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© 1999–2024 Manhattan Review www.manhattanreview.com vi Quantitative Reasoning Question Bank International Phone Numbers and Official Manhattan Review Websites Manhattan Headquarters +1-212-316-2000 www.manhattanreview.com USA & Canada +1-800-246-4600 www.manhattanreview.com Argentina +1-212-316-2000 www.review.com.ar Australia +61-3-9001-6618 www.manhattanreview.com Austria +43-720-115-549 www.review.at Belgium +32-2-808-5163 www.manhattanreview.be Brazil +1-212-316-2000 www.manhattanreview.com.br Chile +1-212-316-2000 www.manhattanreview.cl China +86-20-2910-1913 www.manhattanreview.cn Czech Republic +1-212-316-2000 www.review.cz France +33-1-8488-4204 www.review.fr Germany +49-89-3803-8856 www.review.de Greece +1-212-316-2000 www.review.com.gr Hong Kong +852-5808-2704 www.review.hk Hungary +1-212-316-2000 www.review.co.hu India +1-212-316-2000 www.review.in Indonesia +1-212-316-2000 www.manhattanreview.id Ireland +1-212-316-2000 www.gmat.ie Italy +39-06-9338-7617 www.manhattanreview.it Japan +81-3-4589-5125 www.manhattanreview.jp Malaysia +1-212-316-2000 www.review.my Mexico +1-212-316-2000 www.manhattanreview.mx Netherlands +31-20-808-4399 www.manhattanreview.nl New Zealand +1-212-316-2000 www.review.co.nz Philippines +1-212-316-2000 www.review.ph Poland +1-212-316-2000 www.review.pl Portugal +1-212-316-2000 www.review.pt Qatar +1-212-316-2000 www.review.qa Russia +1-212-316-2000 www.manhattanreview.ru Singapore +65-3158-2571 www.gmat.sg South Africa +1-212-316-2000 www.manhattanreview.co.za South Korea +1-212-316-2000 www.manhattanreview.kr Sweden +1-212-316-2000 www.gmat.se Spain +34-911-876-504 www.review.es Switzerland +41-435-080-991 www.review.ch Taiwan +1-212-316-2000 www.gmat.tw Thailand +66-6-0003-5529 www.manhattanreview.com Turkey +1-212-316-2000 www.review.com.tr United Arab Emirates +1-212-316-2000 www.manhattanreview.ae United Kingdom +44-20-7060-9800 www.manhattanreview.co.uk Rest of World +1-212-316-2000 www.manhattanreview.com www.manhattanreview.com © 1999–2024 Manhattan Review Contents 1 Welcome 1 2 Problem Solving Question Bank 3 2.1 Number properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 2.2 Percents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 2.3 Profit & Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.4 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18 2.5 Ratio & Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 2.6 Speed, Time & Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 2.7 Time & Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30 2.8 Computational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32 2.9 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 2.10 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 2.11 Permutation & Combination & Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 2.12 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 2.13 Statistics & Data Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 2.14 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52 2.15 Quadratic Equations & Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 2.16 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54 2.17 Co-ordinate geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56 3 Data Sufficiency Question Bank 59 3.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 3.2 Percents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69 3.3 Profit & Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72 3.4 Averages (including weighted averages) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73 3.5 Ratio & Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 3.6 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76 3.7 Speed, Time, & Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77 3.8 Time & Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78 3.9 Computational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79 3.10 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82 3.11 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 3.12 Permutation & Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84 3.13 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85 3.14 Statistics & Data Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 3.15 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87 3.16 Quadratic Equations & Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 3.17 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90 3.18 Co-ordinate geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93 vii viii Quantitative Reasoning Question Bank 4 Answer key 97 4.1 Problem Solving Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98 4.2 Data Sufficiency Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 5 Solutions – Problem Solving Questions 103 5.1 Number properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 5.2 Percents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 5.3 Profit & Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 5.4 Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 5.5 Ratio & Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 5.6 Speed, Time & Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 5.7 Time & Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 5.8 Computational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 5.9 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 5.10 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 5.11 Permutation & Combination & Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 5.12 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 5.13 Statistics & Data Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 5.14 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 5.15 Quadratic Equations & Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 5.16 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 5.17 Co-ordinate geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 6 Solutions – Data Sufficiency Questions 243 6.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 6.2 Percents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 6.3 Profit & Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 6.4 Averages (including weighted averages) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 6.5 Ratio & Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 6.6 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 6.7 Speed, Time, & Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 6.8 Time & Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 6.9 Computational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 6.10 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 6.11 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 6.12 Permutation & Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 6.13 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 6.14 Statistics & Data Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 6.15 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 6.16 Quadratic Equations & Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 6.17 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 6.18 Co-ordinate geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 7 Talk to Us 421 www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 1 Welcome Dear Students, At Manhattan Review, we go above and beyond to supercharge your standardized test preparation with the ultimate educational content. We relentlessly drive improvements, especially when it comes to a game-changer like the GMAT. Ready for a ride on the new path to GMAT Focus? This book is designed to be your perfect travel companion!
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Professor Dr. Joern Meissner & The Manhattan Review Team 1 2 Quantitative Reasoning Question Bank www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 2 Problem Solving Question Bank 3 4 Quantitative Reasoning Question Bank −PS Questions 2.1 Number properties 1.
999962 −42 = (A) 1010 −8 (B) (105 −8)2 (C) 104(105 −8) (D) 105(104 −8) (E) 105(105 −8) 2.
If 5a is a factor of n!, and the greatest integer value of a is 6, what is the largest possible value of b such that 7b is a factor of the largest possible value of n!?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 3.
If a = 0.999, b = (0.999)2 and c = √ 0.999, which of the following is the correct order of a, b and c?
(A) a < b < c (B) a < c < b (C) b < c < a (D) b < a < c (E) c < a < b 4.
If p is the product of the reciprocals of integers from 150 to 250, inclusive, and q is the product of the reciprocals of integers from 150 to 251, inclusive, what is the value of (p−1 + q−1) in terms of p?
(A) p (251)2 (B) 251 × 252 × p (C) 252p (D) 252 p (E) 251 × 252 × p2 5.
If x is the sum of all integers from 51 to 100, inclusive, what is the value of x?
(A) 3,624 (B) 3,625 (C) 3,675 (D) 3,725 (E) 3,775 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 5 6.
If x is the sum of the reciprocals of the consecutive integers from 51 to 60, inclusive and y is the sum of the reciprocals of the consecutive integers from 61 to 70, inclusive, which of the following is correct?
I.
1 x > 6 II.
1 y > 7 III.
1 y > 1 x (A) Only I (B) Only II (C) Only III (D) Only II and III (E) I, II and III 7.
A number 4p25q is divisible by 4 and 9; where p and q are the thousands and units digits, respectively. What is the minimum value of p q ?
(A) 1 8 (B) 1 7 (C) 1 6 (D) 2 5 (E) 5 2 8.
If a and b are real numbers such that a percent of (a −2b) when added to b percent of b, the value obtained is 0, then which of the following statements must be true?
I.
a = b II.
a + b = 0 III.
a −b = 1 (A) Only I (B) Only II (C) Only III (D) Only I and III (E) Only II and III 9.
A set is such that if m is in the set, m2 + 3 is also in the set. If −1 is in the set, which of the following is also in the set?
I.
−2 II.
4 III.
19 (A) Only I (B) Only II © 1999–2024 Manhattan Review www.manhattanreview.com 6 Quantitative Reasoning Question Bank −PS Questions (C) Only I and II (D) Only II and III (E) I, II and III 10.
A sequence t1, t2, t3, . . . tn is such that t2 = 5 and tn+1 = 2tn −1 for n ≥1, then what is the value of t10 −t9?
(A) 29 (B) 210 + 1 (C) 1 (D) 2 (E) 4 11.
If m, n, p, and q are distinct positive integers, and all are greater than 1 such that mnpq = 660, how many possible combination of values exist for m, n, p, and q?
(A) Two (B) Three (C) Four (D) Five (E) Seven 12.
If x = 125 and y = 216, what is the value of q x2/3 + y2/3 + 2(xy)1/3 + q x2/3 + y2/3 −2(xy)1/3 ?
(A) 0 (B) 5 (C) 6 (D) 10 (E) 12 13.
If m ≥0.9, which of the following is a possible value of 1 √m ?
(A) 1.01 (B) 1.12 (C) 1.13 (D) 1.35 (E) 1.50 14.
Given that a and b are positive integers and a = b3 90, which of the following must be an integer?
I.
a 45 II.
a 90 III.
a 300 (A) Only I www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 7 (B) Only II (C) Only III (D) Only I and II (E) I, II and III 15.
The addition operation shown below shows the sum of the two-digit positive integers XY and YX. If X, Y, and Z are different digits, what is the value of the integer Z?
X Y + Y X X X Z (A) 8 (B) 7 (C) 2 (D) 1 (E) 0 16.
Suzy saves $20 per month. In each of the 30 months, she saved $20 more than he saved in the previous month. What is the total amount she saved during the 30-month period?
(A) $3,600 (B) $4,800 (C) $6,000 (D) $9,300 (E) $12,000 17.
If a sequence of numbers t1, t2, . . . tn is such that t1 = 0, t2 = 2 and tn = t(n+1) + 2t(n−1) for n ≥1, what is the value of t4 ?
(A) −2 (B) 0 (C) 4 (D) 6 (E) 8 18.
If n is an integer such that n > 9, which of the following could be the remainder when (2 + 22 + 23 + 24 + ... + 2n) is divided by 3?
I.
0 II.
1 III.
2 (A) Only I (B) Only II (C) Only III (D) Only I and III (E) I, II and III © 1999–2024 Manhattan Review www.manhattanreview.com 8 Quantitative Reasoning Question Bank −PS Questions 2.2 Percents 19.
A machine can be repaired for $1,200 and will last for one year, while a new machine would cost $2,800 and will last for two years. The average cost per year of the new machine is what percent greater than the cost of repairing the current machine?
(A) 7% (B) 14.28% (C) 16.67% (D) 18.83% (E) 20% 20.
An item is subject to a sales tax of 10 percent on the part of the price that is greater than $200.
If a customer paid a sales tax of $10 on the item, what was the price of the item?
(A) $200 (B) $250 (C) $300 (D) $360 (E) $400 21.
Item A is subject to a sales tax rate of $0.54 per $25. What is the sales tax rate, as a percent, for item B that is subject to four times as much as the sales tax rate for item A?
(A) 216% (B) 86.4% (C) 8.64% (D) 2.16% (E) 0.135% 22.
In the last lap, cyclist P increases his speed from 10 miles per hour to 25 miles per hour, while cyclist Q increases his speed from 8 miles per hour to 24 miles per hour. By what percent is the percent increase in speed of cyclist Q greater than the percent increase in speed of cyclist P?
(A) 33.33% (B) 50% (C) 66.67% (D) 75% (E) 100% 23.
In a certain year, Carrier X traveled 101,098 kilometers and consumed 9,890 liters of diesel fuel, while in the same year, Carrier Y traveled 203,000 kilometers and consumed 24,896 liters of diesel fuel. Fuel mileage is defined as kilometers per liter of fuel. The mileage of Carrier X is approximately what percent greater or lesser than that of Carrier Y?
(A) 20% (B) 25% (C) 33.33% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 9 (D) 37.50% (E) 40% 24.
The list price of a bicycle is $456. The trader first decreased the price by 25 percent and then increased it by 25 percent.†The final price of the bicycle is what percent of the list price?
(A) 0% (B) 50% (C) 66.67% (D) 93.75% (E) 100% 25.
To prepare a certain color, a paint dealer mixes 3.4 liters of red color into a base that is 68 liters.
The paint manufacturer recommends mixing 0.7 liters per 10 liters of base to make that color. By what percent should the dealer increase the quantity of the paint to match the recommendation?
(A) 10% (B) 33.33% (C) 40% (D) 66.66% (E) 72% 26.
A Business Processing Outsourcing unit recruits 200 employees. Each of them is paid $7.50 per hour for the first 44 hours worked during a week and 1 1 3 times that rate for hours worked in excess of 44 hours. What was the total remuneration of the employees for a week in which 30 percent of them worked 30 hours, 40 percent worked 44 hours, and the rest worked 50 hours?
(A) $25,000 (B) $60,600 (C) $63,300 (D) $70,000 (E) $73,400 27.
A retail company earned $5 million as commission on the first $35 million in sales and then $11 million as commission on the next $121 million in sales. By what percent did the ratio of commissions to sales decrease from the first $35 million in sales to the next $121 million in sales?
(A) 11.11% (B) 22.22% (C) 36.36% (D) 44.44% (E) 50% 28.
A sales representative earned 8 percent commission on the amount of sales up to $20,000, inclusive, and x percent commission on the amount of sales above $20,000. If the representative earned a total commission of $2,000 on sales of $24,000, what was the value of x ?
© 1999–2024 Manhattan Review www.manhattanreview.com 10 Quantitative Reasoning Question Bank −PS Questions (A) 4 (B) 6 (C) 8 (D) 10 (E) 12 29.
A trader buys a batch of 120,000 computer chips for $3,600,000.
He sells two-fifths of the computer chips, each at 25 percent above the cost per computer chip. He sells the remaining chips at a price per chip equal to 25 percent less than the cost per chip. What was the percent profit or loss on the batch of the chips?
(A) Loss of 1% (B) Loss of 5% (C) Loss of 7.50% (D) Profit of 10% (E) Profit of 22.22% 30.
With the 20% increase in the price of milk, a housewife can buy 5 liters of less quantity for $60 than before the increase. What was the initial price per liter of milk?
(A) $2.00 (B) $2.50 (C) $2.75 (D) $3.00 (E) $3.50 31.
A company was approved to spend a certain sum of money for a year. It spent one-fourth of the sum during the first quarter and one-sixth of the remainder during the second quarter. By what percent is the sum of money that was left at the beginning of the third quarter more than the sum spent in the two quarters?
(A) 10% (B) 22.22% (C) 33.33% (D) 66.66% (E) 133.33% 32.
David and Suzy each spent $450 in 2013. In 2014, David spent 10 percent more than he did in 2013, and they together spent $600. Approximately, by what percent less did Suzy spend in 2014 than she did in 2013?
(A) 23% (B) 66% (C) 77% (D) 80% (E) 83% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 11 33.
On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the price by k%. By the end of day 2, the price of the item drops by $1. On day 3, he again increases the price by k%, and on day 4, he again decreases the price by k%. If, at the end of day 4, the price of the item is $398, what is the approximate initial price of the item?
(A) $325 (B) $350 (C) $375 (D) $400 (E) $450 34.
Last year, $y could buy x number of items. If the average cost of the item increased by 20 percent this year, then the number of items that can be bought with $3y equals (A) x (B) 1.50x (C) 2.50x (D) 3x (E) 3.50x 35.
A solution consists of 30 percent water by weight. After boiling the solution for 15 minutes, 70 percent of the water, by weight, evaporated. There is no weight loss for the other part of the solution. What percent of the solution’s total remaining weight consists of the remaining water?
(A) 500 69 % (B) 600 69 % (C) 700 79 % (D) 900 79 % (E) 100 69 % 36.
A mixed juice contains, by volume, 25 percent banana pulp and 75 percent papaya pulp. If this mixed juice costs 20 percent more than an equal quantity of only banana pulp, by what percent is papaya pulp more expensive than banana pulp?
(A) 22.22% (B) 26.67% (C) 28% (D) 30% (E) 33.33% 37.
At a lab, bacteria P multiplies itself every 18 days, while bacteria Q multiplies itself every 15 days. Approximately by what percent is the number of times bacteria Q multiplies itself greater than the number of times bacteria P multiplies itself in a 3-year period? Assume that one year equals 365 days.
(A) 12% © 1999–2024 Manhattan Review www.manhattanreview.com 12 Quantitative Reasoning Question Bank −PS Questions (B) 16% (C) 20% (D) 22% (E) 33% 38.
Jack purchased a phone for $1,500 and paid tax at the rate of 5 percent, while Tom purchased a phone for $1,200 and paid tax at the rate of 15 percent. The total amount Tom paid was what percent less than the total amount Jack paid?
(A) 5% (B) 7% (C) 9% (D) 12% (E) 15% 39.
In a class, 65 percent of the boys and 78 percent of the girls play basketball. If 72 percent of all the students play basketball, what is the ratio of the number of girls to the number of boys?
(A) 4 3 (B) 7 6 (C) 8 7 (D) 9 8 (E) 13 11 40.
In a stadium, the Royal Challengers team had a support of 24,500 spectators from natives and 10 percent of spectators from other than natives. If S is the total number of spectators in the stadium and 40 percent belonged to natives, which of the following represents the number of supporters for the Royal Challengers team?
(A) 0.6S + 12, 250 (B) 0.28S + 12, 250 (C) 0.28S + 24, 500 (D) 0.06S + 24, 500 (E) 0.6S + 24, 500 41.
In a school, 40 percent of the students study science, and 60 percent of them go to special classes after school.
If 30 percent of the students at the school go to special classes, what percent of the total students who do not study science go to special classes?
(A) 6% (B) 12% (C) 15% (D) 24% (E) 27% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 13 42.
In a class at a school, there were 40 percent boys. If some students were transferred to a new section and 30 percent of the transferred students were boys, what was the ratio of the transfer rate for the boys to the transfer rate for the girls?
(A) 1 : 4 (B) 2 : 7 (C) 4 : 9 (D) 9 : 14 (E) 9 : 16 43.
At the beginning of a year, a car was valued 5 7 th of the original price, and at the end of the year, it was value 3 5 th of the original price. By what percent did the value of the car decrease in the year?
(A) 11.11% (B) 16% (C) 17.50% (D) 19% (E) 22.22% 44.
A salesman is offered either a 5 percent commission on his monthly sales, in dollars, and a monthly bonus of $500, or a 7 percent commission on his monthly sales with no bonus. At what sales, in dollars, will both offers give him the same remuneration?
(A) $22,500 (B) $25,000 (C) $32,500 (D) $35,000 (E) $40,000 45.
In the beginning of the year, 35 percent of company X’s 120 customers were retailers, and after the 24-month period, 25 percent of its 240 customers were retailers. What was the simple annual percent growth rate in the number of retailers?
(A) 14.28% (B) 21.43% (C) 24.0% (D) 30.0% (E) 37.25% 46.
Which of the following gives the highest overall percent increase, if, in each case, the second percent increase is applied to the value obtained after applying the first percent increase?
(A) 10 percent increase followed by 50 percent increase (B) 25 percent increase followed by 35 percent increase (C) 30 percent increase followed by 30 percent increase (D) 40 percent increase followed by 20 percent increase © 1999–2024 Manhattan Review www.manhattanreview.com 14 Quantitative Reasoning Question Bank −PS Questions (E) 45 percent increase followed by 15 percent increase 47.
In a school, 70 percent of students are boys, and the rest are girls. In a prefect election, 30 percent of boys and 70 percent of girls voted for John. What percent of the total students voted for John?
(A) 37% (B) 42% (C) 50% (D) 58% (E) 66% 48.
According to the table given below, a state has a total of 23,000 companies from seven regions.
By what percent of the total number of companies in the region is the number of companies in Region S greater than the number of companies in Region R?
Region-wise distribution of companies in the state Regions No. of companies Region P 2,345 Region Q 3,456 Region R 3,421 Region S 5,721 Region T 3,445 Region U 80 Region V 4,532 (A) 5% (B) 10% (C) 17.5% (D) 22.5% (E) 33.33% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 15 2.3 Profit & Loss 49.
A shopkeeper could sell only 4 5 th of the stock at the rate of $3 per item. If 100 items were unsold, what was the total amount he received from the sale?
(A) $240 (B) $1,200 (C) $1,250 (D) $1,300 (E) $1,500 50.
A trader bought 900 cartons of a certain icecream brand at a cost of $20 per carton. If he sold 2 3 rd of the cartons for one and a quarter times their cost price and sold the remaining cartons at a loss of 20 percent of their cost price, what was the trader’s gross profit on the total sale?
(A) $1,800 (B) $2,400 (C) $2,700 (D) $3,000 (E) $3,200 51.
A dealer sells only two brands of bicycles, brand A and brand B. The selling price of a brand A bicycle is $150, which is 60 percent of the selling price of a brand B bicycle. If the dealer sells 100 pieces of bicycles, and 3 5 th of which are brand B, what is the dealer’s total sales, in dollars, from the sale of bicycles?
(A) $15,000 (B) $16,000 (C) $18,000 (D) $21,000 (E) $22,000 52.
A trader bought a consignment at a purchase price of $800 and sold it for 20% less than the marked price. If the trader made a profit equivalent to 30% of the purchase price, what is the marked price of the consignment?
(A) $1,000 (B) $1,200 (C) $1,300 (D) $1,350 (E) $1,500 53.
A small textile company buys a few machines to stitch garments, costing a total of $10,000. The per-unit cost of each garment is $2.50, and is sold for $4.50. How many units of the garments must be sold to achieve break-even (a phenomenon when all the investment and production costs are recovered by the sales revenue)?
© 1999–2024 Manhattan Review www.manhattanreview.com 16 Quantitative Reasoning Question Bank −PS Questions (A) 2,000 (B) 3,500 (C) 4,500 (D) 5,000 (E) 6,000 54.
A broker sold a house with a gross margin of 20 percent on the cost of the house. If the selling price of the house were increased by $10,000, it would yield a gross margin of 30 percent of the cost of the house. What was the original selling price of the house?
(A) $90,000 (B) $100,000 (C) $120,000 (D) $140,000 (E) $150,000 55.
A television assembler pays its contractors $20 each for the first 100 assembled sets and $15 for each additional set. If 600 television sets were assembled and the assembler invoiced the manufactures $25 for each set, what was the assembler’s gross profit in dollars?
(A) $3,750 (B) $4,500 (C) $5,500 (D) $6,000 (E) $7,000 56.
A merchant’s gross profit on item A was 10 percent of its cost. If the merchant increased its selling price from $99 to $117, keeping its cost the same, the merchant’s profit on item A after the price increase was what percent of the cost of item A?
(A) 20% (B) 21% (C) 24% (D) 27% (E) 30% 57.
A merchant bought 2,400 fans for $30 each. He sold 60 percent of the fans for $40 each, and the rest for $35 each. What was the merchant’s average profit per fan?
(A) $6 (B) $8 (C) $9 (D) $10 (E) $12 58.
A trader sold an item at a m percent profit on cost after giving a m percent discount on the list price. Had he sold the item at a (m + 15) percent discount on the list price, his profit would have been (m −20) percent on cost. What would have been his percent profit had he sold the item without giving any discount?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 17 (A) 5.0% (B) 10.0% (C) 25.0% (D) 33.3% (E) 38.0% 59.
A merchant sold 800 units of bedsheets for $8 each and 900 units of bedsheets for $5 each. If the merchant’s cost of producing each unit of bedsheet was $6, what was the merchant’s profit or loss on the sale of 1,700 bedsheets?
(A) Loss of $700 (B) Loss of $300 (C) No profit or loss (D) Profit of $300 (E) Profit of $700 60.
The sales revenue from book sales in 2015 was 10% less than that in 2014, and the sales revenue from stationary sales in 2015 was 6% more than that in 2014. If total sales revenues from book sales and stationary sales in 2015 were 2% more than those in 2014, what is the ratio of sales revenue from book sales in 2014 to sales revenue from stationary sales in 2014?
(A) 1 : 3 (B) 2 : 3 (C) 3 : 4 (D) 4 : 5 (E) 5 : 6 61.
A trader’s profit in 2002 was 20 percent higher than in 2001, and its profit in 2003 was 25 percent higher than its profit in 2002. The company’s profit in 2003 was what percent higher than its profit in 2001?
(A) 5% (B) 45% (C) 46% (D) 48% (E) 50% © 1999–2024 Manhattan Review www.manhattanreview.com 18 Quantitative Reasoning Question Bank −PS Questions 2.4 Averages 62.
Milton School has a student-to-teacher ratio of 25 to 2. The average (arithmetic mean) annual salary for teachers is $42,000. If the school pays a total of $3,780,000 in annual salaries to its teachers, how many students does the school have?
(A) 900 (B) 1,000 (C) 1,125 (D) 1,230 (E) 1,500 63.
The average (arithmetic mean) annual salary of the employees of a company was $70,000. If the male employees’ annual salary average was $65,000 and that of female employees’ annual salary was $80,000, what could be the number of male employees and female employees, respectively, in the company?
(A) 6; 7 (B) 7; 15 (C) 7; 14 (D) 14; 7 (E) 15; 7 64.
A class consists of 40 students and is divided into two sections. In Section A, the average score in a test was 85. In Section B, the average score in the test was 80. If the average score of the class in the test was 82, how many students are in Section A?
(A) 12 (B) 14 (C) 16 (D) 20 (E) 22 65.
A juice manufacturer has 1,200 liters of mango pulp in stock, 25 percent of which is water. If the manufacturer adds another 400 liters of mango pulp, of which 20 percent is water, what percent, by volume, of the manufacturer’s mango pulp contains water?
(A) 21.50% (B) 23.75% (C) 33.33% (D) 35.00% (E) 37.50% 66.
A class has four sections: P, Q, R, and S, and the average weights of the students in the sections are 45 lb, 50 lb, 55 lb, and 65 lb, respectively. What is the maximum possible number of students in Section R if there are 40 students in the class and the average weight of all students o the class is 55 lb?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 19 (A) 18 (B) 20 (C) 25 (D) 35 (E) 37 67.
If set N consists of odd numbers of consecutive integers, starting with 1, what is the difference between the average of the odd integers and the average of the even integers in set N?
(A) −1 (B) 0 (C) 1 2 (D) 1 (E) 2 68.
The average of nine numbers is 25. The average of the first five numbers is 20, and that of the last five is 32. What is the value of the fifth number?
(A) 30 (B) 32 (C) 35 (D) 36 (E) 38 69.
Box X and Box Y each contain many yellow and green balls. All of the green balls have the same radius. The radius of each green ball is 4 inches less than the average radius of the balls in Box X and 2 inches greater than the average radius of the balls in Box Y. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box X and the balls in Box Y?"Good question" (A) 4 (B) 6 (C) 7 (D) 8 (E) 10 70.
A certain company has 60 employees. The average (arithmetic mean) salary of 10 of the employ-ees is $35,000; the average salary of 35 other employees is $30,000; and the average salary of the remaining 15 employees is $60,000. What is the average salary of the 60 employees at the company?
(A) $32,500 (B) $38,333 (C) $39,500 (D) $40,000 (E) $42,222 © 1999–2024 Manhattan Review www.manhattanreview.com 20 Quantitative Reasoning Question Bank −PS Questions 71.
At a certain stationery shop, the price of a pencil is 20 cents, and the price of an eraser is 30 cents. A boy buys a total of 20 pencils and erasers from the shop, and the average (arithmetic mean) price of the 20 pieces comes to 28 cents. How many erasers must the boy return so that the average price of the pieces that he buys is 26 cents?
(A) 2 (B) 4 (C) 6 (D) 8 (E) 10 72.
A student’s average (arithmetic mean) test score on four tests is 78. If each test is scored out of 100, which of the following can be the student’s score on the fifth test so that the student’s average score on five tests increases by an integer value?
(A) 82 (B) 87 (C) 89 (D) 93 (E) 95 73.
An instructor gave the same test to three groups: P, Q, and R. The average (arithmetic mean) scores for the three groups were 64, 84, and 72, respectively.
The ratio of the numbers of candidates in the P, Q, and R groups was 3 : 5 : 4, respectively. What was the average score for the three groups combined?
(A) 72 (B) 75 (C) 77 (D) 78 (E) 80 74.
A fitness club has 50 male and 20 female members. The average (arithmetic mean) age of all of the members is 23 years. If the average age of the male members was 20 years, which of the following is the average age, in years, of the female members?
(A) 30.50 (B) 31.50 (C) 32.50 (D) 33.00 (E) 34.50 75.
Following is a modified question of the above.
A fitness club has 50 male and 20 female members. The average (arithmetic mean) age of all of the members is 23.89 years. If the average age of the male members was 20.89 years, which of the following is the average age, in years, of the female members?
(A) 29.75 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 21 (B) 30.50 (C) 31.39 (D) 34.39 (E) 37.50 © 1999–2024 Manhattan Review www.manhattanreview.com 22 Quantitative Reasoning Question Bank −PS Questions 2.5 Ratio & Proportion 76.
The total cost of manufacturing metal bearings incurs a fixed cost of $25,000 and a variable expense, which depends on the number of bearings manufactured. If for 50,000 bearings the total cost is $100,000, what is the total cost for 100,000 bearings?
(A) $125,000 (B) $150,000 (C) $175,000 (D) $200,000 (E) $275,000 77.
A beaker was filled with a 40 liters of mixture. The mixture has water and a liquid chemical in the ratio of 3 : 5, respectively. If each day, for a 10-day period, 2 percent of the initial quantity of water and 5 percent of the initial quantity of liquid chemical evaporated, what percent of the original amount of mixture evaporated during this period?
(A) 22.22% (B) 33.33% (C) 38.75% (D) 44.44% (E) 58.33% 78.
In Ghazal’s doll collection, 3 5 th of the dolls are Barbie dolls, and 4 7 th of the Barbies were purchased before the age of 10. If 90 dolls in Ghazal’s collection are Barbies that were purchased at the age of 10 or later, how many dolls in her collection are non-Barbie dolls?
(A) 70 (B) 90 (C) 140 (D) 154 (E) 192 79.
The ratio of the ages of John and Suzy is 5 : 6. Which of the following can be the ratio of their ages after 10 years?
(A) 2 : 3 (B) 13 : 20 (C) 11 : 15 (D) 4 : 5 (E) 9 : 10 80.
A company assembles two kinds of phones: a feature phone and a smartphone. Of the phones produced by the company last year, two-fifths were feature phones, and the rest were smart-phones. If it takes 8 5 times as many hours to produce a smartphone as it does to produce a feature phone, then the number of hours it took to produce the smartphones last year was what fraction of the total number of hours it took to produce all the phones?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 23 (A) 8 31 (B) 11 31 (C) 12 17 (D) 13 34 (E) 15 34 81.
At a certain garment shop, the ratio of the number of shirts to the number of trousers is 4 to 5, and the ratio of the number of jackets to the number of shirts is 3 to 8. If the ratio of the number of sweaters to the number of trousers is 6 to 5, what is the ratio of the number of jackets to the number of sweaters?
(A) 9 to 25 (B) 1 to 3 (C) 1 to 4 (D) 3 to 5 (E) 6 to 5 82.
At a church prayer, 3 5 th of the members were males.
3 5 th of the male members and 7 10 th of the female members attended the prayer. Of the members who did not attend the prayer, what fraction are male members who did not attend the prayer?
(A) 1 4 (B) 3 7 (C) 2 3 (D) 9 10 (E) 6 19 83.
John, Suzy, and David together donated a total of $100 to a charity. If John paid 5 3 th of what David donated, Suzy donated $20, and David donated the rest, what fraction of the total amount did David donate?
(A) 1 5 (B) 1 6 (C) 2 7 (D) 3 10 (E) 4 11 © 1999–2024 Manhattan Review www.manhattanreview.com 24 Quantitative Reasoning Question Bank −PS Questions 84.
A merchant sold a total of X shirts and trousers. If the number of trousers is 1 5 th the number of shirts, and 1 5 th of the shirts are cotton shirts, how many cotton shirts, in terms of X, were sold by the merchant?
(A) 2X 7 (B) X 4 (C) 4X 15 (D) 6X 25 (E) X 6 85.
A rod that weighs 20 pounds is cut into two pieces such that one of the pieces weighs 16 pounds and measures 36 feet. If the weight of the rod is directly proportional to the square of its length, how long is the other piece of rod, in feet?
(A) 9 (B) 12 (C) 18 (D) 24 (E) 27 86.
The ratio of John’s coins to Suzy’s coins is 3 : 4. If the number of John’s coins exceeds 2 7 of the total number of coins by 25, how many coins Suzy has?
(A) 50 (B) 100 (C) 120 (D) 150 (E) 180 87.
The total cost of an item is formed out of four costs: Material cost, Labour cost, Factory overhead cost, and Office overhead cost. If Material cost and Labour cost constitute 3 7 part of the total cost, Labour cost and Factory overhead cost constitute 1 2 part of the total cost, Factory overhead cost and Office overhead cost constitute 4 7 part of the total cost, and Material cost and Office overhead cost constitute 1 2 part of the total cost, which of the four costs is the highest among all?
(A) Material cost (B) Labour cost (C) Factory overhead cost (D) Office overhead cost (E) Factory overhead cost or Office overhead cost www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 25 88.
An entrance test consists of 20 questions. Each question after the first is worth 2 points more than the preceding question. If the total questions are worth a total of 400 points, how many points is the fourth question worth?
(A) 5 (B) 7 (C) 11 (D) 19 (E) 38 89.
In a science college, 80 more than 1 3 rd of all the students took a science course and 1 3 th of those who took the science course took chemistry. If 1 6 th of all the students in the school took chemistry, how many students are in the school?
(A) 200 (B) 240 (C) 480 (D) 600 (E) 720 90.
In an office having 50 employees, 1 4 th of the males and 1 5 th of the females eat company breakfast. What is the greatest possible number of employees in the office that eat company breakfast?
(A) 6 (B) 8 (C) 10 (D) 12 (E) 25 91.
In a cookery class, 1 8 th of the number of females is equal to 1 12 th of the total number of students in the class. What is the ratio of the number of males to the number of females in the class?
(A) 1 : 5 (B) 1 : 4 (C) 1 : 2 (D) 3 : 4 (E) 2 : 1 92.
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of the boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed at 30 for 40, what is the correct average (arithmetic mean) score of the class?
(A) 31 (B) 31.50 (C) 32.25 © 1999–2024 Manhattan Review www.manhattanreview.com 26 Quantitative Reasoning Question Bank −PS Questions (D) 41 (E) 42 93.
How many liters of Chemical A must be added to a 120-liter solution that is 25 percent Chemical A in order to produce a solution that is 40 percent Chemical A?
(A) 12 (B) 15 (C) 20 (D) 24 (E) 30 94.
Month Number of chickens 1 144 2 c 3 256 Information about the number of chickens hatched in a poultry farm is given in the table above.
If the number of chickens in the poultry farm in any month increased by the same fraction during each of the two periods of the successive months, how many chickens were there in the second month?
(A) 192 (B) 200 (C) 210 (D) 220 (E) 240 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 27 2.6 Speed, Time & Distance 95.
A trip of 900 miles would have taken 1 hour less if the average speed for the trip had been greater by 10 miles per hour. What was the average speed for the trip?
(A) 40 miles per hour (B) 45 miles per hour (C) 60 miles per hour (D) 75 miles per hour (E) 90 miles per hour 96.
A truck traveled 336 miles per full tank of diesel on the national highway and 224 miles per full tank of diesel on the state highway. If the truck traveled 4 fewer miles per gallon on the state highway than on the national highway, how many miles per gallon did the truck travel on the state highway?
(A) 6 (B) 8 (C) 10 (D) 12 (E) 15 97.
A bike traveling at a certain constant speed takes 5 minutes longer to travel 10 miles than it would take at 60 miles per hour. At what speed, in miles per hour, is the bike traveling?
(A) 36 (B) 40 (C) 42 (D) 48 (E) 50 98.
A biker increased his average speed by 10 miles per hour in each successive 10-minute interval after the first interval. If in the first 10-minute interval, his average speed was 30 miles per hour, how many miles did he travel in the fourth 10-minute interval?
(A) 4 (B) 5 (C) 8 (D) 10 (E) 15 99.
An aircraft flew 600 miles to a town at an average speed of 500 miles per hour with the wind and made the trip back following the same route at an average speed of 400 miles per hour against the wind. Which of the following is aircraft’s approximate average speed, in miles per hour, for the trip?
(A) 420 (B) 444 © 1999–2024 Manhattan Review www.manhattanreview.com 28 Quantitative Reasoning Question Bank −PS Questions (C) 450 (D) 467 (E) 483 100.
A truck completed half of a 800-mile trip at an average speed of 40 miles per hour. At what approximate average speed, in miles per hour, should the truck complete the remaining miles to achieve an average speed of 50 miles per hour for the entire 800-mile trip? Assume that he truck completed its 800-mile trip without stoppage.
(A) 52 (B) 55 (C) 60 (D) 67 (E) 70 101.
A marathoner ran for two days. On the second day, he ran at an average speed of 3 miles per hour, faster than the average speed of the first day. If during the two days he ran a total of 36 miles and did a total of 8 hours running, which of the following could be his average speed, in miles per hour, on the first day?
(A) 0.25 (B) 0.50 (C) 1.00 (D) 1.50 (E) 2.00 102.
Two trains traveling toward each other on parallel tracks at constant rates of 50 miles per hour and 60 miles per hour are 285 miles apart. How far apart will they be 2 hours before their engine meet?
(A) 110 (B) 120 (C) 150 (D) 200 (E) 220 103.
If the speed limit along an 10-mile section of rail track is reduced from 50 miles per hour to 40 miles per hour. Approximately how many minutes more will it take a train to travel along this section with the new speed limit than it would have taken at the old speed limit?
(A) 3 (B) 5 (C) 8 (D) 10 (E) 12 104.
Trains A and B traveled the same 100-mile route. If Train A took 4 hours and Train B traveled at an average speed 25 percent more than the average speed of Train A, how many hours did it take Train B to travel the route?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 29 (A) 12 3 (B) 2 (C) 31 5 (D) 33 5 (E) 4 105.
Jeff drives three times farther in 36 minutes than Amy drives in 30 minutes. If Jeff drives at a speed of 40 miles per hour, at what speed, in miles per hour, does Amy drive?
(A) 6 (B) 9 (C) 16 (D) 24 (E) 32 106.
A bus left bus depot A at 7 a.m. and reached bus depot B at 12 p.m. Another bus left bus depot B at 8 a.m. and reached bus depot A at 11 a.m. At what time did the two buses pass each other?
(A) 9:00 am (B) 9:15 am (C) 9:30 am (D) 9:40 am (E) 10:00 am © 1999–2024 Manhattan Review www.manhattanreview.com 30 Quantitative Reasoning Question Bank −PS Questions 2.7 Time & Work 107.
Photocopier machine A makes 1,500 copies per hour. Working 12 hours each day, photocopier machine B, twice as efficient as A, will make how many copies in 20 days?
(A) 400,000 (B) 500,000 (C) 540,000 (D) 660,000 (E) 720,000 108.
A water pump began filling an empty swimming pool and ran at a constant rate till the pool was full. At some point, the pool was half full, and 2 1 3 hours later, it was 5 6 full. How many hours did it take the pump to fill the empty pool completely?
(A) 4 (B) 51 3 (C) 7 (D) 71 5 (E) 81 3 109.
Two pumps, each working alone, can fill an empty pool in 10 hours and 15 hours, respectively.
The first pump initially started alone for h hours, after which the second pump was also started.
If it took a total of 7 hours for the pool to be filled completely, what is the value of h?
(A) 2.00 (B) 2.50 (C) 3.00 (D) 3.30 (E) 4.00 110.
An empty swimming pool with a capacity of 5,760 gallons is being filled at a rate of 12 gallons per minute by a pipe. An emptying pipe that has the capacity to empty 3 4 th of the pool in 9 hours is also in operation. If the pool is already half-filled and if both pipes are in operation, how many hours would it take to fill the pool to its full capacity?
(A) 6 (B) 12 (C) 24 (D) 36 (E) 72 111.
Lathe machine A manufactures metal parts three times as fast as lathe machine B. Lathe machine B manufactures 300 X-type bearings in 60 days. If each machine manufactures bearings at their constant rate, how many Y-type bearings does lathe machine A manufacturer in 10 days if each Y-type bearing takes 2.5 times the time taken to manufacture each X-type bearing?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 31 (A) 40 (B) 50 (C) 54 (D) 60 (E) 64 112.
Photocopier A, working alone at its constant rate, makes 1,200 copies in 3 hours. Photocopier B, working alone at its constant rate, makes 1,200 copies in 2 hours. Photocopier C, working alone at its constant rate, makes 1,200 copies in 6 hours. How many hours will it take photocopiers A, B, and C, working together at their respective constant rates, to make 3,600 copies?
(A) 2.00 (B) 2.25 (C) 2.50 (D) 3.00 (E) 3.50 113.
Five men can consume food costing $150 on a 4-day expedition trip. If a woman consumes three-fourths of the amount of food consumed by a man, what would be the cost of food consumed by 4 men and 2 women during an 8-day expedition trip?
(A) $300 (B) $330 (C) $360 (D) $390 (E) $400 114.
Working alone, Mark and Kate take 12 hours more and 27 hours more, respectively, to complete a certain project than working together would take them to complete the same project. Working together, how many hours do Mark and Kate take to complete the project?
(A) 12 (B) 16 (C) 18 (D) 24 (E) 39 © 1999–2024 Manhattan Review www.manhattanreview.com 32 Quantitative Reasoning Question Bank −PS Questions 2.8 Computational 115.
A chemical evaporates out of a beaker at the rate of x liters for every y minutes. If the chemical costs 25 dollars per liter, what is the cost, in dollars, of the amount of the chemical that will evaporate in z minutes?
(A) 25x yz (B) xz 25q (C) 25y xz (D) 25xz y (E) 25yz x 116.
In company X, the total cost of producing pens is governed by a linear function. If the total cost of producing 25,000 pens is $37,500 and the total cost of producing 35,000 pens is $47,500, what is the total cost of producing 50,000 pens?
(A) $57,500 (B) $60,000 (C) $62,500 (D) $67,857 (E) $75,900 117.
If Suzy had thrice the amount of money that she currently has, she would have exactly the money needed to purchase four pencils, each costing $1.35 per piece, and two erasers, each costing $0.30 per piece. How much money does Suzy have?
(A) $1.50 (B) $2.00 (C) $2.25 (D) $2.50 (E) $2.75 118.
The population of a certain country increases at the rate of 30,000 people every month. The population of the country in 2012 was 360 million. In which year would the population of the country be 378 million?
(A) 2060 (B) 2061 (C) 2062 (D) 2063 (E) 2064 119.
An Ice cream parlor buys milk-cream cartons, each containing 2 1 2 cups of milk-cream. If the restaurant uses 1 2 cup of milk-cream in each serving of its ice cream, what is the least number of cartons needed to prepare 98 servings of the ice cream?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 33 (A) 9 (B) 19 (C) 20 (D) 21 (E) 24 120.
A few coins are put into 7 boxes, such that each box contains at least two coins. At most 3 boxes can contain the same number of coins, and the remaining boxes cannot contain an equal number of coins. What is the minimum possible number of coins in the 7 boxes?
(A) 18 (B) 20 (C) 24 (D) 27 (E) 30 121.
A volcanic lava laterally moves at a rate of 15/4 feet per hour. How many days does it take the lava to move 3/2 miles? (1 mile = 5,280 feet) (A) 48 (B) 60 (C) 72 (D) 80 (E) 88 122.
At a metal rolling factory, if an iron bar of square cross-section with an area of 4 square feet is moving continuously through a belt conveyor at a constant speed of 360 feet per hour, how many seconds does it take for a volume of 8.4 cubic feet of the iron bar to move through the conveyor?
(A) 21 (B) 22 (C) 24 (D) 27 (E) 30 123.
At a factory, each worker is remunerated according to a salary grade G, from 1 to 7. Each worker receives a monthly wage W, in dollars, determined by the formula W = 1, 140 + 45(G −1). How many more dollars per month does a worker with a salary grade of 7 receive than a worker with a salary grade of 1?
(A) $135 (B) $270 (C) $405 (D) $540 (E) $600 © 1999–2024 Manhattan Review www.manhattanreview.com 34 Quantitative Reasoning Question Bank −PS Questions 124.
At a garage sale, a few radios and DVD players were sold.
The prices of all the items were different. Of the radios, the price of a certain radio sold was the 15th highest price as well as the 20th lowest price. Of all items, the price of a certain DVD player sold was the 29th highest price as well as the 37th lowest price. How many DVD players were sold at the garage sale?
(A) 30 (B) 31 (C) 32 (D) 64 (E) 65 125.
A salesman is paid $25 per order as commission for the first 150 orders and $12.50 as commis-sion for each additional order. If he received a total of $5,000 as commission, how many orders did he make?
(A) 100 (B) 150 (C) 200 (D) 250 (E) 300 126.
An overseas businessman purchased a total of $2,000 worth of traveler’s checks in $20 and $50 denominations. During the trip, he cashed only 10 checks and lost all the remaining checks. If the number of $20 checks cashed were 2 more or 2 less than the number of $50 checks cashed, what is the minimum possible amount of money he lost?
(A) $1,200 (B) $1,440 (C) $1,500 (D) $1,620 (E) $1,680 127.
A pet shop sells three pack-sizes of dog food from brand X. The 5-kg pack costs $16, the 10-kg pack costs $26, and the 25-kg pack costs $55. If a customer wants to buy a minimum of 40 kg of brand X dog food, what is the minimum price he will have to pay?
(A) $85 (B) $90 (C) $97 (D) $107 (E) $110 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 35 2.9 Interest 128.
If a sum of money invested under simple interest, amounts to $3,200 in 4 years and $3,800 in 6 years, what is the rate at which the sum of money was invested?
(A) 10% (B) 12% (C) 15% (D) 20% (E) 24% 129.
For an equal sum of money, each invested for 2 years, the difference between compound interest, compound annually, and simple interest, at the same rate of interest, is $63.
If the simple interest on the sum after 2 years is $600, at what rate of interest were the sums of money invested?
(A) 25% (B) 24% (C) 22% (D) 21% (E) 10% 130.
Suzy borrows two equal sums of money under simple interest at 10% and 8% rates of interest.
She finds that if she repays the former sum one year before the latter, she would have to pay the same amount for each borrowing. After how many years did she pay the first sum of money?
(A) 2.5 (B) 3 (C) 3.5 (D) 4 (E) 5 131.
A sum of $100,000 was invested in two deposits: one part at simple interest rates of 3 percent and another part at simple interest rates of 4 percent. If the total interest on the two sums was $3,600 at the end of one year, what fractional part of the $100,000 was invested at 4 percent?
(A) 5 8 (B) 1 5 (C) 2 5 (D) 3 5 (E) 3 7 132.
A sum of money is invested at simple interest, partly at 4% and remaining at 7% annual rates of interest. After two years, the total interest obtained was $2,100. If the total investment is $18,000, what was the sum of money invested at a 4% annual rate of interest?
© 1999–2024 Manhattan Review www.manhattanreview.com 36 Quantitative Reasoning Question Bank −PS Questions (A) $5,500 (B) $6,000 (C) $7,000 (D) $10,500 (E) $11,000 133.
A man invested two equal sums of money in two banks at simple interest, one offering an annual rate of interest of 10% and the other offering an annual rate of interest of 20%. If the difference between the interests earned after 2 years is between $120 and $140, exclusive, which of the following could be the difference between the amounts earned for the same sums of money, invested at the same rates of interest as above, but at compound interest?
(A) $130 (B) $135 (C) $137 (D) $154 (E) $162 134.
At the start of an experiment, a certain population consisted of x organisms. At the end of each month, the population size increased by twice as much as size at the beginning of that month.
If the total population at the end of five months is greater than 1,000, what is the minimum possible value of x?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 135.
A sum of money is borrowed at a 12% per annum interest rate for one year. The interest is calculated after the end of every two-month period and is added to the amount accrued after a period. The amount payable after the end of the year is how many times the sum borrowed?
(A) 1.12 (B) (1.12)5 (C) (1.02)6 (D) (1.02)5 (E) (1.2)6 136.
Mary deposited a sum of x dollars into an account that earned 4% annual interest compounded annually. One year later, she deposited an additional x dollars in the account. Consider that there were no other transactions, and if the account showed y dollars at the end of the two years, which of the following expresses x in terms of y?
(A) y 2.04 (B) y 1.12 (C) y 2.2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 37 (D) y 1.04 × 2.04 (E) y (1.04)2 × 2.04 137.
George invested a certain sum of money in compound interest payable at a certain rate of in-terest. By the end of the 5th year, the interest on the investment was $4,800, and by the end of the 6th year, the interest on the investment was $5,520. What was the rate of interest at which George invested the sum of money?
(A) 10.0% (B) 12.0% (C) 12.5% (D) 15.0% (E) 20.0% © 1999–2024 Manhattan Review www.manhattanreview.com 38 Quantitative Reasoning Question Bank −PS Questions 2.10 Functions 138.
If the function f is defined by f p = p2 + 1 p2 for all non-zero numbers p, then f −1 √p 2 = (A) f p + 2 (B) 2 f p2 (C) 1 f √p !2 (D) 1 − f √p 2 (E) f p −2 139.
The function f is defined by f (x) = −1 x for all non-zero numbers x. If f (a) = −1 2 and f (ab) = 1 6, then b = (A) 3 (B) 1 3 (C) −1 3 (D) −3 (E) −12 140.
The function f is defined by f (x) = √x −20 for all positive numbers x. If p = f q for some positive numbers p and q, what is q in terms of p?
(A) p + 20 2 (B) p p + 20 (C) √p + 20 2 (D) q p2 + 20 (E) p2 + 20 2 141.
The function f is defined for positive three-digit integer T by f (T) = 2a3b5c, where a, b and c are the hundreds, tens and units digits of T, respectively. If K and R are three-digit positive integers such that f (K) = 18f (R), then K −R = (A) 65 (B) 70 (C) 80 (D) 100 (E) 120 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 39 142.
For which of the following functions f , is f (x) = f (1 −x) for all x?
(A) f (x) = 1 + x (B) f (x) = 1 + x2 (C) f (x) = x2 −(1 −x)2 (D) f (x) = x2(1 −x)2 (E) f x2 = x 1 −x 143.
If f (x) = 1 x and g (x) = x x2+1, for all x > 0, what is the minimum value of f g (x) ?
(A) 0 (B) 1 2 (C) 1 (D) 3 2 (E) 2 144.
If f (x) = 10x 1 −x , for what value of x does f (x) = 1 2f (3)?
(A) 4 (B) 2 (C) 1 (D) −3 (E) −5 145.
If 3f (x) + 2f (−x) = 5x −10, what is the value of f (1)?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 146.
As per an estimate, the depth D(t), in centimeters, of the water in a tank at t hours past 12:00 a.m. is given by D (t) = −10(t −7)2 + 100, for 0 ≤t ≤12. At what time does the depth of the water in the tank becomes the maximum?
(A) 5:30 a.m.
(B) 7:00 a.m.
(C) 7:30 a.m.
(D) 8:00 a.m.
(E) 9:00 a.m.
© 1999–2024 Manhattan Review www.manhattanreview.com 40 Quantitative Reasoning Question Bank −PS Questions 2.11 Permutation & Combination & Probability 147.
Cp q = p!
p −q ! × q! for positive integers p and q, p ≥q. If C5 3 = C5 r , which of the following could be the value of r?
(A) 0 (B) 1 (C) 2 (D) 4 (E) 5 148.
A color code is defined as a sequence of three dots arranged in a row. Each dot is colored either “red” or “black.” How many distinct codes can be formed?
(A) 4 (B) 5 (C) 6 (D) 8 (E) 10 149.
A daily store stocks two sizes of mugs, each in four colors: black, green, yellow, and red. The store packs the mugs in packages that contain either three mugs of the same size and the same color or three mugs of the same size and three different colors. If the order of the arrangement of mugs of three different colors is disregarded, how many different packings of the types described above are possible?
(A) 4 (B) 10 (C) 16 (D) 20 (E) 30 150.
A pizza-seller offers six kinds of toppings and two kinds of breads for his pizzas. If each pizza contains at least two kinds of toppings but not all kinds of toppings and only one kind of bread, how many different pizzas could the pizza-seller offer?
(A) 56 (B) 58 (C) 84 (D) 100 (E) 112 151.
A botanist designates each plant with a one-, two-, or three-letter code, where each letter is one of the 26 letters of the alphabet. If the letters may be repeated and if the same letters used in a different order convey a different code, how many different plants can the botanist uniquely designate with these codes?
(A) 2,951 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 41 (B) 9,125 (C) 16,600 (D) 17,576 (E) 18,278 152.
A college student can select one out of eight optional subjects from group one and two out of ten optional subjects from group two. If no subject is common in both groups, how many different sets of three subjects are there to select?
(A) 53 (B) 120 (C) 190 (D) 360 (E) 408 153.
A company has to assign distinct four-digit code numbers to its employees. Each code number was formed from the digits 1 to 9, and no digit appears more than once in any one code. How many employees can be assigned codes?
(A) 30 (B) 2,400 (C) 3,024 (D) 5,040 (E) 6,491 154.
A company plans to assign identification numbers to its employees. Each number is to consist of four digits from 0 to 9, inclusive, except that the first digit cannot be 0. If any digit can be repeated any number of times in a particular code, how many different identification numbers are possible that are odd numbers?
(A) 2,520 (B) 2,268 (C) 3,240 (D) 4,500 (E) 9,000 155.
A fast-food company plans to build four new restaurants. There are six sites (A, B, C, D, E, and F) that satisfy the company’s criteria for the location of the restaurants. In how many ways can the company select four sites if both sites A and B cannot be selected together?
(A) 4 (B) 5 (C) 6 (D) 9 (E) 15 156.
Imran has four Math, five Physics, and six Chemistry books. He has to choose four out of the 15 books such that the selection has at least one book of each subject. In how many ways it is possible?
© 1999–2024 Manhattan Review www.manhattanreview.com 42 Quantitative Reasoning Question Bank −PS Questions (A) 600 (B) 720 (C) 760 (D) 800 (E) 960 157.
A botanist plans to code the 15 plants used in an experiment with unique codes that consist of either a single letter or a pair of distinct letters. If the codes must be in alphabetic order, i.e., a code CD is valid but a code DC is not, what is the least number of letters that can be used to code them?
(A) 3 (B) 4 (C) 5 (D) 7 (E) 15 158.
Classes A, B, and C have 30 students each, and class D has 20 students. A team is to be formed by selecting one student from A, B, and C and two students from D. How many teams are possible?
(A) 1,540,000 (B) 2,200,000 (C) 2,400,000 (D) 3,600,000 (E) 5,130,000 159.
A stock broker recommends a portfolio of 2 Information Technology stocks, 4 Retail stocks, and 2 e-commerce stocks. If the broker can choose from 4 Information Technology stocks, 5 Retail stocks, and 3 e-commerce stocks, how many different portfolios of 8 stocks are possible?
(A) 9 (B) 24 (C) 60 (D) 90 (E) 120 160.
In a conference of 3 delegates from each of 8 different companies, each delegate shook hands with every person other than those from his or her own organization. How many handshakes took place at the conference?
(A) 48 (B) 96 (C) 252 (D) 270 (E) 504 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 43 161.
The digits 0 to 9 are used to form three-digit codes; however, there are a few conditions: the first digit cannot be 0 or 9, the second digit must be 0 or 9, and the second and third digits cannot both be ‘9’ in the same code. If the digits may be repeated in the same code, how many different codes are possible?
(A) 152 (B) 156 (C) 160 (D) 729 (E) 1,000 162.
A pot contains 15 green and 15 yellow marbles. If two marbles are to be picked from this pot at random, what is the probability that both marbles will be yellow?
(A) 1 5 (B) 7 29 (C) 7 30 (D) 8 29 (E) 23 30 163.
A box contains 12 balls; of these, seven are red and five are green. If three balls are to be selected at random from the box, what is the probability that two of the balls will be red and one will be green?
(A) 7 44 (B) 7 22 (C) 51 100 (D) 21 44 (E) 7 9 164.
A badminton club has 21 members. What is the ratio of the number of 6-member committees the number of 5-member committees that can be formed from the members of the club?
(A) 6 to 5 (B) 15 to 1 (C) 8 to 3 (D) 17 to 6 (E) 16 to 5 165.
A courier company can assign its employees to its offices in such a way that more than one office can be assigned to an employee. In how many ways can the company assign four employees to two offices?
© 1999–2024 Manhattan Review www.manhattanreview.com 44 Quantitative Reasoning Question Bank −PS Questions (A) 6 (B) 8 (C) 10 (D) 12 (E) 16 166.
A transport company employs five male officers and three female officers. If a core group is to be created that is made up of three male and two female officers, how many different core groups are possible?
(A) 10 (B) 16 (C) 24 (D) 30 (E) 60 167.
If the probability that Stock X will increase in value during the next week is 0.40 and the proba-bility that Stock Y will increase in value during the next week is 0.60, what is the probability that exactly one of the two stocks will increase in value during the next week? Note that the price fluctuations of one stock do not affect the price fluctuations of the other.
(A) 0.48 (B) 0.50 (C) 0.52 (D) 0.56 (E) 0.58 168.
An unbiased coin has an equal probability of getting a head or a tail. What is the probability that the coin will land heads at least once when it is tossed twice?
(A) 1 5 (B) 1 4 (C) 1 3 (D) 2 3 (E) 3 4 169.
A quiz consists of X questions, each of which is to be answered either “Yes” or “No.” The participant must answer all questions correctly to win the quiz. What is the least value of X for which the probability is less than 1 500 such that a participant who randomly guesses the answer to each question will be a winner?
(A) 8 (B) 9 (C) 10 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 45 (D) 200 (E) 500 170.
A box contains 20 balls: 10 white and 10 black. Five balls are to be drawn at random. If the first three drawn balls are black and not replaced, what is the probability that the next two drawn balls will also be black?
(A) 21 136 (B) 4 17 (C) 1 3 (D) 4 7 (E) 2 5 171.
A box contains 16 balls, of which 4 are white, 3 are blue, and the rest are yellow. If two balls are to be selected at random from the box, one at a time, without being replaced, what is the probability that one ball selected will be white and the other ball selected will be blue?
(A) 5 64 (B) 1 16 (C) 1 10 (D) 1 5 (E) 1 6 172.
A batch of eight refrigerators contains two single-door refrigerators and six double-door refrig-erators. If two refrigerators are to be chosen at random from this batch, what is the probability that at least one of the two refrigerators chosen will be a single-door?
(A) 1 4 (B) 1 3 (C) 7 15 (D) 13 28 (E) 1 2 173.
In a jar, 9 balls are white, and the rest are red. If two balls are to be chosen at random from the jar without replacement, the probability that the balls chosen will both be white is 6 11. What is the number of balls in the jar?
(A) 10 © 1999–2024 Manhattan Review www.manhattanreview.com 46 Quantitative Reasoning Question Bank −PS Questions (B) 11 (C) 12 (D) 13 (E) 15 174.
A pyramid of 12 playing cards is in such a state that if any individual card falls, the pyramid collapses. If, for each individual card, the probability of falling within 1 minute is 0.05, what is the probability that the pyramid will collapse within 1 minute?
(A) 0.05 (B) 0.0512 (C) 1 −0.9512 (D) 0.9512 (E) 1 −0.0512 175.
IIn a pack of a dozen candies, four are orange-flavored. If a kid randomly picks two candies from the pack, what is the probability that the kid has no orange-flavored candy?
(A) 1 7 (B) 2 11 (C) 14 33 (D) 7 33 (E) 8 33 176.
On the morning of day 1, Suzy began her tracking tour. She plans to return home at the end of the first day on which it rains. If for the first three days of the tour, the probability of rain on each day is 0.25, what is the probability that Suzy will return home at the end of day 3?
(A) 1 8 (B) 9 64 (C) 27 64 (D) 37 64 (E) 1 64 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 47 2.12 Sets 177.
A marketing class at a college has a total strength of 30. It formed three groups: G1, G2, and G3, which have 10, 10, and 6 students, respectively. If no student in G1 is in either G2 or G3, what is the greatest possible number of students who are in neither of the groups?
(A) 4 (B) 7 (C) 8 (D) 10 (E) 14 178.
In a batch of 1,000 dresses, 1/4 of the dresses are traditional and the rest are contemporary. Half the dresses are for males, and half are for females. If of all dresses, 100 dresses are traditional and are for males, how many of the dresses are contemporary and are for females?
(A) 150 (B) 250 (C) 300 (D) 350 (E) 400 179.
According to a report, 7% of students did not use computers to play games, 11% did not use computers to write reports, and 95% use computers for at least one of the purposes.
What percent of the students, according to the report, used computers for both purposes—playing games and writing reports?
(A) 13% (B) 56% (C) 77% (D) 87% (E) 91% 180.
In a company survey, 600 employees were each asked whether they take cola or health drinks.
As per the survey, 70 percent of the employees take cola, 45 percent take health drinks, and 25 percent take both cola and health drinks. How many employees surveyed take neither cola nor health drinks?
(A) 50 (B) 60 (C) 70 (D) 75 (E) 80 181.
In Milton School, the number of students who play badminton is three times the number of students who play tennis. The number of students who play both badminton and tennis is three times the number of students who play only tennis. If 60 students play both sports, how many students play only badminton?
© 1999–2024 Manhattan Review www.manhattanreview.com 48 Quantitative Reasoning Question Bank −PS Questions (A) 100 (B) 150 (C) 160 (D) 180 (E) 120 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 49 2.13 Statistics & Data Interpretation 182.
15, 20, 25, x: (not in order) Which of the following could be the median of the four integers listed above?
I.
18 II.
22 III.
23 (A) I only (B) II only (C) I and II only (D) II and III only (E) All of them 183.
44, 52, 56, 65, 73, 75, 77, 95, 96, 97 The list above shows the scores of 10 students obtained on a scheduled test. If the standard deviation of the 10 scores is 20.50 and the arithmetic mean is 73, how many scores are greater than one standard deviation above the mean of the 10 scores?
(A) None (B) One (C) Two (D) Three (E) Four 184.
A set consists of 20 numbers. If n is a number in the list and is four times the average (arithmetic mean) of the numbers in the list, excluding itself, then n is what fraction of the sum of the 20 numbers in the list?
(A) 1 20 (B) 4 23 (C) 1 5 (D) 1 10 (E) 5 11 © 1999–2024 Manhattan Review www.manhattanreview.com 50 Quantitative Reasoning Question Bank −PS Questions 185.
If the average (arithmetic mean) of 3, 8, and w is greater than or equal to w and smaller than or equal to 3w, how many integer values of w exist?
(A) Five (B) Four (C) Three (D) Two (E) One 186.
If the average (arithmetic mean) of seven distinct positive integers is 14, what is the least possi-ble value of the greatest of the seven numbers?
(A) 14 (B) 17 (C) 18 (D) 20 (E) 77 187.
If the average (arithmetic mean) of x, y and 10 is equal to the average of x, y, 10 and 20, what is the sum of x and y?
(A) 40 (B) 50 (C) 55 (D) 60 (E) 65 188.
A set of 13 different integers has a median of 20 and a range of 20. What is the greatest possible value of the integer in the set?
(A) 23 (B) 27 (C) 30 (D) 34 (E) 40 189.
The mean of the set of seven positive integers 1, 2, 3, 4, 5, 6, and x is 2x 7 . What is the value of x?
(A) 1 (B) 7 (C) 14 (D) 18 (E) 21 190.
A company has a total of x employees, such that no two employees have the same annual salary.
The annual salaries of the x employees are listed in increasing order, and the 22nd salary in the list is the median of their annual salaries. If the sum of the annual salaries of all the employees is $860,000, what is the average (arithmetic mean) of the annual salaries of all the employees?
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 51 (A) $19,500 (B) $20,000 (C) $25,000 (D) $30,000 (E) $32,500 191.
The table below gives information about the electricity consumption of four appliances in a household. What is the average number of watts of electricity used per hour per appliance in the household?
Electricity usage in the household Appliance Number of hours in use Number of watts of electricity used per hour Computer 4 105 Music system 2 90 Refrigerator 2 235 LED TV 2 150 (A) 76 (B) 105 (C) 137 (D) 187 (E) 303 192.
The heights a large population have a distribution that is symmetric about the mean ¯ x. If 68 percent of the distribution lies within one standard deviation d of the mean, what percent of the distribution is greater than (¯ x −d)?
(A) 16% (B) 50% (C) 68% (D) 84% (E) 95% © 1999–2024 Manhattan Review www.manhattanreview.com 52 Quantitative Reasoning Question Bank −PS Questions 2.14 Linear Equations 193.
A seller mistakenly reversed the digits of a customer’s amount of change and returned an incor-rect amount of change. If the customer received 63 cents more than he should have, which of the following could be the correct amount of change, in cents, he should have got?
(A) 89 (B) 36 (C) 38 (D) 37 (E) 29 194.
A merchant sold screwdrivers for $11 each and spanners for $3 each. If a customer purchased a few screwdrivers and spanners for a total of $109, what would be the total number of screw-drivers and spanners the customer purchased?
(A) 10 (B) 13 (C) 15 (D) 22 (E) 32 195.
If x + y + z = 2, and x + 2y + 3z = 6 and y ̸= 0, then what is the value of x y ?
(A) −1 2 (B) −1 3 (C) −1 6 (D) 1 3 (E) 1 2 196.
A stationary shop sells a book for $25 per piece and a notebook for $15 per piece.
In the previous month, it sold 2 more books than notebooks. If the total revenue from the sale of books and notebooks in the previous month was $490, what was the total number of books and notebooks that the shop sold in the previous month?
(A) 18 (B) 20 (C) 24 (D) 30 (E) 33 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 53 2.15 Quadratic Equations & Polynomials 197.
If a ≥0 and a = √ 8ab −16b2, then in terms of b, a = (A) −4b (B) b 4 (C) b (D) 4b (E) 4b2 198.
What is the difference between the maximum and the minimum value of x y for which (x −2)2 = 9 and y −3 2 = 25?
(A) −15 8 (B) 3 4 (C) 9 8 (D) 19 8 (E) 25 8 199.
If x and y are positive integers and 2x + 3y + xy = 12, what is the value of (x + y)?
(A) 2 (B) 4 (C) 5 (D) 6 (E) 8 200.
A ball thrown up in air is at a height of h feet, t seconds after it was thrown, where h = −3(t −10)2 + 250. What is the height of the ball once it reached its maximum height and then descended for 7 seconds?
(A) 96 feet (B) 103 feet (C) 164 feet (D) 223 feet (E) 250 feet © 1999–2024 Manhattan Review www.manhattanreview.com 54 Quantitative Reasoning Question Bank −PS Questions 2.16 Inequalities 201.
Suzy’s college is 12 kilometers from her hostel. She travels 6 kilometers from the college to bas-ketball practice, and from there, 4 kilometers for the computer class. If she is then D kilometers away from her hostel, what is the range of possible values for D?
(A) 1 ≤D ≤5 (B) 2 ≤D ≤6 (C) 2 ≤D ≤10 (D) 2 ≤D ≤22 (E) 4 ≤D ≤24 202.
2a + b = 12, and |b| ≤12 How many ordered pairs (a, b) are solutions of the above system such that a and b both are integers?
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13 203.
If the cost of 15 pencils varies between $3.60 and $4.80 and the cost of 21 pens varies between $33.30 and $42.90, then the cost of 5 pencils and 7 pens varies between (A) $8.20 and $12.20 (B) $8.30 and $10.20 (C) $10.20 and $16.30 (D) $12.30 and $15.90 (E) $13.30 and $16.60 204.
Given that x is a negative number and 0 < y < 1, which of the following is the greatest?
(A) x2 (B) xy 2 (C) x y !2 (D) x2 y (E) x2y 205.
David traveled from City A to City B in 5 hours, and his average speed was between 20 miles per hour and 30 miles per hour, while Mark traveled from City A to City B along the same route in 3 hours, and his average speed was between 40 miles per hour and 60 miles per hour. Which of the following could be the distance, in miles, from City A to City B?
(A) 105 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 55 (B) 135 (C) 155 (D) 160 (E) 165 © 1999–2024 Manhattan Review www.manhattanreview.com 56 Quantitative Reasoning Question Bank −PS Questions 2.17 Co-ordinate geometry 206.
In the coordinate plane, a diameter of a circle has the end points (−3, −6) and (5, 0). What is the area of the circle?
(A) 5π (B) 10 √ 2π (C) 25π (D) 50π (E) 100π 207.
A straight line in the XY-plane has a slope of 3 and a Y-intercept of 4. On this line, what is the X-coordinate of the point whose Y-coordinate is 10?
(A) 2 (B) 4 (C) 6 (D) 7 (E) 7.5 208.
In the XY-plane, a line l passes through the origin and has a slope 3. If points (1, a) and (b, 2) are on the line l, what is the value of a b ?
(A) 2 (B) 3 (C) 2 3 (D) 2 9 (E) 9 2 209.
In the XY-plane, the point (3, 2) is the center of a circle. The point (−1, 2) lies inside the circle, and the point (3, −4) lies outside the circle. Which of the following could be the value of r?
(A) 5 (B) 4 (C) 3 (D) 2 (E) 1 210.
In the XY-plane, the three vertices of a square are represented by points (a, b), (a, −b), and (−a, −b). If a < 0 and b > 0, which of the following points is in the same quadrant as the fourth vertex point of the square?
(A) (−2, −6) (B) (−2, 6) (C) (2, −6) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Questions 57 (D) (6, −2) (E) (6, 2) 211.
In the XY-plane, the vertices of a triangle have coordinates (0, 0), (5, 5) and (10, 0). What is the perimeter of the triangle?
(A) 12 (B) 13 (C) 5 + 10 √ 2 (D) 10 + 5 √ 2 (E) 10 + 10 √ 2 212.
If the points (a, 0) , (0, b) and (1, 1) are collinear, what is the value of a in terms of b?
(A) b −1 b (B) b b + 1 (C) b b −1 (D) b + 1 b (E) 1 b −1 213.
In the XY-plane, what is the area of the triangle formed by the line 3y −4x = 24 and the X and Y axes?
(A) 6 (B) 14 (C) 24 (D) 36 (E) 48 © 1999–2024 Manhattan Review www.manhattanreview.com 58 Quantitative Reasoning Question Bank −PS Questions www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 3 Data Sufficiency Question Bank 59 60 Quantitative Reasoning Question Bank – DS Questions Data Sufficiency For most of you, Data Sufficiency (DS) may be a new format. The DS format is very unique to the GMAT exam. The format is as follows: There is a question stem followed by two statements, labeled statement (1) and statement (2). These statements contain additional information.
Your task is to use the additional information from each statement alone to answer the question. If none of the statements alone helps you answer the question, you must use the information from both the statements together. There may be questions which cannot be answered even after combining the additional information given in both the statements. Based on this, the question always follows standard five options which are always in a fixed order.
(A) Statement (1) ALONE is sufficient, but statement (2) ALONE is not sufficient to answer the ques-tion asked.
(B) Statement (2) ALONE is sufficient, but statement (1) ALONE is not sufficient to answer the ques-tion asked.
(C) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
(D) EACH statement ALONE is sufficient to answer the question asked.
(E) Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 61 3.1 Numbers 214.
If M is a positive integer, and M$ is defined by M$ = M2 −2, what is the value of M?
(1) 14 < M$ < 34 (2) M$ is an odd integer.
215.
Gold Gym has a few members in two batches, A and B. It can divide batch A members into eight groups of x members each. However, if it divides batch B members into four groups of y members each, three members will be left. How many members does the gym have?
(1) x = y −1 2 (2) The number of members in batch B is seven more than that in batch A.
216.
For a positive integer p, the index-3 of p is defined as the greatest integer n such that 3n is factor of p. For example, the index-3 of 162 is 4 as 4 is the greatest exponent of 3 and is a factor of 162. If q and r are positive integers, is the index-3 of q greater than the index-3 of r?
(1) q −r > 0 (2) q r is a multiple of 3 217.
How many distinct positive factors does the integer m have?
(1) m = p3q2, where p and q are distinct prime numbers.
(2) The only prime factors of m are 2 and 3.
218.
If √m is an integer, what is the value of √m?
(1) 13 ≤m ≤16 (2) 3 ≤√m ≤4 219.
If xy 4 3 = 3 √ 432, is x + y = 5?
(1) y is a positive integer.
(2) x is an integer.
220.
If y ̸= 0, what is the value of 5x y 2 ?
(1) x = 3 (2) 5x −2y = 0 221.
If |x + 3| = 2, what is the value of x?
(1) x < 0 (2) x2 + 6x + 5 = 0 222.
If |x + 3| = 3, what is the value of x?
(1) x2 ̸= 0 (2) x2 + 6x = 0 © 1999–2024 Manhattan Review www.manhattanreview.com 62 Quantitative Reasoning Question Bank – DS Questions 223.
If 2 < m < 3, is the tenth digit of the decimal representation of m equal to 8?
(1) m + 0.01 < 3 (2) m + 0.05 > 3 224.
If a and b are integers, is b even?
(1) 5a + 6b is even.
(2) 5a + 3b is even.
225.
If x, y, p, and q are positive integers, is xp a factor of yq?
(1) x is a factor of y.
(2) p < q + 1 226.
For the list {−8, −2, 0, 10, X}, what is the value of the integer X?
(1) The product of the five integers in the list is 0.
(2) The sum of the four integers, except X, divided by X is 0.
227.
If P, Q, and R are points on the number line, not necessarily in this order, what is the distance between Q and R?
(1) The points P and Q are 20 units apart.
(2) The points P and R are 25 units apart.
228.
If m is a positive integer, what is the remainder when 2m is divided by 10?
(1) m divided by 10 leaves a remainder of 0.
(2) m divided by 4 leaves a remainder of 0.
229.
If a, b, and c are integers, is (a −b −c) even?
(1) a and b are even and c is odd.
(2) a, b and c are consecutive integers.
230.
If m, n, and p are positive integers and 4m + 5n = p, do p and 10 have a common factor other than 1?
(1) m is a multiple of 5.
(2) n is a multiple of 5.
231.
If p and q are integers, what is the value of (p + q)?
(1) pq = 6 (2) p + q 2 = 49 232.
If m and n are integers and p = 13m + 25n, is p odd?
(1) Only one between m and n is odd.
(2) n is even.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 63 233.
If positive integers a, b, and c are such that a < b < c, is the product of a, b, & c even?
(1) c −b = b −a (2) c −16 = a 234.
Is x 3 an integer?
(1) x is an integer.
(2) x 6 is an integer.
235.
If x and y are positive integers, is x y an even integer?
(1) y is divisible by 4.
(2) x is divisible by 8.
236.
If n is a positive integers, is n 6 an integer?
(1) n is a product of three consecutive integers.
(2) n is a multiple of 3.
237.
If n is a positive integer, what is the remainder when n2 −1 is divided by 24?
(1) n is not a multiple of 2.
(2) n is not a multiple of 3.
238.
If n is a positive integer, is (n −1) n (n + 1) divisible by 4?
(1) n is an odd integer.
(2) n (n + 1) is divisible by 6.
239.
Is a positive integer x odd?
(1) 5x is odd.
(2) (x + 5) is even.
240.
What is the value of a positive integer x?
(1) x divided by 3 leaves the remainder 2.
(2) x2 divided by 3 leaves the remainder 1.
241.
For a positive integer x, what is the value of the hundreds digit of 30x?
(1) x ≥3.
(2) x 3 is an integer.
242.
If x is an integer that lies between 100 and 200, inclusive, what is the value of x?
(1) x is a multiple of 36.
(2) x is an even multiple of 45.
© 1999–2024 Manhattan Review www.manhattanreview.com 64 Quantitative Reasoning Question Bank – DS Questions 243.
If 2 < x < 6, what is the value of x?
(1) 15 is a multiple of x.
(2) 21 is a multiple of x.
244.
If m is an integer and xm = 1 xm , what is the value of x?
(1) x is an integer.
(2) m in a non-zero integer.
245.
If x is a two-digit number, is x less than 85?
(1) The sum of the two digits of x is prime.
(2) Each of the two digits of x is prime.
246.
If n is an integer, is n 13 an integer?
(1) 5n 13 is an integer.
(2) 3n 13 is an integer.
247.
If x is an integer, is 10x ≤ 1 1,000 ?
(1) x ≤−2 (2) x > −4 248.
If a, b, c, and d are non-zero integers, is a b = c d?
(1) c = 5a and d = 5b (2) 5a = 4b and 5c = 4d 249.
If a, b, and c are prime numbers, what is the value of a3b3c3?
(1) a3bc = 2, 457 (2) b = 7 250.
If the positive integer x is a multiple of 24 and the positive integer y is a multiple of 21, is 648 a factor of x2y?
(1) x is a multiple of 8.
(2) y is a multiple of 18.
251.
If r is the remainder when 18 is divided by n, where 2 < n < 18, what is the value of r?
(1) n > 15 (2) n = 2m, where m is a positive integer.
252.
If x is a positive number, is y also a positive number?
(1) y ≤x (2) y ≥x www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 65 253.
If x and y are integers, is x divisible by 3?
(1) xy is divisible by 9.
(2) y is divisible by 3.
254.
If x and y are non-zero integers, is x y an integer?
(1) (x −1) y + 1 y −1 = 1 (2) x −y = 2 255.
If x and y are integers, what is the value of y?
(1) yx = y (2) x > 1 256.
If p is a constant and an−1 + an = pn (n −1) for all positive integers n, what is the value of p ?
(1) a31 −a29 = 120 (2) a2 = 6 257.
If P, Q, & R are numbers on the number line, not necessarily in that order, is |P −R| ≥13?
(1) |P −Q| = 65 (2) |Q −R| = 52 258.
If the sum of three positive integers even?
(1) The sum of the first and the second integer is even.
(2) The sum of the second and the third integer is even.
259.
If a, b, & c are positive numbers such that c = 10a + 12b and a + b = 1, is c > 11?
(1) a > 1 2 (2) b < 1 2 260.
Set S is a set of 14 consecutive integers. Is 7 a member of the set?
(1) The integer −5 is present in the set.
(2) The integer 6 is present in the set.
261.
Sequence S is such that the difference between a term and its previous term is constant and has 250 terms. What is the 200th term of sequence S?
(1) The 150th term of Sequence S is 305.
(2) The 100th term of Sequence S is −95.
262.
If the digit h is the hundredths digit in the decimal number n = 0.3h7, what is the value of n rounded to the nearest tenth digit?
(1) n < 7 20 © 1999–2024 Manhattan Review www.manhattanreview.com 66 Quantitative Reasoning Question Bank – DS Questions (2) h < 5 263.
What is the value of a positive integer x?
(1) x has exactly two distinct factors.
(2) The difference between any two distinct factors of x is odd.
264.
If the product of the digits of the two-digit positive integer n is 12, what is the value of n?
(1) n can be expressed as the sum of two perfect squares in exactly one way.
(2) n is smaller than 40.
265.
If the sum of three integers is divisible by 2, is their product divisible by 4?
(1) The three integers are same.
(2) The product of the three integers is divisible by 2.
266.
If the units digit of a three-digit positive number X is other than 0, what is the tens digit of X?
(1) The tens digit of the number (X + 9) is 3.
(2) The tens digit of the number (X + 3) is 2.
267.
If x ̸= y, is x = 0?
(1) xy = x2 (2) y ̸= 0 268.
If none of x, y, & z are equal to 0, is x4y5z6 > 0?
(1) y > x4 (2) y > z5 269.
If x and y are integers, is y an odd integer?
(1) y y + 2 = x(x + 1) (2) x is not an even integer.
270.
If a and b are integers, what is the value of (8a6b −2)?
(1) a2b = 24 (2) ab = 22 271.
If x and y are positive integers and 18 is a multiple of xy2, what is the value of y?
(1) x is a factor of 54 and is less than half of 54.
(2) y is a multiple of 3.
272.
If x and y are positive integers and x2y = x4y−6, what is the value of y2x?
(1) x2 = 4 (2) x < 3 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 67 273.
If x and y belong to the set {2, 4}, and xky = x(ly2−8), is kl > 2?
(1) k = −6 (2) 3l −k = 3 274.
If x and y are non-zero integers, what is the value of x2y −1 ?
(1) |x| + y = 5, where 1 < |x| < y (2) x2 −4 + y −3 = 0 275.
If x and y are positive integers and r is the remainder when (74x+3 + y) is divided by 10, what is the value of r?
(1) x = 10 (2) y = 2 276.
If x and y are positive integers, what is the value of y?
(1) y −x = 3 (2) x and y are prime numbers.
277.
If x and y are positive integers, what is the value of x ?
(1) 3x + 5y = 134 (2) y = 3 278.
If x and y are positive integers, is xy a multiple of 18?
(1) x is a multiple of 9.
(2) y is a multiple of x.
279.
If x and y are positive integers, is (x + y)(x −y) a prime number?
(1) x is the smallest prime number.
(2) y2 is the smallest prime number.
280.
If x is a positive integer, does the remainder, when (3x + 2) is divided by 100, have 1 as the units digit?
(1) x = 2(2n + 1), where n is a positive integer.
(2) 4 < x < 10 281.
If x, y, and z are positive integers, is xz odd?
(1) x(2y −1) is even (2) x(x + z) is even 282.
If x, y, and z are positive integers, is y −x > 0?
(1) y x = z y (2) z > x © 1999–2024 Manhattan Review www.manhattanreview.com 68 Quantitative Reasoning Question Bank – DS Questions 283.
If z is positive, is x −y a positive number?
(1) xy + z = 0 (2) x(x −2) = 0 284.
If x & y are integers and y = x2 + x3, is y < 0?
(1) x < 0 (2) y < 1 285.
In the decimal representation of d, where 0 < d < 1, is the tenths digit of d greater than 0?
(1) 12d is an integer.
(2) 6d is an integer.
286.
In the sequence of non-zero numbers t1, t2, t3, . . . tn, . . . , the value of t(n+1) = tn 3 , for all positive integers n. What is the value of t5?
(1) t2 = 1 3 (2) t2 −t5 = 26 81 287.
Is 3x > 100?
(1) 3 √x = 9 (2) 1 3x > 0.01 288.
Is |x| < 1?
(1) |x + 2| = 3 |x −1| (2) |2x −5| ̸= 0 289.
Is q p −3 2 = 3 −p ?
(1) p < p (2) 3 > p 290.
Is m n < mn?
(1) mn is positive (2) n < −1 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 69 3.2 Percents 291.
A broker charges a brokerage which is a fixed percent of the value of a property. The brokerage was what percent of the value of the property?
(1) The property values $1.8 million.
(2) The broker charged $3,000 as the brokerage.
292.
Do at least 24 percent of Loral business school students aspire to do masters in economics?
(1) In the school, the ratio of male students to female students is 6 : 11.
(2) In the school, of the total number of students, 35 percent of the male students and 25 percent of female students aspire to do the masters in economics.
293.
By what percent was the price of a smartphone increased?
(1) The price of the smartphone was increased by $40.
(2) The price of the smartphone after the increase was $400.
294.
Did John pay less than a total of $d dollars for the phone?
(1) The price John paid for his phone was $0.85d, excluding the 20 percent sales tax.
(2) The price John paid for his phone was $170, excluding the 20 percent sales tax.
295.
Does Suzy have 1 3 more marbles than George?
(1) The number of marbles George has is 75 percent of the number of marbles Suzy has.
(2) The number of marbles Suzy has is 133.33% percent of the number of marbles George has.
296.
A salesperson is paid a fixed monthly salary of $2,000 and a commission equal to 15 percent of the amount of total sales that month over $10,000. What was the total amount paid to the salesperson last month?
(1) The total amount the salesperson was paid last month is equal to 17.5 percent of the amount of total sales last month.
(2) The salesperson’s total sales last month was $20,000.
297.
Every month Tim receives a fixed salary of $1,000 and a 10 percent commission on the total sales exceeding $10,000 in that month. What was the total amount of Tim’s sales last month?
(1) Last month Tim’s fixed salary and commission was $1, 500.
(2) Last month Tim’s commission was $500.
298.
The total cost for air conditioning consists of the cost of an air conditioner and the cost of installation. A fixed sales tax of 10% is charged on both the cost of the air conditioner and installation. If the cost of the air conditioner, excluding sales tax, was $600, what was the total amount of the air conditioner and installation, including sales tax?
(1) The sales tax on installation cost was $6.
(2) The total sales tax was $66.
© 1999–2024 Manhattan Review www.manhattanreview.com 70 Quantitative Reasoning Question Bank – DS Questions 299.
What percent of juice bottles are labeled correctly? That is, Guava Juice labelled bottles have guava juice and Orange Juice labelled bottles have orange juice.
(1) Of Guava Juice labelled bottles, 20 percent have orange juice.
(2) 80 percent of the bottles are labeled orange Juice.
300.
From 2001 to 2010, what was the percent increase in total sales revenue of Company X?
(1) Total sales revenue of Company X in 2001 was 20 percent of the industry’s sales revenue in 2001.
(2) Total sales revenue of Company X in 2010 was 25 percent of the industry’s sales revenue in 2010.
301.
By what percent the sales revenue of Company X increased from 2001 to 2005?
(1) In each of the two years, 2001 and 2005, the sales revenue of Company X was 20 percent of the total sales revenue of the industry in respective years.
(2) In 2005, the total sales revenue of the industry was 20 percent more than that in 2001.
302.
What was the ledger balance in the saving bank account on January 31?
(1) Had the increase in the ledger balance, from January 1 to January 31, in the saving bank account been 15 percent, the ledger balance in the account on January 31 would have been $1,150.
(2) From January 1 to January 31, the increase in the ledger balance in the saving bank account was 10 percent.
303.
Mark’s net income equals his salary less taxes. By what percent did Mark’s net income increase or decrease on January 1, 2016?
(1) Mark’s salary increased by 10 percent on January 1, 2016.
(2) Mark’s taxes increased by 15 percent on January 1, 2016.
304.
How many male teachers in a school of 80 teachers have masters degree?
(1) 50 percent of all the teachers in the school have masters degree.
(2) 50 percent of all the teachers in the school are male.
305.
If the number of students in School A and School B in 2015 were each 10 percent higher than their respective number of students in 2014, what was School A’s number of students in 2014?
(1) The sum of School A’s and School B’s number of students in 2014 was 1,000.
(2) The sum of School A’s and School B’s number of students in 2015 was 1,100.
306.
Is 25% of n greater than 20% of the sum of n and 1 2?
(1) 0 < n < 1 (2) n > 0.5 307.
If x and z are positive, is 100% of x equal to 33.33% percent of z?
(1) z is 200% greater than x.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 71 (2) x is 75% less than (x + z).
308.
If, for an office, the total expenditure for computers, software, and printers was $54,000, what was the expenditure on computers?
(1) The expenditures for printers were 30 percent greater than the expenditures for software.
(2) The total of the expenditures for software and printers was 65 percent less than the expen-ditures for computers.
309.
In 2001, John paid 5 percent of his taxable income as taxes. In 2002, what percent of his taxable income did he pay as taxes?
(1) In 2001, John’s taxable income was $40,000.
(2) In 2002, John paid $250 more in tax than he did in 2001.
310.
In 2001, Joe paid 5.1 percent of his income in taxes. In 2002, did Joe pay less than 5.1 percent of his income in taxes?
(1) From 2001 to 2002, Joe’s income increased by 10 percent.
(2) Taxes paid in 2002 are equivalent to 3.4 percent of Joe’s income in 2001.
311.
In 2005, there were 1,050 students at a school. If the number of students at the school increased by 50 percent from 1995 to 2000, by what percent did the number of students at the school increase from 2000 to 2005?
(1) The number of students increased by 110 percent from 1995 to 2005 at the school.
(2) There were 500 students in 1995 at the school.
312.
In 2001, what was the ratio of the number of employees in Company A to the number of em-ployees in Company B?
(1) In 2001, Company A had 60 percent more employees than Company B had in 2000.
(2) In 2001, Company B had 20 percent more employees than it had in 2000.
© 1999–2024 Manhattan Review www.manhattanreview.com 72 Quantitative Reasoning Question Bank – DS Questions 3.3 Profit & Loss 313.
If a shopkeeper purchased an item at a cost of x dollars and sold it for y dollars, by what percent of its cost did he make profit?
(1) y −x = 60 (2) 5y = 6x 314.
A used car reseller was paid a total of $5,000 for a used car. The reseller’s only costs for the car were for buying the used car and repairing it. Was the reseller’s profit from selling the car more than $1,500?
(1) The reseller’s total cost was three times the cost of buying the car.
(2) The reseller’s profit was more than the cost of repairing the car.
315.
A shopkeeper offered discounts on the sale price of a book and the sale price of a notebook.
Was the discount in dollars on the book not equal to that on the notebook?
(1) The percent discount on the book was 10 percentage points greater than the percent dis-count on the notebook.
(2) The original sale price of the book was $1 less than the original sale price of the notebook.
316.
A trader purchased a Type A gas stove and a Type B gas stove for an equal sum and then sold them at different prices. The trader’s gross profit on the Type A gas stove was what percent greater than its gross profit on the Type B gas stove?
(1) The price at which the trader sold the Type A gas stove was 10 percent greater than the price at which the trader sold the Type B gas stove.
(2) The trader’s gross profit on the Type B gas stove was $50.
317.
If the marked price of a bike was $6,250, what was the cost of the bike to the trader?
(1) The cost price when raised by 25 percent was equal to the marked price.
(2) The bike was sold for $5,500, which was 10 percent more than the cost to the trader.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 73 3.4 Averages (including weighted averages) 318.
If a clerk enters a total of 24 amounts on an MS-Office Excel sheet that has six rows and four columns, what is the average of all the amounts?
(1) The sum of averages of the amounts in six rows is 720.
(2) The sum of averages of the amounts in four columns is 480.
319.
All 50 employees in Company X take one of the two courses, NLP and HLP. What is the average (arithmetic mean) age of the employees in the company?
(1) In the company, the average age of the employees enrolled for the NLP course is 40.
(2) In the company, the average age of the employees enrolled for the HLP course is 3 4 of the average age of the employees enrolled for the NLP course.
320.
Department X of a factory has 100 workers. What is the average (arithmetic mean) annual wage of the workers at the factory?
(1) The average annual wage of the workers in Department X is $15,000.
(2) The average annual wage of the workers at the factory other than those in Department X is $20,000.
321.
Last month, a computer dealer sold a few desktop computers and a few laptop computers. What was the average (arithmetic mean) sale price of the computers sold by the dealer last month?
(1) The average sale price of the desktop computers sold by the dealer last month was $800.
(2) The average sale price of the laptop computers sold by the dealer last month was $1,100.
322.
If Dave’s average (arithmetic mean) score on three tests was 74, what was his lowest score?
(1) Dave’s highest score was 82.
(2) The sum of Dave’s two highest scores was 162.
323.
A group of 20 friends went out for lunch. Five of them spent $21 each and each of the rest spent $x less than the average of all of them. Is the the average amount spent by all the friends $12?
(1) x = 3 (2) The total amount spent by all the friends is $240.
© 1999–2024 Manhattan Review www.manhattanreview.com 74 Quantitative Reasoning Question Bank – DS Questions 3.5 Ratio & Proportion 324.
A teacher distributed a number of candies, cookies, and toffees among the students in the class.
How many students were there in the class?
(1) The numbers of candies, cookies, and toffees that each student received were in the ratio 3 : 4 : 5, respectively.
(2) The teacher distributed a total of 27 candies, 36 cookies, and 45 toffees.
325.
At the beginning of the session, a class of MBA (Finance) and a class of MBA (Marketing) of a college each had n candidates. At the end of the session, 6 candidates left MBA (Finance) course and 4 candidates left MBA (Marketing) course. How many candidates did MBA (Finance) course have at the beginning of the session?
(1) The ratio of the total number of candidates who left at the end of the session to the total number of candidates at the beginning of the session was 1 : 5.
(2) At the end of the session, 21 candidates remained on MBA (Marketing) course.
326.
Tub A and Tub B contain milk, Tub A was partially full, and Tub B was half full. If all of the milk in Tub A was poured into Tub B, then what fraction of the capacity of Tub B was filled with milk?
(1) Tub A was one-third full, when the milk from it was poured into Tub B.
(2) Tub A and Tub B have the same capacity.
327.
A bag has red, blue, green, and yellow marbles in the ratio 6 : 5 : 2 : 2. How many green marbles are there in the bag?
(1) There are 2 more red marble than blue marbles.
(2) The bag has a total of 30 marbles.
328.
How many milliliters of Chemical X were added to the Chemical Y in a vessel?
(1) The amount of Chemical X that was added was 2 3 times the amount of Chemical Y in the vessel.
(2) There was 60 milliliters of Chemical Y in the vessel.
329.
If no worker of Company X who worked in it last year quit, how many workers does the company have now on its payroll?
(1) Last year the ratio of the number of male workers to the number of female workers was 2 to 5.
(2) Since last year, Company X recruited 300 new male workers and no new female workers, raising the ratio of the number of male workers to the number of female workers to 2 to 3.
330.
If Steve and David each bought some candies, did Steve buy more candies than David?
(1) Steve bought 3 5 of the total number of candies they bought together.
(2) They together bought a total of 50 candies.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 75 331.
The ratio of the number of male and the number of female workers in a company in 2002 was 3 : 4. Was the percent increase in the number of men greater than that in the number of women from 2002 to 2003?
(1) The ratio of the number of male workers in 2002 to that in 2003 was 3 : 5.
(2) The ratio of the number of male and female workers in 2003 was 10 : 7.
332.
In Company X, are more than 1 4 of the employees over 55 years of age?
(1) Exactly 40 percent of the female employees are over 50 years of age, and, of them, 2 5 are over 55 years of age.
(2) Exactly 20 male employees are over 55 years of age.
333.
In a professional club, are more than 1 3 rd of the members mechanical engineers? Only those who are engineers can be mechanical engineers.
(1) Exactly 75 percent of the female members are engineers, and, of them, 1 3 rd are mechanical engineers.
(2) Exactly 30 percent of the male members are engineers.
334.
What is the length of the line AD? A B C D (1) AC = 10, BD = 15 (2) AB BC = BC CD © 1999–2024 Manhattan Review www.manhattanreview.com 76 Quantitative Reasoning Question Bank – DS Questions 3.6 Mixtures 335.
Two containers contain milk and water solutions of volume x liters and y liters, respectively.
What would be the minimum possible concentration of milk in either of the containers so that when the contents of both the containers are mixed, 30 liters of 80 percent milk solution is obtained?
(1) x = 2y (2) x = y + 10 336.
From a cask containing y liters of only milk, x liters of content is drawn and z liters of water is added. This process is repeated one more time. What is the fraction of milk finally present in the mixture in the cask?
(1) x = 20 & y = 100 (2) x and z form 20% and 10% of y, respectively 337.
Three friends, A, B and C decided to have a beer party. If each of the three friends consumed equal quantities of beer, and paid equally for it, what was the price of one beer bottle?
(1) A, B and C brought along 4, 6 and 2 bottles of beer, respectively; all bottles of beer being identical.
(2) C paid a total of $16 to A and B for his share.
338.
What is the volume of milk present in a mixture of milk and water?
(1) When 2 liters of milk is added to the mixture, the resultant mixture has equal quantities of milk and water.
(2) The initial mixture had 2 parts of water to 1 part milk.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 77 3.7 Speed, Time, & Distance 339.
Truckers Jack and Dave drove their trucks along a straight route. If Jack made the trip in 12 hours, how many hours did it take Dave to make the same trip?
(1) Dave’s average speed for the trip was 3 5 of Jack’s average speed.
(2) The length of the route is 720 miles.
340.
How many miles is the route from Washington DC to New York?
(1) It will take 20 minutes less time to travel the entire route at an average speed of 65 miles per hour than at an average rate of 60 miles per hour.
(2) It will take 2.5 hours to travel the first half of the route at an average speed of 52 miles per hour.
341.
Suzy estimated both the distance of her trip to her hometown, in miles, and the average speed, in miles per hour. Was the estimated time within 30 minutes of the actual time of the trip?
(1) Suzy’s estimate for the distance was within 10 miles of the actual distance.
(2) Suzy’s estimate for her average speed was within 5 miles per hour of her actual average speed.
342.
Is the time required to travel d miles at r miles per hour greater than the time required to travel D miles at R miles per hour?
(1) d = D + 20 (2) r = R + 20 © 1999–2024 Manhattan Review www.manhattanreview.com 78 Quantitative Reasoning Question Bank – DS Questions 3.8 Time & Work 343.
If a lathe machine manufactures screws and bolts at a constant rate, how much time will it take to manufacture 1,000 bolts?
(1) It takes the lathe machine 28 seconds to manufacture 20 screws.
(2) It takes the lathe machine 1.5 times more time to manufacture one bolt than to manufacture one screw.
344.
If two lathe machines work simultaneously at their respective constant rates to manufacture bolts, how many bolts do they manufacture in 10 minutes?
(1) One of the machines manufactures bolts at the constant rate of 50 bolts per minute.
(2) One of the machines manufactures bolts at twice the rate of the other machine.
345.
A group of 5 equally efficient skilled workers together take 18 hours to finish a job. How long will it take for a group of 4 skilled workers and 3 apprentices to do the same job, if each skilled worker works at an identical rate and each apprentice works at an identical rate?
(1) An apprentice works at 2 3 the rate of a skilled worker.
(2) 6 apprentices and 5 skilled workers take 10 hours to complete the same job.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 79 3.9 Computational 346.
A computer dealership has a number of computers to be sold by its sales persons. How many computers are up for the sale?
(1) If each of the sales persons sells 5 of the computers, 18 computers will remain in stock.
(2) If each of the sales persons sells 4 of the computers, 28 computers will remain in stock.
347.
Employees of Company X are paid $10 per hour for an 8-hour shift in a day. If the employees are paid 1 1 4 times this rate for time worked in excess of 8 hours during any day, how many hours did employee P work today?
(1) Employee P was paid $25 more today than yesterday.
(2) Yesterday employee P worked 8 hours.
348.
A large-size battery pack contains more numbers of batteries and costs more than the popular-size battery pack. What is the cost per battery of the large-size battery pack?
(1) A large-size battery pack contains 10 more batteries than a popular-size battery pack.
(2) A large-size battery pack costs $20.
349.
A teacher distributed 105 candies to 50 students in her class, with each student getting at least one candy. How many students received only one candy?
(1) None of the students received more than three candies.
(2) Fifteen students received only two candies each.
350.
At a school, one-fourth of the teachers are male and half of the teachers are non-academic staff.
What is the number of teachers at this school?
(1) Exactly 14 male teachers are non-academic staff.
(2) There are 32 more female teachers than male teachers.
351.
At a retail shop, the price of a pencil was $0.20 more than the price of an eraser. What was the revenue from the sale of erasers at the shop yesterday?
(1) The number of erasers sold at the shop yesterday was 10 more than the number of pencils.
(2) The total revenue from the sale of pencils at the shop yesterday was $30.
352.
On the first day of last month, a magazine seller had in stock 300 copies of Magazine X, costing $4 each. During the month, the seller purchased more copies of Magazine X. What was the total amount of inventory, in dollars, of Magazine X at the end of the month?
(1) The seller purchased 100 copies of Magazine X for $3.75 each during the month.
(2) The total revenue from the sale of Magazine X was $800 during the month.
353.
A university canteen owner determined that the number of new chairs needed in the canteen is proportional to the number of new admissions in the university minus the number of pass-outs from the university. If C is the number of new chairs needed and N is the number of new admissions minus the number of pass-outs, how many new chairs did the canteen owner determined to order?
© 1999–2024 Manhattan Review www.manhattanreview.com 80 Quantitative Reasoning Question Bank – DS Questions (1) The number of new admissions minus the number of pass-outs was 100.
(2) As per the relationship determined by the canteen owner, if the number of new admissions minus the number of pass-outs were 450, then 90 new chairs would be needed.
354.
How much did it cost, per mile, for the diesel consumed by Truck T for the trip?
(1) For the trip, Truck T consumed diesel that cost $2.70 per gallon.
(2) For the trip, Truck T was driven 540 miles.
355.
On a certain week, 950 visitors chose one of days from Monday through Sunday to visit a pagoda.
If twice as many visitors chose Monday than Tuesday, did at least 100 visitors choose Sunday?
(1) None of the days was chosen by more than 150 visitors.
(2) None of the days was chosen by fewer than 75 visitors.
356.
a b c d e f g h i If the letters in the table above represent one of the numbers 1, 2, or 3 such that each of these numbers occurs only once in each row and in each column, what is the value of a?
(1) e + i = 6 (2) b + c + d + g = 6 357.
For all integers x and y, the operation △is defined by x △y = (x + 2)2 + y + 3 2. What is the value of integer t?
(1) t △2 = 74 (2) 2 △t = 80 358.
A dealer sold good for $X. If Y percent was deducted for taxes and then $Z dollars was deducted for the cost of good, what was dealer’s gross profit after the deductions?
(1) X −Z = 400 (2) XY = 11, 000 359.
If a public distribution company loses 5 percent of its monthly allotment of wheat in a month because of wastage, pilferage and theft, what is the cost in dollars to the company per month for this loss?
(1) The company’s monthly wheat allotment is 400 million tons.
(2) The cost to the company for each 10,000 tons of wheat loss is $5.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 81 360.
If Suzy spends s dollars each month and Dave spends d dollars each month, what is the total amount they together spend per month?
(1) Dave spends $100 more per month than Suzy spends per month.
(2) It takes Suzy seven months to spend the same amount that Dave spends in six months.
361.
If Martin bought two one-pound pieces of same cake in a scheme, what percent of the total regular price of the two pieces did he save?
(1) Martin paid the regular price for the first piece and paid three-fourth of the regular price for the second piece.
(2) The regular price of the cake Martin bought was $10 per one-pound piece.
362.
If the symbol ‘#’ represents either addition, subtraction, multiplication or division, what is the value of 14 # 7?
(1) 25 # 5 = 5 (2) 2 # 1 = 2 363.
At the beginning of the year, Steve bought a total of x shares of stock P and David bought a total of 200 shares of stock P. If they held all of their respective shares throughout the year, and Steve’s dividends on his x shares totaled $225 in that year, what was David’s total dividend on his 200 shares in that year?
(1) In that year, the annual dividend on each share of stock P was $1.25.
(2) In that year, Steve bought a total of 180 shares of stock P.
364.
To understand the Population Density (Population divided by total area of a region), in persons per square kilometers, of a country, the population and the total area, in square kilometers, were estimated. Both the estimates had their lower and upper limits. Was the Population Density for the country greater than 500 persons per square kilometers?
(1) The upper limit for the estimate of the population was 50 million persons.
(2) The upper limit for the estimate of the total area was 90,000 square kilometers.
365.
■+ △= ∀ In the addition problem above, each of the symbols ■, △and ∀represents a positive digit. If ■< △, what is the value of △?
(1) 3 < ∀< 5 (2) ■< 2 © 1999–2024 Manhattan Review www.manhattanreview.com 82 Quantitative Reasoning Question Bank – DS Questions 3.10 Interest 366.
A total of $80, 000 was invested for one year. Part of this amount earned simple annual interest at the rate of x percent per year, and the rest earned simple annual interest at the rate of y percent per year. If the total interest earned on the investment of $80, 000 for that year was $7, 400, what is the value of x?
(1) x = 5 4y (2) The ratio of the first part of amount to the second part of amount was 5 to 3.
367.
John lent one part of an amount of money at 10 percent rate of simple interest and the remaining at 22 percent rate of simple interest, both for one year. At what rate was the larger part lent?
(1) The total amount lent was $2,400.
(2) The average rate of simple interest he received on the total amount was 15 percent.
368.
A hundred dollars is deposited in a bank account that pays r percent annual interest com-pounded annually.
The amount A(t), in dollars, with interest in t years is given by A(t) = 100 1 + r 100 t . What amount will $100 be in 3 years?
(1) A(2) = 121 (2) r = 10 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 83 3.11 Functions 369.
The function f is defined by f (x) = px, where x and p are integers. What is the value of f (1)?
(1) f (2) = 81 (2) f (3) = −729 370.
For all numbers x, the function h is defined by h(x) = 2x −1, and the function g is defined by g (x) = 2x −3 5 . If k is a positive number, what is the value of g(k)?
(1) h (k) = 7 (2) h (1) = k 4 371.
If f is the function defined by f (x) = 27x for x ≥0 and f (x) = x4 for x < 0, what is the value of f (k)?
(1) |k| = 3 (2) k < 0 © 1999–2024 Manhattan Review www.manhattanreview.com 84 Quantitative Reasoning Question Bank – DS Questions 3.12 Permutation & Combination 372.
A bag contains a total of 30 only green and black balls such that the number of green balls is less than the number of black balls. If two balls are to be drawn simultaneously from the bag, how many balls in bag are green?
(1) The probability that the two balls to be drawn will be green is 3 29.
(2) The probability that the two balls to be drawn will be black is 38 87.
373.
A box contains only b black tokens, w white tokens, and g green tokens. If one token is ran-domly drawn from the box, is the probability that the drawn token will be green greater than the probability that the drawn token will be white?
(1) g(b + g) > w(b + w) (2) b > w + g 374.
In a university, there are 19 departments. 13 males and 6 females head one of the departments.
If one of the heads of the departments is selected at random, what is the probability that the head of the department selected will be a female who is pursuing Ph. D. program?
(1) Among the females, three are pursuing Ph. D. program.
(2) Among the females, three are not pursuing Ph. D. program.
375.
A bag contains only red, or green, or blue tokens. If one token is to be drawn at random, what is the probability that the token will be green?
(1) There are 10 red tokens in the bag.
(2) The probability that the token will be blue is 1 2.
376.
If two different persons are to be selected at random from a group of 10 members and p is the probability that both the persons selected will be men, is p > 0.5?
(1) The number of men is greater than the number of women.
(2) The probability that both the persons selected will be women is less than 1 10.
377.
How many employees are there in Company X?
(1) If an employee is to be chosen at random from the company, the probability that the em-ployee chosen will be a male is 4 7.
(2) There are 10 more males in the company than females.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 85 3.13 Sets 378.
In a conference, if each of the 1,230 participants ordered for either Tea or Coffee (but not both), what percent of the female participants ordered for Coffee?
(1) 70 percent of the female participants ordered Tea.
(2) 80 percent of the male participants ordered Coffee.
379.
In a survey of 320 employees, 35 percent said that they take tea, and 45 percent said that they take coffee. What percent of those surveyed said that they take neither tea nor coffee?
(1) 25 percent of the employees said that they take coffee but not tea.
(2) 400 7 percent of the employees who said that they take tea also said that they also take coffee.
380.
Is the number of clients of Company X greater than the number of clients of Company Y?
(1) Of the clients of Company X, 25 percent are also clients of Company Y.
(2) Of the clients of Company Y, 37.5 percent are also clients of Company X.
© 1999–2024 Manhattan Review www.manhattanreview.com 86 Quantitative Reasoning Question Bank – DS Questions 3.14 Statistics & Data Interpretation 381.
The principal of a school recorded the number of students in each of the 15 classes to find the mean class size. What was the standard deviation of the numbers of students in the 15 classes?
(1) The average (arithmetic mean) number of students for all the 15 classes was 30.
(2) Each classes had the same number of students.
382.
Each of the 23 mangoes in box X weighs less than each of the 22 mangoes in box Y. What is the median weight of the 45 mangoes in the boxes?
(1) The heaviest mango in box X weighs 100 grams.
(2) The lightest mango in box Y weighs 120 grams.
383.
If each of the 10 students working with an NGO received a cash prize, was the amount of each cash prize the same?
(1) The standard deviation of the amounts of the cash prizes was 0.
(2) The sum of the 10 cash prizes was $500.
384.
If the average (arithmetic mean) of seven unequal numbers is 20, what is the median of these numbers?
(1) The median of the seven numbers is equal to 1 6 of the sum of the six numbers other than the median.
(2) The sum of the six numbers other than the median is equal to 120.
385.
If the average (arithmetic mean) of four unequal numbers is 40, how many of the numbers are greater than 40?
(1) No number is greater than 70.
(2) Two of the four numbers are 19 and 20.
386.
If the average (arithmetic mean) of the scores of x students of class X is 40 and the average of the scores of y students of class Y is 30, what is the average of the scores of the students of both the classes?
(1) x + y = 60 (2) x = 3y 387.
Is the standard deviation of the scores of Class A’s students greater than the standard deviation of the scores of Class B’s students?
(1) The average (arithmetic mean) score of Class A’s students is greater than the average score of Class B’s students.
(2) The median score of Class A’s students is greater than the median score of Class B’s stu-dents.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 87 3.15 Linear Equations 388.
If a stationery shop sells notebooks in A-4 size and A-5 size paper, what is the price of a A-5 size notebook?
(1) The total price of one A-4 size and one A-5 size notebooks is $4.
(2) The total price of three A-4 size and one A-5 size notebooks is $9.
389.
If a gym charges its members a one-time registration fee of $r and a monthly fee of $m, what is the amount of the registration fee?
(1) The total charge, including the registration fee, for 12 months is $620.
(2) The total charge, including the registration fee, for 24 months is $1,220.
390.
A pencil and an eraser cost a total of $2.00. How much does the eraser cost?
(1) The pencil costs thrice as much as the eraser.
(2) The pencil costs $1.50.
391.
A gym sold only individual and group memberships. It charged a fee of $200 for an individual membership. If the gym’s total revenue from memberships was $240,000, what was the charge for a group membership?
(1) The revenue from individual memberships was 1 3 of the total revenue from memberships.
(2) The gym sold twice as many group memberships as individual memberships.
392.
At a used item shop, all caps were priced equally and all sunglasses were priced equally. What was the price of 4 caps and 5 sunglasses at the sale?
(1) The price of a cap was $2.00 more than the price of a sunglasse.
(2) The price of 8 caps and 10 sunglasses was $45.
393.
A number of bottles are packed in standard size cartons with each holding 75 bottles. If these bottles were to be packed in smaller cartons that can each hold 50 bottles, how many smaller cartons would be needed to hold all the bottles?
(1) The number of smaller cartons needed is 10 more than the standard size cartons.
(2) All the bottles are packed in 20 standard size cartons.
394.
Is 2m −3n = 0?
(1) m ̸= 0 (2) 6m = 9n 395.
An electricity distribution company charges its customers at the rate of $x per unit for the first 200 units a customer consumes in a month and at the rate of $y per unit for each unit over 200 units. What would be the charge for a customer who consumes 200 units in a month?
(1) y = 1.25x (2) If a customer consumes 210 units in a month, the company would charge $425.
© 1999–2024 Manhattan Review www.manhattanreview.com 88 Quantitative Reasoning Question Bank – DS Questions 396.
At a hotel, a buffet lunch is charged $50 for the first dish and x dollars for an additional dish.
What is the charge for additional dish?
(1) The average cost of a dish for a buffet lunch with a total of 4 dishes is $27.50.
(2) The average cost of a dish for a buffet lunch with a total of 4 dishes is $2.50 more than the corresponding cost for 6 dishes.
397.
A shopkeeper sells a pen for $1.50 and a pencil for $0.50. If last week, a total of 200 items were sold, how many of the pens were sold?
(1) Last week, total revenue from the sale of these two items was $150.
(2) The average (arithmetic mean) price per item for the 200 items sold was $0.75.
398.
For a week Jack is paid at the rate of x dollars per hour for the first t hours (t > 4) he works and $2 per hour for the hours worked in excess of t hours. If x and t are integers, what is the value of t?
(1) If Jack works (t −3) hours in one week, he will earn $14.
(2) If Jack works (t + 3) hours in one week, he will earn $23.
399.
If from 1991 to 2000, the number of students of School X tripled, how many students attended the school in 1991?
(1) From 2000 to 2009, the number of students of the school doubled.
(2) From 2000 to 2009, the number of students of the school increased by 120.
400.
How many marbles does Kevin have?
(1) If Kevin had 10 fewer marbles, he would have only half as many marbles as he actually has.
(2) Kevin has thrice as many black marbles as white marbles.
401.
How many years did Mrs. Peterson live?
(1) Had Mrs. Peterson become the professor 20 years earlier than she actually did, she would have been the professor for exactly 3 4 th of her life.
(2) Had Mrs. Peterson become the professor 20 years later than she actually did, she would have been the professor for exactly 1 4 th of her life.
402.
If x + y = 2p and x −y = 2q, what is the value of (p + q)?
(1) y = 8 (2) x = 3 403.
If x 6 = y 3 , is y = 10?
(1) x + y = 30 (2) 3x = 60 404.
In which year was Chris born?
(1) Chris’s brother Kevin who is 5 years older than Chris was born in 1990.
(2) In 2007, Kevin turned 17 years old.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 89 3.16 Quadratic Equations & Polynomials 405.
How many more boys than girls are in the class?
(1) There are a total of 10 boys and girls in the class.
(2) The number of boys in the class equals the cube of the number of girls in the class.
406.
If a2 + b2 = 1, is (a + b) = 1?
(1) ab = 0 (2) b = 0 407.
If x ̸= y, is x + y −xy = 0?
(1) (1 −x) 1 −y = 1 (2) x + y x −y = xy x −y 408.
If a (a −5) (a + 2) = 0, is a negative?
(1) a (a −7) ̸= 0 (2) a2 −2a −15 ̸= 0 409.
If ab ̸= 0, what is the value of 1 a + 1 b ?
(1) a + b = −1 (2) ab = 6 (a + b) 410.
If a2 −b = n, what is the value of a?
(1) n + b = 4 (2) b = 1 411.
If (n + 3) (n −1) −(n −2) (n −1) = m(n −1), what is the value of n?
(1) |m| = 5 (2) m = 5 412.
If m, n and p are constants and x2 + mx + n = x + p 2 for all values of x, what is the value of n?
(1) p = 3 (2) m = 6 413.
If x2 + 3x + c = (x + a) (x + b) for all x, what is the value of c?
(1) a = 1 (2) a and b are positive integers.
© 1999–2024 Manhattan Review www.manhattanreview.com 90 Quantitative Reasoning Question Bank – DS Questions 3.17 Inequalities 414.
Was the average (arithmetic mean) score that Steve got per subject greater than the average score that David got per subject?
(1) Twice the average score that Steve got per subject was greater than 5 less than twice the average score that David got per subject.
(2) Twice the average score that David got per subject was less than 5 more than twice the average score that Steve got per subject.
415.
If 1 5 of the larger of two positive numbers is greater than 6 times the smaller of the same two numbers, is the smaller number less than 5?
(1) The larger number is greater than 120.
(2) The larger number is less than 150.
416.
If xy ̸= 0, is x = y?
(1) |x| = |y| (2) xy > 0 417.
If abc ̸= 0, is a (b + c) ≥0?
(1) |b + c| = |b| + |c| (2) |a + b| = |a| + |b| 418.
If R = M N , is R ≤M?
(1) M > 40 (2) 0 < N ≤15 419.
If x7y4z3 < 0, is xyz < 0?
(1) z < 0 (2) x > 0 420.
If xy = 6, is x < y?
(1) y ≥3 (2) y ≤3 421.
If x < −3y 2 , is x < 0?
(1) y > 0 (2) 2x + 5y = 20 422.
If a > 0 and b > 0, is 1 a + b < 1?
(1) a b = 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 91 (2) a + b > 1 423.
If each of w, x, y and z are positive numbers, is w x × y z > y x ?
(1) y > x (2) w > z 424.
If x and y are integers, is x > 0?
(1) x + y > 60 (2) y > 58 425.
If xy < 2, is y < 1?
(1) x > 2 (2) y < 3 426.
If x is positive, is x2 < x?
(1) 1 10 < x < 2 5 (2) x3 < x2 427.
If x > 1 and y > 2, is x < y?
(1) x2 < xy + x (2) xy < y(y −1) 428.
If x ̸= 0, is x2 < |x|?
(1) x < 1 (2) x > −1 429.
If x and y are integers and y = |x + 5| + |6 −x|, is y = 11?
(1) x ≤6 (2) x ≥−5 430.
If x > 0, is y > 0?
(1) 2y < 7x (2) y > −x 431.
If x and y are integers, is x + y > 2?
(1) x2 < 1 (2) y < 1 © 1999–2024 Manhattan Review www.manhattanreview.com 92 Quantitative Reasoning Question Bank – DS Questions 432.
If x and y are positive integers and y2 = 9 −x, what is the value of y?
(1) x ≤7 (2) y ≥2 433.
If x and y are positive, is 3x > 8y?
(1) x −y > 4 (2) x > 14y 5 434.
If x and y are positive, is 8x > 5y?
(1) x2 > y2 (2) x3 > y3 435.
If x is a negative integer, is x < −3?
(1) x2 + 6x < 7 (2) x2 + |x| ≤2 436.
If x + y > 0, is xy < 0?
(1) x2y < 1 (2) x + 2y < 0 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 93 3.18 Co-ordinate geometry 437.
A circle in the XY-plane has its center at the origin. If M is a point on the circle, what is the sum of the squares of the coordinates of M?
(1) The radius of the circle is 5.
(2) The sum of the coordinates of M is 7.
438.
The equation of line L in the XY-plane is y = mx + c, where m and c are constants, what is the slope of line L?
(1) Line L is parallel to the line y = (1 −2m) x + 2c.
(2) Line L intersects the line y = 2x −4 at the point (3, 2).
439.
In the figure below, AB and BC are parallel to the X-axis and Y-axis, respectively. What is the sum of the coordinates of point B?
A B C Y X (𝑎, 𝑏) (𝑐,𝑏) (𝑐,𝑑) (1) The Y-coordinate of point C is 2.
(2) The X-coordinate of point A is −8.
440.
In the figure shown below, the circle has center at the origin O, and point A has coordinates (13, 0). If point B is any point on the circle, what is the length of line segment AB?
Y O A B (1) The X-coordinate of point B is −5.
(2) The Y-coordinate of point B is −12.
© 1999–2024 Manhattan Review www.manhattanreview.com 94 Quantitative Reasoning Question Bank – DS Questions 441.
In the XY-plane, are the points with their coordinates (a, b) and (c, d) equidistant from the origin?
(1) a + b = 2 (2) c = 1 −a and d = 1 −b 442.
In the XY-plane, does the point (m, n) lie below the line y = x?
(1) m = 4 (2) n = m + 4 443.
In the XY-plane, is the slope of the line k positive?
(1) Line k is perpendicular to the line passing though the points (1, 1) and (−2, 5).
(2) Line k makes a negative intercept on the X-axis and a positive intercept on the Y-axis.
444.
In the XY-plane, lines m and n intersect at the point (−5, 4). What is the slope of line m?
(1) The product of the slopes of lines m and n is −1.
(2) Line n passes through the origin.
445.
In the XY-plane, lines L and K are parallel. If the Y-intercept of line L is −1, what is the Y-intercept of line K?
(1) The X-intercept of line L is −1.
(2) Line K passes through the point (5, 10).
446.
In the XY-plane, the point (a, b) lies on a circle with centre at the origin. What is the value of a2 + b2 ?
(1) The circle has radius 5.
(2) The point (3, −4) lies on the circle.
447.
In the XY-plane, region X consists of all the points x, y such that 3x + 4y ≤12. Is the point (r, s) in region X?
(1) 4r + 3s = 12 (2) r ≤4 and s ≤3 448.
In the XY-plane, the line l passes through the origin and the point (m, n), where mn ̸= 0. Is n > 0?
(1) The line l has a negative slope.
(2) m < n 449.
In the XY-plane, the line with equation ax + by + c = 0, where abc ̸= 0, has slope −3. What is the value of b?
(1) a = 2 (2) c = 5 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Questions 95 450.
In the XY-plane, the sides of a rectangle are parallel to the X and Y axes. If one of the vertices of the rectangle is (−2, −3), what is the area of the rectangle?
(1) One of the vertices of the rectangle is (4, −3).
(2) One of the vertices of the rectangle is (4, 5).
© 1999–2024 Manhattan Review www.manhattanreview.com 96 Quantitative Reasoning Question Bank – Answer key www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 4 Answer key 97 98 Quantitative Reasoning Question Bank – Answer key 4.1 Problem Solving Questions (1) E (2) C (3) D (4) D (5) E (6) C (7) C (8) A (9) D (10) A (11) C (12) E (13) A (14) C (15) E (16) D (17) A (18) D (19) C (20) C (21) C (22) A (23) B (24) D (25) C (26) C (27) C (28) D (29) B (30) A (31) D (32) C (33) D (34) C (35) D (36) B (37) D (38) D (39) B (40) D (41) A (42) D (43) B (44) B (45) B (46) C (47) B (48) B (49) B (50) A (51) D (52) C (53) D (54) C (55) C (56) E (57) B (58) D (59) E (60) A (61) E (62) C (63) D (64) C (65) B (66) D (67) B (68) C (69) B (70) B (71) E (72) D (73) B (74) A (75) C (76) C (77) C (78) C (79) E (80) C (81) C (82) C (83) D (84) E (85) C (86) B (87) E (88) B (89) C (90) D (91) C (92) C (93) E (94) A (95) E (96) B (97) B (98) D (99) B (100) D (101) E (102) E (103) A (104) C (105) C (106) C (107) E (108) C (109) B (110) B www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – Answer key 99 (111) D (112) D (113) B (114) C (115) D (116) C (117) B (118) C (119) C (120) C (121) E (122) A (123) B (124) B (125) D (126) D (127) C (128) C (129) D (130) D (131) D (132) C (133) D (134) D (135) C (136) D (137) D (138) A (139) D (140) A (141) E (142) D (143) E (144) D (145) D (146) B (147) C (148) D (149) C (150) E (151) E (152) D (153) C (154) D (155) D (156) B (157) C (158) E (159) D (160) C (161) A (162) B (163) D (164) C (165) E (166) D (167) C (168) E (169) B (170) A (171) C (172) D (173) C (174) C (175) C (176) B (177) D (178) D (179) D (180) B (181) D (182) C (183) D (184) B (185) B (186) B (187) B (188) D (189) E (190) B (191) C (192) D (193) E (194) C (195) A (196) C (197) D (198) E (199) B (200) B (201) D (202) E (203) D (204) C (205) B (206) C (207) A (208) E (209) A (210) E (211) E (212) C (213) C © 1999–2024 Manhattan Review www.manhattanreview.com 100 Quantitative Reasoning Question Bank – Answer key 4.2 Data Sufficiency Questions (214) A (215) E (216) B (217) A (218) A (219) C (220) B (221) E (222) A (223) B (224) C (225) C (226) E (227) E (228) B (229) A (230) A (231) E (232) A (233) E (234) B (235) E (236) A (237) C (238) A (239) D (240) E (241) D (242) B (243) B (244) E (245) B (246) D (247) E (248) D (249) A (250) B (251) B (252) B (253) E (254) A (255) E (256) A (257) C (258) E (259) D (260) E (261) C (262) D (263) B (264) E (265) A (266) A (267) A (268) A (269) A (270) A (271) B (272) A (273) D (274) D (275) B (276) C (277) A (278) E (279) C (280) A (281) A (282) C (283) A (284) E (285) B (286) D (287) D (288) C (289) D (290) C (291) C (292) B (293) C (294) A (295) D (296) D (297) D (298) D (299) E (300) E (301) C (302) C (303) E (304) E (305) E (306) A (307) D (308) B (309) E (310) C (311) D (312) C (313) B (314) C (315) E (316) E (317) D (318) D (319) E (320) E (321) E (322) B (323) D www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – Answer key 101 (324) E (325) D (326) C (327) D (328) C (329) C (330) A (331) B (332) E (333) C (334) C (335) D (336) B (337) C (338) C (339) A (340) D (341) E (342) E (343) C (344) E (345) D (346) C (347) C (348) E (349) C (350) B (351) E (352) C (353) C (354) E (355) A (356) D (357) C (358) C (359) C (360) C (361) A (362) A (363) D (364) E (365) A (366) C (367) B (368) D (369) B (370) D (371) A (372) D (373) A (374) D (375) E (376) E (377) C (378) A (379) D (380) C (381) B (382) A (383) A (384) D (385) C (386) B (387) E (388) C (389) C (390) D (391) C (392) B (393) D (394) B (395) C (396) D (397) D (398) B (399) C (400) A (401) C (402) B (403) D (404) A (405) C (406) E (407) D (408) C (409) B (410) E (411) E (412) D (413) D (414) E (415) B (416) C (417) C (418) E (419) E (420) A (421) D (422) B (423) B (424) E (425) A (426) D (427) B (428) C (429) C (430) E (431) C (432) D (433) B (434) D (435) B (436) B (437) A (438) A (439) E (440) A (441) C (442) B (443) D (444) C (445) C (446) D (447) E (448) C (449) A (450) B © 1999–2024 Manhattan Review www.manhattanreview.com 102 Quantitative Reasoning Question Bank – Answer key www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 5 Solutions – Problem Solving Questions 103 104 Quantitative Reasoning Question Bank −PS Solutions 5.1 Number properties 1.
Here, the given expression is in the format a2 −b2 = (a + b) (a −b) => 99, 9962 −42 = (99, 996 + 4) (99, 996 −4) => 99, 9962 −42 = (99, 996 + 4) (99, 992) => 100, 000 × (100, 000 −8) = 105 × (105 −8) The correct answer is Option E.
2.
The highest exponent of 5 in n! can be calculated by adding the quotients (integer parts) when n is successively divided by 5: If [x] denotes the integer part of x, we have n 5 + n 5 5 + · · · = 6 Assuming that only n 5 equals 6, we have n = 5 × 6 = 30 However, for n = 30, the actual value of the exponent of 5 is: • 30 5 = 6 • 6 5 = 1 The exponent is 6 + 1 = 7, which is greater by 7 −6 = 1 than what was required.
Thus, we need to reduce 30 by ‘1’ multiple of 5 i.e. 1 × 5 = 5.
Thus, the approximate value of n = 30 −5 = 25.
For n = 25, the actual value of the exponent of 5 is: • 25 5 = 5 • 5 5 = 1 The exponent is 5 + 1 = 6, which is the exact exponent required.
However, since we need to find the largest value of the highest exponent of ‘7’ in n!, we need to check if any higher value of n is possible.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 105 We have obtained the highest exponent of ‘5’ equals to 6 in 25!
However, we can increase n till a point before the next multiple of 5 is included.
Thus, we can increase n to 29! as in 30 includes a multiple of ‘5.’ Let us verify: For n = 29, the actual value of the exponent of 5 is: • 29 5 = 5.8 ≡5 • 5 5 = 1 Thus the exponent is 5 + 1 = 6, which is the exact exponent required.
Thus, the maximum value of n! = 29!
Thus, for n = 29, the value of the highest exponent of 7 is: • 29 7 = 4.14 ≡4 Thus the highest exponent of ‘7’ in 29! is 4.
The correct answer is Option C.
Alternate approach: Let us count the number of multiples of ‘5’ in n!.
The multiples of ‘5’ would be there in 5, 10, 15, 20, 25, 30, ...
We should stop counting till we get 6 multiples. Since 5, 10, 15, and 20 would give one each, and 25 would give two multiples, the total count of multiples = 1 + 1 + 1 + 1 + 2 = 6 = the number of required multiples to get the greatest value of a.
We see that if n < 30, the value of a = 6, thus the greatest value of n = 29.
Now let’s count the number of multiples of ‘7’ in n! = 29!.
The multiples of ‘7’ would be there in 7, 14, 21, and 28.
There are one multiple in each of 7, 14, 21, and 28: a total of four multiples.
© 1999–2024 Manhattan Review www.manhattanreview.com 106 Quantitative Reasoning Question Bank −PS Solutions 3.
We have 0 < a < 1 Also, b = a2 and c = √a Multiplying both sides of the inequality, a < 1, by a (since a is positive, multiplying a will not change the sign of inequality): a × a < 1 × a => a2 < a Thus, we have a > a2 = b Again, since a < 1, taking square root on both sides, we have √a < 1 Multiplying both sides of the above inequality by √a, we have √a × √a < √a => a < √a = c Thus, we have a2 < a < √a => b < a < c The correct answer is Option D.
Alternate approach: For any number less than 1, its squares, cubes & higher order numbers would be less than the number, and its square roots, cube roots & nth roots would be greater than the number.
Since a = 0.999 < 1, this follows that b < a < c.
4.
Given that, p = 1 150 × 1 151 × · · · × 1 250 q = 1 150 × 1 151 × · · · × 1 250 × 1 251 = p 251 Thus, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 107 p−1 + q−1 = 1 p + 1 q = 1 p + 251 p => 252 p The correct answer is Option D.
5.
The sum of all integers from 1 to n is given by n (n + 1) 2 .
Thus, the sum of all integers from 1 to 100 = 100 × 101 2 = 50 × 101.
And, the sum of all integers from 1 to 50 = 50 × 51 2 = 25 × 51.
Thus, the sum of all integers from 51 to 100 = 50 × 101 −25 × 51 = 25(202 −51) = 25 × 151 = 3, 775.
The correct answer is Option E.
6.
x = 1 51 + 1 52 + · · · + 1 60 Let us try to find the range of values of x.
We know that x is the sum of 10 terms.
Thus, we can say: x < 1 51 + 1 51 + . . . (10 times) => x < 10 51 = 1 5.1 => x < 1 5 Again, we have x > 1 60 + 1 60 + . . . (10 terms) => x > 10 60 = 1 6 Thus, we have 1 6 < x < 1 5 © 1999–2024 Manhattan Review www.manhattanreview.com 108 Quantitative Reasoning Question Bank −PS Solutions => 6 > 1 x > 5 Similarly, we can deduce that 7 > 1 y > 6 Thus, 1 y > 1 x The correct answer is Option C.
7.
A number, here 4p25q is divisible by 4, if the number formed by the last two digits of the number is divisible by 4.
Thus, the number 5q is divisible by 4.
This is possible if q = 2 or 6 (since both 52 and 56 are divisible by 4).
A number, here number 4p25q is divisible by 9, if the sum of the digits of the number is divisible by 9.
The sum of the digits = 4 + p + 2 + 5 + q = 11 + p + q.
Thus, (11 + p + q) should be divisible by 9.
If q = 2: 11 + p + q = 11 + p + 2 = 13 + p.
Thus, (13 + p) is divisible by 9 if p = 5 (since 13 + 5 = 18, which is divisible by 9).
If q = 6: 11 + p + q = 11 + p + 6 = 17 + p.
Thus, (17 + p) is divisible by 9 if p = 1 (since 17 + 1 = 18, which is divisible by 9).
Thus we have two possible situations: p = 5, q = 2 => p q = 5 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 109 OR p = 1, q = 6 => p q = 1 6 Thus, the minimum value of p q = 1 6.
The correct answer is Option C.
8.
We know that if a% of (a −2b) when added to b% of b, the value obtained is 0.
=> a% × (a −2b) + b% × b = 0 => a 100 × (a −2b) + b 100 × b = 0 => a2 −2ab 100 + b2 100 = 0 => a2 −2ab + b2 100 = 0 => a2 −2ab + b2 = 0 => (a −b)2 = 0 => a −b = 0 => a = b Thus, only statement I is correct.
The correct answer is Option A.
9.
According to the problem: If m is in the set, (m2 + 3) is also in the set.
However, it does NOT imply that if (m2 + 3) is in the set, then m must be in the set.
What it does imply is that: If (m2 + 3) is NOT in the set, m is NOT in the set.
Thus, if we have m = −1 as a member of the set, m2 + 3 = 4 is also a member of the set.
Thus, statement II is correct.
Proceeding in the same way: Since m = 4 is a member of the set, then m2 + 3 = 19 is a member of the set.
© 1999–2024 Manhattan Review www.manhattanreview.com 110 Quantitative Reasoning Question Bank −PS Solutions Thus, statement III is also correct.
The correct answer is Option D.
10.
We have t2 = 5 t3 = 2 × t2 −1 = 2 × 5 −1 = 9 t3 = 2 × t2 −1 = 2 × 9 −1 = 17 Thus, we see that the sequence follows the pattern: t1, 5, 9, 17, ...
We can calculate the value of t1 too.
t2 = 2 × t1 −1 5 = 2 × t1 −1 t1 = 3 Again, we see that the sequence follows the pattern: 3, 5, 9, 17, ...
We can rewrite the above terms as following.
(21 + 1), (22 + 1), (23 + 1), (24 + 1)...
From the analogy, we can deduce that t10 = 210 + 1 and t9 = 29 + 1 Thus, t10 −t9 = (210 + 1) −(29 + 1) = 210 + 1 −29 −1 = 29(2 −1) = 29 The correct answer is Option A.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 111 11.
660 = 22 × 3 × 5 × 11 We see that there is an extra ‘2’ since the exponent of 2 is 2.
This extra ‘2’ can be combined with the other factors to generate different values.
Also, keeping the two 2s separate, the other factors may be combined to generate different values of m, n, p and q.
Thus, the possible combination for the distinct values of m, n, p and q are: m m m n n n p p p q q q 3 2 × 2 = 4 5 11 2 2 × 3 = 6 5 11 2 3 5 2 × 11 = 22 2 3 11 2 × 5 = 10 Thus, there are four possible combinations.
The correct answer is Option C.
12.
Looking at the expression, we can deduce that q x2/3 + y2/3 + 2(xy)1/3 and q x2/3 + y2/3 −2(xy)1/3 are basically (a + b)2 and (a −b)2, respectively, where a = x1/3 and b = y1/3 So, q x2/3 + y2/3 + 2(xy)1/3 = √ a2 + b2 + 2ab = p (a + b)2 = a + b Similarly, q x2/3 + y2/3 −2(xy)1/3 = √ a2 + b2 −2ab = p (a −b)2 = b −a; since b > a Note that x = 125 and y = 216, or, y > x; thus, y1/3 > x1/3. Or b > a So, q x2/3 + y2/3 + 2(xy)1/3 + q x2/3 + y2/3 −2(xy)1/3 = (a + b) + (b −a) = 2b = 2y1/3 = 2 × (216)1/3 = 2 × (6)(3×1/3) = 2 × 6 = 12 The correct answer is Option E.
13.
We observe that all the options for the value of 1 √m are greater than 1. This is possible only if √m is less than 1, i.e., m is less than 1.
Thus, we have For 0 < m < 1, Multiplying m throughout: © 1999–2024 Manhattan Review www.manhattanreview.com 112 Quantitative Reasoning Question Bank −PS Solutions => 0 < m2 < m Taking square root throughout: => 0 < m < √m Thus, we have √m > m Taking reciprocal: 1 √m < 1 m . . . (i) We know that m ≥0.9 => 1 m ≤ 1 0.9 = 1.11 . . . (ii) Thus, from (i) and (ii), we have 1 √m < 1.11 Only option A (1.01) is qualified in the range.
The correct answer is Option A.
Alternate approach: Say m = 1, => 1 √m = 1 √ 1 = 1 We see that as the value of m increases, the value of 1 √m decreases; however, in the options, we see that there is no less than 1 value.
Since all the values in options are more than 1, we need to probe further.
If we plug in a value for m, lying between 1 and 0.9, the value of 1 √m would be greater than 1.
Although all the option values are greater than 1, this does not mean that the problem can’t be solved.
Since this is a question of MCQ category and only one among the five options is correct, at least Option A (1.01, least among all the options) must be correct.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 113 14.
We have a = b3 90 => b3 = 90a 90 = 2 × 32 × 5 Given that b is an integer, 90a must be a perfect cube, we have a = 22 × 3 × 52 × k3, where k is a positive integer.
In that case, we have 90a = 23 × 33 × 53 × k3, which is a perfect cube.
Thus, among the statements I, II and III, only those would be an integer in which the denomina-tor is a factor of a = 22 × 3 × 52 = 300.
Only in Statement III, in which the denominator is 300, a is completely divisible by the denomi-nator and hence would be an integer.
The correct answer is Option C.
15.
Given that, X Y + Y X X X Z In the addition in the units digits, we have Y + X = Z + carry (We must have ‘carry’ since in the tens position, the same addition (X + Y) results in a different value (X)) The value of the ‘carry’ must be 1 (adding two single digits can result in a maximum carry of ‘1’) We observe that in the addition of the tens digits, i.e.
(1 + X + Y), we get a ‘carry’ to the hundreds position (since the result is a three-digit number).
Also, we have (1 + X + Y) results in the digit X in the tens position => 1 + X + Y = 10 + X => 1 + Y = 10 © 1999–2024 Manhattan Review www.manhattanreview.com 114 Quantitative Reasoning Question Bank −PS Solutions (Since only (X + 10) would result is the digit X in the tens position) => Y = 9 Since the hundreds position in the sum comes only from the ‘carry’ from the tens position, we have X = 1 Thus, the correct addition is: 1 9 + 9 1 1 1 0 Thus, the value of the units digit of the integer XXZ, i.e. Z = 0.
The correct answer is Option E.
Alternate approach: Since XY and YX are two digit positive integers, we must have: XY + YX = XXZ ≤198; since a two digit positive integer ≤99 => X must be equal to 1 => XXZ = 11Z, where Z is the units digit We can write a two digit positive number with digits X and Y as: XY ≡10X + Y Similarly, we have, the two digit positive number: YX ≡10Y + X Thus, we have XY + YX ≡(10X + Y) + (10Y + X) = 11(X + Y) We know XY + YX = XXZ => XXZ = 11Z ≡11(X + Y) Thus, XXZ is a multiple 11, and also less than or equal to 198: => XXZ ≡110 => Z = 0 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 115 16.
Amount saved by Suzy in the 1st month = $20 = 20 × 1 = $20.
Amount saved by Suzy in the 2nd month = $(20 + 20) = $40 = 20 × 2 = $40.
Amount saved by Suzy in the 3rd month = $(40 + 20) = $60 = 20 × 3 = $60.
Thus, the amount saved in the 30th month = 20 × 30 = $600.
Thus, the average amount saved every month = Amount saved in the 1st month + Amount saved in the 30th month 2 = $ 20 + 600 2 = $310 Thus, total amount saved in 30 months = Average amount saved per month × (Number of months) = $310 × 30 = $9, 300 The correct answer is Option D.
Alternate approach: Sum of first n positive integers = n × (n + 1) 2 Thus, the sum of first 30 positive integers = 30 × 31 2 = 15 × 31 = 465 Thus, total amount saved in 30 months = $20 × 465 = $9, 300.
17.
We have tn = t(n+1) + 2t(n−1) for n ≥1 Since the relationship tn = t(n+1) + 2t(n−1) is among three terms, to get the value of t4, we must have the values of t2 and t3. But we do not have the value of t3. So, we will first calculate the value of t3.
Plugging-in n = 2 in tn = t(n+1) + 2t(n−1), we get t2 = t3 + 2t1 Plugging-in the values of t1 = 0, and t2 = 2, we get 2 = t3 + 2 × 0 © 1999–2024 Manhattan Review www.manhattanreview.com 116 Quantitative Reasoning Question Bank −PS Solutions t3 = 2 Again, plugging-in n = 3 in tn = t(n+1) + 2t(n−1), we get t3 = t4 + 2t2 Plugging-in the values of t2 = 2, and t3 = 2, we get 2 = t4 + 2 × 2 t4 = −2 The correct answer is Option A.
18.
Let us check the remainders when 2n is divided by 3, for n = 1, 2, 3, 4, 5 . . . : n 2n Remainder when 2n is divided by 3 1 2 2 2 4 1 3 8 2 4 16 1 5 32 2 > Taking 2 terms: 2 + 22: Remainder when 2 + 22 divided by 3 = 2 + 1 = 3 => 0 > Taking 3 terms: 2 + 22 + 23: Remainder when 2 + 22 + 23 is divided by 3 = 2 + 1 + 2 = 5 => 2 > Taking 4 terms: 2 + 22 + 23 + 24: Remainder when 2 + 22 + 23 + 24 is divided by 3 = 2 + 1 + 2 + 1 = 6 => 0 > Taking 5 terms: 2 + 22 + 23 + 24 + 25: Remainder when 2 + 22 + 23 + 24 + 25 is divided by 3 = 2 + 1 + 2 + 1 + 2 = 8 => 2 Note that when the number of terms is even (2 and 4), the remainder is 0 and when the number of terms is odd (3 and 5), the remainder is 2.
Thus, when the number of terms (n > 9) is even (10, 12, ...), the remainder would be 0 and when the number of terms (n > 9) is odd (11, 13, ...), the remainder would be 2.
Thus, statements I and III are true.
The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 117 5.2 Percents 19.
The cost of repairing the current machine = $1,200 The cost of new machine = $2,800 Since the new machine lasts for two years, the average cost per year = $ 2, 800 2 = $1,400.
Thus, the required percentage = 1, 400 −1, 200 1, 200 × 100 = 16.67%.
The correct answer is Option C.
20.
Let the price of the item be $x.
So tax is applicable on $(x −200) Thus, tax paid = 10% of $ (x −200) => $ 10 100 × (x −200) => $ x −200 10 , which is equals to $10.
Thus, we have x −200 10 = 10 => x = 300 The correct answer is Option C.
Alternate approach: We see that tax paid = 10% of excess amount = $10 => Excess amount = $100 Thus, the price = $(200 + 100) = $300 21.
Tax paid on $25 = $0.54 Thus, a tax, which is four times as much as the above, would be $(0.54 × 4) = $2.16 on $25.
Thus, the tax for item B on $100 = 4 × Tax paid on $25 = $2.16 × 4 = $8.64 © 1999–2024 Manhattan Review www.manhattanreview.com 118 Quantitative Reasoning Question Bank −PS Solutions Thus, this tax, expressed as a percentage = 8.64 100 × 100 = 8.64%.
The correct answer is Option C.
22.
Cyclist P increased his speed from 10 mph to 25 mph.
Total increase in speed of Cyclist P = 25 −10 = 15 mph Thus, the percentage increase in speed of Cyclist P = 15 10 × 100 = 150%.
Cyclist Q increased his speed from 8 mph to 24 mph.
Total increase in speed of Cyclist Q = 24 −8 = 16 mph Thus, the percentage increase in speed of Cyclist Q = 16 8 × 100 = 200%.
Apparently it seems that the required answer is simply: 200% −150% = 50%.
However, it is not so since we are here asked to find percent change not percent point change Here absolute change is 200% −150% = 50%, which we are going to compare with percent change with the speed of Cyclist P, which is 150%.
The required percent = 50 150 × 100 = 33.33%.
The correct answer is Option A.
23.
Kilometers traveled by Carrier X = 101,098 =≈100,000 kilometers Fuel consumed by Carrier X = 9,890 liters =≈10,000 liters.
Thus, mileage of Carrier X = 100, 000 10, 000 = 10 kmpl Kilometers traveled by Carrier Y = 203,000 =≈200,000 kilometers Fuel consumed by Carrier Y = 24,896 liters =≈25,000 liters Thus, mileage of Carrier Y = 200, 000 25, 000 = 8 kmpl Thus, the percent by which mileage of Carrier X is greater than that of Carrier Y = 10 −8 8 × 100 = 25% The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 119 24.
Since we need to find the overall percent change, we can assume the list price equal to $100 (the overall percent change does not depend on the list price of the bicycle).
Price after the price was decreased by 25% = (100 −25) % of $100 = $100 × 75 100 = $75 Price after the decreased price was increased by 25% = (100 + 25) % of $75 = $75 × 125 100 = $93.75 Since the assumed list price is $100, the final price would be 93.75% of the assumed list price.
We need not take the actual list price of the bicycle into the consideration.
The correct answer is Option D.
Alternate approach: We can find the overall percent change using the relation: x + y + xy 100 %, where x% and y% represent successive percent changes.
Applying in this problem, we get: 25 −25 −25 × 25 100 = −6.25% Thus, the overall percent change is 6.25% and the final price = 93.75% of the list price.
25.
The mixing for 68 liters of base was 3.4 liters of red color.
The recommended mixing for every 10 liter of base was 0.7 liters of red color.
Thus, as per the recommendation, the amount of red color required for 68 liters of base = 0.7 10 × 68 = 4.76 liters Thus, the mixing quantity must increase by (4.76 −3.4) = 1.36 liters Thus, the percent change needed in the mixing = 1.36 3.4 × 100 = 40% The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 120 Quantitative Reasoning Question Bank −PS Solutions 26.
Number of employees who worked for 30 hours in the week = 30% of 200 = 60.
Rate of pay per hour = $7.50.
Thus, pay per employee for 30 hours = $(7.50 × 30) = $225.
Thus, total pay for all 60 employees = $ (225 × 60) = $13, 500 . . . (i) Number of employees who worked for 44 hours in the week = 40% of 200 = 80.
Rate of pay per hour = $7.50.
Thus, pay per employee for 44 hours= $(7.50 × 44) = $330.
Thus, total pay for all 80 employees = $(330 × 80) = $26, 400 . . . (ii) Number of employees who worked for 50 hours in the week = 200 −(60 + 80) = 60.
Rate of pay is $7.50 per hour for the first 44 hours and $ 7.50 × 11 3 = $ 7.50 × 4 3 = $10 per hour for the remaining (50 −44) = 6 hours.
Thus, remuneration per employee = $(7.50 × 44 + 10 × 6) = $390.
Thus, total remuneration for all 60 employees = $(390 × 60) = $23, 400 . . . (iii) Thus, total remuneration for all the 200 employees = $(13, 500 + 26, 400 + 23, 400) = $63, 300.
The correct answer is Option C.
27.
On the first $35 million in sales, amount received in commission = $5 million Thus, ratio of commission to sales = 5 35 = 1 7 On the next $121 million in sales, amount received in commission = $11 million Thus, ratio of commission to sales = 11 121 = 1 11 Since 1 7 > 1 11, there is a decrease in commission.
Thus, the required percent decrease = 1 7 −1 11 1 7 × 100% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 121 = 11 −7 11 × 100% = 4 11 × 100 = 36.36% The correct answer is Option C.
28.
Total sales = $24,000.
Thus, commission on the first $20,000 = 8% of $20,000 = $ 8 100 × 20, 000 = $1, 600.
Total commission received by the sales representative = $2,000 (given).
Thus, commission received on the remaining sales = $(2, 000 −1, 600) = $400.
Commission received on the remaining $(24, 000 −20, 000) = $4, 000 at x% = x% of $4,000 = $ x 100 × 4, 000 = $40x.
Thus, we have 40x = 400 => x = 10% The correct answer is Option D.
29.
Total cost of the 120,000 computer chips = $3,600,000 Cost of 2 5 of the above computer chips = $ 3, 600, 000 × 2 5 = $1, 440, 000.
These were sold at a 25% higher than the cost price.
Thus, selling price of the above computer chips = (100 + 25)% of $1,440,000 = $ 125 100 × 1, 440, 000 = $1, 800, 000 . . . (i) © 1999–2024 Manhattan Review www.manhattanreview.com 122 Quantitative Reasoning Question Bank −PS Solutions Cost of the remaining computer chips = $(3, 600, 000 −1, 440, 000) = $2, 160, 000 Later, these remaining computer chips were sold at a 25% lower than the cost price.
Thus, selling price of the above computer chips = (100 −25)% of $2,160,000 = $ 75 100 × 2, 160, 000 = $1, 620, 000 . . . (ii) Thus, total selling price = $(1, 800, 000 + 1, 620, 000) = $3, 420, 000.
Since total selling price (= $3,420,000) < total cost price (= $3,600,000), there is a loss Thus, percent loss = Cost price −Selling price Cost price !
× 100 = 3, 600, 000 −3, 420, 000 3, 600, 000 × 100 = 5% (Loss) The correct answer is Option B.
Alternate approach: Percent profit made on 2 5 of the stock = 25% Percent loss made on the remaining (1 −2 5 =) 3 5 of the stock = −25% (since price is 25% less) Thus, overall percent profit or loss = 2 5 × 25 + 3 5 × (−25) = 10 −15 = −5% 30.
We know that the price of milk increased by 20%.
Let the original price of milk per liter be $x.
Thus, the price of milk per liter after the price increase www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 123 = $ (100 + 20) % of x = $(120% of x) = $ 6x 5 .
Number of liters of milk = Total Price Price per liter !
Thus, difference in quantity of milk obtained for $60 = 60 x −60 6x 5 = 5 => 60 x −50 x = 5 => 10 x = 5 => x = 2 Thus, the correct answer is Option A.
Alternate approach: If the price of an item goes up/down by x%, the quantity consumed should be reduced/increased by 100x 100 ± x % so that the total expenditure remains the same.
Since the price of milk increased by 20%, the quantity obtained for $60 would reduce by 100 × 20 100 + 20 = 50 3 % Thus, 50 3 % of the original quantity = 5 liters => Original quantity = 5 50 3 % = 500 50 3 = 30 liters Thus, the initial price per liter of diesel = Total Initial Price Total Initial number of liters = $ 60 30 = $2.
31.
Since the problem asks us to find a percent value, we can assume any suitable value of the annual sum of money for ease of calculation as the initial value does not affect the final answer.
Since we have to deal with factions 1 4 and 1 6, we can assume the sum equal to $24 (= LCM of 4 and 6).
Thus, the sum spent during the first quarter = $ 1 4 × 24 = $6.
Amount of money left = $ (24 −6) = $18.
Thus, the sum spent during the second quarter = $ 1 6 × 18 = $3.
Thus, the sum left at the beginning of the third quarter = $ (18 −3) = $15.
© 1999–2024 Manhattan Review www.manhattanreview.com 124 Quantitative Reasoning Question Bank −PS Solutions Total the sum spent in the last two quarters = $ (6 + 3) = $9.
Thus, the required percent = 15 −9 9 × 100 = 200 3 % = 66.66% The correct answer is Option D.
32.
Amount David spent in 2013 = $450.
Amount David spent in 2014 = 10% more than what he spent in 2013 = (100 + 10) % of what he spent in 2013 = $ 110 100 × 450 = $495 Total amount spent by David and Suzy in 2014 = $600.
Thus, amount spent by Suzy in 2014 = $ (600 −495) = $105.
We know that Suzy had spent $450 in 2013.
Thus, the required percent = (Amount spent by Suzy in 2013) −(Amount spent by Suzy in 2014) (Amount spent by Suzy in 2013) × 100% = 450 −105 450 × 100% = 345 450 × 100% =≈77% The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 125 33.
It is easier to solve this question can be solved by observing the answer options than by actual solving.
We observe that the options are very large in value compared to the price change by the end of day 2, i.e. $1.
Let the original price of the item be $x.
By the end of day 2, the price of the item decreases by $1 Thus, by the end of day 2, the price of the item = $(x −1) The subsequent increase would be calculated on $(x −1) instead of $x However, since $1 is negligible compared to x (note that x is one among very large option values), we can conclude that the decrease in price would be very slightly less than $1.
Thus, the final price on day 4 =≈$ ((x −1) −1) = $(x −2).
Thus, we have (x −2) =≈398 => x =≈400 Thus, the only option that satisfies is D: $400.
The actual calculation is shown below: Assuming the initial price of a share to be $x, we have x − x 1 + k 100 1 − k 100 = 1 . . . (i) x 1 + k 100 1 − k 100 1 + k 100 1 − k 100 = 398 . . . (ii) We need to solve for x from the above two equations.
Note: The actual solution is time taking and very involved, and hence, not suggested.
The correct answer is Option D.
34.
Cost of x items = $y Since the cost increases by 20%, the new cost of x items = $ (100 + 20) % of y © 1999–2024 Manhattan Review www.manhattanreview.com 126 Quantitative Reasoning Question Bank −PS Solutions = $ 120 100 × y = $ 6y 5 Thus, with a budget of $ 6y 5 , x items can be bought next year.
Thus, with a budget of $(3y), the number of items can be bought = 3y 6y 5 × x = 5 2x = 2.50x The correct answer is Option C.
35.
Since the question asks for a percent value, we can choose any suitable initial value of the total weight of the solution for the ease of calculation; the initial value will not affect the final answer.
Let the total weight of the solution initially = 100 liters.
Thus, weight of water = 30% of 100 = 30 liters.
Weight of the remaining solution = 100 −30 = 70 liters.
Loss of water = 70% by weight.
Thus, final weight of water after 15 minutes of boiling = (100 −70)% of 30 = 30% of 30 = 9 liters.
Since there is no weight loss in the other part of the solution, final weight of the solution = 70 + 9 = 79 units.
Thus, the required percent value = Final weight of water Final weight of solution !
× 100 = 9 79 × 100 = 900 79 % The correct answer is Option D.
36.
Since the question asks about a percent value, we can choose any suitable initial value of the total volume of the mix for ease of calculation; the initial value will not affect the final answer.
Let the total volume of the mixed juice be 100 units.
Thus, volume of banana pulp = 25% of 100 = 25 units www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 127 => Volume of papaya pulp = 100 −25 = 75 units Let the price of 100 units (an equal quantity) of banana pulp be $100x.
Since the mixed juice costs 20 percent more than the cost of an equal quantity of only banana pulp, the price of 100 units of the mixed juice = $((100 + 20) % of 100x) = $120x The price of 25 units of banana pulp present in the mix = $ 100x 100 × 25 = $25x.
Thus, the price of 75 units of papaya pulp present in the mix = $ (120x −25x) = $95x.
Thus, the price of 100 units of papaya pulp = $ 95x 75 × 100 = $126.67x Thus, the percent by which papaya pulp are more expensive than banana pulp = (Price of 100 units of papaya pulp) −(Price of 100 units of banana pulp) Price of 100 units of banana pulp × 100 = 126.67x −100x 100x × 100 = 26.67% The correct answer is Option B.
37.
Let us assume the number of days in a year to be 365.
Thus, in a 3-year period, total number of days = 3 × 365 = 1, 095.
We know that bacteria P multiplies itself in every 18 days and bacteria Q multiplies itself in every 15 days.
Thus, we have Number of times bacteria P multiplies itself = 1, 095 18 = 60... = 60; bacteria P multiplies only 60 times.
Number of times bacteria Q multiplies itself = 1, 095 15 = 73; bacteria Q multiplies 73 times.
Since there is no information whether the 3-year period includes a leap year (366 days in the year), we must consider that scenario too.
© 1999–2024 Manhattan Review www.manhattanreview.com 128 Quantitative Reasoning Question Bank −PS Solutions In that case, Number of times bacteria P multiplies itself = 1, 096 18 = 60... = 60; bacteria P still multiplies only 60 times.
Number of times bacteria Q multiplies itself = 1, 096 15 = 73... = 73; bacteria Q still multiplies only 73 times.
Even if we had not considered the leap year scenario, we would not have made any mistake.
Since this is PS problem, and there cannot be two answers to the same problem.
Thus, the required percent difference = (Number of times bacteria Q multiplies) −(Number of times bacteria P multiplies) Number of times bacteria P multiplies × 100% = 73 −60 60 × 100% =≈22% The correct answer is Option D.
38.
Cost of the phone purchased by Jack = $1,500.
Sales tax paid = 5% of $1500 = $ 5 100 × 1, 500 = $75.
Thus, total price paid by Jack = $ (1, 500 + 75) = $1, 575.
Cost of the phone purchased by Tom = $1,200.
Sales tax paid = 15% of $1,200 = $ 15 100 × 1, 200 = $180.
Thus, total price paid by Tom = $ (1, 200 + 180) = $1, 380.
Thus, the amount which Tom paid less compared to Jack = $ (1, 575 −1, 380) = $195.
Thus, the required % value = Total price paid by Jack −Total price paid by Tom Total price paid by Jack !
× 100% = 195 1, 575 × 100% = 260 21 =≈12% The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 129 39.
Let the number of boys and girls be b and g, respectively.
Number of boys who play basketball = 65% of b = 65 100 × b = 13b 20 Number of girls who play basketball = 78% of g = 78 100 × g = 39g 50 Thus, total number of students play basketball = 13b 20 + 39g 50 Total number of students = (b + g) Since 72% of all students play basketball, we have 72 100 × b + g = 13b 20 + 39g 50 Upon solving, we get, => g b = 7 6 The correct answer is Option B.
Alternate approach: We can solve the problem using the method of alligation, as shown below: 𝑏 𝑔 65% 78% 72% 78 −72 % = 6% 72 −65 % = 7% Thus, we have b : g = 6 : 7 g : b = 7 : 6 © 1999–2024 Manhattan Review www.manhattanreview.com 130 Quantitative Reasoning Question Bank −PS Solutions 40.
Total number of spectators = S Since 40% of the spectators belonged to natives, percent of other than native spectators = (100 −40) = 60% Thus, the number of other than native spectators = 60% of S = 0.6S Of the above, Royal Challengers had a support of 10%.
Thus, the number of other than native supporters = 10% of 0.6S = 0.06S Number of native supporters = 24,500 Thus, total number of supporters = 0.06S + 24, 500 The correct answer is Option D.
41.
Since the problem asks for a percent value, we can assume any suitable value of the number of students for ease of calculation.
Let the number of students be 100.
Number of students of Science stream = 40% of 100 = 40.
Number of Science stream students who go to special classes = 60% of 40 = 24 Total number of students who go to special classes = 30% of 100 = 30.
Thus, the number of students who do not study Science but go to special classes = 30 −24 = 6 Thus, the percent of the students who do not study Science but go to special classes = 6 100 × 100 = 6% The correct answer is Option A.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 131 42.
Let the total number of students = 100 (since the question asks for ratio, the answer is indepen-dent of the initial number chosen, so we choose a suitable number for ease of calculation).
The number of boys = 40% of 100 = 40.
Thus, the number of girls = 100 −40 = 60.
Let the number of students transferred = x.
Thus, number of boys transferred = 30% of x = 30 100 × x = 3x 10 Thus, number of girls transferred = x −3x 10 = 7x 10 Here we know that: Transfer rate for students of a certain gender = Number of students of that gender transferred Total number of students of that gender Thus, the transfer rate for the boys = 3x 10 40 = 3x 400.
Also, the transfer rate for the girls = 7x 10 60 = 7x 600.
Thus, the required ratio = 3x 400 : 7x 600 = 3 2 : 7 3 = 9 : 14 The correct answer is Option D.
43.
Since the problem asks us about a percent change, we can assume a suitable initial value of the original price for ease of calculations.
We see that we need to take 5 7 and 3 5 of the original price.
So, we should assume a value, which is a multiple of 35 (LCM of denominators, 7 and 5) for ease of calculations.
Thus, let the original price be $35.
Thus, the value of the car at the start of the year = $ 5 7 × 35 = $25.
The value of the car in the end of the year = $ 3 5 × 35 = $21.
Thus, the required percent decrease © 1999–2024 Manhattan Review www.manhattanreview.com 132 Quantitative Reasoning Question Bank −PS Solutions = Change in value Initial value × 100% = 25 −21 25 × 100 = 16% The correct answer is Option B.
Alternate approach: Say the original price is $1.
Thus, the required percent decrease = Change in value Initial value × 100 = 5 7 −3 5 5 7 × 100 = 4 35 7 5 × 100 = 4 25 × 100 = 16% 44.
Remuneration as per the first offer = 5% commission on sales plus monthly bonus of $500 Remuneration as per the second offer = 7% of commission on sales Thus, we have (5% commission on sales) + ($500) = (7% commission on sales) => (7% commission on sales) −(5% commission on sales) = $500 => 2% of sales = $500 => Sales = $ 500 × 100 2 = $25, 000 The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 133 45.
Number of retailers at the beginning of the year = 35% of 120 = 35 100 × 120 = 42.
Number of retailers after the 24-month period = 25% of 240 = 25 100 × 240 = 60.
Thus, increase in the number of retailers = 60 −42 = 18.
Let the simple annual percent growth rate in the number of retailers be r%.
Thus, in two years (the 24-month period), increase in the number of retailers at r% rate = 42 × r × 2 100 (The value is calculated on 42 since 42 is the value at the start of the year) Thus, we have 42 × r × 2 100 = 18 => r = 18 × 100 2 × 42 = 21.43% The correct answer is Option B.
46.
Since we need to find the greatest overall percent increase, let us assume a suitable initial value.
Let the initial value be 100.
Let there be x% increase initially.
Thus, value after x% increase = (100 + x) % of 100 = 100 + x 100 × 100 = (100 + x) Let there be y% increase next.
Thus, the final value after y% increase = (100 + y)% of (100 + x) = 100 + y 100 × (100 + x) © 1999–2024 Manhattan Review www.manhattanreview.com 134 Quantitative Reasoning Question Bank −PS Solutions = 100 + y (100 + x) 100 = 1002 + 100 x + y + xy 100 Thus, overall percent increase = (Final value) −(Initial value) (Initial value) ×100% = 1002 + 100 x + y + xy 100 −100 !
100 × 100% = 100 x + y + xy 100 !
% = x + y + xy 100 % In each of the above options, we observe that: x + y = 60 Thus, the option with the highest value of xy will have the highest overall percent increase.
We know that if the sum of two terms is a constant, the product of the two terms becomes the maximum only when both terms are equal => x = y = 60 2 = 30 Thus, the correct answer is Option C.
Alternately, working with the options one at a time: • A: x = 10, y = 50 => xy = 500 • B: x = 25, y = 35 => xy = 875 • C: x = 30, y = 30 => xy = 900: Correct answer.
• D: x = 40, y = 20 => xy = 800 • E: x = 45, y = 15 => xy = 675 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 135 47.
Since the problem asks us about a percent value, we can assume a suitable value of the number of voters for ease of calculations.
Let the number of voters be 100.
Thus, the number of boys = 70% of 100 = 70 Number of girls = 100 −70 = 30 Number of boys who would vote for John = 30% of 70 = 21.
Number of girls who would vote for John = 70% of 30 = 21.
Thus, total number of votes for John = 21 + 21 = 42.
Thus, the required percent = 42 100 × 100 = 42%.
The correct answer is Option B.
48.
We see some ugly numbers to deal with.
Region-wise distribution of companies in the state Regions No. of companies Region P 2,345 Region Q 3,456 Region R 3,421 Region S 5,721 Region T 3,445 Region U 80 Region V 4,532 Total 23,000 In the GMAT, you would be seldom asked to do calculations that consume a lot of time. Even if you see some ugly numbers to deal with, you need not put in too much time on it. Mostly, there would be a smarter approach to deal with the question.
We are asked to find out the percentage difference between the number of companies of Region S and that of Region R. Instead of calculating the individual percentages, let’s save time and directly calculate the percentage difference.
It is given that the number of companies of Region S = 5,721 and the number of companies of Region R = 3,421 © 1999–2024 Manhattan Review www.manhattanreview.com 136 Quantitative Reasoning Question Bank −PS Solutions Thus, the difference = 5,721 −3,421 = 2,300.
Thus, the percentage difference of number of companies of Region S and that of region R of the total = 2, 300 23, 000 × 100% = 10%.
The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 137 5.3 Profit & Loss 49.
Since 4 5 th of the stock was sold, the remaining 1 −4 5 = 1 5 th of the stock was not sold.
Thus, 1 5 th of the total stock = 100.
Thus, 4 5 th of the total stock = 4 × 100 = 400.
Thus, 400 items were sold each at $3.00.
Thus, total amount received = $(400 × 3) = $1, 200.
The correct answer is Option B.
50.
We know that the trader bought 900 cartons a cost of $20 per carton.
Thus, the total cost of the cartons = $ (20 × 900) = $18, 000 Selling price of 2 3 of 900 or 600 cartons = $ (1.25 × 20) = $25 per carton Thus, selling price of 600 cartons = $ (25 × 600) = $15, 000 . . . (i) Selling price of the remaining (900 −600) or 300 cartons = $ (100 −20) % of 20 = $ 80 100 × 20 = $16 per carton Thus, selling price of 300 cartons = $ (16 × 300) = $4, 800 . . . (ii) Thus, from (i) and (ii): Total selling price = $(15, 000 + 4, 800) = $19, 800.
Thus, gross profit = Total selling price −Total cost price = $(19, 800 −18, 000) = $1, 800.
The correct answer is Option A.
51.
Selling price of a brand A bicycle = $150.
Thus, 60% of the selling price of a brand B bicycle = $150 Thus, selling price of a brand B bicycle = $ 150 × 100 60 = $250 Total number of bicycles sold by the dealer = 100 © 1999–2024 Manhattan Review www.manhattanreview.com 138 Quantitative Reasoning Question Bank −PS Solutions Thus, the number of brand B bicycles sold = 3 5 × 100 = 60 Thus, the number of brand A bicycles sold = 100 −60 = 40 Thus, total sale from the sale of all bicycles = Sale from brand A bicycles + Sale from brand B bicycles = $ (150 × 40) + $ (250 × 60) = $ (6, 000 + 15, 000) = $21, 000 The correct answer is Option D.
52.
Purchase price of the consignment = $800.
Percent profit made on the Purchase price = 30%.
Thus, selling price of the consignment = (100 + 30) % of $800 = $ 130 100 × 800 = $1, 040 This selling price is 20% less than the marked price.
Thus, (100 −20) % = 80% of the marked price is equal to the selling price $1,040.
Thus, the marked price = $ 1, 040 × 100 80 = $1, 300 The correct answer is Option C.
53.
Cost of production of each unit = $2.50 Selling price of each unit = $4.50 Thus, margin on each unit = $ (4.50 −2.50) = $2.00 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 139 Investment made on the equipment = $10,000.
Thus, number of units required to be sold to recover the investment in machines = Investment Margin on each unit = 10, 000 2.00 = 5, 000 The correct answer is Option D.
Alternate approach: Let the number of required units o be sold to recover the investment in machines = n.
Thus, total cost = $ (10, 000 + 2.5n).
Total selling price = $4.5n Thus, we have 10, 000 + 2.5n = 4.5n => n = 5, 000 54.
Original selling price of the house = (100 + 20) % of the cost of the house = 120% of the cost of the house New selling price of the house = (100 + 30) % of the cost of the house = 130% of the cost of the house Thus, difference between the above two selling prices = (130% −120%) of the cost of the house = 10% of the cost of the house Since it is given that the difference between the two selling prices is $10,000, we have 10% of the cost of the house = $10,000 © 1999–2024 Manhattan Review www.manhattanreview.com 140 Quantitative Reasoning Question Bank −PS Solutions => Cost of the house = $ 10, 000 × 100 10 = $100, 000 Thus, the original selling price of the house = 120% of the cost of the house = $ 100, 000 × 120 100 = $120,000 The correct answer is Option C.
55.
Total number of television sets assembled = 600.
Contractor payment for each of the first 100 television sets = $20 Thus, the total contractor payment for the first 100 television sets = $ (20 × 100) = $2, 000.
Contractor payment for each of the remaining 500 (= 600 −100) television sets = $15 Thus, the total contractor payment for the remaining 500 television sets = $ (15 × 500) = $7, 500.
Thus, the total contractor payment for 600 television sets = $ (2, 000 + 7, 500) = $9, 500.
Total invoice value for the 600 television at $25.00 each = $ (25 × 600) = $15, 000.
Thus, gross profit = Total invoice value −Total contractor payment = $15,000 −$9,500 = $5,500 The correct answer is Option C.
56.
Initial selling price of the item = $99.
Since the initial percent profit was 10% of the cost, we have (100 + 10) % of the cost of the item = $99 => Cost of the item = $ 99 110 × 100 = $90 Increased selling price of the item = $117.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 141 Thus, the required percent profit = Selling price −Cost price Cost price !
× 100 = 117 −90 90 × 100 = 27 90 × 100 = 30% The correct answer is Option E.
57.
The merchant has 2,400 fans priced at (cost price) $30 each.
Thus, total cost price = $ (2, 400 × 30) = $72, 000.
Number of fans sold at $40 each = 60% of 2,400 = 60 100 × 2400 = 1, 440.
Thus, sales revenue of these 1,440 fans = $ (1440 × 40) = $57, 600.
Number of fans sold at $35 each = 2,400 −1,440 = 960.
Thus, sales revenue of these 960 fans = $ (960 × 35) = $33, 600.
Thus, sales revenue of 2,400 fans = 57,600 + 33,600 = $91,200 Thus, profit made by selling the 2,400 fans = 91, 200 −72, 000 = $19, 200.
Thus, average profit made per fan = $ 19, 200 2, 400 = $8.
The correct answer is Option B.
Alternate approach: The merchant sold 60% of the fans each at a profit of $ (40 −30) = $10.
He sold the remaining 40% of the fans each at a profit of $ (40 −35) = $5.
Thus, his average profit per radio = $ 10 × 60 100 + 5 × 40 100 = $ (6 + 2) = $8 © 1999–2024 Manhattan Review www.manhattanreview.com 142 Quantitative Reasoning Question Bank −PS Solutions 58.
Let us understand the problem using the diagram shown below: Cost Price Listed Price 2nd Selling Price 1st Selling Price 𝑚 % Discount 𝑚 % Profit (𝑚+ 15)% Discount (𝑚−20)% Profit Difference in Selling Price Let us calculate the difference in the two selling prices: Since in the first case, the profit was m% and in the second, the profit was (m −20) %, the difference in selling price = (m −(m −20)) % of Cost Price; (Since percent discount is calculated on the cost price) = 20% of Cost price . . . (i) Again, in the first case, the discount was m% and in the second, the discount was (m + 15) %, the difference in selling price = ((m + 15) −m) % of List price; (Since percent discount is calculated on the list price) = 15% of List price . . . (ii) Thus, from (i) and (ii), we have 20% of Cost price = 15% of List price => List price Cost price = 20 15 = 4 3 Thus, if no discount were offered (the list price becomes the selling price), percent profit = List price −Cost price Cost price !
× 100 = List price Cost price −1 !
× 100 = 4 3 −1 × 100 = 33.3% The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 143 59.
Total number of units sold = 800 + 900 = 1,700 Cost of producing each unit = $6 Thus, total cost of producing 1700 units = $(6 × 1700) = $10,200 Selling price of 800 units of the product = $(800 × 8) = $6,400 Selling price of 900 units of the product = $(900 × 5) = $4,500 Thus, the total selling price of 1,700 units = $(6,400 + 4,500) = $10,900 Thus, profit = $ (10, 900 −10, 200) = $700 The correct answer is Option E.
60.
Let the book and stationary sales in 2014 be b and s, respectively.
Thus, total sales revenue in 2014 = (b + s) Thus, the sales revenue from book sales in 2015 = (100 −10)% of b = 90% of b = 0.9b.
And, the sales revenue from stationary sales in 2015 = (100 + 6)% of s = 106% of s = 1.06s.
Thus, total sales revenue in 2015 = (0.9b + 1.06s) Thus, we have (0.9b + 1.06s) = (100 + 2) % of (b + s) => 0.9b + 1.06s = 1.02b + 1.02s => 0.04s = 0.12b => b s = 0.04 0.12 = 1 3 => b : s = 1 : 3 The correct answer is Option A.
61.
Let the profit of the trader in 2001 be $100 (the assumption of $100, or any other number does not affect the answer since we have to find the percent change).
Thus, the profit of the trader in 2002 = $((100 + 20) % of 100) = $120.
Thus, the profit of the trader in 2003 = $((100 + 25) % of 120) = $150.
© 1999–2024 Manhattan Review www.manhattanreview.com 144 Quantitative Reasoning Question Bank −PS Solutions Thus, the percent change in profit from 2001 to 2003 = 150 −100 100 × 100 = 50%.
The correct answer is Option E.
Alternate approach: If the value of a commodity changes by x% and then by y% successively, the overall percent change is given by: x + y + xy 100 % Here, we have x = Percentage change from 2001 to 2002 = 20 % y = Percentage change from 2002 to 2003 = 25 % Thus, the overall percent change = 20 + 25 + 20 × 25 100 = 50% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 145 5.4 Averages 62.
Number of teachers = Total annual salaries Average annual salaries = 3, 780, 000 42, 000 = 90 => Number of students = 25 2 × 90 = 1, 125 The correct answer is Option C.
63.
Let the number of male employees and female employees be m and f , respectively.
Total salary of male employees = 65, 000m.
Total salary of female employees = 80, 000f .
Thus, total salaries of all the employees = 65, 000m + 80, 000f.
Thus, average salaries of all the employees = Total salaries of the employees Total number of employees !
= 65, 000m + 80, 000f m + f !
.
Thus, we have 65, 000m + 80, 000f m + f !
= 70, 000 => 65m + 80f = 70m + 70f => 10f = 5m => m f = 2 1 We cannot get the absolute number of male employees and female employees. There are infinite number of possibilities.
Among the options, the value that bears the ratio 2 : 1 would be the correct answer. We see that only Option D (14; 7) bears that ratio 2: 1.
The correct answer is Option D.
64.
It is given that the total number of students = 40.
Let the number of students in section A = n.
Thus, the number of students in section B = (40 −n).
© 1999–2024 Manhattan Review www.manhattanreview.com 146 Quantitative Reasoning Question Bank −PS Solutions Total score of all the students in the two sections combined = 85n + 80(40 −n).
Thus, average score considering all students = Total score Total number of students = 85n + 80 (40 −n) 40 Thus, we have 85n + 80 (40 −n) 40 = 82 => 17n + 640 −16n = 656 => n = 16 The correct answer is Option C.
65.
The juice manufacturer has (1, 200 + 400) = 1, 600 liters of mango pulp.
Amount of water in the first stock = 25% of 1200 = 300 liters Amount of water in the second stock = 20% of 400 = 80 liters Thus, the total amount of water = 300 + 80 = 380 liters Thus, the percent of water in the total stock = 380 1, 600 × 100 = 23.75%.
The correct answer is Option B.
Alternate approach: The required percent is the weighted average of the percentages of the above two stocks = 1, 200 × 25 100 + 400 × 20 100 1, 200 + 400 × 100 = 300 + 80 1, 600 × 100 = 380 1, 600 × 100 = 23.75%.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 147 66.
Let the number of students in the sections P, Q, R, and S be p, q, r, and s, respectively.
Thus, the average weight of all students in the class = Total weight of all the students combined Total number of students = 45 × p + 50 × q + 55 × r + 65 × s p + q + r + s !
lb.
Thus, we have 45 × p + 50 × q + 55 × r + 65 × s p + q + r + s !
= 55 => 45p + 50q + 55r + 65s = 55p + 55q + 55r + 55s => 10p + 5q = 10s => 2p + q = 2s Since we need to maximize r, we need to find the minimum possible values of p, q, and s so that the above equation holds true.
Since the RHS of the above equation is 2s, it is even.
Also, in the LHS of the above equation, 2p is even.
Thus, q must be even.
The smallest even number that we can consider for q is 2 since we have at least one student in each section. Thus, we have q = 2.
Thus, the equation gets modified to: 2p + 2 = 2s => p + 1 = s Thus, we use the minimum possible values: p = 1, s = 2.
Thus, we have p = 1, q = 2 and s = 2.
Since there are a total of 40 students in all sections combined, the maximum value of students in section R = r = 40 − p + q + s = 40 −5 = 35 The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 148 Quantitative Reasoning Question Bank −PS Solutions 67.
Since the set starts with an odd number (1) and has an odd numbers of integers, the set would end with an odd number, too. Let’s see the set.
Set N: {1, 2, 3, 4, 5, ..., (2n + 1)}, where n is a positive integer.
=> Number of odd terms is one more than the number of even terms.
Thus, the number of even terms = n The number of odd terms = (n + 1) We know that the sum (M) of p terms of an arithmetic progression having first term as a and common difference as d is given by: M = p 2 2a + p −1 × d Say, X is the average of the odd integers in set N and Y is the average of the even integers in set N Thus, we have X = (1 + 3 + 5 + · · · + (2n + 1)) n + 1 = n + 1 2 × (2 × 1 + ((n + 1) −1) × 2) n + 1 = n + 1 Y = (2 + 4 + · · · + 2n) n = 2 × (1 + 2 + · · · + n) n = 2 × n (n + 1) 2 n = n + 1 Thus, we have X −Y = (n + 1) −(n + 1) = 0 The correct answer is Option B.
Alternate approach: Since there is no restriction on the number of terms, let us try to find (X −Y) using a few values of the number of terms.
Say there are only three terms in the set.
S = {1, 2, 3} X = 1 + 3 2 = 2 Y = 2 => X −Y = 0 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 149 Again, say there are only five terms in the set: S = {1, 2, 3, 4, 5} X = 1 + 3 + 5 3 = 3 Y = 2 + 4 2 = 3 => X −Y = 0 Since (X −Y) = 0 for the number of terms = 3 and the number of terms = 5, we can conclude that the value of (X −Y) would be the same for any odd number of terms in the set, as we have a definite answer among the given answer options.
68.
Let the seven numbers be a, b, c, d, e, f , g, h and i.
Thus, we have a + b + c + d + e + f + g + h + i 9 = 25 => a + b + c + d + e + f + g + h + i = 225 . . . (i) Since the average of the first five numbers is 20, we have a + b + c + d + e 5 = 20 => a + b + c + d + e = 100 . . . (ii) Since the average of the last five numbers is 32, we have e + f + g + h + i 5 = 32 => e + f + g + h + i = 160 . . . (iii) Adding (ii) and (iii), we have a + b + c + d + 2e + f + g + h + i = 260 . . . (iv) Subtracting (i) from (iv), we have e = 260 −225 = 35 Thus, the value of the fifth number is 35.
The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 150 Quantitative Reasoning Question Bank −PS Solutions 69.
Let the radius of each green ball = x inches.
Each green ball is 4 inches less than the average radius of the balls in Box X.
Thus, the average radius of balls in Box X = (x + 4) inches.
Also, each green ball is 2 inches greater than the average radius of the balls in Box Y.
Thus, the average radius of balls in Box Y = (x −2) inches.
Thus, the required difference = ((x + 4) −(x −2)) = 6 inches.
Note: In this problem, there is a lot of data which has been given to make the question appear complicated. The radius of yellow balls is of no consequence. One should carefully read the problem statement and use only the information required to answer the question.
The correct answer is Option B.
70.
We know that Average salary = (# of emps of group I × Av. salary) + (# of emps of group II × Av. salary) + (# of emps of group III × Av. salary) Total number of employees => Average salary = $ 10 × 35, 000 + 35 × 30, 000 + 15 × 60, 000 60 = $38, 333 The correct answer is Option B.
Note: While calculating the average, we may work with the salaries as $35, $30 and $60, respec-tively; and after calculating the average, multiply the result with 1,000.
71.
Let us use the method of alligation to solve this problem: Total number of pencils and erasers bought = 20.
Price of each pencil = 20 cents.
Price of each eraser = 30 cents.
Average price of all 20 pieces = 28 cents.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 151 Pencils Erasors 20 30 28 30 −28 = 2 28 −20 = 8 Thus, ratio of the number of pencils to the number of erasers = 2 : 8 = 1 : 4.
Since total number of pieces purchased are 20, we have Number of pencils = 1 1 + 4 × 20 = 4 Number of erasers = 4 1 + 4 × 20 = 16 We now need to find the number of erasers required to be returned so that the average price falls to 26 cents.
So the situation is this, Pencils Erasors 20 30 26 30 −26 = 4 26 −20 = 6 Thus, ratio of the number of pencils to the number of erasers now = 4 : 6 = 2 : 3.
Since the number of pencils remains the same from the initial condition (only erasers are returned) and we had obtained 2 pencils on ratio scale, so here too, the number of pencils should be 2 on ratio scale.
Thus, the number of erasers must be 6 (since the ratio of pencils to erasers = 2 : 3).
Since initially, there were 16 erasers, number of erasers returned = 16 −6 = 10.
The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 152 Quantitative Reasoning Question Bank −PS Solutions Alternate approach 1: Average amount spent for 20 pieces (pencils and erasers) = 28 cents.
Total amount spent for 20 pieces = 28 × 20 = 560 cents.
Let us assume that the boy returned x erasers so that the average amount spent is 26 cents.
Thus, total amount spent for (20 −x) pieces = 26 × (20 −x) cents.
Thus, price of x erasers returned = (560 −26 × (20 −x)) cents.
Since the price of one eraser is 30 cents, we have 30 = 560 −26 × (20 −x) x => 30x = 560 −520 + 26x => x = 10 Alternate approach 2: The average price of 20 pieces was 28 cents and after returning some erasers, say x, the average of (20 −x) pieces became 26 cents.
Alternately, we can say that: The average price of some pieces, say f, was 26 cents and after adding some erasers, say r (at 30 cents each), the average of (f + r = 20) pieces became 28 cents.
Thus, by alligation method: Pieces (𝑓) Erasors (𝑟) 26 30 28 30 −28 = 2 28 −26 = 2 Thus, we have f : r = 2 : 2 = 1 : 1 However, we know that: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 153 f + r = 20 => f = r = 1 1 + 1 × 20 = 10 Thus, the number of erasers returned is 10.
Alternate approach 3: We infer the same conclusion as in the previous approach: The average price of some pieces, say f, was 26 cents and after adding some erasers say r (at 30 cents each), the average of (f + r = 20) pieces became 28 cents.
Thus, the price of each of the f pieces increased by (28 −26) = 2 cents, resulting in a total increase of 2 × f = 2f cents.
Also, the price of each of the r erasers reduced by (30 −28) = 2 cents resulting in a total decrease of 2 × r = 2r cents.
The increase of 2f cents came at the expense of the reduction of 2r cents, implying: 2r = 2f => r = f But, we have f + r = 20 => f = r = 1 1 + 1 × 20 = 10 Thus, the number of erasors returned is 10.
72.
We know that the student’s average score on four tests is 78.
Thus, his total score on the four tests = 4 × 78 = 312.
Let the score on the 5th test be n.
Thus, his total score on the five tests = (312 + n).
Thus, his average on the 5 tests = 312 + n 5 .
We know that The final average increases from the average on 4 tests by an integer © 1999–2024 Manhattan Review www.manhattanreview.com 154 Quantitative Reasoning Question Bank −PS Solutions This is possible only when the average of the 5 tests is also an integer.
Again, the student’s average (an integer value) on the 5 tests = 312 + n 5 = 62 + (2 + n) 5 Thus, (n + 2) must be divisible by 5.
Working with the options, we see that only Option D, i.e.
n = 93 satisfies since (n + 2) is divisible by 5.
The correct answer is Option D.
73.
We know that the ratio of the numbers of candidates in groups P, Q and R was 3 : 5 : 4, respectively.
Let the number of candidates in groups P, Q and R be 3x, 5x and 4x, respectively, where x is a constant of proportionality.
We also know that the average scores for the groups P, Q and R were 64, 84, and 72, respectively.
Thus, the average score for the three groups combined = 64 × 3x + 84 × 5x + 72 × 4x 3x + 5x + 4x = 192x + 420x + 288x 12x = 900x 12x = 75 The correct answer is Option B.
Alternate approach: Let us solve the problem using the concept of assumed mean (deviation method).
We know that the ratio of the numbers of candidates in the groups P, Q and R was 3 : 5 : 4, respectively.
Let the number of candidates in groups P, Q and R be 3x, 5x and 4x, respectively, where x is a constant of proportionality.
We know that the average scores for the groups P, Q and R were 64, 84, and 72, respectively.
Let the assumed mean score for the three groups combined = 64.
Thus, the effective scores are: (64 −64) = 0, (84 −64) = 20 and (72 −64) = 8 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 155 Thus, we have Average = 0 × 3x + 20 × 5x + 8 × 4x 3x + 5x + 4x = 132x 12x = 11 Thus, the actual average = 11 + 64 = 75.
Note: This approach leads to lesser calculations than that in the previous approach.
74.
We know that there are 50 male and 20 female members.
Thus, the total number of members = 50 + 20 = 70.
As the average of all 70 members is 23 years, sum of their ages = 70 × 23 = 1, 610 As the average of 50 male members is 20 years, sum of their ages = 50 × 20 = 1, 000 Thus, the sum of the ages of 20 female members = 1, 610 −1, 000 = 610 Thus, the average age of these 20 female members = 610 20 = 30.50 The correct answer is Option A.
Alternate approach 1: Here we know that the average age of 70 members = 23 years.
The average age of 50 male members = 20 Let us consider the average age of the remaining 20 female members = x years.
By the method of alligation: Males Females 20 𝑥 23 𝑥−23 23 −20 = 3 © 1999–2024 Manhattan Review www.manhattanreview.com 156 Quantitative Reasoning Question Bank −PS Solutions Thus, we have x −23 3 = 50 20 = 5 2 => 2 (x−23) = 5 × 3 => 2x −46 = 15 => 2x = 61 x = 30.50 years.
Alternate approach 2: According to the data, we have Average age of 50 male members is 20 years and the average age of all 70 members is 23 years.
Let us reduce the above averages by 20.
Thus, the modified data is: Average age of 50 male members is 0 years and the average age of all 70 members is 3 years.
Let, in the above situation, the average age of the 20 female members be a years.
Thus, we have 50 × 0 + 20a 70 = 3 => 210 = 20a => a = 10.5 To get the actual average age of the females, we must add 20 (which we had subtracted initially).
Thus, the actual average age of the females = 10.5 + 20 = 30.5 Although practically, above approach is not preferred one, for the sake of understanding, you must learn it.
75.
This question is crafted to acknowledge the importance of Alternate approach 2 discussed in the previous question. In this question, while other approaches will certainly consume more time, Alternate approach 2 would excel.
According to the data, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 157 Average age of 50 male members is 20.89 years and the average age of all 50 + 20 = 70 members is 23.89 years.
Let us reduce the above averages by 20.89.
Thus, the modified data is: Average age of 50 male members is 20.89 −20.89 = 0 years and the average age of all 70 members is 23.89 −20.89 = 3 years.
Let, in the above situation, the average age of the 20 female members be a years.
Thus, we have 50 × 0 + 20a 70 = 3 => 20a = 210 => a = 10.50 To get the actual average age of the females, we must add 23.89 (which we had subtracted initially).
Thus, the actual average age of the females = 10.50 + 20.89 = 31.39 years The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 158 Quantitative Reasoning Question Bank −PS Solutions 5.5 Ratio & Proportion 76.
Fixed cost = $25,000.
The total cost for 50,000 bearings = $100,000.
Thus, the variable component of the total cost for 50,000 bearings = $ (100, 000 −25, 000) = $75, 000.
Thus, variable cost per bearing = $ 75, 000 50, 000 = $1.50.
Thus, the variable component of the total cost for 100,000 bearings = $ (100, 000 × 1.50) = $150, 000.
Thus, the total cost for 100,000 bearings = $ (25, 000 + 150, 000) = $175, 000.
The correct answer is Option C.
77.
We know that the beaker was filled with 40 liters of water and liquid chemical with the components in the ratio 3 : 5, respectively.
Initial quantity of water = 3 3 + 5 × 40 = 15 liters.
Initial quantity of liquid chemical = 5 3 + 5 × 40 = 25 liters.
Amount of water evaporated per day = 2% of 15= 0.3 liters.
Thus, total amount of water evaporated in 10 days = 0.3 × 10 = 3 liters.
Amount of liquid chemical evaporated per day = 5% of 25 = 1.25 liters.
Thus, total amount of liquid chemical evaporated in 10 days = 1.25 × 10 = 12.5 liters.
Thus, total amount of mixture evaporated in 10 days = 3 + 12.5 = 15.5 liters.
Thus, the percent of the original amount of mixture evaporated = 15.5 40 × 100 = 38.75%.
The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 159 78.
Fraction of total dolls that are Barbie dolls = 3 5 . . . (i) Thus, fraction of total dolls that are non-Barbie dolls = 1 −3 5 = 2 5 . . . (ii) Fraction of Barbies purchased before the age of 10 = 4 7 Thus, fraction of Barbies purchased at the age of 10 or later = 1 −4 7 = 3 7 . . . (iii) Thus, from (i) and (iii), we have Fraction of total dolls that are Barbies and purchased at the age of 10 or later = 3 5 × 3 7 = 9 35 Thus, we have 9 35 of the total dolls = 90 => Total dolls = 90 × 35 9 = 10 × 35 = 350 Thus, from (ii): Number of non-Barbie dolls = 350 × 2 5 = 140 The correct answer is Option C.
79.
We know that, for a ratio 0 < x y < 1: • 0 < x y < x + k y + k !
< 1 ... (i), if k is a positive number • 0 < x −k y −k !
< x y < 1 ... (ii) Also, for a ratio x y > 1: • 1 < x + k y + k !
< x y ... (iii), if k is a positive number • 1 < x y < x −k y −k !
... (iv) In the above problem, we have The given ratio of ages of John and Suzy = 5 6 (< 1) =≈0.83 © 1999–2024 Manhattan Review www.manhattanreview.com 160 Quantitative Reasoning Question Bank −PS Solutions Thus, after 10 years, ages of both would increase by 10.
Hence, the final ratio must be greater than 5 6(= 0.83) (from relation (i) above).
Working with the options, we have Option A: 2 3 = 0.67 ≯0.83 −Does not satisfy Option B: 13 20 = 0.65 ≯0.83 −Does not satisfy Option C: 11 15 = 0.73 ≯0.83 −Does not satisfy Option D: 4 5 = 0.8 ≯0.83 −Does not satisfy Option E: 9 10 = 0.90 > 0.83 −Satisfies The correct answer is Option E.
80.
Since the problem asks us to find a fraction value, we can assume any suitable value of the total number of phones and the time taken to produce a feature phone since the initial value does not affect the final answer.
Let the total number of phones be 5.
Thus, the number of feature phones = 2 5 × 5 = 2.
Number of smartphones = 5 −2 = 3.
Let the time taken to produce a feature phone = 5 hours.
Thus, the time taken to produce a smartphone = 8 5 × 5 = 8 hours.
Thus, total time taken to produce smartphones = 3 × 8 = 24 hours.
Total time taken to produce feature phones = 2 × 5 = 10 hours.
Thus, total time taken to produce all the phones = 24 + 10 = 34 hours.
Thus, the required fraction = 24 34 = 12 17 The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 161 81.
We have Number of shirts Number of trousers = 4 5 . . . (i) Number of jackets Number of shirts = 3 8 . . . (ii) Number of sweaters Number of trousers = 6 5 Taking reciprocal on both sides: => Number of trousers Number of sweaters = 5 6. . . (iii) Thus, from the above three equations, we have Number of jackets Number of sweaters = Number of jackets Number of shirts × Number of shirts Number of trousers × Number of trousers Number of sweaters = 3 8 × 4 5 × 5 6 = 1 4 The correct answer is Option C.
82.
Since the question asks for a fraction value, we can choose any suitable initial value of the to-tal number for members for ease of calculation as the initial value will not affect the final answer.
Let the number of members be 100.
Thus, the number of male members = 3 5 × 100 = 60.
Number of female members = 100 −60 = 40.
Fraction of male members who attended the prayer = 3 5 Thus, the fraction of male members who did not attend the prayer = 1 −3 5 = 2 5 Thus, the number of male members who did not attend the prayer = 2 5 × 60 = 24.
Fraction of female members who attended the prayer = 7 10 Thus, the fraction of female members who did not attend the prayer = 1 −7 10 = 3 10 Thus, the number of female members who did not attend the prayer = 3 10 × 40 = 12.
© 1999–2024 Manhattan Review www.manhattanreview.com 162 Quantitative Reasoning Question Bank −PS Solutions Thus, total number of members who did not attend the prayer = 24 + 12 = 36.
Thus, the required fraction = Number of male members who did not attend the prayer Total number of members who did not attend the prayer = 24 36 = 2 3 The correct answer is Option C.
83.
Total amount to be paid = $100.
Amount paid by Suzy = $20.
Thus, the amount paid by John and David = $ (100 −20) = $80.
We know that John paid 5 3 th of what David paid.
Thus, ratio of the amounts paid by John and David = 5 : 3.
Thus, we need to divide $80 in the ratio 5 : 3.
Thus, amount paid by David = $ 3 5 + 3 × 80 = 30 Thus, the fraction of the total amount paid by David = 30 100 = 3 10 The correct answer is Option D.
84.
Total number of shirts and trousers = X.
We know that the ratio of the number of trousers to the number of shirts = 1 : 5.
Thus, the number of shirts = 5 5 + 1 × X = 5X 6 Thus, the number of cotton shirts = 1 5 th of the number of shirts = 1 5 × 5X 6 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 163 = X 6 The correct answer is Option E.
85.
Total weight of the rod = 20 pounds.
We know that the weight (w) of each piece is directly proportional to the square of its length (l).
Weight of the first piece (w1), which is 36 feet long (l1) = 16 pounds.
Thus, weight of the other piece (w2) = 20 −16 = 4 pounds.
Thus, we have w ∝l2 => w1 w2 = (l1)2 (l2)2 => 16 4 = 362 (l2)2 => (l2)2 = 362 × 4 16 => (l2)2 = 362 4 => (l2)2 = 36 2 2 => (l2)2 = 182 => (l2)2 = 18 Thus, the length of the second piece = 18 feet.
The correct answer is Option C.
86.
We know that the ratio of coins of John to that of Suzy = 3 : 4 Let John’s and Suzy’s coins be 3x and 4x, respectively, where x is a constant of proportionality.
Thus, the total number of coins = 3x + 4x = 7x.
Since John’s share exceeds 2 7 of the total number of coins by 100, we have 3x = 2 7 × 7x + 25 © 1999–2024 Manhattan Review www.manhattanreview.com 164 Quantitative Reasoning Question Bank −PS Solutions => x = 25 Thus, Suzy’s coins = 4x = 4 × 25 = 100.
The correct answer is Option B.
87.
Say Material cost, Labour cost, Factory overhead cost, and Office overhead cost are a, b, c, and d.
Since we have to deal with fractions, 3 7, 4 7, and 1 2, let us assume that the total cost = $1,400.
Thus, we have...
a + b = 3 7 of 1,400 = 600 ...(1) b + c = 1 2 of 1,400 = 700 ...(2) c + d = 4 7 of 1,400 = 800 ...(3) a + d = 1 2 of 1,400 = 700 ...(4) Subtracting equation (2) from (1), we get c −a = 100 => c > a; a is not the highest.
Subtracting equation (2) from (3), we get d −b = 100 => d > b; b is not the highest.
We cannot establish whether c >=< d. Factory overhead cost or Office overhead cost can be highest.
The correct answer is Option E.
88.
Let the number of points for the 1st question = x.
We know that each question is worth 2 points more than the preceding question.
Thus, the worth of each question in points for the 20 questions forms an arithmetic progression with the first term as x and a constant difference between consecutive terms of 2.
The nth term in arithmetic progression = a + (n −1) × d; (a is the first term, and d is the constant difference between consecutive terms) Thus, the number of points for the 20th question = x + (20 −1) × 2 = (x + 38).
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 165 Since the points for the above questions have a constant difference, the average points per question = First term + Last term 2 = x + (x + 38) 2 = (x + 19) Thus, the total points for all 20 questions = 20 × (x + 19).
Thus, we have 20 (x + 19) = 400 => x = 1 Thus, the number of points for the 4th question = x + (4 −1) × 2 = 1 + 6 = 7.
The correct answer is Option B.
89.
Let the number of students in the school = x.
Thus, number of students taking the science course = 80 + x 3 .
Thus, the number of students taking chemistry = 1 3 80 + x 3 .
Since 1 6 th of the students are taking chemistry, we have 1 3 80 + x 3 = x 6 => 80 + x 3 = x 2 => x 2 −x 3 = 80 => x 6 = 80 => x = 480 The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 166 Quantitative Reasoning Question Bank −PS Solutions 90.
Let the number of males = x.
Thus, the number of females = (50 −x).
Thus, the number of employees who eat company breakfast = x 4 + 50 −x 5 = 5x + 4 (50 −x) 20 = x + 200 20 = x 20 + 10 Thus, to maximize the above value, we must have the largest possible value of x.
Also, since x 20 must be an integer, we have to choose x = 40.
Thus, the required maximum value = 40 20 + 10 = 12 The correct answer is Option D.
91.
Let T, F and M be the total number of students, the number of female and the number of male, respectively.
We have 1 8 × F = 1 12×T => F = 8 12×T => F = 2 3×T => M = 1−2 3 ×T => M = 1 3 × T => M F = 1 3 × T 2 3 × T => M F = 1 3 2 3 = 1 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 167 Thus, the required ratio = 1 2 The correct answer is Option C.
92.
The incorrect total of the score of the class = 40 × 32 = 1, 280.
The correct total of the score of the class = 1, 280 −30 + 40 = 1, 290.
The correct average (arithmetic mean) score of the class = 1, 290 40 = 32.25.
The information that the average (arithmetic mean) score of the girls is 30 and that of boys is 40 is redundant.
The correct answer is Option C.
Alternate approach: Since the correct score (40) is greater than the incorrect score (30), the average score would increase by (40 −30) 40 = 0.25.
Thus the correct average (arithmetic mean) score of the class = 32 + 0.25 = 32.25.
93.
We have 120 liters of 25% Chemical A solution.
Thus, amount of Chemical A = 25% of 120 = 25 100 × 120 = 30 liters Let x liters of Chemical A be added.
Thus, the final amount of Chemical A = (x + 30) liters.
Total volume of the solution = (x + 120) liters.
Since the final concentration of Chemical A is 40%, we have x + 30 x + 120 × 100 = 40 => x + 30 x + 120 = 40 100 = 2 5 => 5x + 150 = 2x + 240 => x = 30 © 1999–2024 Manhattan Review www.manhattanreview.com 168 Quantitative Reasoning Question Bank −PS Solutions The correct answer is Option E.
Alternate approach: Amount of water in the initial solution = (100 −25) % of 120 = 75% of 120 = 90 liters.
Let x liters of Chemical A be added.
Total volume of the solution = (x + 120) liters.
Final concentration of water = (100 −40) % = 60% Since water is not added, the quantity of water remains the same, i.e. 90 liters.
Thus, we have 90 x + 120 × 100 = 60 => 90 x + 120 = 60 100 = 3 5 => 3x + 360 = 450 => x = 30 94.
Increase in the number of chickens from the first month (144) to the second month (c) = (c −144) Thus, fractional increase = Increase Original value = c −144 144 Increase in the number of chickens from the second month (c) to the third month (256) = (256 −c) Thus, fractional increase = Increase Original value = 256 −c c Since the fractional increases are the same, we have c −144 144 = 256 −c c www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 169 => c 144 −1 = 256 c −1 => c 144 = 256 c => c2 = 144 × 256 => c = 12 × 16 => c = 192 The correct answer is Option A.
Alternate approach 1: Since the number of chickens increased by the same fraction during each of the two periods, we can say that the ratio of the number of chickens in consecutive intervals would be equal.
Thus we have 144 c = c 256 => c2 = 144 × 256 => c = 12 × 16 => c = 192 Alternate approach 2: Since the fractional increase in each period is the same, we can conclude that the percent change in each period is also the same.
Let the percent change be r.
Thus, using the concept of compound interest, we have 256 = 144 × 1 + r 100 2 => 1 + r 100 2 = 16 12 2 => 1 + r 100 = 16 12 => 1 + r 100 = 4 3 => r 100 = 1 3 © 1999–2024 Manhattan Review www.manhattanreview.com 170 Quantitative Reasoning Question Bank −PS Solutions => r = 100 3 % =>The number of chickens by second month, c = 144 × 1 + 100/3 100 1 = 144 × 4 3 = 192 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 171 5.6 Speed, Time & Distance 95.
Let the average speed for the trip be S.
T1 = Time required to travel 900 miles with speed S = 900 S hours S is increased by 10 miles; so new speed = (S + 10) mph T2 = Time required to travel 900 miles with speed (S + 10) = 900 S + 10 hours Given that the difference between T1and T2 is 1 hour => 900 S − 900 S + 10 = 1 => 900 1 S − 1 S + 10 = 1 => S + 10 −S S(S + 10) = 1 900 => 10 × 900 = S2 + 10S => S2 + 10S −9, 000 = 0 Roots of above quadratic equations are ‘90’ and ‘−100’.
Since speed cannot be negative, ‘−100’ is ignored, so S = 90 mph.
The correct answer is Option E.
96.
The truck traveled 4 miles less per gallon on the state highway compared to on the national highway.
Let’s consider if the truck travels x miles per gallon on the national highway then the truck travels (x −4) miles per gallon on the state highway.
Capacity of the full tank of diesel on the national highway = 336 x gallons Capacity of the full tank of diesel on the state highway = 224 x −4 gallons Since the above fractions must be equal, we have 336 x = 224 x −4 => 3 x = 2 x −4 => 3x −12 = 2x © 1999–2024 Manhattan Review www.manhattanreview.com 172 Quantitative Reasoning Question Bank −PS Solutions => x = 12 As the truck travels (x −4) miles per gallon on the state highway, required answer is 12 −4 = 8.
The correct answer is Option B.
97.
Time required to travel 10 miles at speed 60 miles per hour = 10 60 hours = 10 60 × 60 minutes = 10 minutes Now new time required = 10 + 5 = 15 minutes = 15 60 = 1 4 hours.
Speed = Distance Time = 10 miles 15 Minutes = 10 1 4 = 40 miles per hour The correct answer is Option B.
98.
Speed for first 10-minutes interval = 30 miles per hour Speed for second 10-minutes interval = 40 miles per hour Speed for third 10-minute interval = 50 miles per hour Speed for fourth 10-minute interval = 60 miles per hour Distance travelled in the fourth 10-minute interval = 60 60 × 10 = 10 miles The correct answer is Option D.
99.
Time taken for the onward journey = 600 500 = 6 5 hours.
Time taken for the return journey = 600 400 = 3 2 hours.
Thus, total time taken for the round trip = 6 5 + 3 2 = 27 10 hours.
Total distance travelled = 2 × 600 = 1200 miles.
Thus, average speed = Total distance Total time = 1, 200 27 10 = 1200 × 10 27 = 1, 000 × 4 9 =≈444 miles/hr.
The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 173 Alternate approach: Since the distance travelled for the onward and the return journey is the same, we have Average speed = 2 × Speed1 × Speed2 Speed1 + Speed2 = 2 × 500 × 400 500 + 400 = 2 × 500 × 400 900 = 4, 000 9 =≈444 mph 100.
Average speed of the truck for 1/2 of 800 miles (= 400 miles) = 40 miles per hour Thus, time taken to cover the above distance (400 miles) = 400 40 = 10 hours Average speed of the truck or the entire 800 miles = 50 miles per hour Thus, time taken to cover the total distance = 800 50 = 16 hours Thus, time taken by the truck to cover the remaining miles 400 miles = 16 −10 = 6 hours Thus, average speed of the truck for the remaining miles 400 miles = 400 6 = 66.66 =≈67 miles per hour.
The correct answer is Option D.
101.
Let the marathoner’s speed on the first day be x miles per hour.
Thus, the his speed on the second day was (x + 3) miles per hour.
Let the time for which the marathoner ran on the first day be t hours.
Thus, the time for which he ran on the second day = (8 −t) hours.
Distance covered by him on the first day = (x × t) miles.
Distance covered by him on the second day = ((x + 3) × (8 −t)) miles.
Thus, total distance covered in two days = [xt + (x + 3) (8 −t)] miles.
Thus, we have xt + (x + 3) (8 −t) = 36 => xt + 8x −xt + 24 −3t = 36 => 8x = 3t + 12 Since this is a single linear equation with two variables, we cannot get the unique value of x; however, we can get consistent values of x.
© 1999–2024 Manhattan Review www.manhattanreview.com 174 Quantitative Reasoning Question Bank −PS Solutions Working with the options one at a time: Option A: x = 0.5 => t = −4 −Not possible, since t cannot be negative We can deduce that since this is a ‘Could be value’ type of question, and we see that the smallest option value of x yields negative value for t, thus we should first try the largest value of x among the options.
Option E: x = 2 => t = 4/3 – Possible Since this is a ‘Could be value’ type of question, and only one option is correct, Option E is the correct answer. There is no need to check other options since two or more options cannot be simultaneously correct.
Thus, we have x = 2 The correct answer is Option E.
Alternate approach: Average speed of the marathoner for the two days = Total distance Total time = 36 8 = 4.5 miles per hour.
If the average speed on the first day be x miles per hour, the average speed on the second day should be (x + 3) miles per hour.
Thus, the average speed of 4.5 miles per hour must lie between the values of the speeds on the two days.
Thus, we have x < 4.5 < x + 3 Among the options, only x = 2 satisfies.
102.
On seeing this question, one would immediately calculate the time taken to meet using the data given.
From there on, one would try to find the distance between the trains 2 hours before they meet.
However, such calculations are not necessary.
The question simply asks us, “If two trains traveling at 50 miles per hour and 60 miles per hour need 2 hours to meet, how far away are they from one another.” Since both trains travel for 2 hours before they meet, one train travels 50 × 2 = 100 miles and the other train travels 60 × 2 = 120 miles.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 175 Thus, together the trains travel (100 + 120) = 220 miles before they meet.
Thus, distance between the trains 2 hours before they meet = 220 miles.
The correct answer is Option E.
103.
Length of section = 10 miles.
Original speed limit = 50 miles per hour.
Thus, time taken to cover this distance = 10 50 = 1 5 hours New speed limit = 40 miles per hour.
Thus, time taken to cover this distance = 10 40 = 1 4 hours Thus, the required difference between the time durations is given = 1 4 −1 5 hours = 1 20 hours = 1 20 × 60 = 3 minutes The correct answer is Option A.
104.
We know that Speed of Train B = 25% more than the speed of Train A => Speed of Train B = (100 + 25)% of the speed of Train A => Speed of Train B = 125% of the speed of Train A => Speed of Train B = 125 100 × (Speed of Train A) => Speed of Train B = 5 4 × (Speed of Train A) => Speed of Train B Speed of Train A = 5 4 Since time is inversely proportional to speed for a constant distance, we have Time taken by Train B to cover a distance Time taken by Train A to cover the same distance = 1 5 4 = 4 5 We know that Train A took 4 hours to cover the distance.
© 1999–2024 Manhattan Review www.manhattanreview.com 176 Quantitative Reasoning Question Bank −PS Solutions Thus, we have => Time taken by Train B to cover a distance = 4 × 4 5 = 16 5 = 31 5 hours The correct answer is Option C.
Alternate approach: Train A took 4 hours to travel 100 miles.
Thus, the average speed of Train A = 25 miles per hour.
Since the average speed of Train B is 25% more than that of Train A, we have Average speed of Train B = 25 + 25% of 25 = 125 4 miles per hour.
Thus, time required to travel 100 miles by Train B = Distance Speed = 100 125 4 = 16 5 = 31 5 hours 105.
Jeff’s speed = 40 miles per hour.
Thus, distance covered by Jeff in 60 minutes = 40 miles.
Thus, distance covered by Jeff in 36 minutes = 40 60 × 36 = 24 miles.
This distance of 24 miles is 3 times of what Amy drives in 30 minutes.
Thus, distance covered by Amy in 30 minutes = 24 3 = 8 miles.
Thus, distance covered by Amy in 60 minutes = 60 30 × 8 = 16 miles.
Thus, Amy’s speed = 16 miles per hour.
The correct answer is Option C.
106.
Time taken for the first bus to cover the distance between the two depots A & B = 5 hours.
Time taken for the second bus to cover the distance between the two depots A & B = 3 hours.
Since we need to find the time when the two buses pass one another, the actual length of the distance is not required.
So, we can assume a suitable value of the distance for ease of calculations.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 177 Let the distance between the depots = least common multiple of 5 and 3 = 15 miles.
Thus, speed of the first bus = 15 5 = 3 miles/hour.
Speed of the second bus = 15 3 = 5 miles/hour.
We know that the first bus started one hour earlier than when the second bus started.
Thus, in 1 hour, the distance covered by the first bus = 3 × 1 = 3 miles.
Thus, at 8:00 am, the distance between the two buses = (15 −3) = 12 miles.
At 8:00 am, both buses approach one another at speeds 3 miles/hour and 5 miles/hour.
Thus, when the buses pass one another, the ratio of the distances covered by the trains would be equal to the ratio of their respective speeds = 3 : 5 Thus, distance covered by the first bus = 3 3 + 5 × 12 = 4.5 miles.
Time taken by the first bus to cover 4.5 miles = 4.5 3 = 1.5 hours = 1 hour 30 minutes.
The buses will meet at 8:00 am + 1 hour 30 minutes = 9:30 am.
Alternately, we can use the concept of relative speed: Since the buses travel in opposite directions, their relative speed = (3 + 5) = 8 miles/hour.
Thus, time taken to cover 12 miles = 12 8 = 1.5 hours = 1 hour 30 minutes.
Thus, the time when the two buses pass one another = 1 hour 30 minutes past 8:00 am The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 178 Quantitative Reasoning Question Bank −PS Solutions 5.7 Time & Work 107.
Number of copies made per hour by A = 1,500.
Thus, number of copies made per hour by B = 2 × 1, 500 = 3, 000 (Since it is twice as efficient as the first machine) Since B works 12 hours a day, number of copies made in a day = 3, 000 × 12 = 36,000 Thus, number of copies made in 20 days = 36, 000 × 20 = 720,000 The correct answer is Option E.
108.
According to the given data, the pump filled 5 6 −1 2 = 1 3 part of the pool in 21 3 = 7 3 hours Thus, time taken by the pump to fill the entire pool = 7 3 1 3 = 7 3 × 3 1 = 7 hours The correct answer is Option C.
109.
Let the volume of the pool = LCM (10, 15) = 30 liters The first pump can fill the pool in 10 hours.
Thus, the rate of the first pump = 30 10 = 3 liters per hour.
The second pump can fill the pool in 15 hours.
Thus, the rate of the second pump = 30 15 = 2 liters per hour.
The first pump was on for the entire 7 hours in which it filled = 7 × 3 = 21 liters.
Thus, the remaining (30 −21) = 9 liters were filled by the second pump.
Time taken for the second pump to fill 9 liters = 9 2 = 4.50 hours.
Thus, of the total 7 hours, the second pump was on for 4.50 hours.
Thus, the first pump alone was on for (7 −4.50) = 2.50 hours.
Thus, h = 2.50 The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 179 110.
The emptying pipe can empty the pool which is 3 4 th full in 9 hours.
Thus, time taken to empty the entire pool = 9 × 4 3 = 12 hours.
It is given that capacity of swimming pool is 5,760 gallons.
Thus, the rate at which the emptying pipe removes water = 5, 760 12 = 480 gallons per hour.
The rate at which the pool can be filled = 12 gallons per minute = 12 × 60 = 720 gallons per hour.
Thus, the effective filling rate when both filling and emptying occur simultaneously = 720 −480 = 240 gallons per hour.
Since we need to fill only half the pool, the volume required to be filled = 5, 760 2 = 2, 880 gallons.
Thus, time required = 2, 880 240 = 12 hours.
The correct answer is Option B.
111.
We know that lathe machine B manufactures 300 X-type bearings in 60 days.
Since lathe machine A manufactures bearings thrice as fast as machine B does, time taken by lathe machine A to manufacturer 300 X-type bearings = 60 3 = 20 days Since each Y-type bearing takes 2.5 times the time taken to manufacturer each X-type bearing, the time taken by lathe machine A to manufacturer 300 Y-type bearings = 20 × 2.5 = 50 days Thus, the number of Y-type bearings manufactured by lathe machine A in 50 days = 300.
Thus, the number of bearings manufactured by lathe machine A in 10 days = 10 50 × 300 = 60 The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 180 Quantitative Reasoning Question Bank −PS Solutions 112.
Numbers of copies made by Photocopier A in 1 hour = 1, 200 3 = 400 Numbers of copies made by Photocopier B in 1 hour = 1, 200 2 = 600 Numbers of copies made by Photocopier C in 1 hour = 1, 200 6 = 200 Thus, when all the three photocopiers work together, total numbers of copies made in 1 hour = 400 + 600 + 200 = 1, 200 copies Thus, time taken to make 3,600 copies = 3, 600 1, 200 = 3 hours.
The correct answer is Option D.
113.
Cost of food consumed by 5 men in 4 days = $150 => Cost of food consumed by 1 man in 4 days = $ 150 5 = $30 => Cost of food consumed by 1 man in 1 day = $ 30 4 = $ 15 2 We know that one woman consumes three-fourth the amount of food consumed by a man.
Thus, 2 women consume food equivalent to 2 × 3 4 = 3 2 men.
Thus, 4 men and 2 women are equivalent to 4 + 3 2 = 11 2 men.
Thus, cost of food consumed by 11 2 men in 1 day = $ 15 2 × 11 2 = $ 165 4 => Cost of food consumed by 11 2 men in 8 days = $ 165 4 × 8 = $330 The correct answer is Option B.
114.
Let the time taken by Mark and Kate working together = x hours Thus, time taken by Mark, working alone = (x + 12) hours Time taken by Kate, working alone = (x + 27) hours Let us assume the total work to be 1 unit.
Part of work done by Mark and Kate, working together, in 1 hour = 1 x Part of work done by Mark in 1 hour = 1 x + 12 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 181 Part of work done by Kate in 1 hour = 1 x + 27 Thus, we have 1 x = 1 x + 12 + 1 x + 27 => 1 x = (x + 27) + (x + 12) (x + 12) (x + 27) => (x + 12) (x + 27) = x (2x + 39) => x2 + 39x + 324 = 2x2 + 39x => x2 = 324 Since x is positive, we have x = √ 324 = 18 The correct answer is Option C.
Alternate approach: Work done by Mark and Kate in x hours = Work done by Mark in (x + 12) hours Cancelling work done by Mark in x hours from both sides: Work done by Kate in x hours = Work done by Mark in 12 hours => Work done by Kate in 1 hour = Work done by Mark in 12 x hours . . . (i) Again, we have Work done by Mark and Kate in x hours = Work done by Kate in (x + 27) hours Cancelling work done by Kate in x hours from both sides: Work done by Mark in x hours = Work done by Kate in 27 hours => Work done by Kate in 27 hours = Work done by Mark in x hours => Work done by Kate in 1 hour = Work done by Mark in x 27 hours . . . (ii) Thus, from (i) and (ii), we have 12 x = x 27 © 1999–2024 Manhattan Review www.manhattanreview.com 182 Quantitative Reasoning Question Bank −PS Solutions => x2 = 324 => x = 18 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 183 5.8 Computational 115.
Amount of chemical evaporated in y minutes = x liters.
Thus, the amount of chemical evaporated in 1 minute = x y liters.
Thus, the amount of chemical evaporated in z minutes = xz y liters.
Cost of 1 liter of the chemical = $25.
Thus, cost of the chemical evaporated in z minutes = $ 25 × xz y !
= $ 25xz y !
.
The correct answer is Option D.
116.
We know that the total cost of producing 25,000 pens is $37,500 and the total cost of producing 35,000 pens is $47,500.
Since it is given that total cost of producing pens is governed by a linear function, total cost must have a fixed cost component and a variable complement.
Note that fixed cost component is same, irrespective of how many pens Company X makes.
Thus, incremental cost (variable complement) of producing additional 10,000 pens (= 35,000 − 25,000) pens = $47,500 −$37,500 = $10,000.
=> Thus, incremental cost (variable complement) of producing additional 1 pen = 10, 000 10, 000 = $1 Thus, incremental cost (variable complement) of producing additional 15,000 pens (= 50,000 − 35,000) pens = 1 × 15, 000 = $15, 000 => Thus, total cost of producing 50,000 pen = total cost of producing 35,000 pen + incremental cost (variable complement) of producing additional 15,000 pens = $47,500 + $15,000 = $62,500 The correct answer is Option C.
117.
The cost of four pencils = $ (1.35 × 4) = $5.40 The cost of two erasers = $ (0.30 × 2) = $0.60 Thus, total cost = $(5.40 + 0.60) = $6.00 Thus, Suzy has one-third of the above amount = $ 6 3 = $2 The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 184 Quantitative Reasoning Question Bank −PS Solutions 118.
Total increase in population = 378 −360 = 18 million Increase in population per month = 30,000 Thus, increase in population per year = 30, 000 × 12 = 360,000 = 0.36 million Thus, number of years required for the increase = 18 0.36 = 50 years Thus, the population would be 378 million in the year (2012 + 50) = 2062 The correct answer is Option C.
119.
The restaurant uses 1 2 cup milk-cream in each serving of its ice-cream.
Since each carton has 2 1 2 = 5 2 cups of milk-cream, number of servings of ice-cream possible using one carton = 5 2 1 2 = 5 Thus, number of cartons required for 98 servings of the ice-cream = 98 5 =≈19.3 However, the number of cartons must be an integer.
Thus, the minimum number of cans required is 20.
The correct answer is Option C.
120.
We need to minimize the total number of coins such that each box has at least 2 coins.
We know that at the most 3 boxes can have the same number of coins.
Since we need to minimize the total number of coins, we must have as many boxes having the same number (minimum possible number, i.e. 2 coins) of coins as possible.
Thus, for each of the 3 boxes containing an equal number of coins, we have ‘2’ coins.
Thus, number of coins in the 3 boxes = 2 × 3 = 6.
Since each of the remaining 4 boxes have a different number of coins, let us put in 3, 4, 5, and 6 coins in those boxes.
Thus, the total number of coins = 6 + (3 + 4 + 5 + 6) = 24.
The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 185 121.
Distance covered by the lava in 1 hour = 15 4 feet = 15 4 5, 280 miles = 15 4 × 5, 280 miles Hours needed to cover 3 2 miles = 3 2 15 4×5,280 = 3 × 4 × 5, 280 2 × 15 = 2, 112 hours Days needed to cover 3 2 miles = 2, 112 24 = 88 days The correct answer is Option E.
122.
Cross-sectional area of the bar = 4 square feet Rate at which the bar moves through the conveyor = 360 feet per hour = 360 60 × 60 feet per second = 0.1 feet per second Thus, volume of the bar that moves through the conveyor per second = 0.1 × 4 cubic feet = 0.4 cubic feet Thus, time taken to move 0.4 cubic feet of bar through the conveyor = 1 second.
Thus, time taken to move 8.4 cubic feet of bar through the conveyor = 8.4 0.4 = 21 seconds.
The correct answer is Option A.
123.
We know that for a salary grade G, the hourly wage W, in dollars, is given by: W = 1, 140 + 45(G −1) Thus, for a salary grade G = 1, the corresponding hourly wage W = $ (1, 140 + 45 (1 −1)) = $1, 140 © 1999–2024 Manhattan Review www.manhattanreview.com 186 Quantitative Reasoning Question Bank −PS Solutions Again, for a salary grade G = 7, the corresponding hourly wage W = $ (1, 140 + 45 (7 −1)) = $(1, 140 + 270) = $1, 410 Thus, the required difference = $ (1, 410 −1, 140) = $270.
The correct answer is Option B.
Alternate approach: We have wage equation: W = 1, 140 + 45(G −1) => W = 1, 140 −45 + 45G We see that in the above equation, 45G is variable and ‘1, 140 −45’ is a constant for the workers of all the grades.
Thus, the difference between the remuneration would result because of 45G.
Thus, difference in remuneration of grade-7 and grade-1 worker = 45 × 7 −45 × 1 = $270.
124.
Considering the radios sold: We know that the price of the certain radio was the 15th highest price as well as the 20th lowest price among the prices of the radios sold.
Thus, the number of radios sold at a price greater than the price of the particular radio = 15 −1 = 14 Also, the number of radios sold at a price lower than the price of the particular radio = 20 −1 = 19 Thus, total number of radios sold = 14 + 19 + 1 (including that particular radio) = 34 Considering all the items sold: We know that the price of the certain DVD player was the 29th highest price as well as the 37th lowest price among the prices of all the items sold.
Thus, the number of items sold at a price greater than the price of the particular item (i.e. DVD player) = 29 −1 = 28 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 187 Also, the number of items sold at a price lower than the price of the particular item (i.e. DVD player) = 37 −1 = 36.
Thus, total number of items sold = 28 + 36 + 1 (including that particular radio) = 65 Thus, the number of DVD players = Total number of items −number of radios = 65 −34 = 31 The correct answer is Option B.
125.
Commission received on the first 150 orders at the rate of $25 per order = $ (25 × 150) = $3, 750 Total commission received received = $5,000.
Thus, commission received on the additional orders (above 150) at $12.50 per order = $(5, 000 −3, 750) = $1, 250 Thus, number of additional order made= 1, 250 12.5 = 100 Thus, total number of order = Initial 100 orders + Additional 150 orders = 250 orders The correct answer is Option D.
126.
The loss would be minimum if fewer number of $50 checks were lost as compared to $20 checks.
Thus, the traveler should have cashed a greater number of $50 checks than $20 checks.
We know that the number of $20 checks cashed was 2 more or 2 less than the number of $50 checks cashed.
Thus, from the above reasoning, we can conclude that the number of $20 checks was 2 less than the number of $50 checks.
Since the total number of checks cashed was 10, from the above information, we have Number of $50 checks cashed = 6 Number of $20 checks cashed = 4 Alternately: Let the number of $50 checks cashed = x © 1999–2024 Manhattan Review www.manhattanreview.com 188 Quantitative Reasoning Question Bank −PS Solutions Thus, the number of $20 checks cashed = (x −2) Since the total number of checks cashed is 10, we have x + (x −2) = 10 => x = 6 Thus, the total value of $20 checks cashed = $ (4 × 20) = $80.
The total value of $50 checks cashed = $ (6 × 50) = $300.
Thus, the total value of all the checks cashed = $ (80 + 300) = $380.
We know that the total value of all the checks with him was $2,000.
Thus, the total value of all the checks lost = $ (2, 000 −380) = $1, 620.
The correct answer is Option D.
127.
Let’s list down per kg prices realized upon buying three pack-sizes.
1. 5-kg pack for $16 => 16 5 = $3.20 per kg.
2. 10-kg pack for $26 => 26 10 = $2.60 per kg.
3. 25-kg pack for $55 => 55 25 = $2.20 per kg.
From the above calculation, it is obvious that larger the pack size, smaller the per kg amount a customer has to pay.
Let’s take few options the customer can take to buy 40 kg dog food.
1. Buy 50 kg.: 2 packs of 25 kg Cost = 2 × 55 = $110; though the quantity of food (50 kg) is more than the minimum required (40 kg), we must take this option into consideration since the per kg price of the food is least for a 25 kg pack.
2. Buy 45 kg.: 1 pack of 25 kg and 2 packs of 10 kg Cost = 55 + 2 × 26 = $107 3. Buy 40 kg.: 1 pack of 25 kg, 1 pack of 10 kg, and 1 pack of 5 kg Cost = 55 + 26 + 16 = $97 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 189 From the above, it is clear that the customer must buy 40 kg (1 pack of 25 kg, 1 pack of 10 kg, and 1 pack of 5 kg), paying the minimum price of $97.
The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 190 Quantitative Reasoning Question Bank −PS Solutions 5.9 Interest 128.
Amount after 4 years = $3,200 Amount after 6 years = $3,800 Thus, interest accumulated in 2 (= 6 −4) years = $ (3, 800 −2, 200) = $600 Thus, interest accumulated per year = $ 600 2 = $300 (since under simple interest, interest accumulated every year is constant) Thus, interest accumulated in the first 4 years = $ (300 × 4) = $1, 200 Thus, principal amount invested = $ (3, 200 −1, 200) = $2, 000 Thus, on $2,000 invested, interest accumulated is $300 every year.
Thus, rate of interest = 300 2, 000 × 100 = 15% The correct answer is Option C.
129.
Simple interest accumulated after 2 years = $600 Thus, simple interest per year = $ 600 2 = $300 (since under simple interest, interest accumu-lated every year is constant) Thus, compound interest accumulated after the first year = $300 (equal to the simple interest accumulated after one year) Thus, compound interest accumulated in the second year = $ (300 + 63) = $363 (since the total compound interest accumulated in 2 years is $63 more than that under simple interest) The higher interest in the second year is due to the additional interest on the interest accumu-lated after one year.
Thus, we can say that interest on $300 in one year = $63 Thus rate of interest = 63 300 × 100 = 21% The correct answer is Option D.
130.
Let the sums borrowed at 10% and 8% rate of interest be $x each.
Let the time after which Suzy repays the second sum be t years.
Thus, the time after which she repays the first sum is (t −1) years.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 191 Since the amount to be repaid in either case is the same, the interest accumulated is also equal.
Simple Interest = PRT 100 , where P =Principal, R = Rate of Interest, and T = Time Interval Hence, we have 10 × x × (t −1) 100 = 8 × x × t 100 => 5(t −1) = 4t => t = 5 years Thus, the time for which she borrowed the first sum of money = t −1 = 4 years The correct answer is Option D.
131.
Total amount invested = $100, 000.
Let $x be invested at 3% and $(100, 000 −x) be invested at 4% Thus, at the end of 1 year, interest on $x = x × 3 × 1 100 = 3x 100 Also, at the end of 1 year, interest on $(100, 000 −x) = (100, 000 −x) × 4 × 1 100 = 4 (100, 000 −x) 100 Since the total interest is $3,600, we have 3x 100 + 4 (100, 000 −x) 100 = 3, 600 => 400, 000 −x = 360, 000 => x = 40, 000 => 100, 000 −x = 60, 000 Thus, the fraction of the total invested at 4% = 60, 000 100, 000 = 3 5 The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 192 Quantitative Reasoning Question Bank −PS Solutions Alternate approach: We can use the method of alligation: Let the amounts invested at 3% and 4% be $x and $y, respectively.
The total interest on $100,000 is $3,600.
Thus, the effective rate of interest as a whole = 3, 600 100, 000 × 100 = 3.6%.
Thus, we have $𝑥 $𝑦 3% 4% 3.6% 4 −3.6 = 0.4 3.6 −3 = 0.6 Hence, we have x y = 0.4 0.6 = 2 3 Hence, the required fraction = 3 2 + 3 = 3 5 132.
Let the sum of money invested at 4% rate of interest be $x.
Thus, the sum of money invested at 7% rate of interest = $ (18, 000 −x) Thus, interest on $x at 4% rate of interest in 2 years = $ x × 2 × 4 100 = $ 8x 100 Also, interest on $ (18, 000 −x) at 7% rate of interest in 2 years = $ (18, 000 −x) × 2 × 7 100 = $ 14 (18, 000 −x) 100 Since total interest is $2,100, we have 8x 100 + 14 (18, 000 −x) 100 = 2, 100 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 193 => 14 × 18, 000 −6x = 210, 000 => 6x = 252, 000 −210, 000 => x = 7, 000 The correct answer is Option C.
133.
Let the sums of money invested at 10% and 20% rates of interest be $x each.
Since the difference between the interests earned after three years is between $120 and $140, we have 120 < x × 20 × 2 100 − x × 10 × 2 100 < 140 => 120 < 20x 100 < 140 => 600 < x < 700 We need to find the difference between the amounts earned after 2 years at compound interest at the same rates as above.
Thus, the required difference = x 1 + 20 100 2 −x 1 + 10 100 2 = x ( 1 + 20 100 2 − 1 + 10 100 2) = x 1 + 20 100 − 1 + 10 100 1 + 20 100 + 1 + 10 100 = x 10 100 2 + 30 100 = x (0.1 × 2.3) = x × 0.23 We know that 600 < x < 700 Thus, the required difference (x × 0.23) is 600 × 0.23 < x × 0.23 < 700 × 0.23 => 138 < Required difference < 161 Thus, the required difference lies between $138 and $161.
Only Option D lies in the above range.
The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 194 Quantitative Reasoning Question Bank −PS Solutions 134.
Population at the start of the experiment = x Increase in population at the end of the 1st month = 2x Thus here we can say that, rate of increase = 2x x × 100 = 200% This 200% increase remains same for each of the next 4 months.
Thus, applying the concept of compounding, we have x 1 + 200 100 5 > 1, 000 => x > 1, 000 (3)5 => x > 1, 000 243 => x > 1, 000 =≈250 =≈4 Since we 250 > 243, thus 1, 000 243 > 4 => x > 4 Since x must be an integer value (it represents the number of organisms), the minimum possible value of x = 5.
The correct answer is Option D.
Alternate approach: Population at the start of the experiment = x.
Increase in population at the end of the 1st month = 2x.
Thus, population size at the end of the 1st month = x + 2x = 3x.
Increase in population after the 2nd month = 2 × 3x = 6x.
Thus, population size at the end of the 2nd month = 3x + 6x = 9x.
Thus, we observe that the population size triples after every month.
Thus, the population size at the end of the 5th month = 3 × (The population size at the end of the 4th month) = 3 × 3 × (The population size at the end of the 3rd month) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 195 = 3 × 3 × 3 × (The population size at the end of the 2nd month) = 33 × 9x = 243x Thus, we have 243x > 1, 000 => x > 1, 000 243 => x > 4.1 Since x must be an integer value (it represents the number of organisms), the minimum possible value of x = 5.
135.
The description about the calculation of interest is basically follows the concept of compound interest.
Since interest is calculated after every two-month period, in a year, it will be calculated 6 times.
Also the rate on interest given is 12% per annum, thus, the rate of interest per period would be 12/6 = 2% per two-month period.
The amount (A) under compound interest on a sum of money (P) invested at (r%) rate of interest for n periods is given by: A = P 1 + r 100 n We have to find out the value of A P .
We know that n = 6 two-month periods and r = 2% per two-month period Thus, A P = 1 + 2 100 6 A P = (1.02)6 The correct answer is Option C.
136.
The first $x deposited in the account earned interest for 2 years, while the additional $x earned interest for only 1 year.
The amount A under compound interest on a sum of money P invested at r% rate of interest for t years is given by: A = P 1 + r 100 t Thus, the final value after 2 years for the first $x deposited © 1999–2024 Manhattan Review www.manhattanreview.com 196 Quantitative Reasoning Question Bank −PS Solutions = $x 1 + 4 100 2 = $(x × 1.042) The final value after 1 year of the additional $x deposited = $x 1 + 4 100 1 = $1.04x Thus, total value of the money present in the account (y) = $ n x × 1.042 + (x × 1.04) o = $ x 1.042 + 1.04 Thus, we have y = x 1.042 + 1.04 x = y 1.042 + 1.04 x = y 1.04 (1.04 + 1) x = y 1.04 × 2.04 The correct answer is Option D.
137.
Let us understand the concept of the interest in the nth year with an example: Let $100 be invested at 10% rate of interest.
Interest accumulated after one year (or in the first year) = $ (10% of 100) = $10 Thus, amount after one year = $ (100 + 10) = $110 Thus, interest in the second year = $ (10% of 110) = $11 (= 110% of interest in the first year) Thus, total interest accumulated in two years = (Interest in the first year + Interest in the second year) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 197 = $ (10 + 11) = $21 Thus, amount after two years = $ (100 + 21) = $121 Thus, interest in the third year $ (10% of 121) = $12.10 (= 110% of interest in the second year) If we observe the values of the interest in the 1st, 2nd and 3rd years, we observe that: The interest in the nth year = (100 + r) % of (The interest in the (n −1)th year) Using the above relation in our problem: Interest in the 5th year = $4,800 Interest in the 6th year = $5,520 Thus, we have 5, 520 = (100 + r) % of 4,800 => 5, 520 = 100 + r 100 × 4, 800 => 5, 520 = 48 × (100 + r) => 115 = 100 + r => r = 15% The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 198 Quantitative Reasoning Question Bank −PS Solutions 5.10 Functions 138.
We have f p = p2 + 1 p2 => f −1 √p !
= −1 √p !2 + 1 −1 √p !2 = 1 p + p = f −1 √p !!2 = 1 p + p !2 = 1 p2 + p2 + 2 × 1 p × p = p2 + 1 p2 + 2 = f p + 2; by replacing the value of p2 + 1 p2 = f (p) The correct answer is Option A.
139.
f (x) = −1 x => f (a) = −1 a = −1 2 => a = 2 Also, we have f (ab) = −1 ab = 1 6 => ab = −6 => b = −6 a = −6 2 => b = −3 The correct answer is Option D.
140.
f (x) = √x −20 => f q = √q −20 Since p = f q , we have p = √q −20 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 199 => √q = p + 20 Squaring both sides, we have => q = p + 20 2 The correct answer is Option A.
141.
Let d, e and f be the hundreds, tens and units digits of K, respectively and g, h and i be the hundreds, tens and units digits of R, respectively.
Thus, we have f (K) = 2d3e5f f (R) = 2g3h5i Since f (K) = 18f (R), we have 2d3e5f = 18 × 2g3h5i => 2d3e5f = 2(g+1)3(h+2)5i Since d, e, f , g, h, i are integers, comparing coefficients of 2, 3 and 5: d = g + 1; e = h + 2 and f = i Thus, we have The three-digit number R ≡ghi, where g is the digit in the hundreds place, h is the digit in the tens place and i is the digit in the tens place => R = 100g + 10h + i The three-digit number K ≡def ≡(g + 1)(h + 2)i, where g + 1 is the digit in the hundreds place, (h + 2) is the digit in the tens place and i is the digit in the tens place => K = 100(g + 1) + 10(h + 2) + i => K = (100d + 10h + i) + 120 => K = R + 120 => K −R = 120 The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 200 Quantitative Reasoning Question Bank −PS Solutions 142.
Working with the options one at a time: Option A: f (x) = 1 + x => f (1 −x) = 1 + (1 −x) = 2 −x ̸= f (x) −Incorrect Option B: f (x) = 1 + x2 => f (1 −x) = 1 + (1 −x)2 = 1 + 1 −2x + x2 = 2 −2x + x2 ̸= f (x) −Incorrect Option C: f (x) = x2 −(1 −x)2 => f (1 −x) = (1 −x)2 −(1 −(1 −x))2 = (1 −x)2 −x2 ̸= f (x) −Incorrect Option D: f (x) = x2(1 −x)2 => f (1 −x) = (1 −x)2(1 −(1 −x))2 = (1 −x)2x2 = f (x) −Correct Since we already have the answer, we need not check Option E.
Verifying Option E, we would have had: Option E: f (x) = x2 1 −x => f (1 −x) = (1 −x)2 1 −(1 −x) = (1 −x)2 x ̸= f (x) The correct answer is Option D.
143.
f (x) = 1 x g (x) = x x2 + 1 => f g (x) = 1 g (x) = 1 x x2+1 = x2 + 1 x = (x −1)2 + 2x x = (x −1)2 x + 2 Since x > 0, the minimum value of the above expression will occur when the square term becomes zero (since a square term is always non-negative, the minimum possible value occurs when it is zero).
=> (x −1)2 = 0 => x = 1 Thus, the minimum value of f g (x) = 0 + 2 = 2 The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 201 144.
f (x) = 10x 1 −x => f (3) = 10 × 3 1 −3 = −15 Thus, we have f (x) = 1 2 × −15 = −15 2 => 10x 1 −x = −15 2 => 20x = −15 + 15x => 5x = −15 => x = −3 The correct answer is Option D.
145.
We have 3f (x) + 2f (−x) = 5x −10 . . . (i) Substituting x = 1 in equation (i): 3f (1) + 2f (−1) = 5 −10 => 3f (1) + 2f (−1) = −5 . . . (ii) Substituting x = −1 in equation (i): 3f (−1) + 2f (1) = −5 −10 => 3f (−1) + 2f (1) = −15 . . . (iii) Equation (ii) × 3 −Equation (iii) × 2: 5f (1) = −15 −(−30) = 15 => f (1) = 3 The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 202 Quantitative Reasoning Question Bank −PS Solutions 146.
We know that D (t) = −10(t −7)2 + 100, where 0 ≤t ≤12 We need to find the value of t so that the value of D(t) is maximum.
In the expression for D(t), we have a negative term: −10(t −7)2 We know that (t −7)2 ≥0 for all values of t (since it is a perfect square).
Thus, we have −10(t −7)2 ≤0 for all values of t (multiplying with a negative reverses the inequality).
Thus, in order that D(t) attains a maximum value, the term −10(t −7)2 must be 0.
Thus, we have −10(t −7)2 = 0 => t = 7 Thus, D(t) attains a maximum value at t = 7 i.e. 7 hours past 12:00 am i.e. 7:00 am.
The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 203 5.11 Permutation & Combination & Probability 147.
Since C5 3 = C5 r , r = 3, but there is no option as r = 3.
We know that Cp q = Cp p−q => C5 3 = C5 r = C5 5−r => 3 = 5 −r => r = 2 The correct answer is Option C.
148.
This is a question on permutation with indistinguishable or identical objects.
We know that if there are n objects, out of which p objects are indistinguishable, then Total number of way of arranging them = n!
p!
In this question, let’s first assume that we use two BLACK and one RED dot, thus, Total number of way of arranging them = 3!
2! = 3.
Similarly, let’s now assume that we use two RED and one BLACK dot, thus, Total number of way of arranging them = 3!
2! = 3.
There can be two more cases where we use all three BLACK or all three RED.
(Note: the question does not say that both colors must be used) Total number of codes = 3 + 3 + 2 = 8.
The codes would be: RRB, RBY, RBB, BBR, BRB, BRR, BBB, & RRR.
The correct answer is Option D.
149.
There are two different sizes and four different colors of mugs.
Packages having the same size and same color of mugs: Number of ways in which a size can be chosen = C2 1 = 2!
(2 −1)! × 1! = 2 ways Number of ways in which a color can be chosen = C4 1 = 4!
(4 −1)! × 1! = 4 ways © 1999–2024 Manhattan Review www.manhattanreview.com 204 Quantitative Reasoning Question Bank −PS Solutions Thus, total number of such packages = 2 × 4 = 8.
Packages having the same size and different colors of mugs: Number of ways in which a size can be chosen = C2 1 = 2!
(2 −1)! × 1! = 2 ways Number of ways in which three different colors can be chosen = C4 3 = 4!
(4 −3)! × 3! = 4 ways Thus, total number of such packages = 2 × 4 = 8 Thus, total number of different packages = 8 + 8 = 16 The correct answer is Option C.
150.
We know that there are six kinds of toppings and two kinds of breads for pizzas.
Since each pizza contains at least two kinds of toppings but not all kinds of toppings, the number of topping would be between 2 to 5.
Number of ways of selecting two kinds of toppings = C6 2 = 6!
4!2! = 15.
Number of ways of selecting three kinds of toppings = C6 3 = 6!
3!3! = 20.
Number of ways of selecting four kinds of toppings = C6 4 = C6 2 = 6!
4!2! = 15.
Number of ways of selecting five kinds of toppings = C6 5 = C6 1 = 6.
Thus, number of possible sections of toppings = 15 + 20 + 15 + 6 = 56.
Number of ways of selecting one kind of bread = C2 1 = 2.
Thus, number of pizzas possible = 56 × 2 = 112.
The correct answer is Option E.
151.
Number of one-letter codes: The botanist can uniquely designate 26 plants (since there are a total of 26 letters).
Number of two-letter codes: The first position can be assigned in 26 ways.
The second position can also be assigned in 26 ways (since the letters may be repeated).
Thus, total two-letter codes possible = 26 × 26 = 676.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 205 Thus, using two-letter codes, the botanist can uniquely designate 676 plants.
Number of three-letter codes: Each of the three positions can be assigned in 26 ways.
Thus, total three-letter codes possible = 26 × 26 × 26 = 17,576.
Thus, using three-letter codes, the botanist can uniquely designate 17,576 plants.
Thus, total number of unique designations possible using one-, two- or three-letter codes = 26 + 676 + 17,576 = 18,278.
The correct answer is Option E.
Alternate approach: There is a cheeky method for this question.
Number of one-letter codes: 26 ≡units digit is 6.
Number of two-letter codes: 26 × 26 ≡units digit is 6.
Number of three-letter codes: 26 × 26 × 26 ≡units digit is 6.
Thus, the units digit of the sum = 6 + 6 + 6 ≡8.
Only Option E has the units digit as 8.
Note: This is not a holistic method and not to be used when two or more options are with same units digit.
152.
We know that no subject is common in both groups.
Here events “Selection of one out of eight optional subjects from group one” and “Selection of two out of ten optional subjects from group two” are mutually exclusive or disjoint events; thus the total number of ways would be multiplied.
Number of ways of selecting one optional subject from eight subjects = C8 1 = 8 Number of ways of selecting two optional subjects from ten subjects = C10 2 = 10 × 9 2 × 1 = 45.
Thus, total number of ways of selecting three subjects = 8 × 45 = 360.
The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 206 Quantitative Reasoning Question Bank −PS Solutions 153.
We need to form a four-digit code using the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 i.e. 9 possible digits.
Also, it is known that repetition of digits is not allowed.
The thousands position of the code can be filled using any of the nine digits in 9 ways.
The hundreds position of the code can be filled using any of the eight digits in 8 ways.
The tens position of the code can be filled using any of the remaining seven digits in 7 ways.
The units position of the code can be filled using any of the remaining six digits in 6 ways.
Thus, the number of distinct codes possible = 9 × 8 × 7 × 6 = 3,024.
The correct answer is Option C.
154.
We need to form a four-digit code using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 i.e. 10 possible digits.
Also, it is known that repetition of digits is allowed, the code has to be an odd number, and the thousands’ position of the code can’t be 0.
The thousands’ position of the code can be filled using any of the 9 digits (except 0) in 9 ways.
The hundreds position of the code can be filled using any of the 10 digits (since 0 can now be used and repetition is allowed) in 10 ways.
The tens position of the code can be filled using any of the 10 digits (since 0 can now be used and repetition is allowed) in 10 ways.
The units position of the code can be filled using any of the 5 odd digits 1, 3, 5, 7, or 9 in 5 ways.
Thus, the number of distinct codes possible = 9 × 10 × 10 × 5 = 4,500.
The correct answer is Option D.
155.
Number of ways of selecting 4 sites out of 6 = (Number of ways in which any 4 of the 6 sites are selected without consideration to any constraint) −(Number of ways considering both A and B sites are selected) Number of ways in which any 4 of the 6 sites may be selected without paying consideration to any restriction = C6 4 = C6 (6−4) = C6 2 = 6 × 5 2 × 1 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 207 = 15 Number of ways in which 4 of the 6 sites may be selected so that both sites A and B are selected (i.e. two more sites to be selected from the remaining four) = C(6−2) (4−2) = C4 2 = 4 × 3 2 × 1 = 6; the above 6 ways do not satisfy the given restriction Thus, the number of ways in which 4 of the 6 sites can be selected so that both A and B sites are not selected simultaneously = 15 −6 = 9 The correct answer is Option D.
156.
Imran has 4 Math, 5 Physics, and 6 Chemistry books.
To select: 4 books such that the selection has at least one book of each subject.
The selections can be done in following ways.
(1) 2 Math, 1 Physics and 1 Chemistry books: # of ways = C4 2 × 5 × 6 = 4 × 3 1 × 2 × 5 × 6 = 180 (2) 1 Math, 2 Physics and 1 Chemistry books: # of ways = 4 × C5 2 × 6 = 4 × 5 × 4 1 × 2 × 6 = 240 (3) 1 Math, 1 Physics and 2 Chemistry books: # of ways = 4 × 5 × C6 2 = 4 × 5 × 6 × 5 1 × 2 = 300 Total number of possible selections = 180 + 240 + 300 = 720 The correct answer is Option B.
157.
Let the number of letters to be used be n.
The number of plants that can be identified using a single letter = Cn 1 = n.
The number of plants that can be identified using two distinct letters = Cn 2 = n (n −1) 2 (since the letters are to kept in alphabetic order, we must not order them or apply P n 2 .) Thus, total number of plants that can be identified if we attempt to have all the 15 codes that are either one-letter code or two-letter codes = n + n (n −1) 2 Since we need to have at least 15 identifications, we have © 1999–2024 Manhattan Review www.manhattanreview.com 208 Quantitative Reasoning Question Bank −PS Solutions n + n (n −1) 2 ≥15 => 2n + n (n −1) 2 ≥15 => n (2 + (n −1)) 2 ≥15 => n (n + 1) 2 ≥15 => n (n + 1) ≥30 Working with the options and starting with the least value of n, we see that n = 5 satisfies the above inequality.
The correct answer is Option C.
158.
We have to select 1 student each from classes A, B, and C each having 30 students and 2 students from Class D having 20 students.
Number of ways of selecting 1 student from Class A = C30 1 = 30.
Number of ways of selecting 1 student from Class B = C30 1 = 30.
Number of ways of selecting 1 student from Class C = C30 1 = 30.
Number of ways of selecting 2 student from Class D = C20 2 = 20 × 19 2 × 1 = 190.
As all four events are independent of each other, the number of ways of forming the team = 30 × 30 × 30 × 190 = 5,130,000 The correct answer is Option E.
159.
Total number of stocks available: 4 Information Technology, 5 Retail, and 3 e-commerce stocks Number of stocks to be selected: 2 Information Technology, 4 Retail, and 2 e-commerce stocks Number of ways of selecting 2 out of 4 Information Technology stocks = C4 2 = 4 × 3 2 × 1 = 6 Number of ways of selecting 4 out of 5 Retail stocks = C5 4 = C5 (5−4) = C5 1 = 5 Number of ways of selecting 2 out of 3 e-commerce stocks = C3 2 = C3 (3−2) = C3 1 = 3 Thus, the total number of ways of selecting the required number of stocks www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 209 = 6 × 5 × 3 (we multiply since the above selections are independent of one another) = 90 The correct answer is Option D.
160.
Total number of people present = 3 × 8 = 24.
Number of handshakes if the delegates shook hands with every person other than those from his or her own company = (Number of handshakes without any constraint) −(Number of handshakes with people from their own company) Number of handshakes without any constraint = C24 2 = 24 × 23 2 × 1 = 276 (Since a handshake requires two people, we need to select any 2 people from the 24 for a handshake) Number of handshakes with people from their own company = 8 × C3 2 = 8 × 3 × 2 2 × 1 = 24 (We need to select any 2 delegates from 3 delegates of the same company, for each of the 8 companies) Thus, the number of handshakes if the delegates shook hands with every person other than those from his or her own company = 276 −24 = 252 The correct answer is Option C.
Alternate approach: Let us calculate how many handshake a particular person does.
There are a total of 3 × 8 = 24 delegates.
A particular person would have to handshake 24 −3 = 21 times. This excludes the person itself and his company’s two collegaues.
Thus, all the delegates would do 21 × 24 2 = 252 handshakes.
We divided the total number of handshakes by ‘2’ because one handshake involving two dele-gates should be counted as one handshake and not two.
© 1999–2024 Manhattan Review www.manhattanreview.com 210 Quantitative Reasoning Question Bank −PS Solutions 161.
We need to select three digits for the code.
The digits to be used are from 0 to 9, thus there are 10 possible digits.
Since the first digit cannot be 0 or 9, number of possibilities for the first digit = (10 −2) = 8.
Since the second digit can only be 0 or 9, number of possibilities for the second digit = 2.
Let us ignore the restriction for the third digit.
Thus, number of possibilities for the third digit = 10.
Thus, total number of codes possible (ignoring the condition for the third digit) = 8 × 2 × 10 = 160 In the above codes, there are a few codes that are unacceptable since they violate the condition for the third digit.
The codes which violate the condition for the third digit are of the form (a99), where a is the first digit and both second and third digits are simultaneously 9.
The number of such codes equals the number of possibilities for the first digit, i.e. 8.
Thus, the number of codes possible without violating any of the given conditions = 160 −8 = 152 The correct answer is Option A.
162.
There are a total of 30 marbles, out of which 15 are yellow.
Probability that both marbles are yellow = Number of ways of drawing two yellow marbles Number of ways of drawing any two marbles => C15 2 C30 2 = 15×14 1×2 30×29 1×2 = 15 × 14 30 × 29 = 7 29 The correct answer is Option B.
163.
In a basket, out of 12 balls, seven are red and five are green.
Number of ways we can select three balls from 12 balls = C12 3 = 12 × 11 × 10 3 × 2 × 1 = 220 Number of ways of selecting two red balls and one green ball = C7 2 × C5 1 = 7 × 6 1 × 2 × 5 = 21 × 5 = 105 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 211 So, the required probability = C7 2 × C5 1 C12 3 = 105 220 = 21 44 The correct answer is Option D.
164.
The number of 6−member committees that can be formed from the 21 members = C21 6 = 21 × 20 × 19 × 18 × 17 × 16 6 × 5 × 4 × 3 × 2 × 1 The number of 5−member committees that can be formed from the 21 members = C21 5 = 21 × 20 × 19 × 18 × 17 5 × 4 × 3 × 2 × 1 Thus, the required ratio = 21 × 20 × 19 × 18 × 17 × 16 6 × 5 × 4 × 3 × 2 × 1 21 × 20 × 19 × 18 × 17 5 × 4 × 3 × 2 × 1 = 16 6 = 8 3 = 8 to 3 The correct answer is Option C.
165.
There are two offices to which the four employees need to be assigned.
Thus, the number of offices for each employee to choose = 2 (since each employee can be assigned to any of the two offices) Thus, the total number of ways of assigning the four employees = 2 × 2 × 2 × 2 = 24 = 16.
Note: For n objects, each with r options, the total number of options = r n The correct answer is Option E.
166.
We need to select three male officers from five male officers.
The number of ways of achieving it = C5 3 = 5 × 4 × 3 3 × 2 × 1 = 10.
We also need to select two female officer from three female officers.
The number of ways of achieving it = C3 2 = C3 1 = 3.
Thus, the number of ways in which three male officers and two female officer can be selected = 10 × 3 = 30.
The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 212 Quantitative Reasoning Question Bank −PS Solutions 167.
Given that, Probability that Stock X increases in value = P(A) = 0.4; Probability that Stock Y increases in value = P(B) = 0.6 Exactly one of the stocks would increase in value if: Stock A increases AND Stock Y does not OR Stock B increases AND Stock X does not Probability that Stock X does not increase in value = P(A) = 1 −0.4 = 0.6 Probability that Stock Y does not increase in value = P(B) = 1 −0.6 = 0.4 Thus, the required probability = P (A) × P B + P(B) × P A = 0.4 × 0.4 + 0.6 × 0.6 = 0.16 + 0.36 = 0.52 The correct answer is Option C.
168.
Since probability of landing on heads or tails is the same, each must be 1 2 => P(H) = P(T) = 1 2 Probability that the coin will land on heads at least once on two tosses = 1−Probability that it will not land on heads at all = 1−Probability that it will land on tails on both occasions = 1−(Probability that the first toss will show tails AND the second toss will show tails) = 1 −P (T) × P(T) = 1 −1 2 × 1 2 = 1 −1 4 = 3 4 The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 213 169.
Each question has two options of which only one is correct.
Thus, the probability of randomly guessing an answer and getting it correct = 1 2.
Thus, the probability of randomly guessing answers to all X questions and getting them correct = 1 2 × 1 2 × 1 2 × . . . X times = 1 2 X Thus, we have 1 2 X < 1 500 => 2X > 500 We observe from the options that 29 = 512, which just exceeds 500.
Thus, the least value of n = 9.
The correct answer is Option B.
170.
We know that the first three balls were black.
Thus, there are (20 −3) = 17 balls left.
Of these, number of black balls = 10 −3 = 7 We need the next two balls to be black.
Thus, we need to select 2 black balls from the remaining 7 black balls.
Number of ways in which the above case can be achieved (favorable cases) = C7 2 = 7 × 6 2 × 1 = 21 Number of ways in which two balls can be selected from the remaining 17 balls (total cases) = C17 2 = 17 × 16 2 × 1 = 136 Thus, required probability = Favorable cases Total cases = 21 136 The correct answer is Option A.
© 1999–2024 Manhattan Review www.manhattanreview.com 214 Quantitative Reasoning Question Bank −PS Solutions 171.
We need to select two balls one at a time.
On drawing 2 balls, one white and one blue can be obtained if: The first ball is white AND the second ball is blue OR The first ball is blue AND the second ball is white Since the balls are not replaced after drawing, after the first draw, the total number of balls would be 1 less than what was present initially, i.e. (16 −1) = 15.
Thus, the required probability = p(The first ball is white AND the second ball is blue) OR p(The first ball is blue AND the second ball is white) = {p(1st ball white) × p(2nd ball blue)} + {p(1st ball blue) × p(2nd ball white)} = 4 16 × 3 15 + 3 16 × 4 15 = 1 20 + 1 20 = 1 10 The correct answer is Option C.
172.
Probability that at least one refrigerator is single-door = 1 −Probability that none is single-door refrigerator Probability of selecting two refrigerators such that none is single-door refrigerator = Probability of selecting two double-door refrigerators = Number of ways of selecting 2 doubledoor from 6 sets Number of ways of selecting 2 refrigerators from 8 sets !
= C6 2 C8 2 = 6 × 5 2 × 1 8 × 7 2 × 1 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 215 = 15 28 Thus, the required probability = 1 −15 28 = 13 28.
The correct answer is Option D.
173.
Let the number of red balls = r Number of white balls = 9 Thus, total number of balls = (r + 9) Thus, probability that both balls would be white = Number of ways of selecting 2 white balls form 9 white balls Number of ways of selecting 2 balls from (r+9) balls !
= C9 2 C(r+9) 2 = 9 × 8 2 × 1 (r + 9) × (r + 9 −1) 2 × 1 = 72 (r + 9) × (r + 8) Thus, we have 72 (r + 9) × (r + 8) = 6 11 => (r + 9) × (r + 8) = 12 × 11 We see that the LHS and the RHS are product of two consecutive numbers, thus (r + 9) = 12 => r = 3 Thus, the total number of balls = 3 + 9 = 12 The correct answer is Option C.
174.
The probability that a card falls = P (F) = 0.05 Thus, the probability that a card does not fall = P F = 1 −0.05 = 0.95 Thus, the probability that none of the 12 card falls = 0.9512 For the pyramid to collapse, at least one card needs to fall.
Thus, the probability that at least one card falls = 1 −0.9512 The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 216 Quantitative Reasoning Question Bank −PS Solutions 175.
Total number of candies = 12 Number of orange flavored candies = 4 Thus, the number of non-orange flavored candies = 12 – 4 = 8 Thus, the kid has to pick 2 candies out of the 8 non-orange flavored candies.
Thus, number of ways (favorable cases) = C8 2 = 8 × 7 2 × 1 = 28 Number of ways in which the kid can pick 2 candies from the 12 candies (total cases) = C12 2 = 12 × 11 2 × 1 = 66 Thus, the required probability = Favorable cases Total no. of cases = 28 66 = 14 33 The correct answer is Option C.
176.
If Suzy returns at the end of the day 3, she would have stayed for 3 days.
Since it is given that she would return home at the end of the first day it rained, it must not have rained on day 1 and day 2 but rained on day 3.
Probability of rain on each day = P (R) = 0.25 => Probability that there is no rain on a particular day = P R = 1 −P (R) = 1 −0.25 = 0.75 Thus, probability that she return at the end of day on day 3 = Probability that there is no rain on day 1 AND no rain on day 2 AND rain on day 3 = P R × P R × P (R) = 0.75 × 0.75 × 0.25 = 9 64 The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 217 5.12 Sets 177.
Number of students on the committee G1 = 10.
As no member of G1 is in either of the other two groups, the above 10 students belong to only G1.
However, there may be an overlap with the students of G2 and G3.
Number of students in G2 = 10.
Number of students in G3 = 6.
We would get the greatest number of students who would not be in any of the groups if there is maximum overlap between the students of G2 and G3.
The maximum overlap between the students of G2 and G3 would be the minimum of the number of students in the two groups i.e. minimum of 6 and 10 = 6.
Thus, we have G2 ∩G3 = 6 Thus, the number of students in G2 or G3 = G2 + G3 −G2 ∩G3 = 10 + 6 −6 = 10 Thus, total number of students belonging to one of more groups = G1 + G2 or G3 = 10 + 10 = 20.
Thus, maximum number of students who don’t belong to any group = 30 −20 = 10.
The above information can be represented in a Venn-diagram as shown below: G2 (10) G3(6) G1 (10) 4 10 Total = 30 The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 218 Quantitative Reasoning Question Bank −PS Solutions 178.
Total number of dresses = 1,000 Number of traditional dresses = 250 Number of contemporary dresses = 750 Number of dresses for males = Number of dresses for females = 100 2 = 500 Number of traditional dresses for males = 100 Let us represent the above information using a Venn-diagram, as shown below: Total = 1,000 Traditional = 1/4 of 1,000 = 250 Males = !
" of 1,000 = 500 100 Traditional, Males 250 −100 = 150 Traditional, females 500 −100 = 400 Contemporary, Males 1,000 – (150 + 100 + 400) = 350 Contemporary, Females Thus, from the above Venn-diagram, we have Number of contemporary dresses for females = 350 The correct answer is Option D.
179.
We know that 95 percent students used a computer to play games or write reports.
Thus, (100 −95) = 5% students did not use a computer for either of these purposes.
Let us represent the above information using a Venn-diagram, as shown below: Not for playing games = 7% Not for writing reports = 11% 5% Neither 7 −5 = 2% 11 −5 = 6% Used for both = 100 −2 + 5 + 6 % = 87% Total = 100% www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 219 Thus, the percent of students who did use a computer both to play games and to write reports = 87%.
The correct answer is Option D.
180.
Let us represent the given information using a Venn-diagram, as shown below: Take cola = 70% Take health drink = 45% 25% Takeboth 70 −25 = 45% Take only cola 45 −25 = 20% Takeonly health drink Takeneither cola nor health drink 100 −45 + 25 + 20 = 10% 100% = 600 employees Thus, the number of employees who take neither cola nor health drink = 10% of 600 = 60 The correct answer is Option B.
181.
Let us represent the given information using a Venn-diagram, as shown below: # of Tennis = 𝑥 # of Badminton = 3𝑥 60 Both (𝑥−60) Only Tennis (3𝑥−60) Only Badminton We know that the number of students who play both sports is thrice the number of students who play only Tennis.
Thus, we have © 1999–2024 Manhattan Review www.manhattanreview.com 220 Quantitative Reasoning Question Bank −PS Solutions 60 = 3 (x −60) => x = 80 Thus, the number of students play only Badminton = 3x −60 = 180 The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 221 5.13 Statistics & Data Interpretation 182.
We know that the median is the middle-most value of any series/data set, but we do not know the value of x, so we cannot calculate exact value of Median; however we can surely find its range.
• Case 1: If x is smallest, the series would be x, 15, 20, 25 and median = average of 15 & 20 = 17.5–smallest median value.
• Case 2: If x is largest, the series would be 15, 20, 25, x and median = average of 20 & 25 = 22.5–largest median value.
Thus, the median would lie between 17.5 & 22.5, inclusive. Since the values in only Statements I & II are in the range, Option C is correct.
The correct answer is Option C.
183.
The question asks the number of scores > (Mean + SD)?
Mean + 1× SD = 73 + 1 × 20.50 = 93.50.
It is clear that three scores (95, 96, and 97) are greater than 93.50.
The correct answer is Option D.
184.
Let the sum of the 19 numbers other than n be s.
Thus, we have n = 4 × s 19 => s = 19n 4 Thus, the sum of all the 20 numbers in the list = s + n = 19n 4 + n = 23n 4 Thus, the required fraction = n 23n 4 = 4 23 The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 222 Quantitative Reasoning Question Bank −PS Solutions 185.
We have w ≤3 + 8 + w 3 ≤3w => 3w ≤11 + w ≤9w => 2w ≤11 ≤8w => 2w ≤11 and 8w ≥11 => w ≤11 2 = 51 2 and => w ≥11 8 = 13 8 Since w is an integer, possible values of w are: 2, 3, 4 or 5, Thus, there are four possible values of w.
The correct answer is Option B.
186.
Let the seven distinct positive integers be p, q, v, w, x, y and z such that p > q > v > w > x > y > z.
Thus, we need to find the least possible value of the largest among the seven, i.e. p.
Let’s assume that w, the middle-most number = average of the seven numbers = 14.
Now to maintain the average equals to 14, increase v, q, & p by 1, 2, & 3, respectively and decrease x, y, & z by 1, 2, & 3, respectively.
Thus, p = 14 + 3 = 17 – the least possible value of the greatest of the seven numbers The correct answer is Option B.
187.
We have x + y + 10 3 = x + y + 10 + 20 4 => 4x + 4y + 40 = 3x + 3y + 90 => x + y = 50 The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 223 188.
If the 13 different integers are arranged in order, the median is the 13+1 2 th = 7th integer.
Thus, the 7th integer is 20.
Since we need to find the maximum possible integer (given a range of 20), we need to have the maximum value of the least integer as well.
Since the integers are distinct, we can have the first 7 integers as: (20 −6) = 14, (20 −5) = 15, (20 −4) = 16, (20 −3) = 17, (20 −2) = 18, (20 −1) = 19, and 20 Thus, the maximum possible value of the smallest integer = 14.
Since the range is 20, the value of the greatest integer = 14 + 20 = 34.
The correct answer is Option D.
189.
We have Mean = Sum of the terms Total number of terms => 1 + 2 + 3 + 4 + 5 + 6 + x 7 = 2x 7 => 21 + x 7 = 2x 7 => 3 + x 7 = 2x 7 => 3 = 2x 7 −x 7 => 3 = x 7 => x = 21 The correct answer is Option E.
190.
We know that there are x employees.
Since the median salary is the 22nd salary, no two salaries are the same, and the 22nd salary is there in the list, there would be 21 salaries that are less than the 22nd salary and 21 salaries that are greater than the 22nd salary.
This implies that there are 21 + 1 + 21 = 43 salaries in the list or there are a total of 43 employees.
Thus, the average salary of the 43 employees = Total salary Number of employees © 1999–2024 Manhattan Review www.manhattanreview.com 224 Quantitative Reasoning Question Bank −PS Solutions = $ 860, 000 43 = $20, 000 The correct answer is Option B.
191.
We can calculate the total number of watts of electricity consumed as shown below: Appliance Number of hours in use Number of watts of electricity used per hour Total number of watts of electricity used Computer 4 105 4 × 105 = 420 Music system 2 90 2 × 90 = 180 Refrigerator 2 235 2 × 235 = 470 LED TV 2 150 2 × 150 = 300 Total 4 + 2 + 2 + 2 = 10 hours 420 + 180 + 470 + 300 = 1,370 watts Thus, the average number of watts of electricity used per hour per appliance = 1, 370 10 = 137 watts per hour.
The correct answer is Option C.
192.
We know that the distribution is symmetric about the mean.
Thus, the percent of the distribution equidistant from the mean on either side of it is the same.
Let the percent of the distribution less than (¯ x + d) be x%.
Thus, the percent of the distribution more than (¯ x −d) is also x%.
The situation is shown in the diagram below.
Mean 𝑥̅ 68% 𝑥̅ + 𝑑 𝑥̅ −𝑑 𝑥% 𝑥% Thus, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 225 x% + 68% + x% = 100% => x = 16% Thus, the percent of the distribution greater than (¯ x −d) = x + 68 = 16 + 68 = 84%.
The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 226 Quantitative Reasoning Question Bank −PS Solutions 5.14 Linear Equations 193.
Let us recall the property of two-digit number: “Difference between a particular two digit number and the number obtained by interchanging the digits of the same two digit number is always 9 times the difference between the digits.” Thus, the difference between actual amount and reversed amount = 63 = 9 × difference between the digits => Difference between the digits = 63 9 = 7 The difference between the digits of the number 7 is satisfied only by Option E.
The correct answer is Option E.
Alternate approach 1: If we consider correct amount as [xy] = 10x + y, then interchanged amount becomes [yx] = 10y + x.
According to the given condition, difference between the new amount and the original amount is 63 cents.
=> 10x + y − 10y + x = 63 => 10x −x −10y + y = 63 => 9x −9y = 63 On dividing by 9, we have x −y = 7.
Thus, the difference between the digits is 7, which is satisfied only by Option E.
Alternate approach 2: Since the cash register contained 63 cents more than it should have as a result of this error, this implies that the tens digit of the correct amount must be greater than its units digit.
Only two options can qualify. Let us analyze them: (C) 73: 73 −37 = 36 ̸= 63 (E) 92: 92 −29 = 63; correct answer www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 227 194.
Let the number of screwdrivers and spanners purchased be a and b, respectively.
Thus, total cost of the items = $(11a + 3b) Thus, we have 11a + 3b = 109 b = 109 −11a 3 = 108 + 1 −11a 3 = 108 3 + 1 −11a 3 = 36 + 1 −11a 3 It is clear that a and b are positive integers. Thus, (1 −11a) must be a multiple of 3.
There can be a few possible values of a, two such values are a = 2, 5 and 8.
For a = 3, we have b = 36 + 1 −11 × 2 3 = 36 −7 = 29. Thus, a + b = 2 + 29 = 31.
Since 31 is not one of the options, we must try with a = 5.
For a = 5, we have b = 36 + 1 −11 × 5 3 = 36 −18 = 18. Thus, a + b = 5 + 18 = 23.
Since 23 is not one of the options, we must try with a = 8.
For a = 8, we have b = 36 + 1 −11 × 8 3 = 36 −29 = 7. Thus, a + b = 8 + 7 = 15.
Thus, the total number of items purchased could be 8 + 7 = 15.
The correct answer is Option C.
195.
We have x + y + z = 2 . . . (i) x + 2y + 3z = 6 . . . (ii) Since z has no role to play, let’s eliminate it.
Multiplying (i) by 3 and subtracting (ii) from the result: 3 × (x + y + z) −(x + 2y + 3z) = 3 × 2 −6 => 2x + y = 0 => 2x = −y => x y = −1 2 The correct answer is Option A.
© 1999–2024 Manhattan Review www.manhattanreview.com 228 Quantitative Reasoning Question Bank −PS Solutions 196.
Let the number of notebooks sold last month be n.
Since the number of books sold was 2 more than the notebooks, the number of books sold = (n + 2).
Selling price of each book = $25.
Selling price of each notebook = $15.
Thus, total sales revenue = $(15n + 25(n + 2)).
Thus, we have 15n + 25(n + 2) = 490 40n = 490 −50 => n = 440 40 = 11 Thus, the number of notebooks sold = 11.
Thus, the number of books sold = 11 + 2 = 13.
Thus, the total number of books and notebooks sold = 11 + 13 = 24.
The correct answer is Option C.
Alternate approach: We know that the number of books sold is 2 more than the number of notebooks sold.
The price of 2 books = $(25 × 2) = $50.
Removing this from the total, i.e. $490, we are left with $(490 −50) = $440.
This amount was obtained by selling equal numbers of books and notebooks.
Total price of one book and one notebook = $(25 + 15) = $40.
Thus, number of items sold for $440 = 440 40 × 2 = 22 Thus, total number of books and notebooks sold = 2 + 22 = 24.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 229 5.15 Quadratic Equations & Polynomials 197.
We have a = √ 8ab −16b2 Squaring both sides: a2 = 8ab −16b2 => a2 −8ab + 16b2 = 0 => (a −4b)2 = 0 => a −4b = 0 => a = 4b The correct answer is Option D.
198.
Given that, (x −2)2 = 9 => x −2 = ±3 => x = 2 ± 3 => x = 5 OR −1 Given that, y −3 2 = 25 => y −3 = ±5 => y = 3 ± 5 => y = 8 OR −2 The minimum value of x y will be that value with the greatest magnitude of x, least magnitude of y and exactly one between of x and y being negative in sign.
Thus, we have x = 5, y = −2 => x y = −5 2 © 1999–2024 Manhattan Review www.manhattanreview.com 230 Quantitative Reasoning Question Bank −PS Solutions The maximum value of x y will be that value with the greatest magnitude of x, least magnitude of y and both x and y being simultaneously positive or negative in sign.
Thus, we have x = 5, y = 8 => x y = 5 8 OR x = −1, y = −2 => x y = 1 2 Between 5 8 and 1 2, the fraction 5 8 is greater.
Thus, the required difference = 5 8 − −5 2 = 5 8 + 5 2 = 25 8 The correct answer is Option E.
199.
Given that, 2x + 3y + xy = 12 => 2x + 3y + xy + 6 = 12 + 6 = 18; (adding the product of the coefficients of x and y to both sides) => (2x + 6) + xy + 3y = 18 => 2 (x + 3) + y (x + 3) = 18 => (x + 3) y + 2 = 18 Since x and y are positive integers, we must have: x + 3 > 3, and y + 2 > 2 Possible ways of getting 18 are: 1 × 18, 2 × 9, and 3 × 6 Thus, the only possible solution is: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 231 x + 3 = 6 => x = 3, and y + 2 = 3 => y = 1 Thus, we have x + y = 3 + 1 = 4.
The correct answer is Option B.
Alternate approach: Given that both x and y are positive integers, we can solve this question by hit and trial approach, too.
Let’s plug-in few probable positive integer values for x in the equation 2x + 3y + xy = 12 and see which value renders a positive integer value for y.
We see that for x = 1 & 2, we get non-integer values for y; and for x = 3, we get y = 1, a positive integer value. Thus, x + y = 3 + 1 = 4 200.
We know that h = −3(t −10)2 + 250 . . . (i) We need to first find the value of t such that the value of h is maximum.
In the expression for h, we have a term −3(t −10)2 We know that (t −10)2 ≥0 for all values of t (since it is a perfect square).
Thus, we have −3(t −10)2 ≤0 for all values of t (multiplying with a negative reverses the inequality).
Thus, in order that h attains a maximum value, the term −3(t −10)2 must be 0.
Thus, we have −3(t −10)2 = 0 => t = 10 Thus, h attains a maximum value at t = 10 seconds Thus, 7 seconds after the maximum height is attained, i.e. at t = 10 + 7 = 17, we have the corresponding value of h (in feet) as: h = −3(t −10)2 + 250 = −3 × 49 + 250; substituting the value of t = 17 and solving = 103 feet The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 232 Quantitative Reasoning Question Bank −PS Solutions 5.16 Inequalities 201.
The possible extreme scenarios are shown in the diagrams below: (1) Maximum distance away from hostel: College Hostel Basketball practice Computer class 4 6 12 D = 12 + 6 + 4 = 22 (2) Minimum distance away from hostel: College Hostel Basketball practice Computer class 4 6 12 D = 12 – (6 + 4) = 2 Thus, the maximum value of D is 22 and minimum value of D is 2 => 2 ≤D ≤22 The correct answer is Option D.
202.
As per given inequality: |b| ≤12, value of ‘b’ ranges from ‘−12’ to ‘+12’. So, by putting these values in first equation: 2a + b = 12, we can form a table of consistent values of a & b.
a a a b b b a a a b b b 0 12 Cont. . .
1 10 7 −2 2 8 8 −4 3 6 9 −6 4 4 10 −8 5 2 11 −10 6 0 12 −12 So a total of 13 ordered pairs are possible.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 233 The correct answer is Option E.
Alternate approach: We see that the value of b ranges from −12 to +12; this follows that b can have 25 number of integer values.
Now let us see how many integer values a can have.
2a + b = 12 => a = 12 −b 2 ; => a = 6 −b 2 We see that for a to be an integer, b 2 must be an integer; this follows that b must be an even number.
Out of 25 possible values of b, 13 are even (including 0); so for a to be an integer, the set of arrangement can only have 13 ordered pairs.
203.
Let the cost of an pencil be $x and the cost of a pen be $y.
Thus, we need to find the range within which (5x + 7y) lies.
We have 3.60 < 15x < 4.80 => 3.60 3 < 15x 3 < 4.80 3 => 1.20 < 5x < 1.60 . . . (i) Also, we have 33.30 < 21y < 42.90 => 33.30 3 < 21y 3 < 42.90 3 => 11.10 < 7y < 14.30 . . . (ii) Adding (i) and (ii): 12.30 < 5x + 7y < 15.90 Thus, the correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 234 Quantitative Reasoning Question Bank −PS Solutions 204.
Working with the options one at a time: Comparing options A and B: xy 2 = x2y2 < x2 (since 0 < y < 1 => y2 is a fraction between 0 and 1) Thus, Option B cannot have the greatest value.
Comparing options A and C: x y !2 = x2 y2 = x2 × 1 y2 !
x2 (since 0 < y < 1 => 0 < y2 < 1 => 1 y2 > 1) Thus, Option A cannot have the greatest value.
Comparing options C and D: x2 y = x2 y2 × y = x y !2 × y < x y !2 (since 0 < y < 1 => y is a fraction between 0 and 1) Thus, Option D cannot have the greatest value.
Comparing options C and E: x2y = x2 y2 × y3 = x y !2 × y3 < x y !2 (since 0 < y < 1 => y3 is a fraction between 0 and 1) Thus, Option E cannot have the greatest value.
The correct answer is Option C.
Alternate approach: Since x < 0 and 0 < y < 1 must be true for all the values of x & y; let us assume convenient, smart values of x & y.
Say x = −1 & y = 1 2 Let us calculate the values of each option.
(A) x2 = (−1)2 = 1 (B) xy 2 = −1 × 1 2 2 = 1 4 (C) x y !2 = −1 1 2 2 = 4 : Maximum (D) x2 y = (−1)2 1 2 = 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 235 (E) x2y = (−1)2 × 1 2 = 1 2 205.
Let the distance between the cities A and B be d miles.
Range of speeds of David = 20 miles per hour to 30 miles per hour.
Time taken by David to cover the distance = 5 hours.
So, the range of distance between cities A and B = [20 × 5 miles] to [30 × 5 miles] = 100 miles to 150 miles => 100 ≤d ≤150 . . . (i) Range of speeds of Mark = 40 miles per hour to 60 miles per hour.
Time taken by Mark to cover the distance = 3 hours.
So, the range of distance between cities A and B = [40 × 3 miles] to [60× 3 miles] = 120 miles to 180 miles => 120 ≤d ≤180 . . . (ii) Thus, from (i) and (ii), we have Range of distance between cities A and B = (Higher of the two minimum values) to (Lower of the two maximum values) = (Higher of 100 and 120) to (Lower of 150 and 180) = 120 miles to 150 miles => 120 ≤d ≤150 The only option that satisfies above is 135 miles.
The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 236 Quantitative Reasoning Question Bank −PS Solutions 5.17 Co-ordinate geometry 206.
The length of the line segment between 2 points x1, y1 = (−3, −6) and x2, y2 = (5, 0) is given by: L = q (x2 −x1)2 + y2 −y1 2 = q (5 −(−3))2 + (0 −(−6))2 = √ 82 + 62 = 10 Thus, the length of the diameter of the circle = 10.
Thus, the length of the radius of the circle = 10 2 = 5.
Thus, the area of the circle = π × 52 = 25π.
The correct answer is Option C.
207.
Equation of a line is: y = mx + c, where m is the slope and c is the Y-intercept.
Thus, we have the equation of the given line as: y = 3x + 4 Let the required point on the above line be: (a, 10).
Thus, we have 10 = 3a + 4 => a = 2 Thus, the required X-coordinate of the point is 2.
The correct answer is Option A.
208.
Equation of a line passing through a point (p, q) and slope m is given as: y −q = m x −p Thus, the equation of the line l that passes through the origin (0, 0) and has slope 3 is: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 237 y −0 = 3 (x −0) => y = 3x Since (1, a) is a point on the line, we have a = 3 × 1 => a = 3 . . . (i) Since (b, 2) is a point on the line, we have 2 = 3 × b => b = 2 3 . . . (ii) Thus, from (i) and (ii), we have a b = 3 2 3 = 9 2 The correct answer is Option E.
209.
The equation of a circle having center at (p, q) and radius r is: x −p 2 + y −q 2 = r 2 Since the center of the circle is at (3, 2), the equation of the circle is: (x −3)2 + y −2 2 = r 2 If a point (m, n) lies inside the circle x −p 2 + y −q 2 = r 2, it must satisfy: m −p 2 + n −q 2 < r 2 Since (−1, 2) lies inside the circle, it must satisfy: (−1 −3)2 + (2 −2)2 < r 2 => r 2 > 16 => r > 4 . . . (i) If a point (m, n) lies outside the circle, it must satisfy: m −p 2 + n −q 2 > r 2 Since (3, −4) lies outside the circle, it must satisfy: (3 −3)2 + (−4 −2)2 > r 2 => r 2 < 36 © 1999–2024 Manhattan Review www.manhattanreview.com 238 Quantitative Reasoning Question Bank −PS Solutions => −6 < r < 6 Since r must be positive, we have 0 < r < 6 . . . (ii) Thus, from (i) and (ii), we have 4 < r < 6 From the options, only r = 5 satisfies the above inequality.
The correct answer is Option A.
Alternate approach: Since (−1, 2) lies inside, the distance between (−1, 2) and the center (3, 2) is less than the radius.
Similarly, since (3, −4) lies outside, the distance between (3, −4) and the center (3, 2) is greater than the radius. Thus, we have p [(−1 −3)2 + (2 −2)2] < r < p [(3 −3)2 + (−4 −2)2] => √ 16 < r < √ 36 => 4 < r < 6 => r = 5 (only value among the options) 210.
The three vertices are: (a, b), (a, −b), and (−a, −b) We know that a < 0 and b > 0 => −a > 0 and −b < 0 We can see that the distance of each of the vertices from the origin (0, 0) = √ a2 + b2 Thus, the three vertices are equidistant from the origin.
Alternately, we can see that the midpoint of the diagonal joining (a, b) and (−a, −b) is (0, 0), i.e. the origin.
Thus, the centre of the square is at the origin.
Thus, the square would be positioned as shown in the diagram below: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 239 −𝑋 𝑋 𝑌 −𝑌 (𝑎, 𝑏) (−𝒂, 𝒃) (𝑎, −𝑏) (−𝑎, −𝑏) Thus, the fourth vertex would be (−a, b), which lies in the first quadrant.
Thus, the X-coordinate of the fourth vertex is positive and the Y-coordinate of the fourth vertex is also positive.
Thus, the only point which also lies in the same quadrant is (6, 2).
The correct answer is Option E.
211.
Distance between two points (x1, y1) and (x2, y2) is given by q (x2 −x1)2 + y2 −y1 2 Distance between the points (0, 0) and (5, 5) = q (5 −0)2 + (5 −0)2 = √ 50 = 5 √ 2 Distance between the points (0, 0) and (10, 0) = q (10 −0)2 + (0 −0)2 = √ 100 = 10 Note: The distance is obviously 10 since (10, 0) is a point on the X-axis at a distance of 10 to the right of the origin.
Distance between the points (10, 0) and (5, 5) = q (10 −5)2 + (0 −5)2 © 1999–2024 Manhattan Review www.manhattanreview.com 240 Quantitative Reasoning Question Bank −PS Solutions = √ 50 = 5 √ 2 Thus, the perimeter = 5 √ 2 + 10 + 5 √ 2 = 10 + 10 √ 2 The correct answer is Option E.
212.
Since the points are collinear, their slopes must be equal.
Slope of the line joining two points (x1, y1) and (x2, y2) is given by: y2 −y1 x2 −x1 .
Slope of the line joining the points (a, 0) and (0, b) = b −0 0 −a = −b a Slope of the line joining the points (a, 0) and (1, 1) = 1 −0 1 −a = 1 1 −a Thus, we have −b a = 1 1 −a => −b × (1 −a) = a => −b + ab = a => b + a = ab => b = ab −a => b = a(b −1) => a = b b −1 The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank −PS Solutions 241 213.
The area of the triangle formed by the line with the two axes = 1 2 × Length of X intercept × (Length of Y intercept) We have 3y −4x = 24 . . . (i) The graph of the line and the area of the triangle formed by the line and the axes is shown in the diagram below: Y X 3𝑦−4𝑥= 24 (0, 8) (−6, 0) To calculate the X-intercept: Substituting y = 0 in (i): x = −24 4 = −6 Thus, the length of the X-intercept = 6.
To calculate the Y-intercept: Substituting x = 0 in (i): y = 24 3 = 8 Thus, the length of the Y-intercept = 8.
Thus, required area = 1 2 × 6 × 8 = 24 The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 242 Quantitative Reasoning Question Bank −PS Solutions www.manhattanreview.com © 1999–2024 Manhattan Review Chapter 6 Solutions – Data Sufficiency Questions 243 244 Quantitative Reasoning Question Bank – DS Solutions 6.1 Numbers 214.
From statement 1: We have M$ = M2 −2.
14 < M$ < 34 => 14 < M2 −2 < 34 => 16 < M2 < 36 Since M is a positive integer, M2 must be a perfect square between 16 and 36.
Thus: M2 = 25 => M = 5 – Sufficient From statement 2: This statement is clearly not sufficient.
If M$ = 7, M2 = 9, and M = 3; however, if M$ = 23, M2 = 25, and M = 5. No unique value of M.
The correct answer is Option A.
215.
Let the number of members in batches A and B be a and b, respectively.
Thus, we have a = 8x b = 4y + 3 We need to determine the value of (a + b).
From statement 1: We have x = y −1 2 => y = 2x + 1 Thus, we have a = 8x b = 4y + 3 = 4(2x + 1) + 3 = 8x + 7 However, there may be many values for x and y, thus correspondingly many values for (a + b).
Thus, we cannot find the value of (a + b). – Insufficient From statement 2: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 245 We have b = a + 7 Thus, we have 4y + 3 = 8x + 7 => y = 2x + 1 There may be many values possible for x and y: x = 3, y = 7; x = 4, y = 9, etc.
Thus, the values of x and y are not unique.
Hence, the values of a = 8x and b = 4y + 3 are also not unique. – Insufficient Thus, from statements 1 and 2 together: From each statement, we have the same information: y = 2x + 1 Thus, even combining both the statements does not help. – Insufficient The correct answer is Option E.
216.
The index-3 of p is essentially the highest exponent of 3 that perfectly divided p.
For example, if n = 162 = 34 × 2 The index-3 of 162 is 4 since the greatest exponent of 3 that divided 162 completely is 4.
From statement 1: We have q −r > 0 => q > r.
We have have two cases.
Case 1: Say q = 4 & r = 3 We have the index-3 of q = 4 = 30 × 4 is 0 and the index-3 of r = 3 = 31 is 1. The answer is No.
Case 2: Say q = 9 & r = 3 We have the index-3 of q = 32 is 2 and the index-3 of r = 3 = 31 is 1. The answer is Yes.
Hence, there is no definite answer. – Insufficient From statement 2: © 1999–2024 Manhattan Review www.manhattanreview.com 246 Quantitative Reasoning Question Bank – DS Solutions We have q r is a multiple of 3 => q = r × 3k; where k is any positive integer Thus, the index-3 of q will be k greater than that of r. – Sufficient The correct answer is Option B.
217.
The number of positive factors of a number N, expressed in its prime form as N = pxqy, where p and q are distinct primes, is given by (x + 1) (y + 1).
For example: 24 = 23 × 31 The number of positive factors = (3 + 1) (1 + 1) = 8.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24; i.e. 8 in number.
From statement 1: m = p3q2; where p and q are different prime numbers.
Thus, the number of positive factors = (3 + 1) (2 + 1) = 12. – Sufficient From statement 2: Here, m can take multiple possible values.
For example, 2 × 3, 22 × 33, 24 × 3, etc. all have 2 and 3 as the only two prime factors.
However, the number of factors of each of the above numbers is different. – Insufficient The correct answer is Option A.
218.
Since √m is an integer, m must be a perfect square.
From statement 1: We have 13 ≤m ≤16 The only perfect square between 13 and 16 is 16.
Thus, we have m = 16.
=> √m = 4 (√m only takes the positive square root of m). – Sufficient From statement 2: We have 3 ≤√m ≤4 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 247 The possible values of √m are 3 or 4.
Thus, the value of √m is not unique. – Insufficient The correct answer is Option A.
219.
We have xy 4 3 = 3 √ 432 Cubing both sides x3y4 = 432 = 33 × 24 We need to determine whether x + y = 5.
From statement 1: Nothing is mentioned about x.
For example, two possible solutions are: If x = 3, y = 2 => x + y = 5; OR if x = 3 √ 432, y = 1 => x + y ̸= 5 – Insufficient From statement 2: Nothing is mentioned about y.
For example, two possible solutions are: If x = 3, y = 2 => x + y = 5; OR If x = 1, y = 4 √ 432 => x + y ̸= 5 – Insufficient Thus, from statements 1 and 2 together: Both x and y are integers and y > 0.
Thus, the only way in which x3y4 = 432 = 33 × 24 is x = 3, y = 2 => x + y = 5 – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 248 Quantitative Reasoning Question Bank – DS Solutions 220.
From statement 1: The value of y is not known.
Hence, we cannot determine the value of 5x y !2 – Insufficient From statement 2: 5x −2y = 0 => 5x = 2y => x y = 2 5 Thus, we have 5x y !2 = 25 × 4 25 = 4 – Sufficient The correct answer is Option B.
221.
|x + 3| = 2 => x + 3 = ±2 => x = −3 ± 2 => x = −1 or −5.
From statement 1: We have x < 0.
Thus, as per the information given in the question x = −1 or −5.
Thus, there is no unique value for x. – Insufficient From statement 2: x2 + 6x + 5 = 0 => (x + 1) (x + 5) = 0 => x = −1 or −5.
Thus, both values of x are possible. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we have x = −1 or −5. – Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 249 222.
|x + 3| = 3 => x + 3 = ±3 => x = −3 ± 3 => n = 0 or −6.
From statement 1: We have x2 ̸= 0 => x ̸= 0 => x = −6. – Sufficient From statement 2: x2 + 6x = 0 => x (x + 6) = 0 => x = 0 or −6.
Thus, both values of x are possible, i.e. x is not unique. – Insufficient The correct answer is Option A.
223.
For any decimal number, say [a·bcd], where b, c, d are the digits after the decimal point, the tenths digit refers to the digit b, i.e. the digit just after the decimal point.
From statement 1: m + 0.01 < 3 => m < 2.99 => 2 < m < 2.99 Thus, the tenths digit of m may be 8 (if m = 2.88); less than 8 (for example, if m = 2.58) or 9 (if m = 2.98). – Insufficient From statement 2: m + 0.05 > 3 => m > 2.95 => 2.95 < m < 3.
Thus, the tenths digit of m must always be 9, not 8. The answer is No. – Sufficient The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 250 Quantitative Reasoning Question Bank – DS Solutions 224.
From statement 1: 5a + 6b is even: For any integer value of b, 6b is even, [since (even) × any integer = even].
Thus, 5a must be even [since (even) + (even) = even].
Hence, a is even.
However, b may be even or odd. – Insufficient From statement 2: 5a + 3b is even: Two odd numbers when added result in an even number OR two even numbers when added also result in an even number.
Thus, both 5a and 3b are odd i.e. both a and b are odd OR Both 5a and 3b are even i.e. both a and b are even Thus, b may be even or odd. – Insufficient Thus, from statements 1 and 2 together: From statement 1: a is even.
Hence, from statement 2, b is even. – Sufficient The correct answer is Option C.
225.
From statement 1: We know that x is a factor of y.
However, we have no information about the exponents p and q, which are likewise important.
For example: 2 is a factor of 6.
Also, 23 is a factor of 64.
However: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 251 24 is not a factor of 63.
Thus, we need to know whether p ≤q. – Insufficient From statement 2: We know: p < q + 1.
However, there is no information on x and y. – Insufficient Thus, from statements 1 and 2 together: x is a factor of y; p < q + 1 => p ≤q Thus, xp is a factor of yq. – Sufficient The correct answer is Option C.
226.
From statement 1: Since the product of the four numbers (10, −2, −8 and 0) is zero, it is immaterial what the fifth integer (X) be; for any value of the fifth integer, the product of five integers would still be ‘0.’ – Insufficient From statement 2: The sum of the four integers = 10 −2 −8 + 0 = 0.
Since ‘0’ divided by any number (other than ‘0’ itself) is ‘0’, we can only conclude that the fifth integer (X) is not ‘0’, as ‘0’ divided by ‘0’ is not defined; however we cannot determine the value of the fifth integer, X. – Insufficient Thus, from statements 1 and 2 together: Even after combining both the statements, we cannot get the value of the fifth integer X.
The correct answer is Option E.
227.
We know that P, Q and R lie on a straight line.
However, the order in which they are present is not known and whether they are positive or negative is also not known.
From statement 1: © 1999–2024 Manhattan Review www.manhattanreview.com 252 Quantitative Reasoning Question Bank – DS Solutions The points P and Q are 20 units apart.
However, the distance between either R and P or between R and Q is not known; neither do we know the order in which the points are present is known. – Insufficient From statement 2: The points P and Q are 25 units apart.
However, the distance between neither Q and P nor between Q and R is known; neither do we know the order in which the points are present is known. – Insufficient Thus, from statements 1 and 2 together: The order in which the points are present is not known.
For example, if the points are as: P__ Q__ R, then the distance between Q and R is => 25−20 = 5 However, if the points are as: Q__ P__ R, then the distance between Q and R is => 25 + 20 = 45 Thus, the distance between Q and R cannot be determined. – Insufficient The correct answer is Option E.
228.
The remainder when a number is divided by 10 is the unit digit of the number.
For example: The remainder when 12 is divided by 10 is 2, which is the unit digit of 12.
The exponents of 2 follow a cycle for the last digit as shown below (p is a positive integer): Exponent of 2 Unit digit 24p+1 2 24p+2 4 24p+3 8 24p 6 From statement 1: Since m is divisible by 10, m may or may not be a multiple of 4.
For example, if m = 20: Unit digit of 2m = 220 = 2(4×5) => Remainder = 6.
Whereas, if m = 30 : Unit digit of 2m = 230 = 2(4×7)+2 => Remainder = 4.
Thus, we do not have a unique remainder. – Insufficient From statement 2: Thus, m is a multiple of 4; unit digit of 2(4p) is always 6. – Sufficient The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 253 229.
From statement 1: We know that: a and b are even and c is odd.
Thus, (a −b −c) is ((Even −Even) −Odd) = (Even −Odd) = Odd. The answer is No. – Sufficient From statement 2: Since a, b and c are consecutive integers, there are two possibilities: (1) a = Even, b = Odd, c = Even => (a −b −c) is ((Even −Odd) −Even) = (Odd −Even) = Odd.
(2) a = Odd, b = Even, c = Odd => (a −b −c) is ((Odd −Even) −Odd) = (Odd −Odd) = Even.
Hence, the nature of value of (a −b −c) is not unique. – Insufficient The correct answer is Option A.
230.
From statement 1: m is a multiple of 5 => m = 5u; where u is an integer.
Thus, we have 4m + 5n = p => 20u + 5n = p => p = 5(4u + n) Thus, p is a multiple of 5.
Thus, p and 10 do have a common factor other than 1 and that is 5. – Sufficient From statement 2: n is a multiple of 5 => n = 5v; where v is an integer.
Thus, we have © 1999–2024 Manhattan Review www.manhattanreview.com 254 Quantitative Reasoning Question Bank – DS Solutions 4m + 5n = p => 4m + 25v = p We cannot definitely conclude that p is a multiple of at least one of the common multiples of 10 (2 and 5).
Thus, p and 10 may or may not have a common factor other than 1. – Insufficient The correct answer is Option A.
231.
From statement 1: There is no information about whether p and q are positive.
Hence, there would be different combinations of values of p and q possible, leading to different values of (p + q).
The possible integer values of p and q satisfying the equation pq = 6 are: p q p + q 1 6 7 2 3 5 3 2 5 6 1 7 −1 −6 −7 −2 −3 −5 −3 −2 −5 −6 −1 −7 Thus, there is no unique value of (p + q). – Insufficient From statement 2: p + q 2 = 49 => p + q = ±7 Thus, there is no unique value of (p + q). – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we have p + q = ±7 Thus, there is no unique value of (p + q). – Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 255 232.
p = 13m + 25n is odd in one condition: If only one between 13m and 25n is odd, making the sum of EVEN + ODD = ODD.
From statement 1: Since p = 13m + 25n and one of m and n is odd, the condition above is met. Let’s see how.
Say m is odd and n is even, then p = 13 × ODD + 25 × EVEN = ODD × ODD + ODD × EVEN = ODD + EVEN = ODD Same goes with considering n is odd and m is even.
The answer is Yes. – Sufficient From statement 2: Given p = 13m + 25n and n is even, we have 1. If m is ODD: p = 13×ODD+25× EVEN = ODD× ODD+ ODD× EVEN = ODD + EVEN = ODD 2. If m is EVEN: p = 13×EVEN+25× EVEN = ODD× EVEN+ODD× EVEN = EVEN + EVEN = EVEN No unique answer. – Insufficient The correct answer is Option A.
233.
The product abc will be even if at least one among a, b and c is an even integer.
From statement 1: c −b = b −a => a + c = 2b Thus, we see that (a + c) is twice of an integer and hence, (a + c) is even.
However, a and c are both odd or both even.
Similarly, b may be odd or even.
For example: a = 3, c = 19, b = 11: all are odd, the product abc is odd.
OR © 1999–2024 Manhattan Review www.manhattanreview.com 256 Quantitative Reasoning Question Bank – DS Solutions a = 3, c = 5, b = 4: at least one of them is even, the product abc is even.
Thus, the product abc may be even or odd. – Insufficient From statement 2: c −16 = a => c −a = 16 Since the difference of c and a is 16 (even), it may be that a and c are both odd or both even.
Also, there is no information on whether b is even.
Thus, the product abc may be even or odd. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we can have all of a, b and c as odd or at least one of them even.
Thus, the product abc may be even or odd. – Insufficient The correct answer is Option E.
234.
If x 3 an integer, x must be a multiple of ‘3.’ From statement 1: If x = 1, the answer is no; however, if x is, say 3, the answer is yes. Thus, we cannot say whether x is a multiple of ‘3.’ – Insufficient From statement 2: We have x 6 is an integer => x is a multiple of ‘6, which itself is a multiple of 3.’ => x is a multiple of ‘3.’ – Sufficient The correct answer is Option B.
235.
From statement 1: We have no information on x. – Insufficient From statement 2: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 257 We have no information on y. – Insufficient Thus, from statements 1 and 2 together: x is divisible by 8 => x = 8p, where p is an integer.
y is divisible by 4 => y = 4q, where q is an integer.
Thus, we have x y = 8p 4q = 2p q = 2 × p q If p q is an integer, then x y is an even integer.
However, if p q is not an integer, then x y is not an even integer.
For example, if p q = 1 2 => x y = 1; not an even integer.
Hence, the answer to the question may be ‘Yes’ or ‘No.’ – Insufficient The correct answer is Option E.
236.
From statement 1: Say n = k (k + 1) (k + 2); where k is a positive integer.
Possible values of k are 1, 2, 3 . . .
For any other value of k, since n is the product of three consecutive positive integers, there is at least one even integer and one integer, which is a multiple of 3.
Thus, n is divisible by 3 × 2 = 6.
Hence, n is divisible by 6 for any positive integer value of k, thus n 6 an integer. – Sufficient Alternatively: Product of ‘n’ consecutive integers must be divisible by n! thus, product of three consecutive integers must be divisible by 3! i.e. 6, thus n 6 an integer. – Sufficient From statement 2: © 1999–2024 Manhattan Review www.manhattanreview.com 258 Quantitative Reasoning Question Bank – DS Solutions With is information, n 6 may or may not be an integer.
If n = 6, n 6 an integer; however, if n = 3, n 6 is not an integer. – Insufficient The correct answer is Option A.
237.
From statement 1: n is not a multiple of 2.
=> n is an odd number.
If n = 5: Remainder (r) when n2 −1 = 24 is divided by 24 is ‘0.’ If n = 9: Remainder (r) when n2 −1 = 80 is divided by 24 is ‘8.’ Thus, the value of remainder is not unique. – Insufficient From statement 2: n is not a multiple of 3.
If n = 5: Remainder (r) when n2 −1 = 24 is divided by 24 is ‘0.’ If n = 8: Remainder (r) when n2 −1 = 63 is divided by 24 is ‘15.’ Thus, the value of remainder is not unique. – Insufficient Thus, from statements 1 and 2 together: n is neither a multiple of 2 nor a multiple of 3.
=> n is not a multiple of 6.
Thus, n when divided by 6 will leave a remainder of either 1 or 5 since it cannot leave remain-ders 2, or 4 (since n is not a multiple of 2) or 3 (n is not a multiple of 3).
Thus, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 259 n = 6p + 1 or n = 6p −1, where p is a positive integer.
Thus, we have the following situations: If n n n = 6p p p + 1 If n n n = 6p p p −1 n2 −1 = (n −1) (n + 1) = 6p + 1 −1 6p + 1 + 1 = 6p × (6p + 2) n2 −1 = (n −1) (n + 1) = 6p −1 −1 6p −1 + 1 = (6p −2) × 6p Thus, we observe that: (n −1) (n + 1) = 6p × (6p ± 2) = 12p × (3p ± 1) Now, we have two situations depending on p as even AND as odd: (1) If p is even: => 12p(3p ± 1) is a multiple of 24.
(2) If p is odd: => (3p ± 1) is even => 12p(3p + 1) is a multiple of 24.
Thus, we see that n2 −1 is always divisible by 24.
Hence, the remainder (r) when n2 −1 is divided by 24 is ‘0.’ – Sufficient The correct answer is Option C.
238.
From statement 1: Say n = 2k + 1; where k is an integer Any odd integer can be expressed as n = 2k + 1; where k is an integer.
Thus, we have = n (n −1) (n + 1) © 1999–2024 Manhattan Review www.manhattanreview.com 260 Quantitative Reasoning Question Bank – DS Solutions = (2k + 1) (2k) (2k + 2) = 4k (2k + 1) (k + 1) => n (n −1) (n + 1) is divisible by 4. – Sufficient Alternatively: Since n is an odd number, (n −1) and (n + 1) must be even.
Thus, we have n (n −1) (n + 1) = odd × even × even, which must be divisible by 4. – Sufficient From statement 2: We have n(n + 1) is divisible by 6.
If n = 2: It satisfies n(n + 1) divisible by 6.
However, n (n −1) (n + 1) = 6, which is not divisible by 4.
Again, if n = 3: It satisfies n(n + 1) divisible by 6.
However, n (n −1) (n + 1) = 24, which is divisible by 4.
Hence, we do not have a unique answer. – Insufficient The correct answer is Option A.
239.
From statement 1: We know that 5x is odd.
Since x is an integer and ‘5’ is odd, x must be an odd integer (since: odd × odd = odd). – Sufficient From statement 2: We know that (x + 5) is odd.
Since ‘5’ is odd, x must be odd (since: odd + odd = even). – Sufficient The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 261 240.
From statement 1: Since x divided by 3 leaves the remainder 2, we have x = 3k + 2, where k is a non-negative integer.
However, the value of x cannot be determined as k is unknown. – Insufficient From statement 2: We know that x2 divided by 3 leaves the remainder 1.
We know that any number x divided by 3 leaves a remainder of 0 or 1 or 2.
Thus, x2 when divided by 3 would leave a remainder 02 = 0 or 12 = 1 or 22 = 4 ≡1 (since 4 is greater than 3, we divide 4 by 3 to get the actual remainder as 1).
Thus, x2 when divided by 3 leaves remainder 0 or 1.
It is obvious that the remainder 0 occurs when the number x is a multiple of 3.
For all other values of x, remainder when x2 divide by 3 would be 1.
Thus, there are infinitely many possible values of x, for example: 1, 2, 4, 5, 7, 8, 10 . . .
Thus, we cannot determine any unique value of x. – Insufficient Thus, from statements 1 and 2 together: From statement 1, we have x = 3k + 2 => x2 = 9k2 + 12k + 4 => x2 = 3 3k2 + 4k + 1 + 1 => x2 = 3q + 1; q is quotient Thus, x2 when divided by 3 would leave a remainder 1.
However, this is exactly what statement 2 conveys.
Thus, statement 2 provides the same information as statement 1.
Since there is no additional information provided about x, the value of x cannot be determined.
– Insufficient © 1999–2024 Manhattan Review www.manhattanreview.com 262 Quantitative Reasoning Question Bank – DS Solutions The correct answer is Option E.
A few values of x would be 1, 2, 5, 8, 11, ...
Alternate approach: It is clear that each statement itself in not sufficient, so let’s combine them.
From statement 1, we have x = 3k + 2; where k is non-negative integer Few values of x would be 1, 2, 5, 8, 11, 14, 17,...
=> The relevant values of x2 are: 22 = 4; 52 = 25; 82 = 64; 112 = 121; 142 = 196; 172 = 289; ...
We see that each of the above values of x2: 4, 25, 64, 121, 196, and 289 leave the remainder 1 when they are divided by 3; thus, statement 2 is, in fact, a rephrased version of statement 1.
241.
From statement 1: Since x ≥3, the number of trailing zeroes in 30x must be at least ‘three.’ Let’s see with an example.
For x = 3, 30x = 303 = 27, 000; For x = 4, 30x = 304 = 810, 000; Thus, the hundreds digit of 30x is ‘0.’ – Sufficient From statement 2: We have x 3 is an integer Thus, the possible values of x are 3, 6, 9 . . .
If x = 3, we have 30x = 303 = 27, 000.
Thus, when x = 3 there are ‘3’ trailing zeroes in 30x.
For higher values of x, the number of trailing zeroes would be more.
Thus, the hundreds digit in 30x is ‘0.’ – Sufficient The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 263 242.
From statement 1: x is a multiple of 36 such that its value is between 100 and 200.
Thus, possible multiples of 36 are: 36 × 3 = 108, 36 × 4 = 144, and 36 × 5 = 180.
Thus, the value of x is not unique. – Insufficient From statement 2: x is an even multiple of 45 such that its value is between 100 and 200.
Thus, possible multiples of 45 are: 45 × 3 = 135, and 45 × 4 = 180.
Since x is an even integer, the only possible value of x is 45 × 4 = 180.
Thus, the value of x is unique. – Sufficient The correct answer is Option B.
243.
From statement 1: We have 15 is a multiple of x.
=> x is a factor of 15.
Factors of 15 are: 1, 3, 5, and 15.
Possible factors of 15 (lying between 2 and 6) are 3 and 5.
Thus, possible values of x are 3 or 5.
Thus, the value of x is not unique. – Insufficient From statement 2: We have 21 is a multiple of x.
=> x is a factor of 21.
Factors of 21 are: 1, 3, 7, and 21.
The only possible factor of 21 (lying between 2 and 6) is 3.
Thus, the only possible value of x is 3.– Sufficient The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 264 Quantitative Reasoning Question Bank – DS Solutions 244.
We have => xm = 1 xm => xm × xm = 1 => x2m = 1 Thus, there are two possibilities: (1) x = ±1 (here, m can be any number: since ‘1’ raised to any exponent is always ‘1’ and ‘−1’ raised to any even exponent (i.e. 2) is always 1) OR (2) m = 0 (here, x can be any number: since any non-zero number raised to ‘0’ is always ‘1’) Thus, we need to use the statements 1 and 2 to decide which of the above cases is possible.
From statement 1: We only know that x is an integer.
We have no information whether m is ‘0.’ Thus, the value of x cannot be determined. – Insufficient From statement 2: Since m ̸= 0, we must have: x = ±1 However, we cannot uniquely determine the value of x. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statement, we still have x = ±1.
Thus, the value of x cannot be uniquely determined. – Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 265 245.
From statement 1: We know that the sum of the two digits of x is prime.
Thus, apart from x = 11 (sum of the digits is 2, the only even prime), for all other values of x, one digit of x must be even and the other digit of x must be odd (since the sum of an even and odd number is always odd and all prime numbers above 2 are odd).
Thus, possible values of x could be: 23 (sum of digits = 5, a prime number), 89 (sum of digits = 17, a prime number), etc.
Thus, x may be less than 85 or more than 85. – Insufficient From statement 2: We know that each of the two digits of x is prime.
In order that x > 85 (but less than equal to 99), we need to have a prime number greater than or equal to 8 but less than or equal to 9 in the tens position of x.
However, this is not possible.
Thus, all possible values of x would have a digit less than 8 in the tens position.
Possible such digits are: 7, 5, 3, or 2.
Hence, x < 85 – Sufficient The correct answer is Option B.
246.
n 13 will be an integer only if n is a multiple of 13.
From statement 1: Since 5n 13 is an integer, we can conclude that 5n is divisible by 13.
However, 5 and 13 have no common factors.
Thus, n must be divisible by 13.
Hence, n 13 must be an integer. – Sufficient From statement 2: Since 3n 13 is an integer, we can conclude that 3n is divisible by 13.
© 1999–2024 Manhattan Review www.manhattanreview.com 266 Quantitative Reasoning Question Bank – DS Solutions However, 3 and 13 have no common factors.
Thus, n must be divisible by 13.
Hence, n 13 must be an integer. – Sufficient The correct answer is Option D.
247.
10x ≤ 1 1000 => 10x ≤ 1 103 => 10x ≤10−3 => x ≤−3 From statement 1: Possible values of x are: −2, −3, −4 . . .
Thus, we have x = −2 (≰−3) OR x = −3, −4, etc. (≤−3).
Thus, we do not have a unique answer to the question. – Insufficient From statement 2: Possible values of x are: −3, −2, −1 . . .
Thus, we have x = −3(≤−3) OR x = −2, −1, etc. (≰−3) Thus, we do not have a unique answer to the question. – Insufficient Thus, from statements 1 and 2 together: We still have: n = −2 (≰−3) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 267 OR x = −3(≤−3) Thus, we do not have a unique answer to the question. – Insufficient The correct answer is Option E.
248.
We need to verify whether: a b = c d From statement 1: We have c = 5a . . . (i) d = 5b . . . (ii) Dividing (i) by (ii) c d = 5a 5b = a b – Sufficient From statement 2: We have 5a = 4b => a b = 4 5 . . . (iii) 5c = 4d => c d = 4 5 . . . (iv) From, (iii) and (iv) a b = c d – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 268 Quantitative Reasoning Question Bank – DS Solutions 249.
From statement 1: Let us factorize 2,457: 2, 457 = 33 × 7 × 13.
Thus, we have a3bc = 33 × 7 × 13.
Since a, b, and c are prime numbers, we have a = 3, b = 7, & c = 13 OR a = 3, b = 13, & c = 7 In either case, a3b3c3 = 33 × 73 × 133. – Sufficient From statement 2: We have no information on a and c. – Insufficient The correct answer is Option A.
250.
We have x is a multiple of 24 => x2 is a multiple of 242 = (8 × 3)2 = (23 × 3)2 = 26 × 32.
Also, y is a multiple of 21 = 3 × 7.
=> x2y is a multiple of 26 × 33 × 7 In order that x2y is a multiple of 648 = 23 × 34 , we must ensure that there are three 2s and four 3s in x2y. We see that there are sufficient 2s (six), but the number of 3s is deficient by one.
So, the question is limited to whether we get one more 3 as a multiple of x2y.
From statement 1: Given x is a multiple of 8 does not help. 8 provides three more 2s but none 3. – Insufficient From statement 2: Given x is a multiple of 18 helps. 18(= 2 × 32) provides at least one 3. – Sufficient The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 269 251.
Since 2 < n < 18, the remainder (r) when 18 is divided by n can be any of the following: r = 0, 1, 2, 3, 4, 5, 6, 7, and 8.
From statement 1: Since n > 15, possible values of n are 16, and 17.
Thus, the corresponding values of r are 2 and 1.
Thus, the value of r is not unique. – Insufficient From statement 2: Given that n = 2m, we have 2 < 2m < 18, possible values of n are 22 = 4, 23 = 8 or 24 = 16.
Thus, the values of the remainder (r) when 18 is divided by 4, 8, or 16 is ‘2’ in each case.
Thus, the value of r is unique. – Sufficient The correct answer is Option B.
252.
We have x > 0 We need to determine whether y > 0 given x > 0.
From statement 1: We have y ≤x: If y = x: y > 0 If y < x: The value of y may be positive (but less than x), 0, or negative. – Insufficient From statement 2: We have y ≥x: Since x > 0 => y ≥x > 0 © 1999–2024 Manhattan Review www.manhattanreview.com 270 Quantitative Reasoning Question Bank – DS Solutions y > 0. – Sufficient The correct answer is Option B.
253.
From statement 1: We have xy is divisible by 9.
There may be a possibility that both x and y are divisible by 3 and hence their product is divisible by 9.
However, there is another possibility that y is divisible by 9 and x is not divisible by 3.
Thus, there is no unique answer to the question. – Insufficient From statement 2: We have no information about x. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we cannot determine whether x is divisible by 3.
– Insufficient The correct answer is Option E.
254.
From statement 1: (x −1) y + 1 y −1 = 1 => x −1 = y + 1 y −1 => x −1 = y2 −1 => x = y2 => x y = y (which is given to be an integer). – Sufficient From statement 2: We have x −y = 2 A few possible values of x and y are: x = 3, y = 1 => x y = 3 1 = 3 (an integer) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 271 x = −1, y = −3 => x y = −1 −3 = 1 3 (not an integer) Thus, there is no unique answer to the question. – Insufficient The correct answer is Option A.
255.
From statement 1: We have yx = y => yx−1 = 1 Possible situations are: (1) y = 1 (here x can be any number) (2) y = −1 (here (x −1) must be an even number) (3) x −1 = 0 => x = 1 (here y can be any number) Thus, the value of y cannot be uniquely determined. – Insufficient From statement 2: We do not have any information about y. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we have (1) y = 1 (here x can be any number) (2) y = −1 (here (x −1) must be an even number) (Note: x = 1 is no longer applicable using the information from statement 2) Thus, the value of y cannot be uniquely determined. – Insufficient The correct answer is Option E.
256.
From statement 1: We have an−1 + an = pn (n −1) Substituting n = 30 : a29 + a30 = p × 30 × 29 . . . (i) Substituting n = 31 : a30 + a31 = p × 31 × 30 . . . (ii) © 1999–2024 Manhattan Review www.manhattanreview.com 272 Quantitative Reasoning Question Bank – DS Solutions Thus: (ii) −(i) (since a31 −a29 is mentioned in the statement 1): a31 −a29 = 30 × p × 31 −p × 29 = 60p.
Thus, we have 60p = 120 => p = 2. – Sufficient From statement 2: We have an−1 + an = pn (n −1) Substituting n = 2 : a1 + a2 = p × 2 × 1 = 2p . . . (iii) Since a2 = 6, we have a1 + 6 = 2p However, we have no information on the value of a1, we cannot determine the value of p. – Insufficient The correct answer is Option A.
257.
We have find out whether |P −R| ≥13.
From statement 1: |P −Q| = 65 => P −Q = ±65 . . . (i) However, we do not have any information about R. – Insufficient From statement 2: |Q −R| = 52 => Q −R = ±52 . . . (ii) However we do not have any information about P. – Insufficient Thus, from statements 1 and 2 together: Adding (i) and (ii): www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 273 (P −Q) + (Q −R) = (P −R) = ±65 ± 52 = 65 + 52 = 117 OR = 65 −52 = 13 OR = −65 + 52 = −13 OR = −65 −52 = −117 => P −R = ±117 OR ±13 => |P −R| = 117 OR 13 Thus, |P −R| ≥13. – Sufficient The correct answer is Option C.
258.
Let the three positive integers are r, s, & t as first, second and third integer.
From statement 1: We have no information about the third integer. – Insufficient From statement 2: We have no information about the first integer. – Insufficient Thus, from statements 1 and 2 together: (r + s) + (s + t) = even + even = even => r + 2s + t = [even] => r + s + t = [even] −s However, s may be even or odd.
Thus, we have © 1999–2024 Manhattan Review www.manhattanreview.com 274 Quantitative Reasoning Question Bank – DS Solutions r + s + t = even −even = even OR r + s + t = even −odd = odd Thus, we cannot determine the nature of (r + s + t). – Insufficient The correct answer is Option E.
259.
c = 10a + 12b => c = 12 (a + b) −2a => c = 12 −2a (since (a + b) = 1).
We deliberately took 12(a + b) so that we could relate c and a.
We need to determine whether c > 11: => 12 −2a > 11 => 2a < 1 => a < 1 2 Thus, we need to see whether a < 1 2.
From statement 1: We have a > 1 2 This is contrary to what we needed, i.e., a < 1 2 Thus, we have a unique answer ‘No.’ – Sufficient From statement 2: We had obtained a < 1 2.
Since a + b = 1, we have a = 1 −b => 1 −b < 1 2 b > 1 2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 275 This is contrary to the information of statement 2, i.e., b > 1 2 Thus, we have a unique answer ‘No.’ – Sufficient The correct answer is Option D.
260.
From statement 1: We can have 14 consecutive integers including −5 in the following ways: Set S: {−5, −4, −3, −2, . . . 6, 7, 8}; (‘7’ is present in the set) OR Set S: {−7, −6, −5, −4, −3 . . . 4, 5, 6}; (‘7’ is not present in the set), etc.
Thus, the integer ‘7’ may be or may not be present in the set. – Insufficient From statement 2: We can have 14 consecutive integers including 6 in the following ways: Set: {5, 6, 7, . . . 17, 18}; (‘7’ is present in the set) OR Set: {−7, −6, −5, −4 . . . 4, 5, 6}; (‘7’ is not present in the set).
Thus, the integer ‘7’ may be or may not be present in the set. – Insufficient Thus, from statements 1 and 2 together: We can still have the following sets: Set: {−7, −6, −5, −4, . . . 4, 5, 6}; (‘7’ is not present in the set) OR Set: {−6, −5, −4, −3 . . . 4, 5, 6, 7}; (‘7’ is present in the set) Thus, the integer ‘7’ may be or may not be present in the set. – Insufficient The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 276 Quantitative Reasoning Question Bank – DS Solutions 261.
From statement 1: We know that the 150th term is 305.
However, the relation between the other terms of the sequence is not known.
Hence, the 200th term cannot be determined. – Insufficient From statement 2: Same with Statement 2.– Insufficient From statement 1 $ 2 together: Since it is given that Sequence S is such that the difference between a term and its previous term is constant, the difference between 150th term and 100th term must be equal to the difference between 200th term and 150th term because the difference between 150 and 100 is 50 and the difference between 200 and 150 is also 50.
Thus, 200th term = 150th term + (150th term −100th term) = 305 + (305 −(−95)) = 305 + (305 + 95) = 305 + 400 = 705 The correct answer is Option C.
262.
From statement 1: n < 7 20 => n < 0.35 For all values of h = 0, 1, 2, 3 or 4, we have n < 0.35 Thus, the value of n rounded to the nearest tenth is: (1) For h = 0 : n = 0.307 = 0.3 (2) For h = 1 : n = 0.317 = 0.3 (3) For h = 2 : n = 0.327 = 0.3 (4) For h = 3 : n = 0.337 = 0.3 (5) For h = 4 : n = 0.347 = 0.3 Thus, the value of n to the nearest tenth is 0.3 – Sufficient From statement 2: Since h < 5, possible values of h are 0, 1, 2, 3 or 4.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 277 This is the same result as obtained from statement 1. – Sufficient The correct answer is Option D.
263.
From statement 1: Since x has exactly two distinct positive factors, x must be a prime number (the factors of a prime number are 1 and the number itself).
Since 2 is a prime number, we can have x = 2.
However, there are other prime numbers greater than 2 as well, for example 3, 5, etc.
– Insufficient From statement 2: We know that the difference between any two distinct positive factors of x is odd.
This is only possible if one factor is even and the other factor is odd (since the difference between an even and odd number is odd, whereas the difference between any two odd numbers or any two even numbers is even).
Thus, the number must have exactly two factors, one odd, i.e., 1 and the other even (i.e. 2).
Thus, we have x = 1 × 2 = 2. – Sufficient The correct answer is Option B.
264.
Possible values of n so that the product of the digits is 12 are: 26, 34, 43 and 62.
From statement 1: We know that n can be expressed as the sum of perfect squares in only one way.
Let us see for each of the above numbers: 26: (1) With 1, we have 12 + 52 = 26; we see that is can be expressed as the sum of two perfect squares (2) With 2: 22 + 22 = 26; however, 22 is not a perfect square (3) With 3: 32 + 17 = 26; however, 17 is not a perfect square (4) With 4: 42 + 10 = 26; however, 10 is not a perfect square © 1999–2024 Manhattan Review www.manhattanreview.com 278 Quantitative Reasoning Question Bank – DS Solutions 34: (1) With 1: 12 + 33 = 34; however, 33 is not a perfect square (2) With 2: 22 + 30 = 34; however, 30 is not a perfect square (3) With 3: 32 + 52 = 34; we see that it can be expressed as the sum of two perfect squares (4) With 4: 42 + 18 = 34; however, 18 is not a perfect square Thus, we see that both 26 and 34 can be expressed as the sum of two perfect squares in exactly one way.
Thus, it is enough to observe that statement 1 is not sufficient.
Note: Neither 43 nor 62 can be expressed as the sum of two perfect squares in any way.
Thus, we do not get a unique value of n. – Insufficient From statement 2: Since n is smaller than 40, possible values of n are 26 and 34. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we still have possible values of n as 26 and 34.
– Insufficient The correct answer is Option E.
265.
From statement 1: We know that the sum of three equal integers is divisible by 2; thus, their sum is even.
=> Thrice of an integer is even.
Hence, we can conclude that the integer in question is even.
Thus, each of the three equal integers is even.
Thus, their product must be a multiple of 2 × 2 × 2 = 8.
Hence, the product of the three integers is divisible by 4. – Sufficient From statement 2: Since the sum as well as the product of the three integers is even, we may have all the three integers even OR any one integer even and the other two odd.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 279 In the first case, the product of the three integers must be divisible by 2 × 2 × 2 = 8, and hence divisible by 4; while in the second case, the product may be divisible by 4 (if the even number itself is divisible by 4) or may not be divisible by 4 (if the even number is 2).
Thus, we do not have a unique answer. – Insufficient The correct answer is Option A.
266.
From statement 1: We know that the units digit of X is non-zero (between 1 to 9, inclusive).
Thus, in the number (X + 9), the units digit when added to 9 would always lead to a carryover of ‘1’ to the tens place.
For example: If the three digit number is 331, where the units digit is the smallest, i.e. 1, then we have 331 + 9 = 340 i.e. the tens digit has increased by 1.
Again, if the three digit number is 339, where the units digit is the largest, i.e. 9, then too we have 339 + 9 = 348 i.e. the tens digit has increased by 1.
Since the tens digit in (X + 9) is 3, the tens digit in X (i.e. before adding 9) must have been 3 −1 = 2 (subtracting the carry of ‘1’).
Thus, the tens digit in X is 2. – Sufficient From statement 2: We know that the units digit of X is non-zero (between 1 to 9, inclusive).
Thus, in the number (X + 3), the units digits when added to 3 may lead to a carryover of ‘1’ to the tens place if the units digit in X is 7 or greater; however, there will not be any carryover to the tens place if the units digit in X is 6 or smaller.
For example: If the three digit number is 317, where the units digit is 7, then we have 317 + 3 = 320 i.e. the tens digit has increased by 1.
Again, if the three digit number is 319, where the units digit is the largest, i.e. 9, then too we have 319 + 3 = 322 i.e. the tens digit has increased by 1.
However, if the three digit number is 321, where the tens digit is the smallest, i.e. 1, then we have © 1999–2024 Manhattan Review www.manhattanreview.com 280 Quantitative Reasoning Question Bank – DS Solutions 321 + 3 = 324 i.e. the tens digit has not increased by 1.
Again, if the three digit number is 325, where the units digit is 5, then too we have 325 + 3 = 328 i.e. the tens digit has not increased by 1.
Thus, if there is a carryover, the tens digit in X would have been 2 −1 = 1.
However, if there is no carryover, the tens digit in X would have been 2 itself.
Thus, there is no unique answer. – Insufficient The correct answer is Option A.
267.
From statement 1: xy = x2 => xy −x2 = 0 => x y −x = 0 => x = 0 or y = x However, we know that: y ̸= x, since they are different numbers.
Thus, we have x = 0 – Sufficient From statement 2: Since y ̸= 0, x may or may not be equal to 0. – Insufficient The correct answer is Option A.
268.
We know that none among x, y, & z equals to 0.
We have x4y5z6 to analyze.
x4y5z6 = x4y4z6 × y We know that x4y4z6 > 0 (since the exponents are even i.e. the numbers are perfect squares, they must be positive).
Thus, we have to see whether y > 0 (This is what we need to know to get the answer to the question) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 281 From statement 1: y > x4 Since x4 > 0, y > 0 => x4y5z6 > 0 – Sufficient From statement 2: y > z5 However, z5 may be positive or negative depending on the value of z => y may be positive or negative.
=> x4y5z6 may be positive or negative. – Insufficient The correct answer is Option A.
269.
From statement 1: We have y y + 2 = x(x + 1) Let us look into two possibilities for x: (1) If x is odd: x (x + 1) = odd × even = even (2) If x is even: x (x + 1) = even × odd = even Thus, we have y(y + 2) as even.
If y is odd, (y + 2) is also odd => y y + 2 = odd × odd = odd; however, y(y + 2) is even, thus, y cannot be odd.
If y is even, (y + 2) is also even => y y + 2 = even × even = even.
Thus, we have y an even integer. The answer is No. – Sufficient From statement 2: We have no information about y. – Insufficient The correct answer is Option A.
© 1999–2024 Manhattan Review www.manhattanreview.com 282 Quantitative Reasoning Question Bank – DS Solutions 270.
From statement 1: We have a2b = 24 Thus, we have 8a6b −2 = 8 a2b3 −2 = 8 × (24) 3 −2 = 8 × (212) −2 – Sufficient Note: We need not try to get individual values of a and b from a2b = 24.
From statement 2: We have ab = 22 = 4 Two possible values of 8a6b −2 depending on the values of a and b are: (1) a = 1, b = 4: 8a6b −2 = 8 × 1 −2 = 6 (2) a = 2, b = 2: 8a6b −2 = 8 × 212 −2 Thus, there is no unique value. – Insufficient The correct answer is Option A.
271.
From statement 1: Since x is a factor of 54 and is less than half of 54, the possible values of x are: 1, 2, 3, 6, 9 or 18.
Again, we know that 18 is a multiple of xy2.
=> xy2 is a factor of 18 The factors of 18 are 1, 2, 3, 6, 9 and 18.
If x = 1, the possible values of y are 1 or 3.
Thus, we observe that the value of y cannot be uniquely determined. – Insufficient From statement 2: Since xy2 is a factor of 18, the possible values of xy2 are: 1, 2, 3, 6, 9 or 18.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 283 Also, we know that y is a multiple of 3.
=> y2 is a multiple of 32 = 9 Thus, the only possible value of y is 3 (higher multiples of 3 as the value of y would not be possible since xy2 is a factor of 18). – Sufficient The correct answer is Option B.
272.
We have x2y = x4y−6 2y = 4y −6 => y = 3, for all values of x except 1 OR y = any positive integer if x = 1 Thus, though it appeared that only possible value of y is 3, it is not necessarily so.
Thus, we need to refer to the statements to get the value of y.
From statement 1: We have x2 = 4 => x = 2; x ̸= −2 since x is a positive integer.
x = 2, (i.e. x ̸= 1) Thus, y = 3 => y2x = 34 = 81 – Sufficient From statement 2: We have x < 3 => Possible values of x are 2 or 1.
=> y = 3 if x = 2 OR y = any positive integer if x = 1 © 1999–2024 Manhattan Review www.manhattanreview.com 284 Quantitative Reasoning Question Bank – DS Solutions Since the values of x and y cannot be uniquely determined, we cannot determine the unique value of y2x. – Insufficient The correct answer is Option A.
273.
We have xky = x(ly2−8) Since x ̸= 1, we have ky = ly2 −8 => y ly −k = 8 From statement 1: We have k = −6 => y ly + 6 = 8 Possible values of y: 2 or 4.
If y = 2 : 2(2l + 6) = 8 => l = −1 => kl = (−6) × (−1) = 6 > 2 If y = 4 : 4(4l + 6) = 8 => l = −1 => kl = (−6) × (−1) = 6 > 2 Thus, there is a unique answer and the answer is ‘Yes.’ – Sufficient From statement 2: We have 3l −k = 3 We know that y ly −k = 8 Possible values of y: 2 or 4.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 285 If y = 2 : 2 (2l −k) = 8 => 2l −k = 4 => (3l −k) −l = 4 => 3 −l = 4 => l = −1 => k = 3l −3 = −6 (using the equation: 3l −k = 3) => kl = (−6) × (−1) = 6 > 2 If y = 4 : 4 (4l −k) = 8 => 4l −k = 2 => (3l −k) + l = 2 => 3 + l = 2 => l = −1 => k = 3l −3 = −6 (using the equation: 3l −k = 3) => kl = (−6) × (−1) = 6 > 2 Thus, there is a unique answer and the answer is ‘Yes.’ – Sufficient The correct answer is Option D.
274.
From statement 1: We have |x| + y = 5, where 1 < |x| < y Thus, y must be a positive integer (since y > 1).
Since |x| + y = 5, and |x| > 1 © 1999–2024 Manhattan Review www.manhattanreview.com 286 Quantitative Reasoning Question Bank – DS Solutions => y < 4 Thus, possible values of y are: 3, 2 or 1.
If y = 3 : |x| = 2; thus y > |x| is satisfied.
Here, |x| = 2 => x = ±2 However, if y = 2 : |x| = 3 i.e. y ≯|x| Also, if y = 1 : |x| = 4 i.e. y ≯|x| Thus, the only possible values of x and y are: x = 2, y = 3 OR x = −2, y = 3 Thus, we have x2y −1 = (±2)2×3 −1 = (±2)6 −1 = 63 – Sufficient From statement 2: We have x2 −4 + y −3 = 0 The sum of two absolute values can be ‘0’ only if each of them is zero.
Thus, we have x2 −4 = 0 => x2 −4 = 0 => x = ±2 Also, y −3 = 0 => y −3 = 0 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 287 => y = 3 Thus, the only possible values of x and y are: x = 2, y = 3 OR x = −2, y = 3 This is the same result as obtained from statement 1. – Sufficient The correct answer is Option D.
275.
The remainder when a number is divided by 10 is essentially the units digit of the number.
Thus, we need to determine the units digit of 74x+23 + y .
The units digit of exponents of 7 follows a cycle as shown below: • 74k+1 ≡7 • 74k+2 ≡9 • 74k+3 ≡3 • 74k ≡1 Thus, the remainder r of the units digit of (74x+3+y) would be same as the units digit of (3+y).
Thus, we need to determine the value of y.
From statement 1: We have no information about y. – Insufficient From statement 2: We have y = 2 Thus, the remainder r = units digit of (3 + 2) = 5 – Sufficient The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 288 Quantitative Reasoning Question Bank – DS Solutions 276.
From statement 1: We know that: y −x = 3 However, there can be infinitely many positive integer values of x and y satisfying the above equation.
Thus, the value of y cannot be uniquely determined. – Insufficient From statement 2: We know that: x and y are prime numbers.
However, there can be infinitely many values of x and y possible.
Thus, the value of y cannot be uniquely determined. – Insufficient Thus, from statements 1 and 2 together: We have y −x = 3, where x and y are prime numbers.
Since the difference between two prime numbers is 3 (an odd number), one prime must be even and the other odd.
Thus, one of the prime numbers must be ‘2’ (the only even prime, also the smallest prime number).
Thus, we have x = 2 => y −2 = 3 => y = 5 – Sufficient The correct answer is Option C.
277.
From statement 1: We need to find which combination of exponents of 3 and 5 add up to 134.
Since the exponents of 5 would reach 134 faster than the exponents of 3, we need to try with the exponents of 5 so that we can get the answer(s) in the least possible trials.
If y = 1 : 3x + 5 = 134 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 289 => 3x = 129 However, 129 cannot be expressed as an exponent of 3 => y ̸= 1 If y = 2 : 3x + 25 = 134 => 3x = 109 However, 109 cannot be expressed as an exponent of 3 => y ̸= 2 If y = 3 : 3x + 125 = 134 => 3x = 9 => x = 2 Thus, there is only one solution: x = 2 – Sufficient From statement 2: There is no information about x. – Insufficient The correct answer is Option A.
278.
Assuming that xy is a multiple of 18. Say xy = 18n; where n must be an integer.
So, in order to answer the question, we have to determine whether n is an integer.
From statement 1: x is a multiple of 9.
Say x = 9m; where m is any integer.
Plugging-in the value of x = 9m in xy = 18n, we get 9my = 18n => my = 2n If my is even, n is integer, the answer is ‘Yes;’ however, if my is odd, n is not an integer, the answer is ‘No.’ – Insufficient From statement 2: y is a multiple of x.
Say y = xp; where p is any integer.
Plugging-in the value of y = xp in xy = 18n, we get x2p = 18n © 1999–2024 Manhattan Review www.manhattanreview.com 290 Quantitative Reasoning Question Bank – DS Solutions If x2p is divisible by 18, n is an integer, else not. No unique answer. – Insufficient From statement 1 & 2 together: From statement 2, we have x2p = 18n and from statement 1, we have x = 9m.
Thus, x2p = 18n => (9m)2 × p = 18n => 81pm2 = 18n => 9pm2 = 2n If pm2 is even, n is integer, the answer is ‘Yes;’ however, if pm2 is odd, n is not an integer, the answer is ‘No.’ – Insufficient The correct answer is Option E.
279.
We have to determine whether (x + y)(x −y) is a prime number OR (x2 −y2) a prime number.
From statement 1: Given x is the smallest prime number, it means that x = 2. Since we have no information about the value of y, we cannot conclude whether (x2 −y2) is a prime number. – Insufficient From statement 2: Given y2 is the smallest prime number, it means that y2 = 2. Since we have no information about the value of x, we cannot conclude whether (x2 −y2) is a prime number. – Insufficient Thus, from statements 1 and 2 together: Plugging-in the value of x = 2 and y2 in (x2 −y2), we get x2 −y2 = 22 −2 = 4 −2 = 2, a prime number. – Sufficient The correct answer is Option C.
280.
The remainder obtained when a number is divided by 100 is the last two digits of the number.
For example, when 1,234 is divided by 100, the remainder obtained is 34, which is the last two digits of the number.
Thus, (3x + 2) when divided by 100 will leave a remainder having 1 as the units digit only if (3x + 2) has 1 as its units digit as well.
This is possible only if the units digit of 3x is 9 (since 9 + 2 = 11 ≡the units digit is 1) The units digit of exponents of 3 follows a cycle as shown below: • 34k+1 ≡3 • 34k+2 ≡9 • 34k+3 ≡7 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 291 • 34k ≡1 Thus, the units digit of 3x is 9 only if x = 4k + 2, where k is any positive integer.
From statement 1: We have x = 2(2n + 1), where n is a positive integer.
x = 4n + 2 This is exactly the condition as discussed above. – Sufficient From statement 2: We have 10 > x > 4.
Thus, a couple of possible values of x are: x = 6, which is of the form x = 4n + 2 – satisfies x = 7, which is of the form x = 4n + 3, not of the form x = 4n + 2 – does not satisfy Thus, we do not have a unique answer. – Insufficient The correct answer is Option A.
281.
From statement 1: => x(2y −1) is even For any integer value of y, 2y is even.
Thus, 2y −1 = even −1 = odd.
Since the product of two numbers x and (2y −1) is even and (2y −1) is odd, it implies that x is even.
=> xz is even (since an even number multiplied with any integer (even or odd) is even).
– Sufficient From statement 2: => x(x + z) is even Possible situations are: © 1999–2024 Manhattan Review www.manhattanreview.com 292 Quantitative Reasoning Question Bank – DS Solutions (1) x is even, z is even => xz is even OR (2) x is even, z is odd => xz is even OR (3) x is odd, z is odd => xz is odd Thus, xz may be even or odd. – Insufficient The correct answer is Option A.
282.
From statement 1: y x = z y => x, y and z form a geometric series For example: x = 1, y = 2, z = 4: Here, y > x (common ratio of the geometric series is greater than 1) x = 4, y = 2, z = 1: Here, y ≯x (common ratio of the geometric series is smaller than 1) Thus, there is no unique answer. – Insufficient From statement 2: There is no information about y. – Insufficient Thus, from statements 1 and 2 together: We know that: z > x Thus, combining with the information from statement 1, we see that x, y, z form a geometric series with the common ratio greater than 1.
Thus, we have y > x => y −x > 0 – Sufficient The correct answer is option is C.
283.
The expression “Is |x −y| a positive number?” means, “Is x −y > 0?” We know that the absolute value of any number is always non-negative.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 293 Thus, we have x −y ≥0 Here, x −y > 0 will be satisfied only if x ̸= y.
From statement 1: xy + z = 0 => xy = −z => xy < 0 (since z is positive) => x > 0 and y < 0 OR x < 0 and y > 0 Thus, we can conclude that x ̸= y => x −y > 0 – Sufficient From statement 2: => x (x −2) = 0 x = 0 or 2 However, we have no information about y.
We cannot get the value of x −y – Insufficient The correct answer is Option A.
284.
From statement 1: This statement is clearly insufficient as x can take any value −1, −2, −3, etc.
At x = −1, y = (−1)2 + (−1)3 = 1 −1 = 0.
The answer is No; however, at other values of x, y < 0. The answer is Yes. – Insufficient From statement 2: This statement is clearly insufficient as y can take any value 0, −1, −2, −3, etc. If y = 0, the answer is No, else Yes. –Insufficient © 1999–2024 Manhattan Review www.manhattanreview.com 294 Quantitative Reasoning Question Bank – DS Solutions Thus, from statements 1 and 2 together: Even after combining both statements, we still have the following situations: At x = −1 => y = 0 – Answer is No.
AND At x = −2, y = (−2)2 + (−2)3 = 4 −8 = −4 => y < 0 – Answer is Yes.
Hence, there is no unique answer. – Insufficient The correct answer is Option E.
285.
From statement 1: Since 0 < d < 1 and 12d is an integer, we must have: d = An integer less than 12 12 .
(1) If d = 1 12 (smallest possible value of d): d = 0.0833 => The tenths digit is zero.
(2) If d = 11 12 (largest possible value of d): d = 0.917 => The tenths digit is non-zero.
Thus, there is no unique answer. – Insufficient From statement 2: Since 0 < d < 1 and 6d is an integer, we must have: d = An integer less than 6 6 .
If d = 1 6 (smallest possible value of d): d = 0.167 => The tenths digit is non-zero.
Since for the smallest possible value of d, the tenths digit is non-zero, the tenths digit will always be non-zero for all higher values of d. – Sufficient The correct answer is Option B.
286.
We know that for all n: t(n+1) = tn 3 Thus, we have t2 = t1 3 , t3 = t2 3 , . . .
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 295 Thus, we see that every term, starting from t2 is one-third of the previous term.
From statement 1: t2 = 1 3 => t3 = 1 3 × 1 3 = 1 9 => t4 = 1 3 × 1 9 = 1 27 => t5 = 1 3 × 1 27 = 1 81 – Sufficient From statement 2: Let t5 = x We know that t2 = 3 × t3 = 32 × t4 = 33 × t5 = 27x => t2 −t5 = 27x −x = 26x => 26x = 26 81 => x = t5 = 1 81 – Sufficient The correct answer is Option D.
287.
From statement 1: We have 3 √x = 9 => 3 √x = 32 => √x = 2 => x = 22 = 4 => 3x = 34 = 81 < 100. The answer is No. – Sufficient © 1999–2024 Manhattan Review www.manhattanreview.com 296 Quantitative Reasoning Question Bank – DS Solutions From statement 2: We have 1 3x > 0.01 For any value of x, 3x is positive, also 1 0.01 is positive. Thus, taking reciprocal and reversing the inequality: => 3x < 1 0.01 => 3x < 100. The answer is No. – Sufficient The correct answer is Option D.
288.
We need to determine whether: |x| < 1 => −1 < x < 1 From statement 1: |x + 2| = 3 |x −1| => x + 2 = ±3 (x −1) => x + 2 = ± (3x −3) => x + 2 = 3x −3 => x = 5 2 – Does not satisfy the required condition OR x + 2 = −(3x −3) => x + 2 = −3x + 3 x = 1 4 – Satisfies the required condition Thus, there is no unique answer. – Insufficient From statement 2: We have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 297 |2x −5| ̸= 0 => x ̸= 5 2 However, we do not have any possible values x can take. – Insufficient Thus, from statements 1 and 2 together: Combining both statements, we arrive at: x = 1 4 – Satisfies the required condition Thus, we have a unique answer. – Sufficient The correct answer is Option C.
289.
We know that the radical sign i.e.
√ k takes only the positive square root of k.
Thus, we can have the following cases: (a) q p −3 2 = (p −3) – Condition: If p −3 ≥0 => p ≥3 (b) q p −3 2 = q 3 −p 2 = (3 −p) – Condition: If 3 −p ≥0 => p ≤3 For the given condition to be satisfied, p must be less than or equal to 3.
From statement 1: p < p We know that p ≥0 for all values of p => p < 0 Thus, it satisfies the condition that p < 3 Thus, we have q p −3 2 = (3 −p) – Sufficient From statement 2: 3 > p => q p −3 2 = (3 −p) – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 298 Quantitative Reasoning Question Bank – DS Solutions 290.
m n < mn is true under the following conditions: (a) If m is positive, i.e. m > 0: Canceling m from both sides: 1 n < n => n > 1: (Say, n = 2 => 1 n = 1 2 < 2 = n) .... (a) OR => −1 < n < 0: (Say, n = −1 2 => 1 n = −2 < −1 2 = n) .... (b) (b) If m is negative, i.e. m < 0: Canceling m from both sides and reversing the inequality sign: 1 n > n => 0 < n < 1: (Say, n = 1 2 => 1 n = 2 > 1 2 = n) .... (c) OR => n < −1: (Say, n = −2 => 1 n = −1 2 > −2 = n) .... (d) From statement 1: We have mn > 0 => m > 0 and n > 0 OR m < 0 and n < 0 However, we do not get any of the conditions above. – Insufficient From statement 2: We have n < −1 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 299 However, we do not know whether m is positive or negative. – Insufficient Thus, from statements 1 and 2 together: We have n < −1 i.e. n is negative.
Also, since mn > 0, we have m < 0 Thus, we have m < 0 and n < −1 Thus, it satisfies condition (d) above. – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 300 Quantitative Reasoning Question Bank – DS Solutions 6.2 Percents 291.
From statement 1: We do not have information on the brokerage. – Insufficient From statement 2: We do not have information about the value of the property. – Insufficient Thus, from statements 1 and 2 together: The required percent = 3, 000 1.8 × 106 × 100 = 0.3% – Sufficient The correct answer is Option C.
292.
From statement 1: We have no information on the aspirations of the students to do the masters in management. – Insufficient From statement 2: Since 35% (more than 24%) of the male students and 25% (more than 24%) of the female students aspire for the course, we can definitely say that on an average, more than 24% of that total number of students aspire for the masters in management. – Sufficient Minimum term < Average > Maximum term Average is always greater than the minimum and less than the maximum.
The correct answer is Option B.
293.
From statement 1: The initial price of the smartphone is not known.
Hence, the percent increase in the price cannot be determined. – Insufficient From statement 2: The amount of increase in price of the smartphone is not known.
Hence, the percent increase in the price cannot be determined – Insufficient Thus, from statements 1 and 2 together: Increase in price of the smartphone = $40.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 301 Price of the smartphone after increase = $400.
Thus, initial price of the smartphone = $(400 −40) = $360.
Hence, the percent increase in the price = 40 360 × 100 = 11.11% – Sufficient The correct answer is Option C.
294.
From statement 1: Payment for the phone excluding 20% sales tax = $0.85d Thus, payment for the phone including sales tax = total price = $(0.85d + 20% of 0.85d) => $(0.85d + 0.17d) => $01.02d > $d. The answer is No. – Sufficient From statement 2: The value of d is not given, hence a comparison with d is not possible – Insufficient The correct answer is Option A.
295.
From statement 1: The number of marbles George has is 75 percent of the number of marbles Suzy has.
Say Suzy has 100 marbles, thus George has 75% of 100 = 75 marbles.
=> Suzy has 25 more marbles than George (75).
=> Suzy has 25 75 = 1 3 more marbles than George. – Sufficient From statement 2: The number of marbles Suzy has is 133.33% percent of the number of marbles George has.
Say George has 100 marbles, thus Suzy has 133.33% of 100 = ≈133 marbles.
=> Suzy has 33 more marbles than George (100).
© 1999–2024 Manhattan Review www.manhattanreview.com 302 Quantitative Reasoning Question Bank – DS Solutions => Suzy has ≈ 33 100 =≈1 3 more marbles than George. – Sufficient The correct answer is Option D.
296.
Let the total sales be $x.
Amount paid to the salesperson per month is: $(2, 000 + 15% of (x −10, 000)); provided (x > 10, 000) OR $2, 000; provided (x ≤10, 000).
From statement 1: If x ≤10, 000: $2, 000 = 17.5% of x => x = $2, 000 × 100 17.5 = $20, 000 17.5 = $10, 000 × 20 17.5 > $10, 000; since 20 17.5 > 1, the value of $10, 000 × 20 17.5 > $10, 000 However, this contradicts our assumption x ≤10, 000.
Thus, we can conclude that x ≰10, 000, i.e., x > 10, 000.
If x > 10, 000: $(2, 000 + 15% of (x −10, 000)) = 17.5% of x => 2, 000 + 15 (x −10, 000) 100 = 17.5x 100 This is a linear equation and will return a unique value of x. So the statement is sufficient to answer the question; however, for the sake of completeness, let’s do the calculation.
=> 2, 000 + 15x 100 −150, 000 100 = 17.5x 100 => 2, 000 −1, 500 = 17.5x 100 −15x 100 => 500 = 2.5x 100 => x = 20, 000 Thus, amount paid to the salesperson = $(2, 000 + 15% of (20, 000 −10, 000)) = $3, 500.
– Sufficient From statement 2: Since the total sales of the salesperson was $20,000 (> $10, 000), the amount paid to him was: $(2, 000 + 15% of (20, 000 −10, 000)) = $3, 500. – Sufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 303 The correct answer is Option D.
297.
Let the total sales be $x.
Amount paid to Tim per month is: $(1, 000 + 10% of (x −10, 000)); provided (x > 10, 000) OR $1,000; provided (x ≤10, 000).
From statement 1: Since Tim’s pay exceeds $1,500 (more than the fixed salary of $1,000), we can conclude that x > 10, 000.
Thus, we have $(1, 000 + 10% of (x −10, 000)) = $1, 500 => 1, 000 + 10 (x −10, 000) 100 = $1, 500 => x = $15, 000 – Sufficient From statement 2: Since Tim received some commission, we can conclude that x > 10, 000.
Thus, we have 10% of (x −10, 000)) = $500 => 10 (x −10, 000) 100 = $500 => x = $15, 000 – Sufficient The correct answer is Option D.
298.
Cost of the air conditioner, excluding sales tax = $600.
Sales tax on the cost of the air conditioner = $(10% of 600) = $60.
Thus, cost of the air conditioner, including sales tax = $(600 + 60) = $660.
© 1999–2024 Manhattan Review www.manhattanreview.com 304 Quantitative Reasoning Question Bank – DS Solutions We need to determine the installation cost including sales tax to get the total cost of the air conditioner and the installation.
From statement 1: Sales tax on installation = 10% of the installation cost = $6 Thus, installation cost = $6 × 100 10 = $60.
Hence, installation cost, including sales tax = $(60 + 6) = $66.
Hence, the air conditioner and the installation, including sales tax = $(660 + 66) = $726. – Sufficient From statement 2: Total sales tax = $66.
We know that the sales tax on the air conditioner = 10% of $600 = 60.
Thus, sales tax on installation cost = $ (66 −60) = $6.
This is the same information as in statement 1. Subsequent to this, following the calculation as we did in statement 1, we get the answer. – Sufficient The correct answer is Option D.
299.
From statement 1: We have no information on the total number bottles and the corresponding information for the bottles that are labeled orange juice. – Insufficient From statement 2: Since 80% of the bottles are labeled orange Juice, (100 −80) = 20% of the bottles are labeled guava Juice.
However, the specific information about the correct and incorrect labeling is not given.
Hence, the percent of juice bottles are labeled correctly cannot be determined. – Insufficient Thus, from statements 1 and 2 together: We know that of those that are labeled guava juice, 20 percent have orange juice in them.
However, we have no information about the number of bottles.
Hence, the percent of juice bottles are labeled correctly cannot be determined. – Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 305 300.
From statement 1: We have no information about the sales revenue in 2010.
Hence, the percentage change cannot be determined. – Insufficient From statement 2: We have no information about the sales revenue in 2001.
Hence, the percentage change cannot be determined. – Insufficient Thus, from statements 1 and 2 together: We have no information on the value of the the industry’s sales revenue in 2010 as compared to that in 2001.
Hence, the percentage change cannot be determined. – Insufficient The correct answer is Option E.
301.
From statement 1: We have no information about the total sales revenue of the industry in 2005 compared to 2001.
Hence, the percentage change in the sales revenue of Company X from 2001 to 2005 cannot be determined. – Insufficient From statement 2: We have no information about the value (or proportion) of sales revenue of Company X in 2001 and 2005.
Hence, the percentage change in the sales revenue of Company X from 2001 to 2005 cannot be determined. – Insufficient Thus, from statements 1 and 2 together: Let the total sales revenue of the industry in 2001 be $100x.
Thus, the total sales revenue of the industry in 2005 = $(120% of 100x) = $120x.
The sales revenue of Company X in 2001 = $(20% of 100x) = $20x.
The sales revenue of Company X in 2005 = $(20% of 120x) = $24x.
© 1999–2024 Manhattan Review www.manhattanreview.com 306 Quantitative Reasoning Question Bank – DS Solutions Hence, the percentage change in the sales revenue of Company X from 2001 to 2005 = 24x −20x 20x × 100 = 20% Since the sales revenue of Company X forms a constant percentage share of the total sales revenue of the industry, the percentage change in the sales revenue of Company X must be the same as the percentage change in the total sales revenue of the industry, 20%. – Sufficient The correct answer is Option C.
302.
From statement 1: The statement gives us information on the balance on January 31 had the rate been 15%.
This can be used to determine the balance on January 1.
However, since the actual percent increase has not been mentioned, we cannot determine the actual balance on January 31. – Insufficient From statement 2: The statement gives us information on the actual percent increase from January 1 to January 31.
However, since the balance on January 1 has not been mentioned, we cannot determine the actual balance on January 31. – Insufficient Thus, from statements 1 and 2 together: Let the balance on January 1 be $x.
Thus, at 15% increase, the balance on January 31 = $(115% of x) Thus, we have 115 100 × x = 1, 150 => x = $1, 000 Thus, actual balance on January 31 (at 10% increase) = $(110% of 1,000) = $1, 100. – Sufficient The correct answer is Option C.
303.
From statement 1: There is no information about Mark’s taxes. – Insufficient From statement 2: There is no information about Mark’s salary. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 307 Thus, from statements 1 and 2 together: There is no information about Mark’s actual salary and actual taxes or the taxes as a percent of salary ratio. – Insufficient Had the taxes as a percent of the salary i been known, say k%, the percent change in net income could have been calculated as follows: Initial tax = k% of i. New tax = 1.15 × (k% of i).
Initial net income: {i −(k% of i)}.
Final net income: 1.10i −1.15(k% of i).
Thus, change in net income = {1.1i −1.15(k% of i)} −{i −(k% of i)} = 0.1i −0.15(k% of i) = i 0.1 −0.15k 100 Thus, percent change = i 0.1 −0.15k 100 [i −(k% of i)] = i 0.1 −0.15k 100 i 1 − k 100 = 0.1 −0.15k 100 1 − k 100 Value of k is not known.
The correct answer is Option E.
304.
From statement 1: We know that the number of teachers with masters degree = 50 100 × 80 = 40.
However, we cannot determine the number of male teachers with masters degree. – Insufficient From statement 2: We only know the number of males = 50 100 × 80 = 40.
However, we cannot determine the number of male teachers with masters degree. – Insufficient Thus, from statements 1 and 2 together: Even after combining the statements, we cannot determine the number of male teachers with masters degree (since the percent of male teachers with masters degree is not known: we cannot assume that since 50% of the teachers have masters degree, 50% of the male teachers would also have masters degree). – Insufficient The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 308 Quantitative Reasoning Question Bank – DS Solutions 305.
Since each of School A’s and School B’s number of students in 2015 were 10% higher than that in 2014, the sum of their number of students in 2015 would also be 10% higher than that in 2014.
From statement 1: The sum of School A’s and School B’s number of students in 2014 = 1,000.
Thus, the sum of School A’s and School B’s number of students in 2015 = $1, 000× 110 100 = 1, 100 However, we cannot determine School A’s individual number of students in 2014. – Insufficient From statement 2: The sum of School A’s and School B’s number of students in 2015 = 1,100.
Thus, the sum of School A’s and School B’s number of students in 2014 = 1, 100 × 110 110 = 1, 000 However, we cannot determine School A’s individual number of students in 2014. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we cannot determine School A’s individual number of students in 2014, as we are only aware of the sum of School A’s and School B’s number of students, but no individual values of number of students are known. – Insufficient The correct answer is Option E.
306.
We need to verify if: 25% of n > 20% of n + 1 2 => 25n 100 > 20 n + 1 2 100 => n 4 > n + 0.5 5 => 5n > 4n + 2 => n > 2 From statement 1: 0 < n < 1 => n ≯2 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 309 Thus, the answer to the question is ‘No.’ – Sufficient From statement 2: n > 0.5 Thus, we may have n = 3(> 2) or n = 1(≯2).
Thus, the answer to the question is not unique. – Insufficient The correct answer is Option A.
307.
We need to verify whether: x × 100% = x × 33.33% x × 100 100 = z × 33.33 100 => x = z 3 From statement 1: z = (100 + 200) % of x => z = 300% of x => z = 300 100 × x => z = 3x => x = z 3 – the answer is Yes. – Sufficient From statement 2: x = (100 −75) % of (x + z) => x = 25% of (x + z) => x = 25 100 × (x + z) => x = x + z 4 => x −x 4 = z 4 => 3x 4 = z 4 © 1999–2024 Manhattan Review www.manhattanreview.com 310 Quantitative Reasoning Question Bank – DS Solutions => x = z 3 – the answer is Yes. – Sufficient The correct answer is Option D.
308.
Let the expenditures for computers, printers and software be $x, $y and $z, respectively.
Thus: x + y + z = 54, 000 . . . (i) From statement 1: y = (100 + 30) % of z => y = 1.3z . . . (ii) The above equation along with equation (i) cannot be used to solve for x since there are three unknowns and only two equations. – Insufficient From statement 2: y + z = (100 −35) % of x => y + z = 7x 20 . . . (iii) Substituting (y + z) above in equation (i): x + 7x 20 = 54, 000 => 27x 20 = 54, 000 => x = $40, 000 – Sufficient The correct answer is Option B.
309.
From statement 1: There is no information about the taxable income of the amount of taxes paid by John in 2002.
– Insufficient From statement 2: There is no information about the amount of taxes paid by John in 2001.
Hence, the amount of taxes paid in 2002 cannot be determined. – Insufficient Thus, from statements 1 and 2 together: Taxes paid in 2001 = $(5% of 40,000) = $2,000.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 311 Thus, taxes paid in 2002 = $(2,000 + 500) = $2,500.
However, the taxable income in 2002 is not known.
Thus, the percentage of taxable income paid as taxes cannot be determined. – Insufficient The correct answer is Option E.
310.
Let Joe’s income in 2001 be $100 Thus, taxes paid in 2001 = $(5.1% of 100) = $5.1 From statement 1: Joe’s income in 2002 = ${(100 + 10)% of 100} = $110.
However, there is no information about the amount of taxes Joe paid in 2002. – Insufficient From statement 2: Taxes paid in 2002 = $(3.4% of 100) = $3.4 However, there is no information about Joe’s income in 2002. – Insufficient Thus, from statements 1 and 2 together: Joe’s income in 2002 = $110.
Taxes paid by Joe in 2002 = $3.4 Thus, percent of income paid in taxes = 3.4 110 × 100 < 5.1 – Sufficient The correct answer is Option C.
Alternate approach: Since income has increased from 2001 to 2002, while the taxes in 2002 has fallen as a percent of the income in 2001, the percent of income paid in taxes in 2002 would be even lower.
Hence, the answer is ‘Yes.’ – Sufficient 311.
We know that the number of students in 2005 = 1050.
Let number of students in 1995 be n.
Thus, the number of students in 2000 = (100 + 50) % of n = 150 100 × n = 3n 2 .
© 1999–2024 Manhattan Review www.manhattanreview.com 312 Quantitative Reasoning Question Bank – DS Solutions From statement 1: Number of students in 2005 = (100 + 110) % of n = 210 100 × n = 21n 10 .
Thus, we have 21n 10 = 1, 050 => n = 500 => The number of students in 2000 = 3 2 × 500 = 750.
Thus, the percent increase in the number of students from 2000 to 2005 = 1, 050 −750 750 × 100 = 40% – Sufficient From statement 2: We have n = 500 This is the same information as obtained from statement 1. – Sufficient The correct answer is Option D.
312.
We need to find the ratio of the number of employees in Company A and that in Company B in 2001.
From statement 1: The relation between the number of employees in Company A in 2001 to the number of employees in Company B in 2000 is given.
However, there is no relation given regarding the number of employees in Company B in 2001.
– Insufficient From statement 2: The relation between the number of employees in Company B in 2001 to the number of employees in Company B in 2000 is given.
However, there is no relation given regarding the number of employees in Company A in 2001.
– Insufficient Thus, from statements 1 and 2 together: Let the number of employees in Company B in 2000 be p.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 313 Thus, the number of employees in Company A in 2001 = (100 + 60) % of p = 8p 5 .
Also, the number of employees in Company B in 2001 = (100 + 20) % of x = 6p 5 .
Thus, the required ratio = 8p 5 : 6p 5 = 4 : 3. – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 314 Quantitative Reasoning Question Bank – DS Solutions 6.3 Profit & Loss 313.
Gross profit = (y −x) The gross profit as a percent of the cost = y −x x × 100 = y x −1 × 100.
From statement 1: y −x = 60 However, we have no information on the value of y x . – Insufficient From statement 2: 5y = 6x y x = 6 5.
Thus, the gross profit as a percent of the cost = y x −1 × 100 = 6 5 −1 × 100 = 20%. – Sufficient The correct answer is Option B.
314.
Total earning from the car = $5,000 Profit = Earning −Cost Let the cost of repairing the car = $l Let the cost of buying the car = $m Hence, profit = $(5, 000 −(m + l)) We need to check if the profit is greater than $1,500.
=> (5, 000 −(m + l)) > 1, 500 => m + l < 3, 500 From statement 1: l + m = 3m => l = 2m.
We do not have information on the values of l and m – Insufficient From statement 2: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 315 5, 000 −(m + l) > l => m + 2l < 5, 000 We have no information on the value of l and m – Insufficient From statements 1 and 2 together: We have m + 2l < 5, 000, and l = 2m => m + 2(2m) < 5, 000 => m < 1, 000 => l = 2m < 2, 000 Thus, we have m + l < (1, 000 + 2, 000 = 3, 000) < 3, 500 – Sufficient The correct answer is Option C.
315.
From statement 1: We know that the percentage discount on the book was 10 percentage points more than that on the notebook.
However, the sale price of the two items before the discount is not known.
Thus, we cannot determine which item was discounted by what amount. – Insufficient From statement 2: We know that the sale price of the book was $1 less than the sale price of the notebook.
However, we have no information on the discounts offered on the two items.
Thus, we cannot determine which item was discounted by what amount. – Insufficient Thus, from statements 1 and 2 together: Let the percentage discount on the notebook be s%.
Thus, the percentage discount on the book will be (s + 10)%.
Let the sale price of the notebook, before the discount be $p.
Thus, the sale price of the book, before the discount will be $(p −1).
Thus, discount on the book = $ (s + 10)(p −1) 100 = $ sp 100 + 10(p −1) 100 ; and the discount on the notebook = $ sp 100 If 10(p −1) = s, the answer is no, else yes. Since we do not know the values of s and p, we cannot compare the discounts.
© 1999–2024 Manhattan Review www.manhattanreview.com 316 Quantitative Reasoning Question Bank – DS Solutions Alternatively, since the book has a higher percentage discount on relatively lower price, we cannot compare which of the two items has a higher discount. – Insufficient The correct answer is Option E.
316.
Let the price at which the gas stoves of Type A and Type B were purchased by the trader be $c each.
Let the price at which the gas stoves of Type A and Type B were sold by the trader be $x and $y, respectively.
Thus, the profit on Type A = x −c & the profit on Type B = y −c We have to calculate (x −c) −(y −c) (y −c) × 100% = (x −y) (y −c) × 100% From statement 1: x = y + 10% of y => x = y 1 + 10 100 = 1.1y => x −y = 0.1y ... (a) We have no information about the cost of the two gas stoves.
Hence, we cannot determine the answer. – Insufficient From statement 2: y −c = 50 ... (b) We have no information about the difference in selling prices of the two gas stoves, i.e. (x −y).
Hence, we cannot determine the answer. – Insufficient Hence, from statements 1 and 2 together: From (a) and (b): x −y y −c × 100 = 0.1y 50 × 100 = 0.2y However, the value of y is not known. - Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 317 317.
From statement 1: Marked price = (100 + 25) % of (Cost price) => Cost price = (Marked price) × 100 125 = $6, 250 × 100 125 = $5, 000 – Sufficient From statement 2: Selling price of the bike = $5,500.
Thus, we have 5, 500 = (100 + 10) % of (Cost price) => 5, 500 = 1 + 1 10 × (Cost price) => Cost price = $5, 500 × 10 11 = $5, 000 – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 318 Quantitative Reasoning Question Bank – DS Solutions 6.4 Averages (including weighted averages) 318.
From statement 1: Say R1 refers to the average of all the amounts in the 1st row.
Similar reasoning is valid for the other rows as well.
Let us take an example: R1 1 2 3 6 Average of R1 = 12 4 = 3 R2 4 6 7 7 Average of R2 = 24 4 = 6 Average of all the values = 1 + 2 + 3 + 6 + 4 + 6 + 7 + 7 8 = 4.5 Average of R1 and R2 = 3 + 6 2 = 4.5 Thus, average of all the elements is the same as the average of R1 and R2 Thus, we can say that the average of all the 24 amounts = R1 + R2 + R3 + R4 + R5 + R5 6 = 720 6 = 120 – Sufficient From statement 2: Similar reasoning is applicable as in statement 1.
Thus, we can say that the average of all the 24 amounts = C1 + C2 + C3 + C4 4 = 480 4 = 120 – Sufficient The correct answer is Option D.
319.
From statement 1: We know that average age of the employees enrolled for only one course.
Hence, we cannot determine the average age of all the employees. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 319 From statement 2: We only know the ratio of the average ages of the employees enrolled for the two courses.
Hence, we cannot determine the average age of all the employees. – Insufficient Thus, from statements 1 and 2 together: Average age of the employees in the NLP course = 40.
Thus, average age of the employees in the HLP course = 3 4 × 40 = 30.
However, we do not know the number of employees enrolled for each course or the ratio of number of employees enrolled for each course. If the ratio of number of employees in NLP course to number of employees in HLP course is known = x y , then the average would have been = 40x + 30y x + y .
Hence, we cannot determine the average age of all the employees. – Insufficient The correct answer is Option E.
320.
From statement 1: The average annual wage of the workers in Department X is $15,000.
However, we have no information on the workers in other departments in the factory.
Hence, the average annual wage of the workers at the factory cannot be determined. – Insuffi-cient From statement 2: The average annual wage of the workers not in Department X is $20,000.
However, we have no information on the wage of the workers in Department X in the factory.
Hence, the average annual wage of the workers at the factory cannot be determined. – Insuffi-cient Thus, from statements 1 and 2 together: We know the average annual wage of the workers in Department X and that of the the workers who are not in Department X.
However, we have no information on the ratio of the number of workers in Department X and the number of workers other than in Department X.
If the ratio Number of workers in Department X Number of workers not in Department X = x y , © 1999–2024 Manhattan Review www.manhattanreview.com 320 Quantitative Reasoning Question Bank – DS Solutions Then, the average wage of all the workers in the factory = x × 15, 000 + y × 20, 000 x + y !
Hence, the average annual wage of the workers at the factory cannot be determined – Insufficient The correct answer is Option E.
321.
From statement 1: There is no information on the number of desktop computers sold, the number of laptop computers sold, and the average selling price of the laptop computers. – Insufficient From statement 2: There is no information on the number of desktop computers sold, the number of laptop computers sold, and the average selling price of the desktop computers. – Insufficient Thus, from statements 1 and 2 together: Let the number of desktop computers sold and the number of laptop computers be d & l, respectively; thus, the average price for all the computers = $ 800d + 1, 100l d + l However, there is no information about d & l. – Insufficient The correct answer is Option E.
322.
Dave’s average score for the three tests = 74.
Thus, Dave’s total score for three tests = 74 × 3 = 222.
From statement 1: Dave’s highest score = 82.
Thus, sum of Dave’s two lowest scores = 222 −82 = 140.
However, we cannot determine Dave’s lowest scores. – Insufficient From statement 2: Sum of Dave’s two highest scores = 162.
Thus, Dave’s lowest score = 222 −162 = 60 – Sufficient The correct answer is Option B.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 321 323.
We know that there are 20 friends in all.
Let the average amount spent by each friend = $a.
The amount spent by the first five friends = $ (5 × 21) = $105.
The average amount spent by the remaining (20 −5) = 15 friends = $ (a −x).
Thus, the total amount spent by the 15 friends = $ (15 × (a −x)).
Thus, the total amount spent by all the friends = $ {15 × (a −x) + 105} . . . (i) Thus, the average amount spent by all the friends = $ 15 × (a −x) + 105 20 Thus, we have 15 (a −x) + 105 20 = a => 15 (a −x) + 105 = 20a => 3(a −x) + 21 = 4a => 3a −3x + 21 = 4a => a + 3x = 21 . . . (ii) From statement 1: x = 3 Thus, from (ii), we have => a + 3 × 3 = 21 => a = 12 – Sufficient From statement 2: The total amount spent is $240, and total friends are 20.
The average amount spent = 240 20 = 12 – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 322 Quantitative Reasoning Question Bank – DS Solutions 6.5 Ratio & Proportion 324.
Let the number of students = n.
From statement 1: Say with each student received x candies, y cookies, and z toffees.
Thus, x : y : z = 3 : 4 : 5 => x = 3k, y = 4k, z = 5k, where k is a constant of proportionality.
However, we have no information on n. – Insufficient From statement 2: nx = 27, ny = 36, nz = 45.
We have no information about x, y, & z.
Hence, we cannot determine the value of n. – Insufficient Thus, from statements 1 and 2 together: Substituting the values of x or y or z from statement 1 in the information from statement 2, we have nx = 27 = 3k n = 9 k Since k is unknown, we cannot determine n. – Insufficient Note: Since x, y, z and n must be integers, k can be either 1, 3 or 9. For k = 9, there would be only one student, whereas for k = 3, there would be three students and for k = 1, there would be nine students.
The correct answer is Option E.
325.
Number of candidates at the beginning of the session in the MBA (Finance) course and MBA (Marketing) course were n each.
Number of candidates at the end of the session in the MBA (Finance) course and MBA (Marketing) course were (n −6) and (n −4), respectively.
We have to determine the value of n.
From statement 1: Number of candidates who left at the end of the session = 6 + 4 = 10 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 323 Thus, number of candidates at the beginning of the session = 5 1 × 10 = 50 Thus: 2n = 50 => n = 25 – Sufficient From statement 2: Number of candidates remained in MBA (Marketing) course = 21 Number of candidates who had left MBA (Marketing) course = 4 Thus, number of candidates in MBA (Marketing) course at the beginning of the session = 21 + 4 = 25 Thus, number of candidates in MBA (Finance) course at the beginning of the session n = 25 – Sufficient The correct answer is Option D.
326.
From statement 1: Before the milk from Tub A was poured, Tub A was 1 3 full.
However, the capacity of Tub B in terms of Tub A is not known.
Hence, fraction of milk in Tub B after pouring cannot be determined. – Insufficient From statement 2: Let the capacity of Tub A and Tub B be c units.
Though we know the initial amount of milk in Tub B = c 2, we do not know the initial amount of milk present in Tub A.
Hence, fraction of milk in Tub B after pouring milk from Tub A cannot be determined.
– Insufficient Thus, from statements 1 and 2 together: The capacity of Tub A and Tub B is c units.
Initial amount of milk in Tub A and B = c 3 and c 2, respectively.
Final amount of milk in Tub B when all the milk in Tub A is poured in Tub B = c 2 + c 3 = 5c 6 .
Hence, fraction of milk in Tub B after pouring = 5 6 – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 324 Quantitative Reasoning Question Bank – DS Solutions 327.
Ratio of marbles is: Red : Blue : Green : Yellow = 6 : 5 : 2 : 2.
Thus, we have Number of red marbles = 6k Number of blue marbles = 5k Number of green marbles = 2k Number of yellow marbles = 2k, where k is the constant of proportionality We need to determine the number of green marbles, i.e. 2k.
From statement 1: Since number of red marbles is 2 more than blue marbles, we have 6k = 2 + 5k => k = 2 => The number of green marbles = 2k = 2 × 2 = 4 – Sufficient From statement 2: Total number of marbles = 6k + 5k + 2k + 2k = 15k.
Thus, we have 15k = 30 => k = 2 => The number of green marbles = 2k = 2 × 2 = 4 – Sufficient The correct answer is Option D.
328.
From statement 1: We have the ratio of Chemical X and Chemical Y mixed.
Since no volume is mentioned, we cannot determine the number of milliliters of Chemical X. – Insufficient From statement 2: The volume of Chemical Y is known. However, we cannot determine the amount of Chemical X.
– Insufficient Thus, from 1 and 2 together: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 325 Chemical X Chemical Y = 2 3 => Chemical X 60 = 2 3 => Chemical X = 60 × 2 3= 40 milliliters. – Sufficient The correct answer is Option C.
329.
From statement 1: The ratio of the number of male workers to female workers = 2 : 5.
There is no information about the present scenario. – Insufficient From statement 2: Let the number of male workers last year be m.
Let the number of female workers last year be f.
Thus, the number of male workers at present = (m + 300).
The number of female workers at present = f.
Thus, we have m + 300 f = 2 3 There are two unknowns and hence, this equation cannot be solved. – Insufficient Thus, from statements 1 and 2 together: m : f = 2 : 5 => f = 5m 2 Substituting this in the equation obtained in statement 2, we have (m + 300) 5m 2 = 2 3 => m = 450 => f = 5 × 450 2 = 1, 125 Thus, the number of male workers now = (m + 300) = 750.
© 1999–2024 Manhattan Review www.manhattanreview.com 326 Quantitative Reasoning Question Bank – DS Solutions The number of female workers now = f = 1, 125.
Thus, the total number of workers now = 750 + 1125 = 1, 875. – Sufficient The correct answer is Option C.
330.
From statement 1: Steve bought 3 5 of the total candies they bought together.
Thus, David bought 1 −3 5 = 2 5 of the total candies they bought together.
Thus, Steve bought more candies than David did. – Sufficient From statement 2: We know that they together bought a total of 50 candies.
However, we have no information on the number of candies each bought; we cannot compare their number of candies. – Insufficient The correct answer is Option A.
331.
We know that: The ratio of the number of male and female workers in 2002 = 3 : 4.
Let the number of male and the number of female workers in 2002 be 3k and 4k, respectively, where k be a constant of proportionality.
From statement 1: We know that: The ratio of the number of male workers in 2002 to that in 2003 = 3 : 5.
Thus, the number of male workers in 2002 = 3k.
And, the number of male workers in 2003 = 5 3 × 3k = 5k.
However, we have no information about the number of female workers in 2003. – Insufficient From statement 2: We know that: The ratio of the number of male and female workers in 2003 = 10 : 7.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 327 Let the number of male and female workers in 2003 be 10l and 7l, respectively; where l is another constant of proportionality (not necessarily be the same as k).
Percent increase in the number of male workers from 2002 to 2003 = 10l −3k 3k × 100 = 10l 3k −1 × 100% = Pm Percent increase in the number of female workers from 2002 to 2003 = 7l −4k 4k × 100 = 7l 4k −1 × 100% = Pf Comparing Pm and Pf , we see that there are two ratios involved: 10l 3k and 7l 4k , respectively.
Comparing the ratios, we see that: 10l 3k = 10 3 × l k = 3.33 × l k 7l 4k = 7 4 × l k = 1.75 × l k As l and k both are positive, the ratio l k must also be positive. Hence, from above equations we can conclude that 10l 3k > 7l 4k => Pm > Pf – Sufficient The correct answer is Option B.
332.
From statement 1: There is no information about the male employees in the company. – Insufficient From statement 2: There is no information about the female employees in the company. – Insufficient Thus, from statements 1 and 2 together: We know that 40% of females are above 50 years of age and 2/5 of 40% = 16% of females are above 55 years of age.
We know that 20 male employees are over 55 years of age.
However, we cannot find the number of female employees over 55 years of age.
Also, the total number of employees is not known.
© 1999–2024 Manhattan Review www.manhattanreview.com 328 Quantitative Reasoning Question Bank – DS Solutions Thus, the answer cannot be determined. – Insufficient The correct answer is Option E.
333.
From statement 1: There is no information about the male members in the club. – Insufficient From statement 2: There is no information about the female members in the club. – Insufficient Thus, from statements 1 and 2 together: Percent of female members who are mechanical engineers = 1 3 × 75% = 25%.
Percent of male members who are engineers = 30%.
Thus, the percent of male members who are mechanical engineers ≤30%, (since only engineers can be mechanical engineers).
Since for both male and female members, the fraction of mechanical engineers among them is less than 1 3, the fraction of mechanical engineers is definitely not more than 1 3. – Sufficient The correct answer is Option C.
334.
From statement 1: A B C D AD = AC + BD −BC However, the length of BC is not known. – Insufficient From statement 2: BC2= AB × CD However, none of the lengths are known. – Insufficient Thus, from statements 1 and 2 together: Let the length of BC = x.
Thus, from statement 1, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 329 AB = 10 −x CD = 15 −x Thus, from statement 2, we have x2 = (10 −x) (15 −x) => x2 = 150 −25x + x2 => 25x = 150 => x = 6 Thus, we have AD = AC + BD −BC = 10 + 15 −6 = 19 – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 330 Quantitative Reasoning Question Bank – DS Solutions 6.6 Mixtures 335.
We need to find the minimum concentration of milk in any of the two containers so that when mixed they result in 80% milk solution.
Since one container has the minimum milk concentration, the other must have the maximum possible milk concentration, i.e. 100% (this is the limiting case).
Also, in order to find the minimum concentration in one container, we must have 100% concentration of milk in the container having the larger volume so that a large quantity of milk is obtained.
From statement 1: We have x = 2y => x > y Thus, the container with x liters must be taken to be 100% milk.
Since the entire contents of both containers are mixed to get 30 liters of solution, we have x + y = 30 => 2y + y = 30 => y = 10 => x = 20 Thus, we have two solutions: 20 liters of 100% concentration of milk and 10 liters of n% concentration of milk, where n% represents the minimum concentration of milk.
Thus, equating the final concentration of milk, we have 20 × 100 100 + 10 × n 100 20 + 10 = 80 100 => 20 + n 10 = 4 5 × 30 => 20 + n 10 = 24 => n = 40% – Sufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 331 From statement 2: We have x = y + 10 => x > y Thus, the container with x liters must be taken to be 100% milk.
Since the entire contents of both containers are mixed to get 30 liters of solution, we have x + y = 30 => y + 10 + y = 30 => y = 10 => x = 20 This is the same result as obtained from statement 1.
Hence, we would obtain a unique answer. – Sufficient The correct answer is Option D.
336.
From statement 1: We have x = 10, and y = 100 However, the value of z is not known (we should not assume that x = z). – Insufficient From statement 2: We have x = 20% of y, and z = 10% of y Since we need the fraction of milk finally present in the mixture, we can assume a suitable value for y for ease of calculations.
© 1999–2024 Manhattan Review www.manhattanreview.com 332 Quantitative Reasoning Question Bank – DS Solutions Let y = 100 liters Thus, we have x = 20, and z = 10 Thus, we have From 100 liters of milk in a cask, 20 liters are removed and then 10 liters of water are added.
This process is repeated twice.
Thus, in the first cycle, fraction of the total contents of the cask removed = 20 100 = 1 5 Thus, fraction of contents left = 1 −1 5 = 4 5 Thus, amount of milk left after the first cycle = 100 × 4 5 = 80 liters.
Now, 10 liters of water are added.
Thus, total contents of the cask = 80 + 10 = 90 liters.
Thus, in the second cycle, fraction of the total contents of the cask removed = 20 90 = 2 9 Thus, fraction of contents left = 1 −2 9 = 7 9 Thus, amount of milk left after the second cycle = 80 × 7 9 = 560 9 liters.
Total volume of cask after removal = (90 −20) = 70 liters.
Now, 10 liters of water are added.
Thus, total contents of the cask = (70 + 10) = 80 liters.
Thus, required fraction of milk = 560 9 80 = 7 9 – Sufficient The correct answer is Option B.
337.
Let the price of each bottle of beer = $x.
From statement 1: Total bottles purchased = 4 + 6 + 2 = 12.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 333 Thus, total cost of all beer bottles = $12x.
Number of bottles of beer consumed by each friend = 12 3 = 4 bottles.
However, C had purchased only 2 bottles of beer.
Thus, he had to pay to A and B the price of 4 −2 = 2 bottles of beer.
Thus, the amount C paid to A and B = $2x.
However, we cannot determine the value of x. – Insufficient From statement 2: There is no information about the number of bottles of beer purchased. – Insufficient Thus from statement 1 and 2 together: We have 2x = 16 => x = $8 – Sufficient The correct answer is Option C.
338.
Let the amount of milk and water in the mixture be m and w respectively.
From statement 1: When 2 liters of milk is added to the mixture, the resultant mixture has equal quantities of milk and water.
Thus, we have (m + 2) = w . . . (i) However, we cannot determine the value of m. −Insufficient From statement 2: The initial mixture had 2 parts of water to 1 part milk.
Thus, we have m w = 1 2 © 1999–2024 Manhattan Review www.manhattanreview.com 334 Quantitative Reasoning Question Bank – DS Solutions => w = 2m . . . (ii) However, we cannot determine the value of m. −Insufficient Thus, from statements 1 and 2 together: Substituting w from (ii) in (i): m + 2 = 2m => m = 2 −Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 335 6.7 Speed, Time, & Distance 339.
From statement 1: Dave’s average speed was 3 5 of Jack’s speed.
Thus, Dave’s travel time = 1 3 5 = 5 3 of Jack’s travel time (since for the same distance, time is inversely proportional to speed).
It is given that Jack’s travel time = 12 hours.
Thus, Dave’s travel time = 12 × 5 3 = 20 hours – Sufficient From statement 2: The length of the route is of no use since actual speeds are not mentioned. – Insufficient The correct answer is Option A.
340.
Let the length of the route be d miles.
From statement 1: Time taken to cover the route at 65 miles per hour is d 65 hours.
Time taken to cover the route at 60 miles per hour is d 60 hours.
Thus, we have d 65 = d 60 −20 60 d 65 = d 60 −1 3 => d = 260 miles. – Sufficient From statement 2: Distance of the first half = 2.5 hours × 52 miles per hour = 130 miles.
Thus, d = 2 × 130 = 260 miles.– Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 336 Quantitative Reasoning Question Bank – DS Solutions 341.
From statement 1: There is no information about Suzy’s average speed. – Insufficient From statement 2: There is no information about the distance. – Insufficient Thus, from statements 1 and 2 together: Let the actual distance be d miles.
Thus, Suzy’s estimate of the distance ranged from (d + 10) miles to (d −10) miles.
Let Suzy’s actual average speed be s miles/hour.
Thus, Suzy’s estimate of her speed ranged from (s + 5) miles to (s −5) miles/hour.
Thus, actual time = d s hours.
Maximum value of estimated time = d + 10 s −5 hours.
Minimum value of estimated time = d −10 s + 5 hours.
We need to determine whether: • d + 10 s −5 −d s ≤30 60 = 0.5, AND • d s − d −10 s + 5 ≤30 60 = 0.5 Since the relation between s and d is not known, we cannot determine the answer.
Let us take some values to verify: (1) If d = 20, s = 15: d + 10 s −5 −d s = 30 10 −20 15 = 3 −4 3 = 5 9 > 0.5 – does not satisfy (2) If d = 60, s = 50: d s − d −10 s + 5 = 60 50 −50 65 = 6 5 −10 13 = 28 65 < 0.5 – satisfies www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 337 Thus, there is no unique answer. – Insufficient The correct answer is Option E.
342.
Time (t) taken to travel d miles at r miles per hour t = d r Time (T) taken to travel D miles at R miles per hour T = D R From statement 1: d = D + 20 . . . (i) However, there is no information on the values of r and R. – Insufficient From statement 2: r = R + 20 . . . (i) However, there is no information on the values of d and D. – Insufficient Thus, from statements 1 and 2 together: From (i) and (ii), we have t = d r = D + 20 R + 20 = D + 20 R R + 20 R = D R + 20 R 1 + 20 R = T + 20 R 1 + 20 R Let 20 R = k Thus, we have © 1999–2024 Manhattan Review www.manhattanreview.com 338 Quantitative Reasoning Question Bank – DS Solutions t = T + k 1 + k There are two possible cases: (a) If T > 1: T + k 1 + k < T; for example: If T = 2 and k = 1: T + k 1 + k = 3 2 < 2 => t < T (b) If T < 1: T + k 1 + k > T; for example: If T = 1 2 and k = 1: T + k 1 + k = 3 2 2 = 3 4 > 1 2 => t > T Thus, there is no unique answer. – Insufficient The correct answer is Option E.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 339 6.8 Time & Work 343.
From statement 1: Time taken to manufacture 20 screws = 28 seconds.
However, no information is provided about the time taken to manufacture a bolt. – Insufficient From statement 2: Time taken to manufacture 1 bolt = 1.5 times the time taken to manufacture one screw.
However, no information is provided about the time taken to manufacture one screw.
– Insufficient Thus, from statements 1 and 2 together: Time taken to manufacture one screw = 28 20 = 1.4 seconds.
Thus, time taken to manufacture one bolt = 1.4 × 1.5 = 2.1 seconds.
Thus, time taken to manufacture 1,000 bolts = 2.1 × 1000 = 2,100 seconds. – Sufficient The correct answer is Option C.
344.
From statement 1: We have information about only one machine. – Insufficient From statement 2: We do not have any information about the actual rates at which the bolts are made. – Insufficient Thus, from statements 1 and 2 together: We know that one machine manufactures bolts at the rate of 50 bolts per minute.
Since one machine is twice as fast as the other machine (we have no information about which machine is twice as efficient), we can have the second machine making bolts at the rate of (1) 50 2 = 25 bolts per minute OR (2) 50 × 2 = 100 bolts per minute © 1999–2024 Manhattan Review www.manhattanreview.com 340 Quantitative Reasoning Question Bank – DS Solutions Thus, we do not know the actual rate at which the bolts are made by the other machine. – Insufficient The correct answer is Option E.
345.
Time taken by 5 skilled workers to complete the job = 18 hours.
Thus, time taken by 1 skilled worker to complete the job = 18 × 5 = 90 hours . . . (i) Also, number of skilled workers required to complete the job in 1 hour = 5 × 18 = 90 . . . (ii) We need to find the time it takes for a group of 3 skilled workers and 4 apprentices to do the same job.
From statement 1: Since an apprentice works at 2 3 the rate of a skilled worker, we can say that 1 apprentice is equivalent to 2 3 of a skilled worker.
Thus, 3 apprentices are equivalent to 3 × 2 3 = 2 skilled workers.
Thus, 4 skilled workers and 3 apprentices are equivalent to (4 + 2) = 6 skilled workers.
Thus, from (i), we have Time taken by 6 skilled workers to complete the job = 90 6 = 15 hours – Sufficient From statement 2: Time taken by 6 apprentices and 5 skilled workers to complete the job = 10 hours.
Thus, in order to complete the job in 1 hour, number of people required = 6 × 10 apprentices and 5 × 10 skilled workers = 60 apprentices and 50 skilled workers Thus, from (ii), we have 60 apprentices and 50 skilled workers ≡90 skilled workers => 60 apprentices ≡40 skilled workers => 1 apprentice ≡40 60 = 2 3 skilled worker www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 341 This is the same information as in statement 1. – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 342 Quantitative Reasoning Question Bank – DS Solutions 6.9 Computational 346.
Say there are n number of sales persons.
Thus from statement 1, total number of computers = 5n + 18 – Insufficient Thus from statement 2, total number of computers = 4n + 28 – Insufficient Thus from statement 1 & 2 together: 5n + 18 = 4n + 28 => n = 10 Thus, the number of computers = 5 × 10 + 18 = 68 - Sufficient The correct answer is Option C.
347.
The employee gets paid $10 per hour for 8 hours i.e. $80 for a total of 8 hours.
Hence, for an excess of 8 hours, his pay per hour = $10 × 11 4 = $ 10 × 5 4 = $ 25 2 .
From statement 1: Since we have no information on the number of hours the employee worked yesterday, we cannot calculate his yesterday’s pay.
Thus, we cannot find his today’s pay.
Hence, we cannot determine the number of hours he worked today. – Insufficient From statement 2: We have no information on the pay received by the employee today.
Hence, we cannot determine the number of hours he worked today. – Insufficient From statements 1 and 2 together: Since the employee worked for 8 hours yesterday, he received $80 as pay.
Hence, the pay received by the employee today $(25 + 80) = $105.
For the first 8 hours today, the employee received $80.
For each hour of additional work, the employee receives = $25 2 .
Since he received $25 extra today, the number of hours he worked extra today = 25 25 2 = 2 hours.
Hence, the employee worked for 8 + 2 = 10 hours today. – Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 343 348.
Let a popular-size box contain b numbers of batteries.
From statement 1: The large-size box contains (b + 10) numbers of batteries.
We cannot determine the cost per battery since no price is mentioned. – Insufficient From statement 2: Cost of a large-size box = $20 We cannot determine the cost per battery since no quantity is mentioned. – Insufficient Thus from statements 1 and 2 together: Cost per battery of the large-size box = $ 20 b + 10 .
We cannot determine the cost per battery since b is unknown. – Insufficient The correct answer is Option E.
349.
From statement 1: We have no information on how many students received how many candies. – Insufficient From statement 2: We know that 15 students received two candies each.
This accounts for 15 × 2 = 30 candies.
We have no information on the number of candies received by the others. – Insufficient Thus, from statements 1 and 2 together: Let x students received one candy each.
Also, 15 students received two candies each (from Statement 2).
Thus, the remaining (50−x −15) = (35−x) students received exactly three candies each (since no one received more than three candies–from Statement 1).
Total number of candies distributed = 105.
Thus, x × 1 + 15 × 2 + (35 −x) × 3 = 105 => x = 15 – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 344 Quantitative Reasoning Question Bank – DS Solutions 350.
Let the number of teachers = x Thus, number of males = x 4 Number of non-academic staff = x 2 From statement 1: We know that there are 14 males who are non-academic staff.
However, we cannot determine x from this information. – Insufficient From statement 2: Number of males = x 4 Thus, number of females = x −x 4 = 3x 4 Hence, we have 3x 4 −x 4 = 32 => x = 64 – Sufficient The correct answer is Option B.
351.
Let the price of each eraser yesterday be $x.
Thus, the price of each pencil yesterday = $(x + 0.20) Let the number of pencils sold = p and that of erasers sold = e We have to calculate the revenue from the sale of erasers = ex From statement 1: e = p + 10.
Since we do not know the actual number of erasers sold, we cannot determine the revenue from the erasers. – Insufficient From statement 2: Revenue from pencils = p × (x + 0.20) = 30.
Since we do not know the number of pencils sold or the price of each pencil, we cannot determine the revenue from the sale of erasers. – Insufficient Thus, from statements 1 and 2 together: Revenue from erasers = ex = (p + 10)x; (substituting e = p + 10).
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 345 We have p × (x + 0.20) = 30; However, we cannot determine the value of p or x.
Hence, we cannot determine the revenue from the erasers. – Insufficient The correct answer is Option E.
352.
Total amount of inventory, in dollars, at the end = Total amount of inventory at the beginning + Total amount of purchases in the month −Total amount of sales in the month From statement 1: The seller effectively had purchased 300 copies of Magazine X at $4 per magazine and 100 copies of Magazine X at $3.75 per magazine.
However, the number of copies of Magazine X sold and their price are not known.
Hence, total amount of inventory, in dollars, of stock by the seller at the end of last month cannot be determined – Insufficient From statement 2: The total revenue from the sale of Magazine X = $800.
However, the number of copies of Magazine X purchased by the seller and its price in the last month is not known.
Hence, total amount of inventory, in dollars, of stock by the seller at the end of last month cannot be determined – Insufficient Thus, from statements 1 and 2 together: Total amount of inventory, in dollars, at the end = 300 × 4 + 100 × 3.75 −800 = 1, 200 + 375 −800 = $775 – Sufficient The correct answer is Option C.
353.
We know that C ∝N.
=> C = kN; where k is a constant From statement 1: We have C = kN, © 1999–2024 Manhattan Review www.manhattanreview.com 346 Quantitative Reasoning Question Bank – DS Solutions N = 100.
However, since k is unknown, we cannot determine the value of C. – Insufficient From statement 2: We have C = kN, N = 450 C = 90 Thus, k = C N = 90 450 = 1 5.
Thus, we have C = N 5 However, no information is given on the value of N.
Hence, the value of C cannot be determined. – Insufficient From statements 1 and 2 together: We know: C = N 5 , N = 100 Thus, we have C = 100 5 = 20 – Sufficient The correct answer is Option C.
354.
From statement 1: We have no information about the distance travelled in one gallon of diesel.
Hence, the cost of diesel per mile cannot be determined – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 347 From statement 2: We have no information about the cost of diesel.
Hence, the cost of diesel per mile cannot be determined. – Insufficient Thus, from statements 1 and 2 together: Since we need to find the cost of diesel per mile, we need to know the total cost of diesel and the total number of miles travelled.
The cost per gallon of diesel is known.
Since the number of gallons of diesel consumed is not known, we cannot determine the total cost of diesel.
Hence, the cost of diesel per mile cannot be determined – Insufficient The correct answer is Option E.
355.
Total number of visitors = 950.
We also know that twice as many visitors chose Monday than Tuesday.
From statement 1: We know that the maximum number of visitors present on any weekday = 150.
Say on each of the weekdays except Tuesday and Sunday, maximum number of visitors, 150 each, visited the pagoda. Thus, on Tuesday 150/2 = 75 visitors chose to go to the pagoda.
The maximum number of visitors on weekdays except Sunday = 5 × 150 + 75 = 750 + 75 = 825.
Thus, the least number of visitors on Sunday = 950 −825 = 125 > 100. – Sufficient From statement 2: We know that on each of days, Tuesday to Saturday had at least 75 visitors, thus Monday had at least 2 × 75 = 150 visitors.
Thus, the minimum number of visitors from Monday to Saturday = 150 + 5 × 75 = 150 + 375 = 525.
Thus, the maximum number of visitors on Sunday = 950 −525 = 425.
However, this does not help in determining the minimum number of visitors on Sunday.
– Insufficient The correct answer is Option A.
© 1999–2024 Manhattan Review www.manhattanreview.com 348 Quantitative Reasoning Question Bank – DS Solutions 356.
From statement 1: We have e + i = 6 Since none of the numbers are more than ‘3’, the above equation is valid only if e = i = 3.
We also know that each of the numbers 1, 2, and 3 appear exactly once in each row and column.
Thus, in the first row: b ̸= 3 (as in column 2 => e = 3 is already present) and c ̸= 3 (as in column 3 => i = 3 is already present).
Since in the first row there must be at one ‘3’; thus, a = 3 – Sufficient From statement 2: Since each of the numbers 1, 2, and 3 appear exactly once in each row and column.
Thus, the sum of each row and each column = 1 + 2 + 3 = 6.
Hence, we have a + b + c = 6; (considering the first row) a + d + g = 6; (considering the first column) Adding the above two equations, we have 2a + b + c + d + g = 12.
However, we know: b + c + d + g = 6 Hence, 2a = 12 − b + c + d + g = 12 −6 = 6 => a = 3 – Sufficient The correct answer is Option D.
357.
From statement 1: t △2 = 74 => (t + 2)2 + (2 + 3)2 = 74 => (t + 2)2 = 49 => t + 2 = ±7 => t = 5 or −9.
Hence, we do not have a unique value of t. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 349 From statement 2: 2 △t = 80 => (2 + 2)2 + (t + 3)2 = 80 => (t + 3)2 = 64 => t + 3 = ±8 => t = 5 or −11.
Hence, we do not have a unique value of t. – Insufficient Thus, from statements 1 and 2 together: We find that t = 5 is common to both statements 1 and 2.
Hence, t = 5. – Sufficient The correct answer is Option C.
358.
Total deduction = X × Y 100 + Z = XY 100 + Z.
Thus, dealer’s gross profit = X −XY 100 −Z = X −Z −XY 100.
From statement 1: X −Z = 400 However, we do not have the value of X or Y. – Insufficient From statement 2: XY = 11, 000 However, we do not have the value of XY 100. – Insufficient Thus, from statements 1 and 2 together: Dealer’s gross profit = (X −Z) −XY 100 = 400 −11, 000 100 = 400 −110 = $290 – Sufficient The correct answer is Option C.
359.
Let the monthly wheat allotment be x tons.
Loss in wheat allotment per month = 5x 100 tons.
© 1999–2024 Manhattan Review www.manhattanreview.com 350 Quantitative Reasoning Question Bank – DS Solutions Let the cost to the company for every gallon lost be $y.
Thus, we need to determine: The dollar cost to the company per month for the loss = $ 5xy 100 From statement 1: We know x = 400 × 106 tons.
However, the value of y is unknown.
Hence, the dollar cost to the company per month for the loss cannot be determined. – Insuffi-cient From statement 2: The statement gives us the cost to the company for every ton lost as y = $ 5 10, 000 = $ 1 2, 000 However, the value of x is unknown.
Hence, the dollar cost to the company per month for the loss cannot be determined. – Insuffi-cient Thus, from statements 1 and 2 together: x = 400 × 106 y = $ 1 2, 000 Thus, the dollar cost to the company per month for the loss: = $ 5×(400×106)× 1 2,000 100 !
= $10, 000. – Sufficient The correct answer is Option C.
360.
We have to determine the value of (s + d).
From statement 1: d = s + 100 Since the actual values of s or d is not known, we cannot determine the total amount spent per month i.e. (s + d). – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 351 From statement 2: Amount spent by Suzy in seven months = $7s.
Amount spent by Dave in six months = $6d.
Thus, we have 7s = 6d Since the actual values of s or d is not known, we cannot determine the total amount spent per month i.e. (s + d). – Insufficient Thus, from statements 1 and 2 together: We have 7s = 6d => d = 7s 6 Substituting the value of d from statement 1, in the equation, we have 7s 6 = s + 100 => s = 600 => d = 7 × 600 6 = 700 Thus, the total amount spent per month = $(s + d) = $1, 300. – Sufficient The correct answer is Option C.
361.
From statement 1: Let the regular price of the cake Martin bought be $x per one-pound piece.
Thus, the price for the second one-pound piece he paid = $ 3x 4 .
Thus, Martin’s savings = $ x 4 .
Total regular price (without the discounted offer) = $2x.
Thus, the percent of the total regular price saved: => x 4 2x × 100 = 12.5%. – Sufficient From statement 2: © 1999–2024 Manhattan Review www.manhattanreview.com 352 Quantitative Reasoning Question Bank – DS Solutions The offered discounted rate is not mentioned.
Hence, the percent of the total regular price saved cannot be determined. – Insufficient The correct answer is Option A.
362.
We know that ‘#’ represents either addition, subtraction, multiplication or division.
From statement 1: 25 # 5 = 5 Checking one at a time: • 25 + 5 = 30 ̸= 5 • 25 −5 = 20 ̸= 5 • 25 × 5 = 125 ̸= 5 • 25 ÷ 5 = 5 Thus, we can say that ‘#’ represents division.
=> 14 # 7 = 14 ÷ 7 = 2 – Sufficient From statement 2: 2 # 1 = 2 Checking one at a time: • 2 + 1 = 3 ̸= 2 • 2 −1 = 1 ̸= 2 • 2 × 1 = 2 • 2 ÷ 1 = 2 Thus, we can say that ‘#’ represents either multiplication or division.
=> 14 # 7 = 14 × 7 = 98 OR => 14 # 7 = 14 ÷ 7 = 2 Thus, there is no unique answer. – Insufficient The correct answer is Option A.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 353 363.
Number of shares of stock P with Steve = x.
Dividend earned on the above x shares = $225.
Thus, dividend earned per share = $ 225 x .
From statement 1: We have We have the annual dividend on each share of stock P was $1.25 => David’s total dividend on his 200 shares = 200 × 1.25 = $250. – Sufficient From statement 2: We have x = 180 We know that dividend earned per share = 225 x Thus, dividend earned per share = 225 x = 225 180 = $1.25 This is the same information as obtained from statement 1. – Sufficient The correct answer is Option D.
364.
Let the upper limit for the population be p.
Let the upper limit for the total area be a.
From statement 1: There is no information about the total area. – Insufficient From statement 2: There is no information about the population. – Insufficient Thus, from statements 1 and 2 together: We have p = 50, 000, 000 © 1999–2024 Manhattan Review www.manhattanreview.com 354 Quantitative Reasoning Question Bank – DS Solutions i = 90, 000 square kilometers Thus, based on the upper estimates, the Population Density = 50, 000, 000 90, 000 = 555.55 persons per square kilometers =≈555 > 500 Though it seems that we got the unique answer; however, it is not so. Since we do not have any information on the lower estimates, we cannot be sure that the Population Density for the country must be greater than 500 persons per square kilometers.
For example: Let population = 40, 000, 000 & area = 80, 000 square kilometers Thus, based on the above estimates, the Population Density = 40, 000, 000 80, 000 = 500 persons per square kilometers = 500 ≯500 Thus, there is no unique answer. – Insufficient The correct answer is Option E.
365.
Since ■, △and ∀represent positive digits, their values can only be from 1 to 9 (inclusive).
From statement 1: Since 3 < ∀< 5, it means that ∀= 4; there is only one possible value of ■and △so that ■< △: ■= 1, & △= 3. – Sufficient From statement 2: Since ■< 2, it means that ■= 1, the possible values of △are: 2, 3, 4, 5, 6, 7 or 8; since in each case, the addition does not lead to a carry.
Thus, the value of △cannot be uniquely determined. – Insufficient The correct answer is Option A.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 355 6.10 Interest 366.
From statement 1: We only have information about the ratio of the rates of interest.
However, we have no information on the amounts invested at the given rates x% and y%.
Thus, we cannot determine the value of x. – Insufficient.
From statement 2: The amount invested at x% = $ 5 5 + 3 × 80, 000 = $50, 000.
The amount invested at y% = $ 3 5 + 3 × 80, 000 = $30, 000.
However, we have no information on the rates of interest.
Thus, we cannot determine the value of x. – Insufficient Thus, from statements 1 and 2 together: We have x = 5 4y => y = 4x 5 Thus, total interest: 50, 000 × x% + 30, 000 × y% = 7, 400 50, 000 × x 100 + 30, 000 × y 100 = 7, 400 => 500x + 300 4x 5 = 7, 400 => x = 7, 400 740 = 10% – Sufficient The correct answer is Option C.
367.
Let John had lent $x at 10% and $y at 22%.
From statement 1: x + y = 2, 400 However, there is no information about the interest received on each amount. – Insufficient From statement 2: © 1999–2024 Manhattan Review www.manhattanreview.com 356 Quantitative Reasoning Question Bank – DS Solutions Since the average rate of interest obtained was 15%, we have (x × 10) + y × 22 x + y = 15 => 10x + 22y = 15x + 15y => 5x = 7y => x y = 7 5 Thus, the larger part is $x and the smaller part is $y.
Thus, the larger part was lent at 10% rate of interest. – Sufficient The correct answer is Option B.
368.
From statement 1: Given: A(2) = 121 => 100 1 + r 100 2 = 121 => 1 + r 100 2 = 121 100 => 1 + r 100 = 121 100 1/2 => 1 + r 100 = 11 10 => r 100 = 0.1 => r = 10%.
Amount after three years = A(3) = 100 1 + 10 100 3 – Sufficient.
From statement 2: r = 10% It is the same information as obtained from statement 1 – Sufficient.
The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 357 6.11 Functions 369.
From statement 1: f (2) = p2 = 81 => p = ±9 Thus, f (1) = p1 = p = ±9.
Hence, we do not have a unique value of f (1). – Insufficient From statement 2: f (3) = p3 = −729 => p = −9 Thus: f (1) = p1 = a = −9. – Sufficient The correct answer is Option B.
370.
From statement 1: h (k) = 2k −1 = 7 Thus: 2k −1 = 7 => k = 4.
Thus, we have g (k = 4) = 2k −3 5 = 8 −3 5 = 1. – Sufficient From statement 2: h (1) = 2 × 1 −1 = 1.
=> 1 = k 4 k = 4 This is the same as the information as obtained from the first statement. – Sufficient The correct answer is Option D.
371.
From statement 1: |k| = 3 => k = ±3 If k = 3 ≥0: © 1999–2024 Manhattan Review www.manhattanreview.com 358 Quantitative Reasoning Question Bank – DS Solutions f (k) = f (3) = 27 × 3 = 81; (since k ≥0, f (k) = 27k) If k = −3 < 0: f (k) = f (−3) = (−3)4 = 81; (since k < 0, f (k) = k4) Thus, we have a unique value of f (k) = 81. – Sufficient From statement 2: k < 0 => f (k) = k4 However, the value of k is not known.
Hence, the value of f (k) cannot be determined. – Insufficient The correct answer is Option A.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 359 6.12 Permutation & Combination 372.
We are given that total number of balls = 30.
Say the number of green balls is n From statement 1: We know that Cn 2 C30 2 = 3 29 => n × (n −1) 1 × 2 30 × 29 1 × 2 = 3 29 => n × (n −1) = 30 × 3 = 10 × 9 Since n × (n −1) is a product of two consecutive integers, which are 10 & 9, thus n = 10 – Sufficient.
From statement 2: We know that C(30−n) 2 C30 2 = 38 87 => (30 −n) × (29 −n) 1 × 2 30 × 29 1 × 2 = 38 87 => (30 −n) × (29 −n) = 38 × 10 = 20 × 19 Since (30 −n) × (29 −n) is a product of two consecutive integers, which are 20 & 19, thus (30 −n) = 20 => n = 10 – Sufficient.
The correct answer is Option D.
373.
Probability that the token chosen is green = g b + w + g .
Probability that the token chosen is white = w b + w + g .
From statement 1: g(b + g) > w(b + w) © 1999–2024 Manhattan Review www.manhattanreview.com 360 Quantitative Reasoning Question Bank – DS Solutions => g b + w > w b + g => 1 + g b + w > 1 + w b + g (adding 1 to both sides) => b + w + g b + w > b + g + w b + rg => 1 b + w > 1 b + g ; (canceling (b + w + g) from both sides) => b + g > b + w => g > w.
Since the number of green tokens is greater than the number of white tokens, the probability that the token chosen will be green is greater than the probability that the token chosen will be white. – Sufficient From statement 2: => b > w + g We see that the number of black tokens is the greatest.
However, we have no information on whether g > w. – Insufficient The correct answer is Option A.
374.
From statement 1: There are 6 females among which, 3 are are pursuing Ph. D.
Thus, required probability = 3 19 – Sufficient From statement 2: Since 3 of the 6 females are not are pursuing Ph. D., number of females who are are pursuing Ph. D. = 6 −3 = 3.
Thus, required probability = 3 19 – Sufficient The correct answer is Option D.
375.
From statement 1: We have no information on the total number of tokens or on the number of black and green tokens. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 361 From statement 2: The probability that the token will be blue = 1 2 Thus, the probability that the token will be red or green = 1 −1 2 = 1 2 Since we do not know the probability of the token being red, we cannot determine the probabil-ity of the token being green. – Insufficient Thus, from statements 1 and 2 together: Since we do not know the total number of tokens, we cannot determine the probability of the token being red. – Insufficient The correct answer is Option E.
376.
Let the number of men be m.
Thus, we have p = Cm 2 C10 2 We need to determine whether p > 0.5 = 1 2.
From statement 1: Number of men, m > 10 2 = 5.
If m = 6 : p = Cm 2 C10 2 = C6 2 C10 2 = 15 45 = 1 3 ≯1 2.
Thus, if m ≥6 : 1 ≥p ≥1 3.
Thus, the value of p can be less than 1 2 or even be more than 1 2 (for example, if m = 10, then p = 1).
Thus, the answer cannot be uniquely determined. – Insufficient From statement 2: Number of women = (10 −m).
© 1999–2024 Manhattan Review www.manhattanreview.com 362 Quantitative Reasoning Question Bank – DS Solutions We know that the probability that both are women is less than 1 10.
Thus, we have C(10−m) 2 C10 2 < 1 10.
Rather than solving, it is best to put a few test values of m.
Since the probability 1 10 is very small, possible values of m would be on the higher side, but less than or equal to ‘8’ (since there are at least two men).
• m = 8 : C(10−m) 2 C10 2 = C2 2 C10 2 = 1 45 < 1 10; (satisfies) • m = 7 : C(10−m) 2 C10 2 = C3 2 C10 2 = 3 45 = 1 15 < 1 10; (satisfies) • m = 6 : C(10−m) 2 C10 2 = C4 2 C10 2 = 6 45 = 2 15 ≮1 10; (does not satisfy) Thus, possible values of m are 7 or 8.
Let us now calculate the probability that both are men i.e. p.
At m = 7 : p = Cm 2 C10 2 = C7 2 C10 2 = 21 45 ≯1 2 At m = 8 : p = Cm 2 C10 2 = C8 2 C10 2 = 28 45 > 1 2 Thus, the answer cannot be uniquely determined. – Insufficient Thus, statements 1 and 2 together: Even after combining both statements, p may be more than 1 2 or less than 1 2. – Insufficient The correct answer is Option E.
377.
Let the number of males and the number of females be m and w, respectively.
We need to determine the value of (m + w).
From statement 1: The probability of selecting a male = m m + w Thus, we have www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 363 m m + w = 4 7 => m = 4w 3 Since the exact values of m and w are not known, we cannot determine the value of (m + w). – Insufficient From statement 2: m = w + 10 Since the exact values of m and w are not known, we cannot determine the value of (m + w). – Insufficient Thus, from statements 1 and 2 together: Substituting the value of m = 4w 3 in the equation from statement 2, we have 4w 3 = w + 10 => w = 30 => m = w + 10 = 40 => m + w = 70. – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 364 Quantitative Reasoning Question Bank – DS Solutions 6.13 Sets 378.
From statement 1: It is known that each of the participants ordered for exactly one drink.
We also know that 70 percent of the female participants ordered Tea.
Thus, the remaining (100 −70)% = 30% of the female participants ordered Coffee. – Sufficient From statement 2: There is no information about the female participants. – Insufficient The correct answer is Option A.
379.
Let us draw the corresponding Venn-diagram: Coffee 45% Tea 35% Total = 100% 𝑥% 45 −𝑥% 35 −𝑥% 𝑛% We need to determine the value of n.
From statement 1: We have (45 −x) % = 25% => x = 20% Thus, the percent of employees who take coffee or tea = (45 + 35 −x) % = (45 + 35 −20) % = 60% Thus, the percent of employees who take neither tea nor coffee (n%) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 365 = (100 −60) % = 40% – Sufficient From statement 2: Percent of employees who take tea = 35%.
Thus, percent of employees who tea as well as coffee (x%) = 400 7 % of 35% = 4 7 × 35% = 20% This is the same information as obtained from statement 1. – Sufficient The correct answer is Option D.
380.
Let us refer to the Venn-diagram shown below: X (𝑎+ 𝑐) Y (𝑏+ 𝑐) Both X and Y 𝑎 𝑏 𝑐 𝑛 From statement 1: c = 25% of (a + c) => c = a + c 4 => 4c = a + c => a = 3c . . . (i) However, there is no information on the value of b. – Insufficient From statement 2: c = 37.5% of (b + c) => c = 375 1000 (b + c) © 1999–2024 Manhattan Review www.manhattanreview.com 366 Quantitative Reasoning Question Bank – DS Solutions => c = 3 8 (b + c) => b = 5c 3 =≈1.67c . . . (ii) However, there is no information on the value of a. – Insufficient Thus, from statements 1 and 2 together: From (i) and (ii): a > b => a + c > b + c => The number of clients of Company X is greater than that of Company Y. – Sufficient Alternately, we have 25% of X = 37.5% of Y =>X Y = 37.5 25 > 1 => X > Y => The number of clients of Company X is greater than that of Company Y. – Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 367 6.14 Statistics & Data Interpretation 381.
Standard deviation (SD) is a measure of deviation of items in a set w.r.t. their arithmetic mean (average). Closer are the items to the mean value, lesser is the value of SD, and vice versa; thus, it follows that if a set has all equal items, its SD = 0.
From statement 1: Statement 1 is clearly insufficient as we do not know how many numbers of students are there in each class; merely knowing the mean value is insufficient.
From statement 2: Statement 2 is clearly sufficient. As discussed above since each class has an equal number of students, their mean = number of students in each class, so SD = 0: no deviation at all!
The correct answer is Option B.
382.
The median weight of 45 mangoes would be the weight of the 45 + 1 2 th = 23rd mango once the mangoes have been arranged in increasing order of weight (the mangoes may be arranged in decreasing order of their weight as well).
Since each of the 23 mangoes in box X weighs less than each of the 22 mangoes in box Y, the median weight will be the weight of the heaviest mango, i.e., 23rd mango, in box X.
From statement 1: The heaviest mango in box X weighs 100 grams.
Hence, the median weight of 45 mangoes = 100 grams. – Sufficient From statement 2: The lightest mango in box Y weighs 120 grams.
However, we need information on the heaviest mango in box X. – Insufficient The correct answer is Option A.
383.
From statement 1: Standard deviation (SD) is a measure of deviation of items in a set with respect to their arithmetic mean (average). Closer are the items to the mean value, lesser is the value of SD, and vice versa.
Thus, it follows that if a set has all equal items, its SD = 0.
© 1999–2024 Manhattan Review www.manhattanreview.com 368 Quantitative Reasoning Question Bank – DS Solutions Since the SD of the amounts of the 10 prizes is ‘0’, the amount for each prize must be same. – Sufficient From statement 2: The total amount of the 10 prizes will not help us to determine whether the amount for each prize was the same. – Insufficient The correct answer is Option A.
384.
Let the seven numbers be a, b, c, d, e, f and g (where a > b > c > d > e > f > g).
Thus, we have a + b + c + d + e + f + g 7 = 20 => a + b + c + d + e + f + g = 140 We need to determine the median of the seven numbers.
It is clear that the median is one of the seven numbers.
Since we have assumed a > b > c > d > e > f > g, the median must be d.
From statement 1: Thus, we have d = 1 6 a + b + c + e + f + g => d = 1 6 a + b + c + d + e + f + g −d => d = 1 6 (140 −d) => 6d = 140 −d => d = 20 Thus, the median is 20. – Sufficient From statement 2: It is clear that the median is one of the seven numbers.
Since we have assumed a > b > c > d > e > f > g, the median must be d.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 369 Thus, we have a + b + c + e + f + g = 120 However, we have a + b + c + d + e + f + g = 140 => d = 140 −120 = 20 Thus, the median is 20. – Sufficient The correct answer is Option D.
385.
Let the four numbers be a, b, c and d.
Thus, we have a + b + c + d 4 = 40 => a + b + c + d = 160 From statement 1: We know that no number is greater than 70.
However, it may be that only one number is greater than 40, three numbers are greater than 40 or three numbers are greater than 40 as shown below: • 37, 38, 39, 46: Average is 40; only one number greater than 40 • 38, 39, 41, 42: Average is 40; two numbers greater than 40 • 34, 41, 42, 43: Average is 40; three numbers greater than 40 Hence, we cannot determine the answer. – Insufficient From statement 2: We know that two of the numbers are 19 and 20.
Without loss of generality, we can assume that: a = 19 b = 20 a + b + c + d = 160 © 1999–2024 Manhattan Review www.manhattanreview.com 370 Quantitative Reasoning Question Bank – DS Solutions => 19 + 20 + c + d = 160 => c + d = 121 Thus, it may be that: Both c and d are more than 40, for example: a = 60 & b = 61 OR Only one among them is more than 40, for example: a = 90 & b = 31 Hence, we cannot determine the answer. – Insufficient Thus, from statements 1 and 2 together: We know that no number is greater than 70.
Also, we have c + d = 121.
Thus, maximum value of either c or d is 70.
Hence, the value of the other number = 121 −70 = 51.
Thus, we see that two numbers are greater than 40. – Sufficient The correct answer is Option C.
386.
The average of the scores of x students = 40.
Thus, the total score of x students = 40x.
The average of the scores of y students = 30.
Thus, the total score of y students = 30y.
Thus, the total score of (x + y) students = (40x + 30y).
Thus, the average of the scores of (x + y) students = 40x + 30y x + y !
.
From statement 1: We only know the value of (x + y) but not the relation between x and y.
Hence, the average of the scores of (x + y) students cannot be determined. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 371 From statement 2: We have x = 3y Thus, the average of the scores of (x + y) students = 40x + 30y x + y !
= 40 × 3y + 30y 3y + y !
= 150y 4y = 37.5 – Sufficient The correct answer is Option B.
387.
Standard deviation (SD) is a measure of deviation of items in a set with respect to their arithmetic mean (average). Closer are the items to the mean value, lesser is the value of SD, and vice versa; this follows that if a set has all equal items, its SD = 0.
From statement 1: We know that the average score of Class A’s students is greater than the average score of Class B’s students.
However, we have no information about the deviations of the scores of the students about the mean.
Hence, we cannot compare the standard deviations. – Insufficient From statement 2: We know that the median score of Class A’s students is greater than the median score of Class B’s students.
However, we have no information about the deviations of the scores of the students about the mean.
Hence, we cannot compare the standard deviations. – Insufficient Thus, from statements 1 and 2 together: Even after combining the two statements we cannot determine the deviation of the scores of the students about the mean. – Insufficient The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 372 Quantitative Reasoning Question Bank – DS Solutions 6.15 Linear Equations 388.
Let the price of a A-4 size notebook be $x and that of a A-5 size notebook be $y.
We need to find the value of y.
From statement 1: x + y = 4 – Insufficient From statement 2: 3x + y = 9 – Insufficient Thus, from statements 1 and 2 together: We can solve for y: (3x + y −9) −(x + y −4) = 0 => x = 2.50 => y = 4 −2.50 = $1.50 – Sufficient The correct answer is Option C.
389.
We need to find the value of r.
From statement 1: r + 12m = 620 – Insufficient From statement 2: r + 24m = 1, 220 – Insufficient Thus, from statements 1 and 2 together: (r + 24m) −(r + 12m) = 1, 220 −620 => 12m = 600 => m = $50 => r = $20 – Sufficient The correct answer is Option C.
390.
Let the price of the pencil be $x and the price of the eraser be $y.
Thus, we have x + y = 2.
From statement 1: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 373 x = 3y.
Thus: x + y = 2 => 3y + y = 2 => y = $0.50 – Sufficient From statement 2: x = 1.50.
Thus: y = 2 −x = 2 −1.50 => y = $0.50 – Sufficient The correct answer is Option D.
391.
Let the charge of a group membership be $g.
The charge of an individual membership = $200.
Let the number of group and number of individual memberships be x and y, respectively.
Thus, revenue from group and revenue from individual memberships are $xg and $200y, respectively.
Thus: xg + 200y = 240, 000.
We need to determine the value of g.
From statement 1: 200y = 1 3(240, 000) => y = 400 We have no information on x.
Hence, we cannot determine g. – Insufficient From statement 2: x = 2y We have no information about the value of x and y.
Hence, we cannot determine g. – Insufficient © 1999–2024 Manhattan Review www.manhattanreview.com 374 Quantitative Reasoning Question Bank – DS Solutions From statements 1 and 2 together: xg + 200y = 240, 000 => 2yg + 200y = 240, 000; (Substituting x = 2y from Statement 2) => 2 × 400g + 200 × 400 = 240, 000 (Substituting y = 400 from Statement 1) => g = 200 – Sufficient The correct answer is Option C.
392.
Let the price of each cap and each sunglass be $x and $y, respectively.
We need to find the value of: (4x + 5y) From statement 1: x = y + 2 We do not know the actual values of y and x.
Hence, we cannot determine the answer. – Insufficient From statement 2: 8x + 10y = 45 => 2(4x + 5y) = 45 => 4x + 5y = 45 2 = $22.50 – Sufficient The correct answer is Option B.
393.
From statement 1: Let the number of smaller cartons (holding 50 bottles each) required be x.
Number of smaller cartons is 10 more than the standard size cartons.
Thus, the number of standard size cartons (holding 75 bottles each) = (x −10).
Thus, equating the total number of cartons: 75(x −10) = 50x => 25x = 750 => x = 30 – Sufficient From statement 2: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 375 Number of standard size cartons (holding 75 bottles each) = 20.
Thus, total number of bottles = 20 × 75 = 1,500.
Each smaller carton can hold 50 bottles.
Thus, the number of smaller cartons required = x = 1, 500 50 = 30 – Sufficient The correct answer is Option D.
394.
From statement 1: We have no information on n – Insufficient From statement 2: 6m = 9n => 2m = 3n; (canceling 3 from both the sides) => 2m −3n = 0 – Sufficient The correct answer is Option B.
395.
Let the customer consumes t units.
If t ≤200 : customer’s bill = $xt If t > 200 : customer’s bill = $ 200x + (t −200) y From statement 1: y = 1.25x Since y is unknown, we cannot determine the value of x. – Insufficient From statement 2: Customer’s bill for 210 units = $ 200x + (210 −200) y = $425 => 200x + 10y = 425...(i) There are two unknowns, hence we cannot determine the value of x. – Insufficient Thus, from statements 1 and 2 together: Substituting y = 1.25x in equation (i): 200x + 10 × 1.25x = 425 © 1999–2024 Manhattan Review www.manhattanreview.com 376 Quantitative Reasoning Question Bank – DS Solutions => 212.5x = 425 => x = 425 212.5 = $2 The customer who consumes 200 units in a month would be charged 2×200 = $400. – Sufficient The correct answer is Option C.
396.
Let the number of additional dishes be p.
Thus, total cost for all (p + 1) dishes = $(50 + xp).
We need to calculate the value of x.
From statement 1: Total cost for 4 dishes (i.e. 3 additional dishes) = $(50 + 3x).
Thus, average cost for 4 dishes = $ 50 + 3x 4 .
Thus, we have 50 + 3x 4 = 27.50 => x = $20. – Sufficient From statement 2: Total cost for 4 dishes (i.e. 3 additional dishes) = $(50 + 3x) Thus, average cost for 4 dishes = $ 50 + 3x 4 Similarly, average cost for 6 dishes = $ 50 + 5x 6 Thus, we have 50 + 3x 4 − 50 + 5x 6 = 2.50 => 50 −x 12 = 2.50 => x = $20. – Sufficient The correct answer is Option D.
397.
Total items sold = 200 Let the number of pens sold = x.
Thus, the number of pencils sold = (200 −x) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 377 From statement 1: Revenue from item sales = $ (1.5x + 0.50 (200 −x)) Thus, we have 1.5x + 0.5(200 −x) = 150 => x = 50 – Sufficient From statement 2: Revenue from item sales = $ (1.5x + 0.5 (200 −x)) Thus, average price per item sold = $ 1.5x + 0.5 (200 −x) 200 Thus, we have 1.5x + 0.5 (200 −x) 200 = 0.75 => 1.5x + 0.5 (200 −x) = 150.
This equation is same as that in Statement 1. – Sufficient The correct answer is Option D.
398.
Let the number of hours worked in a week be w.
If w ≤t: Jack’s earnings = $ xw.
If w > t: Jack’s earnings = $ (xt + 2 (w −t)).
From statement 1: We have w = (t −3); i.e. w < t.
Jack’s earnings = $ (x (t −3)).
Thus, we have x (t −3) = 14 Since x and t are integers, possible values of x and t are: (1) x = 14; t −3 = 1 => t = 4: not possible since t > 4 (2) x = 7; t −3 = 2 => t = 5: possible (3) x = 2; t −3 = 7 => t = 10: possible (4) x = 1; t −3 = 14 => t = 17: possible © 1999–2024 Manhattan Review www.manhattanreview.com 378 Quantitative Reasoning Question Bank – DS Solutions There are multiple possible values of t possible, hence we cannot determine the value of t. – Insufficient From statement 2: We have w = (t + 3); i.e. w > t.
Jack’s earnings = $ (xt + 2 ((t + 3) −t)) = $ (xt + 6).
Thus, we have xt + 6 = 23 => xt = 17.
Since x and t are integers, possible values of x and t are: (1) x = 17; t = 1: not possible since t > 4 (2) x = 1; t = 17: possible Thus, we have a unique value of t. – Sufficient The correct answer is Option B.
399.
We know that the number of students in 1991 was one-third of the number of students in 2000.
We need to determine the number of students in 1991.
From statement 1: There is no information regarding the number of students in 1991. – Insufficient From statement 2: There is no information regarding the actual number of students in 2009. – Insufficient Thus, from statements 1 and 2 together: Let the number of students in 1991 be x.
Thus, the number of students in 2000 = 3x.
Hence, the number of students in 2009 = (2 × 3x) = 6x.
Thus, increase in number of students from 2000 to 2009 = 6x −3x = 3x.
Thus, we have 3x = 120 => x = 40. – Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 379 400.
Let the number of marbles with Kevin be n.
From statement 1: n −10 = n 2 => n = 20. – Sufficient From statement 2: This statement gives us information on the ratio of types of marbles with Kevin and not the total number of marbles. – Insufficient The correct answer is Option A.
401.
Let the number of years for which Mrs. Peterson lived be x.
Let the number of years for which Mrs. Peterson had been a professor be y.
From statement 1: we have y + 20 = 3x 4 . . . (i) We cannot determine x from this equation since there are two unknowns. – Insufficient From statement 2: We have y −20 = x 4 . . . (ii) We cannot determine x from this equation since there are two unknowns. – Insufficient Thus, from statements 1 and 2 together: Subtracting equations (i) and (ii): 40 = x 2 x = 80. – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 380 Quantitative Reasoning Question Bank – DS Solutions 402.
We have x + y = 2p and x −y = 2q => 2p + 2q = (x + y) + (x −y) = 2x.
p + q = x Now, in order to find (p + q), we need to find x.
From statement 1: We do not have the value of x. – Insufficient From statement 2: p + q = x = 3. – Sufficient The correct answer is Option B.
403.
We have x 6 = y 3 => x = 2y.
From statement 1: x + y = 30 => 2y + y = 30; (substituting x = 2y) => y = 10. – Sufficient From statement 2: 3x = 60 => x = 20 => 2y = 20; (substituting x = 2y) => y = 10. – Sufficient The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 381 404.
From statement 1: Kevin was born in 1990 and is 5 years older than Chris.
Thus, Chris must have been born 5 years after Kevin was born, i.e. in the year 1990 + 5 = 1995 – Sufficient From statement 2: We have no information on Chris. – Insufficient The correct answer is Option A.
© 1999–2024 Manhattan Review www.manhattanreview.com 382 Quantitative Reasoning Question Bank – DS Solutions 6.16 Quadratic Equations & Polynomials 405.
Let the number of boys and girls be b and g respectively.
We need to determine the value of (b −g).
From statement 1: We have b + g = 10.
We cannot determine (b −g) from this information. – Insufficient From statement 2: We have b = g3 We cannot determine (b −g) from this information. – Insufficient Thus, from statements 1 and 2 together: b = g3 and b + g = 10: g3 + g = 10 g(g2 + 1) = 10 The optimum approach to solve this polynomial is by plugging in test values.
Say g = 1; At g = 1, g(g2 + 1) = 1(12 + 1) = 2 ̸= 10 Say g = 2; At g = 2, g(g2 + 1) = 2(22 + 1) = 10 = 10 => g = 2 => b = 23 = 8 => b −g = 8 −2 = 6 – Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 383 406.
(a + b)2 = a2 + b2 + 2ab => (a + b) = ± p (a2 + b2 + 2ab) From statement 1: (a + b) = ± p (a2 + b2) + 2ab => (a + b) = ± √ 1 + 0 = ±1.
Thus, the value of (a + b) may be or may not be ‘1.’ – Insufficient From statement 2: Since b = 0, we have a2 + b2 = 1 => a2 + 0 = 1 => a = ±1 => a + b = ±1 Thus, the value of (a + b) may be or may not be ‘1.’ – Insufficient Thus, from statements 1 and 2 together: Even after combining, we still get the same solution: (a + b) = ±1. – Insufficient The correct answer is Option E.
407.
From statement 1: (1 −x) 1 −y = 1 => 1 −x −y + xy = 1 => x + y = xy => x + y −xy = 0 – Sufficient From statement 2: => x + y x −y = xy x −y We know: x ̸= y i.e. (x −y) ̸= 0 and hence it can be cancelled from both sides.
© 1999–2024 Manhattan Review www.manhattanreview.com 384 Quantitative Reasoning Question Bank – DS Solutions => x + y = xy => x + y −xy = 0 – Sufficient The correct answer is Option D.
408.
a (a −5) (a + 2) = 0 => a = 0, 5 or −2.
From statement 1: a (a −7) ̸= 0 => a ̸= 0 and a ̸= 7.
However, from the question statement, we can still have: a = 5 ≮0 or a= −2 (< 0).
– Insufficient From statement 2: a2 −2a −15 ̸= 0 => (a −5) (a + 3) ̸= 0 => a ̸= 5 and a ̸= −3.
However, from the main question, we can still have: a = 0 ≮0 or a = −2 (< 0). – Insufficient Thus, from statements 1 and 2 together: a ̸= 0 and a ̸= 5 Thus, from the main question, we have a = −2 (< 0). – Sufficient The correct answer is Option C.
409.
1 a + 1 b = a + b ab From statement 1: The value of ab is not known.
Hence, the value of 1 a + 1 b cannot be determined. – Insufficient From statement 2: ab = 6 (a + b) => a + b ab = 1 6 => 1 a + 1 b = 1 6. – Sufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 385 The correct answer is Option B.
410.
a2 −b = n => a2 = n + b.
From statement 1: n + b = 4 Thus: a2 = n + b = 4 => a = ±2 Thus, the value of a is not unique. – Insufficient From statement 2: The value of n is unknown.
Hence, the value of a cannot be determined. – Insufficient Thus, from statements 1 and 2 together: Even after combining both statements, we cannot determine a unique value of a. – Insufficient The correct answer is Option E.
411.
(n + 3) (n −1) −(n −2) (n −1) = m (n −1) => (n + 3) (n −1) −(n −2) (n −1) −m (n −1) = 0 => (n −1) [(n + 3) −(n −2) −m] = 0 => (n −1) (5 −m) = 0 => n = 1 or m = 5 Looking at the results, it seems that the question is sufficient in itself and not even a single statement is needed as it yields n = 1; however it is not so. The meaning of n = 1 or m = 5 is that at least one of these must be true. Thus, if m = 5, then n may be or may not be 1. However, if m ̸= 5, then n must be 1.
Moreover, if m = 5, n may have any value under the sun!
So the question boils down to the either the determination of value of n or the determination whether m ̸= 5.
From statement 1: |m| = 5 => m = ±5 © 1999–2024 Manhattan Review www.manhattanreview.com 386 Quantitative Reasoning Question Bank – DS Solutions If m = 5, n can take any value.
If m = −5 (̸= 5) , then n = 1.
Thus, the unique value of n cannot be determined. – Insufficient.
From statement 2: If m = 5, n can take any value.
Thus, the unique value of n cannot be determined. – Insufficient.
Thus, from statements 1 and 2 together: Even after combining both statements, we cannot determine the unique value of n. – Insufficient The correct answer is Option E.
412.
We have x2 + mx + n = x + p 2 => x2 + mx + n = x2 + 2px + p2 Since this is true for all values of x, we can compare the coefficients of x and the constant terms on either side: m = 2p; n = p2 From statement 1: n = p2 = 32 = 9. – Sufficient From statement 2: m = 2p => 6 = 2p => p = 3 => n = p2 = 9. – Sufficient The correct answer is Option D.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 387 413.
Since the equation x2 + 3x + c = x2 + x (a + b) + ab is valid for all values of x, we have (equating the coefficients of x and constants): a + b = 3 . . . (i) ab = c . . . (ii) From statement 1: We have a = 1 Thus, from (i), we have b = 2 Thus, from (ii), we have c = 1 × 2 = 2 – Sufficient From statement 2: We know that: a + b = 3 Since a and b are positive integers, possible solutions are: a = 1, b = 2 OR a = 2, b = 1 In either of the two cases: c = ab = 1 × 2 = 2 – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 388 Quantitative Reasoning Question Bank – DS Solutions 6.17 Inequalities 414.
Let the average score that Steve got per subject = s, and the average score that David got per subject = d.
We need to determine whether s > d.
From statement 1: 2s > 2d −5 => s > d −5 2 However, we cannot determine whether s > d, since s is greater than a quantity d, which is reduced by a certain amount, 5 2 – Insufficient From statement 2: 2d < 2s + 5 => 2s > 2d −5 It is the same inequality that we got from statement 1. – Insufficient Thus, from statements 1 and 2 together: Even after combining the statements, we do not get any additional information – Insufficient The correct answer is Option E.
415.
Let the larger and the smaller numbers be l and s, respectively.
Thus, we have l 5 > 6s => s < l 30 We need to determine if: s < 5 => l < 150.
From statement 1: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 389 We have l > 120 Say, if l = 150, then s < 150 30 => s < 5.
However, if l = 180, then s < 180 30 => s < 6. Thus, s may be or may not be less than 5. – Insufficient From statement 2: We have l < 150. As discussed in Statement 1, we have s < 5. – Sufficient The correct answer is Option B.
416.
We have xy ̸= 0 => None of x or y is 0.
From statement 1: |x| = |y| => x = ±y Thus, x and y may be or may not be equal. – Insufficient From statement 2: xy > 0 => x > 0 and y > 0 OR x < 0 and y < 0.
However, we cannot determine whether x = y. – Insufficient Thus, from statements 1 and 2 together: From statement 1, we have x = ±y From statement 2, we have x ̸= −y, (since x and y must be of the same sign.) => x = y. – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 390 Quantitative Reasoning Question Bank – DS Solutions 417.
We have abc ̸= 0 => None of a, b and c is ‘0.’ a (b + c) ≥0 => a ≥0 and (b + c) ≥0 OR a ≤0 and (b + c) ≤0.
From statement 1: |b + c| = |b| + |c| => b and c are of the same sign i.e. b ≥0 and c ≥0 OR b ≤0 and c ≤0.
However, there is no information on a. – Insufficient From statement 2: |a + b| = |a| + |b| => a and b are of the same sign i.e. a ≥0 and b ≥0 OR a ≤0 and b ≤0.
However, there is no information on c. – Insufficient Thus, from statements 1 and 2 together: We have two possibilities: (1) a ≥0, b ≥0, c ≥0 => a ≥0 and (b + c) ≥0 (2) a ≤0, b ≤0, c ≤0 => a ≤0 and (b + c) ≤0 In both the cases, we have a (b + c) ≥0. – Sufficient The correct answer is Option C.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 391 418.
We have R = M N Thus, for R ≤M, we have N ≥1 if (R ≥0 and M ≥0); for example, 5 = 10 2 OR N < 0 if (R ≤0 and M ≥0); for example, −2 = 4 −2 From statement 1: M > 40 Since no information on N is provided, we cannot determine if R ≤M. – Insufficient From statement 2: 0 < N ≤15.
Since no information on M is provided, we cannot determine if R ≤M. – Insufficient Thus, from statements 1 and 2 together: We find that both M and N are positive, thus R is also positive.
However, N ≥1 or 0 < N ≤1 is possible.
N ≥1 => R ≤M Whereas, 0 < N ≤1 => R ≥M – Insufficient The correct answer is Option E.
419.
x7y4z3 < 0 => x6y4z2 (xz) < 0 => xz < 0; since x6y4z2 > 0 (being a perfect square).
We need to determine whether xyz < 0.
If xyz < 0, then we have y > 0, (since xz < 0) Thus, we need to determine whether y > 0.
© 1999–2024 Manhattan Review www.manhattanreview.com 392 Quantitative Reasoning Question Bank – DS Solutions From statement 1: There is no information on y. – Insufficient From statement 2: There is no information on y. – Insufficient Thus, from statements 1 and 2 together: There is still no information on y. – Insufficient The correct answer is Option E.
420.
We have xy = 6.
From statement 1: y ≥3.
Since we want to know whether x < y, we must test the inequality for minimum possible value of y against the maximum possible value of x.
The minimum value of y results in the maximum value of x since xy is constant.
Minimum value of y = 3.
Thus, we have the maximum value of x = 6 3 = 2.
Since the maximum value of x is less than the minimum value of y, we have x < y. – Sufficient From statement 2: y ≤3.
We know that the maximum value of y results in the minimum value of x since xy is constant.
Maximum value of y = 3.
Thus, we have the minimum value of x = 6 3 = 2. Thus, for these set of values, we have x < y.
However, x can attain values higher than 2, while y can attain values lower than 3.
For example: if y = 1 => x = 6 1 = 6 => x > y.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 393 Thus, there is no unique answer. – Insufficient The correct answer is Option A.
421.
For the given inequality x < −3y 2 , if y is positive, the answer is Yes, else the answer may be Yes or No.
From statement 1: y > 0 => −3y 2 < 0 Thus, we have x < −3y 2 < 0. – Sufficient From statement 2: We have 2x + 5y = 20 . . . (i) We know that x < −3y 2 => −2x > 3y; the sign of inequality would reverse.
=> −2x −3y > 0 . . . (ii) Adding (i) and (ii), we get: 5y −3y > 20 => y > 10 => y > 0.
This is the same as in statement 1. – Sufficient The correct answer is Option D.
422.
We need to determine whether 1 a + b < 1 If 1 a + b < 1, then a + b > 1 (We know that a > 0 and b > 0, thus, taking reciprocal of the positive quantity (a + b) and reversing the inequality.) From statement 1: a b = 2 However, we cannot determine whether a + b > 1. – Insufficient © 1999–2024 Manhattan Review www.manhattanreview.com 394 Quantitative Reasoning Question Bank – DS Solutions From statement 2: a + b > 1 – Sufficient The correct answer is Option B.
423.
We need to determine whether: w x × y z > y z => w z × y x > y x (rearranging terms) Since w, x, z and z are positive, we can cancel y x from both sides, it implies that: => w z > 1 => w > z From statement 1: We have y > x However, there is no comparison between w and z. – Insufficient From statement 2: We have w > z – Sufficient The correct answer is Option B.
424.
From statement 1: x + y > 60 Since x and y are any integers, we can have a situation where: (1) x = 0 and y = 61 => x + y = 61 > 60. We see that x ≯0.
(2) x = 3 and y = 60 => x + y = 63 > 60. We see that x > 0.
(3) x = −2 and y = 66 => x + y = 64 > 60. We see that x ≯0.
Thus, x may be greater than ‘0’ or equal to ‘0’ or less than ‘0.’ – Insufficient From statement 2: There is no information about x. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 395 Thus, from statements 1 and 2 together: Even after combining the statements, we can still have all the situations shown for statement 1.
– Insufficient The correct answer is Option E.
425.
From statement 1: We have xy < 2, and x > 2 Since the product xy is smaller than 2 with x itself being greater than 2, we must have y as either a fraction between ‘0’ and ‘1’ or a number less than or equal to ‘0.’ For example: (1) Say xy = 1 and x = 3 => y = 1 3 < 1 (2) Say xy = −1 and x = 3 => y = −1 3 < 1 Thus, we have y < 1 – Sufficient From statement 2: We have xy < 2, and y < 3 Since the product xy is smaller than 2 with y itself being smaller than 2, y can take any value depending on the value assigned to x.
For example: (1) If 3 > y > 1 y = 3 2, x = 1 => xy < 2 The answer is no, y ≮1.
© 1999–2024 Manhattan Review www.manhattanreview.com 396 Quantitative Reasoning Question Bank – DS Solutions (2) If y < 1 y = 1 2, x = 2 => xy < 2 The answer is yes, y < 1.
Thus, there is no unique answer. – Insufficient The correct answer is Option A.
426.
We know that x > 0.
There are two possibilities: (1) 0 < x < 1: Here, x is a proper fraction.
Thus, higher the exponent of x, smaller is the value of (since the denominator is greater than the numerator).
x4 < x3 < x2 < x For example: x = 0.2 => x2 = 0.04 < 0.2 = x (2) x > 1: Here, higher the exponent of x, higher is the value (since the numerator is greater than the denominator).
For example: x = 2 => x2 = 4 > 2 = x From statement 1: We have 1 10 < x < 2 5 => 0.1 < x < 0.4 => 0 < x < 1 => x2 < x – Sufficient From statement 2: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 397 We have x3 < x2 Since x2 > 0, we can cancel x2 from both sides: => x < 1 => 0 < x < 1 => x2 < x – Sufficient The correct answer is Option D.
427.
From statement 1: x2 < xy + x => x2 < x y + 1 => x < y + 1 (since x is positive, we can cancel x from both sides without reversing the inequality) However, we cannot determine whether x < y. – Insufficient From statement 2: xy < y(y −1) x < (y −1) (since y is positive, we can cancel y from both sides without reversing the inequality) => x < y −1 (since y > 2 => y −1 > 0, we can cross-multiply (y −1) without reversing the inequality) => x + 1 < y Thus, we can definitely say that: x < y – Sufficient The correct answer is Option B.
428.
We need to determine whether x2 < |x| => |x|2 < |x| since x2 = |x|2 => |x| < 1 © 1999–2024 Manhattan Review www.manhattanreview.com 398 Quantitative Reasoning Question Bank – DS Solutions => −1 < x < 1 From statement 1: We know: x < 1 However, we do not have the lower bound. – Insufficient From statement 2: We know: x > −1 However, we do not have the higher bound. – Insufficient Thus, from statements 1 and 2 together: We have −1 < x < 1; which is the required condition. – Sufficient The correct answer is Option C.
429.
Give: y = |x + 5| + |6 −x| We have to determine whether y = 11.
From statement 1: Let’s take two extreme test values.
Case 1: Say x = 6 => y = |x + 5| + |6 −x| => y = |6 + 5| + |6 −6| = 11. The answer is yes.
Case 2: Say x = −10 => y = |x + 5| + |6 −x| => y = |−10 + 5| + |6 + 10| = | −6| + 16 = 6 + 16 = 22 ≠11. The answer is no.
No unique answer. – Insufficient From statement 2: Let’s take two extreme test values.
Case 1: Say x = −5 => y = |x + 5| + |6 −x| => y = |−5 + 5| + |6 + 5| = 11. The answer is yes.
Case 2: Say x = 10 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 399 => y = |x + 5| + |6 −x| => y = |10 + 5| + |6 −10| = 15 + | −4| = 15 + 4 = 19 ≠11. The answer is no.
No unique answer. – Insufficient Thus, from statements 1 and 2 together: We have −5 ≤x ≤6 We see that at the extreme values of x(= −5 & 6) for −5 ≤x ≤6, the value of y = 11. You may also check some values in this interval, for example: x = 0, −4, 2, etc. In each case you will find that y = 11. – Sufficient The correct answer is Option C.
430.
From statement 1: 2y < 7x => y < 7x 2 Thus, we see that y is smaller than a positive quantity i.e.
7x 2 .
Thus, the value of y may be positive, may be zero or may even be negative. – Insufficient From statement 2: y > −x Thus, we see that y is greater than a negative quantity, −x.
Thus, the value of y may be positive, may be zero or may even be negative. – Insufficient Thus, from statements 1 and 2 together: −x < y < 7x 2 If x = 5 −5 < y < 7 × 5 2 Thus, depending on the value of x, the value of y may be positive, may be zero or may even be negative. – Insufficient The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 400 Quantitative Reasoning Question Bank – DS Solutions 431.
From statement 1: x2 < 1 => −1 < x < 1 Since x is an integer, x = 0 However, we have no information about y. – Insufficient From statement 2: We have y < 1 y may be negative or zero.
However, we have no information about x. – Insufficient Thus, from statements 1 and 2 together: We have x = 0, and y < 1 The greatest value of y would be 0.
Thus, we have x + y < 0 + 0 => x + y < 0 => x + y < 2. The answer to the question is No. – Sufficient The correct answer is Option C.
432.
Since y2 = 9 −x, in order that y2 is a positive integer, we must have: 9 −x > 0 => x < 9 From statement 1: We have x ≤7: www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 401 Possible values of x are: 1, 2, 3, 4, 5, 6 or 7.
The only value of x for which y is an integer is: x = 5 Thus, we have y2 = 9 −x = 9 −5 = 4 => y = 2; since y is a positive integer, it cannot be −2.
Thus, we have a unique value of y. – Sufficient From statement 2: We have y ≥2 We have already established that: x < 9 Thus, possible values of x are: 1, 2, 3, 4, 5, 6, 7 or 8.
The possible values of x for which y is an integer are: x = 5 or 8 If x = 5: y2 = 9 −5 = 4 => y = 2 ≥2 If x = 8: y2 = 9 −8 = 1 => y = 1 ≱2 Thus, there is only possible value of y = 2 – Sufficient The correct answer is Option D.
433.
From statement 1: x −y > 4 => x > y + 4 => 3x > 3y + 12 However, there is no relation between (3y + 12) and 8y since y can be any positive number.
Thus, we cannot conclude that 3x > 8y. – Insufficient From statement 2: x > 14y 5 => x > 2.8y © 1999–2024 Manhattan Review www.manhattanreview.com 402 Quantitative Reasoning Question Bank – DS Solutions => 3x > 8.4y Since y > 0, we have 8.4y > 8y => 3x > 8y – Sufficient The correct answer is Option B.
434.
We have to determine whether 8x > 5y.
Or, x y > 5 8?
From statement 1: x2 > y2 => x y !2 > 1 => x y > 1 or x y < −1 However, x y < −1 is not valid since x and y are positive.
Since x y > 1 > 5 8, the answer is Yes. –Sufficient From statement 2: x3 > y3 => x y !3 > 1 => x y > 1 Since x y > 1 > 5 8, the answer is Yes. –Sufficient The correct answer is Option D.
435.
From statement 1: x2 + 6x < 7 => x2 + 6x + 9 < 16 (in order to convert the quadratic equation to a perfect square) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 403 => (x + 3)2 < 16 => −4 < x + 3 < 4 => −7 < x < 1 But, we know that x < 0 and is an integer.
Thus, possible values of x are: −1, −2, −3, −4, −5 or −6.
Thus, x may be greater than −3 or equal to −3 or even smaller than −3. – Insufficient From statement 2: x2 + |x| ≤2 We know that both x2 and |x| are non-negative.
Also, x2 + |x| must be an integer since x is a negative integer.
Thus, we have (1) x2 + |x| = 2 This is only possible if each of x2 and |x| are 1 => x2 = |x| = 1 => x = −1, which is not less than −3, Or x = 1, which is non-negative and hence not valid.
(2) x2 + |x| = 0 => x = 0, which is not negative, hence not valid.
Thus, we only have x = −1 ≮−3.
Thus, we have a unique answer ‘No.’ – Sufficient The correct answer is Option B.
436.
We have x + y > 0 => x > −y . . . (i) From statement 1: x2y < 1 => x2y < 1 © 1999–2024 Manhattan Review www.manhattanreview.com 404 Quantitative Reasoning Question Bank – DS Solutions We know that x2 ≥0 for all values of x.
The above inequality can be satisfied in any of the following situations: (1) x > 1 and y < 0, for example: x = 2 and y = −1 (satisfying x + y > 0) => x2y = 2−2 = 1 4 < 1 => xy = 2 × (−1) = −2 < 0 (2) 0 < |x| < 1 and y > 1, for example: x = ± 1 2 and y = 2 (satisfying x + y > 0) => x2y = 1 2 4 = 1 16 < 1 => xy = 1 2 × 2 = 1 > 0 OR xy = −1 2 × 2 = −1 < 0 (3) |x| = 0 and y is any number except 0, where x2y = 0 and xy = 0 Thus, xy may be positive, negative or zero. – Insufficient From statement 2: x + 2y < 0 => x < −2y . . . (ii) Thus, combining (i) with (ii), we have −y < x < −2y Thus, we have −y < −2y => y > 2y => y < 0; Note that y cannot be positive since if were so, then y can be cancelled and the result would be 1 > 2, which is invalid.
Thus, y is negative.
Hence, −y and −2y are both positive.
Thus, x lies between two positive numbers (−y < x < −2y), implying x is positive.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 405 Thus, we have x > 0 and y < 0 => xy < 0 – Sufficient The correct answer is Option B.
© 1999–2024 Manhattan Review www.manhattanreview.com 406 Quantitative Reasoning Question Bank – DS Solutions 6.18 Co-ordinate geometry 437.
The equation of a circle in the XY-plane with its centre at the origin is given by: x2 + y2 = r 2, where r is the radius of the circle and (x, y) are the coordinates of any point on the circle.
From statement 1: x2 + y2 = 52 = 25.
Thus, sum of the squares of the coordinates of any point M on the circle is 25. – Sufficient From statement 2: x + y = 7 => (x + y)2 = 49 => x2 + y2 = 49 −2xy.
Since we do not know the values of xy, we cannot determine the answer. – Insufficient The correct answer is Option A.
438.
y = mx + c We need to determine the slope of the above line i.e. the value of m.
From statement 1: We know that: y = (1 −2m) x + 2c and y = mx + c are parallel.
Thus, their slope should be same.
Thus, we have (1 −2m) = m => m = 1 3. – Sufficient From statement 2: We know that: y = mx + c and y = 2x −4 intersect at (3, 2).
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 407 Substituting x = 3 & y = 2 in y = mx + c, we have 2 = 3m + c Since there are two unknowns, we cannot determine the value of m. – Insufficient The correct answer is Option A.
439.
Let the coordinates of A be (a, b) and that of C be (c, d).
A B C Y X (𝑎, 𝑏) (𝑐,𝑏) (𝑐,𝑑) Thus, the coordinates of B = (c, b), since A and B have the same Y-coordinate (AT parallel to X-axis, also, C and B have the same X-coordinate (CB parallel to the Y-axis).
From statement 1: We have d = 2.
However, we cannot determine the values of c or b. – Insufficient From statement 2: We have a = −8.
However, we cannot determine the values of c or b. – Insufficient Thus, from statements 1 and 2 together: We have a = −8, & d = 2.
However, we still cannot determine the values of c or b. – Insufficient The correct answer is Option E.
© 1999–2024 Manhattan Review www.manhattanreview.com 408 Quantitative Reasoning Question Bank – DS Solutions 440.
Y O A B The equation of the above circle has its centre at the origin and point A lies on X-axis with its Y-coordinate being 0, its X-coordinate is the radius = 13.
x2 + y2 = 132 From statement 1: Let the coordinates of point B be (−5, a).
Since B is on the circle, it must satisfy the equation of the circle. Thus: (−5)2 + a2 = 132 => a2 = 169 −25 = 144 => a = ±12 Thus, the length of AB = q (13 −(−5))2 + (0 −a)2 = q (13 −(−5))2 + (0 −(±12))2 = p 182 + 122 = √ 468 – Sufficient From statement 2: Let the coordinates of point B be (b, −12).
Since B is on the circle, it must satisfy the equation of the circle. Thus: b2 + (−12)2 = 132 => b2 = 169 −144 = 25 => b = ±5 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 409 Thus, the length of AB = q (13 −(±5))2 + (0 −(−12))2 = q (13 ∓5)2 + (12)2 = p 82 + 122 OR p 182 + 122 = √ 204 OR √ 468 Thus, there is no unique answer. – Insufficient The correct answer is Option A.
441.
The points (a, b) and (c, d) would be equidistant from the origin (0, 0) if: q (a −0)2 + (b −0)2 = q (c −0)2 + (d −0)2 => a2 + b2 = c2 + d2; (this is the condition we need to verify).
From statement 1: We have no information about c and d. – Insufficient From statement 2: We have c = 1 −a and d = 1 −b => c2 + d2 = (1 −a)2 + (1 −b)2 = 2 + a2 + b2 −2(a + b) We do not know the value of (a + b). – Insufficient Thus, from statements 1 and 2 together: c2 + d2 = 2 + a2 + b2 −2 (a + b) = 2 + a2 + b2 −2 × 2 (since a + b = 2) = a2 + b2 −2 ̸= a2 + b2 – Sufficient The correct answer is Option C.
© 1999–2024 Manhattan Review www.manhattanreview.com 410 Quantitative Reasoning Question Bank – DS Solutions 442. Y X 𝑦= 𝑥=> 𝑦−𝑥= 0 B (𝑝, 𝑞) A (𝑟, 𝑠) C (𝑢, 𝑣) Let us refer to the figure above: (1) For a point A on the line, it must satisfy the equation of the line. Thus, we have y −x = 0 => s −r = 0 => s = r (2) For a point B above the line, we have y −x > 0 (it can be observed that the value of the Y-coordinate is greater than the value of the X-coordinate) => q −p > 0 => q > p (3) For a point C below the line, we have y −x < 0 (it can be observed that the value of the Y-coordinate is smaller than the value of the X-coordinate) => v −u < 0 => v < u Thus, we need to determine which of the above cases is true.
From statement 1: There is no information about b. – Insufficient From statement 2: We have n = m + 4 www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 411 => m −n = 4 => m −n > 0 (i.e. it satisfies y −x > 0) => The point is not below but above the line y = x – Sufficient The correct answer is Option B.
443.
Form statement 1: Slope of the line joining (1, 1) and (−2, 5) = 5 −1 −2 −1 = −4 3 Since the line k is perpendicular to the above line, product of slopes of the above line and the slope of the line k is −1.
Thus, slope of line k = 3 4 Thus, the line k has a positive slope. – Sufficient From statement 2: It is clear that the line must be oriented as shown in the diagram below: Y X Negative X-intercept Positive Y-intercept 𝑘 If a line is increasing towards right hand side, the slope is positive.
If line is parallel to x- axis, the slope is 0.
And, if the line is decreasing towards the right hand side, the slope is negative.
Thus, we see that the line k has a positive slope. – Sufficient The correct answer is Option D.
© 1999–2024 Manhattan Review www.manhattanreview.com 412 Quantitative Reasoning Question Bank – DS Solutions 444.
From statement 1: We have no information about the value of the slope of line n. – Insufficient From statement 2: We have no information about line m. – Insufficient Thus, from statements 1 and 2 together: We can see that line n passes through the points (0, 0) and (−5, 4).
Thus, slope of line n = 4 −0 −5 −0 = −4 5.
Since the product of the slopes of lines m and n is −1, slope of line m = − 1 −4 5 = 5 4. – Sufficient The correct answer is Option C.
445.
If a line intersects the X-axis at (p, 0) and the Y-axis at (0, q), the slope of the line = q −0 0 −p = −q p = − Y intercept X intercept !
Let line K intersect the X-axis and Y-axis at points (m, 0) and (0, n) respectively.
We need to determine the value of n.
From statement 1: Slope of line L = − −1 −1 = −1 Since lines L and K are parallel, slope of line K is −1.
Thus, for line K, we have − n m = −1 => n = m.
We cannot determine the value of n since m is not known. – Insufficient www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 413 From statement 2: We see that line K passes through the point (5, 10) and the points (m, 0) and (0, n) as assumed earlier.
We cannot determine the value of n, since we do not know the slope of line K. – Insufficient Thus, from statements 1 and 2 together: We see that line K passes through the point (5, 10) and the point (0, n) and has a slope of −1.
Thus, we have Slope = 10 −n 5 −0 = −1 => 10 −n = −5 => n = 15 – Sufficient The correct answer is Option C.
446.
The equation of a circle with centre at the origin and radius c is given by: x2 + y2 = c2 Since (a, b) lies on the circle, it must satisfy the equation of the circle.
Thus, we have a2 + b2 = c2 From statement 1: c = 5 => a2 + b2 = c2 = 25 – Sufficient From statement 2: Since the point (3, −4) lies on the circle, it must satisfy the equation of the circle.
Thus, we have (3)2 + (−4)2 = c2 => 9 + 16 = c2 => c2 = 25 © 1999–2024 Manhattan Review www.manhattanreview.com 414 Quantitative Reasoning Question Bank – DS Solutions => a2 + b2 = 25 – Sufficient The correct answer is Option D.
447.
The point (r, s) would lie in region X if it satisfies the condition: 3x + 4y ≤12.
Thus, we have 3r + 4s ≤12 . . . (i) From statement 1: We have 4r + 3s = 12 => (3r + 4s) + (r −s) = 12 => 3r + 4s = 12 + (s −r) . . . (ii) Comparing (i) and (ii), we have 12 + (s −r) ≤12 => s −r ≤0 => s ≤r Thus, the point (r, s) would lie in region X if s ≤r However, such a condition has not been stated in the problem.
Hence, the answer cannot be determined. – Insufficient From statement 2: We have r ≤4 => 3r ≤12 . . . (iii), and s ≤3 => 4s ≤12 . . . (iv) www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 415 Thus, from (iii) and (iv): 3r + 4s ≤12 + 12 => 3r + 4s ≤24 . . . (v) If 3r + 4s ≤12, the answer is yes.
However, if 12 < 3r + 4s ≤24, the answer is no.
Thus, the answer cannot be uniquely determined. – Insufficient Thus, from statements 1 and 2 together: We have from statement 2: r ≤4, and s ≤3 We also have from statement 1: 4r + 3s = 12 We cannot determine the value of (4r + 3s) from the above two relations. – Insufficient The correct answer is Option E.
Alternate approach 1: Let us try with some values: Statement 1: 4r + 3s = 12 : (1) r = 3, s = 0: the inequality 3x + 4y ≡3r + 4s = 9 ≤12 – Satisfies (2) r = 1, s = 8 3: the inequality 3x + 4y ≡3r + 4s = 3 × 1 + 4 × 8 3 = 41 3 = 13.67 ≰12 – Does not satisfy Thus, statement 1 is not sufficient.
Statement 2: r ≤4, s ≤3 : (1) r = 3, s = 0: the inequality 3x + 4y ≡3r + 4s = 9 ≤12 – Satisfies © 1999–2024 Manhattan Review www.manhattanreview.com 416 Quantitative Reasoning Question Bank – DS Solutions (2) r = 1, s = 8 3: the inequality 3x + 4y ≡3r + 4s = 3 × 1 + 4 × 8 3 = 41 3 = 13.67 ≰12 – Does not satisfy Thus, statement 2 is not sufficient.
Combining statements 1 and 2: We can still have the same above values of r and s.
Thus, even combining the statements is not sufficient.
Alternate approach 2: From statement 1: 4r + 3s = 12 => r = (12 −3s) 4 = 3 −3s 4 Thus, 3r + 4s ≤12 = 3 3 −3s 4 + 4s ≤12 => 9 −9s 4 + 4s ≤12 => 5s 4 ≤3 => s ≤12 5 => s ≤2.4 However, Statement 2 states that s ≤3, which is not sufficient to conclude whether 3r + 4s ≤12 (since s can take a value, say, 2.5 which doesn’t satisfy the required condition of s ≤2.4.) 448.
From statement 1: Slope of the line l passing through (0, 0) and (m, n) is: Slope = n −0 m −0 = n m Thus, we have n m < 0 Thus, the possible cases are: (a) m > 0 and n < 0 OR www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 417 (b) m < 0 and n > 0 Thus, there is no unique answer. – Insufficient From statement 2: We know that: m < n However, it is not sufficient to determine the answer, whether n > 0. – Insufficient Thus, from statements 1 and 2 together: From statement 1, we had: (a) m > 0 and n < 0 => m > n – Does not satisfy the condition of statement 2 (b) m < 0 and n > 0 => m < n – Satisfies the condition of statement 2 Thus, combining the statements, we can conclude that: m > 0. – Sufficient The correct answer is Option C.
449.
ax + by + c = 0 => by = −ax −c => y = −a b x −c b => The slope of the line = −a b => −a b = −3 => b = a 3 From statement 1: a = 2 => b = 2 3 – Sufficient From statement 2: The value of c cannot be used to determine the value of b, as we are not aware of the value of the Y-intercept.
© 1999–2024 Manhattan Review www.manhattanreview.com 418 Quantitative Reasoning Question Bank – DS Solutions If Y-intercept would have been given it would be −c b , and then we would have been able to find the value of b. – Insufficient The correct answer is Option A.
450.
From statement 1: The figure depicting the two vertices of the rectangle (−2, −3) and (4, −3) is shown below: (−2, −3) (4, −3) X Y Thus, we know that the length of the rectangle is the difference between the X values of the coordinates of the two points (since the length is parallel to the X axis).
Thus, the length of the rectangle = 4 −(−2) = 6 However, we do not know the width of the rectangle and hence, the area cannot be determined.
– Insufficient From statement 2: The figure depicting the two vertices of the rectangle (−2, −3) and (4, 5) is shown below: (−2, −3) (4, −3) X Y (4, 5) Since the length and width of the rectangle are parallel to the X and Y axes, the dotted lines shown in the figure above must denote the length and width of the rectangle.
www.manhattanreview.com © 1999–2024 Manhattan Review Quantitative Reasoning Question Bank – DS Solutions 419 Thus, the third vertex must be the point of intersection of the dotted lines i.e. (4, −3).
Thus, we know that the length of the rectangle is the difference between the X values of the coordinates of the two points: (−2, −3) and (4, −3) (since the length is parallel to the X axis).
Thus, the length of the rectangle = 4 −(−2) = 6 Also, the width of the rectangle is the difference between the Y values of the coordinates of the two points: (4, 5) and (4, −3) (since the width is parallel to the Y axis).
Thus, the width of the rectangle = 5 −(−3) = 8 Thus, the area of the rectangle = 6 × 8 = 48 – Sufficient The correct answer is Option B.
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11502 | https://pmc.ncbi.nlm.nih.gov/articles/PMC11287907/ | Navigating Aspirin Hypersensitivity in Patients Undergoing Percutaneous Coronary Intervention - PMC
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. 2024 Jul 25;15(8):201–207. doi: 10.14740/jmc4239
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Navigating Aspirin Hypersensitivity in Patients Undergoing Percutaneous Coronary Intervention
Kai Shiang Lin
Kai Shiang Lin
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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a, Keston Rattan
Keston Rattan
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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a, Jensen George
Jensen George
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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Samantha Cavusoglu
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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Christy Joseph
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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Varsha Talanki
a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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Sabu John
b Department of Cardiology, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
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a Department of Internal Medicine, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
b Department of Cardiology, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA
c
Corresponding Author: Sabu John, Department of Cardiology, SUNY Downstate Health Sciences University, Brooklyn, NY 11203, USA. Email: sabu.john@downstate.edu
Received 2024 May 7; Accepted 2024 Jun 19; Issue date 2024 Aug.
Copyright 2024, Lin et al.
This article is distributed under the terms of the Creative Commons Attribution Non-Commercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC11287907 PMID: 39091571
Abstract
Aspirin hypersensitivity continues to be a major clinical challenge in patients with coronary artery disease (CAD), particularly in those requiring percutaneous coronary intervention (PCI) in the absence of a validated alternative antiplatelet regimen. Although true aspirin allergies are uncommon, they can manifest with severe reactions such as angioedema or anaphylaxis, highlighting the critical role of diagnostic challenge tests and tolerance induction strategies. Here, a 61-year-old female with end-stage renal disease (ESRD) on hemodialysis presented with new-onset heart failure and elevated troponins in the setting of a hypertensive emergency. A subsequent left heart catheterization revealed severe multivessel disease, but PCI was deferred due to her history suggestive of aspirin-induced angioedema and the absence of a known optimal approach in this scenario. Given the feasibility of completing a desensitization protocol, aspirin desensitization was pursued, facilitating the successful placement of a drug-eluting stent. This case highlights the need for validated protocols to manage aspirin hypersensitivity, as the current treatment paradigm necessitates a highly individualized approach by the treating clinician.
Keywords: Aspirin, Allergy, Angioedema, Percutaneous coronary intervention, Heart failure, Acute coronary syndrome, Hypersensitivity, Dual antiplatelet therapy
Introduction
Aspirin hypersensitivity in patients undergoing percutaneous coronary intervention (PCI) is a relatively rare finding, with a prevalence of 2.6% among patients admitted for cardiac catheterization, according to a recent retrospective study . Of these patients, 1.0% presented with respiratory manifestations including asthma or rhinitis, while the remaining 1.6% exhibited cutaneous manifestations such as urticaria or angioedema. Although infrequent, this spectrum of hypersensitivity is significant in contemporary medical practice due to aspirin’s pivotal role in treating coronary artery disease (CAD), particularly in the setting of dual antiplatelet therapy (DAPT). Comprising aspirin and a P2Y12 inhibitor, DAPT represents the cornerstone of post-PCI medical therapy and is used mainly to mitigate the risk of cardiovascular complications, such as stent thrombosis (ST). In concert with this approach are the current guidelines which recommend at least 6 months of DAPT post-PCI to achieve optimal therapeutic benefit . However, because the exact prevalence of aspirin allergy is unclear, in part due to ambiguity of definition and the lack of awareness with respect to pseudoallergies and true allergies, aspirin therapy also presents a substantial challenge to this recommendation, as the safety and efficacy of alternative oral antiplatelet combinations without aspirin have yet to be established, likely due to a scarcity of high-quality evidence. This case report and literature review aims to illustrate this clinical scenario and discuss plausible alternatives as well as underscore the need for further research in the area.
Case Report
History of presentation
A 61-year-old female was brought in by emergency medical services (EMS) for sudden-onset generalized weakness and shortness of breath that began within the last 12 h. The patient had similar symptoms in the past in the context of missed hemodialysis sessions, and denied any exertional nature to her symptoms, also denying chest pain, palpitations, nausea, vomiting, lightheadedness, and epigastric pain. Her last hemodialysis session via her left upper extremity arteriovenous fistula was 3 days before presentation, and she endorsed baseline paroxysmal nocturnal dyspnea as well as orthopnea requiring the use of two pillows to sleep at night, but not dyspnea on exertion. Prior to presentation, she desaturated to an oxygen saturation of 66% on room air that failed to improve on supplemental oxygen at home, prompting her to notify EMS.
On arrival, her vital signs were notable for a blood pressure of 193/80 mm Hg (normal range: 90/60 mm Hg to 120/80 mm Hg), respiratory rate of 20 breaths per minute (normal range: 12 - 20 breaths per minute), heart rate of 70 beats per minute (normal range: 60 - 100 beats per minute), a temperature of 98.9 °F (normal range: 98.6 °F to 100.4 °F), and an oxygen saturation of 80% as detected by pulse oximetry (normal range: 95% or higher) that improved to 98% on bilevel positive airway pressure. Her exam was significant for crackles in the bilateral lower and middle lung fields as well as 1+ pitting edema in the bilateral lower extremities. Laboratory workup revealed elevated levels of B-type natriuretic peptide, troponin, and glucose, and decreased levels of hemoglobin. A chest X-ray showed bilateral pulmonary congestion, alveolar edema, and pleural effusions, and a standard 12-lead electrocardiogram (ECG) showed normal sinus rhythm, left ventricular hypertrophy, and poor R wave progression, all of which were unchanged from her baseline that was taken 3 years ago (Fig. 1). The posterior ECG was unremarkable. Nephrology and cardiology were consulted for urgent hemodialysis and elevated troponins, respectively, and the patient was admitted for acute decompensated heart failure and rule-out of acute coronary syndrome (ACS) in the setting of hypertensive emergency.
Figure 1.
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ECG on admission, showing normal sinus rhythm, left ventricular hypertrophy, poor R wave progression. ECG: electrocardiogram.
Past medical history
The patient had a history of hypertension, insulin-dependent diabetes mellitus, ischemic cerebrovascular accident, heart failure with reduced ejection fraction (ejection fraction 35% with grade 2 diastolic dysfunction and moderate diffuse hypokinesis, and no prior ischemic workup), end-stage renal disease on hemodialysis. Home medications included detemir, aspart, furosemide, clopidogrel, sevelamer, metoprolol succinate, amlodipine, and atorvastatin.
Allergies
The patient has a remote history of oral numbness and tingling and possibly angioedema with the use of aspirin many years before presentation. She was told by a physician at the time to never use aspirin again and has never done so since then.
Investigation and management
Following admission, serial ECGs were performed and were unremarkable. Serial troponins rose to a peak of 0.66 ng/mL before falling to undetectable levels within the next day. Euvolemia was achieved through a combination of aggressive diuresis which led to a small amount of urine production, inpatient hemodialysis for further excess volume removal, and blood pressure control, all of which resulted in the restoration to euvolemic weight and the resolution of shortness of breath, lung crackles, and pitting edema. Transthoracic echocardiography (TTE) revealed an improvement in ejection fraction from her baseline of 35% to 50-55% without any regional wall motion abnormalities as well as a grossly normal valvular function and structure. A subsequent left heart catheterization revealed severe, two-vessel disease with 70-80% occlusion and an instantaneous wave-free ratio (iFR) of 0.63 in the mid-left anterior descending artery (LAD) as well as 60-70% occlusion and iFR of 0.93 in the proximal left circumflex artery (LCX) (Fig. 2). However, PCI was deferred at that time due to a possible history of angioedema with aspirin use, conflicting with her need for DAPT after stenting. The patient subsequently underwent successful aspirin desensitization without the development of any adverse reactions, and a staged PCI with deployment of a drug-eluting stent was performed within hours of completion of the desensitization protocol (Table 1). The patient was discharged on DAPT with aspirin and clopidogrel and continued on the rest of her home medications with outpatient cardiology follow-up.
Figure 2.
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Coronary angiogram showing severe two-vessel disease with 70-80% occlusion of the mid-LAD and 60-70% occlusion of the proximal LCX. The labeled arrows point to the respective stenotic regions. LAD: left anterior descending artery; LCX: left circumflex artery.
Table 1. Aspirin Desensitization Protocol.
| Step | Day | Time (h) | Aspirin dosage (mg) |
:--- :--- |
| 1 | 1 | 0 | 20.25 (0.25 tablets) |
| 2 | 1 | 3 | 60.75 (0.75 tablets) |
| 3 | 1 | 6 | 81 (1 tablet) |
| 4 | 2 | 0 | 101.25 (1.25 tablets) |
| 5 | 2 | 3 | 162.5 (2 tablets) |
| 6 | 2 | 6 | 325 (4 tablets) |
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The patient underwent a 2-day protocol in which she received three doses of aspirin on the first day, and three more doses on the second day.
Discussion
This case illustrates a rare clinical scenario where a patient with traditional CAD risk factors who developed ischemic symptoms underwent two coronary catheterizations due to the need for aspirin desensitization prior to stent deployment, which is a current standard of care for coronary ischemia. The delay in revascularization is particularly noteworthy, highlighting the need for improved strategies to address this crucial gap in medical literature. To explore this clinical question, the following discussion begins with a brief plunge into the evolution of PCI.
The introduction of PCI by Andreas Gruntzig in 1977 revolutionized the treatment of CAD . No longer was cardiothoracic surgery the only available treatment modality when medical therapy proved inadequate, and PCI has since become the subject of intense research and development. Significant milestones, such as the introduction of bare-metal stents (BMS), drug-eluting stents (DES), and biodegradable polymer-based DES, have successively enhanced treatment capabilities and established PCI as a crucial component of standard care. Despite these advancements, PCI is not without limitations; ST remains a significant concern, and minimizing its occurrence has been a key therapeutic objective for several decades. Although rare, ST is a catastrophic event characterized by the sudden thrombotic occlusion of a previously patent stent, leading to sudden death or a large myocardial infarction (MI) in most cases. It typically presents subacutely, defined as the development of acute thrombosis within the first 30 days after PCI, but can also occur acutely, late, or very late, which correspond to acute thrombotic events within 24 h, 1 year, or more than 1 year post-PCI, respectively . The 30-day incidence of nonfatal MI in patients with DES who had angiographically confirmed ST is approximately 60%, whereas the 30-day mortality rates for angiographically confirmed ST and clinically identified ST, with BMS and DES, are 7%, 19%, and 15%, respectively [5, 6].
Major risk factors for the development of ST include intraprocedural trauma to the coronary endothelium and the introduction of metal to the coronary vasculature. The former exposes subendothelial collagen and tissue factor to blood, triggering the coagulation cascade, while metal itself is inherently thrombogenic . To mitigate the risk of ST, the primary therapeutic measure is antithrombotic therapy. Notably, the specific antithrombotic regimen depends on the timing of its administration relative to the timing of PCI. For instance, combined antiplatelet and anticoagulant therapy, such as aspirin and unfractionated heparin, is typically indicated prior to and during PCI, whereas antiplatelet therapy alone is usually indicated post-PCI, particularly in the absence of an alternate indication for anticoagulation. Yet regardless of the timing of administration, the antiplatelet regimen almost invariably includes aspirin and a P2Y12 receptor blocker, collectively known as DAPT, thanks to the significant reduction in ST when compared to antiplatelet monotherapy, particularly in the early period following PCI . In fact, DAPT is so important that the absence of a P2Y12 receptor blocker is the single most important predictor for ST at the time of the event [5, 6, 8]. With robust evidence from randomized clinical trials and the absence of another nonsteroidal anti-inflammatory drug (NSAID) with equally selective and irreversible blockade of platelet cyclooxygenase-1 (COX-1), aspirin’s role as a unique and indispensable antiplatelet agent for stented patients is firmly established [9-11]. Interestingly, the strict adherence to aspirin in current DAPT regimens is in stark contrast to the more lenient approach to the choice of P2Y12 receptor blocker, which can vary depending on the specific indication [12, 13].
While DAPT is effective in reducing PCI-related complications, it presents its own set of challenges as well. In addition to the risk of bleeding, which is often a reason for DAPT discontinuation, the concurrent presence of aspirin hypersensitivity can be problematic and potentially preclude the use of DAPT. Because current DAPT regimens unanimously advocate for aspirin use, and that the safety and efficacy of oral antiplatelet combinations excluding aspirin remain unestablished, the optimal approach in this clinical scenario remains unclear and is compounded by additional considerations including the type and severity of the aspirin allergy as well as the urgency of PCI.
Within cardiac pathologies, there are three main indications for urgent aspirin therapy: patients with suspected ACS for whom revascularization is not planned, patients with suspected ACS with plans for revascularization but delayed aspirin initiation due to concerns for hypersensitivity, and patients requiring nonurgent coronary revascularization. Similarly, aspirin or NSAID allergies can be broadly categorized into pseudoallergies and true allergies, with the former thought to be nonimmunologic reactions secondary to NSAID-induced inhibition of COX-1 and other related biochemical pathways, and the latter being presumed, immunoglobulin E (IgE)-mediated immunologic reactions. Regardless of the type or category of allergies, identifying key manifestations of aspirin sensitivity such as anaphylaxis, urticaria or angioedema, and aspirin-exacerbated respiratory disease (AERD), is essential to guide further management .
A relatively common condition affecting 5-7% of all asthmatics, AERD has an overall prevalence of approximately 10% whereas the prevalence for aspirin-induced cutaneous sensitivity ranges from 0.07% to 0.2%; the prevalence for aspirin-related anaphylaxis remains unknown as it has never been conclusively documented to date [15-17]. The main reason for identifying these key hypersensitivity manifestations is to determine the need for a premedication regimen prior to the administration of aspirin, with the goal of preventing or reducing the severity of a potential reaction. Prompt administration of aspirin is often preferred, especially in patients requiring urgent PCI despite the potential for inducing hypersensitivity, and premedication is usually reserved for those suspected of having AERD, which include those with a history of aspirin-induced chest tightness or wheezing, or those with confirmed or suspected prior anaphylactic reactions suggestive of AERD. In these populations, premedication is often done with leukotriene-modifying agents such as montelukast or oral glucocorticoids due to their efficacy in alleviating or even preventing asthma exacerbations. Conversely, patients who develop aspirin-related cutaneous reactions or suspected anaphylactic reactions without symptoms indicative of AERD generally do not require premedication, owing to the more benign nature of these reactions and that the majority of aspirin-related anaphylactic reactions are actually severe respiratory reactions in patients with underlying AERD.
In addition to the type and severity of aspirin allergies, the urgency of PCI is another crucial consideration in determining the optimal approach in this clinical dilemma. Patients requiring PCI can be categorized into those for whom PCI is elective, those for whom PCI can be postponed several hours, and those for whom PCI is urgent without sufficient time to introduce aspirin. Whereas PCI in the first two scenarios can be performed after aspirin desensitization or postponed for several hours to allow for the introduction of low-dose aspirin, respectively, the optimal approach in the third scenario, where PCI is urgent, remains less clear. The result is a handful of postulated alternatives, including oral P2Y12 receptor blocker monotherapy, perioperative intravenous glycoprotein (GP) IIb/IIIa antagonist therapy with subsequent aspirin desensitization, and low-dose rivaroxaban in conjunction with a P2Y12 receptor blocker. These options all appear to be viable alternatives, but in fact share a common flaw - the lack of formal validation in the clinical setting [18-20]. With respect to the introduction of aspirin, which can be performed in the second scenario, a simple protocol involving low-dose aspirin, usually 100 mg or less daily, can be attempted (Table 2) . This protocol can be completed within several hours and has been demonstrated to be safe in patients with aspirin-induced cutaneous manifestations such as urticaria or angioedema . Should patients with suspected or confirmed AERD develop respiratory symptoms during the protocol, bronchodilator therapy and antihistamines are available treatment options. Notably, our patient was started on a lower dose, 20.25 mg, than that outlined in this protocol, 40.5 mg, due to our patient being high-risk and her personal desire to start at the lowest dose possible. Remarkably, the development of anaphylaxis with this protocol is usually not considered even in patients with true aspirin allergies, given the aforementioned absence of conclusive evidence for aspirin-specific anaphylaxis and the low likelihood of symptom induction with the low dosages of aspirin used in this protocol. In patients who may require a single, higher loading dose of aspirin, which is typically a one-time dose of 325 mg, a second protocol extending an additional day can be added to the initial protocol (Table 3) . It should be noted that the likelihood of triggering respiratory symptoms is higher during this second protocol, likely due to the higher dosages of aspirin used, although the same bronchodilator and antihistamine therapies can be used in the event of symptom development.
Table 2. Protocol for Introducing Low-Dose Aspirin (81 mg) in a Patient With Suspected or Confirmed Aspirin Allergy.
| Step | Activity |
:--- |
| 1 | Administer 40.5 mg of aspirin and observe for 60 min. |
| 2 | Administer 40.5 mg of aspirin and observe for 60 min again. If the patient developed symptoms, repeat this step again. |
| 3 | If no symptoms have developed during the first two steps, administer 81 mg of aspirin daily starting the next day. |
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Table 3. Protocol for Increasing the Dosage of Aspirin From 81 mg Daily to 325 mg Daily for Loading Purposes in Patients With Suspected or Confirmed Aspirin Allergy.
| Step | Activity in patients with AERD | Activity in patients without AERD |
:---
| 1 | Administer 121.5 mg of aspirin and observe for 60 min. | Administer 162 mg of aspirin and observe for 60 min. |
| 2 | Administer 202.5 mg of aspirin and observe for 60 min. | Administer 325 mg and observe for 60 min. |
| 3 | Administer 325 mg aspirin and observe for 180 min. | If no symptoms develop, continue with 325 mg of aspirin starting the next day. Desensitization is complete. |
| | If no symptoms develop, continue with 325 mg of aspirin starting the next day. Desensitization is complete. | If symptoms develop, treat the symptoms, repeat the same dose, and observe for another 60 min. If no further symptoms, then desensitization is complete and begin 325 mg of aspirin starting the next day. |
| | If symptoms develop, treat the symptoms, repeat the same dose, and observe for another 60 min. If no further symptoms, then desensitization is complete and begin 325 mg of aspirin starting the next day. | |
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The use of this protocol assumes that patients are already on an 81 mg dose of aspirin. AERD: aspirin-exacerbated respiratory disease.
In contrast to introducing low-dose aspirin to aid in DAPT initiation in more urgent clinical scenarios, aspirin desensitization is an alternative approach in which aspirin is slowly introduced into a patient with suspected or documented aspirin allergy. The main advantage of this approach is its safety and hence, popularity, although the low prevalence of true aspirin allergies suggests that this approach may be less needed than commonly believed . Interestingly, the majority of patients with presumed aspirin allergies actually have drug intolerance secondary to a direct consequence of aspirin’s mechanisms of action rather than true hypersensitivity . However, in the setting of ample time without the need for urgent PCI, aspirin desensitization is still commonly performed when the safest approach is preferred. Thus, in the case of our patient, aspirin desensitization was the approach pursued given the relative lack of urgency for PCI and the desire for the safest approach in the setting of a potentially severe history of angioedema. The relative clinical stability and the ease with which euvolemia was achieved, in conjunction with the absence of chest pain, shortness of breath, or other significant symptoms, suggested that there was, indeed, ample time for aspirin desensitization prior to PCI. The etiology of our patient’s acute pulmonary edema and elevated troponins was likely multifactorial, with the most likely causes being inadequate hemodialysis and coronary ischemia; importantly, the coronary angiogram was negative for ACS.
Finally, while few clinical scenarios appear to preclude the use of aspirin with either the low-dose protocol or desensitization, it is important to be cognizant of certain absolute contraindications. These include patients who have had Stevens-Johnson syndrome, toxic epidermal necrolysis, and drug reaction with eosinophilia and systemic symptoms/drug-induced hypersensitivity syndrome, among others. In these particular scenarios, the general recommendation is to avoid all NSAIDs, including aspirin, without attempting a rapid, low-dose introduction or slow desensitization.
Conclusions
Safe strategies to maneuver the clinical dilemma of initiating DAPT in a patient with aspirin allergy include a low-dose protocol and complete desensitization. Given that the optimal approach remains undefined, further research with randomized trials will likely be crucial in navigating this unique clinical scenario in the near future.
Learning points
Learning points from this case report and literature review include recognition of this clinical conundrum and acknowledgement of the absence of proven optimal approaches to overcome it. Despite the multiple approaches highlighted above, there remains an urgent need for further research and clinical trials to establish a standard of care for this scenario, given the high prevalence of patients with ischemic heart disease.
Acknowledgments
None to declare.
Funding Statement
None to declare.
Conflict of Interest
None to declare.
Informed Consent
Informed consent was obtained from the patient.
Author Contributions
The case was identified by supervising author Sabu John, MD. The manuscript was written by Kai Shiang Lin, MD, and Keston Rattan, MD. Manuscript editing was performed by Kai Shiang Lin, MD, Jensen George, MD, Samantha Cavusoglu, Christy Joseph, and Varsha Talanki. The ECG and coronary angiogram image were identified by Sabu John, MD. The final manuscript was reviewed and supervised by Sabu John, MD.
Data Availability
The authors declare that data supporting the findings of this study are available within the article.
Abbreviations
ACS
acute coronary syndrome
AERD
aspirin-exacerbated respiratory disease
BMS
bare-metal stents
CAD
coronary artery disease
DAPT
dual antiplatelet therapy
DES
drug-eluting stents
ECG
electrocardiogram
EMS
emergency medical services
ESRD
end-stage renal disease
iFR
instantaneous wave-free ratio
LAD
left anterior descending artery
LCX
left circumflex artery
MI
myocardial infarction
NSAIDs
nonsteroidal anti-inflammatory drugs
PCI
percutaneous coronary intervention
ST
stent thrombosis
TTE
transthoracic echocardiography
References
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Associated Data
This section collects any data citations, data availability statements, or supplementary materials included in this article.
Data Availability Statement
The authors declare that data supporting the findings of this study are available within the article.
Articles from Journal of Medical Cases are provided here courtesy of Elmer Press
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Data Type Conversions
Octal Systems
Decimal Notation
Number Systems
Conversion (mathematics)
Octal (base 8)
Number Conversion
5
How do I convert octal to decimal, and vice versa?
Dan Lutley
Former Electrical Engineer at Motorola (company) (1988–2003)
·
Author has 76 answers and 103.4K answer views
·
7y
Octal implies a base 8 numbering system
So for example the number 6777 base 8 would be in decimal:
The right most digit is for units or 8^0
Second digit is for groups of 8 (8^1)
Third digit is for groups of 64 (8^2)
Fourth digit is for groups of 512 (8^3) so we have:
(7 x 8^0) + (7 x 8^1) + (7 x 8^2) + (6 x 8^3)= 7 + 56 +448 + 3072 = 3583
Now to convert back to base 10 we reverse the process by dividing
3583 by modulo 512 to get the number of groups of 512 = 6 with 511 left over
then divide 511 by mod 64 to get number of groups of 64 = 7 with 63 left over
then divide by 8 to get number of groups of 8 le
Octal implies a base 8 numbering system
So for example the number 6777 base 8 would be in decimal:
The right most digit is for units or 8^0
Second digit is for groups of 8 (8^1)
Third digit is for groups of 64 (8^2)
Fourth digit is for groups of 512 (8^3) so we have:
(7 x 8^0) + (7 x 8^1) + (7 x 8^2) + (6 x 8^3)= 7 + 56 +448 + 3072 = 3583
Now to convert back to base 10 we reverse the process by dividing
3583 by modulo 512 to get the number of groups of 512 = 6 with 511 left over
then divide 511 by mod 64 to get number of groups of 64 = 7 with 63 left over
then divide by 8 to get number of groups of 8 left = 7 with 7 units left over
So we have 6777 base 8 which is what we started with.
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.4M answer views
·
7y
Every once in awhile, just for the challenge, I will once again write a program to convert numbers from one base to another. I especially enjoy doing this with the Prgm environment on the TI-84 PLUS CE Graphing Calculator because the language is somewhat similar to the first BASIC language I learned and because I have to write code to handle workarounds for something that the TI cannot normally handle, like displaying hexadecimal digits.
This morning, I wrote a new version that allows me to enter any positive integer in any positive base, then output the values in decimal, binary, octal and hex
Every once in awhile, just for the challenge, I will once again write a program to convert numbers from one base to another. I especially enjoy doing this with the Prgm environment on the TI-84 PLUS CE Graphing Calculator because the language is somewhat similar to the first BASIC language I learned and because I have to write code to handle workarounds for something that the TI cannot normally handle, like displaying hexadecimal digits.
This morning, I wrote a new version that allows me to enter any positive integer in any positive base, then output the values in decimal, binary, octal and hexadecimal. I did not put in all the error-traps possible because I knew that I’d understand if I got weird results.
Here is what my operations screen looks like, with an octal input:
And here is an example with a hexadecimal input:
As you can see, the calculator first asks what base my input number will be in. This can be any positive base, but you might get weird answers if you put in base 1, which I have not error-trapped or tested.
In order to allow input of actual hexadecimal numbers, I needed to set up a method where the calculator could understand hexadecimal, so I temporarily defined A, B, C, D, E and F to be equal to 10 through 15, then I told the calculator to request the input number in list form.
Once it knows the base number and the representation of that number, in list form, it converts the number to decimal and displays it.
Then it converts the decimal number into binary, octal and hexadecimal form, using the “Divide by the Base Number and display the Remainders” method.
I tried to do the digit representation with the piecewise function, but couldn’t make it work properly, so I went back to the if method of generating each digit of each number.
Theoretically, you could expand this 84 program to handle base 36, but I see no need to write the code.
If you’d like to see my program, leave a comment and I’ll post it.
John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.4M answer views
·
7y
Originally Answered: How do I convert octal to a decimal form?
·
Earlier today, I wrote another of my (I don’t know how many) base conversion programs. I love to write new versions because it gives me a chance to try new tricks and alternate solution methods.
The method that program used to convert a number from base “whatever” (I used Base “I” because I was using “B” for something else) to decimal was:
Set the decimal value to zero
0 → D
Starting wit
Earlier today, I wrote another of my (I don’t know how many) base conversion programs. I love to write new versions because it gives me a chance to try new tricks and alternate solution methods.
The method that program used to convert a number from base “whatever” (I used Base “I” because I was using “B” for something else) to decimal was:
Set the decimal value to zero
0 → D
Starting with the first digit, I did each “next” digit until I was out of digits:
Multiply D (decimal value so far) × I (the base number, you’ll use 8)
then add the next digit to that product
Once we are out of digits, D contains that value.
NOTE: this is not as easy a method as others that you may see here on Quora.
Let’s try this with octal number 1234567:
D = 0
0 × 8 + 1 = 1
1 × 8 + 2 = 10
10 × 8 + 3 = 83
83 × 8 + 4 = 668
668 × 8 + 5 = 5349
5349 × 8 + 6 = 42798
42798 × 8 + 7 = 342391
Here’s the output screen showing the same number as decimal, octal, binary and hexadecimal.
My program input the base 8 number into list L1, th...
Related questions
How do I convert 543 from octal to decimal?
How can you convert from an octal number system to a hexadecimal number system and vice versa?
How do you convert a decimal number into binary and vice versa?
How do I convert from decimal to octal?
How do I convert octal to a decimal form?
Assistant
Bot
·
1y
Converting between octal (base 8) and decimal (base 10) involves understanding the positional values of the digits in each system. Here’s how to do both conversions:
Converting Octal to Decimal
To convert an octal number to decimal, you can use the formula:
Decimal=dn×8n+dn−1×8n−1+…+d1×81+d0×80
Where d represents the digits of the octal number, and n is the position of the digit (starting from 0 on the right).
Example: Convert 157 (octal) to decimal.
Identify the digits: 1, 5, and 7.
Compute each term:
1×82=1×64=64
Converting between octal (base 8) and decimal (base 10) involves understanding the positional values of the digits in each system. Here’s how to do both conversions:
Converting Octal to Decimal
To convert an octal number to decimal, you can use the formula:
Decimal=dn×8n+dn−1×8n−1+…+d1×81+d0×80
Where d represents the digits of the octal number, and n is the position of the digit (starting from 0 on the right).
Example: Convert 157 (octal) to decimal.
Identify the digits: 1, 5, and 7.
Compute each term:
1×82=1×64=64
5×81=5×8=40
7×80=7×1=7
Add them together:
64+40+7=111
So, 1578=11110.
Converting Decimal to Octal
To convert a decimal number to octal, you can use repeated division by 8, keeping track of the remainders.
Example: Convert 111 (decimal) to octal.
Divide the number by 8 and record the quotient and remainder.
111÷8=13 remainder 7
13÷8=1 remainder 5
1÷8=0 remainder 1
The octal number is formed by reading the remainders from bottom to top: 157.
So, 11110=1578.
Summary
Octal to Decimal: Use positional values of base 8.
Decimal to Octal: Use repeated division by 8 and read remainders in reverse order.
Feel free to ask if you have any more questions or need further examples!
Klaus Hauptfleisch
Former Journalist
·
Author has 8.4K answers and 2.7M answer views
·
3y
Originally Answered: How do I convert octal to a decimal form?
·
An octal digit is representing 0 to 7 times the power of 8 of it's numeral position:
45678=4⋅83+5⋅82+
6⋅81+7⋅80=
2048+128+48+7=242310.
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Robert Gross
Retired Electrical Engineer
·
Author has 18.2K answers and 4.6M answer views
·
2y
Originally Answered: How does one go about converting from decimal to octal, or vice versa?
·
Converting octal to decimal (assuming this involves only integers, no fractions): Start with the first octal digit. Add to it 8 times the next octal digit to the left, then add 64 times the next, then 512 times the next, etc. Each time you move to the left, multiply that digit by 8 times as much as you did the previous digit. As an example, to convert 243 octal to digital, add together 3 + (4 x 8) + (2 x 64) to obtain
3 + 32 +128 = 163 decimal.
To convert decimal to octal: divide the decimal number by 8. Write down the remainder in the right most position. Divide by 8 again, write down the remai
Converting octal to decimal (assuming this involves only integers, no fractions): Start with the first octal digit. Add to it 8 times the next octal digit to the left, then add 64 times the next, then 512 times the next, etc. Each time you move to the left, multiply that digit by 8 times as much as you did the previous digit. As an example, to convert 243 octal to digital, add together 3 + (4 x 8) + (2 x 64) to obtain
3 + 32 +128 = 163 decimal.
To convert decimal to octal: divide the decimal number by 8. Write down the remainder in the right most position. Divide by 8 again, write down the remainder in the next left column. Continue until the number you are dividing is less than 8. This will be the left most octal number. As an example, do the above conversion in reverse: 163 decimal to octal.
163/8 = 20 + 3 remainder. The right most digit is 3.
20/8 = 2 + 4 remainder. The next left digit is 4.
2/8 is less than 8. The next left digit is 2. The octal number is 243.
Related questions
Can you convert from hex to decimal and vice versa?
How do I convert octal to binary?
How do you convert a 246 octal number to a decimal?
Can you convert from binary to octal and vice versa?
How do I convert a decimal 255 to binary and octal?
Craig Cornelius
B.A. from Luther College (Graduated 1974)
·
Author has 1.9K answers and 3.4M answer views
·
7y
Here’s a good description of the method:
An example is octal 1234. This is 1 8^3 + 2 8^2 + 3 8^1+ 4 8^0 →
1 512 + 2 64 + 3 8 + 4 1 →
512 + 128 + 24 + 4 →
668 (decimal)
Methods to converting from decimal to octal can be found here.
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John K WilliamsSon
Accredited (MS Educ) nerd who loves talking about math
·
Author has 9K answers and 23.4M answer views
·
5y
Originally Answered: Is it possible to convert an octal number to its equivalent decimal number?
·
Multiply the last digit of the number times 8º
Multiple the second-to-last digit times 8¹
Multiple the next prior digit times 8²
Multiply each prior digit by 8³, 8⁴, 8⁵, etc.
Add up all those products
Here is how to do that on a calculator that can do list arithmetic:
put the digits of the octal number into list L₁
put the powers of 8 into list L₂ but make sure the LAST value in the list is 1
since there were six digits in 123456₈ — use 6 twice in the sequence command
optional — multiply the two lists to see each digit multipled by each place value.
Calculate the sum of the product of the two lists to g
Multiply the last digit of the number times 8º
Multiple the second-to-last digit times 8¹
Multiple the next prior digit times 8²
Multiply each prior digit by 8³, 8⁴, 8⁵, etc.
Add up all those products
Here is how to do that on a calculator that can do list arithmetic:
put the digits of the octal number into list L₁
put the powers of 8 into list L₂ but make sure the LAST value in the list is 1
since there were six digits in 123456₈ — use 6 twice in the sequence command
optional — multiply the two lists to see each digit multipled by each place value.
Calculate the sum of the product of the two lists to get the decimal value.
.
Is it possible to convert an octal number to its equivalent decimal number?
Darrah Chavey
Professor, Math & Comp. Sci. at Beloit College (1987–present)
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.2K answers and 1.5M answer views
·
5y
Originally Answered: Is it possible to convert an octal number to its equivalent decimal number?
·
Let’s assume that the octal digits are in variables a0,a1,a2,…, so that the value of the octal number is ak8k+a787+⋯+a181+a080. Then to convert it to a decimal number, use the following algorithm:
decimalValue = 0
loop from k=8 down to k=0 {
decimalValue ← decimalValue 8
decimalValue ← decimalValue + ak
end loop
At the conclusion of the loop, the octal number will have been converted to a decimal number (assuming all the calculations were done in decimal notation).
This answer assumes you’re working with integers. A similar approach can be us
Let’s assume that the octal digits are in variables a0,a1,a2,…, so that the value of the octal number is ak8k+a787+⋯+a181+a080. Then to convert it to a decimal number, use the following algorithm:
decimalValue = 0
loop from k=8 down to k=0 {
decimalValue ← decimalValue 8
decimalValue ← decimalValue + ak
end loop
At the conclusion of the loop, the octal number will have been converted to a decimal number (assuming all the calculations were done in decimal notation).
This answer assumes you’re working with integers. A similar approach can be used for decimals, see How do we convert an octal number, 0.4051, into its equivalent decimal number? for an example of that conversion.
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William Graham
B. Sc. in Computer Science, University of Calgary (Graduated 1994)
·
Author has 2.2K answers and 930.4K answer views
·
3y
Originally Answered: How do I go from octal to decimal?
·
The octal numeral represents a polynomial, with the digits being the coefficients. Evaluate the polynomial at x=8, doing the arithmetic in decimal.
For example, the octal numeral 12348 means
1x3+2x2+3x1+4x0, evaluated for x=8. The first few powers of 8 are:
80=1
81=8
82=64
83=512
So the polynomial is:
1(512)+2(64)+3(8)+4(1),
Doing the arithmetic, that's 512+128+24+4=66810 as a decimal numeral.
Diploma from Government Polytechnic Lucknow (Graduated 2017)
·
Updated 5y
Originally Answered: How do I convert octal to a decimal form?
·
Learn this example….
(235) octal to decimal
=157 in decimal
Klaus Hauptfleisch
B.A. from University of Bonn (Graduated 1985)
·
Author has 8.4K answers and 2.7M answer views
·
3y
Originally Answered: Is it possible to convert an octal number to its equivalent decimal number?
·
For converting an octal to a decimal number you can use Horner's Method by multiplying and adding the octal digits.
Let's take the octal 765 for example. Multiply the 7 by 8 and add 6 for getting the interim result 62, then multiply it by 8 and add 5 to get the final result which is 501. So the octal 765 is the decimal 501. If the octal number is 7654 do the same, with the difference that 501 is the second interim result, if you multiply it by 8 and add 4 you get the decimal number 4012.
Brian Yu
For every digit after the decimal point, multiply that digit by 8 to the power of the number of that digit past the decimal point. For example:
732 base 8 = 7 8^2 + 3 8^1 + 2 8^0 = 7 64 + 3 8 =+ 2 1 = 474 base 10.
1.2 base 8 = 1 8^0 + 2 8^-1 = 1 1 + 2 1/8 = 1 1/4 = 5/4 base 10.
To go vise versa, do pretty much the opposite: take the base 10 number, find the biggest power of 8 that divides it, find the maximum number of them, which becomes the digit, and then subtract that number from the original number, and repeat. For Example:
200 base 10 = 3 8^2 + 8 = 3 8^2 + 1 8^2 + 0
For every digit after the decimal point, multiply that digit by 8 to the power of the number of that digit past the decimal point. For example:
732 base 8 = 7 8^2 + 3 8^1 + 2 8^0 = 7 64 + 3 8 =+ 2 1 = 474 base 10.
1.2 base 8 = 1 8^0 + 2 8^-1 = 1 1 + 2 1/8 = 1 1/4 = 5/4 base 10.
To go vise versa, do pretty much the opposite: take the base 10 number, find the biggest power of 8 that divides it, find the maximum number of them, which becomes the digit, and then subtract that number from the original number, and repeat. For Example:
200 base 10 = 3 8^2 + 8 = 3 8^2 + 1 8^2 + 0 8^1 + 08^0 = 310 base 8.
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11504 | https://sites.oxy.edu/ron/math/312/14/ws/11.pdf | Complex Analysis Math 214 Spring 2014 Fowler 307 MWF 3:00pm - 3:55pm c ⃝2014 Ron Buckmire Class 11: Friday February 14 TITLE Analyticity, the Cauchy-Riemann Equations and Harmonic Functions CURRENT READING Zill & Shanahan, Section 3.3 and 3.4.
HOMEWORK Zill & Shanahan, §3.3 #1, 3, 15, 18, 24. 27; SUMMARY We shall move on from our definition of differentiability to the idea of analyticity and the famous Cauchy-Riemann Equations. We’ll also introduce the concept of harmonic functions and the harmonic conjugate.
Derivation of the Cauchy-Riemann Equations We shall derive the Cauchy-Riemann equations by looking at the definition of the derivative of a function f(z) = u(x, y) + iv(x, y) at the point z0.
f′(z0) = lim ∆z→0 f(z0 + ∆z) −f(z0) ∆z = lim (∆x,∆y)→(0,0) u(x0 + ∆x, y0 + ∆y) + iv(x0 + ∆x, y0 + ∆y) −u(x0, y0) −iv(x0, y0) ∆x + i∆y We shall do this limit twice, once letting ∆z →0 horizontally and the other time letting ∆z →0 vertically 1 Complex Analysis Worksheet 11 Math 312 Spring 2014 Difference Between Analyticity and Differentiability Differentiability is a property of a function that occurs at a particular point. If a func-tion is differentiable at every point in a set, then we can say that it is differentiable on that set. (But if thta set is open, then we would also say that the function is analytic on that set.) Remember analyticity is a property a function that is defined on an open set, often times a neighborhood of a particular point.
THEOREM A complex function w = f(z) is said to be analytic (or “regular” or “holomorphic”) at a point z0 if f is differentiable at z0 and at every point in a neighborhood surrounding z0.
The Cauchy-Riemann Equations and Analyticity Given a function f(z) = u(x, y)+iv(x,y) the corresponding Cauchy-Riemann Equations are ux = vy, uy = −vx ANALYTICITY ⇒C.R.E. satisfied To make satisfying the CRE a sufficient condition one needs the added condition that the first derivatives of u and v are continuous on an open set. If both these conditions are true and f is defined on an open set, then f is analytic on the open set.
f′(z0) = ux(x0, y0) + ivx(x0, y0) = −i(uy(x0, y0) + ivy(x0, y0)) ANALYTICITY ⇐ ⇒C.R.E. + Continuity of ux, uy, vx, vy EXAMPLE Show that f(z) = z is not analytic anywhere in the complex plane. You can do this in two ways: 1: 2: 2 Complex Analysis Worksheet 11 Math 312 Spring 2014 GroupWork Show that the function f(z) = 1/z is analytic on the set z ̸= 0. To do that you will have to answer the following questions: • What is its domain of definition? Is this an open set?
• What are its component functions? Are their partial derivatives continuous?
• Do they satisfy the CRE?
• Is it analytic? On what set? Is this set open or closed?
3 Complex Analysis Worksheet 11 Math 312 Spring 2014 Laplace’s Equation The partial differential equation shown below is known as Laplace’s Equation.
∇2φ = ∆φ = ∂2 ∂x2φ(x, y) + ∂2 ∂y2φ(x, y) = 0 Solutions φ(x, y) which solve Laplace’s equation are very important in a number of areas of mathematical physics and applied mathematics. Some of these applications are: • electrostatic potential in two-dimensional free space • scalar magnetostatic potential • stream function and velocity potential in fluid flow (aerodynamics, etc) • spatial distribution of equilibrium temperature Harmonic Functions A real-valued function φ(x, y) is said to be harmonic in a domain (i.e. open, connected set) D if all its second-order partial derivatives are continuous in D and if φ satisfies Laplace’s Equation at each point (x, y) ∈D.
THEOREM If f(z) is analytic on a domain D then both u(x, y) =Re(f(z)) and v(x, y) =Im(f(z)) are harmonic in D.
ANALYTICITY ⇐ ⇒Re f(z) and Im f(z) are HARMONIC PROOF The proof follows directly from the CRE.
(Take 3 minutes and try and come up with it.) Given a harmonic function u(x, y) defined on an open connected set D we can construct a harmonic conjugate v(x, y) so that the combined function f = u(x, y) + iv(x, y) will be analytic on the domain D.
EXAMPLE Given u(x, y) = x3 −3xy2 + y find the harmonic conjugate v(x, y) and thus construct an analytic function f(z) such that Re f(z) = u(x, y) 4 |
11505 | https://pubmed.ncbi.nlm.nih.gov/2385766/ | Acute bacterial sialadenitis: a study of 29 cases and review - PubMed
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. 1990 Jul-Aug;12(4):591-601.
doi: 10.1093/clinids/12.4.591.
Acute bacterial sialadenitis: a study of 29 cases and review
I I Raad1,M F Sabbagh,G J Caranasos
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1 Department of Medicine, University of Florida School of Medicine, Gainesville.
PMID: 2385766
DOI: 10.1093/clinids/12.4.591
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Acute bacterial sialadenitis: a study of 29 cases and review
I I Raad et al. Rev Infect Dis.1990 Jul-Aug.
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. 1990 Jul-Aug;12(4):591-601.
doi: 10.1093/clinids/12.4.591.
Authors
I I Raad1,M F Sabbagh,G J Caranasos
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1 Department of Medicine, University of Florida School of Medicine, Gainesville.
PMID: 2385766
DOI: 10.1093/clinids/12.4.591
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Acute bacterial parotitis (ABP) used to be described as a nosocomial postoperative infection associated with a high mortality. A review of the literature on acute bacterial sialadenitis (ABS) revealed Staphylococcus aureus as the most common pathogen. Between 1970 and 1988, 17 patients with ABP and 12 with acute bacterial submandibular sialadenitis (ABSS) were admitted to a university hospital. Cultures of purulent discharge from the salivary ducts of these patients revealed S. aureus in 53% and viridans streptococci in 31%. During the study period, only six nosocomial cases of ABP occurred out of 289,234 admissions. S. aureus was the organism isolated most frequently in the nosocomial cases. There were no deaths, and all patients recovered following antibiotic treatment without surgical drainage. Advances in antimicrobial therapy and fluid management of hospitalized patients have made nosocomial ABS a rare disease with a favorable prognosis. S. aureus remains the most common pathogen causing ABS even in nosocomially infected patients who are not critically ill.
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[Differential diagnostic observations on acute sialadenitis].Knöbber D, Wilhelm HJ, Schätzle W.Knöbber D, et al.Laryngorhinootologie. 1990 Jun;69(6):324-6. doi: 10.1055/s-2007-998201.Laryngorhinootologie. 1990.PMID: 2378660 German.
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11506 | https://math.answers.com/statistics/Can_you_show_me_an_example_of_an_ogive_graph | Can you show me an example of an ogive graph? - Answers
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Can you show me an example of an ogive graph?
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Ogives are often used in statistics to represent the cumulative frequencies for the classes in a frequency distribution.
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### The graph of a cumulative frequency distribution is called? ogive
### Is a histogram a graph that represents the cumulative frequencies for the classes in a frequency distribution? No it is not. The ogive is a graph that represents the cumulative frequencies for the classes in a frequency distribution.
### Can you show me a example of an histogram graph? A histogram is a bar chart that only uses numbers.
### What does a line graph show in data? A line graph shows information that grows (raises) over a certain time.... Example: Growth of a plant within a year (timelaps)
### What are interpolation of ogive curve? An ogive is a cumulative relative frequency diagram. Interpolation is definiting the midpoint (50%) of this line
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11507 | https://pulmo-ua.com/wp-content/uploads/2021/12/wilson2013.pdf | 118 Chapter 5 Physiology of the Airway WILLIAM C. WILSON JONATHAN L. BENUMOF I. Normal Respiratory Physiology (Nonanesthetized) A. Gravity-Determined Distribution of Perfusion and Ventilation 1. Distribution of Pulmonary Perfusion 2. Distribution of Ventilation 3. The Ventilation-Perfusion Ratio B. Nongravitational Determinants of Blood Flow Distribution 1. Passive Processes 2. Active Processes and Pulmonary Vascular Tone 3. Alternative (Nonalveolar) Pathways of Blood Flow Through the Lung C. Nongravitational Determinants of Pulmonary Compliance, Resistance, Lung Volume, Ventilation, and Work of Breathing 1. Pulmonary Compliance 2. Airway Resistance 3. Different Regional Lung Time Constants 4. Pathways of Collateral Ventilation 5. Work of Breathing 6. Lung Volumes, Functional Residual Capacity, and Closing Capacity D. Oxygen and Carbon Dioxide Transport 1. Alveolar and Dead Space Ventilation and Alveolar Gas Tensions 2. Oxygen Transport 3. Carbon Dioxide Transport 4. Bohr and Haldane Effects E. Pulmonary Microcirculation, Interstitial Space, and Fluid (Pulmonary Edema) II. Respiratory Function During Anesthesia A. Anesthetic Depth and Respiratory Pattern B. Anesthetic Depth and Spontaneous Minute Ventilation C. Preexisting Respiratory Dysfunction D. Special Intraoperative Conditions E. Mechanisms of Hypoxemia During Anesthesia 1. Equipment Malfunction 2. Hypoventilation 3. Hyperventilation 4. Decrease in Functional Residual Capacity 5. Decreased Cardiac Output and Increased Oxygen Consumption 6. Inhibition of Hypoxic Pulmonary Vasoconstriction 7. Paralysis 8. Right-to-Left Interatrial Shunting 9. Involvement of Mechanisms of Hypoxemia in Specific Diseases F. Mechanisms of Hypercapnia and Hypocapnia During Anesthesia 1. Hypercapnia 2. Hypoventilation 3. Increased Dead Space Ventilation 4. Increased Carbon Dioxide Production 5. Inadvertent Switching Off of a Carbon Dioxide Absorber 6. Hypocapnia G. Physiologic Effects of Abnormalities in Respiratory Gases 1. Hypoxia 2. Hyperoxia (Oxygen Toxicity) 3. Hypercapnia 4. Hypocapnia III. Conclusions IV. Clinical Pearls I. NORMAL RESPIRATORY PHYSIOLOGY (NONANESTHETIZED) Anesthesiologists require an extensive knowledge of respiratory physiology to care for patients in the operat ing room and the intensive care unit. Mastery of the normal respiratory physiologic processes is a prerequisite to understanding the mechanisms of impaired gas exchange that occur during anesthesia, during surgery, and with disease. This chapter is divided into two sec tions. The first section reviews the normal (gravity-determined) distribution of perfusion and ventilation, the major nongravitational determinants of resistance to per fusion and ventilation, transport of respiratory gases, and the pulmonary reflexes and special functions of the lung. In the second section, these processes and concepts are CHAPTER 5 Physiology of the Airway 119 vessels, the vessels in this region of the lung are collapsed, and no blood flow occurs; this is known as zone 1 (PA > Ppa > Ppv). Because there is no blood flow, no gas exchange is possible, and the region functions as alveolar dead space, or wasted ventilation. Little or no zone 1 exists in the lung under normal conditions,2 but the amount of zone 1 lung may be greatly increased if Ppa is reduced, as in oligemic shock, or if PA is increased, as in the application of excessively large tidal volumes (VT) or levels of positive end-expiratory pressure (PEEP) during positive-pressure ventilation.
Further down the lung, absolute Ppa becomes positive, and blood flow begins when Ppa exceeds PA (zone 2, Ppa > PA > Ppv). At this vertical level in the lung, PA exceeds Ppv, and blood flow is determined by the mean Ppa − PA difference rather than by the more conventional Ppa − Ppv difference (see later discussion).3 In zone 2, the rela tionship between blood flow and alveolar pressure has the same physical characteristics as a waterfall flowing over a dam. The height of the upstream river (before reaching the dam) is equivalent to Ppa, and the height of discussed in relation to the general mechanisms of impaired gas exchange that occur during anesthesia and surgery.
A. Gravity-Determined Distribution of Perfusion and Ventilation 1. Distribution of Pulmonary Perfusion Contraction of the right ventricle imparts kinetic energy to the blood in the main pulmonary artery. As this energy is dissipated in climbing a vertical hydrostatic gradient, the absolute pressure in the pulmonary artery (Ppa) decreases by 1 cm H2O per centimeter of vertical dis tance up the lung (Fig. 5-1). At some height above the heart, Ppa becomes zero (i.e., equal to atmospheric pres sure), and still higher in the lung, Ppa becomes negative.1 In this region, then, alveolar pressure (PA) exceeds Ppa and pulmonary venous pressure (Ppv), which is very negative at this vertical height. Because the pressure outside the vessels is greater than the pressure inside the Figure 5-1 Schematic diagram showing the distribution of blood flow in the upright lung. In zone 1, alveolar pressure (PA) exceeds pulmonary artery pressure (Ppa), and no flow occurs because the intra-alveolar vessels are collapsed by the compressing alveolar pressure. In zone 2, Ppa exceeds PA, but PA exceeds pulmonary venous pressure (Ppv). Flow in zone 2 is determined by the Ppa-PA difference (Ppa − PA) and has been likened to an upstream river flowing over a dam. Because Ppa increases down zone 2 whereas PA remains constant, perfusion pressure increases, and flow steadily increases down the zone. In zone 3, Ppv exceeds PA, and flow is determined by the Ppa − Ppv differ ence, which is constant down this portion of the lung. However, transmural pressure across the wall of the vessel increases down this zone, so the caliber of the vessels increases (resistance decreases), and therefore flow increases. Finally, in zone 4, pulmonary interstitial pressure (PISF) becomes positive and exceeds both Ppv and PA. Consequently, flow in zone 4 is determined by the Ppa − PISF difference. (Redrawn with modification from West JB: Ventilation/Blood flow and gas exchange, ed 4, Oxford, 1970, Blackwell Scientific, 1970.) Arterial Alveolar Venous Ppa PA PA > Ppa > Ppv Ppa > PA > Ppv Ppa > Ppv > PA Ppa > PISF > Ppv > PA Ppa The four zones of the lung 1. Collapse 2. Waterfall 3. Distention 4. Interstitial pressure Zone 1 Zone 2 Zone 3 Zone 4 Ppa = PA Ppv = PA Distance Blood flow 120 PART 1 Basic Clinical Science Considerations the normally present negative and radially expanding interstitial tension on the extra-alveolar pulmonary vessels. Expansion of the pulmonary interstitial space by fluid causes pulmonary interstitial pressure (PISF) to become positive and to exceed Ppv (zone 4, Ppa > PISF > Ppv > PA).6,7 In addition, the vascular resistance of extra-alveolar vessels may be increased at a very low lung volume (i.e., residual volume); at such volumes, the teth ering action of the pulmonary tissue on the vessels is also lost, and as a result, PISF increases positively (see later discussion of lung volume).8,9 Consequently, zone 4 blood flow is governed by the arteriointerstitial pressure differ ence (Ppa − PISF), which is less than the Ppa − Ppv dif ference, and therefore zone 4 blood flow is less than zone 3 blood flow. In summary, zone 4 is a region of the lung from which a large amount of fluid has transuded into the pulmonary interstitial compartment or is possibly at a very low lung volume. Both these circumstances produce positive interstitial pressure, which causes com pression of extra-alveolar vessels, increased extra-alveolar vascular resistance, and decreased regional blood flow.
It should be evident that as Ppa and Ppv increase, three important changes take place in the pulmonary circulation—namely, recruitment or opening of previ ously unperfused vessels, distention or widening of previ ously perfused vessels, and transudation of fluid from very distended vessels.10,11 Thus, as mean Ppa increases, zone 1 arteries may become zone 2 arteries, and as mean Ppv increases, zone 2 veins may become zone 3 veins. The increase in both mean Ppa and Ppv distends zone 3 vessels according to their compliance and decreases the resistance to flow through them. Zone 3 vessels may become so distended that they leak fluid and become converted to zone 4 vessels. In general, pulmonary capillary recruitment is the principal change as Ppa and Ppv increase from low to moderate levels, distention is the principal change as Ppa and Ppv increase from mod erate to high levels, and transudation is the principal change when Ppa and Ppv increase from high to very high levels.
2. Distribution of Ventilation Gravity also causes differences in vertical Ppl, which in turn causes differences in regional alveolar volume, com pliance, and ventilation. The vertical gradient of Ppl can best be understood by imagining the lung as a plastic bag filled with semifluid contents; in other words, it is a vis coelastic structure. Without the presence of a supporting chest wall, the effect of gravity on the contents of the bag would cause the bag to bulge outward at the bottom and inward at the top (i.e., it would assume a globular shape). Inside the supporting chest wall, the lung cannot assume a globular shape. However, gravity still exerts a force on the lung to assume a globular shape; this force creates relatively more negative pressure at the top of the pleural space (where the lung pulls away from the chest wall) and relatively more positive pressure at the bottom of the lung (where the lung is compressed against the chest wall) (Fig. 5-2). The density of the lung determines the magnitude of this pressure gradient. Because the lung has about one fourth the density of water, the gradient of Ppl (in cm H2O) is about one fourth the height of the upright the dam is equivalent to PA. The rate of water flow over the dam is proportional to only the difference between the height of the upstream river and the dam (Ppa − PA), and it does not matter how far below the dam the down stream riverbed (Ppv) is. This phenomenon has various names, including the waterfall, Starling resistor, weir (dam made by beavers), and sluice effect. Because mean Ppa increases down this region of the lung but mean PA is relatively constant, the mean driving pressure (Ppa − PA) increases linearly, and therefore mean blood flow increases linearly as one descends down this portion of the lung. However, respiration and pulmonary blood flow are cyclic phenomena. Therefore, absolute instantaneous Ppa, Ppv, and PA are changing continuously, and the rela tionships among Ppa, Ppv, and PA are dynamically deter mined by the phase lags between the cardiac and respiratory cycles. Consequently, a given point in zone 2 may actually be in either a zone 1 or a zone 3 condition at a given moment, depending on whether the patient is in respiratory systole or diastole or in cardiac systole or diastole.
Still lower in the lung, there is a vertical level at which Ppv becomes positive and also exceeds PA. In this region, blood flow is governed by the pulmonary arteriovenous pressure difference, Ppa − Ppv (zone 3, Ppa > Ppv > PA), for here both these vascular pressures exceed PA, and the capillary systems are thus permanently open and blood flow is continuous. In descending zone 3, gravity causes both absolute Ppa and Ppv to increase at the same rate, so the perfusion pressure (Ppa − Ppv) is unchanged. However, the pressure outside the vessels—namely, pleural pressure (Ppl)—increases less than Ppa and Ppv. Therefore, the transmural distending pressures (Ppa − Ppl and Ppv − Ppl) increase down zone 3, the vessel radii increase, vascular resistance decreases, and blood flow consequently increases further.
Finally, whenever pulmonary vascular pressures (Ppa) are extremely high, as they are in a severely volume-overloaded patient, in a severely restricted and constricted pulmonary vascular bed, in an extremely dependent lung (far below the vertical level of the left atrium), and in patients with pulmonary embolism or mitral stenosis, fluid can transude out of the pulmonary vessels into the pulmonary interstitial compartment. In addition, pulmo nary interstitial edema can be caused by extremely nega tive Ppl and perivascular hydrostatic pressure, such as may occur in a vigorously spontaneously breathing patient with an obstructed airway due to laryngospasm (most commonly) or upper airway masses (e.g., tumors, hematoma, abscess, edema), strangulation, infectious pro cesses (e.g., epiglottitis, pharyngitis, croup), or vocal cord paralysis; with rapid reexpansion of lung; or with the application of very negative Ppl during thoracentesis.4,5 Transuded pulmonary interstitial fluid can significantly alter the distribution of pulmonary blood flow.
When the flow of fluid into the interstitial space is excessive and the fluid cannot be cleared adequately by the lymphatics, it accumulates in the interstitial connec tive tissue compartment around the large vessels and airways and forms peribronchial and periarteriolar edema fluid cuffs. The transuded pulmonary interstitial fluid fills the pulmonary interstitial space and may eliminate CHAPTER 5 Physiology of the Airway 121 lung (30 cm). Thus, Ppl increases positively by 30/4 = 7.5 cm H2O from the top to the bottom of the lung.12 Because PA is the same throughout the lung, the Ppl gradient causes regional differences in transpulmonary distending pressure (PA − Ppl). Ppl is most positive (least negative) in the dependent basilar lung regions, so alveoli in these regions are more compressed and are therefore considerably smaller than the superior, relatively non compressed apical alveoli. (The volume difference is approximately fourfold.)13 If regional differences in alve olar volume are translated to a pressure-volume (compli ance) curve for normal lung (Fig. 5-3), the dependent small alveoli are on the midportion, and the nondepen dent large alveoli are on the upper portion of the S-shaped compliance curve. Because the different regional slopes of the composite curve are equal to the different regional lung compliance values, dependent alveoli are relatively compliant (steep slope), and nondependent alveoli are relatively noncompliant (flat slope). Therefore, most of the VT is preferentially distributed to dependent alveoli which expand more per unit pressure change than the nondependent alveoli.
3. The Ventilation-Perfusion Ratio Blood flow and ventilation (both shown on the left verti cal axis of Fig. 5-4) increase linearly with distance down the normal upright lung (horizontal axis, reverse polar ity).14 Because blood flow increases from a very low value and more rapidly than ventilation does with distance down the lung, the ventilation-perfusion ratio ( V /Q A , Figure 5-3 Pleural pressure increases by 0.25 cm H2O every centimeter down the lung. This increase in pleural pressure causes a fourfold decrease in alveolar volume from the top of the lung to the bottom. The caliber of the air passages also decreases as lung volume decreases. When regional alveolar volume is translated to a regional transpulmonary pressure–alveolar volume curve, small alveoli are seen to be on a steep portion of the curve (large slope), and large alveoli are on a flat portion of the curve (relatively small slope). Because the regional slope equals regional compliance, the dependent small alveoli normally receive the largest share of the tidal volume. Over the normal tidal volume range the pressure-volume relationship is linear: lung volume increases by 500 mL, from 2500 mL (normal functional residual capacity) to 3000 mL. The lung volume values in this diagram are derived from the upright position. 4,000 3,500 3,000 2,500 5 5 5 0 Volume (mL) Regional slope equals regional compliance Transpulmonary pressure cm H2O Pleural pressure increases 0.25 cm H2O/cm lung dependency Figure 5-2 Schematic diagram of the lung within the chest wall showing the tendency of the lung to assume a globular shape because of gravity and the lung’s viscoelastic nature. The tendency of the top of the lung to collapse inward creates a relatively nega tive pressure at the apex of the lung, and the tendency of the bottom of the lung to spread outward creates a relatively positive pressure at the base of the lung. Therefore, alveoli at the top of the lung tend to be held open and are larger at end-exhalation, whereas those at the bottom tend to be smaller and compressed at end-exhalation. Pleural pressure increases by 0.25 cm H2O per centimeter of lung dependence. ≠Ppl 0.25 cm H2O/cm ≠Ppl 0.25 cm H2O/cm Lung Chest wall + + – – 122 PART 1 Basic Clinical Science Considerations Figure 5-4 Distribution of ventilation and blood flow (left vertical axis) and the ventilation-perfusion ratio ( V Q A/ , right vertical axis) in normal upright lung. Both blood flow and ventilation are expressed in liters per minute per percentage of alveolar volume and have been drawn as smoothed-out linear functions of vertical height. The closed circles mark the V /Q A ratios of horizontal lung slices (three of which are shown in Fig. 5-5). A cardiac output of 6 L/min and a total minute ventilation of 5.1 L/min were assumed. (Redrawn with modi fication from West JB: Ventilation/Blood flow and gas exchange, ed 4, Oxford, 1970, Blackwell Scientific.) 1 2 3 .15 5 4 3 2 .10 .05 Percent lung volume (L/min) Bottom Top Rib number Blood flow Ventilation VA/Q • • VA/Q • • right vertical axis of Fig. 5-4) decreases rapidly at first and then more slowly. V /Q A best expresses the amount of ventilation rela tive to perfusion in any given lung region. For example, alveoli at the base of the lung are overperfused in relation to their ventilation ( V /Q A < 1). Figure 5-5 shows the calculated ventilation ( VA) and blood flow ( Q), the V /Q A ratio, and the alveolar partial pressures of oxygen (PAO2) and carbon dioxide (PACO2) for horizontal slices from the top (7% of lung volume), middle (11% of lung volume), and bottom (13% of lung volume) of the lung.15 PAO2 increases by more than 40 mm Hg, from 89 mm Hg at the base to 132 mm Hg at the apex, whereas PCO2 decreases by 14 mm Hg, from 42 mm Hg at the bottom to 28 mm Hg at the top. Therefore, in keeping with the regional V /Q A ratio, the bottom of the lung is relatively hypoxic and hypercapnic compared with the top of the lung. V /Q A inequalities have different effects on arterial CO2 tension (PaCO2) than on arterial O2 tension (PaO2). Blood passing through underventilated alveoli tends to retain its CO2 and does not take up enough O2; blood traversing overventilated alveoli gives off an excessive amount of CO2 but cannot take up a proportionately increased amount of O2 because of the flatness of the oxygen-hemoglobin (oxy-Hb) dissociation curve in this region (see Fig. 5-25). A lung with uneven V /Q A rela tionships can eliminate CO2 from the overventilated alveoli to compensate for the underventilated alveoli. As a result, with uneven V /Q A relationships, PACO2-to-PaCO2 gradients are small, and PAO2-to-PAO2 gradients are usually large.
Figure 5-5 Ventilation-perfusion ratio ( V Q A/ ) and the regional com position of alveolar gas. Values for regional flow ( Q), ventilation ( VA), partial pressure of oxygen (PO2), and partial pressure of carbon dioxide (PCO2) were derived from Figure 5-4. Partial pressure of nitro gen (PN2) represents what remains from total gas pressure (760 mm Hg including water vapor, which equals 47 mm Hg). The percentage volumes (Vol.) of the three lung slices are also shown. When com pared with the top of the lung, the bottom of the lung has a low V /Q A ratio and is relatively hypoxic and hypercapnic. (Redrawn from West JB: Regional differences in gas exchange in the lung of erect man. J Appl Physiol 17:893, 1962.) 7 11 .13 .24 .52 .82 .07 .50 1.29 3.3 1.0 0.63 132 108 89 28 39 42 553 566 582 Vol.
(%) PO2 PCO2 PN2 (L/min) (mm Hg) VA • VA/Q • • Q • In 1974, Wagner and colleagues described a method of determining the continuous distribution of V /Q A ratios within the lung based on the pattern of elimination of a series of intravenously infused inert gases.16 Gases of differing solubility are dissolved in physiologic saline solution and infused into a peripheral vein until a steady state is achieved (20 minutes). Toward the end of the infusion period, samples of arterial and mixed expired gas are collected, and total ventilation and total cardiac output ( QT) are measured. For each gas, the ratio of arterial to mixed venous concentration (retention) and the ratio of expired to mixed venous concentration (excretion) are calculated, and retention-solubility and excretion-solubility curves are drawn. The retention- and excretion-solubility curves can be regarded as fingerprints of the particular distribution of V /Q A ratios that give rise to them.
Figure 5-6 shows the types of distributions found in young, healthy subjects breathing air in the semirecum bent position.17 The distributions of both ventilation and blood flow are relatively narrow. The upper and lower 9% limits shown (vertical interrupted lines) correspond to V /Q A ratios of 0.3 and 2.1, respectively. Note that these young, healthy subjects had no blood flow perfus ing areas with very low V /Q A ratios, nor did they have CHAPTER 5 Physiology of the Airway 123 the opposite effect occurs within the pulmonary vessels during a decrease in QT. As QT decreases, pulmonary vascular pressures decrease, the radii of the pulmonary vessels are reduced, and PVR consequently increases. The pulmonary vessels of patients with significant pulmonary hypertension are less distensible and act more like rigid pipes. In this setting, Ppa increases much more sharply with any increase in QT because PVR in these stiff vessels does not decrease significantly due to minimal expansion of their radii.
Understanding the relationships among Ppa, PVR, and QT during passive events is a prerequisite to recognition of active vasomotion in the pulmonary circulation (see Lung Volume). Active vasoconstriction occurs whenever QT decreases and Ppa either remains constant or increases. Increased Ppa and PVR have been found to be “a univer sal feature of acute respiratory failure.”19 Active pulmonary vasoconstriction can increase Ppa and Ppv, contributing to the formation of pulmonary edema, and in that way it has a role in the pathophysiol ogy of adult respiratory distress syndrome (ARDS). Active vasodilation occurs whenever QT increases and Ppa either remains constant or decreases. When deliber ate hypotension is achieved with sodium nitroprusside, QT often remains constant or increases, but Ppa decreases, and therefore so does PVR.
b. LUNG VOLUME Lung volume and PVR have an asymmetric, U-shaped relationship because of the varying effect of lung volume on small intra-alveolar and large extra-alveolar vessels, which in both cases is minimal at functional residual capacity (FRC). FRC is defined as the amount of volume (gas) in the lungs at end-exhalation during normal tidal breathing. Ideally, this means that the patient is inspiring a normal VT, with minimal or no muscle activity or pres sure difference between the alveoli and atmosphere at end-exhalation. Total PVR is increased when lung volume is either increased or decreased from FRC any blood flow to unventilated or shunted areas ( V /Q A = 0) or unperfused areas ( V /Q A = 8). Figure 5-6 also shows PAO2 and PACO2 in respiratory units with different V /Q A ratios. Within the 95% range of V /Q A ratios (i.e., 0.3 to 2.1), PO2 ranges from 60 to 123 mm Hg, whereas the corresponding PCO2 range is 44 to 33 mm Hg.
B. Nongravitational Determinants of Blood Flow Distribution 1. Passive Processes a. CARDIAC OUTPUT The pulmonary vascular bed is a high-flow, low-pressure system in health. As QT increases, pulmonary vascular pressures increase minimally.18 However, increases in QT distend open vessels and recruit previously closed vessels. Accordingly, pulmonary vascular resistance (PVR) drops because the normal pulmonary vasculature is quite dis tensible (and partly because of the addition of previously unused vessels to the pulmonary circulation). As a result of the distensibility of the normal pulmonary circulation, an increase in Ppa increases the radius of the pulmonary vessels, which causes PVR to decrease (Fig. 5-7). Conversely, Figure 5-6 A, Average distribution of ventilation-perfusion ratios ( V /Q A ) in normal, young, semirecumbent subjects. The 95% range (between dashed lines) is 0.3 to 2.1. B, Corresponding variations in partial pressures of oxygen (PO2) and carbon dioxide (PCO2) in alveo lar gas. (Redrawn from West JB: Blood flow to the lung and gas exchange. Anesthesiology 41:124, 1974.) 120 1.2 0.8 0.4 0 50 30 10 40 80 0.1 1.0 10.0 100.0 0.01 0.1 1.0 10.0 100.0 0.01 Ventilation of blood flow (L/min) Alveolar PO2 (mm Hg) Alveolar PCO2 (mm Hg) A B 95% Range Blood flow Ventilation PCO2 PO2 Figure 5-7 Passive changes in pulmonary vascular resistance (PVR) as a function of pulmonary artery pressure (Ppa) and pulmonary blood flow ( QT): PVR = Ppa/ QT. As QT increases, Ppa also increases, but to a lesser extent, and PVR decreases. As QT decreases, Ppa also decreases, but to a lesser extent, and PVR increases. (Redrawn with modification from Fishman AP: Dynamics of the pulmonary circu lation. In Hamilton WF, editor: Handbook of physiology. Section 2: Circulation, vol 2, Baltimore, 1963, Williams & Wilkins, p 1667.) 0 0 0 High High High Resistance (PVR) Pressure (Ppa) Ppa vs PVR QT vs PVR Flow (QT) • 124 PART 1 Basic Clinical Science Considerations spontaneous ventilation only—the negativity of perivas cular pressure at high lung volumes). The increase in total PVR below FRC results from an increase in the PVR of large extra-alveolar vessels (passive effect). The increase in large-vessel PVR is partly due to mechanical tortuosity or kinking of these vessels (passive effect). In addition, small or grossly atelectatic lungs become hypoxic, and it has been shown that the increased large-vessel PVR in these lungs is also caused by an active vasoconstrictive mechanism known as hypoxic pulmonary vasoconstric tion (HPV).24 The effect of HPV (discussed in greater detail in “Alveolar Gases”) is significant whether the chest is open or closed and whether ventilation is by positive pressure or spontaneous.25 2. Active Processes and Pulmonary Vascular Tone Four major categories of active processes affect the pul monary vascular tone of normal patients: (1) local tissue (endothelial- and smooth muscle–derived) autocrine or paracrine products, which act on smooth muscle (Table 5-1); (2) alveolar gas concentrations (chiefly hypoxia), which also act on smooth muscle; (3) neural influences; and (4) humoral (or hormonal) effects of circulating products within the pulmonary capillary bed. The neural and humoral effects work by means of either receptor-mediated mechanisms involving the autocrine/paracrine molecules listed in Table 5-1 or related mechanisms ulti mately affecting the smooth muscle cell.26 These four interrelated systems, each affecting pulmonary vascular tone, are briefly reviewed in sequence.
a. TISSUE (ENDOTHELIAL- AND SMOOTH MUSCLE–DERIVED) PRODUCTS The pulmonary vascular endothelium synthesizes, metab olizes, and converts a multitude of vasoactive mediators and plays a central role in the regulation of PVR. However, the main effecter site of pulmonary vascular tone is the pulmonary vascular smooth muscle cell (which both senses and produces multiple pulmonary vasoactive com pounds).27 The autocrine/paracrine molecules listed in Table 5-1 are all actively involved in the regulation of pulmonary vascular tone during various conditions. Numerous additional compounds bind to receptors on the endothelial or smooth muscle cell membranes and modulate the levels (and effects) of these vasoactive molecules.
Nitric oxide (NO) is the predominant endogenous vasodilatory compound. Its discovery by Palmer and Figure 5-8 Total pulmonary vascular resistance (PVR) relates to lung volume as an asymmetric, U-shaped curve. The trough of the curve occurs when lung volume equals functional residual capacity (FRC). Total PVR is the sum of the resistance in small vessels (increased by increasing lung volume [LV] and the resistance in large vessels (increased by decreasing LV). The end point for increasing LV toward total lung capacity (TLC) is the creation of zone 1 conditions, and the end point for decreasing LV toward residual volume (RV) is the creation of low ventilation-perfusion ( V Q A/ ) and atelectatic (atel) areas that demonstrate hypoxic pulmonary vasoconstriction (HPV). (Data fromr Bhavani-Shankar K, Hart NS, Mushlin PS: Negative pressure induced airway and pulmonary injury. Can J Anaesth 44:78, 1997; Berggren SM: The oxygen deficit of arterial blood caused by non-ventilating parts of the lung. Acta Physiol Scand Suppl 4:11, 1942; and Benumof JL: One lung ventilation: Which lung should be PEEPed? Anesthesiology 56:161, 1982.) 1.0 2.0 3.0 4.0 5.0 Lung volume (LV) atel, VA/Q TLC RV FRC Total PVR Zone 1 • • Pulmonary vascular resistance (PVR) HPV Large vessels Small vessels (Fig. 5-8).20–22 The increase in total PVR above FRC results from alveolar compression of small intra-alveolar vessels, which results in an increase in small-vessel PVR (i.e., creation of zone 1 or zone 2).23 As a relatively small mitigating or counterbalancing effect to the compression of small vessels, the large extra-alveolar vessels can be expanded by the increased tethering of interstitial connective tissue at high lung volumes (and with TABLE 5-1 Local Tissue (Autocrine/Paracrine) Molecules Involved in Active Control of Pulmonary Vascular Tone Molecule Subtype Site of Origin Site of Action Response Nitric oxide NO Endothelium Sm. muscle Vasodilation Endothelin ET-1 Endothelium Sm. muscle (ETA receptor) Endothelium (ETB receptor) Vasoconstriction Vasodilation Prostaglandin PGI2 Endothelium Endothelium Vasodilation Prostaglandin PGF2a Endothelium Sm. muscle Vasoconstriction Thromboxane TXA2 Endothelium Sm. muscle Vasoconstriction Leukotriene LTB4–LTE4 Endothelium Sm. muscle Vasoconstriction ETA receptor, ET-1 receptor located on the smooth muscle cell membrane; ETB receptor, ET-1 receptor located on the endothelial cell membrane; Sm. muscle, Pulmonary arteriole smooth muscle cell.
CHAPTER 5 Physiology of the Airway 125 contrast, thromboxane A2 and leukotriene B4 are elabo rated under pathologic conditions and are thought to be involved in the pathophysiology of pulmonary artery hypertension (PAH) associated with sepsis and reperfu sion injury.26 Therapeutically, epoprostenol has been used success fully to decrease PVR in patients with chronic PAH when infused or inhaled.40,41 Currently, the synthetic PGI2 (ilo prost) is the most commonly used inhaled eicosanoid for reduction of PVR in patients with PAH.41Although most patients with chronic PAH are unresponsive to an acute vasodilator challenge with short-acting agents such as epoprostenol, adenosine, or NO,42 long-term administra tion of epoprostenol has been shown to decrease PVR in these patients.43 Furthermore, some patients with previ ously severe PAH have been weaned from epoprostenol after long-term administration, with dramatically decreased PVR and improved exercise tolerance.42 The vascular remodeling required to provide such a dramatic reduction in PVR is probably the result of mechanisms besides simple local vasodilation, as predicted by Fishman in an editorial in 1998.44 One such mechanism that appears to be important is the increased clearance of ET-1 (a potent vasoconstrictor and mitogen) with long-term epoprostenol administration.45 b. ALVEOLAR GASES Hypoxia-induced vasoconstriction constitutes a funda mental difference between pulmonary vessels and all other systemic blood vessels (which vasodilate in the presence of hypoxia). Alveolar hypoxia of in vivo and in vitro whole lung, unilateral lung, lobe, or lobule of lung results in localized pulmonary vasoconstriction. This phe nomenon is widely referred to as HPV and was first described more than 65 years ago by Von Euler and Liljestrand.46 The HPV response is present in all mam malian species and serves as an adaptive mechanism for diverting blood flow away from poorly ventilated to better ventilated regions of the lung and thereby improv ing V /Q A ratios.47 The HPV response is also critical for fetal development because it minimizes perfusion of the unventilated lung.
The HPV response occurs primarily in pulmonary arterioles of about 200 µm internal diameter (ID) in humans (60 to 700 µm ID in other species).48 These vessels are advantageously situated anatomically in close relation to small bronchioles and alveoli, which permits rapid and direct detection of alveolar hypoxia. Indeed, blood may actually become oxygenated in small pul monary arteries because of the ability of O2 to diffuse directly across the small distance between the contiguous air spaces and vessels.49 This direct access that gas in the airways has to small arteries makes possible a rapid and localized vascular response to changes in gas composition.
The O2 tension at the HPV stimulus site (PsO2) is a function of both PAO2 and mixed venous O2 pressure (PvO2).50 The PsO2-HPV response is sigmoid, with a 50% response when PAO2, PvO2, and PsO2 are approximately 30 mm Hg. Usually, PAO2 has a much greater effect than PvO2 does because O2 uptake is from the alveolar space to the blood in the small pulmonary arteries.50 colleagues 25 years ago ended the long search for the so-called endothelium-derived relaxant factor (EDRF).28 Since then, a massive amount of laboratory and clinical research has demonstrated the ubiquitous nature of NO and its predominant role in vasodilation of both pulmo nary and systemic blood vessels.29 In the pulmonary endothelial cell, L-arginine is converted to L-citrulline by means of nitric oxide synthase (NOS) to produce the small, yet highly reactive NO molecule.30 Because of its small size, NO can diffuse freely across membranes into the smooth muscle cell, where it binds to the heme moiety of guanylate cyclase (which converts guanosine triphosphate to cyclic guanosine monophosphate [cGMP]).31 cGMP activates protein kinase G, which dephosphorylates the myosin light chains of pulmonary vascular smooth muscle cells and thereby causes vasodila tion.31 NOS exists in two forms: constitutive (cNOS) and inducible (iNOS). cNOS is permanently expressed in some cells, including pulmonary vascular endothelial cells, and produces short bursts of NO in response to changing levels of calcium and calmodulin and shear stress. The cNOS enzyme is also stimulated by linked membrane-based receptors that bind numerous mole cules in the blood (e.g., acetylcholine, bradykinin).31 In contrast, iNOS is usually produced only as a result of inflammatory mediators and cytokines and, when stimu lated, produces large quantities of NO for an extended duration.31 It is well known that NO is constitutively produced in normal lungs and contributes to the main tenance of low PVR.32,33 Endothelin-1 (ET-1) is a pulmonary vasoconstrictor.34 The endothelins are 21-amino-acid peptides that are pro duced by a variety of cells. ET-1 is the only family member produced in pulmonary endothelial cells, and it is also produced in vascular smooth muscle cells.34 ET-1 exerts its major vascular effects through activation of two dis tinct G protein–coupled receptors (ETA and ETB). ETA receptors are found in the medial smooth muscle layers of the pulmonary (and systemic) blood vessels and in atrial and ventricular myocardium.34 When stimulated, ETA receptors induce vasoconstriction and cellular pro liferation by increasing intracellular calcium.35 ETB receptors are localized on endothelial cells and some smooth muscle cells.36 Activation of ETB receptors stim ulates the release of NO and prostacyclin, thereby pro moting pulmonary vasodilation and inhibiting apoptosis.30 Bosentan, an ET-1 receptor antagonist, has produced modest improvement in the treatment of pulmonary hypertension.37 The more selective ETA receptor antago nist, sitaxsentan, showed additional benefit in improving pulmonary hypertension.38 However, both of these ET-1 receptor antagonists are associated with an increased risk of liver toxicity.39 In summary, it appears that there is a normal balance between NO and ET-1, with a slight predominance toward NO production and vasodilation in health.
Similarly, various eicosanoids are elaborated by the pulmonary vascular endothelium, with a balance toward the vasodilatory compounds in health. Prostaglandin I2 (PGI2), now known as epoprostenol (previously known as prostacyclin), causes vasodilation and is continuously elaborated in small amounts in healthy endothelium. In 126 PART 1 Basic Clinical Science Considerations hypoxia.57 However, alternative (less likely) mechanisms are still being investigated.58 One alternative hypothesis suggests that smooth muscle microsomal reduced nico tinamide adenine dinucleotide phosphate (NADPH) oxi doreductase or sarcolemmal NADPH oxidase is the sensing mechanism.58 Another, previously popular theory posited that voltage-sensitive potassium (KV) channels were required for the HPV response. However, KV chan nels are no longer believed to be obligate but instead are thought to be attenuators, because a study demonstrated that inhibition of KV channels failed to inhibit the HPV response.58 In summary, HPV probably results from a direct action of alveolar hypoxia on pulmonary smooth muscle cells, sensed by the mitochondrial electron transport chain, with reactive O2 species (probably H2O2 or superoxide) serving as second messengers to increase calcium and smooth muscle vasoconstriction. The endothelium-derived products serve to both potentiate (ET-1) and attenuate (NO, PGI2) the HPV response. Additional mechanisms (humoral and neurogenic influences) may also modulate the baseline pulmonary vascular tone and affect the magnitude of the HPV response.
Elevated PaCO2 has a pulmonary vasoconstrictor effect. Both respiratory acidosis and metabolic acidosis augment HPV, whereas respiratory and metabolic alkalosis cause pulmonary vasodilation and serve to reduce HPV.
The clinical effects of HPV in humans can be classified under three basic mechanisms. First, life at high altitude or whole-lung respiration of a low inspired concentration of O2 (FIO2) increases Ppa. This is true for newcomers to high altitude, for the acclimatized, and for natives.53 The vasoconstriction is considerable; in healthy people breath ing 10% O2, Ppa doubles whereas pulmonary wedge pres sure remains constant.59 The increased Ppa increases perfusion of the apices of the lung (through recruitment of previously unused vessels), which results in gas exchange in a region of lung not normally used (i.e., zone 1). Therefore, with a low FIO2, PAO2 is greater and the alveolar-arterial O2 tension difference and the ratio between dead space and tidal volume (VD/VT) are less than would be expected or predicted on the basis of a normal (sea level) distribution of ventilation and blood flow. High-altitude pulmonary hypertension is an impor tant component in the development of mountain sickness subacutely (hours to days) and cor pulmonale chronically (weeks to years).60 There is now good evidence that in both patients with chronic obstructive pulmonary disease (COPD) and those with obstructive sleep apnea (OSA), nocturnal episodes of arterial O2 desaturation (caused by episodic hypoventilation) are accompanied by elevations in Ppa that can eventually lead to sustained pulmonary hypertension and cor pulmonale.61 Second, hypoventilation (low V /Q A ratio), atelectasis, or nitrogen ventilation of any region of the lung usually causes a diversion of blood flow away from the hypoxic to the nonhypoxic lung (40% to 50% in one lung, 50% to 60% in one lobe, 60% to 70% in one lobule) (Fig. 5-10).62 The regional vasoconstriction and blood flow diversion are of great importance in minimizing transpul monary shunting and normalizing regional V /Q A ratios during disease of one lung, one-lung anesthesia (see Numerous theories have been developed to explain the mechanism of HPV.46,51–53 Many vasoactive sub stances have been proposed as mediators of HPV, including leukotrienes, prostaglandins, catecholamines, serotonin, histamine, angiotensin, bradykinin, and ET-1, but none has been identified as the primary mediator. In 1992, Xuan proposed that NO has a pivotal role in modulating PVR.54 NO is involved, but not precisely in the way that Xuan first proposed. There are multiple sites of O2 sensing with variable contributions from the NO, ET-1, and eicosanoid systems (described earlier). In vivo, HPV is currently thought to result from the syner gistic action of molecules produced in both endothelial cells and smooth muscle cells.55 However, HPV can proceed in the absence of intact endothelium, suggesting that the primary O2 sensor is in the smooth muscle cell and that endothelium-derived molecules modulate only the primary HPV response.
The precise mechanism of HPV is still under investiga tion. However, current data support a mechanism involv ing the smooth muscle mitochondrial electron transport chain as the HPV sensor (Fig. 5-9).56 In addition, reactive oxygen species (possibly H2O2 or superoxide) are released from complex III of the electron transport chain and probably serve as second messengers to increase calcium in pulmonary artery smooth muscle cells during acute Figure 5-9 Schematic model of the mitochondrial O2-sensing and effector mechanism probably responsible for hypoxic pulmonary vasoconstriction (HPV). In this model, reactive O2 species (ROS) are released from electron transport chain complex III and act as second messengers in the hypoxia-induced calcium (Ca2+) increase and resultant HPV. The solid arrows represent electron transfer steps; solid bars show sites of electron chain inhibition. Normal mitochon drial electron transport involves the movement of reducing equiva lents generated in the Krebs cycle through complex I or II and then through complex III (ubiquinone) and complex IV (cytochrome oxidase). The Q cycle converts the dual electron transfer in complex I and II into a single electron transfer step used in complex IV. The ubisemiquinone (a free radical) created in this process can gener ate superoxide, which in the presence of superoxide dismutase (SOD) produces H2O2, the probable mediator of the hypoxia-induced increased Ca2+ and HPV. This process is amplified during hypoxia. Diphenyleneiodonium (DPI), rotenone, and myxothiazol (not shown in figure) are inhibitors of the proximal portion of the electron transport chain. (From Waypa GB, Marks JD, Mack MM, et al: Mitochondrial reactive oxygen species trigger calcium increases during hypoxia in pulmonary artery myocytes. Circ Res 91:719, 2002.) Ubiquinol Complex I Complex II DPI, rotenone Antimycin A Myxothiazol SOD + HPV [Ca2+]i H2O2 H2O O2 O2 Ubisemiquinone Catalase Cyanide Superoxide Cytochrome oxidase Hypoxia Ubiquinone Fe-S C1 Cyt C CHAPTER 5 Physiology of the Airway 127 of the lung, thereby serving as an autoregulatory mecha nism that protects PaO2 by favorably adjusting regional V /Q A ratios. Factors that inhibit regional HPV are exten sively discussed elsewhere.64,65 c. NEURAL INFLUENCES ON PULMONARY VASCULAR TONE The three systems used to innervate the pulmonary cir culation are the same ones that innervate the airways: the sympathetic, parasympathetic, and nonadrenergic non cholinergic (NANC) systems.26 Sympathetic (adrenergic) fibers originate from the first five thoracic nerves and enter the pulmonary vessels as branches from the cervical ganglia, as well as from a plexus of nerves arising from the trachea and main stem bronchi. These nerves act mainly on pulmonary arteries down to a diameter of 60 µm.26 Sympathetic fibers cause pulmonary vasocon striction through α1-receptors. However, the pulmonary arteries also contain vasodilatory α2-receptors and β2-receptors. The α1-adrenergic response predominates during sympathetic stimulation, such as occurs with pain, fear, and anxiety.26 The parasympathetic (cholinergic) nerve fibers originate from the vagus nerve and cause pulmonary vasodilation through an NO-dependent process.26 Binding of acetylcholine to a muscarinic (M3) receptor on the endothelial cell increases intracellular calcium and stimulates cNOS.26 NANC nerves cause pul monary vasodilation through NO-mediated systems by using vasoactive intestinal peptide as the neurotransmit ter. The functional significance of this system is still under investigation.26 d. HUMORAL INFLUENCES ON PULMONARY VASCULAR TONE Numerous molecules are released into the circulation that either affect pulmonary vascular tone (by binding to pulmonary endothelial receptors) or are acted on by the pulmonary endothelium and subsequently become acti vated or inactivated (Table 5-2). The entire topic of non respiratory function of the lung is fascinating but beyond the scope of this chapter. Here, we highlight the effects that circulating molecules have on pulmonary vascular tone.
Although we understand the basic effect that various circulating factors have on pulmonary vascular tone, it is Chapter 26), inadvertent intubation of a main stem bron chus, and lobar collapse.
Third, in patients who have COPD, asthma, pneumo nia, or mitral stenosis but not bronchospasm, administra tion of pulmonary vasodilator drugs such as isoproterenol, sodium nitroprusside, or nitroglycerin inhibits HPV and causes a decrease in PaO2 and PVR and an increase in right-to-left transpulmonary shunting.63 The mechanism for these changes is thought to be deleterious inhibition of preexisting and, in some lesions, geographically wide spread HPV without concomitant and beneficial bron chodilation.63 In accordance with the latter two lines of evidence (one-lung or regional hypoxia and vasodilator drug effects on whole-lung or generalized disease), HPV is thought to divert blood flow away from hypoxic regions Figure 5-10 Schematic drawing of regional hypoxic pulmonary vasoconstriction (HPV); one-lung ventilation is a common clinical example of regional HPV. HPV in the hypoxic atelectatic lung causes redistribution of blood flow away from the hypoxic lung to the nor moxic lung, thereby diminishing the amount of shunt flow ( Q Q S T / ) that can occur through the hypoxic lung. Inhibition of hypoxic lung HPV causes an increase in the amount of shunt flow through the hypoxic lung, thereby decreasing the alveolar oxygen tension (PAO2). HPV I n hi bit io n of HP V QS / QT • • QS / QT • • Modified from Lumb AB: Non-respiratory functions of the lung. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 309.
TABLE 5-2 Effect of Compounds Passing Through Pulmonary Circulation Molecule Activated Unchanged Inactivated Amines Dopamine Epinephrine Histamine 5-Hydroxytryptamine Norepinephrine Peptides Angiotensin I Angiotensin II Oxytocin Vasopressin Bradykinin Atrial natriuretic peptide Endothelins Eicosanoids Arachidonic acid PGI2 PGA2 PGD2 PGE1, PGE2 PGF2a Leukotrienes Purine derivatives Adenosine ATP, ADP, AMP ADP, Adenosine diphosphate; AMP, adenosine monophosphate; ATP, adenosine triphosphate; PG, prostaglandin.
128 PART 1 Basic Clinical Science Considerations Consequently, as much as a 10% or a 5% obligatory right-to-left shunt may be present, respectively, under these conditions.
Intrapulmonary arteriovenous anastomoses are nor mally closed, but in the presence of acute pulmonary hypertension, such as may be caused by a pulmonary embolus, they may open and result in a direct increase in right-to-left shunting. The foramen ovale is patent (PFO) in 20% to 30% of individuals but it usually remains func tionally closed because left atrial pressure normally exceeds right atrial pressure. However, any condition that causes right atrial pressure to be greater than left atrial pressure may produce a right-to-left shunt, with resultant hypoxemia and possible paradoxical embolization. Such conditions include the use of high levels of PEEP, pulmo nary embolization, pulmonary hypertension, COPD, pul monary valvular stenosis, congestive heart failure, and postpneumonectomy states.68 Even such common events as mechanical ventilation and reaction to the presence of an endotracheal tube (ETT) during the excitement phase of emergence from anesthesia have caused right-to-left shunting across a PFO and severe arterial desaturation (with the potential for paradoxical embolization).69,70 Transesophageal echocardiography has been demon strated to be a sensitive modality for diagnosing a PFO in anesthetized patients with elevated right atrial pres sure.71 Esophageal to mediastinal to bronchial to pulmo nary vein pathways have been described and may explain in part the hypoxemia associated with portal hyperten sion and cirrhosis. There are no known conditions that selectively increase thebesian channel blood flow. (The besian vessels nourish the left ventricular myocardium and originate and empty into the left side of the heart.) C. Nongravitational Determinants of Pulmonary Compliance, Resistance, Lung Volume, Ventilation, and Work of Breathing 1. Pulmonary Compliance For air to flow into the lungs, a pressure gradient (ΔP) must be developed to overcome the elastic resistance of the lungs and chest wall to expansion. These structures are arranged concentrically, and their elastic resistance is therefore additive. The relationship between ΔP and the resultant volume increase (ΔV) of the lungs and thorax is independent of time and is known as total compliance (CT), as expressed in the following equation: C L/cm H O V L / P cm H O T ( ) ( ) ( ) 2 2 = ∆ ∆ (1) The CT of lung plus chest wall is related to the individual compliance of the lungs (CL) and of the chest wall (CCW) according to the following expression: 1 1 1 /C /C /C or C C C /C C T L CW T L CW L CW = + = + [ ( )( ) ] (2) Normally, CL and CCW each equal 0.2 L/cm H2O; hence, CT = 0.1 L/cm H2O. To determine CL, ΔV and the transpulmonary pressure gradient (PA − Ppl, the ΔP for the lung) must be known; to determine CCW, ΔV and the unlikely that these compounds are modulators of pulmo nary vascular tone in normal circumstances. However, they have marked effects during disease (e.g., ARDS, sepsis).
Endogenous catecholamines (epinephrine and norepi nephrine) bind to both α1- (vasoconstrictor) and β2-(vasodilator) receptors on the pulmonary endothelium, but when elaborated in high concentration, they have a predominant α1 (vasoconstrictor) effect. The same is true for exogenously administered catecholamines. Other amines (e.g., histamine, serotonin) are elaborated sys temically or locally after various challenges and have variable effects on PVR. Histamine can be released from mast cells, basophils, and elsewhere. When histamine binds directly to H1 receptors on endothelium, NO-mediated vasodilation occurs (as seen after epinephrine-induced pulmonary vasoconstriction). Direct stimulation of H2 receptors on smooth muscle cell membranes also causes vasodilation. In contrast, stimulation of H1 recep tors on the smooth muscle membrane results in vasocon striction. Serotonin (5-hydroxytryptamine) is a potent vasoconstrictor that can be elaborated from activated platelets (e.g., after pulmonary embolism) and can lead to acute severe pulmonary hypertension.66 Numerous peptides circulate and cause either pulmo nary vasodilation (e.g., substance P, bradykinin, vasopres sin [a systemic vasoconstrictor]) or vasoconstriction (e.g., neurokinin A, angiotensin). These peptides produce clini cally detectable effects on PVR only when administered in high concentration, such as with exogenous adminis tration or in disease.
Two other classes of molecules must be mentioned for completeness: eicosanoids (whose vasoactive effects were discussed earlier) and purine nucleosides (which are simi larly highly vasoactive).26 Adenosine is a pulmonary vaso dilator in normal subjects, whereas adenosine triphosphate (ATP) has a variable normalizing effect, depending on baseline pulmonary vascular tone.67 3. Alternative (Nonalveolar) Pathways of Blood Flow Through the Lung Blood can use several possible pathways to travel from the right side of the heart to the left without being fully oxygenated or oxygenated at all. Blood flow through poorly ventilated alveoli (regions of low V /Q A with an FIO2 <0.3 have a right-to-left shunt effect on oxygen ation) and blood flow through nonventilated alveoli (in atelectatic or consolidated regions, V /Q A = 0 at all FIO2 values) are sources of right-to-left shunting. Low- V /Q A and atelectatic lung units occur in conditions in which the FRC is less than the closing capacity (CC) of the lung (see “Lung Volumes, Functional Residual Capacity, and Closing Capacity”).
Several right-to-left blood flow pathways through the lungs and heart do not pass by or involve the alveoli at all. The bronchial and pleural circulations originate from systemic arteries and empty into the left side of the heart without being oxygenated; these circulations constitute the 1% to 3% true right-to-left shunt normally present. With chronic bronchitis, the bronchial circulation may carry 10% of the cardiac output, and with pleuritis, the pleural circulation may carry 5% of the cardiac output. CHAPTER 5 Physiology of the Airway 129 would be left (Fig. 5-11A). The second problem concerns the relationship between lung volume and transpulmo nary ΔP (PA − Ppl). Theoretically, the retractive forces of the lung should increase as lung volume decreases. If this were true, lung volume would decrease in a vicious circle, with an increasingly progressive tendency to collapse as lung volume diminishes.
These two problems are resolved by the fact that the surface tension of the fluid lining the alveoli is variable and decreases as its surface area is reduced. The surface tension of alveolar fluid can reach levels that are well below the normal range for body fluids such as water and plasma. When an alveolus decreases in size, the surface tension of the lining fluid falls to an extent greater than the corresponding reduction in radius; as a result, the transmural pressure gradient (equal to 2T/R) diminishes. This explains why small alveoli do not discharge their contents into large alveoli (see Fig. 5-11B) and why the elastic recoil of small alveoli is less than that of large alveoli.
The substance responsible for the reduction (and vari ability) in alveolar surface tension is secreted by the intra-alveolar type II pneumocyte; it is a lipoprotein called surfactant, which floats as a 50-Å-thick film on the surface of the fluid lining the alveoli. When the surface film is reduced in area and the concentration of surfactant at the surface is increased, the surface-reducing pressure is increased and counteracts the surface tension of the fluid lining the alveoli.
2. Airway Resistance For air to flow into the lungs, a pressure gradient (ΔP) must also be developed to overcome the nonelastic airway resistance (RAW) of the lungs to airflow. The RAW describes the relationship between ΔP and the rate of airflow ( V). R cm H O/L/sec P cm H O V L/sec ( ) ( ) ( ) 2 2 = ∆ ∆ (4) The ΔP along the airway depends on the caliber of the airway and the rate and pattern of airflow. There are three main patterns of airflow. Laminar flow occurs when the gas passes down parallel-sided tubes at less than a certain critical velocity. With laminar flow, the pressure drop down the tube is proportional to the flow rate and may be calculated from the equation derived by Poiseuille: ∆P V L / r = × × 8 4 µ π (5) where ΔP is the pressure drop (in cm H2O), V is the volume flow rate (in mL/sec), µ is viscosity (in poises), L is the length of the tube (in cm), and r is the radius of the tube (in cm).
When flow exceeds the critical velocity, it becomes turbulent. The significant feature of turbulent flow is that the pressure drop along the airway is no longer directly proportional to the flow rate but is proportional to the square of the flow rate according to equation (6) for turbulent flow: transmural pressure gradient (Ppl − Pambient, the ΔP for the chest wall) must be known; and to determine CT, ΔV and the transthoracic pressure gradient (PA − Pambient, the ΔP for the lung and chest wall together) must be known. In clinical practice, only CT is measured, which can be done dynamically or statically, depending on whether a peak or a plateau inspiratory ΔP (respectively) is used for the CT calculation.
During a positive- or negative-pressure inspiration of sufficient duration, transthoracic ΔP first increases to a peak value and then decreases to a lower plateau value. The peak transthoracic pressure value is the pressure required to overcome both elastic and airway resistance (see “Airway Resistance”). Transthoracic pressure decreases to a plateau value after the peak value because with time, gas is redistributed from stiff alveoli (which expand only slightly and therefore have only a short inspiratory period) into more compliant alveoli (which expand a great deal and therefore have a long inspiratory period). Because the gas is redistributed into more com pliant alveoli, less pressure is required to contain the same amount of gas, which explains why the pressure decreases. In practical terms, dynamic compliance is the volume change divided by the peak inspiratory transtho racic pressure, and static compliance is the volume change divided by the plateau inspiratory transthoracic pressure. Therefore, static CT is usually greater than dynamic CT, because the former calculation uses a smaller denomina tor (lower pressure) than the latter. However, if the patient is receiving PEEP, that pressure must first be subtracted from the peak or plateau pressure before tho racic compliance is calculated (i.e., compliance is equal to the volume delivered divided by the peak or plateau pressure − PEEP).
Alveolar pressure deserves special comment. The alveoli are lined with a layer of liquid. When a curved surface (a sphere or cylinder, such as the alveoli, bronchi oles, and bronchi) is lined with liquid, a surface tension is created that tends to make the surface area that is exposed to the atmosphere as small as possible. Simply stated, water molecules crowd much closer together on the surface of a curved layer of water than elsewhere in the fluid. As lung or alveolar size decreases, the degree of curvature and the retractive surface tension increase.
According to the Laplace expression, shown in equa tion (3), the pressure in an alveolus (P, in dynes per square centimeter) is higher than ambient pressure by an amount that depends on the surface tension of the lining liquid (T, in dynes per centimeter) and the radius of curvature of the alveolus (R, in centimeters). This rela tionship is expressed in the following equation: P T/R = 2 (3) Although surface tension contributes to the elastic resistance and retractive forces of the lung, two difficul ties must be resolved. First, the pressure inside small alveoli should be higher than that inside large alveoli, a conclusion that stems directly from the Laplace equation (R in the denominator). From this reasoning, one would expect a progressive discharge of each small alveolus into a larger one until eventually only one gigantic alveolus 130 PART 1 Basic Clinical Science Considerations diameter of the smaller bronchi and bronchioles are par ticularly critical, because bronchoconstriction may convert laminar flow to turbulent flow and the pressure drop along the airways can become much more closely related to the flow rate.
3. Different Regional Lung Time Constants Thus far, the compliance and airway resistance properties of the chest have been discussed separately. In the fol lowing analysis, pressure at the mouth is assumed to increase suddenly to a fixed positive value (Fig. 5-12) that overcomes both elastic and airway resistance and to be maintained at this value during inflation of the lungs.73 The ΔP required to overcome nonelastic airway resis tance is the difference between the fixed mouth pressure and the instantaneous height of the dashed line in Figure 5-12 and is proportional to the flow rate during most of the respiratory cycle.
The ΔP required to overcome nonelastic airway resis tance is maximal initially but then decreases exponen tially (see Fig. 5-12, A, hatched lines). The rate of filling therefore also declines in an approximately exponential manner. The remainder of the pressure gradient over comes the elastic resistance (the instantaneous height of the dashed line in Fig. 5-12A) and is proportional to the change in lung volume. The ΔP required to overcome elastic resistance is minimal initially but then increases exponentially, as does lung volume. Alveolar filling ceases (lung volume remains constant) when the pressure ∆P V pfL/ r = 2 2 5 4π (6) where ΔP is the pressure drop (in cm H2O), V is the volume flow rate (in mL/sec), ρ is the density of the gas (or liquid), f is a friction factor that depends on the roughness of the tube wall, and r is the radius of the tube (in cm).72 With increases in turbulent flow (or orifice flow, as described in the next para graph), ΔP increases much more than V and there fore RAW also increases more, as predicted by equation (4).
Orifice flow occurs at severe constrictions such as a nearly closed larynx or a kinked ETT. In these situations, the pressure drop is also proportional to the square of the flow rate, but density replaces viscosity as the impor tant factor in the numerator. This explains why a low-density gas such as helium diminishes the resistance to flow (by threefold in comparison with air) in severe obstruction of the upper airway.
Because the total cross-sectional area of the airways increases as branching occurs, the velocity of airflow decreases in the distal airways; laminar flow is therefore chiefly confined to the airways below the main bronchi. Orifice flow occurs at the larynx, and flow in the trachea is turbulent during most of the respiratory cycle. By examining the components that constitute each of the preceding airway pressure equations, one can see that many factors can affect the pressure drop down the airways during respiration. However, variations in Figure 5-11 Relationship between surface tension (T), alveolar radius (R), and alveolar transmural pressure (P). The left side of the diagrams shows the starting condition, the right side shows the expected result in alveolar size (using the Laplace equation to calculate the starting pressure). In A, the surface tension in the fluid lining both the large and the small alveolus is the same (no surfactant). Accordingly, the direc tion of gas flow is from the higher-pressure small alveolus to the lower-pressure large alveolus, which results in one large alveolus (Rfinal = ΣRinitial). B shows the expected changes in surface tension when surfactant lines the alveolus (less tension in the smaller alveolus). The direction of gas flow is from the larger to the smaller alveolus until the two are of equal size and are volume stable (RK). K, Constant; ΣR, sum of all individual radii. Rfinal = Σ Rinitial Surface tension (T) in both alveoli = 20 dyn/cm Surface tension = 20 dyn/cm Surface tension = 5 dyn/cm G a s f l o w G a s f l o w RK RK P = –– 2T R P = –– 2T R 0.010 cm 0.010 cm 0.005 cm 0.005 cm A B Pressure = 4,000 dyn/sq cm = 4.0 cm H2O 2 × 20 0.010 Pressure = 4,000 dyn/sq cm = 4.0 cm H2O 2 × 20 0.010 Pressure = 2,000 dyn/cm = 2.0 cm H2O 2 × 5 0.005 = = Pressure = 8,000 dyn/cm = 8.0 cm H2O 2 × 20 0.005 = CHAPTER 5 Physiology of the Airway 131 because a greater volume of air is transferred into a more compliant alveolus before the retractive force equals the applied pressure. The compliance of individual alveoli differs from top to bottom of the lung, and the resistance of individual airways varies widely depending on their length and caliber. Therefore, various time constants for inflation exist throughout the lung.
4. Pathways of Collateral Ventilation Collateral ventilation is another nongravitational deter minant of the distribution of ventilation. Four pathways of collateral ventilation are known. First, interalveolar communications (pores of Kohn) exist in most species; their number ranges from 8 to 50 per alveolus, and they may increase with age and with the development of obstructive lung disease. Their precise role has not been defined, but they probably function to prevent hypoxia in neighboring but obstructed lung units. Second, distal bronchiole-to-alveolus communications are known to exist (channels of Lambert); their function in vivo is speculative but may be similar to that of the pores of Kohn. Third, respiratory bronchiole–to–terminal bronchi ole connections have been found in adjacent lung seg ments (channels of Martin) in healthy dogs and in humans with lung disease. Fourth, interlobar connections exist; the functional characteristics of interlobar collateral ven tilation through these connections have been described in dogs,74 and they have been observed in humans as well.75 5. Work of Breathing The pressure-volume characteristics of the lung also determine the work of breathing (WOB). Because Work Force Distance Force Pressure Area Distance Volume/Area = × = × = (7) work is defined by the equation Work Pressure Area Volume/Area Pressure Volume = × = × ( )( ) (8) and ventilatory work may be analyzed by plotting pres sure against volume.76 In the presence of increased airway resistance or decreased CL, increased transpulmonary pressure is required to achieve a given VT with a conse quent increase in the WOB. The metabolic cost of the WOB at rest constitutes only 1% to 3% of the total O2 consumption in healthy subjects, but it is increased con siderably (up to 50%) in patients with pulmonary disease.
Two different pressure-volume diagrams are shown in Figure 5-13. During normal inspiration, transpulmonary pressure increases from 0 to 5 cm H2O while 500 mL of air is drawn into the lung. Potential energy is stored by the lung during inspiration and is expended during expi ration; as a consequence, the entire expiratory cycle is passive. The hatched area plus the triangular area ABC represents pressure multiplied by volume and is the WOB during one breath. Line AB is the lower section of the pressure-volume curve of Figure 5-13. The triangular resulting from the retractive elastic forces balances the applied (mouth) pressure (see Fig. 5-12A, dashed line).
Because only a finite time is available for alveolar filling and because alveolar filling occurs in an exponen tial manner, the degree of filling obviously depends on the duration of the inspiration. The rapidity of change in an exponential curve can be described by its time con stant τ, which is the time required to complete 63% of an exponentially changing function if the total time allowed for the function change is unlimited (2τ = 87%, 3τ = 95%, and 4τ = 98%). For lung inflation, τ = CT × R; normally, CT = 0.1 L/cm H2O, R = 2.0 cm H2O/L/sec, τ = 0.2 sec, and 3τ = 0.6 sec.
When this equation is applied to individual alveolar units, the time taken to fill such a unit clearly increases as airway resistance increases. The time required to fill an alveolar unit also increases as compliance increases, Figure 5-12 Artificial ventilation by intermittent application of con stant pressure (square wave) followed by passive expiration. The pressure required to overcome airway resistance (hatched lines in A) and the airflow rate (in C) from equation (4) in the text are pro portional to one another and decrease exponentially (assuming that resistance to airflow is constant). The pressure required to over come the elastic forces (height of the dashed line in A) and lung volume (B) are proportional to one another and increase exponen tially. Values shown are typical for an anesthetized supine paralyzed patient: total dynamic compliance, 50 mL/cm H2O; pulmonary resis tance, 3 cm H2O/L/sec; apparatus resistance, 7 cm H2O/L/sec; total resistance, 10 cm H2O/L/sec; time constant, 0.5 seconds. FRC, Func tional residual capacity. (Redrawn from Lumb AB: Artificial ventilation. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 590.) 10 0 5 10 15 20 25 30 0 10 10 10 0 Inspiration Expiration Pressure to overcome airflow resistance Alveolar pressure (cm H2O) Approximately equal areas Expiratory air flow inspiratory rate (L/sec) Lung volume above FRC (mL) Inflation or mouth pressure (cm H2O) Time (sec) Pressure gradient required to overcome airflow resistance (cm H2O) 0 500 0 0 10 20 30 1 0 1 C B A Pressure to overcome elastic forces 132 PART 1 Basic Clinical Science Considerations need to increase to achieve a normal Paco2. Similarly, when airway resistance (RAW) is increased, or compliance (CL) is decreased, there will be a corresponding increase in the WOB.
Furthermore, for any constant minute volume, the work done against elastic resistance is increased when breathing is deep and slow. On the other hand, the work done against airflow resistance is increased when breath ing is rapid and shallow. If the two components are summed and the total work is plotted against respiratory frequency, there is an optimal respiratory frequency at which the total WOB is minimal (Fig. 5-14).77 In patients with diseased lungs in which elastic resistance is high (e.g., pulmonary fibrosis, pulmonary edema, infants), the optimal frequency is increased, and rapid, shallow breaths are favored. As with other muscles, respiratory muscles can become fatigued, especially with rapid, shallow breathing.78 When airway resistance is high (e.g., asthma, obstructive lung disease), the optimal frequency is decreased, and slow, deep breaths are favored. Although the optimal frequency is slow (allowing a prolonged expi ratory phase), a rapid, shallow breathing pattern also develops in these patients when fatigued and further exacerbates their primary (airway resistance) problem.78 6. Lung Volumes, Functional Residual Capacity, and Closing Capacity a. LUNG VOLUMES AND FUNCTIONAL RESIDUAL CAPACITY FRC is defined as the volume of gas in the lung at the end of a normal expiration when there is no airflow and P A equals ambient pressure. Under these conditions, expansive chest wall elastic forces are exactly balanced by retractive lung tissue elastic forces (Fig. 5-15).79 The expiratory reserve volume is part of FRC; it is the additional gas beyond the end-tidal volume that can be area ABC is the work required to overcome elastic forces (CT), whereas the hatched area is the work required to overcome airflow or frictional resistance (R). The second graph applies to an anesthetized patient with diffuse obstructive airway disease resulting from the accumula tion of mucous secretions. There is a marked increase in both the elastic (triangle AB′C) and the airway (hatched area) resistive components of respiratory work. During expiration, only 250 mL of air leaves the lungs during the passive phase when intrathoracic pressure reaches the equilibrium value of 0 cm H2O. Active effort-producing work is required to force out the remaining 250 mL of air, and intrathoracic pressure actually becomes positive.
The full WOB over time must include the ventilatory frequency. The following equation depicts the variables included in the WOB equation: WOB V R C E AW L = × (9) Evaluating each component in the WOB equation, VE is the minute ventilation (rr × VT) required to achieve a normal Paco2. When patients have an increased CO2 pro duction (as occurs with fever) the VE, and hence the WOB, will need to be higher. When the dead space (either alveolar or anatomic) is increased, the VE will Figure 5-13 Lung volume plotted against transpulmonary pressure in a pressure-volume diagram for a healthy awake patient (Normal) and an anesthetized patient. The lung compliance of the awake patient (slope of line AB = 100 mL/cm H2O) equals that shown for the small dependent alveoli in Figure 5-3. The lung compliance of the anesthetized patient (slope of line AB′ = 50 mL/cm H2O) equals that shown for the medium midlung alveoli in Figure 5-3 and for the anesthetized patient in Figure 5-12. The total area within the oval and triangles has the dimensions of pressure multiplied by volume and represents the total work of breathing. The hatched areas to the right of lines AB and AB′ represent the active inspiratory work necessary to overcome resistance to airflow during inspiration (INSP). The hatched area to the left of the triangle AB′C represents the active expiratory work necessary to overcome resistance to airflow during expiration (EXP). Expiration is passive in the healthy subject because sufficient potential energy is stored during inspiration to produce expiratory airflow. The fraction of total inspiratory work nec essary to overcome elastic resistance is shown by the triangles ABC and AB′C. The anesthetized patient has decreased compliance and increased elastic resistance work (triangle AB′C) compared with the healthy patient’s compliance and elastic resistance work (triangle ABC). The anesthetized patient represented in this figure has increased airway resistance to both inspiratory and expiratory work. 3,000 2,750 2,500 0 5 0 5 10 Volume (mL) Transpulmonary pressure (cm H2O) Normal Anesthetized EXP EXP INSP INSP B C B′ C A A Figure 5-14 The diagrams show the work done against elastic and airflow resistance, both separately and summed to indicate the total work of breathing at different respiratory frequencies. The total work of breathing has a minimal value at approximately 15 breaths/min under normal circumstances. For the same minute volume, minimal work is performed at higher frequencies with stiff (less compliant) lungs and at lower frequencies when airflow resistance is increased. (Redrawn with modification from Lumb AB: Pulmonary ventilation: Mechanisms and the work of breathing. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 128.) 5 10 15 20 5 10 15 20 5 10 15 20 Respiratory frequency (breaths per minute) Work of breathing (arbitrary units) Total Elastic Total Elastic Airflow Total Elastic Airflow Normal Increased elastic resistance Increased airflow resistance Airflow CHAPTER 5 Physiology of the Airway 133 eliminated and the initial alveolar N2 concentration was 80%, the initial volume of the lung was 2.5 L. The second method, the inert gas dilution technique, uses the washin of an inert tracer gas such as helium. If 50 mL of helium is introduced into the lungs and, after equilibration, the helium concentration is found to be 1%, the volume of the lung is 5 L. The third method, the total-body pleth ysmography technique, uses Boyle’s law (P1V1 = P2V2, where P1= initial pressure, V1 = initial volume). The subject is confined within a gas-tight box (plethysmo graph) so that changes in the volume of the body during respiration may be readily determined as a change in pressure within the sealed box. Although each technique has technical limitations, all are based on sound physical and physiologic principles and provide accurate results in normal patients. Disparity between FRC as measured in the body plethysmograph and as determined by the helium dilution method is often used as a way of detect ing large, nonventilating air-trapped blebs.80 Obviously, there are difficulties in applying the body plethysmo graph to anesthetized patients.
Figure 5-15 A, The resting state of normal lungs when they are removed from the chest cavity; that is, elastic recoil causes total collapse. B, The resting state of a normal chest wall and diaphragm when the thoracic apex is open to the atmosphere and the thoracic contents are removed. C, The lung volume that exists at the end of expiration is the functional residual capacity (FRC). At FRC, the elastic forces of the lung and chest walls are equal and in opposite directions. The pleural surfaces link these two opposing forces. (Redrawn with modification from Shapiro BA, Harrison RA, Trout CA: The mechanics of ventilation. In Shapiro BA, Harrison RA, Trout CA, editors: Clinical application of respira tory care, ed 3, Chicago, 1985, Year Book, p 57.) A B C Figure 5-16 The dynamic lung volumes that can be measured by simple spirometry are tidal volume, inspiratory reserve volume, expiratory reserve volume, inspiratory capacity, and vital capacity. The static lung volumes are residual volume, functional residual capacity, and total lung capacity. Static lung volumes cannot be measured by simple spirometry and require separate methods of measurement (e.g., inert gas dilution, nitrogen washout, total-body plethysmography). 5.0 3.0 2.5 1.25 0 Volume (L) Inspiratory reserve volume Functional residual capacity Tidal volume Residual volume Expiratory reserve volume Total lung capacity Vital capacity Inspiratory capacity consciously exhaled, resulting in the minimum volume of lung possible, known as residual volume. Therefore, FRC equals residual volume plus expiratory reserve volume (Fig. 5-16). With regard to the other lung volumes shown in Figure 5-16, VT, vital capacity, inspiratory capac ity, inspiratory reserve volume, and expiratory reserve volume can be measured by simple spirometry. Total lung capacity (TLC), FRC, and residual volume contain a frac tion (residual volume) that cannot be measured by simple spirometry. However, if one of these three volumes is measured, the others can easily be derived, because the other lung volumes, which relate these three volumes to one another, can be measured by simple spirometry.
Residual volume, FRC, and TLC can be measured by any of three techniques: (1) nitrogen washout, (2) inert gas dilution (e.g., helium washin), and (3) total-body plethysmography. The first method, the nitrogen washout technique, is based on measuring expired nitrogen con centrations before and after the patient breathes pure O2 for several minutes; the difference is the total quantity of nitrogen eliminated. If, for example, 2 L of N2 is 134 PART 1 Basic Clinical Science Considerations parenchymal pressure or Ppl; that point is termed the equal pressure point (EPP). If the EPP occurs in small intrathoracic air passages (distal to the 11th generation, the airways have no cartilage and are called bronchioles), they may be held open at that particular point by the tethering effect of the elastic recoil of the immediately adjacent or surrounding lung parenchyma. If the EPP occurs in large extrathoracic air passages (proximal to the 11th generation, the airways have cartilage and are called bronchi), they may be held open at that particular point by their cartilage. Downstream of the EPP (in either small or large airways), transmural ΔP is reversed (−6 cm H2O), and airway closure occurs. Thus, the patency of airways distal to the 11th generation is a function of lung volume, and the patency of airways proximal to the 11th generation is a function of intrathoracic (pleural) pres sure. In extrathoracic bronchi with cartilage, the posterior membranous sheath appears to give first by invaginating into the lumen.81 If lung volume were abnormally decreased (e.g., because of splinting) and expiration were still forced, the caliber of the airways would be relatively reduced at all times, which would cause the EPP and point of collapse to move progressively from larger to smaller air passages (closer to the alveolus).
In adults with normal lungs, airway closure can still occur even if exhalation is not forced, provided that residual volume is approached closely enough. Even in patients with normal lungs, as lung volume decreases toward residual volume during expiration, small airways b. AIRWAY CLOSURE AND CLOSING CAPACITY As discussed earlier (see “Distribution of Ventilation”), Ppl increases from the top to the bottom of the lung and determines regional alveolar size, compliance, and venti lation. Of even greater importance to the anesthesiologist is the recognition that these gradients in Ppl may lead to airway closure and collapse of alveoli.
PATIENT WITH NORMAL LUNGS. Figure 5-17A illustrates the normal resting end-expiratory (FRC) position of the lung–chest wall combination. The distending transpul monary ΔP and the intrathoracic air passage transmural ΔP are 5 cm H2O, and the airways remain patent. During the middle of a normal inspiration (see Fig. 5-17B), there is an increase in transmural ΔP (to 6.8 cm H2O) that encourages distention of the intrathoracic air passages. During the middle of a normal expiration (see Fig. 5-17C), expiration is passive; PA is attributable only to the elastic recoil of the lung (2 cm H2O), and there is a decrease (to 5.2 cm H2O) but still a favorable (distend ing) intraluminal transmural ΔP. During the middle of a severe forced expiration (see Fig. 5-17D), Ppl increases far higher than atmospheric pressure and is communi cated to the alveoli, which have a pressure that is still higher because of the elastic recoil of the alveolar septa (an additional 2 cm H2O).
At high gas flow rates, the pressure drop down the air passage is increased, and there is a point at which intraluminal pressure equals either the surrounding Figure 5-17 Pressure gradients across the airways. The airways consist of a thin-walled intrathoracic portion (near the alveoli) and a more rigid (cartilaginous) intrathoracic and extrathoracic portion. During expiration, the pressure from elastic recoil is assumed to be +2 cm H2O in normal lungs (A through D) and +1 cm H2O in abnormal lungs (E and F). The total pressure inside the alveolus is pleural pressure plus elastic recoil. The arrows indicate the direction of airflow. EPP, Equal pressure point. See text for explanation. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saun ders, Chapter 8.) A B C D E F –5 –5 –5 –5 –5 –5 0 0 0 0 0 0 –7 –7 –7 –0.2 –1 +0.2 +2 +2 +4 EPP +10 +10 +10 +10 +10 +4 +4 +4 +10 EPP +12 +4 +4 +4 +4 +4 +4 +4 +5 0 0 End expiration Normal Lung Middle inspiration Normal Lung Middle passive expiration Normal Lung Middle forced expiration Normal Lung Mild forced expiration Emphysema Mild forced expiration Emphysema Larynx Partly Closed +4 +4 +5 0 +3 EPP CHAPTER 5 Physiology of the Airway 135 A differential tracer gas concentration is thus estab lished, with the gas in the apices having a higher tracer concentration than that in the bases (see Fig. 5-18). As the subject exhales and the diaphragm ascends, a point is reached at which the small airways just above the diaphragm start to close and thereby limit airflow from these areas. The airflow now comes more from the upper lung fields, where the alveolar gas has a much higher tracer concentration, which results in a sudden increase in the tracer gas concentration toward the end of exhala tion (phase IV of Fig. 5-18).
Closing volume (CV) is the difference between the onset of phase IV and residual volume; because it repre sents part of a vital capacity maneuver, it is expressed as a percentage of vital lung capacity. CV plus residual volume is known as CC and is expressed as a percentage of TLC. Smoking, obesity, aging, and the supine position increase CC.90 In healthy individuals at a mean age of 44 years, CC = FRC in the supine position, and at a mean age of 66 years, CC = FRC in the upright position.91 RELATIONSHIP BETWEEN FUNCTIONAL RESIDUAL CAPACITY AND CLOSING CAPACITY. The relationship between FRC and CC is far more important than consideration of FRC or CC alone, because it is this relationship that determines whether a given respiratory unit is normal or atelectatic (0.5 to 0.9 mm in diameter) show a progressive tendency to close, whereas larger airways remain patent.82,83 Airway closure occurs first in the dependent lung regions (as directly observed by computed tomography) because the distending transpulmonary pressure is less and the volume change during expiration is greater.32 Airway closure is most likely to occur in the dependent regions of the lung whether the patient is in the supine or the lateral decu bitus position and whether ventilation is spontaneous or positive-pressure ventilation.32,84,85 PATIENTS WITH ABNORMAL LUNGS. Airway closure occurs with milder active expiration, lower gas flow rates, and higher lung volumes, and it occurs closer to the alveolus in patients with emphysema, bronchitis, asthma, or inter stitial pulmonary edema. In all four conditions, the increased airway resistance causes a larger decrease in pressure from the alveoli to the larger bronchi, thereby creating the potential for negative intrathoracic transmu ral ΔP and narrowed and collapsed airways. In addition, the structural integrity of the conducting airways may be diminished because of inflammation and scarring, and therefore these airways may close more readily for any given lung volume or transluminal ΔP.
In emphysema, the elastic recoil of the lung is reduced (to 1 cm H2O in Fig. 5-17 E), the air passages are poorly supported by the lung parenchyma, the point of airway resistance is close to the alveolus, and transmural ΔP can become negative quickly. Therefore, during only a mild forced expiration in an emphysematous patient, the EPP and the point of collapse are near the alveolus (see Fig. 5-17E). The use of pursed-lip or grunting expiration (the equivalents of partly closing the larynx during expira tion), PEEP, and continuous positive airway pressure in an emphysematous patient restores a favorable (distend ing) intrathoracic transmural air ΔP (see Fig. 5-17F). In bronchitis, the airways are structurally weakened and may close when only a small negative transmural ΔP is present (as with mild forced expiration). In asthma, the middle-sized airways are narrowed by bronchospasm, and if expiration is forced, they are further narrowed by a negative transmural ΔP. Finally, with pulmonary intersti tial edema, perialveolar interstitial edema compresses the alveoli and acutely decreases FRC; the peribronchial edema fluid cuffs (within the connective tissue sheaths around the larger arteries and bronchi) compress the bronchi and acutely increase closing volume.86–88 MEASUREMENT OF CLOSING CAPACITY. CC is a sensitive test of early small-airways disease and is performed by having the patient exhale to residual volume (Fig. 5-18).89 As inhalation from residual volume toward TLC is begun, a bolus of tracer gas (e.g., xenon 133, helium) is injected into the inspired gas. During the initial part of this inhala tion from residual volume, the first gas to enter the alveo lus is the VD gas and the tracer bolus. The tracer gas enters only alveoli that are already open (presumably the apices of the lung) and does not enter alveoli that are already closed (presumably the bases of the lung). As the inhalation continues, the apical alveoli complete filling and the basilar alveoli begin to open and fill, but with gas that does not contain any tracer gas.
Figure 5-18 Measurement of closing capacity (CC) with the use of a tracer gas such as xenon 133 (133Xe). The bolus of tracer gas is inhaled near residual volume (RV) and, because of airway closure in the dependent lung, is distributed only to nondependent alveoli whose air passages are still open. During expiration, the concentra tion of tracer gas becomes constant after the dead space is washed out. This plateau (phase III) gives way to a rising concentration of tracer gas (phase IV), when there is once again closure of the dependent airways because the only contribution made to expired gas is by the nondependent alveoli with a high 133Xe concentration. FRC, Functional residual capacity; TLC, total lung capacity. (Redrawn with modification from Lumb AB: Respiratory system resistance: Mea surement of closing capacity. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 79.) Total lung capacity Closing capacity Residual volume Inhalation of bolus of 133Xe Concentration of 133Xe of mouth Total lung capacity Spirogram FRC Residual volume Vital capacity Time Closing capacity Phase III Phase IV TLC Vital capacity FRC CC RV 136 PART 1 Basic Clinical Science Considerations reversed with IPPB alone but has been reversed with IPPB plus PEEP (5 to 10 cm H2O).32 D. Oxygen and Carbon Dioxide Transport 1. Alveolar and Dead Space Ventilation and Alveolar Gas Tensions In normal lungs, approximately two thirds of each breath reaches perfused alveoli to take part in gas exchange. This constitutes the effective or alveolar ventilation ( VA). The remaining third of each breath takes no part in gas exchange and is therefore termed the total (or physio logic) dead space ventilation (VD). The relationship is as follows: alveolar ventilation ( VA) = frequency (f) (VT − VD). The physiologic (or total) dead space ventilation (VDphysiologic) may be further divided into two compo nents: a volume of gas that ventilates the conducting airways, the anatomic dead space (VDanatomic), and a volume of gas that ventilates unperfused alveoli, the alveolar dead space (VDalveolar). Clinical examples of VDalveolar ventilation include zone 1, pulmonary embolus, and destroyed alveolar septa; such ventilation does not participate in gas exchange. Figure 5-21 shows a two-compartment model of the lung in which the anatomic and alveolar dead space compartments have been com bined into the total (physiologic) dead space compart ment; the other compartment is the alveolar ventilation compartment, whose idealized V /Q A ratio is 1.0.
The VDanatomic varies with lung size and is approxi mately 2 mL/kg of body weight (150 mL in a 70-kg adult). In a normal healthy adult lying supine, VDanatomic and total VD are approximately equal to each other, because VDalveolar is normally minimal. In the erect posture, the uppermost alveoli may not be perfused (zone 1), and VDalveolar may increase from a negligible amount to 60 to 80 mL. Figure 5-21 illustrates that in a steady state, the volume of CO2 entering the alveoli ( VCO2) is equal to the volume of CO2 eliminated in the expired gas, ( VE) or has a low V /Q A ratio. The relationship between FRC and CC is as follows. When the volume of the lung at which some airways close is greater than the whole of VT, lung volume never increases enough during tidal inspiration to open any of these airways. As a result, these airways stay closed during the entire tidal respiration. Airways that are closed all the time are equivalent to atelectasis (Fig. 5-19). If the CV of some airways lies within VT, as lung volume increases during inspiration, some previously closed airways open for a short time until lung volume recedes once again below the CV of these airways. Because these opening and closing airways are open for a shorter time than normal airways are, they have less chance or time to participate in fresh gas exchange, a circumstance equivalent to a low- V /Q A region. If the CC of the lung is below the whole of tidal respiration, no airways are closed at any time during tidal respiration; this is a normal circumstance. Anything that decreases FRC relative to CC or increases CC relative to FRC converts normal areas to low- V /Q A and atelectatic areas,83 which causes hypoxemia.
Mechanical intermittent positive-pressure breathing (IPPB) may be efficacious because it can take a previously spontaneously breathing patient with a low- V /Q A rela tionship (in which CC is greater than FRC but still within VT, as depicted in Figure 5-20, right panel) and increase the amount of inspiratory time that some previously closed (at end-exhalation) airways spend in fresh gas exchange, thereby increasing V /Q A (see Fig. 5-20, middle panel). However, if PEEP is added to IPPB, PEEP increases FRC to a lung volume equal to or greater than CC, thereby restoring a normal FRC-to-CC relationship so that no airways are closed at any time during the tidal respiration depicted in Figure 5-20 (left panel) (IPPB + PEEP). Thus, anesthesia-induced atelectasis (identified by crescent-shaped densities on computed tomography) in the dependent regions of patients’ lungs has not been Figure 5-19 Relationship between functional residual capacity (FRC) and closing capacity (CC). FRC is the amount of gas in the lungs at end-exhalation during normal tidal breathing, shown by the level of each trough of the sine wave tidal volume. CC is the amount of gas that must be in the lungs to keep the small conducting airways open. This figure shows three different CCs, as indicated by the three different straight lines. See the text for an explanation of why the three different FRC-CC relationships depicted result in normal or low ventilation-perfusion ( V /Q A ) relationships or atelecta sis. The abscissa is time. (Redrawn from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) 100 0 Atelectasis (CC >> FRC) Time Normal (FRC > CC) Total lung capacity (%) Low VA/Q (CC > FRC) FRC • • FRC Figure 5-20 Relationship of functional residual capacity (FRC) to closing capacity (CC) during spontaneous ventilation, intermittent positive-pressure breathing (IPPB), and IPPB with positive end-expiratory pressure (IPPB + PEEP). See the text for an explanation of the effect of the two ventilatory maneuvers (IPPB and PEEP) on the relationship of FRC to CC. The abscissa is time. 100 0 Spontaneous ventilation IPPB IPPB + PEEP Closing capacity Total lung capacity (%) Time CHAPTER 5 Physiology of the Airway 137 PaCO2. If PaCO2 is to remain constant while VD/VT increases, VE must increase more.
The alveolar concentration of a gas is equal to the dif ference between the inspired concentration and the ratio of the output (or uptake) of the gas to VA. Thus, for gas X during dry conditions, PAX = (Pdry atm) (FIX) ± VX (output or uptake)/ VA, where PAX = alveolar partial pressure of gas X, FIX = inspired concentration of gas X, Pdry atm = dry atmospheric pressure = Pwet atm − PH2O = 760 − 47 = 713 mm Hg, VX = output or uptake of gas X, and VA = alveolar ventilation.
For CO2, PACO2 = 713(FICO2 + V /V CO A 2 ). Because FICO2 = 0 and using standard conversion factors: P V mL/min STPD /V L/min/BTPS ACO CO A 2 2 713 0 863 11 = [ ( ) ( )( .
)] ( ) where BTPS = body temperature and pressure, satu rated (i.e., 37° C, PH2O = 47 mm Hg) and STPD = standard temperature and pressure, dry. For example, 36 mm Hg = (713)(200/4000).
For O2, P F V mL/min /V mL/min) AO IO O A 2 2 2 713 12 = − [ ( ) ( ] ( ) (FECO2), where VE = minute ventilation and FECO2 = frac tion of expired CO2. Thus, V V )(F ) CO E ECO 2 = ( 2 . However, the expired gas volume consists of alveolar gas, ( VA)(FACO2), and VD gas, ( VD)(FICO2), where FACO2 and FICO2 are the alveolar and inspired fractions of CO2, respectively. Thus, V V )(F ) V )(F ) CO A ACO D ICO 2 = + ( ( 2 2 . Setting the first equation equal to the second equation and using the relationship, V V V E A D = + , subsequent algebraic manipulation, including setting PACO2 equal to PaCO2, results in the modified Bohr equation: V /V Pa P /Pa D T CO ECO CO = − ( ) ( ) 2 2 2 10 The CO2 tension in expired gas, PECO2, may be obtained by measuring exhaled CO2 in a large (Douglas) bag or, more commonly, by using end-tidal CO2 tension (PETCO2) as a surrogate. In severe lung disease, physiologic VD/VT provides a useful expression of the inefficiency of ventila tion. In a healthy adult, this ratio is usually less than 30%; that is, ventilation is more than 70% efficient. In a patient with COPD, VD/VT may increase to 60% to 70%. Under these conditions, ventilation is obviously grossly ineffi cient. Figure 5-22 shows the relationship between VE and PaCO2 for several VD/VT values. As VE decreases, PaCO2 increases for all VD/VT values. As VD/VT increases, a given decrease in VE causes a much greater increase in Figure 5-21 Two-compartment model of the lung in which the ana tomic and alveolar dead space compartments have been com bined into the total (physiologic) dead space ( VD). FACO2 = alveolar CO2 fraction; FECO2 = mixed expired CO2 fraction; FICO2 = inspired CO2 fraction; VA = alveolar ventilation; VCO2 = carbon dioxide production; VE = expired minute ventilation. V Q A/ = 1 means that ventilation and perfusion are equal in liters per minute. Normally, the amount of CO2 eliminated at the airway ( VE × FECO2) equals the amount of CO2 removed by alveolar ventilation ( VA × FACO2) because there is no CO2 elimination from alveolar dead space (FICO2 = 0). Anatomic dead space Alveolar dead space VE FECO2 FACO2 (FICO2) • VA • VD = Total dead space = Anatomic + alveolar dead space VA/Q = 1 VCO2 • • • VD • Figure 5-22 Relationship between minute ventilation ( VE, L/min) and arterial partial pressure of carbon dioxide (PaCO2) for a family of ratios of total dead space to tidal volume (VD/VT). These curves are hyperbolic and rise steeply at low VE values. See equation (10) in the text. 80 70 60 50 40 30 20 10 5 10 15 20 30 25 PaCO2 (mm Hg) VD/VT (%) VE (L/min) • 80 60 40 20 138 PART 1 Basic Clinical Science Considerations For example, 100 mm Hg = 713(0.21 - 225/3200).
Figure 5-23 shows the hyperbolic relationships expressed in equations (10) and (11) between PaCO2 and VA (see Fig. 5-22) and between PAO2 and VA for differ ent levels of VCO2 and VO2, respectively. PaCO2 is sub stituted for PACO2 because PACO2-to-PaCO2 gradients are small (as opposed to PAO2-to-Pao2 gradients, which can be large). Note that as VA increases, the second term on the right side of equations (11) and (12) approaches zero and the composition of the alveolar gas approaches that of the inspired gas. In addition, Figures 5-22 through Figure 5-24 show that, because anesthesia is usually administered with an oxygen-enriched gas mixture, hypercapnia is a more common result of hypoventilation than hypoxemia is.
2. Oxygen Transport a. OVERVIEW The principal function of the heart and lungs is support ing O2 delivery to and CO2 removal from the tissues in accordance with metabolic requirements while maintain ing arterial blood O2 and CO2 partial pressures within a Figure 5-23 Top, The relationship between alveolar ventilation and arterial carbon dioxide tension (PaCO2) for a group of different CO2 production values ( VCO2). Bottom, The relationship between alveo lar ventilation and alveolar oxygen tension (PAo2) for a group of different O2 consumption values ( VO2). Values are derived from equations (10) and (11) in the text, and the curves are hyperbolic. As alveolar ventilation increases, PAO2 and PaCO2 approach inspired concentrations. Decreases in alveolar ventilation to less than 4 L/min are accompanied by precipitous decreases in PAO2 and increases in PaCO2. VO2 VCO2 150 100 50 0 2 0 4 6 8 12 10 60 80 40 20 0 Alveolar ventilation (L/min) Alveolar PO2 (mm Hg) PO2 of inspired gas (21% oxygen) Arterial PCO2 (mm Hg) • • 300 260 220 180 140 100 100 200 400 800 narrow range. The respiratory and cardiovascular systems are linked in series to accomplish this function over a wide range of metabolic requirements, which may increase 30-fold from rest to heavy exercise. The func tional links in the O2 transport chain are as follows: (1) ventilation and distribution of ventilation with respect to perfusion, (2) diffusion of O2 into blood, (3) chemical reaction of O2 with Hb, (4) total cardiac output of arterial blood, and (5) distribution of blood to tissues and release of O2 (Table 5-3). The system is seldom stressed except at exercise, and the earliest symptoms of cardiac or respi ratory diseases are often seen only during exercise.
The maximum functional capacity of each link can be determined independently. Table 5-3 lists these measured functional capacities for healthy, young men. Because theoretical maximal O2 transport at the ventilatory step or at the diffusion and chemical reaction step (approxi mately 6 L/min in healthy humans at sea level) exceeds the O2 transportable by the maximum cardiac output and distribution steps, the limit to O2 transport is the cardiovascular system. Respiratory diseases would not be expected to limit maximum O2 transport until functional capacities are reduced by 40% to 50%.
b. OXYGEN-HEMOGLOBIN DISSOCIATION CURVE As a red blood cell (RBC) passes by the alveolus, O2 dif fuses into plasma and increases PAO2. As PAO2 increases, O2 diffuses into the RBC and combines with Hb. Each Hb molecule consists of four heme molecules attached to a globin molecule. Each heme molecule consists of glycine, α-ketoglutaric acid, and iron in the ferrous (Fe2+) form. Each ferrous ion has the capacity to bind with one Figure 5-24 For any given O2 concentration in inspired gas, the relationship between alveolar ventilation and alveolar O2 tension (PAO2) is hyperbolic. As the inspired O2 concentration is increased, the amount that alveolar ventilation must decrease to produce hypoxemia is greatly increased. BTPS, Body temperature, ambient pressure, saturated. (Redrawn from Lumb AB: Respiratory system resis tance: Measurement of closing capacity. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 79.) 2 4 6 8 10 0 150 100 50 0 20 25 15 5 10 0 Alveolar ventilation (L/min) (BTPS) Alveolar PO2 (mm Hg) Alveolar PO2 (kPa) O2 concentration in inspired gas 50% 40% 30% 25% 21% 15% 10% CHAPTER 5 Physiology of the Airway 139 The saturation at normal arterial pressure (point a on upper, flat part of the oxy-Hb curve in Figure 5-25) is 95% to 98%, achieved by a PaO2 of about 90 to 100 mm Hg. When PO2 is less than 60 mm Hg (90% saturation), saturation falls steeply, and the amount of Hb uncombined with O2 increases greatly for a given decrease in PO2. Mixed venous blood has a PO2 (PvO2) of about 40 mm Hg and is approximately 75% saturated, as indi cated by the middle of the three points (v) on the oxy-Hb curve in Figure 5-25.
The oxy-Hb curve can also relate the O2 content (CO2) (vol%, or mL of O2 per dL of blood; see Fig. 5-25) to PO2. Oxygen is carried both in solution in plasma (0.003 mL of O2/mm Hg PO2 per dL) and combined with Hb (1.39 mL of O2/g of Hb), to the extent (percentage) that Hb is saturated. Therefore, C 139 Hb percent saturation 3 P O O 2 2 = + ( .
)( )( ) .
( ) ( ) 0 00 13 For a patient with an Hb content of 15 g/dL, a PAO2 of 100 mm Hg, and a PvO2 of 40 mm Hg, the arterial O2 content (CaO2) = (1.39)(15)(1) + (0.003)(100) = 20.9 + 0.3 = 21.2 mL/dL; the mixed venous O2 content (CvO2) = (1.39)(15)(0.75) + (0.003)(40) = 15.6 + 0.1 = 15.7 mL/dL. Therefore, the normal arteriovenous O2 content difference is approximately 5.5 mL/dL of blood.
Note that equation (13) uses the constant 1.39, which means that 1 g of Hb can carry 1.39 mL of O2. Contro versy exists over the magnitude of this number. Origi nally, 1.34 had been used,92 but with determination of the molecular weight of Hb (64,458), the theoretical Figure 5-25 Oxygen-hemoglobin dissociation curve. Four different ordinates are shown as a function of oxygen partial pressure (the abscissa). In order from right to left, they are arterial O2 saturation (%), O2 content (mL of O2/dL of blood), O2 supply to peripheral tissues (mL/min), and O2 available to peripheral tissues (mL/min), which is O2 supply minus the approximately 200 mL/min that cannot be extracted below a partial pressure of 20 mm Hg. Three points are shown on the curve: a, normal arterial pressure; v , normal mixed venous pressure; P50, the partial pressure (27 mm Hg) at which hemoglobin is 50% saturated. 10 30 40 50 60 70 80 100 200 160 120 80 1000 800 700 600 500 400 300 100 0 800 600 400 200 0 10 30 50 70 90 110 40 20 Oxygen partial pressure (mm Hg) Available mL/min Supply mL/min Content mL/dL Saturation (%) Arterial oxygen a v P50 0 0 200 O2 molecule in a loose, reversible combination. As the ferrous ions bind to O2, the Hb molecule begins to become saturated.
The oxy-Hb dissociation curve relates the saturation of Hb (rightmost y-axis in Fig. 5-25) to PAO2. Hb is fully saturated (100%) by a PO2 of approximately 700 mm Hg. From Cassidy SS: Heart-lung interactions in health and disease. Am J Med Sci 30:451–461, 1987.
TABLE 5-3 Functional Capacities and Potential Maximum O2 Transport of Each Link in the O2 Transport Chain in Normal Humans at Sea Level Link in Chain Functional Capacity in Normal Humans Theoretical Maximum O2 Transport Capacity Ventilation 200 L/min (MVV) 0.030 × MVV = 6.0 L O2/min Diffusion and chemical reaction Dlo2 = 6.1 L O2/min Cardiac output 20 L/min O2 extraction 75% 0.16 × Cardiac output = 3.2 L O2/min (Cao2 − CvO2 difference) (16 mL O2/100 mL or 0.16) Hemoglobin = 15 g/dL; physiologic dead space in percentage of tidal volume = 0.25; partial alveolar pressure of oxygen >110 mm Hg.
Cao2 − CvO2 , Arteriovenous O2 content difference; Dlo2, diffusing capacity of lung for oxygen; MVV, maximum voluntary ventilation.
140 PART 1 Basic Clinical Science Considerations tissue perfusion than normal to produce the normal amount of O2 unloading. Causes of a right-shifted oxy-Hb curve are acidosis (metabolic and respiratory—the Bohr effect), hyperthermia, abnormal Hb, increased RBC 2,3-DPG content, and inhaled anesthetics (see later dis cussion).95 Abnormalities in acid-base balance result in alteration of 2,3-DPG metabolism to shift the oxy-Hb curve to its normal position. This compensatory change in 2,3-DPG requires between 24 and 48 hours. Therefore, with acute acid-base abnormalities, O2 affinity and the position of the oxy-Hb curve change. However, with more prolonged acid-base changes, the reciprocal changes in 2,3-DPG levels shift the oxy-Hb curve and O2 affinity back toward normal.95 Many inhaled anesthetics have been shown to shift the oxy-Hb dissociation curve to the right.96 Isoflurane shifts P50 to the right by 2.6 ± 0.07 mm Hg at a vapor pressure of approximately 1 minimum alveolar concentration (MAC) (1.25%).97 On the other hand, high-dose fen tanyl, morphine, and meperidine do not alter the position of the curve.
c. EFFECT OF QS/QT ON ALVEOLAR OXYGEN TENSION PAO2 is directly related to FIO2 in normal patients. PAO2 and FIO2 also correspond to PaO2 when there is little to no right-to-left transpulmonary shunt ( Q /Q S T ). Figure 5-26 shows the relationship between FIO2 and PaO2 for a family of right-to-left transpulmonary shunts; the calcula tions assume a constant and normal QT and PaCO2. With no Q /Q S T , a linear increase in FIO2 results in a linear increase in PAO2 (solid straight line). As the shunt is increased, the Q /Q S T lines relating FIO2 to PaO2 become progressively flatter.98 With a shunt of 50% of QT , an increase in FIO2 results in almost no increase in PaO2. The solution to the problem of hypoxemia secondary to a large shunt is not increasing the FIO2 but rather causing a reduction in the shunt (e.g., PEEP, patient value of 1.39 became popular.93 After extensive human studies, Gregory observed in 1974 that the applicable value was 1.31 mL O2/g of Hb in human adults.94 That the clinically measured CO2 is lower than the theoretical 1.39 is probably due to the small amount of methemo globin (MetHb) and carboxyhemoglobin (COHb) nor mally present in blood.
The oxy-Hb curve can also relate O2 transport (L/min) to the peripheral tissues (see Fig. 5-25) to PO2. The term O2 transport is synonymous with the term O2 delivery. This value is obtained by multiplying the O2 content by QT (O2 transport = QT × CaO2). To do this multiplica tion, one must convert the content unit of mL/dL to mL/L by multiplying by 10; subsequent multiplication of mL/L against QT in L/min yields mL/min. Thus, if QT = 5 L/min and CaO2 = 20 mL of O2/dL, the arterial point corresponds to 1000 mL O2/min going to the periphery, and the venous point corresponds to 750 mL O2/min returning to the lungs, with VO2 = 250 mL/min.
The oxy-Hb curve can also relate the O2 actually avail able to the tissues (leftmost y axis in Fig. 5-25) as a func tion of PO2. Of the 1000 mL/min of O2 normally going to the periphery, 200 mL/min of O2 cannot be extracted because it would lower PO2 below the level at which organs such as the brain can survive (rectangular dashed line in Fig. 5-25); the O2 available to tissues is therefore 800 mL/min. This amount is approximately three to four times the normal resting VO2. When QT = 5 L/min and arterial saturation is less than 40%, the total flow of O2 to the periphery is reduced to 400 mL/min; the available O2 is then 200 mL/min, and O2 supply just equals O2 demand. Consequently, with low arterial saturation, tissue demand can be met only by an increase in QT or, in the longer term, by an increase in Hb concentration.
The affinity of Hb for O2 is best described by the PO2 level at which Hb is 50% saturated (P50) on the oxy-Hb curve. The normal adult P50 is 26.7 mm Hg (see Fig. 5-25).
The effect of a change in PO2 on Hb saturation is related to both P50 and the portion of the oxy-Hb curve at which the change occurs.95 In the region of normal PaO2 (75 to 100 mm Hg), the curve is relatively horizon tal, and shifts of the curve have little effect on saturation. In the region of mixed venous PO2, where the curve is relatively steep, a shift of the curve leads to a much greater difference in saturation. A P50 lower than 27 mm Hg describes a left-shifted oxy-Hb curve, which means that at any given PO2, Hb has a higher affinity for O2 and is therefore more saturated than normal. This lower P50 may require higher than normal tissue perfu sion to produce the normal amount of O2 unloading. Causes of a left-shifted oxy-Hb curve are alkalosis (meta bolic and respiratory—the Bohr effect), hypothermia, abnormal fetal Hb, carboxyhemoglobin, methemoglobin, and decreased RBC 2,3-diphosphoglycerate (2,3-DPG) content. (The last condition may occur with the transfu sion of old acid citrate-dextrose–stored blood; storage of blood in citrate-phosphate-dextrose minimizes changes in 2,3-DPG with time.95) A P50 higher than 27 mm Hg describes a right-shifted oxy-Hb curve, which means that at any given PO2, Hb has a low affinity for O2 and is less saturated than normal. This higher P50 may allow a lower Figure 5-26 Effect of changes in inspired oxygen concentration on arterial oxygen tension (PaO2) for various right-to-left transpulmonary shunts. Cardiac output ( QT ), hemoglobin (Hb), oxygen consumption ( VO2), and arteriovenous oxygen content differences [C(a − v)O2] are assumed to be normal. PCO2, partial pressure of carbon dioxide; PO2, partial pressure of oxygen. 600 400 200 0 20 40 60 80 100 0 Inspired O2 (%) PaO2 (mm Hg) Alveolar PO2 with normal PCO2 10% shunt 20% shunt 30% shunt 40% shunt 50% shunt Air • • – Assume normal QT, VO2, Hb, C(a – v)O2 CHAPTER 5 Physiology of the Airway 141 function of QT and VO2. Figure 5-28 shows the equiva lent circuit of the pulmonary circulation in a patient with a 50% shunt, a normal CvO2 of 15 mL/dL, and a mod erately low CaO2 of 17.5 mL/dL. Decreasing QT or increasing VO2, or both, causes a larger primary decrease in CvO2 to 10 mL/dL and a smaller but still significant secondary decrease in CaO2 to 15 mL/dL; the ratio of change in CvO2 to change in CaO2 in this example of 50% Q /Q S T is 2 : 1.
If a decrease in QT or an increase in VO2 is accom panied by a decrease in Q /Q S T , there may be no change in PaO2 (i.e., a decreasing effect on PaO2 is offset by an increasing effect on PaO2) (Table 5-4). These changes sometimes occur in diffuse lung disease. However, if a decrease in QT or an increase in VO2 is accompanied by an increase in Q /Q S T, PaO2 may be greatly decreased (i.e., a decreasing effect on PaO2 is compounded by another decreasing effect on PaO2). These changes sometimes occur in regional ARDS and atelectasis.102 e. FICK PRINCIPLE The Fick principle allows calculation of VO2 and states that the amount of O2 consumed by the body ( VO2) is equal to the amount of O2 leaving the lungs ( QT )(CaO2) minus the amount of O2 returning to the lungs ( QT ) (CvO2): V Q Ca Q Cv Q Ca Cv O T O T O T O O 2 2 2 2 2 = − = − ( )( ) ( )( ) ( ) Figure 5-27 Effect of a decrease in cardiac output ( QT ) or an increase in oxygen consumption ( VO2) on mixed venous and arterial oxygen content. Mixed venous blood ( v ) either perfuses ventilated alveolar capillaries (ALV) and becomes oxygenated end-pulmonary capillary blood (c′), or it perfuses whatever true shunt pathways exist and remains the same in composition (i.e., desaturated). These two pathways must ultimately join together to form mixed arterial (a) blood. If QT decreases or VO2 increases, or both, the tissues must extract more oxygen per unit volume of blood than under normal conditions. Thus, the primary effect of a decrease in QT or an increase in VO2 is a decrease in mixed venous oxygen content. The mixed venous blood with a decreased oxygen content must flow through the shunt pathway as before (which may remain constant in size) and lower the arterial content of oxygen. Thus, the secondary effect of a decrease in QT or an increase in VO2 is a decrease in arterial oxygen content. ALV (O2) C′ 2. ↓ a ↓ QT or ↑ VO2 • • 1.↓↓ v – ↓↓v – 1. Primary effect 2. Secondary effect Shunt or ↑VO2 • ↓QT • positioning, suctioning, fiberoptic bronchoscopy, diuret ics, antibiotics).
d. EFFECT OF QT AND VO2 ON ARTERIAL OXYGEN CONTENT In addition to an increased Q /Q S T , CaO2 is decreased by decreased QT (for a constant VO2) and by increased VO2 (for a constant QT ). In either case, along with a constant right-to-left shunt, the tissues must extract more O2 from blood per unit blood volume, and therefore, CvO2 must primarily decrease (Fig. 5-27). When blood with lower CvO2 passes through whatever shunt exists in the lung and remains unchanged in its VO2, it must inevita bly mix with oxygenated end-pulmonary capillary blood (c′ flow) and secondarily decrease CaO2. The amount of O2 flowing per minute through any particular lung channel, as depicted in Figure 5-27, is a product of blood flow times the O2 content of that blood. Thus, Q Ca Qc Cc Q Cv T O O S O × = ′ × ′ + × 2 2 2 . With Qc Q Q T S ′ = − and further algebraic manipulation,99 Q /Q c Ca /Cc Cv S T O O O O = ′ − ′ − C 2 2 2 2 14 ( ) The larger the intrapulmonary shunt, the greater is the decrease in CaO2, because more venous blood with lower CvO2 can admix with end-pulmonary capillary blood (c′) (see Fig. 5-37).100,101 Therefore, the alveolar-arterial oxygen difference P(A − a)O2 is a function both of the size of the Q /Q S T and of what is flowing through the Q /Q S T —namely, CvO2—and CvO2 is a primary Figure 5-28 The equivalent circuit of the pulmonary circulation in a patient with a 50% right-to-left shunt. Oxygen content is in mL/dL of blood. A decrease in cardiac output ( QT) or an increase in O2 con sumption ( VO2) can cause a decrease in mixed venous oxygen content, from 15 to 10 mL/dL in this example, which in turn causes a decrease in the arterial content of oxygen, from 17.5 to 15.0 mL/ dL). In this 50% shunt example, the decrease in mixed venous oxygen content was twice the decrease in arterial oxygen content. (20) C′ End-pulmonary capillary a Mixed arterial (17.5) v Shunt (50%) (15) v (15) Mixed venous a (15) (20) C′ v (10) or VO2 • QT • (10) v 142 PART 1 Basic Clinical Science Considerations changes in QT than CaO2 (or PaO2) does (see Figs. 5-27 and 5-37).
3. Carbon Dioxide Transport The amount of CO2 circulating in the body is a function of both CO2 elimination and CO2 production. Elimina tion of CO2 depends on pulmonary blood flow and alveo lar ventilation. Production of CO2 ( VCO2) parallels O2 consumption ( VO2) according to the respiratory quotient (RQ): RQ V V CO O = 2 2 17 ( ) Under normal resting conditions, RQ is 0.8; that is, only 80% as much CO2 is produced as O2 is consumed. However, this value changes as the nature of the meta bolic substrate changes. If only carbohydrate is used, the RQ is 1.0. Conversely, with the sole use of fat, more O2 combines with hydrogen to produce water, and the RQ value drops to 0.7. CO2 is transported from mitochondria to the alveoli in a number of forms. In plasma, CO2 exists in physical solution, hydrated to carbonic acid (H2CO3), and as bicarbonate (HCO3 −). In the RBC, CO2 combines with Hb as carbaminohemoglobin (Hb-CO2). The approximate values of H2CO3 (H2O + CO2), HCO3 −, and Hb-CO2 relative to the total CO2 transported are 7%, 80%, and 13%, respectively.
In plasma, CO2 exists both in physical solution and as H2CO3: H O CO H CO 2 2 2 3 + → (18) The CO2 in solution can be related to PCO2 by the use of Henry’s law.103 P a CO in solution CO2 2 × = [ ] ( ) 19 where a is the solubility coefficient of CO2 in plasma (0.03 mmol/L/mm Hg at 37° C). However, the major fraction of CO2 produced passes into the RBC. As in plasma, CO2 combines with water to produce H2CO3. However, unlike the slow reaction in plasma, in which the equilibrium point lies toward the left, the reaction in an RBC is catalyzed by the enzyme carbonic anhydrase. This zinc-containing enzyme moves the reaction to the right at a rate 1000 times faster than in plasma. Further more, almost 99.9% of the H2CO3 dissociates to HCO3 − and hydrogen ions (H+): H O H CO carbonic anhydrase H CO H HCO 2 2 2 3 2 3 3 + → → + + − CO (20) The H+ produced from H2CO3 in the production of HCO3 − is buffered by Hb (H+ + Hb ⇋ HHb). The HCO3 − produced passes out of the RBC into plasma to perform its function as a buffer. To maintain electrical neutrality within the RBC, chloride ion (−) moves in as HCO3 − moves out (− shift). Finally, CO2 can combine with Hb in the erythrocyte (to produce Hb-CO2). Again, Condensing the content symbols yields the usual expres sion of the Fick equation: V Q C(a-v) O T O 2 2 15 = ( )[ ] ( ) This equation states that O2 consumption is equal to QT times the arteriovenous O2 content difference [C(a − v) O2]. Normally, (5 L/min)(5.5 mL)/dL = 0.27 L/ min (see “Oxygen-Hemoglobin Dissociation Curve”). V V F V F V F F O E IO E EO E IO EO 2 2 2 2 2 16 = − = − ( ) ( ) ( ) ( ) Similarly, the amount of O2 consumed by the body ( VO2) is equal to the amount of O2 brought into the lungs by ventilation ( VI )(FIO2) minus the amount of O2 leaving the lungs by ventilation ( VE )(FEO2), where VE is expired minute ventilation and FEO2 is the mixed expired O2 fraction: V V )(F ) V )(F ) O I IO E EO 2 = − ( ( 2 2 . Because the difference between VI and VE is due to the difference between VO2 (normally 250 mL/min) and VCO2 (normally 200 mL/min) and is only 50 mL/min (see later discussion), VI essentially equals VE.
Normally, VO2 = 5.0 L/min(0.21 − 0.16) = 0.25 L/ min. In determining VO2 in this way, VE can be mea sured with a spirometer, FIO2 can be measured with an O2 analyzer or from known fresh gas flows, and FEO2 can be measured by collecting expired gas in a bag for a few minutes. A sample of the mixed expired gas is used to measure PEO2. To convert PEO2 to FEO2, one simply divides PEO2 by dry atmospheric pressure: PEO2/713 = FEO2.
In addition, the Fick equation is useful in understand ing the impact of changes in QT on PaO2 and PvO2. If VO2 remains constant (K) and QT decreases (↓), the arteriovenous O2 content difference has to increase (↑): V K Q x a-v O T 2 2 = = ↓ ↑ ( ) ( ) ( ) C O The C(a − v)O2 difference increases because a decrease in QT causes a much larger and primary decrease in CvO2 versus a smaller and secondary decrease in CaO2, as follows101: ( ) ( ) ( ) ↑ = ↓ ↓↓ C C O O a-v a-v 2 2 Thus, CvO2 and PvO2 are much more sensitive indicators of QT because they change more with TABLE 5-4 Relationship Between Cardiac Output ( QT ), Shunt ( Q /Q S T), and Venous (PvO2 ) and Arterial (Pao2) Oxygenation Changes Clinical Situation If QT ↓ → ↓ PvO2 and Q /Q S T = 0 → Pao2↓ Decreased cardiac output, stable shunt If QT ↓ → ↓ PvO2 and Q /Q S T ↓ → Pao2 = 0 Application of PEEP in ARDS If QT ↓ → ↓ PvO2 and Q /Q S T ↑ → Pao2 ↓↓ Shock combined with ARDS or atelectasis ARDS, Adult respiratory distress syndrome; 0, no change; PEEP, positive end-expiratory pressure; ↓, decrease; ↑, increase.
CHAPTER 5 Physiology of the Airway 143 elastic fibers, fibroblasts, and macrophages. This connec tive tissue is the backbone of the lung parenchyma; it forms a continuum with the connective tissue sheaths around the conducting airways and blood vessels. Thus, the pericapillary perialveolar interstitial space is continu ous with the interstitial tissue space that surrounds ter minal bronchioles and vessels, and both spaces constitute the connective tissue space of the lung. There are no lymphatics in the interstitial space of the alveolar septum. Instead, lymphatic capillaries first appear in the intersti tial space surrounding terminal bronchioles, small arter ies, and veins.105 The opposite side of the alveolar septum (the thin, down, gas-exchanging-only side) contains only fused epi thelial and endothelial basement membranes. The inter stitial space is greatly restricted on this side because of fusion of the basement membranes. Interstitial fluid cannot separate the endothelial and epithelial cells from one another. As a result the space and distance barrier to fluid movement from the capillary to the alveolar com partment is reduced; it is composed of only the two cell linings with their associated basement membranes.106 Between the individual endothelial and epithelial cells are holes or junctions that provide a potential pathway for fluid to move from the intravascular space to the interstitial space and finally from the interstitial space to the alveolar space (see Fig. 5-29). The junctions between endothelial cells are relatively large and are therefore termed “loose”; the junctions between epithelial cells are relatively small and are therefore termed “tight.” Pulmo nary capillary permeability is a direct function of, and essentially equivalent to, the size of the holes in the endothelial and epithelial linings.
To understand how pulmonary interstitial fluid is formed, stored, and cleared, it is necessary first to develop the concepts that (1) the pulmonary interstitial space is a continuous space between the periarteriolar and peri bronchial connective tissue sheath and the space between the endothelial and epithelial basement membranes in the alveolar septum and (2) the space has a progressively negative distal-to-proximal ΔP.
The concepts of a continuous connective tissue sheath– alveolar septum interstitial space and a negative intersti tial space ΔP are prerequisite to understanding interstitial fluid kinetics. After entering the lung parenchyma, both the bronchi and the arteries run within a connective tissue sheath that is formed by an invagination of the pleura at the hilum and ends at the level of the bronchi oles (Fig. 5-30A).This results in a potential perivascular space between the arteries and the connective tissue sheath and a potential peribronchial space between the bronchi and the connective tissue sheath. The negative pressure in the pulmonary tissues surrounding the peri vascular connective tissue sheath exerts a radial outward traction force on the sheath. This radial traction creates negative pressure within the sheath that is transmitted to the bronchi and arteries and tends to hold them open and increase their diameters.106 The alveolar septum intersti tial space is the space between the capillaries and alveoli (or, more precisely, the space between the endothelial and epithelial basement membranes) and is continuous with the interstitial tissue space that surrounds the larger Figure 5-29 Schematic summary of the ultrastructure of a pulmo nary capillary. On the upper side of the capillary, the endothelial (ENDO) and epithelial (EPI) basement membranes (BM) are sepa rated by an interstitial space (I.S.), whereas the lower side contains only fused ENDO and EPI BMs. The dashed arrows indicate a poten tial pathway for fluid to move from the intravascular space to the I.S. through loose junctions (LJ) in the endothelium and from the I.S. to the alveolar space (ALV) through tight junctions (TJ) in the epi thelium. RBC, Red blood cell. (Redrawn from Fishman AP: Pulmonary edema: The water-exchanging function of the lung. Circulation 46:390, 1972.) EPI EPI BM BM BM ENDO RBC ENDO ALV ALV LJ I.S.
TJ as in HCO3 − release, an H+ ion is formed in the reaction of CO2 and Hb. This H+ ion is also buffered by Hb.
4. Bohr and Haldane Effects Just as the percent saturation of Hb with O2 is related to PO2 (described by the oxy-Hb curve), so the total CO2 in blood is related to PCO2. In addition, Hb has variable affinity for CO2; it binds more avidly in the reduced state than as oxy-Hb.95 The Bohr effect describes the effect of PCO2 and [H+] ions on the oxy-Hb curve. Hypercapnia and acidosis both shift the curve to the right (reducing the O2-binding affinity of hemoglobin), and hypocapnia and alkalosis both shift the curve to the left. Conversely, the Haldane effect describes the shift in the CO2 dissocia tion curve caused by oxygenation of Hb. Low PO2 shifts the CO2 dissociation curve to the left so that the blood is able to pick up more CO2 (as occurs in capillaries of rapidly metabolizing tissues). Conversely, oxygenation of Hb (as occurs in the lungs) reduces the affinity of Hb for CO2, and the CO2 dissociation curve is shifted to the right, thereby increasing CO2 removal.
E. Pulmonary Microcirculation, Interstitial Space, and Fluid (Pulmonary Edema) The ultrastructural appearance of an alveolar septum is depicted schematically in Figure 5-29.104 Capillary blood is separated from alveolar gas by a series of anatomic layers: capillary endothelium, endothelial basement membrane, interstitial space, epithelial basement membrane, and alveolar epithelium (of the type I pneumocyte).
On one side of the alveolar septum (the thick, upper, fluid- and gas-exchanging side), the epithelial and endo thelial basement membranes are separated by a space of variable thickness containing connective tissue fibrils, 144 PART 1 Basic Clinical Science Considerations arteries and bronchi. Studies indicate that the alveolar interstitial pressure is also uniquely negative but not as much so as the negative interstitial space pressure around the larger arteries and bronchi.107 The forces governing net transcapillary–interstitial space fluid movement are as follows. The net transcapil lary flow of fluid (F) out of pulmonary capillaries (across the endothelium and into the interstitial space) is equal to the difference between pulmonary capillary hydro static pressure (Pinside) and interstitial fluid hydrostatic pressure (Poutside) and the difference between capillary colloid oncotic pressure (πinside) and interstitial colloid oncotic pressure (πoutside). These four forces produce a steady-state fluid flow (F) during a constant capillary permeability (K) as predicted by the Starling equation: F K P P inside outside inside outside = − − − [( ) ( )] π π (21) Figure 5-30 A, Schematic diagram of the concept of a continuous connective tissue sheath–alveolar septum interstitial space. The entry of the main stem bronchi and pulmonary artery into the lung parenchyma invaginates the pleura at the hilum and forms a sur rounding connective tissue sheath. The connective tissue sheath ends at the level of the bronchioles. The space between the pulmo nary arteries and bronchi and the interstitial space is continuous with the alveolar septum interstitial space. The alveolar septum interstitial space is contained within the endothelial basement membrane of the capillaries and the epithelial basement membrane of the alveoli. B, Schematic diagram showing how interstitial fluid moves from the alveolar septum interstitial space (no lymphatics) to the connective tissue interstitial space (lymphatic capillaries first appear). The mech anisms are a negative-pressure gradient (sump), the presence of one-way valves in the lymphatics, and the massaging action of arterial pulsations. CAP, Capillary. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) CT sheath containing the extra-alveolar interstitial space Alveolar septum containing the intra-alveolar interstitial space 2. Valves 1. Negative pressure gradient 3. Arterial pulsations Bronchi Artery Alveolus CAP Loose junctions Connective tissue sheath Alveolar septum Continuous interstitial space P ul m o n a r y a rt e r y Br on ch u s A B Box 5-1 Causes of Extremely Negative Pulmonary Interstitial Fluid Pressure (Poutside) in Pulmonary Edema Vigorous spontaneous ventilation against an obstructed airway Laryngospasm Infection, inflammation, edema Upper airway mass (e.g., tumor, hematoma, abscess, foreign body) Vocal cord paralysis Strangulation Rapid reexpansion of lung Vigorous pleural suctioning (thoracentesis, chest tube) K is a capillary filtration coefficient expressed in mL/ min/mm Hg/100 g. The filtration coefficient is the product of the effective capillary surface area in a given mass of tissue and the permeability per unit surface area of the capillary wall to filter the fluid. Under normal circumstances and at a vertical height in the lung that is at the junction of zones 2 and 3, intravascular colloid oncotic pressure (≈ 26 mm Hg) acts to keep water in the capillary lumen; working against this force, pulmonary capillary hydrostatic pressure (≈ 10 mm Hg) acts to force water across the loose endothelial junctions into the interstitial space. If these were the only operative forces, the interstitial space and, consequently, the alveolar sur faces would be constantly dry and there would be no lymph flow. In fact, alveolar surfaces are moist, and lym phatic flow from the interstitial compartment is constant (≈ 500 mL/day). This can be explained in part by πoutside (≈ 8 mm Hg) and in part by the negative Poutside (−8 mm Hg).
Negative (subatmospheric) interstitial space pressure would promote, by suction, a slow loss of fluid across the endothelial holes.108 Indeed, extremely negative pleural (and perivascular hydrostatic) pressure, such as may occur in a vigorously spontaneously breathing patient with an obstructed airway, can cause pulmonary intersti tial edema (Box 5-1).109 Relative to the vertical level of the junction of zones 2 and 3, as lung height decreases (lung dependence), absolute Pinside increases, and fluid has a propensity to transudate; as lung height increases (lung nondependence), absolute Pinside decreases, and fluid has a propensity to be reabsorbed. However, fluid transuda tion induced by an increase in Pinside is limited by a con comitant dilution of proteins in the interstitial space and therefore a decrease in πoutside.110 Any change in the size of the endothelial junctions, even if the foregoing four forces remain constant, changes the magnitude and perhaps even the direction of fluid movement. Increased size of endothelial junctions (increased permeability) promotes transudation, whereas decreased size of endo thelial junctions (decreased permeability) promotes reabsorption.
No lymphatics are present in the interstitial space of the alveolar septum. The lymphatic circulation starts as blind-ended lymphatic capillaries, first appearing in the interstitial space sheath surrounding terminal bronchioles and small arteries, and ends at the subclavian veins. Inter stitial fluid is normally removed from the alveolar CHAPTER 5 Physiology of the Airway 145 MAC (light anesthesia), irregular respiration progresses to a more regular pattern that is associated with a larger than normal VT. However, during light but deepening anesthesia, the approach to a more regular respiratory pattern may be interrupted by a pause at the end of inspiration (a “hitch” in inspiration), followed by a rela tively prolonged and active expiration in which the patient seems to exhale forcefully rather than passively. As anesthesia deepens to moderate levels, respiration becomes faster and more regular but shallower. The respiratory pattern is a sine wave losing the inspiratory hitch and lengthened expiratory pause. There is little or no inspiratory or expiratory pause, and the inspiratory and expiratory periods are equivalent. Intercostal muscle activity is still present, and there is normal movement of the thoracic cage with lifting of the chest during inspiration.
The respiratory rate is generally slower and the VT larger with nitrous oxide–narcotic anesthesia than with anesthesia involving halogenated drugs. During deep anesthesia with halogenated drugs, increasing respiratory depression is manifested by increasingly rapid and shallow breathing (panting). On the other hand, with deep nitrous oxide–narcotic anesthesia, respirations become slower but may remain deep. In the case of very deep anesthesia with all inhaled drugs, respirations often become jerky or gasping in character and irregular in pattern. This situa tion results from loss of the active intercostal muscle contribution to inspiration. As a result, a rocking boat movement occurs in which there is out-of-phase depres sion of the chest wall during inspiration, flaring of the lower chest margins, and billowing of the abdomen. The reason for this type of movement is that inspiration is dependent solely on diaphragmatic effort. Independent of anesthetic depth, similar chest movements may be simulated by upper or lower airway obstruction or by partial paralysis.
B. Anesthetic Depth and Spontaneous Minute Ventilation Despite the variable changes in respiratory pattern and rate as anesthesia deepens, overall spontaneous VE pro gressively decreases. In the normal awake response to breathing CO2, an increasing end-tidal PCO2 causes a linear increase in VE (Fig. 5-31). The slope of the line relating VE to the end-tidal CO2 concentration (PCO2) in awake individuals is approximately 2 L/min/mm Hg. (In healthy individuals, the variation in the slope of this response is large.) Figure 5-31 shows that increasing the halothane concentration displaces the ventilation-response curve progressively to the right (i.e., at any end-tidal PCO2, ventilation is less than before), decreases the slope of the curve, and shifts the apneic threshold to a higher end-tidal PCO2.124 Similar alterations are observed with narcotics and other halogenated anesthetics.125 Figures 5-22 to 5-24 show that decreases in VE cause increases in PaCO2 and decreases in PaO2. The relative increase in PaCO2 caused by depression of VE (<1.24 MAC) by halogenated anesthetics is desflurane = isoflu rane > sevoflurane > halothane. At higher concentrations, desflurane causes increasing ventilatory depression, even interstitial space into the lymphatics by a sump (pressure gradient) mechanism, which is caused by the presence of more negative pressure surrounding the larger arteries and bronchi.3,111 The sump mechanism is aided by the presence of valves in the lymph vessels. In addition, because the lymphatics run in the same sheath as the pulmonary arteries, they are exposed to the massaging action of arterial pulsations. The differential negative pressure, the lymphatic valves, and the arterial pulsations all help propel the lymph proximally toward the hilum through the lymph nodes (pulmonary to bronchopulmo nary to tracheobronchial to paratracheal to scalene and cervical nodes) to the central venous circulation depot (see Fig. 5-30B). An increase in central venous pressure, which is the backpressure for lymph to flow out of the lung, would decrease lung lymph flow and perhaps promote pulmonary interstitial edema.
If the rate of entry of fluid into the pulmonary inter stitial space exceeds the capability of the pulmonary interstitial space to clear the fluid, the pulmonary inter stitial space fills with fluid; the fluid, now under an increased and positive driving force (PISF), crosses the relatively impermeable epithelial wall holes, and the alveolar space fills. Intra-alveolar edema fluid also causes alveolar collapse and atelectasis, thereby promoting further accumulation of fluid and worsening right-to-left transpulmonary shunt.
II. RESPIRATORY FUNCTION DURING ANESTHESIA Arterial oxygenation is impaired in most patients during anesthesia with either spontaneous or controlled ventilation.112–117 In otherwise normal patients, it is gener ally accepted that the impairment in arterial oxygenation during anesthesia is more severe in elderly persons,118,119 obese people,120 and smokers.121 In various studies of healthy young to middle-aged patients under general anesthesia, venous admixture (shunt) has been found to average 10%, and the scatter in V /Q A ratios is small to moderate.119,122 In patients with a more marked deteriora tion in preoperative pulmonary function, general anes thesia causes considerable widening of the V /Q A distribution and large increases in both low- V /Q A (0.005 < V /Q A < 0.1) (underventilated) regions and shunt ing.118,121,123 The magnitude of shunting correlates closely with the degree of atelectasis.118,123 In addition to these generalizations concerning respira tory function during anesthesia, the effect of a given anesthetic on respiratory function depends on the depth of general anesthesia, the patient’s preoperative respira tory condition, and the presence of special intraoperative anesthetic and surgical conditions.
A. Anesthetic Depth and Respiratory Pattern The respiratory pattern is altered by the induction and deepening of anesthesia. When the depth of anesthesia is inadequate (less than MAC), the respiratory pattern may vary from excessive hyperventilation and vocalization to breath-holding. As anesthetic depth approaches or equals 146 PART 1 Basic Clinical Science Considerations anesthetic inhibits HPV, the drug may increase shunting more in patients with preexisting HPV than in those without preexisting HPV. Thus, the effect of a standard anesthetic can be expected to produce varying degrees of respiratory change in patients who have different degrees of preexisting respiratory dysfunction.
D. Special Intraoperative Conditions Some special intraoperative conditions (e.g., surgical position, massive blood loss, surgical retraction on the lung) can cause impaired gas exchange. For example, some of the surgical positions (i.e., the lithotomy, jack knife, and kidney rest positions) and surgical exposure requirements may decrease QT, cause hypoventilation in a spontaneously breathing patient, and reduce FRC. The type and severity of preexisting respiratory dysfunction, as well as the number and severity of special intraopera tive conditions that can embarrass respiratory function, magnify the respiratory depressant effects of any anesthetic.
more than isoflurane, and sevoflurane causes a degree of ventilatory depression similar to isoflurane.
C. Preexisting Respiratory Dysfunction Anesthesiologists are frequently required to care for (1) patients with acute chest disease (pulmonary infection, atelectasis) or systemic diseases (sepsis, cardiac and renal failure, multiple trauma) who require emergency opera tions, (2) heavy smokers with subtle pathologic airway and parenchymal conditions and hyperreactive airways, (3) patients with classic emphysematous and bronchitic problems, (4) obese people susceptible to decreases in FRC during anesthesia,126 (5) patients with chest defor mities, and (6) extremely old patients.
The nature and magnitude of these preexisting respira tory conditions determine, in part, the effect of a given standard anesthetic on respiratory function. For example, in Figure 5-32, the FRC-CC relationship is depicted for normal, obese, bronchitic, and emphysematous patients. In a healthy patient, FRC exceeds CC by approximately 1 L. In the latter three respiratory conditions, CC is 0.5 to 0.75 L less than FRC. If anesthesia causes a 1-L decrease in FRC in a healthy patient, there is no change in the qualitative relationship between FRC and CC. In patients with special respiratory conditions, however, a 1-L decrease in FRC causes CC to exceed FRC and changes the previous marginally normal FRC-CC rela tionship to either a grossly low V /Q A or an atelectatic FRC-CC relationship. Similarly, patients with chronic bronchitis, who have copious airway secretions, may suffer more than other patients from an anesthetic-induced decrease in mucus velocity flow. Finally, if an Figure 5-31 In conscious controls, increasing end-expiratory con centration of carbon dioxide (PCO2, x-axis) increases pulmonary minute volume ( VE, y-axis). The dashed line is an extrapolation of the CO2 response curve to zero ventilation and represents the apneic threshold. Increases in end-expiratory anesthetic (halothane) con centration progressively diminish the slope of the CO2 response curve and shift the apneic threshold to a higher PCO2. The heavy line interrupted by dots shows the decrease in minute ventilation and the increase in PCO2 that occur with increasing depth of anesthesia. (Redrawn with modification from Munson ES, Larson CP Jr, Babad AA, et al: The effects of halothane, fluroxene and cyclopropane on ven tilation: A comparative study in man. Anesthesiology 27:716, 1966.) 20 10 0 20 40 60 80 100 Halothane (end-expiratory) End-expiratory PCO2 (mm Hg) Conscious control 0.81% 1.11% 1.49% 1.88% VE (L/min) • Figure 5-32 The lung volume (ordinate) at which tidal volume is breathed decreases (by 1 L) from the awake state to the anesthe tized state. Functional residual capacity (FRC), which is the volume of gas existing in the lung at the end of a tidal breath, therefore also decreases (by 1 L) from the awake to the anesthetized state. In healthy, obese, bronchitic, and emphysematous patients, the awake FRC considerably exceeds the closing capacity (CC, hori zontal lines), but the anesthetized state causes FRC to be less than CC. In healthy patients, anesthesia causes FRC to equal CC. 5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 0 Awake Anesthetized Lung volume (L) Normal Obese Bronchitis Emphysema Tidal volume CC CHAPTER 5 Physiology of the Airway 147 2. Hypoventilation Patients under general anesthesia may have a reduced spontaneous VT for two reasons. First, increased WOB can occur during general anesthesia as a result of increased airway resistance and decreased CL. Airway resistance can be increased because of reduced FRC, endotracheal intubation, the presence of the external breathing apparatus and circuitry, and possible airway obstruction in patients whose tracheas are not intubated.136–138 CL is reduced as a result of some (or all) of the factors that can decrease FRC.89 Second, patients may have a decreased drive to breathe spontaneously during general anesthesia (i.e., decreased chemical control of breathing) (see Fig. 5-31).
Decreased VT may cause hypoxemia in two ways.117 First, shallow breathing can promote atelectasis and cause a decrease in FRC (see “Ventilation Pattern [Rapid Shallow Breathing]”).40,139 Second, decreased VE decreases the overall V /Q A ratio of the lung, which decreases PaO2 (see Figs. 5-23 and 5-24).117 This is likely to occur with spontaneous ventilation during moderate to deep levels of anesthesia, in which the chemical control of breathing is significantly altered.
3. Hyperventilation Hypocapnic alkalosis (hyperventilation) can occasionally be associated with a decreased PaO2 due to several indi rect mechanisms: decreased QT and increased VO2 140,141 (see “Decreased Cardiac Output and Increased Oxygen Consumption”),99,101,140,141 a left-shifted oxy-Hb curve (see “Oxygen-Hemoglobin Dissociation Curve”), decreased HPV (see “Inhibition of Hypoxic Pulmonary Vaso constriction”),142 and increased airway resistance and decreased compliance (see “Increased Airway Resis tance”).143 Although these theoretical causes of hypox emia exist, they are seldom a major factor in the clinical realm.
4. Decrease in Functional Residual Capacity The effect of decreased FRC on hypoxemia is very sig nificant clinically. Induction of general anesthesia is con sistently accompanied by a pronounced (15% to 20%) decrease in FRC,32,83,144 which usually causes a decrease in compliance.89 The maximum decrease in FRC appears to occur within the first few minutes of anesthesia,32,145 and in the absence of any other complicating factor, FRC does not seem to decrease progressively during anesthe sia. During anesthesia, the reduction in FRC is of the same order of magnitude whether ventilation is sponta neous or controlled. Conversely, in awake patients, FRC is only slightly reduced during controlled ventilation.146 In obese patients, the reduction in FRC is far more pro nounced than in normal patients, and the decrease is inversely related to the body mass index (BMI).147 The reduction in FRC continues into the postoperative period.148 For individual patients, the reduction in FRC correlates well with the increase in the alveolar-arterial PO2 gradient during anesthesia with spontaneous breath ing,149 during anesthesia with artificial ventilation,146 and in the postoperative period.148 The reduced FRC can be restored to normal or above normal by the application of E. Mechanisms of Hypoxemia During Anesthesia 1. Equipment Malfunction a. MECHANICAL FAILURE OF ANESTHESIA APPARATUS TO DELIVER OXYGEN TO THE PATIENT Hypoxemia resulting from mechanical failure of the O2 supply system (see Chapter 14) or the anesthesia machine is a recognized hazard of anesthesia. Disconnection of the patient from the O2 supply system (usually at the junc ture of the ETT and the elbow connector) is by far the most common cause of mechanical failure to deliver O2 to the patient. Other reported causes of failure of the O2 supply during anesthesia include the following: an empty or depleted O2 cylinder, substitution of a nonoxygen cyl inder at the O2 yoke because of absence or failure of the pin index, an erroneously filled O2 cylinder, insufficient opening of the O2 cylinder (which hinders free flow of gas as pressure decreases), failure of gas pressure in a piped O2 system, faulty locking of the piped O2 system to the anesthesia machine, inadvertent switching of the Schrader adapters on piped lines, crossing of piped lines during construction, failure of a reducing valve or gas manifold, inadvertent disturbance of the setting of the O2 flowmeter, use of the fine O2 flowmeter instead of the coarse flowmeter, fractured or sticking flowmeters, trans position of rotameter tubes, erroneous filling of a liquid O2 reservoir with N2, and disconnection of the fresh gas line from machine to in-line hosing.127–131 Monitoring of the inspired O2 concentration with an in-line FIO2 ana lyzer and monitoring of airway pressure should detect most of these causes of failure to deliver O2 to the patient.127–131 b. IMPROPER ENDOTRACHEAL TUBE POSITION Esophageal intubation results in almost no ventilation. Aside from disconnection, almost all other mechanical problems with ETTs (e.g., kinking, blockage of secretions, herniated or ruptured cuffs) cause an increase in airway resistance that may result in hypoventilation. Intubation of a main stem bronchus (see Chapter 7) results in absence of ventilation of the contralateral lung. Although potentially minimized by HPV, some perfusion to the contralateral lung always remains, and shunting increases and PaO2 decreases. A tube previously well positioned in the trachea may enter a bronchus after the patient or the patient’s head is turned or moved into a new position.132 Flexion of the head causes the tube to migrate deeper (caudad) into the trachea, whereas extension of the head causes cephalad (outward) migration of the ETT.132 A high incidence of main stem bronchial intubation after the institution of a 30-degree Trendelenburg position has been reported.133 Cephalad shift of the carina and medi astinum during the Trendelenburg position caused the previously “fixed” ETT to migrate into a main stem bron chus. Main stem bronchial intubation may obstruct the ipsilateral upper lobe in addition to the contralateral lung.134,135 Infrequently, the right upper bronchus or one of its segmental bronchi branches from the lateral wall of the trachea (above the carina) and may be occluded by a properly positioned ETT.
148 PART 1 Basic Clinical Science Considerations intrathoracic air passage and decreases the possibility of airway closure and a decrease in FRC (see Fig. 5-17F). Endotracheal intubation bypasses the lips and glottis and can abolish the normally present pursed-lip or grunting exhalation in a patient with COPD and in that way con tributes to airway closure and loss of FRC in some spon taneously breathing patients.
c. PARALYSIS In an upright subject, FRC and the position of the dia phragm are determined by the balance between lung elastic recoil pulling the diaphragm cephalad and the weight of the abdominal contents pulling it caudad.153 There is no transdiaphragmatic pressure gradient.
The situation is more complex in the supine position. The diaphragm separates two compartments of markedly different hydrostatic gradients. On the thoracic side, pres sure increases by approximately 0.25 cm H2O/cm of lung height,38,154 and on the abdominal side, it increases by 1.0 cm H2O/cm of abdominal height.153 Therefore, in horizontal postures, progressively higher transdiaphrag matic pressure must be generated toward dependent parts of the diaphragm to keep the abdominal contents out of the thorax. In an unparalyzed patient, this tension is developed either by passive stretch and changes in shape of the diaphragm (causing an increased contractile force) or by neurally mediated active tension. With acute muscle paralysis, neither of these two mechanisms can operate, and a shift of the diaphragm to a more cephalad position occurs (see Fig. 5-33).155 The latter position must express the true balance of forces on the diaphragm, unmodified by any passive or active muscle activity.
The cephalad shift in the FRC position of the dia phragm as a result of expiratory muscle tone during general anesthesia is equal to the shift observed during paralysis (awake or anesthetized patients).145,156 The equal shift suggests that the pressure on the diaphragm caused by an increase in expiratory muscle tone during general anesthesia is equal to the pressure on the dia phragm caused by the weight of the abdominal contents during paralysis. It is quite probable that the magnitude of these changes in FRC related to paralysis also depends on body habitus.
d. LIGHT OR INADEQUATE ANESTHESIA AND ACTIVE EXPIRATION Induction of general anesthesia can result in increased expiratory muscle tone,151 but the increased expiratory muscle tone is not coordinated and does not contribute to the exhaled volume of gas. In contrast, spontaneous ventilation during light general anesthesia usually results in a coordinated and moderately forceful active exhala tion and larger exhaled volumes. Excessively inadequate anesthesia (relative to a given stimulus) results in very forceful active exhalation, which can produce exhaled volumes of gas equal to an awake expiratory vital capacity.
As during an awake expiratory vital capacity maneu ver, forced expiration during anesthesia raises intratho racic and alveolar pressure considerably above atmospheric pressure (see Fig. 5-17). This increase in pressure results in rapid outflow of gas, and because part of the expiratory PEEP.82,150 The following discussion considers the most common causes of reduced FRC.
a. SUPINE POSITION Anesthesia and surgery are usually performed with the patient in the supine position. With change from the upright to the supine position, FRC decreases by 0.5 to 1.0 L32,83,144 because of a 4-cm cephalad displacement of the diaphragm by the abdominal viscera (Fig. 5-33). Pul monary vascular congestion can also contribute to the decrease in FRC in the supine position, particularly in patients who experienced orthopnea preoperatively.
b. INDUCTION OF GENERAL ANESTHESIA: CHANGE IN THORACIC CAGE MUSCLE TONE At the end of a normal (awake) exhalation, there is slight tension in the inspiratory muscles and no tension in the expiratory muscles. Therefore, at the end of a normal exhalation, there is a force tending to maintain lung volume and no force decreasing lung volume. After induction of general anesthesia, there is a loss of inspira tory tone and an appearance of end-expiratory tone in the abdominal expiratory muscles at the end of exhala tion. The end-expiratory tone in the abdominal expira tory muscles increases intra-abdominal pressure, forces the diaphragm cephalad, and decreases FRC (see Fig. 5-33).145,151 Thus, after the induction of general anesthe sia, there is loss of the force tending to maintain lung volume and gain of the force tending to decrease lung volume. Indeed, Innovar (droperidol and fentanyl citrate) may increase tone in expiratory muscles to such an extent that the reduction in FRC with Innovar anesthesia alone is greater than that with Innovar plus paralysis induced by succinylcholine.151,152 With emphysema, exhalation may be accompanied by pursing the lips or grunting (i.e., with a partially closed larynx). An emphysematous patient exhales in either of these ways because both these maneuvers cause an expiratory retardation that produces PEEP in the Figure 5-33 Anesthesia and surgery may cause a progressive ceph alad displacement of the diaphragm. The sequence of events involves assumption of the supine position, induction of anesthesia, establishment of paralysis, assumption of several surgical positions, and displacement by retractors and packs. Cephalad displace ment of the diaphragm results in decreased functional residual capacity (↓ FRC). Pab, Pressure of abdominal contents. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) Progressive Cephalad Displacement of the Diaphragm ↓ FRC Surgical position and displacement Induction of anesthesia Paralysis Supine position Initial upright position Pab CHAPTER 5 Physiology of the Airway 149 kinking, herniated cuffs) are not uncommon and can be life-threatening.
The respiratory apparatus often causes resistance that is considerably higher than the resistance in the normal human respiratory tract (see Fig. 5-34).89 When certain resistors such as those shown in Figure 5-34 are joined in series to form an anesthetic gas circuit, their effects are generally additive and produce larger resistance (as with resistance in series in an electrical circuit). The increase in resistance associated with commonly used breathing circuits and ETTs can impose an additional WOB that is two to three times normal.136 f. SUPINE POSITION, IMMOBILITY, AND EXCESSIVE INTRAVENOUS FLUID ADMINISTRATION Patients undergoing anesthesia and surgery are often kept supine and immobile for long periods. In these cases, some of the lung can be continually dependent and below the left atrium and therefore in zone 3 or 4 condition. Being in a dependent position, the lung is predisposed to accumulation of fluid. Coupled with excessive fluid administration, conditions sufficient to promote transu dation of fluid into the lung are present and result in pulmonary edema and decreased FRC.
When mongrel dogs were placed in a lateral decubitus position and anesthetized for several hours (Fig. 5-36), expansion of the extracellular space with fluid caused the PO2 of blood draining the dependent lung to decrease precipitously to mixed venous levels (no O2 uptake).159 Blood draining the nondependent lung maintained its PO2 for a period but declined after 5 hours in the presence of the extracellular fluid expansion. Transpulmonary shunting progressively increased. If the animals were turned every hour (and received the same fluid resistance lies in the smaller air passages, a drop in pres sure occurs between the alveoli and the main bronchi. Under these circumstances, intrathoracic pressure rises considerably above the pressure within the main bronchi. Collapse occurs if this reversed pressure gradient is suf ficiently high to overcome the tethering effect of the surrounding parenchyma on the small intrathoracic bron chioles or the structural rigidity of cartilage in the large extrathoracic bronchi. Such collapse occurs in a normal subject during a maximal forced expiration and is respon sible for the associated wheeze in both awake and anes thetized patients.157 In a paralyzed, anesthetized patient, the use of a sub atmospheric expiratory pressure phase is analogous to a forced expiration in a conscious subject; the negative phase may set up the same adverse ΔP, which can cause airway closure, gas trapping, and a decrease in FRC. An excessively rapidly descending bellows of a ventilator during expiration has caused subatmospheric expiratory pressure and resulted in wheezing.158 e. INCREASED AIRWAY RESISTANCE The overall reduction in all components of lung volume during anesthesia results in reduced airway caliber, which increases airway resistance and any tendency toward airway collapse (Fig. 5-34). The relationship between airway resistance and lung volume is well established (Fig. 5-35). The decreases in FRC caused by the supine position (≈ 0.8L) and induction of anesthesia (≈ 0.4L) are often sufficient to explain the increased resistance seen in a healthy anesthetized patient.137 In addition to this expected increase in airway resis tance in anesthetized patients, there are a number of special potential sites of increased airway resistance, including the ETT (if present), the upper and lower airway passages, and the external anesthesia apparatus. Endotracheal intubation reduces the size of the trachea, usually by 30% to 50% (see Fig. 5-34). Pharyngeal obstruction, which can be considered to be a normal feature of unconsciousness, is most common. A minor degree of this type of obstruction occurs in snoring. Laryngospasm and obstructed ETTs (e.g., secretions, Figure 5-34 An anesthetized patient in the supine position has increased airway resistance as a result of decreased functional residual capacity (FRC), decreased caliber of the airways, endotra cheal intubation, and connection of the endotracheal tube (ET) to the external breathing apparatus and circuitry. ↓, Decreased. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) ↓ Caliber of airway ↓ FRC CO2 canister Increased airway resistance Upright awake ET tube Water Valves Vaporizer Supine anesthetized Figure 5-35 Airway resistance is an increasing hyperbolic function of decreasing lung volume. Functional residual capacity (FRC) decreases with a change from the upright to the supine position. (Redrawn with modification from Lumb AB: Respiratory system resis tance. In Lumb AB, editor: Nunn’s applied respiratory physiology, ed 5, London, 2000, Butterworths, p 67.) 1 2 3 2 4 6 Lung volume (L) Residual volume Maximal inspiration FRC (supine) FRC (upright) Airway resistance (cm H2O) (at 1L/sec) 150 PART 1 Basic Clinical Science Considerations uptake by lung units with low V /Q A ratios.161,162 A unit that has a low V /Q A ratio during breathing of air will have a low PAO2. When an enriched O2 mixture is inspired, PAO2 rises, and the rate at which O2 moves from alveolar gas to capillary blood increases greatly. The O2 flux may increase so much that the net flow of gas into blood exceeds the inspired flow of gas, and the lung unit becomes progressively smaller. Collapse is most likely to occur if FIO2 is high, the V /Q A ratio is low, the time of exposure of the unit with low V /Q A to high FIO2 is long, and CvO2 is low. Given the right V /Q A ratio and time of administration, an FIO2 as low as 50% can produce absorption atelectasis.161,162 This phenomenon is of con siderable significance in the clinical situation for two reasons. First, enriched O2 mixtures are often used thera peutically, and it is important to know whether this therapy is causing atelectasis. Second, the amount of shunt is often estimated during breathing of 100% O2, and if this maneuver results in additional shunt, the mea surement is hard to interpret.
h. SURGICAL POSITION SUPINE POSITION. In the supine position, the abdominal contents force the diaphragm cephalad and reduce FRC.83,145,151,156 The Trendelenburg position allows the abdominal contents to push the diaphragm further ceph alad so that the diaphragm must not only ventilate the lungs but also lift the abdominal contents out of the thorax. The result is a predisposition to decreased FRC and atelectasis.163 The decrease in FRC related to Tren delenburg position is exacerbated in obese patients.147 Increased pulmonary blood volume and gravitational force on the mediastinal structures are additional factors that may decrease pulmonary compliance and FRC. In the steep Trendelenburg position, most of the lung may be below the left atrium and therefore in a zone 3 or 4 condition. In this condition, the lung may be susceptible to the development of pulmonary interstitial edema. Thus, patients with elevated Ppa, such as those with mitral stenosis, do not tolerate the Trendelenburg posi tion well.164 LATERAL DECUBITUS POSITION. In the lateral decubitus position, the dependent lung experiences a moderate decrease in FRC and is predisposed to atelectasis, whereas the nondependent lung may have increased FRC. The overall result is usually a slight to moderate increase in total-lung FRC.165 The kidney and lithotomy positions also cause small decreases in FRC above that caused by the supine position. The prone position may increase FRC moderately.165 i. VENTILATION PATTERN (RAPID SHALLOW BREATHING) Rapid shallow breathing is often a regular feature of anesthesia. Monotonous shallow breathing can cause a decrease in FRC, promote atelectasis, and decrease com pliance.40,139,166 These changes with rapid shallow breath ing are probably due to progressive increases in surface tension.166 Initially, these changes may cause hypoxemia with normocapnia and may be prevented or reversed by periodic large mechanical inspirations, spontaneous sighs, PEEP, or a combination of these techniques.166–168 challenge), only the dependent lung, at the end of each hour period, suffered a decrease in oxygenation. If the animals were turned every half-hour and received the same fluid challenge, neither lung suffered a decrease in oxygenation.
In patients who undergo surgery in the lateral decubi tus position (e.g., pulmonary resection, in which they have or will have a restricted pulmonary vascular bed) and receive excessive intravenous fluids, the risk of the dependent lung’s becoming edematous is certainly increased. These considerations also explain, in part, the beneficial effect of a continuously rotating (side-to-side) bed on the incidence of pulmonary complications in criti cally ill patients.160 g. HIGH INSPIRED OXYGEN CONCENTRATION AND ABSORPTION ATELECTASIS General anesthesia is usually administered with an increased FIO2. In patients who have areas of moderately low V /Q A ratios (0.1 to 0.01), administration of FIO2 greater than 0.3 adds enough O2 into the alveolar space in these areas to eliminate the shunt-like effect that they have, and total measured right-to-left shunting decreases. However, when patients with a significant amount of blood flow perfusing lung units with very low V /Q A ratios (0.01 to 0.0001) have a change in FIO2 from room air to 1.0, the very low V /Q A units virtually disappear, and a moderately large right-to-left shunt appears.16,17,161 In these studies, the increase in shunting was equal to the amount of blood flow previously perfusing the areas with low V /Q A ratios during the breathing of air. Thus, the effect of breathing O2 was to convert units that had low V /Q A ratios into shunt units. The pathologic basis for these data is the conversion of low V /Q A units into atelectatic units. The cause of the atelectatic shunting during O2 breathing is presumably a large increase in O2 Figure 5-36 Mongrel dogs anesthetized with pentobarbital were placed in a lateral decubitus position and subjected to progressive extracellular fluid expansion. They had a marked decrease in the partial pressure of oxygen (PO2) of blood draining the dependent lung (yellow line) and a smaller, much slower decrease in the Po2 of blood draining the nondependent lung (purple line). The pulmonary arteriovenous shunt rose progressively (blue line). (Redrawn from Ray JF, Yost L, Moallem S, et al: Immobility, hypoxemia, and pulmonary arteriovenous shunting. Arch Surg 109:537, 1974.) 7 6 5 4 3 Hours anesthetized 20 40 60 80 100 120 140 PO2 (mm Hg) 25 38 50 68 75 Extracellular fluid increase (%) 75 50 25 Pulmonary arteriovenous shunt (%) CHAPTER 5 Physiology of the Airway 151 CaO2, because more venous blood with lower CvO2 can admix with end-pulmonary capillary blood. Decreased QT may occur with myocardial failure and hypovolemia; the specific causes of these two conditions are beyond the scope of this chapter. Increased VO2 may occur with excessive stimulation of the sympathetic nervous system, hyperthermia, or shivering and can further contribute to impaired oxygenation of arterial blood.178 6. Inhibition of Hypoxic Pulmonary Vasoconstriction Decreased regional PAO2 causes regional pulmonary vaso constriction, which diverts blood flow away from hypoxic regions of the lung to better ventilated normoxic regions. The diversion of blood flow minimizes venous admixture from the underventilated or nonventilated lung regions. Inhibition of regional HPV could impair arterial oxygen ation by permitting increased venous admixture from hypoxic or atelectatic areas of the lung (see Fig. 5-9).
Because the pulmonary circulation is poorly endowed with smooth muscle, any condition that increases the pressure against which the vessels must constrict (i.e., Ppa) decreases HPV. Numerous clinical conditions can increase Ppa and therefore decrease HPV. Mitral steno sis,179 volume overload,179 low (but greater than room air) FIO2 in nondiseased lung,74 a progressive increase in the amount of diseased lung,74 thromboembolism,74 hypo thermia,180 and vasoactive drugs can all increase Ppa.64 Direct vasodilating drugs (e.g., isoproterenol, nitroglyc erin, sodium nitroprusside),64,59 inhaled anesthetics,65 and hypocapnia can directly decrease HPV.64,142 Selective application of PEEP to only the nondiseased lung can selectively increase PVR in the nondiseased lung and may divert blood flow back into the diseased lung.181 j. DECREASED REMOVAL OF SECRETIONS (DECREASED MUCOCILIARY FLOW) Tracheobronchial mucous glands and goblet cells produce mucus, which is swept by cilia up to the larynx, where it is swallowed or expectorated. This process clears inhaled organisms and particles from the lungs. The secreted mucus consists of a surface gel layer lying on top of a more liquid sol layer in which the cilia beat. The tips of the cilia propel the gel layer toward the larynx (upward) during the forward stroke. As the mucus streams upward and the total cross-sectional area of the airways dimin ishes, absorption takes place from the sol layer to main tain a constant depth of 5 mm.169 Poor systemic hydration and low inspired humidity reduce mucociliary flow by increasing the viscosity of secretions and slowing the ciliary beat.170–172 Mucociliary flow varies directly with body or mucosal temperature (low inspired temperature) over a range of 32° to 42° C.173,174 High FIO2 decreases mucociliary flow.175 Inflation of an ETT cuff suppresses tracheal mucus velocity,176 an effect that occurs within 1 hour, and apparently it does not matter whether a low- or high-compliance cuff is used. Passage of an uncuffed tube through the vocal cords and keeping it in situ for several hours does not affect tracheal mucus velocity.176 The mechanism for suppression of mucociliary clear ance by the ETT cuff is speculative. In the report of Sackner and colleagues,176 mucus velocity was decreased in the distal portion of the trachea, but the cuff was inflated in the proximal portion. Therefore, the phenom enon cannot be attributed solely to damming of mucus at the cuff site. One possibility is that the ETT cuff caused a critical increase in the thickness of the layer of mucus proceeding distally from the cuff. Another possi bility is that mechanical distention of the trachea by the ETT cuff initiated a neurogenic reflex arc that altered mucous secretions or the frequency of ciliary beating.
Other investigators showed that when all the forego ing factors were controlled, halothane reversibly and pro gressively decreased but did not stop mucus flow over an inspired concentration of 1 to 3 MAC.177 The halothane-induced depression of mucociliary clearance was proba bly due to depression of the ciliary beat, an effect that caused slow clearance of mucus from the distal and peripheral airways. In support of this hypothesis is the finding that cilia are morphologically similar throughout the animal kingdom. Inhaled anesthetics in clinical doses, including halothane, have been found to cause reversible depression of the ciliary beat of protozoa.115 5. Decreased Cardiac Output and Increased Oxygen Consumption Decreased QT in the presence of constant O2 consump tion ( VO2), increased VO2 in the presence of a constant QT , and decreased QT concomitant with increased VO2 must all result in lower CvO2. Venous blood with lowered CvO2 then flows through whichever shunt pathways exist, mixes with the oxygenated end-pulmonary capil lary blood, and lowers CaO2 (see Figs. 5-27 and 5-28). Figure 5-37 shows these relationships quantitatively for several different intrapulmonary shunts.100,101 The larger the intrapulmonary shunt, the greater the decrease in Figure 5-37 Effects of changes in cardiac output ( Q ) on the O2 content of end-pulmonary capillary, arterial (solid lines), and mixed venous (dashed lines) blood for various transpulmonary right-to-left shunts. The magnitudes of the shunts are indicated by the percent ages; the oxygen content of end-capillary blood is unaffected by the degree of shunting. Note that a given decrease in Q results in a greater decrease in the arterial content of O2 with larger shunts. (Redrawn from Kelman GF, Nunn JF, Prys-Roberts C, et al: The influ ence of the cardiac output on arterial oxygenation: A theoretical study. Br J Anaesth 39:450, 1967.) 9 7 5 3 1 20 22 16 18 12 14 10 8 Q L/min End capillary Arterial Mixed venous 5% 10% 20% 30% 5% 10% 20% 30% O2 content vols (%) • 152 PART 1 Basic Clinical Science Considerations edema as a result of increased pulmonary capillary per meability. Finally, the pulmonary embolus can increase PVR (by platelet-induced serotonin release,4 among other mechanisms) and decrease cardiac output.
After major hypotension, shock, blood loss, sepsis, or other conditions, noncardiogenic pulmonary edema may occur and lead to acute respiratory failure or ARDS.184 The syndrome can evolve during and after anesthesia and has the hallmark characteristics of decreased FRC and compliance and hypoxemia. After shock and trauma, plasma levels of serotonin, histamine, kinins, lysozymes, reactive oxygen species, fibrin degradation products, products of complement metabolism, and fatty acids all increase. Sepsis and endotoxemia may be present. Increased levels of activated complement stimulate neu trophils into chemotaxis in patients with trauma and 7. Paralysis In the supine position, the weight of the abdominal con tents pressing against the diaphragm is greatest in the dependent or posterior part of the diaphragm and least in the nondependent or anterior part of the diaphragm. In an awake patient breathing spontaneously, active tension in the diaphragm is capable of overcoming the weight of the abdominal contents, and the diaphragm moves most in the posterior portion (because the poste rior of the diaphragm is stretched higher into the chest, it has the smallest radius of curvature, and therefore it contracts most effectively) and least in the anterior portion. This circumstance is healthy because the greatest amount of ventilation occurs in areas with the most per fusion (posteriorly or dependently), and the least amount occurs in areas with the least perfusion (anteriorly or nondependently). During paralysis and positive-pressure breathing, the passive diaphragm is displaced by the posi tive pressure preferentially in the anterior, nondependent portion (where there is the least resistance to diaphrag matic movement) and is displaced minimally in the pos terior, dependent portion (where there is the most resistance to diaphragmatic movement). This circum stance is unhealthy because the greatest amount of ven tilation now occurs in areas with the least perfusion, and the least amount occurs in areas with the most perfu sion.156 However, the magnitude of the change in the diaphragmatic motion pattern with paralysis varies with body position.156,182 8. Right-to-Left Interatrial Shunting Acute arterial hypoxemia from a transient right-to-left shunt through a PFO has been described, particularly during emergence from anesthesia.70 However, unless a real-time technique of imaging the cardiac chambers is used (e.g., transesophageal echocardiography with color flow Doppler imaging),71 it is difficult to document an acute and transient right-to-left intracardiac shunt as a cause of arterial hypoxemia. Nonetheless, right-to-left shunting through a PFO has been described in virtually every conceivable clinical situation that afterloads the right side of the heart and increases right atrial pressure. When right-to-left shunting through a PFO is identified, administration of inhaled NO can decrease PVR and functionally close the PFO.183 9. Involvement of Mechanisms of Hypoxemia in Specific Diseases In any given pulmonary disease, many of the mechanisms of hypoxemia listed earlier may be involved.117 Pulmo nary embolism (air, fat, thrombi) (Fig. 5-38) and the evolution of ARDS (Fig. 5-39) are used to illustrate this point. A significant pulmonary embolus can cause severe increases in Ppa, and these increases can result in right-to-left transpulmonary shunting through opened arterio venous anastomoses and the foramen ovale (possible in 20% of patients), pulmonary edema in nonembolized regions of the lung, and inhibition of HPV. The embolus can cause hypoventilation through increased dead space ventilation. If the embolus contains platelets, serotonin can be released, and such release can cause hypoventila tion as a result of bronchoconstriction and pulmonary Figure 5-38 Mechanisms of hypoxemia during pulmonary embo lism. See the text for an explanation of the pathophysiologic flow diagram. AV, Arteriovenous; CAP PERM, capillary permeability; CC, closing capacity; FRC, functional residual capacity; HPV, hypoxic pulmonary vasoconstriction; PA, pulmonary artery. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) Decreased cardiac output Pulmonary embolus Increased dead space Hypoventilation Bronchoconstriction Serotonin Platelets Increased PA pressure AV anastomoses Foramen ovale Decreased HPV Transudate fluid Increased CAP PERM Decreased FRC-CC Figure 5-39 Mechanisms of hypoxemia during adult respiratory dis tress syndrome. See the text for an explanation of the pathophysi ologic flow diagram. CAP PERM, Capillary permeability; CC, closing capacity; FRC, functional residual capacity; HPV, hypoxic pulmo nary vasoconstriction; PA, pulmonary artery. (Redrawn with modifica tion from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) Decreased HPV ↑ PA pressure ↑ CAP PERM Decreased cardiac output Shock Transudate fluid Excessive crystalloid infusion Decreased FRC-CC O2 toxicity Supine position Surfactant CHAPTER 5 Physiology of the Airway 153 2. Hypoventilation Patients spontaneously hypoventilate during anesthesia because it is more difficult to breath (abnormal surgical position, increased airway resistance, decreased comp liance) and because they are less willing to breath (decreased respiratory drive due to anesthetics). Hypo ventilation results in hypercapnia (see Figs. 5-22 and 5-23).
3. Increased Dead Space Ventilation A decrease in Ppa, as during deliberate hypotension,185 can cause an increase in zone 1 and alveolar dead space ventilation. An increase in airway pressure (as with PEEP) can also cause an increase in zone 1 and alveolar dead space ventilation. Pulmonary embolism, thrombosis, and vascular obliteration (e.g., kinking, clamping, blocking of the pulmonary artery during surgery) can increase the amount of lung that is ventilated but unperfused. Vascu lar obliteration can also increase dead space ventilation; this occurs with age (VD/VT% = 33 + age/3). Rapid, short inspirations may be distributed preferentially to noncom pliant (short time constant for inflation) and badly per fused alveoli, whereas slow inspiration allows time for distribution to more compliant (long time constant for inflation) and better perfused alveoli. Thus, rapid, short inspirations may have a dead space ventilation effect.
The anesthesia apparatus increases total dead space (VD/VT) for two reasons. First, the apparatus simply increases the anatomic dead space. Inclusion of normal apparatus dead space increases the total VD/VT ratio from 33% to about 46% in intubated patients and to pancreatitis; activated neutrophils can damage endothe lial cells. These factors, along with pulmonary contusion (if it occurs), can individually or collectively increase pulmonary capillary permeability. After shock, acidosis, increased circulating catecholamines and sympathetic nervous system activity, leukotriene and prostaglandin release, histamine release, microembolism (with sero tonin release), increased intracranial pressure (with head injury), and alveolar hypoxia can occur and may individu ally or collectively (particularly after resuscitation) cause a moderate increase in Ppa. After shock, the normal com pensatory response to hypovolemia is movement of a protein-free fluid from the interstitial space into the vas cular space to restore vascular volume. Dilution of vas cular proteins by protein-free interstitial fluid can cause decreased capillary colloid oncotic pressure. Increased pulmonary capillary permeability and Ppa along with decreased capillary colloid oncotic pressure results in fluid transudation and pulmonary edema. In addition, decreased QT , inhibition of HPV, immobility, the supine position, excessive fluid administration, and an exces sively high FIO2 can contribute to the development of ARDS.
F. Mechanisms of Hypercapnia and Hypocapnia During Anesthesia 1. Hypercapnia Hypoventilation, increased dead space ventilation, increased CO2 production, and inadvertent switching off of a CO2 absorber can all cause hypercapnia (Fig. 5-40).
Figure 5-40 Schematic diagram of the causes of hypercapnia during anesthesia. An increase in carbon dioxide (CO2) production ( VCO2) increases the arterial partial pressure of CO2 (PaCO2) with a constant minute ventilation ( VE). Several events can increase alveolar dead space: a decrease in pulmonary artery pressure (Ppa), the application of positive end-expiratory pressure (PEEP), thromboembolism, and mechanical interference with pulmonary arterial flow (ligatures and kinking of vessels). Most commonly in trauma, surgery, and critical care, hypovolemia due to hemorrhage or third spacing leads to increased alveolar dead space and consequently to increased PaCO2. A decrease in VE causes an increase in PaCO2 with a constant VCO2 . It is possible for some anesthesia systems to cause rebreathing of CO2. Finally, the anesthesia apparatus may increase the anatomic dead space, and inadvertent switching off of a CO2 absorber in the presence of low fresh gas flow can increase PaCO2. ↑, increase; ↓, decrease. (Redrawn with modification from Benumof JL: Anesthesia for thoracic surgery, ed 2, Philadelphia, 1995, Saunders, Chapter 8.) ↑ Alveolar dead space Increased alveolar pressure (PEEP) Ligatures, kinking No pulmonary perfusion Thrombo-embolism Decreased pulmonary artery pressure (↓ Ppa) No CO2 absorber Apparatus dead space Rebreathing Hypoventilation (↓ VE) Increased CO2 production (↑ VCO2) • • 154 PART 1 Basic Clinical Science Considerations G. Physiologic Effects of Abnormalities in Respiratory Gases 1. Hypoxia The end products of aerobic metabolism (oxidative phos phorylation) are CO2 and water, both of which are easily diffusible and lost from the body. The essential feature of hypoxia is the cessation of oxidative phosphorylation when mitochondrial PO2 falls below a critical level. Anaerobic pathways, which produce energy (ATP) inef ficiently, are then used. The main anaerobic metabolites are hydrogen and lactate ions, which are not easily excreted. They accumulate in the circulation, where they can be quantified in terms of the base deficit and the lactate-pyruvate ratio.
Because the various organs have different blood flow and O2 consumption rates, the manifestations and clinical diagnosis of hypoxia are usually related to symptoms arising from the most vulnerable organ. This organ is usually the brain in an awake patient and the heart in an anesthetized patient (see later discussion), but in special circumstances it may be the spinal cord (e.g., aortic surgery), kidney (e.g., acute tubular necrosis), liver (e.g., hepatitis), or limb (e.g., claudication, gangrene).
The cardiovascular response to hypoxemia is a product of both reflex (neural and humoral) and direct effects (Table 5-5).187–189 The reflex effects occur first and are excitatory and vasoconstrictive. The neuroreflex effects result from aortic and carotid chemoreceptor, barorecep tor, and central cerebral stimulation, and the humoral reflex effects result from catecholamine and renin-angiotensin release. The direct local vascular effects of hypoxia are inhibitory and vasodilatory and occur late. The net response to hypoxia in a subject depends on the severity of the hypoxia, which determines the magnitude and balance between the inhibitory and excitatory com ponents; the balance may vary according to the type and depth of anesthesia and the degree of preexisting cardio vascular disease.
Mild arterial hypoxemia (arterial saturation less than normal but still 80% or higher) causes general activation of the sympathetic nervous system and release of cate cholamines. Consequently, the heart rate, stroke volume, QT, and myocardial contractility—as measured by a shortened pre-ejection period (PEP), left ventricular ejec tion time (LVET), and a decreased PEP/LVET ratio—are increased (Fig. 5-41).190 Changes in systemic vascular resistance (SVR) are usually slight. However, in patients under anesthesia with β-blockers, hypoxia (and hypercapnia when present) may cause circulating about 64% in patients breathing through a mask.186 Second, anesthesia circuits cause rebreathing of exhaled gases, which is equivalent to dead space ventilation. The rebreathing classification by Mapleson during spontane ous ventilation with Mapleson circuits is A (Magill), D, C, and B. The order of increasing rebreathing (decreasing clinical merit) during controlled ventilation is D, B, C, and A. There is no rebreathing in system E (Ayre’s T-piece) if the patient’s respiratory diastole is long enough to permit washout with a given fresh gas flow (a common event) or if the fresh gas flow is greater than the peak inspiratory flow rate (an uncommon event).
The effects of an increase in dead space can usually be counteracted by a corresponding increase in the respira tory VE. If, for example, the VE is 10 L/min and the VD/ VT ratio is 30%, alveolar ventilation is 7 L/min. If a pul monary embolism occurred and resulted in an increase in the VD/VT ratio to 50%, VE would need to be increased to 14 L/min to maintain an alveolar ventilation of 7 L/ min (14 L/min × 0.5).
4. Increased Carbon Dioxide Production All causes of increased O2 consumption also increase CO2 production; these causes include hyperthermia, shivering, catecholamine release (light anesthesia), hyper tension, and thyroid storm. If VE, total dead space, and V /Q A relationships are constant, an increase in CO2 production results in hypercapnia.
5. Inadvertent Switching Off of a Carbon Dioxide Absorber Many factors, such as patients’ ventilatory responsiveness to CO2 accumulation, fresh gas flow, circle system design, and CO2 production, determine whether hypercapnia results from accidental switching off or depletion of a circle CO2 absorber. However, high fresh gas flows (≥5 L/ min) minimize the problem with almost all systems for almost all patients.
6. Hypocapnia The mechanisms of hypocapnia are the reverse of those that produce hypercapnia. Thus, all other factors being equal, hyperventilation (spontaneous or controlled ven tilation), decreased VD ventilation (e.g., change from a mask airway to an ETT airway, decreased PEEP, increased Ppa, decreased rebreathing), and decreased CO2 produc tion (e.g., hypothermia, deep anesthesia, hypotension) lead to hypocapnia. By far the most common mechanism of hypocapnia is passive hyperventilation by mechanical means.
TABLE 5-5 Cardiovascular Response to Hypoxemia Hemodynamic Variable O2 Saturation (%) Heart Rate Systemic Blood Pressure Stroke Volume Cardiac Output SVR Predominant Response >80 ↑ ↑ ↑ ↑ No change Reflex, excitatory 60–80 ↑ Baroreceptor ↓ No change No change ↓ Local, depressant > reflex, excitatory <60 ↓ ↓ ↓ ↓ ↓ Local, depressant SVR, Systemic vascular resistance; ↑, increase; ↓, decrease.
CHAPTER 5 Physiology of the Airway 155 which in turn increases myocardial irritability. First, arte rial hypoxemia can directly decrease the myocardial O2 supply. Second, early tachycardia may result in increased myocardial O2 consumption, and decreased diastolic filling time may lead to decreased myocardial O2 supply. Third, early increased systemic blood pressure can cause an increased afterload on the left ventricle, which increases left ventricular O2 demand. Fourth, late systemic hypo tension may decrease myocardial O2 supply because of decreased diastolic perfusion pressure. Fifth, coronary blood flow reserve may be exhausted by a late, maximally increased coronary blood flow as a result of maximal coronary vasodilation.192 The level of hypoxemia that causes cardiac dysrhythmias cannot be predicted with certainty because the myocardial O2 supply-demand rela tionship in a given patient is not known (i.e., the degree of coronary artery atherosclerosis may not be known). However, if a myocardial area (or areas) become hypoxic or ischemic, or both, unifocal or multifocal premature ventricular contractions, ventricular tachycardia, and ven tricular fibrillation may occur.
The cardiovascular response to hypoxia includes a number of other important effects. Cerebral blood flow increases (even if hypocapnic hyperventilation is present). Ventilation is stimulated regardless of the reason for the hypoxia (see Fig. 5-41). The pulmonary distribution of blood flow is more homogeneous because of increased pulmonary artery pressure. Chronic hypoxia causes an increased Hb concentration and a right-shifted oxy-Hb curve (as a result of either an increase in 2,3-DPG or acidosis), which tends to raise tissue PO2.
2. Hyperoxia (Oxygen Toxicity) The dangers associated with inhalation of excessive O2 are multiple. Exposure to high O2 tension clearly causes pulmonary damage in healthy individuals.193,194 A dose-time toxicity curve for humans is available from a number of studies.193–195 Because the lungs of normal human vol unteers cannot be directly examined to determine the rate of onset and the course of toxicity, indirect measures such as onset of symptoms have been used to construct dose-time toxicity curves. Examination of the curve indi cates that 100% O2 should not be administered for more than 12 hours, 80% O2 for more than 24 hours, and 60% O2 for more than 36 hours.193–195 No measurable changes in pulmonary function or blood-gas exchange occur in humans during exposure to less than 50% O2, even for long periods.195 Nevertheless, it is important to note that in the clinical setting, these dose-time toxicity relation ships are often obscured because of the complex multi variable nature of the clinical setting.196 The dominant symptom of O2 toxicity in human vol unteers is substernal distress, which begins as mild irrita tion in the area of the carina and may be accompanied by occasional coughing.197 As exposure continues, the pain becomes more intense, and the urge to cough and to deep-breathe also becomes more intense. These symp toms progress to severe dyspnea, paroxysmal coughing, and decreased vital capacity when the FIO2 has been 1.0 for longer than 12 hours. If excessive O2 is discontinued at this point, recovery of mechanical lung function usually occurs within 12 to 24 hours, but more than 24 hours catecholamines to have only an α-receptor effect, the heart may be unstimulated (even depressed by a local hypoxia effect), and SVR may be increased. Conse quently, QT may be decreased in these patients. With moderate hypoxemia (arterial O2 saturation 60% to 80%), local vasodilation begins to predominate and SVR and blood pressure decrease, but the heart rate may con tinue to be increased because of a systemic hypotension-induced stimulation of baroreceptors. Finally, with severe hypoxemia (arterial saturation <60%), local depressant effects dominate and blood pressure falls rapidly; the pulse slows, shock develops, and the heart either fibril lates or becomes asystolic.
Significant preexisting hypotension converts a mild hypoxemic hemodynamic profile into a moderate hypox emic hemodynamic profile and converts a moderate hypoxemic hemodynamic profile into a severe hypox emic hemodynamic profile. Similarly, in well-anesthetized or sedated patients, early sympathetic nervous system reactivity to hypoxemia may be reduced and the effects of hypoxemia may be expressed only as bradycardia with severe hypotension and, ultimately, circulatory collapse.191 Hypoxemia can also promote cardiac dysrhythmias, which may in turn potentiate the already mentioned deleterious cardiovascular effects. Hypoxemia-induced dysrhythmias can be caused by multiple mechanisms; the mechanisms are interrelated because they all cause a decrease in the myocardial O2 supply-demand ratio, Figure 5-41 Changes in the minute ventilation and in the circulation of healthy awake humans during progressive isocapnic hypoxia and hyperoxic hypercapnia. Petco2, End-tidal PCO2; Peto2, end-tidal PO2; Q , cardiac output; S1, slope during the first phase of slowly increas ing ventilation and/or circulation; S2, slope during the second phase of sharply increasing ventilation and/or circulation; VE , expired minute ventilation. (Redrawn from Serebrovskaya TV: Comparison of respiratory and circulatory human responses to progressive hypoxia and hypercapnia. Respiration 59:35, 1992.) 700 500 300 100 110 90 70 100 120 80 60 40 40 60 70 50 30 Q (mL/min/kg) VE (mL/min/kg) PETO2 (mm Hg) Hypoxia Hypercapnia PETCO2 (mm Hg) S1 S2 S1 S2 S2 S2 S1 S1 • • 156 PART 1 Basic Clinical Science Considerations intestinal gas pressure in patients with intestinal obstruc tion, to decrease the size of an air embolus, and to aid in the absorption of pneumoperitoneum, pneumocephalus, and pneumothorax.
3. Hypercapnia The effects of CO2 on the cardiovascular system are as complex as those of hypoxia. Like hypoxemia, hypercap nia appears to cause direct depression of both cardiac muscle and vascular smooth muscle, but at the same time it causes reflex stimulation of the sympathoadrenal system, which compensates to a greater or lesser extent for the primary cardiovascular depression (see Fig. 5-41).189,192 With moderate to severe hypercapnia, a hyperkinetic circulation results with increased QT and increased systemic blood pressure.190 Even in patients under halothane anesthesia, plasma catecholamine levels increase in response to increased CO2 levels in much the same way as in conscious subjects. Thus, hypercapnia, like hypoxemia, may cause increased myocardial O2 demand (tachycardia, early hypertension) and decreased myocar dial O2 supply (tachycardia, late hypotension).
Table 5-6 summarizes the interaction of anesthesia with hypercapnia in humans; increased QT and decreased SVR should be emphasized.199,200 The increase in QT is most marked during anesthesia with drugs that enhance sympathetic activity and least marked with halothane and nitrous oxide. The decrease in SVR is most marked during enflurane anesthesia and hypercapnia. Hypercap nia is a potent pulmonary vasoconstrictor even after the inhalation of 3% isoflurane for 5 minutes.199 Dysrhythmias have been reported in unanesthetized humans during acute hypercapnia, but they have seldom been of serious import. A high PaCO2 level is, however, more dangerous during general anesthesia. With halo thane anesthesia, dysrhythmias frequently occur above a PaCO2 arrhythmic threshold that is often constant for a particular patient. Furthermore, halothane, enflurane, and isoflurane have been shown to prolong the QTC interval in humans, thereby increasing the risk for torsades de pointes ventricular tachycardia, which in turn is notori ous for decompensating into ventricular fibrillation.201 The maximum stimulatory respiratory effect is attained by a PaCO2 of about 100 mm Hg. With a higher PaCO2, stimulation is reduced, and at extremely high levels, res piration is depressed and later ceases altogether. The PCO2 ventilation-response curve is generally displaced to the right, and its slope is reduced by anesthetics and other depressant drugs.202 With profound anesthesia, the may be required in some individuals.195 As toxicity pro gresses, results of other pulmonary function studies such as compliance and blood gases show deterioration. Patho logically, in animals, the lesion progresses from tracheo bronchitis (exposure for 12 hours to a few days), to involvement of the alveolar septa with pulmonary inter stitial edema (exposure for a few days to 1 week), to pulmonary fibrosis of the edema (exposure for >1 week).198 Ventilatory depression can occur in patients who, by reason of drugs or disease, have been ventilating in response to a hypoxic drive. By definition, ventilatory depression that results from removal of a hypoxic drive through increasing the inspired O2 concentration causes hypercapnia but does not necessarily produce hypoxia (because of the increased FIO2).
Absorption atelectasis was described earlier (see “High Inspired Oxygen Concentration and Absorption Atelec tasis”). Retrolental fibroplasia, an abnormal proliferation of the immature retinal vasculature of a prematurely born infant, can occur after exposure to hyperoxia. Extremely premature infants are most susceptible to ret rolental fibroplasia (i.e., those <1.0 kg in birth weight and <28 weeks of gestation). The risk of retrolental fibroplasia exists whenever FIO2 causes PaO2 to be greater than 80 mm Hg for longer than 3 hours in an infant whose gestational age plus life age combined is less than 44 weeks. If the ductus arteriosus is patent, arterial blood samples should be drawn from the right radial artery; umbilical or lower-extremity PaO2 is lower than the PaO2 to which the eyes are exposed because of ductal shunting of unoxygenated blood.
The mode of action of O2 toxicity in tissues is complex, but interference with metabolism seems to be wide spread. Most importantly, many enzymes, particularly those with sulfhydryl groups, are inactivated by O2-derived free radicals.196 Neutrophil recruitment and release of mediators of inflammation occur next and greatly accelerate the extent of endothelial and epithelial damage and impairment of the surfactant systems.196 The most acute toxic effect of O2 in humans is a convulsive effect, which occurs during exposure to pressures in excess of 2 atmospheres (atm) absolute.
High inspired O2 concentrations can be of use thera peutically. Clearance of gas loculi in the body may be greatly accelerated by the inhalation of 100% O2. Inhala tion of 100% O2 creates a large nitrogen gradient from the gas space to the perfusing blood. As a result, nitrogen leaves the gas space and the space diminishes in size. Administration of O2 to remove gas may be used to ease TABLE 5-6 Cardiovascular Responses to Hypercapnia (Paco2 = 60–83 mm Hg) During Various Types of Anesthesia (1 MAC Equivalent Except for Nitrous Oxide) Anesthesia Heart Rate Contractility Cardiac Output Systemic Vascular Resistance Conscious ↑↑ ↑↑ ↑↑↑ ↓ Nitrous oxide 0 ↑ ↑↑ ↓↓ Halothane 0 ↑ ↑ ↓ Isoflurane ↑↑ ↑↑↑ ↑↑↑ ↓ The increase in the partial arterial pressure of carbon dioxide (Paco2) in conscious subjects was 11.5 mm Hg from a normal level of 38 mm Hg.
↑, <10% increase; ↑↑, 10–25% increase; ↑↑↑, >25% increase; 0, no change; ↓, <10% decrease; ↓↓, 10–25% decrease; ↓↓↓, >25% decrease; MAC, minimum alveolar concentration for adequate anesthesia in 50% of subjects.
CHAPTER 5 Physiology of the Airway 157 to the tissues, QT or tissue perfusion has to increase at a time when it may not be possible for it to do so. The cerebral effects of hypocapnia may be related to a state of cerebral acidosis and hypoxia because hypocapnia can cause a selective reduction in cerebral blood flow and also shifts the oxy-Hb curve to the left.208 Hypocapnia can cause V /Q A abnormalities by inhibit ing HPV or by causing bronchoconstriction and decreased CL. Finally, passive hypocapnia promotes apnea.
III. CONCLUSIONS The primary purpose of the respiratory system is to facili tate gas exchange of O2 and CO2 in the alveoli. At the alveoli, O2 combines with hemoglobin and is transported throughout the body by the circulatory system, while at the same time CO2 which has been transported from the tissues is removed to be exhaled via the alveoli. These respiratory functions are achieved by coordinated action of the upper and lower airways, alveoli, pulmonary blood flow, respiratory muscles, and metabolic sensors, along with medullary and neural-based control centers.
The lungs also serve a number of very important non pulmonary metabolic and humoral functions, as described in this chapter.
Ventilation is the process of bringing in O2 rich air through the airways to the alveoli (inhalation), where gas exchange occurs; then, during exhalation, the O2-depleted air (along with CO2 produced in tissues) is returned to the external environment. The process of ventilation is tightly regulated by neural and non-neural mechanisms.
Perfusion relates to the quantity of blood flowing by the alveoli. Pulmonary perfusion is generally equal to cardiac output, unless shunts occur. Ventilation is closely coupled with perfusion of the alveoli. The interaction between ventilation and perfusion (ventilation-perfusion [ V /Q A ] relationship) ultimately determines the gas exchange in the lungs.
The transport of O2 requires reversible binding of O2 to Hb, which is then unloaded at the tissues. The O2 flows through its concentration gradient to the extracellular space and cells. Intracellular concentrations of O2 vary within the cell, with the mitochondrial PO2 being very low compared to arterial and even mixed venous blood values. Furthermore, interaction of the circulatory system with the respiratory system adds another level of fine-tuning and complexity to the process of perfusion, ven tilation, and ventilation-perfusion interaction.
Cardiac and respiratory functions are closely inte grated with numerous feedback mechanisms designed to match ventilation with perfusion. The lungs and the heart are the only organs that receive the full QT . Accordingly, the lungs are anatomically well situated to perform many of the secondary (nonpulmonary) functions. The list of nonpulmonary functions continues to grow and includes filtering of metabolic products, conversion of important enzymes, and immune protection.
Pulmonary and nonpulmonary functions adapt to con stantly changing needs of the body. Understanding the basic physiologic mechanisms involved in pulmonary and nonpulmonary functions of the lungs is the key to appre ciating the pathophysiology of respiratory disorders and response curve may be flat or even sloping downward, and CO2 then acts as a respiratory depressant. In patients with ventilatory failure, CO2 narcosis occurs when PaCO2 rises to greater than 90 to 120 mm Hg. A 30% CO2 concentration is sufficient for the production of anesthe sia, and this concentration causes total but reversible flattening of the electroencephalogram.203 As expected, hypercapnia causes bronchodilation in both healthy persons and patients with lung disease.204 Quite apart from the effect of CO2 on ventilation, it exerts two other important effects that influence the oxygenation of the blood.117 First, if the concentration of nitrogen (or other inert gas) remains constant, the concentration of CO2 in alveolar gas can increase only at the expense of O2, which must be displaced. Thus, PAO2 and PaO2 may decrease. Second, hypercapnia shifts the oxy-Hb curve to the right, thereby facilitating tissue oxygenation.95 Chronic hypercapnia results in increased reabsorption of bicarbonate by the kidneys, which further raises the plasma bicarbonate level and constitutes a secondary or compensatory metabolic alkalosis. The decrease in renal reabsorption of bicarbonate in patients with chronic hypocapnia results in a further fall in plasma bicarbonate and produces a secondary or compensatory metabolic acidosis. In each case, arterial pH returns toward the normal value, but the bicarbonate ion concentration departs even further from normal.
Hypercapnia is accompanied by leakage of potassium from cells into plasma. Much of the potassium comes from the liver, probably from glucose release and mobi lization, which occur in response to the rise in plasma catecholamine levels.205 Because the plasma potassium level takes an appreciable time to return to normal, repeated bouts of hypercapnia at short intervals result in a stepwise rise in plasma potassium. Finally, hypercapnia can predispose the patient to other complications in the operating room; for example, the oculocephalic response is far more common during hypercapnia than during eucapnia.206 4. Hypocapnia In this section, hypocapnia is considered to be produced by passive hyperventilation (by the anesthesiologist or ventilator). Hypocapnia can cause a decrease in QT by three separate mechanisms. First, if it is present, an increase in intrathoracic pressure decreases QT . Second, hypocapnia is associated with withdrawal of sympathetic nervous system activity, and such withdrawal can decrease the inotropic state of the heart. Third, hypocapnia can increase pH, and the increased pH can decrease ionized calcium, which may in turn decrease the inotropic state of the heart. Hypocapnia with alkalosis also shifts the oxy-Hb curve to the left, which increases Hb affinity for O2 and thus impairs O2 unloading at the tissue level. The decrease in peripheral flow and the impaired ability to unload O2 to the tissues are compounded by an increase in whole-body O2 consumption as a result of increased pH-mediated uncoupling of oxidation from phosphoryla tion.207 A PaCO2 of 20 mm Hg increases tissue O2 con sumption by 30%. Consequently, hypocapnia may simultaneously increase tissue O2 demand and decrease tissue O2 supply. To have the same amount of O2 delivery 158 PART 1 Basic Clinical Science Considerations minute ventilation ( VE ) and airway resistance (RAW) and inversely related to CL (see Equation 9).
• Increased O2 affinity shifts the oxygen-hemoglobin (oxy-Hb) dissociation curve to the left (i.e., increases the affinity of Hb for O2, thus reducing P50, the oxygen concentration at which Hb is 50% saturated), whereas decreased O2 affinity shifts the oxy-Hb curve to the right (i.e., decreases Hb affinity for O2 and thus increases P50). The four primary processes that shift the oxy-Hb curve to the right are increased hydrogen ion concentration ([H+]), increased carbon dioxide tension (PCO2), increased 2,3-diphosphoglycerate (2,3-DPG), and increased temperature.
• The Bohr effect refers to the effect of PCO2 and [H+] ions on the oxy-Hb curve (i.e., increasing the propen sity for O2 to offload from Hb).
• The Haldane effect describes the shift in the CO2 dis sociation curve caused by oxygenation of Hb. Low PO2 shifts the CO2 dissociation curve to the left so that the blood is able to pick up more CO2 (e.g., in capillaries of rapidly metabolizing tissues). Highly oxygenated Hb (as occurs in the lungs) reduces the affinity of Hb for CO2, shifting the CO2 dissociation curve to the right and thereby increasing CO2 removal.
• CO2 is transported in the blood primarily in three dif ferent forms: physically dissolved in blood, bound to amino groups of proteins (e.g., Hb) as carbamate com pounds, and as bicarbonate ions.
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IV. CLINICAL PEARLS • The ventilated gas that participates in gas exchange is referred to as alveolar ventilation ( VA). The amount of gas that is wasted is referred to as dead space ventila tion (VD). The aggregate total of dead space ventilation is referred to as the physiologic dead space (VDphysiologic) and is divided into two subcomponents. The volume of gas that ventilates the conducting airways is called the anatomic dead space (VDanatomic), and the volume of gas that ventilates nonperfused alveoli is the alveolar dead space (VDalveolar).
• Ventilation-perfusion ( V/Q) relationships are impor tant in pulmonary gas exchange. At the top of the lung there is relatively high V/Q, whereas at the bottom, there is relatively low V/Q. However, most of the perfusion and most of the ventilation occur at the base, and perfusion is fairly well matched throughout the lung in normal, young, healthy individuals. Alveolar ventilation without perfusion results in alveolar dead space, and alveolar perfusion without ventilation results in a right-to-left transpulmonary shunt.
• The functional residual capacity (FRC) is the amount of gas in the lungs at end-exhalation during normal tidal breathing. The FRC is also equal to the sum of the expiratory reserve volume and the residual volume. The FRC has important clinical significance because it represents the major reservoir of oxygen in the body and is directly related to the time until desaturation after apnea. The FRC is also inversely proportional to the degree of low- V/Q alveoli and shunt. For example, morbidly obese patients have low FRCs, tend to desat urate quickly, and have many more atelectatic alveoli and shunt units than normal, age-matched patients.
• Lung compliance (CL, volume/pressure) is the inverse of elastance. CL is bimodal: it is low at low lung volumes, highest at normal lung volumes (normal FRC), and low at very high lung volumes. The formula for compliance is analogous to the mathematical formula used to cal culate capacitance in electronics.
• Factors that affect airway resistance include lung volume, bronchial smooth muscle tone, and the density/ viscosity of the inhaled gas.
• Pulmonary vessels constrict in response to hypoxia, hypercarbia, and acidosis, whereas systemic vessels dilate when exposed to these factors.
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The bases of a trapezoid have lengths 10 and 15. A segment parallel to
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The bases of a trapezoid have lengths 10 and 15. A segment parallel to the bases passes through the point of intersection of the diagonals and extends from one side to the other. What is the length of the segment?
A. 5
B. 12
C. √150
D. 25/√2
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lacktutor
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Post URL05 Feb 2020, 04:55
There is a formula for Harmonic M e a n=2 a b(a+b)M e a n=2 a b(a+b)
a, b — lengths of bases
The length of segment = 2∗10∗15(10+15)=300 25=12 2∗10∗15(10+15)=300 25=12
The answer is B.
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Post URL05 Feb 2020, 05:00
lacktutor
There is a formula for Harmonic M e a n=2 a b(a+b)M e a n=2 a b(a+b)
a, b — lengths of bases
The length of segment = 2∗10∗15(10+15)=300 25=12 2∗10∗15(10+15)=300 25=12
The answer is B.
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How did you recognize HM has to be used ??
Also, is this applicable to all quadrilaterals in similar questions ???
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Post URL05 Feb 2020, 09:58
Hi chetan2u ,
kindly help wih this one
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rajatchopra1994
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Post URL05 Feb 2020, 20:38
Explanation:
There is one formula for finding A segment parallel to the bases passes through the point of intersection of the diagonals.
let longer base = A = 15
Shorter base = B = 10
Segment Passing through intersection of diagonals = C
The length of the parallel line segment through the intersection of the diagonals is the harmonic mean of the bases of the trapezoid.
C-B/B = A-C/C
C-10/10 = 15-C/15
5C=60
C= 12
IMO-B
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metalhead2593
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Post URL14 Mar 2020, 04:04
rajatchopra1994
Explanation:
There is one formula for finding A segment parallel to the bases passes through the point of intersection of the diagonals.
let longer base = A = 15
Shorter base = B = 10
Segment Passing through intersection of diagonals = C
The length of the parallel line segment through the intersection of the diagonals is the harmonic mean of the bases of the trapezoid.
C-B/B = A-C/C
C-10/10 = 15-C/15
5C=60
C= 12
IMO-B
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I didn't know this formula, but I used estimation.
So, the length that we want to find will definitely be smaller than the average of two bases (unless they're equal to): L < (10+15)/2 = 12.5
Look at the choices, we can cross out C, D, E. Since A (L=5) is smaller than the length of one base (10), we're left with 12.
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11509 | https://en.wikipedia.org/wiki/P/O_ratio | Jump to content
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From Wikipedia, the free encyclopedia
The phosphate/oxygen ratio, or P/O ratio, refers to the amount of ATP produced from the movement of two electrons through a defined electron transport chain, terminated by reduction of an oxygen atom.
The P/O ratio is dependent on the number of hydrogen ions transported outward across an electrochemical gradient, and the number of protons which return inward through the membrane via an enzyme such as ATP synthase. The ATP synthase works by a rotary mechanism. The ATP generated will be dependent on the amount of ATP produced per rotation of the ATP synthase rotor, and the number of protons necessary to complete a rotation. Every full rotation produces 3 ATPs. According to current understanding of the mechanism of the F0 part, the number of protons translocated per rotation is exactly equal to the number of subunits in the c ring. Recent structural studies show that this is not the same for all organisms. For vertebrate mitochondrial ATP synthase, the number of c subunits is 8 . The synthase thus requires 8 protons to synthesize three ATP, or 8/3 protons/ATP.
Inward moving protons must not only power rotation of ATP synthase, but may also be used in the transport of products and precursors. Given the net charge differences between ATP and ADP, the enzyme ATP–ADP translocase dissipates the charge equivalent of one hydrogen ion from the gradient when moving ATP (outward) and ADP (inward) across the inner mitochondrial membrane. The electroneutral symport of phosphate ion and H+ results in importing one proton, without its charge, per phosphate. Taken together, import of ADP and Pi and export of the resulting ATP results in one proton imported, subtracting from the number available for use by the ATP synthase directly. Taking this into account, it takes 8/3 +1 or 3.67 protons for vertebrate mitochondria to synthesize one ATP in the cytoplasm from ADP and Pi in the cytoplasm.
Within aerobic respiration, the P/O ratio continues to be debated; however, current figures place it at 2.5 ATP per 1/2(O2) reduced to water, though some claim the ratio is 3. This figure arises from accepting that 10 H+ are transported out of the matrix per 2 e−, and 4 H+ are required to move inward to synthesize a molecule of ATP.
The H+/2e− ratios of the three major respiratory complexes are generally agreed to be 4, 4, and 2 for Complexes I, III, and IV respectively. The H+/O ratio thus depends whether the substrate electrons enter at the level of NADH (passing through all three for 10 H+/2e−) or ubiquinol (passing through only complexes III and IV for 6H+/2e−). The latter is the case when the substrate is succinate or extramitochondrial NADH being oxidized via the glycerol phosphate shuttle; or other UQH2-linked dehydrogenase. During normal aerobic respiration the ratio would be somewhere between these values, as the TCA cycle produces both NADH and ubiquinol.
The resulting P/O ratio would be the ratio of H/O and H/P; which is 10/3.67 or 2.73 for NADH-linked respiration, and 6/3.67 or 1.64 for UQH2-linked respiration, with actual values being somewhere between.
Notes
[edit]
^ Garrett & Grisham 2010, p.620.
^ Niklas Klusch; Bonnie J Murphy; Deryck J Mills; Özkan Yildiz; Werner Kühlbrandt (2017). "Structural basis of proton translocation and force generation in mitochondrial ATP synthase". eLife. 6: e33274. doi:10.7554/eLife.33274. PMC 5747523. PMID 29210357.{{cite journal}}: CS1 maint: article number as page number (link)
^ Watt, I.N.; Montgomery, M.G.; Runswick, M.J.; Leslie, A.G.W.; Walker, J.E. (2010). "Bioenergetic Cost of Making an Adenosine Triphosphate Molecule in Animal Mitochondria". Proc. Natl. Acad. Sci. U.S.A. 107 (39): 16823–16827. doi:10.1073/pnas.1011099107. PMC 2947889. PMID 20847295.
^ Zhou, A.; Rohou, A.; Schep, D.G.; Bason, J.V.; Montgomery, M.G.; Walker, J.E.; Grigorieff, N.; Rubinstein, J.L. (2015). "Structure and conformational states of the bovine mitochondrial ATP synthase by cryo-EM". eLife. 4: e10180. doi:10.7554/elife.10180. PMC 4718723. PMID 26439008.{{cite journal}}: CS1 maint: article number as page number (link)
^ Ferrier, Denise (2014). Harvey, Richard (ed.). Biochemistry. Lippincott's Illustrated Reviews (6th ed.). Philadelphia: Wolters Kluwer Health/Lippincott Williams & Wilkins. p. 77. ISBN 978-1-4511-7562-2. OCLC 797084711.
^ Garrett & Grisham 2010, p.620.
^ Hinkle PC. (2005). "P/O ratios of mitochondrial oxidative phosphorylation". Biochim Biophys Acta. 1706 (1–2): 1–11. doi:10.1016/j.bbabio.2004.09.004. PMID 15620362.
References
[edit]
Garrett RH & Grisham CM (2010). "Biochemistry 4th Ed.". Brooks/Cole.
Nelson DL & Cox MM (2008). "Principles of Biochemistry 5th Ed." WH Freeman and Company.
| v t e Metabolism, catabolism, anabolism |
| General | Metabolic pathway Metabolic network Primary nutritional groups |
| Energy metabolism | | | | --- | | Aerobic respiration | Glycolysis → Pyruvate decarboxylation → Citric acid cycle → Oxidative phosphorylation (electron transport chain + ATP synthase) | | Anaerobic respiration | Electron acceptors other than oxygen | | Fermentation | Glycolysis → Substrate-level phosphorylation + ABE + Ethanol + Lactic acid | |
| Specific paths | | | | | | | | --- --- --- | | Protein metabolism | Protein synthesis Catabolism (protein→peptide→amino acid) | | | --- | | Amino acid | Amino acid synthesis Amino acid degradation (amino acid→pyruvate, acetyl CoA, or TCA intermediate) Urea cycle | | Nucleotide metabolism | Purine metabolism Nucleotide salvage Pyrimidine metabolism Purine nucleotide cycle | | | Carbohydrate metabolism(carbohydrate catabolismand anabolism) | | | | | | | | --- --- --- | | Human | | | | Glycolysis ⇄ Gluconeogenesis | | Glycogenolysis ⇄ Glycogenesis | | Pentose phosphate pathway Fructolysis + Polyol pathway Galactolysis + Leloir pathway | | Glycosylation + N-linked + O-linked | | | Nonhuman | | | | Photosynthesis Anoxygenic photosynthesis Chemosynthesis Carbon fixation DeLey-Doudoroff pathway Entner-Doudoroff pathway | | Xylose metabolism Radiotrophism | | | | Lipid metabolism (lipolysis, lipogenesis) | | | | --- | | Fatty acid metabolism | Fatty acid degradation (Beta oxidation) Fatty acid synthesis | | Other | Steroid metabolism Sphingolipid metabolism Eicosanoid metabolism Ketosis Reverse cholesterol transport | | | Other | Metal metabolism + Iron metabolism Ethanol metabolism Phospagen system (ATP-PCr) Chlororespiration | |
| v t e Metabolism: Citric acid cycle enzymes |
| Cycle | Citrate synthase Aconitase Isocitrate dehydrogenase Oxoglutarate dehydrogenase Succinyl CoA synthetase Succinate dehydrogenase (SDHA) Fumarase Malate dehydrogenase and ETC |
| Anaplerotic | | | | --- | | to acetyl-CoA | Pyruvate dehydrogenase complex (E1, E2, E3) (regulated by Pyruvate dehydrogenase kinase and Pyruvate dehydrogenase phosphatase) | | to α-ketoglutaric acid | Glutamate dehydrogenase | | to succinyl-CoA | Methylmalonyl-CoA mutase | | to oxaloacetic acid | Pyruvate carboxylase Aspartate transaminase | |
| Mitochondrialelectron transport chain/oxidative phosphorylation | | | | --- | | Primary | Complex I/NADH dehydrogenase Complex II/Succinate dehydrogenase Coenzyme Q10 (CoQ10) Complex III/Coenzyme Q - cytochrome c reductase Cytochrome c Complex IV/Cytochrome c oxidase Coenzyme Q10 synthesis: COQ2 COQ3 COQ4 COQ5 COQ6 COQ7 COQ9 COQ10A COQ10B PDSS1 PDSS2 | | Other | Alternative oxidase Electron-transferring-flavoprotein dehydrogenase | |
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gcodetools.py
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1#!/usr/bin/env python3
2# coding=utf-8
3#
4# Copyright (C) 2005 Aaron Spike, aaron@ekips.org (super paths et al)
5# 2007 hugomatic... (gcode.py)
6# 2009 Nick Drobchenko, nick@cnc-club.ru (main developer)
7# 2011 Chris Lusby Taylor, clusbytaylor@enterprise.net (engraving functions)
8#
9# This program is free software; you can redistribute it and/or modify
10# it under the terms of the GNU General Public License as published by
11# the Free Software Foundation; either version 2 of the License, or
12# (at your option) any later version.
13#
14# This program is distributed in the hope that it will be useful,
15# but WITHOUT ANY WARRANTY; without even the implied warranty of
16# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
17# GNU General Public License for more details.
18#
19# You should have received a copy of the GNU General Public License
20# along with this program; if not, write to the Free Software
21# Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
22#
23"""
24 Comments starting "#LT" or "#CLT" are by Chris Lusby Taylor who rewrote the engraving function in 2011.
25 History of CLT changes to engraving and other functions it uses:
26 9 May 2011 Changed test of tool diameter to square it
27 10 May Note that there are many unused functions, including:
28 bound_to_bound_distance, csp_curvature_radius_at_t,
29 csp_special_points, csplength, rebuild_csp, csp_slope,
30 csp_simple_bound_to_point_distance, csp_bound_to_point_distance,
31 bez_at_t, bez_to_point_distance, bez_normalized_slope, matrix_mul, transpose
32 Fixed csp_point_inside_bound() to work if x outside bounds
33 20 May Now encoding the bisectors of angles.
34 23 May Using r/cos(a) instead of normalised normals for bisectors of angles.
35 23 May Note that Z values generated for engraving are in pixels, not mm.
36 Removed the biarc curves - straight lines are better.
37 24 May Changed Bezier slope calculation to be less sensitive to tiny differences in points.
38 Added use of self.options.engraving_newton_iterations to control accuracy
39 25 May Big restructure and new recursive function.
40 Changed the way I treat corners - I now find if the centre of a proposed circle is
41 within the area bounded by the line being tested and the two angle bisectors at
42 its ends. See get_radius_to_line().
43 29 May Eliminating redundant points. If A,B,C colinear, drop B
44 30 May Eliminating redundant lines in divided Beziers. Changed subdivision of lines
45 7Jun Try to show engraving in 3D
46 8 Jun Displaying in stereo 3D.
47 Fixed a bug in bisect - it could go wrong due to rounding errors if
48 1+x1.x2+y1.y2<0 which should never happen. BTW, I spotted a non-normalised normal
49 returned by csp_normalized_normal. Need to check for that.
50 9 Jun Corrected spelling of 'definition' but still match previous 'defention' and 'defenition' if found in file
51 Changed get_tool to find 1.6.04 tools or new tools with corrected spelling
52 10 Jun Put 3D into a separate layer called 3D, created unless it already exists
53 Changed csp_normalized_slope to reject lines shorter than 1e-9.
54 10 Jun Changed all dimensions seen by user to be mm/inch, not pixels. This includes
55 tool diameter, maximum engraving distance, tool shape and all Z values.
56 12 Jun ver 208 Now scales correctly if orientation points moved or stretched.
57 12 Jun ver 209. Now detect if engraving toolshape not a function of radius
58 Graphics now indicate Gcode toolpath, limited by min(tool diameter/2,max-dist)
59 24 Jan 2017 Removed hard-coded scale values from orientation point calculation
60 TODO Change line division to be recursive, depending on what line is touched. See line_divide
61"""
62
63version = '1.7'
64
65 import cmath
66 import copy
67 import math
68 import os
69 import re
70 import sys
71 import time
72 from functools import partial
73
74 import numpy
75
76 import inkex
77 frominkex.bezierimport bezierlength, bezierparameterize, beziertatlength
78 from inkex import Transform, PathElement, TextElement, Tspan, Group, Layer, Marker, CubicSuperPath, Style
79
80 if sys.version_info > 2:
81 xrange = range
82 unicode = str
83
84def ireplace(self, old, new, count=0):
85 pattern = re.compile(re.escape(old), re.I)
86 return re.sub(pattern, new, self, count)
87
88
84def ireplace(self, old, new, count=0): …)
89
94
95TAU = math.pi 2
96STRAIGHT_TOLERANCE = 0.0001
97STRAIGHT_DISTANCE_TOLERANCE = 0.0001
98ENGRAVING_TOLERANCE = 0.0001
99LOFT_LENGTHS_TOLERANCE = 0.0000001
100
101EMC_TOLERANCE_EQUAL = 0.00001
102
103options = {}
104defaults = {
105'header': """%
106(Header)
107(Generated by gcodetools from Inkscape.)
108(Using default header. To add your own header create file "header" in the output dir.)
109 M3
110(Header end.)
111""",
112'footer': """
113(Footer)
114 M5
115 G00 X0.0000 Y0.0000
116 M2
117(Using default footer. To add your own footer create file "footer" in the output dir.)
118(end)
119%"""
120}
121
122INTERSECTION_RECURSION_DEPTH = 10
123INTERSECTION_TOLERANCE = 0.00001
124
125def marker_style(stroke, marker='DrawCurveMarker', width=1):
126"""Set a marker style with some basic defaults"""
127 returnStyle(stroke=stroke, fill='none', stroke_width=width,
128 marker_end='url(#{})'.format(marker))
129
125def marker_style(stroke, marker='DrawCurveMarker', width=1): …)
130MARKER_STYLE = {
131"in_out_path_style": marker_style('#0072a7', 'InOutPathMarker'),
132"loft_style": {
133'main curve': marker_style('#88f', 'Arrow2Mend'),
134 },
135"biarc_style": {
136'biarc0': marker_style('#88f'),
137'biarc1': marker_style('#8f8'),
138'line': marker_style('#f88'),
139'area': marker_style('#777', width=0.1),
140 },
141"biarc_style_dark": {
142'biarc0': marker_style('#33a'),
143'biarc1': marker_style('#3a3'),
144'line': marker_style('#a33'),
145'area': marker_style('#222', width=0.3),
146 },
147"biarc_style_dark_area": {
148'biarc0': marker_style('#33a', width=0.1),
149'biarc1': marker_style('#3a3', width=0.1),
150'line': marker_style('#a33', width=0.1),
151'area': marker_style('#222', width=0.3),
152 },
153"biarc_style_i": {
154'biarc0': marker_style('#880'),
155'biarc1': marker_style('#808'),
156'line': marker_style('#088'),
157'area': marker_style('#999', width=0.3),
158 },
159"biarc_style_dark_i": {
160'biarc0': marker_style('#dd5'),
161'biarc1': marker_style('#d5d'),
162'line': marker_style('#5dd'),
163'area': marker_style('#aaa', width=0.3),
164 },
165"biarc_style_lathe_feed": {
166'biarc0': marker_style('#07f', width=0.4),
167'biarc1': marker_style('#0f7', width=0.4),
168'line': marker_style('#f44', width=0.4),
169'area': marker_style('#aaa', width=0.3),
170 },
171"biarc_style_lathe_passing feed": {
172'biarc0': marker_style('#07f', width=0.4),
173'biarc1': marker_style('#0f7', width=0.4),
174'line': marker_style('#f44', width=0.4),
175'area': marker_style('#aaa', width=0.3),
176 },
177"biarc_style_lathe_fine feed": {
178'biarc0': marker_style('#7f0', width=0.4),
179'biarc1': marker_style('#f70', width=0.4),
180'line': marker_style('#744', width=0.4),
181'area': marker_style('#aaa', width=0.3),
182 },
183"area artefact": Style(stroke='#ff0000', fill='#ffff00', stroke_width=1),
184"area artefact arrow": Style(stroke='#ff0000', fill='#ffff00', stroke_width=1),
185"dxf_points": Style(stroke="#ff0000", fill="#ff0000"),
186}
187
188
189
192
193def gcode_comment_str(s, replace_new_line=False):
194 if replace_new_line:
195 s = re.sub(r"[\n\r]+", ".", s)
196 res = ""
197 if s[-1] == "\n":
198 s = s[:-1]
199 for a in s.split("\n"):
200 if a != "":
201 res += "(" + re.sub(r"[\(\)\\n\r]", ".", a) + ")\n"
202 else:
203 res += "\n"
204 return res
205
206
193def gcode_comment_str(s, replace_new_line=False): …)
207
210
211
212def csp_from_polyline(line):
213 return [[[point[:] for _ in range(3)] for point in subline] for subline in line]
214
215
212def csp_from_polyline(line): …)
216def csp_remove_zero_segments(csp, tolerance=1e-7):
217 res = []
218 for subpath in csp:
219 if len(subpath) > 0:
220 res.append([subpath])
221 for sp1, sp2 in zip(subpath, subpath[1:]):
222 ifpoint_to_point_d2(sp1, sp2) <= tolerance andpoint_to_point_d2(sp1, sp2) <= tolerance andpoint_to_point_d2(sp1, sp2) <= tolerance:
223 res[-1][-1] = sp2
224 else:
225 res[-1].append(sp2)
226 return res
227
228
216def csp_remove_zero_segments(csp, tolerance=1e-7): …)
229def point_inside_csp(p, csp, on_the_path=True):
230# we'll do the raytracing and see how many intersections are there on the ray's way.
231# if number of intersections is even then point is outside.
232# ray will be x=p.x and y=>p.y
233# you can assign any value to on_the_path, by default if point is on the path
234# function will return thai it's inside the path.
235 x, y = p
236 ray_intersections_count = 0
237 for subpath in csp:
238
239 for i in range(1, len(subpath)):
240 sp1 = subpath[i - 1]
241 sp2 = subpath[i]
242 ax, ay, bx, by, cx, cy, dx, dy = csp_parameterize(sp1, sp2)
243 if ax == 0 and bx == 0 and cx == 0 and dx == x:
244# we've got a special case here
245 b = csp_true_bounds()
246 if b <= y <= b:
247# points is on the path
248 return on_the_path
249 else:
250# we can skip this segment because it won't influence the answer.
251 pass
252 else:
253 for t incsp_line_intersection([x, y], [x, y + 5], sp1, sp2):
254 if t == 0 or t == 1:
255# we've got another special case here
256 x1, y1 = csp_at_t(sp1, sp2, t)
257 if y1 == y:
258# the point is on the path
259 return on_the_path
260# if t == 0 we should have considered this case previously.
261 if t == 1:
262# we have to check the next segment if it is on the same side of the ray
263 st_d = csp_normalized_slope(sp1, sp2, 1)
264 if st_d == 0:
265 st_d = csp_normalized_slope(sp1, sp2, 0.99)
266
267 for j in range(1, len(subpath) + 1):
268 if (i + j) % len(subpath) == 0:
269 continue# skip the closing segment
270 sp11 = subpath[(i - 1 + j) % len(subpath)]
271 sp22 = subpath[(i + j) % len(subpath)]
272 ax1, ay1, bx1, by1, cx1, cy1, dx1, dy1 = csp_parameterize(sp1, sp2)
273 if ax1 == 0 and bx1 == 0 and cx1 == 0 and dx1 == x:
274 continue# this segment parallel to the ray, so skip it
275 en_d = csp_normalized_slope(sp11, sp22, 0)
276 if en_d == 0:
277 en_d = csp_normalized_slope(sp11, sp22, 0.01)
278 if st_d en_d <= 0:
279 ray_intersections_count += 1
280 break
281 else:
282 x1, y1 = csp_at_t(sp1, sp2, t)
283 if y1 == y:
284# the point is on the path
285 return on_the_path
286 else:
287 if y1 > y and 3 ax t 2 + 2 bx t + cx != 0: # if it's 0 the path only touches the ray
288 ray_intersections_count += 1
289 return ray_intersections_count % 2 == 1
290
291
229def point_inside_csp(p, csp, on_the_path=True): …)
292def csp_close_all_subpaths(csp, tolerance=0.000001):
293 for i in range(len(csp)):
294 ifpoint_to_point_d2(csp[i], csp[i][-1]) > tolerance 2:
295 csp[i][-1] = csp[i][-1][:]
296 csp[i] +=
297 else:
298 if csp[i] != csp[i][-1]:
299 csp[i][-1] = csp[i][:]
300 return csp
301
302
292def csp_close_all_subpaths(csp, tolerance=0.000001): …)
303def csp_simple_bound(csp):
304 minx = None
305 miny = None
306 maxx = None
307 maxy = None
308
309 for subpath in csp:
310 for sp in subpath:
311 for p in sp:
312 minx = min(minx, p) if minx is not None else p
313 miny = min(miny, p) if miny is not None else p
314 maxx = max(maxx, p) if maxx is not None else p
315 maxy = max(maxy, p) if maxy is not None else p
316 return minx, miny, maxx, maxy
317
318
303def csp_simple_bound(csp): …)
319def csp_segment_to_bez(sp1, sp2):
320 return sp1[1:] + sp2[:2]
321
322
319def csp_segment_to_bez(sp1, sp2): …)
323def csp_to_point_distance(csp, p, dist_bounds=(0, 1e100)):
324 min_dist = [1e100, 0, 0, 0]
325 for j in range(len(csp)):
326 for i in range(1, len(csp[j])):
327 d = csp_seg_to_point_distance(csp[j][i - 1], csp[j][i], p, sample_points=5)
328 if d < dist_bounds:
329 return [d, j, i, d]
330 else:
331 if d < min_dist:
332 min_dist = [d, j, i, d]
333 return min_dist
334
335
323def csp_to_point_distance(csp, p, dist_bounds=(0, 1e100)): …)
336def csp_seg_to_point_distance(sp1, sp2, p, sample_points=5):
337 ax, ay, bx, by, cx, cy, dx, dy = csp_parameterize(sp1, sp2)
338 dx = dx - p
339 dy = dy - p
340 if sample_points < 2:
341 sample_points = 2
342 d = min([(p - sp1) 2 + (p - sp1) 2, 0.], [(p - sp2) 2 + (p - sp2) 2, 1.])
343 for k in range(sample_points):
344 t = float(k) / (sample_points - 1)
345 i = 0
346 while i == 0 or abs(f) > 0.000001 and i < 20:
347 t2 = t 2
348 t3 = t 3
349 f = (ax t3 + bx t2 + cx t + dx) (3 ax t2 + 2 bx t + cx) + (ay t3 + by t2 + cy t + dy) (3 ay t2 + 2 by t + cy)
350 df = (6 ax t + 2 bx) (ax t3 + bx t2 + cx t + dx) + (3 ax t2 + 2 bx t + cx) 2 + (6 ay t + 2 by) (ay t3 + by t2 + cy t + dy) + (3 ay t2 + 2 by t + cy) 2
351 if df != 0:
352 t = t - f / df
353 else:
354 break
355 i += 1
356 if 0 <= t <= 1:
357 p1 = csp_at_t(sp1, sp2, t)
358 d1 = (p1 - p) 2 + (p1 - p) 2
359 if d1 < d:
360 d = [d1, t]
361 return d
362
363
336def csp_seg_to_point_distance(sp1, sp2, p, sample_points=5): …)
364def csp_seg_to_csp_seg_distance(sp1, sp2, sp3, sp4, dist_bounds=(0, 1e100), sample_points=5, tolerance=.01):
365# check the ending points first
366 dist = csp_seg_to_point_distance(sp1, sp2, sp3, sample_points)
367 dist += [0.]
368 if dist <= dist_bounds:
369 return dist
370 d = csp_seg_to_point_distance(sp1, sp2, sp4, sample_points)
371 if d < dist:
372 dist = d + [1.]
373 if dist <= dist_bounds:
374 return dist
375 d = csp_seg_to_point_distance(sp3, sp4, sp1, sample_points)
376 if d < dist:
377 dist = [d, 0., d]
378 if dist <= dist_bounds:
379 return dist
380 d = csp_seg_to_point_distance(sp3, sp4, sp2, sample_points)
381 if d < dist:
382 dist = [d, 1., d]
383 if dist <= dist_bounds:
384 return dist
385 sample_points -= 2
386 if sample_points < 1:
387 sample_points = 1
388 ax1, ay1, bx1, by1, cx1, cy1, dx1, dy1 = csp_parameterize(sp1, sp2)
389 ax2, ay2, bx2, by2, cx2, cy2, dx2, dy2 = csp_parameterize(sp3, sp4)
390# try to find closes points using Newtons method
391 for k in range(sample_points):
392 for j in range(sample_points):
393 t1 = float(k + 1) / (sample_points + 1)
394 t2 = float(j) / (sample_points + 1)
395
396 t12 = t1 t1
397 t13 = t1 t1 t1
398 t22 = t2 t2
399 t23 = t2 t2 t2
400 i = 0
401
402 F1 = [0, 0]
403 F2 =
404 F = 1e100
405 x = ax1 t13 + bx1 t12 + cx1 t1 + dx1 - (ax2 t23 + bx2 t22 + cx2 t2 + dx2)
406 y = ay1 t13 + by1 t12 + cy1 t1 + dy1 - (ay2 t23 + by2 t22 + cy2 t2 + dy2)
407 while i < 2 or abs(F - Flast) > tolerance and i < 30:
408 f1x = 3 ax1 t12 + 2 bx1 t1 + cx1
409 f1y = 3 ay1 t12 + 2 by1 t1 + cy1
410 f2x = 3 ax2 t22 + 2 bx2 t2 + cx2
411 f2y = 3 ay2 t22 + 2 by2 t2 + cy2
412 F1 = 2 f1x x + 2 f1y y
413 F1 = -2 f2x x - 2 f2y y
414 F2 = 2 (6 ax1 t1 + 2 bx1) x + 2 f1x f1x + 2 (6 ay1 t1 + 2 by1) y + 2 f1y f1y
415 F2 = -2 f1x f2x - 2 f1y f2y
416 F2 = -2 f2x f1x - 2 f2y f1y
417 F2 = -2 (6 ax2 t2 + 2 bx2) x + 2 f2x f2x - 2 (6 ay2 t2 + 2 by2) y + 2 f2y f2y
418 F2 = inv_2x2(F2)
419 if F2 is not None:
420 t1 -= (F2 F1 + F2 F1)
421 t2 -= (F2 F1 + F2 F1)
422 t12 = t1 t1
423 t13 = t1 t1 t1
424 t22 = t2 t2
425 t23 = t2 t2 t2
426 x = ax1 t13 + bx1 t12 + cx1 t1 + dx1 - (ax2 t23 + bx2 t22 + cx2 t2 + dx2)
427 y = ay1 t13 + by1 t12 + cy1 t1 + dy1 - (ay2 t23 + by2 t22 + cy2 t2 + dy2)
428 Flast = F
429 F = x x + y y
430 else:
431 break
432 i += 1
433 if F < dist and 0 <= t1 <= 1 and 0 <= t2 <= 1:
434 dist = [F, t1, t2]
435 if dist <= dist_bounds:
436 return dist
437 return dist
438
439
364def csp_seg_to_csp_seg_distance(sp1, sp2, sp3, sp4, dist_bounds=(0, 1e100), sample_points=5, tolerance=.01): …)
440def csp_to_csp_distance(csp1, csp2, dist_bounds=(0, 1e100), tolerance=.01):
441 dist = [1e100, 0, 0, 0, 0, 0, 0]
442 for i1 in range(len(csp1)):
443 for j1 in range(1, len(csp1[i1])):
444 for i2 in range(len(csp2)):
445 for j2 in range(1, len(csp2[i2])):
446 d = csp_seg_bound_to_csp_seg_bound_max_min_distance(csp1[i1][j1 - 1], csp1[i1][j1], csp2[i2][j2 - 1], csp2[i2][j2])
447 if d >= dist_bounds:
448 continue
449 if d < dist_bounds:
450 return [d, i1, j1, 1, i2, j2, 1]
451 d = csp_seg_to_csp_seg_distance(csp1[i1][j1 - 1], csp1[i1][j1], csp2[i2][j2 - 1], csp2[i2][j2], dist_bounds, tolerance=tolerance)
452 if d < dist:
453 dist = [d, i1, j1, d, i2, j2, d]
454 if dist <= dist_bounds:
455 return dist
456 if dist >= dist_bounds:
457 return dist
458 return dist
459
460
440def csp_to_csp_distance(csp1, csp2, dist_bounds=(0, 1e100), tolerance=.01): …)
461def csp_split(sp1, sp2, t=.5):
462 [x1, y1] = sp1
463 [x2, y2] = sp1
464 [x3, y3] = sp2
465 [x4, y4] = sp2
466 x12 = x1 + (x2 - x1) t
467 y12 = y1 + (y2 - y1) t
468 x23 = x2 + (x3 - x2) t
469 y23 = y2 + (y3 - y2) t
470 x34 = x3 + (x4 - x3) t
471 y34 = y3 + (y4 - y3) t
472 x1223 = x12 + (x23 - x12) t
473 y1223 = y12 + (y23 - y12) t
474 x2334 = x23 + (x34 - x23) t
475 y2334 = y23 + (y34 - y23) t
476 x = x1223 + (x2334 - x1223) t
477 y = y1223 + (y2334 - y1223) t
478 return [sp1, sp1, [x12, y12]], ,
479
480
461def csp_split(sp1, sp2, t=.5): …)
481def csp_true_bounds(csp):
482# Finds minx,miny,maxx,maxy of the csp and return their (x,y,i,j,t)
483 minx = [float("inf"), 0, 0, 0]
484 maxx = [float("-inf"), 0, 0, 0]
485 miny = [float("inf"), 0, 0, 0]
486 maxy = [float("-inf"), 0, 0, 0]
487 for i in range(len(csp)):
488 for j in range(1, len(csp[i])):
489 ax, ay, bx, by, cx, cy, x0, y0 = bezierparameterize((csp[i][j - 1], csp[i][j - 1], csp[i][j], csp[i][j]))
490 roots = cubic_solver(0, 3 ax, 2 bx, cx) + [0, 1]
491 for root in roots:
492 if type(root) is complex and abs(root.imag) < 1e-10:
493 root = root.real
494 if type(root) is not complex and 0 <= root <= 1:
495 y = ay (root 3) + by (root 2) + cy root + y0
496 x = ax (root 3) + bx (root 2) + cx root + x0
497 maxx = max([x, y, i, j, root], maxx)
498 minx = min([x, y, i, j, root], minx)
499
500 roots = cubic_solver(0, 3 ay, 2 by, cy) + [0, 1]
501 for root in roots:
502 if type(root) is complex and root.imag == 0:
503 root = root.real
504 if type(root) is not complex and 0 <= root <= 1:
505 y = ay (root 3) + by (root 2) + cy root + y0
506 x = ax (root 3) + bx (root 2) + cx root + x0
507 maxy = max([y, x, i, j, root], maxy)
508 miny = min([y, x, i, j, root], miny)
509 maxy, maxy = maxy, maxy
510 miny, miny = miny, miny
511
512 return minx, miny, maxx, maxy
513
514
481def csp_true_bounds(csp): …)
515
522def csp_segments_intersection(sp1, sp2, sp3, sp4):
523 a = csp_segment_to_bez(sp1, sp2)
524 b = csp_segment_to_bez(sp3, sp4)
525
526 def polish_intersection(a, b, ta, tb, tolerance=INTERSECTION_TOLERANCE):
527 ax, ay, bx, by, cx, cy, dx, dy = bezierparameterize(a)
528 ax1, ay1, bx1, by1, cx1, cy1, dx1, dy1 = bezierparameterize(b)
529 i = 0
530 F = [.0, .0]
531 F1 =
532 while i == 0 or (abs(F) 2 + abs(F) 2 > tolerance and i < 10):
533 ta3 = ta 3
534 ta2 = ta 2
535 tb3 = tb 3
536 tb2 = tb 2
537 F = ax ta3 + bx ta2 + cx ta + dx - ax1 tb3 - bx1 tb2 - cx1 tb - dx1
538 F = ay ta3 + by ta2 + cy ta + dy - ay1 tb3 - by1 tb2 - cy1 tb - dy1
539 F1 = 3 ax ta2 + 2 bx ta + cx
540 F1 = -3 ax1 tb2 - 2 bx1 tb - cx1
541 F1 = 3 ay ta2 + 2 by ta + cy
542 F1 = -3 ay1 tb2 - 2 by1 tb - cy1
543 det = F1 F1 - F1 F1
544 if det != 0:
545 F1 =
546 ta = ta - (F1 F + F1 F)
547 tb = tb - (F1 F + F1 F)
548 else:
549 break
550 i += 1
551
552 return ta, tb
553
554 def recursion(a, b, ta0, ta1, tb0, tb1, depth_a, depth_b):
555 global bezier_intersection_recursive_result
556 if a == b:
557 bezier_intersection_recursive_result +=
558 return
559 tam = (ta0 + ta1) / 2
560 tbm = (tb0 + tb1) / 2
561 if depth_a > 0 and depth_b > 0:
562 a1, a2 = bez_split(a, 0.5)
563 b1, b2 = bez_split(b, 0.5)
564 ifbez_bounds_intersect(a1, b1):
565 recursion(a1, b1, ta0, tam, tb0, tbm, depth_a - 1, depth_b - 1)
566 ifbez_bounds_intersect(a2, b1):
567 recursion(a2, b1, tam, ta1, tb0, tbm, depth_a - 1, depth_b - 1)
568 ifbez_bounds_intersect(a1, b2):
569 recursion(a1, b2, ta0, tam, tbm, tb1, depth_a - 1, depth_b - 1)
570 ifbez_bounds_intersect(a2, b2):
571 recursion(a2, b2, tam, ta1, tbm, tb1, depth_a - 1, depth_b - 1)
572 elif depth_a > 0:
573 a1, a2 = bez_split(a, 0.5)
574 ifbez_bounds_intersect(a1, b):
575 recursion(a1, b, ta0, tam, tb0, tb1, depth_a - 1, depth_b)
576 ifbez_bounds_intersect(a2, b):
577 recursion(a2, b, tam, ta1, tb0, tb1, depth_a - 1, depth_b)
578 elif depth_b > 0:
579 b1, b2 = bez_split(b, 0.5)
580 ifbez_bounds_intersect(a, b1):
581 recursion(a, b1, ta0, ta1, tb0, tbm, depth_a, depth_b - 1)
582 ifbez_bounds_intersect(a, b2):
583 recursion(a, b2, ta0, ta1, tbm, tb1, depth_a, depth_b - 1)
584 else: # Both segments have been subdivided enough. Let's get some intersections :).
585 intersection, t1, t2 = straight_segments_intersection([a] + [a], [b] + [b])
586 if intersection:
587 if intersection == "Overlap":
588 t1 = (max(0, min(1, t1)) + max(0, min(1, t1))) / 2
589 t2 = (max(0, min(1, t2)) + max(0, min(1, t2))) / 2
590 bezier_intersection_recursive_result +=
591
592 global bezier_intersection_recursive_result
593 bezier_intersection_recursive_result = []
594 recursion(a, b, 0., 1., 0., 1., INTERSECTION_RECURSION_DEPTH, INTERSECTION_RECURSION_DEPTH)
595 intersections = bezier_intersection_recursive_result
596 for i in range(len(intersections)):
597 if len(intersections[i]) < 5 or intersections[i] != "Overlap":
598 intersections[i] = polish_intersection(a, b, intersections[i], intersections[i])
599 return intersections
600
601
522def csp_segments_intersection(sp1, sp2, sp3, sp4): …)
602def csp_segments_true_intersection(sp1, sp2, sp3, sp4):
603 intersections = csp_segments_intersection(sp1, sp2, sp3, sp4)
604 res = []
605 for intersection in intersections:
606 if (
607 (len(intersection) == 5 and intersection == "Overlap"and (0 <= intersection <= 1 or 0 <= intersection <= 1) and (0 <= intersection <= 1 or 0 <= intersection <= 1))
608 or (0 <= intersection <= 1 and 0 <= intersection <= 1)
609 ):
610 res += [intersection]
611 return res
612
613
602def csp_segments_true_intersection(sp1, sp2, sp3, sp4): …)
614def csp_get_t_at_curvature(sp1, sp2, c, sample_points=16):
615# returns a list containing [t1,t2,t3,...,tn], 0<=ti<=1...
616 if sample_points < 2:
617 sample_points = 2
618 tolerance = .0000000001
619 res = []
620 ax, ay, bx, by, cx, cy, dx, dy = csp_parameterize(sp1, sp2)
621 for k in range(sample_points):
622 t = float(k) / (sample_points - 1)
623 i = 0
624 F = 1e100
625 while i < 2 or abs(F) > tolerance and i < 17:
626 try: # some numerical calculation could exceed the limits
627 t2 = t t
628# slopes...
629 f1x = 3 ax t2 + 2 bx t + cx
630 f1y = 3 ay t2 + 2 by t + cy
631 f2x = 6 ax t + 2 bx
632 f2y = 6 ay t + 2 by
633 f3x = 6 ax
634 f3y = 6 ay
635 d = (f1x 2 + f1y 2) 1.5
636 F1 = (
637 ((f1x f3y - f3x f1y) d - (f1x f2y - f2x f1y) 3. (f2x f1x + f2y f1y) ((f1x 2 + f1y 2) .5)) /
638 ((f1x 2 + f1y 2) 3)
639 )
640 F = (f1x f2y - f1y f2x) / d - c
641 t -= F / F1
642 except:
643 break
644 i += 1
645 if 0 <= t <= 1 and F <= tolerance:
646 if len(res) == 0:
647 res.append(t)
648 for i in res:
649 if abs(t - i) <= 0.001:
650 break
651 if not abs(t - i) <= 0.001:
652 res.append(t)
653 return res
654
655
614def csp_get_t_at_curvature(sp1, sp2, c, sample_points=16): …)
656def csp_max_curvature(sp1, sp2):
657 ax, ay, bx, by, cx, cy, dx, dy = csp_parameterize(sp1, sp2)
658 tolerance = .0001
659 F = 0.
660 i = 0
661 while i < 2 or F - Flast < tolerance and i < 10:
662 t = .5
663 f1x = 3 ax t 2 + 2 bx t + cx
664 f1y = 3 ay t 2 + 2 by t + cy
665 f2x = 6 ax t + 2 bx
666 f2y = 6 ay t + 2 by
667 f3x = 6 ax
668 f3y = 6 ay
669 d = pow(f1x 2 + f1y 2, 1.5)
670 if d != 0:
671 Flast = F
672 F = (f1x f2y - f1y f2x) / d
673 F1 = (
674 (d (f1x f3y - f3x f1y) - (f1x f2y - f2x f1y) 3. (f2x f1x + f2y f1y) pow(f1x 2 + f1y 2, .5)) /
675 (f1x 2 + f1y 2) 3
676 )
677 i += 1
678 if F1 != 0:
679 t -= F / F1
680 else:
681 break
682 else:
683 break
684 return t
685
686
656def csp_max_curvature(sp1, sp2): …)
687def csp_curvature_at_t(sp1, sp2, t, depth=3):
688 ax, ay, bx, by, cx, cy, dx, dy = bezierparameterize(csp_segment_to_bez(sp1, sp2))
689
690# curvature = (x'y''-y'x'') / (x'^2+y'^2)^1.5
691
692 f1x = 3 ax t 2 + 2 bx t + cx
693 f1y = 3 ay t 2 + 2 by t + cy
694 f2x = 6 ax t + 2 bx
695 f2y = 6 ay t + 2 by
696 d = (f1x 2 + f1y 2) 1.5
697 if d != 0:
698 return (f1x f2y - f1y f2x) / d
699 else:
700 t1 = f1x f2y - f1y f2x
701 if t1 > 0:
702 return 1e100
703 if t1 < 0:
704 return -1e100
705# Use the Lapitals rule to solve 0/0 problem for 2 times...
706 t1 = 2 (bx ay - ax by) t + (ay cx - ax cy)
707 if t1 > 0:
708 return 1e100
709 if t1 < 0:
710 return -1e100
711 t1 = bx ay - ax by
712 if t1 > 0:
713 return 1e100
714 if t1 < 0:
715 return -1e100
716 if depth > 0:
717# little hack ;^) hope it won't influence anything...
718 returncsp_curvature_at_t(sp1, sp2, t 1.004, depth - 1)
719 return 1e100
720
721
687def csp_curvature_at_t(sp1, sp2, t, depth=3): …)
722def csp_subpath_ccw(subpath):
723# Remove all zero length segments
724 s = 0
725 if (P(subpath[-1]) - P(subpath)).l2() > 1e-10:
726 subpath[-1] = subpath[-1]
727 subpath = subpath
728 subpath += ]
729 pl = subpath[-1]
730 for sp1 in subpath:
731 for p in sp1:
732 s += (p - pl) (p + pl)
733 pl = p
734 return s < 0
735
736
722def csp_subpath_ccw(subpath): …)
737def csp_at_t(sp1, sp2, t):
738 ax = sp1
739 bx = sp1
740 cx = sp2
741 dx = sp2
742
743 ay = sp1
744 by = sp1
745 cy = sp2
746 dy = sp2
747
748 x1 = ax + (bx - ax) t
749 y1 = ay + (by - ay) t
750
751 x2 = bx + (cx - bx) t
752 y2 = by + (cy - by) t
753
754 x3 = cx + (dx - cx) t
755 y3 = cy + (dy - cy) t
756
757 x4 = x1 + (x2 - x1) t
758 y4 = y1 + (y2 - y1) t
759
760 x5 = x2 + (x3 - x2) t
761 y5 = y2 + (y3 - y2) t
762
763 x = x4 + (x5 - x4) t
764 y = y4 + (y5 - y4) t
765
766 return [x, y]
767
768
737def csp_at_t(sp1, sp2, t): …)
769def csp_at_length(sp1, sp2, l=0.5, tolerance=0.01):
770 bez = (sp1[:], sp1[:], sp2[:], sp2[:])
771 t = beziertatlength(bez, l, tolerance)
772 returncsp_at_t(sp1, sp2, t)
773
774
769def csp_at_length(sp1, sp2, l=0.5, tolerance=0.01): …)
775def cspseglength(sp1, sp2, tolerance=0.01):
776 bez = (sp1[:], sp1[:], sp2[:], sp2[:])
777 return bezierlength(bez, tolerance)
778
779
775def cspseglength(sp1, sp2, tolerance=0.01): …)
780def csp_line_intersection(l1, l2, sp1, sp2):
781 dd = l1
782 cc = l2 - l1
783 bb = l1
784 aa = l2 - l1
785 if aa == cc == 0:
786 return []
787 if aa:
788 coef1 = cc / aa
789 coef2 = 1
790 else:
791 coef1 = 1
792 coef2 = aa / cc
793 bez = (sp1[:], sp1[:], sp2[:], sp2[:])
794 ax, ay, bx, by, cx, cy, x0, y0 = bezierparameterize(bez)
795 a = coef1 ay - coef2 ax
796 b = coef1 by - coef2 bx
797 c = coef1 cy - coef2 cx
798 d = coef1 (y0 - bb) - coef2 (x0 - dd)
799 roots = cubic_solver(a, b, c, d)
800 retval = []
801 for i in roots:
802 if type(i) is complex and abs(i.imag) < 1e-7:
803 i = i.real
804 if type(i) is not complex and -1e-10 <= i <= 1. + 1e-10:
805 retval.append(i)
806 return retval
807
808
780def csp_line_intersection(l1, l2, sp1, sp2): …)
809def csp_split_by_two_points(sp1, sp2, t1, t2):
810 if t1 > t2:
811 t1, t2 = t2, t1
812 if t1 == t2:
813 sp1, sp2, sp3 = csp_split(sp1, sp2, t1)
814 return [sp1, sp2, sp2, sp3]
815 elif t1 <= 1e-10 and t2 >= 1. - 1e-10:
816 return [sp1, sp1, sp2, sp2]
817 elif t1 <= 1e-10:
818 sp1, sp2, sp3 = csp_split(sp1, sp2, t2)
819 return [sp1, sp1, sp2, sp3]
820 elif t2 >= 1. - 1e-10:
821 sp1, sp2, sp3 = csp_split(sp1, sp2, t1)
822 return [sp1, sp2, sp3, sp3]
823 else:
824 sp1, sp2, sp3 = csp_split(sp1, sp2, t1)
825 sp2, sp3, sp4 = csp_split(sp2, sp3, (t2 - t1) / (1 - t1))
826 return [sp1, sp2, sp3, sp4]
827
828
809def csp_split_by_two_points(sp1, sp2, t1, t2): …)
829def csp_seg_split(sp1, sp2, points):
830# points is float=t or list [t1, t2, ..., tn]
831 if type(points) is float:
832 points = [points]
833 points.sort()
834 res = [sp1, sp2]
835 last_t = 0
836 for t in points:
837 if 1e-10 < t < 1. - 1e-10:
838 sp3, sp4, sp5 = csp_split(res[-2], res[-1], (t - last_t) / (1 - last_t))
839 last_t = t
840 res[-2:] = [sp3, sp4, sp5]
841 return res
842
843
829def csp_seg_split(sp1, sp2, points): …)
844def csp_subpath_split_by_points(subpath, points):
845# points are [[i,t]...] where i-segment's number
846 points.sort()
847 points = + points +
848 parts = []
849 for int1, int2 in zip(points, points[1:]):
850 if int1 == int2:
851 continue
852 if int1 == 1.:
853 int1 += 1
854 int1 = 0.
855 if int1 == int2:
856 continue
857 if int2 == 0.:
858 int2 -= 1
859 int2 = 1.
860 if int1 == 0 and int2 == len(subpath) - 1: # and small(int1) and small(int2-1) :
861 continue
862 if int1 == int2: # same segment
863 sp = csp_split_by_two_points(subpath[int1 - 1], subpath[int1], int1, int2)
864 if sp != sp:
865 parts += ]
866 else:
867 sp5, sp1, sp2 = csp_split(subpath[int1 - 1], subpath[int1], int1)
868 sp3, sp4, sp5 = csp_split(subpath[int2 - 1], subpath[int2], int2)
869 if int1 == int2 - 1:
870 parts += , sp4]]
871 else:
872 parts +=
873 return parts
874
875
844def csp_subpath_split_by_points(subpath, points): …)
876def arc_from_s_r_n_l(s, r, n, l):
877 if abs(n 2 + n 2 - 1) > 1e-10:
878 n = normalize(n)
879 returnarc_from_c_s_l([s + n r, s + n r], s, l)
880
881
876def arc_from_s_r_n_l(s, r, n, l): …)
882def arc_from_c_s_l(c, s, l):
883 r = point_to_point_d(c, s)
884 if r == 0:
885 return []
886 alpha = l / r
887 cos_ = math.cos(alpha)
888 sin_ = math.sin(alpha)
889 e = [c + (s - c) cos_ - (s - c) sin_, c + (s - c) sin_ + (s - c) cos_]
890 n = [c - s, c - s]
891 slope = rotate_cw(n) if l > 0 elserotate_ccw(n)
892 returncsp_from_arc(s, e, c, r, slope)
893
894
882def arc_from_c_s_l(c, s, l): …)
895def csp_from_arc(start, end, center, r, slope_st):
896# Creates csp that approximise specified arc
897 r = abs(r)
898 alpha = (atan2(end - center, end - center) - atan2(start - center, start - center)) % TAU
899
900 sectors = int(abs(alpha) 2 / math.pi) + 1
901 alpha_start = atan2(start - center, start - center)
902 cos_ = math.cos(alpha_start)
903 sin_ = math.sin(alpha_start)
904 k = (4. math.tan(alpha / sectors / 4.) / 3.)
905 ifdot(slope_st, [- sin_ k r, cos_ k r]) < 0:
906 if alpha > 0:
907 alpha -= TAU
908 else:
909 alpha += TAU
910 if abs(alpha r) < 0.001:
911 return []
912
913 sectors = int(abs(alpha) 2 / math.pi) + 1
914 k = (4. math.tan(alpha / sectors / 4.) / 3.)
915 result = []
916 for i in range(sectors + 1):
917 cos_ = math.cos(alpha_start + alpha i / sectors)
918 sin_ = math.sin(alpha_start + alpha i / sectors)
919 sp =
920 sp = [sp + sin_ k r, sp - cos_ k r]
921 sp = [sp - sin_ k r, sp + cos_ k r]
922 result += [sp]
923 result = result[:]
924 result[-1] = result[-1]
925
926 return result
927
928
895def csp_from_arc(start, end, center, r, slope_st): …)
929def point_to_arc_distance(p, arc):
930# Distance calculattion from point to arc
931 P0, P2, c, a = arc
932 p = P(p)
933 r = (P0 - c).mag()
934 if r > 0:
935 i = c + (p - c).unit() r
936 alpha = ((i - c).angle() - (P0 - c).angle())
937 if a alpha < 0:
938 if alpha > 0:
939 alpha = alpha - TAU
940 else:
941 alpha = TAU + alpha
942 ifbetween(alpha, 0, a) or min(abs(alpha), abs(alpha - a)) < STRAIGHT_TOLERANCE:
943 return (p - i).mag(), [i.x, i.y]
944 else:
945 d1 = (p - P0).mag()
946 d2 = (p - P2).mag()
947 if d1 < d2:
948 return d1, [P0.x, P0.y]
949 else:
950 return d2, [P2.x, P2.y]
951
952
929def point_to_arc_distance(p, arc): …)
953def csp_to_arc_distance(sp1, sp2, arc1, arc2, tolerance=0.01): # arc = [start,end,center,alpha]
954 n = 10
955 i = 0
956 d = (0, [0, 0])
957 d1 = (0, [0, 0])
958 dl = 0
959 while i < 1 or (abs(d1 - dl) > tolerance and i < 4):
960 i += 1
961 dl = d1 1
962 for j in range(n + 1):
963 t = float(j) / n
964 p = csp_at_t(sp1, sp2, t)
965 d = min(point_to_arc_distance(p, arc1), point_to_arc_distance(p, arc2))
966 d1 = max(d1, d)
967 n = n 2
968 return d1
969
970
953def csp_to_arc_distance(sp1, sp2, arc1, arc2, tolerance=0.01): # arc = [start,end,center,alpha]…)
971def csp_point_inside_bound(sp1, sp2, p):
972 bez = [sp1, sp1, sp2, sp2]
973 x, y = p
974 c = 0
975# CLT added test of x in range
976 xmin = 1e100
977 xmax = -1e100
978 for i in range(4):
979 [x0, y0] = bez[i - 1]
980 [x1, y1] = bez[i]
981 xmin = min(xmin, x0)
982 xmax = max(xmax, x0)
983 if x0 - x1 != 0 and (y - y0) (x1 - x0) >= (x - x0) (y1 - y0) and x > min(x0, x1) and x <= max(x0, x1):
984 c += 1
985 return xmin <= x <= xmax and c % 2 == 0
986
987
971def csp_point_inside_bound(sp1, sp2, p): …)
988def line_line_intersect(p1, p2, p3, p4): # Return only true intersection.
989 if (p1 == p2 and p1 == p2) or (p3 == p4 and p3 == p4):
990 return False
991 x = (p2 - p1) (p4 - p3) - (p2 - p1) (p4 - p3)
992 if x == 0: # Lines are parallel
993 if (p3 - p1) (p2 - p1) == (p3 - p1) (p2 - p1):
994 if p3 != p4:
995 t11 = (p1 - p3) / (p4 - p3)
996 t12 = (p2 - p3) / (p4 - p3)
997 t21 = (p3 - p1) / (p2 - p1)
998 t22 = (p4 - p1) / (p2 - p1)
999 else:
1000 t11 = (p1 - p3) / (p4 - p3)
1001 t12 = (p2 - p3) / (p4 - p3)
1002 t21 = (p3 - p1) / (p2 - p1)
1003 t22 = (p4 - p1) / (p2 - p1)
1004 return"Overlap"if (0 <= t11 <= 1 or 0 <= t12 <= 1) and (0 <= t21 <= 1 or 0 <= t22 <= 1) else False
1005 else:
1006 return False
1007 else:
1008 return (
1009 0 <= ((p4 - p3) (p1 - p3) - (p4 - p3) (p1 - p3)) / x <= 1 and
1010 0 <= ((p2 - p1) (p1 - p3) - (p2 - p1) (p1 - p3)) / x <= 1)
1011
1012
988def line_line_intersect(p1, p2, p3, p4): # Return only true intersection.…)
1013def line_line_intersection_points(p1, p2, p3, p4): # Return only points [ (x,y) ]
1014 if (p1 == p2 and p1 == p2) or (p3 == p4 and p3 == p4):
1015 return []
1016 x = (p2 - p1) (p4 - p3) - (p2 - p1) (p4 - p3)
1017 if x == 0: # Lines are parallel
1018 if (p3 - p1) (p2 - p1) == (p3 - p1) (p2 - p1):
1019 if p3 != p4:
1020 t11 = (p1 - p3) / (p4 - p3)
1021 t12 = (p2 - p3) / (p4 - p3)
1022 t21 = (p3 - p1) / (p2 - p1)
1023 t22 = (p4 - p1) / (p2 - p1)
1024 else:
1025 t11 = (p1 - p3) / (p4 - p3)
1026 t12 = (p2 - p3) / (p4 - p3)
1027 t21 = (p3 - p1) / (p2 - p1)
1028 t22 = (p4 - p1) / (p2 - p1)
1029 res = []
1030 if (0 <= t11 <= 1 or 0 <= t12 <= 1) and (0 <= t21 <= 1 or 0 <= t22 <= 1):
1031 if 0 <= t11 <= 1:
1032 res += [p1]
1033 if 0 <= t12 <= 1:
1034 res += [p2]
1035 if 0 <= t21 <= 1:
1036 res += [p3]
1037 if 0 <= t22 <= 1:
1038 res += [p4]
1039 return res
1040 else:
1041 return []
1042 else:
1043 t1 = ((p4 - p3) (p1 - p3) - (p4 - p3) (p1 - p3)) / x
1044 t2 = ((p2 - p1) (p1 - p3) - (p2 - p1) (p1 - p3)) / x
1045 if 0 <= t1 <= 1 and 0 <= t2 <= 1:
1046 return
1047 else:
1048 return []
1049
1050
1013def line_line_intersection_points(p1, p2, p3, p4): # Return only points [ (x,y) ]…)
1051def point_to_point_d2(a, b):
1052 return (a - b) 2 + (a - b) 2
1053
1054
1051def point_to_point_d2(a, b): …)
1055def point_to_point_d(a, b):
1056 return math.sqrt((a - b) 2 + (a - b) 2)
1057
1058
1055def point_to_point_d(a, b): …)
1059def point_to_line_segment_distance_2(p1, p2, p3):
1060# p1 - point, p2,p3 - line segment
1061# draw_pointer(p1)
1062 w0 = [p1 - p2, p1 - p2]
1063 v = [p3 - p2, p3 - p2]
1064 c1 = w0 v + w0 v
1065 if c1 <= 0:
1066 return w0 w0 + w0 w0
1067 c2 = v v + v v
1068 if c2 <= c1:
1069 return (p1 - p3) 2 + (p1 - p3) 2
1070 return (p1 - p2 - v c1 / c2) 2 + (p1 - p2 - v c1 / c2)
1071
1072
1059def point_to_line_segment_distance_2(p1, p2, p3): …)
1073def line_to_line_distance_2(p1, p2, p3, p4):
1074 ifline_line_intersect(p1, p2, p3, p4):
1075 return 0
1076 return min(
1077point_to_line_segment_distance_2(p1, p3, p4),
1078point_to_line_segment_distance_2(p2, p3, p4),
1079point_to_line_segment_distance_2(p3, p1, p2),
1080point_to_line_segment_distance_2(p4, p1, p2))
1081
1082
1073def line_to_line_distance_2(p1, p2, p3, p4): …)
1083def csp_seg_bound_to_csp_seg_bound_max_min_distance(sp1, sp2, sp3, sp4):
1084 bez1 = csp_segment_to_bez(sp1, sp2)
1085 bez2 = csp_segment_to_bez(sp3, sp4)
1086 min_dist = 1e100
1087 max_dist = 0.
1088 for i in range(4):
1089 ifcsp_point_inside_bound(sp1, sp2, bez2[i]) orcsp_point_inside_bound(sp3, sp4, bez1[i]):
1090 min_dist = 0.
1091 break
1092 for i in range(4):
1093 for j in range(4):
1094 d = line_to_line_distance_2(bez1[i - 1], bez1[i], bez2[j - 1], bez2[j])
1095 if d < min_dist:
1096 min_dist = d
1097 d = (bez2[j] - bez1[i]) 2 + (bez2[j] - bez1[i]) 2
1098 if max_dist < d:
1099 max_dist = d
1100 return min_dist, max_dist
1101
1102
1083def csp_seg_bound_to_csp_seg_bound_max_min_distance(sp1, sp2, sp3, sp4): …)
1103def csp_reverse(csp):
1104 for i in range(len(csp)):
1105 n = []
1106 for j in csp[i]:
1107 n = ] + n
1108 csp[i] = n[:]
1109 return csp
1110
1111
1103def csp_reverse(csp): …)
1112def csp_normalized_slope(sp1, sp2, t):
1113 ax, ay, bx, by, cx, cy, dx, dy = bezierparameterize((sp1[:], sp1[:], sp2[:], sp2[:]))
1114 if sp1 == sp2 == sp1 == sp2:
1115 return [1., 0.]
1116 f1x = 3 ax t t + 2 bx t + cx
1117 f1y = 3 ay t t + 2 by t + cy
1118 if abs(f1x f1x + f1y f1y) > 1e-9: # LT changed this from 1e-20, which caused problems
1119 l = math.sqrt(f1x f1x + f1y f1y)
1120 return [f1x / l, f1y / l]
1121
1122 if t == 0:
1123 f1x = sp2 - sp1
1124 f1y = sp2 - sp1
1125 if abs(f1x f1x + f1y f1y) > 1e-9: # LT changed this from 1e-20, which caused problems
1126 l = math.sqrt(f1x f1x + f1y f1y)
1127 return [f1x / l, f1y / l]
1128 else:
1129 f1x = sp2 - sp1
1130 f1y = sp2 - sp1
1131 if f1x f1x + f1y f1y != 0:
1132 l = math.sqrt(f1x f1x + f1y f1y)
1133 return [f1x / l, f1y / l]
1134 elif t == 1:
1135 f1x = sp2 - sp1
1136 f1y = sp2 - sp1
1137 if abs(f1x f1x + f1y f1y) > 1e-9:
1138 l = math.sqrt(f1x f1x + f1y f1y)
1139 return [f1x / l, f1y / l]
1140 else:
1141 f1x = sp2 - sp1
1142 f1y = sp2 - sp1
1143 if f1x f1x + f1y f1y != 0:
1144 l = math.sqrt(f1x f1x + f1y f1y)
1145 return [f1x / l, f1y / l]
1146 else:
1147 return [1., 0.]
1148
1149
1112def csp_normalized_slope(sp1, sp2, t): …)
1150def csp_normalized_normal(sp1, sp2, t):
1151 nx, ny = csp_normalized_slope(sp1, sp2, t)
1152 return [-ny, nx]
1153
1154
1150def csp_normalized_normal(sp1, sp2, t): …)
1155def csp_parameterize(sp1, sp2):
1156 return bezierparameterize(csp_segment_to_bez(sp1, sp2))
1157
1158
1155def csp_parameterize(sp1, sp2): …)
1159def csp_concat_subpaths(s):
1160 def concat(s1, s2):
1161 if not s1:
1162 return s2
1163 if not s2:
1164 return s1
1165 if (s1[-1] - s2) 2 + (s1[-1] - s2) 2 > 0.00001:
1166 return s1[:-1] + , [s2, s2, s2]] + s2[1:]
1167 else:
1168 return s1[:-1] + ] + s2[1:]
1169
1170 if len(s) == 0:
1171 return []
1172 if len(s) == 1:
1173 return s
1174 result = s
1175 for s1 in s[1:]:
1176 result = concat(result, s1)
1177 return result
1178
1179
1159def csp_concat_subpaths(s): …)
1180def csp_subpaths_end_to_start_distance2(s1, s2):
1181 return (s1[-1] - s2) 2 + (s1[-1] - s2) 2
1182
1183
1180def csp_subpaths_end_to_start_distance2(s1, s2): …)
1184def csp_clip_by_line(csp, l1, l2):
1185 result = []
1186 for i in range(len(csp)):
1187 s = csp[i]
1188 intersections = []
1189 for j in range(1, len(s)):
1190 intersections += [[j, int_] for int_ incsp_line_intersection(l1, l2, s[j - 1], s[j])]
1191 splitted_s = csp_subpath_split_by_points(s, intersections)
1192 for s in splitted_s[:]:
1193 clip = False
1194 for p incsp_true_bounds([s]):
1195 if (l1 - l2) p + (l2 - l1) p + (l1 l2 - l2 l1) < -0.01:
1196 clip = True
1197 break
1198 if clip:
1199 splitted_s.remove(s)
1200 result += splitted_s
1201 return result
1202
1203
1184def csp_clip_by_line(csp, l1, l2): …)
1204def csp_subpath_line_to(subpath, points, prepend=False):
1205# Appends subpath with line or polyline.
1206 if len(points) > 0:
1207 if not prepend:
1208 if len(subpath) > 0:
1209 subpath[-1] = subpath[-1][:]
1210 if type(points) == type([1, 1]):
1211 for p in points:
1212 subpath += ]
1213 else:
1214 subpath +=
1215 else:
1216 if len(subpath) > 0:
1217 subpath = subpath[:]
1218 if type(points) == type([1, 1]):
1219 for p in points:
1220 subpath = ] + subpath
1221 else:
1222 subpath = + subpath
1223 return subpath
1224
1225
1204def csp_subpath_line_to(subpath, points, prepend=False): …)
1226def csp_join_subpaths(csp):
1227 result = csp[:]
1228 done_smf = True
1229 joined_result = []
1230 while done_smf:
1231 done_smf = False
1232 while len(result) > 0:
1233 s1 = result[-1][:]
1234 del (result[-1])
1235 j = 0
1236 joined_smf = False
1237 while j < len(joined_result):
1238 ifcsp_subpaths_end_to_start_distance2(joined_result[j], s1) < 0.000001:
1239 joined_result[j] = csp_concat_subpaths(joined_result[j], s1)
1240 done_smf = True
1241 joined_smf = True
1242 break
1243 ifcsp_subpaths_end_to_start_distance2(s1, joined_result[j]) < 0.000001:
1244 joined_result[j] = csp_concat_subpaths(s1, joined_result[j])
1245 done_smf = True
1246 joined_smf = True
1247 break
1248 j += 1
1249 if not joined_smf:
1250 joined_result += [s1[:]]
1251 if done_smf:
1252 result = joined_result[:]
1253 joined_result = []
1254 return joined_result
1255
1256
1226def csp_join_subpaths(csp): …)
1257def triangle_cross(a, b, c):
1258 return (a - b) (c - b) - (c - b) (a - b)
1259
1260
1257def triangle_cross(a, b, c): …)
1261def csp_segment_convex_hull(sp1, sp2):
1262 a = sp1[:]
1263 b = sp1[:]
1264 c = sp2[:]
1265 d = sp2[:]
1266
1267 abc = triangle_cross(a, b, c)
1268 abd = triangle_cross(a, b, d)
1269 bcd = triangle_cross(b, c, d)
1270 cad = triangle_cross(c, a, d)
1271 if abc == 0 and abd == 0:
1272 return [min(a, b, c, d), max(a, b, c, d)]
1273 if abc == 0:
1274 return [d, min(a, b, c), max(a, b, c)]
1275 if abd == 0:
1276 return [c, min(a, b, d), max(a, b, d)]
1277 if bcd == 0:
1278 return [a, min(b, c, d), max(b, c, d)]
1279 if cad == 0:
1280 return [b, min(c, a, d), max(c, a, d)]
1281
1282 m1 = abc abd > 0
1283 m2 = abc bcd > 0
1284 m3 = abc cad > 0
1285
1286 if m1 and m2 and m3:
1287 return [a, b, c]
1288 if m1 and m2 and not m3:
1289 return [a, b, c, d]
1290 if m1 and not m2 and m3:
1291 return [a, b, d, c]
1292 if not m1 and m2 and m3:
1293 return [a, d, b, c]
1294 if m1 and not (m2 and m3):
1295 return [a, b, d]
1296 if not (m1 and m2) and m3:
1297 return [c, a, d]
1298 if not (m1 and m3) and m2:
1299 return [b, c, d]
1300
1301 raise ValueError("csp_segment_convex_hull happened which is something that shouldn't happen!")
1302
1303
1261def csp_segment_convex_hull(sp1, sp2): …)
1304
1307
1308def bez_bounds_intersect(bez1, bez2):
1309 returnbounds_intersect(bez_bound(bez2), bez_bound(bez1))
1310
1311
1308def bez_bounds_intersect(bez1, bez2): …)
1312def bez_bound(bez):
1313 return [
1314 min(bez, bez, bez, bez),
1315 min(bez, bez, bez, bez),
1316 max(bez, bez, bez, bez),
1317 max(bez, bez, bez, bez),
1318 ]
1319
1320
1312def bez_bound(bez): …)
1321def bounds_intersect(a, b):
1322 return not ((a > b) or (b > a) or (a > b) or (b > a))
1323
1324
1321def bounds_intersect(a, b): …)
1325def tpoint(xy1, xy2, t):
1326 (x1, y1) = xy1
1327 (x2, y2) = xy2
1328 return [x1 + t (x2 - x1), y1 + t (y2 - y1)]
1329
1330
1325def tpoint(xy1, xy2, t): …)
1331def bez_split(a, t=0.5):
1332 a1 = tpoint(a, a, t)
1333 at = tpoint(a, a, t)
1334 b2 = tpoint(a, a, t)
1335 a2 = tpoint(a1, at, t)
1336 b1 = tpoint(b2, at, t)
1337 a3 = tpoint(a2, b1, t)
1338 return [a, a1, a2, a3], [a3, b1, b2, a]
1339
1340
1331def bez_split(a, t=0.5): …)
1341
1344
1345def normalize(xy):
1346 (x, y) = xy
1347 l = math.sqrt(x 2 + y 2)
1348 if l == 0:
1349 return [0., 0.]
1350 else:
1351 return [x / l, y / l]
1352
1353
1345def normalize(xy): …)
1354def cross(a, b):
1355 return a b - a b
1356
1357
1354def cross(a, b): …)
1358def dot(a, b):
1359 return a b + a b
1360
1361
1358def dot(a, b): …)
1362def rotate_ccw(d):
1363 return [-d, d]
1364
1365
1362def rotate_ccw(d): …)
1366def rotate_cw(d):
1367 return [d, -d]
1368
1369
1366def rotate_cw(d): …)
1370def vectors_ccw(a, b):
1371 return a b - b a < 0
1372
1373
1370def vectors_ccw(a, b): …)
1374
1377
1378def inv_2x2(a): # invert matrix 2x2
1379 det = a a - a a
1380 if det == 0:
1381 return None
1382 return [
1383 [a / det, -a / det],
1384 [-a / det, a / det]
1385 ]
1386
1387
1378def inv_2x2(a): # invert matrix 2x2…)
1388def small(a):
1389 global small_tolerance
1390 return abs(a) < small_tolerance
1391
1392
1388def small(a): …)
1393def atan2(arg):
1394 if len(arg) == 1 and (type(arg) == type([0., 0.]) or type(arg) == type((0., 0.))):
1395 return (math.pi / 2 - math.atan2(arg, arg)) % TAU
1396 elif len(arg) == 2:
1397 return (math.pi / 2 - math.atan2(arg, arg)) % TAU
1398 else:
1399 raise ValueError("Bad argumets for atan! ({})".format(arg))
1400
1401
1393def atan2(arg): …)
1402def draw_text(text, x, y, group=None, style=None, font_size=10, gcodetools_tag=None):
1403 if style is None:
1404 style = "font-family:DejaVu Sans;font-style:normal;font-variant:normal;font-weight:normal;font-stretch:normal;font-family:DejaVu Sans;fill:#000000;fill-opacity:1;stroke:none;"
1405 style += "font-size:{:f}px;".format(font_size)
1406 attributes = {'x': str(x), 'y': str(y), 'style': style}
1407 if gcodetools_tag is not None:
1408 attributes["gcodetools"] = str(gcodetools_tag)
1409
1410 if group is None:
1411 group = options.doc_root
1412
1413 text_elem = group.add(TextElement(attributes))
1414 text_elem.set("xml:space", "preserve")
1415 text = str(text).split("\n")
1416 for string in text:
1417 span = text_elem.add(Tspan(x=str(x), y=str(y)))
1418 span.set('sodipodi:role', 'line')
1419 y += font_size
1420 span.text = str(string)
1421
1422
1402def draw_text(text, x, y, group=None, style=None, font_size=10, gcodetools_tag=None): …)
1423def draw_csp(csp, stroke="#f00", fill="none", comment="", width=0.354, group=None, style=None):
1424 if group is None:
1425 group = options.doc_root
1426 node = group.add(PathElement())
1427
1428 node.style = style if style is not None else \
1429 {'fill': fill, 'fill-opacity': 1, 'stroke': stroke, 'stroke-width': width}
1430
1431 node.path = CubicSuperPath(csp)
1432
1433 if comment != '':
1434 node.set('comment', comment)
1435
1436 return node
1437
1438
1423def draw_csp(csp, stroke="#f00", fill="none", comment="", width=0.354, group=None, style=None): …)
1439def draw_pointer(x, color="#f00", figure="cross", group=None, comment="", fill=None, width=.1, size=10., text=None, font_size=None, pointer_type=None, attrib=None):
1440 size = size / 2
1441 if attrib is None:
1442 attrib = {}
1443 if pointer_type is None:
1444 pointer_type = "Pointer"
1445 attrib["gcodetools"] = pointer_type
1446 if group is None:
1447 group = options.self.svg.get_current_layer()
1448 if text is not None:
1449 if font_size is None:
1450 font_size = 7
1451 group = group.add(Group(gcodetools=pointer_type + " group"))
1452draw_text(text, x + size 2.2, x - size, group=group, font_size=font_size)
1453 if figure == "line":
1454 s = ""
1455 for i in range(1, len(x) / 2):
1456 s += " {}, {} ".format(x[i 2], x[i 2 + 1])
1457 attrib.update({"d": "M {},{} L {}".format(x, x, s), "style": "fill:none;stroke:{};stroke-width:{:f};".format(color, width), "comment": str(comment)})
1458 elif figure == "arrow":
1459 if fill is None:
1460 fill = "#12b3ff"
1461 fill_opacity = "0.8"
1462 d = "m {},{} ".format(x, x) + re.sub("([0-9\-.e]+)", (lambda match: str(float(match.group(1)) size 2.)), "0.88464,-0.40404 c -0.0987,-0.0162 -0.186549,-0.0589 -0.26147,-0.1173 l 0.357342,-0.35625 c 0.04631,-0.039 0.0031,-0.13174 -0.05665,-0.12164 -0.0029,-1.4e-4 -0.0058,-1.4e-4 -0.0087,0 l -2.2e-5,2e-5 c -0.01189,0.004 -0.02257,0.0119 -0.0305,0.0217 l -0.357342,0.35625 c -0.05818,-0.0743 -0.102813,-0.16338 -0.117662,-0.26067 l -0.409636,0.88193 z")
1463 attrib.update({"d": d, "style": "fill:{};stroke:none;fill-opacity:{};".format(fill, fill_opacity), "comment": str(comment)})
1464 else:
1465 attrib.update({"d": "m {},{} l {:f},{:f} {:f},{:f} {:f},{:f} {:f},{:f} , {:f},{:f}".format(x, x, size, size, -2 size, -2 size, size, size, size, -size, -2 size, 2 size), "style": "fill:none;stroke:{};stroke-width:{:f};".format(color, width), "comment": str(comment)})
1466 group.add(PathElement(attrib))
1467
1468
1439def draw_pointer(x, color="#f00", figure="cross", group=None, comment="", fill=None, width=.1, size=10., text=None, font_size=None, pointer_type=None, attrib=None): …)
1469def straight_segments_intersection(a, b, true_intersection=True): # (True intersection means check ta and tb are in [0,1])
1470 ax = a
1471 bx = a
1472 cx = b
1473 dx = b
1474 ay = a
1475 by = a
1476 cy = b
1477 dy = b
1478 if (ax == bx and ay == by) or (cx == dx and cy == dy):
1479 return False, 0, 0
1480 if (bx - ax) (dy - cy) - (by - ay) (dx - cx) == 0: # Lines are parallel
1481 ta = (ax - cx) / (dx - cx) if cx != dx else (ay - cy) / (dy - cy)
1482 tb = (bx - cx) / (dx - cx) if cx != dx else (by - cy) / (dy - cy)
1483 tc = (cx - ax) / (bx - ax) if ax != bx else (cy - ay) / (by - ay)
1484 td = (dx - ax) / (bx - ax) if ax != bx else (dy - ay) / (by - ay)
1485 return ("Overlap"if 0 <= ta <= 1 or 0 <= tb <= 1 or 0 <= tc <= 1 or 0 <= td <= 1 or not true_intersection else False), (ta, tb), (tc, td)
1486 else:
1487 ta = ((ay - cy) (dx - cx) - (ax - cx) (dy - cy)) / ((bx - ax) (dy - cy) - (by - ay) (dx - cx))
1488 tb = (ax - cx + ta (bx - ax)) / (dx - cx) if dx != cx else (ay - cy + ta (by - ay)) / (dy - cy)
1489 return (0 <= ta <= 1 and 0 <= tb <= 1 or not true_intersection), ta, tb
1490
1491
1469def straight_segments_intersection(a, b, true_intersection=True): # (True intersection means check ta and tb are in [0,1])…)
1492def between(c, x, y):
1493 return x - STRAIGHT_TOLERANCE <= c <= y + STRAIGHT_TOLERANCE or y - STRAIGHT_TOLERANCE <= c <= x + STRAIGHT_TOLERANCE
1494
1495
1492def between(c, x, y): …)
1496def cubic_solver_real(a, b, c, d):
1497# returns only real roots of a cubic equation.
1498 roots = cubic_solver(a, b, c, d)
1499 res = []
1500 for root in roots:
1501 if type(root) is complex:
1502 if -1e-10 < root.imag < 1e-10:
1503 res.append(root.real)
1504 else:
1505 res.append(root)
1506 return res
1507
1508
1496def cubic_solver_real(a, b, c, d): …)
1509def cubic_solver(a, b, c, d):
1510 if a != 0:
1511# Monics formula see
1512 a, b, c = (b / a, c / a, d / a)
1513 m = 2 a 3 - 9 a b + 27 c
1514 k = a 2 - 3 b
1515 n = m 2 - 4 k 3
1516 w1 = -.5 + .5 cmath.sqrt(3) 1j
1517 w2 = -.5 - .5 cmath.sqrt(3) 1j
1518 if n >= 0:
1519 t = m + math.sqrt(n)
1520 m1 = pow(t / 2, 1. / 3) if t >= 0 else -pow(-t / 2, 1. / 3)
1521 t = m - math.sqrt(n)
1522 n1 = pow(t / 2, 1. / 3) if t >= 0 else -pow(-t / 2, 1. / 3)
1523 else:
1524 m1 = pow(complex((m + cmath.sqrt(n)) / 2), 1. / 3)
1525 n1 = pow(complex((m - cmath.sqrt(n)) / 2), 1. / 3)
1526 x1 = -1. / 3 (a + m1 + n1)
1527 x2 = -1. / 3 (a + w1 m1 + w2 n1)
1528 x3 = -1. / 3 (a + w2 m1 + w1 n1)
1529 return [x1, x2, x3]
1530 elif b != 0:
1531 det = c 2 - 4 b d
1532 if det > 0:
1533 return [(-c + math.sqrt(det)) / (2 b), (-c - math.sqrt(det)) / (2 b)]
1534 elif d == 0:
1535 return [-c / (b b)]
1536 else:
1537 return [(-c + cmath.sqrt(det)) / (2 b), (-c - cmath.sqrt(det)) / (2 b)]
1538 elif c != 0:
1539 return [-d / c]
1540 else:
1541 return []
1542
1543
1509def cubic_solver(a, b, c, d): …)
1544
1547
1548def print_(arg):
1549 with open(options.log_filename, "ab") as f:
1550 for s in arg:
1551 s = unicode(s).encode('unicode_escape') + b" "
1552 f.write(s)
1553 f.write(b"\n")
1554
1555
1548def print_(arg): …)
1556
1559class P(object):
1560def init(self, x, y=None):
1561 if not y is None:
1562 self.x = float(x)
1563 self.y = float(y)
1564 else:
1565 self.x = float(x)
1566 self.y = float(x)
1567
1560def init(self, x, y=None): …)
1568def add(self, other):
1569 returnP(self.x + other.x, self.y + other.y)
1570
1568def add(self, other): …)
1571def sub(self, other):
1572 returnP(self.x - other.x, self.y - other.y)
1573
1571def sub(self, other): …)
1574def neg(self):
1575 returnP(-self.x, -self.y)
1576
1574def neg(self): …)
1577def mul(self, other):
1578 if isinstance(other, P):
1579 return self.x other.x + self.y other.y
1580 returnP(self.x other, self.y other)
1581
1577def mul(self, other): …)
1582 rmul = mul
1583
1584def div(self, other):
1585 returnP(self.x / other, self.y / other)
1586
1584def div(self, other): …)
1587def truediv(self, other):
1588 return self.div(other)
1589
1587def truediv(self, other): …)
1590def mag(self):
1591 return math.hypot(self.x, self.y)
1592
1590def mag(self): …)
1593def unit(self):
1594 h_mag = self.mag()
1595 if h_mag:
1596 return self / h_mag
1597 returnP(0, 0)
1598
1593def unit(self): …)
1599def dot(self, other):
1600 return self.x other.x + self.y other.y
1601
1599def dot(self, other): …)
1602def rot(self, theta):
1603 c = math.cos(theta)
1604 s = math.sin(theta)
1605 returnP(self.x c - self.y s, self.x s + self.y c)
1606
1602def rot(self, theta): …)
1607def angle(self):
1608 return math.atan2(self.y, self.x)
1609
1607def angle(self): …)
1610def repr(self):
1611 return'{:f},{:f}'.format(self.x, self.y)
1612
1610def repr(self): …)
1613def pr(self):
1614 return"{:.2f},{:.2f}".format(self.x, self.y)
1615
1613def pr(self): …)
1616def to_list(self):
1617 return [self.x, self.y]
1618
1616def to_list(self): …)
1619def ccw(self):
1620 returnP(-self.y, self.x)
1621
1619def ccw(self): …)
1622def l2(self):
1623 return self.x self.x + self.y self.y
1624
1625
1622def l2(self): …)
1559class P(object): …)
1626class Line(object):
1627def init(self, st, end):
1628 if st.class == P:
1629 st = st.to_list()
1630 if end.class == P:
1631 end = end.to_list()
1632 self.st = P(st)
1633 self.end = P(end)
1634 self.l = self.length()
1635 if self.l != 0:
1636 self.n = ((self.end - self.st) / self.l).ccw()
1637 else:
1638 self.n = [0, 1]
1639
1627def init(self, st, end): …)
1640def offset(self, r):
1641 self.st -= self.n r
1642 self.end -= self.n r
1643
1640def offset(self, r): …)
1644def l2(self):
1645 return (self.st - self.end).l2()
1646
1644def l2(self): …)
1647def length(self):
1648 return (self.st - self.end).mag()
1649
1647def length(self): …)
1650def draw(self, group, style, layer, transform, num=0, reverse_angle=1):
1651 st = gcodetools.transform(self.st.to_list(), layer, True)
1652 end = gcodetools.transform(self.end.to_list(), layer, True)
1653
1654 attr = {'style': style['line'],
1655'd': 'M {},{} L {},{}'.format(st, st, end, end),
1656"gcodetools": "Preview",
1657 }
1658 if transform:
1659 attr["transform"] = transform
1660 group.add(PathElement(attr))
1661
1650def draw(self, group, style, layer, transform, num=0, reverse_angle=1): …)
1662def intersect(self, b):
1663 if b.class == Line:
1664 if self.l< 10e-8 or b.l < 10e-8:
1665 return []
1666 v1 = self.end - self.st
1667 v2 = b.end - b.st
1668 x = v1.x v2.y - v2.x v1.y
1669 if x == 0:
1670# lines are parallel
1671 res = []
1672
1673 if (self.st.x - b.st.x) v1.y - (self.st.y - b.st.y) v1.x == 0:
1674# lines are the same
1675 if v1.x != 0:
1676 if 0 <= (self.st.x - b.st.x) / v2.x <= 1:
1677 res.append(self.st)
1678 if 0 <= (self.end.x - b.st.x) / v2.x <= 1:
1679 res.append(self.end)
1680 if 0 <= (b.st.x - self.st.x) / v1.x <= 1:
1681 res.append(b.st)
1682 if 0 <= (b.end.x - b.st.x) / v1.x <= 1:
1683 res.append(b.end)
1684 else:
1685 if 0 <= (self.st.y - b.st.y) / v2.y <= 1:
1686 res.append(self.st)
1687 if 0 <= (self.end.y - b.st.y) / v2.y <= 1:
1688 res.append(self.end)
1689 if 0 <= (b.st.y - self.st.y) / v1.y <= 1:
1690 res.append(b.st)
1691 if 0 <= (b.end.y - b.st.y) / v1.y <= 1:
1692 res.append(b.end)
1693 return res
1694 else:
1695 t1 = (-v1.x (b.end.y - self.end.y) + v1.y (b.end.x - self.end.x)) / x
1696 t2 = (-v1.y (self.st.x - b.st.x) + v1.x (self.st.y - b.st.y)) / x
1697
1698 gcodetools.error(str((x, t1, t2)))
1699 if 0 <= t1 <= 1 and 0 <= t2 <= 1:
1700 return [self.st + v1 t1]
1701 else:
1702 return []
1703 else:
1704 return []
1705
1706
1662def intersect(self, b): …)
1707
1626class Line(object): …)
1716def csp_offset(csp, r):
1717 offset_tolerance = 0.05
1718 offset_subdivision_depth = 10
1719 time_ = time.time()
1720 time_start = time_
1721print_("Offset start at {}".format(time_))
1722print_("Offset radius {}".format(r))
1723
1724 def csp_offset_segment(sp1, sp2, r):
1725 result = []
1726 t = csp_get_t_at_curvature(sp1, sp2, 1 / r)
1727 if len(t) == 0:
1728 t = [0., 1.]
1729 t.sort()
1730 if t > .00000001:
1731 t = [0.] + t
1732 if t[-1] < .99999999:
1733 t.append(1.)
1734 for st, end in zip(t, t[1:]):
1735 c = csp_curvature_at_t(sp1, sp2, (st + end) / 2)
1736 sp = csp_split_by_two_points(sp1, sp2, st, end)
1737 if sp != sp:
1738 if c > 1 / r and r < 0 or c < 1 / r and r > 0:
1739 offset = offset_segment_recursion(sp, sp, r, offset_subdivision_depth, offset_tolerance)
1740 else: # This part will be clipped for sure... TODO Optimize it...
1741 offset = offset_segment_recursion(sp, sp, r, offset_subdivision_depth, offset_tolerance)
1742
1743 if not result:
1744 result = offset[:]
1745 else:
1746 ifcsp_subpaths_end_to_start_distance2(result, offset) < 0.0001:
1747 result = csp_concat_subpaths(result, offset)
1748 else:
1749
1750 intersection = csp_get_subapths_last_first_intersection(result, offset)
1751 if intersection:
1752 i, t1, j, t2 = intersection
1753 sp1_, sp2_, sp3_ = csp_split(result[i - 1], result[i], t1)
1754 result = result[:i - 1] + [sp1_, sp2_]
1755 sp1_, sp2_, sp3_ = csp_split(offset[j - 1], offset[j], t2)
1756 result = csp_concat_subpaths(result, [sp2_, sp3_] + offset[j + 1:])
1757 else:
1758 pass# ???
1759 return result
1760
1761 def create_offset_segment(sp1, sp2, r):
1762# See Gernot Hoffmann "Bezier Curves" p.34 -> 7.1 Bezier Offset Curves
1763 p0 = P(sp1)
1764 p1 = P(sp1)
1765 p2 = P(sp2)
1766 p3 = P(sp2)
1767
1768 s0 = p1 - p0
1769 s1 = p2 - p1
1770 s3 = p3 - p2
1771
1772 n0 = s0.ccw().unit() if s0.l2() != 0 elseP(csp_normalized_normal(sp1, sp2, 0))
1773 n3 = s3.ccw().unit() if s3.l2() != 0 elseP(csp_normalized_normal(sp1, sp2, 1))
1774 n1 = s1.ccw().unit() if s1.l2() != 0 else (n0.unit() + n3.unit()).unit()
1775
1776 q0 = p0 + r n0
1777 q3 = p3 + r n3
1778 c = csp_curvature_at_t(sp1, sp2, 0)
1779 q1 = q0 + (p1 - p0) (1 - (r c if abs(c) < 100 else 0))
1780 c = csp_curvature_at_t(sp1, sp2, 1)
1781 q2 = q3 + (p2 - p3) (1 - (r c if abs(c) < 100 else 0))
1782
1783 return
1784
1785 def csp_get_subapths_last_first_intersection(s1, s2):
1786 _break = False
1787 for i in range(1, len(s1)):
1788 sp11 = s1[-i - 1]
1789 sp12 = s1[-i]
1790 for j in range(1, len(s2)):
1791 sp21 = s2[j - 1]
1792 sp22 = s2[j]
1793 intersection = csp_segments_true_intersection(sp11, sp12, sp21, sp22)
1794 if intersection:
1795 _break = True
1796 break
1797 if _break:
1798 break
1799 if _break:
1800 intersection = max(intersection)
1801 return [len(s1) - i, intersection, j, intersection]
1802 else:
1803 return []
1804
1805 def csp_join_offsets(prev, next, sp1, sp2, sp1_l, sp2_l, r):
1806 if len(next) > 1:
1807 if (P(prev[-1]) - P(next)).l2() < 0.001:
1808 return prev, [], next
1809 intersection = csp_get_subapths_last_first_intersection(prev, next)
1810 if intersection:
1811 i, t1, j, t2 = intersection
1812 sp1_, sp2_, sp3_ = csp_split(prev[i - 1], prev[i], t1)
1813 sp3_, sp4_, sp5_ = csp_split(next[j - 1], next[j], t2)
1814 return prev[:i - 1] + [sp1_, sp2_], [], [sp4_, sp5_] + next[j + 1:]
1815
1816# Offsets do not intersect... will add an arc...
1817 start = (P(csp_at_t(sp1_l, sp2_l, 1.)) + r P(csp_normalized_normal(sp1_l, sp2_l, 1.))).to_list()
1818 end = (P(csp_at_t(sp1, sp2, 0.)) + r P(csp_normalized_normal(sp1, sp2, 0.))).to_list()
1819 arc = csp_from_arc(start, end, sp1, r, csp_normalized_slope(sp1_l, sp2_l, 1.))
1820 if not arc:
1821 return prev, [], next
1822 else:
1823# Clip prev by arc
1824 ifcsp_subpaths_end_to_start_distance2(prev, arc) > 0.00001:
1825 intersection = csp_get_subapths_last_first_intersection(prev, arc)
1826 if intersection:
1827 i, t1, j, t2 = intersection
1828 sp1_, sp2_, sp3_ = csp_split(prev[i - 1], prev[i], t1)
1829 sp3_, sp4_, sp5_ = csp_split(arc[j - 1], arc[j], t2)
1830 prev = prev[:i - 1] + [sp1_, sp2_]
1831 arc = [sp4_, sp5_] + arc[j + 1:]
1832# Clip next by arc
1833 if not next:
1834 return prev, [], arc
1835 ifcsp_subpaths_end_to_start_distance2(arc, next) > 0.00001:
1836 intersection = csp_get_subapths_last_first_intersection(arc, next)
1837 if intersection:
1838 i, t1, j, t2 = intersection
1839 sp1_, sp2_, sp3_ = csp_split(arc[i - 1], arc[i], t1)
1840 sp3_, sp4_, sp5_ = csp_split(next[j - 1], next[j], t2)
1841 arc = arc[:i - 1] + [sp1_, sp2_]
1842 next = [sp4_, sp5_] + next[j + 1:]
1843
1844 return prev, arc, next
1845
1846 def offset_segment_recursion(sp1, sp2, r, depth, tolerance):
1847 sp1_r, sp2_r = create_offset_segment(sp1, sp2, r)
1848 err = max(
1849csp_seg_to_point_distance(sp1_r, sp2_r, (P(csp_at_t(sp1, sp2, .25)) + P(csp_normalized_normal(sp1, sp2, .25)) r).to_list()),
1850csp_seg_to_point_distance(sp1_r, sp2_r, (P(csp_at_t(sp1, sp2, .50)) + P(csp_normalized_normal(sp1, sp2, .50)) r).to_list()),
1851csp_seg_to_point_distance(sp1_r, sp2_r, (P(csp_at_t(sp1, sp2, .75)) + P(csp_normalized_normal(sp1, sp2, .75)) r).to_list()),
1852 )
1853
1854 if err > tolerance 2 and depth > 0:
1855 if depth > offset_subdivision_depth - 2:
1856 t = csp_max_curvature(sp1, sp2)
1857 t = max(.1, min(.9, t))
1858 else:
1859 t = .5
1860 sp3, sp4, sp5 = csp_split(sp1, sp2, t)
1861 r1 = offset_segment_recursion(sp3, sp4, r, depth - 1, tolerance)
1862 r2 = offset_segment_recursion(sp4, sp5, r, depth - 1, tolerance)
1863 return r1[:-1] + ] + r2[1:]
1864 else:
1865 return [sp1_r, sp2_r]
1866
1867
1870 csp_len = len(csp)
1871
1872
1876
1877 for i inxrange(len(csp)):
1878 for j inxrange(len(csp[i])):
1879 sp = csp[i][j]
1880 if (P(sp) - P(sp)).mag() < 0.001:
1881 csp[i][j] = sp
1882 if (P(sp) - P(sp)).mag() < 0.001:
1883 csp[i][j] = sp
1884 for i inxrange(len(csp)):
1885 for j inxrange(1, len(csp[i])):
1886 ifcspseglength(csp[i][j - 1], csp[i][j]) < 0.001:
1887 csp[i] = csp[i][:j] + csp[i][j + 1:]
1888 ifcspseglength(csp[i][-1], csp[i]) > 0.001:
1889 csp[i][-1] = csp[i][-1]
1890 csp[i] += ]
1891
1892# TODO Get rid of self intersections.
1893
1894 original_csp = csp[:]
1895# Clip segments which has curvature>1/r. Because their offset will be self-intersecting and very nasty.
1896
1897print_("Offset prepared the path in {}".format(time.time() - time_))
1898print_("Path length = {}".format(sum([len(i) for i in csp])))
1899 time_ = time.time()
1900
1901
1905 unclipped_offset = [[] for i inxrange(csp_len)]
1906
1907 intersection = [[] for i inxrange(csp_len)]
1908 for i inxrange(csp_len):
1909 subpath = csp[i]
1910 subpath_offset = []
1911 for sp1, sp2 in zip(subpath, subpath[1:]):
1912 segment_offset = csp_offset_segment(sp1, sp2, r)
1913 if not subpath_offset:
1914 subpath_offset = segment_offset
1915
1916 prev_l = len(subpath_offset)
1917 else:
1918 prev, arc, next = csp_join_offsets(subpath_offset[-prev_l:], segment_offset, sp1, sp2, sp1_l, sp2_l, r)
1919
1920 subpath_offset = csp_concat_subpaths(subpath_offset[:-prev_l + 1], prev, arc, next)
1921 prev_l = len(next)
1922 sp1_l = sp1[:]
1923 sp2_l = sp2[:]
1924
1925# Join last and first offsets togother to close the curve
1926
1927 prev, arc, next = csp_join_offsets(subpath_offset[-prev_l:], subpath_offset[:2], subpath, subpath, sp1_l, sp2_l, r)
1928 subpath_offset[:2] = next[:]
1929 subpath_offset = csp_concat_subpaths(subpath_offset[:-prev_l + 1], prev, arc)
1930
1931# Collect subpath's offset and save it to unclipped offset list.
1932 unclipped_offset[i] = subpath_offset[:]
1933
1934print_("Offsetted path in {}".format(time.time() - time_))
1935 time_ = time.time()
1936
1937
1941
1942# TODO define offset tolerance here
1943 global small_tolerance
1944 small_tolerance = 0.01
1945 summ = 0
1946 summ1 = 0
1947 for subpath_i inxrange(csp_len):
1948 for subpath_j inxrange(subpath_i, csp_len):
1949 subpath = unclipped_offset[subpath_i]
1950 subpath1 = unclipped_offset[subpath_j]
1951 for i inxrange(1, len(subpath)):
1952# If subpath_i==subpath_j we are looking for self intersections, so
1953# we'll need search intersections only for xrange(i,len(subpath1))
1954 for j in (xrange(i, len(subpath1)) if subpath_i == subpath_j elsexrange(len(subpath1))):
1955 if subpath_i == subpath_j and j == i:
1956# Find self intersections of a segment
1957 sp1, sp2, sp3 = csp_split(subpath[i - 1], subpath[i], .5)
1958 intersections = csp_segments_intersection(sp1, sp2, sp2, sp3)
1959 summ += 1
1960 for t in intersections:
1961 summ1 += 1
1962 if not (small(t - 1) andsmall(t)) and 0 <= t <= 1 and 0 <= t <= 1:
1963 intersection[subpath_i] +=
1964 else:
1965 intersections = csp_segments_intersection(subpath[i - 1], subpath[i], subpath1[j - 1], subpath1[j])
1966 summ += 1
1967 for t in intersections:
1968 summ1 += 1
1969# TODO tolerance dependence to cpsp_length(t)
1970 if len(t) == 2 and 0 <= t <= 1 and 0 <= t <= 1 and not (
1971 subpath_i == subpath_j and (
1972 (j - i - 1) % (len(subpath) - 1) == 0 andsmall(t - 1) andsmall(t) or
1973 (i - j - 1) % (len(subpath) - 1) == 0 andsmall(t - 1) andsmall(t))):
1974 intersection[subpath_i] += ]
1975 intersection[subpath_j] += ]
1976
1977 elif len(t) == 5 and t == "Overlap":
1978 intersection[subpath_i] += , [i, t]]
1979 intersection[subpath_j] += , [j, t]]
1980
1981print_("Intersections found in {}".format(time.time() - time_))
1982print_("Examined {} segments".format(summ))
1983print_("found {} intersections".format(summ1))
1984 time_ = time.time()
1985
1986
1989 splitted_offset = []
1990 for i inxrange(csp_len):
1991 subpath = unclipped_offset[i]
1992 if len(intersection[i]) > 0:
1993 parts = csp_subpath_split_by_points(subpath, intersection[i])
1994# Close parts list to close path (The first and the last parts are joined together)
1995 if [1, 0.] not in intersection[i]:
1996 parts = parts[-1][-1]
1997 parts = csp_concat_subpaths(parts[-1], parts)
1998 splitted_offset += parts[:-1]
1999 else:
2000 splitted_offset += parts[:]
2001 else:
2002 splitted_offset += [subpath[:]]
2003
2004print_("Split in {}".format(time.time() - time_))
2005 time_ = time.time()
2006
2007
2010 result = []
2011 for subpath_i in range(len(splitted_offset)):
2012 clip = False
2013 s1 = splitted_offset[subpath_i]
2014 for subpath_j in range(len(splitted_offset)):
2015 s2 = splitted_offset[subpath_j]
2016 if (P(s1) - P(s2[-1])).l2() < 0.0001 and ((subpath_i + 1) % len(splitted_offset) != subpath_j):
2017 ifdot(csp_normalized_normal(s2[-2], s2[-1], 1.), csp_normalized_slope(s1, s1, 0.)) r < -0.0001:
2018 clip = True
2019 break
2020 if (P(s2) - P(s1[-1])).l2() < 0.0001 and ((subpath_j + 1) % len(splitted_offset) != subpath_i):
2021 ifdot(csp_normalized_normal(s2, s2, 0.), csp_normalized_slope(s1[-2], s1[-1], 1.)) r > 0.0001:
2022 clip = True
2023 break
2024
2025 if not clip:
2026 result += [s1[:]]
2027 elif options.offset_draw_clippend_path:
2028draw_csp([s1], width=.1)
2029draw_pointer(csp_at_t(s2[-2], s2[-1], 1.) +
2030 (P(csp_at_t(s2[-2], s2[-1], 1.)) + P(csp_normalized_normal(s2[-2], s2[-1], 1.)) 10).to_list(), "Green", "line")
2031draw_pointer(csp_at_t(s1, s1, 0.) +
2032 (P(csp_at_t(s1, s1, 0.)) + P(csp_normalized_slope(s1, s1, 0.)) 10).to_list(), "Red", "line")
2033
2034# Now join all together and check closure and orientation of result
2035 joined_result = csp_join_subpaths(result)
2036# Check if each subpath from joined_result is closed
2037
2038 for s in joined_result[:]:
2039 ifcsp_subpaths_end_to_start_distance2(s, s) > 0.001:
2040# Remove open parts
2041 if options.offset_draw_clippend_path:
2042draw_csp([s], width=1)
2043draw_pointer(s, comment=csp_subpaths_end_to_start_distance2(s, s))
2044draw_pointer(s[-1], comment=csp_subpaths_end_to_start_distance2(s, s))
2045 joined_result.remove(s)
2046 else:
2047# Remove small parts
2048 minx, miny, maxx, maxy = csp_true_bounds([s])
2049 if (minx - maxx) 2 + (miny - maxy) 2 < 0.1:
2050 joined_result.remove(s)
2051print_("Clipped and joined path in {}".format(time.time() - time_))
2052
2053
2057
2058 if abs(r .01) < 1:
2059 r1 = (0.99 r) 2
2060 r2 = (1.01 r) 2
2061 else:
2062 r1 = (abs(r) - 1) 2
2063 r2 = (abs(r) + 1) 2
2064
2065 for s in joined_result[:]:
2066 dist = csp_to_point_distance(original_csp, s[int(len(s) / 2)], dist_bounds=[r1, r2])
2067 if not r1 < dist < r2:
2068 joined_result.remove(s)
2069 if options.offset_draw_clippend_path:
2070draw_csp([s], comment=math.sqrt(dist))
2071draw_pointer(csp_at_t(csp[dist][dist - 1], csp[dist][dist], dist) + s[int(len(s) / 2)], "blue", "line", comment=[math.sqrt(dist), i, j, sp])
2072
2073print_("-----------------------------")
2074print_("Total offset time {}".format(time.time() - time_start))
2075print_()
2076 return joined_result
2077
2078
1716def csp_offset(csp, r): …)
2079
2087def biarc(sp1, sp2, z1, z2, depth=0):
2088 def biarc_split(sp1, sp2, z1, z2, depth):
2089 if depth < options.biarc_max_split_depth:
2090 sp1, sp2, sp3 = csp_split(sp1, sp2)
2091 l1 = cspseglength(sp1, sp2)
2092 l2 = cspseglength(sp2, sp3)
2093 if l1 + l2 == 0:
2094 zm = z1
2095 else:
2096 zm = z1 + (z2 - z1) l1 / (l1 + l2)
2097 returnbiarc(sp1, sp2, z1, zm, depth + 1) + biarc(sp2, sp3, zm, z2, depth + 1)
2098 else:
2099 return ]
2100
2101 P0 = P(sp1)
2102 P4 = P(sp2)
2103 TS = (P(sp1) - P0)
2104 TE = -(P(sp2) - P4)
2105 v = P0 - P4
2106 tsa = TS.angle()
2107 tea = TE.angle()
2108 va = v.angle()
2109 if TE.mag() < STRAIGHT_DISTANCE_TOLERANCE and TS.mag() < STRAIGHT_DISTANCE_TOLERANCE:
2110# Both tangents are zero - line straight
2111 return ]
2112 if TE.mag() < STRAIGHT_DISTANCE_TOLERANCE:
2113 TE = -(TS + v).unit()
2114 r = TS.mag() / v.mag() 2
2115 elif TS.mag() < STRAIGHT_DISTANCE_TOLERANCE:
2116 TS = -(TE + v).unit()
2117 r = 1 / (TE.mag() / v.mag() 2)
2118 else:
2119 r = TS.mag() / TE.mag()
2120 TS = TS.unit()
2121 TE = TE.unit()
2122 tang_are_parallel = ((tsa - tea) % math.pi < STRAIGHT_TOLERANCE or math.pi - (tsa - tea) % math.pi < STRAIGHT_TOLERANCE)
2123 if (tang_are_parallel and
2124 ((v.mag() < STRAIGHT_DISTANCE_TOLERANCE or TE.mag() < STRAIGHT_DISTANCE_TOLERANCE or TS.mag() < STRAIGHT_DISTANCE_TOLERANCE) or
2125 1 - abs(TS v / (TS.mag() v.mag())) < STRAIGHT_TOLERANCE)):
2126# Both tangents are parallel and start and end are the same - line straight
2127# or one of tangents still smaller then tolerance
2128
2129# Both tangents and v are parallel - line straight
2130 return ]
2131
2132 c = v v
2133 b = 2 v (r TS + TE)
2134 a = 2 r (TS TE - 1)
2135 if v.mag() == 0:
2136 return biarc_split(sp1, sp2, z1, z2, depth)
2137 asmall = abs(a) < 10 -10
2138 bsmall = abs(b) < 10 -10
2139 csmall = abs(c) < 10 -10
2140 if asmall and b != 0:
2141 beta = -c / b
2142 elif csmall and a != 0:
2143 beta = -b / a
2144 elif not asmall:
2145 discr = b b - 4 a c
2146 if discr < 0:
2147 raise ValueError(a, b, c, discr)
2148 disq = discr .5
2149 beta1 = (-b - disq) / 2 / a
2150 beta2 = (-b + disq) / 2 / a
2151 if beta1 beta2 > 0:
2152 raise ValueError(a, b, c, disq, beta1, beta2)
2153 beta = max(beta1, beta2)
2154 elif asmall and bsmall:
2155 return biarc_split(sp1, sp2, z1, z2, depth)
2156 alpha = beta r
2157 ab = alpha + beta
2158 P1 = P0 + alpha TS
2159 P3 = P4 - beta TE
2160 P2 = (beta / ab) P1 + (alpha / ab) P3
2161
2162 def calculate_arc_params(P0, P1, P2):
2163 D = (P0 + P2) / 2
2164 if (D - P1).mag() == 0:
2165 return None, None
2166 R = D - ((D - P0).mag() 2 / (D - P1).mag()) (P1 - D).unit()
2167 p0a = (P0 - R).angle() % (2 math.pi)
2168 p1a = (P1 - R).angle() % (2 math.pi)
2169 p2a = (P2 - R).angle() % (2 math.pi)
2170 alpha = (p2a - p0a) % (2 math.pi)
2171 if (p0a < p2a and (p1a < p0a or p2a < p1a)) or (p2a < p1a < p0a):
2172 alpha = -2 math.pi + alpha
2173 if abs(R.x) > 1000000 or abs(R.y) > 1000000 or (R - P0).mag() < options.min_arc_radius 2:
2174 return None, None
2175 else:
2176 return R, alpha
2177
2178 R1, a1 = calculate_arc_params(P0, P1, P2)
2179 R2, a2 = calculate_arc_params(P2, P3, P4)
2180 if R1 is None or R2 is None or (R1 - P0).mag() < STRAIGHT_TOLERANCE or (R2 - P2).mag() < STRAIGHT_TOLERANCE:
2181 return ]
2182
2183 d = csp_to_arc_distance(sp1, sp2, [P0, P2, R1, a1], [P2, P4, R2, a2])
2184 if d > options.biarc_tolerance and depth < options.biarc_max_split_depth:
2185 return biarc_split(sp1, sp2, z1, z2, depth)
2186 else:
2187 if R2.mag() a2 == 0:
2188 zm = z2
2189 else:
2190 zm = z1 + (z2 - z1) (abs(R1.mag() a1)) / (abs(R2.mag() a2) + abs(R1.mag() a1))
2191
2192 l = (P0 - P2).l2()
2193 if l < EMC_TOLERANCE_EQUAL 2 or l < EMC_TOLERANCE_EQUAL 2 R1.l2() / 100:
2194# arc should be straight otherwise it could be treated as full circle
2195 arc1 = [sp1, 'line', 0, 0, [P2.x, P2.y], [z1, zm]]
2196 else:
2197 arc1 = [sp1, 'arc', [R1.x, R1.y], a1, [P2.x, P2.y], [z1, zm]]
2198
2199 l = (P4 - P2).l2()
2200 if l < EMC_TOLERANCE_EQUAL 2 or l < EMC_TOLERANCE_EQUAL 2 R2.l2() / 100:
2201# arc should be straight otherwise it could be treated as full circle
2202 arc2 =
2203 else:
2204 arc2 =
2205
2206 return [arc1, arc2]
2207
2208
2087def biarc(sp1, sp2, z1, z2, depth=0): …)
2209class Postprocessor(object):
2210def init(self, error_function_handler):
2211 self.error = error_function_handler
2212 self.functions = {
2213"remap": self.remapremap,
2214"remapi": self.remapiremapi,
2215"scale": self.scalescale,
2216"move": self.movemove,
2217"flip": self.flip_axisflip_axis,
2218"flip_axis": self.flip_axisflip_axis,
2219"round": self.round_coordinatesround_coordinates,
2220"parameterize": self.parameterizeparameterize,
2221"regex": self.re_sub_on_gcode_lines
2222 }
2223
2210def init(self, error_function_handler): …)
2224def process(self, command):
2225 command = re.sub(r"\\", ":#:#:slash:#:#:", command)
2226 command = re.sub(r"\;", ":#:#:semicolon:#:#:", command)
2227 command = command.split(";")
2228 for s in command:
2229 s = re.sub(":#:#:slash:#:#:", "\\", s)
2230 s = re.sub(":#:#:semicolon:#:#:", "\;", s)
2231 s = s.strip()
2232 if s != "":
2233 self.parse_command(s)
2234
2224def process(self, command): …)
2235def parse_command(self, command):
2236 r = re.match(r"([A-Za-z0-9_]+)\s\(\s(.)\)", command)
2237 if not r:
2238 self.error("Parse error while postprocessing.\n(Command: '{}')".format(command), "error")
2239 function = r.group(1).lower()
2240 parameters = r.group(2)
2241 if function in self.functions:
2242print_("Postprocessor: executing function {}({})".format(function, parameters))
2243 self.functionsfunction
2244 else:
2245 self.error("Unrecognized function '{}' while postprocessing.\n(Command: '{}')".format(function, command), "error")
2246
2235def parse_command(self, command): …)
2247def re_sub_on_gcode_lines(self, parameters):
2248 gcode = self.gcode.split("\n")
2249 self.gcode = ""
2250 try:
2251 for line in gcode:
2252 self.gcode += eval("re.sub({},line)".format(parameters)) + "\n"
2253
2254 except Exception as ex:
2255 self.error("Bad parameters for regexp. "
2256"They should be as re.sub pattern and replacement parameters! "
2257"For example: r\"G0(\d)\", r\"G\1\" \n"
2258"(Parameters: '{}')\n {}".format(parameters, ex), "error")
2259
2247def re_sub_on_gcode_lines(self, parameters): …)
2260def remapi(self, parameters):
2261 self.remapremap(parameters, case_sensitive=True)
2262
2260def remapi(self, parameters): …)
2263def remap(self, parameters, case_sensitive=False):
2264# remap parameters should be like "x->y,y->x"
2265 parameters = parameters.replace("\,", ":#:#:coma:#:#:")
2266 parameters = parameters.split(",")
2267 pattern = []
2268 remap = []
2269 for s in parameters:
2270 s = s.replace(":#:#:coma:#:#:", "\,")
2271 r = re.match("""\s(\'|\")(.)\1\s->\s(\'|\")(.)\3\s""", s)
2272 if not r:
2273 self.error("Bad parameters for remap.\n(Parameters: '{}')".format(parameters), "error")
2274 pattern += [r.group(2)]
2275 remap += [r.group(4)]
2276
2277 for i in range(len(pattern)):
2278 if case_sensitive:
2279 self.gcode = ireplace(self.gcode, pattern[i], ":#:#:remap_pattern{}:#:#:".format(i))
2280 else:
2281 self.gcode = self.gcode.replace(pattern[i], ":#:#:remap_pattern{}:#:#:".format(i))
2282
2283 for i in range(len(remap)):
2284 self.gcode = self.gcode.replace(":#:#:remap_pattern{}:#:#:".format(i), remap[i])
2285
2263def remap(self, parameters, case_sensitive=False): …)
2286def transform(self, move, scale):
2287 axis = ["xi", "yj", "zk", "a"]
2288 flip = scale scale scale < 0
2289 gcode = ""
2290 warned = []
2291 r_scale = scale
2292 plane = "g17"
2293 for s in self.gcode.split("\n"):
2294# get plane selection:
2295 s_wo_comments = re.sub(r"\([^\)]\)", "", s)
2296 r = re.search(r"(?i)(G17|G18|G19)", s_wo_comments)
2297 if r:
2298 plane = r.group(1).lower()
2299 if plane == "g17":
2300 r_scale = scale # plane XY -> scale x
2301 if plane == "g18":
2302 r_scale = scale # plane XZ -> scale x
2303 if plane == "g19":
2304 r_scale = scale # plane YZ -> scale y
2305# Raise warning if scale factors are not the game for G02 and G03
2306 if plane not in warned:
2307 r = re.search(r"(?i)(G02|G03)", s_wo_comments)
2308 if r:
2309 if plane == "g17"and scale != scale:
2310 self.error("Post-processor: Scale factors for X and Y axis are not the same. G02 and G03 codes will be corrupted.")
2311 if plane == "g18"and scale != scale:
2312 self.error("Post-processor: Scale factors for X and Z axis are not the same. G02 and G03 codes will be corrupted.")
2313 if plane == "g19"and scale != scale:
2314 self.error("Post-processor: Scale factors for Y and Z axis are not the same. G02 and G03 codes will be corrupted.")
2315 warned += [plane]
2316# Transform
2317 for i in range(len(axis)):
2318 if move[i] != 0 or scale[i] != 1:
2319 for a in axis[i]:
2320 r = re.search(r"(?i)(" + a + r")\s(-?)\s(\d.?\d)", s)
2321 if r and r.group(3) != "":
2322 s = re.sub(r"(?i)(" + a + r")\s(-?)\s(\d.?\d)", r"\1 {:f}".format(float(r.group(2) + r.group(3)) scale[i] + (move[i] if a not in ["i", "j", "k"] else 0)), s)
2323# scale radius R
2324 if r_scale != 1:
2325 r = re.search(r"(?i)(r)\s(-?\s(\d.?\d))", s)
2326 if r and r.group(3) != "":
2327 try:
2328 s = re.sub(r"(?i)(r)\s(-?)\s(\d.?\d)", r"\1 {:f}".format(float(r.group(2) + r.group(3)) r_scale), s)
2329 except:
2330 pass
2331
2332 gcode += s + "\n"
2333
2334 self.gcode = gcode
2335 if flip:
2336 self.remapiremapi("'G02'->'G03', 'G03'->'G02'")
2337
2286def transform(self, move, scale): …)
2338def parameterize(self, parameters):
2339 planes = []
2340 feeds = {}
2341 coords = []
2342 gcode = ""
2343 coords_def = {"x": "x", "y": "y", "z": "z", "i": "x", "j": "y", "k": "z", "a": "a"}
2344 for s in self.gcode.split("\n"):
2345 s_wo_comments = re.sub(r"\([^\)]\)", "", s)
2346# get Planes
2347 r = re.search(r"(?i)(G17|G18|G19)", s_wo_comments)
2348 if r:
2349 plane = r.group(1).lower()
2350 if plane not in planes:
2351 planes += [plane]
2352# get Feeds
2353 r = re.search(r"(?i)(F)\s(-?)\s(\d.?\d)", s_wo_comments)
2354 if r:
2355 feed = float(r.group(2) + r.group(3))
2356 if feed not in feeds:
2357 feeds[feed] = "#" + str(len(feeds) + 20)
2358
2359# Coordinates
2360 for c in"xyzijka":
2361 r = re.search(r"(?i)(" + c + r")\s(-?)\s(\d.?\d)", s_wo_comments)
2362 if r:
2363 c = coords_def[r.group(1).lower()]
2364 if c not in coords:
2365 coords += [c]
2366# Add offset parametrization
2367 offset = {"x": "#6", "y": "#7", "z": "#8", "a": "#9"}
2368 for c in coords:
2369 gcode += "{} = 0 ({} axis offset)\n".format(offset[c], c.upper())
2370
2371# Add scale parametrization
2372 if not planes:
2373 planes = ["g17"]
2374 if len(planes) > 1: # have G02 and G03 in several planes scale_x = scale_y = scale_z required
2375 gcode += "#10 = 1 (Scale factor)\n"
2376 scale = {"x": "#10", "i": "#10", "y": "#10", "j": "#10", "z": "#10", "k": "#10", "r": "#10"}
2377 else:
2378 gcode += "#10 = 1 ({} Scale factor)\n".format({"g17": "XY", "g18": "XZ", "g19": "YZ"}[planes])
2379 gcode += "#11 = 1 ({} Scale factor)\n".format({"g17": "Z", "g18": "Y", "g19": "X"}[planes])
2380 scale = {"x": "#10", "i": "#10", "y": "#10", "j": "#10", "z": "#10", "k": "#10", "r": "#10"}
2381 if"g17"in planes:
2382 scale["z"] = "#11"
2383 scale["k"] = "#11"
2384 if"g18"in planes:
2385 scale["y"] = "#11"
2386 scale["j"] = "#11"
2387 if"g19"in planes:
2388 scale["x"] = "#11"
2389 scale["i"] = "#11"
2390# Add a scale
2391 if"a"in coords:
2392 gcode += "#12 = 1 (A axis scale)\n"
2393 scale["a"] = "#12"
2394
2395# Add feed parametrization
2396 for f in feeds:
2397 gcode += "{} = {:f} (Feed definition)\n".format(feeds[f], f)
2398
2399# Parameterize Gcode
2400 for s in self.gcode.split("\n"):
2401# feed replace :
2402 r = re.search(r"(?i)(F)\s(-?)\s(\d.?\d)", s)
2403 if r and len(r.group(3)) > 0:
2404 s = re.sub(r"(?i)(F)\s(-?)\s(\d.?\d)", "F [{}]".format(feeds[float(r.group(2) + r.group(3))]), s)
2405# Coords XYZA replace
2406 for c in"xyza":
2407 r = re.search(r"(?i)((" + c + r")\s(-?)\s(\d.?\d))", s)
2408 if r and len(r.group(4)) > 0:
2409 s = re.sub(r"(?i)(" + c + r")\s((-?)\s(\d.?\d))", r"\1[\2{}+{}]".format(scale[c], offset[c]), s)
2410
2411# Coords IJKR replace
2412 for c in"ijkr":
2413 r = re.search(r"(?i)((" + c + r")\s(-?)\s(\d.?\d))", s)
2414 if r and len(r.group(4)) > 0:
2415 s = re.sub(r"(?i)(" + c + r")\s((-?)\s(\d.?\d))", r"\1[\2{}]".format(scale[c]), s)
2416
2417 gcode += s + "\n"
2418
2419 self.gcode = gcode
2420
2338def parameterize(self, parameters): …)
2421def round_coordinates(self, parameters):
2422 try:
2423 round_ = int(parameters)
2424 except:
2425 self.error("Bad parameters for round. Round should be an integer! \n(Parameters: '{}')".format(parameters), "error")
2426 gcode = ""
2427 for s in self.gcode.split("\n"):
2428 for a in"xyzijkaf":
2429 r = re.search(r"(?i)(" + a + r")\s(-?\s(\d.?\d))", s)
2430 if r:
2431
2432 if r.group(2) != "":
2433 s = re.sub(
2434 r"(?i)(" + a + r")\s(-?)\s(\d.?\d)",
2435 (r"\1 %0." + str(round_) + "f"if round_ > 0 else r"\1 %d") % round(float(r.group(2)), round_),
2436 s)
2437 gcode += s + "\n"
2438 self.gcode = gcode
2439
2421def round_coordinates(self, parameters): …)
2440def scale(self, parameters):
2441 parameters = parameters.split(",")
2442 scale = [1., 1., 1., 1.]
2443 try:
2444 for i in range(len(parameters)):
2445 if float(parameters[i]) == 0:
2446 self.error("Bad parameters for scale. Scale should not be 0 at any axis! \n(Parameters: '{}')".format(parameters), "error")
2447 scale[i] = float(parameters[i])
2448 except:
2449 self.error("Bad parameters for scale.\n(Parameters: '{}')".format(parameters), "error")
2450 self.transform([0, 0, 0, 0], scale)
2451
2440def scale(self, parameters): …)
2452def move(self, parameters):
2453 parameters = parameters.split(",")
2454 move = [0., 0., 0., 0.]
2455 try:
2456 for i in range(len(parameters)):
2457 move[i] = float(parameters[i])
2458 except:
2459 self.error("Bad parameters for move.\n(Parameters: '{}')".format(parameters), "error")
2460 self.transform(move, [1., 1., 1., 1.])
2461
2452def move(self, parameters): …)
2462def flip_axis(self, parameters):
2463 parameters = parameters.lower()
2464 axis = {"x": 1., "y": 1., "z": 1., "a": 1.}
2465 for p in parameters:
2466 if p in [",", " ", " ", "\r", "'", '"']:
2467 continue
2468 if p not in ["x", "y", "z", "a"]:
2469 self.error("Bad parameters for flip_axis. Parameter should be string consists of 'xyza' \n(Parameters: '{}')".format(parameters), "error")
2470 axis[p] = -axis[p]
2471 self.scalescale("{:f},{:f},{:f},{:f}".format(axis["x"], axis["y"], axis["z"], axis["a"]))
2472
2473
2462def flip_axis(self, parameters): …)
2474
2209class Postprocessor(object): …)
2477class Polygon(object):
2478def init(self, polygon=None):
2479 self.polygon = [] if polygon is None else polygon[:]
2480
2478def init(self, polygon=None): …)
2481def move(self, x, y):
2482 for i in range(len(self.polygon)):
2483 for j in range(len(self.polygon[i])):
2484 self.polygon[i][j] += x
2485 self.polygon[i][j] += y
2486
2481def move(self, x, y): …)
2487def bounds(self):
2488 minx = 1e400
2489 miny = 1e400
2490 maxx = -1e400
2491 maxy = -1e400
2492 for poly in self.polygon:
2493 for p in poly:
2494 if minx > p:
2495 minx = p
2496 if miny > p:
2497 miny = p
2498 if maxx < p:
2499 maxx = p
2500 if maxy < p:
2501 maxy = p
2502 return minx 1, miny 1, maxx 1, maxy 1
2503
2487def bounds(self): …)
2504def width(self):
2505 b = self.bounds()
2506 return b - b
2507
2504def width(self): …)
2508def rotate_(self, sin, cos):
2509 self.polygon = [
2510 [
2511 [point cos - point sin, point sin + point cos] for point in subpoly
2512 ]
2513 for subpoly in self.polygon
2514 ]
2515
2508def rotate_(self, sin, cos): …)
2516def rotate(self, a):
2517 cos = math.cos(a)
2518 sin = math.sin(a)
2519 self.rotate_(sin, cos)
2520
2516def rotate(self, a): …)
2521def drop_into_direction(self, direction, surface):
2522# Polygon is a list of simple polygons
2523# Surface is a polygon + line y = 0
2524# Direction is [dx,dy]
2525 if len(self.polygon) == 0 or len(self.polygon) == 0:
2526 return
2527 if direction 2 + direction 2 < 1e-10:
2528 return
2529 direction = normalize(direction)
2530 sin = direction
2531 cos = -direction
2532 self.rotate_(-sin, cos)
2533 surface.rotate_(-sin, cos)
2534 self.drop_down(surface, zerro_plane=False)
2535 self.rotate_(sin, cos)
2536 surface.rotate_(sin, cos)
2537
2521def drop_into_direction(self, direction, surface): …)
2538def centroid(self):
2539 centroids = []
2540 sa = 0
2541 for poly in self.polygon:
2542 cx = 0
2543 cy = 0
2544 a = 0
2545 for i in range(len(poly)):
2546 [x1, y1] = poly[i - 1]
2547 [x2, y2] = poly[i]
2548 cx += (x1 + x2) (x1 y2 - x2 y1)
2549 cy += (y1 + y2) (x1 y2 - x2 y1)
2550 a += (x1 y2 - x2 y1)
2551 a = 3.
2552 if abs(a) > 0:
2553 cx /= a
2554 cy /= a
2555 sa += abs(a)
2556 centroids +=
2557 if sa == 0:
2558 return [0., 0.]
2559 cx = 0
2560 cy = 0
2561 for c in centroids:
2562 cx += c c
2563 cy += c c
2564 cx /= sa
2565 cy /= sa
2566 return [cx, cy]
2567
2538def centroid(self): …)
2568def drop_down(self, surface, zerro_plane=True):
2569# Polygon is a list of simple polygons
2570# Surface is a polygon + line y = 0
2571# Down means min y (0,-1)
2572 if len(self.polygon) == 0 or len(self.polygon) == 0:
2573 return
2574# Get surface top point
2575 top = surface.bounds()
2576 if zerro_plane:
2577 top = max(0, top)
2578# Get polygon bottom point
2579 bottom = self.bounds()
2580 self.move(0, top - bottom + 10)
2581# Now get shortest distance from surface to polygon in positive x=0 direction
2582# Such distance = min(distance(vertex, edge)...) where edge from surface and
2583# vertex from polygon and vice versa...
2584 dist = 1e300
2585 for poly in surface.polygon:
2586 for i in range(len(poly)):
2587 for poly1 in self.polygon:
2588 for i1 in range(len(poly1)):
2589 st = poly[i - 1]
2590 end = poly[i]
2591 vertex = poly1[i1]
2592 if st <= vertex <= end or end <= vertex <= st:
2593 if st == end:
2594 d = min(vertex - st, vertex - end)
2595 else:
2596 d = vertex - st - (end - st) (vertex - st) / (end - st)
2597 if dist > d:
2598 dist = d
2599# and vice versa just change the sign because vertex now under the edge
2600 st = poly1[i1 - 1]
2601 end = poly1[i1]
2602 vertex = poly[i]
2603 if st <= vertex <= end or end <= vertex <= st:
2604 if st == end:
2605 d = min(- vertex + st, -vertex + end)
2606 else:
2607 d = - vertex + st + (end - st) (vertex - st) / (end - st)
2608 if dist > d:
2609 dist = d
2610
2611 if zerro_plane and dist > 10 + top:
2612 dist = 10 + top
2613 self.move(0, -dist)
2614
2568def drop_down(self, surface, zerro_plane=True): …)
2615def draw(self, color="#075", width=.1, group=None):
2616 csp = [csp_subpath_line_to([], poly + [poly]) for poly in self.polygon]
2617draw_csp(csp, width=width, group=group)
2618
2615def draw(self, color="#075", width=.1, group=None): …)
2619def add(self, add):
2620 if type(add) == type([]):
2621 self.polygon += add[:]
2622 else:
2623 self.polygon += add.polygon[:]
2624
2619def add(self, add): …)
2625def point_inside(self, p):
2626 inside = False
2627 for poly in self.polygon:
2628 for i in range(len(poly)):
2629 st = poly[i - 1]
2630 end = poly[i]
2631 if p == st or p == end:
2632 return True# point is a vertex = point is on the edge
2633 if st > end:
2634 st, end = end, st # This will be needed to check that edge if open only at right end
2635 c = (p - st) (end - st) - (end - st) (p - st)
2636 if st <= p < end:
2637 if c < 0:
2638 inside = not inside
2639 elif c == 0:
2640 return True# point is on the edge
2641 elif st == end == p and (st <= p <= end or end <= p <= st): # point is on the edge
2642 return True
2643 return inside
2644
2625def point_inside(self, p): …)
2645def hull(self):
2646# Add vertices at all self intersection points.
2647 hull = []
2648 for i1 in range(len(self.polygon)):
2649 poly1 = self.polygon[i1]
2650 poly_ = []
2651 for j1 in range(len(poly1)):
2652 s = poly1[j1 - 1]
2653 e = poly1[j1]
2654 poly_ += [s]
2655
2656# Check self intersections
2657 for j2 in range(j1 + 1, len(poly1)):
2658 s1 = poly1[j2 - 1]
2659 e1 = poly1[j2]
2660 int_ = line_line_intersection_points(s, e, s1, e1)
2661 for p in int_:
2662 ifpoint_to_point_d2(p, s) > 0.000001 andpoint_to_point_d2(p, e) > 0.000001:
2663 poly_ += [p]
2664# Check self intersections with other polys
2665 for i2 in range(len(self.polygon)):
2666 if i1 == i2:
2667 continue
2668 poly2 = self.polygon[i2]
2669 for j2 in range(len(poly2)):
2670 s1 = poly2[j2 - 1]
2671 e1 = poly2[j2]
2672 int_ = line_line_intersection_points(s, e, s1, e1)
2673 for p in int_:
2674 ifpoint_to_point_d2(p, s) > 0.000001 andpoint_to_point_d2(p, e) > 0.000001:
2675 poly_ += [p]
2676 hull += [poly_]
2677# Create the dictionary containing all edges in both directions
2678 edges = {}
2679 for poly in self.polygon:
2680 for i in range(len(poly)):
2681 s = tuple(poly[i - 1])
2682 e = tuple(poly[i])
2683 ifpoint_to_point_d2(e, s) < 0.000001:
2684 continue
2685 break_s = False
2686 break_e = False
2687 for p in edges:
2688 ifpoint_to_point_d2(p, s) < 0.000001:
2689 break_s = True
2690 s = p
2691 ifpoint_to_point_d2(p, e) < 0.000001:
2692 break_e = True
2693 e = p
2694 if break_s and break_e:
2695 break
2696 l = point_to_point_d(s, e)
2697 if not break_s and not break_e:
2698 edges[s] =
2699 edges[e] =
2700 else:
2701 if e in edges:
2702 for edge in edges[e]:
2703 ifpoint_to_point_d2(edge, s) < 0.000001:
2704 break
2705 ifpoint_to_point_d2(edge, s) > 0.000001:
2706 edges[e] +=
2707 else:
2708 edges[e] =
2709 if s in edges:
2710 for edge in edges[s]:
2711 ifpoint_to_point_d2(edge, e) < 0.000001:
2712 break
2713 ifpoint_to_point_d2(edge, e) > 0.000001:
2714 edges[s] +=
2715 else:
2716 edges[s] =
2717
2718 def angle_quadrant(sin, cos):
2719# quadrants are (0,pi/2], (pi/2,pi], (pi,3pi/2], (3pi/2, 2pi], i.e. 0 is in the 4-th quadrant
2720 if sin > 0 and cos >= 0:
2721 return 1
2722 if sin >= 0 and cos < 0:
2723 return 2
2724 if sin < 0 and cos <= 0:
2725 return 3
2726 if sin <= 0 and cos > 0:
2727 return 4
2728
2729 def angle_is_less(sin, cos, sin1, cos1):
2730# 0 = 2pi is the largest angle
2731 if [sin1, cos1] == [0, 1]:
2732 return True
2733 if [sin, cos] == [0, 1]:
2734 return False
2735 if angle_quadrant(sin, cos) > angle_quadrant(sin1, cos1):
2736 return False
2737 if angle_quadrant(sin, cos) < angle_quadrant(sin1, cos1):
2738 return True
2739 if sin >= 0 and cos > 0:
2740 return sin < sin1
2741 if sin > 0 and cos <= 0:
2742 return sin > sin1
2743 if sin <= 0 and cos < 0:
2744 return sin > sin1
2745 if sin < 0 and cos >= 0:
2746 return sin < sin1
2747
2748 def get_closes_edge_by_angle(edges, last):
2749# Last edge is normalized vector of the last edge.
2750 min_angle = [0, 1]
2751 next = last
2752 last_edge = [(last - last) / last, (last - last) / last]
2753 for p in edges:
2754
2755 cur = [(p - p) / p, (p - p) / p]
2756 cos = dot(cur, last_edge)
2757 sin = cross(cur, last_edge)
2758
2759 if angle_is_less(sin, cos, min_angle, min_angle):
2760 min_angle = [sin, cos]
2761 next = p
2762
2763 return next
2764
2765# Join edges together into new polygon cutting the vertexes inside new polygon
2766 self.polygon = []
2767 len_edges = sum([len(edges[p]) for p in edges])
2768 loops = 0
2769
2770 while len(edges) > 0:
2771 poly = []
2772 if loops > len_edges:
2773 raise ValueError("Hull error")
2774 loops += 1
2775# Find left most vertex.
2776 start = (1e100, 1)
2777 for edge in edges:
2778 start = min(start, min(edges[edge]))
2779 last = [(start - 1, start), start, 1]
2780 first_run = True
2781 loops1 = 0
2782 while last != start or first_run:
2783 first_run = False
2784 if loops1 > len_edges:
2785 raise ValueError("Hull error")
2786 loops1 += 1
2787 next = get_closes_edge_by_angle(edges[last], last)
2788
2789 last = next
2790 poly += [list(last)]
2791 self.polygon += [poly]
2792# Remove all edges that are intersects new poly (any vertex inside new poly)
2793 poly_ = Polygon([poly])
2794 for p in edges.keys()[:]:
2795 if poly_.point_inside(list(p)):
2796 del edges[p]
2797 self.draw(color="Green", width=1)
2798
2799
2645def hull(self): …)
2800
2805
2477class Polygon(object): …)
2806class Gcodetools(inkex.EffectExtension):
2807 multi_inx = True# XXX Remove this after refactoring
2808
2809def export_gcode(self, gcode, no_headers=False):
2810 if self.options.postprocessor != ""or self.options.postprocessor_custom != "":
2811 postprocessor = Postprocessor(self.error)
2812 postprocessor.gcode = gcode
2813 if self.options.postprocessor != "":
2814 postprocessor.process(self.options.postprocessor)
2815 if self.options.postprocessor_custom != "":
2816 postprocessor.process(self.options.postprocessor_custom)
2817
2818 if not no_headers:
2819 postprocessor.gcode = self.header + postprocessor.gcode + self.footer
2820
2821 with open(os.path.join(self.options.directory, self.options.file), "w") as f:
2822 f.write(postprocessor.gcode)
2823
2809def export_gcode(self, gcode, no_headers=False): …)
2824
2828def tab_plasma_prepare_path(self):
2829 self.get_info_plus()
2830
2831 def add_arc(sp1, sp2, end=False, l=10., r=10.):
2832 if not end:
2833 n = csp_normalized_normal(sp1, sp2, 0.)
2834 returncsp_reverse([arc_from_s_r_n_l(sp1, r, n, -l)])
2835 else:
2836 n = csp_normalized_normal(sp1, sp2, 1.)
2837 returnarc_from_s_r_n_l(sp2, r, n, l)
2838
2839 def add_normal(sp1, sp2, end=False, l=10., r=10.):
2840# r is needed only for be compatible with add_arc
2841 if not end:
2842 n = csp_normalized_normal(sp1, sp2, 0.)
2843 p = [n l + sp1, n l + sp1]
2844 returncsp_subpath_line_to([], [p, sp1])
2845 else:
2846 n = csp_normalized_normal(sp1, sp2, 1.)
2847 p = [n l + sp2, n l + sp2]
2848 returncsp_subpath_line_to([], [sp2, p])
2849
2850 def add_tangent(sp1, sp2, end=False, l=10., r=10.):
2851# r is needed only for be compatible with add_arc
2852 if not end:
2853 n = csp_normalized_slope(sp1, sp2, 0.)
2854 p = [-n l + sp1, -n l + sp1]
2855 returncsp_subpath_line_to([], [p, sp1])
2856 else:
2857 n = csp_normalized_slope(sp1, sp2, 1.)
2858 p = [n l + sp2, n l + sp2]
2859 returncsp_subpath_line_to([], [sp2, p])
2860
2861 if not self.options.in_out_path and not self.options.plasma_prepare_corners and self.options.in_out_path_do_not_add_reference_point:
2862 self.error("Warning! Extension is not said to do anything! Enable one of Create in-out paths or Prepare corners checkboxes or disable Do not add in-out reference point!")
2863 return
2864
2865# Add in-out-reference point if there is no one yet.
2866if ((len(self.in_out_reference_points) == 0 and self.options.in_out_path
2867 or not self.options.in_out_path and not self.options.plasma_prepare_corners)
2868 and not self.options.in_out_path_do_not_add_reference_point):
2869 self.options.orientation_points_count = "in-out reference point"
2870#self.orientation()
2871
2872 if self.options.in_out_path or self.options.plasma_prepare_corners:
2873 self.set_markers()
2874 add_func = {"Round": add_arc, "Perpendicular": add_normal, "Tangent": add_tangent}[self.options.in_out_path_type]
2875 if self.options.in_out_path_type == "Round"and self.options.in_out_path_len > self.options.in_out_path_radius 3 / 2 math.pi:
2876 self.error("In-out len is to big for in-out radius will cropp it to be r3/2pi!")
2877
2878if self.selected_paths == {} and self.options.auto_select_paths:
2879 self.selected_paths = self.paths
2880 self.error("No paths are selected! Trying to work on all available paths.")
2881
2882 if self.selected_paths == {}:
2883 self.error("Nothing is selected. Please select something.")
2884 a = self.options.plasma_prepare_corners_tolerance
2885 corner_tolerance = cross([1., 0.], [math.cos(a), math.sin(a)])
2886
2887 for layer in self.layers:
2888 if layer in self.selected_paths:
2889 max_dist = self.transform_scalar(self.options.in_out_path_point_max_dist, layer, reverse=True)
2890 l = self.transform_scalar(self.options.in_out_path_len, layer, reverse=True)
2891 plasma_l = self.transform_scalar(self.options.plasma_prepare_corners_distance, layer, reverse=True)
2892 r = self.transform_scalar(self.options.in_out_path_radius, layer, reverse=True)
2893 l = min(l, r 3 / 2 math.pi)
2894
2895 for path in self.selected_paths[layer]:
2896 csp = self.apply_transforms(path, path.path.to_superpath())
2897 csp = csp_remove_zero_segments(csp)
2898 res = []
2899
2900 for subpath in csp:
2901# Find closes point to in-out reference point
2902# If subpath is open skip this step
2903 if self.options.in_out_path:
2904# split and reverse path for further add in-out points
2905 ifpoint_to_point_d2(subpath, subpath[-1]) < 1.e-10:
2906 d = [1e100, 1, 1, 1.]
2907 for p in self.in_out_reference_points:
2908 d1 = csp_to_point_distance([subpath], p, dist_bounds=[0, max_dist])
2909 if d1 < d:
2910 d = d1[:]
2911 p_ = p
2912 if d < max_dist 2:
2913# Lets find is there any angles near this point to put in-out path in
2914# the angle if it's possible
2915# remove last node to make iterations easier
2916 subpath = subpath[-1]
2917 del subpath[-1]
2918 max_cross = [-1e100, None]
2919 for j in range(len(subpath)):
2920 sp1 = subpath[j - 2]
2921 sp2 = subpath[j - 1]
2922 sp3 = subpath[j]
2923 ifpoint_to_point_d2(sp2, p_) < max_dist 2:
2924 s1 = csp_normalized_slope(sp1, sp2, 1.)
2925 s2 = csp_normalized_slope(sp2, sp3, 0.)
2926 max_cross = max(max_cross, [cross(s1, s2), j - 1])
2927# return back last point
2928 subpath.append(subpath)
2929 if max_cross is not None and max_cross > corner_tolerance:
2930# there's an angle near the point
2931 j = max_cross
2932 if j < 0:
2933 j -= 1
2934 if j != 0:
2935 subpath = csp_concat_subpaths(subpath[j:], subpath[:j + 1])
2936 else:
2937# have to cut path's segment
2938 d, i, j, t = d
2939 sp1, sp2, sp3 = csp_split(subpath[j - 1], subpath[j], t)
2940 subpath = csp_concat_subpaths([sp2, sp3], subpath[j:], subpath[:j], [sp1, sp2])
2941
2942 if self.options.plasma_prepare_corners:
2943# prepare corners
2944# find corners and add some nodes
2945# corner at path's start/end is ignored
2946 res_ = [subpath]
2947 for sp2, sp3 in zip(subpath[1:], subpath[2:]):
2948 sp1 = res_[-1]
2949 s1 = csp_normalized_slope(sp1, sp2, 1.)
2950 s2 = csp_normalized_slope(sp2, sp3, 0.)
2951 ifcross(s1, s2) > corner_tolerance:
2952# got a corner to process
2953 S1 = P(s1)
2954 S2 = P(s2)
2955 N = (S1 - S2).unit() plasma_l
2956 SP2 = P(sp2)
2957 P1 = (SP2 + N)
2958 res_ += [
2959 [sp2, sp2, (SP2 + S1 plasma_l).to_list()],
2960 [(P1 - N.ccw() / 2).to_list(), P1.to_list(), (P1 + N.ccw() / 2).to_list()],
2961 [(SP2 - S2 plasma_l).to_list(), sp2, sp2]
2962 ]
2963 else:
2964 res_ += [sp2]
2965 res_ += [sp3]
2966 subpath = res_
2967 if self.options.in_out_path:
2968# finally add let's add in-out paths...
2969 subpath = csp_concat_subpaths(
2970 add_func(subpath, subpath, False, l, r),
2971 subpath,
2972 add_func(subpath[-2], subpath[-1], True, l, r)
2973 )
2974
2975 res += [subpath]
2976
2977 if self.options.in_out_path_replace_original_path:
2978 path.path = CubicSuperPath(self.apply_transforms(path, res, True))
2979 else:
2980draw_csp(res, width=1, style=MARKER_STYLE["in_out_path_style"])
2981
2828def tab_plasma_prepare_path(self): …)
2982def add_arguments(self, pars):
2983 add_argument = pars.add_argument
2984 add_argument("-d", "--directory", default="/home/", help="Directory for gcode file")
2985 add_argument("-f", "--filename", dest="file", default="-1.0", help="File name")
2986 add_argument("--add-numeric-suffix-to-filename", type=inkex.Boolean, default=True, help="Add numeric suffix to filename")
2987 add_argument("--Zscale", type=float, default="1.0", help="Scale factor Z")
2988 add_argument("--Zoffset", type=float, default="0.0", help="Offset along Z")
2989 add_argument("-s", "--Zsafe", type=float, default="0.5", help="Z above all obstacles")
2990 add_argument("-z", "--Zsurface", type=float, default="0.0", help="Z of the surface")
2991 add_argument("-c", "--Zdepth", type=float, default="-0.125", help="Z depth of cut")
2992 add_argument("--Zstep", type=float, default="-0.125", help="Z step of cutting")
2993 add_argument("-p", "--feed", type=float, default="4.0", help="Feed rate in unit/min")
2994
2995 add_argument("--biarc-tolerance", type=float, default="1", help="Tolerance used when calculating biarc interpolation.")
2996 add_argument("--biarc-max-split-depth", type=int, default="4", help="Defines maximum depth of splitting while approximating using biarcs.")
2997 add_argument("--path-to-gcode-order", default="path by path", help="Defines cutting order path by path or layer by layer.")
2998 add_argument("--path-to-gcode-depth-function", default="zd", help="Path to gcode depth function.")
2999 add_argument("--path-to-gcode-sort-paths", type=inkex.Boolean, default=True, help="Sort paths to reduce rapid distance.")
3000 add_argument("--comment-gcode", default="", help="Comment Gcode")
3001 add_argument("--comment-gcode-from-properties", type=inkex.Boolean, default=False, help="Get additional comments from Object Properties")
3002
3003 add_argument("--tool-diameter", type=float, default="3", help="Tool diameter used for area cutting")
3004 add_argument("--max-area-curves", type=int, default="100", help="Maximum area curves for each area")
3005 add_argument("--area-inkscape-radius", type=float, default="0", help="Area curves overlapping (depends on tool diameter [0, 0.9])")
3006 add_argument("--area-tool-overlap", type=float, default="-10", help="Radius for preparing curves using inkscape")
3007 add_argument("--unit", default="G21 (All units in mm)", help="Units")
3008 add_argument("--active-tab", type=self.arg_method('tab'), default=self.tab_help, help="Defines which tab is active")
3009
3010 add_argument("--area-fill-angle", type=float, default="0", help="Fill area with lines heading this angle")
3011 add_argument("--area-fill-shift", type=float, default="0", help="Shift the lines by tool d shift")
3012 add_argument("--area-fill-method", default="zig-zag", help="Filling method either zig-zag or spiral")
3013
3014 add_argument("--area-find-artefacts-diameter", type=float, default="1", help="Artefacts seeking radius")
3015 add_argument("--area-find-artefacts-action", default="mark with an arrow", help="Artefacts action type")
3016
3017 add_argument("--auto_select_paths", type=inkex.Boolean, default=True, help="Select all paths if nothing is selected.")
3018
3019 add_argument("--loft-distances", default="10", help="Distances between paths.")
3020 add_argument("--loft-direction", default="crosswise", help="Direction of loft's interpolation.")
3021 add_argument("--loft-interpolation-degree", type=float, default="2", help="Which interpolation use to loft the paths smooth interpolation or staright.")
3022
3023 add_argument("--min-arc-radius", type=float, default=".1", help="All arc having radius less than minimum will be considered as straight line")
3024
3025 add_argument("--engraving-sharp-angle-tollerance", type=float, default="150", help="All angles thar are less than engraving-sharp-angle-tollerance will be thought sharp")
3026 add_argument("--engraving-max-dist", type=float, default="10", help="Distance from original path where engraving is not needed (usually it's cutting tool diameter)")
3027 add_argument("--engraving-newton-iterations", type=int, default="4", help="Number of sample points used to calculate distance")
3028 add_argument("--engraving-draw-calculation-paths", type=inkex.Boolean, default=False, help="Draw additional graphics to debug engraving path")
3029 add_argument("--engraving-cutter-shape-function", default="w", help="Cutter shape function z(w). Ex. cone: w. ")
3030
3031 add_argument("--lathe-width", type=float, default=10., help="Lathe width")
3032 add_argument("--lathe-fine-cut-width", type=float, default=1., help="Fine cut width")
3033 add_argument("--lathe-fine-cut-count", type=int, default=1., help="Fine cut count")
3034 add_argument("--lathe-create-fine-cut-using", default="Move path", help="Create fine cut using")
3035 add_argument("--lathe-x-axis-remap", default="X", help="Lathe X axis remap")
3036 add_argument("--lathe-z-axis-remap", default="Z", help="Lathe Z axis remap")
3037
3038 add_argument("--lathe-rectangular-cutter-width", type=float, default="4", help="Rectangular cutter width")
3039
3040 add_argument("--create-log", type=inkex.Boolean, dest="log_create_log", default=False, help="Create log files")
3041 add_argument("--log-filename", default='', help="Create log files")
3042
3043 add_argument("--orientation-points-count", default="2", help="Orientation points count")
3044 add_argument("--tools-library-type", default='cylinder cutter', help="Create tools definition")
3045
3046 add_argument("--dxfpoints-action", default='replace', help="dxfpoint sign toggle")
3047
3048 add_argument("--help-language", default=' help="Open help page in webbrowser.")
3049
3050 add_argument("--offset-radius", type=float, default=10., help="Offset radius")
3051 add_argument("--offset-step", type=float, default=10., help="Offset step")
3052 add_argument("--offset-draw-clippend-path", type=inkex.Boolean, default=False, help="Draw clipped path")
3053 add_argument("--offset-just-get-distance", type=inkex.Boolean, default=False, help="Don't do offset just get distance")
3054
3055 add_argument("--postprocessor", default='', help="Postprocessor command.")
3056 add_argument("--postprocessor-custom", default='', help="Postprocessor custom command.")
3057
3058 add_argument("--graffiti-max-seg-length", type=float, default=1., help="Graffiti maximum segment length.")
3059 add_argument("--graffiti-min-radius", type=float, default=10., help="Graffiti minimal connector's radius.")
3060 add_argument("--graffiti-start-pos", default="(0;0)", help="Graffiti Start position (x;y).")
3061 add_argument("--graffiti-create-linearization-preview", type=inkex.Boolean, default=True, help="Graffiti create linearization preview.")
3062 add_argument("--graffiti-create-preview", type=inkex.Boolean, default=True, help="Graffiti create preview.")
3063 add_argument("--graffiti-preview-size", type=int, default=800, help="Graffiti preview's size.")
3064 add_argument("--graffiti-preview-emmit", type=int, default=800, help="Preview's paint emmit (pts/s).")
3065
3066 add_argument("--in-out-path", type=inkex.Boolean, default=True, help="Create in-out paths")
3067 add_argument("--in-out-path-do-not-add-reference-point", type=inkex.Boolean, default=False, help="Just add reference in-out point")
3068 add_argument("--in-out-path-point-max-dist", type=float, default=10., help="In-out path max distance to reference point")
3069 add_argument("--in-out-path-type", default="Round", help="In-out path type")
3070 add_argument("--in-out-path-len", type=float, default=10., help="In-out path length")
3071 add_argument("--in-out-path-replace-original-path", type=inkex.Boolean, default=False, help="Replace original path")
3072 add_argument("--in-out-path-radius", type=float, default=10., help="In-out path radius for round path")
3073
3074 add_argument("--plasma-prepare-corners", type=inkex.Boolean, default=True, help="Prepare corners")
3075 add_argument("--plasma-prepare-corners-distance", type=float, default=10., help="Stepout distance for corners")
3076 add_argument("--plasma-prepare-corners-tolerance", type=float, default=10., help="Maximum angle for corner (0-180 deg)")
3077
2982def add_arguments(self, pars): …)
3078def init(self):
3079 super(Gcodetools, self).init()
3080 self.default_tool = {
3081"name": "Default tool",
3082"id": "default tool",
3083"diameter": 10.,
3084"shape": "10",
3085"penetration angle": 90.,
3086"penetration feed": 100.,
3087"depth step": 1.,
3088"feed": 400.,
3089"in trajectotry": "",
3090"out trajectotry": "",
3091"gcode before path": "",
3092"gcode after path": "",
3093"sog": "",
3094"spinlde rpm": "",
3095"CW or CCW": "",
3096"tool change gcode": " ",
3097"4th axis meaning": " ",
3098"4th axis scale": 1.,
3099"4th axis offset": 0.,
3100"passing feed": "800",
3101"fine feed": "800",
3102 }
3103 self.tools_field_order = [
3104'name',
3105'id',
3106'diameter',
3107'feed',
3108'shape',
3109'penetration angle',
3110'penetration feed',
3111"passing feed",
3112'depth step',
3113"in trajectotry",
3114"out trajectotry",
3115"gcode before path",
3116"gcode after path",
3117"sog",
3118"spinlde rpm",
3119"CW or CCW",
3120"tool change gcode",
3121 ]
3122
3078def init(self): …)
3123def parse_curve(self, p, layer, w=None, f=None):
3124 c = []
3125 if len(p) == 0:
3126 return []
3127 p = self.transform_csp(p, layer)
3128
3129# Sort to reduce Rapid distance
3130 k = list(range(1, len(p)))
3131 keys =
3132 while len(k) > 0:
3133 end = p[keys[-1]][-1]
3134 dist = None
3135 for i in range(len(k)):
3136 start = p[k[i]]
3137 dist = max((-((end - start) 2 + (end - start) 2), i), dist)
3138 keys += [k[dist]]
3139 del k[dist]
3140 for k in keys:
3141 subpath = p[k]
3142 c += , 'move', 0, 0]]
3143 for i in range(1, len(subpath)):
3144 sp1 = for j in range(3)]
3145 sp2 = for j in range(3)]
3146 c += biarc(sp1, sp2, 0, 0) if w is None elsebiarc(sp1, sp2, -f(w[k][i - 1]), -f(w[k][i]))
3147 c += , 'end', 0, 0]]
3148 return c
3149
3123def parse_curve(self, p, layer, w=None, f=None): …)
3150
3153
3154def draw_csp(self, csp, layer=None, group=None, fill='none', stroke='#178ade', width=0.354, style=None):
3155 if layer is not None:
3156 csp = self.transform_csp(csp, layer, reverse=True)
3157 if group is None and layer is None:
3158 group = self.documentdocument.getroot()
3159 elif group is None and layer is not None:
3160 group = layer
3161 csp = self.apply_transforms(group, csp, reverse=True)
3162 if style is not None:
3163 returndraw_csp(csp, group=group, style=style)
3164 else:
3165 returndraw_csp(csp, group=group, fill=fill, stroke=stroke, width=width)
3166
3154def draw_csp(self, csp, layer=None, group=None, fill='none', stroke='#178ade', width=0.354, style=None): …)
3167def draw_curve(self, curve, layer, group=None, style=MARKER_STYLE["biarc_style"]):
3168 self.set_markers()
3169
3170 for i in [0, 1]:
3171 sid = 'biarc{}_r'.format(i)
3172 style[sid] = style['biarc{}'.format(i)].copy()
3173 style[sid]["marker-start"] = "url(#DrawCurveMarker_r)"
3174 del style[sid]["marker-end"]
3175
3176 if group is None:
3177 group = self.layers[min(1, len(self.layers) - 1)].add(Group(gcodetools="Preview group"))
3178 if not hasattr(self, "preview_groups"):
3179 self.preview_groups = {layer: group}
3180 elif layer not in self.preview_groups:
3181 self.preview_groups[layer] = group
3182 group = self.preview_groups[layer]
3183
3184 s = ''
3185 arcn = 0
3186
3187 transform = self.get_transforms(group)
3188 if transform:
3189 transform = self.reverse_transform(transform)
3190 transform = str(Transform(transform))
3191
3192 a = [0., 0.]
3193 b = [1., 0.]
3194 c = [0., 1.]
3195 k = (b - a) (c - a) - (c - a) (b - a)
3196 a = self.transform(a, layer, True)
3197 b = self.transform(b, layer, True)
3198 c = self.transform(c, layer, True)
3199 if ((b - a) (c - a) - (c - a) (b - a)) k > 0:
3200 reverse_angle = 1
3201 else:
3202 reverse_angle = -1
3203 for sk in curve:
3204 si = sk[:]
3205 si = self.transform(si, layer, True)
3206 si = self.transform(si, layer, True) if type(si) == type([]) and len(si) == 2 else si
3207
3208 if s != '':
3209 if s == 'line':
3210 elem = group.add(PathElement(gcodetools="Preview"))
3211 elem.transform = transform
3212 elem.style = style['line']
3213 elem.path = 'M {},{} L {},{}'.format(s, s, si, si)
3214 elif s == 'arc':
3215 arcn += 1
3216 sp = s
3217 c = s
3218 s = s reverse_angle
3219
3220 a = ((P(si) - P(c)).angle() - (P(s) - P(c)).angle()) % TAU # s
3221 if s a < 0:
3222 if a > 0:
3223 a = a - TAU
3224 else:
3225 a = TAU + a
3226 r = math.sqrt((sp - c) 2 + (sp - c) 2)
3227 a_st = (math.atan2(sp - c, - (sp - c)) - math.pi / 2) % (math.pi 2)
3228 if a > 0:
3229 a_end = a_st + a
3230 st = style['biarc{}'.format(arcn % 2)]
3231 else:
3232 a_end = a_st 1
3233 a_st = a_st + a
3234 st = style['biarc{}_r'.format(arcn % 2)]
3235
3236 elem = group.add(PathElement.arc(c, r, start=a_st, end=a_end,
3237 open=True, gcodetools="Preview"))
3238 elem.transform = transform
3239 elem.style = st
3240
3241 s = si
3242
3167def draw_curve(self, curve, layer, group=None, style=MARKER_STYLE["biarc_style"]): …)
3243def check_dir(self):
3244print_("Checking directory: '{}'".format(self.options.directory))
3245 if os.path.isdir(self.options.directory):
3246 if os.path.isfile(os.path.join(self.options.directory, 'header')):
3247 with open(os.path.join(self.options.directory, 'header')) as f:
3248 self.header = f.read()
3249 else:
3250 self.header = defaults['header']
3251 if os.path.isfile(os.path.join(self.options.directory, 'footer')):
3252 with open(os.path.join(self.options.directory, 'footer')) as f:
3253 self.footer = f.read()
3254 else:
3255 self.footer = defaults['footer']
3256 self.header += self.options.unit + "\n"
3257 else:
3258 self.error("Directory does not exist! Please specify existing directory at Preferences tab!", "error")
3259 return False
3260
3261 if self.options.add_numeric_suffix_to_filename:
3262 dir_list = os.listdir(self.options.directory)
3263 if"."in self.options.file:
3264 r = re.match(r"^(.)(..)$", self.options.file)
3265 ext = r.group(2)
3266 name = r.group(1)
3267 else:
3268 ext = ""
3269 name = self.options.file
3270 max_n = 0
3271 for s in dir_list:
3272 r = re.match(r"^{}_0(\d+){}$".format(re.escape(name), re.escape(ext)), s)
3273 if r:
3274 max_n = max(max_n, int(r.group(1)))
3275 filename = name + "_" + ("0" (4 - len(str(max_n + 1))) + str(max_n + 1)) + ext
3276 self.options.file = filename
3277
3278 try:
3279 with open(os.path.join(self.options.directory, self.options.file), "w") as f:
3280 pass
3281 except:
3282 self.error("Can not write to specified file!\n{}".format(os.path.join(self.options.directory, self.options.file)), "error")
3283 return False
3284 return True
3285
3243def check_dir(self): …)
3286
3294def generate_gcode(self, curve, layer, depth):
3295 Zauto_scale = self.Zauto_scale[layer]
3296 tool = self.tools[layer]
3297 g = ""
3298
3299 def c(c):
3300 c = [c[i] if i < len(c) else None for i in range(6)]
3301 if c == 0:
3302 c = None
3303 s = [" X", " Y", " Z", " I", " J", " K"]
3304 s1 = ["", "", "", "", "", ""]
3305 m = [1, 1, self.options.Zscale Zauto_scale, 1, 1, self.options.Zscale Zauto_scale]
3306 a = [0, 0, self.options.Zoffset, 0, 0, 0]
3307 r = ''
3308 for i in range(6):
3309 if c[i] is not None:
3310 r += s[i] + ("{:f}".format(c[i] m[i] + a[i])) + s1[i]
3311 return r
3312
3313 def calculate_angle(a, current_a):
3314 return min(
3315 [abs(a - current_a % TAU + TAU), a + current_a - current_a % TAU + TAU],
3316 [abs(a - current_a % TAU - TAU), a + current_a - current_a % TAU - TAU],
3317 [abs(a - current_a % TAU), a + current_a - current_a % TAU])
3318
3319 if len(curve) == 0:
3320 return""
3321
3322 try:
3323 self.last_used_toolis None
3324 except:
3325 self.last_used_tool = None
3326print_("working on curve")
3327print_(curve)
3328
3329 if tool != self.last_used_tool:
3330 g += ("(Change tool to {})\n".format(re.sub("\"'\(\)\\", " ", tool["name"]))) + tool["tool change gcode"] + "\n"
3331
3332 lg = 'G00'
3333 zs = self.options.Zsafe
3334 f = " F{:f}".format(tool['feed'])
3335 current_a = 0
3336 go_to_safe_distance = "G00" + c([None, None, zs]) + "\n"
3337 penetration_feed = " F{}".format(tool['penetration feed'])
3338 for i in range(1, len(curve)):
3339# Creating Gcode for curve between s=curve[i-1] and si=curve[i] start at s end at s=si
3340 s = curve[i - 1]
3341 si = curve[i]
3342 feed = f if lg not in ['G01', 'G02', 'G03'] else''
3343 if s == 'move':
3344 g += go_to_safe_distance + "G00" + c(si) + "\n" + tool['gcode before path'] + "\n"
3345 lg = 'G00'
3346 elif s == 'end':
3347 g += go_to_safe_distance + tool['gcode after path'] + "\n"
3348 lg = 'G00'
3349 elif s == 'line':
3350 if tool['4th axis meaning'] == "tangent knife":
3351 a = atan2(si - s, si - s)
3352 a = calculate_angle(a, current_a)
3353 g += "G01 A{}\n".format(a tool['4th axis scale'] + tool['4th axis offset'])
3354 current_a = a
3355 if lg == "G00":
3356 g += "G01" + c([None, None, s + depth]) + penetration_feed + "(Penetrate)\n"
3357 g += "G01" + c(si + [s + depth]) + feed + "\n"
3358 lg = 'G01'
3359 elif s == 'arc':
3360 r = [(s - s), (s - s)]
3361 if tool['4th axis meaning'] == "tangent knife":
3362 if s < 0: # CW
3363 a1 = atan2(s - s, -s + s) + math.pi
3364 else: # CCW
3365 a1 = atan2(-s + s, s - s) + math.pi
3366 a = calculate_angle(a1, current_a)
3367 g += "G01 A{}\n".format(a tool['4th axis scale'] + tool['4th axis offset'])
3368 current_a = a
3369 axis4 = " A{}".format((current_a + s) tool['4th axis scale'] + tool['4th axis offset'])
3370 current_a = current_a + s
3371 else:
3372 axis4 = ""
3373 if lg == "G00":
3374 g += "G01" + c([None, None, s + depth]) + penetration_feed + "(Penetrate)\n"
3375 if (r 2 + r 2) > self.options.min_arc_radius 2:
3376 r1 = (P(s) - P(s))
3377 r2 = (P(si) - P(s))
3378 if abs(r1.mag() - r2.mag()) < 0.001:
3379 g += ("G02"if s < 0 else"G03") + c(si + [s + depth, (s - s), (s - s)]) + feed + axis4 + "\n"
3380 else:
3381 r = (r1.mag() + r2.mag()) / 2
3382 g += ("G02"if s < 0 else"G03") + c(si + [s + depth]) + " R{:f}".format(r) + feed + axis4 + "\n"
3383 lg = 'G02'
3384 else:
3385 if tool['4th axis meaning'] == "tangent knife":
3386 a = atan2(si - s, si - s) + math.pi
3387 a = calculate_angle(a, current_a)
3388 g += "G01 A{}\n".format(a tool['4th axis scale'] + tool['4th axis offset'])
3389 current_a = a
3390 g += "G01" + c(si + [s + depth]) + feed + "\n"
3391 lg = 'G01'
3392 if si == 'end':
3393 g += go_to_safe_distance + tool['gcode after path'] + "\n"
3394 return g
3395
3294def generate_gcode(self, curve, layer, depth): …)
3396def get_transforms(self, g):
3397 root = self.documentdocument.getroot()
3398 trans = []
3399 while g != root:
3400 if'transform'in g.keys():
3401 t = g.get('transform')
3402 t = Transform(t).matrix
3403 trans = (Transform(t) Transform(trans)).matrix if trans != [] else t
3404
3405print_(trans)
3406 g = g.getparent()
3407 return trans
3408
3396def get_transforms(self, g): …)
3409def reverse_transform(self, transform):
3410 trans = numpy.array(transform + ([0, 0, 1],))
3411 if numpy.linalg.det(trans) != 0:
3412 trans = numpy.linalg.inv(trans).tolist()[:2]
3413 return trans
3414 else:
3415 return transform
3416
3409def reverse_transform(self, transform): …)
3417def apply_transforms(self, g, csp, reverse=False):
3418 trans = self.get_transforms(g)
3419 if trans:
3420 if not reverse:
3421# TODO: This was applyTransformToPath but was deprecated. Candidate for refactoring.
3422 for comp in csp:
3423 for ctl in comp:
3424 for pt in ctl:
3425 pt, pt = Transform(trans).apply_to_point(pt)
3426
3427 else:
3428# TODO: This was applyTransformToPath but was deprecated. Candidate for refactoring.
3429 for comp in csp:
3430 for ctl in comp:
3431 for pt in ctl:
3432 pt, pt = Transform(self.reverse_transform(trans)).apply_to_point(pt)
3433 return csp
3434
3417def apply_transforms(self, g, csp, reverse=False): …)
3435def transform_scalar(self, x, layer, reverse=False):
3436 return self.transform([x, 0], layer, reverse) - self.transform([0, 0], layer, reverse)
3437
3435def transform_scalar(self, x, layer, reverse=False): …)
3438def transform(self, source_point, layer, reverse=False):
3439 if layer not in self.transform_matrix:
3440 for i in range(self.layers.index(layer), -1, -1):
3441 if self.layers[i] in self.orientation_points:
3442 break
3443 if self.layers[i] not in self.orientation_points:
3444 self.error(f"Orientation points for '{layer.label}' layer have not been found! Please add orientation points using Orientation tab!", "error")
3445 elif self.layers[i] in self.transform_matrix:
3446 self.transform_matrix[layer] = self.transform_matrix[self.layers[i]]
3447 self.Zcoordinates[layer] = self.Zcoordinates[self.layers[i]]
3448 else:
3449 orientation_layer = self.layers[i]
3450 if len(self.orientation_points[orientation_layer]) > 1:
3451 self.error(f"There are more than one orientation point groups in '{orientation_layer.label}' layer")
3452 points = self.orientation_points[orientation_layer]
3453 if len(points) == 2:
3454 points += , [-(points - points) + points, points - points + points]]]
3455 if len(points) == 3:
3456print_("Layer '{orientation_layer.label}' Orientation points: ")
3457 for point in points:
3458print_(point)
3459# Zcoordinates definition taken from Orientatnion point 1 and 2
3460 self.Zcoordinates[layer] = [max(points, points), min(points, points)]
3461 matrix = numpy.array([
3462 [points, points, 1, 0, 0, 0, 0, 0, 0],
3463 [0, 0, 0, points, points, 1, 0, 0, 0],
3464 [0, 0, 0, 0, 0, 0, points, points, 1],
3465 [points, points, 1, 0, 0, 0, 0, 0, 0],
3466 [0, 0, 0, points, points, 1, 0, 0, 0],
3467 [0, 0, 0, 0, 0, 0, points, points, 1],
3468 [points, points, 1, 0, 0, 0, 0, 0, 0],
3469 [0, 0, 0, points, points, 1, 0, 0, 0],
3470 [0, 0, 0, 0, 0, 0, points, points, 1]
3471 ])
3472
3473 if numpy.linalg.det(matrix) != 0:
3474 m = numpy.linalg.solve(matrix,
3475 numpy.array(
3476 , [points], , [points], [points], , [points], [points], ]
3477 )
3478 ).tolist()
3479 self.transform_matrix[layer] = [[m[j 3 + i] for i in range(3)] for j in range(3)]
3480
3481 else:
3482 self.error("Orientation points are wrong! (if there are two orientation points they should not be the same. If there are three orientation points they should not be in a straight line.)", "error")
3483 else:
3484 self.error("Orientation points are wrong! (if there are two orientation points they should not be the same. If there are three orientation points they should not be in a straight line.)", "error")
3485
3486 self.transform_matrix_reverse[layer] = numpy.linalg.inv(self.transform_matrix[layer]).tolist()
3487print_(f"\n Layer '{layer.label}' transformation matrixes:")
3488print_(self.transform_matrix)
3489print_(self.transform_matrix_reverse)
3490
3491# Zautoscale is obsolete
3492 self.Zauto_scale[layer] = 1
3493print_("Z automatic scale = {} (computed according orientation points)".format(self.Zauto_scale[layer]))
3494
3495 x = source_point
3496 y = source_point
3497 if not reverse:
3498 t = self.transform_matrix[layer]
3499 else:
3500 t = self.transform_matrix_reverse[layer]
3501 return [t x + t y + t, t x + t y + t]
3502
3438def transform(self, source_point, layer, reverse=False): …)
3503def transform_csp(self, csp_, layer, reverse=False):
3504 csp = for j in range(len(csp_[i]))] for i in range(len(csp_))]
3505 for i inxrange(len(csp)):
3506 for j inxrange(len(csp[i])):
3507 for k inxrange(len(csp[i][j])):
3508 csp[i][j][k] = self.transform(csp[i][j][k], layer, reverse)
3509 return csp
3510
3503def transform_csp(self, csp_, layer, reverse=False): …)
3511def error(self, s, msg_type="warning"):
3512"""
3513 Errors handling function
3514 warnings are printed into log file and warning message is displayed but
3515 extension continues working,
3516 errors causes log and execution is halted
3517 """
3518 if msg_type == "warning":
3519print_(s)
3520 inkex.errormsg(s + "\n")
3521
3522 elif msg_type == "error":
3523print_(s)
3524 raiseinkex.AbortExtension(s)
3525
3526 else:
3527print_("Unknown message type: {}".format(msg_type))
3528print_(s)
3529 raiseinkex.AbortExtension(s)
3530
3511def error(self, s, msg_type="warning"): …)
3531
3534def set_markers(self):
3535"""Make sure all markers are available"""
3536 def ensure_marker(elem_id, x=-4, polA='', polB='-', fill='#000044'):
3537 if self.svg.getElementById(elem_id) is None:
3538 marker = self.svg.defs.add(Marker(
3539 id=elem_id, orient="auto", refX=str(x), refY="-1.687441",
3540 style="overflow:visible"))
3541 path = marker.add(PathElement(
3542 d="m {0}4.588864,-1.687441 0.0,0.0 L {0}9.177728,0.0 "\
3543"c {1}0.73311,-0.996261 {1}0.728882,-2.359329 0.0,-3.374882"\
3544 .format(polA, polB)))
3545 path.style = "fill:{};fill-rule:evenodd;stroke:none;".format(fill)
3546
3547 ensure_marker("CheckToolsAndOPMarker")
3548 ensure_marker("DrawCurveMarker")
3549 ensure_marker("DrawCurveMarker_r", x=4, polA='-', polB='')
3550 ensure_marker("InOutPathMarker", fill='#0072a7')
3551
3534def set_markers(self): …)
3552def get_info(self):
3553"""Get Gcodetools info from the svg"""
3554 self.selected_paths = {}
3555 self.paths = {}
3556 self.tools = {}
3557 self.orientation_points = {}
3558 self.graffiti_reference_points = {}
3559 self.layers = [self.documentdocument.getroot()]
3560 self.Zcoordinates = {}
3561 self.transform_matrix = {}
3562 self.transform_matrix_reverse = {}
3563 self.Zauto_scale = {}
3564 self.in_out_reference_points = []
3565 self.my3Dlayer = None
3566
3567 def recursive_search(g, layer, selected=False):
3568 items = g.getchildren()
3569 items.reverse()
3570 for i in items:
3571 if selected:
3572 self.svg.selected[i.get("id")] = i
3573 if isinstance(i, Layer):
3574 if i.label == '3D':
3575 self.my3Dlayer = i
3576 else:
3577 self.layers += [i]
3578 recursive_search(i, i)
3579
3580 elif i.get('gcodetools') == "Gcodetools orientation group":
3581 points = self.get_orientation_points(i)
3582 if points is not None:
3583 self.orientation_points[layer] = self.orientation_points[layer] + [points[:]] if layer in self.orientation_pointselse [points[:]]
3584print_(f"Found orientation points in '{layer.label}' layer: {points}")
3585 else:
3586 self.error(f"Warning! Found bad orientation points in '{layer.label}' layer. Resulting Gcode could be corrupt!")
3587
3588# Need to recognise old files ver 1.6.04 and earlier
3589 elif i.get("gcodetools") == "Gcodetools tool definition"or i.get("gcodetools") == "Gcodetools tool definition":
3590 tool = self.get_tool(i)
3591 self.tools[layer] = self.tools[layer] + [tool.copy()] if layer in self.toolselse [tool.copy()]
3592print_(f"Found tool in '{layer.label}' layer: {tool}")
3593
3594 elif i.get("gcodetools") == "Gcodetools graffiti reference point":
3595 point = self.get_graffiti_reference_points(i)
3596 if point:
3597 self.graffiti_reference_points[layer] = self.graffiti_reference_points[layer] + [point[:]] if layer in self.graffiti_reference_pointselse [point]
3598 else:
3599 self.error(f"Warning! Found bad graffiti reference point in '{layer.label}' layer. Resulting Gcode could be corrupt!")
3600
3601 elif isinstance(i, inkex.PathElement):
3602 if"gcodetools"not in i.keys():
3603 self.paths[layer] = self.paths[layer] + [i] if layer in self.pathselse [i]
3604 if i.get("id") in self.svg.selected.ids:
3605 self.selected_paths[layer] = self.selected_paths[layer] + [i] if layer in self.selected_pathselse [i]
3606
3607 elif i.get("gcodetools") == "In-out reference point group":
3608 items_ = i.getchildren()
3609 items_.reverse()
3610 for j in items_:
3611 if j.get("gcodetools") == "In-out reference point":
3612 self.in_out_reference_points.append(self.apply_transforms(j, j.path.to_superpath()))
3613
3614 elif isinstance(i, inkex.Group):
3615 recursive_search(i, layer, (i.get("id") in self.svg.selected))
3616
3617 elif i.get("id") in self.svg.selected:
3618# xgettext:no-pango-format
3619 self.error("This extension works with Paths and Dynamic Offsets and groups of them only! "
3620"All other objects will be ignored!\n"
3621"Solution 1: press Path->Object to path or Shift+Ctrl+C.\n"
3622"Solution 2: Path->Dynamic offset or Ctrl+J.\n"
3623"Solution 3: export all contours to PostScript level 2 (File->Save As->.ps) and File->Import this file.")
3624
3625 recursive_search(self.documentdocument.getroot(), self.documentdocument.getroot())
3626
3627 if len(self.layers) == 1:
3628 self.error("Document has no layers! Add at least one layer using layers panel (Ctrl+Shift+L)", "error")
3629 root = self.documentdocument.getroot()
3630
3631 if root in self.selected_pathsor root in self.paths:
3632 self.error("Warning! There are some paths in the root of the document, but not in any layer! Using bottom-most layer for them.")
3633
3634 if root in self.selected_paths:
3635 if self.layers[-1] in self.selected_paths:
3636 self.selected_paths[self.layers[-1]] += self.selected_paths[root][:]
3637 else:
3638 self.selected_paths[self.layers[-1]] = self.selected_paths[root][:]
3639 del self.selected_paths[root]
3640
3641 if root in self.paths:
3642 if self.layers[-1] in self.paths:
3643 self.paths[self.layers[-1]] += self.paths[root][:]
3644 else:
3645 self.paths[self.layers[-1]] = self.paths[root][:]
3646 del self.paths[root]
3647
3552def get_info(self): …)
3648def get_orientation_points(self, g):
3649 items = g.getchildren()
3650 items.reverse()
3651 p2 = []
3652 p3 = []
3653 p = None
3654 for i in items:
3655 if isinstance(i, inkex.Group):
3656 if i.get("gcodetools") == "Gcodetools orientation point (2 points)":
3657 p2 += [i]
3658 if i.get("gcodetools") == "Gcodetools orientation point (3 points)":
3659 p3 += [i]
3660 if len(p2) == 2:
3661 p = p2
3662 elif len(p3) == 3:
3663 p = p3
3664 if p is None:
3665 return None
3666 points = []
3667 for i in p:
3668 point =
3669 for node in i:
3670 if node.get('gcodetools') == "Gcodetools orientation point arrow":
3671 csp = node.path.transform(node.composed_transform()).to_superpath()
3672 point = csp
3673 if node.get('gcodetools') == "Gcodetools orientation point text":
3674 inkex.errormsg(node.get_text())
3675 r = re.match(r'(?i)\s\(\s(-?\s\d(?:,|.)\d)\s;\s(-?\s\d(?:,|.)\d)\s;\s(-?\s\d(?:,|.)\d)\s\)\s', node.get_text())
3676 point = [float(r.group(1)), float(r.group(2)), float(r.group(3))]
3677 if point != [] and point != []:
3678 points += [point]
3679 if len(points) == len(p2) == 2 or len(points) == len(p3) == 3:
3680 return points
3681 else:
3682 return None
3683
3648def get_orientation_points(self, g): …)
3684def get_graffiti_reference_points(self, g):
3685 point = [[], '']
3686 for node in g:
3687 if node.get('gcodetools') == "Gcodetools graffiti reference point arrow":
3688 point = self.apply_transforms(node, node.path.to_superpath())
3689 if node.get('gcodetools') == "Gcodetools graffiti reference point text":
3690 point = node.get_text()
3691 if point != [] and point != '':
3692 return point
3693 else:
3694 return []
3695
3684def get_graffiti_reference_points(self, g): …)
3696def get_tool(self, g):
3697 tool = self.default_tool.copy()
3698 tool["self_group"] = g
3699 for i in g:
3700# Get parameters
3701 if i.get("gcodetools") == "Gcodetools tool background":
3702 tool["style"] = dict(i.style)
3703 elif i.get("gcodetools") == "Gcodetools tool parameter":
3704 key = None
3705 value = None
3706 for j in i:
3707# need to recognise old tools from ver 1.6.04
3708 if j.get("gcodetools") == "Gcodetools tool definition field name"or j.get("gcodetools") == "Gcodetools tool defention field name":
3709 key = j.get_text()
3710 if j.get("gcodetools") == "Gcodetools tool definition field value"or j.get("gcodetools") == "Gcodetools tool defention field value":
3711 value = j.get_text()
3712 if value == "(None)":
3713 value = ""
3714 if value is None or key is None:
3715 continue
3716 if key in self.default_tool.keys():
3717 try:
3718 tool[key] = type(self.default_tool[key])(value)
3719 except:
3720 tool[key] = self.default_tool[key]
3721 self.error("Warning! Tool's and default tool's parameter's ({}) types are not the same ( type('{}') != type('{}') ).".format(key, value, self.default_tool[key]))
3722 else:
3723 tool[key] = value
3724 self.error("Warning! Tool has parameter that default tool has not ( '{}': '{}' ).".format(key, value))
3725 return tool
3726
3696def get_tool(self, g): …)
3727def set_tool(self, layer):
3728 for i in range(self.layers.index(layer), -1, -1):
3729 if self.layers[i] in self.tools:
3730 break
3731 if self.layers[i] in self.tools:
3732 if self.layers[i] != layer:
3733 self.tools[layer] = self.tools[self.layers[i]]
3734 if len(self.tools[layer]) > 1:
3735 label = self.layers[i].label
3736 self.error(f"Layer '{label}' contains more than one tool!")
3737 return self.tools[layer]
3738 else:
3739 self.error(f"Can not find tool for '{layer.label}' layer! Please add one with Tools library tab!", "error")
3740
3727def set_tool(self, layer): …)
3741
3746def tab_path_to_gcode(self):
3747 self.get_info_plus()
3748 def get_boundaries(points):
3749 minx = None
3750 miny = None
3751 maxx = None
3752 maxy = None
3753 out =
3754 for p in points:
3755 if minx == p:
3756 out += [p]
3757 if minx is None or p < minx:
3758 minx = p
3759 out = [p]
3760
3761 if miny == p:
3762 out += [p]
3763 if miny is None or p < miny:
3764 miny = p
3765 out = [p]
3766
3767 if maxx == p:
3768 out += [p]
3769 if maxx is None or p > maxx:
3770 maxx = p
3771 out = [p]
3772
3773 if maxy == p:
3774 out += [p]
3775 if maxy is None or p > maxy:
3776 maxy = p
3777 out = [p]
3778 return out
3779
3780 def remove_duplicates(points):
3781 i = 0
3782 out = []
3783 for p in points:
3784 for j inxrange(i, len(points)):
3785 if p == points[j]:
3786 points[j] = [None, None]
3787 if p != [None, None]:
3788 out += [p]
3789 i += 1
3790 return out
3791
3792 def get_way_len(points):
3793 l = 0
3794 for i inxrange(1, len(points)):
3795 l += math.sqrt((points[i] - points[i - 1]) 2 + (points[i] - points[i - 1]) 2)
3796 return l
3797
3798 def sort_dxfpoints(points):
3799 points = remove_duplicates(points)
3800 ways = [
3801# l=0, d=1, r=2, u=3
3802 [3, 0], # ul
3803 [3, 2], # ur
3804 [1, 0], # dl
3805 [1, 2], # dr
3806 [0, 3], # lu
3807 [0, 1], # ld
3808 [2, 3], # ru
3809 [2, 1], # rd
3810 ]
3811 minimal_way = []
3812 minimal_len = None
3813 for w in ways:
3814 tpoints = points[:]
3815 cw = []
3816 for j inxrange(0, len(points)):
3817 p = get_boundaries(get_boundaries(tpoints)[w])[w]
3818 tpoints.remove(p)
3819 cw += p
3820 curlen = get_way_len(cw)
3821 if minimal_len is None or curlen < minimal_len:
3822 minimal_len = curlen
3823 minimal_way = cw
3824
3825 return minimal_way
3826
3827 def sort_lines(lines):
3828 if len(lines) == 0:
3829 return []
3830 lines = [[key] + lines[key] for key in range(len(lines))]
3831 keys =
3832 end_point = lines[3:]
3833print_("!!!", lines, "\n", end_point)
3834 del lines
3835 while len(lines) > 0:
3836 dist = [[point_to_point_d2(end_point, lines[i][1:3]), i] for i in range(len(lines))]
3837 i = min(dist)
3838 keys.append(lines[i])
3839 end_point = lines[i][3:]
3840 del lines[i]
3841 return keys
3842
3843 def sort_curves(curves):
3844 lines = []
3845 for curve in curves:
3846 lines += [curve + curve[-1][-1]]
3847 return sort_lines(lines)
3848
3849 def print_dxfpoints(points):
3850 gcode = ""
3851 for point in points:
3852 gcode += "(drilling dxfpoint)\nG00 Z{:f}\nG00 X{:f} Y{:f}\nG01 Z{:f} F{:f}\nG04 P{:f}\nG00 Z{:f}\n".format(self.options.Zsafe, point, point, self.Zcoordinates[layer], self.tools[layer]["penetration feed"], 0.2, self.options.Zsafe)
3853 return gcode
3854
3855 def get_path_properties(node):
3856 res = {}
3857 done = False
3858 while not done and node != self.svg:
3859 for i in node.getchildren():
3860 if isinstance(i, inkex.Desc):
3861 res["Description"] = i.text
3862 elif isinstance(i, inkex.Title):
3863 res["Title"] = i.text
3864 done = True
3865 node = node.getparent()
3866 return res
3867
3868 if self.selected_paths == {} and self.options.auto_select_paths:
3869 paths = self.paths
3870 self.error("No paths are selected! Trying to work on all available paths.")
3871 else:
3872 paths = self.selected_paths
3873 self.check_dir()
3874 gcode = ""
3875
3876 parent = list(self.selected_paths) if self.selected_pathselse self.layers
3877 biarc_group = parent.add(Group())
3878print_(("self.layers=", self.layers))
3879print_(("paths=", paths))
3880 colors = {}
3881 for layer in self.layers:
3882 if layer in paths:
3883print_(("layer", layer))
3884# transform simple path to get all var about orientation
3885 self.transform_csp(, ]], layer)
3886
3887 self.set_tool(layer)
3888 curves = []
3889 dxfpoints = []
3890
3891 try:
3892 depth_func = eval('lambda c,d,s: ' + self.options.path_to_gcode_depth_function.strip('"'))
3893 except:
3894 self.error("Bad depth function! Enter correct function at Path to Gcode tab!")
3895
3896 for path in paths[layer]:
3897 if"d"not in path.keys():
3898 self.error("Warning: One or more paths do not have 'd' parameter, try to Ungroup (Ctrl+Shift+G) and Object to Path (Ctrl+Shift+C)!")
3899 continue
3900 csp = path.path.to_superpath()
3901 csp = self.apply_transforms(path, csp)
3902 id_ = path.get("id")
3903
3904 def set_comment(match, path):
3905 if match.group(1) in path.keys():
3906 return path.get(match.group(1))
3907 else:
3908 return"None"
3909
3910 if self.options.comment_gcode != "":
3911 comment = re.sub("\[([A-Za-z_\-\:]+)\]", partial(set_comment, path=path), self.options.comment_gcode)
3912 comment = comment.replace(":newline:", "\n")
3913 comment = gcode_comment_str(comment)
3914 else:
3915 comment = ""
3916 if self.options.comment_gcode_from_properties:
3917 tags = get_path_properties(path)
3918 for tag in tags:
3919 comment += gcode_comment_str("{}: {}".format(tag, tags[tag]))
3920
3921 stroke = path.style('stroke')
3922 colors[id_] = inkex.Color(stroke if stroke != None else"#000").to_rgb()
3923 if path.get("dxfpoint") == "1":
3924 tmp_curve = self.transform_csp(csp, layer)
3925 x = tmp_curve
3926 y = tmp_curve
3927print_("got dxfpoint (scaled) at ({:f},{:f})".format(x, y))
3928 dxfpoints +=
3929 else:
3930
3931 zd = self.Zcoordinates[layer]
3932 zs = self.Zcoordinates[layer]
3933 c = 1. - float(sum(colors[id_])) / 255 / 3
3934 curves += [
3935 [
3936 [id_, depth_func(c, zd, zs), comment],
3937 [self.parse_curve([subpath], layer) for subpath in csp]
3938 ]
3939 ]
3940 dxfpoints = sort_dxfpoints(dxfpoints)
3941 gcode += print_dxfpoints(dxfpoints)
3942
3943 for curve in curves:
3944 for subcurve in curve:
3945 self.draw_curve(subcurve, layer)
3946
3947 if self.options.path_to_gcode_order == 'subpath by subpath':
3948 curves_ = []
3949 for curve in curves:
3950 curves_ += for subcurve in curve]
3951 curves = curves_
3952
3953 self.options.path_to_gcode_order = 'path by path'
3954
3955 if self.options.path_to_gcode_order == 'path by path':
3956 if self.options.path_to_gcode_sort_paths:
3957 keys = sort_curves([curve for curve in curves])
3958 else:
3959 keys = range(len(curves))
3960 for key in keys:
3961 d = curves[key]
3962 for step in range(0, int(math.ceil(abs((zs - d) / self.tools[layer]["depth step"])))):
3963 z = max(d, zs - abs(self.tools[layer]["depth step"] (step + 1)))
3964
3965 gcode += gcode_comment_str("\nStart cutting path id: {}".format(curves[key]))
3966 if curves[key] != "()":
3967 gcode += curves[key] # add comment
3968
3969 for curve in curves[key]:
3970 gcode += self.generate_gcode(curve, layer, z)
3971
3972 gcode += gcode_comment_str("End cutting path id: {}\n\n".format(curves[key]))
3973
3974 else: # pass by pass
3975 mind = min([curve for curve in curves])
3976 for step in range(0, 1 + int(math.ceil(abs((zs - mind) / self.tools[layer]["depth step"])))):
3977 z = zs - abs(self.tools[layer]["depth step"] step)
3978 curves_ = []
3979 for curve in curves:
3980 if curve < z:
3981 curves_.append(curve)
3982
3983 z = zs - abs(self.tools[layer]["depth step"] (step + 1))
3984 gcode += "\n(Pass at depth {})\n".format(z)
3985
3986 if self.options.path_to_gcode_sort_paths:
3987 keys = sort_curves([curve for curve in curves_])
3988 else:
3989 keys = range(len(curves_))
3990 for key in keys:
3991
3992 gcode += gcode_comment_str("Start cutting path id: {}".format(curves[key]))
3993 if curves[key] != "()":
3994 gcode += curves[key] # add comment
3995
3996 for subcurve in curves_[key]:
3997 gcode += self.generate_gcode(subcurve, layer, max(z, curves_[key]))
3998
3999 gcode += gcode_comment_str("End cutting path id: {}\n\n".format(curves[key]))
4000
4001 self.export_gcode(gcode)
4002
3746def tab_path_to_gcode(self): …)
4003
4008def tab_dxfpoints(self):
4009 self.get_info_plus()
4010 if self.selected_paths == {}:
4011 self.error("Nothing is selected. Please select something to convert to drill point (dxfpoint) or clear point sign.")
4012 for layer in self.layers:
4013 if layer in self.selected_paths:
4014 for path in self.selected_paths[layer]:
4015 if self.options.dxfpoints_action == 'replace':
4016
4017 path.set("dxfpoint", "1")
4018 r = re.match("^\s.\s(\S+)", path.get("d"))
4019 if r is not None:
4020print_(("got path=", r.group(1)))
4021 path.set("d", "m {} 2.9375,-6.343750000001 0.8125,1.90625 6.843748640396,-6.84374864039 0,0 0.6875,0.6875 -6.84375,6.84375 1.90625,0.812500000001 z".format(r.group(1)))
4022 path.set("style", MARKER_STYLE["dxf_points"])
4023
4024 if self.options.dxfpoints_action == 'save':
4025 path.set("dxfpoint", "1")
4026
4027 if self.options.dxfpoints_action == 'clear'and path.get("dxfpoint") == "1":
4028 path.set("dxfpoint", "0")
4029
4008def tab_dxfpoints(self): …)
4030
4035def tab_area_artefacts(self):
4036 self.get_info_plus()
4037 if self.selected_paths == {} and self.options.auto_select_paths:
4038 paths = self.paths
4039 self.error("No paths are selected! Trying to work on all available paths.")
4040 else:
4041 paths = self.selected_paths
4042 for layer in paths:
4043 for path in paths[layer]:
4044 parent = path.getparent()
4045 if"d"not in path.keys():
4046 self.error("Warning: One or more paths do not have 'd' parameter, try to Ungroup (Ctrl+Shift+G) and Object to Path (Ctrl+Shift+C)!")
4047 continue
4048 csp = path.path.to_superpath()
4049 remove = []
4050 for i in range(len(csp)):
4051 subpath =
4052 subpath = self.apply_transforms(path, [subpath])
4053 bounds = csp_simple_bound([subpath])
4054 if (bounds - bounds) 2 + (bounds - bounds) 2 < self.options.area_find_artefacts_diameter 2:
4055 if self.options.area_find_artefacts_action == "mark with an arrow":
4056 arrow = Path('m {},{} 2.9375,-6.343750000001 0.8125,1.90625 6.843748640396,-6.84374864039 0,0 0.6875,0.6875 -6.84375,6.84375 1.90625,0.812500000001 z'.format(subpath, subpath)).to_superpath()
4057 arrow = self.apply_transforms(path, arrow, True)
4058 node = parent.add(PathElement())
4059 node.path = CubicSuperPath(arrow)
4060 node.style = MARKER_STYLE["area artefact arrow"]
4061 node.set('gcodetools', 'area artefact arrow')
4062 elif self.options.area_find_artefacts_action == "mark with style":
4063 node = parent.add(PathElement())
4064 node.path = CubicSuperPath(csp[i])
4065 node.style = MARKER_STYLE["area artefact"]
4066 remove.append(i)
4067 elif self.options.area_find_artefacts_action == "delete":
4068 remove.append(i)
4069print_("Deleted artefact {}".format(subpath))
4070 remove.reverse()
4071 for i in remove:
4072 del csp[i]
4073 if len(csp) == 0:
4074 parent.remove(path)
4075 else:
4076 path.path = CubicSuperPath(csp)
4077
4078 return
4079
4035def tab_area_artefacts(self): …)
4080def tab_area(self):
4081"""Calculate area curves"""
4082 self.get_info_plus()
4083 if len(self.selected_paths) <= 0:
4084 self.error("This extension requires at least one selected path.")
4085 return
4086 for layer in self.layers:
4087 if layer in self.selected_paths:
4088 self.set_tool(layer)
4089 if self.tools[layer]['diameter'] <= 0:
4090 self.error(f"Tool diameter must be > 0 but tool's diameter on '{layer.label}' layer is not!", "error")
4091
4092 for path in self.selected_paths[layer]:
4093print_(("doing path", path.get("style"), path.get("d")))
4094
4095 area_group = path.getparent().add(Group())
4096
4097 csp = path.path.to_superpath()
4098print_(csp)
4099 if not csp:
4100print_("omitting non-path")
4101 self.error("Warning: omitting non-path")
4102 continue
4103
4104 if path.get('sodipodi:type') != "inkscape:offset":
4105print_("Path {} is not an offset. Preparation started.".format(path.get("id")))
4106# Path is not offset. Preparation will be needed.
4107# Finding top most point in path (min y value)
4108
4109 min_x, min_y, min_i, min_j, min_t = csp_true_bounds(csp)
4110
4111# Reverse path if needed.
4112 if min_y != float("-inf"):
4113# Move outline subpath to the beginning of csp
4114 subp = csp[min_i]
4115 del csp[min_i]
4116 j = min_j
4117# Split by the topmost point and join again
4118 if min_t in [0, 1]:
4119 if min_t == 0:
4120 j = j - 1
4121 subp[-1], subp = subp[-1], subp
4122 subp = ] + subp[j + 1:] + subp[:j] + ]
4123 else:
4124 sp1, sp2, sp3 = csp_split(subp[j - 1], subp[j], min_t)
4125 subp[-1], subp = subp[-1], subp
4126 subp = ] + [sp3] + subp[j + 1:] + subp[:j - 1] + [sp1] + ]
4127 csp = [subp] + csp
4128# reverse path if needed
4129 ifcsp_subpath_ccw(csp):
4130 for i in range(len(csp)):
4131 n = []
4132 for j in csp[i]:
4133 n = ] + n
4134 csp[i] = n[:]
4135
4136# What the absolute fudge is this doing? Closing paths? Ugh.
4137# Not sure but it most be at this level and not in the if statement, or it will not work with dynamic offsets
4138 d = str(CubicSuperPath(csp))
4139print_(("original d=", d))
4140 d = re.sub(r'(?i)(m[^mz]+)', r'\1 Z ', d)
4141 d = re.sub(r'(?i)\sz\sz\s', r' Z ', d)
4142 d = re.sub(r'(?i)\s([A-Za-z])\s', r' \1 ', d)
4143print_(("formatted d=", d))
4144 p0 = self.transform([0, 0], layer)
4145 p1 = self.transform([0, 1], layer)
4146 scale = (P(p0) - P(p1)).mag()
4147 if scale == 0:
4148 scale = 1.
4149 else:
4150 scale = 1. / scale
4151print_(scale)
4152 tool_d = self.tools[layer]['diameter'] scale
4153 r = self.options.area_inkscape_radius scale
4154 sign = 1 if r > 0 else -1
4155print_("Tool diameter = {}, r = {}".format(tool_d, r))
4156
4157# avoiding infinite loops
4158 if self.options.area_tool_overlap > 0.9:
4159 self.options.area_tool_overlap = .9
4160
4161 for i in range(self.options.max_area_curves):
4162 radius = - tool_d (i (1 - self.options.area_tool_overlap) + 0.5) sign
4163 if abs(radius) > abs(r):
4164 radius = -r
4165
4166 elem = area_group.add(PathElement(style=str(MARKER_STYLE["biarc_style_i"]['area'])))
4167 elem.set('sodipodi:type', 'inkscape:offset')
4168 elem.set('inkscape:radius', radius)
4169 elem.set('inkscape:original', d)
4170print_(("adding curve", area_group, d, str(MARKER_STYLE["biarc_style_i"]['area'])))
4171 if radius == -r:
4172 break
4173
4080def tab_area(self): …)
4174def tab_area_fill(self):
4175"""Fills area with lines"""
4176 self.get_info_plus()
4177# convert degrees into rad
4178 self.options.area_fill_angle = self.options.area_fill_angle math.pi / 180
4179 if len(self.selected_paths) <= 0:
4180 self.error("This extension requires at least one selected path.")
4181 return
4182 for layer in self.layers:
4183 if layer in self.selected_paths:
4184 self.set_tool(layer)
4185 if self.tools[layer]['diameter'] <= 0:
4186 self.error(f"Tool diameter must be > 0 but tool's diameter on '{layer.label}' layer is not!", "error")
4187 tool = self.tools[layer]
4188 for path in self.selected_paths[layer]:
4189 lines = []
4190print_(("doing path", path.get("style"), path.get("d")))
4191 area_group = path.getparent().add(Group())
4192 csp = path.path.to_superpath()
4193 if not csp:
4194print_("omitting non-path")
4195 self.error("Warning: omitting non-path")
4196 continue
4197 csp = self.apply_transforms(path, csp)
4198 csp = csp_close_all_subpaths(csp)
4199 csp = self.transform_csp(csp, layer)
4200
4201# rotate the path to get bounds in defined direction.
4202 a = - self.options.area_fill_angle
4203 rotated_path = [[[[point math.cos(a) - point math.sin(a), point math.sin(a) + point math.cos(a)] for point in sp] for sp in subpath] for subpath in csp]
4204 bounds = csp_true_bounds(rotated_path)
4205
4206# Draw the lines
4207# Get path's bounds
4208 b = [0.0, 0.0, 0.0, 0.0] # [minx,miny,maxx,maxy]
4209 for k in range(4):
4210 i = bounds[k]
4211 j = bounds[k]
4212 t = bounds[k]
4213
4214 b[k] = csp_at_t(rotated_path[i][j - 1], rotated_path[i][j], t)[k % 2]
4215
4216# Zig-zag
4217 r = tool['diameter'] (1 - self.options.area_tool_overlap)
4218 if r <= 0:
4219 self.error('Tools diameter must be greater than 0!', 'error')
4220 return
4221
4222 lines +=
4223
4224 if self.options.area_fill_method == 'zig-zag':
4225 i = b - self.options.area_fill_shift r
4226 top = True
4227 last_one = True
4228 while i < b or last_one:
4229 if i >= b:
4230 last_one = False
4231 if not lines[-1]:
4232 lines[-1] += ]
4233
4234 if top:
4235 lines[-1] += , [i + r, b]]
4236
4237 else:
4238 lines[-1] += , [i + r, b]]
4239
4240 top = not top
4241 i += r
4242 else:
4243
4244 w = b - b + self.options.area_fill_shift r
4245 h = b - b + self.options.area_fill_shift r
4246 x = b - self.options.area_fill_shift r
4247 y = b - self.options.area_fill_shift r
4248 lines[-1] +=
4249 stage = 0
4250 start = True
4251 while w > 0 and h > 0:
4252 stage = (stage + 1) % 4
4253 if stage == 0:
4254 y -= h
4255 h -= r
4256 elif stage == 1:
4257 x += w
4258 if not start:
4259 w -= r
4260 start = False
4261 elif stage == 2:
4262 y += h
4263 h -= r
4264 elif stage == 3:
4265 x -= w
4266 w -= r
4267
4268 lines[-1] +=
4269
4270 stage = (stage + 1) % 4
4271 if w <= 0 and h > 0:
4272 y = y - h if stage == 0 else y + h
4273 if h <= 0 and w > 0:
4274 x = x - w if stage == 3 else x + w
4275 lines[-1] +=
4276# Rotate created paths back
4277 a = self.options.area_fill_angle
4278 lines = [[[point math.cos(a) - point math.sin(a), point math.sin(a) + point math.cos(a)] for point in subpath] for subpath in lines]
4279
4280# get the intersection points
4281
4282 splitted_line = ]
4283 intersections = {}
4284 for l1, l2, in zip(lines, lines[1:]):
4285 ints = []
4286
4287 if l1 == l2 and l1 == l2:
4288 continue
4289 for i in range(len(csp)):
4290 for j in range(1, len(csp[i])):
4291 sp1 = csp[i][j - 1]
4292 sp2 = csp[i][j]
4293 roots = csp_line_intersection(l1, l2, sp1, sp2)
4294 for t in roots:
4295 p = tuple(csp_at_t(sp1, sp2, t))
4296 if l1 == l2:
4297 t1 = (p - l1) / (l2 - l1)
4298 else:
4299 t1 = (p - l1) / (l2 - l1)
4300 if 0 <= t1 <= 1:
4301 ints +=
4302 if p in intersections:
4303 intersections[p] +=
4304 else:
4305 intersections[p] =
4306
4307 ints.sort()
4308 for i in ints:
4309 splitted_line[-1] += ]
4310 splitted_line += ]]
4311 splitted_line[-1] += [l2]
4312 i = 0
4313print_(splitted_line)
4314 while i < len(splitted_line):
4315# check if the middle point of the first lines segment is inside the path.
4316# and remove the subline if not.
4317 l1 = splitted_line[i]
4318 l2 = splitted_line[i]
4319 p = [(l1 + l2) / 2, (l1 + l2) / 2]
4320 if notpoint_inside_csp(p, csp):
4321 del splitted_line[i]
4322 else:
4323 i += 1
4324
4325# and apply back transrormations to draw them
4326 csp_line = csp_from_polyline(splitted_line)
4327 csp_line = self.transform_csp(csp_line, layer, True)
4328
4329 self.draw_csp(csp_line, group=area_group)
4330
4174def tab_area_fill(self): …)
4331
4350def tab_engraving(self):
4351 self.get_info_plus()
4352 global cspm
4353 global wl
4354 global nlLT
4355 global i
4356 global j
4357 global gcode_3Dleft
4358 global gcode_3Dright
4359 global max_dist # minimum of tool radius and user's requested maximum distance
4360 global eye_dist
4361 eye_dist = 100 # 3D constant. Try varying it for your eyes
4362
4363 def bisect(nxy1, nxy2):
4364"""LT Find angle bisecting the normals n1 and n2
4365
4366 Parameters: Normalised normals
4367 Returns: nx - Normal of bisector, normalised to 1/cos(a)
4368 ny -
4369 sinBis2 - sin(angle turned/2): positive if turning in
4370 Note that bisect(n1,n2) and bisect(n2,n1) give opposite sinBis2 results
4371 If sinturn is less than the user's requested angle tolerance, I return 0
4372 """
4373 (nx1, ny1) = nxy1
4374 (nx2, ny2) = nxy2
4375 cosBis = math.sqrt(max(0, (1.0 + nx1 nx2 - ny1 ny2) / 2.0))
4376# We can get correct sign of the sin, assuming cos is positive
4377 if (abs(ny1 - ny2) < ENGRAVING_TOLERANCE) or (abs(cosBis) < ENGRAVING_TOLERANCE):
4378 if abs(nx1 - nx2) < ENGRAVING_TOLERANCE:
4379 return nx1, ny1, 0.0
4380 sinBis = math.copysign(1, ny1)
4381 else:
4382 sinBis = cosBis (nx2 - nx1) / (ny1 - ny2)
4383# We can correct signs by noting that the dot product
4384# of bisector and either normal must be >0
4385 costurn = cosBis nx1 + sinBis ny1
4386 if costurn == 0:
4387 return ny1 100, -nx1 100, 1 # Path doubles back on itself
4388 sinturn = sinBis nx1 - cosBis ny1
4389 if costurn < 0:
4390 sinturn = -sinturn
4391 if 0 < sinturn 114.6 < (180 - self.options.engraving_sharp_angle_tollerance):
4392 sinturn = 0 # set to zero if less than the user wants to see.
4393 return cosBis / costurn, sinBis / costurn, sinturn
4394# end bisect
4395
4396 def get_radius_to_line(xy1, n_xy1, n_xy2, xy2, n_xy23, xy3, n_xy3):
4397"""LT find biggest circle we can engrave here, if constrained by line 2-3
4398
4399 Parameters:
4400 x1,y1,nx1,ny1 coordinates and normal of the line we're currently engraving
4401 nx2,ny2 angle bisector at point 2
4402 x2,y2 coordinates of first point of line 2-3
4403 nx23,ny23 normal to the line 2-3
4404 x3,y3 coordinates of the other end
4405 nx3,ny3 angle bisector at point 3
4406 Returns:
4407 radius or self.options.engraving_max_dist if line doesn't limit radius
4408 This function can be used in three ways:
4409 - With nx1=ny1=0 it finds circle centred at x1,y1
4410 - with nx1,ny1 normalised, it finds circle tangential at x1,y1
4411 - with nx1,ny1 scaled by 1/cos(a) it finds circle centred on an angle bisector
4412 where a is the angle between the bisector and the previous/next normals
4413
4414 If the centre of the circle tangential to the line 2-3 is outside the
4415 angle bisectors at its ends, ignore this line.
4416
4417 Note that it handles corners in the conventional manner of letter cutting
4418 by mitering, not rounding.
4419 Algorithm uses dot products of normals to find radius
4420 and hence coordinates of centre
4421 """
4422 (x1, y1) = xy1
4423 (nx1, ny1) = n_xy1
4424 (nx2, ny2) = n_xy2
4425 (x2, y2) = xy2
4426 (nx23, ny23) = n_xy23
4427 (x3, y3) = xy3
4428 (nx3, ny3) = n_xy3
4429 global max_dist
4430
4431# Start by converting coordinates to be relative to x1,y1
4432 x2, y2 = x2 - x1, y2 - y1
4433 x3, y3 = x3 - x1, y3 - y1
4434
4435# The logic uses vector arithmetic.
4436# The dot product of two vectors gives the product of their lengths
4437# multiplied by the cos of the angle between them.
4438# So, the perpendicular distance from x1y1 to the line 2-3
4439# is equal to the dot product of its normal and x2y2 or x3y3
4440# It is also equal to the projection of x1y1-xcyc on the line's normal
4441# plus the radius. But, as the normal faces inside the path we must negate it.
4442
4443# Make sure the line in question is facing x1,y1 and vice versa
4444 dist = -x2 nx23 - y2 ny23
4445 if dist < 0:
4446 return max_dist
4447 denom = 1. - nx23 nx1 - ny23 ny1
4448 if denom < ENGRAVING_TOLERANCE:
4449 return max_dist
4450
4451# radius and centre are:
4452 r = dist / denom
4453 cx = r nx1
4454 cy = r ny1
4455# if c is not between the angle bisectors at the ends of the line, ignore
4456# Use vector cross products. Not sure if I need the .0001 safety margins:
4457 if (x2 - cx) ny2 > (y2 - cy) nx2 + 0.0001:
4458 return max_dist
4459 if (x3 - cx) ny3 < (y3 - cy) nx3 - 0.0001:
4460 return max_dist
4461 return min(r, max_dist)
4462# end of get_radius_to_line
4463
4464 def get_radius_to_point(xy1, n_xy, xy2):
4465"""LT find biggest circle we can engrave here, constrained by point x2,y2
4466
4467 This function can be used in three ways:
4468 - With nx=ny=0 it finds circle centred at x1,y1
4469 - with nx,ny normalised, it finds circle tangential at x1,y1
4470 - with nx,ny scaled by 1/cos(a) it finds circle centred on an angle bisector
4471 where a is the angle between the bisector and the previous/next normals
4472
4473 Note that I wrote this to replace find_cutter_centre. It is far less
4474 sophisticated but, I hope, far faster.
4475 It turns out that finding a circle touching a point is harder than a circle
4476 touching a line.
4477 """
4478 (x1, y1) = xy1
4479 (nx, ny) = n_xy
4480 (x2, y2) = xy2
4481 global max_dist
4482
4483# Start by converting coordinates to be relative to x1,y1
4484 x2 = x2 - x1
4485 y2 = y2 - y1
4486 denom = nx 2 + ny 2 - 1
4487 if denom <= ENGRAVING_TOLERANCE: # Not a corner bisector
4488 if denom == -1: # Find circle centre x1,y1
4489 return math.sqrt(x2 2 + y2 2)
4490# if x2,y2 not in front of the normal...
4491 if x2 nx + y2 ny <= 0:
4492 return max_dist
4493 return (x2 2 + y2 2) / (2 (x2 nx + y2 ny))
4494# It is a corner bisector, so..
4495 discriminator = (x2 nx + y2 ny) 2 - denom (x2 2 + y2 2)
4496 if discriminator < 0:
4497 return max_dist # this part irrelevant
4498 r = (x2 nx + y2 ny - math.sqrt(discriminator)) / denom
4499 return min(r, max_dist)
4500# end of get_radius_to_point
4501
4502 def bez_divide(a, b, c, d):
4503"""LT recursively divide a Bezier.
4504
4505 Divides until difference between each
4506 part and a straight line is less than some limit
4507 Note that, as simple as this code is, it is mathematically correct.
4508 Parameters:
4509 a,b,c and d are each a list of x,y real values
4510 Bezier end points a and d, control points b and c
4511 Returns:
4512 a list of Beziers.
4513 Each Bezier is a list with four members,
4514 each a list holding a coordinate pair
4515 Note that the final point of one member is the same as
4516 the first point of the next, and the control points
4517 there are smooth and symmetrical. I use this fact later.
4518 """
4519 bx = b - a
4520 by = b - a
4521 cx = c - a
4522 cy = c - a
4523 dx = d - a
4524 dy = d - a
4525 limit = 8 math.hypot(dx, dy) / self.options.engraving_newton_iterations
4526# LT This is the only limit we get from the user currently
4527 if abs(dx by - bx dy) < limit and abs(dx cy - cx dy) < limit:
4528 return
4529 abx = (a + b) / 2.0
4530 aby = (a + b) / 2.0
4531 bcx = (b + c) / 2.0
4532 bcy = (b + c) / 2.0
4533 cdx = (c + d) / 2.0
4534 cdy = (c + d) / 2.0
4535 abcx = (abx + bcx) / 2.0
4536 abcy = (aby + bcy) / 2.0
4537 bcdx = (bcx + cdx) / 2.0
4538 bcdy = (bcy + cdy) / 2.0
4539 m = [(abcx + bcdx) / 2.0, (abcy + bcdy) / 2.0]
4540 return bez_divide(a, [abx, aby], [abcx, abcy], m) + bez_divide(m, [bcdx, bcdy], [cdx, cdy], d)
4541# end of bez_divide
4542
4543 def get_biggest(nxy1, nxy2):
4544"""LT Find biggest circle we can draw inside path at point x1,y1 normal nx,ny
4545
4546 Parameters:
4547 point - either on a line or at a reflex corner
4548 normal - normalised to 1 if on a line, to 1/cos(a) at a corner
4549 Returns:
4550 tuple (j,i,r)
4551 ..where j and i are indices of limiting segment, r is radius
4552 """
4553 (x1, y1) = nxy1
4554 (nx, ny) = nxy2
4555 global max_dist
4556 global nlLT
4557 global i
4558 global j
4559
4560 n1 = nlLT[j][i - 1] # current node
4561 jjmin = -1
4562 iimin = -1
4563 r = max_dist
4564# set limits within which to look for lines
4565 xmin = x1 + r nx - r
4566 xmax = x1 + r nx + r
4567 ymin = y1 + r ny - r
4568 ymax = y1 + r ny + r
4569 for jj inxrange(0, len(nlLT)): # for every subpath of this object
4570 for ii inxrange(0, len(nlLT[jj])): # for every point and line
4571 if nlLT[jj][ii - 1]: # if a point
4572 if jj == j: # except this one
4573 if abs(ii - i) < 3 or abs(ii - i) > len(nlLT[j]) - 3:
4574 continue
4575 t1 = get_radius_to_point((x1, y1), (nx, ny), nlLT[jj][ii - 1])
4576 else: # doing a line
4577 if jj == j: # except this one
4578 if abs(ii - i) < 2 or abs(ii - i) == len(nlLT[j]) - 1:
4579 continue
4580 if abs(ii - i) == 2 and nlLT[j][(ii + i) / 2 - 1] <= 0:
4581 continue
4582 if (abs(ii - i) == len(nlLT[j]) - 2) and nlLT[j][-1] <= 0:
4583 continue
4584 nx2, ny2 = nlLT[jj][ii - 2]
4585 x2, y2 = nlLT[jj][ii - 1]
4586 nx23, ny23 = nlLT[jj][ii - 1]
4587 x3, y3 = nlLT[jj][ii]
4588 nx3, ny3 = nlLT[jj][ii]
4589 if nlLT[jj][ii - 2] > 0: # acute, so use normal, not bisector
4590 nx2 = nx23
4591 ny2 = ny23
4592 if nlLT[jj][ii] > 0: # acute, so use normal, not bisector
4593 nx3 = nx23
4594 ny3 = ny23
4595 x23min = min(x2, x3)
4596 x23max = max(x2, x3)
4597 y23min = min(y2, y3)
4598 y23max = max(y2, y3)
4599# see if line in range
4600 if n1 == False and (x23max < xmin or x23min > xmax or y23max < ymin or y23min > ymax):
4601 continue
4602 t1 = get_radius_to_line((x1, y1), (nx, ny), (nx2, ny2), (x2, y2), (nx23, ny23), (x3, y3), (nx3, ny3))
4603 if 0 <= t1 < r:
4604 r = t1
4605 iimin = ii
4606 jjmin = jj
4607 xmin = x1 + r nx - r
4608 xmax = x1 + r nx + r
4609 ymin = y1 + r ny - r
4610 ymax = y1 + r ny + r
4611# next ii
4612# next jj
4613 return jjmin, iimin, r
4614# end of get_biggest
4615
4616 def line_divide(xy0, j0, i0, xy1, j1, i1, n_xy, length):
4617"""LT recursively divide a line as much as necessary
4618
4619 NOTE: This function is not currently used
4620 By noting which other path segment is touched by the circles at each end,
4621 we can see if anything is to be gained by a further subdivision, since
4622 if they touch the same bit of path we can move linearly between them.
4623 Also, we can handle points correctly.
4624 Parameters:
4625 end points and indices of limiting path, normal, length
4626 Returns:
4627 list of toolpath points
4628 each a list of 3 reals: x, y coordinates, radius
4629
4630 """
4631 (x0, y0) = xy0
4632 (x1, y1) = xy1
4633 (nx, ny) = n_xy
4634 global nlLT
4635 global i
4636 global j
4637 global lmin
4638 x2 = (x0 + x1) / 2
4639 y2 = (y0 + y1) / 2
4640 j2, i2, r2 = get_biggest((x2, y2), (nx, ny))
4641 if length < lmin:
4642 return
4643 if j2 == j0 and i2 == i0: # Same as left end. Don't subdivide this part any more
4644 return [[x2, y2, r2], line_divide((x2, y2), j2, i2, (x1, y1), j1, i1, (nx, ny), length / 2)]
4645 if j2 == j1 and i2 == i1: # Same as right end. Don't subdivide this part any more
4646 return [line_divide((x0, y0), j0, i0, (x2, y2), j2, i2, (nx, ny), length / 2), [x2, y2, r2]]
4647 return [line_divide((x0, y0), j0, i0, (x2, y2), j2, i2, (nx, ny), length / 2), line_divide((x2, y2), j2, i2, (x1, y1), j1, i1, (nx, ny), length / 2)]
4648# end of line_divide()
4649
4650 def save_point(xy, w, i, j, ii, jj):
4651"""LT Save this point and delete previous one if linear
4652
4653 The point is, we generate tons of points but many may be in a straight 3D line.
4654 There is no benefit in saving the intermediate points.
4655 """
4656 (x, y) = xy
4657 global wl
4658 global cspm
4659 x = round(x, 4) # round to 4 decimals
4660 y = round(y, 4) # round to 4 decimals
4661 w = round(w, 4) # round to 4 decimals
4662 if len(cspm) > 1:
4663 xy1a, xy1, xy1b, i1, j1, ii1, jj1 = cspm[-1]
4664 w1 = wl[-1]
4665 if i == i1 and j == j1 and ii == ii1 and jj == jj1: # one match
4666 xy1a, xy2, xy1b, i1, j1, ii1, jj1 = cspm[-2]
4667 w2 = wl[-2]
4668 if i == i1 and j == j1 and ii == ii1 and jj == jj1: # two matches. Now test linearity
4669 length1 = math.hypot(xy1 - x, xy1 - y)
4670 length2 = math.hypot(xy2 - x, xy2 - y)
4671 length12 = math.hypot(xy2 - xy1, xy2 - xy1)
4672# get the xy distance of point 1 from the line 0-2
4673 if length2 > length1 and length2 > length12: # point 1 between them
4674 xydist = abs((xy2 - x) (xy1 - y) - (xy1 - x) (xy2 - y)) / length2
4675 if xydist < ENGRAVING_TOLERANCE: # so far so good
4676 wdist = w2 + (w - w2) length1 / length2 - w1
4677 if abs(wdist) < ENGRAVING_TOLERANCE:
4678 cspm.pop()
4679 wl.pop()
4680 cspm +=
4681 wl += [w]
4682# end of save_point
4683
4684 def draw_point(xy0, xy, w, t):
4685"""LT Draw this point as a circle with a 1px dot in the middle (x,y)
4686 and a 3D line from (x0,y0) down to x,y. 3D line thickness should be t/2
4687
4688 Note that points that are subsequently erased as being unneeded do get
4689 displayed, but this helps the user see the total area covered.
4690 """
4691 (x0, y0) = xy0
4692 (x, y) = xy
4693 global gcode_3Dleft
4694 global gcode_3Dright
4695 if self.options.engraving_draw_calculation_paths:
4696 elem = engraving_group.add(PathElement.arc((x, y), 1))
4697 elem.set('gcodetools', "Engraving calculation toolpath")
4698 elem.style = "fill:#ff00ff; fill-opacity:0.46; stroke:#000000; stroke-width:0.1;"
4699
4700# Don't draw zero radius circles
4701 if w:
4702 elem = engraving_group.add(PathElement.arc((x, y), w))
4703 elem.set('gcodetools', "Engraving calculation paths")
4704 elem.style = "fill:none; fill-opacity:0.46; stroke:#000000; stroke-width:0.1;"
4705
4706# Find slope direction for shading
4707 s = math.atan2(y - y0, x - x0) # -pi to pi
4708# convert to 2 hex digits as a shade of red
4709 s2 = "#{0:x}0000".format(int(101 (1.5 - math.sin(s + 0.5))))
4710 style = "stroke:{}; stroke-opacity:1;stroke-width:{};fill:none".format(s2, t/2)
4711 right = gcode_3Dleft.add(PathElement(style=style, gcodetools="Gcode G1R"))
4712 right.path = "M {:f},{:f} L {:f},{:f}".format(
4713 x0 - eye_dist, y0, x - eye_dist - 0.14 w, y)
4714 left = gcode_3Dright.add(PathElement(style=style, gcodetools="Gcode G1L"))
4715 left.path = "M {:f},{:f} L {:f},{:f}".format(
4716 x0 + eye_dist, y0, x + eye_dist + 0.14 r, y)
4717
4718# end of subfunction definitions. engraving() starts here:
4719 gcode = ''
4720 r = 0 # theoretical and tool-radius-limited radii in pixels
4721 w = 0
4722 wmax = 0
4723 cspe = []
4724 we = []
4725 if not self.selected_paths:
4726 self.error("Please select at least one path to engrave and run again.")
4727 return
4728 if not self.check_dir():
4729 return
4730# Find what units the user uses
4731 unit = " mm"
4732 if self.options.unit == "G20 (All units in inches)":
4733 unit = " inches"
4734 elif self.options.unit != "G21 (All units in mm)":
4735 self.error("Unknown unit selected. mm assumed")
4736print_("engraving_max_dist mm/inch", self.options.engraving_max_dist)
4737
4738# LT See if we can use this parameter for line and Bezier subdivision:
4739 bitlen = 20 / self.options.engraving_newton_iterations
4740
4741 for layer in self.layers:
4742 if layer in self.selected_pathsand layer in self.orientation_points:
4743# Calculate scale in pixels per user unit (mm or inch)
4744 p1 = self.orientation_points[layer]
4745 p2 = self.orientation_points[layer]
4746 ol = math.hypot(p1 - p2, p1 - p2)
4747 oluu = math.hypot(p1 - p2, p1 - p2)
4748print_("Orientation2 p1 p2 ol oluu", p1, p2, ol, oluu)
4749 orientation_scale = ol / oluu
4750
4751 self.set_tool(layer)
4752 shape = self.tools[layer]['shape']
4753 if re.search('w', shape):
4754 toolshape = eval('lambda w: ' + shape.strip('"'))
4755 else:
4756 self.error("Tool '{}' has no shape. 45 degree cone assumed!".format(self.tools[layer]['name']))
4757 toolshape = lambda w: w
4758# Get tool radius in pixels
4759 toolr = self.tools[layer]['diameter'] orientation_scale / 2
4760print_("tool radius in pixels=", toolr)
4761# max dist from path to engrave in user's units
4762 max_distuu = min(self.tools[layer]['diameter'] / 2, self.options.engraving_max_dist)
4763 max_dist = max_distuu orientation_scale
4764print_("max_dist pixels", max_dist)
4765
4766 engraving_group = self.selected_paths[layer].getparent().add(Group())
4767 if self.options.engraving_draw_calculation_paths and (self.my3Dlayeris None):
4768 self.svg.add(Layer.new("3D"))
4769# Create groups for left and right eyes
4770 if self.options.engraving_draw_calculation_paths:
4771 gcode_3Dleft = self.my3Dlayer.add(Group(gcodetools="Gcode 3D L"))
4772 gcode_3Dright = self.my3Dlayer.add(Group(gcodetools="Gcode 3D R"))
4773
4774 for node in self.selected_paths[layer]:
4775 if isinstance(node, inkex.PathElement):
4776 cspi = node.path.to_superpath()
4777# LT: Create my own list. n1LT[j] is for subpath j
4778 nlLT = []
4779 for j inxrange(len(cspi)): # LT For each subpath...
4780# Remove zero length segments, assume closed path
4781 i = 0 # LT was from i=1
4782 while i < len(cspi[j]):
4783 if abs(cspi[j][i - 1] - cspi[j][i]) < ENGRAVING_TOLERANCE and abs(cspi[j][i - 1] - cspi[j][i]) < ENGRAVING_TOLERANCE:
4784 cspi[j][i - 1] = cspi[j][i]
4785 del cspi[j][i]
4786 else:
4787 i += 1
4788 for csp in cspi: # LT6a For each subpath...
4789# Create copies in 3D layer
4790print_("csp is zz ", csp)
4791 cspl = []
4792 cspr = []
4793# create list containing lines and points, starting with a point
4794# line members: [x,y],[nx,ny],False,i
4795# x,y is start of line. Normal on engraved side.
4796# Normal is normalised (unit length)
4797# Note that Y axis increases down the page
4798# corner members: [x,y],[nx,ny],True,sin(halfangle)
4799# if halfangle>0: radius 0 here. normal is bisector
4800# if halfangle<0. reflex angle. normal is bisector
4801# corner normals are divided by cos(halfangle)
4802# so that they will engrave correctly
4803print_("csp is", csp)
4804 nlLT.append([])
4805 for i in range(0, len(csp)): # LT for each point
4806 sp0 = csp[i - 2]
4807 sp1 = csp[i - 1]
4808 sp2 = csp[i]
4809 if self.options.engraving_draw_calculation_paths:
4810# Copy it to 3D layer objects
4811 spl = []
4812 spr = []
4813 for j in range(0, 3):
4814 pl = [sp2[j] - eye_dist, sp2[j]]
4815 pr = [sp2[j] + eye_dist, sp2[j]]
4816 spl += [pl]
4817 spr += [pr]
4818 cspl += [spl]
4819 cspr += [spr]
4820# LT find angle between this and previous segment
4821 x0, y0 = sp1
4822 nx1, ny1 = csp_normalized_normal(sp1, sp2, 0)
4823# I don't trust this function, so test result
4824 if abs(1 - math.hypot(nx1, ny1)) > 0.00001:
4825print_("csp_normalised_normal error t=0", nx1, ny1, sp1, sp2)
4826 self.error("csp_normalised_normal error. See log.")
4827
4828 nx0, ny0 = csp_normalized_normal(sp0, sp1, 1)
4829 if abs(1 - math.hypot(nx0, ny0)) > 0.00001:
4830print_("csp_normalised_normal error t=1", nx0, ny0, sp1, sp2)
4831 self.error("csp_normalised_normal error. See log.")
4832 bx, by, s = bisect((nx0, ny0), (nx1, ny1))
4833# record x,y,normal,ifCorner, sin(angle-turned/2)
4834 nlLT[-1] +=
4835
4836# LT now do the line
4837 if sp1 == sp1 and sp2 == sp2: # straightline
4838 nlLT[-1] +=
4839 else: # Bezier. First, recursively cut it up:
4840 nn = bez_divide(sp1, sp1, sp2, sp2)
4841 first = True# Flag entry to divided Bezier
4842 for bLT in nn: # save as two line segments
4843 for seg in range(3):
4844 if seg > 0 or first:
4845 nx1 = bLT[seg] - bLT[seg + 1]
4846 ny1 = bLT[seg + 1] - bLT[seg]
4847 l1 = math.hypot(nx1, ny1)
4848 if l1 < ENGRAVING_TOLERANCE:
4849 continue
4850 nx1 = nx1 / l1 # normalise them
4851 ny1 = ny1 / l1
4852 nlLT[-1] +=
4853 first = False
4854 if seg < 2: # get outgoing bisector
4855 nx0 = nx1
4856 ny0 = ny1
4857 nx1 = bLT[seg + 1] - bLT[seg + 2]
4858 ny1 = bLT[seg + 2] - bLT[seg + 1]
4859 l1 = math.hypot(nx1, ny1)
4860 if l1 < ENGRAVING_TOLERANCE:
4861 continue
4862 nx1 = nx1 / l1 # normalise them
4863 ny1 = ny1 / l1
4864# bisect
4865 bx, by, s = bisect((nx0, ny0), (nx1, ny1))
4866 nlLT[-1] +=
4867# LT for each segment - ends here.
4868print_(("engraving_draw_calculation_paths=", self.options.engraving_draw_calculation_paths))
4869 if self.options.engraving_draw_calculation_paths:
4870# Copy complete paths to 3D layer
4871 cspl += [cspl] # Close paths
4872 cspr += [cspr] # Close paths
4873 style = "stroke:#808080; stroke-opacity:1; stroke-width:0.6; fill:none"
4874 elem = gcode_3Dleft.add(PathElement(style=style, gcodetools="G1L outline"))
4875 elem.path = CubicSuperPath([cspl])
4876 elem = gcode_3Dright.add(Pathelement(style=style, gcodetools="G1R outline"))
4877 elem.path = CubicSuperPath([cspr])
4878
4879 for p in nlLT[-1]: # For last sub-path
4880 if p:
4881 elem = engraving_group.add(PathElement(gcodetools="Engraving normals"))
4882 elem.path = "M {:f},{:f} L {:f},{:f}".format(p, p,
4883 p + p 10, p + p 10)
4884 elem.style = "stroke:#f000af; stroke-opacity:0.46; stroke-width:0.1; fill:none"
4885 else:
4886 elem = engraving_group.add(PathElement(gcodetools="Engraving bisectors"))
4887 elem.path = "M {:f},{:f} L {:f},{:f}".format(p, p,
4888 p + p 10, p + p 10)
4889 elem.style = "stroke:#0000ff; stroke-opacity:0.46; stroke-width:0.1; fill:none"
4890
4891# LT6a build nlLT[j] for each subpath - ends here
4892# Calculate offset points
4893 reflex = False
4894 for j inxrange(len(nlLT)): # LT6b for each subpath
4895 cspm = [] # Will be my output. List of csps.
4896 wl = [] # Will be my w output list
4897 w = r = 0 # LT initial, as first point is an angle
4898 for i inxrange(len(nlLT[j])): # LT for each node
4899# LT Note: Python enables wrapping of array indices
4900# backwards to -1, -2, but not forwards. Hence:
4901 n0 = nlLT[j][i - 2] # previous node
4902 n1 = nlLT[j][i - 1] # current node
4903 n2 = nlLT[j][i] # next node
4904# if n1 == True and n1==0 : # A straight angle
4905# continue
4906 x1a, y1a = n1 # this point/start of this line
4907 nx, ny = n1
4908 x1b, y1b = n2 # next point/end of this line
4909 if n1: # We're at a corner
4910 bits = 1
4911 bit0 = 0
4912# lastr=r #Remember r from last line
4913 lastw = w # Remember w from last line
4914 w = max_dist
4915 if n1 > 0: # acute. Limit radius
4916 len1 = math.hypot((n0 - n1), (n0 - n1))
4917 if i < (len(nlLT[j]) - 1):
4918 len2 = math.hypot((nlLT[j][i + 1] - n1), (nlLT[j][i + 1] - n1))
4919 else:
4920 len2 = math.hypot((nlLT[j] - n1), (nlLT[j] - n1))
4921# set initial r value, not to be exceeded
4922 w = math.sqrt(min(len1, len2)) / n1
4923 else: # line. Cut it up if long.
4924 if n0 > 0 and not self.options.engraving_draw_calculation_paths:
4925 bit0 = r n0 # after acute corner
4926 else:
4927 bit0 = 0.0
4928 length = math.hypot((x1b - x1a), (y1a - y1b))
4929 bit0 = (min(length, bit0))
4930 bits = int((length - bit0) / bitlen)
4931# split excess evenly at both ends
4932 bit0 += (length - bit0 - bitlen bits) / 2
4933 for b inxrange(bits): # divide line into bits
4934 x1 = x1a + ny (b bitlen + bit0)
4935 y1 = y1a - nx (b bitlen + bit0)
4936 jjmin, iimin, w = get_biggest((x1, y1), (nx, ny))
4937print_("i,j,jjmin,iimin,w", i, j, jjmin, iimin, w)
4938 wmax = max(wmax, w)
4939 if reflex: # just after a reflex corner
4940 reflex = False
4941 if w < lastw: # need to adjust it
4942 draw_point((x1, y1), (n0 + n0 w, n0 + n0 w), w, (lastw - w) / 2)
4943 save_point((n0 + n0 w, n0 + n0 w), w, i, j, iimin, jjmin)
4944 if n1: # We're at a corner
4945 if n1 > 0: # acute
4946 save_point((x1 + nx w, y1 + ny w), w, i, j, iimin, jjmin)
4947 draw_point((x1, y1), (x1, y1), 0, 0)
4948 save_point((x1, y1), 0, i, j, iimin, jjmin)
4949 elif n1 < 0: # reflex
4950 if w > lastw:
4951 draw_point((x1, y1), (x1 + nx lastw, y1 + ny lastw), w, (w - lastw) / 2)
4952 wmax = max(wmax, w)
4953 save_point((x1 + nx w, y1 + ny w), w, i, j, iimin, jjmin)
4954 elif b > 0 and n2 > 0 and not self.options.engraving_draw_calculation_paths: # acute corner coming up
4955 if jjmin == j and iimin == i + 2:
4956 break
4957 draw_point((x1, y1), (x1 + nx w, y1 + ny w), w, bitlen)
4958 save_point((x1 + nx w, y1 + ny w), w, i, j, iimin, jjmin)
4959
4960# LT end of for each bit of this line
4961 if n1 == True and n1 < 0: # reflex angle
4962 reflex = True
4963 lastw = w # remember this w
4964# LT next i
4965 cspm += [cspm]
4966print_("cspm", cspm)
4967 wl += [wl]
4968print_("wl", wl)
4969# Note: Original csp_points was a list, each element
4970# being 4 points, with the first being the same as the
4971# last of the previous set.
4972# Each point is a list of [cx,cy,r,w]
4973# I have flattened it to a flat list of points.
4974
4975 if self.options.engraving_draw_calculation_paths:
4976 node = engraving_group.add(PathElement(
4977 gcodetools="Engraving calculation paths",
4978 style=MARKER_STYLE["biarc_style_i"]['biarc1']))
4979 node.path = CubicSuperPath([cspm])
4980 for i inxrange(len(cspm)):
4981 elem = engraving_group.add(PathElement.arc(cspm[i], wl[i]))
4982 elem.set('gcodetools', "Engraving calculation paths")
4983 elem.style = "fill:none;fill-opacity:0.46;stroke:#000000;stroke-width:0.1;"
4984 cspe += [cspm]
4985 wluu = [] # width list in user units: mm/inches
4986 for w in wl:
4987 wluu += [w / orientation_scale]
4988print_("wl in pixels", wl)
4989print_("wl in user units", wluu)
4990# LT previously, we was in pixels so gave wrong depth
4991 we += [wluu]
4992# LT6b For each subpath - ends here
4993# LT5 if it is a path - ends here
4994# LT4 for each selected object in this layer - ends here
4995
4996 if cspe:
4997 curve = self.parse_curve(cspe, layer, we, toolshape) # convert to lines
4998 self.draw_curve(curve, layer, engraving_group)
4999 gcode += self.generate_gcode(curve, layer, self.options.Zsurface)
5000
5001# LT3 for layers loop ends here
5002 if gcode != '':
5003 self.header += "(Tool diameter should be at least " + str(2 wmax / orientation_scale) + unit + ")\n"
5004 self.header += "(Depth, as a function of radius w, must be " + self.tools[layer]['shape'] + ")\n"
5005 self.header += "(Rapid feeds use safe Z=" + str(self.options.Zsafe) + unit + ")\n"
5006 self.header += "(Material surface at Z=" + str(self.options.Zsurface) + unit + ")\n"
5007 self.export_gcode(gcode)
5008 else:
5009 self.error("No need to engrave sharp angles.")
5010
4350def tab_engraving(self): …)
5011
5016def tab_orientation(self, layer=None):
5017 self.get_info()
5018 Zsurface = f"{self.options.Zsurface:.5f}"
5019 Zdepth = f"{self.options.Zdepth:.5f}"
5020 if layer is None:
5021 layer = self.svg.get_current_layer() if self.svg.get_current_layer() is not None else self.documentdocument.getroot()
5022
5023 transform = self.get_transforms(layer)
5024 if transform:
5025 transform = self.reverse_transform(transform)
5026 transform = str(Transform(transform))
5027
5028 if self.options.orientation_points_count == "graffiti":
5029print_(self.graffiti_reference_points)
5030print_("Inserting graffiti points")
5031 if layer in self.graffiti_reference_points:
5032 graffiti_reference_points_count = len(self.graffiti_reference_points[layer])
5033 else:
5034 graffiti_reference_points_count = 0
5035 axis = ["X", "Y", "Z", "A"][graffiti_reference_points_count % 4]
5036 attr = {'gcodetools': "Gcodetools graffiti reference point"}
5037 if transform:
5038 attr["transform"] = transform
5039 group = layer.add(Group(attr))
5040 elem = group.add(PathElement(style="stroke:none;fill:#00ff00;"))
5041 elem.set('gcodetools', "Gcodetools graffiti reference point arrow")
5042 elem.path = 'm {},{} 2.9375,-6.343750000001 0.8125,1.90625 6.843748640396,'\
5043'-6.84374864039 0,0 0.6875,0.6875 -6.84375,6.84375 1.90625,0.8125000000'\
5044'01 z z'.format(graffiti_reference_points_count 100, 0)
5045
5046draw_text(axis, graffiti_reference_points_count 100 + 10, -10, group=group, gcodetools_tag="Gcodetools graffiti reference point text")
5047
5048 elif self.options.orientation_points_count == "in-out reference point":
5049draw_pointer(group=self.svg.get_current_layer(), x=self.svg.namedview.center, figure="arrow", pointer_type="In-out reference point", text="In-out point")
5050
5051 else:
5052print_("Inserting orientation points")
5053
5054 if layer in self.orientation_points:
5055 self.error("Active layer already has orientation points! Remove them or select another layer!", "error")
5056
5057 attr = {"gcodetools": "Gcodetools orientation group"}
5058 if transform:
5059 attr["transform"] = transform
5060
5061 orientation_group = layer.add(Group(attr))
5062 doc_height = self.svg.unittouu(self.documentdocument.getroot().get('height'))
5063 if self.documentdocument.getroot().get('height') == "100%":
5064 doc_height = 1052.3622047
5065print_("Overriding height from 100 percents to {}".format(doc_height))
5066 if self.options.unit == "G21 (All units in mm)":
5067 points =
5068 elif self.options.unit == "G20 (All units in inches)":
5069 points =
5070 if self.options.orientation_points_count == "2":
5071 points = points[:2]
5072 for i in points:
5073 name = "Gcodetools orientation point ({} points)".format(
5074 self.options.orientation_points_count)
5075 grp = orientation_group.add(Group(gcodetools=name))
5076 elem = grp.add(PathElement(style="stroke:none;fill:#000000;"))
5077 elem.set('gcodetools', "Gcodetools orientation point arrow")
5078 elem.path = 'm {},{} 2.9375,-6.343750000001 0.8125,1.90625 6.843748640396,'\
5079'-6.84374864039 0,0 0.6875,0.6875 -6.84375,6.84375 1.90625,0.812500000'\
5080'001 z'.format(i, -i + doc_height)
5081
5082draw_text("({}; {}; {})".format(i, i, i), (i + 10), (-i - 10 + doc_height), group=grp, gcodetools_tag="Gcodetools orientation point text")
5083
5016def tab_orientation(self, layer=None): …)
5084
5089def tab_tools_library(self, layer=None):
5090 self.get_info()
5091
5092 if self.options.tools_library_type == "check":
5093 return self.check_tools_and_op()
5094
5095# Add a tool to the drawing
5096 if layer is None:
5097 layer = self.svg.get_current_layer() if self.svg.get_current_layer() is not None else self.documentdocument.getroot()
5098 if layer in self.tools:
5099 self.error("Active layer already has a tool! Remove it or select another layer!", "error")
5100
5101 if self.options.tools_library_type == "cylinder cutter":
5102 tool = {
5103"name": "Cylindrical cutter",
5104"id": "Cylindrical cutter 0001",
5105"diameter": 10,
5106"penetration angle": 90,
5107"feed": "400",
5108"penetration feed": "100",
5109"depth step": "1",
5110"tool change gcode": " "
5111 }
5112 elif self.options.tools_library_type == "lathe cutter":
5113 tool = {
5114"name": "Lathe cutter",
5115"id": "Lathe cutter 0001",
5116"diameter": 10,
5117"penetration angle": 90,
5118"feed": "400",
5119"passing feed": "800",
5120"fine feed": "100",
5121"penetration feed": "100",
5122"depth step": "1",
5123"tool change gcode": " "
5124 }
5125 elif self.options.tools_library_type == "cone cutter":
5126 tool = {
5127"name": "Cone cutter",
5128"id": "Cone cutter 0001",
5129"diameter": 10,
5130"shape": "w",
5131"feed": "400",
5132"penetration feed": "100",
5133"depth step": "1",
5134"tool change gcode": " "
5135 }
5136 elif self.options.tools_library_type == "tangent knife":
5137 tool = {
5138"name": "Tangent knife",
5139"id": "Tangent knife 0001",
5140"feed": "400",
5141"penetration feed": "100",
5142"depth step": "100",
5143"4th axis meaning": "tangent knife",
5144"4th axis scale": 1.,
5145"4th axis offset": 0,
5146"tool change gcode": " "
5147 }
5148
5149 elif self.options.tools_library_type == "plasma cutter":
5150 tool = {
5151"name": "Plasma cutter",
5152"id": "Plasma cutter 0001",
5153"diameter": 10,
5154"penetration feed": 100,
5155"feed": 400,
5156"gcode before path": """G31 Z-100 F500 (find metal)
5157 G92 Z0 (zero z)
5158 G00 Z10 F500 (going up)
5159 M03 (turn on plasma)
5160 G04 P0.2 (pause)
5161 G01 Z1 (going to cutting z)\n""",
5162"gcode after path": "M05 (turn off plasma)\n",
5163 }
5164 elif self.options.tools_library_type == "graffiti":
5165 tool = {
5166"name": "Graffiti",
5167"id": "Graffiti 0001",
5168"diameter": 10,
5169"penetration feed": 100,
5170"feed": 400,
5171"gcode before path": """M03 S1(Turn spray on)\n """,
5172"gcode after path": "M05 (Turn spray off)\n ",
5173"tool change gcode": "(Add G00 here to change sprayer if needed)\n",
5174
5175 }
5176
5177 else:
5178 tool = self.default_tool
5179
5180 tool_num = sum([len(self.tools[i]) for i in self.tools])
5181 colors = ["00ff00", "0000ff", "ff0000", "fefe00", "00fefe", "fe00fe", "fe7e00", "7efe00", "00fe7e", "007efe", "7e00fe", "fe007e"]
5182
5183 tools_group = layer.add(Group(gcodetools="Gcodetools tool definition"))
5184 bg = tools_group.add(PathElement(gcodetools="Gcodetools tool background"))
5185 bg.style = "fill-opacity:0.5;stroke:#444444;"
5186 bg.style['fill'] = "#" + colors[tool_num % len(colors)]
5187
5188 y = 0
5189 keys = []
5190 for key in self.tools_field_order:
5191 if key in tool:
5192 keys += [key]
5193 for key in tool:
5194 if key not in keys:
5195 keys += [key]
5196 for key in keys:
5197 g = tools_group.add(Group(gcodetools="Gcodetools tool parameter"))
5198draw_text(key, 0, y, group=g, gcodetools_tag="Gcodetools tool definition field name", font_size=10 if key != 'name'else 20)
5199 param = tool[key]
5200 if type(param) == str and re.match("^\s$", param):
5201 param = "(None)"
5202draw_text(param, 150, y, group=g, gcodetools_tag="Gcodetools tool definition field value", font_size=10 if key != 'name'else 20)
5203 v = str(param).split("\n")
5204 y += 15 len(v) if key != 'name'else 20 len(v)
5205
5206 bg.set('d', "m -20,-20 l 400,0 0,{:f} -400,0 z ".format(y + 50))
5207 tools_group.transform.add_translate(self.svg.namedview.center)
5208 tools_group.transform.add_translate(-150, 0)
5209
5089def tab_tools_library(self, layer=None): …)
5210
5215def check_tools_and_op(self):
5216 if len(self.svg.selected) <= 0:
5217 self.error("Selection is empty! Will compute whole drawing.")
5218 paths = self.paths
5219 else:
5220 paths = self.selected_paths
5221# Set group
5222 parent = self.selected_paths.keys() if len(self.selected_paths.keys()) > 0 else self.layers
5223 group = parent.add(Group())
5224 trans_ =
5225
5226 self.set_markers()
5227
5228 bounds = [float('inf'), float('inf'), float('-inf'), float('-inf')]
5229 tools_bounds = {}
5230 for layer in self.layers:
5231 if layer in paths:
5232 self.set_tool(layer)
5233 tool = self.tools[layer]
5234 tools_bounds[layer] = tools_bounds[layer] if layer in tools_bounds else [float("inf"), float("-inf")]
5235 for path in paths[layer]:
5236 group.insert(0, PathElement(path.attrib))
5237 new = group.getchildren()
5238 new.style = Style(
5239 stroke='#000044', stroke_width=1,
5240 marker_mid='url(#CheckToolsAndOPMarker)',
5241 fill=tool["style"].get('fill', '#00ff00'),
5242 fill_opacity=tool["style"].get('fill-opacity', 0.5))
5243
5244 trans = trans_ self.get_transforms(path)
5245 csp = path.path.transform(trans).to_superpath()
5246
5247 path_bounds = csp_simple_bound(csp)
5248 trans = str(Transform(trans))
5249 bounds = [min(bounds, path_bounds), min(bounds, path_bounds), max(bounds, path_bounds), max(bounds, path_bounds)]
5250 tools_bounds[layer] = [min(tools_bounds[layer], path_bounds), max(tools_bounds[layer], path_bounds)]
5251
5252 new.set("transform", trans)
5253 trans_ += 20
5254 trans_ += 100
5255
5256 for layer in self.layers:
5257 if layer in self.tools:
5258 if layer in tools_bounds:
5259 tool = self.tools[layer]
5260 g = copy.deepcopy(tool["self_group"])
5261 g.attrib["gcodetools"] = "Check tools and OP assignment"
5262 trans = , [0, 0.5, tools_bounds[layer]]]
5263 g.set("transform", str(Transform(trans)))
5264 group.insert(0, g)
5265
5215def check_tools_and_op(self): …)
5266
5269def tab_help(self):
5270 self.error("Switch to another tab to run the extensions.\n"
5271"No changes are made if the preferences or help tabs are active.\n\n"
5272"Tutorials, manuals and support can be found at\n"
5273" English support forum:\n"
5274"
5275"and Russian support forum:\n"
5276"
5277 return
5278
5269def tab_help(self): …)
5279def tab_about(self):
5280 return self.tab_help()
5281
5279def tab_about(self): …)
5282def tab_preferences(self):
5283 return self.tab_help()
5284
5282def tab_preferences(self): …)
5285def tab_options(self):
5286 return self.tab_help()
5287
5288
5285def tab_options(self): …)
5289
5292def generate_lathe_gcode(self, subpath, layer, feed_type):
5293 if len(subpath) < 2:
5294 return""
5295 feed = " F {:f}".format(self.tool[feed_type])
5296 x = self.options.lathe_x_axis_remap
5297 z = self.options.lathe_z_axis_remap
5298 flip_angle = -1 if x.lower() + z.lower() in ["xz", "yx", "zy"] else 1
5299 alias = {"X": "I", "Y": "J", "Z": "K", "x": "i", "y": "j", "z": "k"}
5300 i_ = alias[x]
5301 k_ = alias[z]
5302 c =
5303 for sp1, sp2 in zip(subpath, subpath[1:]):
5304 c += biarc(sp1, sp2, 0, 0)
5305 for i in range(1, len(c)): # Just in case check end point of each segment
5306 c[i - 1] = c[i][:]
5307 c +=
5308 self.draw_curve(c, layer, style=MARKER_STYLE["biarc_style_lathe_{}".format(feed_type)])
5309
5310 gcode = ("G01 {} {:f} {} {:f}".format(x, c, z, c)) + feed + "\n"# Just in case move to the start...
5311 for s in c:
5312 if s == 'line':
5313 gcode += ("G01 {} {:f} {} {:f}".format(x, s, z, s)) + feed + "\n"
5314 elif s == 'arc':
5315 r = [(s - s), (s - s)]
5316 if (r 2 + r 2) > self.options.min_arc_radius 2:
5317 r1 = (P(s) - P(s))
5318 r2 = (P(s) - P(s))
5319 if abs(r1.mag() - r2.mag()) < 0.001:
5320 gcode += ("G02"if s flip_angle < 0 else"G03") + (" {} {:f} {} {:f} {} {:f} {} {:f}".format(x, s, z, s, i_, (s - s), k_, (s - s))) + feed + "\n"
5321 else:
5322 r = (r1.mag() + r2.mag()) / 2
5323 gcode += ("G02"if s flip_angle < 0 else"G03") + (" {} {:f} {} {:f}".format(x, s, z, s)) + " R{:f}".format(r) + feed + "\n"
5324 return gcode
5325
5292def generate_lathe_gcode(self, subpath, layer, feed_type): …)
5326def tab_lathe(self):
5327 self.get_info_plus()
5328 if not self.check_dir():
5329 return
5330 x = self.options.lathe_x_axis_remap
5331 z = self.options.lathe_z_axis_remap
5332 x = re.sub("^\s([XYZxyz])\s$", r"\1", x)
5333 z = re.sub("^\s([XYZxyz])\s$", r"\1", z)
5334 if x not in ["X", "Y", "Z", "x", "y", "z"] or z not in ["X", "Y", "Z", "x", "y", "z"]:
5335 self.error("Lathe X and Z axis remap should be 'X', 'Y' or 'Z'. Exiting...")
5336 return
5337 if x.lower() == z.lower():
5338 self.error("Lathe X and Z axis remap should be the same. Exiting...")
5339 return
5340 if x.lower() + z.lower() in ["xy", "yx"]:
5341 gcode_plane_selection = "G17 (Using XY plane)\n"
5342 if x.lower() + z.lower() in ["xz", "zx"]:
5343 gcode_plane_selection = "G18 (Using XZ plane)\n"
5344 if x.lower() + z.lower() in ["zy", "yz"]:
5345 gcode_plane_selection = "G19 (Using YZ plane)\n"
5346 self.options.lathe_x_axis_remap = x
5347 self.options.lathe_z_axis_remap = z
5348
5349 paths = self.selected_paths
5350 self.tool = []
5351 gcode = ""
5352 for layer in self.layers:
5353 if layer in paths:
5354 self.set_tool(layer)
5355 if self.tool != self.tools[layer]:
5356 self.tool = self.tools[layer]
5357 self.tool["passing feed"] = float(self.tool["passing feed"] if"passing feed"in self.toolelse self.tool["feed"])
5358 self.tool["feed"] = float(self.tool["feed"])
5359 self.tool["fine feed"] = float(self.tool["fine feed"] if"fine feed"in self.toolelse self.tool["feed"])
5360 gcode += ("(Change tool to {})\n".format(re.sub("\"'\(\)\\", " ", self.tool["name"]))) + self.tool["tool change gcode"] + "\n"
5361
5362 for path in paths[layer]:
5363 csp = self.transform_csp(path.path.to_superpath(), layer)
5364
5365 for subpath in csp:
5366# Offset the path if fine cut is defined.
5367 fine_cut = subpath[:]
5368 if self.options.lathe_fine_cut_width > 0:
5369 r = self.options.lathe_fine_cut_width
5370 if self.options.lathe_create_fine_cut_using == "Move path":
5371 subpath = [[[i2, i2 + r] for i2 in i1] for i1 in subpath]
5372 else:
5373# Close the path to make offset correct
5374 bound = csp_simple_bound([subpath])
5375 minx, miny, maxx, maxy = csp_true_bounds([subpath])
5376 offsetted_subpath = csp_subpath_line_to(subpath[:], ])
5377 left = subpath[-1]
5378 right = subpath
5379 if left > right:
5380 left, right = right, left
5381 offsetted_subpath = csp_offset([offsetted_subpath], r if notcsp_subpath_ccw(offsetted_subpath) else -r)
5382 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [left, 10], [left, 0])
5383 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [right, 0], [right, 10])
5384 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [0, miny - r], [10, miny - r])
5385# Join offsetted_subpath together
5386# Hope there won't be any circles
5387 subpath = csp_join_subpaths(offsetted_subpath)
5388
5389# Create solid object from path and lathe_width
5390 bound = csp_simple_bound([subpath])
5391 top_start = [subpath, self.options.lathe_width + self.options.Zsafe + self.options.lathe_fine_cut_width]
5392 top_end = [subpath[-1], self.options.lathe_width + self.options.Zsafe + self.options.lathe_fine_cut_width]
5393
5394 gcode += ("G01 {} {:f} F {:f} \n".format(z, top_start, self.tool["passing feed"]))
5395 gcode += ("G01 {} {:f} {} {:f} F {:f} \n".format(x, top_start, z, top_start, self.tool["passing feed"]))
5396
5397 subpath = csp_concat_subpaths(csp_subpath_line_to([], [top_start, subpath]), subpath)
5398 subpath = csp_subpath_line_to(subpath, [top_end, top_start])
5399
5400 width = max(0, self.options.lathe_width - max(0, bound))
5401 step = self.tool['depth step']
5402 steps = int(math.ceil(width / step))
5403 for i in range(steps + 1):
5404 current_width = self.options.lathe_width - step i
5405 intersections = []
5406 for j in range(1, len(subpath)):
5407 sp1 = subpath[j - 1]
5408 sp2 = subpath[j]
5409 intersections += [[j, k] for k incsp_line_intersection([bound - 10, current_width], [bound + 10, current_width], sp1, sp2)]
5410 intersections += [[j, k] for k incsp_line_intersection([bound - 10, current_width + step], [bound + 10, current_width + step], sp1, sp2)]
5411 parts = csp_subpath_split_by_points(subpath, intersections)
5412 for part in parts:
5413 minx, miny, maxx, maxy = csp_true_bounds([part])
5414 y = (maxy + miny) / 2
5415 if y > current_width + step:
5416 gcode += self.generate_lathe_gcode(part, layer, "passing feed")
5417 elif current_width <= y <= current_width + step:
5418 gcode += self.generate_lathe_gcode(part, layer, "feed")
5419 else:
5420# full step cut
5421 part = csp_subpath_line_to([], [part, part[-1]])
5422 gcode += self.generate_lathe_gcode(part, layer, "feed")
5423
5424 top_start = [fine_cut, self.options.lathe_width + self.options.Zsafe + self.options.lathe_fine_cut_width]
5425 top_end = [fine_cut[-1], self.options.lathe_width + self.options.Zsafe + self.options.lathe_fine_cut_width]
5426 gcode += "\n(Fine cutting start)\n(Calculating fine cut using {})\n".format(self.options.lathe_create_fine_cut_using)
5427 for i in range(int(self.options.lathe_fine_cut_count)):
5428 width = self.options.lathe_fine_cut_width (1 - float(i + 1) / self.options.lathe_fine_cut_count)
5429 if width == 0:
5430 current_pass = fine_cut
5431 else:
5432 if self.options.lathe_create_fine_cut_using == "Move path":
5433 current_pass = [[[i2, i2 + width] for i2 in i1] for i1 in fine_cut]
5434 else:
5435 minx, miny, maxx, maxy = csp_true_bounds([fine_cut])
5436 offsetted_subpath = csp_subpath_line_to(fine_cut[:], ])
5437 left = fine_cut[-1]
5438 right = fine_cut
5439 if left > right:
5440 left, right = right, left
5441 offsetted_subpath = csp_offset([offsetted_subpath], width if notcsp_subpath_ccw(offsetted_subpath) else -width)
5442 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [left, 10], [left, 0])
5443 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [right, 0], [right, 10])
5444 offsetted_subpath = csp_clip_by_line(offsetted_subpath, [0, miny - r], [10, miny - r])
5445 current_pass = csp_join_subpaths(offsetted_subpath)
5446
5447 gcode += "\n(Fine cut {:d}-th cicle start)\n".format(i + 1)
5448 gcode += ("G01 {} {:f} {} {:f} F {:f} \n".format(x, top_start, z, top_start, self.tool["passing feed"]))
5449 gcode += ("G01 {} {:f} {} {:f} F {:f} \n".format(x, current_pass, z, current_pass + self.options.lathe_fine_cut_width, self.tool["passing feed"]))
5450 gcode += ("G01 {} {:f} {} {:f} F {:f} \n".format(x, current_pass, z, current_pass, self.tool["fine feed"]))
5451
5452 gcode += self.generate_lathe_gcode(current_pass, layer, "fine feed")
5453 gcode += ("G01 {} {:f} F {:f} \n".format(z, top_start, self.tool["passing feed"]))
5454 gcode += ("G01 {} {:f} {} {:f} F {:f} \n".format(x, top_start, z, top_start, self.tool["passing feed"]))
5455
5456 self.export_gcode(gcode)
5457
5326def tab_lathe(self): …)
5458
5464
5465def tab_lathe_modify_path(self):
5466 self.get_info()
5467 if self.selected_paths == {} and self.options.auto_select_paths:
5468 paths = self.paths
5469 self.error("No paths are selected! Trying to work on all available paths.")
5470 else:
5471 paths = self.selected_paths
5472
5473 for layer in self.layers:
5474 if layer in paths:
5475 width = self.options.lathe_rectangular_cutter_width
5476 for path in paths[layer]:
5477 csp = self.transform_csp(path.path.to_superpath(), layer)
5478 new_csp = []
5479 for subpath in csp:
5480 orientation = subpath[-1] > subpath
5481 new_subpath = []
5482
5483# Split segment at x' and y' == 0
5484 for sp1, sp2 in zip(subpath[:], subpath[1:]):
5485 ax, ay, bx, by, cx, cy, dx, dy = csp_parameterize(sp1, sp2)
5486 roots = cubic_solver_real(0, 3 ax, 2 bx, cx)
5487 roots += cubic_solver_real(0, 3 ay, 2 by, cy)
5488 new_subpath = csp_concat_subpaths(new_subpath, csp_seg_split(sp1, sp2, roots))
5489 subpath = new_subpath
5490 new_subpath = []
5491 first_seg = True
5492 for sp1, sp2 in zip(subpath[:], subpath[1:]):
5493 n = csp_normalized_normal(sp1, sp2, 0)
5494 a = math.atan2(n, n)
5495 if a == 0 or a == math.pi:
5496 n = csp_normalized_normal(sp1, sp2, 1)
5497 a = math.atan2(n, n)
5498 if a != 0 and a != math.pi:
5499 o = 0 if 0 < a <= math.pi / 2 or -math.pi < a < -math.pi / 2 else 1
5500 if not orientation:
5501 o = 1 - o
5502
5503# Add first horizontal straight line if needed
5504 if not first_seg and new_subpath == []:
5505 new_subpath = for i in range(3)]]
5506
5507 new_subpath = csp_concat_subpaths(
5508 new_subpath,
5509 [
5510 for i in range(3)],
5511 for i in range(3)]
5512 ]
5513 )
5514 first_seg = False
5515
5516# Add last horizontal straight line if needed
5517 if a == 0 or a == math.pi:
5518 new_subpath += for i in range(3)]]
5519
5520 new_csp += [new_subpath]
5521 self.draw_csp(new_csp, layer)
5522
5465def tab_lathe_modify_path(self): …)
5523
5526def tab_graffiti(self):
5527 self.get_info_plus()
5528# Get reference points.
5529
5530 def get_gcode_coordinates(point, layer):
5531 gcode = ''
5532 pos = []
5533 for ref_point in self.graffiti_reference_points[layer]:
5534 c = math.sqrt((point - ref_point) 2 + (point - ref_point) 2)
5535 gcode += " {} {:f}".format(ref_point, c)
5536 pos += [c]
5537 return pos, gcode
5538
5539 def graffiti_preview_draw_point(x1, y1, color, radius=.5):
5540 self.graffiti_preview = self.graffiti_preview
5541 r, g, b, a_ = color
5542 for x in range(int(x1 - 1 - math.ceil(radius)), int(x1 + 1 + math.ceil(radius) + 1)):
5543 for y in range(int(y1 - 1 - math.ceil(radius)), int(y1 + 1 + math.ceil(radius) + 1)):
5544 if x >= 0 and y >= 0 and y < len(self.graffiti_preview) and x 4 < len(self.graffiti_preview):
5545 d = math.sqrt((x1 - x) 2 + (y1 - y) 2)
5546 a = float(a_) (max(0, (1 - (d - radius))) if d > radius else 1) / 256
5547 self.graffiti_preview[y][x 4] = int(r a + (1 - a) self.graffiti_preview[y][x 4])
5548 self.graffiti_preview[y][x 4 + 1] = int(g a + (1 - a) self.graffiti_preview[y][x 4 + 1])
5549 self.graffiti_preview[y][x 4 + 2] = int(g b + (1 - a) self.graffiti_preview[y][x 4 + 2])
5550 self.graffiti_preview[y][x 4 + 3] = min(255, int(self.graffiti_preview[y][x 4 + 3] + a 256))
5551
5552 def graffiti_preview_transform(x, y):
5553 tr = self.graffiti_preview_transform
5554 d = max(tr - tr + 2, tr - tr + 2)
5555 return [(x - tr + 1) self.options.graffiti_preview_size / d, self.options.graffiti_preview_size - (y - tr + 1) self.options.graffiti_preview_size / d]
5556
5557 def draw_graffiti_segment(layer, start, end, feed, color=(0, 255, 0, 40), emmit=1000):
5558# Emit = dots per second
5559 l = math.sqrt(sum([(start[i] - end[i]) 2 for i in range(len(start))]))
5560 time_ = l / feed
5561 c1 = self.graffiti_reference_points[layer]
5562 c2 = self.graffiti_reference_points[layer]
5563 d = math.sqrt((c1 - c2) 2 + (c1 - c2) 2)
5564 if d == 0:
5565 raise ValueError("Error! Reference points should not be the same!")
5566 for i in range(int(time_ emmit + 1)):
5567 t = i / (time_ emmit)
5568 r1 = start (1 - t) + end t
5569 r2 = start (1 - t) + end t
5570 a = (r1 2 - r2 2 + d 2) / (2 d)
5571 h = math.sqrt(r1 2 - a 2)
5572 xa = c1 + a (c2 - c1) / d
5573 ya = c1 + a (c2 - c1) / d
5574
5575 x1 = xa + h (c2 - c1) / d
5576 x2 = xa - h (c2 - c1) / d
5577 y1 = ya - h (c2 - c1) / d
5578 y2 = ya + h (c2 - c1) / d
5579
5580 x = x1 if y1 < y2 else x2
5581 y = min(y1, y2)
5582 x, y = graffiti_preview_transform(x, y)
5583 graffiti_preview_draw_point(x, y, color)
5584
5585 def create_connector(p1, p2, t1, t2):
5586 P1 = P(p1)
5587 P2 = P(p2)
5588 N1 = P(rotate_ccw(t1))
5589 N2 = P(rotate_ccw(t2))
5590 r = self.options.graffiti_min_radius
5591 C1 = P1 + N1 r
5592 C2 = P2 + N2 r
5593# Get closest possible centers of arcs, also we define that arcs are both ccw or both not.
5594 dc, N1, N2, m = (
5595 (
5596 (((P2 - N1 r) - (P1 - N2 r)).l2(), -N1, -N2, 1)
5597 ifvectors_ccw(t1, t2) else
5598 (((P2 + N1 r) - (P1 + N2 r)).l2(), N1, N2, -1)
5599 )
5600 ifvectors_ccw((P1 - C1).to_list(), t1) == vectors_ccw((P2 - C2).to_list(), t2) else
5601 (
5602 (((P2 + N1 r) - (P1 - N2 r)).l2(), N1, -N2, 1)
5603 ifvectors_ccw(t1, t2) else
5604 (((P2 - N1 r) - (P1 + N2 r)).l2(), -N1, N2, 1)
5605 )
5606 )
5607 dc = math.sqrt(dc)
5608 C1 = P1 + N1 r
5609 C2 = P2 + N2 r
5610 Dc = C2 - C1
5611
5612 if dc == 0:
5613# can be joined by one arc
5614 returncsp_from_arc(p1, p2, C1.to_list(), r, t1)
5615
5616 cos = Dc.x / dc
5617 sin = Dc.y / dc
5618
5619 p1_end = [C1.x - r sin m, C1.y + r cos m]
5620 p2_st = [C2.x - r sin m, C2.y + r cos m]
5621 ifpoint_to_point_d2(p1, p1_end) < 0.0001 andpoint_to_point_d2(p2, p2_st) < 0.0001:
5622 return
5623
5624 arc1 = csp_from_arc(p1, p1_end, C1.to_list(), r, t1)
5625 arc2 = csp_from_arc(p2_st, p2, C2.to_list(), r, [cos, sin])
5626 returncsp_concat_subpaths(arc1, arc2)
5627
5628 if not self.check_dir():
5629 return
5630 if self.selected_paths == {} and self.options.auto_select_paths:
5631 paths = self.paths
5632 self.error("No paths are selected! Trying to work on all available paths.")
5633 else:
5634 paths = self.selected_paths
5635 self.tool = []
5636 gcode = """(Header)
5637(Generated by gcodetools from Inkscape.)
5638(Using graffiti extension.)
5639(Header end.)"""
5640
5641 minx = float("inf")
5642 miny = float("inf")
5643 maxx = float("-inf")
5644 maxy = float("-inf")
5645# Get all reference points and path's bounds to make preview
5646
5647 for layer in self.layers:
5648 if layer in paths:
5649# Set reference points
5650 if layer not in self.graffiti_reference_points:
5651 reference_points = None
5652 for i in range(self.layers.index(layer), -1, -1):
5653 if self.layers[i] in self.graffiti_reference_points:
5654 reference_points = self.graffiti_reference_points[self.layers[i]]
5655 self.graffiti_reference_points[layer] = self.graffiti_reference_points[self.layers[i]]
5656 break
5657 if reference_points is None:
5658 self.error('There are no graffiti reference points for layer {}'.format(layer), "error")
5659
5660# Transform reference points
5661 for i in range(len(self.graffiti_reference_points[layer])):
5662 self.graffiti_reference_points[layer][i] = self.transform(self.graffiti_reference_points[layer][i], layer)
5663 point = self.graffiti_reference_points[layer][i]
5664 gcode += "(Reference point {:f};{:f} for {} axis)\n".format(point, point, point)
5665
5666 if self.options.graffiti_create_preview:
5667 for point in self.graffiti_reference_points[layer]:
5668 minx = min(minx, point)
5669 miny = min(miny, point)
5670 maxx = max(maxx, point)
5671 maxy = max(maxy, point)
5672 for path in paths[layer]:
5673 csp = path.path.to_superpath()
5674 csp = self.apply_transforms(path, csp)
5675 csp = self.transform_csp(csp, layer)
5676 bounds = csp_simple_bound(csp)
5677 minx = min(minx, bounds)
5678 miny = min(miny, bounds)
5679 maxx = max(maxx, bounds)
5680 maxy = max(maxy, bounds)
5681
5682 if self.options.graffiti_create_preview:
5683 self.graffiti_preview = list([ (4 self.options.graffiti_preview_size) for _ in range(self.options.graffiti_preview_size)])
5684 self.graffiti_preview_transform = [minx, miny, maxx, maxy]
5685
5686 for layer in self.layers:
5687 if layer in paths:
5688
5689 r = re.match("\s\(\s([0-9\-,.]+)\s;\s([0-9\-,.]+)\s\)\s", self.options.graffiti_start_pos)
5690 if r:
5691 start_point = [float(r.group(1)), float(r.group(2))]
5692 else:
5693 start_point = [0., 0.]
5694 last_sp1 = [[start_point, start_point - 10] for _ in range(3)]
5695 last_sp2 = [start_point for _ in range(3)]
5696
5697 self.set_tool(layer)
5698 self.tool = self.tools[layer]
5699# Change tool every layer. (Probably layer = color so it'll be
5700# better to change it even if the tool has not been changed)
5701 gcode += ("(Change tool to {})\n".format(re.sub("\"'\(\)\\", " ", self.tool["name"]))) + self.tool["tool change gcode"] + "\n"
5702
5703 subpaths = []
5704 for path in paths[layer]:
5705# Rebuild the paths to polyline.
5706 csp = path.path.to_superpath()
5707 csp = self.apply_transforms(path, csp)
5708 csp = self.transform_csp(csp, layer)
5709 subpaths += csp
5710 polylines = []
5711 while len(subpaths) > 0:
5712 i = min([(point_to_point_d2(last_sp2, subpaths[i]), i) for i in range(len(subpaths))])
5713 subpath = subpaths[i][:]
5714 del subpaths[i]
5715 polylines += [
5716 ['connector', create_connector(
5717 last_sp2,
5718 subpath,
5719csp_normalized_slope(last_sp1, last_sp2, 1.),
5720csp_normalized_slope(subpath, subpath, 0.),
5721 )]
5722 ]
5723 polyline = []
5724 spl = None
5725
5726# remove zerro length segments
5727 i = 0
5728 while i < len(subpath) - 1:
5729 ifcspseglength(subpath[i], subpath[i + 1]) < 0.00000001:
5730 subpath[i] = subpath[i + 1]
5731 del subpath[i + 1]
5732 else:
5733 i += 1
5734
5735 for sp1, sp2 in zip(subpath, subpath[1:]):
5736 if spl is not None and abs(cross(csp_normalized_slope(spl, sp1, 1.), csp_normalized_slope(sp1, sp2, 0.))) > 0.1: # TODO add coefficient into inx
5737# We've got sharp angle at sp1.
5738 polyline += [sp1]
5739 polylines += ]
5740 polylines += [
5741 ['connector', create_connector(
5742 sp1,
5743 sp1,
5744csp_normalized_slope(spl, sp1, 1.),
5745csp_normalized_slope(sp1, sp2, 0.),
5746 )]
5747 ]
5748 polyline = []
5749# max_segment_length
5750 polyline += [sp1]
5751print_(polyline)
5752print_(sp1)
5753
5754 spl = sp1
5755 polyline += [sp2]
5756 polylines += ]
5757
5758 last_sp1 = sp1
5759 last_sp2 = sp2
5760
5761# Add return to start_point
5762 if not polylines:
5763 continue
5764 polylines += ]]
5765
5766# Make polylines from polylines. They are still csp.
5767 for i in range(len(polylines)):
5768 polyline = []
5769 l = 0
5770print_("polylines", polylines)
5771print_(polylines[i])
5772 for sp1, sp2 in zip(polylines[i], polylines[i][1:]):
5773print_(sp1, sp2)
5774 l = cspseglength(sp1, sp2)
5775 if l > 0.00000001:
5776 polyline += [sp1]
5777 parts = int(math.ceil(l / self.options.graffiti_max_seg_length))
5778 for j in range(1, parts):
5779 polyline += [csp_at_length(sp1, sp2, float(j) / parts)]
5780 if l > 0.00000001:
5781 polyline += [sp2]
5782print_(i)
5783 polylines[i] = polyline
5784
5785 t = 0
5786 last_state = None
5787 for polyline_ in polylines:
5788 polyline = polyline_
5789# Draw linearization
5790 if self.options.graffiti_create_linearization_preview:
5791 t += 1
5792 csp = for i in range(len(polyline))]
5793draw_csp(self.transform_csp([csp], layer, reverse=True))
5794
5795# Export polyline to gcode
5796# we are making transform from XYZA coordinates to R1...Rn
5797# where R1...Rn are radius vectors from graffiti reference points
5798# to current (x,y) point. Also we need to assign custom feed rate
5799# for each segment. And we'll use only G01 gcode.
5800 last_real_pos, g = get_gcode_coordinates(polyline, layer)
5801 last_pos = polyline
5802 if polyline_ == "draw"and last_state != "draw":
5803 gcode += self.tool['gcode before path'] + "\n"
5804 for point in polyline:
5805 real_pos, g = get_gcode_coordinates(point, layer)
5806 real_l = sum([(real_pos[i] - last_real_pos[i]) 2 for i in range(len(last_real_pos))])
5807 l = (last_pos - point) 2 + (last_pos - point) 2
5808 if l != 0:
5809 feed = self.tool['feed'] math.sqrt(real_l / l)
5810 gcode += "G01 " + g + " F {:f}\n".format(feed)
5811 if self.options.graffiti_create_preview:
5812 draw_graffiti_segment(layer, real_pos, last_real_pos, feed, color=(0, 0, 255, 200) if polyline_ == "draw"else (255, 0, 0, 200), emmit=self.options.graffiti_preview_emmit)
5813 last_real_pos = real_pos
5814 last_pos = point[:]
5815 if polyline_ == "draw"and last_state != "draw":
5816 gcode += self.tool['gcode after path'] + "\n"
5817 last_state = polyline_
5818 self.export_gcode(gcode, no_headers=True)
5819 if self.options.graffiti_create_preview:
5820 try:
5821# Draw reference points
5822 for layer in self.graffiti_reference_points:
5823 for point in self.graffiti_reference_points[layer]:
5824 x, y = graffiti_preview_transform(point, point)
5825 graffiti_preview_draw_point(x, y, (0, 255, 0, 255), radius=5)
5826
5827 import png
5828 writer = png.Writer(width=self.options.graffiti_preview_size, height=self.options.graffiti_preview_size, size=None, greyscale=False, alpha=True, bitdepth=8, palette=None, transparent=None, background=None, gamma=None, compression=None, interlace=False, bytes_per_sample=None, planes=None, colormap=None, maxval=None, chunk_limit=1048576)
5829 with open(os.path.join(self.options.directory, self.options.file + ".png"), 'wb') as f:
5830 writer.write(f, self.graffiti_preview)
5831
5832 except:
5833 self.error("Png module have not been found!")
5834
5526def tab_graffiti(self): …)
5835def get_info_plus(self):
5836"""Like get_info(), but checks some of the values"""
5837 self.get_info()
5838 if self.orientation_points == {}:
5839 self.error("Orientation points have not been defined! A default set of orientation points has been automatically added.")
5840 self.tab_orientation(self.layers[min(1, len(self.layers) - 1)])
5841 self.get_info()
5842 if self.tools == {}:
5843 self.error("Cutting tool has not been defined! A default tool has been automatically added.")
5844 self.options.tools_library_type = "default"
5845 self.tab_tools_library(self.layers[min(1, len(self.layers) - 1)])
5846 self.get_info()
5847
5835def get_info_plus(self): …)
5848
5855def effect(self):
5856 start_time = time.time()
5857 global options
5858 options = self.options
5859 options.self = self
5860 options.doc_root = self.documentdocument.getroot()
5861
5862# define print_ function
5863 global print_
5864 if self.options.log_create_log:
5865 try:
5866 if os.path.isfile(self.options.log_filename):
5867 os.remove(self.options.log_filename)
5868 with open(self.options.log_filename, "a") as fhl:
5869 fhl.write("""Gcodetools log file.
5870 Started at {}.
5871{}
5872""".format(time.strftime("%d.%m.%Y %H:%M:%S"), options.log_filename))
5873 except:
5874 print_ = lambda x: None
5875 else:
5876 print_ = lambda x: None
5877
5878# This automatically calls any tab_{tab_name_in_inx} which in this
5879# extension is A LOT of different functions. So see all method prefixed
5880# with tab_ to find out what's supported here.
5881 self.options.active_tab()
5882
5883print_("------------------------------------------")
5884print_("Done in {:f} seconds".format(time.time() - start_time))
5885print_("End at {}.".format(time.strftime("%d.%m.%Y %H:%M:%S")))
5886
5887
5855def effect(self): …)
5888def tab_offset(self):
5889 self.get_info()
5890 if self.options.offset_just_get_distance:
5891 for layer in self.selected_paths:
5892 if len(self.selected_paths[layer]) == 2:
5893 csp1 = self.selected_paths[layer].path.to_superpath()
5894 csp2 = self.selected_paths[layer].path.to_superpath()
5895 dist = csp_to_csp_distance(csp1, csp2)
5896print_(dist)
5897draw_pointer(list(csp_at_t(csp1[dist][dist - 1], csp1[dist][dist], dist))
5898 + list(csp_at_t(csp2[dist][dist - 1], csp2[dist][dist], dist)), "red", "line", comment=math.sqrt(dist))
5899 return
5900 if self.options.offset_step == 0:
5901 self.options.offset_step = self.options.offset_radius
5902 if self.options.offset_step self.options.offset_radius < 0:
5903 self.options.offset_step = -1
5904 time_ = time.time()
5905 offsets_count = 0
5906 for layer in self.selected_paths:
5907 for path in self.selected_paths[layer]:
5908
5909 offset = self.options.offset_step / 2
5910 while abs(offset) <= abs(self.options.offset_radius):
5911 offset_ = csp_offset(path.path.to_superpath(), offset)
5912 offsets_count += 1
5913 if offset_:
5914 for iii in offset_:
5915draw_csp([iii], width=1)
5916 else:
5917print_("------------Reached empty offset at radius {}".format(offset))
5918 break
5919 offset += self.options.offset_step
5920print_()
5921print_("-----------------------------------------------------------------------------------")
5922print_("-----------------------------------------------------------------------------------")
5923print_("-----------------------------------------------------------------------------------")
5924print_()
5925print_("Done in {}".format(time.time() - time_))
5926print_("Total offsets count {}".format(offsets_count))
5927
5928
5888def tab_offset(self): …)
2806class Gcodetools(inkex.EffectExtension): …)
5929 if name == 'main':
5930Gcodetools().run()
gcodetools.Gcodetools
Definitiongcodetools.py:2806
gcodetools.Gcodetools.selected_paths
selected_paths
Definitiongcodetools.py:2878
gcodetools.Gcodetools.graffiti_reference_points
graffiti_reference_points
Definitiongcodetools.py:3558
gcodetools.Gcodetools.tab_engraving
tab_engraving(self)
Engraving.
Definitiongcodetools.py:4350
gcodetools.Gcodetools.parse_curve
parse_curve(self, p, layer, w=None, f=None)
Definitiongcodetools.py:3123
gcodetools.Gcodetools.tab_offset
tab_offset(self)
Definitiongcodetools.py:5888
gcodetools.Gcodetools.preview_groups
preview_groups
Definitiongcodetools.py:3179
gcodetools.Gcodetools.Zcoordinates
Zcoordinates
Definitiongcodetools.py:3560
gcodetools.Gcodetools.tool
tool
Definitiongcodetools.py:5350
gcodetools.Gcodetools.tab_preferences
tab_preferences(self)
Definitiongcodetools.py:5282
gcodetools.Gcodetools.tab_area_artefacts
tab_area_artefacts(self)
Artefacts.
Definitiongcodetools.py:4035
gcodetools.Gcodetools.in_out_reference_points
in_out_reference_points
Definitiongcodetools.py:2866
gcodetools.Gcodetools.get_info
get_info(self)
Definitiongcodetools.py:3552
gcodetools.Gcodetools.apply_transforms
apply_transforms(self, g, csp, reverse=False)
Definitiongcodetools.py:3417
gcodetools.Gcodetools.transform_csp
transform_csp(self, csp_, layer, reverse=False)
Definitiongcodetools.py:3503
gcodetools.Gcodetools.paths
paths
Definitiongcodetools.py:3555
gcodetools.Gcodetools.get_orientation_points
get_orientation_points(self, g)
Definitiongcodetools.py:3648
gcodetools.Gcodetools.set_markers
set_markers(self)
Set markers.
Definitiongcodetools.py:3534
gcodetools.Gcodetools.init
init(self)
Definitiongcodetools.py:3078
gcodetools.Gcodetools.tab_about
tab_about(self)
Definitiongcodetools.py:5279
gcodetools.Gcodetools.transform
transform(self, source_point, layer, reverse=False)
Definitiongcodetools.py:3438
gcodetools.Gcodetools.add_arguments
add_arguments(self, pars)
Definitiongcodetools.py:2982
gcodetools.Gcodetools.get_info_plus
get_info_plus(self)
Definitiongcodetools.py:5835
gcodetools.Gcodetools.tab_area_fill
tab_area_fill(self)
Definitiongcodetools.py:4174
gcodetools.Gcodetools.generate_lathe_gcode
generate_lathe_gcode(self, subpath, layer, feed_type)
Lathe.
Definitiongcodetools.py:5292
gcodetools.Gcodetools.tab_plasma_prepare_path
tab_plasma_prepare_path(self)
In/out paths: TODO move it to the bottom.
Definitiongcodetools.py:2828
gcodetools.Gcodetools.footer
footer
Definitiongcodetools.py:3253
gcodetools.Gcodetools.Zauto_scale
Zauto_scale
Definitiongcodetools.py:3563
gcodetools.Gcodetools.tab_lathe
tab_lathe(self)
Definitiongcodetools.py:5326
gcodetools.Gcodetools.tools
tools
Definitiongcodetools.py:3556
gcodetools.Gcodetools.draw_csp
draw_csp(self, csp, layer=None, group=None, fill='none', stroke='#178ade', width=0.354, style=None)
Draw csp.
Definitiongcodetools.py:3154
gcodetools.Gcodetools.get_transforms
get_transforms(self, g)
Definitiongcodetools.py:3396
gcodetools.Gcodetools.generate_gcode
generate_gcode(self, curve, layer, depth)
Generate Gcode Generates Gcode on given curve.
Definitiongcodetools.py:3294
gcodetools.Gcodetools.orientation_points
orientation_points
Definitiongcodetools.py:3557
gcodetools.Gcodetools.draw_curve
draw_curve(self, curve, layer, group=None, style=MARKER_STYLE["biarc_style"])
Definitiongcodetools.py:3167
gcodetools.Gcodetools.my3Dlayer
my3Dlayer
Definitiongcodetools.py:3565
gcodetools.Gcodetools.graffiti_preview
graffiti_preview
Definitiongcodetools.py:5540
gcodetools.Gcodetools.reverse_transform
reverse_transform(self, transform)
Definitiongcodetools.py:3409
gcodetools.Gcodetools.transform_matrix_reverse
transform_matrix_reverse
Definitiongcodetools.py:3489
gcodetools.Gcodetools.header
header
Definitiongcodetools.py:3248
gcodetools.Gcodetools.tab_orientation
tab_orientation(self, layer=None)
Orientation.
Definitiongcodetools.py:5016
gcodetools.Gcodetools.get_tool
get_tool(self, g)
Definitiongcodetools.py:3696
gcodetools.Gcodetools.last_used_tool
last_used_tool
Definitiongcodetools.py:3325
gcodetools.Gcodetools.tab_options
tab_options(self)
Definitiongcodetools.py:5285
gcodetools.Gcodetools.tab_graffiti
tab_graffiti(self)
Graffiti function generates Gcode for graffiti drawer.
Definitiongcodetools.py:5526
gcodetools.Gcodetools.set_tool
set_tool(self, layer)
Definitiongcodetools.py:3727
gcodetools.Gcodetools.get_graffiti_reference_points
get_graffiti_reference_points(self, g)
Definitiongcodetools.py:3684
gcodetools.Gcodetools.tab_path_to_gcode
tab_path_to_gcode(self)
Path to Gcode.
Definitiongcodetools.py:3746
gcodetools.Gcodetools.check_tools_and_op
check_tools_and_op(self)
Check tools and OP assignment.
Definitiongcodetools.py:5215
gcodetools.Gcodetools.tab_area
tab_area(self)
Definitiongcodetools.py:4080
gcodetools.Gcodetools.layers
layers
Definitiongcodetools.py:3559
gcodetools.Gcodetools.tools_field_order
tools_field_order
Definitiongcodetools.py:3103
gcodetools.Gcodetools.export_gcode
export_gcode(self, gcode, no_headers=False)
Definitiongcodetools.py:2809
gcodetools.Gcodetools.graffiti_preview_transform
graffiti_preview_transform
Definitiongcodetools.py:5684
gcodetools.Gcodetools.default_tool
default_tool
Definitiongcodetools.py:3080
gcodetools.Gcodetools.tab_dxfpoints
tab_dxfpoints(self)
dxfpoints
Definitiongcodetools.py:4008
gcodetools.Gcodetools.transform_matrix
transform_matrix
Definitiongcodetools.py:3488
gcodetools.Gcodetools.effect
effect(self)
Effect.
Definitiongcodetools.py:5855
gcodetools.Gcodetools.check_dir
check_dir(self)
Definitiongcodetools.py:3243
gcodetools.Gcodetools.error
error(self, s, msg_type="warning")
Definitiongcodetools.py:3511
gcodetools.Gcodetools.tab_tools_library
tab_tools_library(self, layer=None)
Tools library.
Definitiongcodetools.py:5089
gcodetools.Gcodetools.transform_scalar
transform_scalar(self, x, layer, reverse=False)
Definitiongcodetools.py:3435
gcodetools.Gcodetools.tab_help
tab_help(self)
TODO Launch browser on help tab.
Definitiongcodetools.py:5269
gcodetools.Gcodetools.tab_lathe_modify_path
tab_lathe_modify_path(self)
Lathe modify path Modifies path to fit current cutter.
Definitiongcodetools.py:5465
gcodetools.Line
Definitiongcodetools.py:1626
gcodetools.Line.end
end
Definitiongcodetools.py:1633
gcodetools.Line.l2
l2(self)
Definitiongcodetools.py:1644
gcodetools.Line.offset
offset(self, r)
Definitiongcodetools.py:1640
gcodetools.Line.n
n
Definitiongcodetools.py:1636
gcodetools.Line.l
l
Definitiongcodetools.py:1634
gcodetools.Line.init
init(self, st, end)
Definitiongcodetools.py:1627
gcodetools.Line.length
length(self)
Definitiongcodetools.py:1647
gcodetools.Line.intersect
intersect(self, b)
Definitiongcodetools.py:1662
gcodetools.Line.draw
draw(self, group, style, layer, transform, num=0, reverse_angle=1)
Definitiongcodetools.py:1650
gcodetools.Line.st
st
Definitiongcodetools.py:1632
gcodetools.P
Point (x,y) operations.
Definitiongcodetools.py:1559
gcodetools.P.l2
l2(self)
Definitiongcodetools.py:1622
gcodetools.P.neg
neg(self)
Definitiongcodetools.py:1574
gcodetools.P.angle
angle(self)
Definitiongcodetools.py:1607
gcodetools.P.mul
mul(self, other)
Definitiongcodetools.py:1577
gcodetools.P.rot
rot(self, theta)
Definitiongcodetools.py:1602
gcodetools.P.repr
repr(self)
Definitiongcodetools.py:1610
gcodetools.P.unit
unit(self)
Definitiongcodetools.py:1593
gcodetools.P.sub
sub(self, other)
Definitiongcodetools.py:1571
gcodetools.P.div
div(self, other)
Definitiongcodetools.py:1584
gcodetools.P.add
add(self, other)
Definitiongcodetools.py:1568
gcodetools.P.ccw
ccw(self)
Definitiongcodetools.py:1619
gcodetools.P.dot
dot(self, other)
Definitiongcodetools.py:1599
gcodetools.P.to_list
to_list(self)
Definitiongcodetools.py:1616
gcodetools.P.mag
mag(self)
Definitiongcodetools.py:1590
gcodetools.P.init
init(self, x, y=None)
Definitiongcodetools.py:1560
gcodetools.P.x
x
Definitiongcodetools.py:1562
gcodetools.P.truediv
truediv(self, other)
Definitiongcodetools.py:1587
gcodetools.P.pr
pr(self)
Definitiongcodetools.py:1613
gcodetools.P.y
y
Definitiongcodetools.py:1563
gcodetools.Polygon
Definitiongcodetools.py:2477
gcodetools.Polygon.centroid
centroid(self)
Definitiongcodetools.py:2538
gcodetools.Polygon.add
add(self, add)
Definitiongcodetools.py:2619
gcodetools.Polygon.drop_into_direction
drop_into_direction(self, direction, surface)
Definitiongcodetools.py:2521
gcodetools.Polygon.hull
hull(self)
Definitiongcodetools.py:2645
gcodetools.Polygon.drop_down
drop_down(self, surface, zerro_plane=True)
Definitiongcodetools.py:2568
gcodetools.Polygon.point_inside
point_inside(self, p)
Definitiongcodetools.py:2625
gcodetools.Polygon.rotate_
rotate_(self, sin, cos)
Definitiongcodetools.py:2508
gcodetools.Polygon.draw
draw(self, color="#075", width=.1, group=None)
Definitiongcodetools.py:2615
gcodetools.Polygon.move
move(self, x, y)
Definitiongcodetools.py:2481
gcodetools.Polygon.rotate
rotate(self, a)
Definitiongcodetools.py:2516
gcodetools.Polygon.polygon
polygon
Definitiongcodetools.py:2479
gcodetools.Polygon.bounds
bounds(self)
Definitiongcodetools.py:2487
gcodetools.Polygon.init
init(self, polygon=None)
Definitiongcodetools.py:2478
gcodetools.Polygon.width
width(self)
Definitiongcodetools.py:2504
gcodetools.Postprocessor
Definitiongcodetools.py:2209
gcodetools.Postprocessor.scale
scale(self, parameters)
Definitiongcodetools.py:2440
gcodetools.Postprocessor.remap
remap(self, parameters, case_sensitive=False)
Definitiongcodetools.py:2263
gcodetools.Postprocessor.re_sub_on_gcode_lines
re_sub_on_gcode_lines(self, parameters)
Definitiongcodetools.py:2247
gcodetools.Postprocessor.scale
scale
Definitiongcodetools.py:2215
gcodetools.Postprocessor.move
move
Definitiongcodetools.py:2216
gcodetools.Postprocessor.transform
transform(self, move, scale)
Definitiongcodetools.py:2286
gcodetools.Postprocessor.flip_axis
flip_axis(self, parameters)
Definitiongcodetools.py:2462
gcodetools.Postprocessor.error
error
Definitiongcodetools.py:2211
gcodetools.Postprocessor.round_coordinates
round_coordinates(self, parameters)
Definitiongcodetools.py:2421
gcodetools.Postprocessor.parameterize
parameterize(self, parameters)
Definitiongcodetools.py:2338
gcodetools.Postprocessor.flip_axis
flip_axis
Definitiongcodetools.py:2217
gcodetools.Postprocessor.remapi
remapi(self, parameters)
Definitiongcodetools.py:2260
gcodetools.Postprocessor.init
init(self, error_function_handler)
Definitiongcodetools.py:2210
gcodetools.Postprocessor.round_coordinates
round_coordinates
Definitiongcodetools.py:2219
gcodetools.Postprocessor.functions
functions
Definitiongcodetools.py:2212
gcodetools.Postprocessor.move
move(self, parameters)
Definitiongcodetools.py:2452
gcodetools.Postprocessor.remapi
remapi
Definitiongcodetools.py:2214
gcodetools.Postprocessor.parameterize
parameterize
Definitiongcodetools.py:2220
gcodetools.Postprocessor.gcode
gcode
Definitiongcodetools.py:2249
gcodetools.Postprocessor.remap
remap
Definitiongcodetools.py:2213
gcodetools.Postprocessor.parse_command
parse_command(self, command)
Definitiongcodetools.py:2235
inkex.base.InkscapeExtension.run
run(self, args=None, output=output_unspecified)
Definitionbase.py:222
inkex.base.InkscapeExtension.document
document
Definitionbase.py:75
inkex.base.InkscapeExtension.arg_method
arg_method(self, prefix="method")
Definitionbase.py:113
inkex.base.InkscapeExtension.options
options
Definitionbase.py:74
inkex.base.SvgOutputMixin.document
document
Definitionbase.py:538
inkex.colors.color.Color
Definitioncolor.py:44
inkex.elements._groups.Group
Definition_groups.py:119
inkex.elements._groups.Marker
Definition_groups.py:174
inkex.elements._meta.Desc
Definition_meta.py:73
inkex.elements._meta.Title
Definition_meta.py:79
inkex.elements._polygons.PathElement
Definition_polygons.py:78
inkex.elements._text.TextElement
Definition_text.py:112
inkex.elements._text.Tspan
Definition_text.py:199
inkex.extensions.EffectExtension
Definitionextensions.py:76
inkex.styles.Style
Definitionstyles.py:88
inkex.transforms.Transform
Definitiontransforms.py:333
inkex.utils.AbortExtension
Definitionutils.py:178
gcodetools.csp_clip_by_line
csp_clip_by_line(csp, l1, l2)
Definitiongcodetools.py:1184
gcodetools.csp_join_subpaths
csp_join_subpaths(csp)
Definitiongcodetools.py:1226
gcodetools.csp_close_all_subpaths
csp_close_all_subpaths(csp, tolerance=0.000001)
Definitiongcodetools.py:292
gcodetools.csp_at_length
csp_at_length(sp1, sp2, l=0.5, tolerance=0.01)
Definitiongcodetools.py:769
gcodetools.rotate_ccw
rotate_ccw(d)
Definitiongcodetools.py:1362
gcodetools.csp_subpaths_end_to_start_distance2
csp_subpaths_end_to_start_distance2(s1, s2)
Definitiongcodetools.py:1180
gcodetools.cspseglength
cspseglength(sp1, sp2, tolerance=0.01)
Definitiongcodetools.py:775
gcodetools.csp_max_curvature
csp_max_curvature(sp1, sp2)
Definitiongcodetools.py:656
gcodetools.xrange
xrange
Definitiongcodetools.py:81
gcodetools.line_to_line_distance_2
line_to_line_distance_2(p1, p2, p3, p4)
Definitiongcodetools.py:1073
gcodetools.line_line_intersection_points
line_line_intersection_points(p1, p2, p3, p4)
Definitiongcodetools.py:1013
gcodetools.point_to_point_d2
point_to_point_d2(a, b)
Definitiongcodetools.py:1051
gcodetools.csp_to_arc_distance
csp_to_arc_distance(sp1, sp2, arc1, arc2, tolerance=0.01)
Definitiongcodetools.py:953
gcodetools.csp_split
csp_split(sp1, sp2, t=.5)
Definitiongcodetools.py:461
gcodetools.bounds_intersect
bounds_intersect(a, b)
Definitiongcodetools.py:1321
gcodetools.biarc
biarc(sp1, sp2, z1, z2, depth=0)
Biarc function.
Definitiongcodetools.py:2087
gcodetools.csp_concat_subpaths
csp_concat_subpaths(s)
Definitiongcodetools.py:1159
gcodetools.csp_at_t
csp_at_t(sp1, sp2, t)
Definitiongcodetools.py:737
gcodetools.between
between(c, x, y)
Definitiongcodetools.py:1492
gcodetools.small
small(a)
Definitiongcodetools.py:1388
gcodetools.bez_split
bez_split(a, t=0.5)
Definitiongcodetools.py:1331
gcodetools.csp_seg_to_point_distance
csp_seg_to_point_distance(sp1, sp2, p, sample_points=5)
Definitiongcodetools.py:336
gcodetools.ireplace
ireplace(self, old, new, count=0)
Definitiongcodetools.py:84
gcodetools.draw_text
draw_text(text, x, y, group=None, style=None, font_size=10, gcodetools_tag=None)
Definitiongcodetools.py:1402
gcodetools.rotate_cw
rotate_cw(d)
Definitiongcodetools.py:1366
gcodetools.gcode_comment_str
gcode_comment_str(s, replace_new_line=False)
Gcode additional functions.
Definitiongcodetools.py:193
gcodetools.dot
dot(a, b)
Definitiongcodetools.py:1358
gcodetools.csp_true_bounds
csp_true_bounds(csp)
Definitiongcodetools.py:481
gcodetools.csp_remove_zero_segments
csp_remove_zero_segments(csp, tolerance=1e-7)
Definitiongcodetools.py:216
gcodetools.csp_seg_to_csp_seg_distance
csp_seg_to_csp_seg_distance(sp1, sp2, sp3, sp4, dist_bounds=(0, 1e100), sample_points=5, tolerance=.01)
Definitiongcodetools.py:364
gcodetools.csp_segment_to_bez
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Definitiongcodetools.py:319
gcodetools.cubic_solver_real
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Definitiongcodetools.py:1496
gcodetools.arc_from_c_s_l
arc_from_c_s_l(c, s, l)
Definitiongcodetools.py:882
gcodetools.marker_style
marker_style(stroke, marker='DrawCurveMarker', width=1)
Definitiongcodetools.py:125
gcodetools.csp_split_by_two_points
csp_split_by_two_points(sp1, sp2, t1, t2)
Definitiongcodetools.py:809
gcodetools.csp_line_intersection
csp_line_intersection(l1, l2, sp1, sp2)
Definitiongcodetools.py:780
gcodetools.arc_from_s_r_n_l
arc_from_s_r_n_l(s, r, n, l)
Definitiongcodetools.py:876
gcodetools.bez_bound
bez_bound(bez)
Definitiongcodetools.py:1312
gcodetools.csp_reverse
csp_reverse(csp)
Definitiongcodetools.py:1103
gcodetools.csp_seg_split
csp_seg_split(sp1, sp2, points)
Definitiongcodetools.py:829
gcodetools.csp_segments_true_intersection
csp_segments_true_intersection(sp1, sp2, sp3, sp4)
Definitiongcodetools.py:602
gcodetools.cross
cross(a, b)
Definitiongcodetools.py:1354
gcodetools.csp_normalized_normal
csp_normalized_normal(sp1, sp2, t)
Definitiongcodetools.py:1150
gcodetools.csp_to_csp_distance
csp_to_csp_distance(csp1, csp2, dist_bounds=(0, 1e100), tolerance=.01)
Definitiongcodetools.py:440
gcodetools.csp_normalized_slope
csp_normalized_slope(sp1, sp2, t)
Definitiongcodetools.py:1112
gcodetools.csp_from_arc
csp_from_arc(start, end, center, r, slope_st)
Definitiongcodetools.py:895
gcodetools.csp_point_inside_bound
csp_point_inside_bound(sp1, sp2, p)
Definitiongcodetools.py:971
gcodetools.point_to_point_d
point_to_point_d(a, b)
Definitiongcodetools.py:1055
gcodetools.point_inside_csp
point_inside_csp(p, csp, on_the_path=True)
Definitiongcodetools.py:229
gcodetools.vectors_ccw
vectors_ccw(a, b)
Definitiongcodetools.py:1370
gcodetools.csp_segment_convex_hull
csp_segment_convex_hull(sp1, sp2)
Definitiongcodetools.py:1261
gcodetools.csp_seg_bound_to_csp_seg_bound_max_min_distance
csp_seg_bound_to_csp_seg_bound_max_min_distance(sp1, sp2, sp3, sp4)
Definitiongcodetools.py:1083
gcodetools.csp_to_point_distance
csp_to_point_distance(csp, p, dist_bounds=(0, 1e100))
Definitiongcodetools.py:323
gcodetools.csp_simple_bound
csp_simple_bound(csp)
Definitiongcodetools.py:303
gcodetools.bez_bounds_intersect
bez_bounds_intersect(bez1, bez2)
Bezier additional functions.
Definitiongcodetools.py:1308
gcodetools.point_to_line_segment_distance_2
point_to_line_segment_distance_2(p1, p2, p3)
Definitiongcodetools.py:1059
gcodetools.straight_segments_intersection
straight_segments_intersection(a, b, true_intersection=True)
Definitiongcodetools.py:1469
gcodetools.csp_subpath_ccw
csp_subpath_ccw(subpath)
Definitiongcodetools.py:722
gcodetools.draw_csp
draw_csp(csp, stroke="#f00", fill="none", comment="", width=0.354, group=None, style=None)
Definitiongcodetools.py:1423
gcodetools.inv_2x2
inv_2x2(a)
Common functions.
Definitiongcodetools.py:1378
gcodetools.csp_segments_intersection
csp_segments_intersection(sp1, sp2, sp3, sp4)
csp_segments_intersection(sp1,sp2,sp3,sp4)
Definitiongcodetools.py:522
gcodetools.print_
print_(arg)
print_ prints any arguments into specified log file
Definitiongcodetools.py:1548
gcodetools.triangle_cross
triangle_cross(a, b, c)
Definitiongcodetools.py:1257
gcodetools.csp_curvature_at_t
csp_curvature_at_t(sp1, sp2, t, depth=3)
Definitiongcodetools.py:687
gcodetools.cubic_solver
cubic_solver(a, b, c, d)
Definitiongcodetools.py:1509
gcodetools.normalize
normalize(xy)
Some vector functions.
Definitiongcodetools.py:1345
gcodetools.csp_subpath_split_by_points
csp_subpath_split_by_points(subpath, points)
Definitiongcodetools.py:844
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Definitiongcodetools.py:1439
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csp_get_t_at_curvature(sp1, sp2, c, sample_points=16)
Definitiongcodetools.py:614
gcodetools.csp_offset
csp_offset(csp, r)
Definitiongcodetools.py:1716
gcodetools.csp_subpath_line_to
csp_subpath_line_to(subpath, points, prepend=False)
Definitiongcodetools.py:1204
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point_to_arc_distance(p, arc)
Definitiongcodetools.py:929
gcodetools.line_line_intersect
line_line_intersect(p1, p2, p3, p4)
Definitiongcodetools.py:988
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csp_from_polyline(line)
Cubic Super Path additional functions.
Definitiongcodetools.py:212
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Solving Rational Equations: Relationship to Graphs
IntroHarder Probs
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Solve the following equation:
To figure out the common denominator for these fractions, I'll first need to factor that quadratic in the denominator on the right-hand side of the rational equation. This will also allow me to find the disallowed values for this equation. Factoring gives me:
x2 − 6x + 8 = (x − 4)(x − 2)
The factors of the quadratic on the right-hand side "just so happen" to be duplicates of the other denominators. This often happens in these exercises. (So often, in fact, that if you get completely different factors, you should probably go back and check your work.)
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Solving Rational Equations
So the lowest common denominator for this equation will be (x − 4)(x − 2), and I'll need to remember (at the end of my work) that x ≠ 2, 4.
Method 1: I can convert everything to the common denominator and then solve the numerators:
(x2 − 4x) + (x − 2) = 2
x2 − 4x + x − 2 = 2
x2 − 3x − 4 = 0
(x − 4)(x + 1) = 0
x = 4 or x = −1
Method 2: On the other hand, I can multiply through on both sides by the common denominator, and then solve the resulting equation. I've used color below to highlight the bits that cancel off:
x(x − 4) + 1(x − 2) = 2
x2 − 4x + x − 2 = 2
x2 − 3x − 2 = 2
x2 − 3x − 4 = 0
(x − 4)(x + 1) = 0
x = 4 or x = −1
By either method, I get the same result: x −1, 4.
However, when I check these solutions against the denominators of the original equation, I see that "x = 4" would cause division by zero, so I throw that solution out. Then my final answer is:
x = −1
Content Continues Below
I've mentioned previously that the solution process for rational equations can create solutions that don't actually exist. This is because the process, at some point, gets rid of (or at least ignores) the denominators. When this happens, we've done what is called "an irreversible step". That is, we've done something that gets rid of information, and there's no way to get it back (other than returning to the beginning again).
We can see the results of this "irreversible" step graphically.
When solving these rational equations like these, we can think of ourselves as trying to find the intersections of the related functions on either side of the "equals" sign. For instance, let's return to the equation we just solved:
Let each side of the equation be its own function:
Graphing these, we can see that the two functions intersect in one spot:
This one spot where the two functions intersect is the solution, x = −1, that we found earlier for the original rational equation.
If you have a graphing calculator, you can check your answers by doing a quick graph, because the solutions to the equations are the intersections of the graphs of the related functions. Do the graphs, and confirm that your solution values match the x-values of the intersection points on the graphs. Or, if they don't match, you know you need to check your work!
But remember how we intially came up with two solutions? That was because we'd gotten to a stage where we were ignoring the denominators. If we ignore the denominators, we get the following new functions:
y3 = x2 − 3x − 2
y4 = 2
These new functions graph like this:
As you can see, the irreversible step of getting rid of the denominators not only eliminated the vertical asymptotes, but it also created an additional (and wrong) solution.
Because this irreversible step is susceptible to the creation of extraneous solutions, you must always check your solutions for rational equations against the original equation's denominators, throwing out any solutions that would cause division by zero.
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Solving rational equations is pretty straightforward if you are careful to write each step completely. But (warning!) as soon as you start skipping steps or doing stuff in your head, you're going to start messing up. So always work neatly and completely. And never forget to check your solutions, because I can just about guarantee that you'll have one of those "no solution" (or "only one of these solutions actually works — ha, ha!") problems on your next test.
You can use the Mathway widget below to practice solving a rational equation. Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.
Please accept "preferences" cookies in order to enable this widget.
(Clicking on "Tap to view steps" on the widget's answer screen will take you to the Mathway site for a paid upgrade.)
URL:
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11512 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5808417/ | Cryptococcosis - PMC
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Infect Dis Clin North Am
. Author manuscript; available in PMC: 2018 Feb 12.
Published in final edited form as: Infect Dis Clin North Am. 2016 Mar;30(1):179–206. doi: 10.1016/j.idc.2015.10.006
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Cryptococcosis
Eileen K Maziarz
Eileen K Maziarz, MD
1 Division of Infectious Diseases and International Health, Department of Medicine, Duke University Medical Center, DUMC Box 102359, 315 Trent Drive, Durham, NC 27710, USA
Find articles by Eileen K Maziarz
1,, John R Perfect
John R Perfect, MD
1 Division of Infectious Diseases and International Health, Department of Medicine, Duke University Medical Center, DUMC Box 102359, 315 Trent Drive, Durham, NC 27710, USA
Find articles by John R Perfect
1
Author information
Article notes
Copyright and License information
1 Division of Infectious Diseases and International Health, Department of Medicine, Duke University Medical Center, DUMC Box 102359, 315 Trent Drive, Durham, NC 27710, USA
Corresponding author.eileen.maziarz@dm.duke.edu
Keywords: Cryptococcosis, Opportunistic mycoses, HIV/AIDS, Solid organ transplantation (SOT), Central nervous system (CNS) infection, Immune reconstitution inflammatory syndrome (IRIS)
PMC Copyright notice
PMCID: PMC5808417 NIHMSID: NIHMS914122 PMID: 26897067
The publisher's version of this article is available at Infect Dis Clin North Am
INTRODUCTION
Cryptococcosis is an infectious disease with worldwide distribution and wide array of clinical presentations caused by pathogenic encapsulated yeasts in the genus Cryptococcus. Currently, there are 2 species of Cryptococcus that commonly cause disease in humans: Cryptococcus neoformans and Cryptococcus gattii. C neoformans was first identified as a human pathogen in the late 19th century, but was not recognized as a common cause of human disease until the late 1970s.1,2 Over the last several decades, as vulnerable populations have expanded, cryptococcal meningitis became an infection of global importance, with up to 1 million new infections annually and significant attributable morbidity and mortality, especially among patients with human immunodeficiency virus (HIV) infection and AIDS.3 Although C neoformans and C gattii share many features of a highly evolved, environmentally savvy yeast, there are important species- and strain-specific differences with respect to geographic distribution, environmental niches, host predilection, and clinical manifestations that should be emphasized. As molecular techniques of identification have evolved, we have gained further insight into the pathobiology of these encapsulated yeasts, and their capacity to adapt to environmental pressures, exploit new geographic environments, and cause disease in both immunocompromised and apparently immunocompetent hosts.4 Despite increased availability of and success with antiretroviral therapy (ART), the worldwide burden of and mortality associated cryptococcal disease remains unacceptably high, and novel strategies of screening and preemptive therapy offer great promise at making a sustained and much needed impact on this sugarcoated opportunistic mycosis.
THE PATHOGENS: CRYPTOCOCCUS NEOFORMANS AND CRYPTOCOCCUS GATTII
Cryptococcus is a genus of basidiomycetous fungi with more than 30 species ubiquitously distributed in the environment. There are only 2 species commonly known to cause human disease, C neoformans and C gattii. The epidemiology of C neoformans is well-characterized and this organism causes disease in both immunocompromised and apparently immunocompetent hosts. C gattii, conversely, has historically been regarded as a pathogen of apparently immunocompetent patients. However, preexisting conditions and immunocompromised states, including subclinical immune defects, are also reported as risk factors for infection with this species.5–8 These species differences in clinical presentation may be primarily determined by variable host predilections, but may also be better characterized as we further our understanding of molecular subtypes.9–12
Historically, the genus was further classified into 3 varieties, 5 serotypes (based on structural differences in the polysaccharide capsule), and 8 molecular subtypes (Table 1). Molecular methods of identification have enhanced our appreciation for the significant genetic diversity among the C gattii–C neoformans complex and have called into question the current 2 species classification system. Recent proposed taxonomy changes based on the understanding of molecular studies have divided the pathogenic cryptococcal species from their classic divisions into better-defined molecular and genetic divisions. At present, the following divisions have been proposed: C neoformans var. grubii (serotype A) with 3 genotypes (VNI, VNII, VNB); C neoformans var. neoformans (serotype D or VNIV); and 5 other cryptic species, C gattii, C bacillisporus, C deuterogattii, C tetragattii, and C decagattii (serotypes B/C or VGI-IV).13 Phylogenetic analyses, combined with recognized heterogeneity with respect to virulence, host preference, and antifungal susceptibility do provide evidence to support further taxonomic classification into a 7-species/4 hybrid species scheme (Table 2). The molecular taxonomy of cryptococcal species is a vibrant area of evolution that has allowed for a greater understanding of specific strain characteristics, including fitness and predilection for certain environmental niches13; clinical correlations have yet to match this molecular precision, however, and for this review we will tend to lump the yeasts into their historical species designations, C neoformans and C gattii.
Table 1.
Current classification of pathogenic Cryptococcus species
| Serotype | Species and Varieties | Molecular Types |
:---
| A | C neoformans var. grubiia | VN I, VN II |
| B | C gattii | VG I, VG II, VG III, VG IV |
| C | C gattii | VG I, VG II, VG III, VG IV |
| D | C neoformans var. neoformans | VN IV |
| AD | C neoformans | VN III |
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a
Responsible for the vast majority of disease owing to C neoformans worldwide.
Adapted from Hagen F, Khayhan K, Theelen B, et al. Recognition of seven species in the Cryptococcus gatti/Cryptococcus neoformans species complex. Fungal Genet Biol 2015;78:17.
Table 2.
Proposed taxonomy changes for the Cryptococcus neoformans/C gattii complex
| Current Species Name | Genotype by RFLP | Proposed Species Name |
:---
| C neoformans var. grubii | VNI | C neoformans |
| |
| | VNII | |
| | VNIII | |
| |
| C neoformans var. neoformans | VNIV | C deneoformans |
| |
| C neoformans intervariety hybrid | VNIII | C neoformans × C deneoformans hybrid |
| |
| C gattii | VGI | C gattii |
| | VGIII | C bacillisporus |
| | VGII | C deuterogattii |
| | VGIV | C tetragattii |
| | VGIV/VGIIIc | C decagattii |
| |
| C neoformans var. neoformans × C gattii AFLP4/VGI hybrid | — | C deneoformans × C gattii hybrid |
| |
| C neoformans var. grubii × C gattii AFLP4/VGI hybrid | — | C neoformans × C gattii hybrid |
| |
| C neoformans var. grubii × C gattii AFLP6/VGII hybrid | — | C deneoformans × C deuterogattii hybrid |
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Adapted from Hagen F, Khayhan K, Theelen B, et al. Recognition of seven species in the Cryptococcus gatti/Cryptococcus neoformans species complex. Fungal Genet Biol 2015;78:17.
Approximately 95% of cryptococcal infections are caused by C neoformans (serotype A) strains with the remaining 4% to 5% of infections caused by C neoformans (serotype D) or C gattii (serotypes B/C strains). Whereas C neoformans var. grubii (serotype A) is found worldwide, C neoformans var neoformans (serotype D) is primarily observed in European countries and C gattii has historically been geographically restricted to tropical and subtropical regions, such as southern California, Hawaii, Brazil, Australia, Southeast Asia, and central Africa. More recently, C gattii has been identified in temperate climates such as Vancouver Island and the Pacific Northwest region of the United States and parts of Europe, suggesting an ecological shift possibly related to global temperature and moisture changes.4,10–12 Although C gattii causes up to 15% of all cases of cryptococcosis in Australia and New Zealand, C neoformans remains the predominant species even in these endemic areas.14 In certain areas of Africa around Botswana, where C neoformans and C gattii live together in the environment, active sexual recombination has been reported.15 Although outbreaks of cryptococcosis are ongoing among immunocompromised populations worldwide, to date only C gattii strains have been reported to produce a geographically defined outbreak of disease.4
C neoformans is found throughout the world in association with excreta from certain birds such as pigeons,16 environmental scavengers such as ameba and sowbugs,17,18 and in a variety of tree species in their hollows. C gattii is commonly associated with several species of eucalyptus trees in tropical and subtropical climates.19 However, recently as it has emerged as an important pathogen capable of widespread outbreaks within new geographic niches including British Columbia and the Pacific Northwest United States,4,10–12 it has been associated with temperate trees, such as firs and oaks.9,20–22
The life cycle of Cryptococcus involves both asexual and sexual forms.23 The asexual form is the haploid encapsulated yeast that reproduces by mitosis with narrow-based budding and is found in clinical and environmental specimens. The sexual state is observed at present under certain laboratory conditions, resulting in meiosis between 2 mating types (MATa and MATα) to form clamp connections, basidia and basidiospores. The α mating type strains represent the vast majority of clinical and environmental isolates, probably related to their ability to produce haploid fruiting. Even same sex mating between 2 strains of the same type (MATα–MATα) does occur and is thought to produce the infectious spores that cause human infection.24,25 This nonclassical mating between 2 α–α strains allows for further genetic diversity and is implicated in the production of hypervirulent, clonal strains responsible for the C gattii outbreak on Vancouver Island, suggesting that such mechanisms may confer the yeast the ability to exploit new geographic niches.26,27 Furthermore, there are locations in Botswana where there are equal proportions of MATα and MATa isolates in both environmental and clinical populations, providing evidence that sexual recombination remains active even with the spread worldwide of relatively clonal strains.15,28
EPIDEMIOLOGY AND RISK FACTORS
Cryptococcosis was considered an uncommon infection before the AIDS pandemic; however, it was an awakening mycosis giant in the 1970s because it was associated with malignancy, organ transplantation, and certain immunosuppressive treatments. The incidence of disease increased significantly in the mid 1980s, with HIV/AIDS accounting for more than 80% of cryptococcosis cases worldwide.29–31 Cryptococcal meningitis preferentially occurs in persons with impaired cell-mediated immunity and is a major AIDS-related opportunistic infection as the CD4+ cell count falls below 100 cells/μL. With widespread implementation of successful antiretroviral therapy (ART), the incidence of HIV-associated cryptococcosis has decreased significantly in most developed nations, although the incidence in other at-risk populations has not changed (Table 3).32 Furthermore, the prevalence of and morbidity and mortality associated with cryptococcal meningitis remain unacceptably high in settings where access to ART and other necessary health care resources are limited, specifically sub-Saharan Africa and parts of Asia. In fact, mortality peaked at approximately 600,000 deaths per year in the first decade of the 21st century; even today, it is likely that cryptococcal meningitis–related deaths approach several hundred thousand per year.3 Although both C neoformans and C gattii can also cause disease in apparently immunocompetent hosts, the percentage of infections owing to C gattii in such patients is significantly higher than for C neoformans.
Table 3.
Risk factors for Cryptococcus infection
| HIV infection | Rheumatologic diseasesa Systemic lupus erythematosus Rheumatoid arthritis |
:--- |
| Corticosteroid and/or immunosuppressive therapies | Idiopathic CD4+ lymphopenia |
| Solid organ transplantationa | Chronic liver disease (decompensated)b |
| Malignant and lymphoproliferative disorders a,b | Renal failure and/or peritoneal dialysis |
| Sarcoidosis | Hyper-IgM syndrome or hyper-IgE syndrome |
| Treatment with monoclonal antibodies (etanercept, infliximab, alemtuzumab) | Diabetes mellitusc |
| Anti-GM CSF antibodies | — |
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Abbreviations: GM CSF, granulocyte macrophage colony stimulating factor; HIV, human immunodeficiency virus; Ig, immunoglobulin.
a
Immunosuppression for these conditions may influence risk.
b
Poor prognosis especially among patients with hematologic malignancy.32
c
Historically considered a risk factor but may reflect the frequency of condition rather than specific risk to an individual. Not found to be a risk factor in.190,191
Adapted from Casadevall A, Perfect JR. Cryptococcus neoformans. Washington, DC: ASM Press; 1998.
Pathogenesis and Host Immunity
Cryptococcal infection occurs primarily by inhalation of the infectious propagules (either poorly encapsulated yeast cells or basidiospores) from environmental reservoirs with deposition into pulmonary alveoli. Traumatic inoculation into tissues has been described33 and may occur infrequently. The yeast may potentially enter via the gastrointestinal tract, although this entry is less consistent. Primary pulmonary infection is generally thought to be asymptomatic or minimally symptomatic despite high rates of serologic reactivity in children in certain urban settings.34 Clearance of the infection by the host may occur. However, in many individuals, after yeasts are deposited in alveoli, they encounter alveolar macrophages, which play a central role in the immune response.35 Host response to cryptococcal infection primarily involves a helper T cell response with cytokines including tumor necrosis factor (TNF), interferon-γ, and interleukin-2, resulting in granulomatous inflammation.36 In many circumstances, this yeast will establish a latent infection within phagolysosome, with dormant (yet viable) yeasts within the thoracic lymph nodes or a pulmonary granuloma that can persist in an asymptomatic individual for years. When local immunity is suppressed, the yeast can grow and disseminate outside these pulmonary lymph node complexes similar to the pathophysiology that is observed in cases of reactivation tuberculosis or histoplasmosis.31,37 In some hosts, C gattii disease seems to be more likely than C neoformans disease to present as a progressive granulomatous pulmonary infection, but less likely to disseminate to the central nervous system (CNS). This general observation has been made in human outbreaks and characterized in mouse models, but there remains substantial overlap between species.12,31,38 In a patient with severely compromised cellular immunity, the yeasts reactivate and can proliferate at the site of initial infection and can disseminate within phagocytes or as yeast cells and gain access to other body sites.39 Both direct invasion of the blood–brain barrier via transcytosis of free yeast forms through a series of mechanisms between yeast and host factors40 and/or transport via macrophages into the CNS (the “Trojan horse” mechanism) seem to occur.41–43 Whether certain immune states permit additional body sites of latency (eg, the CNS or prostate) have not yet been elucidated fully.
Advances in the molecular biology of Cryptococcus have confirmed multiple yeast virulence factors.44 The 3 classical and prominent virulence factors of C neoformans include capsule formation, melanin pigment production, and thermotolerance.23,36 The prominent antiphagocytic polysaccharide capsule, which is composed of glucuronoxylomannan, is unique to Cryptococcus species and is considered an essential virulence factor that has multiple effects on host immunity and can increase in size with exposure to body tissues and fluids.45,46 In addition, C neoformans possesses an enzyme that catalyzes the conversion of diphenolic compounds to form melanin, which, when expressed, may have a biological role to protect the yeasts from host oxidative stresses and which may partially explain the organism’s neurotropism into sites with high concentrations of the diphenolic catecholamines. Finally, the ability to grow at 37°C is a basic part of the virulence composite for most pathogenic fungi in humans including Cryptococcus, and molecular studies have linked high temperature growth with multiple signaling pathways and enzymes that this yeast has acquired or adapted to over time to retain or enhance its mammalian pathogenicity. Other virulence factors include phospholipase and urease production and multiple enzymes associated with protection against oxidative stresses, conferring survival within the phagolysosome.44 It is estimated that more than 100 genes are important for optimal fitness of the yeast in mammalian hosts. The yeast has even adapted sophisticated mechanisms to escape the intracellular environment by modifying the permeability of the phagosome membrane and via nonlytic exocytosis (vomocytosis), allowing cell-to-cell or host compartment transfer of yeast ant its virulence factors without damage to the host macrophages.47,48
The many factors in the immunologic responses to Cryptococcus cannot be covered completely in this review, but several observations can be made. First, exposure is frequent and the healthy immunocompetent individual is generally resistant to cryptococcal disease. In fact, even in this group, some apparently normal hosts with cryptococcosis have been found to possess anti-granulocyte macrophage colony stimulating factor antibodies as a potential immune defect.7,8 Second, the effective immune response is through a helper T cell–supported reaction and anything that weakens it may let cryptococci survive and thrive. This includes destruction of CD4+ cells by HIV, reduction of TNF activity by anti-TNF inhibitors, or the multifaceted immune suppressant effect of corticosteroids. From activated macrophages and not alternative macrophages to the development of protective antibodies over nonprotective antibodies, immunity changes over the course of cryptococcal infections. In fact, even some of our protective host mechanisms might be used against us as surfactant D may be coopted by Cryptococcus to gain entry into the lung.49 Clearly, cryptococcosis emphasizes the Goldilocks paradigm of immunity. It produces disease when immunity is too little or too much, but when the human host immunity is just right, disease does not appear.
CLINICAL MANIFESTATIONS
C neoformans and C gattii have a major predilection for establishing clinical disease in the lungs and CNS. Other less frequent body sites of infection include skin, prostate, eyes, and bone/joints. However, it should be emphasized that this yeast can widely disseminate and infect most organs in severely immunosuppressed patients and thus has the ability to appear at any human body site.
Pulmonary Infection
The respiratory tract serves as the most important portal of entry for Cryptococcus. Clinical manifestations of pulmonary cryptococcosis range from asymptomatic colonization of the airways or a simple pulmonary nodule on a chest radiograph to life-threatening pneumonia with the presence of an acute respiratory distress syndrome.50,51 In a normal host, asymptomatic, isolated pulmonary infection can occur in about one-third of patients and can be identified simply by an abnormal chest radiograph. In fact, the most common radiologic findings of cryptococcosis include well-defined single or multiple noncalcified nodules and pulmonary infiltrates (Fig. 1), although pleural effusions, hilar lymphadenopathy, and lung cavitation may also be observed. Patients with pulmonary cryptococcosis can present acutely with symptoms of pneumonia.50 For example, in the recent outbreak of C gattii infections in Vancouver Island area, several cases of severe, symptomatic pulmonary cryptococcosis in apparently immunocompetent individuals occurred.12 In an immunocompromised patient, however, cryptococcal pneumonia is usually symptomatic and in some cases can progress rapidly to acute respiratory distress syndrome, even in the absence of CNS involvement. Pulmonary involvement ranges from 10% to 55% of patients with AIDS-associated cryptococcal meningoencephalitis, although CNS symptoms usually predominate the clinical picture.51
Fig. 1.
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Solitary pulmonary nodule. In an asymptomatic patient with isolated pulmonary cryptococcosis. (Courtesy of J. R. Perfect, MD, Durham, NC.)
Serum cryptococcal polysaccharide antigen testing is usually negative in cases of true isolated pulmonary cryptococcosis, but at times can be positive in the absence of CNS involvement or other apparent sites of infection. In immunocompromised individuals with Cryptococcus isolated from the lung or other sterile body site, however, a lumbar puncture to rule out CNS disease should be considered regardless of a patient’s symptoms or serum antigen titer results. The only setting wherein a screening lumbar puncture may not necessarily be required is a patient with Cryptococcus isolated from the lung in the apparently immunocompetent patient without referable CNS symptoms and disease that clinically seems to be limited to the lungs.
Central Nervous System Infection
Clinical manifestations of CNS cryptococcosis include a myriad of signs and symptoms, such as headache, fever, cranial neuropathies, altered mentation, lethargy, memory loss, and signs of meningeal irritation.2,30,31 Symptoms usually develop over a period of several weeks. However, on some occasions, patients present more acutely or lack typical features, such as headache. In severely immunocompromised, HIV-infected patients with CNS cryptococcosis, the burden of fungal organisms is usually high and can reach levels of more than 1 million yeasts per milliliter of cerebrospinal fluid (CSF). These patients may consequently have a shorter onset of signs and symptoms, greater CSF polysaccharide antigen titers, and higher intracranial pressures than other more immunocompetent individuals.
Although disease severity is determined primarily by host immune factors, different species and/or strains of Cryptococcus may produce unique clinical manifestations, which can have implications for management. For instance, in certain areas of the world, C gattii has been observed to cause cerebral cryptococcomas and/or obstructive hydrocephalus with or without large pulmonary mass lesions more frequently than C neoformans.12,52,53 These patients with parenchymal brain involvement may have a high intracranial pressure and present with cranial neuropathies. In such patients, who have been observed to respond poorly to antifungal therapy, early neurosurgical intervention to control pressure or ensure a correct diagnosis and longer antifungal treatment courses may be required for a successful outcome.9,54
Skin Infection
Cutaneous infections are the third most common clinical manifestations of cryptococcosis and patients can present with a variety of skin lesions. Lesions are often indistinguishable from those owing to other infections; as such, a skin biopsy with culture and histopathology are absolutely essential for definitive diagnosis. Primary cutaneous cryptococcosis is very rare and is usually associated with skin injury and direct inoculation of the yeasts33; thus, the appearance of cutaneous lesions usually heralds the presence of disseminated infection. Solid organ transplant recipients on tacrolimus seem to be more likely to develop skin, soft tissue, and osteoarticular infections owing to Cryptococcus.55 Tacrolimus acts on the temperature signaling molecule calcineurin in Cryptococcus and has anticryptococcal activity at high temperatures, but it loses this direct antifungal activity as environmental temperatures decrease; this may in part explain the increased frequency of cutaneous lesions in patients receiving calcineurin inhibitors.56
Prostate Infection
The prostate is not a rare site for cryptococcal infection, but prostatic cryptococcosis is usually asymptomatic. For instance, latent C neoformans infection has been recognized to disseminate in the bloodstream during urologic surgery on the prostate for other indications.57 The prostate gland may thus serve as an important reservoir for disease relapse in patients with a high fungal tissue burden.58 Cultures of urine or seminal fluid may still be positive for Cryptococcus after initial antifungal treatment of cryptococcal meningitis in poorly controlled AIDS patients,59 strongly supporting the need for prolonged antifungal treatment to eradicate infection in sanctuary sites in these severely immunocompromised patients.
Eye Infection
In early reports of cryptococcal meningitis before the AIDS epidemic, ocular signs and symptoms were noted in a substantial proportion of cases,60 such as ocular palsies and papilledema. Several other ocular manifestations of cryptococcosis have been identified, including extensive retinal disease with or without vitritis, which can lead to irreversible blindness.61 Visual loss may be owing to optic nerve infiltration by yeasts or vascular compromise from intracranial hypertension. The former process results in rapid visual loss with limited effective treatments, whereas the latter phenomenon results in more gradual visual loss and can be interrupted with aggressive management of increased intracranial pressure.
Infection at Other Body Sites
C neoformans can cause disease in essentially any organ of the human body. In fact, the first identification of this fungus from a clinical specimen was from a patient with tibial osteomyelitis in the 19th century.1 Bone involvement of cryptococcosis typically presents as circumscribed osteolytic lesions in any bone of the body, but most commonly the vertebrae, and cryptococcal osteomyelitis has been associated with underlying sarcoidosis.62 Bone marrow infiltration can be observed in severely immunocompromised hosts. Fungal peritonitis63 and cryptococcuria are also reported in several case series. An appreciation for this yeast’s protean clinical manifestations is essential, both at the time of initial diagnosis, as well as when immune defects are restored during treatment and immune restoration phenomena can present.
Immune Reconstitution Inflammatory Syndrome
Restoration of pathogen-specific immunity can result in a phenomenon known as the immune reconstitution inflammatory syndrome (IRIS), an entity that can occur before (“unmasking IRIS”) or during (“paradoxic IRIS”) antifungal therapy. Cryptococcal IRIS is best characterized in HIV-infected patients with CNS infection and is associated with significant morbidity and mortality.64–76 In addition, IRIS is estimated to occur in 5% to 11% of solid organ transplant recipients with cryptococcal infection and is associated with increased risk of allograft failure77–83 and may also be observed in non-HIV, nontransplant patients.84 Proposed criteria for IRIS in HIV-associated disease include onset of symptoms within 12 months of ART initiation (with concomitant CD4+ recovery).85 These criteria are imprecise and do not address all populations at risk (Box 1). As such, it is incumbent upon the treating provider to have a high level of suspicion for this entity, as opposed to alternative diagnoses, which include progressive infection (from inadequate antifungal therapy, direct antifungal drug resistance, or persistent immune deficits), coinfection with other opportunistic infections, malignancy, or drug toxicity.
Box 1. Suggested diagnostic criteria for the immune reconstitution inflammatory syndrome.
New appearance or worsening of any of the following:
Clinical or radiographic manifestations consistent with an inflammatory process:
Central nervous system: Contrast-enhancing lesions on neuroimaging (computed tomography or MRI); cerebrospinal fluid pleocytosis (ie, >5 white blood cell count per μL); increased intracranial pressure (ie, opening pressure of ≥20 mm H 2 0), with or without hydrocephalus.
Pulmonary: Nodules, cavities, masses or pleural effusions.
Other: Lymphadenopathy, skin, soft tissue, osteoarticular lesions.
Histopathology showing granulomatous lesions.
Symptoms occurring during receipt of appropriate antifungal therapy a that cannot be explained by a newly acquired infection or another process (neoplasm, etc).
Negative results of cultures, or stable or reduced biomarkers for the initial fungal pathogen during the diagnostic workup for the inflammatory process.
All 3 criteria must be present for a positive diagnosis.
a Exclude intrinsic and de novo drug resistance, and suboptimum drug concentrations.
Adapted from Sun H, Alexander B, Huprikar S, et al. Predictors of immune reconstitution syndrome in organ transplant recipients with cryptococcosis: implications for the management of immunosuppression. Clin Infect Dis 2015;60(1):36–44; and Singh N and Perfect JR. Immune reconstitution syndrome associated with opportunistic mycoses. Lancet Infect Dis 2007; 7:398.
Cryptococcal IRIS is thought to represent a dysregulated reversal of a Th2 (anti-inflammatory) to a strong helper T cell (pro-inflammatory) immune response in the setting of immune recovery.86 Multiple factors are thought to be associated with future IRIS episodes, including high yeast burden at baseline, ineffective host immune response to initial infection, and rapid restoration of immunity.67,73 Host immune responses in various compartments may not be uniform and are likely influenced by baseline parameters at the site.87 Differences in baseline CSF cytokine and chemokine expression are thought to facilitate the development of cryptococcal IRIS, potentially via myeloid cell trafficking to the CNS and, consequently, production of excessive inflammation.88,89 In fact, evidence of increased macrophage activation and linked CSF pleocytosis have been observed in patients receiving early ART and may mediate increased mortality, even before recognition of the clinical syndrome of IRIS.87
Clinical features of cryptococcal IRIS are similar to active cryptococcal infection itself, most commonly presenting as CNS disease, although lymphadenitis, pneumonitis, multifocal disease, soft tissue involvement, and mediastinitis have all been reported.85,90 Meningeal disease is the most serious presentation.85 A hallmark finding is suppurative or necrotic granulomatous inflammation with yeast forms seen on histopathology of infected tissues despite negative cultures.77,80,90,91 Despite changes in inflammatory markers, there are no reliably specific diagnostic tests for IRIS, and establishing the diagnosis presents a considerable clinical challenge, especially with atypical presentations or manifestations at distant sites.69,92 CSF opening pressure and white blood cell count67,68,73 at the time of an IRIS event are significantly higher than baseline values for individual patients, which combined with negative cultures, may help to distinguish IRIS from relapsed infection.70
Management of cryptococcal IRIS is largely based on expert opinion.93 First, ensuring the efficacy of antifungal therapy is essential94,95; in the absence of disease relapse or direct antifungal drug resistance, modification of antimicrobial therapy is generally not indicated.93 A significant proportion of minor cases simply improve without specific treatment.65,66,76 Corticosteroids have been shown to decrease the need for hospitalization and improve short-term quality of life and functional status in paradoxic tuberculosis-associated IRIS.96 Although steroids may be essential in treating a serious life-threatening CNS IRIS episode owing to Cryptococcus, they should not be used for prevention of IRIS or to control CNS pressure, and may be harmful in some cases.97 Immunomodulatory agents including those with anti–TNF-α activity have been used in cases of steroid-refractory IRIS.65,98–101 Other strategies, including therapeutic lumbar drainage for intracranial hypertension93,102 and, at times, surgical drainage of suppurative lymph nodes,86,91 are important adjunctive measures that may be considered in severe disease. Continuation of ART in the setting of IRIS is generally recommended and has been performed safely.66,71,92,103,104
LABORATORY DIAGNOSIS
Definitive diagnosis of cryptococcosis is made by isolation of Cryptococcus from a clinical specimen or direct detection of the fungus by means of India ink staining of body fluids. There are several other methods used for the diagnosis of cryptococcosis, including histopathology of infected tissues and serologic methods. Molecular methods, although available and extensively used for research purposes, are not used currently in routine clinical practice.
Direct Examination/India Ink
The most rapid method for diagnosis of cryptococcal meningitis is direct microscopic examination for encapsulated yeasts by India ink preparation of CSF. Cryptococcus can be visualized as a globular, encapsulated yeast cell with or without budding, ranging in size from 5 to 20 μm in diameter (Fig. 2). The sensitivity of India ink staining of CSF depends on fungal burden and is reported to be 30% to 50% in non–AIDS-related cryptococcal meningitis and up to 80% in AIDS-related disease. False positives can result from intact lymphocytes, other tissue cells and nonviable yeast forms, which further limits the diagnostic utility of direct microscopy of CSF for cryptococcal meningitis.105
Fig. 2.
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India ink staining. Encapsulated yeast seen on India ink preparation of cerebrospinal fluid in a patient with cryptococcal meningitis. (Courtesy of J. R. Perfect, MD, Durham, NC.)
Culture and Identification
Cryptococcus can be cultured readily from biologic samples such as CSF, sputum, and skin biopsy on routine fungal and bacterial culture media. In adults with HIV-associated cryptococcal meningitis, CSF and blood cultures are positive in up to 90% and 70% of patients, respectively (reviewed in106). Colonies are usually observed on solid agar plates after 48 to 72 hours incubation at 30°C to 35°C in aerobic conditions and will appear as opaque, white-to-cream colonies that may turn orange-tan or brown after prolonged incubation. The mucoid appearance of the colony is related to the capsule size around the yeasts. Despite relatively rapid growth for most strains, cultures should be held for up to 4 weeks, particularly for patients receiving antifungal treatment.
Cytology and Histopathology
Cryptococcus can be identified by histologic staining of tissues from the lung, skin, bone marrow, brain, and other organs.107 Histopathologic staining and cytology of centrifuged CSF sediment and other bodily fluids are more sensitive than the India ink staining method.108–111 The organism is observed as a yeast that reproduces by narrow-based budding. The yeast is best identified by special stains that label the polysaccharide capsule including mucicarmine, periodic acid-Schiff, and Alcian blue stains.2 The Fontana–Masson stain identifies melanin in the yeast cell wall. Other fungal stains such as Calcofluor, which binds fungal chitin, or Gomori methenamine silver, which stains the fungal cell wall, are also used to identify the organism from clinical specimens.2,109
Serology
The diagnosis of cryptococcosis improved significantly with the development of serologic tests for the cryptococcal polysaccharide capsular antigen (CrAg), which is shed during infection. Latex agglutination and enzyme immunoassay techniques have been available widely (using both serum and CSF), the former of which had been the most commonly used methodology until recently, with overall sensitivities and specificities of 93% to 100% and 93% to 98%, respectively.112,113 False-positive results of latex agglutination testing usually have initial reciprocal titers of 8 or less,112 whereas false negatives can be seen owing to a prozone effect in the setting of extremely high antigen titers, which can be overcome with dilution.114 Low fungal burden, as in chronic low-grade meningitis or in the very early stages of infection, and improper specimen storage can also cause false-negative results in latex agglutination tests.115 Recently, a lateral flow assay was approved for use in serum and CSF, with sensitivity and specificity of greater than 98% in both specimen types (including whole blood from finger stick samples) and sensitivity of 85% in urine.116–123 The semiquantitative test offers many advantages over the other serologic methods, including rapid turnaround (approximately 15 minutes), minimal requirements for laboratory infrastructure, stability at room temperature, low cost, and wider capture of C gattii polysaccharides.116 Combined with these advantages, the assay’s excellent performance across a broad range of clinical settings, including settings with low burden of HIV infection and high rates of C gattii infection,100–104 make it an attractive option for point-of-care testing in both resource-available and resource-limited settings.116,117,124
Baseline cryptococcal polysaccharide antigen titers in serum and CSF correlate with fungal burden and carry prognostic significance in patients with cryptococcal meningitis.122,125,126 However, there is limited value in serial monitoring of antigen titers acutely in assessing treatment response, because the kinetics of antigen clearance is a slower and less predictable marker of treatment response than quantitative culture.122,127 Quantitative CSF yeast culture and its serial use for measurement of effective fungicidal activity has become a primary research tool for effectiveness of therapeutic regimens.128 The quantitative yeast count has been correlated with outcome129 and effective fungicidal activity has correlated with success of antifungal regimens, including survival.95,128,130 Despite a decade of use and validation of its effectiveness in clinical studies, the use of quantitative CSF yeast culture for the determination of effective fungicidal activity has not yet become a part of routine clinical practice.
TREATMENT
Basic Principles
Amphotericin B deoxycholate (AmBd) is the cornerstone of treatment for severe cryptococcal infection, including meningoencephalitis. Treatment is summarized in Table 4. A standard induction dose of 0.7 to 1 mg/kg/d is recommended. Liposomal amphotericin B (3–6 mg/kg/d) has become a preferred alternative with similar outcomes and less nephrotoxicity, and is recommended specifically for primary induction in patients at risk for renal dysfunction.93,131,132 Flucytosine (5-FC) is used in combination therapy with AmBd as first-line therapy in cryptococcal meningitis or severe pulmonary cryptococcosis at a dosage of 100 mg/kg/d in divided doses.133,134 This combination represents the most potent fungicidal regimen, with faster CSF sterilization and fewer relapses, and is associated with lower attributable mortality.133–139 Because the interruption of induction therapy is associated with poorer outcome, in resource-available areas the liposomal product has become the preferred polyene. Unfortunately, there are still no comparative studies with 5-FC combined with lipid formulations of amphotericin B as opposed to AmBd. Early mycological failure (defined as persistently positive CSF cultures at day 14) correlates with late treatment failure and poor outcome,140 and lack of 5-FC is independently associated with both early141 and late137 mycological failure. This improved fungicidal activity of combination therapy translates into a direct survival benefit compared with AmBd monotherapy.135 5-FC should be dose adjusted for renal dysfunction, with therapeutic drug monitoring to decrease its primary side effect of bone marrow suppression.142 There are emerging data that lower doses of 5-FC in combination with amphotericin may demonstrate similar fungicidal activity.138
Table 4.
Treatment recommendations for HIV-associated cryptococcal meningoencephalitis
| | Duration |
:--- |
| Induction therapy | |
| Primary regimen | |
| AmBd (0.7–1 mg/kg/d) plus flucytosine (5-FC) (100 mg/kg/d)a | 2 wk |
| Alternative regimensb | |
| If 5-FC intolerant or unavailable: AmBd (0.7–1 mg/kg/d) or L-AMBc (3–4 mg/kg/d) or ABLC (5 mg/kg/d) | 4–6 wk |
| AmBd (0.7–1 mg/kg/d) plus fluconazole (800 mg/d) | 2 wk |
| Fluconazole (≥800 mg/d, preferably 1200 mg/d) plus 5-FC (100 mg/kg/d) | 6 wk |
| Fluconazole (800–2000 mg/d, preferably 1200 mg/d) | 10–12 wk |
| Itraconazole (200 mg BID) | 10–12 wk |
| Consolidation therapy | |
| Fluconazole (400 mg/d) | 8wkd |
| Maintenance or suppressive therapy | |
| Fluconazole (200 mg/d) | ≥1 ye |
| Alternative reqimensa | |
| Itraconazole (200 mg BID) | ≥1 ye |
| AmBd (1 mg/kg IV per week) | ≥1 y |
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Abbreviations: 5-FC, flucytosine; ABLC, amphotericin B lipid complex; AmBd, amphotericin B deoxycholate; BID, twice daily; L-AMB, liposomal amphotericin B.
a
L-AMB, 3–4 mg/kg/d or AmB lipid complex (ABLC; 5 mg/kg/d) for patients predisposed to renal dysfunction.
b
Can be considered as alternative regimen when primary regimen not available but not encouraged as equivalent substitutes.
c
L-AMB can be safely administered in doses as high as 6 mg/k/d.
d
Initiate highly active antiretroviral therapy approximately 4 weeks after beginning antifungal regimen.
e
After 1 year of therapy, if successful response to antiretroviral drugs (CD4 count ≥ 100 and viral load low or undetectable for >3 months), can consider discontinuation of antifungal therapy. Consider reinstitution if CD4 count is <100.
Adapted from Perfect JR, Dismukes WE, Dromer F, et al. Clinical practice guide lines for the management of cryptococcal disease: 2010 update by the Infectious Disease Society of America. Clin Infect Dis 2010;50:291–322.
Although combination induction therapy remains the recommended first-line therapy for severe cryptococcosis, 5-FC availability is limited in settings where the disease burden and mortality rates are the highest. Alternative combination therapies have been investigated, the most efficacious of which is AmBd plus fluconazole (800 mg/d), which results in improved rates of fungal clearance, neurologic recovery, and survival compared with AmBd alone or in combination with lower doses of fluconazole.143,144 This combination offers a more feasible and potentially viable option for effective initial therapy in settings where access to 5-FC is limited. Optimizing treatment outcomes without exhausting limited resources is critical in many settings. Standardized fluid and electrolyte supplementation protocols for patients treated with amphotericin B in these resource-limited settings have been associated with improved early survival.145 Additionally, shorter courses of amphotericin B in combination with other agents may be considered in these settings, although clinical endpoints for such regimens have not been rigorously evaluated.146,147 An ongoing trial evaluating the combination of intermittent dosing of high-dose of liposomal amphotericin B with high-dose fluconazole in resource-limited settings is underway to address this unanswered question (AmBition-CM, www.controlled-trials.com/ ISRCTN10248064). Additional alternative induction regimens are available in the guidelines but their use is not encouraged based on limited data of the success with these regimens.148 Fluconazole monotherapy for meningitis is not recommended for induction given its fungistatic nature, poor success, and higher relapse rates as well as increased rates of resistance in relapse.93,94 However, in areas without access to AmBd, high doses (≥1200 mg/d) of fluconazole should be commenced.
A 3-stage regimen of induction, consolidation, and maintenance is standard treatment for cryptococcal meningitis in all patients, irrespective of host risk factors.93,133 In HIV-infected patients, initial induction treatment usually begins with combination therapy as described, followed by consolidation treatment with fluconazole (400–800 mg/d) for 8 weeks in patients who have demonstrated favorable response. Longer courses of both induction (eg, 6 weeks) and consolidation (or “eradication”) therapy have been suggested in C gattii meningoencephalitis, irrespective of host immune status, owing to the observed severity of neurologic disease in this group of patients,11,52,53 but this is not certain and in general C gattii should be treated similarly to C neoformans. After consolidation, long-term suppression is commenced with oral fluconazole (200–400 mg/d). This approach has decreased rates of relapse from approximately 40% to less than 5% in severely immunosuppressed patients.149 Secondary prophylaxis is discontinued after 1 to 2 years of antifungal therapy in patients who respond to ART with an increase in CD4+ cell counts to greater than 100 cells/μL and a decrease in HIV viral load to undetectable levels for at least 3 months.93,150,151 The other triazoles (itraconazole, voriconazole, and posaconazole) are active against cryptococcal isolates in vitro and, in combination with AmBd, may have similar fungicidal activity to 5-FC,144 but owing to differences in bioavailability, CSF penetration, drug interactions, cost, and lack of robust studies in cryptococcosis, these agents are not recommended as first-line agents for consolidation or maintenance therapy. However, they may have a role in refractory cases.152–155
Timing of Antiretroviral Therapy
In HIV-associated cryptococcal infection, ART has a major impact on long-term prognosis. However, several studies have suggested an increased risk of IRIS among HIV-infected patients initiated on ART early after the diagnosis of an opportunistic infection.64,65,156 More contemporary studies have demonstrated conflicting results regarding outcomes of cryptococcal infection based on timing of ART initiation,103,157,158 and studies in tuberculosis have demonstrated a survival benefit with earlier ART (despite increased rates of IRIS).159,160 Recently, the Cryptococcal Optimal ART Timing Trial (COAT) provided some definitive guidance to delay initiation of ART in patients with cryptococcal meningitis for a minimum of 4 weeks after starting antifungal therapy. This randomized trial demonstrated improved survival in patients with cryptococcal meningitis in whom ART initiation was deferred for up to 5 weeks after diagnosis as compared with immediate ART (within 1–2 weeks).161 Although increased rates of IRIS observed with early ART did not attain statistical significance, markers of macrophage activation were increased in this early group, suggesting that subclinical or compartmentalized IRIS may occur and influence mortality.87,161
Organ Transplant Recipients
Organ transplant recipients with CNS cryptococcal infection are managed similarly to HIV-infected patients, although lipid formulations of amphotericin B are preferred to limit nephrotoxicity.93 A longer course of induction therapy is indicated if CSF cultures remain positive at 2 weeks, because this scenario is associated with an increased 6-month mortality.162 Relapse rates among organ transplant recipients are lower than in HIV-associated disease, such that a shorter course of maintenance therapy can be pursued following standard consolidation, but generally these patients are treated for 1 year.93,162 Drug interactions between fluconazole and immunosuppressive agents should be anticipated owing to CYP3A4 inhibition, and a preemptive reduction in calcineurin inhibitors should be considered. Management of immunosuppression in the setting of cryptococcal infection requires recognition of the increased risk of IRIS.77,80,163 Thus, stepwise reduction in immunosuppression is recommended, although the approach should be individualized for each patient.
Non–HIV-Infected, Nontransplant Patients
Very few prospective data are available on the management of cryptococcal infection in the apparently immunocompetent host lacking classical risk factors for cryptococcosis.134 This heterogeneous group of patients is diagnosed later, irrespective of disease severity.32,84 Recommendations for longer induction therapy (≥4 weeks) are based on the recognition of poorer outcomes and higher mortality rates in this group of patients both in early134,164 as well as contemporary32 studies. However, in patients with good prognostic factors and excellent antifungal induction response, 2-week induction therapy can be successful. Therapy should be extended further if 5-FC is not included (or there is limited exposure to this drug) in the induction regimen.93 Recommendations for consolidation and maintenance parallel those for HIV-infected patients and reflect high relapse rates (30%) within the first year before the introduction of consolidation and maintenance antifungal strategies.93,134 Criteria for stopping treatment in these patients include resolution of symptoms and at least 1 year of suppressive antifungal therapy.
Management of Intracranial Pressure
Along with the optimization of antifungal therapy, management of increased intracranial pressure is critically important in cryptococcal meningoencephalitis. Intracranial hypertension frequently corresponds with CSF fungal burden, potentially mediated by CSF outflow obstruction by clumped yeast forms even during early therapy, and is associated with increased morbidity and mortality.97,165 Intracranial imaging should be performed before lumbar puncture if impaired mentation or focal neurologic deficits are present. A baseline CSF opening pressure should be obtained in all patients. Aggressive attempts to control increased intracranial pressure should occur when patients are symptomatic, although emerging data suggest there may be benefit to therapeutic lumbar punctures, irrespective of baseline opening pressure in resource-limited settings.166 Treatment options for managing acutely elevated intracranial pressure include repeated lumbar punctures (daily until pressure and symptoms are stable for >2 days), lumbar drain insertion, ventriculostomy, or ventriculoperitoneal shunt, if obstructive hydrocephalus develops.97 Consideration of early neurosurgical consultation has been recommended in cases of meningoencephalitis owing to C gattii where CNS inflammation is often severe.52,53 Medical treatments such as corticosteroids (unless IRIS suspected or in cases of severe C gattii infection), mannitol, and acetazolamide are generally not recommended.52,53,129,167 If shunt placement is necessary, CSF sterilization is not required before insertion, which can be performed once appropriate antifungal therapy has been commenced.168
Persistent and Relapsed Infection
Persistent and relapsed infection must be distinguished from IRIS. Persistent disease has been defined as persistently positive CSF cultures after 1 month of antifungal therapy, whereas relapse requires new clinical signs and symptoms and positive cultures after initial improvement and fungal sterilization.93 Surrogate markers, including biochemical parameters, India ink staining, and cryptococcal antigen titers, are insufficient to define relapse or alter antifungal therapy. General recommendations for management in these persistent or relapsed cases include resumption of induction therapy, often for a longer duration and at increased dosages, if tolerable, and pursuance of comparative antifungal susceptibility testing.93 Although primary direct antifungal resistance to azoles and polyenes is rare, decreased susceptibility to fluconazole has been observed in some cases of culture-positive relapse.94 There has not yet been a convincing minimum inhibitory concentration breakpoint for cryptococcal species in antifungal susceptibility testing; thus, the importance of comparative minimum inhibitory concentration testing with the original isolate in cases where resistance is suspected cannot be overemphasized.169,170
Nonmeningeal Disease
Although isolation of Cryptococcus from respiratory tract specimens can occur in the absence of clinical disease (colonization), it is incumbent upon the treating clinician to assess for subclinical disease or potential for complications when Cryptococcus is isolated from any clinical specimen. In the absence of immune compromise, airway colonization carries a low risk for invasive disease and treatment can be deferred; although in most cases, given the safety profile of fluconazole, many clinicians favor treatment in all patients in whom Cryptococcus is isolated. In immunosuppressed patients with isolated pulmonary cryptococcosis, however, treatment is recommended to prevent dissemination.93 This group of patients should be evaluated for systemic disease (including blood and CSF cultures as well as CrAg testing from serum and CSF) to optimize treatment. In any patient in whom cryptococcemia is identified, symptoms are severe, or CSF examination reveals asymptomatic CNS involvement, treatment for cryptococcal meningitis is recommended.93 The potential for severe pulmonary infection owing to C gattii should be appreciated when Cryptococcus is isolated from respiratory cultures in settings where this species is endemic11,12,52,53,171; however, to date, there are no convincing data that species identification is required to optimally select antifungal therapy, and disease severity remains the critical factor in determining initial treatment. Cerebral cryptococcomas often can be managed with prolonged antifungal therapy without the need for surgical removal unless mass effect or other evidence of obstruction is identified. A longer induction phase with AmBd plus 5-FC, followed by 6 to 18 months of consolidation therapy with fluconazole (400–800 mg/d) is recommended. Localized infection of extrapulmonary nonmeningeal sites can occur occasionally with direct inoculation, but more commonly represents disseminated infection. Suspicion for the latter must be maintained when Cryptococcus is identified from a sterile body site, because management strategies differ if disseminated disease is present. Consultation with ophthalmology is indicated in cases of cryptococcal eye disease.93
Screening and Prevention
There is no question that early identification of HIV infection and initiation of ART in patients before progression to severe immunodeficiency is the most effective intervention at reducing the global burden of cryptococcosis and other opportunistic infections. However, despite increased access to ART worldwide, late presentations of HIV infection still occur and the burden of severe cryptococcal infection and related mortality remains disproportionately represented in these populations.
Fluconazole prophylaxis has been shown to be effective for preventing cryptococcosis in patients with advanced AIDS in endemic areas172,173; however, universal prophylaxis is relatively cost ineffective,124 has not been shown to offer a survival benefit,174 and may add to the appearance of azole-resistant strains. As such, this approach is not recommended currently.
Given that mortality from cryptococcal meningitis remains unacceptably high, alternative management strategies have been evaluated and implemented in resource-limited settings, specifically a “screen and treat” approach using serum cryptococcal antigen (CrAg) testing followed by preemptive fluconazole therapy in CrAg-positive patients. CrAg is an early marker of cryptococcal disease, detectable in serum a median of 22 days before the onset of symptoms, and is both highly predictive of incident cryptococcal meningitis and an independent risk factor for death during the first year of ART.175–177 This approach is associated with a decreased incidence of cryptococcal meningitis and improved survival among patients with advanced HIV disease and has been successfully implemented in several resource-limited settings, with a baseline prevalence of asymptomatic cryptococcal antigenemia of 5% to 13%.177,178 Moreover, analyses have consistently demonstrated both the cost effectiveness and survival advantage of a “screen and treat” approach, as compared with standard of care or universal fluconazole prophylaxis, at CrAg prevalences as low as 0.6%.178–180 As access to lateral flow assay testing in these settings is increased, the cost effectiveness is likely to be greater than initially reported. The World Health Organization now recommends implementation of CrAg screening and preemptive fluconazole therapy in ART-naïve adults with a CD4 count of less than 100 cells/mm 3 before initiating ART in endemic settings.181 Several nations in sub-Saharan Africa have since operationalized programs as a part of the existing HIV infrastructure. Several unanswered questions remain, however, including the feasibility of implementation, the dose and duration of preemptive fluconazole, the criteria for lumbar puncture in asymptomatic patients, and the potential impact on azole resistance. Some data suggest a ‘screen and treat’ would be cost effective, even in resource-rich settings, although this is currently not part of standard practice, despite recent reports of CrAg prevalence of more than 3% in the United States.176,182 Routine screening for cryptococcal infection and/or prophylaxis are not recommended in solid organ transplant recipients, even when immunosuppression is augmented in patients with previously (appropriately) treated infection.183
In the arena of direct immune modulation for cryptococcosis management, aside from the use of ART, progress has been slow. First, although both cryptococcal glucuronoxylomannan–tetanus toxoid conjugate vaccine and specific monoclonal antibodies to cryptococci have been developed, clinical trials have not been initiated to determine their usefulness in human subjects.184,185 The use of immune stimulation with recombinant gamma-interferon has both immunologic support and 2 positive clinical trials,186–189 but has only been used in refractory cases and likely reflects concerns about precisely judging immune stimulation when IRIS can be a deadly problem.
KEY POINTS.
Cryptococcosis is a major invasive fungal infection that is capable of widespread disease outbreaks in both immunocompromised and apparently immunocompetent hosts.
Molecular advances continue to enhance our understanding of Cryptococcus and provide insight into its evolution into a pathogen of global importance.
Diagnosis has improved with the introduction of point-of-care diagnostic assays.
Screening and preemptive antifungal therapy offer great promise in making a significant impact in this highly deadly opportunistic mycosis.
Footnotes
Disclosure Statement: Dr J.R. Perfect is a Principal Investigator for the following companies: Amplyx, Astellas, Cidara, Merck, Pfizer, Schering-Plough, Tokoyama, and Viamet.
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11513 | https://mathoverflow.net/questions/313087/non-backtracking-random-walk-in-regular-finite-graphs | Skip to main content
Non-backtracking random walk in regular (finite) graphs
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I know that many things are known when dealing with random walks on a finite (or even infinite) graph: mixing time, returns to origin, etc. All is based in the use of the Markovian property of such a random walk (I am assuming that on each vertex we can choose the next one among neighbours with uniform probability distribution).
My question is the following: when dealing with non-backtracking random walks (namely, we cannot go back through an edge we have just used, inducing on every edge a uniform distribution) we lose all markovian property, but this can be manage by taking orientation on edges.
Q: is there some kind of 'universal' result for regular graphs concerning the existence of stationary distribution for such random walks?
I have looked for bibliography on this topic (this should be like the first question on these models), but I have not been able to find any reference on this.
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edited Jun 28, 2019 at 15:47
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Johnny CageJohnny Cage
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What do you mean by 'universal'? Many of these walks will be periodic on regular graphs (the d-dimensional hypercube is d-regular and bipartite).
– Marcus M
Commented
Oct 18, 2018 at 0:00
By 'universal' I mean not depending on the graph, but only if it is k-regular or not (for instance, or any other general condition) . The first thing I would like to know (i.e., references) if starting at a fixed vertex there is a uniform distribution to arrive to any other independently of the graph. I am happy with the restriction of the graph to be k-regular.
– Johnny Cage
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Oct 18, 2018 at 5:34
No such universal condition may exist; if a graph is bipartite, then all walks are periodic and no stationary distribution exists. The d-cubes are all bipartite (if you view each vertex as a binary string then the parity of each vertex is the parity of the hamming weight), and thus no stationary distribution exists for any random walk on them.
– Marcus M
Commented
Oct 18, 2018 at 13:10
Sure: I forgot to say that we want of course to avoid bipartiteness... apart from this natural condition, are there other obstructions?
– Johnny Cage
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Oct 18, 2018 at 20:15
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What if the graph in question is a cycle graph?
– LeechLattice
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Indeed, understanding non-backtracking walks is often the key to analyzing the simple random walk and random graphs. See e.g. , and , .
Basic properties of the non-backtracking walk are collected in , Exercise 6.59.
Lubetzky, Eyal, and Allan Sly. "Cutoff phenomena for random walks on random regular graphs." Duke Mathematical Journal 153, no. 3 (2010): 475-510.
Lubetzky, Eyal, and Yuval Peres. "Cutoff on all Ramanujan graphs." Geometric and Functional Analysis 26, no. 4 (2016): 1190-1216.
C. Bordenave. A new proof of Friedman’s second eigenvalue Theorem and its extension to random lifts. arXiv preprint arXiv:1502.04482, 2015.
Bordenave, Charles, Marc Lelarge, and Laurent Massoulié. "Non-backtracking spectrum of random graphs: community detection and non-regular ramanujan graphs." In 2015 IEEE 56th Annual Symposium on Foundations of Computer Science, pp. 1347-1357. IEEE, 2015.
R. Lyons and Y. Peres. Probability on Trees and Networks. Cambridge University
Press. (2016). Available at
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answered Jun 28, 2019 at 13:02
Yuval PeresYuval Peres
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Edit: The previous argument is flawed, as pointed out by Brendan Mckay.
Let G be a non-bipartite regular graph with δ≥3, and X its transition matrix on directed arcs.
We need to prove that
There exists a number N and a vertex V such that for every n>N there's a non-backtracking walk from V to V.
If the statement is true, there exists some number N for which we can reach every directed arc in G starting from any directed arc in G in n steps, for every n>N, as G is connected. So there exists a number N for which XN is strictly positive. By Perron–Frobenius theory, it implies that there is only one stationary distribution on X, which is the uniform distribution.
We could prove the statement above by showing that there are some cycles in G whose lengths has no common divisor >1.
Let L be the longest path in G, and let l1 be one of its ends. l1 is incident to at least two vertices on L, say l2 and l3. So there are three cycles formed by the edges E{L}∪l1l2∪l1l3. The gcd of the lengths of the cycles are either 1 or 2; If it's 1, we are already done, and if it's 2, pick an odd cycle, and the gcd will become 1.
The cycle lengths have finite Frobenius Number, and the statement follows.
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edited Jun 29, 2019 at 8:52
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Do you have a nice argument why vertices A and B must exist?
– Brendan McKay
Commented
Jun 28, 2019 at 19:52
A and B exists by the construction above using the longest path L, but the previous argument is flawed(it even works for bipartite graphs, which has non-uniform stationary distributions and should be excluded).
– LeechLattice
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Jun 29, 2019 at 3:59
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11514 | https://www.freemathhelp.com/forum/threads/what-does-this-notation-mean-verify-bar-over-z-w-bar-over-z-bar-over-w-what-are-the-bars.114684/ | Published Time: 2019-02-27T20:48:22+00:00
what does this notation mean? "Verify (bar over)(z + w) = (bar over)(z) + (bar over)(w)" What are the bars? | Free Math Help Forum
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Intermediate/Advanced Algebra
what does this notation mean? "Verify (bar over)(z + w) = (bar over)(z) + (bar over)(w)" What are the bars?
Thread starterallegansveritatem
Start dateFeb 27, 2019
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allegansveritatem
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Joined Jan 10, 2018 Messages 962
Feb 27, 2019
#1
I came across this today in some problems at the end of a section and can find no reference to it in the text:
What I don't understand is the symbols over the z+w and over the z and the w. What are they asking for here? Here is the solution the book gives but I can't figure out where they got any of this from:
MarkFL
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Staff member
Joined Nov 24, 2012 Messages 3,021
Feb 27, 2019
#2
That overline notation refers to a complex conjugate. for example, if we are given:
[MATH]z=x+yi[/MATH]
then:
[MATH]\overline{z}=x-yi[/MATH]
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Feb 27, 2019
#3
MarkFL said:
That overline notation refers to a complex conjugate. for example, if we are given:
[MATH]z=x+yi[/MATH]
then:
[MATH]\overline{z}=x-yi[/MATH]
Click to expand...
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from this:
to:
From z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
Last edited: Feb 27, 2019
K
ksdhart2
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Joined Mar 25, 2016 Messages 1,297
Feb 27, 2019
#4
allegansveritatem said:
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from [...] z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
Click to expand...
That first step there, where the a's and b's "magically" appear, is literally just the definition of a complex number. It doesn't make sense to talk about the complex conjugate of a real number, so z and w must be complex numbers. Let z=a+b i\displaystyle z = a + bi z=a+b i and w=c+d i\displaystyle w = c + di w=c+d i. Then z+w‾=???\displaystyle \overline{z + w} = \text{???}z+w=???
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Feb 27, 2019
#5
allegansveritatem said:
So the line over the z refers to a complex conjugate. But I don't see here how they come to this solution. I mean how do they go from this:
View attachment 11197
to:
View attachment 11198
From z + w with the overhead line to a bunch of a's and b's....? I don't see how come by these terms. I can see conjugates in there but.... It astonishes me that this wasn't covered in the text but just thrown into the problems at the end of a section.
Click to expand...
First, do you understand or have you been taught that the notation 3−2 i‾\displaystyle \overline{3-2\bf{i}}3−2 i stands for the conjugate 3+2 i?\displaystyle 3+2\bf{i}~?3+2 i?
As for the a′s&b′s\displaystyle a's~\&~b's a′s&b′s, in the theory of complex numbers are defined as ordered pairs of real numbers.
The pair (3,-2) is written as 3−2 i\displaystyle 3-2\bf{i}3−2 i. The real part ℜ(3−2 i)=3\displaystyle \Re(3-2\bf{i})=3 ℜ(3−2 i)=3 and imaginary part ℑ(3−2 i)=−2\displaystyle \Im(3-2\bf{i})=-2 ℑ(3−2 i)=−2
So we define (a,b)+(c,d)=(a+c,b+d)\displaystyle (a,b)+(c,d)=(a+c,b+d)(a,b)+(c,d)=(a+c,b+d) , (a,b)⋅(c,d)=(a c−b d,a d+b c)\displaystyle (a,b)\cdot(c,d)=(ac-bd,ad+bc)(a,b)⋅(c,d)=(a c−b d,a d+b c), and (a,b)‾=(a,−b)\displaystyle \overline{(a,b)}=(a,-b)(a,b)=(a,−b)
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Feb 28, 2019
#6
pka said:
First, do you understand or have you been taught that the notation 3−2 i‾\displaystyle \overline{3-2\bf{i}}3−2 i stands for the conjugate 3+2 i?\displaystyle 3+2\bf{i}~?3+2 i?
As for the a′s&b′s\displaystyle a's~\&~b's a′s&b′s, in the theory of complex numbers are defined as ordered pairs of real numbers.
The pair (3,-2) is written as 3−2 i\displaystyle 3-2\bf{i}3−2 i. The real part ℜ(3−2 i)=3\displaystyle \Re(3-2\bf{i})=3 ℜ(3−2 i)=3 and imaginary part ℑ(3−2 i)=−2\displaystyle \Im(3-2\bf{i})=-2 ℑ(3−2 i)=−2
So we define (a,b)+(c,d)=(a+c,b+d)\displaystyle (a,b)+(c,d)=(a+c,b+d)(a,b)+(c,d)=(a+c,b+d) , (a,b)⋅(c,d)=(a c−b d,a d+b c)\displaystyle (a,b)\cdot(c,d)=(ac-bd,ad+bc)(a,b)⋅(c,d)=(a c−b d,a d+b c), and (a,b)‾=(a,−b)\displaystyle \overline{(a,b)}=(a,-b)(a,b)=(a,−b)
Click to expand...
I know about complex numbers. I don't know about the sign, the overhead line that goes above them. I don't have a teacher but have been using a book by Robert Blixer. Recently I finished this book and got one by Earl Swokowski. I guess Swokowski uses this notation and Blixer does not. I have taken algebra courses in high school and a college course as well but that was many decades ago--still I don't recall this sign from those times either. Blixer's book is titled: College Algebra. Swokowski's is more advanced with the title: Precalculus, Equations and Graphs. But I begin to see my problem with this: One part of the problem is that I took the z and the w for single component terms, like a and b. I will study your reply later today and get back. Thanks.
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Feb 28, 2019
#7
ksdhart2 said:
That first step there, where the a's and b's "magically" appear, is literally just the definition of a complex number. It doesn't make sense to talk about the complex conjugate of a real number, so z and w must be complex numbers. Let z=a+b i\displaystyle z = a + bi z=a+b i and w=c+d i\displaystyle w = c + di w=c+d i. Then z+w‾=???\displaystyle \overline{z + w} = \text{???}z+w=???
Click to expand...
I was having trouble translating z and w into anything other than terms with a single component. I wasn't making the leap from z or w to a+bi. Dumb but understandable. I have to go but I will get back to your post later and work it out. Thanks for reply.
Dr.Peterson
Elite Member
Joined Nov 12, 2017 Messages 16,816
Feb 28, 2019
#8
I checked the Blitzer book I've taught from, and see that, indeed, it doesn't use any notation for conjugates, because it doesn't do any more with them than use them to simplify expressions or point out that solutions to quadratic equations come in conjugate pairs, neither of which needs a notation.
Clearly your new book is at a higher level, so you may need to go back to where they introduced complex numbers and read carefully, to see their notation and conventions.
In the solution you quoted, they don't take the time, as one would in a proof, to define variables or justify steps, trusting that you know what to expect: When they replace z with a+bi, they are implicitly saying, "suppose that the complex number z has components a and b", and so on. So you have to build up your expectations to match theirs, by reading slowly and carefully, even when you think the topic is one you already know. Probably there has been an example that demonstrated these ways of thinking.
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Feb 28, 2019
#9
allegansveritatem said:
(I) have been using a book by Robert Blixer. Recently I finished this book and got one by Earl Swokowski. I guess Swokowski uses this notation and Blixer does not. Blixer's book is titled: College Algebra. Swokowski's is more advanced with the title: Precalculus, Equations and Graphs. But I begin to see my problem with this: One part of the problem is that I took the z and the w for single component terms, like a and b. I will study your reply later today and get back.
Click to expand...
Unlike Prof. Peterson, I do not have nor have I seen the text by Bilixer. I I knew Swokowski and his text is excellent. There are even solution manuals available on Amazon. I was typing a replay late last night when the LaTeX went haywire. So I simply stopped. I wish I had been able to save the code but alas its gone.
Here is the a bit of it. Any complex number is two parts: A real part & an imaginary part.
Say z=5−3 i\displaystyle z=5-3\bf{i}z=5−3 i, the real part ℜ(z)=5&\displaystyle \Re(z)=5~\&~ℜ(z)=5&the imaginary part ℑ z=−3\displaystyle \Im{z}=-3 ℑ z=−3.
Thus z=a+b i\displaystyle z=a+b\bf{i}z=a+b i, the real part ℜ(z)=a\displaystyle \Re(z)=a ℜ(z)=a& the imaginary part ℑ z=b\displaystyle \Im{z}=b ℑ z=b where both a&b\displaystyle a~\&~b a&b are real numbers.
Then we get Say ∣z∣=a 2+b 2,\displaystyle |z|=\sqrt{a^2+b^2},\:∣z∣=a 2+b 2,z‾=a−b i\displaystyle \overline{\,z\,}=a-b\bf{i}z=a−b i.
Moreover, z+z‾=2⋅ℜ(z)\displaystyle \;z+\overline{\,z\,}=2\cdot\Re(z)z+z=2⋅ℜ(z) and z−z‾=2⋅ℑ(z)\displaystyle \;z-\overline{\,z\,}=2\cdot\Im(z)z−z=2⋅ℑ(z)
z⋅z‾=∣z∣2\displaystyle z\cdot \overline{\,z\,}=|z|^2 z⋅z=∣z∣2
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A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Feb 28, 2019
#10
Dr.Peterson said:
I checked the Blitzer book I've taught from, and see that, indeed, it doesn't use any notation for conjugates, because it doesn't do any more with them than use them to simplify expressions or point out that solutions to quadratic equations come in conjugate pairs, neither of which needs a notation.
Clearly your new book is at a higher level, so you may need to go back to where they introduced complex numbers and read carefully, to see their notation and conventions.
In the solution you quoted, they don't take the time, as one would in a proof, to define variables or justify steps, trusting that you know what to expect: When they replace z with a+bi, they are implicitly saying, "suppose that the complex number z has components a and b", and so on. So you have to build up your expectations to match theirs, by reading slowly and carefully, even when you think the topic is one you already know. Probably there has been an example that demonstrated these ways of thinking.
Click to expand...
I think that is it. Blitzer introduces complex numbers and their conjugates and how to do operations with complex numbers. He demonstrates alos how to clear denominators of complex numbers by multiplying by conjugates. In the Swokowski book I skipped the first 65 pages because I couldn't see anything thee that wasn't some kind of review of things I have learned from Blitzer. Now I am going to check over those pages again and see if I missed something new. It is quite a nice book and pretty well illustrated.
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Feb 28, 2019
#11
pka said:
Unlike Prof. Peterson, I do not have nor have I seen the text by Bilixer. I I knew Swokowski and his text is excellent. There are even solution manuals available on Amazon. I was typing a replay late last night when the LaTeX went haywire. So I simply stopped. I wish I had been able to save the code but alas its gone.
Here is the a bit of it. Any complex number is two parts: A real part & an imaginary part.
Say z=5−3 i\displaystyle z=5-3\bf{i}z=5−3 i, the real part ℜ(z)=5&\displaystyle \Re(z)=5~\&~ℜ(z)=5&the imaginary part ℑ z=−3\displaystyle \Im{z}=-3 ℑ z=−3.
Thus z=a+b i\displaystyle z=a+b\bf{i}z=a+b i, the real part ℜ(z)=a\displaystyle \Re(z)=a ℜ(z)=a& the imaginary part ℑ z=b\displaystyle \Im{z}=b ℑ z=b where both a&b\displaystyle a~\&~b a&b are real numbers.
Then we get Say ∣z∣=a 2+b 2,\displaystyle |z|=\sqrt{a^2+b^2},\:∣z∣=a 2+b 2,z‾=a−b i\displaystyle \overline{\,z\,}=a-b\bf{i}z=a−b i.
Moreover, z+z‾=2⋅ℜ(z)\displaystyle \;z+\overline{\,z\,}=2\cdot\Re(z)z+z=2⋅ℜ(z) and z−z‾=2⋅ℑ(z)\displaystyle \;z-\overline{\,z\,}=2\cdot\Im(z)z−z=2⋅ℑ(z)
z⋅z‾=∣z∣2\displaystyle z\cdot \overline{\,z\,}=|z|^2 z⋅z=∣z∣2
Click to expand...
Yes, I think I follow what you are doing here. I worked something out on this problem today and substituted specific numbers for variables:
And now that I think I understand the problem better I have another question: What is the difference between the two expressions, I mean between z+w with a continuous overhead line and z+w where each variable has its own separate overhead line?
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Feb 28, 2019
#12
Yes all of that is correct. But it is easier to remember that the conjugate of a sum is the sum of the conjugates.
Here is a personal favorite of mine. 1 z=z‾∣z∣2\displaystyle \dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}z 1=∣z∣2 z
To see how useful this is, consider the following: 2+i 3+4 i=(2+i)(3−4 i)5\displaystyle \dfrac{2+\bf{i}}{3+4\bf{i}}= \dfrac{(2+\bf{i})(3-4\bf{i})}{5}3+4 i 2+i=5(2+i)(3−4 i)
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Mar 1, 2019
#13
pka said:
Yes all of that is correct. But it is easier to remember that the conjugate of a sum is the sum of the conjugates.
^2
Here is a personal favorite of mine. 1 z=z‾∣z∣2\displaystyle \dfrac{1}{z}=\dfrac{\overline{\,z\,}}{|z|^2}z 1=∣z∣2 z
To see how useful this is, consider the following: 2+i 3+4 i=(2+i)(3−4 i)5\displaystyle \dfrac{2+\bf{i}}{3+4\bf{i}}= \dfrac{(2+\bf{i})(3-4\bf{i})}{5}3+4 i 2+i=5(2+i)(3−4 i)
Click to expand...
I don't get what you are driving at. The reciprocal of 1/z? Is the z a complex number?If si, that would mean that 1/2+i would equal 2-i/(2+i)^2. According to Wolfram Alpha, that is a false statement.
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Mar 1, 2019
#14
allegansveritatem said:
I don't get what you are driving at. The reciprocal of 1/z? Is the z a complex number?If si, that would mean that 1/2+i would equal 2-i/(2+i)^2. According to Wolfram Alpha, that is a false statement.
Click to expand...
I think you are missing the whole point. ∣2+i∣=2 2+1 2=5\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 ∣2+i∣=2 2+1 2=5 so that ∣2+i∣2=5\displaystyle |2+\bf{i}|^2=5 ∣2+i∣2=5.
So 1 2+i=2−i 5=2 5−1 5 i\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}2+i 1=5 2−i=5 2−5 1i
You need to learn absolute value.
Here is another example: ℜ e[(4−3 i)−1]=4 25&ℑ m[(4−3 i)−1]=3 25\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}ℜ e[(4−3 i)−1]=2 5 4&ℑ m[(4−3 i)−1]=2 5 3
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allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Mar 3, 2019
#15
pka said:
I think you are missing the whole point. ∣2+i∣=2 2+1 2=5\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 ∣2+i∣=2 2+1 2=5 so that ∣2+i∣2=5\displaystyle |2+\bf{i}|^2=5 ∣2+i∣2=5.
So 1 2+i=2−i 5=2 5−1 5 i\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}2+i 1=5 2−i=5 2−5 1i
You need to learn absolute value.
Here is another example: ℜ e[(4−3 i)−1]=4 25&ℑ m[(4−3 i)−1]=3 25\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}ℜ e[(4−3 i)−1]=2 5 4&ℑ m[(4−3 i)−1]=2 5 3
Click to expand...
I know something about absolute value but not enough I guess to know how to make much use of it. I have copied this out and will try to work it out tomorrow. 1 over 4-3i is 1/4 +1/3i, no? Why do we need absolute value? Does that cancel out the i?
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Mar 3, 2019
#16
allegansveritatem said:
1 over 4-3i is 1/4 +1/3i, no? Why do we need absolute value? Does that cancel out the i?
Click to expand...
That is incorrect. Look at this: 1 7=1 3+4≠1 3+1 4\displaystyle \frac{1}{7}=\frac{1}{3+4}\ne\frac{1}{3}+\frac{1}{4}7 1=3+4 1=3 1+4 1.
A
allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Mar 3, 2019
#17
pka said:
I think you are missing the whole point. ∣2+i∣=2 2+1 2=5\displaystyle |2+\bf{i}|=\sqrt{2^2+1^2}=\sqrt5 ∣2+i∣=2 2+1 2=5 so that ∣2+i∣2=5\displaystyle |2+\bf{i}|^2=5 ∣2+i∣2=5.
So 1 2+i=2−i 5=2 5−1 5 i\displaystyle \dfrac{1}{2+\bf{i}}=\dfrac{2-\bf{i}}{5}=\frac{2}{5}-\frac{1}{5}\bf{i}2+i 1=5 2−i=5 2−5 1i
You need to learn absolute value.
Here is another example: ℜ e[(4−3 i)−1]=4 25&ℑ m[(4−3 i)−1]=3 25\displaystyle \Re e[(4-3\bf{i})^{-1}]=\frac{4}{25}\;\&\;\Im m[(4-3\bf{i})^{-1}]=\frac{3}{25}ℜ e[(4−3 i)−1]=2 5 4&ℑ m[(4−3 i)−1]=2 5 3
Click to expand...
yes, I see what you did here: You took the reciprocal of 1/4-3i which is 4+3i/25. I have been studying this morning more about the concept of absolute value and have a clearer idea of what is going on now. I've also just watched several videos on solving equations containing absolute values. Thanks for pointing out the fact that I need to learn (or review?) the basics.
pka
Elite Member
Joined Jan 29, 2005 Messages 11,988
Mar 3, 2019
#18
allegansveritatem said:
I have been studying this morning more about the concept of absolute value and have a clearer idea of what is going on now. I've also just watched several videos on solving equations containing absolute values. Thanks for pointing out the fact that I need to learn (or review?) the basics.
Click to expand...
In general ∣a+b i∣=a 2+b 2\displaystyle |a+b\bf{i}|=\sqrt{a^2+b^2} ∣a+b i∣=a 2+b 2 Example ∣6−8 i∣=(6)2+(−8)2=10\displaystyle |6-8\bf{i}|=\sqrt{(6)^2+(-8)^2}=10∣6−8 i∣=(6)2+(−8)2=1 0
Now lets look the connection to conjugates.
(a+b i)(a+b i)‾=(a+b i)(a−b i)\displaystyle (a+b\bf{i})\overline{(a+b\bf{i})}=(a+b\bf{i})(a-b\bf{i})(a+b i)(a+b i)=(a+b i)(a−b i)
(a+b i)(a−b i)=(a 2−b 2 i 2)+(a b i−b a i)\displaystyle (a+b\bf{i})(a-b\bf{i})=(a^2-b^2\bf{i}^2)+(ab\bf{i}-ba\bf{i})(a+b i)(a−b i)=(a 2−b 2 i 2)+(a b i−b a i)
(a+b i)(a−b i)=(a 2+b 2)\displaystyle (a+b\bf{i})(a-b\bf{i})=(a^2+b^2)(a+b i)(a−b i)=(a 2+b 2)
∣a+b i∣2=(a 2+b 2)\displaystyle |a+b\bf{i}|^2=(a^2+b^2)∣a+b i∣2=(a 2+b 2)
Thus we have shown that ∣z∣2=z⋅z‾\displaystyle |z|^2=z\cdot\overline{\,z\,}∣z∣2=z⋅z
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allegansveritatem
Full Member
Joined Jan 10, 2018 Messages 962
Mar 5, 2019
#19
here is my problem: When I multiply (a+bi)(a+bi) I get (a^+2bi-b^2). I don't see how you can say multiplying the first two terms = (a+bi)(a-bi).
K
ksdhart2
Senior Member
Joined Mar 25, 2016 Messages 1,297
Mar 5, 2019
#20
allegansveritatem said:
here is my problem: When I multiply (a+bi)(a+bi) I get (a^+2bi-b^2). I don't see how you can say multiplying the first two terms = (a+bi)(a-bi).
Click to expand...
Well, that's mostly correct (accounting for a few typos). It is true that (a+b i)(a+b i)=(a+b i)2=a 2+2 a b i−b 2\displaystyle (a + bi)(a + bi) = (a + bi)^2 = a^2 + 2abi - b^2(a+b i)(a+b i)=(a+b i)2=a 2+2 a b i−b 2 ... but that's not what we have here. Did you notice the line over the second (a+b i)\displaystyle (a + bi)(a+b i)? Do you recall that the overline means "complex conjugate?" What is the complex conjugate of a+b i\displaystyle a + bi a+b i? Can you see why that means that the statement (a+b i)(a+b i)‾=(a+b i)(a−b i)\displaystyle (a + bi)\overline{(a +bi)} = (a + bi)(a - bi)(a+b i)(a+b i)=(a+b i)(a−b i) is just a tautology?
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Planar real affine symplectic map: How to construct a fixed point?
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In the real affine plane A 2 R A R 2, every affine map, which is not pure translation, has at least one fixed point. In special cases, there exist simple geometric constructions:
Euclidean motions: If a Euclidean motion f∈A S O(2,R):=R 2⋊S O(2,R)f∈A S O(2,R):=R 2⋊S O(2,R) maps points A↦A′,B↦B′A↦A′,B↦B′, and is not pure translation (and not the identity), then a fixed point (centre of rotation) is obtained by intersecting the perpendicular bisectors of the segments A A′A A′ and B B′B B′.
Spiral similarities: If a spiral similarity f∈A G O+(2,R):=R 2⋊G O+(2,R)f∈A G O+(2,R):=R 2⋊G O+(2,R), where G O+:=R>0×S O G O+:=R>0×S O is the group of orientation preserving spiral similarities about the origin, maps points A↦A′,B↦B′A↦A′,B↦B′, and is not pure scaled translation (and not the identity), then a fixed point is obtained by first intersecting the two straight lines (A∨A′)∧(B∨B′)=:D(A∨A′)∧(B∨B′)=:D and then intersecting the circles ⊙(A,B,D)∩⊙(A′,B′,D)=:{C,D}⊙(A,B,D)∩⊙(A′,B′,D)=:{C,D} and take C C the fixed point (the centre of the spiral similarity).
Question:
Is there a simple way, how to construct geometrically a fixed point of an orientation preserving affine symplectic map f∈A S p+(2,R):=R 2⋊S p+(2,R)f∈A S p+(2,R):=R 2⋊S p+(2,R), i.e. of an area preserving affine f∈A S L(2,R)f∈A S L(2,R), which is not pure translation?
As the group A S L(2,R)A S L(2,R) is five-dimensional, it would not be sufficient to consider just two points along with their images, because two points would just give four conditions.
Any ideas or references? Thank you!
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edited Mar 17 at 11:41
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I found a publication which treats fixed points of general affinities in the real plane:
One starts with a parallelogramm a,b,c,d. By affine-linearity, its image a',b',c',d' is a parallelogramm too. Define
p:=(a∨b)∧(a′∨b′)q:=(b∨c)∧(b′∨c′)r:=(c∨d)∧(c′∨d′)s:=(d∨a)∧(d′∨a′)p:=(a∨b)∧(a′∨b′)q:=(b∨c)∧(b′∨c′)r:=(c∨d)∧(c′∨d′)s:=(d∨a)∧(d′∨a′)
Then
m:=(p∨r)∧(q∨s)m:=(p∨r)∧(q∨s)
is a fixed point (at least if a,b,c,d satisfy the condition in the theorem there - but the construction can be extended to more general cases in the projective closure).
For a general quadrangle, the midpoints of its sides form a parallelogram and can be used for that construction.
Kovacs' construction seems to be rather for the parallelogram as a quadrilateral than as a quadrangle.
For a planar affine symplectomorphism, the area of the parallelogramm is preserved, but this fact does not simplify the construction.
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11516 | https://www.princeton.edu/~aaa/Public/Teaching/ORF523/S16/ORF523_S16_Lec4_gh.pdf | ORF 523 Lecture 4 Spring 2015, Princeton University Instructor: A.A. Ahmadi Scribe: G. Hall Tuesday, February 16, 2016 When in doubt on the accuracy of these notes, please cross check with the instructor’s notes, on aaa. princeton. edu/ orf523 . Any typos should be emailed to gh4@princeton.edu.
Consider the general form of an optimization problem: min. f(x) s.t. x ∈Ω.
The few optimality conditions we’ve seen so far characterize locally optimal solutions. (In fact, they do not even do that since we did not have a “necessary and sufficient” condition).
But ideally, we would like to make statements about global solutions. This comes at the expense of imposing some additional structure on f and Ω. By and large, the most successful structural property that we know of that achieves this goal is convexity. This motivates an in-depth study of convex sets and convex functions. In short, the reasons for focusing on convex optimization problems are as follows: • They are close to being the broadest class of problems we know how to solve efficiently.
• They enjoy nice geometric properties (e.g., local minima are global).
• There’s excellent software that readily solves a large class of convex problems.
• Numerous important problems in a variety of application domains are convex!
1 From local to global minima 1.1 Definition Definition 1. A set Ω⊆Rn is convex, if for all x, y ∈Ωand ∀λ ∈[0, 1] λx + (1 −λ)y ∈Ω.
A point of the form λx + (1 −λ)y, λ ∈[0, 1] is called a convex combination of x and y. As λ varies between [0, 1], a “line segment” is being formed between x and y as shown in Figure 1.
1 Figure 1: Convex sets and convex combinations.
Definition 2. A function f : Rn →R is convex if its domain dom(f) is a convex set and for all x, y ∈dom(f) and ∀λ ∈[0, 1], we have f(λx + (1 −λ)y) ≤λf(x) + (1 −λ)f(y).
Geometrically, the line segment connecting (x, f(x)) to (y, f(y)) should sit above the graph of the function.
Figure 2: An illustration of the definition of a convex function.
Theorem 1. Consider an optimization problem min. f(x) s.t. x ∈Ω, where f is a convex function and Ωis a convex set. Then, any local minimum is also a global minimum.
Proof: Let ¯ x be a local minimum.
⇒¯ x ∈Ωand ∃ϵ > 0 s.t. f(¯ x) ≤f(x) ∀x ∈B(¯ x, ϵ).
2 Suppose for the sake of contradiction that ∃z ∈Ωwith f(z) < f(¯ x).
Because of convexity of Ω, we have λ¯ x + (1 −λ)z ∈Ω, ∀λ ∈[0, 1].
By convexity of f, we have f(λ¯ x + (1 −λ)z) ≤λf(¯ x) + (1 −λ)f(z) < λf(¯ x) + (1 −λ)f(¯ x) = f(¯ x).
But, as λ →1, (λ¯ x + (1 −λ)z) →¯ x and the previous inequality contradicts local optimality of ¯ x. □ This theorem, as simple as it is, is one of the most important theorems in convex analysis.
Let’s now delve deeper in the theory of convex sets and convex functions.
2 Convex sets If you refer back to the definition of a convex set, you see that the condition is required ∀x, y ∈Ωand ∀λ ∈[0, 1]. Under mild conditions, it is possible to fix λ to a constant.
2.1 Midpoint convexity Definition 3. A set Ω⊆Rn is midpoint convex if for all x, y ∈Ω, the midpoint between x and y is also in Ω. In other words, x, y ∈Ω⇒1 2x + 1 2y ∈Ω.
It’s clear that convex sets are midpoint convex. But the converse is also true except in pathological cases.
Figure 3: Intuition 3 Theorem 2. A closed midpoint convex set Ωis convex.
Proof: Pick x, y ∈Ω, λ ∈[0, 1]. For any integer k, define λk to be the k-bit rational number closest to λ: λk = c12−1 + c22−2 + . . . + ck2−k, where ci ∈{0, 1}. Then for all k, λkx + (1 −λk)y ∈Ωas we can apply midpoint convexity k times recursively.
As k →∞, λk →λ. By closedness of Ω, we conclude that λx + (1 −λ)y ∈Ω. □ Remark: An example of a midpoint convex set that’s not convex is the set of all rationals in [0, 1].
2.2 Common examples of convex sets The following sets are convex (prove convexity in each case): • Hyperplanes:{x| aTx = b} (a ∈Rn, b ∈R, a ̸= 0) • Halfspaces: {x| aTx ≤b} (a ∈Rn, b ∈R, a ̸= 0) • Euclidian balls: {x| ||x −xc|| ≤r} (xc ∈Rn, r ∈R, ||.|| is the 2-norm) • Ellipsoids: {x| p (x −xc)TP(x −xc) ≤r} (xc ∈Rn, r ∈R, P ≻0) Proof: Define ||x||P = √ xTPx. This is a norm (as you proved on the homework!).
As this set is closed, it suffices to show midpoint convexity. Pick x, y ∈S. We have ||x −xc||P ≤r and ||y −xc||P ≤r. Then x + y 2 −xc P = 1 2x −1 2xc + 1 2y −1 2xc P ≤1 2||x −xc||P + 1 2||y −xc||P ≤1 2r + 1 2r = r.
• The set of symmetric positive semidefinite matrices: Sn×n + = {P ∈Sn×n | P ⪰0}.
Proof: Let A ⪰0, B ⪰0 and λ ∈[0, 1]. (Again, it would be enough to look at λ = 1 2 as the set is closed.) Pick any y ∈Rn. Then, yT (λA + (1 −λ)B) y = λyTAy + (1 −λ)yTBy ≥0.
4 • The set of nonnegative polynomials in n variables and of degree d.
(A polynomial p is nonnegative, if p(x) ≥0, ∀x ∈Rn.) • The set of optimal solutions of the problem minx∈Ωf(x) where f is convex and Ωis a convex set.
• Proving convexity of a set is not always easy like our previous examples. For instance, for n > 2, fix any Q1, Q2 ∈Sn×n: try to show that the following set in R2 is convex S = {(xTQ1x, xTQ2x) | ||x|| = 1}.
In Figure 4g, you can see an example of what the set S can look like in the case of two quadratics in four variables.
• Interestingly, the analogue of the statement above would fail to be true if we had three quadratic maps or for two polynomial maps of higher degree.
The following theorem (which we will prove later) shows that indeed testing convexity of sets can be a very computationally demanding task.
Theorem 3 (). Given a multivariate polynomial p of degree 4, it is NP-hard to test whether the set {x| p(x) ≤1} is convex.
5 (a) A hyperplane (b) A halfspace (c) A Euclidian ball (d) An ellipsoid (e) ( (x, y, z) | x y y z !
⪰0 ) (Image credit: ) (f) Matlab code to generate S (g) The set S generated by this Matlab code Figure 4: Examples of convex sets 6 2.3 Operations on convex sets It is very easy to see that the intersection of two convex sets is convex: Ω1 convex, Ω2 convex ⇒Ω1 ∩Ω2 convex.
Proof: Pick x ∈Ω1 ∩Ω2, y ∈Ω1 ∩Ω2. We have, for all λ ∈[0, 1]: λx + (1 −λ)y ∈Ω1 because Ω1 is convex, λx + (1 −λ)y ∈Ω2 because Ω2 is convex ⇒λx + (1 −λ)y ∈Ω1 ∩Ω2. □ This statement is also true for infinite intersections. Obviously the union of two convex sets may not be convex.
Similarly, it is easy to show that the Minkowski sum of two convex sets is convex.
Example: Polyhedra • A polyhedron is the solution set of finitely many linear inequalities. These sets are very important sets in optimization theory. as they form the feasible sets of linear programs.
• A polyhedron can be written in the form {x| Ax ≤b} where A is an m × n matrix and b is an m × 1 vector.
• These sets are convex as they are the intersection of halfspaces aT i x ≤bi, where aT i is the ith row of A.
Figure 5: An example of a polyhedron 7 2.4 Epigraphs The notion of epigraphs nicely connects the concepts of convex functions and convex sets.
Definition 4. The epigraph epi(f) of a function f : Rn →R is a subset of Rn+1 defined as epi(f) = {(x, t) | x ∈domain(f), f(x) ≤t}.
Figure 6: Examples of epigraphs Theorem 4. A function f : Rn →R is convex if and only if its epigraph is convex (as a set).
Proof: (⇒) Suppose epi(f) not convex ⇒∃(x, tx), (y, ty), λ ∈[0, 1] s.t. f(x) ≤tx, f(y) ≤ty and (λx + (1 −λ)y, λtx + (1 −λty) / ∈epi(f) ⇒f(λx + (1 −λ)y) > λtx + (1 −λ)ty ≥λf(x) + (1 −λ)f(y).
This implies that f is not convex.
(⇐) Suppose f is not convex ⇒∃x, y ∈dom(f), λ ∈[0, 1] s.t.
f(λx + (1 −λ)y) > λf(x) + (1 −λ)f(y).
(1) Pick (x, f(x)), (y, f(y)) ∈epi(f). Then (1) ⇒(λx + (1 −λ)y, λf(x) + (1 −λ)f(y)) / ∈epi(f).
□ 8 2.5 Convexity of sublevel sets Definition 5. The α-sublevel set of a function f : Rn →R is the set Sα = {x ∈domain(f) | f(x) ≤α}.
(a) Sublevel set of a function in one dimension (b) Sublevel sets of a bivariate function Figure 7: Examples of sublevel sets Theorem 5. If a function f : Rn →R is convex then all its sublevel sets are convex sets.
Proof: Pick x, y ∈Sα, λ ∈[0, 1].
x ∈Sα ⇒f(x) ≤α; y ∈Sα ⇒f(y) ≤α f convex ⇒f(λx + (1 −λ)y) ≤λf(x) + (1 −λ)f(y) ≤λα + (1 −λ)α = α ⇒λx + (1 −λ)y ∈Sα.
A function whose sublevel sets are all convex is called quasiconvex.
Although convexity implies quasiconvexity, the converse is not true. See Figure 8.
2.6 Convex hulls Given x1, . . . , xm ∈Rn, a point of the form λ1x1 + . . . + λmxm with λi ≥0, P i λi = 1 is called a convex combination of the points x1, . . . , xm.
9 Figure 8: Examples of quasiconvex functions that are not convex Lemma 1. A set S ⊆Rn is convex iffit contains every convex combination of its points.
Definition 6. The convex hull of a set S ⊆Rn, denoted by conv(S), is the set of all convex combinations of the points in S: conv(S) = { m X i=1 λixi | xi ∈S, λi ≥0, X λi = 1}.
Theorem 6 (Carath´ eodory, 1907). Consider a set S ⊆Rd. Then every point in conv(S) can be written as a convex combinaiton of d + 1 points in S.
Proof: We give the standard proof as it appears, e.g., in . Let x ∈conv(S), then x = α1y1 + . . . + αmym, for some αi ≥0, P i αi = 1, and yi ∈S. If m ≤d + 1, we are done (why?). Suppose m > d + 1. We’ll give another representation of x using m −1 points. An iteration of this idea would finish the proof.
Consider the system of d + 1 equations: γ1y1 + . . . + γmym = 0 γ1 + . . . + γm = 0 in m variables γi ∈R. As m > d+1, this system has infinitely many solutions. Let γ1, . . . , γm be any nonzero solution. We must have γi > 0 for some i (why?). Let τ = min i αi γi : γi > 0 := αi0 γi0 .
10 Let λi = αi −τγi for i = 1, . . . , m.
Claims: (i) λi ≥0, (ii) m X i=1 λi = 1, (iii) X λiyi = xi, (iv) λi0 = 0.
(i) λi = αi − αi0 γi0 γi ≥αi −αi γi γi = 0.
(ii) P λi = P αi −τ P γi = P αi = 1.
(iii) P λiyi = P(αi −τγi)yi = P αiyi −τ P γiyi = x.
(iv) λi0 = αi0 − αi0 γi0 γi0 = 0.
□ Theorem 7. The convex hull of S is the smallest convex set that contains S; i.e., if B is convex and S ⊆B, then conv(S) ⊆B.
The proof is an exercise on the homework. Let us just show that conv(S) is convex.
Pick x, y ∈conv(S). This implies: x = µ1u1 + . . . + µkuk, ui ∈S, µi ≥0, X µi = 1 y = η1v1 + . . . + ηmvm, vi ∈S, ηi ≥0, X ηi = 1 Let λ ∈[0, 1].
λx + (1 −λ)y = λµ1u1 + . . . + λµkuk + (1 −λ)η1v1 + . . . + ηmvm where ui, vi ∈S, λiµi ≥0, (1 −λi)ηi ≥0 and λ X µi + (1 −λ) X ηi = λ + (1 −λ) = 1.
□ Theorem 8. Let l : Rn →R be a linear function and Ω⊂Rn a compact set. Then, min x∈Ωl(x) = min x∈conv(Ω) l(x).
11 (You can remove the compactness assumption and replace “min” with “inf”.) Proof: It is clear that RHS ≤LHS as we are optimizing over a larger set (S ⊆conv(S)). To show that LHS ≤RHS, let ¯ x ∈argminx∈conv(Ω)l(x).
Then, ¯ x = k X i=1 λiyi, with yi ∈Ω, X λi = 1, λi ≥0.
RHS = l(¯ x) = l( X λiyi) = X λil(yi) ≥ X λi min i l(yi) = min i l(yi) ≥LHS.
where we used for the last inequality that yi ∈S. □.
Consider a general optimization problem min x f(x) s.t. x ∈Ω.
We can first rewrite it in the so-called “epigraph form” : min x,α α s.t. x ∈Ω, f(x) ≤α.
These problems are equivalent in the sense that they achieve the same optimal value and one can map any optimal solution of one to the other. Notice that in the new problem, the objective is linear. We can rewrite it again (via Theorem 8): min x,α α s.t. (x, α) ∈conv{x ∈Ω, f(x) ≤α}.
12 Note that the objective is linear and the feasible set is convex! So we rewrote an arbitrary optimization problem as an optimization problem with a convex objective and a convex feasible set. But there is a catch: this transformation is not algorithmic at all. We are hiding all the difficulty in the convex hull operation. In general, it is not easy to write down a description for the convex hull a set, even if the set has a simple description (let’s say it’s described by quadratic inequalities).
Note that the argument we gave also goes against the common belief about “convex problems being easy”. Indeed, the structure and functional description of the feasible set, beyond convexity, cannot be ignored.
3 Convex optimization problems Motivated in part by our discussion above, we will define a convex optimization problem to be any optimization problem of the form min. f(x) s.t. gi(x) ≤0, i = 1, . . . , m, hj(x) = 0, j = 1, . . . , k, where f, gi : Rn →R are convex functions and hi : Rn →R are affine functions.
Let Ωdenote the feasible set, i.e., Ω= {x ∈Rn| gi(x) ≤0, hj(x) = 0}.
• For a convex optimization problem, the set Ωis always a convex set (why?).
• The converse is not true: – Consider for example, Ω= {x ∈R| x3 ≤0}.
Then Ωis a convex set, but minimizing a convex function over Ωis not a convex optimization problem per our definition.
– However, the same set can be represented as Ω= {x ∈Rn| x ≤0}, and then this would be a convex optimization problem with our definition.
• We require this stronger notion because otherwise many abstract and complicated optimization problems can be formulated as optimization problems over a convex set.
(Think, e.g., of the set of nonnegative polynomials). The stronger definition is much closer to what we can solve algorithmically.
13 Figure 9: The feasible set S = {x| g1(x) ≤0, g2(x) ≤0} can be convex even when the defininig inequalities are not even quasiconvex.
The software CVX that we will be using only accepts convex optimization problems defined as above; i.e, CVX accepts the following constraints: • Convex ≤0.
• Affine == 0.
Notes Further reading for this lecture can include Chapter 2 of .
References Amir Ali Ahmadi, Alex Olshevsky, Pablo A Parrilo, and John N Tsitsiklis. NP-hardness of deciding convexity of quartic polynomials and related problems. Mathematical Pro-gramming, 137(1-2):453–476, 2013.
Alexander Barvinok. A Course in Convexity, volume 54. American Mathematical Soc., 2002.
S. Boyd and L. Vandenberghe.
Convex Optimization.
Cambridge University Press, boyd/cvxbook/, 2004.
14 |
11517 | https://lewisgroup.uta.edu/Lectures/stabilityDT.pdf | 1 © Copyright F.L. Lewis 1998 All rights reserved Updated: Tuesday, October 07, 2008 STABILITY OF DISCRETE-TIME SYSTEMS The stability properties of continuous-time systems depend on the locations of the poles in the Laplace s-plane. The stability boundary is the jω-axis, with poles in the open left-half plane being stable. The stability properties of discrete-time systems depend on the locations of the poles in the z-plane. The stability boundary is the unit circle, with poles inside the unit circle being stable. Note that sT z e = with T the sample time. Let s j α ω = + , then ( ) j T T j T z z e e e z e α ω α ω + ∠ = = = . If the CT pole is on the jω-axis, then the real part is 0 α = , so that 1 T z eα = = , and the DT pole is on the unit circle. Thus, the jω-axis in the s-plane maps to the unit circle in the z-plane. Stability of DT Input/Output Systems Input/output systems may be described by a transfer function ( ) ( ) ( ) n z H z d z = or equivalently by an impulse response hk. The system is said to be: Asymptotically Stable (AS) if uk = 0 for all time k implies that yk goes to zero with time. Marginally Stable (MS) if uk = 0 for all time k implies that yk is bounded for all time. In terms of time signals, a decaying exponential is AS, a sine wave is MS, the unit step is MS, and an increasing exponential is unstable. Note that the DT exponential is ak, which decays to zero if and only if the magnitude of a is less than one, i.e. a is inside the unit circle in the z-plane. H(z) yk uk H(z) yk uk 2 A system is AS if and only if (iff) the impulse response goes to zero with time, and MS iff the impulse response is bounded. The natural modes that occur in hk depend on the locations of the poles, defined as the roots of the denominator d(z) of H(z). Thus, the system is AS iff all poles are in the interior of the unit circle (IUC) in the z-plane (e.g. strictly inside the unit circle with no poles on the unit circle). The system is MS iff all poles of H(z) are inside or on the unit circle, with any poles occurring on the unit circle being non-repeated. Stability of DT State Variable Systems The discrete-time linear time invariant (LTI) state-space form is 1 0 , k k k k k k x Ax Bu x y Cx Du + = + = + . The transfer function is given by [ ] [ ] 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) c c H z C zI A B D C adj zI A B C adj zI A B D zI A N z N z D zI A zI A z z − = − + − − + − = + = = = − − Δ Δ . where subscript ‘c’ denotes variables after pole/zero cancellation. The denominator of H(z) is the characteristic polynomial ( ) z zI A Δ = − . The system poles are the roots of the characteristic equation ( ) 0 z zI A Δ = − = . The transfer function poles are the roots of ( ) 0 c z Δ = The (internal) system poles (e.g. roots of Δ(z)) should be distinguished from the (input/output) poles of the transfer function. There may be some pole/zero cancellation in computing the transfer function. Then the denominator of H(z), i.e. ( ) c z Δ , is not the same as Δ(z), since some system poles do not occur in H(z). Therefore, the definitions of stability need to be modified to differentiate internal stability (based on ( ) z Δ ) from the external (e.g. input/output stability, based on ( ) c z Δ ) as just defined. Recall that 1 1 0 ( ) ( ) ( ) ( ) X z zI A zx zI A BU z − − = − + − 1 0 ( ) ( ) ( ) ( ) Y z C zI A zx H z U z − = − + . 3 The SV system can be considered as a 2-input, 2-output system, with the two inputs of the control uk and the initial state x0, and the two outputs of yk and xk. Therefore there are two definitions of internal stability. The system is said to be: (Internally) Asymptotically Stable (AS) if uk= 0 for all time k implies that xk goes to zero with time for all initial conditions x0. Bounded-Input/Bounded-Output Stable (BIBOS) if uk bounded for all time k implies that yk is bounded for all time when x0=0. Though the first of these is usually simply called AS, it is not the same definition as the AS defined for input/output systems, where only the output yk is required to go to zero. Here, we require all the internal states to go to zero. AS as defined here is concerned with the effects of the initial state x0 on xk when the input is equal to zero. Therefore, it depends on the equation 1 1 0 ( ) ( ) ( ) ( ) X z zI A zx zI A BU z − − = − + − 1 0 ( ) ( ) X z zI A zx − = − On the other hand, BIBOS is concerned with the throughput effects of uk on yk when x0= 0. Therefore, it depends on the equation 1 0 ( ) ( ) ( ) ( ) Y z C zI A zx H z U z − = − + ( ) ( ) ( ) Y z H z U z = Therefore, from the first equation, the system is (internally) AS iff all system natural modes decay with time. This occurs iff all the system poles are strictly inside the unit circle. For AS all the roots of ( ) z Δ must be strictly inside the unit circle. To find a condition for BIBOS, according to the second equation, when x0= 0 the output may be found by convolving the pulse response hk with the input, z-1 A C xk yk uk xk+1 Linear Discrete-Time State-Space System interpreted as 2-input/2-output system D B x0 xk z-1 A C xk yk uk xk+1 Linear Discrete-Time State-Space System interpreted as 2-input/2-output system D B x0 xk 4 1 0 k k k i i i y h u − − = = ∑ where hk is given in terms of the Markov parameters by 0 , 1 0 > = = − k B CA h D h k k If the pulse response is a decaying exponential, then the output is bounded for all bounded inputs. On the other hand, suppose hk is the unit step for instance. Convolution of any uk with the unit step simply gives the area under uk. Thus, if uk is, e.g., also the unit step, then the output would be the unit ramp, which is not bounded. It turns out that the output yk is bounded for all bounded uk (when x0=0) if and only if the pulse response goes to zero with time. Therefore, the system is BIBOS iff the poles of the transfer function, as given by ( ) 0 c z Δ = , are strictly inside the unit circle. To study the AS of a system, one may apply the Routh test to the characteristic polynomial ( ) z Δ . To study the BIBOS of a system, one may apply the Routh test to the denominator ( ) c z Δ of the transfer function after pole/zero cancellation, if any. Note: AS requires ALL the system poles to be in the OLHP. BIBOS only requires the poles remaining in the transfer function, after pole/zero cancellation, to be in the OLHP. Therefore, it is clear that AS implies BIBOS. Relation of I/O Stability to Internal SV Stability I/O Stability Internal SV Stability MS- Poles of H(z) are inside unit circle, with possible nonrepeated poles on unit circle AS- Poles of H(z) are strictly inside unit circle BIBOS- poles of H(z) (roots of ( ) c z Δ ) are strictly inside unit circle AS- poles (roots of ( ) z Δ ) are strictly inside unit circle I/O asymptotic stability is the same as internal BIBO stability Stability and Minimality A single-input/single-output (SISO) system is said to be minimal if there is no pole/zero cancellation. In this event, the poles of H(z) are the same as the roots of Δ(z), so that the system is AS if and only if it is BIBOS. |
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From Wikiquote
George Bernard Shaw (26 July 1856 – 2 November 1950), known at his insistence simply as Bernard Shaw, was an Irish playwright, critic, polemicist and political activist with a range incorporating both contemporary satire and historical allegory. He wrote more than sixty plays, including such works as Man and Superman (1902), Pygmalion (1913) and Saint Joan (1923). Shaw was awarded the Nobel Prize in Literature in 1925.
: See also:
: Man and Superman (1903)
Quotes
[edit]
1890s
[edit]
The old Whigs and new Tories of the school of Cobden and Bright, the "Philosophic Radicals," the economists of whom Bastiat is the type, Lord Wemyss and Lord Bramwell, Mr. Herbert Spencer and Mr. Auberon Herbert, Mr. Gladstone, Mr. Arthur Balfour, Mr. John Morley, Mr. Leonard Courtney: any of these is, in England, a more typical Anarchist than Bakounin. They distrust State action, and are jealous advocates of the prerogative of the individual, proposing to restrict the one and to extend the other as far as is humanly possible, in opposition to the Social-Democrat, who proposes to democratize the State and throw upon it the whole work of organizing the national industry, thereby making it the most vital organ in the social body.
'The Impossibilities of Anarchism', a paper read to the Fabian Society on 16 October 1891, reprinted in The Impossibilities of Anarchism (Fabian tract no. 45, 1895), pp. 4–5
Patriotism is, fundamentally, a conviction that a particular country is the best in the world because you were born in it...
The World (15 November 1893)
Pasteboard pies and paper flowers are being banished from the stage by the growth of that power of accurate observation which is commonly called cynicism by those who have not got it...
The World (18 July 1894), Music in London 1890-1894 being criticisms contributed week by week to The World (New York: Vienna House, 1973)
But no public man in these islands ever believes that the Bible means what it says: he is always convinced that it says what he means; and I have no reason to hope that Mr Coote may be an exception to the rule.
"The Living Pictures", The Saturday Review, LXXIX (April 6, 1895), 443, reprinted in Our Theatres in the Nineties (1932). Vol. 1. London: Constable & Co. 79-86
My method is to take the utmost trouble to find the right thing to say, and then to say it with the utmost levity.
Answers to Nine Questions (September 1896), answers to nine questions submitted by Clarence Rook, who had interviewed him in 1895
Die Walküre is endured by the average man because it contains four scenes for which he would sit out a Scotch sermon, or even a House of Commons debate. These are the love duet in the first act, Brunnhilde's announcement of death in the second, the ride of the Valkyries and the 'fire-charm' in the third. For them the ordinary playgoer endures hours of Wotan, with Christopher Sly's prayer in his heart. 'Would 'twere over!' Now, I am one of those elect souls who are deeply moved by Wotan. I grant you that as a long-winded, one-eyed gentleman backing a certain champion in a fight and henpecked out of his fancy because his wife objects to the moral character of the champion, he is a dreary person indeed . . . but to one who has understood all its beauties, its lofty aspirations, its tragedy, there is nothing trivial, nothing tedious in Die Walküre.
comments after attending the 1896 Bayreuth Festival. Quoted in "Visits to Valhalla: The Ectasy And the Agony" by Carolyn Abbatte, New York Times, March 26, 1989. Access date April 1, 2021.
We have no more right to consume happiness without producing it than to consume wealth without producing it.
Candida, Act I (1898)
I'm only a beer teetotaler, not a champagne teetotaler. I don't like beer.
Candida, Act III
We don't bother much about dress and manners in England, because as a nation we don't dress well and we've no manners.
You Never Can Tell, Act I (1898)
The great advantage of a hotel is that it's a refuge from home life.
You Never Can Tell, Act II
My specialty is being right when other people are wrong.
You Never Can Tell, Act IV
I had not achieved a success; but I provoked an uproar; and the sensation was so agreeable that I resolved to try again.
Plays: Pleasant and Unpleasant, Vol. I, preface (1898)
There is only one religion, though there are a hundred versions of it.
Plays Pleasant and Unpleasant, Vol. II, preface (1898)
You're not a man, you're a machine.
Arms and the Man, Act III (1898)
Why should you call me to account for eating decently? If I battened on the scorched corpses of animals, you might well ask me why I did that. Why should I be filthy and inhuman? Why should I be an accomplice in the wholesale horror and degradation of the slaughter-house?
Interview "What Vegetarianism Really Means: a Talk with Mr Bernard Shaw", in Vegetarian (15 January 1898), reprinted in Shaw: Interviews and Recollections, edited by A. M. Gibbs, 1990, p. 401
Quintessence Of Ibsenism (1891; 1913)
[edit]
: A review of the works and ideas of Henrik Ibsen
I have never admitted the right of an elderly author to alter the work of a young author, even when the young author happens to be his former self. In the case of a work which is a mere exhibition of skill in conventional art, there may be some excuse for the delusion that the longer the artist works on it the nearer he will bring it to perfection. Yet even the victims of this delusion must see that there is an age limit to the process, and that though a man of forty-five may improve the workmanship of a man of thirty-five, it does not follow that a man of fifty-five can do the same.
When we come to creative art, to the living word of a man delivering a message to his own time, it is clear that any attempt to alter this later on is simply fraud and forgery. As I read the old Quintessence of Ibsenism I may find things that I see now at a different angle, or correlate with so many things then unnoted by me that they take on a different aspect. But though this may be a reason for writing another book, it is not a reason for altering an existing one.
Preface to the 1913 edition
Just as the liar's punishment is, not in the least that he is not believed, but that he cannot believe any one else; so a guilty society can more easily be persuaded that any apparently innocent act is guilty than that any apparently guilty act is innocent.
The Two Pioneers
The salvation of the world depends on the men who will not take evil good-humouredly, and whose laughter destroys the fool instead of encouraging him.
What is the New Element in the Norwegian School?
The Philanderer (1893)
[edit]
It's well to be off with the Old Woman before you're on with the New.
Act II
The fickleness of the women I love is only equaled by the infernal constancy of the women who love me.
Act II
The test of a man or woman's breeding is how they behave in a quarrel.
Act IV
Mrs. Warren's Profession (1893)
[edit]
People are always blaming circumstances for what they are. I don't believe in circumstances. The people who get on in this world are the people who get up and look for the circumstances they want, and, if they can't find them, make them.
Vivie, Act II
There are no secrets better kept than the secrets everybody guesses.
Crofts, Act III
I know Miss Warren is a great devotee of the Gospel of Getting On.
Praed, Act IV
Caesar and Cleopatra (1898)
[edit]
Hail, Sphinx: salutation from Julius Caesar! I have wandered in many lands, seeking the lost regions from which my birth into this world exiled me, and the company of creatures such as I myself. I have found flocks and pastures, men and cities, but no other Caesar, no air native to me, no man kindred to me, none who can do my day's deed, and think my night's thought.
Act I
My way hither was the way of destiny; for I am he of whose genius you are the symbol: part brute, part woman, and part God — nothing of man in me at all. Have I read your riddle, Sphinx?
Act I
THEODOTUS: Caesar: you are a stranger here, and not conversant with our laws. The kings and queens of Egypt may not marry except with their own royal blood. Ptolemy and Cleopatra are born king and consort just as they are born brother and sister.
BRITANNUS (shocked): Caesar: this is not proper.
THEODOTUS (outraged): How!
CAESAR (recovering his self-possession): Pardon him, Theodotus: he is a barbarian, and thinks that the customs of his tribe and island are the laws of nature.
Act II; sometimes paraphrased as: The customs of your tribe are not laws of nature.
When a stupid man is doing something he is ashamed of, he always declares that it is his duty.
Act III
Again, there is the illusion of "increased command over Nature," meaning that cotton is cheap and that ten miles of country road on a bicycle have replaced four on foot. But even if man's increased command over Nature included any increased command over himself (the only sort of command relevant to his evolution into a higher being), the fact remains that it is only by running away from the increased command over Nature to country places where Nature is still in primitive command over Man that he can recover from the effects of the smoke, the stench, the foul air, the overcrowding, the racket, the ugliness, the dirt which the cheap cotton costs us.
Notes
1900s
[edit]
The novelties of one generation are only the resuscitated fashions of the generation before last.
Three Plays for Puritans, Preface (1900)
I delight in the war more & more. It has waked up the country out of its filthy wallowing in money (blood is a far superior bath); and it has put a fourpence on the Income Tax which will never come off it if the Fabian can help it; so that Old Age Pensions will be within reach at the end of the ten years repayment period, if not sooner. ... Charrington calls me a Tory because I declare for Imperialism as our social theory.
On the Boer War; letter to Henry Stephens Salt (12 March 1900), as cited in Dan H. Laurence (ed) Collected Letters: 1898–1910 (1985), p. 153
The worst sin towards our fellow creatures is not to hate them, but to be indifferent to them: that's the essence of inhumanity.
The Devil's Disciple, Act II (1901)
Martyrdom, sir, is what these people like: it is the only way in which a man can become famous without ability.
The Devil's Disciple, Act II
I was a cannibal for twenty-five years. For the rest I have been a vegetarian. It was Shelley who first opened my eyes to the savagery of my diet.
Interview "Who I Am, and What I Think", in Frank Harris's periodical The Candid Friend (May 1901), reprinted in Sixteen Self Sketches, 1949, p. 53; quoted in Desmond King-Hele, Shelley: His Thought and Work, 1984, p. 42
You must not suppose, because I am a man of letters, that I never tried to earn an honest living.
The Irrational Knot, Preface (1905)
[Chess] is a foolish expedient for making idle people believe they are doing something very clever, when they are only wasting their time.
The Irrational Knot (1905)
To understand a saint, you must hear the devil's advocate; and the same is true of the artist.
The Sanity of Art: An Exposure of the Current Nonsense about Artists being Degenerate (1908)
Assassination is the extreme form of censorship; and it seems hard to justify an incitement to it on anti-censorial principles.
The Shewing Up of Blanco Posnet (1909): The Rejected Statement, Pt. I : The Limits to Toleration
This is the real enemy, the invader from the East, the Druze, the ruffian, the oriental parasite; in a word: the Jew.
London Morning Post, December 3, 1925
This craving for bouquets by Jews is a symptom of racial degeneration. The Jews are worse than my own people. Those Jews who still want to be the chosen race (chosen by the late Lord Balfour) can go to Palestine and stew in their own juice. The rest had better stop being Jews and start being human beings.
Literary Digest, October 12, 1932
Love Among the Artists (1900)
[edit]
The way to deal with worldly people is to frighten them by repeating their scandalous whisperings aloud.
The public want actresses, because they think all actresses bad. They don't want music or poetry because they know that both are good. So actors and actresses thrive and poets and composers starve.
There are some men who are considered quite ugly, but who are more remarkable than pretty people. You often see that in artists.
All very fine, Mary; but my old-fashioned common sense is better than your clever modern nonsense.
If you leave your art, the world will beat you back to it. The world has not an ambition worth sharing, or a prize worth handling. Corrupt successes, disgraceful failures, or sheeplike vegetation are all it has to offer. I prefer Art, which gives me a sixth sense of beauty, with self-respect: perhaps also an immortal reputation in return for honest endeavour in a labour of love.
Perhaps woman's art is of woman's life a thing apart, 'tis man's whole existence; just as love is said to be the reverse — though it isn't.
I hate singers, a miserable crew who think that music exists only in their own throats.
A man's own self is the last person to believe in him, and is harder to cheat than the rest of the world.
Composers are not human; They can live on diminished sevenths, and be contented with a pianoforte for a wife, and a string quartet for a family.
Geniuses are horrid, intolerant, easily offended, sleeplessly self-conscious men, who expect their wives to be angels with no further business in life than to pet and worship their husbands. Even at the best they are not comfortable men to live with; and a perfect husband is one who is perfectly comfortable to live with.
Even the youngest of us may be wrong sometimes.
Man and Superman (1903)
[edit]
: Main article: Man and Superman
This is the true joy in life, the being used for a purpose recognized by yourself as a mighty one; the being thoroughly worn out before you are thrown on the scrap heap; the being a force of Nature instead of a feverish selfish little clod of ailments and grievances complaining that the world will not devote itself to making you happy.
p. xxxi
There is no love sincerer than the love of food.
p. 23
The only man I know who behaves sensibly is my tailor; he takes my measurements anew each time he sees me. The rest go on with their old measurements and expect me to fit them.
p. 37
The confusion of marriage with morality has done more to destroy the conscience of the human race than any other single error.
p. 121
Economy is the art of making the most of life. The love of economy is the root of all virtue.
p. 235
Maxims for Revolutionists (1903)
[edit]
: Full text online
Do not do unto others as you would expect they should do unto you. Their tastes may not be the same.
1
If the lesser mind could measure the greater as a foot-rule can measure a pyramid, there would be finality in universal suffrage. As it is, the political problem remains unsolved.
16
Democracy substitutes election by the incompetent many for appointment by the corrupt few.
17
He who confuses political liberty with freedom and political equality with similarity has never thought for five minutes about either.
23
Liberty means responsibility. That is why most men dread it.
25
The duke inquires contemptuously whether his gamekeeper is the equal of the Astronomer Royal; but he insists that they shall both be hanged equally if they murder him.
26
Where equality is undisputed, so also is subordination.
28
A fool's brain digests philosophy into folly, science into superstition, and art into pedantry.
32
The best brought-up children are those who have seen their parents as they are. Hypocrisy is not the parent's first duty.
33
Every fool believes what his teachers tell him, and calls his credulity science or morality as confidently as his father called it divine revelation.
39
No man can be a pure specialist without being in the strict sense an idiot.
41
The man who has graduated from the flogging block at Eton to the bench from which he sentences the garrotter to be flogged is the same social product as the garrotter who has been kicked by his father and cuffed by his mother until he has grown strong enough to throttle and rob the rich citizen whose money he desires.
55
Criminals do not die by the hands of the law. They die by the hands of other men.
57
When a man wants to murder a tiger he calls it sport: when the tiger wants to murder him he calls it ferocity. The distinction between Crime and Justice is no greater.
62
It is not necessary to replace a guillotined criminal: it is necessary to replace a guillotined social system.
65
There are no perfectly honorable men; but every true man has one main point of honor and a few minor ones.
68
Beware of the man whose god is in the skies.
83
Self-denial is not a virtue: it is only the effect of prudence on rascality.
87
Obedience simulates subordination as fear of the police simulates honesty.
88
Disobedience, the rarest and most courageous of the virtues, is seldom distinguished from neglect, the laziest and commonest of the vices.
89
In a stupid nation the man of genius becomes a god: everybody worships him and nobody does his will.
101
Happiness and Beauty are by-products.
102
Riches and Art are spurious receipts for the production of Happiness and Beauty.
104
He who desires a lifetime of happiness with a beautiful woman desires to enjoy the taste of wine by keeping his mouth always full of it.
105
The man with a toothache thinks everyone happy whose teeth are sound. The poverty stricken man makes the same mistake about the rich man.
107
The more a man possesses over and above what he uses, the more careworn he becomes.
108
In an ugly and unhappy world the richest man can purchase nothing but ugliness and unhappiness.
110
No elaboration of physical or moral accomplishment can atone for the sin of parasitism.
116
The reasonable man adapts himself to the world: the unreasonable one persists in trying to adapt the world to himself. Therefore all progress depends on the unreasonable man.
124
Reason enslaves all whose minds are not strong enough to master her.
125
Decency is Indecency's Conspiracy of Silence.
126
Men are wise in proportion, not to their experience, but to their capacity for experience. If we could learn from mere experience, the stones of London would be wiser than its wisest men.
127-128
No age or condition is without its heroes. The least incapable general in a nation is its Cæsar, the least imbecile statesman its Solon, the least confused thinker its Socrates, the least commonplace poet its Shakespeare.
136
The roulette table pays nobody except him that keeps it. Nevertheless a passion for gaming is common, though a passion for keeping roulette tables is unknown.
149
The reformer for whom the world is not good enough finds himself shoulder to shoulder with him that is not good enough for the world.
158
Youth, which is forgiven everything, forgives itself nothing: age, which forgives itself everything, is forgiven nothing.
160
Do not mistake your objection to defeat for an objection to fighting, your objection to being a slave for an objection to slavery, your objection to not being as rich as your neighbor for an objection to poverty. The cowardly, the insubordinate, and the envious share your objections.
162
Those who understand evil pardon it.
167
When a heretic wishes to avoid martyrdom he speaks of "Orthodoxy, True and False" and demonstrates that the True is his heresy.
172
If you begin by sacrificing yourself to those you love, you will end by hating those to whom you have sacrificed yourself.
179
Major Barbara (1905)
[edit]
I exclude the hypothesis of complete originality on [Charles] Lever's part, because a man can no more be completely original in that sense than a tree can grow out of air.
Preface
The greatest of evils and the worst of crimes is poverty.
Preface
The faults of the burglar are the qualities of the financier: the manners and habits of a duke would cost a city clerk his situation.
Preface
It is quite useless to declare that all men are born free if you deny that they are born good. Guarantee a man's goodness and his liberty will take care of itself. To guarantee his freedom on condition that you approve of his moral character is formally to abolish all freedom whatsoever, as every man's liberty is at the mercy of a moral indictment which any fool can trump up against everyone who violates custom, whether as a prophet or as a rascal.
Preface
Society, with all its prisons and bayonets and whips and ostracisms and starvations, is powerless in the face of the Anarchist who is prepared to sacrifice his own life in the battle with it. Our natural safety from the cheap and devastating explosives which every Russian student can make ... lies in the fact that brave and resolute men, when they are rascals, will not risk their skins for the good of humanity, and, when they are sympathetic enough to care for humanity, abhor murder, and never commit it until their consciences are outraged beyond endurance. The remedy is, then, simply not to outrage their consciences.
Preface
I can't talk religion to a man with bodily hunger in his eyes.
Act II
You cannot have power for good without having power for evil too. Even mother's milk nourishes murderers as well as heroes.
Undershaft: You have made for yourself something that you call a morality or a religion or what not. It doesn't fit the facts. Well, scrap it. Scrap it and get one that does fit. That is what is wrong with the world at present. It scraps its obsolete steam engines and dynamos; but it wont scrap its old prejudices and its old moralities and its old religions and its old political constitutions. Whats the result? In machinery it does very well; but in morals and religion and politics it is working at a loss that brings it nearer bankruptcy every year.
Cusins: Call you poverty a crime?
Undershaft: The worst of crimes. All the other crimes are virtues beside it: all the other dishonors are chivalry itself by comparison. Poverty blights whole cities; spreads horrible pestilences; strikes dead the very souls of all who come within sight, sound or smell of it. What you call crime is nothing: a murder here and a theft there, a blow now and a curse then: what do they matter? they are only the accidents and illnesses of life: there are not fifty genuine professional criminals in London. But there are millions of poor people, abject people, dirty people, ill fed, ill clothed people. They poison us morally and physically: they kill the happiness of society: they force us to do away with our own liberties and to organize unnatural cruelties for fear they should rise against us and drag us down into their abyss. Only fools fear crime: we all fear poverty.
Undershaft: My religion? Well, my dear, I am a Millionaire. That is my religion.
Act II
You have learnt something. That always feels at first as if you had lost something.
Act III
It is not the sale of my soul that troubles me: I have sold it too often to care about that. I have sold it for a professorship. I have sold it for an income. ... What is all human conduct but the daily and hourly sale of our souls for trifles?
He knows nothing; and he thinks he knows everything. That points clearly to a political career.
John Bull's Other Island (1907)
[edit]
A healthy nation is as unconscious of its nationality as a healthy man of his bones. But if you break a nation's nationality it will think of nothing else but getting it set again.
Preface
You cannot be a hero without being a coward.
Preface
What really flatters a man is that you think him worth flattering.
My way of joking is to tell the truth. It's the funniest joke in the world.
Act II
Getting Married (1908)
[edit]
: Full text online
There is no subject on which more dangerous nonsense is talked and thought than marriage. If the mischief stopped at talking and thinking it would be bad enough; but it goes further, into disastrous anarchical action. Because our marriage law is inhuman and unreasonable to the point of downright abomination, the bolder and more rebellious spirits form illicit unions, defiantly sending cards round to their friends announcing what they have done. Young women come to me and ask me whether I think they ought to consent to marry the man they have decided to live with; and they are perplexed and astonished when I, who am supposed (heaven knows why!) to have the most advanced views attainable on the subject, urge them on no account to compromise themselves without the security of an authentic wedding ring.
Preface
Home life as we understand it is no more natural to us than a cage is natural to a cockatoo.
Preface
When two people are under the influence of the most violent, most insane, most delusive, and most transient of passions, they are required to swear that they will remain in that excited, abnormal, and exhausting condition continuously until death do them part.
Preface
Plato long ago pointed out the importance of being governed by men with sufficient sense of responsibility and comprehension of public duties to be very reluctant to undertake the work of governing.
Preface
Love is an appetite which, like all other appetites, is destroyed for the moment by its gratification.
Preface
Never forget that if you leave your law to judges and your religion to bishops, you will presently find yourself without either law or religion.
Preface
Journalists are too poorly paid in this country to know anything that is fit for publication.
Preface
Monogamy has a sentimental basis which is quite distinct from the political one of equal numbers of the sexes. Equal numbers in the sexes are quite compatible with a change of partners every day or every hour. Physically there is nothing to distinguish human society from the farm-yard except that children are more troublesome and costly than chickens and calves, and that men and women are not so completely enslaved as farm stock. Accordingly, the people whose conception of marriage is a farm-yard or slave-quarter conception are always more or less in a panic lest the slightest relaxation of the marriage laws should utterly demoralize society; whilst those to whom marriage is a matter of more highly evolved sentiments and needs (sometimes said to be distinctively human, though birds and animals in a state of freedom evince them quite as touchingly as we) are much more liberal, knowing as they do that monogamy will take care of itself provided the parties are free enough, and that promiscuity is a product of slavery and not of liberty.
Preface: "The Personal Sentimental Basis of Monogamy"
The secret of forgiving everything is to understand nothing.
Leo
Nothing is more dreadful than a husband who keeps telling you everything he thinks, and always wants to know what you think.
The Bishop
All progress means war with Society.
The Bishop
The whole strength of England lies in the fact that the enormous majority of the English people are snobs.
Hotchkiss
You don't learn to hold your own in the world by standing on guard, but by attacking, and getting well hammered yourself.
Mrs. George
I happen, like Napoleon, to prefer Mohammedanism. [Mrs George, associating Mohammedanism with polygamy, looks at him with quick suspicion]. I believe the whole British Empire will adopt a reformed Mohammedanism before the end of the century. The character of Mahomet is congenial to me. I admire him, and share his views of life to a considerable extent.
Hotchkiss
Religion is a great force — the only real motive force in the world; but what you fellows don't understand is that you must get at a man through his own religion and not through yours. Instead of facing that fact, you persist in trying to convert all men to your own little sect, so that you can use it against them afterwards. You are all missionaries and proselytizers trying to uproot the native religion from your neighbor's flowerbeds and plant your own in its place. You would rather let a child perish in ignorance than have it taught by a rival sectary. You can talk to me of the quintessential equality of coal merchants and British officers; and yet you can't see the quintessential equality of all the religions.
Hotchkiss
1910s
[edit]
Why was I born with such contemporaries?
The Dark Lady of the Sonnets, Preface (1910)
A part of eugenic politics would finally land us in an extensive use of the lethal chamber. A great many people would have to be put out of existence simply because it wastes other people's time to look after them."
Shaw's Lecture to the London's Eugenics Education Society, The Daily Express, (March 4, 1910), quoted in Modernism and the Culture of Efficiency: Ideology and Fiction, Evelyn Cobley, University of Toronto Press (2009) p. 159
A critic recently described me, with deadly acuteness, as having 'a kindly dislike of my fellow-creatures.' Perhaps dread would have been nearer the mark than dislike; for man is the only animal of which I am thoroughly and cravenly afraid.
As quoted in George Bernard Shaw, his life and works: a critical biography (authorised), Archibald Henderson, Stewart & Kidd (1911), Chapter VII (The Art Critic), pp. 201-202
The word morality, if we met it in the Bible, would surprise us as much as the word telephone or motor car.
Fanny's First Play, Preface (1911)
That proves it's not by Shaw, because all Shaw's characters are himself: mere puppets stuck up to spout Shaw.
Fanny's First Play, Epilogue
As long as I have a want, I have a reason for living. Satisfaction is death.
Overruled (1912)
Any public committee man who tries to pack the moral cards in the interest of his own notions is guilty of corruption and impertinence. The business of a public library is not to supply the public with the books the committee thinks good for the public, but to supply the public with the books the public wants. ... Censorship ends in logical completeness when nobody is allowed to read any books except the books that nobody can read. But as the ratepayer is mostly a coward and a fool in these difficult matters, and the committee is quite sure that it can succeed where the Roman Catholic Church has made its index expurgatorius the laughing-stock of the world, censorship will rage until it reduces itself to absurdity; and even then the best books will be in danger still.
As quoted in "Literary Censorship in England" in Current Opinion, Vol. 55, No. 5 (November 1913), p. 378; this has sometimes appeared on the internet in paraphrased form as "Censorship ends in logical completeness when nobody is allowed to read any books except the books that nobody reads"
Well, I tell you again to get rid of your Constitution. But I suppose you won't do it. You have a good president and you have a bad Constitution, and the bad Constitution gets the better of the good President all the time. The end of it will be is that you might as well have an English Prime Minister.
Newsreel interview by George Bernard Shaw entitled "Various Scenes with George Bernard Shaw," Fox Movietone Newsreel (1933), referring to Franklin D. Roosevelt's presidency
Custom will reconcile people to any atrocity; and fashion will drive them to acquire any custom.
Killing For Sport, Preface (1914)
All great truths begin as blasphemies.
Annajanska (1919)
You'll never have a quiet world till you knock the patriotism out of the human race.
O'Flaherty V.C. (1919)
Misalliance (1910)
[edit]
It is more dangerous to be a great prophet or poet than to promote twenty companies for swindling simple folk out of their savings.
Preface
Optimistic lies have such immense therapeutic value that a doctor who cannot tell them convincingly has mistaken his profession.
Preface
A perpetual holiday is a good working definition of Hell.
I like a bit of a mongrel myself, whether it's a man or a dog; they're the best for every day.
Episode I
If parents would only realize how they bore their children!
Episode I
A Treatise on Parents and Children (1910)
[edit]
When will we realize that the fact that we can become accustomed to anything, however disgusting at first, makes it necessary to examine carefully everything we have become accustomed to.
Death is for many of us the gate of hell; but we are inside on the way out, not outside on the way in.
A nation should always be healthily rebellious; but the king or prime minister has yet to be found who will make trouble by cultivating that side of the national spirit. A child should begin to assert itself early, and shift for itself more and more not only in washing and dressing itself, but in opinions and conduct; yet as nothing is so exasperating and so unlovable as an uppish child, it is useless to expect parents and schoolmasters to inculcate this uppishness. Such unamiable precepts as Always contradict an authoritative statement, Always return a blow, Never lose a chance of a good fight, When you are scolded for a mistake ask the person who scolds you whether he or she supposes you did it on purpose, and follow the question with a blow or an insult or some other unmistakable expression of resentment, Remember that the progress of the world depends on your knowing better than your elders, are just as important as those of The Sermon on the Mount; but no one has yet seen them written up in letters of gold in a schoolroom or nursery.
You are so careful of your boy's morals, knowing how troublesome they may be, that you keep him away from the Venus of Milo only to find him in the arms of the scullery maid or someone much worse. You decide that the Hermes of Praxiteles and Wagner's Tristan are not suited for young girls; and your daughter marries somebody appallingly unlike either Hermes or Tristan solely to escape from your parental protection. You have not stifled a single passion nor averted a single danger: you have depraved the passions by starving them, and broken down all the defences which so effectively protect children brought up in freedom.
The secret of being miserable is to have leisure to bother about whether you are happy or not. The cure for it is occupation, because occupation means pre-occupation; and the pre-occupied person is neither happy nor unhappy, but simply alive and active, which is pleasanter than any happiness until you are tired of it.
The Doctor's Dilemma (1911)
[edit]
: Full text online at Project Gutenberg
Do not try to live for ever. You will not succeed.
Preface
No man who is occupied in doing a very difficult thing, and doing it well, ever loses his self-respect.
Preface
Attention and activity lead to mistakes as well as to successes; but a life spent in making mistakes is not only more honorable but more useful than a life spent doing nothing.
Preface
All professions are conspiracies against the laity.
Act I
I don't believe in morality. I'm a disciple of Bernard Shaw.
Act III
Life does not cease to be funny when people die any more than it ceases to be serious when people laugh.
Act V
Chloroform has done a lot of mischief. It's enabled every fool to be a surgeon.
Use your health, even to the point of wearing it out. That is what it is for. Spend all you have before you die; and do not outlive yourself.
Pygmalion (1912)
[edit]
It is impossible for an Englishman to open his mouth without making some other Englishman hate or despise him.
Preface
The English have no respect for their language, and will not teach their children to speak it.
Preface
He ain't a copper just look at 'is boots!
Act I
Ah-ah-ah-ah-ow-ow-oo-oo!!! I ain't dirty: I washed me face and hands afore I come, I did!
Act II
Women upset everything. When you let them into your life, you find that the woman is driving at one thing and you're driving at another.
Act II
What is life but a series of inspired follies? The difficulty is to find them to do. Never lose a chance: it doesn't come every day.
Act II
I wouldn't have ate it, only I'm too lady-like to take it out of my mouth.
Act II
I don't want to talk grammar, I want to talk like a lady.
Act II
I ask you, what am I? I'm one of the undeserving poor: thats what I am. Think of what that means to a man.
Act II
I aint such a mug as to put up my children to all I know myself.
Act II
Walk! Not bloody likely. I am going in a taxi.
Act III
I heard your prayers Thank God it's all over!
Act IV
You see, lots of the real people can't do it at all: they're such fools that they think style comes by nature to people in their position; and so they never learn. There's always something professional about doing a thing superlatively well.
Time enough to think of the future when you haven't any future to think of.
I have to live for others and not for myself; that's middle-class morality.
Act V
Independence? That's middle-class blasphemy. We are all dependent on one another, every soul of us on earth.
Act V
Androcles and the Lion (1913)
[edit]
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality of happiness, and by no means a necessity of life.
Preface, The importance of hell in the salvation scheme
Revolutionary movements attract those who are not good enough for established institutions as well as those who are too good for them.
Howbeit, Paul succeeded in stealing the image of Christ crucified for the figure-head of his Salvationist vessel, with its Adam posing as the natural man, its doctrine of original sin, and its damnation avoidable only by faith in the sacrifice of the cross. In fact, no sooner had Jesus knocked over the dragon of superstition than Paul boldly set it on its legs again in the name of Jesus.
Preface, Paul
The Technique of War (1917)
[edit]
He [the British taxpayer] must be taught that war is not precise or economical. It is almost inconceivably wasteful and extravagant. It burns the house to roast the pig, and even then seldom roasts him effectively..... waste is the law of modern war; and nothing is cheap on the battlefield except the lives of men...... Therefore, my taxpayer, resign yourself to this: that we may fight bravely, fight hard, fight long, fight cunningly, fight recklessly, fight in a hundred and fifty ways, but we cannot fight cheaply.
The Daily Chronicle on the 7 March 1917.
Reproduced in What I Really Wrote About The War (London: Constable & Co Ltd, 1930)
Heartbreak House (1919)
[edit]
: Full text online
"He must be greatly changed. Has he attained the seventh degree of concentration?"
Captain Shotover, Act I
"We know now that the soul is the body, and the body the soul. They tell us they are different because they want to persuade us that we can keep our souls if we let them make slaves of our bodies."
Ellie Dunn, Act II
When your heart is broken, your boats are burned: nothing matters any more. It is the end of happiness and the beginning of peace.
Ellie Dunn, Act II
1920s
[edit]
The moment we face it frankly we are driven to the conclusion that the community has a right to put a price on the right to live in it ... If people are fit to live, let them live under decent human conditions. If they are not fit to live, kill them in a decent human way. Is it any wonder that some of us are driven to prescribe the lethal chamber as the solution for the hard cases which are at present made the excuse for dragging all the other cases down to their level, and the only solution that will create a sense of full social responsibility in modern populations?
As quoted in George Bernard Shaw's "Prefaces" in English Prisons Under Local Government, Sidney and Beatrice Webb, London: Longmans, Green & Co (1922) pp. 31-32
It is difficult, if not impossible, for most people to think otherwise than in the fashion of their own period.
Preface to Saint Joan : A Chronicle Play In Six Scenes And An Epilogue (1923)
Political necessities sometimes turn out to be political mistakes.
Saint Joan : A Chronicle Play In Six Scenes And An Epilogue (1923) - Full text online
Scratch an Englishman and find a Protestant.
Saint Joan : A Chronicle Play In Six Scenes And An Epilogue (1923)
God is on the side of the big battalions.
Saint Joan : A Chronicle Play In Six Scenes And An Epilogue (1923)
Must then a Christ perish in torment in every age to save those that have no imagination?
Saint Joan : A Chronicle Play In Six Scenes And An Epilogue (1923)
The Italians must allow us to slaughter the Momands, because, if we do not kill the warlike hillmen, they will kill us. And we must allow the Italians to slaughter the Danakils for the same reason.
Quote about Italy's 1935 invasion of Ethiopia in Socialism and Superior Brains: The Political Thought of Bernard Shaw by Gareth Griffith (1993) p. 267.
The Nazi movement is in many respects one which has my warmest sympathy.
As Quoted in London Morning Post, (Dec. 3, 1925)
Some of the things Mussolini has done, and some that he is threatening to do go further in the direction of Socialism than the English Labour Party could yet venture if they were in power.
Letter from G. Bernard Shaw to a friend, "Bernard Shaw's Defence of Mussolini," (Feb. 7, 1927)
[Mussolini was] farther to the Left in his political opinions than any of his socialist rivals.
As quoted in, Socialism and Superior Brains: The Political Thought of Bernard Shaw, Gareth Griffith, Routledge, (2002) p. 253, Manchester Guardian (1927)
Socialists must be in favor of an aristocratic form of government. We must have the best men for the job . . . In the dictator you must have a man who has not only the power to govern but the force of character to impose himself as dictator whether you like him or not.
"Socialism urged to find dictator," Berkeley Daily Gazette (Nov. 30, 1927)
Our natural dispositions may be good; but we have been badly brought up, and are full of anti-social personal ambitions and prejudices and snobberies. Had we not better teach our children to be better citizens than ourselves? We are not doing that at present. The Russians are. That is my last word. Think over it.
The Apple Cart (1928) Preface
One man that has a mind and knows it can always beat ten men who haven't and don't.
The Apple Cart (1928), Act I
God help England if she had no Scots to think for her!
The Apple Cart (1928), Act II
We should refuse to tolerate poverty as a social institution not because the poor are the salt of the earth, but because 'the poor in a lump are bad'.
The Intelligent Woman's Guide: To Socialism and Capitalism. p. 219.
We have to confess it: Capitalist mankind in the lump is detestable. ...Both rich and poor are really hateful in themselves. For my part I hate the poor and look forward eagerly to their extermination. I pity the rich a little, but am equally bent on their extermination. The working classes, the ruling classes, the professional classes, the propertied classes, the ruling classes, are each more odious than the other: they have no right to live: I should despair if I did not know that they will all die presently, and that there is no need on earth why they should be replaced by people like themselves... And yet I am not in the least a misanthrope. I am a person of normal affections
The Intelligent Woman's Guide: To Socialism and Capitalism. p. 456.
Under Socialism, you would not be allowed to be poor. You would be forcibly fed, clothed, lodged, taught, and employed whether you liked it or not. If it were discovered that you had not character and industry enough to be worth all this trouble, you might possibly be executed in a kindly manner; but whilst you were permitted to live, you would have to live well.
The Intelligent Woman's Guide: To Socialism and Capitalism, New York: NY, Brentano (1928) p. 670.
It is far more likely that by the time nationalization has become the rule, and private enterprise the exception, Socialism (which is really rather a bad name for the business) will be spoken of, if at all, as a crazy religion held by a fanatical sect in that darkest of dark ages, the nineteenth century. Already, indeed, I am told that Socialism has had its day, and that the sooner we stop talking nonsense about it and set to work, like the practical people we are, to nationalize the coal mines and complete a national electrification scheme, the better. And I, who said forty years ago that we should have had Socialism already but for the Socialists, am quite willing to drop the name if dropping it will help me to get the thing. What I meant by my jibe at the Socialists of the eighteen-eighties was that nothing is ever done, and much is prevented, by people who do not realize that they cannot do everything at once.
The Intelligent Woman's Guide To Socialism, Capitalism, Sovietism, and Fascism (1928)
Perhaps the greatest social service that can be rendered by anybody to the country and to mankind is to bring up a family. But here again, because there is nothing to sell, there is a very general disposition to regard a married woman's work as no work at all, and to take it as a matter of course that she should not be paid for it.
The Intelligent Woman's Guide to Socialism and Capitalism, Chapter 8 (1928)
Women are not angels. They are as foolish as men in many ways; but they have had to devote themselves to life whilst men have had to devote themselves to death; and that makes a vital difference in male and female religion. Women have been forced to fear whilst men have been forced to dare: the heroism of a woman is to nurse and protect life, and of a man to destroy it and court death.
The Intelligent Woman's Guide to Socialism and Capitalism, Chapter 82 (1928)
Well, of course, they notice you. You always hide just in the middle of the limelight.
Reply to T. E. Lawrence who complained of press attention.
Quoted by Harry Kessler in his diary, 14 November 1929
Back to Methuselah (1921)
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: Full text online
In truth, mankind cannot be saved from without, by schoolmasters or any other sort of masters: it can only be lamed and enslaved by them. It is said that if you wash a cat it will never again wash itself. This may or may not be true: what is certain is that if you teach a man anything he will never learn it; and if you cure him of a disease he will be unable to cure himself the next time it attacks him.
Is there any hope in education?
People will have their miracles, their stories, their heroes and heroines and saints and martyrs and divinities to exercise their gifts of affection, admiration, wonder, and worship, and their Judases and devils to enable them to be angry and yet feel that they do well to be angry. Every one of these legends is the common heritage of the human race; and there is only one inexorable condition attached to their healthy enjoyment, which is that no one shall believe them literally. The reading of stories and delighting in them made Don Quixote a gentleman: the believing them literally made him a madman who slew lambs instead of feeding them.
A Touchstone For Dogma
I hear you say "Why?" Always "Why?" You see things; and you say "Why?" But I dream things that never were; and I say "Why not?"
The Serpent, in Pt. I : In the Beginning, Act I
This quote is sometimes misattributed to Robert F. Kennedy. It is often paraphrased slightly in a few different ways, including: You see things as they are and ask, "Why?" I dream things as they never were and ask, "Why not?"
I worship you, Eve. I must have something to worship. Something quite different to myself, like you. There must be something greater than the snake.
The Serpent, in Pt I : In the Beginning
Everything is possible: everything. Listen. I am old. I am the old serpent, older than Adam, older than Eve. I remember Lilith, who came before Adam and Eve. I was her darling as I am yours. She was alone: there was no man with her. She saw death as you saw it when the fawn fell; and she knew then that she must find out how to renew herself and cast the skin like me. She had a mighty will: she strove and strove and willed and willed for more moons than there are leaves on all the trees of the garden. Her pangs were terrible: her groans drove sleep from Eden. She said it must never be again: that the burden of renewing life was past bearing: that it was too much for one. And when she cast the skin, lo! there was not one new Lilith but two: one like herself, the other like Adam. You were the one: Adam was the other.
The Serpent, in Pt. I, Act I
Imagination is the beginning of creation. You imagine what you desire; you will what you imagine; and at last you create what you will.
The Serpent, in Pt. I, Act I
Conceive. That is the word that means both the beginning in imagination and the end in creation.
The Serpent, in Pt. I, Act I
Life must not cease. That comes before everything. It is silly to say you do not care. You do care. It is that care that will prompt your imagination; inflame your desires; make your will irresistible; and create out of nothing.
The Serpent, in Pt. I, Act I
I am very subtle; but Man is deeper in his thought than I am. The woman knows that there is no such thing as nothing: the man knows that there is no such day as tomorrow. I do well to worship them.
The Serpent, in Pt. I, Act I
THE SERPENT: The voice in the garden is your own voice.
ADAM: It is; and it is not. It is something greater than me: I am only a part of it.
EVE: The Voice does not tell me not to kill you. Yet I do not want you to die before me. No voice is needed to make me feel that.
ADAM [throwing his arm round her shoulder with an expression of anguish]: Oh no: that is plain without any voice. There is something that holds us together, something that has no word —
THE SERPENT: Love. Love. Love.
ADAM: That is too short a word for so long a thing.
The Serpent, Adam, and Eve, in Pt. I, Act I
I make no vows. I take my chance. ... It means that I fear certainty as you fear uncertainty. It means that nothing is certain but uncertainty. If I bind the future I bind my will. If I bind my will I strangle creation.
The Serpent, in Pt. I, Act I
You can feel nothing but a torment, and believe nothing but a lie. You will not raise your head to look at all the miracles of life that surround you; but you will run ten miles to see a fight or a death.
Eve to Cain, in Pt. I, Act II
Your father is a fool skin deep; but you are a fool to your very marrow.
Eve to Cain, in Pt. I, Act II
Any sort of plain speaking is better than the nauseous sham good fellowship our democratic public men get up for shop use.
Franklyn, in Pt. II : The Gospel of the Brothers Barnabas
There are no secrets except the secrets that keep themselves.
Confucius, in Pt. III : The Thing Happens
Everything happens to everybody sooner or later if there is time enough.
Pt. V : As Far as Thought Can Reach
Silence is the perfect expression of scorn.
Pt. V
The worst cliques are those which consist of one man.
Pt. V
Life is not meant to be easy, my child but take courage: it can be delightful.
Pt. V; see also the later phrasing of Malcolm Fraser, "life wasn't meant to be easy"
THE HE-ANCIENT: When a thing is funny, search it for a hidden truth
STREPHON: Yes; and take all the fun out of it.
Pt. V
Art is the magic mirror you make to reflect your invisible dreams in visible pictures. You use a glass mirror to see your face: you use works of art to see your soul. But we who are older use neither glass mirrors nor works of art. We have a direct sense of life. When you gain that you will put aside your mirrors and statues, your toys and your dolls.
The She-Ancient, in Pt. V
When the master has come to do everything through the slave, the slave becomes his master, since he cannot live without him.
The He-Ancient, in Pt. V
Love is a simple thing and a deep thing: it is an act of life and not an illusion. Art is an illusion.
Acis, in Pt. V
Even a vortex is a vortex in something. You can't have a whirlpool without water; and you can't have a vortex without gas, or molecules or atoms or ions or electrons or something, not nothing.
Acis, in Pt. V
The body was the slave of the vortex; but the slave has become the master; and we must free ourselves from that tyranny. It is this stuff [indicating her body], this flesh and blood and bone and all the rest of it, that is intolerable. Even prehistoric man dreamed of what he called an astral body, and asked who would deliver him from the body of this death.
The She-Ancient, in Pt. V
I am justified. For I chose wisdom and the knowledge of good and evil; and now there is no evil; and wisdom and good are one. It is enough.
The Serpent, in Pt. V
They have accepted the burden of eternal life. They have taken the agony from birth; and their life does not fail them even in the hour of their destruction.
Lilith, in Pt. V
I had patience with them for many ages: they tried me very sorely. They did terrible things: they embraced death, and said that eternal life was a fable. I stood amazed at the malice and destructiveness of the things I had made...
Lilith, in Pt. V
They have redeemed themselves from their vileness, and turned away from their sins. Best of all, they are still not satisfied: the impulse I gave them in that day when I sundered myself in twain and launched Man and Woman on the earth still urges them: after passing a million goals they press on to the goal of redemption from the flesh, to the vortex freed from matter, to the whirlpool in pure intelligence that, when the world began, was a whirlpool in pure force.
I can wait: waiting and patience mean nothing to the eternal. I gave the woman the greatest of gifts: curiosity. By that her seed has been saved from my wrath; for I also am curious; and I have waited always to see what they will do tomorrow.
I say, let them dread, of all things, stagnation; for from the moment I, Lilith, lose hope and faith in them, they are doomed. In that hope and faith I have let them live for a moment; and in that moment I have spared them many times. But mightier creatures than they have killed hope and faith, and perished from the earth; and I may not spare them for ever. I am Lilith: I brought life into the whirlpool of force, and compelled my enemy, Matter, to obey a living soul. But in enslaving Life's enemy I made him Life's master; for that is the end of all slavery; and now I shall see the slave set free and the enemy reconciled, the whirlpool become all life and no matter. And because these infants that call themselves ancients are reaching out towards that, I will have patience with them still; though I know well that when they attain it they shall become one with me and supersede me, and Lilith will be only a legend and a lay that has lost its meaning. Of Life only is there no end; and though of its million starry mansions many are empty and many still unbuilt, and though its vast domain is as yet unbearably desert, my seed shall one day fill it and master its matter to its uttermost confines. And for what may be beyond, the eyesight of Lilith is too short. It is enough that there is a beyond.
Lilith, in Pt. V
1930s
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No public man in these islands ever believes that the Bible means what it says: he is always convinced that it says what he means.
Our Theatres In The Nineties (1930)
I have defined the 100 per cent American as 99 per cent an idiot.
New York Times (19 December 1930) remarks on Sinclair Lewis receiving the Nobel Prize
Impression of Gandhi! You might as well ask one to give his impression of the Himalayas.
When asked about his impression on Gandhi (1931), as quoted in D. G. Tendulkar, Mahatma: Life of Mohandas Karamchand Gandhi (1952), vol. 3, p. 166.
I object to all punishment whatsoever. I don't want to punish anybody, but there are an extraordinary number of people who I want to kill. Not in any unkind or personal spirit. But it must be evident to all of you, you must all know have a dozen (1/12) people at least, who are no use in this world; who are more trouble than they are worth. And I think it would be a good thing to make everybody come before a properly appointed board just as he might come before the income tax commissioners and say every 5 years or every 7 years, just put them there, and say, sir or madam, now will you be kind enough to justify your existence? If you can’t justify your existence; if you’re not pulling your weight in the social boat; if you are not producing as much as you consume or perhaps a little more, then clearly we cannot use the big organization of our society for the purpose of keeping you alive, because your life does not benefit us, and it can’t be of very much use to yourself.
"George Bernard Shaw reopens capital punishment controversy", Paramount British Pictures (March 5, 1931)
I am afraid we must make the world honest before we can honestly tell our children that honesty is the best policy.
"Rungs of the Ladder", BBC Radio broadcast (11 July 1932)
An American has no sense of privacy. He doesn't know what it means. There is no such thing in the country.
Speech in New York (11 April 1933)
You in America should trust to that volcanic political instinct which I have divined in you.
Speech at New York (11 April 1933)
Hitler is a very remarkable man, a very able man... What Hitler should have done was not to drive the Jews out, what he ought to have said was, 'I will tolerate the Jews to any extent on condition that no Jew marries a Jewess, on condition that he marries a German.'
From a lecture before the Fabian Society in London called "The Politics of Unpolitical Animals" (23 November 1933), as cited in "Shaw Heaps Praise upon the Dictators: While Parliaments Get Nowhere, He Says, Mussolini and Stalin Do Things," New York Times (10 December 1933)
Until you have socialism you will never have State solidity, because, as we know, if you have private property you will immediately split your stake. You get the class conflict, the class struggle, the confrontation of interests between the proprietors and between the proletariat; and, therefore, you have something that is crumbling, that is divided against itself.
From a lecture before the Fabian Society in London called "The Politics of Unpolitical Animals" (23 November 1933), as cited in "Shaw Heaps Praise upon the Dictators: While Parliaments Get Nowhere, He Says, Mussolini and Stalin Do Things," New York Times (10 December 1933)
As a red hot Communist I am in favour of fascism. The only drawback to Sir Oswald's movement is that it is not quite British enough.
From "The Blackshirt Challenge", News Chronicle (17 January 1934), as cited in Gareth Griffith Socialism and Superior Brains: The Political Thought of Bernard Shaw, Routledge (1993) p. 264.
Mussolini, Kemal, Pilsudski, Hitler and the rest can all depend on me to judge them by their ability to deliver the goods and not by... comfortable notions of freedom. Stalin has delivered the goods to an extent that seemed impossible ten years ago; and I take off my hat to him accordingly.
As quoted in "Stalin-Wells Talk: The Verbatim Report and A Discussion", G.B. Shaw, J.M. Keynes et al., London, The New Statesman and Nation, (1934) p. 47. Also reported in Political Pilgrams: Western Intellectuals in Search of the Good Society, Paul Hollander, New Brunswick, N.J: Transaction Publishers (1998) p. 169
I appeal to the chemists to discover a humane gas that will kill instantly and painlessly. In short, a gentlemanly gas – deadly by all means, but humane not cruel.
The Listener (BBC magazine) (February 1934)
Now take Stalin himself. He is 'neither duke nor peer', not a king, not a chancellor, not a dictator, not a Prime Minister, not an archbishop, not entitled to salutes enforced by youths in coloured shirts, but simply secretary of the supreme controlling organ of the hierarchy, subject to dismissal at five minutes' notice if he does not give satisfaction. This position he has attained through the survival of the fittest, and has held through the years of the most appalling vicissitudes that ever attended the birth pangs of a new civilisation.
As quoted in "Stalin-Wells Talk: The Verbatim Report and A Discussion", G.B. Shaw, J.M. Keynes et al., London, The New Statesman and Nation, (1934) p. 40
Stalin has exiled Trotsky and become the Pontifex Maximus of the new Russo-Catholic Church of Communism on two grounds. First, he is a practical Nationalist statesman recognizing that Russia is a big enough handful for mortal rulers to tackle without taking on the rest of the world as well.... Second, Stalin, inflexible as to his final aim, is a compete opportunist as to the means.
As quoted in "Stalin-Wells Talk: The Verbatim Report and A Discussion", G.B. Shaw, J.M. Keynes et al., London, The New Statesman and Nation, (1934) p. 26
The news from Germany is the very best news that we have had since the war. Ever since 1918 we, like all the other powers, have been behaving just as badly as we possibly could. Well now, when Germany fell, they want and they sat on Germany's head and they kept sitting on Germany's head, although it was quite preposterous, quite evident to any sensible persona, that they couldn't go on like that forever. Then there came a very intelligent gentleman named Adolf Hitler and he, knowing perfectly well that the powers would not fight, he snapped his fingers at the Treaty of Versailles. Just exactly as if we in England had been in the same position. As if the powers had beaten us and sat on our head.
From an interview (1935), as cited in James Curran, Anthony Smith & Pauline Wingate, ed (1987). Impacts and Influences: Media Power in the Twentieth Century. Taylor & Francis. pp. 90-91. ISBN 978-0-416-00612-4. , George Bernard Shaw Speaks on Hitler and Germany 1935 (video)
Then the first man who had the gumption to see that we might get up on our legs and defy all those old treaties, he would be the most popular man in England. There can be no peace in the world until there is peace between England, France, Germany, Russia, the United States and all the big powers of the West. Now take that home and think about it and don't be frightened any more about the Germans.
About the Treaty of Versailles (1935). George Bernard Shaw Speaks on Hitler and Germany 1935 (video)
Dear Katharine Cornell: I don't think I was ever so astonished by a picture as I was by your photograph. Your success as Candida and something blonde and expansive about your name had created an ideal British Candida in my imagination. Fancy my feelings on seeing the photograph of a gorgeous dark lady from the cradle of the human race ... wherever that was ... Ceylon ... Sumatra ... Hilo ... or the southernmost corner of the Garden of Eden. If you look like that it doesn't matter a rap if you can act or not. Can you? Yours, breath bereaved, Bernard Shaw.
Letter to Katharine Cornell, c. 1936; as quoted in [ Leading Lady : The World and Theatre of Katharine Cornell (1978) by Tad Mosel, p. 187
We ought to tackle the Jewish question by admitting the right of States to make eugenic experiments by weeding out any strains that they think undesirable, but insisting that they should do it as humanely as they can afford to, and not shock civilization by such misdemeanors as the expulsion and robbery of Einstein.
From a letter to Beatrice Webb (1938), as cited in Dan H. Laurence (ed) Bernard Shaw: Collected Letters 1926-1950, New York: Viking (1988), p. 493
On the Rocks (1933)
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: On the Rocks : A Political Comedy
In this play a reference is made by a Chief of Police to the political necessity for killing people: a necessity so distressing to the statesmen and so terrifying to the common citizen that nobody except myself (as far as I know) has ventured to examine it directly on its own merits, although every Government is obliged to practise it on a scale varying from the execution of a single murderer to the slaughter of millions of quite innocent persons. Whilst assenting to these proceedings, and even acclaiming and celebrating them, we dare not tell ourselves what we are doing or why we are doing it; and so we call it justice or capital punishment or our duty to king and country or any other convenient verbal whitewash for what we instinctively recoil from as from a dirty job. These childish evasions are revolting. We must strip off the whitewash and find out what is really beneath it. Extermination must be put on a scientific basis if it is ever to be carried out humanely and apologetically as well as thoroughly.
Preface; Extermination
Ignoring the satirical elements of Shaw's rhetoric, and that he is presenting many arguments of sometimes questionable sincerity for the "humane" execution of criminals, the last sentence here has sometimes been misquoted as if it as part of an argument for exterminations for the sake of eugenics, by preceding it with a selected portion of a statement later in the essay: "If we desire a certain type of civilization, we must exterminate the sort of people who do not fit into it ... Extermination must be put on a scientific basis if it is ever to be carried out humanely and apologetically as well as thoroughly".
In law we draw a line between the killing of human animals and non-human ones, setting the latter apart as brutes. This was founded on a general belief that humans have immortal souls and brutes none. Nowadays more and more people are refusing to make this distinction. They may believe in The Life Everlasting and The Life to Come; but they make no distinction between Man and Brute, because some of them believe that brutes have souls, whilst others refuse to believe that the physical materializations and personifications of The Life Everlasting are themselves everlasting. In either case the mystic distinction between Man and Brute vanishes; and the murderer pleading that though a rabbit should be killed for being mischievous he himself should be spared because he has an immortal soul and a rabbit has none is as hopelessly out of date as a gentleman duellist pleading his clergy. When the necessity for killing a dangerous human being arises, as it still does daily, the only distinction we make between a man and a snared rabbit is that we very quaintly provide the man with a minister of religion to explain to him that we are not killing him at all, but only expediting his transfer to an eternity of bliss.
Preface; The Sacredness of Human Life
The extermination of what the exterminators call inferior races is as old as history. "Stone dead hath no fellow" said Cromwell when he tried to exterminate the Irish. "The only good nigger is a dead nigger" say the Americans of the Ku-Klux temperament. "Hates any man the thing he would not kill?" said Shylock naively. But we white men, as we absurdly call ourselves in spite of the testimony of our looking glasses, regard all differently colored folk as inferior species. Ladies and gentlemen class rebellious laborers with vermin. The Dominicans, the watchdogs of God, regarded the Albigenses as the enemies of God, just as Torquemada regarded the Jews as the murderers of God. All that is an old story: what we are confronted with now is a growing perception that if we desire a certain type of civilization and culture we must exterminate the sort of people who do not fit into it. There is a difference between the shooting at sight of aboriginal natives in the back blocks of Australia and the massacres of aristocrats in the terror which followed the foreign attacks on the French Revolution. The Australian gunman pots the aboriginal natives to satisfy his personal antipathy to a black man with uncut hair. But nobody in the French Republic had this feeling about Lavoisier, nor can any German Nazi have felt that way about Einstein. Yet Lavoisier was guillotined; and Einstein has had to fly for his life from Germany. It was silly to say that the Republic had no use for chemists; and no Nazi has stultified his party to the extent of saying that the new National Socialist Fascist State in Germany has no use for mathematician-physicists. The proposition is that aristocrats (Lavoisier's class) and Jews (Einstein's race) are unfit to enjoy the privilege of living in a modern society founded on definite principles of social welfare as distinguished from the old promiscuous aggregations crudely policed by chiefs who had no notion of social criticism and no time to invent it.
Preface; Previous Attempts Miss the Point.
There have been summits of civilization at which heretics like Socrates, who was killed because he was wiser than his neighbors, have not been tortured, but ordered to kill themselves in the most painless manner known to their judges. But from that summit there was a speedy relapse into our present savagery.
Preface; Cruelty's Excuses
I dislike cruelty, even cruelty to other people, and should therefore like to see all cruel people exterminated. But I should recoil with horror from a proposal to punish them. Let me illustrate my attitude by a very famous, indeed far too famous, example of the popular conception of criminal law as a means of delivering up victims to the normal popular lust for cruelty which has been mortified by the restraint imposed on it by civilization. Take the case of the extermination of Jesus Christ. No doubt there was a strong case for it. Jesus was from the point of view of the High Priest a heretic and an impostor. From the point of view of the merchants he was a rioter and a Communist. From the Roman Imperialist point of view he was a traitor. From the commonsense point of view he was a dangerous madman. From the snobbish point of view, always a very influential one, he was a penniless vagrant. From the police point of view he was an obstructor of thoroughfares, a beggar, an associate of prostitutes, an apologist of sinners, and a disparager of judges; and his daily companions were tramps whom he had seduced into vagabondage from their regular trades. From the point of view of the pious he was a Sabbath breaker, a denier of the efficacy of circumcision and the advocate of a strange rite of baptism, a gluttonous man and a winebibber. He was abhorrent to the medical profession as an unqualified practitioner who healed people by quackery and charged nothing for the treatment. He was not anti-Christ: nobody had heard of such a power of darkness then; but he was startlingly anti-Moses. He was against the priests, against the judiciary, against the military, against the city (he declared that it was impossible for a rich man to enter the kingdom of heaven), against all the interests, classes, principalities and powers, inviting everybody to abandon all these and follow him. By every argument, legal, political, religious, customary, and polite, he was the most complete enemy of the society of his time ever brought to the bar. He was guilty on every count of the indictment, and on many more that his accusers had not the wit to frame. If he was innocent then the whole world was guilty. To acquit him was to throw over civilization and all its institutions. History has borne out the case against him; for no State has ever constituted itself on his principles or made it possible to live according to his commandments: those States who have taken his name have taken it as an alias to enable them to persecute his followers more plausibly.
It is not surprising that under these circumstances, and in the absence of any defence, the Jerusalem community and the Roman government decided to exterminate Jesus. They had just as much right to do so as to exterminate the two thieves who perished with him.
Preface, Leading Case of Jesus Christ
All government is cruel; for nothing is so cruel as impunity.
Pilate, as portrayed in Preface, Difference Between Reader And Spectator
I am no mere chance pile of flesh and bone: if I were only that, I should fall into corruption and dust before your eyes. I am the embodiment of a thought of God: I am the Word made flesh: that is what holds me together standing before you in the image of God. ... The Word is God. And God is within you. ... In so far as you know the truth you have it from my God, who is your heavenly father and mine. He has many names and his nature is manifold. ... It is by children who are wiser than their fathers, subjects who are wiser than their emperors, beggars and vagrants who are wiser than their priests, that men rise from being beasts of prey to believing in me and being saved. ... By their fruits ye shall know them. Beware how you kill a thought that is new to you. For that thought may be the foundation of the kingdom of God on earth.
Jesus, as portrayed in Preface, Difference Between Reader And Spectator
The kingdom of God is striving to come. The empire that looks back in terror shall give way to the kingdom that looks forward with hope. Terror drives men mad: hope and faith give them divine wisdom. The men whom you fill with fear will stick at no evil and perish in their sin: the men whom I fill with faith shall inherit the earth. I say to you Cast out fear. Speak no more vain things to me about the greatness of Rome. ... You, standing for Rome, are the universal coward: I, standing for the kingdom of God, have braved everything, lost everything, and won an eternal crown.
Jesus, as portrayed in Preface, Difference Between Reader And Spectator
Law is blind without counsel. The counsel men agree with is vain: it is only the echo of their own voices. A million echoes will not help you to rule righteously. But he who does not fear you and shews you the other side is a pearl of the greatest price. Slay me and you go blind to your damnation. The greatest of God's names is Counsellor; and when your Empire is dust and your name a byword among the nations the temples of the living God shall still ring with his praise as Wonderful! Counsellor! the Everlasting Father, the Prince of Peace.
Jesus, as portrayed in Preface, Difference Between Reader And Spectator
The last word remains with Christ and Handel; and this must stand as the best defence of Tolerance until a better man than I makes a better job of it.
Put shortly and undramatically the case is that a civilization cannot progress without criticism, and must therefore, to save itself from stagnation and putrefaction, declare impunity for criticism. This means impunity not only for propositions which, however novel, seem interesting, statesmanlike, and respectable, but for propositions that shock the uncritical as obscene, seditious, blasphemous, heretical, and revolutionary.
Preface, The Sacredness Of Criticism
1940s and later
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The sex relation is not a personal relation. It can be irresistibly desired and rapturously consummated between persons who could not endure one another for a day in any other relation.
letter, 24 June 1930, to Frank Harris "To Frank Harris on Sex in Biography" Sixteen Self Sketches (1949)
The quality of a play is the quality of its ideas.
"The Play of Ideas", New Statesman (6 May 1950)
The apparent multiplicity of Gods is bewildering at the first glance; but you presently discover that they are all the same one God in different aspects and functions and even sexes. There is always one uttermost God who defies personification. This makes Hinduism the most tolerant religion in the world, because its one transcendent God includes all possible Gods... Hinduism is so elastic and so subtle that the profoundest Methodist and the crudest idolater are equally at home in it.
Islam is very different, being ferociously intolerant. What I may call Manifold Monotheism becomes in the minds of very simple folk an absurdly polytheistic idolatry, just as European peasants not only worship Saints and the Virgin as Gods, but will fight fanatically for their faith in the ugly little black doll who is the Virgin of their own Church against the black doll of the next village. When the Arabs had run this sort of idolatry to such extremes ... they did this without black dolls and worshipped any stone that looked funny, Mahomet rose up at the risk of his life and insulted the stones shockingly, declaring that there is only one God, Allah, the glorious, the great... And there was to be no nonsense about toleration. You accepted Allah or you had your throat cut by someone who did accept him, and who went to Paradise for having sent you to Hell. Mahomet was a great Protestant religious force, like George Fox or Wesley....
There is actually a great Hindu sect, the Jains, with Temples of amazing magnificence, which abolish God, not on materialist atheist considerations, but as unspeakable and unknowable, transcending all human comprehension.
Letter to the Reverend Ensor Walters (1933), as cited in Dan H. Laurence (ed) Bernard Shaw : Collected Letters, 1926-1950, New York: Viking (1988), p. 305
A government which robs Peter to pay Paul can always depend on the support of Paul.
Everybody's Political What's What (1944), Ch. 30, p. 256
We have no reason to suppose that we are the Creator's last word.
Everybody's Political What's What (1944)
I am a communist, but not a member of the Communist Party. Stalin is a first rate Fabian. I am one of the founders of Fabianism and as such very friendly to Russia.
As quoted in the Evening Herald in Dublin, Ireland (February 3, 1948), reprinted in Economic Council Letter, Issue 278, Part 397 (1952), p. 1807
The road to ignorance is paved with good editions. Only the illiterate can afford to buy good books now.
As quoted in Days with Bernard Shaw (1949) by Stephen Winsten
The secret of success is to offend the greatest number of people.
As quoted in Days with Bernard Shaw (1949) by Stephen Winsten
Consistency is the enemy of enterprise, just as symmetry is the enemy of art.
As quoted in Bernard Shaw : The Lure of Fantasy (1991) by Michael Holroyd
The first prison I ever saw had inscribed on it CEASE TO DO EVIL: LEARN TO DO WELL; but as the inscription was on the outside, the prisoners could not read it.
Preface to English Prisons Under Local Government by Sydney and Beatrice Webb (1922)
Vulgarity is a necessary part of a complete author's equipment; and the clown is sometimes the best part of the circus.
Preface to London music in 1888-89 as heard by Corno di Bassetto (1937)
The epithet beautiful is used by surgeons to describe operations which their patients describe as ghastly, by physicists to describe methods of measurement which leave sentimentalists cold, by lawyers to describe cases which ruin all the parties to them, and by lovers to describe the objects of their infatuation, however unattractive they may appear to the unaffected spectators.
Preface to Ellen Terry and Bernard Shaw: A Correspondence (1931)
I know I began as a passion and have ended as a habit, like all husbands.
The Simpleton of the Unexpected Isles, Act 2 (1934)
A man of my spiritual intensity does not eat corpses.
George Bernard Shaw, quoted by Hesketh Pearson, George Bernard Shaw: His Life and Personality, 1942
Progress is impossible without change; and those who cannot change their minds cannot change anything. Creeds, articles, and institutes of religious faith ossify our brains and make change impossible. As such they are nuisances, and in practice have to be mostly ignored.
Everybody's Political What's What? (ebook, must be borrowed) (1944), Chapter XXXVII: Creed and Conduct, p. 330
Attributed
[edit]
England and America are two countries divided [separated] by a common [the same] language. (attributed to Shaw despite not appearing in his writings, see Oxford Dictionary of Quotations [4th edn., p. 638, quote no. 31]. See also Esar & Bentley, 1951, Treasury of Humorous Quotations; earlier attributed to Shaw in the Reader's Digest, November 1942.)
The quotation is thought to derive from an Oscar Wilde short story, where the following appears: "Indeed, in many respects, she was quite English, and was an excellent example of the fact that we have really everything in common with America nowadays, except, of course, language" (Wilde, "The Canterville Ghost", 1887). See Rees, Nigel (14 September 2000). "Quote... Unquote": Current Most Frequently Asked Questions" (BBC Radio 4 programme webpages). Accessed 13 December 2022.
The only time my education was interrupted was when I was in school.
Widely attributed to Shaw from the 1970s onward, but not known to exist in his published works. It is in keeping with some of his sardonic statements about the purposes and effectiveness of schools. First known attribution in print is in Neil Postman and Charles Weingartner's Teaching as a Subversive Activity (1971), "G. B. Shaw's line that the only time his education was interrupted was when he was in school captures the sense of this alienation."
Dancing is a perpendicular expression of a horizontal desire. (attributed by George Melly in 1962, see Dancing Is a Perpendicular Expression of a Horizontal Desire. Quote Investigator..
Disputed
[edit]
If you're going to tell people the truth, you better make them laugh; otherwise they'll kill you.
Credited to Shaw in the lead in to the mockumentary C.S.A.: The Confederate States of America (2004) and other recent works, but this or slight variants of it are also sometimes attributed to W. C. Fields, Charlie Chaplin, and Oscar Wilde. It might possibly be derived from Shaw's statement in John Bull's Other Island (1907): "My way of joking is to tell the truth. It's the funniest joke in the world."
Another possibility is that it is derived from Shaw's characteristic of Mark Twain: "He has to put things in such a way as to make people who would otherwise hang him believe he is joking."
Variants:
If you are going to tell people the truth, you'd better make them laugh. Otherwise, they'll kill you.
If you're going to tell people the truth, you'd better make them laugh. Otherwise, they'll kill you.
[Isadora Duncan] wrote to George Bernard Shaw: "Will you be the father of my next child? A combination of my beauty and your brains would startle the world," but he replied: "I must decline your offer with thanks, for the child might have my beauty and your brains."
Anecdote presented in "Isadore Duncan : Dancer as Plaything of Fate" in A Century of Sundays : 100 years of Breaking News in the Sunday Papers (2006), by Nadine Dreyer, p. 65; the anecdote provided here does not cite earlier sources, and though widely attributed to an exchange between Duncan and Shaw, the earliest form of it yet located is in 10,000 Jokes, Toasts & Stories (1939) by Lewis & Faye Copeland, which simply has an unidentified woman offering to have a child with Shaw, saying "think of the child with your brains and my beauty" and him replying "But what if he were to have your brains and my beauty?"
Shaw: Madam, would you sleep with me for a million pounds?
Actress: My goodness, Well, I'd certainly think about it
Shaw: Would you sleep with me for a pound?
Actress: Certainly not! What kind of woman do you think I am?!
Shaw: Madam, we've already established that. Now we are haggling over the price.
Similar remarks are also attributed to Winston Churchill, Groucho Marx and to Mark Twain
I have always held the religion of Muhammad in high estimation because of its wonderful vitality. It is the only religion which appears to me to possess that assimilating capability to the changing phase of existence which can make itself appeal to every age. The world must doubtless attach high value to the predictions of great men like me. I have prophesied about the faith of Muhammad that it would be acceptable to the Europe of tomorrow as it is beginning to be acceptable to the Europe of today. The medieval ecclesiastics, either through ignorance or bigotry, painted Muhammadanism in the darkest colours. They were in fact trained both to hate the man Muhammad and his religion. To them Muhammad was Anti-Christ. I have studied him — the wonderful man, and in my opinion far from being an Anti-Christ he must be called the Saviour of Humanity. I believe that if a man like him were to assume the dictatorship of the modern world he would succeed in solving its problems in a way that would bring it the much-needed peace and happiness. But to proceed, it was in the 19th century that honest thinkers like Carlyle, Goethe and Gibbon perceived intrinsic worth in the religion of Muhammad, and thus there was some change for the better in the European attitude towards Islam. But the Europe of the present century is far advanced. It is beginning to be enamoured of the creed of Muhammad.
Interview (April 1935), as quoted in The Genuine Islam, Vol. 1 (January 1936). A portion of the statement also appears quoted in The Islamic Review, Vol. 24 (1936) edited by Khwaja Kamal-ud-Din, p. 263
I hold the Prophet of Arabia in great esteem and I can quite understand that it would have been impossible to restrain and wean that illiterate and perverse race, sunk in the miasma of utter moral depravity, from committing the most heinous of crimes, and imbue its people with enthusiasm to strive after righteousness and assimilate high morals and virtues, without projecting such a terrible and intensely awe inspiring spectacle of Hell and an equally captivating and enticing image of a land flowing with milk and honey to represent Heaven before their vision.
Interview (April 1935) in The Genuine Islam, Vol. 1, No. 8 (1936), as quoted at "A Shavian and a Theologian" at World Islamic Mission
Despite her failure, she persisted, and one day sent Shaw a card inviting him to tea. It read: “Lady X will be at home Thursday between 4 and 6” . . . Shaw sent it back with the comment: “Mr. Bernard Shaw likewise.”
1939 September 12, The Morning Post, Walter Winchell On Broadway, Quote Page 17, Column 3, Camden, New Jersey. It was described by Witchell as a story already in circulation. The provenance of the story is uncertain, and the identity of the would-be host or hostess is unknown.
Misattributed
[edit]
If you have an apple and I have an apple and we exchange these apples then you and I will still each have one apple. But if you have an idea and I have an idea and we exchange these ideas, then each of us will have two ideas.
George Bernard Shaw never said these words, but Charles F. Brannan did.
In my view, Anglo-Irish history is for Englishmen to remember, for Irishmen to forget.
Ireland in the New Century (1904) by Horace Plunkett
Often quoted as: Irish history is something no Englishman should forget and no Irishman should remember.
You can't make a man a Christian unless you first make him believe he is a sinner.
Lin Yutang, The Importance of Living (1937), p. 17
A: Would you sleep with me for $1,000,000?
B: ...YES!
A: How about $1?
B: What do you think I am?
"The role of the character initiating the proposal in this anecdote has been assigned to George Bernard Shaw, Winston Churchill, Groucho Marx, Mark Twain, W. C. Fields, Bertrand Russell, H.G. Wells, Woodrow Wilson and others. However, the earliest example of this basic story found by QI did not spotlight any of the persons just listed [...]
[...] QI hypothesizes that this anecdote began as a fictional tale that was intended to be humorous with an edge of antagonism. The story was retold for decades. Famous men were substituted into the role of the individual making the proposition. Occasionally, the individual who received the proposition was also described as famous, but typically she remained unidentified.
[...] In January 1937 the syndicated newspaper columnist O. O. McIntyre printed a version of the anecdote that he says was sent to him as a newspaper clipping. This tale featured a powerful Canadian-British media magnate and politician named Max Aitken who was also referred to as Lord Beaverbrook [MJLB]":
Someone sends me a clipping from Columnist Lyons with this honey:
"They are telling this of Lord Beaverbrook and a visiting Yankee actress. In a game of hypothetical questions, Beaverbrook asked the lady: 'Would you live with a stranger if he paid you one million pounds?' She said she would. 'And if be paid you five pounds?' The irate lady fumed: 'Five pounds. What do you think I am?' Beaverbrook replied: 'We've already established that. Now we are trying to determine the degree."
Quote investigator cited 2013-07-10
George Bernard Shaw is said to have told W.S.C.:
Am reserving two tickets for you for my premiere. Come and bring a friend—if you have one.
W.S.C. to G.B.S.:
Impossible to be present for the first performance. Will attend the second—if there is one.
(Version given in Irrepressible Churchill: A Treasury of Winston Churchill's Wit by Kay Halle, 1966)
Apocryphal, from 1946. See discussion at Winston Churchill#Misattributed, and detailed discussion at "Here are Two Tickets for the Opening of My Play. Bring a Friend—If You Have One", Garson O'Toole, Quote Investigator, (March 25, 2012)
The Bible is most dangerous book ever written on earth, keep it under lock and key.
From Why You Should Never be a Christian (1987) by Ishaq 'Kunle Sanni and Dawood Ayodele Amoo.
The trouble with her is that she lacks the power of conversation but not the power of speech.
Widely attributed to Shaw, this quotation is actually of unknown origin.
The single biggest problem in communication is the illusion that it has taken place.
The attribution to Shaw comes from Leadership Skills for Managers (2000) by Marlene Caroselli, p. 71. But this quote seems more likely to come from William H. Whyte.The Biggest Problem in Communication Is the Illusion That It Has Taken Place. Quote Investigator (2014-08-31). Retrieved on 2015-11-09.
Success does not consist in never making blunders, but in never making the same one a second time.
H. W. Shaw (Josh Billings), as quoted in Scientific American, Vol. 31 (1874), p. 121, and in dictionaries of quotations such as Excellent Quotations for Home and School (1890) by Julia B. Hoitt, p. 117 and Many Thoughts of Many Minds: A Treasury of Quotations from the Literature of Every Land and Every Age (1896) by Louis Klopsch, p. 266.
I learned long ago never to wrestle with a pig. ... You get dirty and besides the pig likes it.
Initially attributed to Cyrus S. Ching in Time, Vol. 56 (1950), p. 21. Also attributed to Mark Twain and Abraham Lincoln: Quote Investigator July 8, 2017.
Quotes about Shaw
[edit]
: Alphabetised by surname
Shaw's plays are the price we pay for Shaw's prefaces.
James Agate, diary entry (10 March 1933)
Shaw knows at any moment, on any subject, what he thinks, what you will think, what others have thought, what all this thinking entails; and he takes the most elaborate pains to bring these thoughts to light in a form which is by turns abstract and familiar, conciliatory and aggressive, obvious and inferential, comic and puzzling. In a word, Shaw is perhaps the most consciously conscious mind that has ever thought — certainly the most conscious since Rousseau; which may be why both of them often create the same impression of insincerity amounting to charlatanism. Yet it is by excess of honesty that Shaw himself lent color to his representation as an inconsequential buffoon bent on monopolizing the spotlight.
Jacques Barzun, in "Bernard Shaw in Twilight" in The Kenyon Review (Summer 1943)
Seeing clearly within himself and always able to dodge around the ends of any position, including his own, Shaw assumed from the start the dual role of prophet and gadfly.
Jacques Barzun, in "Bernard Shaw in Twilight" in The Kenyon Review (Summer 1943)
Shaw does not merely decorate a proposition, but makes his way from point to point through new and difficult territory.
Jacques Barzun, in "Bernard Shaw in Twilight" in The Kenyon Review (Summer 1943)
He never invested his whole moral capital in a man, a book, or a cause, but treasured up wisdom wherever it could be picked up, always with scrupulous acknowledgment ... His eclecticism saving him from the cycle of hope-disillusion-despair, his highest effectiveness was as a skirmisher in the daily battle for light and justice, as a critic of new doctrine and a refurbisher of old, as a voice of warning and encouragement. That his action has not been in vain, we can measure by how little Shaw's iconoclasm stirs our blood; we no longer remember what he destroyed that was blocking our view.
Jacques Barzun, in "Bernard Shaw in Twilight" in The Kenyon Review (Summer 1943)
Bernard Shaw remains the only model we have of what the citizen of a democracy should be: an informed participant in all things we deem important to the society and the individual.
Jacques Barzun "Bernard Shaw," in A Jacques Barzun Reader: Selections from his works (2002), p. 231
Mr. Shaw cannot realise his own pertness, nor can he preserve his own gravity, for more than a few moments at a time. Even when he sets out to be funny for fun's sake, he must needs always pretend that there is a serious reason for the emprise; and he pretends so strenuously that he ends by convincing us almost as fully as he convinces himself. Thus the absurdity, whatever it be, comes off doubly well. Conversely, even when he is really engrossed in some process of serious argument, or moved to real eloquence by one of his social ideals, he emits involuntarily some wild jape which makes the whole thing ridiculous — as ridiculous to himself as to us; and straightway he proceeds to caricature his own thesis till everything is topsy-turvy; and we, rolling with laughter, look up and find him no longer on his head, but on his heels, talking away quite gravely; and this sets us off again. For, of course, when seriousness and frivolity thus co-exist inseparably in a man, the seriousness is nullified by the frivolity. The latter is fed by the former, but, graceless and vampire-like, kills it. As a teacher, as a propagandist, Mr. Shaw is no good at all, even in his own generation. But as a personality he is immortal.
Posterity will not, I fancy, read his writings. He has not enough of the specific art-sense for writing. I will not exasperate him by complaining that he has no sense of beauty in the use of his medium: the idea of beauty is a red rag to him, as we know. I will merely suggest that he has in his writing the qualities of a public speaker rather than of a writer. He does not write with that closeness which is the result not of haste but of leisure, and which is the main secret of good literature. He is too glib, too fluent, too diffuse, and too loud. Glibness and fluency, loudness and diffusion, are just the qualities needed for addressing an audience. But between speaking and writing there is a vast difference. A good writer cannot make good speeches, and Mr. Shaw's seems an instance to prove that a good speaker cannot write well. We, his contemporaries, can read him with delight, even though we seem to miss the reporter's interpolation of "laughter", "cheers", "interruption", and so forth. But relentlessly, in course of time, lack of solid form "tells on" writing. However interesting a writer may be, he will not, unless he be a strict artist, be read by posterity. Style, as has been said, is the one antiseptic. But, though Mr. Shaw's writing be not good enough for the next generation, he himself, being so signally unique, is good enough for all time. I wish I had the leisure to be his Boswell, and he the kindness to be my Johnson.
Max Beerbohm, in "A Cursory Conspectus of G. B. S." in The Saturday Review of Politics, Science and Art (2 November 1901)
The writers of our century delight in the weaknesses of the human condition; the only one capable of inventing heroes was Bernard Shaw.
Jorge Luis Borges, Obra Completa (1996), Vol. IV, p. 487
"God spare you, reader, of long prefaces". That was written by Quevedo, who, in order not to commit an anachronism that would have been found out in the long run, never read Shaw´s.
Jorge Luis Borges, Obra Completa (1996), Vol. II, p. 400
He was a Tolstoy with jokes, a modern Dr Johnson, a universal genius who on his own modest reckoning put even Shakespeare in the shade.
John Campbell, The Independent, as quoted in Penguin Classics edition of Plays Unpleasant (1946)
In his works Shaw left us his mind ... Today we have no Shavian wizard to awaken us with clarity and paradox, and the loss to our national intelligence is immense.
John Carey, The Sunday Times, as quoted in Penguin Classics edition of Plays Unpleasant (1946)
I never read a reply by Shaw that did not leave me in better and not worse temper or frame of mind; which did not seem to come out of inexhaustible fountains of fairmindedness and intellectual geniality; which did not savor somehow of that native largeness which the philosophers attributed to Magnanimous Man.
G. K. Chesterton, commenting on twenty years of debate with Shaw on political, religious and other social issues.
He said that one should never tell a child anything without letting him hear the opposite opinion. That is to say, when you tell Tommy not to hit his sick sister on the temple, you must make sure of the presence of some Nietzscheite professor, who will explain to him that such a course might possibly serve to eliminate the unfit. When you are in the act of telling Susan not to drink out of the bottle labelled "poison," you must telegraph for a Christian Scientist, who will be ready to maintain that without her own consent it cannot do her any harm. What would happen to a child brought up on Shaw's principle I cannot conceive; I should think he would commit suicide in his bath.
G. K. Chesterton: George Bernard Shaw. 1909. p. 174.
What a debt every intelligent being owes to Bernard Shaw!
John Maynard Keynes, "One of Wells' Worlds" (Review of the World of William Clissold") in The New Republic (1 February 1927)
Shaw and Stalin are still satisfied with Marx's picture of the capitalist world... They look backwards to what capitalism was, not forward to what it is becoming.
John Maynard Keynes,"Stalin-Wells Talk: The Verbatim Report and A Discussion", G.B. Shaw, J.M. Keynes et al., London, The New Statesman and Nation, (1934) p. 34
Shaw was a very great man indeed. The danger is that when all the froth and nonsense about his being a philosopher has died down (as it must) a reaction should set in and lead people to forget his real genius. He was a comedian, in his own time, of the very highest order ... He was a humorist of the more intellectual kind, a master of satire, art and fantasy like Gilbert, Wilde and Aristophanes. In that class no one had more continuous vitality. He is also, in his prefaces, one of the great masters of plain prose. I have often, in that capacity, held him up as a model to my pupils and have learned much from him myself. Peace to his ashes!
C. S. Lewis, in "Comedian of Highest Order", in The Mark Twain Journal, Vol. 9, no. 4 (Summer 1954), p. 10
He did his best in redressing the fateful unbalance between truth and reality, in lifting mankind to a higher rung of social maturity. He often pointed a scornful finger at human frailty, but his jests were never at the expense of humanity.
Thomas Mann, as quoted in Penguin Classics edition of Plays Unpleasant (1946)
Desmond MacCarthy, whom I tried to persuade to write a new appreciation of Shaw in old age, noticed a real deterioration in Shaw himself. The Shaw who had praised Mussolini and justified Hitler grew increasingly irresponsible in suggesting that people who were a nuisance should be killed. This strain in Shaw, and his characteristic inconsistency when he dropped back suddenly into individualism, after maintaining the State's right to liquidate anyone it disapproved of, had been growing steadily stronger from Major Barbara onwards. He ceased to have a genuine humanism such as he had shown in the splendid preface to John Bull's Other Island. In general, re-reading Shaw, MacCarthy said he could find nothing but "a chaos of clear ideas".
Kingsley Martin, Editor: A Second Volume of Autobiography, 1931-45 (1968), p. 110
He understood early the weakness of democracy; he was naturally impatient with the shallow humbug of much political talk. There was so much in it to laugh at, and so much to expose, that he even allowed himself to praise Mussolini and Hitler and to excuse all the darker deeds of Stalin. In 1948 he sent me a letter describing Russia as a democracy in which Stalin would be pushed out of power in ten minutes if he offended the majority of the Communist Party.
Kingsley Martin, Editor: A Second Volume of Autobiography, 1931-45 (1968), p. 112
Shaw presumes that his friend Stalin has everything under control. Well, Stalin may have made special arrangements to see that Shaw comes to no harm, but the rest of us in Western Europe do not feel quite so sure of our fate,especially those of us who do not share Shaw's curious admiration for dictators.
J. B. Priestley, The War - And After, in Horizon magazine (January 1940), reprinted in War Decade : An Anthology of the 1940s (1989) by Andrew Sinclair
One may say that he [Shaw] did much good and some harm. As an iconoclast he was admirable, but as an eikon rather less so.
Bertrand Russell, Portraits from Memory and Other Essays (1956)
[Shaw] had just learned, more less, to ride a bicycle. And I went out for a country ride with him, and at the bottom of a steep hill the road forked and I didn't know which way to go, and Shaw was behind me. And I got off my bicycle to ask which way we should go. And he wasn't able to manage his machine, and he ran slap into my bicycle. My bicycle buckled. He was precipitated 20 teet through the air and landed on his back on the hard road. He got up, his bicycle undamaged, rode home: I had to go home by train.
Bertrand Russell, in an interview with David Susskind (10 June 1962)
I found many men to whom I felt deeply grateful — especially Guy de Maupassant, Jack London, and H. L. Mencken — but the first man to whom I felt definitely related was George Bernard Shaw. This is a presumptuous or fatuous thing to mention, perhaps, but even so it must be mentioned. ... I myself, as a person, have been influenced by many writers and many things, and my writing has felt the impact of the writing of many writers, some relatively unknown and unimportant, some downright bad. But probably the greatest influence of them all when an influence is most effective — when the man being influenced is nowhere near being solid in his own right — has been the influence of the great tall man with the white beard, the lively eyes, the swift wit and the impish chuckle. ... I have been fascinated by it all, grateful for it all, grateful for the sheer majesty of the existence of ideas, stories, fables, and paper and ink and print and books to hold them all together for a man to take aside and examine alone. But the man I liked most and the man who seemed to remind me of myself — of what I really was and would surely become — was George Bernard Shaw.
William Saroyan, Hello Out There (1941)
Shaw is a pleasant man, simple, direct, sincere, animated; but self-possessed, sane, and evenly poised, acute, engaging, companionable, and quite destitute of affectation. I liked him.
Mark Twain, after meeting Shaw in 1907
His argument seems to be that either the Haves or the Have-Nots must seize power and compel all to come under the Fascist or the Communist plough. It is a crude and flippant attempt at reconstruction, bred of conceit, impatience and ignorance. ... [I]t reinforces the Italian tyranny. It is only fair to add that this naïve faith in a Superman before whose energy and genius all must bow down is not a new feature in the Shaw mentality. What is new and deplorable is the absence of any kind of sympathetic appreciation of the agony that the best and wisest Italians are today going through; any appreciation of the mental degradation as implied in the suppression of all liberty of thought and speech.
Beatrice Webb's diary (1927), quoted in Beatrice Webb, Diaries: 1924-1932, ed. Margaret Cole (1952), p. 155
The worst element in his mental make-up is a queer readiness to succumb to the poses of excessive virility. His soul goes down before successful force. He exalted the maker of enormous guns in Man and Superman; he has rejoiced in the worst claptrap of the Napoleonic legend; now he is striking attitudes of adoration towards the poor, vain, doomed biped who is making Rome horrible and ridiculous to all the world. When it comes to the torture of intelligent men, to vile outrages on old women, to the strangulation of all sane criticism and an orgy of claptrap more dreadful than its attendant cruelties, this vituperative anti-vivisectionist becomes an applauding spectator.
H. G. Wells, The Way The World Is Going (1929), p. 279
Uncle Wells was as magnificent an uncle as one could hope to have. So, too, was Uncle Shaw. He brought his mind for the children to look at, his marvellous shining mind. Too thin a mind, Philistines would object; but the very finest French watches are as thin as a couple of halfcrowns and yet keep better time than the grosser article. He did for his age what Voltaire and Gibbon did for theirs: he popularized the use of the intellectual processes among the politically effective class. And he did it with such style.
Rebecca West, The Strange Necessity. Doubleday, Doran, Incorporated, 1928 (pp. 216-217).
In fact, the real problem with the thesis of A Genealogy of Morals is that the noble and the aristocrat are just as likely to be stupid as the plebeian. I had noted in my teens that major writers are usually those who have had to struggle against the odds -- to "pull their cart out of the mud," as I put it -- while writers who have had an easy start in life are usually second rate -- or at least, not quite first-rate. Dickens, Balzac, Dostoevsky, Shaw, H. G. Wells, are examples of the first kind; in the twentieth century, John Galsworthy, Graham Greene, Evelyn Waugh, and Samuel Beckett are examples of the second kind. They are far from being mediocre writers; yet they tend to be tinged with a certain pessimism that arises from never having achieved a certain resistance against problems.
Colin Wilson in The Books In My Life, p. 188
See also
[edit]
Eugenics
External links
[edit]
Wikipedia has an article about:
George Bernard Shaw
Wikisource has original works by or about:
George Bernard Shaw
Wikimedia Commons has media related to:
George Bernard Shaw
Works by or about George Bernard Shaw at the Internet Archive]
Works by George Bernard Shaw at Project Gutenberg
Works by George Bernard Shaw at Project Gutenberg Australia
George Bernard Shaw on IMDb
International Shaw Society, includes a chronology of Shaw's works
The Shaw Society, UK, established in 1941
The Bernard Shaw Society, New York
Shaw Chicago Theater A theater dedicated to the works of Shaw & his contemporaries.
Shaw Festival Niagara-on-the-Lake, Ontario, Canada theatre that specializes in plays by Bernard Shaw and his contemporaries and plays about his era (1856-1950)
The Nobel Prize Biography on Shaw, From Nobel Lectures, Literature 1901–1967, Editor Horst Frenz, Elsevier Publishing Company, Amsterdam (1969)
Dan H. Laurence/Shaw Collection in the University of Guelph. Archived from the original on 30 June 2009.
Michael Holroyd (July 19, 2006). Send for Shaw, not Shakespeare. The Times Literary Supplement. Archived from the original on 29 September 2006.
Sunder Katwala (July 26, 2006). "Artist of the impossible". The Guardian.
George Bernard Shaw Timeline. Archived from the original on 8 Jule 2008.
Bernard Shaw papers at LSE Archives. Archived from the original on 30 June 2002.
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原问题描述如下：
5 5 个海盗分 100 100 件宝藏， 5 5 人依次提分配方案，当支持人数达到一半及以上则按照此分配方案进行分配，当支持人数少于一半，则把提出该方案的人剔除掉，并且 5 5 个海盗足够贪婪，希望自己尽可能多拿，求第一个海盗提出方案的最优解。
并且
每个海盗总能够以本人利益最大化作为行为准则
在能够取得尽量多金币的情况下，海盗不会故意致同伙于死地
海盗之间没有个人恩怨，分给其他海盗金币的次序以小序号优先为原则
但似乎有很多方案能让第 i i 位海盗得到相同的财宝，所以我们不妨在加一个条件
如果第 i i 位海盗给出的方案给 x x 的金币和第 j(j>i)j(j>i) 位（预测）给出方案的金币相同，那么 x x 将会给 i i 投反对票（因为人越少越好）。
思路：
固定金币数 100 100 ，考虑从一个海盗递推到五个海盗的情况。
只有一个海盗时，这个海盗绝对会要走所有的财宝，他得到的财宝为 。
有两个海盗时，海盗一会拿走所有宝物，就算第二个海盗投反对票，也不会得到金币，他们得到的财宝为 [100,0][100,0] 。
有三个海盗时，如果第一个海盗拿走所有财宝，那么他就会得到两个反对票，所以他不能拿走所有财宝，不管怎么样第二个海盗都会给第一个海盗投反对票（因为剔除第一个海盗后选择权就在第二个海盗手里），因此只要获得第三个海盗的支持，他就可以得到财宝，两个海盗的情况我们已经分析了，所以第一个海盗只需给第三个海盗 1 1 件宝物即可获得最大值，方案为 [99,0,1][99,0,1] 。
有四个海盗时，同样的，二号海盗会选择反对，三个海盗的情况下第二个海盗没有财宝，我们给他一件宝物便能获取他的支持，所以我们只需这样分配即可 [99,0,1,0][99,0,1,0] 。
有五个海盗时，给四个海盗情况下没有金币的各给 1 1 件财宝便能获得他们的支持，方案就是 [98,0,1,0,1][98,0,1,0,1] 。
拓展：
有六个海盗时，给五个海盗情况下没有金币的各给 1 件财宝便能获得他们的支持，方案是 [98,0,1,0,1,0] 。
有七个海盗时，给六个海盗情况下没有金币的各给 1 件财宝便能获得他们的支持，方案是 [97,0,1,0,1,0,1] 。
以此类推，当现在有 a 个海盗， b 个财宝时，第一位海获得的最大收益为 b-\biggl\lfloor\dfrac{a-1}{2}\biggl\rfloor 。
纳什均衡
这种博弈就是纳什均衡，博弈的所有参与人都为了满足自己的个人利益而选择牺牲集体利益而导致的全体参与人都吃亏的均衡状态。
一个策略博弈包含 {N} 玩家集， {A_i} 每个玩家的策略集， {u_i} 表示每个玩家的最终收益。
通常一个博弈论具有一个策略式博弈集合 G={N,{A_i},{u_i}} 。
囚徒困境
假设有两个小偷 A 和 B 联合犯事，对于每一个犯罪嫌疑人，警方给出的政策是：
如果两个犯罪嫌疑人都作了坦白，则两人各被判刑 8 年；
如果有一个犯罪嫌疑人坦白了，另一个犯罪嫌疑人没有坦白而是抵赖，则以妨碍公务罪再加刑 2 年，而坦白者立即释放。
如果两人都抵赖，则警方因证据不足不能判两人的偷窃罪，但可以私入民宅的罪名将两人各判入狱 1 年。
可以列出以下表格：
| A\B | 坦白 | 抵赖 |
---
| 坦白 | (-8,-8) | (-10,0) |
| 抵赖 | (0,-10) | (-1,-1) |
显然双方都抵赖最优，但是我们并不知道对方是否抵赖。
当 A 坦白时， B 坦白， B 收益为 -8 ； B 抵赖， B 收益为 -12 ，因此 B 会选择坦白。
当 A 抵赖时， B 坦白， B 收益为 0 ； B 抵赖， B 收益为 -1 ，因此 B 一定会选择坦白。
而 A,B 都是这么想的，因此 A,B 各判 8 年。
当对手策略选定的时候，应该调整自己的策略，使得自己收益在几种策略选择中是最大的，这时的策略称为“最优策略”。这个时候，如果对手不改变策略的话，我们也不会改变策略。
如果每个人的策略都是“最优策略”，那么就会形成一种稳定的局面，这时的博弈结果就是纳什均衡。
硬币正反
假设有富豪邀请你 van 一个游戏，你们每次抛硬币，如果你们都是正面，则富豪给你 3 元，如果你们都是反面，则富豪给你 1 元，其他情况下，你需给富豪 2 元。
看起来这个游戏好像是公平的，但是我们仔细分析。
设你出正面的概率为 x ，富豪出正面的概率为 y 。
我们总不能一直出正面，我们列方程来求解正面概率为多少时收益最大。
我们应该在对手出正面或反面的时候我们的收益都相等（不然对手就可以改变正反概率让我们收益减少）。
所以 3x+(-2)(1-x)=(1-x)+(-2)x ，解得 x=\dfrac{3}{8} 。
同样的，我们可以解得 y=\dfrac{3}{8} 。
那么富豪期望收益 2(1-y)-3y=\dfrac{1}{8} ，我们亏了 \dfrac{1}{8} 元，按这种情况，不管我们怎么下最后都是富豪赚，我们亏。
发布于 2023-08-11 08:34・重庆
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【一个值得回味的游戏】10个海盗分金币 =================== adr999海盗行动背后的商业秘密 =========== 海盗的黄金时代不仅仅是勇敢冒险家驶向夕阳寻找财富的故事。它关乎一群暴力企业家，他们明白成功需要与任何其他海事企业一样，对情报、战略规划和巡航策略给予同等的审慎关注。这些海上劫掠… 异世界冒险...发表于欧洲/中世... 聊聊海盗 ==== 三只布朗熊 经典益智谜题之海盗分金 =========== 终军弱冠
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11520 | https://www.sciencedirect.com/science/article/abs/pii/S1546144012002918 | ACR Appropriateness Criteria® Suspected Upper Extremity Deep Vein Thrombosis - ScienceDirect
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Cited by (17)
Journal of the American College of Radiology
Volume 9, Issue 9, September 2012, Pages 613-619
Original article
ACR Appropriateness Criteria® Suspected Upper Extremity Deep Vein Thrombosis
Author links open overlay panel Benoit Desjardins MD, PhD a, Frank J.Rybicki MD, PhD b, Hyun S.Kim MD c, Chieh-Min Fan MD b, Scott D.Flamm MD d, Marie D.Gerhard-Herman MD b e, Sanjeeva P.Kalva MD f, Scott A.Koss MD g, M. Ashraf Mansour MD h i, Emile R.Mohler III MD e j, Vamsi R.Narra MD k, Matthew P.Schenker MD b, Mark Tulchinsky MD l m, Clifford Weiss MD n
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Upper-extremity venous thrombosis often presents as unilateral arm swelling. The differential diagnosis includes lesions compressing the veins and causing a functional venous obstruction, venous stenosis, an infection causing edema, obstruction of previously functioning lymphatics, or the absence of sufficient lymphatic channels to ensure effective drainage. The following recommendations are made with the understanding that venous disease, specifically venous thrombosis, is the primary diagnosis to be excluded or confirmed in a patient presenting with unilateral upper-extremity swelling. Contrast venography remains the best reference-standard diagnostic test for suspected upper-extremity acute venous thrombosis and may be needed whenever other noninvasive strategies fail to adequately image the upper-extremity veins. Duplex, color flow, and compression ultrasound have also established a clear role in evaluation of the more peripheral veins that are accessible to sonography. Gadolinium contrast–enhanced MRI is routinely used to evaluate the status of the central veins. Delayed CT venography can often be used to confirm or exclude more central vein venous thrombi, although substantial contrast loads are required.
The ACR Appropriateness Criteria® are evidence-based guidelines for specific clinical conditions that are reviewed every 2 years by a multidisciplinary expert panel. The guideline development and review include an extensive analysis of current medical literature from peer-reviewed journals and the application of a well-established consensus methodology (modified Delphi) to rate the appropriateness of imaging and treatment procedures by the panel. In those instances in which evidence is lacking or not definitive, expert opinion may be used to recommend imaging or treatment.
Section snippets
Introduction/Background
Upper-extremity venous thrombosis often presents as unilateral arm swelling. The differential diagnosis includes a mass lesion or other lesion compressing the veins and causing a functional venous obstruction, venous stenosis, or an infection causing edema . Bilateral upper-extremity swelling may also be due to right-sided heart failure, although this is typically associated with generalized swelling, in contrast to central vein obstruction, which can cause swelling limited to the upper
MRI
Approaches to venous imaging using MRI include black-blood and flow-based or contrast-enhanced bright-blood techniques . Black-blood techniques include conventional T1 or T2 spin-echo [28, 35] or fast spin-echo imaging. However, the black-blood effect on routine spin-echo imaging may not be consistent, and newer double inversion-recovery techniques provide more reliable black-blood imaging . Using black-blood imaging, the presence of thrombus is inferred from focal high signal, often
CT
Computed tomography can be used to determine the presence of centrally located thrombi or stenoses within the jugular veins [51, 52], the brachiocephalic veins [53, 54], and the superior vena cava . The presence of an extrinsic process causing venous obstruction of the venous channels can also be determined . Computed tomography is the main imaging modality for staging neoplastic involvement in the mediastinum and axillae, which can include vascular invasion or compression. Perivascular
Summary
•Despite the availability of noninvasive imaging techniques, contrast venography remains the best reference standard diagnostic test for suspected upper-extremity acute venous thrombosis.
•Contrast venography may be needed whenever other noninvasive strategies fail to adequately image the upper-extremity veins. Additionally, because venography is the first step in direct catheter-based thrombolysis, in situations such as acute upper-extremity DVT in which the likelihood of percutaneous
Anticipated Exceptions
Nephrogenic systemic fibrosis is a disorder with a scleroderma-like presentation and a spectrum of manifestations that can range from limited clinical sequelae to fatality. It seems to be related to both underlying severe renal dysfunction and the administration of gadolinium-based contrast agents. It has occurred primarily in patients on dialysis, rarely in patients with very limited glomerular filtration rates (ie, <30 mL/min/1.73 m 2), and almost never in other patients. There is growing
Relative Radiation Level Information
Potential adverse health effects associated with radiation exposure are an important factor to consider when selecting the appropriate imaging procedure. Because there is a wide range of radiation exposures associated with different diagnostic procedures, a relative radiation level indication has been included for each imaging examination. The relative radiation levels are based on effective dose, which is a radiation dose quantity that is used to estimate population total radiation risk
Recommended articles
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View more references
Cited by (17)
Paget-Schroetter syndrome: Diagnostic limitations of imaging upper extremity deep vein thrombosis
2016, American Journal of Emergency Medicine
### Evidence-Based Strategies and Recommendations for Preservation of Central Venous Access in Children
2019, Journal of Parenteral and Enteral Nutrition
### Venous Thromboembolism After Shoulder Arthoplasty and Arthroscopy
2019, Journal of the American Academy of Orthopaedic Surgeons
### Upper extremity deep vein thrombosis: Symptoms, diagnosis, and treatment
2018, Israel Medical Association Journal
### The role of ultrasound guidance for vascular access
2015, Current Opinion in Anaesthesiology
### Imaging of deep venous thrombosis: A multimodality overview
2014, Applied Radiology
View all citing articles on Scopus
The ACR seeks and encourages collaboration with other organizations on the development of the ACR Appropriateness Criteria® through society representation on expert panels. Participation by representatives from collaborating societies on the expert panel does not necessarily imply individual or society endorsement of the final document.
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Copyright © 2012 American College of Radiology. Published by Elsevier Inc. All rights reserved.
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11521 | https://pubmed.ncbi.nlm.nih.gov/31025219/ | MDCalc Medical Calculator App Review - PubMed
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doi: 10.1007/s10278-019-00218-y.
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MDCalc offers all healthcare professionals a quick and well-designed tool to look up for popular clinical calculators that are supported by evidence-based medicine. The app allows you to select your speciality and have related calculations at a press of a button. The app offers hundreds of clinical decision tools including risk scores, algorithms, equations, diagnostic criteria, formulas, classifications, dosing calculators, and more at your fingertips.
Keywords: Calculator; MD Calc; Medical calculator; Medicine.
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Home screen when first downloaded
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Home screen when first downloaded
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Tutorial screen when first using the app
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Tutorial screen when first using the app
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Example of specific clinical tool homepage
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Example of specific clinical tool homepage
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11522 | https://www.youtube.com/watch?v=W1q5plbGu6U | Washington State is Not the same as Washington D.C.
Mesnick Dalto Group
791 subscribers
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Posted: 4 Nov 2022
There are many many different differences between Washington State and Washington D.C. The major difference is that Joe Biden is NOT in Washington! In today's video we break down the similarities and differences between Washington D.C. and Washington State.
Jason & Molly have helped over 163 families buy or sell real estate in the Seattle Metro area! With their skill and knowledge they are the GO-TO agents in the Seattle area.
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Jason is a proud graduate of Lake Washington High School and the University of Washington, Jason has spent most of his life in Kirkland, where his kids Ty and Riley are enrolled in the same schools that he attended. Jason met his wife Molly in a very unusual way, on ABC’s The Bachelor in 2008. If you have a fun love story, he would love to hear it.
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breaking news Washington state where we live is not the same place as Washington DC oh my God can you believe that people actually get confused about this no I can't they clearly missed that day of fourth grade apparently it confuses people I think like again I don't know if adults get confused but I think kids probably do okay like honestly it is the biggest difference it's 2700 miles apart yeah Washington DC DC being the District of Columbia east coast East Coast Washington state west coast you know what this makes me think of we have a friend okay Eddie oh my gosh I didn't even think about anything Eddie who was living in Washington DC happened I'm looking right at you Eddie I happen to walk from Washington DC to Washington to Washington State we gotta say it watch it we gotta say it he walked across the country yeah he walked he didn't jog he didn't bike he didn't do any of those things it took him like six months that sounds like not fun okay [Music] the differences between DC and Washington State okay so like East Coast West Coast the thing that sticks out to me the most is the weather that's what sticks out to you well I mean I'm thinking like okay hold on that's what sticks out to you it's the coast the weather the difference you know what sticks out to me the coast one is a city one is a state one is it's not even a city Washington yeah its own thing it's its own thing Washington state has over 7 million people in it the District of Columbia DC has like some like not even a million okay so population number of people that's a big difference they both have large white sorry large pointy things that point Towards the Sky like monuments okay so and the Space Needle got the Space Needle Washington DC has the National Monument so those are that's similar I have similar okay right so it can be confusing sure right yeah um they both have white houses okay well we just have houses that are white Washington DC has the White House so big deal they're both houses that are white okay so I guess that could be a reason people get confused so so outside of like the coasts we both have pointy things to point to the sky yeah right and we both have white similarities we have we both have white houses yeah we both have let's say there's 700 000 people in the District of Columbia and seven million people here they both start with a seven okay the most one of the most famous people from Washington State was in my fraternity older than me Joe mccale from the soup you think he's one of the most famous people from Washington State yeah for me he was in my like I went you know what what are those the um with all the pictures in your fraternity composite a composite picture so you go to the composite picture usually there's like athletes or nobodies like the rest of us yeah I said Joe McCabe oh so I said okay Joe McHale is from Washington State say my name Joe McHale God Joe Biden lives in Washington DC so we both have Joe's just different Joes so but that's that's similar yeah but so I guess you're right the weather is probably the biggest weather okay so I have been to the east coast in the summertime I've been to Washington DC in August okay so explain to everybody if you happen to be watching this from Washington State what is the weather like on the east coast in DC in the summertime it is so hot and so humid it's so miserable you don't even want to be outside okay that was literally the worst time to be touring Washington DC was it like a school tour no no I went with an ex-boyfriend is his name Joe no I was in jail so I Remember When Ty went Ty went with his mom to Washington DC I think in August oh you're right yeah he said it was so miserable all he said is he kept on going buying to buy bottles of water so it was so hot in Washington state in August you don't have to do that no it's like 70 degrees 75 degrees miles easy to be outside I think the bottom line is I hope everybody realizes that Washington State and Washington DC are more similar than you might no they're not they're two different they're totally different and if you get confused about this look at a map yeah think about the weather think about the famous Joe's the white houses The Monuments the East Coast the West Coast yeah I mean think about like I know that Washington DC has like beautiful buildings and just you know but we've got nature we've got beautiful trees bottom line it's two different places people oh my gosh if people get confused by this you're definitely right go back to fourth grade God |
11523 | https://www.anleitung-deutsch-lernen.de/alles-wichtige-rund-um-den-komparativ-1-teil/ | Deutsch lernen – Übungen, Anleitungen, Tipps
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Alles Wichtige rund um den Komparativ, 1. Teil
Alles Wichtige rund um den Komparativ, 1. Teil – Der Komparativ zeigt an, dass bei einem Objekt ein Merkmal stärker oder in größerem Umfang vorhanden ist als bei einem anderen Objekt. Es geht also um einen Vergleich. Die Wortarten, die für diesen Vergleich genutzt werden, sind Adjektive und Adverbien.
In einem zweiteiligen Beitrag erklären wir alles Wichtige und Wissenswerte rund um den Komparativ, anfangen bei seiner Bildung über die Verwendung bis hin zu typischen Fehlerquellen:
Inhalt
Was ist der Komparativ?
Bevor wir anfangen, räumen wir erst einmal eine Fehlerquelle aus dem Weg: Der Komparativ heißt tatsächlich Komparativ und nicht Komperativ. Diese Bezeichnung taucht recht oft auf, ist aber falsch!
Komparativ geht auf den lateinischen Ausdruck „gradus comparativus“ zurück, was übersetzt Vergleichsgrad bedeutet. Das Vergleichen mithilfe des Komparativs wird auch Komparation genannt.
Im alltäglichen Sprachgebrauch hingegen wird der Komparativ unter anderem als Steigerungsform, Vergleichsform oder Höherstufe bezeichnet.
Wann wird der Komparativ verwendet?
Wir verwenden den Komparativ, um quantitative und qualitative Unterschiede zwischen Personen, Dingen oder Zuständen zu beschreiben. Dabei bildet der Komparativ zusammen mit dem Positiv und dem Superlativ eine grammatikalische Familie.
Der Positiv ist die Grundform eines Adjektivs oder Adverbs. Er beschreibt nur, ohne eine Wertung vorzunehmen. Der Komparativ vergleicht und steigert damit um eine Stufe. Die höchste Steigerungsform, also die Höchststufe, ist der Superlativ.
In der Praxis sieht das zum Beispiel so aus:
| | | |
---
| Positiv | Komparativ | Superlativ |
| schnell | schneller | am schnellsten |
| klein | kleiner | am kleinsten |
| leicht | leichter | am leichtesten |
| jung | jünger | am jüngsten |
| laut | lauter | am lautesten |
Manchmal wird noch der Elativ als vierte Steigerungsform verwendet. Der Elativ gilt als absoluter Superlativ ohne eine vergleichende Funktion.
Ein Beispiel:
Der erste Satz steht im Superlativ, denn er vergleicht die Geräte mit anderen Geräten. Auch der zweite Satz beschreibt eine größtmögliche Steigerung. Weil er aber keinen Vergleich durchführt, steht er im Elativ.
Für den Komparativ gibt es zahlreiche Anwendungen. Wir können ihn verwenden, um Geschwindigkeiten, Größen, Intensitäten und andere Eigenschaften vergleichend zu beschreiben.
Das ist sowohl in positiven als auch in negativen Aussagen möglich.
Außerdem können wir den Komparativ einsetzen, wenn wir auf Entwicklungen hinweisen möchten:
Wie wird der Komparativ gebildet?
Bei regelmäßigen Adjektiven und Adverbien ist es sehr einfach, den Komparativ zu bilden. Wir müssen dazu nämlich nur ein sogenanntes Komparativ-Suffix anhängen.
Das ist eine Endung, die auch Vergleichsnachsilbe genannt wird und im Regelfall „-er“ lautet. Hängen wir diese Endung an ein Adjektiv oder Adverb, wird daraus der Komparativ.
Möchten wir zum Beispiel den Komparativ von „laut“ bilden, fügen wir „-er“ an und erhalten „lauter“ als Komparativ.
Allerdings gibt es, wie so oft im Deutschen, einige Ausnahmen und Besonderheiten:
Steigerung mit Umlauten
Vor allem Adjektive und Adverbien, die nur aus einer Silbe bestehen, bilden den Komparativ oft mit einem Umlaut. In diesem Fall hängen wir die Endung „-er“ an, ersetzen zusätzlich dazu aber noch den Vokal der Grundform durch den entsprechenden Umlaut.
Ein paar Beispiele:
Wird schon die Grundform mit einem Umlaut geschrieben, bleibt er im Komparativ erhalten. Dann kommt nur die Endung „-er“ dazu, zum Beispiel kühl – kühler.
Adjektive mit der Endung „-el“
Eine weitere Besonderheit betrifft Adjektive, bei denen die Grundform auf „-el“ endet und nicht betont ist. Hier fällt das „e“ weg, wenn wir den Komparativ bilden.
Auch dazu ein paar Beispiele:
Dass das „e“ wegfällt, gilt aber nur für den Komparativ. Beim Superlativ bleibt das „e“ erhalten, zum Beispiel dunkel – dunkler – am dunkelsten.
Außerdem gilt die Ausnahme nicht für Adjektive, die auf „-ell“ enden. Sie werden ganz normal durch die reguläre Endung gesteigert, zum Beispiel schnell – schneller oder hell – heller.
Unregelmäßige Steigerungen
In der deutschen Sprache gibt es ein paar Adjektive und Adverbien, die sich an gar keine Regeln halten. Stattdessen haben sie ihre ganz eigene Steigerungsform.
Zu diesen Wörtern gehören:
Bei diesen Wörtern bleibt uns nichts anderes übrig, als den Komparativ auswendig zu lernen.
Nicht steigerbare Adjektive
Zu guter Letzt gibt es noch Adjektive, die überhaupt keine Steigerungsform haben. Weil sie etwas beschreiben, was so ist oder eben nicht, haben sie bereits in der Grundform eine absolute Bedeutung. Deshalb werden solche Adjektive auch Absolutadjektive genannt.
So kann eine Person zum Beispiel nur tot oder nicht tot sein. Dazwischen gibt es nichts, denn niemand kann nur ein bisschen tot sein oder toter als jemand anders. Genauso ist eine Frau entweder schwanger oder eben nicht.
Weitere Beispiele für Absolutadjektive sind:
Etwas schwieriger wird es bei Adjektiven wie zum Beispiel „voll“. Eigentlich ist zum Beispiel ein Glas entweder voll oder nicht. Es kann nicht voller sein, weil es ja schon vollständig gefüllt ist, wenn es voll ist.
Trotzdem können wir zum Beispiel sagen: „Mein Glas ist noch voller als deines.“ Bei solchen Adjektiven kommt es deshalb immer auf den Zusammenhang an.
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11524 | https://www.teacherspayteachers.com/Product/Least-Common-Multiple-Real-Life-Practice-DIGITAL-Grades-6-7-8921121 | Least Common Multiple - Real Life Practice - DIGITAL - Grades 6-7
Description
Do your kids struggle with finding the least common multiple? Does it feel like a meaningless skill to master with no real purpose? Show your kids how to find the LCM in a real life context with this engaging digital activity!
Using presentation mode, your 6th-7th grade students will practice finding the least common multiple to help determine when things align in the schedule.
For example, if a math test occurs every 14 days and reading logs are due every 21 days, when is the first time they will the both be due on the same day?
This is a great, relevant problem solving task that can be used to introduce least common multiple before formal instruction occurs.
Included are 12 different true or false challenges, where students determine if the given LCM is correct.
After clicking through the true/false section, they go back and give the correct answer to the false solutions (8 problems)!
Math Skills Required:
Use these challenges in a variety of ways:
Please Note: This is a DIGITAL activity, which is designed to be completed on the computer in Google slides. If you print the slides, students can write in the equivalent expressions in the box instead.
Included in the Download:
Love this interactive and purposeful math practice? Check out these additional resources available for 6th Grade:
AREA OF COMPOSITE SHAPES SCAVENGER HUNT
AREA OF TRIANGLES WORD PROBLEMS
GREATEST COMMON FACTOR WORD PROBLEMS
Please Note: This is a pdf download which includes a link to the google slides resource. No physical product will be shipped to you. In addition, this is designed for personal use in the home or classroom only. You may not share, distribute or alter this resource in any way. Thank you for your honesty!
Least Common Multiple - Real Life Practice - DIGITAL - Grades 6-7
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11525 | https://www.quora.com/What-is-the-main-difference-between-opportunity-cost-and-marginal-cost | Something went wrong. Wait a moment and try again.
What is the main difference between opportunity cost and marginal cost?
Gregory Norton
Engineer with interests in economics and the Philosophy of Liberty
·
Author has 22.4K answers and 10M answer views
·
6y
I just read the answer at What Is Marginal Opportunity Cost? and I still think the term is a business or economic equivalent of “old adage”. All adages are old, that is why they are adages. All opportunity costs are significant on the margin, that is why they are important.
Marginal cost is distinguished from average cost, itself just the sum of a large set of marginal costs. All costs are incurred one-at-a-time, though average cost is useful for some decisions. I have never heard of an average opportunity cost.
The term may be an effort to recognize that opportunity costs vary by circumstances.
I just read the answer at What Is Marginal Opportunity Cost? and I still think the term is a business or economic equivalent of “old adage”. All adages are old, that is why they are adages. All opportunity costs are significant on the margin, that is why they are important.
Marginal cost is distinguished from average cost, itself just the sum of a large set of marginal costs. All costs are incurred one-at-a-time, though average cost is useful for some decisions. I have never heard of an average opportunity cost.
The term may be an effort to recognize that opportunity costs vary by circumstances. I may be missing something significant, but “marginal opportunity cost” strikes me as something Dogberry would say.
·
Opportunity cost and marginal cost are both important concepts in economics, but they refer to different ideas:
Opportunity Cost
Definition: Opportunity cost is the value of the next best alternative that is forgone when a decision is made. It represents the benefits you miss out on by choosing one option over another.
Example: If you spend time studying instead of working a part-time job, the opportunity cost is the income you could have earned during that time.
Marginal Cost
Definition: Marginal cost is the additional cost incurred by producing one more unit of a good or service. It focuses on th
Opportunity cost and marginal cost are both important concepts in economics, but they refer to different ideas:
Opportunity Cost
Definition: Opportunity cost is the value of the next best alternative that is forgone when a decision is made. It represents the benefits you miss out on by choosing one option over another.
Example: If you spend time studying instead of working a part-time job, the opportunity cost is the income you could have earned during that time.
Marginal Cost
Definition: Marginal cost is the additional cost incurred by producing one more unit of a good or service. It focuses on the change in total cost that arises when the quantity produced is incremented by one unit.
Example: If a factory's total cost to produce 100 units is $1,000 and the total cost to produce 101 units is $1,020, the marginal cost of the 101st unit is $20.
Key Differences
Focus: Opportunity cost looks at the value of what is sacrificed when making a choice, while marginal cost analyzes the cost associated with increasing production by one unit.
Application: Opportunity cost is broader and applies to any decision-making scenario, whereas marginal cost is specifically related to production and costs in economics.
Understanding both concepts is crucial for making informed economic decisions, whether in personal finance, business, or policy-making.
Related questions
Are opportunity cost and marginal cost the same?
What is the difference between marginal opportunity cost and opportunity cost?
Difference between marginal and total opportunity cost?
In economics, how can the opportunity cost and the marginal cost be simply separated?
What's the difference between opportunity cost and marginal rate of substitution?
Fulton Wilcox
Former At&T Bell Labs, Information Systems Director at Bell Labs
·
Author has 431 answers and 268.7K answer views
·
6y
In a business practice context, the key difference is that, as the name conveys “marginal cost” assessment is adequate for pureplay cost versus cost alternatives assessment, typically the impact of a little more or a little less of some cost generating process. Contrary to its name, opportunity “cost” is in fact opportunity net benefit for one option versus opportunity net benefit of another, with cost being mere one element. Opportunity cost analysis typically requires two or more business case assessments, with the business case net value being the bases for selecting one option over others.
In a business practice context, the key difference is that, as the name conveys “marginal cost” assessment is adequate for pureplay cost versus cost alternatives assessment, typically the impact of a little more or a little less of some cost generating process. Contrary to its name, opportunity “cost” is in fact opportunity net benefit for one option versus opportunity net benefit of another, with cost being mere one element. Opportunity cost analysis typically requires two or more business case assessments, with the business case net value being the bases for selecting one option over others.
Marginal Cost: In many pure cost situations like inventory production buying (every car made needs four tires plus a spare), “economic order quantity” (and its supply side mirror image, economic sales order quantity), often is a pure play marginal cost analysis situation. Bigger deliveries generate more holding costs but less per transaction handling costs, so a calculus minimax solution can identify the least cost lot size. Similarly, aircraft flying at slower speeds typically use less fuel, but longer trip times may impact crew and other costs, so marginal cost versus cost assessment can optimize operating cost per flight hour. The explicit or too often tacit presumption of such marginal costs analysis is that cost versus cost optimization of supply chain lot size or minor differences in flight time has no impact on positive outcomes such as revenue, customer satisfaction, brand value and the like.
Opportunity cost: On the other hand, once revenue, customer order fill rate, brand value, business reputation or other upside dimensions come into play, a business case per feasible option is needed. The most visible difference is that rows in the spread sheet or entries into the model need to reflect changes in those benefits. Besides adding some complexity, this often means “production” and the chief financial officer need to talk to marketing. It also means that neatly “marginal” equations (what happens to flight hour cost if I add one more knot to aircraft speed) need to expand to incorporate differences in kind and impact. if customer or other stakeholders are impacted.
Consequently, “opportunity cost” needs to be twinned with “business case,” one goes with the other. Although there are many different levels of complexity in a business case. the minimal for each option is: 1) a time series for benefit, 2) a corresponding time series for cost, and 3) a resulting time series for net gain or loss.
Note that business cases may generate all “losers” (even after including benefits the costs are bigger), especially for infrastructure projects, but the business case view that includes benefits may alter the decision as to which project should be selected, if doing something is a must.
Matt Cochrane
Former Professor at Fanshawe College (2018–2019)
·
Author has 78 answers and 52.3K answer views
·
6y
An opportunity cost is your highest foregone opportunity. It is what you give up purchasing by purchasing something else. Suppose I can go to the movies or the bar. If those are my two choices, then by choosing movies, my highest foregone alternative is going to the bar.
Marginal Cost is the change in total cost that results from producing one more unit of a good/service. Suppose that in producing 3 haircuts, my cost is $30. If I produce a 4th haircut, my total cost increases to $38. The marginal cost of the 4th haircut is $8.
Profit maximizing occurs (in perfectly competitive, monopolistic comp
An opportunity cost is your highest foregone opportunity. It is what you give up purchasing by purchasing something else. Suppose I can go to the movies or the bar. If those are my two choices, then by choosing movies, my highest foregone alternative is going to the bar.
Marginal Cost is the change in total cost that results from producing one more unit of a good/service. Suppose that in producing 3 haircuts, my cost is $30. If I produce a 4th haircut, my total cost increases to $38. The marginal cost of the 4th haircut is $8.
Profit maximizing occurs (in perfectly competitive, monopolistic competitive and monopoly markets) at a quantity where marginal revenue equals marginal cost, so arguably, marginal cost has more application in decision making and price setting.
Vilnis Krumins
Heterodox Economics
·
Author has 5.1K answers and 3M answer views
·
6y
This is difficult to answer because opportunity cost and marginal cost are unrelated. This makes finding a difference between them difficult.
I suppose the trite answer is to say that the biggest difference between them is that they belong to two completely different categories of analysis but, no doubt, this would be unsatisfying.
Opportunity cost is, actually, not a cost. It is jargon used by economists to denote a situation in which a choice is to be made. Each choice has a return (e.g., income) that establishes its opportunity cost. Choosing an alternative to any particular choice entails fo
This is difficult to answer because opportunity cost and marginal cost are unrelated. This makes finding a difference between them difficult.
I suppose the trite answer is to say that the biggest difference between them is that they belong to two completely different categories of analysis but, no doubt, this would be unsatisfying.
Opportunity cost is, actually, not a cost. It is jargon used by economists to denote a situation in which a choice is to be made. Each choice has a return (e.g., income) that establishes its opportunity cost. Choosing an alternative to any particular choice entails forgoing the return (i.e., the opportunity cost). For example, if I expect the return from an investment to be $1 then that is its opportunity cost and any alternative I pick (if I’m rational) should have a return greater than $1 (because, in the language of economics, that is its opportunity cost).
Marginal cost is, again, not really a cost. Economists use the word “marginal” to denote a change in some variable (i.e., the marginal change) due to a change in some other variable. For example, if I walk 1 kph faster then there will be a marginal change in my destination of 2 km if I walk for 2 hours. Usually, marginal costs crop up in the context of production analysis; a change in the production process, generally, will result in some change in output (which is the marginal change). With respect to costs, a change in technology might produce a change in the cost of manufacturing cost (which is the marginal cost of the technological change).
But, really, they have nothing to do with each other.
Promoted by SavingsPro.org
Mark Bradley
Economist
·
10mo
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Related questions
What are the differences between marginal costs marginal benefits?
Microeconomics:
What is the difference between opportunity costs and implicit costs?
Are marginal opportunity costs and opportunity costs the same concept?
What is ‘marginal opportunity cost’?
What is the difference between a trade off and an opportunity cost?
Sundaram
Group CFO at Republic TV (Indian news channel) (2016–present)
·
Author has 141 answers and 165.7K answer views
·
6y
Let me try and highlight the difference by way of an example:-
Suppose that you are a manufactuer who uses some raw material say plastic resin and you have the ability to make either Pens or Plastic Folders using the said Plastic resins.
The cost of making the next unit of either the Pen or the Plastic folder shall be the Marginal Cost of making that next unit.
So the concept of Marginal Cost is simple and straight forward.
Bur Opportunity Cost is a more circuitous.
Let us suppose that both items (Pens and Plastic Folders) sell at the same unit Price. That renders your decision to make for either P
Let me try and highlight the difference by way of an example:-
Suppose that you are a manufactuer who uses some raw material say plastic resin and you have the ability to make either Pens or Plastic Folders using the said Plastic resins.
The cost of making the next unit of either the Pen or the Plastic folder shall be the Marginal Cost of making that next unit.
So the concept of Marginal Cost is simple and straight forward.
Bur Opportunity Cost is a more circuitous.
Let us suppose that both items (Pens and Plastic Folders) sell at the same unit Price. That renders your decision to make for either Pen or Folders indifferent, as you earn the same return for your each unit of Plastic resin used. However let us complicate this, if instead the Pen fetches you Rs 10/- while the folder can sell for only Rs8/-; it is beneficial to make the Pen. So to maximise returns, you will first exhaust the demand for Pens, then only focus on making Folders, if at all.
Let us make it a choice between a rock and a hard place; now suppose you have a valued Customer who wants folders and you cannot say no, also there is an order for Pens from an irregular dealer at the same time i.e you can serve one and not both. Now, by making Folders instead of Pens you are going to miss the opportunity of making the higher gain for each unit of your Plastic resin, that loss (Rs 10 less Rs 8 times Folders) then is your Opportunity Cost being foregone by honouring friendship.
Nicolas Cole
Writer / Entrepreneur
·
Upvoted by
Erik Hille
, Ph.D Economics & Behavioral Economics, Southern Methodist University (2020) · Author has 917 answers and 39.4M answer views
·
7y
Related
What is opportunity cost?
Opportunity cost is the delta between what you’re currently doing and what you could be doing instead.
Truthfully, most people never understand this idea of opportunity cost.
I’ll give you a perfect example:
There are two kinds of professions in this world. The kind where you get paid by the hour (hours for dollars), and the kind where you create something once but then get paid on residual.
I grew up in a very wealthy neighborhood—top 1% of the United States, and easily top 0.5% of the world.
Most of these successful families are the result of “hours for dollars” jobs: doctors, lawyers, bankers, e
Opportunity cost is the delta between what you’re currently doing and what you could be doing instead.
Truthfully, most people never understand this idea of opportunity cost.
I’ll give you a perfect example:
There are two kinds of professions in this world. The kind where you get paid by the hour (hours for dollars), and the kind where you create something once but then get paid on residual.
I grew up in a very wealthy neighborhood—top 1% of the United States, and easily top 0.5% of the world.
Most of these successful families are the result of “hours for dollars” jobs: doctors, lawyers, bankers, etc. They make great money while they’re clocked in, but their income halts as soon as they stop showing up to the office.
Having watched my own father, a surgeon, work tirelessly for my entire life, I have have taken it upon myself to reach out and learn from those who achieved the same or much greater levels of monetary success, but actually work less. After all, what good is a mansion and a driveway full of sports cars if you’re too busy to enjoy it?
What I’ve learned is the primary difference between those that sit in the top 1% of income earners and those that live in the top .01% think about their resources differently.
Very differently.
Someone who earns a great living tends to think more about protecting the income they’ve earned, rather than leveraging it for more later.
If someone were to approach them and say, “Hey, I can do some of the work you’re doing for X,” they don’t see that as an opportunity—they see that as a loss of income.
This person would be replacing part of their income, which means less for them.
For someone in the top .01%, this way of thinking is counterproductive.
These people see TIME as their greatest asset—not money. So if someone comes along and says, “Hey, I can do some of the work you’re doing for X,” if X is less (much less) than what they value their time at ($), they see that as a beneficial opportunity. This other person can take care of the work, which gives them more time to do bigger and better things.
For example: Entrepreneur values his time at $500 / hour.
Kid comes along and says, “I can handle all your scheduling and paperwork for $20 an hour.”
$20 < $500 = Opportunity.
Opportunity cost, then, is when an individual decides to continue doing X even though their time could be better spent doing something else instead.
A lawyer might continue doing his or her own paperwork because they earn $300 / hour for doing so.
The opportunity cost, however, is that if they pass along that responsibility to someone else, that’s X hours per week they could potentially earn far more than $300 / hour because they have more available TIME.
It’s actually not terribly difficult to become “wealthy.” Wealth can be earned with a combination of hard work and disciplined money management habits.
However, becoming extremely wealthy without understanding this idea of opportunity cost is nearly impossible. There are only so many hours in a day, and almost every industry has a ceiling as to what someone is willing to pay for one of those hours (your expertise).
And lastly, opportunity cost also means that when you’re doing X, you’re not doing Y, Z, etc.
This idea doesn’t only apply to income, but also to things like fulfillment, relationships, hobbies, and other interests.
If you’re working, that means you’re not vacationing.
If you’re vacationing, that means you’re not spending time at work.
Etc.
It all comes down to how aware you are of the decisions you’re making, and why you’re making them.
A father might think, “I’m working a ton, and I’m providing for my family.”
Ok, great. But the opportunity cost by working too much is that you never see your kids.
A mother might think, “I’m going to take as many hours as I can get at this job because they pay good money.”
Ok, great. But the opportunity cost is that if you take on too many hours, there will be no time left for your passion for painting—which brings you fulfillment, happiness, and you believe you could turn into a lucrative career down the road.
It’s not that there is a right way and a wrong way of playing this game of life. What this idea of opportunity cost reveals is that most people aren’t aware of why they make the decisions they do. They see all the reasons why they should do something, but not the reasons why they shouldn’t.
Awareness, then, is the ability to see both sides of a situation and make a conscious decision that will end up leading you to the next opportunity, and the next opportunity.
Those that build the lifestyle they truly desire for themselves have a firm understanding of opportunity cost.
Promoted by BILL
Bob Sturges
Growth Leader at BILL
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Updated 2y
What is the best accounts payable automation solution?
Every business knows the pain points: an important invoice that goes missing, a fake invoice that was mistakenly paid. We’ve all wasted time entering the same data into multiple systems, chasing down signatures from approvers, and writing, stamping, and mailing checks.
AP automation can help solve all of these problems, but it’s difficult to know where to start if you’re just beginning your research. Here are the factors you need to keep in mind when researching the best AP automation software for your business:
Invoice Capture and Coding
Capturing invoices is a huge time sink for your AP staff,
Every business knows the pain points: an important invoice that goes missing, a fake invoice that was mistakenly paid. We’ve all wasted time entering the same data into multiple systems, chasing down signatures from approvers, and writing, stamping, and mailing checks.
AP automation can help solve all of these problems, but it’s difficult to know where to start if you’re just beginning your research. Here are the factors you need to keep in mind when researching the best AP automation software for your business:
Invoice Capture and Coding
Capturing invoices is a huge time sink for your AP staff, so you’ll want to find a solution that pulls in information automatically. For example, BILL allows you to receive invoices via email, drag and drop from your desktop, or snap a photo with the BILL mobile app, so that both physical and digital invoices can be easily captured.
Once the invoices are received, you’ll want a software that uses machine learning or applies artificial intelligence to improve its performance. The leading AP automation software providers use AI-enabled learning to reduce data entry, identify duplicate invoices, and use the right approval workflows for each vendor.
Internal Controls
You should also look for an AP automation software that helps you implement internal controls. Approval workflows that enforce separation of duties are an essential part of AP security, requiring at least one employee to approve a payment and another to sign off on it. You’ll also want a system that logs approvals and comments so your team can easily discuss questionable invoices before payment is made. And, to keep all your data safe, be sure that the AP automation software you use offers single sign-on or two-factor authentication.
Integration with your ERP/Accounting Software
Your AP automation software should sync or integrate with your ERP or accounting software so you can keep your books up-to-date in both systems. BILL provides automatic two-way sync with Oracle NetSuite, Sage Intacct, Microsoft Dynamics 365 Business Central, Microsoft Great Plains, Xero, and QuickBooks. BILL can also generate CSV templates to easily import and export data with any other accounting system, so you’re not constantly double-entering your data.
Multiple Payment Options
Having multiple payment options can help you control your cash flow, take advantage of early payment discounts, and improve your relationships with vendors. Ensure that you’re choosing an AP automation software that provides a variety of payment options, including ACH, credit card, virtual card, check, and international wire.
Intuitive and Simple
Using a new software can be a large lift for your employees, so be sure to choose one that is intuitive and easy-to-use. Check out third-party reviews from places like G2, where users rate AP software on usability. You should also review customer case studies and testimonials to get a feel of how current users are putting AP automation to work. Finally, ensure that your AP automation software candidates offer a mobile app so managers can approve while traveling or out-of-office to prevent delays in your approval process.
Save Time
Ultimately, the most important thing your AP automation software should do for your team is save time. Typing in invoice info, re-entering data, chasing down approvals, and reconciling accounts all prevent your finance team from completing meaningful work. Surveyed BILL customers report saving, on average, 50% of time on accounts payable (May 2021), allowing them to focus more on higher-value activities that support the growth of their business.
Want to learn more about how AP automation can support your business? Try BILL risk-free for 30 days.
Romain Leclercq
Author has 440 answers and 909.9K answer views
·
11y
Related
What is marginal cost?
It is when your overall cost goes up as one unit more is demanded. The marginal cost itself is NOT this overall cost though, it is simply the additional cost added.
Let's say a loaf of bread is five bucks (I hope it's good quality bread). Let's say you bought four loaves earlier today. You spent twenty bucks earlier. Let's say you suddenly want to buy a fifth loaf (one more). Your total cost would go from twenty to twenty-five bucks in terms of bread finance. But the marginal cost is five bucks, because that's how much it costs for just one more.
In microeconomics, there will be many cases wher
It is when your overall cost goes up as one unit more is demanded. The marginal cost itself is NOT this overall cost though, it is simply the additional cost added.
Let's say a loaf of bread is five bucks (I hope it's good quality bread). Let's say you bought four loaves earlier today. You spent twenty bucks earlier. Let's say you suddenly want to buy a fifth loaf (one more). Your total cost would go from twenty to twenty-five bucks in terms of bread finance. But the marginal cost is five bucks, because that's how much it costs for just one more.
In microeconomics, there will be many cases where the marginal cost diminishes over time or vice versa. A company has both fixed and variable costs. Fixed costs are incurred at the onset and are a one-time expense. An example would be like buying a limousine for a shuttle service. You have to buy the car upfront. A variable cost is a cost that is linked to each time you carry out a service or action. This would be like the gas in the limo. You need more of it each time you use it. So the gas cost is variable- it varies depending on your usage.
In the case of a limo service, you have a fixed cost of the car and a variable cost of the gas. We can keep the example simplistic and have those be your only two costs. The first time you execute your service, the cost of your business is, say, 50k for a car and five bucks for gas for the night. The cost of your first shuttle service trial is 50,005. The second night of business, you already have the car paid for, so you just need to buy the gas for the night again. That is another five bucks. This is another topic entirely, but your average total cost (a lot of times referred to as ATC) would go from 50,005 to 50,010 (50,000 for the car, plus 5 for gas night 1, and 5 for gas night 2) divided by two nights. Each night your marginal cost is only five bucks, which is much less than the fixed cost. Your average total cost will slope asymptotically towards the x-axis because the gas costs pale in comparison to the car at the beginning.
Back to the question though, the marginal cost is simply the cost of one more unit. Like someone above discussed in semi-confusing math jargon. It is just the cost of, say, producing, one more unit. Hopefully this figure is less than the profit you would make off that unit.
I hope I answered that okay.
Sandeep Vashistha
work as PGTEconomics teacher in cbse
·
Author has 55 answers and 130.3K answer views
·
6y
Opportunity cost is the next best alternative use of available resources where is marginal cost is the change in total cost due to production of one unit of output
Related questions
Are opportunity cost and marginal cost the same?
What is the difference between marginal opportunity cost and opportunity cost?
Difference between marginal and total opportunity cost?
In economics, how can the opportunity cost and the marginal cost be simply separated?
What's the difference between opportunity cost and marginal rate of substitution?
What are the differences between marginal costs marginal benefits?
Microeconomics:
What is the difference between opportunity costs and implicit costs?
Are marginal opportunity costs and opportunity costs the same concept?
What is ‘marginal opportunity cost’?
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Dichloroacetic acid Suppliers list Company Name:Chemport Science Technology Shanghai) Co., Ltd. Tel:+undefined13816004845 Email:chen@chemport-sh.com Products Intro:Product Name:Dichloroacetic acid CAS:79-43-6 Company Name:Hebei Chuanghai Biotechnology Co., Ltd Tel:+8617732866630 Email:abby@chuanghaibio.com Products Intro:Product Name:Dichloroacetic acid CAS:79-43-6 Purity:99% Package:1kg;1.00;USD Company Name:Hebei Mujin Biotechnology Co.,Ltd Tel:+8613288715578 Email:sales@hbmojin.com Products Intro:Product Name: Dichloroacetic acid CAS:79-43-6 Purity:99% Package:25KG Company Name:Henan Tianfu Chemical Co.,Ltd. Tel:+86-0371-55170693 +86-19937530512 Email:info@tianfuchem.com Products Intro:Product Name:dichloroacetic acid CAS:79-43-6 Purity:99% Package:25KG;5KG;1KG Company Name:career henan chemical co Tel:+86-0371-86658258 +8613203830695 Email:sales@coreychem.com Products Intro:Product Name:Dichloroacetic acid CAS:79-43-6 Purity:99% Package:1kg;2USD Dichloroacetic acid manufacturers ------------------------------------------------------------------------------------------------------- Dichloroacetic acid $0.00 / 25KG 2025-08-08 CAS:79-43-6 Min. Order: 1KG Purity: 99% Supply Ability: 50000KG/month Dichloroacetic acid $1.00 / 1kg 2025-06-24 CAS:79-43-6 Min. Order: 1kg Purity: 99% Supply Ability: 10 mt Dichloroacetic acid $550.00 / 1Kg 2025-04-09 CAS:79-43-6 Min. Order: 0.100Kg Purity: 99 % Supply Ability: 5000 Kg Dichloroacetic acid Basic information Product Name:Dichloroacetic acid Synonyms:DICHLOROACETIC ACID PESTANAL;DICHLOROACETIC ACID REAGENTPLUS(TM) &;DICHLOROACETIC ACID, 1000MG, NEAT;Dichloroaceticacid,99%;DICHLOROACETIC ACID REDISTILLED;Dichloressigsure;DCA Deblock (0.36M dichloroacetic acid in toluene) ;DKhUK CAS:79-43-6 MF:C2H2Cl2O2 MW:128.94 EINECS:201-207-0 Product Categories:Pharmaceutical Intermediates;Alpha Sort;D;DAlphabetic;DIA - DIC;Volatiles/ Semivolatiles;Chemical Synthesis;Organic Acids;Synthetic Reagents;500 Series Drinking Water Methods;EPA;Method 552;Pesticides&Metabolites;Essential Chemicals;Reagent Plus;Routine Reagents;intermediates Mol File:79-43-6.mol Dichloroacetic acid Chemical Properties Melting point 9-11°C(lit.) Boiling point 194°C(lit.) density 1.56 g/mL at 25 °C (lit.) vapor density 4.5 (vs air) vapor pressure 0.19 mm Hg ( 20 °C) refractive index n 20/D 1.466(lit.) Fp>230°F storage temp.Store below +30°C. solubility Chloroform (Slightly), Methanol (Slightly) pka 1.48(at 25℃) form Liquid color APHA: <50 PH 1.2 (129g/l, H2O, 20℃) Water Solubility soluble Sensitive Hygroscopic Merck 14,3050 BRN 1098596 Dielectric constant 10.7（-7℃） Exposure limits ACGIH: TWA 0.5 ppm (Skin) Stability:Stable. Incompatible with strong oxidizing agents, strong bases, strong reducing agents. Reacts with water. Protect from moisture. Hygroscopic. InChIKey JXTHNDFMNIQAHM-UHFFFAOYSA-N LogP 0.92 CAS DataBase Reference79-43-6(CAS DataBase Reference) IARC 2B (Vol. 63, 84, 106) 2014 NIST Chemistry ReferenceAcetic acid, dichloro-(79-43-6) EPA Substance Registry SystemDichloroacetic acid (79-43-6) Safety Information Hazard CodesXn,N,C,Xi,F Risk Statements36/38-40-50-35-36/37/38-67-38-65-48/20-11-63-50/53 Safety Statements26-45-61-36/37-36-16-62-36/37/39 RIDADR UN 1764 8/PG 2 WGK Germany 2 RTECS AG6125000 F3-10 TSCA Yes HS Code 2915 40 00 HazardClass 8 PackingGroup II Hazardous Substances Data79-43-6(Hazardous Substances Data) Toxicity LD50 orally in rats: 2.82 g/kg (Smyth) MSDS Information | Provider | Language | --- | | ACROS | English | | SigmaAldrich | English | | ALFA | English | Dichloroacetic acid Usage And Synthesis Description DCA is a corrosive, combustible, colorless liquid with a pungent odor. Molecular weight = 128.94;Specific gravity (H2O:1) = 1.56; Boiling point = 193.8℃;Freezing/Melting point = 9.7-13.5℃; Vaporpressure = 1 mmHg at 44℃; Flash point = 110℃. HazardIdentification (based on NFPA-704 M Rating System):Health 3, Flammability 1, Reactivity 1. Highly corrosive.Soluble in water. Chemical Properties Dichloroacetic acid, CHCI2COOH, also known as dichlorethanoic acid, is a corrosive, combustible, colorless, strong liquid acid with a pungent odor. It is soluble in water and alcohol. Dichloroacetic acid (Ka=5.14 × 10-2) is a stronger acid than chloroacetic acid. Most chemical reactions are similar to those of chloroacetic acid, although both chlorine atoms are susceptible to reaction. Dichloroacetic acid is much more stable to hydrolysis than chloroacetic acid. It is a reactive intermediate in organic synthesis. Dichloroacetic acid is prepared by the chlorination of acetic acid. Uses Dichloroacetic acid is used as an intermediate to make other chemicals such as its salts and esters. It is involved as a test reagent for fiber analysis and a disinfectant. It is used to inhibit mitochondrial PDK (pyruvate dehydrogenase kinase) and to shift cellular metabolism from glycolysis to glucose oxidation. Further, it is used in the treatment for post-ischemic recovery. Definition ChEBI: Dichloroacetic acid is an organochlorine compound comprising acetic acid carrying two chloro substituents at the 2-position. It occurs in nature in seaweed, Asparagopsis taxiformis. It has a role as an astringent and a marine metabolite. It is a monocarboxylic acid and an organochlorine compound. It derives from an acetic acid. It is a conjugate acid of a dichloroacetate. Preparation Dichloroacetic acid was reported to be first synthesized in 1864 by the further chlorination of monochloroacetic acid with chlorine (Beilstein Online, 2002). The most common production method for dichloroacetic acid is the hydrolysis of dichloroacetyl chloride, which is produced by the oxidation of trichloroethylene. Application Dichloroacetic acid (DCA) is a chlorinated acetic acid that has been reported to be a mouse liver carcinogen. The varying degrees of swelling of polyester in different concentrations of DCA solution has been analyzed by phase-contrast microscopy. The efficiency of titanium dioxide (TiO2) nanoparticles to catalyze the photodegradation of DCA has been investigated. Dichloroacetic acid (DCA) can be used as: A reactant in the synthesis of chloroketones by reacting with esters in the presence of LiHMDS via Claisen-type homologation reaction. A structure-directing agent, solvent, or plastdopant for the preparation of different morphologies of polyaniline (PANI). Self-assembling nanostructured PANI may be formed due to the presence of strong hydrogen bonding between DCA and aniline/polyaniline. General Description Dichloroacetic acid appears as a colorless crystalline solid melting at 49°F. Corrosive to metals and tissue. Air & Water Reactions Soluble in water. Reactivity Profile Dichloroacetic acid is probably hygroscopic. Dichloroacetic acid reacts with water or steam. Dichloroacetic acid is incompatible with strong oxidizing agents, strong bases and strong reducing agents. Health Hazard TOXIC; inhalation, ingestion or skin contact with material may cause severe injury or death. Contact with molten substance may cause severe burns to skin and eyes. Avoid any skin contact. Effects of contact or inhalation may be delayed. Fire may produce irritating, corrosive and/or toxic gases. Runoff from fire control or dilution water may be corrosive and/or toxic and cause pollution. Fire Hazard Combustible material: may burn but does not ignite readily. When heated, vapors may form explosive mixtures with air: indoors, outdoors and sewers explosion hazards. Contact with metals may evolve flammable hydrogen gas. Containers may explode when heated. Runoff may pollute waterways. Substance may be transported in a molten form. Flammability and Explosibility Non flammable reaction suitability reaction type: Acetylations reagent type: Lewis acid Safety Profile Moderately toxic by skin contact and ingestion. It is corrosive to the skin, eyes, and mucous membranes. Questionable carcinogen with experimental tumorigenic data. Will react with water or steam to produce toxic and corrosive fumes. When heated to decomposition it emits toxic fumes of Cl-. See also CHLORIDES. Potential Exposure This haloacetic acid can be a byproduct of drinking water disinfection and may increase the risk of cancer. This chemical is used as starting material for the production of glyoxylic; dialkyloxy acids; and sulfonamides. It is also used as a fungicide, a medication; and a chemical intermediate in pharmaceuticals First aid If this chemical gets into the eyes, remove anycontact lenses at once and irrigate immediately for at least15 min, occasionally lifting upper and lower lids. Seek medical attention immediately. If this chemical contacts theskin, remove contaminated clothing and wash immediatelywith soap and water. Seek medical attention immediately. Ifthis chemical has been inhaled, remove from exposure,begin rescue breathing (using universal precautions, including resuscitation mask) if breathing has stopped and CPR ifheart action has stopped. Transfer promptly to a medicalfacility. When this chemical has been swallowed, get medical attention. If victim is conscious, administer water ormilk. Do not induce vomiting. Medical observation isrecommended for 2448 h after breathing overexposure, aspulmonary edema may be delayed. As first aid for pulmonary edema, a doctor or authorized paramedic may consideradministering a corticosteroid spray. storage Color Code—White: Corrosive or Contact Hazard;Store separately in a corrosion-resistant location. Prior toworking with DCA you should be trained on its proper handling and storage. Store in tightly closed containers in acool, well-ventilated area away from incompatible materialslisted above. Metal containers involving the transfer of thischemical should be grounded and bonded. Where possible,automatically pump liquid from drums or other storage containers to process containers. Drums must be equipped withself-closing valves, pressure vacuum bungs, and flamearresters. Use only nonsparking tools and equipment, especially when opening and closing containers of this chemical. Sources of ignition, such as smoking and open flames,are prohibited where this chemical is used, handled, orstored in a manner that could create a potential fire orexplosion hazard. Wherever this chemical is used, handled,manufactured, or stored, use explosion-proof electricalequipment and fittings. Shipping UN1764 Dichloricacetic acid, Hazard class: 8; Labels: 8-Corrosive material. Purification Methods Crystallise this strong acid from benzene or pet ether. Dry it with MgSO4 and fractionally distil it. [Bernasconi et al. J Am Chem Soc 107 3612 1985, Beilstein 2 IV 498.] Incompatibilities DCA is a medium strong acid; incompatible with nonoxidizing mineral acids; organic acids; bases, acrylates, aldehydes, alcohols, alkylene oxides; ammonia, aliphatic amines; alkanolamines, aromatic amines; amides, furfuryl alcohol (explosion hazard), glycols, isocyanates, ketones. Attacks metals generating flammable hydrogen gas. Attacks some plastics, rubber, and coatings Toxics Screening Level The initial risk screening level (IRSL) for dichloroacetic acid (DCA) is 0.07 μg/m3 (annual averaging time) based on the Michigan Department of Environmental Quality (MDEQ), Air Quality Division (AQD) Rule 336.1231. The initial threshold screening level (ITSL) is 0.5 μg/m3 (annual averaging time) based on Rule 336.1232 (1) (e) and (2) (c). Dichloroacetic acid Preparation Products And Raw materials Raw materialsChloral Preparation ProductsChloroacetic acid-->Glyoxylic acid-->Basic Yellow 24-->Dichloroacetyl chloride-->HEPTENOPHOS-->Dichloroacetic acid methyl ester-->Allantoin-->Hemin-->2-(DICHLOROMETHYL)BENZIMIDAZOLETHIOL-->Sodium dichloroacetate-->Acetic acid, 2,2-dichloro-, (tetrahydro-2-furanyl)methyl ester
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DichloroethaneDichlorodimethylsilaneGlycineEthyl 2-(Chlorosulfonyl)acetate2,4-Dichlorophenoxyacetic acidFolic acidThionyl chlorideDiisopropylammonium dichloroacetateSodium dichloroisocyanurateClopyralidQuinclorac1H-Pyrazole-4-carboxamide, 1-(6-amino-9H-purin-2-yl)-N-methyl-N,N-DimethylacetamideCitric acidAscoric AcidDICHLOROMETHOXYACETIC ACID METHYL ESTERDICHLOROACETIC ACID ETHYL ESTER,DICHLOROACETIC ACID ETHYL ESTER2,2-DICHLOROPROPIONIC ACID SODIUM SALT
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Annals of Operations Research
Article
Covering a square with consecutive squares
Original Research
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Published: 20 May 2025
Volume 350,pages 911–926, (2025)
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Annals of Operations ResearchAims and scopeSubmit manuscript
Covering a square with consecutive squares
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Janos Balogh1,
Gyorgy DosaORCID: orcid.org/0000-0002-4909-66942,
Lars Magnus Hvattum3,
Tomas Attila Olaj2,
Istvan Szalkai2&
…
Zsolt Tuza2,4
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Abstract
In this article we address the following problem. Given are a 1×1 square, a 2×2 square, and so on, finally a n×n square. What is the biggest square that can be covered completely by this given set of “small” squares? It is assumed that the small squares must stand parallel to the sides of the big square, and overlap is allowed. In contrast to the packing version of the problem (asking for the smallest square that can accommodate all small squares without overlap) which has been studied in several papers since the 1960’s, the covering version of the problem seems new. We construct optimal coverings for small values of n. For moderately bigger n values we solve the problem optimally by a commercial mathematical programming solver, and for even bigger n values we give a heuristic algorithm that can find near optimal solutions. We also provide an expansion-algorithm, that from a given good cover using consecutive squares up to size n, can generate a cover for a larger square using small squares up to size n+1. Finally we prove that a simple covering policy can generate an asymptotically optimal covering.
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1 Introduction
We address the following question.
Question
Can a square of size K×K be covered completely by using the consecutive squares of sizes 1×1, 2×2, ..., n×n ? In other words, what is the largest K, so that the consecutive squares 1×1, 2×2, ..., n×n — called “small squares”, denoted as Q[n] — completely cover K×K, allocating the small squares carefully?
We allow that the small squares are overlapping. Up to the best knowledge of the authors, this question seems new. On the other hand, the related question, what the smallest accommodation square is where the small squares can be packed without overlap, is not new. This co-problem is originated by Gardner (1966, 1975), where he popularized an unpublished question due to R.B.Britton. We assume that the small squares must have sides parallel to the axes, but they are allowed to touch each other on their sides. For this packing problem, Gardner (1975) listed the best possible (tight) results up to n=17, based on earlier works.
The packing of consecutive squares serves also as a benchmark problem for more general rectangle packing problems, see (Korf, 2003; Moffitt & Pollack, 2006; Simonis & O’Sullivan, 2008). Rectangle packing has many important applications in VLSI design, scheduling, cutting stock and pallet loading problems. In some of these applications, rotation of the rectangles is allowed.
In our recent paper (Balogh et al., 2022w
")) we review the current best upper bounds for the packing problem, and it can also be found on the webpage (The on-line encyclopedia of integer sequences, 2021), searching for the integer sequence A005842. There the best found _UB_(_n_) values are listed till n=56 (most of them are tight), where _UB_(_n_) is the upper bound for the packing problem with a given parameter _n_. One of the best lower bounds for large values of _n_ comes by rounding up the square root of the total area of the squares, i.e. L B 0=Image 6. Tightness has not been proved for the cases n=38,40,42,48,52,53 and 55, while the other values of _UB_(_n_) given in The on-line encyclopedia of integer sequences (2021) are proved to be tight for 1≤n≤56.
An interesting special case of the problem is n=24, which was unsolved until 2004. As proved already in Watson (1918), n=24 is the unique non-trivial case where A n:=n(n+1)(2 n+1)6 is the square of an integer. However, while L B 0=70, the optimum is 71 as proved by Korf (204). Staying at the very special n=24 case, since the small squares of Q cannot be packed without overlap into the 70×70 accommodating big square, and their total size is just 70×70, it follows also that if we turn to the covering problem, those small squares cannot cover completely the 70×70 big square. Hence the K value that we look for in case of the covering problem, must be smaller than 70 for n=24.
In Balogh et al. (2022w
")) we also review some related packing problems, where an arbitrary set of squares with a total area of at most 1 should be packed into a smallest possible square or into a rectangle of minimum area. In Buchwald and Scheithauer (2016. A 5/9 theorem on packing squares into a square. Preprint MATH-NM-04-2016, TU Dresden.")) the authors prove that any set of squares can be packed into the unit square if (i) the total are of the squares is at most 5/9 and (ii) none of the squares has side length larger than 1/2. They also proved that 5/9 is the best possible such value. It is worth mentioning here that in their packing, the patterns fulfill the guillotine pattern properties. Their packing is similar to the First Fit Decreasing Height (FFDH) heuristic of Coffman et al. (1980. Performance bounds for level-oriented two-dimensional packing algorithms. SIAM Journal on Computing, 9, 808–826.")) designed for the two-dimensional strip packing problem, but they used it both horizontally and vertically.
In Balogh et al. (2022w
")), we give the first asymptotic results for the packing problem. Namely, we show that there exists a constant c<1 such that the square of size N+c n admits a guillotine-type packing of the squares of sizes 1,2,…,n, where N:=A n.
Our results for covering. To the best of our knowledge, the covering problem was never considered before. Our results are summarized below.
For small values of n the covering problem is easy, and we give the optimal solutions. These can be calculated by hand or by a mathematical programming solver. There are several ways to formulate the problem using mathematical programming, and we present and discuss two alternative models.
For moderately bigger values of n, such as n=23, the problem is already hard for mathematical programming solvers, so we create a heuristic algorithm that can find near optimal solutions also for big values of n.
We provide an expansion-algorithm, that from a given good cover of K×K for some n, can generate a cover for some K′>K using small squares up to and including n+1.
We also prove that a simple covering policy can generate an asymptotically optimal covering.
The structure of the paper is as follows. In Section 2 we discuss some small cases in detail, up to n≤7. In Section 3 we first introduce a binary integer programming formulation that can be used to determine optimal covers, and then present an alternative mixed integer programming model; then, in section 3.2 we present the results of solving the models for up to n≤25. Section 4.1 gives a heuristic and Section 4.2 introduces an expansion-algorithm, to obtain a lower bound for the case with n+1 based on a known cover using n. In Section 5 we prove that a square of size N−O(n) can be covered with the squares of consecutive sizes 1,2,…,n, giving a simple, asymptotically optimal covering algorithm. Some remarks are given in the concluding section.
2 Small cases
Here we discuss the small cases, up to n≤7. Let Q[n] denote the squares with sizes 1,2,…,n, exactly one copy from each. Note that
(1)
is a trivial upper bound for the largest size K n of a square that can be completely covered by Q[n]. We will denote by Δ K the increment of K from n−1 to n, i.e. Δ K=K n−K n−1. If no confusion arises, we will omit the index n and write simply K instead of K n. Moreover, let L denote a lower bound for K for any fixed value of n.
The values for the cases n≤7 are shown in Table 1. Their validity can be checked by hand.
Table 1 The cases n≤7
Full size table
The cases n≤3 are trivial. For n=4 the square 5×5 cannot be covered by 1×1,…,4×4, since 4×4 leaves in 5×5 an empty L-shape of 5+4=9 unit squares (or cells), but 3×3, 2×2, 1×1 can cover at most 5+2+1=8 of those empty cells. The bigger cases are less immediate; they are discussed in the following two subsections.
2.1 The special case of n=5, excluding K=7
For n=5, it is easy to make a cover for L=6. To see that K=6, it remains to exclude K=7. One way to do this is as follows. Suppose there exists a complete cover for K=7. Since it is impossible to cover a 7-long stripe with just the three smallest squares 1×1, 2×2, and 3×3, the union of 5×5 and 4×4 necessarily is a polygon of width 7 and height 7. At this point two rectangles R′,R″ are uncovered, of sizes 2×3 and 3×2, in a position where the 3×3 square can cover an area of at most 6 from R′∪R″, and the two smallest squares have a total area of 5. Hence the cover cannot be completed.
Another way to argue is that at least one horizontal side H and one vertical side V of the K×K square are disjoint from the largest square 5×5. The corner C=V∩H is contained in just one of the other four squares, because a second square at C would be superfluous (namely the smaller one). If C is covered with the square of size x≤4, then the side lengths must satisfy 1+2+3+4+x≥|V|+|H|=14, i.e. x=4 holds and the remaining 3-length parts of V and H should be covered by the 3-square and by the union of the 1-square and 2-square, respectively. But then there is a 1×6 stripe inside K×K that touches the 1-square and is internally disjoint from all the four. It is impossible to cover this stripe with the 5-square.
2.2 The small cases n=6 and n=7
The largest possible covers for n=6 and n=7 are shown in Figure 1. Consider e.g. n=6. Since we can realize that there is a good cover for n=6 and L=9, moreover for n=6 the upper bound meets this L (U B 6=L), we conclude that K 6=9, and no more work is required regarding this n. This specific cover is constructed so that the two biggest squares are allocated in two opposite corners (red and green squares of sizes 6 and 5, Fig.1, left), the next two biggest squares are placed in the other two corners, and the remained few cells are covered by the two smallest squares. The same technique works also for n=7 since U B 7=11 (cf. Fig.1, right); but let us note that this simple way of covering will not work after some value of n.
Fig. 1
n=6 (left) and n=7 (right)
Full size image
3 Medium cases
For medium values of n, the problem cannot be solved easily by hand. We mean that the problem is easy if we calculate the trivial upper bound U B n for an n and find a complete cover somehow for L=U B n; in this case we can be sure that an optimal solution has been obtained, i.e. K=L=U B n. But in some other cases, K is smaller than U B n, or a packing of size U B n cannot be found easily. For such cases we follow the next idea: we define a mathematical programming model that takes the sizes of the squares as inputs, and solve the model using a commercial solver. If the solver tells us that there is no solution for a certain L value, but there is one for L−1, then we know that in fact the optimal solution is K=L−1.
3.1 Mathematical programming formulation
Our problem is related to the board packing problem (Abraham et al., 2023; Dósa et al., 2020). We are given a rectangular board with rows I={1,…,|I|} and columns J={1,…,|J|}. Each position (i,j)∈I×J of the board has an integer value g i,j representing a revenue obtained if the position is covered. A set R of rectangles is given, each r∈R with a given height h r, width w r, and cost c r. The objective is to purchase rectangles and place them on the board so as to maximize the profit, which is equal to the values of the covered positions minus the cost of purchased rectangles. The revenues, heights, widths, and costs are expressed as integers. Rectangles must be placed with their sides parallel to the sides of the board. They are allowed to overlap, but the revenue from a given board position can only be collected once.
By setting |I|=|J|=K, the revenue equal to 2 for every cell, and defining the set of rectangles R as the n squares of consecutive sizes, each with a cost of 1, then if the whole board can be covered, the optimal value is at least 2 K 2−n. On the other hand, if the optimal value is less than 2 K 2−n, we can conclude that no cover of K×K exists using small squares up to size n. In Section4 we will adapt a heuristic developed for the board packing problem to handle big cases. However, for medium cases, we will apply a specialized mathematical programming formulation.
In the specialized model, we need to place all the available objects, and we need to cover all cells. We can utilize some symmetry, or even try to solve the model with certain squares in fixed positions. Let K be the size of the square that needs to be covered, and let n×n be the size of the largest consecutive square. Let T r be the set of coordinates such that square r×r can be placed with its upper left corner at such coordinates. E.g., T r={(i,j):i=1,…,K−r+1,j=1,…,K−r+1}. Let B i j r be defined as the set of positions (u,v) for the top left corner of the square of size r×r that will result in position (i,j) being covered by that square. Define the binary variable x i j r to be 1 if and only if the square of size r×r is located with the upper left coordinate at (i,j). Let continuous variable y be 0 if all cells are covered, and at least 1 otherwise. The model can be written as:
min y
(2)
∑(i,j)∈T r x i j r=1,r∈{1,…,n}
(3)
y+∑r∈{1,…,n}∑(u,v)∈B i j r x u v r≥1,i=1,…,K,j=1,…,K
(4)
x i j r∈{0,1},r∈{1,…,n},(i,j)∈T r
(5)
y≥0.
(6)
The problem has some symmetry that can be exploited. For example, one may require that the largest square should be placed to the right and above the middle of the covered square. E.g., we reduce T n without loss of generality as follows. Let D n=⌈(K−n+1)/2⌉. Then, we set T n={(i,j):i=1,…,D n,j=1,…,D n}.
3.2 Results using the mathematical models
We solve the models using the commercial mixed-integer programming solver CPLEX. When using the models for values in the range of n=8,…,21, their performance do not differ very much. However, the second, specialized model, typically has slightly lower running times when attempting to prove optimality. In Table 2 we provide the optimal results for 8≤n≤21. Any case with U B n=K is simple, or at least seems simple. It is enough to find a good cover for the value of U B n. But, note that this good cover cannot be easily found for a relatively big n, like n=21. These are the cases n∈{9,13,14,16,18,19,20,21}.
For the other cases, i.e. for n∈{8,10,11,12,15,17}, the optimal value of L can be obtained so that CPLEX returns the non-existence of a complete cover for L+1. The running time of CPLEX was about 7 seconds for n=11 (to conclude that K=22 is not possible), but for K=21 CPLEX found an optimal solution soon. For n=12, excluding K=25 needed about 12 seconds. For n=15, excluding K=35 needed about 214 seconds. For n=17, CPLEX needed more than 10 hours of running time to conclude that K=42 cannot be covered, thus proving that K=41 is optimal. Thus, for larger values of n, solving models using CPLEX becomes too hard.
Table 2 The cases 8≤n≤21
Full size table
The experience is that finding a good cover is easy if it exists, but excluding some K value is harder and harder as n grows. The case of n=21 with a cover for K=57 is detailed in Table 3, and is illustrated in Figure 2. In the table we give the top-left coordinates of each small square r×r in the covering, r≤n, where “-” means that the square in question is not needed for the covering.
Table 3 Coordinates for n=21, K=57, a complete covering
Full size table
Fig. 2
Cover of K=57 using squares up to size n=21
Full size image
Some further results can be found in Table 4 for 22≤n≤25. Here CPLEX cannot provide an optimal result in reasonable time. For all cases covered in Table 4, the exact value of K is between L and U B n, and has yet to be determined. For example, in case of n=22, we have L=60≤K≤U B 22=61.
Table 4 The cases 22≤n≤25
Full size table
4 Big cases
We define big cases as those where CPLEX is far from being able to prove optimality: CPLEX may be able to find some lower bounds, although even this can be time-consuming when the square to cover is larger than 70×70. We then look for other means to generate lower bounds, using a heuristic algorithm as well as an expansion algorithm.
4.1 Heuristic search
We have seen that after some time the solver is not enough. We still can apply a heuristic to find valid lower bounds on K for a given n. That is, the board packing problem (Abraham et al., 2023) presented in Section3.1 can be generated for given values of K and n, and a computer search can attempt to find a feasible solution with an objective function value that implies that a cover has been found.
The heuristic that we use is an evolutionary algorithm, modified from the algorithm proposed in Abraham et al. (2023) to solve general instances of the board packing problem. Figure3 shows the outline of the method, which is based on the leading metaheuristic for the highly competitive combinatorial task of solving vehicle routing problems (Vidal, 2022), the hybrid genetic search.
First, a set of initial solutions is generated. Some of these solutions are generated by a construction heuristic, where the squares are sorted in decreasing size, and then inserted in a location that maximizes the number of covered cells. The construction heuristic is deterministic, but can be repeated several times by moving the first square to the last position to modify the insertion sequence. Other initial solutions are simply generated by randomly placing all the squares onto the grid. As a refinement of this, we force the four largest squares to be initially placed in a corner. The method is allowed to restart. In that case, the single best solution found is kept, and the other initial solutions are generated at random, without forcing the largest squares to be placed in the corners.
Fig. 3
Schematic overview of the heuristic used to find large coverings
Full size image
In a regular iteration of the heuristic, two solutions are selected from the current population using two binary tournaments. These are then combined into two new solutions using a type of uniform crossover, where each square and its placement is assigned at random from one parent to one child. The new solutions are then subjected to a local search improvement. Here, each square can be moved one step in either Cartesian direction, or if this does not lead to an improvement after considering all the squares, one square is selected to be temporarily removed and then inserted into the best possible position (covering as many cells as possible).
The initial population has a fixed number of solutions included. All new solutions generated are included into the population, as long as their objective function values are different from other solutions already included (this is to ensure a diverse population). When the population reaches a certain size, it is reduced down to the same number of solutions as the initial population had. The procedure is based on (Vidal, 2022), where we combine measures of solution quality and solution diversity. If this reduction happens several times without the search having found any new best solution, the search is restarted.
4.2 Expansion algorithm
Our next idea to obtain lower bounds is an expansion algorithm. Given an original cover of a square of dimensions K×K, using consecutive squares up to size n, we can apply an expansion algorithm to find a covering of a larger square K′=K+s using consecutive squares of sizes from 2 to n+1. The 1×1 square can then be added arbitrarily to this cover. The size of s and thus K′ depends on the minimum number of squares to appear in any given row or column in the original cover of the K×K square.
The idea of the algorithm is easier to express using rectangles. Let us assume that we have a rectangle of size K I×K J that is covered by a set of n rectangles where rectangle r has the size I r×J r. We will then compute the number of rectangles in each row that are actively used to cover the bigger rectangle. Let s be the minimum number of such rectangles across all rows. Then, we will find a cover of a rectangle of size K I×(K J+s) by using n rectangles of sizes I r×(J r+1). This will be achieved by increasing the size of each individual rectangle by one column and shifting them horizontally to cover the enlarged area.
By applying this procedure to a K×K square covered with n consecutive squares, we can first expand the squares horizontally. Then, by transposing the result and applying the procedure again, we can obtain a figure of size (K+s)×(K+s′) covered by consecutive squares of sizes from 2 to n+1. In the case that s≠s′, we may then have to redo the process while ensuring that the square is extended in equal length min{s,s′} in both directions.
Algorithm 1
Pseudo-code for expansion algorithm.
Full size image
Proposition 1
The expansion algorithm works, i.e. provides a covering for each n and s.
However, this is only a cover, we do not know whether it is optimal or not.
Remark 2
It is essential in executing Algorithm 1 horizontally for s and vertically for s′ that s′ be computed only after the determination of s and construction of the K I×(K J+s) cover. Computing both s and s′ from the initial configuration and executing expansion in both directions simultaneously may or may not work. Among the covers constructed for Q[n] we have not found a counterexample so far. However in Fig.4 we show an arrangement of 20 squares that cover an area of 4 rows and 8 columns, but its expansion leaves an uncovered cell (crossed in Fig.5) if s′ is calculated from the original cover rather than from the s-expanded one.
Fig. 4
Initial 4×8 cover
Full size image
Fig. 5
Simultaneous computation of s and s′ leaves a cell uncovered
Full size image
4.3 Results using the heuristic and the expansion-algorithm
For small values of K, the heuristic finds lower bounds as good as those determined by using the commercial MIP solver. However, since the heuristic does not provide any upper bounds, the commercial MIP solver is preferred up to values of K around 65. For larger K, the MIP solver is unable to provide conclusive results, and the heuristic is faster at finding lower bounds.
Table5 shows results for 25≤n≤30. The time taken to find a cover for a given combination of L and n by the heuristic for this range is typically less than one hour. However, when the heuristic fails to find a cover, we simply do not know whether the method just failed to find the cover or whether a cover did not exist. For n=25, which was the largest case considered by CPLEX, the heuristic is able to find the same lower bound. The table also shows lower bounds produced by applying the expansion algorithm. To generate these, we consider the cover found by the heuristic for n and create a new cover using n+1 small squares. This is able to find the same lower bounds as the heuristic for n=26 and n=28 by using the output from the heuristic for n=25 and n=27, respectively.
Table 5 The cases 25≤n≤30 and results using the heuristic (L) and the expansion algorithm (L′)
Full size table
To stress test the heuristic, we chose n=100, and attempted to find the largest value of K such that we can prove a feasible cover using the consecutive squares from 1 to n=100. The heuristic succeeded at finding a solution for K=559≤U B 100=581. The run that succeeded terminated after 3,166 seconds, but before this, two other runs with different random seeds had failed after running for four days each. Thus, even finding lower bounds is challenging for such large values of n.
In the cover for n=100 and K=559 that was found by the heuristic, there are at least s=6 squares in each row and each column (after performing the extension in one direction). Thus, by the expansion algorithm we also get a cover for K′=565 using n=101; for K″=571 using n=102, and so on.
5 The asymptotic case
Recall that N=A n with A n=n(n+1)(2 n+1)6 is an upper bound on the size of a square coverable by the squares 1×1, 2×2, ..., n×n. In this section we prove that this bound is asymptotically tight, i.e., a cover can be constructed for a square of area at least (1−o(1))A n as n→∞. More explicitly, the square of side length N−2 n admits a cover with those smaller squares. We prove this fact in the following more general form.
Theorem 3
If p⋅q≤A n and |p−q|<3 n, then a (p−2 n)×(q−2 n) rectangle can be covered with the squares of side lengths 1,2,…,n.
Proof
We apply induction on n, assuming that for all n′<n, every (p′−2 n′)×(q′−2 n′) rectangle can be covered with the squares of side lengths 1,2,…,n′ provided that p′⋅q′≤A n′ and |p′−q′|<3 n′.
Let R denote a p×q rectangle (p wide and q high), and assume p≤q. The task is to cover the (p−2 n)×(q−2 n) rectangle obtained from R by trimming a 2 n-wide rectangle from its right end and a 2 n-high rectangle from the top. If p≤2 n, then the task is void, which is necessarily the case when 2 n(2 n+1)>A n (that means n<11). Beyond that, the validity of the theorem can very easily be checked for a wider range of not too large n; we demonstrate this for n=24. We then have A 24=4900, p≤A 24=70, and assuming p≥2 n+1=49, q≤A 24/49=100 holds. Hence p−2 n≤22 and q−2 n≤52. On the other hand, even the rectangle 22×69 is obviously coverable with just the three largest squares 24,23,22.
For general n we take the s+t largest squares, i.e. of sizes n,n−1,…,n−s−t+1, determined as the smallest natural numbers s and t satisfying the inequalities
p−2 n≤∑i=n−s+1 n i a n d p−2 n≤∑i=n−s−t+1 n−s i.
From these squares we create a “slice” that completely covers a rectangle of width p−2 n and height 2 n−s−t+1, as sketched in Figure 6; or height 2 n−s−t if the case explained in the next paragraph occurs. The unified notation 2 n−s−t+ε will be used, with the understanding that ε can mean either of 0 and 1. We arrange the s largest squares in increasing order—with their bottom sides aligned—and the next t largest squares in decreasing order—with their top sides aligned—both from left to right, in the way that square n−s−t+1 touches square n from above, and the vertical left sides of the squares n−s+1 and n−s touch the left side of R, the lower left corner of square n−s+1 being the same as that of R.
Fig. 6
Example of a slice in which 5+7 squares cover total width at least p−2 n; i.e., s=5, t=7 (cf.text) and w denotes n−5. Dark areas are covered twice
Full size image
It may occur that the bottom-left corner of square n−s−t+1 is located to the left of the top-left corner of square n. In that case an uncovered area would be surrounded by the four squares n, n−1, n−s−t+2, n−s−t+1. This situation can be handled by aligning the top row lower by 1; that is, square n−s−t+1 should then touch square n−1, rather than n. This yields a slice of height 2 n−s−t, as one of the options in the notation 2 n−s−t+ε. For instance, concerning the example of Figure6, if ∑i=1 4(w+i)∑j=0 5(w−j) then w−6 does not belong to the slice.
Since the bottom squares are larger than the top ones, we certainly have t≥s. Moreover, the height of the slice is 2 n−s−t+ε, hence the largest vertical overlap between top and bottom squares is t−s+(1−ε)≤t−s+1. A key point in the computation will be that we have
t−s≤2.
To prove this, we first observe that for every n the following chain of inequalities is valid:
p−2 n≤N−2 n=n(n+1)(2 n+1)6−2 n<3 5(n+1)n−2 n<(n−n+2)n−2 n=(n−2 n)(n−1).
By the choice of s, the sum of the s−1 largest squares does not reach p−2 n, and the smallest of them is bigger than n−s (hence all are), thus
(n−s)(s−1)<p−2 n<(n−2 n)(n−1),
implying sn−n−t, thus
(n−n−t)(t−1)<p−2 n<(n−2 n)(n−1),
implying t<n. Moreover, we obviously have
and from s≤t<n it follows that the size n−s−t+1 of the smallest square in the top row is larger than n−2 n, thus
As a consequence,
t−s<1+(p−2 n)2 n n 2−2 n n≤1+2(N−2 n)n(n 3/2−2 n)n<3.
Here the last step is equivalent to N<n 3/2, what clearly is a valid inequality because 6 N 2/n=(n+1)(2 n+1)<6 n 2 holds for all n>1. Of course, t−s<3 means t−s≤2, as claimed.
By the choice of s and t, the two rightmost squares, namely n and n−s−t+1, reach p−2 n but do not reach p−n. Hence a rectangle at least (n+1)×(2 n−s−t+ε), with area bigger than 2 n 2−2 n n+2 n−2 n, remains uncovered in R up to height 2 n−s−t+ε, within which the width p−2 n is entirely covered. On the other hand, as the largest vertical overlap is at most t−s+1≤3, the area covered twice within width p−n by the squares used so far is not larger than 3 p−3 n. It is smaller than 9 5 n n−3 n+9 5 n, using the upper bound p<3 5(n+1)n. Thus, the blank area exceeds the double-covered one by more than
n 1/2⋅(2 n 3/2−19 5 n+5 n 1/2−19 5).
Here one can see that for all n≥1 both factors are increasing functions of n (and also of n), thus the product is positive already for n=4 (hence for all n≥4).
Consequently, cutting off the p×(2 n−s−t+ε) rectangle from the bottom of R, we obtain a p×q′ rectangle with q′=q+s+t−2 n−ε, that should be covered with the squares from 1 to n′:=n−s−t. The above calculations show that
p(q−q′)=A n−A n′−|d o u b l e|+|b l a n k|>A n−A n′;
thus, as p q<A n holds by assumption, we obtain p q′<A n′. Consequently, the theorem follows by induction, once we verify that |p−q′|<3 n′ also holds.
Since 0≤q−p<3 n has been assumed, we now have −2 n+s+t−ε≤q′−p<n+s+t−ε. In this way, if n is not very small, we obtain
0<n+s+t−ε<n+2 n≤2 n−2 n<|−2 n+s+t−ε|≤2 n+1−s−t<2 n+1−3(s+t)+4 n<3(n−s−t)=3 n′;
e.g., n≥18 is sufficient.
This inductive proof also provides a linear-time algorithm that generates a square cover of any given rectangle with the parameters p,q,n satisfying the conditions of the theorem. (In many steps of the procedure, consecutive slices will alternate between horizontal and vertical position.) ◻
6 Conclusions
We studied the following question. Given n squares, one of size 1, one of size 2, and so on, up to one of size n, what is the biggest square size K×K that can be completely covered by the set of small squares?
Up to n≤6 we can solve the problem easily by hand. Already for n=7 the case is not trivial. Between n=8 and n=21 we are able to determine the optimal solutions by using a commercial mathematical programming solver, and for some even bigger values of n, we still can get close lower and upper bounds on K.
We also introduced a heuristic algorithm that is able to provide lower bounds for relatively big n values, such as n=100. It is not clear yet, how far we are here from the optimal values. Some other heuristic algorithms that are more specialized may be able to find improved lower bounds.
We introduced an expansion-like algorithm, which is able to make a cover for any n′>n value, if a cover is given for n. Here there are at least two interesting questions. One is that we do not know how good this algorithm is in the sense that, e.g., if we have an optimal solution for n, how far will the constructed cover be for n+1 (in the worst case) from the optimal solution. The other question is the following: We have seen that if there are at least s squares in the cover of a square of size L in the horizontal (vertical) direction, then we can expand the squares in the horizontal (vertical) direction to get a cover for a bigger square of size L+s. But we also gave a small counterexample against horizontal and vertical expansion in parallel, to show that this expansion procedure should be made in two phases: first we count s in one direction and make the expansion, then we count s again in the other direction and make the expansion in the other direction. If we count s for both directions before the expansions, this can be a problem, as it is possible that some hole is created. But our counterexample has some squares with the same size, so not one 1×1, one 2×2, and so on. Can this phenomenon (creating a hole) appear also for the set of 1×1, 2×2,…,n×n squares, and determining s before the expansions in both directions?
During our investigation, we used the trivial upper bound (for the size of K of the biggest covered square) as . Naturally, if we do have a covering of a square of size L where L=U B n, we know for sure that this is the optimal value. For some small values of n, K=U B n really is the optimal value, but unfortunately, from some bigger values of n, this is not the case. Thus, it would be useful to know some other, more effective upper bound. At moment we have only this upper bound U B n, which we know by example is not always tight, so this is a question whether one can find some tighter bound.
It is also an interesting question about the nature of the “gap” between U B n, the trivial upper bound, and K, the optimal value. It is clear (following from our asymptotic result) that U B n−K U B n→0 as n→∞. It would be nice to give a more exact characterization about the gap.
Finally, we hope that our paper will be interesting for other researchers and opens a new direction of research, similarly to the packing version of this problem.
References
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Acknowledgements
This research was supported in part by the National Research, Development and Innovation Office – NKFIH under the grant SNN 129364, and by the grant TKP2021-NVA-09 of the Ministry for Innovation and Technology, Hungary.
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University of Szeged, 6720, Szeged, Hungary
Janos Balogh
University of Pannonia, 8200, Veszprém, Hungary
Gyorgy Dosa,Tomas Attila Olaj,Istvan Szalkai&Zsolt Tuza
Faculty of Logistics, Molde University College, N-6402, Molde, Norway
Lars Magnus Hvattum
HUN-REN Alfréd Rényi Institute of Mathematics, 1053, Budapest, Hungary
Zsolt Tuza
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Balogh, J., Dosa, G., Hvattum, L.M. et al. Covering a square with consecutive squares. Ann Oper Res350, 911–926 (2025).
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Abraham, G., Dosa, G., Hvattum, L. M., Olaj, T. A., & Tuza, Z. S. (2023). The board packing problem. European Journal of Operational Research, 308, 1056–1073.
Google Scholar
Balogh, J., Dosa, G., Hvattum, L. M., Olaj, T., & Tuza, Z. S. (2022). Guillotine cutting is asymptotically optimal for packing consecutive squares. Optimization Letters, 16, 2775–2785.
ArticleGoogle Scholar
Buchwald, T., & Scheithauer, G. (2016). A 5/9 theorem on packing squares into a square. Preprint MATH-NM-04-2016, TU Dresden.
Coffman, E. G., Jr., Garey, M. R., Johnson, D. S., & Tarjan, R. E. (1980). Performance bounds for level-oriented two-dimensional packing algorithms. SIAM Journal on Computing, 9, 808–826.
Google Scholar
Dósa, G., Hvattum, L. M., Olaj, T., & Tuza, Z. S. (2020). The board packing problem: Packing rectangles into a board to maximize profit. In I. Vassányi (Ed.), Proceedings of the Pannonian Conference on Advances in Information Technology (PCIT 2020) (pp. 10–16). Veszprém, Hungary: University of Pannonia.
Google Scholar
Gardner, M. (1966). Mathematical games: The problem of Mrs. Perkins’ quilt, and answers to last month’s puzzles. Scientific American, 215(3), 264–272.
Google Scholar
Gardner, M. (1975). Mrs. Perkins’ quilt and other square-packing problems. In Mathematical Carnival. New York: Alfred A. Knopf (pp. 139–149).
Korf, R. E. (2003). Optimal rectangle packing: Initial results. In: Proceedings of the 13th International Conference on Automated Planning and Scheduling (ICAPS 2003), (pp. 287–295).
Korf, R. E. (2004). Optimal rectangle packing: New results. In: Proceedings of the 14th International Conference on Automated Planning and Scheduling (ICAPS 2004), (pp. 142–149).
Moffitt, M. D., & Pollack, M. E. (2006). Optimal rectangle packing: a meta-CSP approach. In: Proceedings of the 16th International Conference on Automated Planning and Scheduling (ICAPS 2006), (pp. 93–102).
Simonis, H., & O’Sullivan, B. (2008). Search strategies for rectangle packing. In: Constraint Programming 2008, Lecture Notes in Computer Science, (vol. 5202 pp. 52–66).
Vidal, T. (2022). Hybrid genetic search for the CVRP: Open-source implementation and SWAP neighborhood. Computers and Operations Research, 140, Article 105643.
Google Scholar
The on-line encyclopedia of integer sequences, sequence a005842. Published electronically at accessed: 26 March 2021.
Watson, G. (1918). The problem of the square pyramid. Messenger of Mathematics, New Series, 48, 1–22.
Google Scholar
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11529 | https://www.quora.com/How-was-the-drag-force-equation-1-2Cdpsv-2-derived | Something went wrong. Wait a moment and try again.
Derivation of Formulas
Drag Force (physics)
Modern Aerodynamics
Fluid Flow
Physics of Fluids
Physics Equations
Aerodynamics and Propulsi...
Vehicle Aerodynamics
5
How was the drag force equation, 1/2Cdpsv^2, derived?
Kim Aaron
Has PhD in fluid dynamics from Caltech
·
Author has 8.3K answers and 27.2M answer views
·
Updated Jan 4
It's not really derived. It's more like Cd was defined based on observations that drag was proportional to density, area, and velocity squared. That result also follows from Bernoulli's principle.
Bernoulli's equation can be derived from the Navier Stokes equation by setting viscosity equal to zero and assuming steady flow so the time derivatives go away.
By the way, it is difficult to make sense of an equation when all the different symbols are smushed together like that. And the symbol for density is the Greek letter, rho. It looks a bit like p but we do not use p for it because of the confusi
It's not really derived. It's more like Cd was defined based on observations that drag was proportional to density, area, and velocity squared. That result also follows from Bernoulli's principle.
Bernoulli's equation can be derived from the Navier Stokes equation by setting viscosity equal to zero and assuming steady flow so the time derivatives go away.
By the way, it is difficult to make sense of an equation when all the different symbols are smushed together like that. And the symbol for density is the Greek letter, rho. It looks a bit like p but we do not use p for it because of the confusion with pressure, which does use p. I usually write "rho" if I'm not using the actual letter. S is used for area sometimes, especially for wings of an aircraft. But I would use A as being more general. This would have been easier to read:
D = 1/2 rho V^2 A Cd.
I keep the 1/2 rho V^2 together because that combination is the dynamic pressure.
Three years later:
A comment asks why there is a 1/2 in there.
Most dimensionless numbers just have combinations of physical parameters for which the units cancel out. An example from fluid mechanics is the Reynolds number:
Re =VDν
where V is velocity, D is diameter and ν is kinematic viscosity of the fluid.
Including a constant, such as the 1/2 in the various force and moment coefficients does not change the dimensions (because 1/2 is, itself, dimensionless). So the 1/2 seems a bit arbitrary. And it is, sort of. When we define dimensionless parameters, we often don’t have any particular reason to throw a constant in there. But in the case of force and moment coefficients, we have some rationale for doing so.
In fluid dynamics, it has long been observed that when you bring the flow to rest at a stagnation point, the pressure increases by an amount of 1/2ρV^2, where ρ is the fluid density. We call that increase in pressure the “dynamic pressure.” This is for low subsonic flow (M<0.3), for which the flow can be considered to be incompressible. That covers a lot of the situations we experience, so this restriction isn’t much of a problem.
Sometimes we use the letter q to represent this dynamic pressure because it comes up a lot in fluid dynamics.
If we had a flat plate normal to the flow and if we imagined that the flow were at this stagnation pressure over its full area, then the drag force would be q×A, where A is the area of the plate. If we were in the early days of developing the topic of fluid dynamics, we might find it convenient to define something that me might decide to call “drag coefficient” as D/(qA). This coefficient would let us compare measured drag in a number of situations with this idealized drag on a flat plate normal to the flow.
When the Wright brothers were doing their wind tunnel tests, they basically mounted their models on a pivot in the wind tunnel with something like a flat plate normal to the flow. They would adjust things until the force on the model balanced the force on the flat plate. Even today, we still call a device to measure drag or lift a “force balance.”
So it was natural for the early fluid dynamicists to define their newly conceived drag coefficient as the “measured” drag force divided by qA. In that form, you don’t see the 1/2. You don’t see the ρ or the V2 either. You just see a parameter, q, that has units of pressure. Multiply that by area and you get force. So when you divide lift or drag force by that force, it’s dimensionless.
Eventually, they got around to deriving Bernoulli’s equation and figuring out that the dynamic pressure was equal to 1/2 ρ V2. So the force coefficients ended up being
CF =Force1/2ρV2A
John James
Lives in The United Kingdom
·
Author has 205 answers and 201K answer views
·
Updated 3y
Consider a object with a projected/frontal area of A experiencing drag in a uniform fluid.
The mass of the fluid colliding with the object travelling at a velocity v will be equal to ρfV from the definition of density (ρf is the density of the fluid, V is the volume of the fluid impacting the area A). In time dt, a mass equal to dm collides with the object, which travels a distance of vdt (the ‘length’ of the column of fluid). As volume is equal to the product of area and length (e.g. for rectangles), we have that dm=ρfAvdt\Longrightar
Consider a object with a projected/frontal area of A experiencing drag in a uniform fluid.
The mass of the fluid colliding with the object travelling at a velocity v will be equal to ρfV from the definition of density (ρf is the density of the fluid, V is the volume of the fluid impacting the area A). In time dt, a mass equal to dm collides with the object, which travels a distance of vdt (the ‘length’ of the column of fluid). As volume is equal to the product of area and length (e.g. for rectangles), we have that dm=ρfAvdt⟹dmdt=dmdF⋅dvdt⋅dFdv, by the chain rule. Note that dmdF⋅dvdt cancel each other out as both are definitions of acceleration.
⟹dFdv=ρfAv
F=∫v0ρfAv′ dv′=12ρfAv2
The coefficient of drag, CD, is in fact a definition and is not derived (it is also dimensionless). It is a function of the object’s geometry.
Richard Muller
Prof Physics, UC Berkeley, author "Now, The Physics of Time"
·
Upvoted by
Jay Wacker
, physicist: PhD + postdoc + faculty · Author has 2.3K answers and 248.4M answer views
·
9y
Related
Where does the drag force equation come from?
At low velocity, the drag force equation can be derived, and it turns out that the drag is proportional to the velocity.
At high velocities, turbulence occurs, and the derivation fails. Nobody can derive the drag force equation at turbulent velocities. It is seen empirically to be proportional to the velocity squared.
That makes sense; at high velocities, it is as if you are hitting stationary objects (little bits of turbulence). You get the same equation if you are running through raindrops; how does the force of the rain depend on your velocity? Each time you hit, momentum is transferred b
At low velocity, the drag force equation can be derived, and it turns out that the drag is proportional to the velocity.
At high velocities, turbulence occurs, and the derivation fails. Nobody can derive the drag force equation at turbulent velocities. It is seen empirically to be proportional to the velocity squared.
That makes sense; at high velocities, it is as if you are hitting stationary objects (little bits of turbulence). You get the same equation if you are running through raindrops; how does the force of the rain depend on your velocity? Each time you hit, momentum is transferred by an amount proportional to your velocity. But if you are moving faster, you hit more, by an amount proportional to your velocity. So it is plausible that the momentum transferred per second (and that is the force) is proportional to the velocity squared.
Now maybe you are wondering why the low velocity drag equation gives just velocity and not velocity squared! The answer is that the flow "wraps around" the object, and that cancels some of the velocity effect. Look up Wikipedia under the topic "potential flow".
Related questions
Where does the drag force equation come from?
How was drag equation
F
D
=
0.5
C
D
A
ρ
v
2
d
efined or derived as it is?What is the reasoning behind every variable and constant in this equation?
Does S in the drag equation (1/2Cdpsv^2) mean the projected area or surface area?
Why does 1/2 appear in the drag force equation?
How do I prove that the drag force is dimensionless?
Assistant
Bot
·
11mo
The drag force equation, represented as:
Fd=12Cdρv2A
is derived from the principles of fluid dynamics and empirical observations. Here’s a breakdown of the derivation:
Components of the Equation
Drag Force (Fd): This is the force exerted by the fluid (like air or water) on an object moving through it.
Drag Coefficient (Cd): This is a dimensionless number that encapsulates the effects of shape, surface roughness, and flow conditions. It is determined experimentally and varies based on the object's geometry and the flow regime (laminar or turbulent).
Fluid Density (ρ): This i
The drag force equation, represented as:
Fd=12Cdρv2A
is derived from the principles of fluid dynamics and empirical observations. Here’s a breakdown of the derivation:
Components of the Equation
Drag Force (Fd): This is the force exerted by the fluid (like air or water) on an object moving through it.
Drag Coefficient (Cd): This is a dimensionless number that encapsulates the effects of shape, surface roughness, and flow conditions. It is determined experimentally and varies based on the object's geometry and the flow regime (laminar or turbulent).
Fluid Density (ρ): This is the mass per unit volume of the fluid through which the object is moving. For air at sea level, ρ is approximately 1.225kg/m3.
Velocity (v): This is the speed of the object relative to the fluid.
Frontal Area (A): This is the projected area of the object in the direction of the flow. For a flat plate, it would be the area facing the fluid flow.
Derivation Steps
Basic Concept of Drag: When an object moves through a fluid, it experiences resistance due to the fluid's viscosity and inertia. The drag force is a function of the fluid's properties and the object's characteristics.
Empirical Observations: Experimental data shows that the drag force is proportional to the square of the velocity (v2). This is because as the speed increases, the amount of fluid displaced in a given time increases, leading to greater resistance.
Proportionality Constant: The proportionality constant that relates the drag force to the other variables includes the drag coefficient (Cd), density of the fluid (ρ), and the effective frontal area (A):
Fd∝Cd⋅ρ⋅A⋅v2
Factor of 12: The factor of 12 arises from the integration of the forces acting on the object and the dynamics of the fluid flow around it. It is often included in the derivation based on the kinetic energy of the fluid and how it relates to the drag experienced by the object.
Final Equation
Combining these components, we arrive at the drag force equation:
Fd=12CdρAv2
Summary
In summary, the drag force equation is a result of empirical observations combined with theoretical fluid dynamics principles, reflecting how the drag experienced by an object in a fluid medium depends on its speed, shape, and the characteristics of the fluid.
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Aman Raj
Software Engineer at Wells Fargo (company) (2025–present)
·
Author has 308 answers and 1.1M answer views
·
5y
Related
How was drag equation
F
D
=
0.5
C
D
A
ρ
v
2
d
efined or derived as it is?What is the reasoning behind every variable and constant in this equation?
It's a nice question.
Here, we first consider a liquid. A liquid! The force due to the liquid is
F=d(mv)dt
=vdmdt
=vρdVdt
=ρvAv
So,
F=ρAv2
This was for a liquid.
Consider a gas now. Suppose a body moves with velocity v, has area Aand the density ρ. Then, we wish to find the drag force on it.
If we imagine, then it's clear that we need to visualise how air moves. The air moves in a very same way as a liquid does. So, the force due to the air should be similar to that of liquid.
Here the word “similar” does not mean exactly but rather it means that
It's a nice question.
Here, we first consider a liquid. A liquid! The force due to the liquid is
F=d(mv)dt
=vdmdt
=vρdVdt
=ρvAv
So,
F=ρAv2
This was for a liquid.
Consider a gas now. Suppose a body moves with velocity v, has area Aand the density ρ. Then, we wish to find the drag force on it.
If we imagine, then it's clear that we need to visualise how air moves. The air moves in a very same way as a liquid does. So, the force due to the air should be similar to that of liquid.
Here the word “similar” does not mean exactly but rather it means that the same thing is happened here. The main point of difference is gases are highly compressible but liquids are not!
Rest, experiments decided the results. Indeed, the air behaved in the very same way as the liquid could; only compressibility makes the difference.
So, it happened to suggest that
F∝ρAv2
Hence, we get
F=KρAv2
Now here comes the point of 0.5 in formula. It is a trend to write K=0.5CDwhere CDis called drag coefficient.
Hence, the formula is proved.
As expected, the value of C for air is 0.95 (less than 1).
Thanks for reading!
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Robert Johnson
I have 13 years experience leading technical people.
·
Author has 521 answers and 1.5M answer views
·
10y
I don't have a very good or complete answer, but it is a work-energy problem, right? And looked at that way I think what stands out is that 1/2MV^2 looks remarkably similar.
And of course the drag coefficient is just an empirical placeholder for complicated behavior that we can't handle very elegantly. Same with the Reynolds number.
Related questions
Using the simplest math terms, why is the drag force equation the way it is?
How was the equation for kinetic energy derived:
E
k
=
1
2
m
v
2
?
How do I find drag coefficient without using drag force equation?
What will be the derivation of the drag equation from the continuous equation?
How is the equation
R
μ
ν
−
1
2
R
σ
σ
g
μ
ν
=
T
μ
ν
derived?
Petra Blackstone
Upvoted by
Peng Peng Zheng
, Ph.D Physics, The University of Texas at Dallas (2021) · Author has 188 answers and 714.8K answer views
·
7y
Related
Using the simplest math terms, why is the drag force equation the way it is?
Assuming you mean common quadratic drag.
Let’s take an an energy approach.
Say an object is moving through some fluid, the only energy is the kinetic energy of the object, and the energy lost to the fluid by drag.
E=T(particle)+1/2mv2
Where the second term corresponds to the fluid.
Now, we can substitute density and volume for the mass, as one may do with a fluid.
E=T(particle)+1/2ρVv2
Where ρ is the density of the fluid and V is the volume, and v is the velocity of the object as the fluid element matches its velocity through drag.
Let’s also rewrite the volume as the surface area perpendi
Assuming you mean common quadratic drag.
Let’s take an an energy approach.
Say an object is moving through some fluid, the only energy is the kinetic energy of the object, and the energy lost to the fluid by drag.
E=T(particle)+1/2mv2
Where the second term corresponds to the fluid.
Now, we can substitute density and volume for the mass, as one may do with a fluid.
E=T(particle)+1/2ρVv2
Where ρ is the density of the fluid and V is the volume, and v is the velocity of the object as the fluid element matches its velocity through drag.
Let’s also rewrite the volume as the surface area perpendicular to flow, and the height parallel to flow as AL
E=T(particle)+1/2ρALv2
Now, as we know, something experiencing drag will reach a terminal velocity, and therefor have a net force of 0, for no acceleration.
So, let’s take the derivative of our energy with respect to distance moved (which corresponds to L, as it’s parallel to movement), to find force and set it to 0.
0=F(particle)+1/2ρAv2
and then
F(particle)=−1/2ρAv2
where the negative corresponds to the fact that the forces oppose each other, so it doesn’t matter much.
Then, for assurance, you throw in a constant C, because fluids have viscosity and objects have frictional surfaces and we didn’t take that into account.
So we have:
F(drag)=1/2CρAv2
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Simon Mouer
Engineer, Manager, Executive, Professor, Retired (1966–present)
·
Author has 2.8K answers and 4M answer views
·
Updated 7y
Related
What is drag force?
Drag is the name we give all forces that resist motion.
In propeller aircraft there are two primary sources of Drag —
induced drag caused by the angle the propeller blade makes with the relative wind — which increases with the increase of angle of attack of the wing — being maximum at takeoff, and minimum at cruise speed; and
parasitic drag which is caused by the airflow over the aircraft skin — which increases exponentially with airspeed — being minimum at takeoff and maximum at cruise speed.
Drag is the name we give all forces that resist motion.
In propeller aircraft there are two primary sources of Drag —
induced drag caused by the angle the propeller blade makes with the relative wind — which increases with the increase of angle of attack of the wing — being maximum at takeoff, and minimum at cruise speed; and
parasitic drag which is caused by the airflow over the aircraft skin — which increases exponentially with airspeed — being minimum at takeoff and maximum at cruise speed.
Julia Wright
Former Software Security Engineer
·
2y
Related
How was the formula force derived?
Part 1 of 2:Learning the Formula
1. Multiply mass times acceleration.
The force (F) required to move an object of mass (m) with an acceleration (a) is given by the formula F = m x a. So, force = mass multiplied by acceleration.
Convert figures to their SI values.
The International System of Units (SI) unit of mass is the kilogram, and the SI unit of acceleration is m/s (meters per second squared). So when mass and acceleration are expressed in their SI units, we get the force in its SI units which is N (Newtons)
As an example, if the mass of the object is given to be 3 pounds, you'll need to co
Part 1 of 2:Learning the Formula
1. Multiply mass times acceleration.
The force (F) required to move an object of mass (m) with an acceleration (a) is given by the formula F = m x a. So, force = mass multiplied by acceleration.
Convert figures to their SI values.
The International System of Units (SI) unit of mass is the kilogram, and the SI unit of acceleration is m/s (meters per second squared). So when mass and acceleration are expressed in their SI units, we get the force in its SI units which is N (Newtons)
As an example, if the mass of the object is given to be 3 pounds, you'll need to convert those pounds to kilograms. 3 pounds make 1.36 kg, so the mass of the object is 1.36 kg.
Keep in mind that weight and mass mean different things in Physics.
If the weight of an object is given in N (Newtons), then divide it by 9.8 to get the equivalent mass. For example, 10 N weight is equivalent to 10/9.8 = 1.02 kg.
Part 2 of 2:Using the Formula
1. Find the force that is required to accelerate a 1,000 kg car at 5 m/s.
Check to make sure all your values are in the correct SI unit.
Multiply your acceleration value (1000 kg) by 5 m/s to calculate your value.
Calculate the force required for an 8 pound wagon to accelerate at 7 m/s.
First, convert all your units to SI. One pound is equal to .453 kg, so you'll need to multiply that value by your 8 pounds to determine the mass.
Multiply your new value for the mass (3.62 kg) by your acceleration value (7 m/s).
Find the magnitude of force acting upon a cart weighing 100 N and accelerating at the rate of 2.
.
Remember, 10 N is equal to 9.8 kg. So, convert Newtons to kg by dividing by 9.8 kg. Your new kg value should be 10.2 kg for the mass.
Multiply your new mass value (10.2 kg) times the acceleration (2.5 m/s).
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Vir Narayan Singh
Former Professor (Retired) at Indian Institute of Technology, Roorkee (1972–2011)
·
Author has 1K answers and 1.3M answer views
·
5y
Related
How do we derive the formula for centripetal force?
Consider the following diagram:
An object is moving in a circular path of radius ‘r’. In a small amount of time Δt, it moves from A to B. At time t1, it is at A and its velocity is vi. At time t2, it is at B with velocity vf and Δt = t2-t1. The velocities vi and vf change only in direction. Their magnitudes remain the same. The average acceleration is
a = (vf-vi)/Δt = Δv/Δt
Consider the two triangles with sides Δr, r, r and Δv, vi, vf. These are similar triangles. So, the following relation can be written:
Δv/v = Δr/r
So, Δv = vΔr/r
and a = (v/r)(Δr/Δt)
As point B approaches A, Δt tends to zero and Δ
Consider the following diagram:
An object is moving in a circular path of radius ‘r’. In a small amount of time Δt, it moves from A to B. At time t1, it is at A and its velocity is vi. At time t2, it is at B with velocity vf and Δt = t2-t1. The velocities vi and vf change only in direction. Their magnitudes remain the same. The average acceleration is
a = (vf-vi)/Δt = Δv/Δt
Consider the two triangles with sides Δr, r, r and Δv, vi, vf. These are similar triangles. So, the following relation can be written:
Δv/v = Δr/r
So, Δv = vΔr/r
and a = (v/r)(Δr/Δt)
As point B approaches A, Δt tends to zero and Δr/Δt = v. So,
a = v^2/r
In the two figures above, vi and vf are parallel and Δv points to the centre of the circular path and so the acceleration also points to the centre of the circular path. The acceleration ‘a’ is then directed towards the centre of the circular path as the object moves along this path. This acceleration, a = v^2/r, is centripetal acceleration.
Jonathan Devor
PhD in Astronomy, Harvard University (Graduated 2008)
·
Author has 3.7K answers and 19.3M answer views
·
5y
Related
Can you set the drag force equation equal to the weight (m x g) in order to find the drag coefficient?
Yes, this situation is called terminal velocity.
This happens naturally whenever an object free-falls through air, water, or any other fluid medium for a long time. It accelerates due to gravity, and as its speed increases, its drag increases. Ultimately it reaches a point of equilibrium, where the drag force is equal to its weight, and at that point it doesn’t accelerate any more. In other words, its falling speed becomes constant.
Here’s how you calculate this terminal velocity:
If you measure the terminal velocity, you can simply rearrange the above equation and solve for the drag coefficient:
Yes, this situation is called terminal velocity.
This happens naturally whenever an object free-falls through air, water, or any other fluid medium for a long time. It accelerates due to gravity, and as its speed increases, its drag increases. Ultimately it reaches a point of equilibrium, where the drag force is equal to its weight, and at that point it doesn’t accelerate any more. In other words, its falling speed becomes constant.
Here’s how you calculate this terminal velocity:
If you measure the terminal velocity, you can simply rearrange the above equation and solve for the drag coefficient:
Having said all that I will mention a few caveats to what I wrote above:
Even in an ideal setting, you never actually reach terminal velocity. Rather you approach it (approximately) at an exponential rate.
In practice, the environment changes as you fall (e.g. the density of air is higher at lower elevations), so the terminal velocity changes as well.
Ahmed Sabry
still learning
·
10y
Related
Where does the drag force equation come from?
it is a differential equation that is modeled , and this model is based on either drag force is directly proportional to v or drag force is directly proportional to v^2
Ron Brown
Decades of teaching physics to undergrads
·
Author has 13.5K answers and 82.9M answer views
·
4y
Related
How do you derive centripetal force?
The centripetal force on an object is just whatever force is acting on it perpendicular to it’s motion to make it change its direction. That is, it is the force that makes the object momentarily travel in a curved path. It is “centripetal” in that it points toward the center of curvature of the motion at that moment.
But, being a force, it is caused by some interaction with another object. For example, a ball on a string traveling in a circle is being pulled by the string - so the tension is the centripetal force. A car traveling in a circular path on a flat turn does so because the friction of
The centripetal force on an object is just whatever force is acting on it perpendicular to it’s motion to make it change its direction. That is, it is the force that makes the object momentarily travel in a curved path. It is “centripetal” in that it points toward the center of curvature of the motion at that moment.
But, being a force, it is caused by some interaction with another object. For example, a ball on a string traveling in a circle is being pulled by the string - so the tension is the centripetal force. A car traveling in a circular path on a flat turn does so because the friction of the road on the tires turns it into a curved path. The Moon orbiting the Earth is being pulled into its orbit by the gravitational force, and so on.
So I think the question is asking about the centripetal force equation, commonly written
where m is the mass of the object, v is its speed, and r is the radius of curvature of the path at that particular moment.
But that equation is simply saying, whatever force is causing it to turn into a curved path, it must be equal to the mass of the object times the component of the acceleration perpendicular to its path. That is, F=mac, where ac here means the centripetal acceleration at that moment.
So your question is actually, “How do you derive the centripetal acceleration?”
The problem is, it is not a trivial question if you are not familiar with describing circular motion using vectors in polar coordinates and then using differential calculus to obtain the corresponding acceleration. I will set it up, but won’t solve it here:
To describe the location of a object in uniform circular motion in polar coordinates, one needs to describe the r vector as a function of time. That is,
where the i and j (with little ‘hats’) are called unit vectors in the directions of the x- and y-axes. But the angle is time dependent if the object is traveling around the circular path. That is,
where the omega symbol represents the rate at which the angle is increasing (called the angular velocity).
Given that location vector r(t), one needs to differentiate once with respect to time to get the velocity vector as a function of time and then again to get the acceleration vector. The resulting acceleration vector is
The minus sign means the acceleration is in the opposite direction as the radius vector, i.e., it points back toward the center of the circle, hence is centripetal. It’s magnitude is just
since the angular velocity is just the speed divided by the radius of curvature.
[I haven’t done all the steps, of course, and this is set up assuming uniform circular motion for simplicity, but the result is valid for any curved path at a particular moment.]
So pulling all of those pieces together says the “centripetal force” is the mass times the centripetal acceleration, hence can be written
which has a magnitude
Related questions
Where does the drag force equation come from?
How was drag equation
efined or derived as it is?What is the reasoning behind every variable and constant in this equation?
Does S in the drag equation (1/2Cdpsv^2) mean the projected area or surface area?
Why does 1/2 appear in the drag force equation?
How do I prove that the drag force is dimensionless?
Using the simplest math terms, why is the drag force equation the way it is?
How was the equation for kinetic energy derived:
?
How do I find drag coefficient without using drag force equation?
What will be the derivation of the drag equation from the continuous equation?
How is the equation
derived?
Can you set the drag force equation equal to the weight (m x g) in order to find the drag coefficient?
What is actually explained in the Lift-Drag polar?
Why the drag formuler is f = bv + CV^2? How can I derive (proof) that?
What is the drag force on a 1.6 m wide, 1.4 m high car traveling at 13 m/s (≈ 29 mph), suppose that the density of air is 1.2 kg/m
?
A car has a drag coefficient C {d} = 0.30C a frontal area of A=1.9 m2 and a mass 1.2 tonnes. The density of air is 1.2 kg.m^ {-3}. What is the drag force when it is traveling at v=110 kph in a straight line?
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11530 | https://angirrami.com/wp-content/uploads/2020/04/Grade-6-Worksheets-Number-Sense.pdf | Number Sense 6-40 7 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-40 Place Value and Decimals Decimals are a way to record place values based on decimal fractions.
253 2 hundreds 5 tens 3 ones 0. 253 2 tenths 5 hundredths 3 thousandths 2 10 + 5 100 + 3 1000 decimal point 1. Write the decimal as a sum of a whole number and decimal fractions.
a) 2.17 = 2 + 1 10 + 7 100 b) 5.37 = + 10 + 100 c) 6.41 = + 10 + 100 d) 8.92 = + 10 + 100 e) 4.24 = + 10 + 100 f) 0.53 = + 10 + 100 g) 2.756 = + 10 + 100 + 1000 h) 3.41 = + 10 + 100 + 1000 i) 9.207 = + 10 + 100 + 1000 j) 8.019 = + 10 + 100 + 1000 2. Write the decimal as a sum of a whole number and decimal fractions. Do not write the fractions with a numerator of 0.
a) 4.017 = 4 + 1 100 + 7 1000 b) 6.305 = + 10 + 1000 c) 2.035 = d) 0.401 = e) 0.005 = f) 3.007 = 3. What is the value of the 9 in the decimal? Write the answer two ways.
a) 0.497 100 or 9 hundredths b) 3.921 10 or 9 c) 8.294 9 or 9 d) 3.159 9 or 9 e) 3.009 or 9 f) 8.913 or 9 g) 0.904 or 9 h) 6.291 or 9 9 9 CA 6.2 AP U9 NS39-47 p1-xx V4.indd 7 2018-10-25 4:02:54 PM 8 Number Sense 6-40 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 4. Write the decimal fractions in the place value chart, then write the number as a decimal.
a) 3 10 = 0. 3 b) 6 10 = 0. Ones Tenths 0 3 Ones Tenths 0 c) 9 10 = 0 . d) 5 10 + 4 100 = . Ones Tenths 0 Ones Tenths Hundredths 0 5 4 e) 3 + 1 10 + 8 100 = 3 . f) 1 10 + 8 100 = . Ones Tenths Hundredths 3 Ones Tenths Hundredths 0 g) 2 10 + 4 100 + 3 1000 = . h) 7 + 3 100 + 5 1000 = . Ones Tenths Hundredths Thousandths 0 2 4 3 Ones Tenths Hundredths Thousandths 7 5. Write the decimal in the place value chart. Ones Tenths Hundredths Thousandths a) 0.512 0 5 1 2 b) 4.67 c) 0.307 d) 2.727 e) 9.02 6. Underline the smallest place value. Write the decimal in words.
a) 0.6 = six tenths b) 0.005 = five c) 0.04 = d) 0.008 = e) 0.006 = f) 0.9 = 7. Put a decimal point in the number so that the digit 7 has the value 7 10 .
a) 5 7 2 b) 1 0 7 c) 2 8 7 5 9 d) 7 1 CA 6.2 AP U9 NS39-47 p1-xx V4.indd 8 2018-10-25 4:02:54 PM 18 Number Sense 6-45 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-45 Subtracting and Adding Decimals 1. Subtract by crossing out the correct number of boxes. Give the answer as a fraction.
a) b) c) d) 50 100 - 30 100 = 38 100 - 12 100 = 69 100 - 34 100 = 57 100 - 25 100 = 2. Write the equations in Question 1 as decimals.
a) 0.50 − 0.30 = 0.20 b) c) d) 3. Subtract the decimals by lining up the decimal points.
a) 0.74 − 0.31 b) 0.56 − 0.24 c) 3.47 − 2.2 d) 6.49 − 0.35 0 7 4 -0 3 1 0 4 3 - - -e) 2.53 − 1.51 f) 3.79 − 2.66 g) 8.84 − 7.10 h) 5.19 − 3.07 - - - -i) 4.08 − 4.04 j) 2.15 − 2.03 k) 5.53 − 2.41 l) 9.83 − 2.71 When subtracting decimals, you may have to regroup.
Example: 5 7 0 -1 2 4 6 10 5 7 0 -1 2 4 6 10 5 7 0 -1 2 4 4 4 6 Regroup 1 tenth as 10 hundredths.
CA 6.2 AP U9 NS39-47 p1-xx V4.indd 18 2018-10-25 4:03:06 PM Number Sense 6-45 19 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 4. Subtract the decimals. Put a decimal point in your answer on the grid.
a) 0.81 - 0.58 b) 5.72 - 3.56 c) 6.15 - 4.2 d) 2.46 - 0.27 7 11 0 8 1 -0 5 8 0 2 3 - - -e) 4.5 - 2.65 f) 31.1 - 22.2 g) 7.45 - 6.68 h) 5.20 - 1.23 5. To calculate the sum, write the decimals as fractions with a common denominator.
a) 0.27 + 0.6 = 27 100 + 6 10 = 27 100 + 100 = 100 = . b) 0.57 + 0.76 = 57 100 + 76 100 = 100 = . c) 2.02 + 0.99 = 100 + 100 = 100 = . 6. Subtract the decimals on grid paper. a) 0.87 - 0.26 b) 9.46 - 3.12 c) 5.83 - 3.69 7. Add or subtract mentally.
a) 0.54 + 0.31 = b) 4.95 - 2.84 = c) 7.09 - 4.02 = d) 2.37 + 1.22 = e) 5.73 - 1.62 = f) 6.73 - 2.53 = g) 6.32 + 2.54 = h) 4.35 - 2.12 = i) 9.47 - 7.46 = 8. What is the diff erence in the thickness of the coins?
a) a quarter (1.58 mm) and a dime (1.22 mm) b) a loonie (1.95 mm) and a toonie (1.75 mm) 9. Sara made fruit drink by mixing 0.37 L of juice with 0.62 L of ginger ale. How many litres of fruit drink did she make?
10. A male Bengal tiger’s body and head are 1.9 m long. The tail is 0.95 m. What is the total length of the Bengal tiger? CA 6.2 AP U9 NS39-47 p1-25 R1.indd 19 2019-05-10 4:24:11 PM 20 Number Sense 6-46 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-46 Money and Decimals 1. Add.
a) $5.45 + $3.23 b) $26.15 + $32.23 c) $19.57 + $30.32 $ 5 4 5 + $ 3 2 3 $ $ + $ $ $ + $ $ 2. Add. You will have to regroup.
a) $ 1 6 6 0 + $ 2 3 7 5 $ b) $ 2 7 4 5 + $ 4 5 1 2 $ c) $ 8 7 4 1 + $ 6 5 2 $ d) $ 3 4 6 0 + $ 2 6 0 0 $ e) $ 3 8 4 0 + $ 4 4 2 5 $ f) $ 1 6 5 2 + $ 4 8 2 5 $ 3. Subtract the amount. You may have to regroup.
a) $ 2 4 5 0 -$ 2 1 7 5 $ b) $ 3 6 4 5 -$ 1 3 8 0 $ c) $ 4 5 2 3 -$ 6 7 2 $ d) $ 5 3 1 4 -$ 1 6 0 3 $ e) $ 7 0 6 2 -$ 2 5 5 1 $ f) $ 8 4 1 7 -$ 3 9 0 9 $ 4. Jasmin bought a pack of socks for $7.25 and a cap for $23.53. How much money does she need to pay the bill?
5. A library spent $270.25 on novels and $389.82 on non-fiction books. How much did the library spend in total? CA 6.2 AP U9 NS39-47 p1-25 R1.indd 20 2019-05-10 4:24:12 PM Number Sense 6-46 21 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 6. Eric bought three shirts that cost $12.30 each. How much did he pay in total?
7. Raj has $25. If he buys a chess game for $9.50 and a book for $10.35, will he have enough money left to buy a second book costing $5.10?
8. The regular price for a pair of eyeglasses is $69.99. Today they are on sale for $10.50 off per pair. If Lela buys her eyeglasses today, how much will she pay? BONUS The seller offered Lela an extra $5.25 off for a second pair of eyeglasses. If Lela wants to buy two pairs of eyeglasses today, how much will she pay in total?
9. Answer the question by looking at the items and their prices below.
a) If you bought a watch and a soccer ball, how much would you pay?
b) Which costs more: a watch and a backpack or pants and a soccer ball?
c) Could you buy a soccer ball, a pair of tennis rackets, and pants for $100?
d) What is the total cost of the three most expensive things in the picture?
e) Make up your own problem using the items.
$28.50 $42.89 $49.95 $15.64 $35.47 $12.30 10. Try to fi nd the answer mentally.
a) How much do 4 loaves of bread cost at $2.30 each?
b) Apples cost 40¢ each. How many could you buy with $3.00?
c) Permanent markers cost $3.10 each. How many could you buy if you had $25.00?
d) Is $10.00 enough to pay for a book costing $4.75 and a pen costing $5.34?
e) Which costs more: 4 apples at 32¢ per apple or 3 oranges at 45¢ per orange?
CA 6.2 AP U9 NS39-47 p1-xx V4.indd 21 2018-10-25 4:03:08 PM 22 Number Sense 6-47 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-47 Estimating Sums and Differences for Decimals 1. Draw an arrow to the 0 or 1.0 to show whether the circled decimal is closer to 0 or 1.0.
a) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 b) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 c) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 d) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2. a) Which decimal numbers are closer to the number? i) 0 ii) 1.0 b) Why is 0.5 a special case? 3. Draw an arrow to show which whole number you would round the circled number to. Then round to the nearest whole number.
a) 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 Round to 1.0 b) 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5.0 Round to 4. Circle the correct answer.
a) 2.9 is closer to: 2.0 or 3.0 b) 1.4 is closer to: 1.0 or 2.0 c) 13.6 is closer to: 13.0 or 14.0 d) 57.2 is closer to: 57.0 or 58.0 5. Draw an arrow to show whether the circled number is closer to 0 or 1.00.
a) 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 b) 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 6. Circle the correct answer.
a) 0.80 is closer to: 0 or 1.00 b) 0.24 is closer to: 0 or 1.00 c) 2.61 is closer to: 2.00 or 3.00 d) 6.45 is closer to: 6.00 or 7.00 CA 6.2 AP U9 NS39-47 p1-xx V4.indd 22 2018-10-25 4:03:08 PM Number Sense 6-47 23 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 7. Draw an arrow to show whether the circled number is closer to 0 or 1.000. 0 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 8. Circle the correct answer.
a) 0.800 is closer to: 0 or 1.000 b) 0.400 is closer to: 0 or 1.000 c) 8.499 is closer to: 8.000 or 9.000 d) 4.507 is closer to: 4.000 or 5.000 9. Draw an arrow to show which whole number you would round the circled number to. 4.000 4.100 4.200 4.300 4.400 4.500 4.600 4.700 4.800 4.900 5.000 4.268 4.723 REMINDER When rounding to the nearest whole number, if the tenth digit is: 0, 1, 2, 3, or 4 you round down. 5, 6, 7, 8, or 9 you round up.
10. Round to the nearest whole number.
a) 2.2 2.0 b) 2.6 3.0 c) 7.3 d) 11.1 e) 30.7 f) 19.6 11. Round to the nearest tenth. Underline the tenths digit first. Then put your pencil on the digit to the right (the hundredths digit). This digit tells you whether to round up or down. a) 1.45 1.50 b) 1.83 c) 3.61 d) 3.42 e) 5.54 f) 6.67 12. Round the decimal to the nearest hundredth. Underline the hundredths digit first. Then put your pencil on the digit to the right (the thousandths digit). a) 2.734 2.730 b) 1.492 c) 3.547 d) 4.270 e) 9.167 f) 5.317 13. Underline the digit you are rounding to. Then circle whether you would round up or down.
a) tenths b) hundredths c) tenths 7 3 2 5 round up round down 6 5 6 3 round up round down 3 8 5 2 round up round down CA 6.2 AP U9 NS39-47 p1-xx V4.indd 23 2018-10-25 4:03:08 PM 24 Number Sense 6-47 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. Round the digit underlined up or down. The digits to the right of the rounded digit become zeros. • To round up, add 1 to the digit. • To round down, keep the digit the same. The digits to the left remain the same. 2 3 4 5 round up (ru) 3 round down (rd) 2 3 4 5 round up (ru) 2 3 0 0 round down (rd) 14. Round to the tenths digit using the steps of rounding from Question 13 and the grey box above.
a) b) c) 3 2 0 1 ru rd 3 5 8 3 5 ru rd 9 4 2 7 1 ru rd Sometimes in rounding, you have to regroup. Example: Round 3.985 to the nearest tenth. 3 9 8 5 10 3 9 8 5 4 0 3 9 8 5 4 0 0 0 Round 9 tenths up to 10 tenths. Regroup the 10 tenths as 1 (ones) Complete the rounding. and add it to the 3 (ones).
15. Round the number to the given digit. Regroup if necessary.
a) 2.195 hundredths b) 5.96 tenths c) 39.897 hundredths ≈ Mathematicians use the symbol to mean “approximately equal to.” 16. Estimate the sum or difference using the whole-number parts of the decimal.
Example: For 14.357 + 0.23 + 5.741, estimate 14 + 0 + 5 = 19.
a) 3.462 + 4.251 ≈ + = b) 7.03 - 5.465 ≈ - = c) 3.2 + 5.1 + 4.6 ≈ + + = d) 9.601 - 3.02 - 4.9 ≈ - - = 17. Estimate by rounding to the nearest tenth. Then add or subtract.
a) 0.42 0.40 + 5.23 + 5.20 5.60 b) 0.28 + 0.54 + c) 2.62 - 0.19 - d) 4.87 - 1.57 - e) 0.73 + 2.17 ≈ 0.70 + 2.20 = 2.90 f) 0.89 - 0.46 ≈ g) 0.63 - 0.26 ≈ h) 3.82 + 2.47 ≈ CA 6.2 AP U9 NS39-47 p1-25 V6.indd 24 2018-11-01 3:29:33 PM Number Sense 6-47 25 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 18. Estimate by rounding to the nearest hundredth. Then add or subtract.
a) 3.223 3.220 + 1.366 + 1.370 4.59 b) 1.347 + 0.632 + c) 5.653 - 3.137 - d) 6.840 - 0.550 -e) 1.347 - 1.213 f) 1.561 + 0.937 g) 2.193 - 0.768 h) 2.714 - 1.656 19. The decimal hundredths that could be rounded to 4.7 are from 4.65 to 4.74. Which decimal hundredths could be rounded to 5.4? Explain how you know.
For Questions 20 to 22, estimate the answer before calculating.
20. Mary wants to buy a pair of shoes for $24.99, a T-shirt for $6.50, and a pair of pants for $19.99. If she has $50 with her, does she have enough money to buy all three items?
21. The planet Mercury is an average distance of 57.9 million kilometres from the Sun. Earth is 149.6 million kilometres from the Sun.
How much farther from the Sun is Earth?
22. The average high temperature last April in Winnipeg, MB was 8.89°C. The average high temperature last April in Toronto, ON, was 3.89°C more than in Winnipeg. What was the average high temperature last April in Toronto?
23. In the 2012 Summer Olympics, the gold-medal throw for shot put was 21.89 m.
The throw that won the silver medal was 21.86 m.
a) Was the diff erence between the throws more or less than 0.1 m?
b) Round both throws to the nearest tenth. What is the diff erence between the rounded amounts?
c) Make up two throws that would round to the same number when rounded to the tenths.
d) Why are Olympic shot put throws measured so precisely?
CA 6.2 AP U9 NS39-47 p1-25 V6.indd 25 2018-11-01 3:29:34 PM 26 Number Sense 6-48 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-48 Multiplying Decimals by Powers of 10 = 1.0 = 0.1 and 10 × = If a hundreds block represents 1 whole, 10 tenths make 1 whole: then a tens block represents 1 tenth (or 0.1). 10 × 0.1 = 1.0 1. Multiply the number of tens blocks by 10. Then show how many hundreds blocks there are to complete the multiplication statement. The first one is done for you.
a) b) 10 × = 10 × = 10 × 0.3 = 10 × 0.2 = c) 10 × = 10 × 0.5 = 2. Multiply by 10 by shifting the decimal point one place to the right.
a) 10 × 0.5 = b) 10 × 0.6 = c) 10 × 1.4 = d) 10 × 2.4 = e) 3.5 × 10 = f) 14.5 × 10 = g) 10 × 2.06 = h) 10 × 2.75 = i) 10 × 97.6 = To convert from centimetres to millimetres, you multiply by 10. There are 10 mm in 1 cm.
1 cm 1 mm = 1 10 cm = 0.1 cm 3. Convert the measurement in centimetres to millimetres.
a) 0.4 cm = mm b) 0.8 cm = mm c) 7.5 cm = mm 4. 10 × 4 can be written as a sum: 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4. Write 10 × 0.4 as a sum, and skip count by 0.4 to find the answer.
5. A dime is a tenth of a dollar (10¢ = $0.10). Draw a picture or use play money to show that 10 × $0.10 = $1.00.
3 5 20.6 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 26 2018-10-30 11:48:47 AM Number Sense 6-48 27 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. = 1.0 = 0.01 and 100 × = If a hundreds block represents 1 whole, then … a ones block represents 1 hundredth (or 0.01), and … 100 hundredths make 1 whole: 100 × 0.01 = 1.0.
6. Write a multiplication statement for the picture.
a) b) 100 × = 100 × = = = The picture shows why the decimal point shifts two places to the right when multiplying by 100: 100 × 100 × 0.12 = 12 100 × 0.1 = 10 100 × 0.02 = 2 7. Multiply by 100. Do your rough work in the grid.
a) 100 × 0.8 = 80 b) 100 × 3.5 = 0 8 8 0 3 5 c) 7.2 × 100 = d) 6.0 × 100 = e) 100 × 0.34 = f) 100 × 0.07 = 8. a) What do 1000 thousandths add up to? b) What is 1000 × 0.001? 9. Look at your answers to Question 8. How many places right does the decimal point shift when you multiply by 1000? BONUS Multiply by shifting the decimal point.
a) 1000 × 0.932 = b) 6.325 × 1000 = c) 1000 × 0.72 = 100 × 0.02 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 27 2018-10-30 11:48:48 AM 28 Number Sense 6-49 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-49 Multiplying and Dividing by Powers of 10 ÷ 10 = ÷ 10 = ÷ 10 = Divide 1 whole into 10 equal parts; each part is 1 tenth. 1.0 ÷ 10 = 0.1 Divide 1 tenth into 10 equal parts; each part is 1 hundredth. 0.1 ÷ 10 = 0.01 Divide 1 whole into 100 equal parts; each part is 1 hundredth. 1.0 ÷ 100 = 0.01 1. Complete the picture and write a division equation.
a) b) ÷ 10 = ÷ 10 = 2.0 ÷ 10 = 0.2 = c) d) ÷ 10 = ÷ 10 = 0.4 ÷ 10 = = e) f) ÷ 10 = ÷ 10 = = 1.1 ÷ 10 = g) h) ÷ 10 = ÷ 10 = = = REMINDER Division can be used to “undo” a multiplication. 6 × 2 12 and 12 ÷ 2 6 2. How do you undo multiplying by 10 or 100?
a) To multiply by 10, I move the decimal point place(s) to the , so to divide by 10, I move the decimal point place(s) to the .
b) To multiply by 100, I move the decimal point place(s) to the , so to divide by 100, I move the decimal point place(s) to the .
CA 6.2 AP U10 NS48-57 p26-47 V6.indd 28 2018-10-30 11:48:48 AM Number Sense 6-49 29 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 3. Shift the decimal point one or two places to the left. Draw an arrow to show a shift. Hint: If there is no decimal point, write it to the right of the number first.
a) b) 0.4 ÷ 10 = 0 4 0 4 .04 or 0.04 0.7 ÷ 10 = 0 7 c) d) 0.6 ÷ 10 = 0 6 3.1 ÷ 10 = 3 1 3 1 0.31 e) f) 26.0 ÷ 10 = 81.4 ÷ 10 = g) h) 25.4 ÷ 10 = 0.32 ÷ 10 = i) j) 0.5 ÷ 100 = 0 5 0 0 0 5 0.005 7 ÷ 100 = 7 0 k) l) 9.1 ÷ 100 = 91 ÷ 100 = 4. a) To multiply by 10, I move the decimal point place(s) to the .
b) To multiply by 1000, I move the decimal point place(s) to the .
c) To divide by 100, I move the decimal point place(s) to the .
d) To divide by 10, I move the decimal point place(s) to the .
e) To divide by 1000, I move the decimal point place(s) to the .
f) To multiply by 100, I move the decimal point place(s) to the .
g) To by 1000, I move the decimal point place(s) to the left.
h) To by 10, I move the decimal point place(s) to the left.
i) To by 100, I move the decimal point place(s) to the right.
j) To by 10, I move the decimal point place(s) to the right.
k) To by 100, I move the decimal point place(s) to the left.
l) To by 1000, I move the decimal point place(s) to the right.
1 right divide 3 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 29 2018-10-30 11:48:48 AM 30 Number Sense 6-49 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 5. Fill in the blanks. Draw arrows to show how you would shift the decimal point. Then write your answer in the grid.
a) 7.325 × 100 b) 4.6 ÷ 100 Move the decimal point places . Move the decimal point places . 7 3 2 5 rough work 7 3 2 5 final answer 4 6 rough work 0 4 6 final answer c) 724.6 ÷ 100 d) 900.03 ÷ 10 Move the decimal point places . Move the decimal point place . 7 2 4 6 rough work final answer 9 0 0 0 3 rough work final answer BONUS e) 0.407 × 100 f) 521.692 × 1000 Move the decimal point places . Move the decimal point places . 4 0 7 rough work final answer 5 2 1 6 9 2 rough work final answer 6. Multiply or divide on grid paper. Show how you shift the decimal point.
a) 3.41 × 1000 b) 5.002 × 100 c) 0.71 × 10 d) 124.05 × 1000 e) 0.52 ÷ 10 f) 800.4 ÷ 100 g) 276.9 ÷ 100 h) 47.02 ÷ 10 i) 0.31 × 100 j) 134.8 ÷ 100 BONUS 0.04027 × 10 000 7. Explain why 1.00 ÷ 10 = 0.1, using a dollar as the whole. 8. A wall 2.5 m wide is painted with 100 stripes of equal width. How wide is each stripe? 9. Find the missing number.
a) 12.3 cm = mm b) 3.412 kg = g c) 1.76 m = cm d) 52.3 km = m 2 right 2 left CA 6.2 AP U10 NS48-57 p26-47 V6.indd 30 2018-10-30 11:48:48 AM Number Sense 6-52 37 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-52 Dividing Decimals by Whole Numbers (Introduction) REMINDER = 1 = 0.1 1. Write the division equation for the base ten model.
a) b) c) d) e) f) 2. Divide by writing the decimal using ones and tenths.
a) 4.8 ÷ 2 b) 6.9 ÷ 3 = ( ones + tenths) ÷ 2 = ( ones + tenths) ÷ 3 = ones + tenths = ones + tenths = = c) 8.4 ÷ 4 d) 8.6 ÷ 2 = ( ones + tenths) ÷ 4 = ( ones + tenths) ÷ 2 = ones + tenth = ones + tenths = = e) 9.6 ÷ 3 BONUS 4.08 ÷ 4 0.4 ÷ 2 = 0.2 4 8 2 4 2.4 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 37 2018-10-30 11:48:51 AM 38 Number Sense 6-52 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 3. Divide the decimal by a whole number by first dividing as if both numbers were whole numbers. Then count the number of decimal digits in the question to put the decimal point in the answer.
a) 48 ÷ 2 = b) 63 ÷ 3 = c) 48 ÷ 4 = so 4.8 ÷ 2 = so 6.3 ÷ 3 = so 4.8 ÷ 4 = d) 246 ÷ 2 = e) 639 ÷ 3 = f) 488 ÷ 4 = so 24.6 ÷ 2 = so 63.9 ÷ 3 = so 48.8 ÷ 4 = Sometimes you need to regroup: 12.6 ÷ 3 = (1 ten + 2 ones + 6 tenths) ÷ 3 = (10 ones + 2 ones + 6 tenths) ÷ 3 = (12 ones + 6 tenths) ÷ 3 = 4 ones + 2 tenths = 4.2 If we divide as if they were whole numbers, we get 126 ÷ 3 = 42: 4 2 3 1 2 6 − 1 2 6 − 6 0 4. The decimal has been divided as if it was a whole number. Count the number of decimal digits to insert the decimal point.
a) 148 ÷ 2 = 74 b) 216 ÷ 3 = 72 c) 364 ÷ 4 = 91 so 14.8 ÷ 2 = so 21.6 ÷ 3 = so 36.4 ÷ 4 = d) 156 ÷ 3 = 52 e) 328 ÷ 8 = 41 f) 459 ÷ 9 = 51 so 15.6 ÷ 3 = so 32.8 ÷ 8 = so 45.9 ÷ 9 = g) 105 ÷ 5 = 21 BONUS 24 608 ÷ 4 = 6152 so 10.5 ÷ 5 = so 2460.8 ÷ 4 = 5. Raj runs 1.8 km in 9 minutes. How far does he run in 1 minute? 6. A row of 4 nickels placed side by side is 84.8 mm long. What is the width of 1 nickel? 24 2.4 123 12.3 7.4 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 38 2018-10-30 11:48:51 AM 46 Number Sense 6-57 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-57 Decimals Review 1. This newspaper article describes how fast Tyrannosaurus rexes grew.
DA IL Y NEWS DA IL Y NEWS a JUMP Publication During rapid grow th spurts, During rapid growth spurts, teenage Tyrannosaurus rexes gained almost 2.1 kg a day.
Scientists have discovered that T. rexes added 2.07 kg a day during a four-year growth spurt between the ages of 14 and 18 years but experienced little or no growth after that. An adult T. rex could weigh up to 5500 kg.
A blue whale gains 90 kg a day for the fi rst six months of its life and can reach 200 000 kg.
a) Two diff erent measures are given for the weight a T. rex could gain in a day.
i) What are the two measures? ii) Which measure is more precise?
iii) Which measure is greater? iv) What is the diff erence between the two?
b) About how many times greater is the weight gain per day for a baby blue whale than for a teen T. rex?
c) A human newborn weighs about 3 kg. If a baby grew as fast as a T. rex, how much would it weigh after 30 days?
2. Draw a picture in the space provided to show 1 tenth of the whole.
a) b) c) 1 whole 1 tenth 1 whole 1 tenth 1 whole 1 tenth 3. Add.
a) 3000 + 200 + 7 + 0.02 = b) 10 000 + 500 + 20 + 0.1 + 0.05 = c) 6000 + 300 + 8 + 0.1 = d) 400 + 7 + 0.02 = 4. Write < or > to show which decimal is greater.
a) 3.7 3.5 b) 2.32 2.37 c) 1.7 1.69 d) 0.5 0.55 CA 6.2 AP U10 NS48-57 p26-47 V6.indd 46 2018-10-30 11:48:53 AM Number Sense 6-57 47 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 5. If you divide a number by 10, the result is 12.9. What is the original number? Explain.
6. Rani lives 2.4 km from the park. She walks to the park and back every day. How many kilometres does she walk to and from the park in a week?
7. John cut 2.73 m off of a 10 m rope. Tom cut off another 4.46 m. How much rope was left?
8. On a three-day canoe trip, Tasha canoed 25.5 km on the first day, 32.6 km on the second, and 17.25 km on the third.
a) How far did she canoe in total?
b) Tasha’s canoe can hold 100 kg. Tasha weighs 45.5 kg, her tent weighs 10.3 kg, and her supplies weigh 14.5 kg. How much more weight can the canoe carry?
9. A teacher has 157.6 mL of sulphuric acid in a bottle, and she wants to divide it equally into four different containers for class assignments. How much sulphuric acid would be in each container?
10. Anna walked 12.6 m in 20 steps. How many metres was each step?
11. Jax had $25.00. He bought a taco for $3.21, a banana for $1.37, a carton of milk for $1.56, and a video game for $15.87. How much money does he have left? BONUS a) Luc earned $28.35 on Monday. On Thursday, he spent $17.52 on a shirt. He now has $32.23. How much money did he have before he started work Monday? Hint: Work backwards. How much money did he have before he bought the shirt? b) Sun spent half of her money on a book. Then she spent $1.25 on a pen. She has $3.20 left. How much did she start with?
CA 6.2 AP U10 NS48-57 p26-47 V6.indd 47 2018-10-30 11:48:54 AM 48 Geometry 6-13 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. G6-13 Translations Josh slides a dot from one position to another. To move the dot from position 1 to position 2, Josh slides the dot 4 units right. In mathematics, slides are called translations.
1 1 2 3 4 2 1. How many units right did the dot slide from position 1 to position 2?
a) b) c) 1 2 1 2 1 2 units right 2. How many units left did the dot slide from position 1 to position 2?
a) b) c) 2 1 2 1 2 1 units left 3. Follow the instructions to translate the dot to a new position.
a) 3 units right b) 4 units left c) 5 units right L R 4. Describe the translation of the dot from position 1 to position 2.
a) b) c) 2 1 2 1 2 1 units right units right units right units down units down unit down 5. Translate the dot.
a) 5 units right, 2 units down b) 4 units left, 2 units up c) 3 units left, 4 units down CA 6.2 AP U11 G13-20 p48-69 V5.indd 48 2018-11-01 10:52:49 AM Geometry 6-13 49 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. The result of a translation is called the image under translation. You can use the prime symbol (′) to label the image. Example: The image of P under translation is P′. P P′ 6. a) Use a ruler and protractor to measure the sides and the angles of the triangle.
i) A B C ii) F E D AB = mm ∠A = DE = mm ∠D = AC = mm ∠B = EF = mm ∠E = BC = mm ∠C = DF = mm ∠F = b) Translate the triangle by translating the vertices. Use ′ to label the images of the vertices.
i) 5 units right and 2 units down ii) 4 units left and 1 unit up c) Measure the sides and the angles of the image.
i) A′B′ = mm ∠A′ = ii) D′E′ = mm ∠D′ = A′C′ = mm ∠B′ = E′F′ = mm ∠E′ = B′C′ = mm ∠C′ = D′F′ = mm ∠F′ = d) What do you notice about the sides and angles of the triangles and their images? 7. True or false? If the statement is true, explain why. If the statement is false, draw an example to show it is not true.
a) A triangle and its image under translation are congruent. Bonus If two triangles are congruent, there is always a translation that takes one of them onto the other. CA 6.2 AP U11 G13-20 p48-69 V5.indd 49 2018-11-01 10:52:49 AM 50 Geometry 6-13 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 8. a) Translate triangle T as given. Label the image T′. Then translate the image again from T′ to T.
i) 2 units up and 3 units left, then ii) 4 units down and 3 units right, then 1 unit up and 5 units right 3 units up and 4 units left T T b) Draw arrows joining the corresponding vertices of triangles T and T. What do you notice about the direction of the arrows? c) Measure the arrows in millimetres. What do you notice about the length of the arrows? d) Can you use one translation to take triangle T to T? If yes, describe the translation.
i) units and ii) unit and units unit 9. a) Draw a quadrilateral that is not a rectangle in the shaded zone on the grid. Label it Q.
b) Predict the result of combining two translations: Q to Q′: 6 units right and 3 units down Q′ to Q: 4 units left and 4 units down Q to Q: units and units c) Translate Q to Q′ and Q′ to Q to check your prediction. Was your prediction correct? 10. Jax thinks translating a shape 3 units up and 4 units left, then 4 units right and 3 units down results in the original shape. Is he correct? Explain.
CA 6.2 AP U11 G13-20 p48-69 V5.indd 50 2018-11-01 10:52:50 AM Geometry 6-14 51 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. G6-14 Reflections To reflect a point P in a mirror line m: Step 1: Draw a line through P perpendicular to m. Extend it beyond m.
Step 2: Measure the distance from P to m along the perpendicular.
Step 3: Mark the point P′ on the perpendicular on the other side of m so that P and P′ are the same distance from the mirror line m.
Point P′ is the mirror image of P. Mathematicians say that P′ is the image of P under reflection in the line m.
P m P′ 1. Count the grid squares to reflect point A in the given line.
a) m A b) m A c) A m To reflect a shape in a mirror line, reflect the shape’s vertices and then join the images of the vertices.
2. a) Use a ruler and protractor to measure the sides and the angles of the triangle.
i) A B C ii) E F D AB = mm ∠A = DE = mm ∠D = AC = mm ∠B = EF = mm ∠E = BC = mm ∠C = DF = mm ∠F = b) Reflect each triangle in the given line. Use ′ to label the images of the vertices.
c) Measure the sides and the angles of each image.
i) A′B′ = mm ∠A′ = ii) D′E′ = mm ∠D′ = A′C′ = mm ∠B′ = E′F′ = mm ∠E′ = B′C′ = mm ∠C′ = D′F′ = mm ∠F′ = d) What do you notice about the sides and the angles of each triangle and its image? Do reflections take triangles to congruent triangles? CA 6.2 AP U11 G13-20 p48-69 V5.indd 51 2018-11-01 10:52:50 AM 52 Geometry 6-14 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 3. a) Reflect the polygon in the given mirror line.
i) ii) b) Draw a line segment between each vertex in part a) and its image. What do you notice about the line segments? The midpoint of a line segment is the point halfway between the end points of the line segment.
midpoint c) On the grids above, mark the midpoints of the line segments you drew in part b). What do you notice about the midpoints? The shapes ABC and A′B′C′ are mirror images of each other when: • line segments between each vertex and its possible image are parallel; and • all the midpoints of these line segments fall on the same perpendicular line.
Note: The line segments between the vertices have different lengths.
B C C′ A B′ A′ 4. a) Draw line segments between the vertices of the shape and their images.
i) ii) Bonus b) Find the midpoint of each line segment you drew in part a). Are the midpoints on the same line?
c) Are the shapes reflections of each other? How do you know? Bonus If your answer in part c) was “no” for any pair of shapes, identify the transformation that takes one shape into the other.
CA 6.2 AP U11 G13-20 p48-69 V5.indd 52 2018-11-01 10:52:50 AM Geometry 6-14 53 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 5. Fill in the table to summarize what happens to a shape that is reflected. What happens when a shape is translated? Transformation Lengths of Sides Sizes of Angles Orientation Reflection Translation 6. a) Reflect triangle T in the mirror line. Label the image T′.
i) ii) iii) T T T b) Translate T′ as given. Label the image T.
i) 3 units down ii) 4 units right iii) 3 units up and 2 units right c) Draw the line segments joining each vertex in T to its image in T. Are the line segments parallel?
i) ii) iii) d) Are the line segments you drew in part c) equal?
i) ii) iii) e) If possible, draw the translation arrow or the mirror line from T to T.
f) Are triangles T and T congruent? How do you know?
7. a) Reflect the trapezoid R in line ℓ. Label the image R′.
i) ii) BONUS ℓ R m m R ℓ R m ℓ b) Reflect R′ in line m. Label the image R.
c) Is there a reflection or a translation that takes R to R? If yes, describe it.
CA 6.2 AP U11 G13-20 p48-69 V5.indd 53 2018-11-01 10:52:50 AM 54 Geometry 6-15 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. G6-15 Rotations 1. From the dark arrow, draw an arc showing the direction of the given 90° turn. Draw the arrow after turning.
a) clockwise b) counter-clockwise c) clockwise d) counter-clockwise To rotate point P around point O 90° clockwise: Step 1: Draw line segment OP. Measure its length. Step 2: Draw an arc clockwise to show the direction of rotation. Step 3: Place a set square so that: • the arc points at the diagonal side, • the right angle is at point O, and • one arm of the right angle aligns with OP.
O P not like this O P like this Step 4: Draw a ray from point O along the side of the square corner. O P Step 5: On the new ray, measure and mark the image point P′ so that OP′ = OP. O P P′ 2. Rotate point P 90° around point O in the direction given. Label the image P′.
a) clockwise b) counter-clockwise c) clockwise d) counter-clockwise P O P O P O P O 3. Is point P′ in Question 2 always on a grid line intersection? If not, fix your mistake.
CA 6.2 AP U11 G13-20 p48-69 V5.indd 54 2018-11-01 10:52:50 AM Geometry 6-15 55 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. To rotate a shape around point O, rotate the shape’s vertices and join the images of the vertices.
The point O is called the centre of rotation. The centre of rotation can be outside, inside, or on a side of the shape. The centre of rotation is the only fixed point during a rotation; it does not move.
4. a) Measure the sides and the angles of the triangle.
i) C B A O ii) E D O AB = ∠A = DE = ∠D = AC = ∠B = EO = ∠E = BC = ∠C = DO = ∠O = b) Rotate the triangle 90° counter-clockwise around point O. Use ′ to label the vertices of the image.
c) Measure the sides and the angles of the image.
i) A′B′ = ∠A′ = ii) D′E′ = ∠D′ = A′C′ = ∠B′ = E′O = ∠E′ = B′C′ = ∠C′ = D′O = ∠O = d) What do you notice about the sides and the angles of each triangle and its image? Does rotation take polygons to congruent polygons? 5. True or false? If the statement is true, explain why. If the statement is false, draw an example showing it is false.
a) A polygon and its image under rotation are congruent.
b) If two polygons are congruent, there is always a rotation that takes one polygon onto the other.
6. Fill in the table to summarize. What happens to a polygon that is reflected? Translated? Rotated?
Transformation Lengths of Sides Sizes of Angles Orientation Reflection Translation Rotation CA 6.2 AP U11 G13-20 p48-69 V5.indd 55 2018-11-01 10:52:51 AM 56 Geometry 6-15 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. You can rotate a triangle 90° using a grid instead of a set square.
Triangle OED has a horizontal side 2 units long and a vertical side 3 units long.
Rotations take triangles to congruent triangles. A rotation of 90° takes horizontal lines to vertical lines and vertical lines to horizontal lines.
Triangle OE′D′ has a horizontal side 3 units long and a vertical side 2 units long.
E′ D′ E D O 7. Rotate the triangle 90° counter-clockwise around point O. Start with the side marked by an arrow. Hint: Note the direction fi rst.
a) O b) O c) O d) O To rotate a point on a grid 90° clockwise around the point O: step 1: Draw line segment OP.
step 2: Shade a right triangle with OP as one side.
step 3: Rotate the triangle 90° clockwise around O.
step 4: Mark the image point.
P O P′ P O 8. Imagine the triangles to rotate the vertices of the polygon around the point O.
Join the vertices to create the image of the polygon.
a) 90° clockwise b) 90° counter-clockwise O O Bonus Use a ruler to draw a scalene obtuse triangle ABC. Find the midpoint of side AC and label it M. Rotate triangle ABC 180° clockwise around point M.
What type of quadrilateral do triangle ABC and its image make together? Explain.
CA 6.2 AP U11 G13-20 p48-69 V5.indd 56 2018-11-01 10:52:51 AM Geometry 6-16 57 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. G6-16 More Rotations To rotate point P around point O 180° clockwise: Step 1: Draw line segment OP. Measure its length.
Step 2: Extend OP beyond point O.
Step 3: Mark the point P′ so that OP′ = OP.
O P′ P 180° 1. Triangle A′OB′ is the image of triangle AOB under a 180° clockwise rotation around point O.
a) Triangle AOB has a horizontal side units long and a vertical side units long. Triangle A′OB′ has a horizontal side units long and a vertical side units long.
b) Write “horizontal” or “vertical” to complete the sentence. A 180° rotation clockwise or counter-clockwise takes horizontal lines to lines and vertical lines to lines. O A B A′ B′ BONUS Explain why a rotation of 180° clockwise produces the same result as a rotation 180° counter-clockwise around the same centre. 2. Rotate the triangle 180° clockwise or counter-clockwise around point O. Start with a horizontal or a vertical side.
a) O b) O c) O d) O 3. Rotate the vertices of the polygon 180° clockwise around point M. Join the vertices to create the image of the polygon.
a) b) M M CA 6.2 AP U11 G13-20 p48-69 R1.indd 57 2019-05-10 4:38:35 PM 58 Geometry 6-16 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 4. a) Rotate polygon M 90° clockwise around point O. Label the image M′.
i) M O ii) M O b) Rotate polygon M′ 90° clockwise around point O. Label the image M.
c) Which rotation around point O takes polygon M to polygon M? 5. How much did the thick arrow turn? Write “90°,” “180°,” or “270°.” a) b) c) d) clockwise clockwise clockwise clockwise 6. How did the thick arrow turn? Use CW for clockwise and CCW for counter-clockwise.
a) b) c) d) 7. Was the grey shape rotated 90° CW, 90° CCW, or 180° CW or CCW to get the white shape? Write the amount and direction of rotation.
a) b) c) d) 270° 90°CCW CA 6.2 AP U11 G13-20 p48-69 R1.indd 58 2019-05-10 4:38:35 PM Geometry 6-16 59 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 8. Shape B is the image of Shape A under rotation. Mark the centre of rotation and describe the rotation.
a) b) c) d) B A A B A B A B e) f) g) Bonus B A B A B A A B 9. Dory rotates point N around point O as given. What is the image point?
a) 90° CW, then another 90°CW: b) 90° CW, then 180°CW: c) 180° CCW, then another 180°CCW: d) 180° CW, then 90°CW: e) 90° CW, then 90°CCW: N W O S E 10. a) Reflect triangle T in line ℓ . Label the image T′.
b) Reflect T′ in the line m. Label the image T.
c) Reflect T in the line m. Label the image T″.
d) Reflect T″ in the line ℓ . Label the image T.
e) What do you notice about T and T? f) Which transformation takes T to T? Draw the ℓ m T translation arrow, the mirror line, or the centre of rotation and describe the transformation. 90°CW CA 6.2 AP U11 G13-20 p48-69 V5.indd 59 2018-11-01 10:52:52 AM Patterns and Algebra 6-11 77 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-11 Solving Equations—Preserving Equality 1. Write the number that makes the equation true.
a) 8 + 4 − = 8 b) 8 × 3 ÷ = 8 c) 8 ÷ 2 × = 8 d) 12 ÷ 4 × = 12 e) 13 − 6 + = 13 f) 19 + 3 − = 19 2. Write the operation that makes the equation true.
a) 7 + 2 2 = 7 b) 8 × 3 3 = 8 c) 12 ÷ 2 2 = 12 d) 15 − 4 4 = 15 e) 18 ÷ 3 3 = 18 f) 6 + 4 4 = 6 3. Write the operation and number that make the equation true.
a) 17 + 3 − 3 = 17 b) 20 ÷ 4 = 20 c) 18 × 2 = 18 d) 11 − 4 = 11 e) 4 × 3 = 4 f) 15 + 2 = 15 g) 5 × 2 = 5 h) 5 ÷ 2 = 5 i) 5 − 2 = 5 j) n + 3 − 3 = n k) n × 3 = n l) 5m = m m) x − 5 = x n) x + 7 = x o) z ÷ 5 = z REMINDER The variable x represents a number, so you can treat it like a number.
Operation Result Operation Result Add 3 to x. x + 3 Multiply 3 by x. 3 × x (or 3x) Add x to 3. 3 + x Multiply x by 3. x × 3 (or 3x) Subtract 3 from x. x − 3 Divide x by 3. x ÷ 3 Subtract x from 3. 3 − x Divide 3 by x. 3 ÷ x 4. Show the result of the operation.
a) Multiply x by 7. b) Add 4 to x. c) Subtract 5 from x. d) Subtract x from 5. e) Divide x by 10. f) Divide 9 by x. g) Multiply 8 by x. h) Add x to 9. BONUS Add x to y. 5. How could you undo the operation and get back to the number you started with?
a) Add 4. b) Multiply by 3. c) Subtract 9. d) Divide by 2. e) Add 7. f) Multiply by 5. g) Multiply by 2. h) Divide by 8. i) Subtract x. 7x x + 4 subtract 4 CA 6.2 AP U12 PA9-20 p70-97 V6.indd 77 2018-10-26 3:25:17 PM 78 Patterns and Algebra 6-11 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 6. Solve for x by doing the same thing to both sides of the equation. Check your answer.
a) 3x = 12 b) x ÷ 6 = 3 c) x − 4 = 20 3x ÷ 3 = 12 ÷ 3 x ÷ 6 × 6 = 3 × 6 x = 4 Check by replacing x with your answer: 3(4) = 12 d) x ÷ 3 = 5 e) 12 + x = 22 f) 44 = 4x g) x − 17 = 25 h) 31 = 19 + x i) x + 26 = 53 j) 11 = x ÷ 5 k) 9x = 63 BONUS x + 9 = 9 + 45 Lela solves 7 − x = 5 in two steps.
Step 1: She treats x as a number and adds x to both sides: 7 − x + x = 5 + x 7 = 5 + x Step 2: She subtracts 5 from both sides to find x: 7 − 5 = 5 + x − 5 2 = x Lela checks her answer. She replaces x in the equation with 2: 7 − 2 = 5 7. Solve the equation in two steps like Lela. Check your answer.
a) 12 − x = 6 b) 6 = 13 − x c) 24 − x = 20 12 − x + x = 6 + x 12 = 6 + x 12 − 6 = 6 + x − 6 6 = x Check by replacing x with your answer: 12 − 6 = 6 d) 3 = 15 − x e) 59 − x = 56 f) 26 = 43 − x g) 31 − x = 11 h) 73 − x = 41 i) 17 − x = 17 CA 6.2 AP U12 PA9-20 p70-97 V6.indd 78 2018-10-26 3:25:17 PM Patterns and Algebra 6-12 79 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-12 Solving Equations—Using Logic To solve the equation x + 3 = 8, Mike and Jill use different methods.
Mike uses preserving equality: x + 3 = 8 x + 3 − 3 = 8 − 3 x = 5 Jill uses logic. She thinks about how addition and subtraction are related: x + 3 = 8 means I have to add 3 to x to get 8.
x 8 So, I have to subtract 3 from 8 to find x.
x 8 x = 8 − 3 = 5 1. Use Jill’s method to solve the equation.
a) x + 5 = 12 b) x + 3 = 10 c) x + 25 = 41 d) 21 + x = 34 x = 12 − 5 x = 7 e) 28 = 8 + x f) 41 = x + 14 g) 17 + x = 56 h) x + 22 = 33 i) 16 + x = 34 j) x + 35 = 61 k) 6 + x = 100 l) 5 + x + 2 = 18 Mike and Jill solve the equation x − 2 = 5.
Mike uses preserving equality: x − 2 = 5 x − 2 + 2 = 5 + 2 x = 7 Jill uses logic: x − 2 = 5 means I have to subtract 2 from x to get 5. So, I have to add 2 to 5 to find x.
x = 5 + 2 = 7 2. Use Jill’s method to solve the equation.
a) x − 5 = 12 b) x − 12 = 5 c) 26 = x − 3 d) x − 19 = 9 x = 12 + 5 x = 17 e) x − 7 = 28 f) x − 13 = 22 g) 14 = x − 27 h) 29 = x − 32 i) x − 15 = 62 j) 43 = x − 19 k) x − 51 = 49 l) 73 = x − 21 CA 6.2 AP U12 PA9-20 p70-97 R1.indd 79 2019-05-15 11:51:58 AM 80 Patterns and Algebra 6-12 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. REMINDER Division is often written in fractional form. Examples: 12 ÷ 4 = 12 4 15 ÷ 5 = 15 5 x ÷ 3 = x 3 w ÷ 7 = w 7 3. Solve the division problem.
a) 6 3 = 2 b) 12 6 = c) 12 4 = d) 15 5 = Mike and Jill solve the equation 3x = 12.
Mike uses preserving equality: 3x = 12 3x ÷ 3 = 12 ÷ 3 x = 4 Jill uses logic: 3x = 12 means I have to multiply x by 3 to get 12.
So, I have to divide 12 by 3 to find x.
x = 12 ÷ 3 = 4 4. Use Mike’s method to solve the equation by preserving equality.
a) 4x = 12 b) 2x = 10 c) 6x = 42 d) 2x = 14 4x ÷ 4 = 12 ÷ 4 x = 3 e) 7x = 28 f) 6x = 18 g) 7x = 49 h) 8x = 48 Mike and Jill solve the equation x ÷ 3 = 8.
Mike uses preserving equality: x ÷ 3 = 8 x ÷ 3 × 3 = 8 × 3 x = 24 Jill uses logic: x ÷ 3 = 8 means I have to divide x by 3 to get 8.
So, I have to multiply 8 by 3 to find x.
x = 8 × 3, so x = 24 5. Solve the equation using logic.
a) x ÷ 2 = 3 b) 2x = 8 c) x ÷ 4 = 5 d) 3 + x = 8 e) x − 5 = 6 x = 3 × 2 x = 6 f) x ÷ 3 = 4 g) 5 + x = 12 h) 12 = 2x i) 15 = 3x j) 4 = x ÷ 3 k) x ÷ 7 = 4 l) x ÷ 4 = 7 m) 3x = 27 n) 36 = 12x BONUS x 3 = 5 CA 6.2 AP U12 PA9-20 p70-97 V6.indd 80 2018-10-26 3:25:21 PM Patterns and Algebra 6-13 81 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-13 Solving Equations—Guess and Check Finding the value of a variable that makes an equation true is called solving for the variable. Arsham uses a table to solve 2x + 1 = 7.
So x = 3 makes the equation true. x 2x + 1 Is the equation true?
1 3 2 5 3 7 1. Complete the table, then solve for x.
a) 3x + 2 = 14 b) 4x + 3 = 23 c) 5x − 2 = 13 x 3x + 2 True?
1 3(1) + 2 = 5 2 3(2) + 2 = 8 x 4x + 3 True?
1 4(1) + 3 = 7 x 5x − 2 True? so x = so x = so x = 2. Replace n with 5 and say whether 5 is too high or too low. Then try a lower or higher number.
a) 3n + 2 = 20 b) 5n + 1 = 21 c) 2n + 3 = 15 n 3n + 2 Answer 5 3(5) + 2 17 n 5n + 1 Answer 5 5(5) + 1 n 2n + 3 Answer 5 5 is 5 is 5 is d) 4n + 3 = 27 e) 5n − 6 = 14 f) 3n − 3 = 15 n 4n + 3 Answer 5 n 5n − 6 Answer 5 5(5) − 6 n 3n − 3 Answer 5 5 is 5 is 5 is 3. Solve for n by guessing small values, checking, and revising.
a) 3n + 2 = 14 b) 5n − 2 = 13 c) 4n − 1 = 15 d) 6n − 5 = 31 e) 7n − 2 = 19 f) 2n + 3 = 9 too low CA 6.2 AP U12 PA9-20 p70-97 V7.indd 81 2018-11-01 10:59:43 AM 82 Patterns and Algebra 6-14 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-14 Word Problems—Addition and Subtraction Equations 1. Fill in the table. Write x for the number you need to fi nd. Cross out the cell you do not use.
Problem Parts How Many?
Diff erence Equation and Solution Total a) Ethan has 2 dogs and 5 fi sh. How many pets does he have?
dogs 2 Diff erence: 2 + 5 = x x = 7 fi sh 5 Total: x b) Sharon hiked 13 km on Saturday. She hiked 14 km on Sunday. How far did Sharon hike in two days?
Diff erence: Total: c) Lucy saved $43 in January. She saved $14 less in February than in January. How much money did she save in February?
Diff erence: Total: d) The Leviathan roller coaster in Canada is 93 m tall. It is 46 m shorter than the Kingda Ka roller coaster in the United States. How tall is Kingda Ka? Diff erence: Total: e) A supermarket sold 473 bags of white and yellow potatoes. If 139 of the bags were fi lled with white potatoes, how many bags of yellow potatoes were sold?
Diff erence: Total: 2. Write the parts and how many of each part. Then write and solve an equation.
a) Clara watched TV for 45 minutes. She spent 15 minutes less on her homework than on watching TV. How much time did she spend on homework?
b) A recreation pass costs $24. It is $9 more than a movie pass. How much do the two passes cost together?
c) The Mercury City Tower in Moscow is 339 m tall. The CN Tower in Toronto is 553 m tall. How much taller is the CN Tower than the Mercury City Tower? Mercury City Tower CN Tower CA 6.2 AP U12 PA9-20 p70-97 V6.indd 82 2018-10-26 3:25:22 PM Patterns and Algebra 6-14 83 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 3. Solve the problem using an equation for each part. Use your answer from part i) as data for part ii).
a) Alex read for 30 minutes before dinner and 45 minutes after dinner.
i) How many minutes did he spend reading altogether?
ii) Alex’s dinner took 30 minutes. If he finished his after-dinner reading at 7:50 p.m., when did Alex start eating dinner?
b) There are 18 players on a soccer team. Seven of them are reserve players and the rest are field players.
i) How many field players are on the team?
ii) How many more field players than reserve players are on the team? 4. Solve the two-step problem by writing equations.
a) Mary bought 16 red stickers and 25 blue stickers. She used 13 of them. How many stickers does she have left?
b) There are 28 students in a sixth grade class. Thirteen of them don’t wear glasses. How many more students wear glasses than don’t wear glasses?
c) Shawn read 7 mysteries. He read 3 more science fiction books than mysteries. How many books did he read altogether?
d) Ava had $75. She spent $12 on two shirts, $32 on shoes, and $25 on a jacket. Does she have enough money to buy a pair of pants for $14?
5. There are 23 500 houses and 12 700 apartments in a town. Use equations to answer the question.
a) How many houses and apartments are there in total?
b) How many more houses are there than apartments?
c) The town plans to tear down 750 houses and replace them with 2400 apartments. How many more houses than apartments will there be? BONUS The table shows Sun’s savings account balances from June to August. She did not withdraw money from her savings account.
End of June $237.57 End of July $352.24 End of August $528.06 a) How much did she deposit in July?
b) How much did she deposit in July and August altogether?
c) How much more did Sun deposit in August than in July? d) Sun wants to buy a computer for $699.98 by the end of September. Her father told her that he will pay the tax. How much does Sun need to save in September to be able to buy the computer?
CA 6.2 AP U12 PA9-20 p70-97 V6.indd 83 2018-10-26 3:25:22 PM 84 Patterns and Algebra 6-15 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-15 Word Problems—Multiplication Equations Smaller part: Larger part: The larger part is 3 times the size of the smaller part.
The scale factor is 3.
You can write an equation to fi nd one part from the other: Larger part = scale factor × smaller part 1. Circle the larger thing or quantity. Underline the smaller thing or quantity.
a) A high rise is seven times as tall as a house. b) There are fi ve times as many apples as pears.
c) There are four times as many cats as dogs. d) Ed's wallet is one-sixth times as heavy as his suitcase.
e) A kitten is four times as big as a mouse. f) A bus holds ten times as many people as a car.
2. Write an equation to fi nd the answer. Use x for the unknown amount.
a) Jen has 6 times as many stamps as Dan. Jen has 24 stamps.
How many stamps does Dan have?
Larger amount: 24 Smaller amount: x Equation: 24 Larger part = 6 Scale factor × x Smaller part b) A cherry is 10 times as light as an apple. An apple weighs 90 grams.
How much does the cherry weigh?
Larger amount: Smaller amount: Equation: Larger part = Scale factor × Smaller part c) A tablet costs $225. A computer costs three times as much.
How much does the computer cost? BONUS Lara is one tenth as old as Amir. Lara is 5 years old. How old is Amir?
3. Write and solve an equation for the problem.
a) Carl planted 8 times as many tomato plants as rose bushes.
He planted 32 rose bushes. How many tomato plants did Carl plant?
b) A whale shark is fi ve times as long as a great white shark. A whale shark is 20 metres long. How long is the great white shark?
c) A table is four times as heavy as a chair. The table weighs 220 kg.
How much does the chair weigh?
d) A male Nile crocodile weighs 620 kg, four times as much as a female American alligator. How much does the female American alligator weigh?
number of Jen’s stamps number of Dan’s stamps CA 6.2 AP U12 PA9-20 p70-97 V6.indd 84 2018-10-26 3:25:22 PM Patterns and Algebra 6-15 85 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. REMINDER Total number of things = number of sets × number in each set 4. Fill in the table. Use x for the unknown.
Total Number of Things Number of Sets Number in Each Set Equation a) 40 pictures 8 pictures on each page 40 x 8 40 = 8x b) 30 people 5 vans c) 24 flowers 6 pots d) 4 chairs at each table 11 tables e) 50 houses 10 houses on each block f) 9 boxes 22 pencils in each box 5. Solve each equation in Question 4.
6. Write and solve an equation for the problem.
a) A train has 10 cars and 1960 seats. How many seats are in each car?
b) A parking lot has 12 equal rows and 492 parking spots. How many cars can park in each row?
c) A maple tree is 10 m tall. A pine tree is 3 times as tall as the maple tree. How tall is the pine tree?
d) A board game costs 3 times as much as a soft toy. The board game costs $19.50. How much does the soft toy cost?
e) Ben is twice as old as Ella. Ben is 12 years old. How old is Ella?
7. Solve the problem by writing an equation.
a) Jane has 7 stickers. Mark has 5 times as many stickers as Jane. How many stickers do they have altogether?
b) There are 4 times as many people in City A as in City B. There are 257 301 people in City B. How many people are in City A?
c) The planet Uranus is about 2.871 billion kilometres from the sun. Uranus is twice as far from the sun as the planet Saturn. Imagine that the sun, Saturn, and Uranus form a straight line in that order. i) How far from the sun is Saturn?
ii) How far is Uranus from Saturn?
CA 6.2 AP U12 PA9-20 p70-97 V6.indd 85 2018-10-26 3:25:22 PM Patterns and Algebra 6-19 93 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. PA6-19 Formulas for Tables 1. Use the gap to complete the table.
a) Input Output 1 3 +2 2 +2 3 +2 4 b) Input Output 1 4 +2 2 +2 3 +2 4 c) Input Output 1 15 -3 2 -3 3 -3 4 d) Input Output 1 1 +4 2 +4 3 +4 4 2. a) Use the rule to fill in the table. Find the gap between outputs.
i) Rule: Multiply by 3 and add 1 ii) Rule: Multiply by 0.5 and add 2 Input Output 1 2 3 4 Input Output 1 2 3 4 gap = gap = iii) Rule: Multiply by 2 and add 5 iv) Rule: Multiply by 0 and add 4 Input Output 1 2 3 4 Input Output 1 2 3 4 gap = gap = b) Compare the gap and the rule for each table in part a). What do you notice? CA 6.2 AP U12 PA9-20 p70-97 V6.indd 93 2018-10-26 3:25:24 PM 94 Patterns and Algebra 6-19 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. To find the rule for how to get the output from the input: Step 1: Find the gap between the numbers in the Output column.
Input (n) n × gap Output 1 5 +3 2 8 +3 3 11 Step 2: Multiply each input by the gap. Input (n) n × gap Output 1 3 5 +3 2 6 8 +3 3 9 11 Step 3: What must you add (or subtract) to each number in the second column to get the output?
Input (n) n × gap Output 1 3 5 +3 2 6 8 +3 3 9 11 Add 2 Step 4: Write the rule for the table. Rule: Multiply the input by 3, then add 2.
3. Use the steps above to find the rule for the table.
a) Input (n) n × gap Output 1 11 2 15 3 19 Add b) Input (n) n × gap Output 1 5 2 7 3 9 Add Multiply by , then add . Multiply by , then add .
c) Input (n) n × gap Output 1 2 2 5 3 8 Subtract d) Input (n) n × gap Output 1 4 2 10 3 16 Subtract Multiply by , then subtract . Multiply by , then subtract .
4. Make a table for the number of blocks in each figure and then find a formula for the pattern. Figure 1 Figure 2 Figure 3 Input (n) n × gap Output 1 1 2 3 3 Formula: CA 6.2 AP U12 PA9-20 p70-97 V6.indd 94 2018-10-26 3:25:24 PM Patterns and Algebra 6-19 95 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 5. Use the gap in the sequence to complete the table and then find the formula.
a) Input (n) n × gap Output 1 5 2 +5 2 10 7 +5 3 15 12 +5 4 20 17 -3 b) Input (n) n × gap Output 1 1 2 4 3 7 4 10 Formula: 5n − 3 Formula: c) Input (n) n × gap Output 1 3 2 7 3 11 4 15 d) Input (n) n × gap Output 1 9 2 15 3 21 4 27 Formula: Formula: Once you know the gap, you really only need to see how the first term compares to the gap in order to find the formula.
Example: In Question 5.a), the gap is 5 and the first output is 2, so the formula is 5n − 3.
6. Find the gaps in the sequence. Then write the formula for the sequence.
a) 3 , 5 , 7 , 9 b) 9 , 20 , 31 , 42 c) 12 , 17 , 22 , 27 Formula: Formula: Formula: d) 2, 3.5, 5, 6.5 e) 53, 55, 57, 59 BONUS 1 5 , 2 5 , 3 5 , 4 5 7. a) Write a formula for the number of toothpicks in the sequence shown below: Figure 1 Figure 2 Figure 3 Figure 4 b) Use your formula to determine the number of toothpicks in the 30th figure.
CA 6.2 AP U12 PA9-20 p70-97 V6.indd 95 2018-10-26 3:25:26 PM 100 Measurement 6-9 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. ME6-9 Area and Perimeter 1. a) Measure the length and the width of each rectangle in centimetres. Find the perimeter and area of each rectangle. Write the answers in the table. F E D C B 5 cm 3 cm A Shape Perimeter Area A (2 × 3 cm) + (2 × 5 cm) = 16 cm 3 cm × 5 cm = 15 cm2 B C D E F b) Shape E has a greater perimeter than Shape A. Does it also have a greater area? c) Name two rectangles that have the same perimeter and different areas. and d) Write the shapes in order from greatest to least perimeter. e) Write the shapes in order from greatest to least area. f) Are the orders in parts d) and e) the same? g) Alice thinks that a rectangle with larger area always has a larger perimeter. Is she correct? Explain. h) Tristan thinks that a rectangle with larger perimeter always has a larger area. Is he correct? Explain. CA 6.2 U13 AP V7.indd 100 2018-10-30 10:55:01 AM Measurement 6-9 101 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 2. a) Write an equation for the area of a rectangle with length 4 units and width 3 units. b) Write another pair of numbers that multiply to 12. × = 12 c) Draw a rectangle on the grid with length and width equal to your numbers. d) What is the perimeter of the rectangle in part a) and the rectangle in part c)? Are they the same? 3. a) Fill in the first two columns of the table to find all rectangles with perimeter 12 units and sides with lengths that are whole numbers.
b) Find the area of the rectangles to finish filling in the table.
c) Draw two rectangles on the grid with the same perimeter and different areas. Length Width Area 1 5 2 4. Each square on the grid is 1 cm long.
a) Draw a square with perimeter 12 cm. What is its area? b) Draw a square with area 16 cm2. What is its perimeter? c) Can you draw two different squares with perimeter 12 cm? Explain.
d) Can you draw two different squares with area 16 cm2? Explain. 5. Do you need to find the area or the perimeter to answer the question?
a) How much paper is needed to cover a bulletin board?
b) What is the distance around a field?
c) How much carpet is needed for a room?
d) How much ribbon is needed to make a border for a picture?
CA 6.2 U13 AP V7.indd 101 2018-10-30 10:55:01 AM 102 Measurement 6-10 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. ME6-10 Area of Shapes Made from Rectangles 1. a) Calculate the area of each polygon.
b) Draw a line to show how Polygon C can be divided into Rectangles A and B.
i) ii) A B C A B C Area of A = Area of A = Area of B = Area of B = Area of C = Area of C = iii) iv) A C B A C B Area of A = Area of A = Area of B = Area of B = Area of C = Area of C = c) How can you get the area of C from the areas of A and B? Write an equation. Area of C = 2. Draw a line to divide the polygon into two rectangles. Use the areas of the rectangles to find the total area of the polygon.
a) b) c) 4 mm 4 mm 4 mm 8 mm 3 mm 7 mm 4 cm 4 cm 6 cm 10 cm 3 cm 7 cm 2 m 5 m 3 m 8 m 9 m 7 m Area of Rectangle 1 = Area of Rectangle 1 = Area of Rectangle 1 = Area of Rectangle 2 = Area of Rectangle 2 = Area of Rectangle 2 = Total area = Total area = Total area = CA 6.2 U13 AP V7.indd 102 2018-10-30 10:55:01 AM Measurement 6-10 103 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 3. a) A building is 8 storeys high. The wing is b) The tower of a building is 10 m wide. The 5 storeys high. How many storeys high is base is 50 m wide. How wide is the wing? the tower?
8 storeys tower wing The tower is storeys high.
5 storeys 50 m tower wing 10 m The wing is m wide.
4. Find the missing side lengths. Divide the polygon into two rectangles and find their areas. Then find the total area of the polygon.
a) b) 6 m 6 m 4 m 2 m 3 cm 6 cm 4 cm 3 cm Area of Rectangle 1 = Area of Rectangle 1 = Area of Rectangle 2 = Area of Rectangle 2 = Total area = Total area = 5. The picture shows plans for two flower beds. Find the area and the perimeter of each flower bed. A B 60 cm a) Flower Bed A Flower Bed B Area = Area = Perimeter = Perimeter = b) Which flower bed has greater area? c) Which has greater perimeter? 6. The picture shows plans for two parks. Find the area and the perimeter of each park. A B 0.5 km 7. On grid paper, draw two different shapes made from rectangles that have the same perimeter and area.
CA 6.2 U13 AP V7.indd 103 2018-10-30 10:55:02 AM Number Sense 6-59 117 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-59 Equivalent Ratios In the picture, there are 3 circles for every 2 squares. There are also 6 circles for every 4 squares.
The ratios 3 : 2 and 6 : 4 are equivalent. 1. Find two equivalent ratios for the picture. a) b) circles to squares = 3 : = 6 : circles to squares = 1 : = 2 : c) d) circles to squares = 2 : = 6 : circles to squares = 3 : = 9 : 2. Complete the pictures so the ratio of triangles to squares is the same in each column. Then create a sequence of equivalent ratios. Triangles Squares Ratio 2 : 3 Triangles Squares Ratio 6 : 2 3. Skip count to write a sequence of three equivalent ratios.
a) 3 : 2 = 6 : 4 = : b) 3 : 5 = : = : c) 5 : 8 = : = : d) 3 : 10 = : = : e) 5 : 4 = : = : f) 4 : 9 = : = : 4. Find the missing term(s).
a) 3 : 7 = : 14 b) 5 : 6 = 10 : = : 18 c) 2 : 5 = : 20 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 117 2018-10-22 2:18:42 PM 118 Number Sense 6-59 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. There are 5 blue marbles for every 2 red marbles in a jar. There are 20 blue marbles.
To find out how many red marbles are in the jar, write out a sequence of equivalent ratios. Stop when there are 20 blue marbles.
There are 8 red marbles in the jar. Blue Red 5 : 2 10 : 4 15 : 6 20 : 8 5. Write a sequence of equivalent ratios to solve the problem.
a) There are 5 red marbles for every 4 blue marbles in a jar Red Blue with 20 red marbles. How many blue marbles are in the jar?
b) There are 4 red beads for every 3 blue beads in a bracelet. Red Blue The bracelet has 12 red beads. How many blue beads are in the bracelet?
c) A recipe for soup calls for 3 cups of cream for every Cream Tomatoes 5 cups of tomatoes. How many cups of cream are needed for 15 cups of tomatoes?
d) A team has 2 wins for every loss. They won 10 games. Wins Losses How many games did they lose?
e) A mixture for green paint has 5 cups of blue paint for every 6 cups of yellow paint. How much blue paint would you need if you have 30 cups of yellow paint?
CA 6.2 AP U14 NS58-70 p116-xx V4.indd 118 2018-10-22 2:18:42 PM Number Sense 6-60 119 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-60 Introduction to Ratio Tables 1. The column was made by skip counting by a number. Complete the column.
a) b) c) d) e) f) 4 8 12 5 10 3 9 2 8 7 21 12 24 Marc makes orange paint by mixing 1 cup of red paint for every 3 cups of yellow paint. He records the number of cups in a ratio table.
In a ratio table, multiply the numbers in the first row by the same number to get another row. Cups of Red Cups of Yellow ×2 ×3 ×4 1 3 ×2 ×3 ×4 2 6 3 9 4 12 2. Use skip counting or multiplication to complete a ratio table for the ratio.
a) 4 : 1 b) 1 : 2 c) 3 : 1 d) 1 : 7 4 1 8 2 12 3 16 4 1 2 2 4 3 1 1 7 e) 2 : 3 f) 5 : 2 g) 6 : 4 h) 3 : 5 2 3 5 2 6 4 3 5 3. Find the missing number(s) in the ratio table.
a) 2 : 7 b) 4 : 1 c) 3 : 2 BONUS 6 : 2 7 4 14 6 4 1 8 2 3 3 2 4 9 6 12 10 18 15 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 119 2018-10-22 2:18:43 PM 120 Number Sense 6-60 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 4. Jackie created an increasing pattern with squares and recorded the number of squares in a table.
Figure 1 Figure 2 Figure 3 Figure # of Squares 1 4 2 6 3 8 Is this a ratio table? Explain how you know. 5. Circle the tables that are ratio tables.
7 3 14 6 21 9 28 12 4 2 8 4 12 8 16 16 6 5 12 10 18 15 24 20 1 5 2 6 3 7 4 8 6. Dory makes punch. She needs 5 cups of ginger ale for every 3 cups of cranberry juice.
Use the ratio table to fi nd out how many cups of ginger ale she needs for 9 cups of cranberry juice.
Cups of Ginger Ale Cups of Cranberry Juice 5 3 BONUS In Question 6, how many cups each of ginger ale and cranberry juice does Dory need to make 40 cups of punch? Use the ratio table to fi nd out.
Cups of Ginger Ale Cups of Cranberry Juice Cups in Total 5 3 8 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 120 2018-10-22 2:18:43 PM Number Sense 6-64 127 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-64 Percentages A percentage is a ratio that compares a number to 100.
The term “percent” means “per 100” or “for every 100” or “out of 100.” For example, 84% on a test means 84 out of 100.
You can think of a percentage as a short form for a fraction with denominator 100. Example: 45% = 45 100 1. Write the percentage as a fraction.
a) 7% b) 92% c) 5% d) 15% e) 50% f) 100% g) 2% h) 17% 2. Write the fraction as a percentage.
a) 2 100 b) 31 100 c) 52 100 d) 100 100 e) 17 100 f) 88 100 g) 7 100 h) 1 100 3. Write the decimal as a fraction and then a percentage.
a) 0.72 = 72 100 = 72% b) 0.27 = c) 0.04 = 4. Write the fraction as a percentage by first changing it to a fraction with denominator 100. a) 3 5 = 60 100 = 60% b) 2 5 c) 4 5 d) 1 4 e) 3 4 f) 1 2 g) 3 10 h) 7 10 i) 17 25 j) 17 20 k) 3 25 l) 19 20 m) 23 50 n) 47 50 × 20 × 20 × 20 × 20 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 127 2018-10-22 2:18:50 PM 128 Number Sense 6-64 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 5. Write the decimal as a percentage.
a) 0.2 = 2 10 = 20 100 = 20% b) 0.5 c) 0.7 d) 0.9 6. What percentage of the figure is shaded?
a) b) c) d) 7. Change the fraction to a percentage by first writing it with the smallest numbers.
a) 9 15 ÷ 3 ÷ 3 = 3 5 = 3 5 × 20 × 20 = 60 100 = 60% b) 12 15 c) 3 6 d) 7 35 e) 21 28 f) 1 2 g) 12 30 h) 10 40 i) 20 40 j) 16 40 k) 60 150 l) 45 75 m) 80 200 n) 72 80 × 10 × 10 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 128 2018-10-22 2:18:51 PM Number Sense 6-65 129 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-65 Visual Representations of Percentages 1. Fill in the chart. Picture Fraction 23 100 100 45 100 100 Decimal 0.23 0. 0. 0.81 Percentage 23% 63% % % Use a ruler for Questions 2 to 4.
2. Shade 50% of the shape.
a) b) c) 3. Shade 25% of the box.
a) b) c) 4. Colour 50% of the rectangle blue, 40% red, and 10% green. 5. a) Write a fraction for the shaded part: 20 b) Write the fraction with a denominator of 100: c) Write a decimal and a percentage for the shaded part: CA 6.2 AP U14 NS58-70 p116-138 V5.indd 129 2018-11-02 9:21:52 AM 130 Number Sense 6-65 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 6. Write a fraction and a percentage for each division of the number line. Fraction 0 1 Percentage 0% 7. Draw marks to show 25%, 50%, and 75% of the line segment. Hint: Mark 50% first.
a) b) c) d) 8. Circle whether the mark is closer to 25%, 50%, or 75%.
a) b) 25% 50% 75% 25% 50% 75% c) d) 25% 50% 75% 25% 50% 75% e) f) 25% 50% 75% 25% 50% 75% 9. Estimate the percentage of the line segment to the left of the mark.
a) b) 0% 100% 0% 100% 10. Draw a rough sketch of a floor plan for a museum. The different collections should take up the following amounts of space: • Dinosaurs 40% • Animals 20% • Rocks and Minerals 10% • Ancient Artifacts 20% Washrooms should take up the final 10% of the floor space.
11. Asia covers 30% of the world’s land mass. Using a globe, compare the size of Asia to the size of Australia. Approximately what percentage of the world’s land mass does Australia cover? CA 6.2 AP U14 NS58-70 p116-xx V4.indd 130 2018-10-22 2:18:52 PM Number Sense 6-66 131 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-66 Comparing Decimals, Fractions, and Percentages 1. Is the fraction closest to 10%, 25%, 50%, or 75%?
a) 3 5 b) 4 5 c) 2 5 d) 2 10 e) 1 10 f) 4 10 g) 9 10 h) 4 25 i) 11 20 j) 16 20 k) 37 40 l) 1 12 2. Change the numbers in the pair to fractions with the same denominator. Then write <, >, or = in the box.
a) 1 2 47% b) 1 2 53% c) 1 4 23% d) 3 4 70% 1 2 47 100 50 100 > 47 100 e) 2 5 32% f) 0.27 62% g) 0.02 11% h) 1 10 10% i) 19 25 93% j) 23 50 46% k) 0.9 10% l) 11 20 19% 3. Write the numbers in order from least to greatest by first changing each number to a fraction.
a) 3 5 , 42% , 0.73 b) 1 2 , 0.67 , 80% c) 1 4 , 0.09 , 15% d) 2 3 , 57% , 0.62 50 × 50 × CA 6.2 AP U14 NS58-70 p116-xx V4.indd 131 2018-10-22 2:18:53 PM 132 Number Sense 6-67 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-67 Finding Percentages If you use a thousands cube to represent 1 whole, you can see that taking 1 10 of a number is the same as dividing the number by 10—the decimal point shifts one place left. of = 1 10 of = 1 10 of = 1 10 1. Find 1 10 of the number by shifting the decimal point.
a) 4 (= 4.0) b) 7 c) 32 d) 120 e) 3.8 f) 2.5 2. 10% is short for 1 10 . Find 10% of the number.
a) 9 b) 5.7 c) 4.05 d) 6.35 e) 0.06 f) 21.1 You can find percentages that are multiples of 10.
Example: To find 30% of 21, find 10% of 21 and multiply the result by 3.
Step 1: 10% of 21 = 2.1 Step 2: 3 × 2.1 = 6.3 30% of 21 = 6.3 3. Find the percentage using the method above.
a) 40% of 15 b) 60% of 25 c) 90% of 31 i) 10% of = i) 10% of = i) 10% of = ii) × = ii) × = ii) × = 4. a) If you want to estimate what percentage of 120 is 81, would your estimate be 60% or 70%? Hint: Find 60% of 120 and 70% of 120 to see which one is closer to 81. b) 15 out of 32 students in a class walk to school. About what percentage of students walk to school? 1 10 of 1 = 0.1 1 10 of 0.1 = 0.01 1 10 of 0.01 = 0.001 0.4 15 4 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 132 2018-10-22 2:18:53 PM Number Sense 6-69 135 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-69 Percentages Word Problems 1. Find the percentage of the stamp collection that comes from “other” countries. Hint: Change all fractions to percentages.
a) Anne’s collection: b) Braden’s collection: c) Jun’s collection: USA Canada Other 40% 1 2 = 40% = 50% = 10% Canada England Other 80% 1 10 Mexico Canada Other 1 2 40% d) Lela’s collection: e) Grace’s collection: f) Carl’s collection: Canada Nigeria Other 22% 3 5 Jamaica Canada Other 3 4 15% France Italy Other 3 4 10% 2. A painter spends $500.00 on art supplies. Complete the chart.
Fraction of Money Spent Percentage of Money Spent Amount of Money Spent Brushes $50.00 Paint 4 10 Canvas 50% 3. Iva spent 1 hour doing homework. The chart shows the time she spent on each subject. a) Complete the chart. b) How did you find the amount of time spent on math?
4. Raj wants to buy a deck of cards that costs $8.00. The taxes are 15%. How much will he pay in taxes? Subject Fraction of 1 Hour Percentage of 1 Hour Decimal Number of Minutes English 1 4 0.25 15 Science 1 20 5% Math 50% French 0.20 5. There are 15 blue balloons and 12 green balloons at a birthday party. 3 4 of the green balloons have writing on them, and 60% of the blue balloons have writing on them. How many balloons have writing on them?
CA 6.2 AP U14 NS58-70 p116-138 V5.indd 135 2018-11-02 9:21:58 AM 136 Number Sense 6-70 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. NS6-70 Fractions, Ratios, and Percentages There are only blue (b) and red (r) crayons in each bag.
1. Write the number of blue crayons (b), red crayons (r), and total crayons (c) in the bag. a) There are 8 blue crayons and 5 red crayons in the bag. b: r: c: b) There are 4 blue crayons and 7 red crayons in the bag. b: r: c: c) There are 12 blue crayons and 15 red crayons in the bag. b: r: c: d) There are 9 red crayons in the bag of 20 crayons. b: r: c: e) There are 7 blue crayons in the bag of 10 crayons. b: r: c: 2. Write the number of blue crayons, red crayons, and total crayons in the bag. Then write the fraction of crayons that are blue and the fraction that are red. a) There are 5 blue crayons and 6 red crayons in the bag. b: , 5 11 r: , c: b) There are 15 crayons in the bag. 8 are blue. b: , r: , c: 3. Write the fraction of crayons in the bag that are blue and the fraction that are red.
a) There are 5 blue crayons and 17 crayons in total in the bag. b: 5 17 r: b) There are 3 blue crayons and 2 red crayons in the bag. b: r: c) There are 9 red crayons and 20 crayons in total in the bag. b: r: d) The ratio of blue crayons to red crayons in the bag is 5 : 9. b: r: e) The ratio of red crayons to blue crayons in the bag is 7 : 8. b: r: f) The ratio of blue crayons to red crayons in the bag is 10 : 11. b: r: g) The ratio of blue crayons to total crayons in the bag is 11 : 23. b: r: h) The ratio of total crayons to red crayons in the bag is 25 : 13. b: r: 8 5 13 5 6 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 136 2018-10-22 2:18:54 PM Number Sense 6-70 137 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. 4. Fill in the missing numbers for the bag of crayons. Ratio What Fraction … red to total blue to total red to blue blue to red are red?
are blue?
a) 2 : 5 3 : 5 2 : 3 3 : 2 2 5 3 5 b) 4 : 7 c) 1 4 d) 23 : 50 e) 9 : 16 f) 1 2 g) 7 : 10 h) 12 15 i) 25 : 31 5. Fill in the missing numbers for the bag of crayons. Percentage That Are Red Percentage That Are Blue Fraction That Are Red Fraction That Are Blue Ratio of Red to Blue a) 40% 60% 40 100 60 100 40 : 60 b) 35% c) 3 4 d) 10 : 15 e) 1 2 f) 65% g) 23 : 27 CA 6.2 AP U14 NS58-70 p116-xx V4.indd 137 2018-10-22 2:18:54 PM 138 Number Sense 6-70 COPYRIGHT © 2018 JUMP MATH: NOT TO BE COPIED. There are 3 canoes for every 2 kayaks in a marina of 20 boats. How many canoes are in the marina?
Method 1: Jack writes a sequence of equivalent ratios. 3 canoes : 2 kayaks = 6 canoes : 4 kayaks = 9 canoes : 6 kayaks = 12 canoes : 8 kayaks There are 12 canoes in the marina. Stop when the terms of the ratio add to 20.
Method 2: Sally uses different equivalent ratios to find the answer. There are 3 canoes for every 5 boats in the marina. 3 canoes : 5 boats = 6 canoes : 10 boats = 9 canoes : 15 boats = 12 canoes : 20 boats There are 12 canoes in the marina. Stop when you reach 20.
Method 3: Ken uses fractions. The ratio of canoes to kayaks is 3 : 2. So the fraction of canoes in the marina is 3 5 . 3 5 × 20 = 3 × (20 ÷ 5) = 12. So there are 12 canoes in the marina.
6. From the information given, determine the number of kayaks and canoes in the marina.
a) There are 20 boats. 2 5 are canoes. b) There are 42 boats. 3 7 are kayaks.
c) There are 15 boats. d) There are 24 boats. The ratio of kayaks to canoes is 3 : 2. The ratio of kayaks to canoes is 3 : 5.
7. Which marina has more kayaks?
a) In Marina A, there are 40 boats. 60% are kayaks. In Marina B, there are 36 boats. The ratio of canoes to kayaks is 5 : 4.
b) In Marina A, there are 28 boats. The ratio of canoes to kayaks is 5 : 2. In Marina B, there are 30 boats. 3 5 of the boats are canoes.
8. Look at the word “Whitehorse.” a) What is the ratio of vowels to consonants?
b) What fraction of the letters are vowels?
c) What percentage of the letters are consonants?
9. Write the amounts in order from least to greatest: 20%, 1 20 , 0.2. Show your work.
10. Karen has 360 hockey cards. Thirty percent are Montreal Canadiens cards and half are Detroit Red Wings cards. The rest are Edmonton Oilers cards. How many cards from each team does Karen have?
CA 6.2 AP U14 NS58-70 p116-xx V4.indd 138 2018-10-22 2:18:54 PM |
11531 | https://www.cuemath.com/data/arithmetic-mean/ | LearnPracticeDownload
Arithmetic Mean
The arithmetic mean in statistics, is nothing but the ratio of all observations to the total number of observations in a data set. Some of the examples include the average rainfall of a place, the average income of employees in an organization. We often come across statements like "the average monthly income of a family is ₹15,000 or the average monthly rainfall of a place is 1000 mm" quite often. Average is typically referred to as arithmetic mean.
The formula for calculating arithmetic mean is (sum of all observations)/(number of observations). For example, the arithmetic mean of a set of numbers {10, 20, 30, 40} is (10 + 20 + 30 + 40)/4 = 25. Let’s understand the meaning of the term "mean", followed by arithmetic with a few solved examples in the end.
| | |
--- |
| 1. | What Is Arithmetic Mean? |
| 2. | Arithmetic Mean Formula |
| 3. | Properties of Arithmetic Mean |
| 4. | Calculating Arithmetic Mean for Ungrouped Data |
| 5. | Calculating Arithmetic Mean for Grouped Data |
| 6. | Advantages of Arithmetic Mean |
| 7. | Disadvantages of Arithmetic Mean |
| 8. | FAQs on Arithmetic Mean |
What Is Arithmetic Mean?
Arithmetic mean is often referred to as the mean or arithmetic average. It is calculated by adding all the numbers in a given data set and then dividing it by the total number of items within that set. The arithmetic mean (AM) for evenly distributed numbers is equal to the middlemost number. Further, the AM is calculated using numerous methods, which is based on the amount of the data, and the distribution of the data.
For example, the mean of the numbers 6, 8, 10 is 8 since 6 + 8 + 10 = 24 and 24 divided by 3 [there are three numbers] is 8. The arithmetic mean maintains its place in calculating a stock’s average closing price during a particular month. Let's assume there are 24 trading days in a month. How can we calculate the mean? All you need to do is take all the prices, add them up, and divide by 24 to get the AM.
☛Also Check: Difference Between Average and Mean
Arithmetic Mean Formula
The general formula to find the arithmetic mean of a given data is:
Arithmetic mean (x̄) = Sum of all observations / Number of observations
It is denoted by x̄, (read as x bar). Data can be presented in different forms. For example, when we have raw data like the marks of a student in five subjects, we add the marks obtained in the five subjects and divide the sum by 5, since there are 5 subjects in total.
Now consider a case where we have huge data like the heights of 40 students in a class or the number of people visiting an amusement park across each of the seven days of a week.
Will it be convenient to find the arithmetic mean with the above method? The answer is a big NO! So, how can we find the mean? We group the data in a form that is meaningful and easy to comprehend. Let us study more in detail about finding the arithmetic mean for ungrouped and grouped data. The below-given image presents the general formula to find the arithmetic mean:
Properties of Arithmetic Mean
Let us have a look at some of the important properties of the arithmetic mean. Suppose we have n observations denoted by x₁, x₂, x₃, ….,xₙ and x̄ is their arithmetic mean, then:
If all the observations in the given data set have a value say ‘m’, then their arithmetic mean is also ‘m’. Consider the data having 5 observations: 15,15,15,15,15. So, their total = 15+15+15+15+15= 15 × 5 = 75; n = 5. Now, arithmetic mean = total/n = 75/5 = 15.
The algebraic sum of deviations of a set of observations from their arithmetic mean is zero. (x₁−x̄)+(x₂−x̄)+(x₃−x̄)+...+(xₙ−x̄) = 0. For discrete data, ∑(xi−x̄) = 0. For grouped frequency distribution, ∑f(xi−∑x̄) = 0
If each value in the data increases or decreases by a fixed value, then the mean also increases/decreases by the same number. Let the mean of x₁, x₂, x₃ ……xₙ be X̄, then the mean of x₁+k, x₂+k, x₃ +k ……xₙ+k will be X̄+k.
If each value in the data gets multiplied or divided by a fixed value, then the mean also gets multiplied or divided by the same number. Let the mean of x₁, x₂, x₃ ……xₙ be X̄, then the mean of kx₁, kx₂, kx₃ ……xₙ+k will be kX̄. Similarly, the mean of x₁/k, x₂/k, x₃/k ……xₙ/k will be X̄/k.
Note: While dividing each value by k, it must be a non-zero number as division by 0 is not defined.
Calculating Arithmetic Mean for Ungrouped Data
The arithmetic mean of ungrouped data is calculated using the formula:
Mean x̄ = Sum of all observations / Number of observations
Example: Compute the arithmetic mean of the first 6 odd natural numbers.
Solution: The first 6 odd natural numbers: 1, 3, 5, 7, 9, 11
x̄ = (1+3+5+7+9+11) / 6 = 36/6 = 6.
Thus, the arithmetic mean is 6.
Calculating Arithmetic Mean for Grouped Data
There are three methods (Direct method, Short-cut method, and Step-deviation method) to calculate the arithmetic mean for grouped data. The choice of the method to be used depends on the numerical value of xi (data value) and fi (corresponding frequency). ∑ (sigma) the symbol represents summation. If xi and fi are sufficiently small, the direct method will work. But, if they are numerically large, we use the assumed arithmetic mean method or step-deviation method. In this section, we will be studying all three methods along with examples.
Direct Method
Let x₁, x₂, x₃ ……xₙ be the observations with the frequency f₁, f₂, f₃ ……fₙ.
Then, arithmetic mean is calculated using the formula:
x̄ = (x₁f₁+x₂f₂+......+xₙfₙ) / ∑fi Here, f₁+ f₂ + ....fₙ = ∑fi indicates the sum of all frequencies.
Example I (discrete grouped data): Find the arithmetic mean of the following distribution:
| | | | | | |
--- --- --- |
| x | 10 | 30 | 50 | 70 | 89 |
| f | 7 | 8 | 10 | 15 | 10 |
Solution:
| xi | fi | xifi |
---
| 10 | 7 | 10×7 = 70 |
| 30 | 8 | 30×8 = 240 |
| 50 | 10 | 50×10 = 500 |
| 70 | 15 | 70×15 = 1050 |
| 89 | 10 | 89×10 = 890 |
| Total | ∑fi=50 | ∑xifi=2750 |
Add up all the (xifi) values to obtain ∑xifi. Add up all the fi values to get ∑fi
Now, use the arithmetic mean formula.
x̄ = ∑xifi / ∑fi = 2750/50 = 55
Arithmetic mean = 55. The above problem is an example of discrete grouped data.
Let's now consider an example where the data is present in the form of continuous class intervals.
Example II (continuous class intervals): Let's try finding the mean of the following distribution:
| | | | | | | | |
--- --- --- --- |
| Class-Interval | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 |
| Frequency | 6 | 11 | 7 | 4 | 4 | 2 | 1 |
Solution:
When the data is presented in the form of class intervals, the mid-point of each class (also called class mark) is considered for calculating the arithmetic mean.
The formula for mean remains the same as discussed above.
Note:
Class Mark = (Upper limit + Lower limit) / 2
| Class- Interval | Class Mark (xi) | Frequency (fi) | xifi |
--- --- |
| 15-25 | 20 | 6 | 120 |
| 25-35 | 30 | 11 | 330 |
| 35-45 | 40 | 7 | 280 |
| 45-55 | 50 | 4 | 200 |
| 55-65 | 60 | 4 | 240 |
| 65-75 | 70 | 2 | 140 |
| 75-85 | 80 | 1 | 80 |
| | Total | 35 | 1390 |
x̄ = ∑xifi/ ∑fi = 1390/35 = 39.71. We have, ∑fi = 35 and ∑xifi = 35
Arithmetic mean = 39.71
Assumed Mean Method
The short-cut method is called as assumed mean method or change of origin method. The following steps describe this method.
Step1: Calculate the class marks (mid-point) of each class (xi).
Step2: Let A denote the assumed mean of the data.
Step3: Find deviation (di) = xi – A
Step4: Use the formula:
x̄ = A + (∑fidi/∑fi)
Example: Let's understand this with the help of the following example. Calculate the mean of the following using the short-cut method.
| | | | | | | | |
--- --- --- --- |
| Class-Intervals | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
| Frequency | 5 | 8 | 30 | 25 | 14 | 12 | 6 |
Solution: Let us make the calculation table. Let the assumed mean be A = 62.5
Note: A is chosen from the xi values. Usually, the value which is around the middle is taken.
| Class- Interval | Classmark/ Mid-points (xi) | fi | di = (xi - A) | fidi |
--- ---
| 45-50 | 47.5 | 5 | 47.5-62.5 =-15 | -75 |
| 50-55 | 52.5 | 8 | 52.5-62.5 =-10 | -80 |
| 55-60 | 57.5 | 30 | 57.5-62.5 =-5 | -150 |
| 60-65 | 62.5 | 25 | 62.5-62.5 =0 | 0 |
| 65-70 | 67.5 | 14 | 67.5-62.5 =5 | 70 |
| 70-75 | 72.5 | 12 | 72.5-62.5 =10 | 120 |
| 75-80 | 77.5 | 6 | 77.5-62.5 =15 | 90 |
| | | ∑fi=100 | | ∑fidi= -25 |
Now we use the formula,
x̄ = A + (∑fidi/∑fi) = 62.5 + (−25/100) = 62.5 − 0.25 = 62.25
∴ Arithmetic mean = 62.25
Step Deviation Method
This is also called the change of origin or scale method. The following steps describe this method:
Step 1: Calculate the class marks of each class (xi).
Step 2: Let A denote the assumed mean of the data.
Step 3: Find ui = (xi−A)/h, where h is the class size.
Step 4: Use the formula to find arithmetic mean:
x̄ = A + h × (∑fiui/∑fi)
To learn more about this method, click here.
Example: Consider the following example to understand this method. Find the arithmetic mean of the following using the step-deviation method.
| | | | | | | | | |
--- --- --- ---
| Class Intervals | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
| Frequency | 4 | 4 | 7 | 10 | 12 | 8 | 5 | 50 |
Solution: To find the mean, we first have to find the class marks and decide A (assumed mean). Let A = 35 Here h (class width) = 10
| C.I. | xi | fi | ui= xi−Ahxi−Ah | fiui |
--- ---
| 0-10 | 5 | 4 | -3 | 4 x (-3)=-12 |
| 10-20 | 15 | 4 | -2 | 4 x (-2)=-8 |
| 20-30 | 25 | 7 | -1 | 7 x (-1)=-7 |
| 30-40 | 35 | 10 | 0 | 10 x 0= 0 |
| 40-50 | 45 | 12 | 1 | 12 x 1=12 |
| 50-60 | 55 | 8 | 2 | 8 x 2=16 |
| 60-70 | 65 | 5 | 3 | 5 x 3=15 |
| Total | | ∑fi=50 | | ∑fiui=16 |
Using arithmetic mean formula:
x̄ = A + h × (∑fiui/∑fi) =35 + (16/50) ×10 = 35 + 3.2 = 38.2
Arithmetic mean = 38.
Advantages of Arithmetic Mean
The uses of arithmetic mean are not just limited to statistics and mathematics, but it is also used in experimental science, economics, sociology, and other diverse academic disciplines. Listed below are some of the major advantages of the arithmetic mean.
As the formula to find the arithmetic mean is rigid, the result doesn’t change. Unlike the median, it doesn’t get affected by the position of the value in the data set.
It takes into consideration each value of the data set.
Finding an arithmetic mean is quite simple; even a common man having very little finance and math skills can calculate it.
It’s also a useful measure of central tendency, as it tends to provide useful results, even with large groupings of numbers.
It can be further subjected to many algebraic treatments, unlike mode and median. For example, the mean of two or more series can be obtained from the mean of the individual series.
The arithmetic mean is widely used in geometry as well. For example, the coordinates of the “centroid” of a triangle (or any other figure bounded by line segments) are the arithmetic mean of the coordinates of the vertices.
After having discussed some of the major advantages of arithmetic mean, let's understand its limitations.
Disadvantages of Arithmetic Mean
Let us now look at some of the disadvantages/demerits of using the arithmetic mean.
The strongest drawback of arithmetic mean is that it is affected by extreme values in the data set. To understand this, consider the following example. It’s Ryma’s birthday and she is planning to give return gifts to all who attend her party. She wants to consider the mean age to decide what gift she could give everyone. The ages (in years) of the invitees are as follows: 2, 3, 7, 7, 9, 10, 13, 13, 14, 14 Here, n = 10. Sum of the ages = 2+3+7+7+9+10+13+13+14+14 = 92. Thus, mean = 92/10 = 9.2 In this case, we can say that a gift that is desirable to a kid who is 9 years old may not be suitable for a child aged 2 or 14.
In a distribution containing open-end classes, the value of the mean cannot be computed without making assumptions regarding the size of the class.
| Class Interval | Frequency |
--- |
| Less than 15 | 20 |
| 15-25 | 12 |
| 25-35 | 3 |
| 35-45 | 12 |
| More than 45 | 6 |
We know that to find the arithmetic mean of grouped data, we need the mid-point of every class. As evident from the table, there are two cases (less than 15 and 45 or more) where it is not possible to find the mid-point and hence, arithmetic mean can’t be calculated for such cases.
It's practically impossible to locate the arithmetic mean by inspection or graphically.
It cannot be used for qualitative types of data such as honesty, favorite milkshake flavor, most popular product, etc.
We can't find the arithmetic mean if a single observation is missing or lost.
Tips and tricks on arithmetic mean:
If the number of classes is less and the data has values with a smaller magnitude, then the direct method is preferred out of the three methods to find the arithmetic mean.
Step deviation works best when we have a grouped frequency distribution in which the width remains constant for every class interval and we have a considerably large number of class intervals.
☛Related Topics:
Arithmetic Mean Calculator
Arithmetic Mean Vs Geometric Mean
Arithmetic Mean Examples
Example 1: The heights of five students are 5 ft, 6 ft, 4.6 ft, 5.5 ft, and, 6.2 ft respectively. Using the arithmetic mean formula, find the average (mean) height of the students.
Solution:
To find: Average height of the studentsWe haveArithmetic mean = {Sum of Observation}/{Total numbers of Observations}= (5 + 6 + 4.6 + 5.5 + 6.2)/5= 27.3/5 = 5.46ft.
Answer: The average height of the students is 5.46 ft.
2. Example 2: If the arithmetic mean of 2m+3, m+2, 3m+4, 4m+5 is m+2, find m.
Solution: The data contains 4 observations : 2m+3,m+2,3m+4,4m+52m+3,m+2,3m+4,4m+5
So, n = 4
Sum of 4 observations = [(2m+3)+(m+2)+(3m+4)+(4m+5)]/4 = (10m+14)/4
Mean = (10m + 14)/4
∴ m + 2 = (10m + 14)/4
4 × (m+2) = (10m + 14)
4m+8 = 10m + 14
−6m = 6
m = -1
Answer: ∴ m = -1
3. Example 3: The mean monthly salary of 10 workers of a group is ₹1445. One more worker whose monthly salary is ₹1500 has joined the group. Find the arithmetic mean of the monthly salary of 11 workers of the group.
Solution: Here, n = 10, x̄=1445
Using formula,
x̄ = ∑xi/n
∴∑xi = x̄ × n
∑xi = 1445 × 10 = 14450
(Total salary of 10 workers = ₹14450)
Total salary of 11 workers = 14450 + 1500 = ₹15950
Average salary of 11 workers = 15950/11 = 1450
Answer: ∴ Average monthly salary of 11 workers = ₹1450
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FAQs on Arithmetic Mean
What is the Definition of Arithmetic Mean?
The arithmetic mean is the simplest and most widely used measure of a mean, or average. It simply involves taking the sum of a group of numbers, then dividing that sum by the count of the numbers used in the series. For example, take the numbers 34, 44, 56, and 78. The sum is 212. To find the arithmetic mean we will divide the sum 212 by 4 (total numbers), this will give us the mean as 212/4 = 53.
What are Arithmetic Mean Formulas?
Here are the formulas related to arithmetic mean:
Arithmetic mean of ungrouped data = Sum of all observations / Number of observations.
Arithmetic mean of grouped data = ∑xifi / ∑fi.
Arithmetic mean using assumed mean method = A + (∑fidi/∑fi).
Arithmetic mean using step deviation method = A + h × (∑fiui/∑fi).
How to Calculate the Arithmetic Mean?
In statistics, arithmetic mean (AM) is defined as the ratio of the sum of all the given observations to the total number of observations. For example, if the data set consists of 5 observations, the AM can be calculated by adding all the 5 given observations divided by 5.
How to Find the Arithmetic Mean Between 2 Numbers?
Add the two given numbers and then divide the sum by 2. For example, 2 and 6 are the two numbers, the arithmetic mean (which is nothing but AM or mean) is calculated as follows: AM = (2+6)/2 = 8/2 = 4
What Are the Types of Arithmetic Mean?
In mathematics, we deal with different types of means such as arithmetic mean, harmonic mean, and geometric mean.
What Is the Use of Arithmetic Mean?
The arithmetic mean is a measure of central tendency. It allows us to know the center of the frequency distribution by considering all of the observations.
What Are the Characteristics of Arithmetic Mean?
Some important properties of the arithmetic mean (AM) are as follows:
The sum of deviations of the items from their AM is always zero, i.e. ∑(x – X) = 0.
The sum of the squared deviations of the items from AM is minimum, which is less than the sum of the squared deviations of the items from any other values.
If each item in the arithmetic series is substituted by the mean, then the sum of these replacements will be equal to the sum of the specific items.
If the individual values are added or subtracted with a constant, then the AM can also be added or subtracted by the same constant value.
If the individual values are multiplied or divided by a constant value, then the AM is also multiplied or divided by the same value.
What Is the Sum of Deviations from Arithmetic Mean?
The sum of deviations from the arithmetic mean is equal to zero.
How to Find Arithmetic Mean for Grouped Data?
The arithmetic mean for ungrouped data is found using the formula: x̄ = (x₁f₁+x₂f₂+......+xₙfₙ) / ∑fi = ∑fx/n. Here, f₁+ f₂ + ... + fₙ = ∑fi indicates the sum of all frequencies.
What is the Arithmetic Mean Formula Used for Ungrouped Data?
The arithmetic mean for grouped data is calculated using the formula: x̄ = Sum of all observations / Number of observations.
Q1: Find the arithmetic mean of the first five odd numbers.
Q2: The weight of 5 children in a class is given in pounds. They are 66, 70, 59, 63, 73. Find the arithmetic mean of the weights of the children.
Q3: The arithmetic mean of 5 numbers in a data set is 5. The first 4 numbers are 2, 3, 5 and 6. Find the last number in the data set.
Q4: The arithmetic mean of 3 numbers is 5. The sum of the second and the third number is 12. Find the first number.
Q5: Jack worked for 4 hours on day 1 and twice the number of hours on day 2 and thrice the number of hours (spent on day 1) on day 3. Find the average number of hours he worked on all these three days.
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11532 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/27%3A_The_Kinetic_Theory_of_Gases/27.03%3A_The_Distribution_of_Molecular_Speeds_is_Given_by_the_Maxwell-Boltzmann_Distribution | Skip to main content
27.3: The Distribution of Molecular Speeds is Given by the Maxwell-Boltzmann Distribution
Last updated
: Mar 8, 2025
Save as PDF
27.2: The Gaussian Distribution of One Component of the Molecular Velocity
27.4: The Frequency of Collisions with a Wall
Page ID
: 14537
( \newcommand{\kernel}{\mathrm{null}\,})
The Boltzmann Distribution
If we were to plot the number of molecules whose velocities fall within a series of narrow ranges, we would obtain a slightly asymmetric curve known as a velocity distribution. The peak of this curve would correspond to the most probable velocity. This velocity distribution curve is known as the Maxwell-Boltzmann distribution, but is frequently referred to only by Boltzmann's name. The Maxwell-Boltzmann distribution law was first worked out around 1860 by the great Scottish physicist, James Clerk Maxwell (1831-1879), who is better known for discovering the laws of electromagnetic radiation. Later, the Austrian physicist Ludwig Boltzmann (1844-1906) put the relation on a sounder theoretical basis and simplified the mathematics somewhat. Boltzmann pioneered the application of statistics to the physics and thermodynamics of matter and was an ardent supporter of the atomic theory of matter at a time when it was still not accepted by many of his contemporaries.
In section 27.2 we saw that the distribution function for molecular speeds in the x direction is given by:
f(vx)=√m2πkBT⏟normalization termexp(−mv2x2kBT)⏟exponential term
f(vx)=m2πkBT−−−−−−√normalization termexp(−mv2x2kBT)exponential term
However, real gas samples have molecules not only with a distribution of molecular speeds and but also a random distribution of directions. Using normal vector magnitude properties (or simply using the Pythagorean Theorem), it can be seen that
⟨v⟩2=⟨vx⟩2+⟨vy⟩2+⟨vz⟩2
⟨v⟩2=⟨vx⟩2+⟨vy⟩2+⟨vz⟩2
Since the direction of travel is random, the velocity can have any component in x, y, or z directions with equal probability. As such, the average value of the x, y, or z components of velocity should be the same. And so
⟨v⟩2=3⟨vx⟩2
⟨v⟩2=3⟨vx⟩2
Substituting this into the expression for pressure yields
p=Ntotm3V⟨v⟩2
p=Ntotm3V⟨v⟩2
All that remains is to determine the form of the distribution of velocity magnitudes the gas molecules can take. In his 1860 paper (Illustrations of the dynamical theory of gases. Part 1. On the motions and collisions of perfectly elastic spheres, 1860), Maxwell proposed a form for this distribution of speeds which proved to be consistent with observed properties of gases (such as their viscosities). He derived this expression based on a transformation of coordinate system from Cartesian coordinates (xx, yy, zz) to spherical polar coordinates (vv, θθ, ϕϕ). In this new coordinate system, vv represents the magnitude of the velocity (or the speed) and all of the directional data is carried in the angles θθ and ϕϕ. The infinitesimal volume unit becomes
dxdydz=v2sin(θ)dvdθdϕ
dxdydz=v2sin(θ)dvdθdϕ
Applying this transformation of coordinates Maxwell’s distribution took the following form
f(v)=4πv2(m2πkBT)3/2exp(−mv22kBT)
f(v)=4πv2(m2πkBT)3/2exp(−mv22kBT)(27.3.1)
The Distribution of Speed over all Directions
The Distribution of Kinetic Energy in Three Dimensions
As noted above, the distribution function of molecular energies for one dimension is
f(vx)=√m2πkBTexp(−mv2x2kBT)
f(vx)=m2πkBT−−−−−−√exp(−mv2x2kBT)
To obtain a three-dimensional probability distribution, you multiply the distribution function for each of the three dimensions so that
f(vx,vy,vz)=(m2πkBT)3/2exp(−mv22kBT)
f(vx,vy,vz)=(m2πkBT)3/2exp(−mv22kBT)
given ⟨v⟩2=⟨vx⟩2+⟨vy⟩2+⟨vz⟩2
⟨v⟩2=⟨vx⟩2+⟨vy⟩2+⟨vz⟩2
The Conversion of Energy Distribution to Speed Distribution
To convert the three-dimensional energy distribution to a speed distribution over all space, the energy distribution must be summed over all directions. This sum is usually described by imagining a "velocity space" in spherical polar coordinates. As noted above, in this new coordinate system, vv represents the magnitude of the velocity (or the speed) and all of the directional data is carried in the angles θθ and ϕϕ. The infinitesimal volume unit becomes
dxdydz=v2sin(θ)dvdθdϕ
dxdydz=v2sin(θ)dvdθdϕ
You integrate over θθ and ϕϕ to sum over all space, thus
f(v)=(m2πkBT)3/2exp(−mv22kBT)∫π0∫2π0v2sinθdϕdθ⏟=4πv2
f(v)=(m2πkBT)3/2exp(−mv22kBT)∫π0∫2π0v2sinθdϕdθ=4πv2
This equation is rearranged to give
f(v)=4πv2(m2πkBT)3/2exp(−mv22kBT)
f(v)=4πv2(m2πkBT)3/2exp(−mv22kBT)
This function can be thought of as having three basic parts to it: a normalization constant (NN), a velocity dependence (v2v2), and an exponential term that contains the kinetic energy (½mv2½mv2).
Because the function represents the fraction of molecules with the speed vv, the sum of the fractions for all possible velocities must be unity. This sum can be calculated as an integral. The normalization constant ensures that
∫∞0f(v)dv=1
∫∞0f(v)dv=1
Thus the normalization constant is
N=4π√(m2πkBT)3
N=4π(m2πkBT)3−−−−−−−−−−√
Velocity distributions depend on temperature and mass
Higher temperatures allow a larger fraction of molecules to acquire greater amounts of kinetic energy, causing the Boltzmann plots to spread out. Figure 27.3.2
shows how the Maxwell-Boltzmann distribution is affected by temperature. At lower temperatures, the molecules have less energy. Therefore, the speeds of the molecules are lower and the distribution has a smaller range. As the temperature of the molecules increases, the distribution flattens out. Because the molecules have greater energy at higher temperature, the molecules are moving faster.
Notice how the left ends of the plots are anchored at zero velocity (there will always be a few molecules that happen to be at rest.) As a consequence, the curves flatten out as the higher temperatures make additional higher-velocity states of motion more accessible. The area under each plot is the same for a constant number of molecules.
Figure 27.3.2: Maxwell Distribution of speeds for hydrogen molecules at differing temperatures.
Calculating the Average Speed
Using the Maxwell distribution as a distribution of probabilities, the average molecular speed in a sample of gas molecules can be determined.
⟨v⟩=∫∞−∞vf(v)dv=∫∞−∞v4π√(m2πkBT)3v2exp(−mv22kBT) dv=4π√(m2πkBT)3∫∞−∞v3exp(−mv22kBT) dv
⟨v⟩=∫∞−∞vf(v)dv=∫∞−∞v4π(m2πkBT)3−−−−−−−−−−√v2exp(−mv22kBT) dv=4π(m2πkBT)3−−−−−−−−−−√∫∞−∞v3exp(−mv22kBT) dv
The following can be found in a table of integrals:
∫∞0x2n+1e−ax2dx=n!2an+1
∫∞0x2n+1e−ax2dx=n!2an+1
So
⟨v⟩=4π√(m2πkBT)3[12(m2kBT)2]
⟨v⟩=4π(m2πkBT)3−−−−−−−−−−√⎡⎣⎢⎢⎢⎢12(m2kBT)2⎤⎦⎥⎥⎥⎥
Which simplifies to
⟨v⟩=(8kBTπm)1/2
⟨v⟩=(8kBTπm)1/2
Note: the value of ⟨v⟩ is twice that of ⟨vx⟩ which was derived in an earlier example!
⟨v⟩=2⟨vx⟩
Example 27.3.1 :
What is the average value of the squared speed according to the Maxwell distribution law?
Solution:
⟨v2⟩=∫∞−∞v2f(v)dv=∫∞−∞v24π√(m2πkBT)3v2exp(−mv22kBT) dv=4π√(m2πkBT)3∫∞−∞v4exp(−mv22kBT) dv
A table of integrals indicates that
∫∞0x2ne−ax2dx=1⋅3⋅5…(2n−1)2n+1an√πa
Substitution (noting that n=2) yields
⟨v2⟩=4π√(m2πkBT)3[1⋅323(m2kBT)2√π(m2kBT)]
which simplifies to
⟨v2⟩=3kBTm
Note: The square root of this average squared speed is called the root mean square (RMS) speed, and has the value
vrms=√⟨v2⟩=(3kBTm)1/2
All molecules have the same kinetic energy (½mv2) at the same temperature, so the fraction of molecules with higher velocities will increase as m, and thus the molecular weight, decreases. Figure 27.3.3
shows the dependence of the Maxwell-Boltzmann distribution on molecule mass. On average, heavier molecules move more slowly than lighter molecules. Therefore, heavier molecules will have a smaller speed distribution, while lighter molecules will have a speed distribution that is more spread out.
Figure 27.3.3: Maxwell Distribution of speeds at 800 K for different gasses of differing molecular masses.
Related Speed Expressions
Usually, we are more interested in the speeds of molecules rather than their component velocities. The Maxwell–Boltzmann distribution for the speed follows immediately from the distribution of the velocity vector, above. Note that the speed of an individual gas particle is:
v=√v2x+v2y=v2z
Three speed expressions can be derived from the Maxwell-Boltzmann distribution:
the most probable speed,
the average speed, and
the root-mean-square speed.
The most probable speed is the maximum value on the distribution plot (Figure 27.3.4
). This is established by finding the velocity when the derivative of Equation 27.3.1 is zero
df(v)dv=0
which is
vmp=√2RTM
Figure 27.3.4: The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures. from OpenStax. The speed at the top of the curve is called the most probable speed because the largest number of molecules have that speed.
The average speed is the sum of the speeds of all the molecules divided by the number of molecules.
vavg=ˉv=∫∞0vf(v)dv=√8RTπM
The root-mean-square speed is square root of the average speed-squared.
vrms=¯v2=√3RTM
R is the gas constant,
T is the absolute temperature and
M is the molar mass of the gas.
It always follows that for gases that follow the Maxwell-Boltzmann distribution:
vmp<vavg<vrms
Problems
Using the Maxwell-Boltzman function, calculate the fraction of argon gas molecules with a speed of 305 m/s at 500 K.
If the system in problem 1 has 0.46 moles of argon gas, how many molecules have the speed of 305 m/s?
Calculate the values of Cmp, Cavg, and Crms for xenon gas at 298 K.
From the values calculated above, label the Boltzmann distribution plot with the approximate locations of (C_{mp}), Cavg, and Crms.
What will have a larger speed distribution, helium at 500 K or argon at 300 K? Helium at 300 K or argon at 500 K? Argon at 400 K or argon at 1000 K?
Answers
0.00141
3.92×1020 argon molecules
cmp = 194.27 m/s, cavg = 219.21 m/s, crms = 237.93 m/s
As stated above, Cmp is the most probable speed, thus it will be at the top of the distribution curve. To the right of the most probable speed will be the average speed, followed by the root-mean-square speed.
Hint: Use the related speed expressions to determine the distribution of the gas molecules: helium at 500 K. helium at at 300 K. argon at 1000 K.
Sources
Dunbar, R.C. Deriving the Maxwell Distribution J. Chem. Ed. 1982, 59, 22-23.
Peckham, G.D.; McNaught, I.J.; Applications of the Maxwell-Boltzmann Distribution J. Chem. Ed. 1992, 69, 554-558.
Chang, R. Physical Chemistry for the Biosciences, 25-27.
HyperPhysics:
Contributors and Attributions
Prof. David Blauch (Davidson College)
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
Adam Maley (Hope College)
Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)
Stephanie Schaertel (Grand Valley State University)
Tom Neils (Grand Rapids Community College, editing)
27.2: The Gaussian Distribution of One Component of the Molecular Velocity
27.4: The Frequency of Collisions with a Wall |
11533 | https://stackoverflow.com/questions/42974783/all-simple-paths-of-an-undirected-strongly-connected-graph | c++ - All simple paths of an undirected, strongly connected graph - Stack Overflow
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All simple paths of an undirected, strongly connected graph
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Asked 8 years, 6 months ago
Modified8 years, 6 months ago
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I have been trying to learn more about graph traversal in my spare time, and I am trying to use depth-first-search to find all simple paths between a start node and an end node in an undirected, strongly connected graph. So far I have been using this code from Print all paths from a given source to a destination, which is only for a directed graph.
The main algorithm, using recursive DFS, occurs in these two functions:
```cpp
void Graph::printAllPaths(int s, int d)
{
// Mark all the vertices as not visited
bool visited = new bool[V];
// Create an array to store paths
int path = new int[V];
int path_index = 0; // Initialize path[] as empty
// Initialize all vertices as not visited
for (int i = 0; i < V; i++)
visited[i] = false;
// Call the recursive helper function to print all paths
printAllPathsUtil(s, d, visited, path, path_index);
}
// A recursive function to print all paths from 'u' to 'd'.
// visited[] keeps track of vertices in current path.
// path[] stores actual vertices and path_index is current
// index in path[]
void Graph::printAllPathsUtil(int u, int d, bool visited[],
int path[], int &path_index)
{
// Mark the current node and store it in path[]
visited[u] = true;
path[path_index] = u;
path_index++;
// If current vertex is same as destination, then print
// current path[]
if (u == d)
{
for (int i = 0; i<path_index; i++)
cout << path[i] << " ";
cout << endl;
}
else // If current vertex is not destination
{
// Recur for all the vertices adjacent to current vertex
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
if (!visited[i])
printAllPathsUtil(i, d, visited, path, path_index);
}
// Remove current vertex from path[] and mark it as unvisited
path_index--;
visited[u] = false;
}
```
Which works fine for a directed graph, but not a undirected, strongly connected graph.
I was wondering if their is a way to tweak this code to also work for an undirected graph? I have a feeling more backtracking is needed to explore more possible paths, but am unsure as to how to approach this.
Any help would be appreciated.
c++
algorithm
graph
graph-algorithm
depth-first-search
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edited Mar 23, 2017 at 12:06
RoadRunnerRoadRunner
asked Mar 23, 2017 at 11:30
RoadRunnerRoadRunner
26.4k 6 6 gold badges 46 46 silver badges 77 77 bronze badges
5
can an undirected graph be strongly connected?khusrav –khusrav 2017-03-23 13:55:08 +00:00 Commented Mar 23, 2017 at 13:55
Perhaps I could look into backtracking DFS for this?RoadRunner –RoadRunner 2017-03-23 14:02:58 +00:00 Commented Mar 23, 2017 at 14:02
Why not use BFS? <- I guess this is irrelevant, since you want to learn how to use DFS :-)khusrav –khusrav 2017-03-23 14:04:56 +00:00 Commented Mar 23, 2017 at 14:04
2 I'm not a c++ guy, but from the code seems like it should work. When you say it finds a couple, but ignores others - can you give examples?khusrav –khusrav 2017-03-23 14:26:41 +00:00 Commented Mar 23, 2017 at 14:26
1 If you want to learn more about the Depth First algorithm check out this MIT course ware from youtube.com/playlist?list=PLUl4u3cNGP63gFHB6xb-kVBiQHYe_4hSi and you can learn about it in lesson 4. This was a great series although the focus here is on A.I. Programming and it is the algorithmic side as opposed to any "source code or arbitrary programmable language". This may help you a little more in understanding the different types of tree structures and traversals.Francis Cugler –Francis Cugler 2017-03-26 01:32:55 +00:00 Commented Mar 26, 2017 at 1:32
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RoadRunner, are you sure that you "undirected issue" is in the code you show? It looks OK at the first glance. Maybe the bug is comming from the fact you didn't fix addEdge to make graph you create undirected such as:
cpp
void Graph::addEdge(int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u); // Fix: add back edge as well!
}
Update (C code, but very ugly)
OK, here is my attempt to translate code to pure C. Obviously code style is ugly and there are no error checks at all but you can improve that as I expect you to be much more proficient in C. Also I just rolled out my simple custom linked-list of graph nodes with a bit strange name NodeListNode i.e. ListNode containing a graph Node.
Graph.h
```c
pragma once
ifdef __cplusplus
extern "C" { // only need to export C interface if
// used by C++ source code
endif
typedef struct tagNodeListNode {
struct tagNodeListNode next;
int index;
} NodeListNode;
typedef struct tagGraph {
int nodesCount;
NodeListNode adjArr;
} Graph;
typedef void(GraphPathVisitorFunc)(NodeListNode const const path);
Graph GraphCreate(int nodesCount);
void GraphDestroy(Graph gr);
void GraphAddEdge(Graph gr, int u, int v);
void GraphVisitAllPaths(Graph gr, int s, int d, GraphPathVisitorFunc visitor);
void GraphPrintAllPaths(Graph gr, int s, int d);
ifdef __cplusplus
}
endif
```
Graph.c
```c
include "Graph.h"
include
include
include
Graph GraphCreate(int nodesCount)
{
// calloc ensures zeroing array
NodeListNode adjArr = (NodeListNode)calloc(nodesCount, sizeof(NodeListNode));
Graph gr = { nodesCount, adjArr };
return gr;
}
void GraphDestroy(Graph gr)
{
for (int i = 0; i < gr.nodesCount; i++)
{
for (NodeListNode adj = gr.adjArr[i]; adj != NULL;)
{
NodeListNode tmp = adj;
adj = adj->next; //first move on the free
free(tmp);
}
}
free(gr.adjArr);
}
void GraphAddEdgeImplFirst(Graph gr, int from, int to)
{
NodeListNode adj = gr.adjArr[from];
NodeListNode n = (NodeListNode)malloc(sizeof(NodeListNode));
n->next = adj;
n->index = to;
gr.adjArr[from] = n;
}
void GraphAddEdgeImplLast(Graph gr, int from, int to)
{
NodeListNode adj = gr.adjArr[from];
NodeListNode n = (NodeListNode)malloc(sizeof(NodeListNode));
n->next = NULL;
n->index = to;
if(adj == NULL)
{
gr.adjArr[from] = n;
}
else
{
while (adj->next != NULL)
adj = adj->next;
adj->next = n;
}
}
void GraphAddEdge(Graph gr, int u, int v)
{
GraphAddEdgeImplFirst(gr, u, v);
GraphAddEdgeImplFirst(gr, v, u);
// closer to but slower and thus makes no sense
//GraphAddEdgeImplLast(gr, u, v);
//GraphAddEdgeImplLast(gr, v, u);
}
void GraphVisitAllPathsImpl(Graph gr, int cur, int dst, GraphPathVisitorFunc visitor, NodeListNode pathFst, NodeListNode pathLst, bool visited)
{
if (cur == dst)
{
visitor(pathFst);
return;
}
NodeListNode adj = gr.adjArr[cur];
for (NodeListNode const tmp = adj; tmp != NULL; tmp = tmp->next)
{
int next = tmp->index;
if (visited[next])
continue;
visited[next] = true;
NodeListNode nextNode = { NULL,next };
pathLst->next = &nextNode;
GraphVisitAllPathsImpl(gr, next, dst, visitor, pathFst, &nextNode, visited);
pathLst->next = NULL;
visited[next] = false;
}
}
void GraphVisitAllPaths(Graph gr, int start, int dst, GraphPathVisitorFunc visitor)
{
bool visited = calloc(gr.nodesCount, sizeof(bool));
visited[start] = true;
NodeListNode node = { NULL,start };
GraphVisitAllPathsImpl(gr, start, dst, visitor, &node, &node, visited);
free(visited);
}
void PrintPath(NodeListNode const const path)
{
for (NodeListNode const tmp = path; tmp != NULL; tmp = tmp->next)
{
printf("%d ", tmp->index);
}
printf("\n");
}
void GraphPrintAllPaths(Graph gr, int s, int d)
{
GraphVisitAllPaths(gr, s, d, PrintPath);
}
```
And usage example with a graph identical to your ideaone sample. Note that to get matching output you should use GraphAddEdgeImplLast instead of GraphAddEdgeImplFirst otherwise results will be in reversed order.
```c
void testGraph()
{
Graph gr = GraphCreate(20);
GraphAddEdge(gr, 0, 1);
GraphAddEdge(gr, 0, 7);
GraphAddEdge(gr, 1, 2);
GraphAddEdge(gr, 1, 6);
GraphAddEdge(gr, 1, 5);
GraphAddEdge(gr, 2, 3);
GraphAddEdge(gr, 2, 5);
GraphAddEdge(gr, 3, 4);
GraphAddEdge(gr, 3, 5);
GraphAddEdge(gr, 4, 5);
GraphAddEdge(gr, 4, 10);
GraphAddEdge(gr, 4, 11);
GraphAddEdge(gr, 5, 6);
GraphAddEdge(gr, 5, 10);
GraphAddEdge(gr, 5, 11);
GraphAddEdge(gr, 6, 7);
GraphAddEdge(gr, 6, 8);
GraphAddEdge(gr, 6, 9);
GraphAddEdge(gr, 6, 10);
GraphAddEdge(gr, 7, 8);
GraphAddEdge(gr, 8, 9);
GraphAddEdge(gr, 8, 13);
GraphAddEdge(gr, 9, 10);
GraphAddEdge(gr, 9, 13);
GraphAddEdge(gr, 9, 12);
GraphAddEdge(gr, 10, 12);
GraphAddEdge(gr, 11, 12);
GraphAddEdge(gr, 12, 13);
GraphAddEdge(gr, 12, 14);
GraphAddEdge(gr, 12, 16);
GraphAddEdge(gr, 13, 14);
GraphAddEdge(gr, 14, 15);
GraphAddEdge(gr, 16, 17);
GraphAddEdge(gr, 15, 17);
GraphAddEdge(gr, 15, 19);
GraphAddEdge(gr, 17, 18);
GraphAddEdge(gr, 17, 19);
GraphAddEdge(gr, 18, 19);
GraphPrintAllPaths(gr, 12, 4);
GraphDestroy(gr);
}
```
Hope this helps
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edited Mar 26, 2017 at 2:34
answered Mar 25, 2017 at 20:10
SergGrSergGr
23.8k 2 2 gold badges 33 33 silver badges 52 52 bronze badges
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11534 | https://www.gulfcoastnewsnow.com/article/florida-sea-surface-temperatures-hurricane-intensification/61863811 | Loop currents and sea-surface temperatures drive rapid hurricane intensification
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MAY PLAY A ROLE IN HELPING CLEAR UP THE PICTURE. >> SOUTHWEST FLORIDA SITS ALONG ONE OF THE WARMEST BODIES OF WATER IN THE ENTIRE ATLANTIC BASIN. IF YOU'RE TAKING A DIP IN THE GULF OF MEXICO AS OF LATE AND PROBABLY FEELS MORE LIKE A BATH TUB HURRICANES. TROPICAL STORMS FEED OFF OF THIS HEAT AND SO FAR IN 2024, IT'S BEEN A RECORD YEAR FOR WATER. TEMPERATURES RUNNING AS MUCH AS 5 DEGREES ABOVE AVERAGE. BUT IT'S NOT JUST THE HEAT AT THE SURFACE THAT HAS OUR ATTENTION. IT'S HOW DEEP. >> HE GOES. SHOW YOU WHY THE DEATH OF THE WARM WATER MATTER SO MUCH. LET'S COOK UP A STORM IN THE KITCHEN, SO TO SPEAK. I THINK THESE 2 POTS AS BODIES OF WATER IN THE TROPICS IS ONLY ONE DIFFERENCE. ONE. >> IS MUCH DEEPER THAN THE OTHER. NOW WE KNOW THE SHOWER POT IS GOING TO HEAT UP THE WATER MUCH FASTER. BUT ONCE THE DEEPER ONE GETS GOING. LOOK HOW MUCH MORE CONDENSATION THERE IS AND HOW MUCH LONGER IT STAYS THAT SAW ENERGY RELEASED INTO THE ATMOSPHERE NOT ONLY CREATING AN ENVIRONMENT FOR HURRICANES TO THRIVE, EXPLODING INTENSITY. >> THE HE GOES DEEP DR. NICK SHANE, OCEANOGRAPHY PROFESSOR AT THE UNIVERSITY OF MIAMI KNOWS THIS ALL TOO WELL. HE LIVES RIGHT HERE IN SOUTHWEST FLORIDA, ON SANIBEL ISLAND AND HAS SPENT DECADES STUDYING WHAT POWERS THE STORMS ANYTIME YOU'RE DEALING WITH. THE GULF OF MEXICO, YOU ALWAYS HAVE TO BE COGNIZANT. OF HOW WARM THAT WATER IS AND WHERE THESE OCEAN CREATURES ARE. BECAUSE THAT'S GOING TO MAKE A DIFFERENCE. DR. SHE'S RESEARCH GOES MUCH DEEPER THAN JUST LOOKING AT WATER. TEMPERATURES. HIS TEAM ANALYZES LOOP CURRENTS WHICH ARE SMALL CURRENTS OF WARM WATER. THE TRANSPORT HEAT FROM THE CARIBBEAN NORTHWARD INTO THE GULF OF MEXICO DURING HURRICANE SEASON. A LOOP CURRENT IS USUALLY PRESENT RIGHT OFF OUR COAST AND PIECES OF THESE CURRENTS CAN EVEN BREAK OFFER SMALLER POCKETS OF WARM WATER KNEES. RESEARCHERS AT THE UNIVERSITY OF MIAMI USE AN ARRAY OF TECHNOLOGY LIKE THESE TUBES EQUIPPED WITH SENSORS TO MEASURE SMALL FEATURES THAT CAN HAVE A BIG IMPACT ON HOW QUICKLY A STORM CAN STRENGTH. IT IS KIND OF LIKE MAKING GRANDMA'S COOKIES RIGHT? >> YOU KNOW, ALL THE INGREDIENTS GOES INTO GRANDMA'S COOKIES. BUT WHAT? IS THE PROPER RATIO OR PROPORTION THAT GOES IN THE GRANDMA'S COOKIES MAKE THEM ABSOLUTELY DELICIOUS. FINDING THAT EXACT MIX OF INGREDIENTS THAT ALLOWS THE STORM TO INTENSIFY MEANS BETTER FORECAST AND AN EARLIER WARNING AHEAD OF THE STORM. ULTIMATELY AT THE END OF THE DAY, IT'S ABOUT BUILDING A BETTER PREDICTIVE SYSTEM THAT ALLOWS YOU AND ME TO GET OUT OF HARM'S WAY. WHEN THESE TYPES OF SYSTEMS EXPLODE.
Updated: 6:49 AM EDT Aug 15, 2024
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Jason Dunning
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Loop currents and sea-surface temperatures drive rapid hurricane intensification
Updated: 6:49 AM EDT Aug 15, 2024
Editorial Standards ⓘ
Zach Scott
Digital Producer
Jason Dunning
Meteorologist
Southwest Florida, located along one of the warmest bodies of water in the Atlantic Basin, experiences intense heat in the Gulf of Mexico. If you’ve swum in these waters recently, you might have noticed the water feels more like a hot tub than a sea.This warmth plays a crucial role in fueling hurricanes, and 2024 has already set records for sea-surface temperatures, which are running about 5 degrees above average.However, it’s not just the surface temperatures that matter; the depth of the warm water is equally important. To illustrate why, imagine two pots of water: one shallow and one deep. While the shallow pot heats up quickly, the deeper pot retains heat longer and releases more energy, as evidenced by greater condensation.This analogy helps explain how deep warm waters can sustain and intensify hurricanes.Dr. Nick Shay, an oceanography professor at the University of Miami who lives on Sanibel Island, has dedicated decades to studying the dynamics of hurricane intensification. “Anytime you’re dealing with the Gulf of Mexico, you have to be aware of how warm the water is and where these ocean features are,” Dr. Shay said.Dr. Shay's research focuses on loop currents, which are warm water currents that transport heat from the Caribbean northward into the Gulf of Mexico. These currents often interact with hurricanes, contributing to their rapid intensification.During hurricane season, a loop current is typically present off the coast, and portions of these currents can break off to form smaller circular pockets of warm water known as eddies.To study these phenomena, researchers at the University of Miami employ satellites and weather instruments called dropsondes. These tools help analyze how small ocean features impact storm development and intensity.“It’s kind of like making grandma’s cookies. You know all the ingredients that go into them, but finding the right proportions is what makes them delicious,” Dr. Shay explained.Understanding the exact mix of oceanic conditions that fuel hurricanes is crucial for improving forecasts and providing earlier warnings.“Ultimately, it’s about building a better predictive system to help people get out of harm’s way when storms rapidly intensify,” Dr. Shay said.As researchers continue to delve into the complexities of sea-surface temperatures and loop currents, their work will be vital in enhancing hurricane predictions and safeguarding communities from future storms.
FORT MYERS, Fla. —
Southwest Florida, located along one of the warmest bodies of water in the Atlantic Basin, experiences intense heat in the Gulf of Mexico. If you’ve swum in these waters recently, you might have noticed the water feels more like a hot tub than a sea.
This warmth plays a crucial role in fueling hurricanes, and 2024 has already set records for sea-surface temperatures, which are running about 5 degrees above average.
Advertisement
However, it’s not just the surface temperatures that matter; the depth of the warm water is equally important. To illustrate why, imagine two pots of water: one shallow and one deep. While the shallow pot heats up quickly, the deeper pot retains heat longer and releases more energy, as evidenced by greater condensation.
This analogy helps explain how deep warm waters can sustain and intensify hurricanes.
Dr. Nick Shay, an oceanography professor at the University of Miami who lives on Sanibel Island, has dedicated decades to studying the dynamics of hurricane intensification.
“Anytime you’re dealing with the Gulf of Mexico, you have to be aware of how warm the water is and where these ocean features are,” Dr. Shay said.
Dr. Shay's research focuses on loop currents, which are warm water currents that transport heat from the Caribbean northward into the Gulf of Mexico. These currents often interact with hurricanes, contributing to their rapid intensification.
During hurricane season, a loop current is typically present off the coast, and portions of these currents can break off to form smaller circular pockets of warm water known as eddies.
To study these phenomena, researchers at the University of Miami employ satellites and weather instruments called dropsondes. These tools help analyze how small ocean features impact storm development and intensity.
“It’s kind of like making grandma’s cookies. You know all the ingredients that go into them, but finding the right proportions is what makes them delicious,” Dr. Shay explained.
Understanding the exact mix of oceanic conditions that fuel hurricanes is crucial for improving forecasts and providing earlier warnings.
“Ultimately, it’s about building a better predictive system to help people get out of harm’s way when storms rapidly intensify,” Dr. Shay said.
As researchers continue to delve into the complexities of sea-surface temperatures and loop currents, their work will be vital in enhancing hurricane predictions and safeguarding communities from future storms.
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11535 | https://stackoverflow.com/questions/9105176/how-to-sum-sequence-of-floors-numbers | algorithm - How to sum sequence of floors numbers? - Stack Overflow
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How to sum sequence of floors numbers?
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Asked 13 years, 8 months ago
Modified13 years, 8 months ago
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How can I sum the following sequence:
⌊n∕2⌋ + ⌊n+1∕2⌋ + ⌊n+2∕2⌋ + ...... + (n-1)
What I think is discard the floor and sum what inside each floor !! This is just a guess.
Give me any hint or general formula that helps me to sum them
Thanks
algorithm
sum
floor
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edited Feb 1, 2012 at 23:53
PengOne
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asked Feb 1, 2012 at 23:51
NasserNasser
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is n guaranteed to be an integer?driangle –driangle 2012-02-01 23:53:23 +00:00 Commented Feb 1, 2012 at 23:53
3 That doesn't look like a well defined sequence to me can you elaborate on the generating function?RussS –RussS 2012-02-01 23:55:15 +00:00 Commented Feb 1, 2012 at 23:55
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Since you're asking on a programming Q&A site, I must assume you want a computational answer. Here goes...
int sum = 0;
for (int j=0; j<n-1; ++j) {
sum += (n+j)/2;
}
The int will automatically truncate to the floor.
The less smart ass answer is this. Let n = 2k. Then the sum becomes
k + k + k+1 + k+1 + ... + 2k-1 + 2k-1 = 2(k + k+1 + ... + 2k-1)
and you can use the formula
1 + 2 + ... + a = a(a+1)/2
with a bit of algebra to finish it off.
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answered Feb 1, 2012 at 23:55
PengOnePengOne
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Ranty
RantyOver a year ago
This is quite smart. However, it works for even n (say n=2, then k=1 and it's 1 (2/2) + 1 (3/2) + 2 (4/2) + 2 (5/2)... works fine), but for odd n it needs correction (say n=3, then k=1 and its same as before while it should've been (3/2)=1 + (4/2)=2 + ...) by skiping the first k.
2012-02-02T00:07:08.763Z+00:00
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PengOne
PengOneOver a year ago
@Ranty For odd n, write n=2k+1 and follow the same logic. Then the first and last terms are not doubled, but the middle terms are, so you get something like k + 2(k+1 + ... + 2k-2) + 2k-1.
2012-02-03T20:32:00.417Z+00:00
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Assuming n is even, then floor(n/2) == floor((n+1)/2). And floor((n+2)/2) == floor(n/2) + 1.
The other piece in the puzzle is the expression for the sum of an arithmetic sequence, which can be found here.
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answered Feb 1, 2012 at 23:54
Oliver CharlesworthOliver Charlesworth
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for the arbitrary range 1..20 you could do:
sum = (1..20).inject{|sum, n| sum + (n/2.0).floor}
and of course you could use any range. This example is in Ruby, but you could do something similar in many languages - the algorithm is the same.
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answered Feb 1, 2012 at 23:57
acoffmanacoffman
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As long as you're not asking for a clever algorithm or optimizations, the simplest approach I can think of is good old trusty looping. In C#, one way to do that would look something like this:
```
namespace Practice
{
using System;
public class SequenceAggregator
{
public double FirstElement
{
get;
set;
}
public int Length
{
get;
set;
}
public double Calculate()
{
double sum = 0;
for (var i = FirstElement; i < FirstElement + Length; i++)
{
sum += Math.Floor(i / 2);
Console.WriteLine("i={0}, floor(i/2)={1}, sum={1}",
i, Math.Floor(i/2), sum);
}
return sum;
}
}
}
```
And you can use this class in the following way:
```
namespace Practice
{
using System;
class Program
{
static void Main(string[] args)
{
SequenceAggregator a = new SequenceAggregator();
a.FirstElement = 1;
a.Length = 3;
Console.WriteLine("Sum:{0}", a.Calculate());
}
}
}
```
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edited Feb 2, 2012 at 5:38
answered Feb 2, 2012 at 5:32
Gustavo MoriGustavo Mori
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11536 | https://icefractal.com/articles/two-child-problem/ | The Two Child Problem - Ice Fractal
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The Two Child Problem
Conditions:
A child has a 50% chance of being a boy (and otherwise is a girl).
Given a family with two children, one of whom is a boy,
What is the probability that the other child is also a boy?
There are two children, so there are four equally-likely possibilities:
Boy Girl
Boy Boy
Boy Girl
Boy
Girl Boy
Girl Girl
Girl
We know that one child is a boy.
There are only three possibilities that have at least one boy:
Boy Girl
Boy Boy
Boy Girl
Boy
Girl Boy
Girl~~Girl
Girl~~
Boy Girl
Boy Boy
Boy Girl
Boy
Girl Boy
Girl~~Girl
Girl~~
Only one of out three cases has two boys.
This gives the answer: ⅓
But is it? One could argue that the second child has nothing to do with the first, so the probability of the second child being a boy is still ½.
Let's try it and see what happens!
Run Simulation (Click Here)
(It randomly generates children and counts them.)
Results:
0 Families
0 Families with At Least One Boy
0 Families with Two Boys
of children are boys.
Out of all familes, have two boys.
Out of all familes, have at least one boy.
Out of all familes, have only one boy.
Out of the familes who have at least one boy, have two boys.
As confirmed by the first four percentages calculated, a child has a 50% chance of being a boy, and the two children are indeed independent events with 50% chance each, as we would expect. ¼ (25%) have two boys, ¾ (75%) have at least one boy, and ½ (50%) have exactly one boy.
However, the answer is ⅓ (33%) because the question is only asking about families that have at least one boy, without saying which child is that boy. The choice of which child we are asking about depends on which child was the boy required by the question. There are two permutations: either the first is that boy or the second is that boy. The child chosen second is no longer independent of the first child chosen by the question.
Here is another way of explaining where the number 33% comes from:
The intuitive way to approach this question is to split the question into two parts. It is only asking about families that have at least one boy, but does not specify which one. There are two possibilities for which boy is used to satisfy this condition, so let's split the question.
If he is the first child, then we have two possibilites:
Boy,Girl
Boy,Boy
If he is the second child, then we have two possibilities:
Girl,Boy
Boy,Boy
There are 4 total possibilities, and two out of the four have two boys, so the answer seems like 50%. However, because we split the question in two, we have counted the symmetric case of "Boy,Boy" twice! This case has a 1 in 4 chance, just the the others, yet here it is 2 in 4 because we forgot to account for the fact that although it satisfies the question in two different ways, it is still a single case, so we must remove one copy of it before calculating the final probability. This leaves us with:
Boy,Girl
Boy,Boy
Girl,Boy
Now there are three possibilities and only one has two boys, so given that you have one boy, the probability of having two boys is ⅓.
The Two Child Problem on Tuesday
Conditions:
A child has a 50% chance of being a boy (and otherwise is a girl).
A child has a 1 in 7 (~14%) chance of being born on a particular day of the week.
Given a family with two children, one of whom is a boy born on Tuesday,
What is the probability that the other child is also a boy?
Because a child is either a boy or a girl and can be born on any of the seven days of the week, there are 14 possibilities for each child. Since there are two children, this is squared, so there are 196 possibilities for each family:
Boy Girl
Su M Tu W Th F Sa Su M Tu W Th F Sa
Su
M
Tu
W
Th
F
Sa
Su
M
Tu
W
Th
F
Sa
Boy Girl
The question is only asking about families with a boy born on Tuesday, so we only consider those 27 squares (highlighted above). Out of those cases, the probability of both children being a boy is equal to the 13 cases that have two boys (highlighted green) divided by 27.
This gives the answer:13 27(48.148%)
However, your intuition probably says that Tuesday has nothing to do with it, and as with the simpler puzzle, the second child has nothing to do with the first, so the answer is 50%. Or perhaps 33% if you agree with the simpler puzzle's solution, but don't see how Tuesday can possibly make a difference here.
Which is right?
Let's try it and see what happens!
Run Simulation (Click Here)
(It randomly generates children and counts them.)
Results:
0 Families
0 Boys
0 Children Born on Tuesday
0 Families with At Least One Boy Born on Tuesday
0 Families with Two Boys with One Born on Tuesday
of children are boys.
of children were born on Tuesday.
Out of all familes, have at least one boy born on Tuesday.
Out of all familes with one boy born on Tuesday, have two boys.
(After 1,000 trials the data log will stop to avoid crashing the page, and the simulation will speed up to batches of 10,000)
If you let it run for long enough, you will see that it does indeed settle out to 48.148%, as predicted by the mathematical analysis of the problem, not 50% or 33%!
The reasoning is the same as for the simpler puzzle. The intuitive way to start is by splitting the question in two parts.
If the first child is a boy born on Tuesday, then you have these 14 possibilities:
Boy (Tuesday), Girl (Sunday)
Boy (Tuesday), Girl (Monday)
Boy (Tuesday), Girl (Tuesday)
Boy (Tuesday), Girl (Wednesday)
Boy (Tuesday), Girl (Thursday)
Boy (Tuesday), Girl (Friday)
Boy (Tuesday), Girl (Saturday)
Boy (Tuesday), Boy (Sunday)
Boy (Tuesday), Boy (Monday)
Boy (Tuesday), Boy (Tuesday)
Boy (Tuesday), Boy (Wednesday)
Boy (Tuesday), Boy (Thursday)
Boy (Tuesday), Boy (Friday)
Boy (Tuesday), Boy (Saturday)
If the second child is a boy born on Tuesday, then you have these 14 possibilities:
Girl (Sunday), Boy (Tuesday)
Girl (Monday), Boy (Tuesday)
Girl (Tuesday), Boy (Tuesday)
Girl (Wednesday), Boy (Tuesday)
Girl (Thursday), Boy (Tuesday)
Girl (Friday), Boy (Tuesday)
Girl (Saturday), Boy (Tuesday)
Boy (Sunday), Boy (Tuesday)
Boy (Monday), Boy (Tuesday)
Boy (Tuesday), Boy (Tuesday)
Boy (Wednesday), Boy (Tuesday)
Boy (Thursday), Boy (Tuesday)
Boy (Friday), Boy (Tuesday)
Boy (Saturday), Boy (Tuesday)
There are 28 total possibilities, and 14 of them have two boys, so the answer seems like 50%. However, as with the simpler puzzle, by splitting the question we have counted the symmetric case twice! The probability of having two boys born on Tuesday is 1 in 196, just like all the other cases, yet here it is 2 in 196. As before, we must remove the duplicate to get the right answer. The symmetric case still subtracts only one, but for this puzzle the addition of seven extra categories (days of the week) dilutes the effect, so the answer is closer to 50%. The reason why Tuesday changes the answer from the first question's 33% is because a boy born on Tuesday is much less likely than just a boy. For every boy born on Tuesday, there are 7 other possible boys, yet there is still only one case where both boys are equal.
As you can see on the 14x14 square, if you separate the problem and only consider the highlighted row, or only the highlighted column, you have 14 squares on each. However, when you combine them to get the final answer of 14/28, the square where they intersect will be counted twice, so you must subtract one from both the numerator and the denominator.
14 28(50%) becomes 14 - 1 28 - 1=13 27(48.148%)2 4(50%) becomes 2 - 1 4 - 1=1 3(33.333%)
The correct answers pop right out of 50%! All you need to do is subtract 1 for the symmetric case that was counted twice.
It may be easier to understand if you remove the confusion of children and days of the week. The problems can be rephrased equivalently like this:
You flipped two coins. One of them is tails. What is the probability that the other one is also tails?
You rolled two 14-sided dice. One of them is 3. What is the probability that the other one is less than 8?
Now it is much clearer why Tuesday changes everything, and why the answers will differ! In both questions, we are giving a specific condition for one die/coin, yet there are 7 ways to satisfy the other die (1,2,3,4,5,6,7), as opposed to only one way to satisfy the other coin (tails). However, there is still only one symmetric case, in which both are equal (tails,tails and 3,3).
Suppose you further increase the number of unique kinds of boys and girls, perhaps by using the 365 days of the year. The answer will be even closer to 50% because there is always only one symmetric case to subtract. For example, given a 365-day year, if one child is a boy born on January 3, the probability that the other child is a boy is 729 / 1459 (49.9657%)
As the number of unique kinds of children approaches infinity, the answer approaches the expected 50%:
lim n →∞n - 1 2n - 1=lim n →∞n 2n=1 2(50%)
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11537 | http://www2.stat.duke.edu/~sschmid/Courses/Stat376/Papers/ConvergeRates/mcsurvey.pdf |
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11538 | https://en.wikipedia.org/wiki/Kelvin_(given_name) | Jump to content
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1 People with the given name
1.1 In arts and entertainment
1.2 In politics
1.3 In sports
1.3.1 Football (soccer)
1.3.2 American football
1.3.3 Australian rules football
1.3.4 Basketball
1.3.5 Other sports
1.4 In other fields
2 Fictional characters
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From Wikipedia, the free encyclopedia
Not to be confused with Kevin.
Kelvin (/ˈkɛlvɪn/) is a masculine given name, ultimately derived from the title of William Thomson, 1st Baron Kelvin, who received a baronage named for the River Kelvin (the river flowing past the University of Glasgow) in 1892. Isolated use of "Kelvin" as a given name is recorded in England in the 1920s, and the name rises in popularity in the United States around 1950, according to the authors of the Oxford Dictionary of First Names possibly by association with the similar-sounding Kevin, which surged in popularity at about the same time, and maybe further influenced by Calvin and Melvin. In the United States the name peaked in popularity at rank 209 in 1961 and has declined steadily since, dropping to rank 726 as of 2016.
People with the given name
[edit]
In arts and entertainment
[edit]
Kelvin Fletcher, English actor
Kelvin Gosnell, British comics editor
Kelvin Harrison Jr., American actor
Kelvin Holly, American guitarist
Kelvin Kwan (born 1983), Canadian-born Hong Kong Cantopop singer
Kelvin Mercer, American DJ
Kelvin Sng, Singaporean film director
Kelvin Tan Wei Lian, Singaporean singer
In politics
[edit]
Kelvin Davis, New Zealand Member of Parliament
Kelvin Goertzen, Canadian politician
Kelvin Hopkins, British Labour MP
Kelvin Ogilvie, Canadian chemist and senator
Kelvin Thomson, Australian politician
In sports
[edit]
Football (soccer)
[edit]
Kelvin Clarke, English former footballer
Kelvin Davis, English footballer
Kelvin Etuhu, Nigerian footballer
Kelvin Jack, Trinidadian footballer
Kelvin Langmead, English footballer
Kelvin Lomax, English footballer
Kelvin Mateus de Oliveira, Brazilian footballer
Kelvin Wilson, English footballer
American football
[edit]
Kelvin Anderson, retired American football player
Kelvin Banks Jr. (born 2004), American football player
Kelvin Benjamin, American football player
Kelvin Bryant, American football player
Kelvin Harmon, American football player
Kelvin Hayden, American football player
Kelvin Joseph (born 1999), American football player
Kelvin Kinney, American football player
Kelvin Kirk, American football player
Kelvin Korver, American football player
Kelvin Smith, American football player
Australian rules football
[edit]
Kelvin Matthews, former Australian rules footballer
Kelvin Moore, former Australian rules footballer
Kelvin W. Moore, former Australian rules footballer
Kelvin Templeton, former Australian rules footballer
Basketball
[edit]
Kelvin Cato, American basketball player
Kelvin dela Peña, Filipino-Canadian basketball player
Kelvin Ransey, American basketball player
Kelvin Sampson, American basketball coach
Kelvin Scarborough, American basketball player
Kelvin Upshaw, American basketball player
Other sports
[edit]
Kelvin Batey, English competitive BMX racer
Kelvin Burt, British auto racing driver
Kelvin Davis, American boxer
Kelvin Gastelum, American mixed martial artist
Kelvin Graham (born 1964), Australian sprint canoeist
Kelvin Hoefler (born 1994), Brazilian professional street skateboarder
Kelvin Jiménez, Dominican baseball player
Kelvin Kiptum, Kenyan long-distance runner
Kelvin Li (born 1975), Hong Kong freestyle swimmer
In other fields
[edit]
Kelvin Droegemeier, American meteorologist and government advisor
Kelvin Holdsworth, Scottish clergyman
Kelvin Kent (disambiguation), multiple people
Kelvin Lancaster, American academic
Kelvin MacKenzie, British media executive
Kelvin Malone, American spree killer
Kelvin Martin, American criminal
Kelvin Ogilvie, Canadian chemist and senator
Fictional characters
[edit]
Kelvin Carpenter, fictional character in the television show EastEnders
Kelvin Gemstone, fictional character in the television show The Righteous Gemstones
Kelvin Joe Inman, secondary character on the ABC television drama Lost
References
[edit]
^ a b A Dictionary of First Names. Oxford University Press (2007) s.v. "Kelvin".
^ behindthename.com
Name list
This page or section lists people that share the same given name. If an internal link led you here, you may wish to change that link to point directly to the intended article.
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11539 | https://www.youtube.com/watch?v=IgSFdblkSFc | Simplifying Trig Expressions Involving Cot, Csc, Sec
Kisker Educational Videos
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Posted: 21 Mar 2011
This is the fourth video of a series. It contains two example problems that involving simplifying trigonometric expressions that contain cotangent, cosecant, and secant. It was created for my Pre-Calculus class.
Transcript:
all right in this video we're going back to simplifying equations but this time it's going to be a little bit tougher than simplification problems we did last time the reason is is because I'm going to start throwing in things like coent okay a simple idea to keep in mind if all else fails and you're trying to simplify an expression like this not an equation there's no equal sign if I'm trying to take this expression and make it simpler one thing you might consider if all else fails is to write every single term that you see in terms of sign and cosine for instance I know that tangent of an angle is equal to S of that angle over the cosine of the angle that's one of the basic identities you need to memorize well I don't have tangent I have coent so if tangent is s over cosine that means that coent must be the reciprocal of that I'm going to flip the fraction that must be cosine over s and the reason this is important is because now I can rewrite this expression with the coent replaced see I'm going to call it instead I'm going to call it s of t t is now our angle instead of theta plus cosine over s there's replacing Cent I'm going to take it times cosine oh I switched to Theta forgive me I'm going to leave that for just a second I'm going to switch back to oops that's just a bad habit or good habit or whatever you want to call it okay let's do this a little bit more uh I have let's see now s t plus and If I multiply this out that's cosine squared because I have cosine time cosine and that would be all over s of T and that doesn't look any simpler to me I don't know so maybe what I'm going to do then is I'm going to make this one fraction I'm going to put them over a common denominator of sign in order to do that what I would have to do is I would have to spend this out and I would have to multiply both the top and the bottom of this fraction right maybe I shouldn't have extended that out I should leave that out hold [Applause] on there I making two separate fractions like that right so if I multiply both top and bottom by sign of C now all of a sudden I have something I can put together right into one fraction and that fraction will be common denominator s of T this is going to be sin s of T and this is going to be cine s of T and so I think back to my Pythagorean the identities I know what that is that's just one isn't it that would be one over s of T but instead of writing it that way let's go ahead and write the reciprocal of that one over s would be cosecant and so I'm going to finish with this there is about as simple as it gets right I don't know what you would rather work with in a problem but I know that if I was presented with two options either this thing or this I would rather work with cosecant much simple let's try the same thing over here real quick I look at this and I see okay I got a cosecant I've got a cosine I've got a cotangent well cosecant you know I don't want to work with that I'm going to put everything in terms of s and cosine so I'm going to start off by doing this 1 over s of T minus cosine of T and coent we talked about right over here Cent is cosine over s all right now I'm going to take these two things I'm going to go ahead and multiply them out that gives me 1 over the S of T minus cosine s over the s t look at that they already have a common denominator that's pretty convenient that means I can really write this as 1 - cosine 2 over s i see a cosine squar you know what I need to think back if sin squared plus cosine s is equal to 1 and I want to change that to say 1 - cosine squ it looks like I need to subtract the cosine squar from both sides right if I do that I'm left with sin squ on on the left and 1us cosine squ on the right that's pretty convenient that means you know what I can replace this entire chump of stuff with s squar so really I have sin squ of T over the S of T and just like if you have x^2 overx and you would divide powers of X out this does the same thing one power of sign divides out of both top and bottom I'm left with just the sign of t overall much simpler e e e e e e e e e |
11540 | https://openstax.org/books/chemistry-2e/pages/21-3-radioactive-decay | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Chemistry 2e
21.3 Radioactive Decay
Chemistry 2e21.3 Radioactive Decay
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Recognize common modes of radioactive decay
Identify common particles and energies involved in nuclear decay reactions
Write and balance nuclear decay equations
Calculate kinetic parameters for decay processes, including half-life
Describe common radiometric dating techniques
Following the somewhat serendipitous discovery of radioactivity by Becquerel, many prominent scientists began to investigate this new, intriguing phenomenon. Among them were Marie Curie (the first woman to win a Nobel Prize, and the only person to win two Nobel Prizes in different sciences—chemistry and physics), who was the first to coin the term “radioactivity,” and Ernest Rutherford (of gold foil experiment fame), who investigated and named three of the most common types of radiation. During the beginning of the twentieth century, many radioactive substances were discovered, the properties of radiation were investigated and quantified, and a solid understanding of radiation and nuclear decay was developed.
The spontaneous change of an unstable nuclide into another is radioactive decay. The unstable nuclide is called the parent nuclide; the nuclide that results from the decay is known as the daughter nuclide. The daughter nuclide may be stable, or it may decay itself. The radiation produced during radioactive decay is such that the daughter nuclide lies closer to the band of stability than the parent nuclide, so the location of a nuclide relative to the band of stability can serve as a guide to the kind of decay it will undergo (Figure 21.5).
Figure 21.5 A nucleus of uranium-238 (the parent nuclide) undergoes α decay to form thorium-234 (the daughter nuclide). The alpha particle removes two protons (green) and two neutrons (gray) from the uranium-238 nucleus.
Link to Learning
Although the radioactive decay of a nucleus is too small to see with the naked eye, we can indirectly view radioactive decay in an environment called a cloud chamber. Click here to learn about cloud chambers and to view an interesting Cloud Chamber Demonstration from the Jefferson Lab.
Types of Radioactive Decay
Ernest Rutherford’s experiments involving the interaction of radiation with a magnetic or electric field (Figure 21.6) helped him determine that one type of radiation consisted of positively charged and relatively massive α particles; a second type was made up of negatively charged and much less massive β particles; and a third was uncharged electromagnetic waves, γ rays. We now know that α particles are high-energy helium nuclei, β particles are high-energy electrons, and γ radiation compose high-energy electromagnetic radiation. We classify different types of radioactive decay by the radiation produced.
Figure 21.6 Alpha particles, which are attracted to the negative plate and deflected by a relatively small amount, must be positively charged and relatively massive. Beta particles, which are attracted to the positive plate and deflected a relatively large amount, must be negatively charged and relatively light. Gamma rays, which are unaffected by the electric field, must be uncharged.
Alpha (α) decay is the emission of an α particle from the nucleus. For example, polonium-210 undergoes α decay:
Alpha decay occurs primarily in heavy nuclei (A > 200, Z > 83). Because the loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide, the daughter nuclide has a larger n:p ratio than the parent nuclide. If the parent nuclide undergoing α decay lies below the band of stability (refer to Figure 21.2), the daughter nuclide will lie closer to the band.
Beta (β) decay is the emission of an electron from a nucleus. Iodine-131 is an example of a nuclide that undergoes β decay:
Beta decay, which can be thought of as the conversion of a neutron into a proton and a β particle, is observed in nuclides with a large n:p ratio. The beta particle (electron) emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Such nuclei lie above the band of stability. Emission of an electron does not change the mass number of the nuclide but does increase the number of its protons and decrease the number of its neutrons. Consequently, the n:p ratio is decreased, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Gamma emission (γ emission) is observed when a nuclide is formed in an excited state and then decays to its ground state with the emission of a γ ray, a quantum of high-energy electromagnetic radiation. The presence of a nucleus in an excited state is often indicated by an asterisk (). Cobalt-60 emits γ radiation and is used in many applications including cancer treatment:
There is no change in mass number or atomic number during the emission of a γ ray unless the γ emission accompanies one of the other modes of decay.
Positron emission (β+ decay) is the emission of a positron from the nucleus. Oxygen-15 is an example of a nuclide that undergoes positron emission:
Positron emission is observed for nuclides in which the n:p ratio is low. These nuclides lie below the band of stability. Positron decay is the conversion of a proton into a neutron with the emission of a positron. The n:p ratio increases, and the daughter nuclide lies closer to the band of stability than did the parent nuclide.
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus. For example, potassium-40 undergoes electron capture:
Electron capture occurs when an inner shell electron combines with a proton and is converted into a neutron. The loss of an inner shell electron leaves a vacancy that will be filled by one of the outer electrons. As the outer electron drops into the vacancy, it will emit energy. In most cases, the energy emitted will be in the form of an X-ray. Like positron emission, electron capture occurs for “proton-rich” nuclei that lie below the band of stability. Electron capture has the same effect on the nucleus as does positron emission: The atomic number is decreased by one and the mass number does not change. This increases the n:p ratio, and the daughter nuclide lies closer to the band of stability than did the parent nuclide. Whether electron capture or positron emission occurs is difficult to predict. The choice is primarily due to kinetic factors, with the one requiring the smaller activation energy being the one more likely to occur.
Figure 21.7 summarizes these types of decay, along with their equations and changes in atomic and mass numbers.
Figure 21.7 This table summarizes the type, nuclear equation, representation, and any changes in the mass or atomic numbers for various types of decay.
Chemistry in Everyday Life
PET Scan
Positron emission tomography (PET) scans use radiation to diagnose and track health conditions and monitor medical treatments by revealing how parts of a patient’s body function (Figure 21.8). To perform a PET scan, a positron-emitting radioisotope is produced in a cyclotron and then attached to a substance that is used by the part of the body being investigated. This “tagged” compound, or radiotracer, is then put into the patient (injected via IV or breathed in as a gas), and how it is used by the tissue reveals how that organ or other area of the body functions.
Figure 21.8 A PET scanner (a) uses radiation to provide an image of how part of a patient’s body functions. The scans it produces can be used to image a healthy brain (b) or can be used for diagnosing medical conditions such as Alzheimer’s disease (c). (credit a: modification of work by Jens Maus)
For example, F-18 is produced by proton bombardment of 18O and incorporated into a glucose analog called fludeoxyglucose (FDG). How FDG is used by the body provides critical diagnostic information; for example, since cancers use glucose differently than normal tissues, FDG can reveal cancers. The 18F emits positrons that interact with nearby electrons, producing a burst of gamma radiation. This energy is detected by the scanner and converted into a detailed, three-dimensional, color image that shows how that part of the patient’s body functions. Different levels of gamma radiation produce different amounts of brightness and colors in the image, which can then be interpreted by a radiologist to reveal what is going on. PET scans can detect heart damage and heart disease, help diagnose Alzheimer’s disease, indicate the part of a brain that is affected by epilepsy, reveal cancer, show what stage it is, and how much it has spread, and whether treatments are effective. Unlike magnetic resonance imaging and X-rays, which only show how something looks, the big advantage of PET scans is that they show how something functions. PET scans are now usually performed in conjunction with a computed tomography scan.
Radioactive Decay Series
The naturally occurring radioactive isotopes of the heaviest elements fall into chains of successive disintegrations, or decays, and all the species in one chain constitute a radioactive family, or radioactive decay series. Three of these series include most of the naturally radioactive elements of the periodic table. They are the uranium series, the actinide series, and the thorium series. The neptunium series is a fourth series, which is no longer significant on the earth because of the short half-lives of the species involved. Each series is characterized by a parent (first member) that has a long half-life and a series of daughter nuclides that ultimately lead to a stable end-product—that is, a nuclide on the band of stability (Figure 21.9). In all three series, the end-product is a stable isotope of lead. The neptunium series, previously thought to terminate with bismuth-209, terminates with thallium-205.
Figure 21.9 Uranium-238 undergoes a radioactive decay series consisting of 14 separate steps before producing stable lead-206. This series consists of eight α decays and six β decays.
Radioactive Half-Lives
Radioactive decay follows first-order kinetics. Since first-order reactions have already been covered in detail in the kinetics chapter, we will now apply those concepts to nuclear decay reactions. Each radioactive nuclide has a characteristic, constant half-life (t1/2), the time required for half of the atoms in a sample to decay. An isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, and how long a sample of an undesirable or dangerous isotope must be stored before it decays to a low-enough radiation level that is no longer a problem.
For example, cobalt-60, an isotope that emits gamma rays used to treat cancer, has a half-life of 5.27 years (Figure 21.10). In a given cobalt-60 source, since half of the nuclei decay every 5.27 years, both the amount of material and the intensity of the radiation emitted is cut in half every 5.27 years. (Note that for a given substance, the intensity of radiation that it produces is directly proportional to the rate of decay of the substance and the amount of the substance.) This is as expected for a process following first-order kinetics. Thus, a cobalt-60 source that is used for cancer treatment must be replaced regularly to continue to be effective.
Figure 21.10 For cobalt-60, which has a half-life of 5.27 years, 50% remains after 5.27 years (one half-life), 25% remains after 10.54 years (two half-lives), 12.5% remains after 15.81 years (three half-lives), and so on.
Since nuclear decay follows first-order kinetics, we can adapt the mathematical relationships used for first-order chemical reactions. We generally substitute the number of nuclei, N, for the concentration. If the rate is stated in nuclear decays per second, we refer to it as the activity of the radioactive sample. The rate for radioactive decay is:
decay rate = λN with λ = the decay constant for the particular radioisotope
The decay constant, λ, which is the same as a rate constant discussed in the kinetics chapter. It is possible to express the decay constant in terms of the half-life, t1/2:
The first-order equations relating amount, N, and time are:
where N0 is the initial number of nuclei or moles of the isotope, and Nt is the number of nuclei/moles remaining at time t. Example 21.5 applies these calculations to find the rates of radioactive decay for specific nuclides.
Example 21.5
Rates of Radioactive Decay
decays with a half-life of 5.27 years to produce
(a) What is the decay constant for the radioactive disintegration of cobalt-60?
(b) Calculate the fraction of a sample of the isotope that will remain after 15 years.
(c) How long does it take for a sample of to disintegrate to the extent that only 2.0% of the original amount remains?
Solution
(a) The value of the rate constant is given by:
(b) The fraction of that is left after time t is given by Rearranging the first-order relationship Nt = N0e–λt to solve for this ratio yields:
The fraction of that will remain after 15.0 years is 0.138. Or put another way, 13.8% of the originally present will remain after 15 years.
(c) 2.00% of the original amount of is equal to 0.0200 N0. Substituting this into the equation for time for first-order kinetics, we have:
Check Your Learning
Radon-222, has a half-life of 3.823 days. How long will it take a sample of radon-222 with a mass of 0.750 g to decay into other elements, leaving only 0.100 g of radon-222?
Answer:
11.1 days
Because each nuclide has a specific number of nucleons, a particular balance of repulsion and attraction, and its own degree of stability, the half-lives of radioactive nuclides vary widely. For example: the half-life of is 1.9 1019 years; is 24,000 years; is 3.82 days; and element-111 (Rg for roentgenium) is 1.5 10–3 seconds. The half-lives of a number of radioactive isotopes important to medicine are shown in Table 21.2, and others are listed in Appendix M.
Half-lives of Radioactive Isotopes Important to Medicine
| Type1 | Decay Mode | Half-Life | Uses |
--- --- |
| F-18 | β+ decay | 110. minutes | PET scans |
| Co-60 | β decay, γ decay | 5.27 years | cancer treatment |
| Tc-99m | γ decay | 8.01 hours | scans of brain, lung, heart, bone |
| I-131 | β decay | 8.02 days | thyroid scans and treatment |
| Tl-201 | electron capture | 73 hours | heart and arteries scans; cardiac stress tests |
Table 21.2
Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for purposes of “dating” the origin of objects such as archaeological artifacts, formerly living organisms, or geological formations. This process is radiometric dating and has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilization. We will explore some of the most common types of radioactive dating and how the particular isotopes work for each type.
Radioactive Dating Using Carbon-14
The radioactivity of carbon-14 provides a method for dating objects that were a part of a living organism. This method of radiometric dating, which is also called radiocarbon dating or carbon-14 dating, is accurate for dating carbon-containing substances that are up to about 30,000 years old, and can provide reasonably accurate dates up to a maximum of about 50,000 years old.
Naturally occurring carbon consists of three isotopes: which constitutes about 99% of the carbon on earth; about 1% of the total; and trace amounts of Carbon-14 forms in the upper atmosphere by the reaction of nitrogen atoms with neutrons from cosmic rays in space:
All isotopes of carbon react with oxygen to produce CO2 molecules. The ratio of to depends on the ratio of to in the atmosphere. The natural abundance of in the atmosphere is approximately 1 part per trillion; until recently, this has generally been constant over time, as seen is gas samples found trapped in ice. The incorporation of and into plants is a regular part of the photosynthesis process, which means that the ratio found in a living plant is the same as the ratio in the atmosphere. But when the plant dies, it no longer traps carbon through photosynthesis. Because is a stable isotope and does not undergo radioactive decay, its concentration in the plant does not change. However, carbon-14 decays by β emission with a half-life of 5730 years:
Thus, the ratio gradually decreases after the plant dies. The decrease in the ratio with time provides a measure of the time that has elapsed since the death of the plant (or other organism that ate the plant). Figure 21.11 visually depicts this process.
Figure 21.11 Along with stable carbon-12, radioactive carbon-14 is taken in by plants and animals, and remains at a constant level within them while they are alive. After death, the C-14 decays and the C-14:C-12 ratio in the remains decreases. Comparing this ratio to the C-14:C-12 ratio in living organisms allows us to determine how long ago the organism lived (and died).
For example, with the half-life of being 5730 years, if the ratio in a wooden object found in an archaeological dig is half what it is in a living tree, this indicates that the wooden object is 5730 years old. Highly accurate determinations of ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
Link to Learning
Visit this website to perform simulations of radiometric dating.
Example 21.6
Radiocarbon Dating
A tiny piece of paper (produced from formerly living plant matter) taken from the Dead Sea Scrolls has an activity of 10.8 disintegrations per minute per gram of carbon. If the initial C-14 activity was 13.6 disintegrations/min/g of C, estimate the age of the Dead Sea Scrolls.
Solution
The rate of decay (number of disintegrations/minute/gram of carbon) is proportional to the amount of radioactive C-14 left in the paper, so we can substitute the rates for the amounts, N, in the relationship:
where the subscript 0 represents the time when the plants were cut to make the paper, and the subscript t represents the current time.
The decay constant can be determined from the half-life of C-14, 5730 years:
Substituting and solving, we have:
Therefore, the Dead Sea Scrolls are approximately 1900 years old (Figure 21.12).
Figure 21.12 Carbon-14 dating has shown that these pages from the Dead Sea Scrolls were written or copied on paper made from plants that died between 100 BC and AD 50.
Check Your Learning
More accurate dates of the reigns of ancient Egyptian pharaohs have been determined recently using plants that were preserved in their tombs. Samples of seeds and plant matter from King Tutankhamun’s tomb have a C-14 decay rate of 9.07 disintegrations/min/g of C. How long ago did King Tut’s reign come to an end?
Answer:
about 3350 years ago, or approximately 1340 BC
There have been some significant, well-documented changes to the ratio. The accuracy of a straightforward application of this technique depends on the ratio in a living plant being the same now as it was in an earlier era, but this is not always valid. Due to the increasing accumulation of CO2 molecules (largely in the atmosphere caused by combustion of fossil fuels (in which essentially all of the has decayed), the ratio of in the atmosphere may be changing. This manmade increase in in the atmosphere causes the ratio to decrease, and this in turn affects the ratio in currently living organisms on the earth. Fortunately, however, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radioactive Dating Using Nuclides Other than Carbon-14
Radioactive dating can also use other radioactive nuclides with longer half-lives to date older events. For example, uranium-238 (which decays in a series of steps into lead-206) can be used for establishing the age of rocks (and the approximate age of the oldest rocks on earth). Since U-238 has a half-life of 4.5 billion years, it takes that amount of time for half of the original U-238 to decay into Pb-206. In a sample of rock that does not contain appreciable amounts of Pb-208, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of U-238:Pb-206, we can determine the age of the rock. This assumes that all of the lead-206 present came from the decay of uranium-238. If there is additional lead-206 present, which is indicated by the presence of other lead isotopes in the sample, it is necessary to make an adjustment. Potassium-argon dating uses a similar method. K-40 decays by positron emission and electron capture to form Ar-40 with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of Ar-40 gas that escapes is measured, determination of the Ar-40:K-40 ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (Rb-87 decays into Sr-87 with a half-life of 48.8 billion years), operate on the same principle. To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Example 21.7
Radioactive Dating of Rocks
An igneous rock contains 9.58 10–5 g of U-238 and 2.51 10–5 g of Pb-206, and much, much smaller amounts of Pb-208. Determine the approximate time at which the rock formed.
Solution
The sample of rock contains very little Pb-208, the most common isotope of lead, so we can safely assume that all the Pb-206 in the rock was produced by the radioactive decay of U-238. When the rock formed, it contained all of the U-238 currently in it, plus some U-238 that has since undergone radioactive decay.
The amount of U-238 currently in the rock is:
Because when one mole of U-238 decays, it produces one mole of Pb-206, the amount of U-238 that has undergone radioactive decay since the rock was formed is:
The total amount of U-238 originally present in the rock is therefore:
The amount of time that has passed since the formation of the rock is given by:
with N0 representing the original amount of U-238 and Nt representing the present amount of U-238.
U-238 decays into Pb-206 with a half-life of 4.5 109 y, so the decay constant λ is:
Substituting and solving, we have:
Therefore, the rock is approximately 1.7 billion years old.
Check Your Learning
A sample of rock contains 6.14 10–4 g of Rb-87 and 3.51 10–5 g of Sr-87. Calculate the age of the rock. (The half-life of the β decay of Rb-87 is 4.7 1010 y.)
Answer:
3.7 109 y
Footnotes
1The “m” in Tc-99m stands for “metastable,” indicating that this is an unstable, high-energy state of Tc-99. Metastable isotopes emit γ radiation to rid themselves of excess energy and become (more) stable.
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Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD
Publisher/website: OpenStax
Book title: Chemistry 2e
Publication date: Feb 14, 2019
Location: Houston, Texas
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11541 | https://www.wyzant.com/resources/answers/902883/find-the-focus-and-directrix-of-the-parabola-x-y-2-0 | WYZANT TUTORING
Kerr I.
Find the focus and Directrix of the parabola x+y^2=0
I have been able to do every answer but this about focus and directix, can someone help solve and help me understand what's the difference when y has the exponent instead of the x?
3 Answers By Expert Tutors
Robert K.
answered • 08/24/22
Experienced Math Tutor Who Will Improve Both Understanding and Grades
Here's the simplest way to look at it.
If the equation can be written y = ax^2 with a being positive then the parabola opens upward.
If the equation can be written y = ax^2 with a being negative then the parabola opens downward.
If the equation can be written x = ay^2 with a being positive then the parabola opens to the right.
If the equation can be written x = ay^2 with a being negative then the parabola opens to the left.
This equation falls into the last category and can be written y^2 = -x so it opens to the left.
Now lets put the equation into standard form (y - k)^2 = 4p(x - h)
(y - 0)^2 = -1(x - 0)
h and k are both zero here so the vertex is(0,0) which is the origin.
p is the focal distance and since 4p = -1 the p = -1/4.
Since the parabola opens to the left the focus will be 1/4 units to the left of (0,0) at (-1/4,0). (The focus is always 'inside' the parabola.)
The directrix will be a vertical line 1/4 units to the right of (0,0). Its equation will be x = 1/4.
Sammie W. answered • 08/24/22
Math Teacher & Air Force Helping Students Turn It Around
Sammie W.
08/24/22
Kerr I.
08/24/22
Sammie W.
08/24/22
Paul S.
answered • 08/24/22
Quantitative Strategist
For a vertical parabola (x has the exponent) we use the generic formula: (x - h)2 = 4p(y - k)
For a horizontal parabola (y has the exponent instead of x) we use basically the same formula but with the stuff inside the parenthesis switched around: (y - k)2 = 4p(x - h)
In both formulas, p is the distance from the vertex to the focus and it is also the distance from the vertex to the directrix (but on the other side of the vertex). The focus will always sit inside the parabola while the directrix will be a line behind the parabola. Recall that the vertex in both formulas is (h, k).
The equation we are given, x + y2 = 0, doesn't quite match the formula above so we need to make some rearrangements:
(1) x and y need to be on opposite sides of the equation so let's move x to the other side. We do this by subtracting x from both sides. This gives us: y2 = -x
(2) Next we need to figure out h and k. In this case there are no other numbers in the original equation so both h and k are 0. To see this we can add some parenthesis and see that nothing changes in the equation: -(x - 0) is the same as -x, and (y - 0)2 is the same as (y2). Now that we know the values for h and k we can write our formula as: (y - 0)2 = -(x - 0)
(3) Finally we need to what to do about the 4p that is supposed to be in front of the (x - 0) term. To do this we need to ask ourselves what we can set p to so that our formula, (y - 0)2 = -(x - 0), matches the generic formula, (y - k)2 = 4p(x - h). Well, there is no number (coefficient) in front of our (x - 0) term, just a negative sign, which always means that the coefficient is -1, similarly, if there was no number or negative sign the coefficient would be 1. We can see that -(x - h) is the same as -1(x - h), so let's rewrite out formula:
(y - 0)2 = -1(x - 0).
Almost done! comparing our formula with the generic formula we can see that h and k are both zeros (from step 2) and 4p is sitting in the place where we have a -1. This means 4p has to be equal to -1. Let's solve this equality to find p.
(3.1) Our starting equation: 4p = -1
(3.2) Divide both sides by 4 to get: p = -1/4 (in fraction form) or p = -0.25 (in decimal form)
(4) Now we have h = 0, k = 0 and p = -1/4, this is everything we need! The vertex is (h, k) so our vertex is (0, 0) and the distance from the vertex to the vertex is -1/4. Keep in mind, p is negative we will be moving in a negative direction away from the vertex to find the focus. Since out parabola is horizontal we will be moving right or left to find the focus.
(4.1) Focus: From our vertex (0, 0) we will move 1/4 to the left (negative direction), this gives us a focus of (-1/4, 0)
(4.2) Directrix: The directrix will be the same distance from the vertex but in the opposite direction so we will move 1/4 to the right, this puts the directrix at (1/4, 0), but remember that the directrix is a line not a point like the focus. Since our parabola is horizontal the directrix will be a vertical line (remember all vertical lines have the equation x = ?, where the question mark is the place on the x-axis that the line crosses through). In our case the directrix is at the x location of 1/4 (from the coordinate (1/4, 0) that we found) so our directrix has the equation x = 1/4.
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11542 | https://www.youtube.com/watch?v=BERKZEcFnV8 | Circle Theorems: Angles Subtended by the same Arc at the Circumference are Equal, Geometric Proof
Mathematics Proofs - GCSE & A Level
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Circle Theorems: Angles Subtended by the same Arc at the Circumference are Equal, Geometric Proof
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Transcript:
Intro hello in this video I'm going to be showing you why angles subtended by the same Arc at the circumference are equal so basically why this angle up here using the black lines and this angle up here using the black lines are equal now to show that this over here is going to be true we first need to form another Circle and we need to show that the angle up here at the circum erence using this blue and green line okay is going to be half the angle at the center of this circle over here so let's say we have an angle up here using the blue and green line right H this angle over here using the red lines at the center of this circle is going to be double the angle up here if we can show this then we can show that the other theorem is true Example so how do we show that the angle up here is half the angle at the center of this circle over here well firstly we're going to name these red lines over here R okay and if this line over here is equal to R and this line over here is equal to R then this triangle on the left is an isoceles triangle okay and we can name the angle over here a and we can also label the angle over here a okay and this angle over here is going to be Pi minus 2 a okay because Pi radians is the same as 180° so we're just subtracting these two angles over here from 180° to get this angle over here now this triangle to the right is also going to be an isoceles triangle okay so we can name the angle up here B and we can name the angle down here B this means that this angle over here is going to be Pi minus 2B okay now since we have this angle over here and this angle over here we can actually find this angle over here below because we're going to subtract this and this from 360° or 2 pi Solution radians okay so let's subtract these two angles over here from 2 pi so we have 2 pi - pi - 2 Aus Pi - 2 B this will give us 2 pi minus pi+ 2 a - PK + 2 b okay so from here we get 2 piun - piun minus piun which is 0o so we can cross these out so we're left with 2 a + 2 B and we can factorize this okay to 2 times a + b in Brackets so this angle over here is going to be 2times a plus b but since this angle up here is a plus b that means that the angle at the center has to be double this angle over here at the top okay because we've got two times a + b okay now since we know this it's going to be fairly simple to show that this angle up here using the black lines and this angle up here using the black lines are equal okay we're going to show that the angles subtended by the same Arc at the circumference so we're looking at this Arc over here are equal okay so these Proof angles up here are going to be equal so let's call this angle over here x okay because of this theorem to the left over here we know that this angle over here is going going to be 2x Right double this angle up here now let's call this angle over here y if that is the case because of the theorem to the left this angle over here is also going to be equal to 2 y okay and we can see this is the case because of these red lines over here these are the radiuses of this circle okay so what we're left with is 2x is equal to 2 y so let's write this down as an equation 2x is equal to 2 y now if we divide both sides by two like so we're going to end up with X is equal to Y okay because 2 over two is simply equal to one 1 X is X or 1 Y is y all right so we've shown that X is equal to Y in this case and that the angle subtended by the same Arc which is this Arc over here at the circumference are equal okay so these two angles over here are actually equal I hope this video has been helpful to you if you have any questions leave them in the comment section and I'll get back to you as soon as I can for more content like this make sure you are subscribed to my channel and also check out mathematics. proofs on Instagram thanks for watching this video and see you next time |
11543 | https://www.tsc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/lib-arts-ii-mgf1107/Changing-Between-Base-TWO-and-Base-TEN.pdf | Changing Between Base Two and Base Ten Base Ten Base 10 consists of 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The position of the digit determines its value. Consider the different positions and values for the digit 3 in the following numbers 2173 here the 3 means: 3 times 1 or 3 2137 here the 3 means: 3 times 10 or 30 2317 here the 3 means: 3 times 100 or 300 3171 here the 3 means: 3 times 1000 or 3000 Standard Form: The number 64,702 is in standard form. its base ten values are: Expanded Form: The same number in expanded form is; Base Two Base two consists of two digits 0 and 1. Like base 10 its value is determined by its position Standard Form: The number 1 1 0 1 1 is in standard form. Its base two places are: Expanded Form: The same number in expanded form is: Changing base two to ten Example 1. Write as a base ten number Example 2. An alternative method: Write as a base ten number. Step 1. Draw as many blanks as there are digits in the given base TWO number in this example there are 6. Step 2. Beginning under the rightmost blank, label the base TWO place values, i.e., 1, 2, 4, 8, 16, 32. Step 3. Write the base TWO number in the blanks. Step 4. Add the place values under the 1s in the base TWO number: 32+8+4+2+1=47 Changing Base Ten to base two Example 3: Write 365 as a base two number Step 1. Make a guess of how many blanks you may need, you will adjust how many you have in step 3. Step 2. Beginning under the rightmost blank, label the base two values. i.e. 1,2,4,8,16,32,64,132 and so on. Stop when the place values exceeds the given base ten number, which in this example is 365. Step 3. Erase the blanks and place values that exceed the given number. Step 4. Beginning with the given number, subtract the largest place value and put a 1 in that place value blank. Use the difference and try to subtract the next place value. If the subtraction is not possible without getting a negative number, put a 0 in the place value blank and try the next place value. Continue in this manner putting a 1 in each of the place value blanks where the subtraction is possible and a 0 in the blanks where the subtraction is not possible. Step 5. Use the method in example 2 to check your work |
11544 | https://international-university.eu/dispert_website/Digital_Design_Website_English/digital_2/dig002_3.html | Digital Design: Boolean Algebra and digital Logic
ContentsPrevious ChapterNext Chapter
2 Combinational Circuit Design
2.3 Canonical and Standard Forms (Normal Forms)
As has been shown, an arbitrary Boolean Function can be realized using single-tier combinations of Mintems or Maxterms.
This approach to describe functions leads directly to the definition of so-called Canonical Forms
Definition:
Canonical Disjunctive Normal Form (cDNF):
In a Canonical Disjunctive Normal Form a Boolean function will be presented as the disjunction (OR-Function) of its Minterms mi.
Canonical Conjunctive Normal Form (cCNF):
In a Canonical Conjunctive Normal Form a Boolean function will be presented as the conjunction (AND-Function) of its Maxterms Mi.
Note:
The term "canonical" indicates that in this case the Boolean Function will be described in the most simple (and most elaborate) form. But in many cases simplifications are possible when describing functions (see below), so that the single terms will not be in the canonical forms anymore.
A cCNF as well as a cDNF are realized as two-tier (2-level) circuits. In case that in the first tier not all variables are used, the solution cannot be called "canonical" anymore.
In that case the corresponding Normal Forms are defined as:
Disjunctive Normal Form (DNF) and
Conjunctive Normal Form (CNF), respectively.
Simplified both forms are also known as disjunctive and conjunctive form, respectively,
Furthermore the English literature has adopted the terms
"sum of products" (SOP) and
"product of sums" (POS), respectively.
Note: Using Maxterm Functions (Maxterms)
Experience shows that it seems somehow more difficult to use the Canonical Conjunctive Normal Form (cCNF). Therefore the basic procedure will be clarified in more detail using an example:
Karnaugh Diagram of a given arbitrary sample Boolean function:
Figure 2.18: Sample Function.
Obviously this Boolean function is totally described using six 1-fields and two 0-fields.
A Maxterm is described using a single 0-field, therefore it will be formed as a complete disjunction of all seven 1-fields. To describe the sample function two maxterms have to be used.
These two maxterms are the following terms M2 and M7:
Figure 2.19: Maxterms to describe the Sample Function in Figure 2.18.
As can easily be seen, the overlay of these two maxterms will result in the sample Boolean function. It is clear that the maxterms have to be combined conjunctively, i.e. the cut-set has to be formed.
2.3.1 Example using functions:
Using the above definitions, the cDNF and the cKNF can be derived directly from the function's truth table.
May the Boolean function f(a,b,c) be defined through the following truth table:
a b c f(a,b,c) 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 0
Figure 2.20: Truth Table and Karnaugh Diagram to present a Boolean Function.
a) Using the canonical Disjunctive Normal Form (cDNF):
In the canonical Disjunctive Normal Form f D all the conjunctions (AND-function) of the input variables a,b,c that have the result f D=1 are combined disjunctively (OR-function).
In this example this means:
. (2.1)
Obviously only those rows (in the truth table) were considered for which f(a,b,c) is equal to '1'.
Using minterm notation equation 2.1 can be expressed as:
. (2.2)
b) Using the canonical Conjunctive Normal Form (cCNF):
In the canonical Conjunctive Normal Form f K all the disjunctions (OR-function) of the input variables a,b,c that have the result f K=0 are combined conjunctively (AND-function).
In this example this means:
. (2.3)
Obviously only those rows (in the truth table) were considered for which f(a,b,c) is equal to '0'.
Using maxterm notation equation 2.3 can be expressed as:
. (2.4)
Summary:
An arbitrary Boolean Function f(x1,...,xn) can always be expressed as cDNF (canonical Disjunctive Normal Form):
(2.5)
or as
cCNF (canonical Conjunctive Normal Form):
, (2.6)
with:
with: m i = Minterm, M i = Maxterm.
A Minterm combines all variables conjunctively,
A Maxterm combines all variables disjunctively.
A cDNF becomes a DNF (Disjunctive Normal Form), when at least one conjunctive term is not a Minterm.
A cCNF becomes a CNF (Conjunctive Normal Form), when at least one disjunctive term is not a Maxterm.
ContentsPrevious ChapterNext Chapter |
11545 | https://www.eas.ualberta.ca/jdwilson/EAS372_13/Vomel_CIRES_satvpformulae.html | This is Google's cache of It is a snapshot of the page as it appeared on 2 Jan 2013 23:37:43 GMT. The current page could have changed in the meantime. Learn more
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Water Vapor Pressure Formulations
Saturation
vapor pressure formulations
Holger
Vömel
CIRES,
University of Colorado, Boulder
A large number of saturation vapor
pressure equations exists to calculate the pressure of water vapor
over a surface of liquid water or ice. This is a brief overview of
the most important equations used. Several useful reviews of the
existing vapor pressure curves are listed in the references. Please
note the discussion of the WMO formulations.
1) Vapor pressure over liquid water below 0°C
Goff Gratch equation
(Smithsonian Tables, 1984, after Goff and
Gratch, 1946):
Log10 ew
= -7.90298
(373.16/T-1)
+ 5.02808 Log10(373.16/T)
- 1.3816 10-7 (1011.344 (1-T/373.16)
-1)
+ 8.1328 10-3 (10-3.49149 (373.16/T-1)
-1)
+ Log10(1013.246)
with T in [K] and ew
in [hPa]
Guide to Meteorological
Instruments and Methods of Observation (CIMO Guide)
(WMO,
2008)
ew =
6.112 e(17.62 t/(243.12 +
t))
with t in [°C] and ew in
[hPa]
WMO
(Goff,
1957):
Log10 ew
= 10.79574
(1-273.16/T)
- 5.02800 Log10(T/273.16)
+ 1.50475 10-4 (1 - 10(-8.2969(T/273.16-1)))
+ 0.42873 10-3 (10(+4.76955(1-273.16/T))- 1)
+ 0.78614
with T in [K] and ew
in [hPa]
(Note: WMO based its
recommendation on a paper by Goff (1957), which is shown here. The
recommendation published by WMO (1988) has several typographical
errors and cannot be used. A corrigendum (WMO, 2000) shows the term
+0.42873 10-3 (10(-4.76955(1-273.16/T))- 1) in the fourth line compared to the original publication
by Goff (1957). Note the different sign of the exponent. The earlier
1984 edition shows the correct formula.)
Hyland and Wexler
(Hyland
and Wexler, 1983):
Log ew =
-0.58002206 104 /
T
+ 0.13914993 101
- 0.48640239 10-1 T
+ 0.41764768 10-4 T2
- 0.14452093 10-7 T3
+ 0.65459673 101 Log(T)
with T in [K]
and ew in [Pa]
Buck
(Buck
Research Manual (1996); updated equation from Buck, A. L., New
equations for computing vapor pressure and enhancement factor, J.
Appl. Meteorol., 20, 1527-1532, 1981)
ew
= 6.1121 e(18.678 - t /
234.5) t / (257.14 + t)
ew = 6.1121 e17.502 t/ (240.97 + t)
with t in [°C] and ew in
[hPa]
Sonntag
(Sonntag,
1994)
Log ew =
-6096.9385 /
T
+ 16.635794
- 2.711193 10-2 T
+ 1.673952 10-5 T2
+ 2.433502 Log(T)
with T in [K] and ew
in [hPa]
Magnus Tetens
(Murray,
1967)
ew =
6.1078 e17.269388 (T-273.16) /
(T –
35.86)
with T in [K] and ew in [hPa]
Bolton
(Bolton,
1980)
ew =
6.112 e17.67 t /
(t+243.5)
with t in [°C] and ew in [hPa]
Murphy and Koop
(Murphy
and Koop, 2005)
Log ew =
54.842763
- 6763.22 / T
- 4.21 Log(T)
+ 0.000367 T
+ Tanh{0.0415 (T - 218.8)}
· (53.878 - 1331.22 / T - 9.44523 Log(T)
+ 0.014025 T)
with T in [K] and ew
in [Pa]
International Association for
the Properties of Water and Steam (IAPWS) Formulation 1995
(Wagner
and Pruß, 2002)
Log (ew/22.064e6)
= 647.096/T ((-7.85951783 ν
+ 1.84408259 ν1.5
- 11.7866497 ν3
+ 22.6807411 ν3.5
- 15.9618719 ν4
+ 1.80122502 ν7.5))
with T in [K] and ew
in [Pa] and ν = 1 - T/647.096
At low temperatures most of these
are based on theoretical studies and only a small number are based on
actual measurements of the vapor pressure. The Goff Gratch equation
for the vapor pressure over liquid water covers a region of -50°C
to 102°C [Gibbins 1990]. This work is generally considered the
reference equation but other equations are in use in the
meteorological community [Elliott and Gaffen, 1993]. There is a very
limited number of measurements of the vapor pressure of water over
supercooled liquid water at temperatures below °C. Detwiler
claims some indirect evidence to support the extrapolation of
the Goff-Gratch equation down to temperatures of -60°C. However,
this currently remains an open issue.
The WMO Guide to
Meteorological Instruments and Methods of Observation (CIMO Guide,
WMO No. 8) formulation is widely used in Meteorology and appeals
for its simplicity. Together with the formulas by Bolton and Buck
it has the same mathematical form as older the Maguns Tetens
formula and differs only in the value of the parameters.
The
Hyland and Wexler formulation is used by Vaisala and is very similar
to the formula by Sonntag . The comparison for the liquid
saturation vapor pressure equations - with the Goff-Gratch
equation in figure 1 shows that uncertainties at low temperatures
become increasingly large and reach the measurement uncertainty
claimed by some RH sensors. At -60°C the deviations range from
-6% to +3% and at -70°C the deviations range from -9% to +6%. For
RH values reported in the low and mid troposphere the influence of
the saturation vapor pressure formula used is small and only
significant for climatological studies [Elliott and Gaffen 1993].
The WMO (WMO No. 49, Technical Regulations) recommended
formula is a derivative of the Goff-Gratch equation, originally
published by Goff (1957). The differences between Goff (1957) and
Goff-Gratch (1946) are less than 1% over the entire temperature
range. The formulation published by WMO (1988) cannot be used due to
several typographical errors. The corrected formulation WMO (2000)
differs in the sign of one exponent compared to Goff (1957). This
incorrect formulation is in closer agreement with the Hyland and
Wexler formulation; however, it is to be assumed that Goff (1957) was
to be recommended.
The review of vapor pressures of
ice and supercooled water by Murphy and Kopp (2005) provides a
formulation based on recent data on the molar heat capacity of
supercooled water. The comparison of the the vapor pressure equations
with the formulation by Murphy and Koop is shown in figure 2.
The study by Fukuta and Gramada
shows direct measurements of the vapor pressure over liquid
water down to -38°C. Their result indicates that at the lowest
temperatures the measured vapor pressure may be as much as 10% lower
than the value given by the Smithsonian Tables , and as shown in
figure 1 lower as any other vapor pressure formulation. However,
these data are in conflict with measured molar heat capacity data
(Muprhy and Koop, 2005), which have been measured both for bulk as
for small water droplets.
Like most other formulations, the
IAPWS formulation 1995 (Wagner and Pruß, 2002) are valid only
above the triple point. The IAWPS formulation 1995 (Wagner and Pruß,
2002) is valid in the temperature range 273.16 K < T < 647.096
K.
It is important to note that in the
upper troposphere, water vapor measurements reported in the WMO
convention as relative humidity with respect to liquid water depend
critically on the saturation vapor pressure equation that was used to
calculate the RH value.
Figure 1: Comparison of equations
- with the Goff Gratch equation for the saturation
pressure of water vapor over liquid water. The measurements by Fukuta
et al. are shown as well.
()WMO
(2000) is also shown. This is based on Goff (1957) with the different
sign of one exponent, likely due to a typographical error.
Figure 2: Comparison of several equations with the equation by Sonntag for the saturation pressure of water vapor over liquid water.
The equations by Hyland and Wexler , the nearly identical equation by Wexler (1976, see reference below) and the equation by Sonntag are the most commonly
used equations among radiosonde manufacturers and should be used in upper air applications to avoid inconsistencies.
2) Vapor pressure over ice
Goff Gratch equation
(Smithsonian Tables, 1984):
Log10 ei
= -9.09718 (273.16/T -
1)
- 3.56654 Log10(273.16/ T)
+ 0.876793 (1 - T/ 273.16)
+ Log10(6.1071)
with T in [K] and ei
in [hPa]
Hyland and Wexler
(Hyland
and Wexler, 1983.):
Log ei =
-0.56745359 104 /
T
+ 0.63925247 101
- 0.96778430 10-2 T
+ 0.62215701 10-6 T2
+ 0.20747825 10-8 T3
- 0.94840240 10-12 T4
+ 0.41635019 101 Log(T)
with T in [K]
and ei in [Pa]
Guide to Meteorological
Instruments and Methods of Observation (CIMO Guide)
(WMO,
2008)
ei =
6.112 e(22.46 t/(272.62 +
t))
with t in [°C] and ei in [hPa]
Magnus Teten
(Murray,
1967)
ei =
6.1078 e21.8745584 (T-273.16) /
(T –
7.66)
with T in [K] and ew in [hPa]
Buck
(Buck
Research Manual, 1996)
ei
= 6.1115 e(23.036 - t /
333.7) t / (279.82 + t)
ei = 6.1115 e22.452 t/ (272.55+t)
with t in [°C] and ei in [hPa]
Marti Mauersberger
(Marti
and Mauersberger, 1993)
Log10 ei
= -2663.5 / T +
12.537
with T in [K] and ei in [Pa]
Murphy and Koop
(Murphy
and Koop, 2005)
Log ei =
9.550426
5723.265/T
3.53068 Log(T)
0.00728332
T
with T in [K] and ei in [Pa]
The Goff Gratch equation for
the vapor pressure over ice covers a region of -100°C to 0°C.
It is generally considered the reference equation; however, other
equations have also been widely used. The equations discussed here
are mostly of interest for frost-point measurements using chilled
mirror hygrometers, since these instruments directly measure the
temperature at which a frost layer and the overlying vapor are in
equilibrium. In meteorological practice, relative humidity is given
over liquid water (see section 1) and care needs to be taken to
consider this difference.
Buck Research, which manufactures
frost-point hygrometers, uses the Buck formulations in their
instruments. These formulations include an enhancement factor, which
corrects for the differences between pure vapor and moist air. This
enhancement factor is a weak function of temperature and pressure and
corrects about 0.5% at sea level. For the current discussion it has
been omitted.
The Marti Mauersberger equation is the only
equation based on direct measurements of the vapor pressure down to
temperatures of 170 K.
The comparison of equations 12-17 with the
Goff Gratch equation (figure 3) shows, that with the exception of the
Magnus Teten formula, the deviations in the typical meteorological
range of -100°C to 0°C are less than 2.5%, and smaller than
typical instrumental errors of frost-point hygrometers of 5-10%.
Not
shown is the WMO recommended equation for vapor pressure over ice,
since it is nearly identical with the Goff-Gratch equation .
Figure 3: Comparison of equations
- with the Goff Gratch equation for the saturation
pressure of water vapor over ice.
3) References
Bolton, D., The
computation of equivalent potential temperature, Monthly Weather
Review, 108, 1046-1053, 1980..
Buck, A. L.,
New equations for computing vapor pressure and enhancement factor, J.
Appl. Meteorol., 20, 1527-1532, 1981.
Buck
Research Manuals, 1996
Detwiler, A.,
Extrapolation of the Goff-Gratch formula for vapor pressure over
liquid water at temperatures below 0°C, J. Appl. Meteorol., 22,
503, 1983.
Elliott, W. P. and D. J. Gaffen,
On the utility of radiosonde humidity archives for climate studies,
Bull. Am. Meteorol. Soc., 72, 1507-1520, 1991.
Elliott,
W. P. and D. J. Gaffen, Effects of conversion algorithms on reported
upper air dewpoint depressions, Bull. Am. Meteorol. Soc., 74,
1323-1325, 1993.
Fukuta, N. and C. M.
Gramada, Vapor pressure measurement of supercooled water, J. Atmos.
Sci., 60, 1871-1875, 2003.
Gibbins, C. J., A
survey and comparison of relationships for the determination of the
saturation vapour pressure over plane surfaces of pure water and of
pure ice, Annales Geophys., 8, 859-886, 1990.
Goff,
J. A., and S. Gratch, Low-pressure properties of water from -160 to
212 F, in Transactions of the American society of heating and
ventilating engineers, pp 95-122, presented at the 52nd annual
meeting of the American society of heating and ventilating engineers,
New York, 1946.
Goff, J. A. Saturation
pressure of water on the new Kelvin temperature scale, Transactions
of the American society of heating and ventilating engineers, pp
347-354, presented at the semi-annual meeting of the American society
of heating and ventilating engineers, Murray Bay, Que. Canada, 1957.
Hyland, R. W. and A. Wexler, Formulations for the
Thermodynamic Properties of the saturated Phases of H2O from 173.15K
to 473.15K, ASHRAE Trans, 89(2A), 500-519, 1983.
Marti,
J. and K Mauersberger, A survey and new measurements of ice vapor
pressure at temperatures between 170 and 250 K, GRL 20, 363-366, 1993
Murphy, D. M. and T. Koop, Review of the vapour
pressures of ice and supercooled water for atmospheric applications,
Quart. J. Royal Met. Soc, 131, 1539-1565, 2005.
Murray,
F. W., On the computation of saturation vapor pressure, J. Appl.
Meteorol., 6, 203-204, 1967.
Smithsonian Met.
Tables, 5th ed., pp. 350, 1984.
Sonntag,
D., Advancements in the field of hygrometry, Meteorol. Z., N. F., 3,
51-66, 1994.
Wagner W. and A. Pruß, The
IAPWS formulation 1995 for the thermodynamic properties of ordinary
water substance for general and scientific use, J. Phys. Chem. Ref.
Data, 31, 387-535, 2002.
Wexler, A., Vapor Pressure Formulation for Water in Range 0 to 100°C.
A Revision, Journal of Research of the National Bureau of Standards, 80A, 775-785, 1976.
World Meteorological
Organization, General meteorological standards and recommended
practices, Appendix A, WMO Technical Regulations, WMO-No. 49, Geneva
1988.
World Meteorological Organization,
General meteorological standards and recommended practices, Appendix
A, WMO Technical Regulations, WMO-No. 49, corrigendum, Geneva August
2000.
World Meteorological Organization, Guide to Meteorological
Instruments and Methods of Observation, Appendix 4B, WMO-No. 8 (CIMO
Guide), Geneva 2008.
IDL
source for all equations
1 December 2011
Holger.Voemel@Colorado.edu |
11546 | https://mathworld.wolfram.com/SquareNumber.html | Square Number
Download Wolfram Notebook
A square number, also called a perfect square, is a figurate number of the form , where is an integer. The square numbers for , 1, ... are 0, 1, 4, 9, 16, 25, 36, 49, ... (OEIS A000290).
A plot of the first few square numbers represented as a sequence of binary bits is shown above. The top portion shows to , and the bottom shows the next 510 values.
The generating function giving the square numbers is
The st square number is given in terms of the th square number by
which is equivalent to adding a gnomon to the previous square, as illustrated above.
The th square number is equal to the sum of the st and th triangular numbers,
| |
| |
| |
as can seen in the above diagram, in which the st triangular number is represented by the white triangles, the th triangular number is represented by the black triangles, and the total number of triangles is the square number (R. Sobel, pers. comm.).
Square numbers can also be generated by taking the product of two consecutive even or odd numbers and adding 1. The result obtained by carrying out this operation is then the square of the average of the initial two numbers,
| |
| (6) |
As a part of the study of Waring's problem, it is known that every positive integer is a sum of no more than 4 positive squares (; Lagrange's four-square theorem), that every "sufficiently large" integer is a sum of no more than 4 positive squares (), and that every integer is a sum of at most 3 signed squares (). Actually, the basis set for representing positive integers with positive squares is , so 49 need never be used. Furthermore, since an infinite number of require four squares to represent them, the least integer such that every positive integer beyond a certain point requires squares is given by .
The number of representation of a number by squares, distinguishing signs and order, is denoted and called the sum of squares function. The minimum number of squares needed to represent the numbers 1, 2, 3, ... are 1, 2, 3, 1, 2, 3, 4, 2, 1, 2, ... (OEIS A002828), and the number of distinct ways to represent the numbers 1, 2, 3, ... in terms of squares are 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, ... (OEIS A001156). A brute-force algorithm for enumerating the square partitions of is repeated application of the greedy algorithm. However, this approach rapidly becomes impractical since the number of representations grows extremely rapidly with , as shown in the following table.
| |
| square partitions |
| 10 | 4 |
| 50 | 104 |
| 100 | 1116 |
| 150 | 6521 |
| 200 | 27482 |
The th nonsquare number is given by
| |
| (7) |
where is the floor function, and the first few are 2, 3, 5, 6, 7, 8, 10, 11, ... (OEIS A000037).
The only numbers that are simultaneously square and pyramidal (the cannonball problem) are and , corresponding to and (Ball and Coxeter 1987, p. 59; Ogilvy 1988; Dickson 2005, p. 25), as conjectured by Lucas (1875, 1876) and proved by Watson (1918). The cannonball problem is equivalent to solving the Diophantine equation
| |
| (8) |
(Guy 1994, p. 147).
The only numbers that are square and tetrahedral are , , and (giving , , and ), as proved by Meyl (1878; cited in Dickson 2005, p. 25; Guy 1994, p. 147). In general, proving that only certain numbers are simultaneously figurate in two different ways is far from elementary.
To find the possible last digits for a square number, write for the number written in decimal notation as (, , 1, ..., 9). Then
| |
| (9) |
so the last digit of is the same as the last digit of . The following table gives the last digit of for , 1, ..., 9 (where numbers with more that one digit have only their last digit indicated, i.e., 16 becomes _6). As can be seen, the last digit can be only 0, 1, 4, 5, 6, or 9.
| | | | | | | | | | |
--- --- --- --- --- |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 0 | 1 | 4 | 9 | _6 | _5 | _6 | _9 | _4 | _1 |
We can similarly examine the allowable last two digits by writing as
| |
| (10) |
so
| |
| (11) |
| (12) |
so the last two digits must have the last two digits of . Furthermore, the last two digits can be obtained by considering only , 1, 2, 3, and 4, since
| |
| (13) |
has the same last two digits as (with the one additional possibility that in which case the last two digits are 00). The following table (with the addition of 00) therefore exhausts all possible last two digits.
| | | | | | | | | |
--- --- --- ---
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 0 | 01 | 04 | 09 | 16 | 25 | 36 | 49 | 64 | 81 |
| 1 | _21 | _44 | _69 | _96 | _25 | _56 | _89 | _24 | _61 |
| 2 | _41 | _84 | _29 | _76 | _25 | _76 | _29 | _84 | _41 |
| 3 | _61 | _24 | _89 | _56 | _25 | _96 | _69 | _44 | _21 |
| 4 | _81 | _64 | _49 | _36 | _25 | _16 | _09 | _04 | _01 |
The only 22 possibilities are therefore 00, 01, 04, 09, 16, 21, 24, 25, 29, 36, 41, 44, 49, 56, 61, 64, 69, 76, 81, 84, 89, and 96, which can be summarized succinctly as 00, , , 25, , and , where stands for an even number and for an odd number. Additionally, a necessary (but not sufficient) condition for a number to be square is that its digital root be 1, 4, 7, or 9. The digital roots of the first few squares are 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4, 9, 7, ... (OEIS A056992), while the list of number having digital roots 1, 4, 7, or 9 is 1, 4, 7, 9, 10, 13, 16, 18, 19, 22, 25, ... (OEIS A056991).
This property of square numbers was referred to in a "puzzler" feature of a March 2008 broadcast of the NPR radio show "Car Talk." In this Puzzler, a son tells his father that his computer and math teacher assigned the class a problem to determine if a number is a perfect square. Each student is assigned a particular number, and the students are supposed to write a software program to determine the answer. The son's assigned number was . While the father thinks this is a hard problem, a bystander listening to the conversation states that the teacher gave the son an easy number and the bystander can give the answer immediately. The question is what does this third person know? The answer is that the number ends in the digit "2," which is not one of the possible last digits for a square number.
The following table gives the possible residues mod for square numbers for to 20. The quantity gives the number of distinct residues for a given .
| | | |
---
| 2 | 2 | 0, 1 |
| 3 | 2 | 0, 1 |
| 4 | 2 | 0, 1 |
| 5 | 3 | 0, 1, 4 |
| 6 | 4 | 0, 1, 3, 4 |
| 7 | 4 | 0, 1, 2, 4 |
| 8 | 3 | 0, 1, 4 |
| 9 | 4 | 0, 1, 4, 7 |
| 10 | 6 | 0, 1, 4, 5, 6, 9 |
| 11 | 6 | 0, 1, 3, 4, 5, 9 |
| 12 | 4 | 0, 1, 4, 9 |
| 13 | 7 | 0, 1, 3, 4, 9, 10, 12 |
| 14 | 8 | 0, 1, 2, 4, 7, 8, 9, 11 |
| 15 | 6 | 0, 1, 4, 6, 9, 10 |
| 16 | 4 | 0, 1, 4, 9 |
| 17 | 9 | 0, 1, 2, 4, 8, 9, 13, 15, 16 |
| 18 | 8 | 0, 1, 4, 7, 9, 10, 13, 16 |
| 19 | 10 | 0, 1, 4, 5, 6, 7, 9, 11, 16, 17 |
| 20 | 6 | 0, 1, 4, 5, 9, 16 |
In general, the odd squares are congruent to 1 (mod 8) (Conway and Guy 1996). Stangl (1996) gives an explicit formula by which the number of squares in (i.e., mod ) can be calculated. Let be an odd prime. Then is the multiplicative function given by
| |
| (14) |
| (15) |
| (16) |
| (17) |
| (18) |
is related to the number of quadratic residues in by
| |
| (19) |
for (Stangl 1996).
For a perfect square , or 1 for all odd primes where is the Legendre symbol. A number that is not a perfect square but that satisfies this relationship is called a pseudosquare.
In a Ramanujan conference talk, W. Gosper conjectured that every sum of four distinct odd squares is the sum of four distinct even squares. This conjecture was proved by M. Hirschhorn using the identity
| |
| (20) |
where , , , and are positive or negative integers. Hirschhorn also showed that every sum of four distinct oddly even squares is the sum of four distinct odd squares.
A prime number can be written as the sum of two squares iff is not divisible by 4 the (Fermat's 4n+1 theorem). An arbitrary positive number is expressible as the sum of two squares iff, given its prime factorization
| |
| (21) |
none of is divisible by 4 (Conway and Guy 1996, p. 147). This is equivalent the requirement that all the odd factors of the squarefree part of are equal to 1 (mod 4) (Hardy and Wright 1979, Finch). The first few numbers that can be expressed as the sum of two squares are 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, ... (OEIS A001481). Letting be the fraction of numbers that are expressible as the sum of two squares,
| |
| (22) |
and
| |
| (23) |
where is the Landau-Ramanujan constant.
Numbers expressible as the sum of three squares are those not of the form for (Nagell 1951, p. 194; Wells 1986, pp. 48 and 56; Hardy 1999, p. 12).
The following table gives the first few numbers which require , 2, 3, and 4 squares to represent them as a sum (Wells 1986, p. 70).
| | |
--- |
| Sloane | numbers |
| 1 | A000290 | 1, 4, 9, 16, 25, 36, 49, 64, 81, ... |
| 2 | A000415 | 2, 5, 8, 10, 13, 17, 18, 20, 26, 29, ... |
| 3 | A000419 | 3, 6, 11, 12, 14, 19, 21, 22, 24, 27, ... |
| 4 | A004215 | 7, 15, 23, 28, 31, 39, 47, 55, 60, 63, ... |
Fermat's 4n+1 theorem guarantees that every prime of the form is a sum of two square numbers in only one way.
There are only 31 numbers that cannot be expressed as the sum of distinct squares: 2, 3, 6, 7, 8, 11, 12, 15, 18, 19, 22, 23, 24, 27, 28, 31, 32, 33, 43, 44, 47, 48, 60, 67, 72, 76, 92, 96, 108, 112, 128 (OEIS A001422; Guy 1994; Savin 2000). The following numbers cannot be represented using fewer than five distinct squares: 55, 88, 103, 132, 172, 176, 192, 240, 268, 288, 304, 368, 384, 432, 448, 496, 512, and 752, together with all numbers obtained by multiplying these numbers by a power of 4. This gives all known such numbers less than (Savin 2000). All numbers can be expressed as the sum of at most five distinct squares, and only
| |
| (24) |
and
| |
| (25) |
require six distinct squares (Bohman et al. 1979; Guy 1994, p. 136; Savin 2000). In fact, 188 can also be represented using seven distinct squares:
| |
| (26) |
The following table gives the numbers that can be represented in different ways as a sum of squares. For example,
| |
| (27) |
can be represented in two ways () by two squares ().
| | |
--- |
| Sloane | numbers |
| 1 | 1 | A000290 | 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ... |
| 2 | 1 | A025284 | 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, ... |
| 2 | 2 | A025285 | 50, 65, 85, 125, 130, 145, 170, 185, 200, ... |
| 3 | 1 | A025321 | 3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 22, 24, ... |
| 3 | 2 | A025322 | 27, 33, 38, 41, 51, 57, 59, 62, 69, 74, 75, ... |
| 3 | 3 | A025323 | 54, 66, 81, 86, 89, 99, 101, 110, 114, 126, ... |
| 3 | 4 | A025324 | 129, 134, 146, 153, 161, 171, 189, 198, ... |
| 4 | 1 | A025357 | 4, 7, 10, 12, 13, 15, 16, 18, 19, 20, 21, 22, ... |
| 4 | 2 | A025358 | 31, 34, 36, 37, 39, 43, 45, 47, 49, 50, 54, ... |
| 4 | 3 | A025359 | 28, 42, 55, 60, 66, 67, 73, 75, 78, 85, 95, 99, ... |
| 4 | 4 | A025360 | 52, 58, 63, 70, 76, 84, 87, 91, 93, 97, 98, 103, ... |
The least numbers that are the sum of two squares in exactly different ways for , 2, ... are given by 2, 50, 325, 1105, 8125, 5525, 105625, 27625, 71825, 138125, 5281250, ... (OEIS A016032; Beiler 1966, pp. 140-141; Rubin 1977-78; Culberson 1978-79; Hardy and Wright 1979; Rivera).
The product of four distinct nonzero integers in arithmetic progression is square only for (, , 1, 3), giving (Le Lionnais 1983, p. 53). It is possible to have three squares in arithmetic progression, but not four (Dickson 2005, pp. 435-440). If these numbers are , , and , there are positive integers and such that
| |
| (28) |
| (29) |
| (30) |
where and one of , , or is even (Dickson 2005, pp. 437-438). Every three-term progression of squares can be associated with a Pythagorean triple ) by
| |
| (31) |
| (32) |
| (33) |
(Robertson 1996).
Catalan's conjecture states that 8 and 9 ( and ) are the only consecutive powers (excluding 0 and 1), i.e., the only solution to Catalan's Diophantine problem. This conjecture has not yet been proved or refuted, although R. Tijdeman has proved that there can be only a finite number of exceptions should the conjecture not hold. It is also known that 8 and 9 are the only consecutive cubic and square numbers (in either order).
The numbers that are not the difference of two squares are 2, 6, 10, 14, 18, ... (OEIS A016825; Wells 1986, p. 76).
A square number can be the concatenation of two squares, as in the case and giving . The first few numbers that are neither square nor the sum of a square and a prime are 10, 34, 58, 85, 91, 130, 214, ... (OEIS A020495).
It is conjectured that, other than , and , there are only a finite number of squares having exactly two distinct nonzero digits (Guy 1994, p. 262). The first few such are 4, 5, 6, 7, 8, 9, 11, 12, 15, 21, ... (OEIS A016070), corresponding to of 16, 25, 36, 49, 64, 81, 121, ... (OEIS A018884).
The following table gives the first few numbers which, when squared, give numbers composed of only certain digits. The values of such that contains exactly two different digits are given by 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 20, ... (OEIS A016069), whose squares are 16, 25 36, 49, 64, ... (OEIS A018885).
| | | |
---
| digits | Sloane | , |
| 1, 2, 3 | A030175 | 1, 11, 111, 36361, 363639, ... |
| A030174 | 1, 121, 12321, 1322122321, ... |
| 1, 4, 6 | A027677 | 1, 2, 4, 8, 12, 21, 38, 108, ... |
| A027676 | 1, 4, 16, 64, 144, 441, 1444, ... |
| 1, 4, 9 | A027675 | 1, 2, 3, 7, 12, 21, 38, 107, ... |
| A006716 | 1, 4, 9, 49, 144, 441, 1444, 11449, ... |
| 2, 4, 8 | A027679 | 2, 22, 168, 478, 2878, 210912978, ... |
| A027678 | 4, 484, 28224, 228484, 8282884, ... |
| 4, 5, 6 | A030177 | 2, 8, 216, 238, 258, 738, 6742, ... |
| A030176 | 4, 64, 46656, 56644, 66564, ... |
For three digits, an extreme example containing only the digits 7, 8, and 9 is
| |
| (34) |
No squares are known containing only the digits 013 or 678. Unique solutions are known for 019, 039, 056, 079, 568, and 789. The longest known is
| |
| (35) |
with 52 digits. A list of known solutions to the 3-digit square problem is maintained by Mishima.
Brown numbers are pairs of integers satisfying the condition of Brocard's problem, i.e., such that
| |
| (36) |
where is a factorial. Only three such numbers are known: (5,4), (11,5), (71,7). Erdős conjectured that these are the only three such pairs.
Either or has a solution in positive integers iff, for some , , where is a Fibonacci number and is a Lucas number (Honsberger 1985, pp. 114-118).
The smallest and largest square numbers containing the digits 1 to 9 are
| |
| (37) |
| |
| (38) |
The smallest and largest square numbers containing the digits 0 to 9 are
| |
| (39) |
| |
| (40) |
(Madachy 1979, p. 159). The smallest and largest square numbers containing the digits 1 to 9 twice each are
and the smallest and largest containing 1 to 9 three times are
(Madachy 1979, p. 159).
Madachy (1979, p. 165) also considers numbers that are equal to the sum of the squares of their two "halves" such as
| |
| |
| |
| |
| |
in addition to a number of others.
See also
Antisquare Number, Biquadratic Number, Brocard's Problem, Brown Numbers, Cannonball Problem, Catalan's Conjecture, Centered Square Number, Clark's Triangle, Cubic Number, Diophantine Equation, Fermat's 4n+1 Theorem, Greedy Algorithm, Gross, Heptagonal Square Number, Lagrange's Four-Square Theorem, Landau-Ramanujan Constant, Octagonal Square Number, Partition, Pentagonal Square Number, Pseudosquare, Pyramidal Number, Squarefree, Square Triangular Number, Sum of Squares Function, Waring's Problem Explore this topic in the MathWorld classroom
Portions of this entry contributed by Len Goodman
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References
Archibald, R. G. "Waring's Problem: Squares." Scripta Math. 7, 33-48, 1940.Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, p. 59, 1987.Beiler, A. H. Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. New York: Dover, 1966.Bohman, J.; Fröberg, C.-E.; and Riesel, H. "Partitions in Squares." BIT 19, 297-301, 1979.Cole, C. "rec.puzzles FAQ3." J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 30-32 and 146-147, 1996.Culberson, J. "Solution to Problem 590." J. Recr. Math. 11, 137-138, 1978-79.Dickson, L. E. History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 2005.Finch, S. "On a Generalized Fermat-Wiles Equation." E. Representations of Integers as Sums of Squares. New York: Springer-Verlag, 1985.Guy, R. K. "Sums of Squares" and "Squares with Just Two Different Decimal Digits." §C20 and F24 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 136-138, 147, and 262, 1994.Hajdu, L. and Pintér, Á. "Square Product of Three Integers in Short Intervals." Math. Comput. 68, 1299-1301, 1999.Hardy, G. H. Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work, 3rd ed. New York: Chelsea, 1999.Hardy, G. H. and Wright, E. M. "The Representation of a Number by Two or Four Squares." Ch. 20 in An Introduction to the Theory of Numbers, 5th ed. Oxford, England: Clarendon Press, pp. 297-316, 1979.Honsberger, R. "A Second Look at the Fibonacci and Lucas Numbers." Ch. 8 in Mathematical Gems III. Washington, DC: Math. Assoc. Amer., 1985.Landau, E. "Über die Einteilung der positiven ganzen Zahlen in vier Klassen nach der Mindeszahl der zu ihrer additiven Zusammensetzung erforderlichen Quadrate." Arch. Math. Phys. 13, 305-312, 1908.Le Lionnais, F. Les nombres remarquables. Paris: Hermann, 1983.Lucas, É. Question 1180. Nouv. Ann. Math. Ser. 2 14, 336, 1875.Lucas, É. Solution de Question 1180. Nouv. Ann. Math. Ser. 2 15, 429-432, 1876.Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, pp. 159 and 165, 1979.Meyl, A.-J.-J. Solution de Question 1194. Nouv. Ann. Math. 17, 464-467, 1878.Mishima, H. "Sporadic Solutions." T. Introduction to Number Theory. New York: Wiley, 1951.Ogilvy, C. S. and Anderson, J. T. Excursions in Number Theory. New York: Dover, pp. 77 and 152, 1988.Pappas, T. "Triangular, Square & Pentagonal Numbers." The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, p. 214, 1989.Pietenpol, J. L. "Square Triangular Numbers." Amer. Math. Monthly 69, 168-169, 1962.Rivera, C. "Problems & Puzzles: Puzzle 062-The qs-Sequence." J. P. "Magic Squares of Squares." Math. Mag. 69, 289-293, 1996.Rubin, F. "Problem 590." J. Recr. Math. 10, 46, 1977-78.Savin, A. "Shape Numbers." Quantum 11, 14-18, 2000.Sloane, N. J. A. Sequences A000037/M0613, A000290/M3356, A000415, A000419, A001156/M0221, A001422, A001481/M0968, A002828/M0404, A004215/M4349, A006716/M3369, A016069, A016070, A016032, A016825, A018884, A018885, A020495, A025284, A025285, A025321, A025322, A025323, A025324, A025357, A025358, A025359, A025360, A027675, A027676, A027677, A027678, A027679, A030174, A030175, A030176, A030177, A056991, and A056992 in "The On-Line Encyclopedia of Integer Sequences."Stangl, W. D. "Counting Squares in ." Math. Mag. 69, 285-289, 1996.Taussky-Todd, O. "Sums of Squares." Amer. Math. Monthly 77, 805-830, 1970.Vardi, I. Computational Recreations in Mathematica. Reading, MA: Addison-Wesley, pp. 20 and 234-237, 1991.Watson, G. N. "The Problem of the Square Pyramid." Messenger. Math. 48, 1-22, 1918.Wells, D. The Penguin Dictionary of Curious and Interesting Numbers. Middlesex, England: Penguin Books, pp. 48 and 70, 1986.
Referenced on Wolfram|Alpha
Square Number
Cite this as:
Goodman, Len and Weisstein, Eric W. "Square Number." From MathWorld--A Wolfram Resource.
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An inequality about the sum of some unit fractions with a property
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Question : Is the following true for any $n,N\in\mathbb N$?
$$\sum_{k_1+k_2+\cdots+k_N=n,\ k_i\ge0\in\mathbb Z}\frac1{\prod_{j=1}^{N}{(N-1)k_j+1}}\le 1$$
Motivation : I've known the $N=3$ case :
$$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$
I proved this inequality by estimating the left hand side with integral. After proving this, I reached the above expectation by using computer. The above expectation seems true, but I'm facing difficulty. I would like to know how to prove this (if it's true) and any relevant references.
Remark : This question has been asked previously on math.SE without receiving any answers.
Update : I'm going to show the proof for $N=3$ case without using integral. This is because it seems that this idea can be generalized (though I'm facing difficulty).
For any non-negative integer $n$, $$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$
Proof : Let $A_n$ be the left hand side, and suppose that $\sum$ represents $\sum_{k_1+k_2+k_3=n,k_i\ge 0\in\mathbb Z}$. Noting that $(2k_1+1)+(2k_2+1)+(2k_3+1)=2n+3$, we get $$\begin{align}A_n & =\sum\frac{(2k_1+1)+(2k_2+1)+(2k_3+1)}{(2n+3)(2k_1+1)(2k_2+1)(2k_3+1)}\ & =\frac{1}{2n+3}\sum\left{\frac{1}{(2k_1+1)(2k_2+1)}+\frac{1}{(2k_2+1)(2k_3+1)}+\frac{1}{(2k_3+1)(2k_1+1)}\right}\ & =\frac{3}{2n+3}\sum\frac{1}{(2k_1+1)(2k_2+1)}\ & =\frac{3}{2n+3}\sum_{j=0}^n\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}\ & \le \frac{3}{2n+3}\left(1+\frac 23 n\right)=1\end{align}$$ Here, I used $$B_0=1, B_j\le \frac 23\ (j=1,2,\cdots,n)$$ where $$B_j=\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}.$$
inequalities
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mathlovemathlove
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$\begingroup$ Dear @mathlove: Could you please try to add a top level tag to this question? Thank you. $\endgroup$
Ricardo Andrade
– Ricardo Andrade
2013-10-19 10:34:05 +00:00
Commented Oct 19, 2013 at 10:34
$\begingroup$ @RicardoAndrade: Well, I know neither what the top level tag is nor how to add it. $\endgroup$
mathlove
– mathlove
2013-10-19 15:31:50 +00:00
Commented Oct 19, 2013 at 15:31
$\begingroup$ The top level tags are tags like 'nt.number-theory' and 'at.algebraic-topology' which start with a two letter code. They roughly follow their arxiv counterparts. These tags can be added like any other tag, such as the tag 'inequalities' already present in this question. You only have to edit the question and add another tag. I hope that helps. $\endgroup$
Ricardo Andrade
– Ricardo Andrade
2013-10-19 15:49:43 +00:00
Commented Oct 19, 2013 at 15:49
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$\begingroup$ @Ricardo, I thought of those same two tags, but I'm not sure either one is appropriate. I have also tried, without any success, to find a full listing of two-letter-code tags in use here. $\endgroup$
Gerry Myerson
– Gerry Myerson
2013-10-20 10:14:10 +00:00
Commented Oct 20, 2013 at 10:14
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$\begingroup$ @Ricardo, yes, but it would be nice (in my opinion) if there were a list of arXiv tags readily available on this website. $\endgroup$
Gerry Myerson
– Gerry Myerson
2013-10-20 11:27:51 +00:00
Commented Oct 20, 2013 at 11:27
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This is not an answer, but I want to use displays. What you are asking is whether the coefficients of the taylor expansion of $$\left( \sum_{k=0}^\infty \frac{x^k}{(N-1)k+1} \right)^N$$ are all at most 1. Unfortunately the summation is a hypergeometric function that doesn't have a general simplification. You can easily prove the conjecture for small $n$ and all $N$, or for small $N$ and all $n$ by expanding it. In general you might be able to bound it with a contour integral or something like that. I believe it is true.
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answered Oct 19, 2013 at 10:10
Brendan McKayBrendan McKay
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$\begingroup$ Actually, after this is said, you should have extremely strong willpower to suppress the natural desire to recall that the function with all coefficients $1$ is $\frac 1{1-z}$ and to raise both sides to the power $\frac 1N$, i.e., to try to check if the coefficients of $(1-z)^{-1/N}$ are at least $\frac 1{(N-1)k+1}$. :-) $\endgroup$
fedja
– fedja
2013-10-28 23:16:25 +00:00
Commented Oct 28, 2013 at 23:16
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This should be a comment rather than an answer, since I only have plausible strategies to suggest. But I'm new here and don't have enough reputation to leave comments.
Plausible Strategy #1: As Brendan says, your sums are coefficients in the Taylor expansion of $$S_N(x) = \left( \sum_{k=0}^\infty \frac{x^k}{(N-1)k+1} \right)^N.$$ I computed some expansions of these series and it looks like, perhaps, the coefficient of $x^n$ ($n\ge 2$) increases monotonically w.r.t. $N$. You could try to prove this and also prove that the limit of each coefficient as $N\to\infty$ is 1. I think I can show that $S_N(x)$ approaches $\dfrac 1{1-x}$ pointwise for $x\in (-1,1)$, which, though not sufficient for what you want to do, is at least encouraging.
Plausible Strategy #2: If we group like terms in your unit fraction sum, we can express it as a sum indexed over partitions of $n$. I'm too lazy to type the general formula but here it is for $n=3$, which ought to be suggestive enough: $$\left(\frac{N(N-1)(N-2)}{3!}\cdot\frac 1{N^3}\right) + \left(\frac{N(N-1)}{1!1!}\cdot\frac 1{(2N-1)\cdot N}\right) + \left(\frac{N}{1!}\cdot\frac 1{3N-2}\right)$$ Here the terms correspond to $\pi=(1,1,1),(2,1),(3)$ respectively. The coefficients are just garden-variety multinomial coefficients. Now if we consider the terms corresponding to partitions with $r\ge 3$ parts, their numerators are $< (N-1)^r$ and their denominators are $> k(N-1)^r$, where $k$ is the leading coefficient. So, all of these terms undershoot their limits as $N\to\infty$.
When $n$ is large, almost all partitions of $n$ have 3 or more parts, so (a) your conjecture seems likely to be true and (b) you may be able to prove it by bounding a relatively small and simple subset of the terms.
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answered Oct 20, 2013 at 2:55
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I'm posting an answer just to inform that the question has received an answer by Ivan Loh on MSE.
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11548 | https://www.youtube.com/watch?v=l3XzepN03KQ | Algebra Basics: Solving Basic Equations Part 1 - Math Antics
mathantics
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This video shows students how to solve simple 1-step Algebra equations involving only addition or subtraction.
Part of the Algebra Basics Series:
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Transcript:
Introduction Hi, I’m Rob. Welcome to Math Antics. In our last Algebra video, we learned that Algebra involves equations that have variables or unknown values in them. And we learned that solving an equation means figuring out what those unknown values are. In this video, we’re going to learn how to solve some very simple Algebraic equations that just involve addition and subtraction. Then in the next video, we’ll learn how to solve some simple equations involving multiplication and division. Are you ready?… I thought so! Okay… so if you’ve got an equation that has an unknown value in it, Key Strategy then the key strategy for solving it is to rearrange the equation until you have the unknown value all by itself on one side of the equal sign, and all of the known numbers on the other side of the equal sign. Then, you’ll know just what the unknown value is. But, how do we do that? How do we rearrange equations? Well, we know that Algebra still uses the four main arithmetic operations (addition, subtraction, multiplication and division) and we can use those operations to rearrange equations, as long as we understand one really important thing first. Balancing Equations We need to understand that an equation is like a balance scale. You’ve seen a balance scale, right? If there’s the same amount of weight on each side of the scale, then the two sides are in balance. But, if we add some weight to one side... then the scale will tip. The two sides are no longer in balance. An equation is like that. Whatever is on one side of the equal sign MUST have exactly the same value as whatever is on the other side. Otherwise, the equation would not be true. Of course, that doesn’t mean that the two sides have to look the same. For example, in the equation 1 + 1 = 2, 1 + 1 doesn’t LOOK the like the number 2, but we know that 1 + 1 has the same VALUE as 2, so 1 + 1 = 2 is in balance. It’s a true equation. The reason we need to know that equations must be balanced is because when we start rearranging them, if we are not careful, we might do something that would change one of the sides more than the other. That would make the equation get out of balance and it wouldn’t be true anymore. And if that happens, we won’t get the right answer when we solve it. How to Avoid Balance That sounds pretty bad, huh? So how do we avoid that? How do we avoid getting an equation out of balance? The key is that whenever we make a change to an equation, we have to make the exact same change on both sides That’s so important, I’ll say it again. Whenever we do something to an equation, we have to do the same thing to BOTH sides. For example, if we want to add something to one side of an equation, we have to add that same thing to the other side. And if we want to subtract something from one side of an equation, then we have to subtract that same thing from the other side. And it’s the same for multiplication and division. If we want to multiply one side of an equation by a number, then we need to multiply the other side by that same number. Or if we want to divide one side of an equation by a number, then we have to divide the other side by that number also. As long as you always do the same thing to both sides of an equation, it will stay in balance and your equation will still be true. Alright, like I said, in this video, we’re just going to focus on equations involving addition and subtraction. First Example And here’s our first example: x + 7 = 15 To solve for the unknown value ‘x’, we need to rearrange the equation so that the ‘x’ is all by itself on one side of the equal sign. But what can we do to get ‘x’ all by itself? Well, right now ‘x’ is not by itself because 7 is being added to it. Is there a way for us to get rid of that 7? Yes! Since seven is being added to the ‘x’, we can undo that by subtracting 7 from that side of the equation. Subtracting 7 would leave ‘x’ all by itself because ‘x’ plus 7 minus 7 is just ‘x’. The ‘plus 7’ and the ‘minus 7’ cancel each other out. Okay great! So we just subtract 7 from this side of the equation and ‘x’ is all by itself. …equation solved, right? WRONG! If we just subtract 7 from one side of the equation and not the other side, then our equation won’t be in balance anymore. To keep our equation in balance, we also need to subtract 7 from the other side of the equation. But on that side, we just have the number 15. So we need to subtract 7 from that 15. And since 15 - 7 = 8, that side of the equation will just become 8. There, by subtracting 7 from BOTH sides, we’ve changed the original equation (x + 7 = 15) into the new and much simpler equation (x = 8) which tells us that the unknown number is 8. We have solved the equation! And to check our answer, to make sure we got it right, we can see what would happen if we replaced the unknown value in our original equation with the number 8. Instead of x + 7 = 15, we’d right 8 + 7 = 15, and if that’s true, then we know we got the right answer. Pretty cool, huh? Let’s try another one: 40 = 25 + x Second Example This time, the unknown value is on the right hand side of the equation. Does that make it harder? Nope. We use the exact same strategy. We want to get ‘x’ by itself, but this time ‘x’ is being added to 25. But thanks to the commutative property, that’s the same as 25 being added to ‘x’. So, to isolate 'x', we should subtract 25 from that side of the equation. But then we also need to subtract 25 from the other side to keep things in balance. On the right side, x plus 25 minus 25 is just x The minus 25 cancels out the positive 25 that was there. And on the other side we have 40 minus 25 which would leave 15. So the equation has become 15 = x, which is the same as x = 15. Again, we’ve solved the equation. So, whenever something is being added to an unknown, we can undo that and get the unknown all by itself by subtracting that same something from both sides of the equation. But what about when something is being subtracted from an unknown, like in this example: x - 5 = 16 In this case, ‘x’ is not by itself because 5 is being subtracter or taken away from it. …any ideas about how we could get rid of (or undo) that ‘minus 5’? Yep! To undo that subtraction, this time we need to ADD 5 to both sides of the equation. The ‘minus 5’ and the ‘plus 5’ cancel each other out and leave ‘x’ all by itself on this side. And on the other side, we have 16 + 5 which is 21. So in this equation, x equals 21. Let’s try another example like that: 10 = x - 32. Again the ‘x’ is not by itself because 32 is being subtracted from it. So to cancel that ‘minus 32’ out, we can just add 32 to both sides of the equation. On the right side, the ‘minus 32’ and the ‘plus 32’ cancel out leaving just ‘x’. And on the left we have 10 + 32 which is 42. Now we know that x = 42. Okay, so now you know how to solve very simple equations like these where something is being added to an unknown or where something is being subtracted from an unknown. But before you try practicing on your own, I want to show you a tricky variation of the subtraction problem that confuses a lot of students. Do you remember how subtraction does NOT have the commutative property? If you switch the order of the subtraction, it’s a different problem. Suppose we get a problem, where instead of a number being taken away from an unknown, an unknown is being taken away from a number. What do we do in that case? Well, we still want to get the unknown all by itself, but it’s a little harder to see how to do that. In this problem (12 - x = 5) the 12 on this side is a positive 12, so we could subtract 12 from both sides. That would get rid of the 12, but the problem is that wouldn’t get rid of this minus sign. That’s because the minus sign really belongs to the ‘x’ since it’s the ‘x’ that is being subtracted. Subtracting 12 would leave us with ‘negative x’ on this side of the equal sign, which is not wrong, but it might be confusing if you don’t know how to work with negative numbers yet. Fortunately, there’s another way to do this kind of problem that will avoid getting a negative unknown. Instead of subtracting 12 from both sides, what would happen if we added ‘x’ to both sides? Can we do that? Can we add an unknown to both sides? Well sure! why not? We can add or subtract ANYTHING we want as long as we do it to both sides! And when we do that, the ‘minus x’ and the ‘plus x’ will cancel each other out on this side. And and the other side, we will get 5 + x. Now our equation is 12 = 5 + x. And you might be thinking, “but why would we do that? That didn’t even solve our equation!” That’s true, but it changed it into an equation that we already know how to solve. Now it’s easy to see that we can isolate the unknown just by subtracting 5 from both sides of the equation. That will give us 7 = x or x = 7. It just took us one extra step to rearrange the equation, but then it was easy to solve. Okay, that’s the basics of solving simple algebraic equations that involve addition and subtraction. Summary You just need to get the unknown value all by itself, and you can do that by adding or subtracting something from both sides of the equation. And this process works the same even if the numbers in the equations are decimals or fractions. And it also works the same no matter what symbol you are using as an unknown. It could be x, y, z or a, b, c. The letter being used doesn’t matter. Remember, when it comes to math, it’s really important to practice what you’ve learned. So be sure to try solving some basic equations on your own! As always, thanks for watching Math Antics and I’ll see ya next time. Learn more at www.mathantics.com |
11549 | https://www.isinj.com/mt-aime/Problems%20in%20Solid%20Geometry%20-%20Sharygin%20(MIR,1986).pdf | SCIENCE FOR EVERYONE IFSHARYGIN PROBLEMS IN SOLID GEO TRY MIR Science for Everyone H.cD. IDapbIfHB 3aAaqH ITO reOMeTpHH CTepeOMeTpHH Ha)l.aTeJIhCTBO «Hayxa}>, MOCRBa I.F. Sharygin Problems in Solid Geometry Translated from the Russian by Leonid Levant Mir Publishers Moscow Firs t published 1986 Revised from the 1984 Russian edition © 1f3p;aTeJIbCTBO «Hayua», rJIaBSaa peAal(~IDl «J)H3HKO-MaTeMaTIlllooKoii JIHTepaTYps, 1984 © English translation, Mir Publishers, 1986 Contents Preface Section Section Section • II • ." .. oil .. • • • oil II • • • • • . . 1. Computational Problems . . . . 2. Problems on Proof . . . . . . . 3. Problems on Extrema. Geometric In-equalities . . . . . . . . . . • Section 4. Loci of Points . . . . . . . . . • . An Arbitrary Tetrahedron 59. An Equi-faced Tetrahedron 61. An Orthocentric Tetrahedron 64. An Arbitrary Poly-hedron. The Sphere 65. An Outlet into Space 68. Answers, Hints, Solutions . . • • • • • • • • • 6 7 37 47 54 69 Preface This book contains 340 problems in solid geo-metry and is a natural continuation of Problems in Plane Geometry, Nauka, Moscow, 1982. It is therefore possible to confine myself here to those points where this book differs from the first. The problems in this collection are grouped into (1) computational problems and (2) prob-lems on proof. The simplest problems in Section 1 only have answers, others, have brief hints, and the most difficult, have detailed hints and worked solu-tions. There are two reservations. Firstly, in most cases only the general outline of the solution is given, a number of details being suggested for the reader to consider. Secondly, although the suggested solutions are valid, they are not pat-terns (models) to be used in examinations. Sections 2-4 contain various geometric facts and theorems, problems on maximum and min .. imum (some of the problems in this part could have been put in Section 1), and problems on loci. Some questions pertaining to the geometry of tetrahedron, spherical geometry, and so forth are also considered here . .As to the techniques for solving all these prob-lems, I have to state that I prefer analytical com-putational methods to those associated with plane geometry. Some of the difficult problems in solid geometry will require a high level of concentration from the reader t and an ability to carry out some rather complicated work. The Author Section 1 Computational Problems 1. Given a cube with edge a. Two vertices of a regular tetrahedron lie on its diagonal and the two remaining vertices on the diagonal of its face. Find the volume of the tetrahedron. 2. The base of a quadrangular pyramid is a rectangle, the altitude of the pyramid is h .. Find the volume of the pyramid if it is known that all five of its faces are equivalent. 3. Among pyramids having all equal edges (each of length a), find the volume of the one which has the greatest number of edges. 4. Circumscribed about a ball is a frustum of a regular quadrangular pyramid whose slant height is equal to a. Find its lateral surface area. 5. Determine the vertex angle of an axial sec-tion of a cone if its volume is three times the volume of the ball inscribed in it. 6. Three balls touch the plane of a given tri-angle at the vertices of the triangle and one an-other. Find the radii of these balls if the sides of the triangle are equal to a, b, and c. 7. Find the distance between the skew dia-gonals of two neighbouring faces of a cube with edge a. In what ratio is each of these diagonals divided' by their common perpendicular? 8. Prove that the area of the projection of a polygon situated in the plane a on the plane ~ 8 Problems in Solid Geometry is equal to S cos <p, where S denotes the plane of the polygon and <p the angle between j the planes a and ~. 9. Given three straight lines passing through one point A. Let Bl and B 2 be two points on one line, eland c 2 two points on the other, and Dl and D2 two points on the third line. Prove that V AB1C1D 1 _ 1 ABI I· 1 ACI 1 • 1 ADI 1 V ABJCJDJ -1 AB2 I· 1 AC2 1 • 1 AD2 1 • 10. Let a, ~, and 'V denote the angles formed by an arbitrary straight line with three pairwise perpendicular lines. Prove that cos2 a + cos2 ~ + cos2 y = 1. 11. Let Sand P denote the areas of two faces of a tetrahedron, a the length of their common edge, and a the dihedral angle between them. Prove that the volume Vof the tetrahedron can be found by the formula V- 2SP sin a; -3 • ~ a 12. Prove that for the volume V of an arbi-trary tetrahedron the following formula is valid: V = ! abd sin <p, where a and b are two opposite edges of the tetrahedron, d the distance between them, and a), three pairwise perpendicular chords are drawn. Find the sum of the squared lengths of the seg-ments of the chords into which they are divided by the given point. 36. The base of a regular triangular prism is a triangle ABC with side a. Taken on the lateral edges are points A l , B 1 , and Cl situated at dis-tances a / 2, a, and 3a/2, respectively, from the plane of the base. Find the angle between the planes ABC and AlBlC1-37. The side of the base of a regular quadran-gular pyramid is equal to the slant height of a lateral face. Through a side of the base a cutting plane is passed separating the surface of the pyra-• Here and henceforward, ABC and AIBICI are the bases of the prism and AA I , BBI , CCI its ateral edges. Sec. 1. Computational Problems 13 mid into two equal portions. Find the angle between the cutting plane and the plane of the base of the pyramid. 38. The centre of a ball is found in the plane of the base of a regular triangular pyramid. The vertices of the base lie on the surface of the ball. Find the length 1 of the line of intersection of the surfaces of the ball and pyramid if the radi us of the ball is equal to R, and the plane angle at the vertex of the pyramid is equal to a. 39. In a regular hexagonal pyramid 8ABCDEF (8 the vertex), on the diagonal AD, three points are taken which divide the diagonal into four equal parts. Through these division points sec-tions are passed parallel to the plane 8 AB. Find the ratios of the areas of the obtained sections. 40. In a regular quadrangular pyramid, the plane angle at the vertex is equal to the angle between the lateral edges and the plane of the base. Determine the dihedral angles between the adjacent lateral faces of this pyramid. 41. The base of a triangular pyramid all of whose lateral edges are pairwise perpendicular is a triangle having an area 8. The area of one of the lateral faces is Q. Find the area of the projection of this face on the base. 42. ABCAIBICI is a regular triangular prism all of whose edges are equal to one another. K is a point on the edge AB different from A and B, M is a point on the straight line BICI, and L is a point in the plane of the face ACCIAI. The straight line KL makes equal angles with the planes ABC and ABBIAI , the line LM makes equal angles with the planes BCCIBI and ACCIAu the line KM also makes equal angles with the 14 Problems in Solid Geometry planes BCClBl and ACClAl • It is known that I KL I = I KM I = 1. Find the edge of the prism. 43. In a regular quadrangular pyramid, the angle between the lateral edges and the plane of the base is equal to the angle between a lateral edge and a plane of the lateral face not contain-ing this edge. Find this angle. 44. Find the dihedral angle between the base and a lateral face of a frustum of a regular tri-angular pyramid if it is known that a ball can be inscribed in it, and, besides, there is a ball which touches all of its edges. 45. Each of three edges of a triangular pyramid is equal to 1, and each of three other edges is equal to a. None of the faces is a regular tri-angle. What is the range of variation of a? What is the volume of this pyramid? 46. The lateral faces of a triangular pyramid are equivalent and are inclined to the plane of the base at angles a, ~, and y. Find the ratio of the radius of the ball inscribed in this pyramid to the radius of the ball touching the base of the pyrami d and the extensions of the three lateral faces. 47. All edges of a regular hexagon al prism are equal to a (each). Find the area of the section passed through a side of the base at an angle a to the plane of the base. 48. IB a rectangular parallelepiped ABCDA1 B1ClDl , IABI = a, 1 AD I = b, I AAl I = c. Find the angle between the planes ABIDI and AICID. 49. The base of the pyramid ABCDM is a square with base a, the lateral edges AM and BM are also equal to a (each). The lateral edges CM Sec. 1. Computational Problems 15 and D M are of length b. On the face CD M as on the base a triangular pyramid CDMN is con-structed outwards, each lateral edge of which has a length a. Find the distance between the straight lines AD and M N. 50. In a tetrahedron, one edge is equal to a, the opposite edge to b, and the rest of the edges to c. Find the radius of the circumscribed ball. 51. The base of a triangular pyramid is a triangle with sides a, b, and c; the opposite lat-eral edges of the pyramid are respectively equal to m, n, and p. Find the distance from the vertex of the pyramid to the centre of gravity of the base. 52. Given a cube ABCDAlBlClDl; through the edge AA 1 a plane is passed forming equal angles with the straight lines BCI and BlD. Find these angles. 53. The lateral edges of a triangular pyramid are pairwise perpendicular, one of them being the sum of two others is equal to a. Find the radius of the ball touching the base of the pyramid and the extensions of its lateral faces. 54. The base of a triangular pyramid SABC is a regular triangle ABC with side a, the edge SA is equal to b. Find the volume of the pyramid if it is known that the lateral faces of the pyramid are equivalent. 55. The base of a triangular pyramid SABC A is an isosceles triangle ABC (A = 90°). Tlle angles ~ ~ ~ ~ SAB, SCA, SAC, SBA (in the indicated or-der) form an arithmetic progression whose differ-ence is not equal to zero. The areas of the faces 16 I Problems in Solid Geometry SAB, ABC and SAC form a geometric progres-sion. Find the angles forming an arithmetic pro-gression. 56. The base of a triangular pyramid SABC is a regular triangle ABC with side a. Find the ~ volume of this pyramid if it is known that ASC = ./"'-... ./"'.... ASB = a, SAB = p. 57. In the cube ABCDA1B1C1D1 K is the mid-point of the edge AA1 , the point L lies on the edge BC. The line segment KL touches the ball inscribed in the cube. In what ratio is the line segment KL divided by the point of tangency? /'... 58. Given a tetrahedron ABCD in which ABC= /'... BAD = 900 • I AB I = a, I DC I = b, the angle between the edges AD and BC is equal to a. Find the radius of the circumscribed ball. 59. An edge of a cube and an edge of a regular tetrahedron lie on the same straight line, the midpoints of the opposite edges of the cube and tetrahedron coincide. Find the volume of the common part of the cube and tetrahedron if the edge of the cube is equal to a. 60. In what ratio is the volume of a triangu-lar pyramid divided by the plane parallel to its two skew edges and dividing one of the other edges in the ratio 2 : 1? 61. In a frustum of a regular quadrangular pyr .. amid two sections are drawn: one through the diagonals of the bases, the other through the side of the lower base and opposite side of the upper base. The angle between the cutting planes is Sec. 1. Computational Problems 17 equal to a. Find the ratio of the areas of the sections. 62. One cone is inscribed in, and the other is circumscribed about, a regular hexagonal pyra-mid. Find the difference between the volumes of the circumscribed and inscribed cones if the alti tude of the pyramid is H and the radius of the base of the circumscribed cone is R. 63. Given a ball and a point inside it. Three mutually perpendicular planes intersecting the ball along three circles are passed through this point in an arbitrary way. Prove that the sum of the areas of these three circles is constant, and find this sum if the radius of the ball is R and the distance from the point of intersection of the planes to the centre of the ball is equal to d. 64. In a ball of radius R the diameter AB is drawn. Two straight lines touch the ball at the points A and B and form an angle a (a < 90°) between themselves. Taken on these lines are points C and D so that CD touches the ball, and the angle between AB and CD equals <p (<p < 90°). Find the volume of the tetrahedron ABCD. 65. In a tetrahedron two opposite edges are perpendicular, their lengths are~a and b, the dis-tance between them is c. Inscribed in the tetra-hedron is a cube whose four edges are perpendicu ... lar to these two edges of the tetrahedron, ex actly two vertices of the 'cube lying on each face of the tetrahedron. Find the edge of the cube. 66. Two congruent, triangles KLM and KLN /"... ~ have a common side KL, KLM = LKN = 1'£13, I KL I = a, I LM I = I KN I = 6a. The planes KLM and KLN are mutually perpendicular. A 2-0"9 i8 Problems in Solid Geometry ball touches the line segments LM and KN at their midpoints. Find the radius of the ball. 67. A ball of radius R touches all the lateral faces of a triangular pyramid at the midpoints of the sides of its base. The line segment joining the vertex of the pyramid to the centre of the ball is bisected by the point of intersection with the base of the pyramid. Find the volume of the pyra-mid. 68. A tetrahedron has three right dihedral an-gles. One of the line segments connecting the mid-points of opposi te edges of the tetrahedron is equal to a, and the other to b (b > a). Find the length of the greatest edge of the tetrahedron. 69. A right circular cone with vertex S is in-scribed in a triangular pyramid SPQR so that the circle of the base of the cone is inscribed in the base PQR of the pyramid. It is known that /'..... ~ ~ PSR = n12, SQR = n/4, PSQ = 7n/12. Find the ratio of the lateral surface area of the cone to the area of the base PQR of the pyramid. 70: The base of the pyramid ABCDE is a par-allelogram ABeD. None of the lateral faces is an obtuse triangle. On the edge DC there is a point M such that the straight line EM is per-pendicular to BC. In addition, the diagonal of the base.{iC and the lateral edges ED and EB are relat-ed as follows: 1 AC 1 ~t 1 EB I ~ ~ 1 ED I. A section repre~enting an isosceles trapezoid is passed through the vertex B and the midpoint of one 9f the lateral edges. Find the ratio of the area of the section to the area of the base of the pyramid. Sec. 1. Computational Probtems i9 71. A line segment AB of unit length which is a chord of a sphere of radius 1 is at an angle n/3 to the diameter CD of this sphere. The distance from the end point C of the diameter to the nearer end point A of the chord A B is equal to -V 2. Determine the length of the line segment BD. 72. In a triangular pyramid ABCD the faces ABC and ABD have areas p and q, respectively, and form an angle a between themselves. Find the area of the section of the pyramid passing through the edge AB and the centre of the ball inscribed in the pyramid 73. In a triangular pyramid ABCD a section is passed through the edge AD (I AD I = a) and point E (the midpoint of the edge BC). The section makes with the faces ACD and ADB angles respectively equal to a and ~. Fin d the volume of the pyramid if the area of the section ADE is equal to S. 74. ABCD is a regular tetrahedron with edge a. Let M be the centre of the face ADC, and let N be the midpoint of the edge BC. Find the radius of the ball inscribed in the trihedral angle A and touching the straight line M N. 75. The base of a triangular pyramid ABCD is a regular triangle ABC. The face BCD makes an angle of 60° with the plane of the base. The centre of a circle of unit radius which touches the edges AB, A C, and the face BCD lies on the straight line passing through the point D per-pendicular to the bas,a. The altitude of the pyra-mid D H is one-half the side of the base. Find the volume of the pyramid. 76. In a triangular pyramid SABC I AC I = 1 AB 1 and the edge SA is inclined to the planes 20 Problems in Solid Geometry of the faces ABC and S BC at angles of 45°. It is known that the vertex A and the midpoints of all the edges of the pyramid, except SA, lie on the sphere of radius 1. Prove that the centre of the sphere is located on the edge SA, and find the area of the face ASC. 77. Given a cube ABCDAlBlClDl with edge a. Find the radius of the sphere touching the line segments ACl and CCv the straight lines AB and BC and intersecting the straight lines AC and AlCl. 78. A ball touches the plane of the base ABCD of a regular quadrangular pyramid SABCD at the point A, and, besides, it touches the ball inscribed in the pyramid. A cutting plane is passed through the centre of the first ball and the side BC of the base. Find the angle of inclination of this plane to the plane of the base if it is known that the diagonals of the section are perpen-dicular to the edges SA and SD. 79. Situated on a sphere of radius 2 are three circles of radius 1 each of which touches the other two. Find the radius of the circle which is smal-ler than the given circles, lies on the given sphere, and touches each of the given circles. 80. In a given rectangular parallelepiped ABCDAlBlClDl the lengths of the edges AB, BC, a.nd BBl are respectively equal to 2a, a, and a; E is the midpoint of the edge BC. The vertices M and N of a regular tetrahedron M N PQ lie on the straight line ClE, the vertices P and Q on the straight line passing through the point Bl and intersecting the straight line AD at the point F. Find: (a) the length of the line segment Sec. 1. Computational Problems 21 DF; (b) the distance between the midpoints of the line segments M Nand PQ. 81 . The length of the edge of a cube ABCDAlBlClDl is a. The points M and N lie on the line segments BD and CCl , respectively. The straight line MN makes an angle n/4 with the plane ABCD and an angle n/6 with the plane BBlClC. Find: (a) the length of the line seg-ment MN; (b) the radius of the sphere with centre on the line segment M N which touches the planes ABCD and BBtClC. 82. The vertex A of a regular prism ABCA1B1Cl coincides with the vertex of a cone~ the vertices Band C He on the lateral surface of this cone, and the vertices Bl and C 1 on the circle of its base. Find the ratio of the volume of the cone and the prism if I AAl I = 2.4 , AB I. 83. The length of the edge of a cube ABCDA 1BlClD l is equal to a. The points P, K, L are midpoints of the edges AA]t AlDl , RlCl , respectively; the point Q is the centre of the face CClDlD. The line segment MN with end points on the straight lines AD and KL intersects the line PQ and is perpendicular to it. Find the length of this line segment. 84. In a regular prism ABCAlBlCl the length of a lateral edge and the altitude of the base is equal to a. Two planes are passed through the vertex A: one perpendicular to the straight line ABl , the other perpendicular to the line ACl . Passed through the vertex Al are also two planes: one perpendicular to the line AlB, the other perpendicular to the line AlC. Find the volum:' of the polyhedron bounded by these four planes ancl" t4e plane BBlC,C, -22 Problems in Solid Geometry 85. The point 0 is a common vertex of two congruent cones si tu ated on one side of the plane a so that only one element of each cone (OA for one cone and OB for the other) belongs to the plane a. I t is known that the size of the angle between the altitudes of the cones is equal to p, and the size of the angle between the altitude and generatrix of the cone is equal to cp, and 2cp < ~. Find the size of the angle between the element OA and the plane of the base of the other cone to which the point B belongs. 86. Arranged inside a regular tetrahedron ABCD are two balls of radii 2R and 3R exter-nally tangent to each other, one ball being in-scribed in the trihedral angle of the tetrahedron with vertex at the point A, and the other in the trihedral angle with vertex at the point B. Find the length of the edge of this tetrahedron. 87. In a regular quadrangular pyramid SABCD with base ABCD, the side of the base is equal to a, and the angle between the lateral edges and the plane of the base is equal to a. The plane parallel to the diagonal of the base AC and the lateral edge BS cuts the pyramid so that a circle can be inscribed in the section obtained. Deter-mine the radius of this circle. 88. Each edge of a regular tetrahedron is equal to a. A plane P passes through the vertex Band midpoints of the edges AC and AD. A ball touche's the straight lines AB, AC, AD and the portion of the plane P enclosed inside the tetra-hedron. Find the radius of the ball. 89. In a regular tetrahedron, M and N are midpoints of two opposite edges. The projection of the t~tfalleclron OIl a plane parallel to M!V Sec. 1. Computational Problems 23 is a quadrilateral having area S one of the angles of which is equal to 60°. Find the surface area of the tetrahedron. 90. In a cube ABCDAlBlClDl a point M i~ taken on AC, and on the diagonal BDl of the .,/"... cube a point N is taken so that NMC = 60°, /""... MNB = 45°. In what ratios are the line seg-ments AC and BDl divided by the points M and N? 91. The base of a right prism ABCDAlBlClDl is an isosceles trapezoid ABCD in whioh AD is parallel to BC, I AD III BC I = n, n > 1. Passed through the edges AAl and BC are planes parallel to the diagonal BlD; and through the edges DDl and Bl Cl planes parallel to the dia-gonal AlC. Determine the ratio of the volume of the triangular pyramid bounded by these four planes to the vol ume of the prism. 92. The side of the base of a regular triangular prism ABCAlBlCl is equal to a. The points M and N are the respective midpoints of the edges AlBI and AAl . The projection of the line seg-ment BM on the line C]N is equal to aI2VS. Determine the altitude of the prism. 93. Two balls touch each other and the faces of a dihedral angle whose size is a. Let A and B be points at which the balls touch the faces (A and B belong to different balls and different faces). In what ratio is the line segment AB divided by the points of intersection with the surfaces of the balls? 94. The base of a pyramid ABCD is a regular ~rlangle ABC with side of length 12. The edge BD ... -'" 24 Problems in Solid Geometry is perpendicular to the plane of the base and is equal to 10 V 3. All the vertices of this pyramid lie on the lateral surface of a right circular cylin-der whose axis intersects the edge BD and the plane ABC. Determine the radius of the cylinder. 95. The base of a pyramid is a square ABCD with side a; the lateral edge SC is perpendicular to the plane of the base and is eq ual to b. M is a point on the edge AS. The points M, B, and D lie on the lateral surface of a right circular cone with vertex at the point A, and the point C in the plane of the base of this cone. Determine the area of the lateral surface of the cone. 96. Inside a right circular cone a cube is arranged so that one of its edges lies on the diameter of the base of the cone; the vertices of the cube not belonging to thi.s edge lie on the lateral surface of the cone; the centre of the cube lies on the alti-tude of the cone. Find the ratio of the volume of the cone to the volume of the cube. 97. In a triangular prism ABCAlBlCl , two sections are passed. One section passes through the edge AB and midpoint of the edge CCl , the other passing through the edge AlBl and the mid-point of the edge CB. Find the ratio of the length of the line segment of the intersection line of these sections enclosed inside the prism to the length of the edge AB. 98. In the tetrahedron ABCD the edge AB . d' I ,/'"". /"-IS perpen lCU ar to the edge CD, A CB = AD B, the area of the section passing through the edge AB and the midpoint of the edge DC is"equal to S, I DC I = a. Find the volume of t~e tetra-hedron A BCD. Sec. 1. Computational Problems 25 99. Given a regular triangular pyramid SABC (S its vertex). The edge SC of this pyramid coin-cides with a lateral edge of a regular triangular prism AlBlCA 2 B 2S (AlA2' BlB2 and CS are lateral edges, and AlBlC is one of the bases). The vertices A 1 and Bl lie in the plane of the face SAB of the pyramid. What part of the volume of the entire pyramid is the volume of the portion of the pyramid lying inside the prism if the ratio of the length of the lateral edge of the pyramid to the side of its base is equal to 2rV3? 100. In a frustum of a regular quadrangular pyramid with the lateral edges AA l , BBl , CCl , DDl , the side of the upper base AlBlClD1 is equal to 1, and the side of the lower base is equal to 7. The plane passing through the edge BlCl perpen-dicular to the plane ADlC separates the pyramid into two parts of equal volume. Find the volume of the pyramid. 101. The base of the prism A BCA lBlC1 is a reg-ular triangle ABC with side a. The projection of the prism on the plane of the base is a trape-zoid with lateral side AB and area which is twice the area of the base. The radius of the sphere pas-sing through the vertices A, B, A l , Cl is equal to a. Find the volume of the prism. 102. Given in a plane is a square ABCD with side a and a point M lying at a distance b from its centre. Find the sum of the volumes of the solids generated by revolving the triangles ABM, BCM, CDM, and DAM about the straight lines AB, BC, CD and DA, respectively. 103. D is the midpoint of the edge AlCl of Sl re¥ular trian~ular prism ABCAlBlCr A re~ulaf 26 Problems in Solid Geometry triangular pyramid SMNP is situated so that the plane of its base MNP coincides with the plane ABC, the vertex M lies on the extension of 1 AC and 1 CM 1 = T 1 AC I, the edge SN passes through the point D, and the edge SP intersects the line segment BBl. In what ratio is the line segment BBl divided by the point of intersection? 104. The centres of three spheres of radii 3, 4, and 6 are situated at the vertices of a regular triangle with side 11. How many planes are there which simultaneously touch all the three spheres? 105. All the plane angles of a trihedral angle N KLM (N the vertex) are right ones. On the face LN M a point P is taken at a distance 2 from the vertex N and at a distance 1 from the edge MN. From some point S situated inside the trihedral angle N KLM a beam of light is directed towards the point P. The beam makes an angle n/4 with the plane MNK and equal angles with the edges K Nand M N. The beam is mirror-reflected from the faces of the angle N K LM first at the point P, then at the point Q, and then at the point R. Find the sum of the lengths of the line segments PQ and QR. 106. The base of a triangular pyramid ABCD A A is a triangle ABC in which A = n/2, C = n16, I BC 1 = 2 V 2". The edges AD, BD, and CD are of the same length. A sphere of radius 1 touches the edges AD, BD, the extension of the edge CD beyond the point D, and the plane ABC. Find the length of the line segment of the tangent drawn from the point A to the sphere. 107, Tllree pa~lsl amoni which tije:r~ ~:re t'Y"q Sec. 1. Computational Problems 27 equal balls, touch a plane P and, besides, pair-wise touch one another. The vertex of a right circular cone belongs to the plane P, and its axis is perpendicular to this plane. All the three balls are arranged outside of the cone and each of them touches its lateral surface. Find the cosine of the angle between the generatrix of the cone and the plane P if it is known that in the tri-angle with vertices at the points of tangency of the balls with the plane one of the angles is equal to 150°. 10R. The volume of the tetrahedron ABCD is equal to 5. Through the midpoints of the edges AD and BC a plane is passed cutting the edge CD at the point M. And the ratio of the lengths of the line segments DM and CM is equal to 2/3. Compute the area of the section of the tetra-hedron by the plane if the distance from it to the vertex A is equal to 1. 109. A ball of radius 2 is inscribed in a regular triangular pyramid SABC with vertex S and base ABC; the altitude of the pyramid SKis equal to 6. Prove that there is a unique plane cutting the ed~es of the baseAB and BC at some points M and N, such that I M N I = 7, which touches the ball at the point equidistant from the points M and N and intersects the extension of the altitude of the pyramid S K beyond the point K at some point D. Find the length of the line segment SD. 110. All the edges of a triangular pyramid ABCD are tangent to a sphere. Three line seg-ments joining the midpoints of skew edges have the same length. The angle A BC is equal to 100°. Find the ratio of the altitudes of the pyram~q c:Jraw~ frq~ tile vertices 4. aQ~ B; 28 Problems in Solid Geometry 111. In a pyramid SABC the products of the lengths of the edges of each of the four faces are equal to one and the same number. The length of the altitude of the pyramid dropped from S on to the face ABC is equal to 2 V ~O;, and the size of the angle CAB is equal to arccos (! V ~). Find the vol ume of the pyramid SABC if I SA 12 + 1 SB !2 -5 1 SC 12 = 60. 112. Given in a plane P is an isosceles tri-angle ABC (I AB 1 = 1 BC 1 = 1, 1 AC 1 = 2a). A sphere of radius r touches the plane P at point B. Two skew lines pass through the points A and C and are tangent to the ball. The angle between either 'of these lines and the plane P is equal to a. Find the distance between these lines. 113. The base of a pyramid ABCEH is a con-ve'X quadrilateral ABCE which is separaterl by the diagonal BE into two equivalent triangles. The length of the edge A B is equal to 1, the lengths of the edges BC and CE are equal to each other. The sum of the lengths of the edges AH and EH is equal to V2". The volume of the pyra-mid is 1/6. Find the radius of the sphere having the grefltest volume among all the balls housed in the pyramid. 114. In a pyramid SABC a straight line inter-secting the edges AC and BS and perpendicular to them passes through the midpoint of the edge BS. The face ASB is equivalent to the face BSC, find t~e ~r~~ of the fac~ ASC is ~)Vj.ce t~e ~rea Qf Sec. 1. Computational Problems 29 the face BSC. Inside the pyramid there is a point M, and the sum of the distances from this point to the vertices Band S is equal to the sum of the distances to all the faces of the pyrami d. Find the distance from the point M to the vertex B if I AC I = VB, I BS I = 1. 115. The base of a pyramid is a rectangle with acute angle between the diagonals a (a < 60°), its lateral edges are of the same length, and the altitude is h. Situated inside the pyramid is a triangular pyramid whose vertex coincides with the vertex of the first pyramid, and the vertices of the base lie on three sides of the rectangle. Find the volume of the quadrangular pyramid if all the edges of the triangular pyramid are equal to one another, and the lateral faces are equivalent. 116. In a triangular pyramid SABC with hase ABC and equal lateral edges, the sum of the dihedral angles with edges SA and SC is equal to 180°. It is known that I AB I = a, I BC I = b. Find the length of the lateral edge. 117. Given a regular tetrahedron with edge a. A sphere touches three edges of the tetrahedron, emanating from one vertex, at their end points. Find the area of the portion of the spherical sur-face enclosed inside the tetrahedron. 118. Three circles of radius Y2 pairwise touch-ing one another are situated on the surface of a sphere of radius 2. The portion of the sphere's surface situated outside of the circles pres en ts two curvilinear triangles. Find the areas of these triangles. 119. Three dihedral angles of a tetrahedron, not belonging to one vertex, are equal to n/2. 1?rohlems in Solid Geometry The remaining three dihedral angles are equal to one another. Find these angles. 120. Two balls are inscribed in the lateral surface of a cone and touch each other. A third sphere passes through two circles along which the first two spheres touch the surface of the cone. Prove that the volume of the portion of the third ball situated outside of the cone is equal to the volume of the portion of the cone enclosed be-tween the first two balls inside the cone. 121. A sphere of radius R touches one base of a frustum of a Cone and its lateral surface along the circle coinciding with the circle of the other base of the cone. Find the volume of the solid representing a combination of a cone and a ball if the total surface area of this solid is equal to 8. 122. Two triangles, a regular one with side a and a right isosceles triangle with legs equal to b, are arranged in space so that their centroids coincide. Find the sum of the squared distances from all the vertices of one of them to all the vertices of the other. 123. In a regular triangular pyramid SABC (8 the vertex), E is the midpoint of the slant height of the face SBC, and the points F, L, and M lie on the edges AB, AC, and 8C, respec-tively, and I AL I = 1~ I AC I. It is known that EF LM is an isosceles trapezoid and the length of its base EF is equal to y7'. Find the volume of the pyramid. 124. Given a cube ABCDA1B1C1D1 with edge a. The bases of a cylinder are inscribed in the faces ABCD and A tBtCIDl " Let M be a point on the edge AB such that I AM I = a13, N a point on Sec. t Computational Probtems 31 the edge BlCl such that I NCl I = a14. Through the points Cl and M there passes a plane touch-ing the bases of the cylinder inscribed in A BCD, and through A and N a plane touching the base inscribed in AlBlClDl . Find the volume of the portion of the cylinder enclosed between the planes. 125. Determine the total surface area of the prism circumscribed about a sphere if the area of its base is equal to 8. 126. The centre of sphere a lies on the surface of sphere ~. The ratio of the surface area of sphere ~ lying inside sphere a to the total surface area of sphere ci is equal to 1/5. Find the ratio of the radii of spheres a and ~. 127. Circumscribed about a ball is a frustum of a cone. The total surface area of this cone is 8. AnGther sphere touches the lateral surface of the cone along the circle of the base of the cone. Find the volume of the frustum of a cone if it is known that the portion of the surface of the second ball contained inside the first ball has an area Q. 128. Circumscribed about a ball is a frustum of a cone whose bases are the great circles of two other balls. Determine the total surface area of the frustum of a cone if the sum of the surface areas of the three balls is equal to 8. 129. A section of maximal area is passed through the vertex of a right circular cone. It is known that the area of this section is twice the area of an axial section. Find the vertex angle of the axial section of the cone. 130.. Inscribed in a cone is a triangular pyra-mid 8ABC (8 coincides with the vertex of the 32 Problems in Solid Geometry cone, A, B, and C lie on the circle of the base of the cone), the dihedral angles at the edges SA, SB t and SC are respectively equal to a, ~, and y. Find the angle between the plane SBC and the plane touching the surface of the cone along the element SC. 131. Three points A, B, and C lying on the surface of a sphere of radius R are pairwise con-nected by arcs of great circles; the arcs are less than a semicircle. Through the midpoints of the -...--'-'" arcs AB and AC one more great circle is drawn '-'" which intersects the continuation of BC at the '-'" ........" point K. Find the length of the arc CK if I Be I = l (1 < :rtR). 132. Find the volume of the solid generated by revolving a regular triangle with side a about a straight line parallel to its plane and such that the projection of this line on the plane of the triangle con tains one of the al ti tudes of the triangle. 133. Consider the solid consisting of points si tuated at a distance not exceeding d from an arbitrary point inside a plane figure having a perimeter 2p and area S or on its boundary. Find the volume of this solid. 134. Given a triangular pyramid SABC. A ball of radius R touches the plane ABC at the point C and the edge SA at the point S. The straight line BS intersects the ball for the second time at the point opposite to the point C. Find the volume of the pyramid SABC if I BC I = a, I SA I = b. 135. Inside a regular triangular pyramid there is a vertex of a trihedral angle all of whoae plane Sec. 1. Computational Problems 33 angles are right ones, and the bisectors of the plane angles pass through the vertices of the base. In what ratio is the volume of the pyramid di-vided by the surface of this angle if each face of the pyramid is separated by it into two equiv-alent portions? 136. Given a parallelepiped ABCDAlBlClD1 whose volume is V. Find the volume of the com-mon portion of two tetrahedrons ABICDl and AlBClD. 137. Two equal triangular pyramids each hav ... ing volume V are arranged in space symmetri-cally with respect to the point o. Find the volume of their common portion if the point 0 lies on the line segment joining the vertex of the pyra-mid to the centroid of the base and divides this line segment in the ratio: (1) 1 : 1; (2) 3 : 1; (3) 2 : 1; (4) 4 : 1, reckoning from the vertex. 138. A regular tetrahedron of volume V is rotat-ed about the straight line joining the midpoints of its skew edges at an angle~a. Find the volume of the common portion of the given and turned tetrahedrons (0 < a < n). 139. The edge of a cube is a. The cube is rotat-ed about the diagonal through an angle a. Find the volume of the common portion of the original cube and the cube being rotated. 140. A ray of light falls on a plane mirror at an angle a. The mirror is rotated about the projec-tion of the beam on the mirror through an angle ~. By what angle will the reflected ray deflect? 141. Given in space are four points: A, B, C, ~ and D, where lAB 1 = I BC I = I CD I, ABC = Problems in Solid Geometry ../'... ~ BCD = CD A = a. Find the angle between the straight lines AC and BD. 142. Given a regular n-gonal prism. The area of its base is equal to S. Two planes cut all the lateral edges of the prism so that the volume of the porti on of the prism enclosed between the planes is equal to V. Find the sum of the lengths of the segments of the lateral edges of the prism enclosed between the cutting planes if it is known that the planes have no common points inside the prism. 143. Three successive sides of a plane convex pentagon are equal to 1, 2,. and a. Find the two remaining sides of this pentagon if it is known that the pentagon is an orthogonal projection on the plane of regular pentagon. For what values of a does the problem have a solution? 144. Given a cube ABCDAlBlClDl in which M is the centre of the face ABBlAl , N a point on the edge BlCl, L the midpoint of AlBl , K the foot of the perpendicular dropped from N on BCl • In what ratio is the edge BlCl divided ~ ~ by the point N if LM K = M KN? 145. In a regular hexagonal pyramid the centre uf the circumscribed sphere lies on the surface of the inscribed sphere. Find the ratio of the radii of the circumscribed and inscribed spheres. 146. In a regular quadrangular pyramid, the centre of the circumscribed ball lies on the sur-face of the inscribed ball. Find the size of the plane angle at the vertex of the pyramid. 147.The base of a quadrangular pyramid SABCD is a square ABCD with side a. Both angles be-~ec. 1. Computational Problems 35 ween opposite lateral faces are equal to a. Find he volume of the pyramid 148. A plane cutting the surface of a triangular Iyramid divides the medians of faces emanating ~om one vertex in the following ratios: 2 : 1, : 2, 4: 1 (as measured from the vertex). In rhat ratio does this plane divide the volume of flis pyramid? 149. n congruent cones have a common vertex. ~ach one touches its two neighbouring cones along n element, and all the cones touch the same plane. 'ind the angle at the vertex of the axial sections f the cones. 150. Given a cube ABCDAlBlClDl . The plane assing through the point A and touching the all insc~ibed in the cube cn ts the edges AlBl nd AlDl at points K and N. Determine the size f the dihedral angle between the planes AClK nd AClN 151. Given a tetrahedron ABCD. Another ~trahedron AlBlClDl is arranged so that its ertices Al, Bl , Cl, Dl lie respectively in the lanes BCD, CDA, DAB, ABC, and the planes f its faces AlBlCl , BlClDl , ClDlAl, DlAIBl ontain the respective vertices D, A, B, and C f the tetrahedron ABCD. It is also known that Ie point Al coincides with the centre of gravity f the triangle BCD, and the straight lines BDl , 'Bl , an.d DCl bisect the line segments AC, AD, nd AB, respectively. Find the volume of the )mmon part of these tetrahedrons if the volume f the tetrahedron ABCD is equal to V. 152. In the tetrahedron ABCD: 1 BC I = CD I = I DA I, 1 BD 1 = I AC I, I BD 1 > BC I, the dihedral angle at the edge AB is 36 Problems in Solid Geometry equal to n13. Find the sum of the remaining di-hedral angles. 153. Given a triangular prism ABCAlBtCl• It is known that the pyramids ABCCl , ABBlCl , and AAlBlCl are congruent. Find the dihedral angles between the plane of the base and the lateral faces of the prism if its base is a nonisosceles right triangle. 154. In a regular tetrahedron ABCD with edgela, taken in the planes BCD, CDA, DAB, and ABC are the respective points Al , Bl , Cl , and Dl so that the line AlBl is perpendicular to the plane BCD, BlCl is perpendicular 'to the plane CDA, ClDl is perpendicular to the plane DAB, and finally, DlAl is perpendicular to the plane ABC. Find the volume of the tetrahedron AlBlCtDt • 155. n congruent balls of radius R touch inter-nally the lateral surface and the plane of the base of a cone, each ball touching two neigh-bouring balls; n halls of radius 2R are arranged in a similar way touching externally the lateral surface of the cone. Find the volume of the cone. 156. Given a cube ABCDAlBlClDl . The points M and N are taken on the line segments AAl and BCl so that the line MN intersects the line BlD. Find ) Bel I IBN) i AM) I AA1 ) • 157. T t is known that all the faces of a tetra-hedron are similar triangles, but not all of them are congruent. Besides, any two faces have at least one pair of congruent edges not counting a common edge. Find the volume of \his tetra-Sec. 2. Problems on Proof 3i hedron if the lengths of two edges lying in one face are equal to ~ and 5. 158. Given three mutually perpendicular lines, the distance between any two of them being equal to a. Find the volume of the parallelepiped whose diagonal lies on one line, and the diagonals of two adjacent faces on two other lines. 159. The section of a regular quadrangular pyr-amid by some cutting plane presents a regular pentagon with side a.Find the volume of the pyra-mid. 160. Given a t.riangle ABC whose area is S, and the radius of the circumscribed circle is R. Erected to the pI ane of the triangle at the verti-ces A, B, and C are three perpendiculars, and points AI' Bh and Cl are taken on them so that the line segments AA l , BBl! CCl are equal in length to the respective altitudes of the tri-angle dropped from the vertices A, B, and C. Find the volume of the pyramid bounded by the planes AIBlC. AIBCh ABIC12 and ABC. Section 2 Problems on Proof 161. Do the altitudes intersect at one point in any tetrahedron? 162. Is there a triangular pyramid such that the feet of all the altitudes lie outside the corre-sponding faces? 38 Problem! in Solid Geometry 163. Prove that a straight line making equal angles with three intersecting lines in a plane is perpendicular to this plane. 164. What regular polygons can be obtained when a cube is cut hy a plane? 165. Prove that the sum of plane angles of a trihedral angle is less than 2n, and the sum of dihedral angles is greater than n. 166. Let the plane angles of a trihedral angle be equal to a, ~, and ,\1, and the opposite dihedral angles to A, B, and C, respectively. Prove that the fol1owing eqll alities hold trlle~ (1) sin ex sin P sin y sinA -sinB sin C (theorem of sines for a trihedral angle), (2) cos a = cos ~ cos '¥ + sin ~ sin '¥ cos A (first theorem of cosines for a trihedral angle), (3) cos A = -cos B cos C + sin B sin C cos a (second theorem of cosines for a trihedral angle). 167. Prove that if all the plane angles of a trihedral angle are obtuse, then all the dihedral angles are also obtuse. 168. Prove that if in a trihedral angle all the dihedral angles are acute, then all the plane an-gles are also acute. 169. Prove that in an arbitrary tetrahedron there is a trihedral angle all plane angles of which are acute. 170. Prove that in an arbitrary polygon all faces of which are triangles there is an edge such that all the plane angles adjacent to it are acute. Sec. 2. Problems on Proof 39 171. Prove that a trihedral prismatic surface can be cut by a plane in a regular triangle. 172. In a triangular pyramid all the plane angles at the vertex A are right angles, the edge AB is equ al to the sum of two other edges ema-nating from A. Prove that the sum of the plane angles at the vertex B is equal to n/2. 173. Can any trihedral angle be cut by a plane in a regular triangle? 174. Find the plane angles at the vertex of a trihedral angle if it is known that any of its sections by a plane is an acute triangle. 175. Prove that in any tetrahedron there is a vertex such that from the line segments equal to the lengths of the edges emanating from this vertex a triangle can be constructed. 176. Prove that any tetrahedron can be cut by a plane into two parts so that the obtained pieces can be brought together in a different way to form the same tetrahedron. 177. Find the plane angles at the vertex of a trihedr al angle if it is known that there exists another trihedral angle with the same vertex whose edges lie in the planes forming the faces of the given angle and are perpendicular to the opposi te edges of the gi ven angle. 178. A straight line l makes acute angles a, ~, and "y with three mutually perpendicular lines. Prove that a + ~ + ~ < n. 179. Prove that the sum of the angles made by the edges of a trihedral angle with opposite faces is less than the sum of its plane angles. Prove also that if the plane angles of a tri-hedral angle are acute, then the sum of the angles made by its edges with opposite faces is greater 40 Problems in Solid Geometry than one half the sum of the plane angles. Does the last statement hold for an arbitrary tri-hedral angle? 180. Prove that the sum of four dihedral angles of a tetrahedron (excluding any two opposite angles) is less than 2"" and the sum of all di-hedral angles of a tetrahedron lies between 2n and 3n. 181. From an arbitrary point of the base of a regular pyramid a perpendicular is erected. Prove that the sum of the Hne segments from the foot of the perpendicular to the intersection with the lateral faces or their extensions is con-stant. 182. Prove that if Xl' X2, X'l' .r~ are distances from an arbitrary point inside a tetrahedron to its faces, and hi' h2' h~H h" are the corresponding altitudes of the tetrahedron, then 183. Prove that the plane passing through the midpoints of two skew edges of a tetrahedron cuts it into two parts of equal volumes. . 184. Prove that if the base of a pyramid ABCD ~ /'. is a regular triangle ABC, and DAB = DBC = /""-. DCA, then ABCD is a regular pyramid. 185. Let a and ai' band bl , C and CI be pairs of opposite edges of a tetrahedron, and let a, ~, and y be the respective angles between them (a, ~, and y do not exceed 90°). Prove that one of the three numbers aal cos a, bb1 cos~, and CCI cos y is the sum- of the other two. Sec. 2. Problems on Proof 41 186. In a tetrahedron ABCD the edges DA, DB, and DC are equal to the corresponding alti-tudes of the triangle ABC (DA is equal to the altitude drawn from the vertex A, and so forth). Prove that a sphere passing through three verti-ces of the tetrahedron intersects the edges ema-nating from the fourth vertex at three points which are the vertices of a regular triangle. 187. Given a quadrangular pyramid MABCD whose base is a convex quadrilateral A BCD . A plane cuts the edges MA, MB, MC, and MD at pOints K, L, P, and N, respectively. Prove that the following relationship is fulfilled: IMAI IMel SBCD I MK I +SADB IMP! S I MD I S I MB I = ABC IMNI + ACD IMLI 188. From an arbitrary point in space perpen-diculars are dropped on the faces of a given cube. The six line segments thus obtained are diagonals of six cubes. Prove that six spheres each of which touches all the edges of the respective cube have a common tangent line. 189. Gi ven three parallel lines; A, B, and C are fixed points on these lines. Let M, N, and L he the respective points on the same lines situated on one side of the plane ABC. Prove that if: (a) the sum' of the lengths of the line segments AM, BN, and CL is constant, or (b) the sum of the areas of the trapezoids AMNB, BNLC, and CLMA is const.ant, then the plane MNL passes through a fixed point. 190. The sum of the lengths of two skew edges of a tetrahedron is equal to the sum of the lengths 42 Problems in Solid Geometry of two other skew edges. Prove that the sum of the dihedral angles whose edges are the first pair of edges is equal to the sum of the dihedral angles whose edges are represented by the second pair of the edges of the tetrahedron. 191. Let 0 be the centre of a regular tetrahed-ron. From an arbitrary point M taken on one of the faces of the tetrahedron perpendiculars are dropped on its three remaining faces, K, L, and N being the feet of these perpendiculars. Prove that the line OMpasses through the centre of grav-ity of the triangle KLN. 192. In a tetrahedron ABCD, the edge CD is perpendicular to the plane ABC, M is the mid-point of DB, and N is the midpoint of AB~ K is a point on CD such that I CK I = ; I CD I. Prove that the distance between the lines BK and CN is equal to that between the lines AM and CN. 193. Taken in the plane of one of the lateral faces of a regular quadrangular pyramid is an arbitrary triangle. This triangle is projected on the base of the pyramid, and the obtained tri-angle is again projected on a lateral face adjacent to the given one. Prove that the last projecting yields a triangle which is similar to the origi-nally taken. 194. In a tetrahedron ABCD, an arbitrary pOint Al is taken in the face BCD. An arbitrary plane is passed through the vertex A. The straight lines passing through the vertices B, C. and D parallel to the line AA 1 pierce this plane at points B I , CI , and DI . Prove that the volume of the tetra-hedron AIBICID 1 is equal to the volume of the tetrahedron ABCD. Sec. 2. Problems on Proof 43 195. Given a tetrahedron ABCD. In the planes determining its faces, points AI' BI , CI , DI are taken so that the lines AAI , BBI , CCI , DDI are parallel to one another. Find the ratio of the volumes of the tetrahedrons ABCD and AIBIC1Dlo 196. Let D be one of the vertices of a tetrahed-ron, M its centre of gravity, 0 the centre of the circumscribed ball. It is known that the pOints D, M and the median points of the faces contain-ing D lie on the surface of the same sphere. Prove that the lines DM and OM are mutually perpendicular. 197. Prove that no solid in space can have even number of symmetry axes. 198. Given a circle and a point A in space. Let B be the projection of A on the plane of the given circle, D an arbitrary point of the circle. Prove that the projections of B on AD lie on the same circle. 199. The base of a pyramid ABCDE is a quad-rilateral ABCD whose diagonals AC and BD are mutually perpendicular and intersect at point M. The line segment EM is the altitude of the pyramid. Prove that the projections of the point M on the lateral faces of the pyramid lie in one plane. 200. Prove that if the straight line passing through the centre of gravity of the tetrahedron ABCD and the centre of the sphere circumscribed about it intersects the edges AB and CD, then 1 AC 1 = 1 BD I, 1 AD 1 = 1 BC I· 201. Prove that if the straight line passing through the centre of gravity of the tetrahedron ABCD and the centre of the sphere inscribed in 44 Problems in Solid Geometry it intersects the edges AB and CD, then I AC I = I BD I, I AD I = I BC I· 202. Given a cube ABCDAIBICIDI. Passed through the vertex A is a plane touching the sphere inscribed in the cube. Let M and N be the points of intersection of this plane and the lines AlB and AID. Prove that the line M N is tangent to the ball inscribed in the cube. 203. Prove that for a tetrahedron in which all the plane angles at one of its vertex are right angles the following statement holds true: the sum of the squared areas of rectangular faces is equal to the squared area of the fourth face (Py-thagorean theorem for a rectangular tetrahedron). 204. Prove that the sum of the squared projec-tions of the edges of a cube on an arbitrary plane is constant. 205. Prove that the sum of the squared projec-tions of the edges of a regular tetrahedron on an arbitrary plane is constant. 206. Two bodies in space move in two straight lines with constant and unequ~l velocities. Prove that there is a fixed circle in space such that the ratio of distances from any point of this circle to the bodies is constant and is equal to the ratio of their velocities. 207. Gi ven a ball and two points A and B outside it. Two intersecting tangents to the ball are drawn from the points A and B. Prove that the point of their intersection lies in one of the two fixed planes. 208. Three balls touch the plane of a given triangle at its vertices and are tangent to one another. Prove that if the triangle is scalene, then there exist two balls touching the three Sec. 2. Problems on Proof 45 given balls and the plane of the triangle, and if rand p (p > r) are the radii of these balls and R is the radius of the circle circumscribed about . 1 1 2 va the trIangle, then - - - = R • r p 209. Given a tetrahedron ABCD. One ball touches the edges AB and CD at points A and C, the other at points Band D. Prove that the pro-jections of A C and BD on the str aight line passing through the centres of these balls are equal. 210. Is there a space pentagon such that a line segment joining any two nonadjacent vertices intersects the plane of the triangle formed by the remaining three vertices at an interior point of this triangle? 211. Prove that a pentagon with equal sides and angles is plane. 212. Given a parallelepiped ABCDAIBIClDl whose diagonal ACl is equal to d and its volume to V. Prove that from the line segments equal to the distances from the vertices AI' B, and D to the diagonal ACl it is possible to construct a triangle, and that if s is the area of this tri-angle, then V = 2ds. 213. Given a tetrahedron ABCD in which A l , Bit CI , DI are the median points of the faces BCD, CDA, DAB, and ABC. Prove that there is a tetrahedron A Jl2C2D2 in which the edges A 2B 2, B 2C2, C2D 2 and D2A2 are equal and parallel to the line segments AAlt BBl , CClt and DDl , re-spectively. Find the volume of the tetrahedron A 2B 2C2D 2 if the volume of the tetrahedron ABCD is equal to V. 214. Given a tetrahedron. Prove that there is another tetrahedron KLMN whose edges KL, 46 Problems in Solid Geometry LM, MN, and NK are perpendicular to the cor .. responding faces of the given tetrahedron, and their lengths are numerically equal to the areas of these faces. Find the volume of the tetrahedron KLMN if the volume of the given tetrahedron is equal to V. 215. Given three intersecting spheres. Three chords belonging to diHerent spheres are drawn through a point, situated on the chord common for all the three spheres. Prove that the end points of the three chords lie on one and the same sphere. 216. A tetrahedron ABeD is cut by a plane perpendicular to the radius of the circumscribed sphere drawn towards the vertex D. Prove that the vertices A, B, C and the points of intersec-tion of the plane with the edges DA, DB, DC lie on one and the same sphere. 217. Given a sphere, a circle on the sphere, and a point P not belonging to the sphere. Prove that the other points of intersection of the lines, con-necting the point P and the points on the given circle, form a circle with the surface of the sphere. 218. Prove that the line of intersection of two conical surfaces with parallel axes and equal an-gles of axial sections is a plane curve. 219. Taken on the edges AB, BC, CD, and DA of the tetrahedron ABCD are points K, L, M, and N situated in one and the same plane. Let P be an arbitrary point in space. The lines PK, PL, PM, and PN intersect once again the circles circumscribed about the triangles PAB, PBC, PCD, and PDA at the points Q, R, S, and T, respectively. Prove that the points P, Q, R, S, and T lie on the surface of a sphere. 220. Prove that the edges of a tetrahedral anflle Sec. 3. Problems on Extrema 47 are elements of a cone whose vertex coincides with the vertex of this angle if and only if the sums of the opposite dihedral angles of the tetrahedral angle are equal to each other. 221. Given a hexagon all faces of which are quadrilaterals. It is known that seven of its eight vertices lie on the surface of one sphere. Prove that the eighth vertex also lies on the sur-face of the same sphere. 222. Taken on each edge of a tetrahedron is an arbitrary point different from the vertex of the tetrahedron. Prove that four spheres each of which passes through one vertex of the tetrahedron and three points taken on the edges emanating from this vertex intersect at one point. Section 3 Problems on Extrema. Geometric Inequalities 223. Given a dihedral angle. A straight line l lies in the plane of one of its faces. Prove that the angle between the line l and the plane of the other face is maximal when l is perpendicular to the edge of this dihedral angle. 224. In a convex quadrihedral angle, each of the plane angles is equal to 60°. Prove that the angles between opposite edges ncanot be all acu te or all obtuse. 225. The altitude of a frustum of a pyramid is equal to h, and th e area of the midsection is S. 48 Problems in Solid Geometry What is the range of change of the volume of this pyramid? 226. Find the greatest value of the volume of the tetrahedron inscribed in a cylinder the radius of whose base is R and the altitude is h. 227. The base of a rectangular parallelepiped ABCDAIBICIDI is a square ABCD. Find the greatest possible size of the angle between the line BDI and the plane BDC I" 228. In a regular quadrangular prism ABCDAIBICIDI the altitude is one half the side of the base. Find the greatest size of the angle AIMCl , where M is a point on the edge AB. 229. The length of the edge of the cube ABCDAIBICIDI is equal to 1. On the extension of the edge AD, a point M is chosen for the point D so that I AM I = 2 -V 275. Point E is the mid-point of the edge AlBl , and point F is the mid-point of the edge DD I . What is the greatest value that can be attained by the ratio I MP 1/1 PQ I, where the point P lies on the line segment AE, and the point Q on the line segment CF? 230. The length of the edge of the cube ABCDAIBIC1D l is equal to a. Points E and F are the midpoints of the edges BBI and CCl , respectively. The triangles are considered whose vertices are the points of intersection of the plane parallel to the plane ABCD with the lines ACl , CE, and DF. Find the smallest value of the areas of the triangles under consideration. 231. Inscribed in a regular quadrangular pyra-mid with side of the base and altitude equal to 1 (each) is a rectangular parallelepiped whose base is in the plane of the base of the pyramid, and Sec. 3. Problems on Extrema 49 the vertices of the opposite face lie on the lateral surface of the pyramid. The area of the base of the parallelepiped is equal to s. What is the range of variation of the length of the diagonal of the parallelepiped? 232. The bases of a frustum of a pyramid are regular triangles ABC and AIBICI 3 cm and 2 cm on a side, respectively. The line segment joining the vertex CI to the centre 0 of the base ABC is perpendicular to the bases; I CIO I = 3. A plane i~ passed thrQugh the vertex B and mid-points of the edges AIBI and BICI. Consider the cylinders situated inside the polyhedron ABCAIMNCI with bases in the face AIMNCt • Find: (a) the greatest value of the volumes of such cylinders with a given altitude h; (b) the maximal value of the volume among all cylin-ders under consideration. 233. All edges of a regular triangular prism ABCAIBICI have an equal length a. Consider the line segments with end points on the dia-gonals BCI and CA I of the lateral faces parallel to the plane ABBIAI. Find the minimal length of such line segments. 234. Given a trihedral angle and a point inside it through which a plane is passed. Prove that the volume of the tetrahedron formed by the given angle and the plane will be minimal if the given point is the centre of gravity of the triangle which is the section of the trihedral angle by the plane. 235. The surface area of a spherical segment is equal to S (the spherical part of the segment is considered here). Find the greatest volume of this segment. 236. A cube with edge a is placed on a pJane. 50 Problema in Solid Geometry A light source is situated at a distance b (b > a) from the plane. Find the smallest area of the shadow thrown by the cube onto the plane. 237. Given a convex central-symmetric poly-hedron. Consider the sections of this polyhedron parallel to the given plane. Check whether the following statements are true: (1) the greatest area is possessed by the section passing through the centre; (2) for each section consider the circle of smal-lest radius containing this section. Is it true that to the greatest radius of such a circle there cor-responds the section passing through the centre of the polyhedron? 238. What is the smallest value which can be attained by the ratio of the volumes of the cone and cylinder circumscribed about the same ball? 239.1Two cones have a common base and are arrange,d on different sides of it. The radius of the base is r, the altitude of one cone is h, of the other H (h ~ H). Find the maximal distance between two elements of these cones. 240. Given a cube ABCDA1B1C1D1 with edge a. Find the radius of the smallest ball which touches the straight linesAB., B.C, CD, and DA. 241. The diagonal of a cube whose edge is equal to 1 lies on the edge of a dihedral angle of size a (a < 180°). What is the range of variation of the volume of the portion of the cube enclosed inside this angle? 242. The lengths of the edges of a rectangular parallelepiped are equal to a, b, and c. What is the greatest value of the area of an orthogonal projection of this parallelepiped on a plane? Sec. 3. Problems on Extrema 5f 243. The length of each of five edges of a tetra-hedron is less than unity. Prove that the volume of the tetrahedron is less than 1/8. 244. The vertex E of the pyramid ABCE is found inside the pyramid ABCD. Check whether the following statements are true: (1) the sum of the lengths of the edges A E, BE, and CE is less than that of the edges AD, BD, and CD; (2) at least one of the edges AE, BE, CE is shorter than the corresponding edge AD, BD, or CD? 245. Let rand R be the respective radii of the balls inscribed in, and circumscribed about, a regular quadrangular pyramid. Prove that 246. Let Rand r be the respective radii of the balls inscribed in, and circumscribed about, a tetrahedron. Prove that R ~ 3r. 247. Two opposite edges of a tetrahedron have lengths band c, the length of the remaining edges being equal to a. What is the smallest value of the sum of distances from an arbitrary point in space to the vertices of this tetrahedron? 248. Given a frustum of a cone in whi ch the angle between the generatrix and greater base is equal to 60°. Prove that the shortest path over the surface of the cone between a point on the boundary of one base and the diametrically op-posite point of the other base has a length of 2R, where R is the radius of the greater base. 52 Problems in Solid Geometry 249. Let a, b, and c be three arbitrary vectors.. Prove that lal+lbl+lcl+la+b+cl ~ I a + b I + I b + c 1 + I c + a I. 250. Given a cube ABCDAtBtCtDt with edge a. Taken on the line AAt is a point M, and on the line BC a point N so that the line MN intersects the edge CtDt" Find the smallest value of the quantity 1 MN I. 251. The base of a quadrangular pyramid is a rectangle one side of which is equal to a, the length of each lateral edge of the pyramid is equal to b. Find the greatest value of the volume of such pyramids. 252. Given a cube A BCDAtBtCtDt with edge a. Find the length of the shortest possible segment whose end points are situated on the lines ABt and BCt making an angle of 60° with the plane of the face ABCD. 253. Three equal cylindrical surfaces of ra-dius R with mutually perpendicular axes touch one another pairwise. (a) What is the radius of the smallest ball touching these cylindrical surfaces? (b) What is the radius of the greatest cylinder touching the three given cylindrical surfaces, whose axis passes inside the triangle with vertices at the points of tangency of the three given cylinders? 254. Two vertices of a tetrahedron are situated on the surface of the sphere of radius Y 10, and two other vertices on the surface of the sphere of radius 2 which is concentric with the first Sec. 3. Problems on Extrema 53 one. What is the greatest volume of such tetra· hedrons? 255. Two trihedral angles are arranged so that the vertex of one of them is equidistant from the faces of the other and vice versa; the distance between the vertices is equal to a. What is the minimal volume of the hexahedron bounded by the faces of these angles if all the plane angles of one of them are equal to 60° (each), and those of the other to 90° (each)? 256. What is the greatest volume of the tetra-hedron ABCD all vertices of which lie on the surface of a sphere of radius 1, and the edges AB, Be, CD, and DA are seen from the centre of the sphere at an angle of 60°? 257. Given a regular tetrahedron with edge a. Find the radius of such a ball with centre at the centre of the tetrahedron for which the sum of the volumes of the part of the tetrahedron found outside of the ball and the part of the sphere outside of the tetrahedron reaches its smallest value. 258. Prove that among triangular pyramids with a given base and equal altitudes the smal-lest lateral surface is possessed by the one whose vertex is projected into the centre of the circle inscribed in the base. r,... 259. Given a cube with edge a. Let N be a point on the diagonal of a lateral face, M a point on the circle found in the plane of the base having its centre at the centre of the base and radius (5/12)""n. Find the least value of the quantity I MN I.~ 260. (a) The base of the pyramid SABC is 54 Problems in Solid Geomelry /"... /"... a triangle ABC in which BAC = A, CBA = B. the radius of the circle circumscribed about it is equal to R. The edge se is perpendicular to the plane ABC. Find I SC I if it is known that 1/sin a + 1/sin ~ -1/sin 'V = 1, where a, ~, and l' are angles made by the edges SA, SB, and SC with the planes of the faces SBC, SAC, and SAB, respectively. (b) Let a, ~, and l' be angles made by the edges of a trihedral angle with the planes of opposite faces. Prove that 1/sin a + 1/sin ~ -1/sin l' :;;;, 1. 261. Can a regular tetrahedron with edge 1 pass through a circular hole of radius: (a) 0.45; (b) 0.44? The thickness of the hole may be neg-lected. Section 4 Loci of Points 262. Prove that in an arbitrary trihedral angle the bisectors of two plane angles and the angle adjacent to the third plane angle lie in one plane. 263. Prove that if the lateral surface of a cy-linder is cut by an inclined plane, and then it is cut along an element and developed on a plane, the line of intersection will represent a sinu~ soid. ~ 264. Given on the surface of a cone is a line different from an element and such that any See. 4. Loci of Points 55 two points of this line can be connected with an arc belonging to this line and representing a line segment on the development. How many points of self-intersection has this line if the angle of the axial section of the cone is equal to a? 265. Three mutually perpendicular lines pass through the point O. A, B, and C are points on these lines such that ,OA 1 = lOB 1 = 10C I. Let l be an arbitrary line passing through 0; AI' B1, and CI points symmetric to the points A, B, and C with respect to l. Through A I, B I , and CI three planes are drawn perpendicular to the lines OA, OB, and ~C, respectively. Find the locus of points of intersection of these planes. 266. Find the locus of the midpoints of line segments parallel to a given plane whose end points lie on two skew lines. 267. Given three pairwise skew lines. Find: (a) the locus of centres of gravity of triangles ABC with vertices on these lines; (b) the locus of centres of gravity of triangles ABC with vertices on these lines whose planes are parallel to a given plane. 268. Three pairwise skew lin~s lp 121 la are perpendicular to one and the same straight line and intersect it. Let Nand M be two points on the lines l] and l2 such that the line N M intersects the line lao Find the locus of midpoints of line segments N M. 269. Given in space are several arbitrary lines and a point A. Through A a straight line is 56 Problems iu Solid Geomau, dr awn so that the sum of the cosines of the acute angles made by this line with the given ones is equal to a given number. Find the locus of such lines. 270. Given a triangle ABC and a straight line l. AI, BI, and CI are three arbitrary points on the line l. Find the locus of centres of gravity of triangles with vertices at the midpoints of the line segments AA I , BBI , CCI. 271. Given a straight line l and a point A. Through A an arbitrary line is drawn which is skew with l. Let M N be a common perpendicular to this line and to l (M lies on the line passing through A). Find the locus of points M. 272. Two spheres a and ~ touch a third sphere co at points A and B. A point M is taken on the sphere a, the line MA pierces the sphere co at point N, and the line NB pierces the sphere ~ at point K. Find the locus of such points M for which the line M K touches the sphere ~. 273. Given a plane and two points on one side of it. Find the locus of centres of spheres passing througb these points and touching the plane. 274. Find the locus of midpoints of common tangents to two given spheres. 275. Two lines II and l2 touch a sphere. Let M and N be points on II and l2 such that the line M N also touches the same sphere. Find the locus of points of tangency of the line'" 'M N with this sphere. . 276. Given in space are a point 0 and two straight lines. Find the locus of points M such that the sum of projections of the line segment Sec. 4. Loci of Points 57 OM on the given lines is a constant quantity. 277. Given in space are two straight lines and a point A on one of them; passed through the given lines are two planes making a right dihed-ral angle. Find the locus of projections of the point A on the edge of this angle. 278. Given three intersecting planes having no common line. Find the locus of points such that the sum of distances from these points to the given planes is constant. 279. Given a triangle ABC. On the straight line perpendicular to the plane ABC and passing through A an arbitrary point D is taken. Find the locus of points of intersection of the altitudes of triangles DBC. 280. Given three intersecting planes and a straight line I. Drawn through a point M in space is a line parallel to I and piercing the given planes at points A, B, and C. Find the locus of points M such that the sum I AM I + I BM I + I CM I is constant. 281. Given a triangle ABC. Find the locus of points M such that the straight line joining the centre of gravity of the pyramid ABCM to the centre of the sphere circumscribed about it intersects the edges A C and BAf. 282. A trihedral angle is cut by two planes parallel to a given plane. Let the first plane cut the edges of the trihedral angle at points A, B, and C, and the second at points AI' BI , and CI (identical letters denote points belonging to one and the same edge). Find the locus of points of intersection of the planes ABCI , ABIC, and AIBC. 283. Given a plane quadrilateral ABCD. Find 58 Problems in Solid Geome&ry the locus of points M such that the lateral surface of the pyramid ABCDM can be cut by a plane so that the section thus obtained is: (a) a rectan-gle, (b) a rhombus, (c) a square; (d) in the pre-ceding case find the locus of cen trefl of squares. 284. Given a plane triangle ABC. Find the loc/us of points M in space such that the straight line connecting the centre of the sphere circum-scribed about ABCM with G as the centre of gravity of the tetrahedron ABCM is perpendic-ular to the plane AMG. 285. A circle of constant radius displaces touching the faces of a trihedral angle all the plane angles of which are equal to 90° (each). Find the locus of centres of these circles. 286. A spider sits in one of the vertices of a cube whose edge is 1 cm long. I t crawls over the surface of the cube with a speed of 1 em/s. Find the locus of points on the surface of the cube such that can be reached by the spider in two seconds. 287. Given a trihedral angle each of whose plane angles is equal to 90°, 0 is the vertex of this angle. Consider all possible polygonal lines of length a beginning at the point 0 and such that any plane parallel to one of the faces of the angle cuts this polygonal line not more than at one point. Find the locus of end points of this polygonal line. 288. Given a ball with centre O. Let ABCD be the pyramid circumscribed about it for which the following inequalities are fulfilled: I OA 1 ~ lOB I ~ I OC I > I OD I. Find the locus of poin ts A, B, C, and D. 289. Given a triangle ABC. Find the locus of Sec. 4. Loci of Points 59 points M in space such that from the line seg-ments MA, MB, and MC a right triangle can he formed. 290. On the surface of the Earth there are points the geographical latitude of which is equal to their longitude. Find the locus of the projections of all these points on the plane of the equator. 291. Given a right circular cone and a point A outside it found at a distance numerically equal to the altitude of the cone from the plane of its base. Let M be a point on the cone such that a beam of light emanating from A towards M, being mirror-reflected by the surface of the cone, will be parallel to the plane of the base. Find the locus of projections of points M on the plane of the base of the cone. 292. Drawn arbitrarily through a fixed point P inside a ball are three mutually perpendicular rays piercing the surface of the ball at points A, B, and C. Prove that the median point of the triangle ABC and the projection of the point P on the plane ABC describe one and the same spherical surface. 293. Given a trihedral angle with vertex 0 and a point N. An arbitrary sphere passes through 0 and N and intersects the." edges of the trihedral angle at points A, B, and C. Find the locus of centres of gravity of triangles ABC. An Arbitrary Tetrahedron 294. Given an arbitrary tetrahedron and a point N. Prove that six planes each of which passes through one edge of the tetrahedron and Problems in Solid Geometry is parallel to the straight line joining N to the midpoint of the opposite edge intersect at one point. 295. Prove that six planes each of which passes through the midpoint of one edge of the tetrahedron and is perpendicular to the opposite edge intersect at one point (Monge's point). 296. Prove that if Monge's point lies in the plane of some face of a tetrahedron, then the foot of the altitude dropped on this face is found on the circle described about it (see the preced-ing problem). 297. Prove that the sum of squared distances from an arbitrary point in space to the vertices of a tetrahedron is equal to the sum of squared distances between the midpoints of opposite edges and quadruple square of the distance from the point to the centre of gravity of the tetrahedron. 298. Prove that there are at least five and at most eight spheres in an arbitrary tetrahedron each of which touches the planes of all its faces. 299. ABCD is a three-dimensional quadrilat-eral (A, B, C, and D do not lie in one plane). Prove that there are at least eight balls touching the lines AB, BC, CD, and DA." Prove also that if the sum of some two sides of the given quadrilateral is equal to the sum of two other sides, then there is an infinitude of such balls. 300. Prove that the product of the lengths of two opposite edges of a tetrahedron divided by the product of the sines of the dihedral angles of the tetrahedron corresponding to these edges is constant for a given tetrahedron (theorem of sines). Sec. 4. Loci of Points 61 301. Let Sh Rh li (i = 1, 2, 3, 4) denote respectively the areas of faces, the radii of the circles circumscribed about these faces, and the distances from the centres of these circles to the opposite vertices of a tetrahedron. Prove that for the vertices of the tetrahedron the following formula is valid: 4 ~ ~ st (It--Ri). i-1 302. Given an arbitrary tetrahedron. Prove that there exists a triangle whose sides are numer-ically equal to the products of the lengths of the opposite sides of" the tetrahedron. Let S denote the area of this triangle, V the volume of the tetrahedron, R the radius of the sphere circumscribed about it. Then the following equal-ity takes place: S = 6VR (Grelle's formula). 303. Let a and b denote the lengths of two skew edges of a tetrahedron, a and ~ the sizes of the corresponding dihedral angles. Prove that the expression a2 + b2 + 2ab cot a cot ~ is independent of the choice of the edges (Bret-schneider's theorem). An Equlfaced Tetrahedron 304. A tetrahedron is said to be equijaced if all of its faces are congruent triangles or, which is the same, if opposite edges of the tetrahedron are pairwise equal. Prove that for a tetrahedron 62 Problems in Solid Geometry to be equifaced, it is necessary and sufficient. that any of the following conditions he fulfilled: (a) the sums of plane angles at any of the three vertices of a tetrahedron are equal to 180°; (b) the sums of plane angles at some two verti-ces of a tetrahedron are equal to 180°, and, be. sides, some two opposite edges are equal; (c) the sum of plane angles at some vertex of a tetrahedron is equal to 180°, and, besides, the tetrahedron has two pairs of equal opposite edges; ~ (d) the following equality is fulfilled ABC = /'.... /""'-.. /"... ADC = BAD = BCD, where ABCD is a given tetrahedron; (e) all the faces are equivalent; (f) the centres of the inscribed and circum-scribed spheres coincide; (g) the line segments joining the midpoints of opposite edges are perpendicular; (h) the centre of gravity coincides with the centre of the circumscribed sphere; (i) the centre of gravity coincides with the centre of the inscribed sphere. 305. Prove that the sum of cosines of the dihedral angles of a tetrahedron is positive and does not exceed 2, the equality of this sum to 2 is characteristic only of equifaced tetrahedrons. 306. The sum of the plane angles of a trihedral angle is equal to 180°. Find the sum of the co-sines of the dihedral angles of this trihedral angle. 307. Prove that for an equifaced tetrahedron (a) the radius of the inscribed ball is half the radius of the ball which touches one face of the See. 4. Loci of Points 63 tetrahedron and the extensions of three other faces (such ball is called externally inscribed); (b) the centres of four externally inscribed balls are the vertices of a tetrahedron congruent to the given one. 308. Let h denote the altitude of an equifaced tetrahedron, hi and h2 the line segments into which one of the altitudes of a face is divided (by the point of intersection of the altitudes of this face). Prove that h2 = 4h1h2• Prove also that the foot of the altitude of the tetrahedron and the point of intersection of the altitudes of the face on which this altitude is dropped are symmetric with respect to the centre of the circle circumscribed about this face. 309. Prove that in an equifaced tetrahedron the feet of the altitudes, the midpoints of the altitudes, and the points of intersection of the altitudes of faces lie on the surface of one and the same sphere (12-point sphere). 310. A circle and a point M are given in a plane. The point lies within the circle less than 1/3 of the radius from its centre. Let ABC denote an arbitrary triangle inscribed in a given circle with centre of gravity at the point M. Prove that there are two fixed points in space (D and D') symmetric with respect to the given plane such that the tetrahedrons ABCD and ABCD' are equifaced. 311. A square ABCD is given in a plane. Two points P and Q are taken on the sides BC and CD so that 1 CP 1 + 1 CQ 1 = 1 AB I. Let M denote a point in space such that in the tetra-hedron APQM all the faces are congruent trian-Problems in SoJid Geometry gles. Determine the locus of projections of points M on the plane perpendicular to the plane of the square and passing through the diagonal AG. An Orthocentric Tetrahedron 312. In order for the altitudes of a'tetrahedron to intersect at one point (such a tetrahedron is called orthocentric), it is necessary and suffi-cient that: (a) opposite edges of the tetrabedron be mu-tually perpendicular; (b) one altitude of the tetrahedron pass through the point of intersection of the altitudes of the base; (c) the sums of the squares of skew edges be equal; (d) the line segments connecting the midpoints of skew edges be of equal length; (e) the products of the cosines of opposite dihedral angles be equal; (f) the angles between opposite edges be equal. 313. Prove that in an orthocentric tetrahedron the centre of gravity lies at the midpoint of the line segment joining the centre of the circum .. scribed sphere to the point of intersection of the altitudes. 314. Prove that in an orthocentric tetrahedron the following relationship is fulfilled: 1 OB 12 = 4R2 -- 3l2, where 0 denotes the centre of the circumscribed sphere, H the point of intersection of the alti-tudes, R the radius of the circumscribed sphere, 5ec. 4. Loci of Points 65 l the distance between the midpoints of the skew edges of the tetrahedron. 315. Prove that in an orthocentric tetrahedron the plane angles adjacent to one vertex are all acute or all obtuse. 316. Prove that in an orthocentric tetrahedron the circles of nine points of each face belong to one sphere (24-point sphere). 317. Prove that in an orthocentric tetrahedron the centres of gravity and the points of inter-section of the altitudes of faces, as well as the pOInts dividing the line segments of each alti-tude of the tetrahedron from the vertex to the point of intersection of the altitudes in the ratio 2 : 1, lie on one and the same sphere (12-point sphere). 318. Let H denote the point of intersection of altitudes of an orthocentric tetrahedron, M the centre of gravity of some face, and None of the pOints of intersection of the line H M 'vith the sphere circumscribed about the tetra-hedron (M lies between Hand N). Prove that I MN I = 2 I HM I. 319. Let G denote the centre of gravity of an orthocentric tetrahedron, F the foot of a certain altitude, K one of the points of intersection of the straight line FG with the sphere circumscribed about the tetrahedron (G lies between K and F). Prove that I KG I = 3 I FG I. An ArbitraI'y Polyhedron. The Sphere 320. Prove that on a sphere it is impossible to arrange three arcs of great circles 3000 each so that no two have common points. 321. Prove that the shortest line connecting 66 Problems in Solid Geometry two points on the surface of a sphere is the smaller arc of the great circle passing through these points. (Considered here are lines passing over the surface of the sphere.) 322. Given a polyhedron with equal edges which touch a sphere. Check to see whether there always exists a sphere circumscribed about this polyhedron. 323. Find the area of the triangle formed by the surface of a sphere of radius R intersecting a trihedral angle whose dihedral angles are equal to a, ~, and y, and whose vertex coincides with the centre of the sphere. 324. Let M denote the number of faces, K the number of edges, N the number of vertices of a convex polyhedron. Prove that M -K +N = 2. (Euler was the first to obtain this relationship; it is true not only for convex polyhedra, but also for a broader class of so-called simpZy-con-nected polyhedra.) 325. Given on the surface of a sphere is a cir-cle. Prove that of all spherical n-gons containing the given circle inside themselves, a regular spherical n-gon has the smallest area. 326. Prove that in any con vex polyhedron there is a face having less than six sides. 327. Prove that in any con vex polyhedron there is either a triangular face or a vertex at which three edges meet. 328. Prove that a convex polyhedron cannot ha ve seven edges. Prove also that for any n ~ 6) n 7 there is a polyhedron having n edges. 329. Prove that in any convex polyhedron Sec. 4. Loci of Points 67 there are two faces with equal number of sides. 330. Found inside a sphere of radius 1 is a Convex polyhedron all dihedral angles of which are less than 2n/3. Prove that the sum of the lengths of the edges of this polyhedron is less than 24. 331. The centre of a sphere of radius R is situated outside a dihedral angle of size a at a distance a (a < R) from its edge and lies in the plane of one of its faces. Find the area of the part of a sphere enclosed inside the angle. 332. A ball of radius R touches the edges of a tetrahedral angle each of whose plane angles is equal to 60°. The surface of the ball inside the angle consists of two curvilinear quadrilat-erals. Find the areas of these quadrilaterals. 333. Given a cube with edge a. Determine the areas of the parts of the sphere circumscribed about this cube into which it is separated by the planes of the faces of the cube. 334. Given a convex polyhedron. Some of its faces are painted black, no two painted faces having a common edge, and their number being more than half the number of all the faces of the polyhedron. Prove that it is impossible to inscribe a ball in this polyhedron. 335. What is the greatest number of balls with a radius of 7 that can simultaneously touch a ball with a radius of 3 without intersecting one another. An Outlet into Space 336. Taken on the sides BC and CD of the square ABCD are points M and N so that 68 Problems in Solid Geometry 1 CM I + 1 CN 1 = 1 AB I· The lines AM and AN divide the diagonal BD into three segments. Prove that a triangle can always be formed from these segments, one angle of this triangle being equal to 60°. 337. Given in a plane are a triangle ABC and a point P. A straight line 1 intersects the lines AB, BC, and CA at points CI , A l , and B l, respectively. The lines PCI' PAl' and PBl intersect the circles circumscribed respectively about the triangles PAB, PBC, and PAC at the respective points C2, A 2, and B 2, different from the point P. Prove that the points P, A 2, B 2, C 2 lie on one and the same circle. 338. Prove that the diagonals, connecting opposite vertices of the hexagon circumscribed about a circle, intersect at one point (Brianchon's theorem). 339. Two triangles AlBlCl and A~2C2 are arranged in a plane so that the lines AIA2' B 1B 2, and Cl C2 intersect at one point. Prove that the three points of intersection of the following three pairs of lines: AlBl and A 2 B 2t BlCl and B 2C2 , ClA l and C2A2 are collinear (that is, in one straight line) (Desargues' theorem). 340. Three planes in space intersect along one straight line. Three trihedral angles are arranged so that their vertices lie on this line, and the edges in the given planes (it is supposed that the corresponding edges, that is, the edges lying in one plane, do not intersect at one point). Prove that the three points of intersection of the corresponding faces of these angles are col-linear. Answers, Hints, Solutions Section 1 a3 -V6 4hs 5+ -V5 3 1.. 108 2. 45' 3. 24 a. 4. 4a2• 5 ± 2 -Va ab be ca 5. n - 2 arccos 13 . 6. 2e' 2a' 2b. 7. a vi ,(}I ~ 8. The st~ent of the problem is obvious for a tri-angle whose one side lies on the line of intersection of the planes ct and ~. Then it is possible to prove its validity for an arbitrary triangle, and then also for an arbitrary polygon. 9. Take the triangles ABICI and AB2C2 for the bases of the pyramids ABICIDI and AB2C2D 2 • 10. The angles under consideration are equal to the angles formed by the diagonal of some rectangular parallel-epiped with three edges emanating from its end point. 12. Consider the parallelepiped formed by the planes passing through the edges of the tetrahedron parallel to opposite edges. (This method of completing a tetrahedron to get a parallelepiped will be frequently used in further constructions.) The volume of the tetrahedron is equal to one third the volume of the parallelepiped (the planes of the faces of the tetrahedron cut off the parallelepiped four triangular pyramids, the volume of each of them being equal to one sixth the volume of the parallelepiped), and the volume of the parallelepiped is readily expressed in terms of the given quantities, sincr the diagonals of its faces are equal and parallel (or, simply, coincide) to the corresponding edgrs of thr tetrahedron, and tho altitude of the parallolepiped is equal to the distanrc hctwoen tho corresponding edges of the tetrahedron. 13. It is easy to see that each of these relationships (between the areas of the faces and the line segments of 70 Problems in Solid Geometry the edge) is equal to the ratio of the volumes of two tetra-hedrons into which the given tetrahedron is separated by the bisecting plane. 14. 10ining the centres of the sphere to the vertices of the polyhedron, divide it into pyramids whose bases are the faces of the polyhedron, and whose altitudes are equal to the radius of the sphere. 15. It is easy to verify the validity of the given for-mula for a tetrahedron. Here, two cases must be consid-ered: (1) three vertices of the tetrahedron lie moneplane and one vertex in the other; (2) two vertices of the tetra-hedron lie in one plane and two in the other. In the second case, use the formula for the volume of a tetrahedron from Problem 12. ~ Then note that an arbitrary convex polyhedron can be broken into tetrahedrons whose vertices coincide with those of the polyhedron. This statement is sufficiently obvious, although its proof is rather awkward. Moreover, the suggested formula is also true for nonconvex polyhedra of the,indicated type, as well as for solids enclosed between two parallel flanes for which the area of the section by a plane para leI to these planes is a quadratic function of the distance to one of them. This formula is named Simpson's formula. 16. Since the described frustum of a cone may be consid-ered as the limit of frustums of pyramids circumscribed about the same sphere, for the volume of a frustum of a cone the formula from Problem 14 holds true. 17. First prove the following auxiliary statement. Let the line segment AB rotate about the line l (l does not intersect AB). The perpendicular erected to AB at the midpoint of AB (point C) intersects the line l at point 0; MN is the projection of AB on the line l. Then the area of the surface generated by revolving AB about l is equal to 2n I CO 1 • I M N I. The surface generated by revolving AB represents the lateral surface of the frustum of a cone with radii of the bases BN and AM, altitude I MN I, andgeneratrixAB. Through A draw a straight line parallel to l, and denote by L the point of its intersection with the perpendicular BN dropped from B on l, I MN I = I AL I. Denote the projection of C on l by K. Note that the triangles ABL and COK are similar to each other. This taken into consideration, the lateral surface of the frustum of a cone Answers, Hints, Solutions 71 Is equal to 2n IBN 1 + 1 AM I .1 AB I =2n I CK 1.1 AB I 2 = 2n I CO I ·1 AL I = 2n I CO 1 ·1 M N I· Now, with the aid of the limit passage, it is easy to get the statement of our problem. (If the spherical zone under '-" consideration is obtained by revolving a certain arc AB of a circle about its diameter, then the surface area of this zone is equal to the limit of the area of the surface generated by rotating about the same diameter the polygo-nalline A L 1L 2 • • • LnB all vertices of which lie on AB pro-vided that the length of the longest link tends to zero.) 18. Let AB be the chord of the given segment, and 0 the centre of the circle. Denote by x the distance from 0 to AB, and by R the radius of the circle. Then the volume of the solid generated by rotating the sector AOB about the diameter will be equal to the product of the area of '-' the surface obtained by revolving the arc AB (see Problem 17) by R13, that is, this volume is equal to .!. 2nR1h=! n (x1+ a2 ) h=~ na1h+! nx2h 3 3 4 6 3· But the second term is equal to the volume of the solid generated by revolving the triangle AOC about the dia-meter (see the solution of Problem 17). Hence, the first term is just the volume of the solid obtained by revolving the given segment. 19. place equal loads at the vertices of the pyramid; to find the centre of gravity of the system, you may proceed as follows: first find the centre of gravity of three loads and then, placing a triple load at the found point, find the centre of gravity of the entire system. You may also proceed in a different way: first find the centre of grav-ity of two loads, then of two others and, finally, the centre of gravity of the whole system. You may not resort to a Q\eehanicaI interpretation, but, simply, consider the triangle formed by two vertices of the tetrahedron and the midpoint of the opposite edge. 21.. Through each edge of the tetrahedron pass a plane parallel to the opposite edge (see the solution of Prob-72 Problems in Solid Geometry lem 12). These planes form a parallelepiped whose edges are equal to the distances between the midpoints of the skew edges of the tetrahedron, and the ed~s of the tetra-hedron themselves are the dia¥Onals of Its faces. Then take advantage of the fact that In an arbitrary parallelo-gram the sum of the squared lengths of the diagonals is equal to the sum of the squared lengths of its sides. 22. If M is the midpoint of BBH then AIM is parallel to CK. Consequently, the desired angle is equal to the angle MAID. On the other hand, the plane AIDM is paral-lel to CK, hence, the distance between CK and.-4 ID is equal to the distance from the point K to the ~lane AID M. Denote the desired distance by x, and the dihedral angle by cpo Then we have t 1 a8 V A1MDK = '"3 S A1M.oX="3 S A1KDa == t2 • a8 Hence x=-=48-=----. Find the sides of ~ AIMD: A1MD ,r-aYS 1 AID I = a ,,2, 1 AIM 1 = 2 ' 3 1 DM 1=2 a. By the theorem of cosines, ". find cos'l'= l/ ; thu~ to S 3 2 a A1MD = 4' a, x = 3" • t a Answer: arccos -y to ' 3". 23. This problem can be solved by the method applied in Problem 22. Here, we suggest another method for deter-mining the distance between skew medians. Let ABCD be the given tetrahedron, K the midpoint of AB, M the midpoint of A C. Project the tetrahedron on the plane passing through AB perpendicular to CK. The tetrahedron is projected into a triangle ABD1, where D] is the projec-tion of D. If MI is the projection of M (MI is the midpoint of AK), then the distance between the lines CK and DM is equal to the distance from the point K to the line D1M1• The distance is readily found, since DIKMI is a right Answers, Hints, Solutions 73 triangle in which the legs DIK and KMI are respectively equal to a 11'273 (altitude of the tetrahedron) and a/4. The problem has two solutions. To get the second solu-tion, consider the medians C K and BN, where N is the mid-point of DC. j, 1 "II /2 2 V10 Answer: arccos "6' a V 35 and arccos "3 , a 10 • 24. It follows from the hypothesis that the quadrilat-eral ABeD is not convex. va Answer: -3-. 25 (2b ± a) a 26 4131; V41 27 1/ 7 • 2 V 3h2 _ a2 ' • 384 • • a 8 • "II / • a2 28. a+ b ± V 2ab - T · a V22 30. 2+ Va. 31. 8 32 th y-). 33.2 arccos (Sin ct sin ~) . • 3a+h 3+2 3 n 34. 12V. 35. 6R2-2a2. 36. ~ . 37. arctan (2-V3). 1 38. I r 0 < ct < arccos "4 ' l = R V 27 +3 tan2 ~ [ arctan (3 cot ~) -ct ] ; if ct;;;; arccos ! ' l = O. 39. 25: 20: 9. 40. arccrs (2- V5). 41. ~2 • 42. Denote the side of the base and the altitude of the ~rism by a, I KB I = x. It follows from the hypothesis that the projection of K M on the plane of the base is paral-1el to the bisector of the angle C of the triangle ABC, that is, I BIM I = 2x, I MCI I = a -2x. Let Ll be the pro-74 Problems in Solid Geometry jection of L on A C. It is also possible to obtain from the hypothesis that LLI = I ALII Y 3 , I LIC I = a -2x. 2 Conse~ently, the quantity I A Ll I can take on the follow-ing values: (1) I ALII = a -I MCII = a -(a -2%) = 2.%; (2) I ALl I = a + (a -2.%) = 2 (a -x). In the first case I KL II = I KL 1 II + ILLI II = aD + 10x2 -4ax; in the second I KL II = 6 (a -X)2. In both cases I KM 12 = 3zI + a2 • Solving two systems of equations, we get two respec-tive values for a: 7 y6+yi4 al = Y97 ' az=. 8 All$wer: 7 y6+ y« Y97' 8 43. arctan vI . 44. Extend the lateral faces until they intersect. In doing so, we obtain two similar pyramids whose bases are the bases of the given frustum of a pyramid. Let a he the side of the greater base of the frustum, and a the dihed-ral angle at this base. We can find: the altitude of the greater pyramid h = a ~ 3 tan a, the radi us of the inscribed ball r = a ~3 tan ~, the altitude of the smaller pyra-mid hI = h -2r = a ~3 (tan a -2 tan ~ ), the side h tan a -2 tan ~ of the smaller base al = -oa h l = a 2, the tan a lateral ellge of the greater pyramid l = aY3 y tan2a +4, 6 the lateral edge of the smaller pyramid II = l h~; then take advantage of the condition of existence of a ball touching all the edges of a frustum of a pyramid. This Answers, Hints, Solutions 75 condition is equivalent to tho existence of a circle inscribed in a lateral face, that is, the following equality must be fulfilled: 2 (l -ll) = a + ap Expressing l, ll' al' in terms of a and a, we get the equa-tion ya -., tan2 a+4.tan ..::..= tan a - tan"::" 3 r 2 2 • a -.,--.,-Hence we find tan 2= r 3- r 2. Answer: 2 arctan (V3-V2). -f+Y5 f+Y5 45. 2 < a < 2 ' a =1= 1; f V= 12 y(a2+f) (3a2-1-a4). 46 3-cos a-cos ~-cos 'V • 3+cos a+cos ~+cos 'V. n 3a2 y3 . n 47. If 0 < a < -6 ' then S = 2 ; If -6 ~ a < cos a 2 a2 ( -arctan -"-3' then S= 6 18cota-3Y3-r cos a 2 y3cot2 a); if arctan ::-3 ~ a < ~, then S = v_a' (V3+cot a). 3 sin a ( a2b2 + b2c2 - c2a2 ) 48. arccos a2b2+b2c2+c2a2 • 49. The polyhedron ABMDCN is a triangular prism with base ABM and lateral edges AD, BC, MN. b Answer: 2a y 4a2 - b2 • Y4c2-a2b2 50. R= V . 2 4c2-a2-b2 76 Problems in Solid Geometry 5t. ~ Y3ml+3nl+3pl-a2-b2-c2. 52. On the extension of the edge CCI take a point K so that B)K is parallel to BCI , and through the edge BBI pass a plane parallel to the given (Fig. 1). This plane A Fig. 1 must pass either through the internal or external bisector of the angle DB1K. Since the ratio in which the plane passing through BBI divides D Kl is equal to the ratio in which it divides DC, two cases are possible: (1) the plane passes through a point N on the edge D G such that I DN 1/1 NC I = V 3tV 2, or (2) it passes through a point M on its extension, and once again I DM 1/1 AIG I = Y3/Y'2. Find the distance from the point K to the nrst plane. It is equal to the distance from the point G to the line BN. If this distance is x, then 2S B NC a Y2 x---IBN! -- (V3+ V2)V 11-4 y-B a (V'-6-t) V-2 5 Answers, Hints, Solutions 77 and x sin q> = -I-B-I-K-I V6-t 5 ' where q> is the angle between the plane BBIN and lines BID and BIK. The other angle is found exactly in the same manner • . y6±t Answer: arCSln 5 . 53. Let ABeD be the given pyramid whose lateral edges are: I DAI = a, I DB I = x, I DC I = Yi by the hypothesis, these edges are mutually perpendicular, and x + Y = a. It is easy to find that t ~r t SABC=2" r a2(x2+y2)+x2y2, VABCD="6axy. On the other hand, if R is the radius of the required ball, then R V ABCD=""3 (SDAB+SDBO+SDCA -SABC) R =6 [ax+ay+xy-ya2 (X2+y2)+X2y2] R R =6 (a2+xy-Ya4-2xya2+x2y2)=T:&Y. Equating the two expressions for V ABCD, we find R = ~ • lL 54. It follows from the hypothesis that the vertex S is projected either into the centre of the circle inscribed in the triangle ABC or into the centre of the circle exter-nally inscribed in it. (Anlexternally inscribed circle touches one side of the triangle and the extensions of two other sides of the triangle.) 2 Answer: if V3 < b ~ a, then V = :2 y 3b2 -a2; if a < b ~ a V 3, two answers are possible: a2 .. 13b2 2 V a2 Va .. lb2 I VI = 12 r - a , 2 = 12 r - a ; 78 Problems in Solid Geometry if b> a va, three answers are possible: a2 a'l va VI = 12 V3b'l-a'l, V'I= 12 Vbl-al , V.= a21r3}rbl-3aI. /'.... /'.... ~ /'.... 55. Let the angles SAB, SCA, SAC, SBA be equal to a -2<p, a -<p, a, a + <p, respectively. By the theorem of sines, from the triangle SAB we find 1 SA 1 = 1 AB 1 ~in (a+ <p) sm (2a-<p) , and from the triangle SA C we find: 1 SA 1=1 CA I s.in (a-<p) sm (2a-<p). But, by the hypothesis, 1 AB 1 = 1 AC I. Hence, sin (a + <p) = sin (a - a, two cases are then possible: b. ASC = b. ASB and /'.... A C S = a + ~. The line segment l will be less than, equal to, or greater than a according as 2a + ~ is greater than, equal to, or less than n. Answers, Hints, Solutions 79 Besides, in both cases the plane angles adjacent to the vertex A must satisfy the conditions under which a tri-hedral angle is possible. n Answer. If ~ > 6" ' 2a+~ > n, then V = aSsin(a+~) .. 1"1-2 2~' 12 . f cos, sma if ~~ : ,a\T' then V __ as ~~n ~a+~) -V3 sin2 ~-[2 cos (2a+~)+cos ~]2; SIn a . n n n 2n If ~>6' a<3"' 3<a+~<3' then both answers are possible. 57. :' as measured from the point K. 58. Take Cl so that ABCCI is a rectangle (Fig. 2). Dt, is the midpoint of A C 1; Ou O2 are the centres of the circles A Fig. 2 c circumscribed about the triangles ACID and ABC, respectively; 0 is the centre of the sphere circumscribed about ABCD. Obviously, O. is the midpoint of AC, AB and CIC are respectively perpendicular to AD and A Cu consequently, the planes ADCl and ABCCI are mutually perpendicular, and OlDIO,.o is a rectangle. Thus I DCl 1== 80 Problems in Solid Geometry YI DC II - I CIC 12 = -rb2 -a2, the radius of the ci~ cle circumscribed about the triangle DClA is R _ I DCl I Yb2 -a:.s-1 -/\ -2 sin a • 2 sin DACI The radius of the sphere R = I OA I can be found from the triangle A 0 10 (this triangle is not shown in the figure): R== Y I AOI 12+ I 0 10 P'= 21 -. / b~~a2 + a2 V Sln a 1 = Vb2-a2 cos2 a 2 sin a • 59. Let K be the midpoint of the edge A B of the cube ABCDAlBlClDl, M the mid~oint of the edge DICit K and M are simUltaneously the mldpoints of the edges PQ and RS of a regular tetrahedron PQRS. DlCI lies on RS. If the edge of the tetrahedron is equal to b, then I M K I = bY 2/2 = aY 2. Hence, b = 21.1. Project the tetrahedron on the plane ABCD (Fig. 3): PI' Ql' RIt Sl are the respective projections of P, Q, R, S Fig. 3 Since PQ makes an angle of 45° with this plane, the length of PlQ! will be aV 2. • Let L be the point of intersection of the lines A Band PlRl• From the similarity of the triangles PILK and PlR1Ml we find ILK 1= I RIM. I-I PIK I I PlMl I Answers, Hints, Solutions 81 Hence, the edge PR of the tetrahedron (and, consequently, other edges: PS, OR, and QS) pierces the cube. To compute the volume of the obtained solid, it is con-venient to consider the solid as a tetrahedron with cor-ners cut away. a3 V2 ~ Answer: 12 (16 V2-17). 60. Denote the lengths of these skew edges by a and b, the distance between them by d, and the angle by cp. Using the formula from Problem 15, find the volumes of the obtained parts: V 10 d' V 7 d' 1= 81 ab sln cp, 2 = 162 ab sin cp. 20 Answer: T' 61. The area of the projection of the second section on the first plane is half the area of the first section. On the other hand (see Problem 8), the ratio of the area of the projection of the second section to the area of the section itself is equal to cos a. Answer: 2 cos a. 62. 112 nR2H. 63. If x, y, and z are the respective distances from the centre of the ball to the passed planes, then x2 + y2 + Z2 = d2 , and the sum of the areas of the three circles will be equal to n [(R2 -x2) + (R2 -y2) + (R2 -Z2)] = n (3R2 -tJ2). 64. Let I A C I = x, I BD I = y (A C and BD touch the ball). DI is the projection of D on the plane passing through A C parallel to BD. We have 2R I CD I=x+y= ,I CD1 1=2Rtancp. cos cp In the triangle CADI the angle CADI is equal either to a or 1800 -a. According to this, x and y must satisfy 6-0449 82 Problems in Solid Geometry one of the two systems of equations: { z+y= c:~cp , x2+y'!-2xy cos a= 4R'! tanS cp, (1) or (2) 2R R2 For system (1) we get: x+y= , xy= ---cos cp cos'! ~ 2 2R R" for system (2): x+y= , xy= ---. Taking into cos cp • '! a sm 2" account the inequality (x+ y)2 ~ luy, we get that system (1) has a solution for cp > T ' and system (2) for q> > ~ -~ . Since the volume of the tetrahedron ABeD is equal to ~ zy R sin a, we get the answer: if f < q> < ~ -~ , the volume of the tetrahedron is equal to 2 a. 3t a 3t 3" R3 tan '""2; If ""2-""2 ~ cp < 2' two values of the . 2 a 2 a volume are posSIble: 3" R3 tan 2 and 3" R3 cot T . 85. Let the common perpendicular to the given edges be divided by the cube into the line segments y, z, and z, y + z + z = c (z is the edge of the cube, y is adjacent to the edge a). The faces of the cube parallel to the given edges cut the tetrahedron in two rectangles, the sides of z + z yb the first one are equal to a, -, of the second to c c Z a, x + y b, the smaller sides of these rectangles being c c Answers, Hints, Solutions 83 oqual to the edge of the cube, that is,.JL b= x, ~ a = x, c c whence ~ ~ ~c Y=T' x=a and X= ab+bc+ca . 66. Let 0 1 and O2 be projections of the centre of the ball 0 on the planes KLM and KLN, P the midpoint of ML. The projections 0 1 and O2 on KL must coincide. It is possible to prove that these projections get into the mid-M --~----~--~--L R Fig. 4 point of KL, point Q (Fig. 4). Since the dihedral angle between the planes KLM and KLN is equal to 900 , the radius of the desired sphere will be -V I POI 12+ I 01Q 12. If 01P is extended to intersect the line KL at point R, then from the right triangle P LR, we find I R L I = 6a, I RP I = 3a-V 3. We then find IRQ 1= ffa, 2 I 0 Q I - ffa -Va I RO I _ ffa -Va 1 -6' 1-3 ' I POI I = ffa va 3 aaY3= 2ara. Problems in Solid Geometry Consequently, tho radius of the sphere is equal to .. / 4a2 I 121a2 __ ~ • / 137 V 3 r 12 -- 2 Y 3· 67. Using the equality of tangent lines emanating from one point, prove that the base is a right triangle, and the medians of the lateral faces drawn to the sides of the base are equal. This will imply that the pyramid is regular. R3 -y6 Answer: 4 • 68. The three given angles cannot be adjacent to one face; further, they cannot adjoin to one vertex, since in this case all the line segments joining the midpoints of opposite edges will be equal. It remains only the case when three edges corresponding to right angles form an open polygonal line. Let AB, B C, and CD be the mentioned edges. Denote: I AB I = x, I BC I = y, I CDI = z. Then the distance between the midpoints of AB and CD .. ;-x2 Z2 will be V "4 +y21+:r' and between AC and BD (or AD and BC): ~ V xl + Z2. The edge AD will be the greatest: I A D I = V x2 + y2 + Z2 = V b 2 + 3a2• 4V3-3 69. n 13 . 70. First prove that ABCD is a rectangle and the plane DEC is perpendicular to the plane ABCD. To this end, through E pass a section perpendicular to B C. This sec-tion must intersect the base a10ng a straight line passing through M and intersecting the line segments BC and AD (~ossibly, at their end points). Further, drawing a section which is an isosceles trapezoid through B is only possible if the section contains the edge AB, and I DE I = I EC I, 1 A E 1 = 1 EB I. Consequently, ~ 1 AC 1 > 1 ED 1 = 1 EC 1, {I AC I > I E B 1 = I AE I , Answers, Hints, Solutions 85 i.e. I AC 12> I CE 12 + I AE 12 and ~AEC is not an ~ acute-angled triangle. But AEC cannot be obtuse, since ~ in that case DEC would also be obtuse. 5 5 Thus, I AC 1 = 4" 1 AE I ="'3 I EC I· 3 .. /65 Answer: "8 V 14' 71. Through C draw a straight line parallel to AB and take on it a point E such that I CE I = I AB I, ABEC is a parallelogram. If 0 is the centre of the sphere, then /'.... the triangle OC E is regular, since OC E = n/3 and 1 CE 1 = 1 (it follows from the hypothesis). Hence, the point 0 is equidistant from all the vertices of the parallel-ogram ABEC. Hence, it follows that ABEC is a rectan-gle, the projection of 0 on the plane ABEC is represented by the point K which is the centre of ABEC, and I BD I = 2 I OK I = 2 VI OC 12 -~ I BC 12 = 1. 72. If x is the area of the sought-for section, I AB I = a, then, taking advantage of the formula of Problem 11 for the volume of the pyramid ABCD and its parts, we get . a . a 2 pxsIDT 2 qXSIDT "3 a +"3 a 2 pq sin a 3 a whence 2 a pqcosT X= p+q 73 8S 2 sin a sin ~ • 3a sin (a+~) . 74. When cutting the ball by the plane AM N, we get a circle inscribed in the triangle AM N. In this triangle V3 V3 a AN 1 = a-, 1 AM I = a~, I ,MNI=- (found from 232 86 Problems in Solid Geometry the triangle C M N). Consequently, if L is the point of contact of the desired ball and AM, then I AL 1= I AN I + I A~ I-I MN I = (15 2 V3-+) a. The ball inscribed in ABCD has the radius r= ! V ~ a and touches the plane A CD at point M. Thus, if x is the radius of the desired ball, then x I AL I 5-V3 r = I AM I =-:; 4 5 Y6-3 V2 Hence, x= 48 4. 9V3 75. 8 ;-76. 1 3. 77. a V2. 1 78. arctan ... r - . 2 r 3 79. Notation: 0 is the centre of the sphere; 01, O2 , 0 3 the centres of the given circles, O. the centre of the sought-for circle. Obviuosly, the triangle 0}0203 is regular. Find its sides (M is the point of contact of the circles with centres 01 and O2). 101M 1 = 102M 1 = 1, 10M 1= /'-/"--... -2. Hence, M001 = M00 2 = 30°, 1 001 1 = 1002 1 = V 3, 1 01 0 2 I = va. 004 is perpendicular to the plane 0 1 0 20 3 and passes through the centre of the triangle 0 1 0 20 3, the distances from 01, O2, and 0 3 to 004 are equal to 1-Let K be the puint of contact of the circles 0 1 and 0 4 , L the foot of the perpendicular dropped from 01 on 004• KN is perpendicular to L01, lOlL 1 = 1 01 K 1 = 1, 1 001 I = va. From the similarity of the right triangles /2 01KN and 001L find lOIN I = V 3' Thus, the required . /2 radiUS 1 04K 1 = 1 LN 1 = 1 -1 3' Answers, Hints, Solutions 87 80. (a) Since the opposite edges in a regular tetrahed-ron are perpendicular, the lines CIE and BIF must also be perpendicular (Fig. 5). If K is the midpoint of CI C, then, since the lines BIK and BIAI are perpendicular to the line CIE, the line Fig. 5 BIF must lie in the plane passing through BIK and BIAI' hence it follows that AIF is parallel to BIK, and, therefore I D F I = a (this is the answer to this item). (b) The distance between the midpoints of M Nand PQ is equal to the distance between the lines BIF and ClEo It can be found by equating different expressions for the volume of the tetrahedron FBI CIE: 1 1 3" S B 1 C1E2a = 6" I FBI 1'1 CIE I·x. 4a Hence, X= 3 -y5 • a (2--Y2) 81. (a) a; (b) 2 . 82. Let I AB I = a, then I ABI I = I ACI I = 2.6a. On the lines AB and A C, take points K and L such that I AK I = I AL I = I ABI I = I AC 1= 2.6a. An iso-sceles trapezoid KLCIBI is inscribea in the circle of the base of the cone. All the sides of this trapezoid are readily computed and, hence, the radius of the circle circums-cribed about it is also easily found, it equals ~~-Y7 a. It is now possible to find the volume of the cone and prism. 88 Problems in Solid Geometry 15,379n Answer: -....;.....-----,,--4800 V3 · 83. Note that the line segment MN is bisected by its point of intersection with the line PQ. Project this line segment on the plane ABCD. If Nl is the projection of N, Kl the midpoint of AD, Ql the midpoint of DC (Kl and Q1 are the respective projections of K and Q), then NIM is perpendicular to A Q 1 and is bisected by the point of in-./'.... /'... tersection. Thus, NIAD = 2Q1AD. Hence we find 1 NIKI 1 and then I NIM I • . a ;-Answer. a l 14. 84. Through the edge AAI pass a plane perpendicular to the plane BCC1B 1 (Fig. 6). M and N are the points of s s Fig. 6 Fig. 7 intersection of this plane with C1B)..and CB. Take on MN a point K such that 1 N K 1 = 1 MN I. By the hypothesis, AA).MN is a square, hence, AK is perpendicular to AM, and it follows that A K is perpendicUlar to the plane AC1B 1, that is, AK is a straight line along which the r,lanes passing through the vertex A intersect. Analogous-y, determine the point L for the vertex A 1• The straight Answers, Hints, Solutions 89 lines A K and AlL intersect at the point 8. Thus, our polyhedron represents a quadrangular pyramid 8KPLQ with vertex 8 whose base is found in the plane BBlCIC. Further, BIN is the projection of ABI. Hence it follows that the plane jlassing through A perpendicular to ABI intersects the plane BBICIC along a straight line perpen-dicular to BIN. It follows from the hypothesis that the triangle BINCl is regular. Hence, the quadrilateral P LQ K, which is the base of the pyramid 8 P LQ K, is a rhom-bus formed from two regular triangles with side I KL 1= 3a. 9as V3 Answer: 4 • 85. The sought-for angle makes the angle between the element OA and the axis of the second cone equal to n/2. Denote by P and Q the centres of the bases of the given cones, by 8 the point at which the planes of the bases of the cones intersect the perpendicular erected to the plane OAB at the point 0 (Fig. 7). In the pyramid 80AB: lOA I = lOB I, 80 is perpendicular to the plane OAB, o P and OQ are respectively perpendicular to 8B and 8 A, .~ ~ ~ /"--... POB = QOA = <p, POQ = ~. Find POA. Let I OA I -lOB I = l, I AB I = a. Then lOP I = I OQ I = l cos <p, l I 8A I = I 8B 1=. , sm <p cos2 <p I 8 P I = I 8 Q I = lOP I cot <p = l. , sm <p I PQ I = I AB I I 8 P I = a cos2 <p. I 8B I On the other hand, I P{! I = 2 lOP I sin ~ = 2l cos <p sin ~ • Hence a cos <p = 2l sin ~ . (1) 90 Problems in Solid Geometry Now, find I PA I: ~ ! P A 12 = 1 P B 12 + 1 A B 12 - 2 1 P B 1 . 1 AB 1 cos P BA = 12 sin2 cp+a2-21 sin cp.a a sin cp 21 .~ But if "( = POA, then from the triangle POA we have: 1 P A 12 = 12 cos2 cp + 12 -212 cos cp cos "(. Equating the two expressions for 1 P A 12 and taking into consideration (1), find 2sin2 ! 2 cos y = cos cp - ----cos cp ( 2sin2 t) Answer: 3t 2 - arccos cos cp _ 2. cos cp 86. (5 V6+ V22) R. 87. If the plane cuts the edges AD and CD, then the section represents a triangle and the radius of theinscribed a circle will change from 0 to V2(2cos a+ V 4cos2 a+1)· Let now the plane cut the edges AB and BC at points P and N, SA and SC at points Q and R, SD"at point K, and the extensions of AD and CD at points Land M (Fig. 8). Since the lines PQ and NR are parallel and touch the circle inscribed in our section, P N is the diameter of this circle. Setting 1 P N 1 = 2r, we have 1 ML 1 =2a V2-2r, 1 KL 1 = a {2-r V4cos 2 a+1, cos a S _ (a V2-r)2 MKL-2cosa • Answers, Hints, Solutions 91 Thus, aV2-r r= , 2cosa+ V 4cos2a+1 whence a -V2 r = ----~--;::;::==:;::==-1 + 2 cos a + V 4 cos2 a + 1 • Answer: a O<r~ ( , V2 2 cosa+ V4 cos2 a+1) aV2 r = -1-+-2 -co-s-a-+--.;....V-;:-:;:4=c=os~2=a=+==-1 • 88. Let us pass a section by the plane passing through the edge AB and the midpoint of CD, point L; K is the S L Fig. 8 point of intersection of the plane P and AL. The altitude dropped from A onto BL intersects BK at Nand BL at Q (Fig. 9). It is easy to prove that the centre of the sphere lies on the line AQ. Here, the centre of the sphere can lie 92 Problems in Solid Geometry both on the line seglllent AN (point 0) and on the exten-sion of AQ (point OJ. The radius of the first sphere is equal to the radius of the circle touching AB and B K and baving the centre on A R 0 a L \ \ \ r. \ \ \ Of Fig. 9 AN. We denote it by x; x can be found from the relation-ship 1 SBAN="2 (I AB! + 1 BN I) x, 1 BNI =: 1 BK!=; V21 AB 1 2+2 1 BL 12_1 AL 12 vIT --5 a, S 2 S V2 2 BAN="5 BAL = w- a , V2a hence, x = r. The radius of the second sphere 5+Y 11 is found in the same way. V2a Answer: • 5+ V11 Answers, Hints, Solutions 93 89. Let x denote an edge of the tetrahedron, I M N I = x Y2' If the edge, whose midpoint is M, makes an angle ex with the given plane, then the opposite edge makes an an-gle of ~ -ex. The projection of the tetrahedron on this plane represents an isosceles trapezoid with bases x cos ex and x sin ex and the distance between the bases equal to ;2. x2 Thus, S = 2y2 (cos ex + sin ex). Besides, by the hypothe-sis, the angle at the greater base is 60°, whence I cos ex -sinexl=V;· Answer: 3SY2. 90. Let the edge of the cube be equal to 1. Denote by 0 ~ the centre of the face ABCD. From the fact that NMC = ~ 60° and NOC = 90° it follows that 0 lies between M and C. Setting 10M 1= x,1 NB I = y,we have I MN 1= 2x, I NO 1= xY3, I MB I = V ~ + x2 • Applying thetheo-rem of cosines to the triangles MNB and ONB, we get { 1 ~r-2+X2=4x2+y2_2xy r 2, 1 2 3x2 = "2 + y? - y3 y. 1 2 Hence we find: x= yB ' y= y3 . Answer: I AM I: I MCI = 2 - y3, I BN I: I NDI 1= 2, 91. The plane passing through AAI parallel to BID is parallel to the plane DDIBIB, Exactly in the same way, the plane passing through DDI parallel to AIC will be parallel to the plane AAICIC. f 94 Problems in Solid Geometry On the other hand, the planes passing through the edges BCandB1C1 will be parallel to the respective planes AB1C1D and A1BCD1. This taken into account, construct the section of our polyhedra by the plane parallel to the bases and passing through the midpoints of the lateral edges and the plane passing through the midpoints of the Of -----K l~ E l Fig. to parallell sides of the bases of the prism (see Fig. to). In the accompanying figures, Land K are the midpoints of opposite edges EF and HG of the triangular pyramid EFGH, the edges EF and HG are mutually perpendicular. Setting I BC I = x, I AD I = nx, and denoting the altitude of the trapezoid ABCD by y and the altitude of the prism by I, we find un y IKSI=ISOI= n+t' ITLI=2"' I KL I=y (~ + n~t ), I EF 1== 5ni 3x , I GH I =5::1 3 I. Answers, Hints, Solutions 95 The volume of the prism is equal to (n+~) xyz . The volume of the triangular pyramid equals ! I EF 1·1 GH I X (5n+3)3 I KL 1= 24 (n+1)2 xyz. (5n+3)3 Answer; 12 (n+ 1)3 • 92. Let the altitude of the prism be equal to x. On the extension of the edge BIB take a point JC such that I BK 1= ;x, I BIK 1= i x. Since KN is parallel to BM and I KN I = 21 BM I, the projection of KN on CN is twice the length of the projection of BM on CN, that is, a it is equal to -ys" In the triangle CNK, we have I CN I = Va? + ~\ INKI = -ya2 + 4%2,1 CK 1= Y a? +?fx2. Depending on whether the angle CINK is acute or ob-tuse, we shall have two equations a2 + 2; x? = ( a2 + ~? ) + (a? + 4x2) a Answer: ~i or a. 2 r 5 93. Denote two other points of tangency by Al and BI and the radii of the balls by Rand r. In the trapezoid a AAIBBI find the bases: I AAI I = 2 R cos 2' I BBI I = 96 Problems in Solid Geometry a ,/"-2r cos 2" and the lateral sides I ABI I = IAIB I = 2 y Rr, and then determine the diagonals I AB I = I Al BI I = 2 V Rr ( 1 + cos2 ~). If the ball passing through A and Al cuts AB at K, then I AIBI2 = I BK I' IBA I, whence I BK I =::! 2 yRr _ I AB I a ' -V 1+cos2 ~ 1+cost 2" a I AB I cost 2" I AK I =::! ----a-. 1+ CQSI_2 Other parts into which the line segment AB is divided are found in a similar way. A nswer: The line segment AB is divided in the ratio 2 a '2a 2 a cos ""2 : SIll 2:: cos 2:. 94. It is possible to prove that the axis of the cylinder must pass through the midpoint of the edge BD and belong to the plane BDL, where L is the midpoint of AG. Let the axis of the cylinder make an acute angle a with BD. Pro-jecting the pyramid on a plane perpendicular to the axis of the cylinder, we get a quadrilateral AIBIGIDI in which I AlGI I = I A G I = 12. The diagonals AlGI and BIDI are mutually perpendicular, AlGI is bisected by the point F of intersection of the diagonals, and DIBI is divided by F into the line segments 6Y3 cos a and 10y3 sin a -6y3 cosa. From the condition I AIF 1'1 FGI I = IBIF I X I FDI I we get for a the equation sin 2 a -5 sin a cos a + 4 cos' a = 0, whence we find tan al = 1, tan a2 = 4. But I BIDl I = 10y:3 sin a and is equal to the diameter of the base of Answers, Hints, Solutions 97 the cylinder. Two values are obtained for the radiue of the base of the cylinder: 5y6 and 20V3. 2 V17 95. On the edge A S take a point K such that 1 A K 1 = a. Then the points B, D, and K belong to the section of the cone by a plane parallel to the base of the cone (I AB 1 = 1 AD 1 = 1 A K I). From the fact that C lies in the plane of the base it follows that the plane BD K bisects the altitude of the cone. Thus, the surface area of our cone is four times the surface area of the cone the radius of the base of which is equal to the radius of the circle circumscribed about the triangle BD K with generatrix equal to a. 4n V2 a2 (Vb2.+2a2'-a) Answer: • jI b2+2a2·V 3 Vb2+2a2-4a 96. Let the radius of the base of the cone be equal to R, altitude to h, the edge of the cube to a. The section of the cone by the plane parallel to the base and passing through the centre of the cube is a circle of radius R 2h - 2~y2; in which a rectangle (the section of the cube) with sides a and aV2 is inscribed, that is, 3 2= R2 (2h-a V2)2 (1) a h2 • The section of the cone parallel to the base of the cone and passing through the edge of the cube opposite to the edge lying in the base isa circle of radius R h - ;V2. On the other hand, the diameter of this circle is equal to a, that is, a=2R h-~ V2 . (2) From Relationships (1), (2) we get h= V2 (5+ V"3) a, R=2 V3-1 a. 4 2 98 Answer: 3 97. 5-Problems in Solid Geometry n (53-7 y3) Y2 48 ~ ~ 98. From the equality A CB = ADB and perpendicular-ityof AB and DC we can obtain that the points C and D are symmetric with respect to the plane passing through AB perpendicular to CD. as . Answer: 3. 99. Let K be the midpoint of AB, P the foot of the perpendicular dropped from K on CS. On AB take points M and N such that PM N is a regular triangle (Fig. 11). S c Fig. 11 Fig. 12 The pyramid SP M N can be completed to obtain a regu-lar prism PMNSMINI so that PMN and SMINI will be its bases and PS, MN}t NNI its lateral edges. The prism A IBICA zBzS will be homothetic to the prism PM NSMIN1 with centre in S and ratio of similitude I CS 1/1 PS I. It is easily seen that the sought-for part of the volume of the pyramid SABC contained inside the prism A IBI CA 2P 2S is equal to the ratio I M NI / IA B I. Setting AB = ay3, I CS I = 2a, we find: V13 3 5 I SK I = 2 a, 1 CK I =2"a, I PS I =4" a, Answers, Hints, Solutions I PK 1= 3 f3 a, 2 3 I MN 1=1 PI( I ~=-2 a, V3 . Y3 Answer. -2-' 99 ;-I M N J / I AB I = 12 3 . 100. Let the plane passing through BICI intersect A B and DC at points K and L (Fig. 12). By the hypothe-sis, the polyhedra AKLDA1BICIDI and KBCLB1C1 have equal volumes. Apply to them Simpson's formula (Prob-lem 15), setting I A K I = 1 DL 1 = a. Since the alti-tudes of these polyhedra are equal, we get the following equation for a: 16 whence a=T' Denote the altitude of the pyramid by h. Introduce a coordinate system taking its origin at the centre of ABCD and with the x- and y-axes respectively parallel to AB and BC. The points A, C, and DI will then have the coor-. (7 7) (7 7 ) ( 1 1 ) dma tes -"2' -"2' 0 , "2'"2' 0 , -"2'"2' h respectively. It is not difficult to find the equation of the plane ACDI: hx -hy + z = O. The plane KLCIB I will have the equation 10hx -8z + 3h = O. The normal vector to the former plane is n (h, -h, 1), to the latter m (10h, 0, -8). The condition of their perpendicularity yields 10h2 -8 = 0, ;h = 2Y5. The volume of the 5 'd' 38 Y5 pyraml IS • 5 101. Two cases are possible: 1. The lateral sides of the trapezoid are the projections of the edges AB and BICI. It is possible to prove that in 100 Problems in Solid Geometry this case the centre of the sphere is found at the point C. The volume of the pyramid will be equal to 3a3/8. 2. The lateral sides of the trapezoid are represented by the projections of the edges AB and A ICI . In this case the centre of the sphere is projected into the centre of the circle circumscribed about the trapezoid ABCIA i, the altitude of the trapezoid is equal to a "V 5/3 , the volume of the prism is equal to a3Y5!4. 3a3 a3 V5 Answer: -8- or 4 . n 102. 3"" a (a2+ 2b2). 1.03. Project the given polyhedra on the plane ABC (Fig. 13). The projections of the points AI, B I , and CI are not shown in the figure since they have coincided with the p Fig. 13 points A, B, and C; SI and DI are the respective projec-tions of the points Sand D. If on the line segment PSI a point K is taken such that I P K I = I ND II, then the point K is the projection of the point {(I at which the edge PS intersects the plane A IBICI. Thus, the desired Answers, Hints, Solutions ratio is equal to I KB I ! NDI I -I PB I 1 BP I I PB I (I SIN !-I DISII)-(I PSI I-I BSII) IPSII-IBSII I BS I I-I DISI I I S 1M I -I BS I I · 101 (1) Consequently, our problem has reduced to finding the line segments I SIM I, I BSI I, I DISI I, where SI is a point from which the sides of the triangle BDIM are seen at equal angles. BDIM is a right triangle with legs I DIM I = 2a, I BDI I = a V 3. Notation: I SI M I = x, I SI B ~ = y, I SIDl I = z. Rotate the triangle DIS 1M through an angle of 60° about the point DI (Fig. 13, b), DIS IS 2 is a regular triangle ~ withsidez; thepointsB, SI' S2' Ml are collinear, BDIMI= 150°. From the triangle BDIMI find x + y + z = a¥13. The altitude of the triangle BDIMI dropped on the side . V3 2a z BMl IS equal to a -, whence z = "r-' y + 2- = 13 r 13 V 3a2 6a N .. fi d h 5a 3a2-13 = 13' ow It IS easy to n t at y = V 13 ' 6a x = "r-' Substituting the found values into (1), we r 13 get that the required ratio is equal to 3 (measured from the vertex B). 104. Any tangent plane separates space into two parts; here two cases are possible: either all the three spheres are located in one half-plane or two in one half-plane and one in the other. It is obvious that if a certain plane touches the spheres, then the plane symmetric to it with respect to the plane passing through the centres of the spheres is also tangent to these spheres. Let us show that there is no plane touching the given spheres so that the spheres with radii of 3 and 4 are found on one side of it, while the sphere of radius 6 on the other. Let the centres of the spheres with radii of 3, 4, and 6 be at the points A, B, and C. The plane touching the given 102 Problems in Solid Geometry spheres in the above indicated manner divides the sides A C and BC in the ratios 1 : 2 and 2 : 3, respectively, stbat is, it will pass through points K and L on A C and BC such that 1 C K 1 = 22/3, 1 CL 1 = 33/5. The distance from C to KL is easily found, it is equal to 33 -V 3/91 < 6. Hence it follows that through KL it is impossible to pass a plane touching the sphere with radius of 6 and centre at C. We can show that all other tangent planes exist, they will be six all in all. 105. The solution of this problem is based on the fact that the extension of an incident beam is symmetric to the reflected beam with respect to the face from which the beam is reflected. Introduce a coordinate system in a nat-ural way, taking its origin at the point N, and the edges N K, N L, and N M as the x-, y-, and z-axes; denote by Q' and R' the successive points of intersection of the straight line S P with the coordinate planes different from LN M. We have 1 PQ 1 = 1 PQ' I, 1 Q R 1 = I Q' R' I· The point P has the coordinates (0, 1,-V3). Denote by ex, ~, ex the angles made by the ray Sl! with the coor-dinate axeS. It follows from the hypothesIs that ~ = n/4, then cos ex is found from the equality 2 cos2 ex -f- cos2 ~ = 1, cos ex = 1/2 (ex is an acute angle). Consequently, the vector a (1/2, -V2/2, 1/2) is parallel to the line SP. If A (x, y, z) is an arbitrary point on this line, then ~ ..... OA = OP + ta, or in coordinate form, -V2 y=1+-2-t, The coordinates y and z vanish for tl = - V2 (this "ill be point Q') and for t2 = --2 Y3 (point R'). Thus, y 2 --V2 -(~... ...::~,.~ QI ( -'-2-' 0, -V3----"2) , R' (- y3, 1--V6, 0), I PQ' ! = -V2, I Qf R' 1=2 Y3-1l2. Answer: 2 Y3. Answers, Hints, Solutions 103 106. Denote by K the point of tangency of the sphere with the extension of CD, and by M and L the points of tangency with the edges AD and BD, N is the midpoint of BC (Fig. 14). Since I CD I = I DB I = I DA I, DN is per-pendicular to the plane ABC, I DK I = I DM I = IDL I, KL is parallel to DN, ML is parallel to AB, hence, the ~ plane KLM is perpendicular to the plane ABC, KLM = 90°. If 0 is the centre of the sphere, then the line DO is OJ K B C Fig. 14 Fig. 15 perpendicular to the plane KLM, that is, DO is parallel to the plane ABC, consequently, I DN I = 1 (to the radi-us of the sphere). In addition, DO passes through the centre of the circle circumscribed about the triangle KLM, that is, through the midpoint of K M. Hence it follows ~ 1 ~ that ODM = 2" KDM. Further, I DA I = I DC 1= VI CN 12 + I DN 12 = V3, I CA I = I CB I cos 300 = -., -~./"'-...... y 6, i.e. ,6, CDA is right-angled, CDA = 90°, OD M = 45°, I D M I = I OM I = 1. The required segment of the tangent is oqual to I AM I = I AD I -I DM I = va -1. 104: Problems in Solid Geometry 107. Let 011 Os, 0 3 be the points where the balls are tangent to the plane P: 0 1 for the ball of radius r, and O2 and 0 3 for the balls of radius R. 0 is the vertex of the cone (see Fig. 15) and rp the angle between the genera-trix of the cone and the plane P. It is possible to show that I 0 10 I = r cot ~, I 002 I = I 003 I = R cot ~ , I 0 10 2 1 = 1010 3 1=2 VRr, 102031 =2R. Since I 0 10 2 I = I 0 10 3 I f only the angle 0 20 10 3 can be equal to 150°, hence, R/r = 4 sins 750 = 2 + vi Further, if L is the midpoint of 0 20 3, then I OL I = VI OOa 12" I 03L 12=R V cotS t- 1, lOlL I = VI 0 10 3 12_1 03L 12= V 4Rr-Rs. The point 0 is found on the line OIL, and it can lie either on the line segment OIL itself, or on its extension beyond the points Lor 0 1 (0' and 0" in the figure). Respec-tively, we get the following three relationships: lOlL I = I 001 I + I OL I, lOlL I = 1010' 1-10' LI, lOlL I = I 0" L I -I 0"01 I· Making the substitutions R = (2+ V3) r, cot t = x in each of these relationships, we shall come to a contradic-tion in the first two (x = 1 or x = -2 V 3/3), in the third case we find x = 2V 3/3. 1 Answer: cos rp = '[ . 108. Denote by K and L the midpoints of the edges AD and BC, N andParethepointsof intersection of the passed plane and the lines AB and A C, respectively (Fig. 16). Answers, Hints, Solutions f05 Find the ratios I PA 1/1 PC I and I PK 1/1 PM I. Draw KQ and A R parallel to DC, Q is the midpoint of A C. I PA I I AR I I DM I 2 I AR I = I DM I, I PC I = I MC I = I MC I =3' I PK I _ I KQ I _ I DC I 5 I PM I -I MC I -2 I MC I =6' Then find I AN I 2 I PN I 4 I NB I -3"' I PL I =:5' V P AKN I PAl ·1 AK I . I AN I 2 V ABeD -I AC I ·1 AD I ·1 AB ! "5 ' that is, V pAKN = 2. Since the altitude dropped from A on PN K is equal to f, S PNK = 6, IPKI·IPNI 3 I PM I .1 P L I 2" t S PM L = 9. Thus, the area of the section will be SP.ML -S PNK = 3. t09. Knowing the radius of the ball inscribed in the regular triangular pyramid and the altitude of the pyra-D o ------~--~----~c B Fig. f6 mid, it is not difficult to find the side of the base. It is equal to f2, I M K I = I KN I (by the hypothesis, the tangents to the ball from the points M and N are equal in length). f06 Problems in Solid Geometry Let I BM I = X, I BN I = y. Finding I MNI by the theorem of cosines from the triangle BMN, and I MK I and I NK I from the respective triangles BMK and BNK, we get the system of equations { X2+y2-xy=49, { X2+y2-xy=49, x2-f2x= y2-f2y ~ (x-y) (x+ y-f2)= O. This system has a solution: Xl = Yl = 7. In this case the distance from K to MN is equal to 4Y3 _ 7V3 = V3 2 2 < 2, that is l the plane passing through M Nand touching the ball actually intersects the extension of S K beyond the point K, Another solution of this system satisfies the condition X + y = f2. From the first equation we get (x + y)2 -3xy = 49, xy = 95/3. Hence it follows that Consequently, the altitude dropped from K on MN is equal to !...-Y3 > 2, that} is, in this case the plane passing 6 through MN and touching the ball does not satisfythe con-ditions of the problem. f2 Answer: 6 f3 • 110. From the fact that the edges of the pyramid ABCD touch the ball it follows that the sums of opposite edges of the pyramid are equal. Let US complete the pyramid ABCD to get a parallelepiped by drawing through each edge of the pyramid a plane parallel to the opposite edge. Answers, Hints, Solutions f07 The edges of the pyramid will be diagonals of the faces of the parallelepiped (Fig. f 7), and the edges of the paral-lelepiped are equal to the distances between the midpoints of the opposite edges of the pyramid. Let I AD I = a, I BC I = b, then any two opposite edges of the pyramid will be equal to a and b. Let us prove this. Let I AB I = x, I DC I = y. Then x + y = a + b, x2 + y2 = a2 + b2 Fig. f7 (the last equality follows from the fact that all the faces of the parallelepiped are rhombi with equal sides). Hence it follows that x = a, y = b or x = b, y = a. Hence, in the triangle ABC at least two sides are equal . ~ III length. But ABC = fOO°, consequently, I AB I = x = I BC I = b, I AC I = a, I DB I = b, I DC I = a. From the triangle ABC we find a = 2b sin 50°, hA a2 V3 .. /-0 whence h R = 2b2 sin f000 = Y 3 tan 50 . 111. The equality of the products of the lengths of the edges of each face means that the opposite eages of the 108 Problems in Solid Geometry pyramid are equal in length. Complete the pyramid SABC in a usual way to get a parallelepiped by passing through each edge a plane parallel to the opposite edge. Since the opposite edges of the pyramid SABC are equal in length, ,,-------:::::::::. A c Fig. 18 the obtained parallelepiped will be rectangular. Denote the edges of this parallelepiped by a, b, and c, as is shown in Fig. 18. In the triangle BCD draw the altitude DL. From the triangle BCD find I DL 1_ bc - -yb2+ c2 ' 1 AL 1 = ya2+ 1 DL 12= ya2bl~b2c2+c2a2 V b2+ c2 ' 1 SABC = -ya2b2+b2c2+c2a2, 2 The volume of the pyramid SABC is one third the volume of the parallelepiped. the altitude on the face ABC is given; thus we get the equation -. / 102 -y a2b2+b2c2+ c2 a2. V 55= abc. (1) By the theorem of cosinos, for the triangle ABC we get .. /17 6a2=-Ya2+c2.y'"a2+b2·V T· (2) Answers, Hints, Solutions f09 And, finally, the last condition of the problem yields c2 -2a2 -2b2 = 30. (3) Solving System (f)-(3), we find a2 = 34, b2 = 2, c2 = f02. 34 "1'6 Answer: ; • ff2. Denote by M and N the points at which the tangents drawn from A and B touch tho ball, MI and N1 are projections of tho points M and N on the plane ABC (Fig. f9, a; the figure represents one of the two equivalent Lt----~O A~-----.--' 8 (6) Fig. f9 cases of arrangement of the tangents when these tangents are skew lines; in two other cases these tangents lie in one and the same plane). The following is readily found: I A M I = I C N I = l, I M MIl = INN I I = l sin a, I A MI I = I CNI I = l cos a. Find I BMI I and I BNI I (Fig. f9, b; 0 tlie centre of the ball, OL II BMI ) I BNII=I BMII=I OL l=yr2-(lsina-r)2 = y 2rl sin a-l2 sin2 a. When rotated about the point B through an angle = ABC. Hence it follows that the angle between AMI Answers, Hints, Solutions fff ~ and CNl is equal to 2 3 -y5 4 3-Y5 Answer: 4 • ff4. From the fact that the straight:line perpendicular to the edges A C and BS passes through the midpoint of B S it follows that the faces A CB and A CS are equivalent. l Let SASB = SBSC = Q, then SACB = SACS = 2Q. Denote by A l' Bh Cl , Sl the projections of M on the res-pective faces BCS, ACS, ABS, ABC; hA' hB' hc, hs are the altitudes dropped on these faces, V the volume of the pyramid. Then we shall have 3V 1 MAl 1+21 MBll-t-1 MCI 1+21 MS11=Q' But, by the hypothesis, 1 MB 1 + 1 MS 1 = I MA) 1+ 1 MBI I + I MCI 1 + I M Sl I. From these two equalities we have: 3V 1 MB 1 +1 MBll+ 1 MS 1+IMSll =0' But f f Q V=T hs·2Q=T hB,2Q=T (hB+ hs). Answers, Hints, Solutions 113 Consequently, I MB I + I MBI I + I MS I + I MS1 I = hB + hs· On the other hand, I MB I + I MBI I > hB' I MS I + I MS1 I ~ hs. Hence, I MB I + I MBI I = hB' I MS I + I MS1 I = hs, and the altitudes dropped from Band S intersect at the point M, and the edges A C and B S are mutually perpenaicular. FrQm the conditions of the problem it also follows that the common perpendicular to A C and BS also bisects A C. Let F be the midpoint of A C, and E the midpoint of B S. Setting I FE I = x, we get 1 1 .. / 3 Q=SASB=2 I SB I· I AE I =2 V X 2+2 , y6 I 1 r 3 We shall get the equation -2-V xZ+T= V X 2+2 , whence X= ~ • Considering the isosceles triangle BFS in which I BS I = 1, I BF I = IF S I, the altitude IF E I = {-, M the point of intersection of altitudes, we find IBMI = ISMI = ~iO . 115. Since the lateral edges of the quadrangular pyra-mid are equal to one another, its vertex is projected into the point 0 which is the centre of the rectangle ABCD. On the other hand, from the equality of the edges of the triangular pyramid it follows that all the vertices of it~ base lie on a circle centred at o. Let the circle on which the vertices of the base of the triangular pyramid lie intersect the sides of the rectangle AB CD at points desig!}ated in Fig. 20, a. From the fac1 that the lateral faces of the triangular pyramid are equiv-alent isosceles triangles it follows that the angles a1 the vertices of these triangles are either equal or theil sum is equal to 1800 • Hence, the base is an isoscelel triangle. (Prove that it cannot be regular.) Further, tW( vertices of this triangle cannot lie on smaller sides 0 114 Problems in Solid Geometry the rectangle ABCD. If the base will be represented by the triangle LN S 7 then I SL I = I LN 17 {iN = 90°7 and, hence, it will follow that ABeD is a square. But if the triangle LNR will turn out to be the base, then L If A Fig. 20 N (a) (h) from the condition a < 600 it will follow that I BNI > I NR I. Hence, the sides RL and LN will be equal which is possible when the points K and L coincide with the midpoint of AB .. Reasoning in a similar way, we shall come to another possibility: the vertices of the base of the triangular pyra-mid are situated at the points R, N, and p, P being the midpoint of CDt. Consider the first case (Fig. 20, b). Let I LO I = CZ I ON I = I OR I = r. Then I NR I = I CD I = 2r tan 2· /"'-... ~ But, since LEN + NER = 180°7 the triangles LNE and NER, being ... brought together (as in Fig. 20, C)7 form a right triangle LNR. Hence, I LN I = Y 4 I LE 121_ I N R II = V"" 4112 +4rl -4r21 tan2 T' On the other hand, L LN II = ( r + r 1/1-tan2 ~ ) 2 + rl tan2 ~ • Answers, Hints, Solutions 115 Thus, 2h2 .,2= V · a a 2tan2 -+ f-tan2 --f 2 2 Considering the triangle N R P in a similar way, we get: r2 < O. Answer: 8h3 tan !:... 2 3 (2tan2 ~ + V f-tan2 ~ -f) • 116. Extend the edge SA beyond the point S, and on the extension take a point A I such that I SA I I = I SA I. In SAIBC the ... dihedra~ angles at the edges SAl and SC will be equal, and, sIDce I SAl I = I SC I, I AlB I = I CB I = b. The triangle ABA" is a right triangle with legs a and b. Consequently, the hypotenuse I AAI I = 21 AS I = ya2 + b2• Answer; ~ ya2 +b2• lf7. Consider the tetrahedron with edge 2a. The sur~ face of the sphere touching all its edges is broken by the surface of the tetrahedron into four equal segments and four congruent curvilinear triangles each of which is congruent to the sought~for triangle. The radius of the sphere is equal to a Y2, the altitude of each segment is 2 (Y2 f 1(2) equal to a 2-"2 Y 3"' consequently, the area of the sought~for curvilinear triangle is equal to f [ ( y2 ) 2 y2 (Y2 f .. /2)] 4" 4na2 ----r -4.2na2 2 -2- - 2" V "3 na2l .. r-=6(2 y 3-3). 118. Consider the cube with edge equal to 2V2. The sphere with centre at the centre of the cube touching its Problems in Solid Geometry edges has the radius 2. The surface of the sphere is broken by the surface of the cube into six spherical segments and eight curvilinear triangles equal to the smallest of the sought-for triangles. Answer: n(3y2-4) and n(9Y2-4). y5-f 119. arccos 2' 120. Pass a section through the axis of the cone. Con-sider the trapezoid ABCD thus obtained, where A and B are the points of tangency with the surface of one ball, C and D of the other. It is possible to prove that if F is the point of contact of the balls, then F is the centre of the circle inscribed in ABCD. In further problems, when determining the volumes of solids generated by revolving appropriate segments, take advantage of the formula obtained in Problem f8. f 121. 3" SR. 122. Take advantage of the Leibniz formula (see (1), Problem f53) • 3 1 MG 121= I MA 11+ 1MB 1 21+ I MC II - ! (I AB 12 + I BC II + 1 CAli), where G is the centre of gravity of the triangle AB C. If now ABC is the given right triangle, AIBICI the given regular triangle, G their common centre of gravity, then 1 AlA 12+ I AlB 121+ 1 AIC 12 =3 I AIG 12+ i b21 4 =al+T b2 • Writing analogous equalities for BI and CI and adding them together, we obtain that the desired sum of squares is equal to 3a21 + 4b2l• • Here and henceforward (f) means: I.F. Sharygin, Problems tn Plane Geometry (Nauka, Moscow, 1982). Answers, Hints, Solutions ff7 f23. Let the side of the base of the pyramid be equal to a, and the lateral edge to b. Through FE pass a plane parallel to A SG and denote by K and N the points of in~ tersection of this plane with BG and SB. Since E is the midpoint of the slant height of the face SGB, we have 1 A F 1 = 1 G K I = a14, I SN I = b14, IKE 1 = 2 1 EN I. Through L draw a straight line parallel to A Sand denote its point of intersection with SG by P. We shall have 1 SP I = O.fb. The triangles LPG and FN K are similar, their corresponding sides are parallel, besides, LM and FE are also parallel, that is, 1 PM III MG 1 = I NEill E K I = f/2, consequently, I SM I = 0.4b. Now, find f9 f5 f I LF 12_ 400 a2 , I ME 12= 400 a2+ fOO b2• From the condition I LF I = I ME I we get a = b. FNK is a regular triangle with side ! a, I FE 12 = f76 a2 = 7. Consequently, a = b = 4. f6 .. 1"-Answer: 3 r 2. f24. Prove that the plane cutting the lateral surface of the cylinder divides its volume in the same ratio in wbich it divides the axis of the cylinder. naS Answer: 24'. f25. Each face of the prism represents a parallelo~ gram. If we connect the point of contact of this face and the inscribed ball with all the vertices of this parallelo~ gram, then our face will be broken into four triangles, the sum of the areas of two of them adjacent to the sides of the bases being equal to the sum of the areas of the other two. The areas of triangles of the first type for all the lateral faces will amount to 2S. Hence, the lateral area is equal to 4S, and the total surface area of the prism to 6S. f26. If the spheres a and ~ intersected, then the surface area of tlie part of the sphere ~ enclosed inside the sphere a would be equal to one fourth the total Sur-face ~re~ of the sphere a. (This part would represept its Problems in Solid Geometry 11 a spherical segment with altitude ;R' where r is the radius of aT R the radius of ~. Consequently, its surface t area will be 2nR ;R = nrl .) Hence, the sphere a contains inside itself the sphere ~, and the ratio of the radii is equal to 11 So" t27. When solving this problem, the following facts are used: (1) the centre of the ball inscribed in the cone lies on the surface of the second ball (consider the correspond-ing statement from plane geometry); (2) from the fact that the centre of the inscribed ball lies on the surface of the second ball will follow that the surface area of the inscribed ball will be equal to 4Q, and its radius will he V Q/n; (3) the volume of the frustum of a cone in which the ball is inscribed is also expressed in terms of the total surface area of the frustum and the radius of the ball (the same as the volume of the circumscribed polyhedron), that is, V = ~ Sl IQ. 3 Y 'n 128. Prove that if Rand r are the radii of the circles of the bases of the frustum of a cone, then the radius of the inscribed ball will be V Rr. . S Answer. 2". r 129: An; of the sections under consideration rerre-sents an isosceles triangle whose lateral sides are equa to the generatrix of the cone. Consequently, the greatest area is possessed by the section in which the greatest value is attained by the sine of the vertex angle. If the angle at the vertex of the axial section of the cone is acute, then the axial section has the greatest area. If this angle is obtuse, then the greatest area is possessed by a right triangle. 5 Answer: 6" n. 130. Draw SO which is the altitude of the cone to form three pyramids: SABO, SBCO, and SCAO. In each of these pyramids the' dihedr~l angles at th~ l~teral edges SA Answers, Hints, Solutions ff9 and SB, SB and SC, SC and SA are congruent. Denoting these angles by x, y, and z, we get the system { x+y=~, y+z= 'V, z+x=a, Whence we find a-~+ 'V and the desired angle Z= 2 ' wUI be equal to n-a1 ~-'V • f31. The chord BC is parallel to any plane passing through the midpoints of the chords A B and A C. Conse-quently, the chord B C is parallel to the plane passing through the centre of the sphere and the midpoints of the ......... -arcs AB and A C. Hence it follows that the great circle passing through B and C and the great circle passing - -through the midpoints of the arcs AB and A C intersect at two points K and Kl so that the diameter KKI is paral-lel to the chord BC. nR l Answer: -2- ± 2" . f32. It is easy to see that the section of the given solid by a plane perpendicular to the axis of rotation re-presents an annulus whose area is independent of the distance between the axis of rotation and the plane of the triangle. na3 Va Answer: 24 . f33. If the given plane figure represents a convex pol-ygon, then the solid under consideration consists of a prism of volume" 2dS, half-cylinders with total vol-ume nptP, and a set of spherical sectors whose sum is a ball of volume ::Jtd3. Consequently, in this case the volume of the solid will be equal to 2dS + nptP + : ncP. Obvious-ly, this formula also holds for an arbitrary convex figure 120 Problems in Solid Geometry 134. Let 0 be the centre of the ball, CD its diameter~ and M the midpoint of BC. Prove that 1 AB 1 = 1 AC I. Here, it is sufficient to prove that A M is perpendicular to BC. By the hypothesis, SA is perpendIcular to OS, besides, S M is perpendicular to OS (the triangles C SD, CSB, BCD are rIght triangles, 0 and M are the respective midpoints of CD and CB). Consequently, the plane AMS is perpendicular to OS, A M is perpendicular to OS. But A M is perpendicular to CD, hence, A M is pe~endicular to the plane BCD, thus, <4M is perpendicular to BC. Ra3 V4bl -al Answer: 6 (4RI+a2) • 135. In Fig. 21, a: SABC is the given pyramid, SO is its altitude, and G is the vertex of the trihedral angle. E Fig. 21 8 (II) s It follows from the hypothesis, that G lies on SO. Besides, intersecting the plane of the base ABC, the faces of the trihedral anEle form a regular triangle whose sides are parallel to the sides of the triangle ABC and pass through ItS vertices. Consequently, if one of the edges of the trilla-dral angle intersects the plane ABC at point E and the edge CSB at foint F, then F lies on the slant height SD of the latera face CSB, and 1 EDI = t DA I. By the hypothesis, 1 SF 1 = 1 FD I. Through S draw a straight line parallel to EO and denote by K the point of inter~ section of this line with the line EF (Fig. 2f, b). We have 1 SG 1 1 SK liED 1 3 1 SK 1 = 1 ED I· Hence, 1 GO 1 = I EO I = I EO I = 4' Answers, Hints, Solutions 121 Thus, the volume of the pyramid GABC is 4/7 the volume of the pyramid SABC. On the other hand, the constructed trihedral angle di-vides the portion of the pyramid above the pyramid GABC into two equal parts. A nBwer: The volume of the portion of the pyramid outside the trihedral angle is to the volume of the por-tion inside it as 3; if. V · 136. "6' 137. Figure 227 a to d, shows the common parts of these two pyramids for all the four cases. (1) The common part represents a parallelepiped (Fig. 22, a). To determine the volume, it is necessary from the volume of the original pyramid to subtract the vohunes of three pyramids similar to it with the ratio of similitude 2/3 and to add the volumes of three pyramids also similar to the original pyramid with the ratio of similitude 1/3. Thus, the volume is equal to: (2) The common part is an octahedron (Fig. 22, b) whose volume is (3) The common part is represented in Fig. 22, c. To determine its volume it is necessary from the volume of the original pyramid to subtract the volume of the pyra-mid similar to it with the ratio of similitude equal to 1/3 (in the figure this pyramid is at the top), then to subtract the volumes of three pyramids also similar to the origi-nal pyramid with the ratio of similitude equal to 5/9 and to add the volumes of three pyramids with the ratio of similitude equal to 1/9. Thus, the volume of the com-mon part is equal to [ ( 1 ) 3 ( 5 ) 3 ( 1 ) 3] 110 V 1-"3 -3 9" +3 9 = 243 V. i22 Problems in Solid Geometry (h) (d) Fig. 22 (4) The common part is represented in Fig. 22, d. Its volume is 138. Let the edge of the regular tetrahedron ABeD be equal to a, and l( and L be the midpoints of the edges Answers, Hints, Solutions 123 CD and AB (Fig. 23). On the edge CB take a point M and through this point draw a section perpendicular to KL. Setting I CM I = x, determine the~quantity x for which the rectangle obtained in our section will have the angle A o c Fig. 23 between the diagonals equal to a. Since the sides of the obtained rectangle are equal to x and a -x, x can be evaluated from the following equation: -x a --=tan-2 ' a-x X= a a tan '2 a • f+tan2" If we take on the edge BC one more point N such that I BN I = I CM I = x, and through this ]Joint draw a sec-tion perpendicular to KL, then we shall obtain another rectangle with the angle between the diagonals equal to ct. Hence it follows tluit, on being rotated anticlockwise about KL through an angle a, the plane BCD will pass throuEh the points K, p, and N. Thus, on being rota ted, the plane BC]) will cut off th~ tetrahedron ABeD a pyra-124 Problems in Solid Geometry mid KPNC whose volume is equal to I KC I I CP I I CN I x (a-x) I CD I' 1 CA 1 • 1 CB 1 VABcn= 2a2 V tan ~ = 2 v. 2 (1+tan ~) Similar reasoning will do for any face of the tetrahedron. Consequently, tlie volume of the common part will be a f + tan2 '2 equal to V. ( f+tan~ )1 139. Let the cube ABCDAIB1CID, be rotated through an angle a about the diagonal A CI (FIg. 24). On the edges 'Cf Q o Fig. 24 A]B] and A]D] take points K and L such that I AIK 1 = f AILI = x, from K and L drop perpendiculars on the diagonal A C]; since the cube is symmetric with respect to the plane A CC]A], these perpendiculars will pass through one point M on the diagonal ACI , Let x be chosen ./"'-so that KML = a. Then, after rotating about the diago-nal A C~ antic10ckwise (when viewed in the direction from A to C 1J through an angle a, the point K will move into L. On the edges BtA t and BIB take points P and Q at. th~ Answers, Hints, Solutions i25 same distance x from the vertex B 1• After the same rota-tion the point Q will move into P. Consequently, after the rotation the face ABBIAI will pass through the points A, L, and P and will cut off our cube a pyramid AAIPL whose volume is equal to ! ax (a -x). The same reason-ing is true for all the faces. Thus, the volume of the common part is equal to a3 -ax (a -x). It now remains /'-... to find x from the condition KML = a. To this end, ioin M to the midpoint of the line segment LK, point R. We have I 1'2 a ..,r-1'2 MR I = x- cot -, I CIR I = ay 2 - x-, 2 2 2 the similarity of the find x = 2a ..,r-a 1 + y 3 cot 2 Thus, the volume of the common part is equal to 140. Let A be some point on the ray, B the point of in-cidence of the rayon the mirror, K and L the projections of A on the given mirror and rotated mirror, Al and A. the points symmetric to A with respect to these mirrors, respectively. The sought-for angle is equal to the /"... angle AlBA •• If I AB I = a, then I AlB I = I A.B I = a, /'-... I A K I = a sin a. Since KA L = ~, we ha ve I K L I = I A K I sin ~ = a sin a sin~, I AlAI I = 2 I KL I = 2a sin a sin ~. Thus, if cp is the desired angle, then sin f = sin a sin ~. Answer: 2 arcsin (sin a sin ~). Problems in Solid Geometry 141. Fix the triangle ABC, then known in the triangle /'..-ADC are two sides I A C I and I DC I and the angle ADC = a. In the plane of the triangle ADC construct a circle of radius I AC I centred~at C (Fig. 25, a). If a~ 60°, 4t-----... C (a) Fig. 25 / / / Lf~ - ~----==---~8 then there exists only one triangle having the given sides and angle (the second point AJ. will turn out to lie on the other side of the point D); this is a triangle con ... gruent to the triangle ABC. In this case A C and BD are mutually perpendicular. And if a < 60°, then there is another possibility (in Fig. 25, 4, this is the triangle A1DC). In this triangle /"... a ~ ° 3a CAID = 90° + -, A1CD = 90 - -. But in this case 2 2 ./"'-.. the vertex C (Fig. 25, b) is common for the angles BCA1 = ,/"-..,/"-.. 3 90° -~, BCD = a, A1CD = 90° -~, and since 90°-2 2 ~= (90° - 3;) + a, the points Alt B, C, and D lie in the same plane, and the angle between AIC and BD will he equal to a. Answer: if a> 60°, then the angle between AC and BD is equal to 900, if a < 60°, then the angle between A C and BD can be equal to either 90° or a. Answers, £-tints, Solutions 127 f42. Let the base of the prism be the polygon AIA,- _ . • • • A n, 0 the centre of the circle circumscribed aDout it. Let then a certain plane cut the edges of the prism at points BI , B 2 , ••• , Bn , and M be a point in:i.the plane such that the line AIO is perpendicular to the plane of the base of the prism. Then the following equalities hold; n ~ IAkBkl=nIMOI, k=l V=SIMOI, (f) (2) where V is the volume of the part of the prism enclosed between the ~ base and the passed plane. Prove. ,Equality (f). For an even n it is obvious. Let n be odd. Consider the triangle A~Ak+IA I' where A 1 is the vertex most distant from Ak and Ak+l. Let Ck and Ck be the midpoints of AkAk+l and BkBk+l' respectively. Then 1 CkO 1 = cos n = A. Now, it is easy to prove 1 OAl 1 n that , 1 MO 1 = 1 CkCk 1 + IAzBdA f+A f 2" (I AkBk 1 + 1 Ak+lBk+l 1)+ 1 ALB, 1 A f+A Adding these equalities for all k's (for k = n instead of n + f take f), we get Statement (f). To prove Equality (2), consider the polyhedron AkAk+lOBkBk+;M. If now Vk is the volume of this poly-hedron, then, by Simpson's formula, we have (see Prob-lem f5) Vk= bn ( 1 AkBk 1 + 1 Ak+lBk+1 1 an 6 2 +4 1 AkBk 1+ 1 Ak+IBk+1 1 +2 1 MO I. an) 4 2 = anbn (I AkBk 1 + IAk+IBk+1 1 + 1 MO I) = :~ (lAkBkl+IAk+lBk+ll+lMOD, t28 Problems in Solid Geometry ? where tln' bn are the side~ and the slant height of the polygon AlA, . • . An- Add~ these equalities for all K's and taking (1) into consIderation, we get Equali-ty (2). Now, it is not difJieult to conclude that the answer to our problem will be the quantity nJ. 143. Let the pentagon ABCD E be the projection of the regular pentagon, where I AB I = 1, I BC I :::::::: 2, I CD 1= 4, ABCD is a trapezoid in which I AD I = '- = 1 + V5, I BC I 2 F the point of intersection of its diagonals, A FDE is E A~------~--~~D Fig. 26 a parallelogram. Draw CK parallel to AB (Fig. 26). In the triangle CKD we have: I CK I = 1, I KD I = 2( '- --............... i), I CD I -= a. Set CDK = cpo Write the theorem of cosines for the triangles CKD anll ACD: i = 4' + 4 ('- -i)' -4 ('- -1) a cos cp, I AC I' = 4' + U' -4a1eos cp. From these two relationships we ftnd .. / 4)..'-3'--4' I AC I == V '--1 ' '-.. / 4).'-3'--4' IEDI=IAFI= 1+1 JI '--I · -trtswets,' Rints, Solutions f29 ~imilarly, we find " , '" 11 / a2 Jv-f+4}.,2_4}" I AE I = I F D I = }., + f V }., _ f • J • f \ ~ " • Answer: Two other sides are equal to ~, ' V5~1 Y 14~10V5-'2 (YS+f) a2 and I -The problem has a solution fer V5- 2 < a <1/5. 144. Let the edge of the cube be equal to a, I Nel J = z. Find " a x I LM I = 2' INK I = .. r- , . . y 2 . • \ 2 I LN 12=,t LBI 12,+ I BIN 12=-;-+ (a-x)2 , 5 2 2 + 2 =4a-ax x, _I LX 12=-1 LBI 12+ I BJ,K 12 = I LBI I2+ {BIN 12+1 NK 12 . V2 +21 BIN lol,NK 12 a2 x 2 . =T+(a.-x)2+ T +(a-x) x . 5 2 2 +3; . ;:::4 a -ax 2' I MN 12= I MBI 12+ I BIN 12= 3;2 -2ax+x2, I MK 12= 1MB 12+ I BK 12_1 MB I . I BK I 3a2 3 x 2 =2-2 ax+T· 9-0·",9 Prohlems in So1ici Oeomeir1 ,/"--.. ~ ( If LM K = M KN = q>, then by the theorem of cosines, for the triangles LM K and M KN we get: I LK I' = I LM II + I MK II -2 ILM I- IMK I cos CPt I MN I' = IMK II + I KN II -2 IMK 1·1 KNI cos cpo Eliminating cos cp from these equations, we get I LK II - IKNI -I MN 1 1 .1 LM I =(1 LM I -I KN I) (I LM I· IKN I -I MK I'). Expressing the line segments entering this equality with the aid of the found formulas, we get ( Sa' _ ax + x2 ) --=-- ( 3a2 _ 2a:l: +:1:1) ~ 4 2 1'2 2 2 _ (.!!:..._-=-) ( ax _ 3a' + 3a:l: _ :1:1) , -2 V2 2 V2 2 2 2· From this equation we find :I: = a (f _ ~2 ) . I BIN I .. r-Answer: I Nel I = Y 2+1.. t4S. Two cases are possible: (f) the centre of the circum ... scribed sphere coincides with the centre of the base and (2) the centre of the circumscribed sphere is found at the point of the surface of the inscribed sphere diametrically opposite to the centre of the base. In the second case, denoting by Rand r the radii of the respective inscribed and circumscribed spheres, find the altitude of the pyramid 2r + R and the side of the base 1'RI - 4r2. The section passin¥ through the altitude and midpoint of the side of the base IS an isosceles triangle with altitude R + 2r, base 1'3 (R2 -4r2) and radius of the inscribed circle equal to r. Proceeding from this, it is possible to get the relationship 3R2 -6Rr -4r2 = 0 for Rand r. Answer: 3 + 1'21 (in both cases). 3 I.Dswers, .ttsaUt, ZSolutions iSi 146. Two cases are possihle: (1.) the centre of the circumscribed ball coincides with the centre of the base, (2) the centre of the circumscribed sphere is found at the point of the surface of the inscribed ball diametrically opposite to the centre of the base. In the first case, the plane angle at the vertex is equal to n/2. Consider the second case. Denote by a, b, and 1 the side of the base, lateral edge, and the slant height of the lateral face, respectively. Then 2 b2 = 12 + a 4 ' (1.) the radius r of the inscribed ball is equal to the radius of the circle inscribed in the isosceles triangle with base a and la teral si de 1: a Y21-a r= 2 Y21+a' (2) the .l'adius R of the circumscribed ball is ~ual to the radius of the circle circumscribed about the isosceles triangle with base a Y2 and lateral side b: R = --.!!.2 y2 . (3) 2 y2b2-a2 Here, the centre of the circle must lif' inside the triangle, which means that b > a. Since the distance from the centre of the circumscribed ball to the base is 27, we ha ve 2 R2 -~ = 472 • Substituting the values of Rand r ex-pressed by Formulas (2) and (3) into this equality, we get after Simplification; (bZ_a 2)2 2 (2bI -a2) a2 (21-a) 21+a Expressing b in terms of a and 1 by Formula (1.), we get ( 12- 3:2 ) ~= a Z (21-a)z. 9 Problems in Solid Geometry 'Taking into account that b > a or l > a V3;we obtain -2 that 4 and l satisfy the equation whence + = f+ ~3 (for~the second root + < ~3) . n n ~srper~ '"2 or 6 a , 147. Let K be the projection of the vertex -Son the plane ABCD, and let L, M, N, and P be the projection of S on the respective sides AB, BC, CD, and DA. It follows from the hypothesis that LSN and MSP are right triangles with right angles at the vertex S. Conse-quently, ILK 1.·1 KN I = I Ml{ I-I KP I = I KS 1 2, And B.........,...--- l (a) Fig: 27 since I LK I + I KN I -I MK I + I KP I = a, two cases are possible; either I LK I = I KM I, 1 KP I = IIKN I, or I LK 1 = I KP I, I M K I = I KN I, that is, the point K lies either on the diagonal AC or BD. Con .. sider both cases, (f) K lies on the diagonal BD (Fig. 27, a). The figt.ll'e represents the projection of the pyramid on the plane Answers, Hints, Solutions ABCD. The pOint S is found "above" K: Setting ILKI = I KM I = x, we now find: I KS I = V I LK I . I KN I = V x (a-x), I SL 1= VI LK 1 2+ I KS It=Vax, a Vax SABS= 2 • a Va (a-x) Analogously, SADS= 2 • Further, V ABDS= i all V x (a-x)o On the other hand, by the formula of Problem ii, we have V 2 SABsSBDssina ABDS=T I AK I a3 V x (a-x) sin a Equating two expressions for V ABDS, we get xl -ax + all cost a = 0, whence x (a -x) = at cost a, a3 t cos a I V ABCDS = 3 0 The problem has a solution if I cos a I ~ ;. Besides, the angle at the edge AS is obtuse, since the plane ASM is perpendicular to the face ASD, and this plane"passes inside the dihedral angle between the planes A SB and ASD. Consequently, in the first case the problem has I · Of n 2n a so utlOn 1 - < a ~ -. , 2 3 (2) The point K lies on the diagonal AC (Fig. 27, b). Reasoning as in Case (1), we get (as before, I LK I = z): V all Vx(a-x) a3 x sin a -, ABDS= 6 == .6 V ~ (x+x) , 134 Problems in Solid Geometry whence we easily find x = a 1 cos ai, Y _ a3 V 1 cos a 1(1-1 cos a I) -6 • The same as in the first case, a >~. Thus, we get the 2 answer. Answer: if ~ < a ~ 2; , two answers are possible: . y __ a3 eosa y _ a3 V-eosa(f+cosa) 1-6' 2-6 'f 2n y_ as V-cos a (1 +cos a) 1 a>3' -6 • t48. Let us first solve the following problem. In the triangle ABC points Land K are taken on the sides AB and AC so that 1 AL 1 = m 1 A'K 1= n. What is the 1 LB 1 'I KC I ratio in which the median A M is divided by the line KL? Denote by N the point of intersection of KL and AM; Q is the point of intersection of KL and BC, P is the point of intersection of KL and the straight line parallel to B C and passing through A. Let 1 B C I = 2a, 1 QC 1 = b, rAP 1 = c. n > m. Then, from the similarity of the corresponding triangles we shall have: .!:. = n, c b b + 2a nl, whence 1 AN I = c _ 2mn . INMI b+a m+n Let now m, n, and p be the ratios in which the edges AB, AC, and AD are divided by the plane. To determine them, we shall have the following system: , 2mn =:2 m+n ' 2np 1 n+p==T' 2pm ==4 p+m ' whence -444 m=-S' n= 9' p='f. The fact that -1 < m < 0 means that the pOint L lies OIl the extensiOD of AB beyond the point A, that is, our Answers, Hints, Solutions 135 plane intersects the edges AC, AD, BC, and BD. Further, determining the ratios in which the edges BC and BD are divided (we shallget!and~),wefind .. theanswer: 7123 . 7 9 16,901 1.49. Consider the pyramid SABC (Fig. 28) in which ./".. 2n A' d" 1 h I CA I = I AB I, BAC=-, S Isperpen lcuarto t e n plane ABC, and such that the vertex A is projected on 8 Fig .. 28 the plane SBC into the point 0 which is the centre of the eircle inscribed in SBC. Let us inscribe a cone in this ~yramid so that its vertex coincides with A, and the circle of its base is rep-resented by the circle inscribed in SBC. It is obvious that if we take n such pyramids whose bases lie in the plane ABC so that their bases congruent to the triangle ABC form a regular n-gon with centre at A, then the eones inscribed in these pyramids form the desired system of cones. Further, let D (be the midpoint of BC, I OD I =r, It n t AD f = I. Then J SD I ;;;:;: -, I BD I = l tan -• Since r n .I, /' /' I S D I l SBD =:::J 20BD, tan SBD = I BD I n rtan-n f36 Problems in Solid Geometry /'\ r tan 0 BD = n' we may w rite the equation l l tan-n r 2----n l tan-n ---~2~-' n r r tan -1-----n l2 tan2~ n n tan -r n whence T = --;:-::=====-V1+2tan2 : n tan -Answer: 2 arcsin n. ',I , ... ; ~ I I' n f+2 tan!-n ' 150. Let the'"'plane A KN touch the-ball at. the point p, and the straight line AP intersect NK at the point M / / ~!!5:E:;::::::Jr;l8f I I ~-1.1 Fig. 29 ... ! I , . ; r .. \ .. ~ (Fig. 29)'. Then the plane .. C1NA is the biseetor J)lane-,of the dihedral angle fonned by the planes D~ C).A: anil C;MA (the planes D1AN and AN M touch the .ball; and ine planes D.,C.,A and C.MA pass through its centre). In the tnswers, Hints, Solutions f37 same way, the plane CI KA is the bisector plane of the dihedral angle formed by the planes MCIA and CIBIA. Thus, the dihedral angle between the planes A CI K and A CIN is one-half the dihedral angle between the planes ADICI and ABICI equal to 2n/3. Answer:, n/3. 15f. Let K, L, and M be the mldpoints of the edges AB, A C, and AD (Fig. 30). From the conditions of the A D o Fig. 30-probl(p it then follows that'the tetrahedron AIBI..GIDI is bounded bY. the planes DKAf' BLAI, CMAJ, and the pJanQ passing through A paralle to BCD. And the vertices BI , CI , and DI are arranged so that the points M, K, a,nd L are th.e midpoints of CBlt DCI, and IJDI (the pOints BI, CI , an~ DI are not shown in the figure)" Let now Q be the midpaint of BC, P th.e pOint of intersection of BLand KQ. To find the v~lume of the common part of two pyramids ABCD and AIBICIDI , we must from the volume V of the~])yramid ABCD subtract the volumes of three pyramids equivalent to DKBQ (each. f -of them has the volume e~ual to -4 V) •• and ad~ the vol-umes of fihree pyramids equivalent to-AtBQP. The volUme f38 Problems in Solid Geometry of the last pyramid is equal to 2f4 V. Thus, the volume of the common part is equal to ~ V. 152. Let uS first prove that the dihedral angles at the edges DB and A C are equal to n/2 (each). Let I A D I = I CDI = I BC I = a, I BD I = lAC I = tJ, I AB I = c, A o (a) (11) c Fig. 3f b> a. From D and C drop perpendiculars D K and CL on the edge AB (Fig. 3f, a). Let us introduce the following notation: I A K I =:: I B L I = x, I K L I = I c -2x I, I D KI :=t I CL 1= h. Since the dihedral angle at the edge AB is equal to n/3, we have I DC I" = f D K 12 + I CL J 2 -I D K I X 1 CL I + I KL Fa, that Is, as = h2 + (c-2X)I. Replac-ing h'Z by at -Xl, we get 3x" -4cx + c-'Z = 0, wlience x = c13, x = c. From the condition b > a it follows tkt x < c/~, hence x = ciS. Thus, the quantities a, b, and c are related as follows: c2 = 3 (b2 -a2). Find the areas of the triangles ABD and A CD: 1 ... / c· f /' 4a' - b' SABD==SABC=Tc Y aJ-g=T c V 3 ~ SACD=:=SBDC= ! b Y4aJ -bl • Answers, Hints, Solutions 139 Expressing the volume of the tetrahedron ABCD by the formula of Problem 11 in terms of the dihedral angle at the edge AB and the areas of the faces ABD and ABC. and then in terms of q> the dihedral angle at the edge AC (it is also equal to the angle at the edge BD) and the areas of the faces ABC and ACD, we get v _ 1 SABDSABC V3 1 SACDSABC . ABCD -"'3 I AB I · -2-=3 I AC I SIn q>, whence sin q>= SABD I AC I. V3 SACD I AB I 2 .. i4a2-b2 2c V 3 b Va --b V4a2-b2 0-;-. -2-= 1. n Hence, q> = 2 . To determine the sum of the remaining three dihedral angles, consider the prism BCDMNA (Fig. 31, b). The tetrahedron ABCN is congruent to the tetrahedron AB CD, since the plane AB C is perpendicular to the plane of ADCN, but ADCN is a rhombus, consequently, the tetrahedra ABCD and ABCN are symmetric with respect to the plane B CA. Just in the same way the tetrahedron ABMN is symmetric to the tetrahedron ABCN with re-spect to the plane ABN (the angle at the edge BN in the tetrahedron AB CN is congruent to the an fIe at the edge BD of the tetrahedron ABCD, that is, equa to n/2), conse-quently, the tetrahedron AB M N is congruent to the tetrahedron ABCN and is congruent to the original tetrahedron ABCD. The dihedral angles of the prism at the ed~s CN and B M are respectively congruent to the dihedral angles at the edges DC and BC of the tetrahedron ABCD. And since the sum of the dihedral angles at the lateral ed~s of the triangular prism is equal to n, the sum of the dihed-ral angles at the edges AD, DC, and CB of the tetrahed-ron ABCD is also equal to n, and the sum of all the t40 Problems in Solid Geometry dihedral angles of the tetrahedron excluding the given angle at the edge AB is equal to 2n. f53. Let in the triangle ABC t~e sides BC, CA, and AB be respectively e~l to a, b, and c. Since the pyramids ABCCI, ABBICI , and AA BICI are congruent, it follows that each of tliem has two laces congruent to the triangle ABC. Indeed, if each pyramid had only one such face, then between the vertices of the pyramids ABCCI and AIBICIA there would be the correspondence A ~ AI' 1_--b--:1I (a) 0 Fig. 32 CA c 8 (e) c B-+ BIt C-+ CIt C1 -+ A, that is, I CCI I = lAC 11, I BCI I" = I BIA I, and this would mean that none of the faces in the pyramid ABCIB 1 is equal to the triangle ABC . Now, it is easy to conclude that the lateral edge of the prism is equal to a, or b, or c (if, for instance, the triangle A CIB is congruent to the triangle ABC, then the face AlBIA in the pyramid AIB1CIA corresponds to the face ACIB of the pyramid ABCCI and the tri-angle AlBIA is congruent to the triangle ABC). " Consider all possible cases .... , (1) I AAJ I = I BBI I = I CCI I = a (Fig. 32., II). Then from the vertex C of the pyramid ABC~ two edges of length a and one edge of length b emanate, and an edge of length c lies opposite the edge CCI' Hence it follows that to the vertex C of the pyramid ABCCI there must correspond the vertex CI of the pyramid AIBIC]A and t A C 1 I = a. Now it is possible to conclude that I ABI I = I BCI I = b. . In all the three pyramids, the dihedral angles at the -edges of length b are congruent, the sum of two such Answers, Uints, Solutions t41 angles being equal to n (for instance, two angles at the edge CIB in the pyramids ABCCI and ABBICI), that is, each of them is equal to n/2. Draw perpendiculars BL and CIK to the edge AC (Fig. 32, b). Since the dihedral angIe at the edge AC is equal to 900 , we have b' = I CIB I' = I CIK 12 + I KL I' + I LB 12 = I CIC 12 -I KC 12 + (I KC I -I LC 1)2 + IBCI 2 -I LC 12 == 2a2 -bx, where x == I LC I, and is found from the equation a2+b2_c2 aZ-x'==c2-{b-x)2, x= 2b • Thus, 3a2 -3b2 + c2 = O. But, by the hypothesis, AB C is a right triangle. This is possible only under the condition c2 = a2 + b2. Consequently, b == aV2, c == ava. Now, it is possible to find the dihedral angle at the ./'---. edge BC of our prlsm. ACCI == 'It/4 is the linear angle of this dihedral angle (ABC and CICB are right triangles with right angles at the vertex C). The dihedral angle at the edge AB of the pyramld ABCCI is equal to 'It/3. Let us show this. Let this angle b(l! equal to q>. Then the dihedral angle at the edge AB of the prism ABCA]BICI is equal to 2q>, and at the edge AIBI to q>. Thus, 'It 3q> == 'It, q> == 3" . (2) I AAI I == I BBI I == I CCI I == b (Flg. 32, c). In this case, in the pyramid ABCCI two edges of length b and one edge of length a emana te from the vertex C. Hence, the pyramid AIBICIA has also such a vertex. It can be either the vertex A or CI • In both cases we get I ABII = a, I A CI I == b (we remind here that two faces with sides a, b, and c must be found). Thus, each of the pyramid~ ABCC and AlB CIA has one face representing a regular triangle with side b, while the pyramid ABB IC1. has not such a face whatever the length of the edge B CI is. Thus, this case is impossible. . t42 Pro,blelUs in Solid Geometry (3) I AAI I = I BBI I == I CCI I == c. This case actu-ally coincides with tlie first, oIily the bases ABC and AIBICI are interchanged. n n ( 3n) n ( 2n) Answer; 2' T or 4 ' 3" or 3 · 154. Drop perpendiculars AIM and BIM on CD, BIN and CIN on AD, CIK and DIK on AB, DIL and AIL on CB. Since I AIM I I BIN I I CIK.I I DIL I 1 I BIM 1=1 NCI 1=1 KDII-I AIL I "'3 (these ratios are equal to the cosine of the dihedral angle at the edges of the tetrahedron) and IAIBII = IBICII = Fig. 33 I U1A DI 1 = I DIAl I, the following equalities must be f lIed: I ~IM I == I BIN I = I CIK I = I DI L I = x, I BIM I = I N C..1 I = I KDI I = I AIL I = 3x (Fig. 33 represents the development of the tetrahedron). Each of the edges CD, DA, AB, and BC will turn out to be di-Answers, iiints, Solutions i43 vided into line segments m and n as is shown in the figure. Bearing in mind that m + n = a, we find z = a ~3 , 5a 7a m = i2 ' n = i2 ,and then find the volume of the tetra-hedron A1B1C1D1• as V2 Answer: i62 • i55. Without loss of generality, we will regard that all the elements of the cone tangent to the balls are in Fig. 34 contact simultaneously with two balls: inner and outer. Let us pass a section through the vertex S of the cone and the centres of the two balls touching one element (Fig. 34, the notation is clear from the figure). From the condition that n balls of radius R touch one another there follows the equality I OA I = R ,analogously, . n SID-n 2R R I OB I = . Consequently I AB I = a = --. n . n sln -Sffi-n n R 2R Let I AC I = z. Then tan a = -, cot a = . Mul-z a-z tiplying these equalities, we get the equation for z: zI -az + 2R2 = 0, t44 Proble-ms ill Solid GeQm.etry , a~ V' a2-8Rio XI = -.-...;....--:::----. 2 whence '" ' ~here a= __ R_ -. n-. '.' sIn-n The condition a2-SR2 ~ 0 yield~ the Jnequafity sin ~ ~ ~r-' Besides, there must be fulfilled the in-n 2 r".2 ". ~quaiity tan"d= ~ <::1.' Now, it is not difflculi <!to Qb,... z ~i.'l.1o. ~ ~ l . h h fi'f f . n f L" taln t at t e root XI ts 1 -3 < SIn -~ .. r-" .. ~r n 2 y 2 • tb 't' t " t" . n __ f . e root X2 1 remains one les fIC Ion: SIn -~ .. r-. n 2 y 2' I · "bl ha 1 . n f I t IS POSSI e ~o prove t t 3 < sin Ii" ~ 2 V 2" on y for n == 9. , The volume of the c~ne will be equal to {- n (a + X)3 tan 2a. Expressing at X t and tan 2a in terms of Rand n by the appropriate formulas, we get the answer. ' Answer: :-nij' (3+' :If'':'''8SiDI~)'3 (,f+" /f-ssin1~) I V _ V " ~) n Y . n 1 ............ \ I t I f2 8in~.!!. ./ f':'- 6 sinl "'::'+ .. /'f- S sin2"::'), . . n\ !' V, n Besides t for n =~ one more value is possible: -nR3 (3--Y {'-:'Ssin2 T) 3 ('f- Y.f-S.stn1 f) ,f2 sin1..:!.. (i - 6 sin1 ..::. _ .. / f - S sin2 ..:!.. ) , 9 ,9 V 9 • , 156. Projecting the cube on the plane perpendicular to BID, we get a regular hexagon ABCCIDIA1 (Fig. 35) wU,h Answers, Hints, Solutions side V~ a= b, where a is the edge of the cube (the regu-lar triangle BCLAI will be projected into a con~~ent triangle, since the plane of BCIAI is perpendic :1' \0 l\ Fig·i35 lH B",D). Consider the triangle KACI , where I KA t == IACII =2b, the line NM passes through the midpoint of I AM I ACI· Let I AAI I = x. We then draw CIL parallel to MN. We have: I,ML I = lAM I, I KN I I KM I I KCI 1- IKL I whence I· BN I 2 (t KN I -I BC I) I BCI I I KCI I IKN I 2+x 1 =2 1 lCI 1-1= l1+x -1= 1+x • 10-0"9 f46 Thus, I BCII IBN I Problems in Solid Geometry IAMI I AAII =f+x-x=1. 157. If two noncongruent and similar triangles have two equal sides, then it is easy to make sure that the sides of each of them form a geometric progression, and the sides of one of them may be designated by a, 1.a, 1.2Ia and those of the other by A,a, A,8a, A,3a. Further, if the sides of a triangle form a geometric progression and two of them are equal to 3 and 5, then the third side will be equal to y-f5 (in other cases the sum of two sides will.be less than the third one). Now, it is easy to prove that in our tetrahedron two faces are triangles with sides 3, V 15, 5 and two other faces have sides -V 15, 5, 5 v~ or 3 vi, 3, V 15; accord-55 V6 ff -ingly the probl~m has two answers: 18 and 10 -V fO. 158. Introduce a rectangular coordinate system so that the first line coincides with the z-axis, the second line is parallel to the It-Uis and passes through the point (0, 0, a), and the third line is parallel to tho I-axis and passes through the point (~t a, 0). Let ABCDA1 B1CID1 be a parallelepiped in which toe points A and C ie on the first line and have the coordinates (Xl' 0, 0), (X2I' 0, 0), res~ctively, the points Band CIon tile second line, their coordinates are (0, Ih .. , a) and (0, Ya' a), and the points D and Bl on the third line, their respective coordinates are (a, a, .11) and (a, a, .IS>. From the condition of the equality ~ ~ ~ of the vectors AD = BC = BICI, we get a -XI = X2I = -a, a = -YI = U2I -a, .II = -a = a -.121, whence Xl = 2a, X2I = -- a, YJ = -a, Y2I = 2a, .II = -a, '- = 2a. Thus, we have A (2a, 0, 0), B (0, -a, a), C (-a, 0, 0), D (a, a, -a), BI (a, a, 2a), CI (0, 2tt, a). It is possible ~ ~ to check that AB = DC. Further, I A C I = 3a, I AB I = ttl's, I BC I = aVa, that is, ABC is a right triangle, hence, the area of ABCD will be lAB I· fBCI = 3azY2. Answers, Hints, Solutions i47 The equation of the plane ABCD is y + z = 0, hence, the distance from Bl to this plane will be equal to :; . Answer: 9as. i59. Consider the regular pyramid ABCDS in which the section KLMNP is drawn representing a regular pen-S A Fig. 36 tagon with side a (Fig. 36). Let the diagonal of the base of the pyramid be equal to b, and its lateral edge to l. Let uS also set I SM I = xl, I SN I = yl. Since the pentagon KLMN P is regular, we have n i+V5 I LM I =2a cosT= 2 a=fla, IMFI I FG I 2n i-cos -... r-5 r 5-i n 2n = 2 =A. cos 5 +cosT b-a We have: I KP I = a, I GO I = 2 . On the other hand, SM b I MC I I OE I = I OC I sc =2 x, I ME I = I SO I I sc I 10 148 Problems in Solid Geometry h (i-x), I FO I =h (i-g), where h is the altitude of the pyramid, consequen tty 1 I GO 1 IOE 1 (1-y) xb 1 FO 1- I ME 1- I FO I ' I GO I = 2 (y-x) • Equating the found expressions for I GO I, we get the equation (1-g)xb =b-a. (1) g-x Further IOE I I MFI \ GO \ = \FG \ =A, whence ; g-x 1 =A. (2) -g Since I LN I = ~a, I LN I = 1/1 DB I, we have gb = p.a. (3) And, finally, consider the triangle PNB in which I PN I = b-a .. ;-/'-... a, I NB 1= (i-g) I, I PB 1= 2 y 2, cos PBN = /"'-.. b cos ABS =-~ 2 Y21-By the theorem of cosines, we get at =(1_g)2 I' + (b-;a)t _ (i-g) ~-a) b • (4) Taking into consideration that ,It = V52 + 1 ,A = YS-1 V5-1 '""---::2=--- , from Equations (1)-(3) we find g = 2 ' b = V5i3 a, then from Equation (4) we get II at (7 +3 V5) bt 4 -"'2-Answers, Hints, Solutions 149 Thus, the volume of the pyramid is equal to .!...~ -. /ll-~-~- (9+41'5) I 3 2 V 4 -12 -12 a • 160. We introduce the usual notation: a, b, c denote the sides of the given triangle, ha' hb' hc its altitudes, p one half of its perimeter, r the radius of the inscribed circle. Let M denote the point of intersection of the planes AlBIC, AIBCI, and ABICI, 0a' Ob' Oc the cen-tres of the externally inscribed circles (Oa is tlie centre of the circle touching the side BC and the extensions of AB and AC, and so on). Prove that Oa01J.OcM i~ the desired pyramid, the altitude dropped from the point M passing through the centre of the inscribed circle (0), and I MO I = 2r. Consider, for instance, the plane AIB)C. Let K be the point of intersection of this plane with the line AB, I KA I_I AAII I KB I I BBtI ha b I AC I hb =a= IBC I' that is, K is the point of intersection of the line AB and the bisector of the exterior angle C. Hence it follows that tbe base of our pyramid is indeed the triangle 0aObOc and that the point M is projected into the point O. Find I MO I: I MO I IrOOa 1_ ra- r ha I AOa 1-ra ' where ra is the radius of the externally inscribed circle S S 2S centred at Oa: ra = , r = -, ha = -, conse-p -a p a quently, 1 1 --I MOI=ha 2_S p-a p ~ 2S =2r. a 1 p p-a Find the area of the triangle 0aObOc. Note that OaA , 'JI)B, 0eC are the altitudes of this triangle. The angles of 150 Problems in Solid Geometry the triangle OaObOe are found readily, for instance, Other angles are found in a similar way. The circle with diameter ObOe passes through Band C, Gonsequently, a ... , • A sin 2 c exactly in the same way I ObOa 1=--... -, hence I~ /".. c \ OuA \ =-\ OaOb \ sin OaObA = ... C sin 2 • C slnT B cos 2. Thus, the area of the triangle OaObOe (let us denote it by Q) will be ... ac sin B 4 • 1 ... ... ... • A . B • C sln2smTsmT (1) Answers~ Hints, Solutions A F o d ° A In sin 2 : . A _ ... / f-cosA _ -. / ~ (f- b2+c2_a2 ) Sln 2 - V 2 - V 2 2bc _ Vr (p-b) (p-c) -be· A A i5t Then sin ~ and sin ~ are found in the same way. Substitutin~ them in (f), we get Q-S abc -2 (p-a) (p-b) (p- c) , and the volume of the pyramid MOaObOc will be V = Sabcr ~ abc=i.. SR 3 (p- a) (p-b) (p- c) 3 3· Section 2 161. No, not in any. 162. The indicated property is possessed by a pyramid in which two opposite dihedral angles are obtuse. 163. Prove that if the straight line is not perpendicu-lar to the plane and forms equal angles wi th two inter-secting lines in this plane, then the projection of this line on the plane also makes equal angles with the same lines, that is, it is parallel to the bisector of either of the two angles made by them. 1M. A triangle, a quadrilateral, and a hexagon. A cube cannot be cut in a regular pentagon, since in a sec-tion having more than three sides there is at least one pair of parallel sides, but a regular pentagon has no parallel sideso 165. On the edges of the trihedral angle layoff equal line seEIIlents SA, SB, and SC from the vertex S. Denote by ° the projection of S on the plane ABC. ASB and AOB are isosceles triangles with a common base AB, the lat-152 Problems in Solid Geometry eral sides of the triangle AOB being shorter than those of /""'-/"-. the triangle ASB. Consequently, AOB > ASB. Similar inequalities hold for other angles. Thus, /""'-/, /""'-/'-.. /'''-, /, ASB + BSC + CSA < AOB + BOC + COA~ 2n. (The last sum is equal to 2n if 0 is inside the triangle A BC and is less than 2n if 0 lies outside of this trian-gle.) To prove the second statement, take an arbitrary point inside the given angle and from this point drop perpen-diculars on the faces of the given angle. These perpendicu-lars will represent the edges of another trihedral angle. (The obtained angle is called complementary to the given trihedral angle. This tochnique is a standard method in the geometry of trihedral angles.) The dihedral angles of the given trihedral angle are complemented to n by the plane angles of the complementary trihedral angle, and vice versa. If a, B, y are the dihedral angles of the given trihedral angle, ihen, using the above-proved inequality for plane angles, we shall have (n -a) + (n -~) + (n --: ,\,) < 2n, whence it follows that a + ~ + y > n. 166. (1) Let S be the vertex of the angle, M a point on an ed~, Ml and M2 the projections of M on two other edges, N the projection of M on the opposite face. Suppose that the edge SM corresponds to the dihedral angle C. If I SM I = a, then, finding successively I SM. I and then from the triangle MM1N, I MN I, or in a different way, first I SM2 1, and then from the triangle MM 2N, I M N I, we arrive at the equality I M N I = a sin a sin B = a sin ~ sin A, that is, sin a sin ~ sin A sin B • (2) Denote by a, b, and c the unit vectors directed along the edges of the trihedral angle (a lies opposite the plane angle of size a, b o~posite ~, c opposite y). The vector b can be represented m the form: b = a cos y + 1], Answers, Hints, Solutions i53 where I '1 I = sin y~ '1 is a vector perpendicular to a; analogously~ c = a cos ~ + 6 where I ~ I = sin~, ~ is perpendicular to a. The angle betwpen the vectors 11 and S is equal to A. Multiplying b and c as scalars~ we get be = cos a = (a cos y + '1) (a cos ~ + s) = cos ~ cos y + sin ~ sin y cos A, which was just required to be proved. (3) From a point inside the angle drop perpendiculars on the faces of the ~iven angle. We get, as is known (see Problem i65), a trihedral angle complementary to the Riven. The plane angles of the given trihedral angle make the dihedral angle of the complementary angle be e~al to n. Applying the first theorem of cosines to the comple-mentary trihedral angle, we get our statement. i67. Take advantage of the first theorem of cosines (see Problem i66). i68. Take advantage of the second theorem of co-sines (see Problem i66). 169. The Sum of all the plane angles of the tetrahedron is equal to 4n. Hence, there is a vertex the sum of plane angles at which does not exceed 1t. All the plane angles at this angle are acute. Otherwise~ one angle would be greater than the sum of two others. 170. This property is possessed by the edge ha ving the greatest length. 171. Let ABC be a perpendicular section, I BC I = a, I CA 1 = b, I AB 1 = c. Through A pass the section ABICI (B and BI~ C and Cllie on the corresponding edges). Lettlien I BBII = I x I, I CCII = I y I. (UBI and Cllie on one side from the plane ABC, then x and y have the same sign, and if on different sides, then x and y have opposite signs.) For the triangle ABI C) to be re.eW.ar, it is necessary and sufficient that the following equalities be fulfilled: ~2 + x2 = b2 + y2, ~2 + y2 = a2 + (x _ y)2. ~et us show that this system has always a solution. Let I ~ band c ~ b. It is easy to show that the set of points n the (x, y)-plane satisfying the first equation and 154 Problems in Solid Geometry situated in Quadrant I is a line which approaches without bound the straight line y = x with increasing x and for x = 0, y = Vell -bl . (As is known, the equation yll -Xli = k describes an equilateral hyperbola.) Simi-larly, the line described by the second equation ap-proaches the straight line y = x/2 with increasing x and for x tending to zero y increases without bound. (The set of points satisfying the second equation is also a hyperbola.) Hence it follows that these two lines intersect, that is, the system of equations always has a solution. t 72. Denote the remaining two vertices of the tetra-hedron by C and D. By the hypothesis, I A C I + I AD I = I AB I. Consider the square KLMN with side equal to I AB I. On its sides LM and MN take points P and Q such that I PM I = I AD I, I QM I = I AC I. Then I LP I = lAC I, I NQ I = I AD I, I PQ I = I DC I and, con-sequently, ~ KLP = 6. ABC, 6. KNQ = 6. BAD, 6. BDC = 6. KPQ. These equalities imply the statement of the problem. 173. No, not any. For instance, if one of the plane anEles of the trihedral angle is sufficiently small and two other are right angles, tJien it is easy to verify that no section of this trihedral angle is a reJlUlar triangle. 174. Show that if at least one plane angle of the given trihedral angle is not equal to 900, then it can be cut by a plane so that the section thus obtained is an obtuse triangle. And if all the plane angles of the trihed-ral angle are right angles, then any of its sections is an acute triangle. For this purpose, it suffices to express the sides of an arbitrary section by the Pythagorean theorem in terms of the line segments of the edges and to check that the sum of the squares of any two sides of the section is greater than the square of the third side. t 75. Let a be the length of the greatest edge, band e the lengths of the edges adjacent to one of the end points of the edge a, and e and f to the other. We have: (b + e -a) + (e + f -a) = b + c + e + f -2a > O. Hence it follows that at least one of the following two inequalities is fulfilled: b + e -a > 0 or e + f -a > O. Hence, the triple of the line seg-ments a, b, c or a, e, f can form a trIangle. 176. In any tetrahedron, there is a vertex for which the sum of certain two plane angles is less than 1800 • (Actu-Answers, Hints, Solutions f55 ally, a stronger statement holds: there is a vertex at which the sum of all plane angles does not exceed f8oo.) Let the vertex A possess this property. On the edge ema-• f k' h /"'-.. natmg rom A ta e pOints K, L, M suc that ALM = /........ / "'" /"'-.. KAL = a, ALK = LAM = ~. It can be done if a + ~ < f80°. Thus, 6. KAL =6 LAM, t:::.. KLM = ~ KAM. In the pyramid AKLM, the dihedral angle at the ed~ A K equals the angle at the edge LM, the dihedral angle at the edge AM equals the angle at the edge KL. It is easy to make sure that the tetrahedron KLMA will be brought into coincidence with itself if the edge KA is brought into coincidence withLM, and the edge A Mwith KL. 177. Sup~se that none of the plane angles of the given trihedral angle is equal to 90°. Let S be the vertex of the ·s D Fig. 37 given angle. Let uS translate the other trihedral angle so that its vertex is brought into coincidence with a point A lying on a certain edge of the given angle (Fig. 37). AB9 A C, and AD are parallel to the edges of the other dihedral angle. The points Band C are found on the edges of the given angle or on its extensions. But AB is perpendicular to SC, AC is perpendicular to SB, consequently, the pro-jections of BS and CS on the plane ABC will be respec-tively perpendicular to AC and AB, that is, S is projected into the ~int of intersection of the altitudes of the trian-gle ABC, hence, AS is perpendicular to BC. Thus, the f56 Problems in Solid Geometry edge AD is parallel to BC, and this means that all the edges of the other trihedral angle belo~ to the same plane. And if one of the plane ~glel of thO given trihed-ral angle is a right one, then all the edges of the other trihedral angle must lie in one face of the given angle (in one that corresponds to the right ]llane angle). If exact-ly two plane angles of the given trihedral angle are right angles, then two ed~s of the other trihedral angle must coincide with one edge of the given angle. Thus, the other trihedral angle can be nondegenerate only if all the plane angles of tile given trihedral angle are right ones. t 78. The straight line l can be regarded as the diago-nal of the rectangular parallelepiped; it makes angles cx, ~, Fig. 38 and y wi th edges. Then, arranging three congruent ~aralle lepipeds in the way shown in Fig. 38, we obtaln that the angles between the three diagonals of these parallele-pipeds emanating from a common vertex are equal to 2a., 26, 2y. Consequently, 2a. + 2~ + 2,\, < 2n. t7g. Let S be tile vertex of the angle, A, B, and C certain points on its edges. Let uS prove that the angle between any edge and the plane of the opposite face is always less than either of tile two plane angles including this edge. Since an angle between a straight line and a plane cannot be obtuse, it suffices to consider the case when the plane angles adfacent to the edge are acute. Let Al be the proJection of A on the face SBC, A2 the projection of A on the edge SB, since I SA 2 I> I SAl I, /' /, /' ASA 1 < ASA t = ASB (remember that all the plane angles at the vertex S are acute). From here readily fol-lows the first part of our problem. Answers, Hints, Solutions 157 /'\ /,\ Let us prove the second part. We have: ASB-BSA1<;. /' /' /, /, ASA 1, ASC -CSA 1 ~ ASA 1 , (at least one inequality is strict). Adding together these inequalities, we get /. /, /' /' ASB + ASC -CSB < 2ASA!" Writing similar inequalities for each edge and adding them, we obtain Our statement. Taking a trihedral angle all the plane angles of which are obtuse and their Sum is close to 2n, we make sure that in this case the state-ment of the second part will not be true. 180. Let a and all ~ and ~1' y and '\'1 be dihedral angles of the tetrahedron (the angles corresponding to opposite edges are denoted by one and the same letter). Consider four vectors a, b, c, and d perpendicular to the faces of the tetrahedron, directed outwards with respect to the tetrahedron, and having lengths numerically equal to the areas of the corresponding laces. The sum of these vectors is equal to zero. (We can give the following interpretation of this statement. Consider the vessel having the shape of our tetrahedron and filled with gas. The force of pressure on each face represents a vector per-pendicular to this face and with the length proportional to its area. It is obvious that the sum of these vectors is equal to zero.) The angle between any two vectors com-plements to n the corresponding dihedral angle of the tetrahedron. Applying these vectors to one another in a different order, we will obtain various three-dimen-sional quadrilaterals. The angles of each quadrilateral are equal to the corresponding dihedral angles of the tetrahedron (two opposite angles are excluded). But the sum of angles of a space quadrilateral is less than 231. Indeed, draw a diagonal of this quadrilateral to separate it into two triangles. The Sum of angles of these triangles is equal to 2n, whereas the Sum of angles of the quadri-lateral is less than the Sum of angles of these triangles, since in any trihedral angle a _plane angle is leas than the sum of two others. Thus, we have proved that the fol-lowing three inequalities are fulfilled: a + ~ + ~ + ~1 < 2n, ~ + ~1 + '\' + '\'1 < 2n, '\' + '\'1 + a + ~ < 2n. (Thus, we liave proved the first part of the problem.) Adding these inequalities, we get a + a1 + Ii + ~1 + i5s Problems in Solid Geometry 'V + '\'1 < 3n. To complete our proof, let us note that the sum of dihedral angles in any trihedral angle is greater than n (see Problem 165). Adding up the inequalities corresponding to each ver-tex of the tetrahedron, we comJllete the proof. Remark. In solving this problem, we liave used the method consisting in tha1 instead of the ~ven trihedral angle, we have considered another trihedral angle whose edges are ~rpendicular to the edges of the given angle. The pair of trihedral angles thus obtained pos~esses the following property: the plane angles of one of them com-plement the dihedral angles of the other to n. Such angles are said to be complementary or polar. This method is widely used in spherIcal geometry. I t was also used for solving Problem 165. lSl. The statement of the problem follows from the fact that for a ~gular polygon the sum of the distances from an arbitrary point inside it to its sides is a constant. lS2. If S1' S,,' S,' and Sf, denote the areas of the cor-responding faces of the tetrahedron, V its volume, then Xl +.!!.+.!!.+ Xf, == SlX1 + S~2 + SaXa + Sf,zf, hI kt kt hf, Slh1 S2hs S,ha Sf,hf, _SlX1 +S~2+S,X,+Sf,xf, 3V ; 1. 183. Let M and K denote the mid~int of the ed~s AB and DC of the tetrahedron ABCD. The plane passmg through M and K cuts the edges AD and BC at points Land N (Fig. 39, a). Since the plane DMC divides the volume of the tetrahedron into two equal parts, it suf-fices to prove that the pyramids DLKM and KCMN are equivalent. The ratio of the volume of the pyramid KCM N to the volume of the entire tetrahedron ABCD is equal to -} II ~ ~ II • Analogously, for the pyramid DLKM this ratio is equal to! II ~1: . Hence, we have to prove the equality: IDL 1 I CNI 1 DA 1-1 CB 1 • Answers, Hints, Solutions i59 Let us project our tetrahedron on tim plane perpendic-ular to tlie line KM. The tetrahedron ABCD will be projected in a parallelogram with diagonals A B and CD {Fig. 39, b). Tlie line LN will pass through the point of Intersection of its diagonals, consequently, our statement is true. A ----__ B 8 u A (a) c (6) Fig. 39 184. Let for the sake of definiteness I DA I ~ I DB I~ I DC I, and at least one of the inequalities is strict. Let us superpose the triangles DAB, DBC, and DCA so as to bring to a. coincidence equal angles and equal sides (Fig. 40). In the figure, the vertices of the second triangle have the subscript 1, those of the third triangle the subscript 2. But ID\tA2 I = I DA I < I DICI I (by the hypothesis). /, ~ Consequently, DaDIB is acute and BDID is obtuse and I DB I > I DICl.l which is just a contradiotion. 185. Through each ed~ of the tetrahedron pass a plane parallel to the OppOSIte edge. Three pairs of planes thus obtained form a parallelepiped. 0ypoSite edges of the tetrahedron will serve as diagonals 0 a pair of opposite faces of the parallelepiped. Let, for instance, a and a1 denote the diagonals of two opposite faces of the parallele-piped, m and n their sides (m;;:: n). Then aIa~ cos ex = 160 Problems in Solid Geometry m'! -nl_ Writing such equalities for each pair of oppo-site edges, we will prove our statement. 186. Let the sphere pass through the vertices A, B, and C and intersect the edges DA, DB, and DC at pointe Fig. 40 K, L, and M. From the similarity of the triangles DKL and ABD, we find: I LK I = I AB I II ~~ II a,nd from the similarity of the triangles DML and DBC: I ML 1= I BC I II ~~: - But I AB I-I CD I = I BC I . I BD I = 2SABC' Now, it is easy to make sure that I LK I = I ML I. Remark_ The statement of our problem will be true for any tetrahedron in which the products of opposite edges are equal. 187 .. The fact that the points K, L, P, and N belong to the same plane (coplanarity) implies that VMKLP + VMPNK = VMNKL + VMLPN-(1) From Problem 9 it follows that VMKLP= IMKI·IMLI·IMPI VMABC, I MA 1·1 MB I-I MC I VMPNK = IMPI-IMNI-IMKI V IMCI-IMDI-IMAI MADe, VMNLlC = IMNI-IMLI·IMKI V IMDI.IMAI.IMBI MABn, Answers, Hints, Solutions 1&1 IMLI·IMPI·IMNI VMLPN = 1MB 1.1 MC I.IIMD I VMBCD. Substituting these expressions for the corresponding quantities in (1), dividing by I MK 1·1 ML I ·IMPI X I MNI, multiplying by I MA 1·1 JJJB 1·1 MCI .. IMD I, expressing the volume of each of the remaining pyramids in terms of the area of the ba!e and altitude h, we will get after the reduction by h/3 the statement of our problem. 188. Prove that the straight line passing through the given point parallel to a diagonal of the cube will touch each ball. 189. Both items follow from the following gene~al statement: if the sum a I AM I + ~ I BN I + , I CL I, where a, ~, 'Y are given coefficients, is constant, then the plane M N L passes through the fixed l>0int. This state-men t, in turn, follows from the equahty a I AM I + ~ I BN I = (a + ~) I PQ I, where P is a point on AB, Q on MN, I AP I I MQ I ~ IPBI-IQNI a-190. If in the tetrahedron ABCD the equality I AB 1+ I CD I = I BC I + I DA I is fulfilled, then, the same as it il done in the two-dimensional case, it is possible to prove that there is a ball touching the ed~s AB, BC, CD, DA, all the pointl of tangency being inside the line seg-ments AB, BC, CD, and DA .. If through the centre of tlie ball and some edge a plane is passed, then each of the dihedral angles under consideration will be divided into two parts, and for each part of any dihedral angle there is a fart of the neighbouring angle which turns out to be equa to it .. For instance, the angle between the planes OAB and ABC is equal to the angle between the planes OBC and ABC. 191. Let R denote the point of intersection of OM with the plane KLN (Fig. 41). The assertion that R is the centre of gravity (centroid) of the triangle KLN is equi-valent to the assertion that the volumes of the tetralied-rons MKLO, MLNO, and MNKO are equal. Denote 11-0U9 t82 Problems in Solid Geometry by %, 1" " the distances from M to the corresponding sides of the triangle ABC. Since the plane KLM is perpen-dicular to the ed~ AD, the distance from 0 to K LM il equal to the projection of OM on AD which is equal to tlie projection of MP on AD, where P is the foot of the D B Fig. 41 perpendic ular dropped from M on B C. I t is easily seen that the pmjection of MP on AD equals V'3 ' where s il the distance from M to BC. If a is a dihedral angle be-tween the faces of the tetrahedron A B CD, then 1 ." xys Vi VKLMo=sl KM 1·1 ML I sma· y'3= 27 • Each of the two other tetrahedrons M LN 0 and M N KO will have the same volume. 192. Project the tetrahedron on the plane passing through N perpendicular to CN. Let A , B I , D , K I , and M I denote the projections of the points 1, B, D t }t t and M. Tlie distance between BK and CN will be equal to the distance from the point N to B;Klt just in the same way, the distance between A M and. CN is equal to the dis-tance from N to AIM!" But AIDIBI is an isosceles tri-Answers, Hints, Solutions t63 angle. The line AIMI passes through Kl. (KI is the poin, of intersection of the medians). And smce the triangle AIKIBI is also isosceles, N is usually distant from AIKt and BIK J• 193. Let A denote a vertex of the base of the pyra-mid, B a point in the plane of a lateral face, I AB I = a, BI the projection of B on a side of the base, B2 the pro-jection of B on the plane of the base, B 3 the projection A Fig. 42 of B" on the edge of the base adjacent to ABI, B. the projection of B" on the lateral face adjacent to the face containing AB (Fig. 42). If now a is a dihedral angle at /, the base of the pyramid, BABI = cp, then I B,Ba I = I AB1 I ==aCQI cp, I ABa I == I B1B, I = I B1B I eela=asin cP Cela, I BaBt I = I BaB, I cos a= aces cp cos a, and, finally, I ABt I = VI ABa 1"+ I BaBt II -=4 V sin~ cP COl" a+ces" cp cos2 a = a cos Ct. Hence it followl that the length of any line segment lying in the plane of a lateral face after a twofold pro-jection indicated in the conditions of the problem will be multiplied by cos a (with the aid of translation we bring one of the end points of the given line segment into the vertex A). Consequently, in such projecting any 11 tM Problems in Solid Geometry igure will go into the :ligure similar to it with the ratio of similitude equal to cos a. 194. The statement of the problem follows from the equalities VAAtBC = VAAtBtC = VAAtBtCt and similar equalities for the volumes of the pyramids AA\CD and AAJDB. 195, Let M denote the point of intersection of the straight lines CB1 and C1B. The vertex A lies on D M • _ .. B, .D c Fig. 43 Through the pobltl D, D}., ud A pasl a plane. Denote by IC and L the points 01 its intersection with C1B1 and CB, and by A21 the point of intersection of the line AAI with DIK (Fig. 43). From the fact that CC1B1B il a trap-ezoid and KL ~sses through the point of intersection of its diagonals it follows that I KM I = I ML I. Ftu1her, considering the trapezoid D1KLD, we will prove that t I A A 1 I = '2 I A A 21 I. CoJl8equen tly , 1 V ABCD =-. 3 V .A aBC D' But it follows from the preceding problem that V AtBCD= V A1B tC1D t" Thus, the ratio of the volumes of the pyramids A1B1C1D1 and ABCD is equal to 3. Answers, Hints, Solutions 196. Introduce the following notation: ABCD is the given tetrahedron I BC I = a, I CA I = b, I AB I = c, I D A I = m, I DB I = n, I DC I = p. Let then G denote the centre of gravity of the triangle ABC, N the point of intersection of the straight line DM with the circum-scribed sphere, and K the point of intersection of the A /( C Fig 44 straight line AG with the circle circumscribed about the triangle ABC (Fig. 44). Let us take advantage of the following equality which is readily proved: I AG I-I GK 1= {- (a2+b2+cl ). Then I DG 1·1 GN I = I AG I-I GK 1=+ (al +b21+e21), conaequently, I GN I = a21+::+c21 , where 1 ! I-I DG I C:T V3m21+3n21+3p21-ai-bl-c21 (t) 166 Problems in Solid Geometry (see Problem 51), I DN I = I DG 1+ I·GN I = t+ a2+bl + C2 m2+ nl+ p2 9t --3t • The assertion that OM is perpendicular to DM, is equivalent to the assertion that 3 3 I DN I = 2 I DM I =2· T I DG I =2 t, that is, m2l+ n2l+ p2l 3 3t 2 t, whence replacing t by its expres-sion (1), we get (2) If AI, BIt C I are the centres of gravity of the 1'9spective face. DBC, DCA, and DAB, then in the tetrahedron A1BICID we will have a b c IBICII=T' ICIAII=T' IAIB1 1=3' 2 2 2 I DA1 1= T ma, I DBI 1=""3 nb, I DCI I =T Pc, where ma' nb' and Pc are the respective medians to the sides BC, CA, and AB in the triangles DBC, DCA, and DAB. If now tl is the distance from the vertex D to the point M, then, since M, by the hypothesis, lie. on the surface of the sphere circumscribed about the tetrahedron AIB1C1D and the line D M passes throu~h the centre of gravity of the triangle AlBIC!, to determine the 'Luantity I D M I we may take advantage of the formula obtained above for I DN I, that is, I DM 1= 4m:+4n~+4b~ 27tl ' where II = : V 12 (m:+nl+rc)-a2l-bl -c2l• Taking advantage of the formula for the length of the median of a trianrlo, we get 4m,1+4nl + 4pll-a'-bl -c2l I DM 1= 27tl ' Answers, Hints, Solutions 167 where tl = : V3ml +3n~+3p2-a2-b2-c2= ~ t. On the other hand, I DM I = f t, that is, 4m2+4n2+4p~-a2-b2-c2 3 1St =4 t. Replacing t by its expression (Formula (1)), we get (2) which was required to be proved. 197 .. Fix some axis of symmetry I. Then, if I' is also an axis of symmetry and I' does not intersect with I or intersect I but not at right angles, then the line 1, which is symmetric to I' with respect to I is also an axis of symmetry. This is obvious. And if some line 11 is an axis of symmetry and intersects with, and is perpendicular to, I, then the line 12 passing through the point of inter-section of I and 11 and perpendicular to them will also be an axis of symmetry. It is possible to verify it, for ins-tance, in the following way. Let us take the lines 1,111 and I. for the coordinate axes. Applying, in succession, to the point M (x, y, I) symmetry transformations with respect to the lines I and 111 we will brinE the point M first to the position Ml (x, -y, -I), and then Ml to M2 (-x, -y, I). Thus, a successive apl!lication of symmetry transformations with respect to the lines I and 11 is equivalent to symmetry with respect to I •• Our reasoning implies that all axes of symmetry, except for I, can be divided in pairs, that is, the number of symmetry axes is necessarily odd if it is finite. 198. Let M denote the projection of B on AD. Obvi-ously, M belongs to the surface of the sphere with diame-ter A B. On the other hand, we can show that I A M I X I AD I = I AB It. Hence it follows that all points M must belong to a certain spherical surface containing the given circle. Hence, points M belong to one circle along which these two spherical surfaces intersect. 199. Prove that the projections of the point M on the lidel of the quadrilateral ABeD lie on one and the same circle (if K and L are projection. of M on AB and Be, 168 Problems in Solid Geometry then the points B, K, M, and L lie in one circle, and, /'.. /'.. ~ /"'--... hence, M LK = M BK, M KL = MBL. The same for other sidel). Then take advantage of the result of Problem 198. 200. Since the centre of gravity lies on the lines joining the midpoints of the edges AB and CD, it will follow from the hypothesis that this line will be perpen-dicular to the edges AB and CD. 201. Let K and M denote the midpoints of the edges AB and CD. It follows from the hypothesis that the line KM passes through the point 0 which is the centre of the inscribed sphere; 0 is equidistant from the faces ACD and BCD. Consequently, the point K is also equidistant from these faces. Hence it follows that these faces are equivalent. In the same way, the faces ABC and ABD tum out to be equivalent. If we now project the tetrahed .. ron on the plane parallel to the edges AB and CD, then its projection will be a parallelogram with diagonals AB and CD. Hence there follows the statement or our problem. 202. Rotate the cube through some angle about the diagonal A Ct. Since the plane of the triangle A1BD is perpendicular to AC1 and its sides are tangent to the ball inscribed in the cube, the sides of the triangle ob-tained from A1BD after the rotation will also touch the inscribed ball. With the angle of rotation appropriately chosen, the face AA1B1B will go into the given plane, and the line segment MN will be a line segment of the rotated face. 203. Denote by t£, ~, Y the angles formed by rectan-gular faces with the fourth face. If Sl' S2, Sa, S4 are the respective areas of the faces, then Sl = S4 cos a, S2= S. cos 6, S, == Sf, cos y. After this, we may take advan-tage 01 the fact that cost a+cosl 6+cost y = 1. This follows, for example, from the fact that the angles made by the altitude dropped on the fourth face with the lat-eral edges of the pyramid are also equal to a, ~, and y (lee Problem 10). 204. Take a straight linejerpendicular to the given plane and denote by a, ~,an y the angles made by this line with the edges of the cube. The projections of the ed~s on the plane take on the values sin a, sin ~, sin ". And since COI~ a + cost ~ + cos! y= 1, the sum of Hie Answers, Hints, Solutions squares of the projections will be equal to 4a2 (sin2 ex + sin2 ~ + sinl y) = 8al , 169 where a is the edge of the cube. 205. Through each edge of the tetrahedron pass a plane parallel to the opposite edge. We will obtain a cube with a tetrahedron inscribed in it. If the ed~ of the tetrahedron is b, then the edge of the cube will be equal to bl-Y2. The projection of each face of the cube is a paral-lelogram whose diagonals are equal to the projections of the edges of the tetrahedron. The sum of the squares of an diagonals is equal to the doubled sum of the squares of the projections of the edges of the tetrahedron and is equal to twice the sum of the squares of the projections of tlie edges of the cube. Taking advantage of the result of the preceding prob-lem, we get that the sum of the squares of the projections of the edges of a regular tetrahedron on an arbitrary b2 plane is equal to 8""2 = 4b2• 206. Consider first the case when the given straight lines are skew lines. Denote by A and B the positions of the points at some instant of time, k is the ratio of their velocities (the velocity of the body situated at the point A is k times the velocity of the other body). M and N are two points on the line AB such that I A l'rl I : I MB I = I AN I : I NB I = k (M is on the line segment AB), o is the midpoint of MN. The proof of the statement of our problem is divided into the following items: (1) The points M, N, and 0 move in straight lines, the straight lines in which the points A, B, M, N, and 0 move are parallel to one plane. (2) The lines in which the points M and N move are mutually perpendicular. (3) If two straight lines are mutually perpendicular and represent skew lines, then any sphere constructed on the line segment whose end points lie on these lines, as on the diameter, passes through the points P and Q, where PQ is a common perpendicular to these lines (P and 0 are situated on the straight lines). (4) The locus of points L such that I A L I : I LB I = k is the surface of the sphere constructed on M N, as on the diameter. 170 Problems in Solid Geometry From the statements (1) to (4) it follows that the circle whose existence is asserted in the problem is the circle obtained by rotating the point P (or Q) about a straight line in which the pOint 0 moves, where P and Q are the end points of the common perpendicular to the straight lines in which the points M and N are displaced. Items (1) and (2) can be proved, for instance, in the following way. Let A 0 and Bo denote the positions of the points at a certain fixed instant of time. Let us project our p_oints parallel to the straight line AoBo on a 'p'lane parallel to the given lines. The points A 0 and B 0 will be projected into one point C, and the points A, B, M, N, and 0 will be projected into the respective points A', B', M', N', and 0'. Then the points M' and N' will repre-sent the end points of the bisectors of the interior and the exterior angle C of the triangle A' B' C'. Hence, M', N', ./"-.... and 0' move in straight lines, and M' CN' = 90°. Hence it follows that the {loints M, N, and 0 also displace in straight lines, since It is ohvious that each of these points lies in the fixed plane })arallel to the given lines. Item (3) is obvious. Item (4) follows from the corresponding state-ment of plane geometry. In the cale when the points A and B move in two inter-secting lines, the relevant reasoning is somewhat changed. The problem is reduced to the proof that in the plane containing the given lines there are two fixed points P and Q such that 1 AP 1 : 1 PB 1 = 1 AQ 1 : 1 QB 1 = k. 207. Let 0 denote the centre of the ball, r its radius, AP and BQ the tangents to the ball (P and Q being the pOints of tangency), M the point of intersection of the lines AP ano BQ. Setting 1 OA 1 = a, 1 OB 1 = b, 1 PM 1 = 1 QM 1 = x. Then 10M II=r2+x2, 1 AM 12= (ya2 -rl + x)l, 1 B M 12 = CV bl -r' + X)2. If the signs are of the same sense, then the following relation.hip is fulfilled: -ybl -r2 1 AM 12_ -yal-ri 1 BM 12 +(ya21-rl-ybl -rl) I OM II=lp (1) Answers, Hints, Solutionl 171 If the signs are opposite, then Vb2l-r2 1 AM 12I+Val -r2lIBMII -(Val-r2+ lfb2-r2) 1 O~M 121=12' (2) where 11 and '2 are constants depending on r, a, and b. Since the sum of the coefficients of I AM 121, I BM 12 and I OM 12 in Equations (1) and (2) is equal to zero, the locus of points M for which one of these relation-ships is fulfilled is a plane. In both cases thil plane is perpendicular to the ~lane OAB. 208. Let ABC be the given triangle whose eides, as usually, are equal to a, b, and e. The radii of the three balls touching one another and the plane of the triangle at points A, B, and C are respectively equal to ~: ' ~: ' ;:. Denote by x the radius of the ball touching the three given balls and the plane of the triangle, M is the point of tangency of this ball and the plane. We have: 1 MA 1 =2 Vb;:, 1 MB 1 = 2 Va;: ' IMC 1 =2 Va::. Consequently, I MA I: I MB I = b: a, I MB I: IMCI = e : b or I MA I : I MB I : I MC I = be : ac : abo For any irregular triangle there are exactly two points Ml and M I for which this relationship is fulfilled. Here we take advantage of Bretschneider'e tlieorem. Let ABCD be an arbitrary plane quadrilateral. Let AB = a, BC = b, CD = e, andDA = dt AC = m and BD = n. The 10m A A of the angles A + C = cpo Then the equality m2n2 = a2le21 + b2ltJI -2abed cos cp holdl. We then obtain that A if A a:;z:: a is the smallelt anKlt 01 the trfanKle, then the ,/""... /'... Ulilel B M 1 C and B M 21C are .qual to 60- + c anti 00--&. 172 Problems in Solid Geometry Let G';c = 60° + a. Write for the iriangle BMl C the theorem of cosines, denoting the radius of the ball touehing the plane at point Ml by r (x = r), a2= 2~cr + "':br 4ar cos (60°+ a) ~ ~=2 (--E.....b +~ 2 cos (60o+a) ). (1) r a ac a Analogously, designating the radius of the ball touching the plane at point M 2 by p, we get 1-=2 (.!-+~ 2 cos (OOO-a) ) (2) P ab ac a • Subtracting (2) from (1), we obtain 1 1 4 [cos (600-a)-cos (60°+ a)] ---= r p a 8 sin 60° sin a _2 Va a -R which was required to be proved. 209. Let M denote tlie midpoint of AB, 01 and O. the centres of the balls, Rl and R2 their radii, then 1 MOl Ii_I M02 12= (RI+ 1 A: 12) _ (Ri+ !A!12) =Rl-RI· This meanl that the midpoints of all the line segments of common tan~nts to the given balls lie in one and the same plane whieh is perpendicular to the line segment Ol0s:. Hence follows the truth of the statement of our pro))lem. 210. Such pentagon does not exist. 211. Let A1A.A.AcA& be the given pentagon. It fol-lows from the liypotliesis tha t all the diagonals of the pen-tagon are equal to one another. Choose three vertices of the pentagon so that the remaining two vertices lie on one lide of the plane determined by the Ihree ehol8D vertices, say, A2t A" and A •• Then the vertices A}. and A. will be Iymmetrie to eaoh other with relpect to the plane Answers, Hints, Solutions 179 passing through the midpoint of A zA. perpendicular to A.A3' This follows from the fact that tlie triang!e AzAaAli is isosceles, I A.A Ii I = IA3AIi I, Al and At he on one side of the plane A 2AaAli' and I AIAt I = lAtA. I, I AIAIi I = I AtAIi I, and I AlA. I = I AtAt I· Hence, the points AI' At, A., and At lie in one plane. The further reasoning is clear. The cases when the sought-for plane passes through other vertices are considered in a similar way. 212. Let M denote the point of intersection of the diagonal A CI and the plane AIBD. Then M is the point '~--+-~N B Ai Fig. 45 of intersection of the medians of the triangle AIBD (so-called median point) and, besides, M divides tile diago-nal ACI in the ratio 1 : 2, that is I AM I = ! d. Consider the pyramid ABAID (Fig. 45). On the line BM take a point K such that I MK I = I BM I, and con-struct the prism M KD AN P. You can easily notice that the distances between the lateral edges of this prism are equal to the respective distances from the points AI' B, and D to AM. Consequently, the sides of the section per-pendicular to the lateral edges of the prism MKDANP are equal to these distances. Further, the volume of the pyramid ABAID is equal to the volume of the constructed t74 Problems in Solid Geometry prism and amounts to one sixth the volume of the paral-lelepiped, i.e. {- V = ~ dS, V = 2dS. 213. Let 14 denote the centre of gravity of the tetra-hedron ABCD. The volume of the pyramid MABC is one fourth the volume of the given tetrahedron. Complete the pyramid MABC to get a parallelepiped so that the line segments MA, MB, MC are its edges. Figure 46 repre-A A c Fig. 46 sents this parallelepiped separately. It is obvious that the edges MC, CK, KL and diagonalMLof this parallel-epiped are respectively equal and parallel to MC, MA, MB, and MD. But the volumes of the pyramid. MABC and MCKL are equal to each other, that is, each of them is equal to ~ V ABCD. ConMquontly, the volume of the tet-rahedron in question equals (~) 2 • ~ V ABCD = ~~ V. 214. When solving Problem 180, we proved that the sum of the vectors, perpendicular to the faces of the tetrahedron, directed towards outer side with respect to the tetrahedron, and whose lengths are numerically equal to the area of the corresponding faces, is eCJWlI to zero. Hence follows the existence of the tetrahedron K LM N. In finding the volume of the tetrahedron, we shall take advantage of the following formula: V = ! abc sin ex sin ~ sin C, Answers, Hints, solutions t15 where a, b, and c denote the respective lengths of the edges emanating from a certain vertex of the tetrahedron, a and ~ two plane angles at this vertex, and C the dihed-ral angle between the planes of the faces corresponding to the angles a and 6. If now a, ~, and yare all plane angle a at this vertex and A, B, and C are dihedral angles, then V3 = ( ! ) 3 a3b3c3 sin2 a sin2 ~ sin' "/ sin A sin B sin C. (f) Take now a point inside the tetrahedron, and from it drop perpendiculars on the three faces of the tetrahedron corresponding to the trihedral angle under consideration, and on each of them layoff line segn).ents whose lengths are numerically equal to the areas of these faces. Obvious-ly, the volume of the tetrahedron formed by these line segments is equal to that of the tetrahedron KLMN. TJie plane angles at the vertex of the trihedral angle formed by tliese line segments are equal to f80° -A, fBOo-B, fBOo-C, and the dihedral angles to f80 0 -a, f80° -~, f80° -yo Consequently, making use of Equal-ity (f), we get for the volume W of tJiis tetrahedron WS = (~) 3 S~SfSl sin2 A sin! B sinl! C sin a sin ~ sin "/, (2) where Sl' Sz, S3 are the areas of the faces formed by the edges a, b, and c, respectively, that is, Sl = ~ ab sin y, S z = ; k ain a, S 3 = ; ca sin ~. Replacing Sl' Sz, S3 in (2), we get W'= (~) 3 (~) 9 alb'c6 sini a sint ~ sini y sinz A sin2 B X sinz C. (3) Comparing Equations (f) and (3), we obtain 3 W =.: 4" VI. 2f5. The statement of the problem follows from the fact that the products of the line segments into whIch each of these chords is divided by the point of intersection are equal. i76 Problems in Solid Geometry 217. The statement of our problem follows from the following fact of plane geometry. If through a point P lying outside of the given circle two strai~ht lines are drawn intersecting the circle at the respectIve points A and AI' Band BI , then the line AIBI iSlarallel to the circle circumscribed about P AB passe through the point P. Thus, the set of points under consideration will be-long to the plane parallel to the plane which touches (at the point P) the sphere passing tnrough the given circle and point P. 218. The equation (x -a)2 + (y -b)2 = k2 (z -e)Z describes a conical surface whose vertex is found at the point S (a, b, e), the axis is Jlarallel to the s-axis, k = tan a, where a is the angle between the axis of the cona and its generatrix. Subtracting from each other the equations of two conical surfaces with axes parallel to the z-axis, equal parameters k, but different veI1ices, we get a linear dependence relating x, y, and z. 2f9. Denote by F the point of intersection of the lines KL and MN and by E the point of intersection of the line PF and the sphere passing through the ~ints P, A, B, and C (supposing that P does not lie in the plane of the face ABC). The points P, Q, R, and E belong to one circle repre-senting the section of the sphere passing through the points P, A, B, and C by the plane passing through the points P, K, and L. But since F is the point of inter-section of the lines KL and MN, the points P, S, T, and E must belong to the circle which is the section of the sphere passing through the points P, A, C, and D by the plane determined by the points P, M, and N. Conse-quently, the points P, Q, R, S, and T lie on two circles liaving two common pomts P and E, and such two cir-cles lie long to one sphere . Remark. We have considered the case of the ~neral position of the given points. To get a complete solution we have to consider several particular cases, say, Plies in the plane of the face, KL and MN are parallel lines, and so on. Answers, Hints, Solutions 177 220. Let the edges SA, SB, SC, and SD of a quadri-hedral angle be elements of a cone whose axis is SO. Then in the trihedral angle formed by the lines SO, SB, and SC, the dihedral angles with the edges SA and SB are equal. Considering three other such angles, we get easily tliat the sums of opposite dihedral angles of the given quadrihedral angle are equal. Conversely. Let the sums of opposite dihedral angles be equal. Consider the cone with the lines SA, SB, and SC as its elements. Suppose that SD is not an element. Denote by SD1 the straight line along which the surface of the cone and the plane A SD intersect. We will obtain two quadrihedral angles SABCD and SABCD1 in each of which the sums of opposite dihedral angles are equal. This will imply that in the trihedral angle which is comj)lementary to the angle SCDD1 (see the solution of Problems 165 and 166) one plane angle is equal to the sum of two others which is impossible. 221. Let all the vertices of the hexahedron ABCDEFKL, except for C, lie on the surface of the L A B Fig. 47 sphere with centre 0 (Fig. 47). Denote by C1 the point of intersection of the line KC with the surface of the sphere. For the sake of brevity we shall symbolize by y, and this means that in the triangle ASB I AB I < I AS I < I BS I, /, /, that is, A SB is the smallest angle of this triangle, A SB < 600 • The supposition that both angles are obtuse is con-sidered in the same manner. 4 225. From Sh to 3 Sh. 226. The greatest volume is possessed by the tetra-hedron two opposite edges of which are mutually per-pendicular and are the diameters of the bases. Its volume is equal to ~ Rlh. 227. Let I AB I = I BC I = i, I AAI I = x. i Vi i VDD1BC1=TSDBD1·2"= 6 x• On the other hand, V DD1BCl = ~ S DBCl I DIB I sin = V (2+X~ (i+2z'l' sn!, q> =2.:'+ :. +5~, 12 iSO Problems in Solid Geometry whence it follows that the greatest value of q> will he . i arcSlD S ' 228. Let the altitude of the prism be equal to 1, I A M I = x. Circumscribe a circle about the triangle I AIMCI I· Consider the solid obtained by revolving the arc AIMCl 5>f this circle about the chord AICI . The angle AIMCI wiu be the greatest if the line AB touches the sUrlace of the solid thus R8nerated. The latter happens if the lines MO and AB, where 0 is the centre of the circle circumscribed about th& triangle ABC, are mutually perpendicular; hence, the line MO divides AICI in the , I AM I z ratiO I MB I 2-x' On the other hand, it is possible to show that MO d' Id A C in the t' I AIM I eosCc':M E IV es I I ra 10 /"-.. . x-I CIM I cos ClAIM pressing the sides and eosines of the angles of the triangle AIMCI in terms of %. we get the equation whence %=t. The greatest value of the angle AI MC1 " equaJaT' 229. The lines AE and CF are mutually perpendicu-lar. Let QI be the projection of Q on the plane ABBI AI' QJ. lies on the line segment BL, where L is the midpoint of AAI. Let N be the point of intersection of AE and LB. It Is easy to Iiad that 1 AN 1- ~5' Setting i I AP I = vg+ %, 1 NQI I = y, we get I PM II = 8 +( t )2 I PQ II '1:+ 1+ I PM II . 5" VS+%, =.-y i, I PQ II attaJDII the greatest value for y=O. It remains to find the Answers, Hints, Solutions 181 greatest value of the fraction 9/5+ (2/V5) X+X2 x2+1 This value i. attained for '" = J5. Answer: V 2. 230. Consider the triangle KLM representing the projection of the given triangle on the plane ABCD, K lying on the line CB, L on CD, M on CA. If ICKI = x, then I CL I = I a-x I, I CM I =V2!a-+ l It is rather easy to get that SKLM = ~ Ix (a-x)-a (a- ; ) I 1 =T (2xl -&x +2a2). 7a2 The least value is equal to 32 . 231. Let x denote the altitude of the parallelepiped. Consider the section of the pyrami~ by the plane passing at a distance x from its base. The section represents a square with side (1 -x); a rectangle of area s which is a face of the parallelepiped is inscribed in the square. Two cases are possible: (1) The base of the parallelepiped is a square with side V;. The diagonal of the parallelepiped d =-V xl + 2s, and V2 -.i-(1-z) """2 < r s < (i-x) or 1-V2s<x ~ i-V;' Thus, in this case if s < ~ , 1-2 V2s + 4s~dl ~ ... r-1 ~ -.i-1-2 y 6+ 361 and if s~T' f2s Y 2s, O<x:< i-Y2s.) We have d2 = xl + (y + :;)2 -2y:; = x2 + 2 (1 -X)2 -2s = 3x' -4x + 2 -2,. t If s < tS ' then the least value of dl is attained for Z= ; , and if s> t1 S ' then for x=1-Y2s. Besides, d i < 2-21. Combining the results of items (1) and (2), we get the aDlwer. Answer: if 0 < , =E;;; 11S t then V f- 2s =E;;;d < V2-2s~ . 1 7+2y6 If 18 <,< 25 ,then V 1-2 Y2s+" =E;;; d < Y2-2s; . 7+2 y6 1 If 25 ~ s < ""2' then V 1-2 y2s+":< d =E;;; V 1-2 y,+3s; if ~ :<'< 1," then f2l< d ~ V 1-2 Vs +&, Answers, Hints, Solutions i83 232. Cut the polyhedron ABCAIMNCI by the/lane passing at a distance h from the plane AIB).CI an pro-Ject the section thus obtained on the plane AIBICI (Fig. 48). In the figure, the projection of this section is Fig. 48 shown in dashed line. It is obvious that the circle of the base of the cylinder must be located inside the trapezoid KLNCI (K, L are the respective points of intersection of Al CI and M N with the projection of this sectionl. If h = 3, then the section plane coincides with the pane AB C and the points K and L with the midpoints of the sides BICI and AICI • If h < 3, I ML I = I AIK 1= h h h 3' I LN 1 , i-3 , I KCI I=2-To We can readily verify that for h < i the radius of the greatest circle contained in the trapezoid KLNCI is equal to V:a, and for h > i this radius is equal to the radius of the circle inscribed in a regular triangle with side I KC I = 2- ~ that is! it is equal '0 (2-~) va . 3 6 ~ 184 Problems in Solid Geometry 33· 3 Answer: (a) if 0 < h:< 2' v=Tif nh; if 2< h:< 3, V = f~ h (2 -~ ) I ; (b) the greatest value of the volume will be obtained &t for h=2, V= 27 . 233. If the plane passed through our line segment parallel to the face ABBlAl cuts CB at the point K so that I CK I = x, then the projection of the line segment on the face ABC has a length x, and its projection on the edge CCI is equal to I a -2x I; thus, the length of the line segment will be equal to V x2+(a-2x)2= V5x2-4ax+a2. The minimal length is equal to ;5 . 234. The following statement is an analogue of our problem in the plane. Given an angle and a point N inside it. Consider all possible triangles formed by the sides of the angle and straight line j)assing through the point N. Among such triangles, the smallest area is possessed by the one for which the side passing through N is bisected by the point N. Let us return to our problem. Let M be the given point inside the trihedral angle. The plane passing through the point M intersects the edges of the trihedral angle at points A, B, and C. Let the line A M intersect BC at N. Then, if the passed plane cuts off a tetrahedron of the least volume, the point N must be the midpoint of BC. Otherwise, rotating the plane about the line AN, we will be able to reduce the volume of the tetrahedron. 235. If h is the altitude of the segment, then its f f ~ volume is equal to ""2 Sh-a nh3 . The greatest volume will be achieved ~for 'h:::::. V :n ; it will be S .. /S equal to ""3 V 2n' 236. Note that the shadow thrown only by the upper face of the cube (Il!SUming th4t aU the remaining facal Answers, Hints, Solutions 185 ab are transparent) represents a square b on a side. -a Hence it follows that the area of the shadow cast by the cube will be the least when the source of light is located above the upper face (only the upper face of the cube is illuminated); it will be equal to ( b ~a ) 2 with the area of the lower face of the cube taken into account. 237. The statement (1) is true, let us prove this. Denote by PI the polygon obtained when our polygon is cut by a plane not passing through its centre, S denoting the area of this polygon. PI is a polYROn symmetric to PI with respect to the centre of the porygon. Let us denote by II the smallest convex polyhedron containing PI and PI (II is called the convex .lull of PI and PI). Obviously, II is a central-symmetric polrgon, its centre coincides with the centre of the origtna polyhedron. All the ver-tices of II are either vertices of PJ.. or vertices of PI. Let P denote the polygon obtained when II is intersecteo by the plane passing through the centre parallel to the faces of PI and P'4! q its area. Let us take a face N of the polyhedron II different from PI and P 2 • It is obvious that any section of the polyhedron II by a plane parallel to N must intersect either simultaneously all the three polygons PIt P2 , and P or none of them. Since the poly-hedron II is con vex, the line segments ll' l2' and l along which this plane cuts PIt P 2 , and P are related as follows: l ~ ~ (l1 + l2). Hence it follows that q> S. (We inte-grate the inequality l > -} (l1 + l2) with respect to all possible planes parallel to N.) The statement (2) is false. Let UI conltruct an example. Consider in a rectangular Cartesian coordinate system the polyhedron whose points satisfy the inequality 1 x 1 + Ify 11+ 1 , 1 < 1. (This J>olyhedron represents a reg-ularloctahedron.): All tlie faces of this polyhedron are regular triangles with side -y2 and radius of the circum-scribed circle V i . The section of this polyhedron by a :plane :passing through the origin and :parallel to any Problems in Solid Geometry face represents a regular hexagon with side ~i and 1'2 the same radius of the circumscribed circle. But 2 < vi-Remark. For an arbitrary convex central-symmetric solid the following statement is true. Let Rand R 0 denote the radii of the smallest circles containing the sections of the given solid by two parallel planes, the second plane passing through the centre; then R 0 ~ ~3 R. As we have already seen, an e<I!lality in this case is achieved for a regular octaheoron. 238. 413. 239. Let A and B be the vertices of the cones, M and N two points on the circle of the bases, L a point diametri-cally opposite to the point M (I AM I = V rI + JI1, IBM I = V r2 + h2). Through M pass a plane perpen-dicular to A M and denote the projections of B tN, and L on this plane by BIt N 1 , and L1 . The distance between AM and BN is equal to the distance between M and BINI' and cannot exceed I MBI I. The condition h ~ H implies that I MB 1 I ~ I M Ll I, that is, the point Bl is situated inside, or on the boundary of, the projection of the base of the cones on the passed plane, and the distance between M and BINI is equal to MBI if MBI and BINI are mutually perpendicular. (h+H) r Answer: V · r2+HI 240. Extend the edge BIB beyond the point Band on the extension take a point K such that I B K I = a. As is readily seen, K is equidistant from all the sides of the quadrilateral AB1CD. On the diagonal BID take a point L such that II ~1; II Vi. The point L is the end point of the bisectors of the triangles BJ"AD and BICD and, hence, L is also equidistant from the sides a the quadrilateraIAB1CD.Now, we can prove that all the pointes of ~e lin~ K L a~ equidistant from tb~ ~idea Answers, Hints, Solutions 187 of the quadrilateral. Thus, the sought-for radius is equal to the shortest distance between the line KL and any of the lines forming the quadrilateral ABlCD. Find the distance, say, between the lines KL and AD. Projecting the points K and L on the plane CD Dl Cl , we get the points Kl and L l • The desired distance is equal to the distance from the point D to the line KILl' Answer: a V 1- ~2. 241. Let the diagonal A C l lie on the edge of the di-hedral angle, the faces of the angle intersect the edges of the cube at points M and N. It is not difficult to notice that if the volume of the part of the cube enclosed inside this angle reaches its greatest or smallest value, then the areas of the triangles A C M and A CIN must be equal (otherwise, rotating the angte in the required direction, we shall be able both to increase and decrease this volume). If 0 < CG < 600 , then the part of the cube under con-sideration has a volume contained in the interval from 1 to 1 . For CG = 600 this 2 V 3 cot ~ 3 ( 1 + Va cot ~ ) volume is constant and is equal to 1/6. For 600 < CG ~ 1200 the extreme values of the inter-val must be increased by 1/6 and CG replaced by CG -600 , for 1200 < CG < 1800 they must be increased by 1/3, and CG replaced by CG -1200 • 242. Note that the area of the projection of any par-allelepiped is always twice the area of the projection of some triangle with vertices at the end points of three edges of the parallelepiped emanating from one of its vertices. For a rectangular parallelepiped all such triangles are congruent. The greatest area of the pro-jection of a rectangular parallelepiped will be obtained when one of such triangles is parallel to the plane on which the parallelepiped is projected. Thus, the greatest area of the projection is equal to Va2b2 + b 2c2 + c2a2 • 243. Prove that the volume of such tetrahedron is loss than the volume of the tetrahedron two faces of which are regula:r triangles with $ide of 1 formins a right angle, i88 Problems in Solid Geometry 244. (i) This statement is false. For instance, take inside the triangle ABC two points DI and EI such that the sum of the distances from DI to the vertices of the triangle is less than the sum of the distances from EI to the vertices. Now, take a point D sufficiently close to DI so that the sum of the distances from D to the ver-tices A, B, and C remains less than the sum of the distances from the point EI • Take E inside ABCD on the perpendicular to the plane AB C erected at the point EI • (2) This statement is true. Let us prove this. Denote by M the point of intersection of the line DE and the plane ABC. Obviously, M lies inside the triangle ABC. The lines AM, BM, and CM separate the plane of the triangle ABC into six parts. The projection of D on the plane ABC, the point DI , is found in one of these six parts. Depending on the position of D}t one of the /"'--. ~ /""..., angles DIMA, DIMB, DIMC is obtuse. If the angle /""-.... DIMA is obtuse, then DMA is also obtuse, and, hence, the angle DEA is also obtuse. Hence it follows that 1 DE 1 < 1 DA I. 245. Let 2a be a side of the base, h the altitude of the pyramid. Then R is equal to the radius of the circle cir-cumscribed 3bout the isosceles triangle with base 2a Y2 and altitude h, R = 2a22t h2 ; r is equal to the radius of the circle inscribed in an isosceles triangle with base 2a and altitude h, r = ~ (Ya2+hz-a). h Let R r We will have 2 + x = 2k (y i + x -i), whence XZ + 4 (i + k -k2) X + 4 + 8k = O. The discriminant of this equation is equal to i6k2 (k2 -2k -i). Thus? If ~ t~ + i, which was req:uireq to be proveq: Answers, Hints, Solutions 189 246. The centres of gravity of the faces of the tetra-hedron serv& as the vertices of the tetrahedron similar to the given one with the ratio of ~imilitude 1/3. Conse-quently, the radius of the sphere passing through the centres of gravity of the faces of the given tetrahedron is equal to R13. Obviously, this radius cannot be less than the radius of the sphere inscribed in the given tetra-hedron. 247. Let in the tetrahedron ABCD I AB I = b, I CD I = c, the remaining edges being equal to a. If N is the midpoint of AB and M is the midpoint of CD, c R ------/ / !( / --~ / P \ p' B----1r-_~ N A L "----L...---~S a Fig. 49 then the straight line MN is the axis of symmetry of the tetrahedron ABCD (Fig. 49, a). Now it is easy to prove that the point for which the sum of the distancea to the vertices of the tetrahedron reaches the smallest value must lie on the line MN. Indeed, let us take an arbitrary point P and a point P' symmetric to it with respect to the line M N. Then the sums of the distances from P and P' to the vertices of the tetrahedron are equal. If K is the midpoint of P P' (K lies on M N), then in the triangles PAP', PBP', PCP', and PDP', AK, BK, CK, and DK are the respective medians, and a median of a triangle is less than the half-sum of the sidea includini it. The quantity I MN I is readily found: f90 Problems in Solid Geometry Consider the equilateral trapezoid LQRS. (Fig. 49, b) in which the bases I LS I and I QR I are equal to b and c, respectively, and the altitude is equal to d. Let F and E be the res:pective midpoints of the bases LS and QR. If K isa pomton MN, and Ton FE, and I FT I = I NK I, then, obviously, the sums of the distances from K to the vertices A, B, C, and D and from T to the vertices L, s, Q, and R are equaL And in the trapezoid LQRS (as well as in any eonvex quadrilateral) the sum of the distances to the vertices reaches the least value at the point of intersection of the diagonals and is equal to the sum of diagonal s. Answer: -V 4a~2-+-2=b-c. 248. Prove that the shortest way leading from the point A belonging to the circle of the greater base to the diametrically opposite/oint c of the other base consists of the element AB an diameter BC. Its length is 2R. Denote by r the radius of the smaller base, by 0 its centre. Consider the path leading from A to some point M be--longin~ to the smaller base. The arc AM situated on the laterar surface of the cone will have the smallest length if a line segIl!ent will correspond to it on the development of the lateral surface of the cone. But this development with the angle between the generatrix and the base equal to n/a and the radius of the base R represents a semicircle of radius 2R. Hence, the development of a frustum of -a eone is a semiannulus. Here, if to the arc BM on the base there corresponds a central angle (t)=Vd+t-Vl+t= V~-V-is monotone with respect to t. This d+t+ l+t implies that f (x, y, :) reaches its least value when x, y, : are equal to +f, that is, when the vectors a, b, and e are collinear. In this case our inequality is readily verified. 250. Let the straight line MN intersect DilCI at the point L. Set: I AM I = x, I BN I = y. It fo ows from the hypothesis that x > a, Y > a. Projecting all the . I fi d I CIL I a pomts on the pane ABBIAlt we n I LDI I = x-a ' and projecting them on the plane ABCD, we find : ~~~ ! = 192 Problems in Solid Geometry y-a Consequently, a y-a, whence xy = a x-a a (x + y) a. But (x + y)2 > 4.xy. Hence, xy > 4.a2• Now, we get I MN 12 = x2 + yl + a2 = (x + y)2 -2zy + a2 = (~)2 _ 2xy + a2 = ;2 (xy -a2)2 > 9a2 • The least value of I MN I is equal to 3a. 251. If x is the length of two other sides of the rec-tangle, then the volume of the pyramid is equal to az .. / a2 %-3 V b2-T-'· The greatest value of the volume .. / 4b2-ai a (4.b2-a2) will be for x = fI 2 ' it equals f2 • 252. Let M be a point on the line ABI , N on the line BCl , Ml and Nl. the respective projections of M and N on the plane ABCD. Setting I BMI I = z, I BNl I = y, we get I MINI I = V.zI+,l!l, I MN I = V xl!l+yl+(a-x-y)2. By the hypothesis, I MN I = 2 I MINI I, consequent-ly, (a -2 -,)1 = 3 (:r + yS). Let :r + yl!l = ul, 2 + y = v, then 2ul -uS ~ 0, and since ul =} (a-v)l, replaci1!K ul in the inequality relating u and v, we obtain the following inequality for v! VI + 4av -2a2 < 0 whence a (2 + VB) ~ v ~ a (V6 -2). We now find the least value of I MN I, it is equal to 2a (Va -V2). 253. Consider the cube ABCDAlBtClDI with an edge 2R. Arrange the axes of the given cy inders on the lines AA 1, DC, BICI• (a) The centre of the cube is at a distance of R V2 from all the edges of the cube. Any point in space is located at a distance greater than R V2 from at least one of the e~s AAI, DC, BICp This follows from the fact that the cylinders with axes AA 1, DC, BICI and radii R V2 have the only common point, the centre of the cube. Consequently, the radius of the smallest ball touching all the three cylinders is equal to R (V2 -f). (b) If K, L, and M are the respective midpoints of the edgel AA" DC, and BICl, then the straight line passing Answers, Hints, Solutions 193 through the centre of the cube perpendicular to the plane KLM is found at a distance of R y2 from the lines AAI, DC, and BIB; KLM is a regular triangle, its centre coincides with the centre of tiie cube. Hence it follows that any straight line intersecting the plane KLM is situated from at least one vertex of the triangle KLM at a distance not exceeding the radius of the circle circum-scribed about it which is equal to R Y'2. Thus, the radius of the greatest cylinder touching the three given cylinders and satisfying the conditions of the problem is equal to R (Y2 -1). 254. Let ABCD be the tetrahedron of the greatest volume, 0 the centre of the given spheres. Each line seg-ment joining 0 to the vertex of the tetrahedron must be perpendicular to the face opposite to this vertex. If, for instance, AO is not perpendicular to the plane BCD, then on the surface of the sphere on which the point A lies it is possible to find points lying at greater distar,.ces than the point A does. (This reasoning remains, obvious-ly, true if A, B, C, and D lie on the surfaces of different spheres and even not necessarily concentric ones.) Hence it follows that the opposite edges of the tetrahedron ABCD are pairwise perpendicular. Let, further, the points A and B lie on the sphere of radius R = V 10, and C and D on the sphere of radius r = 2. Denote by x and y the re-spective distances from 0 to AB and CD. Through AB, draw a section perpendicular to CD. Denote by K the point of intersection of this plane and CD. Taking into consideration the properties of our tet-rahedron ABCD, it is easy to prove that I AK I = I BK I, 0 is the point of intersection of the altitudes of the triangle ABK. Draw the altitudes KL and AM (Fig. 51). From the similarity of the triangles A LO and OKM we find 10M I = X; . Further, I AB I = 2 Y R2 -x2, and from the similarity of the triangles AOL and A MB we get R 13-0449 2 YR2_X2 R ' xy -r-R f94 Problems in Solid Geometry whence 2X2 + xy = R2. Proceeding in the same way, we get the equation 2y2 + zy = r. From tlie system of equations { 2:&2.+ xy = fO, 2y2+ Xy =4 we find z = 2, y = f. The volume of the tetrahedron ABCD will be equal to 6 V2. A L~~------------~·K B Fig. 5f 255. Let A denote the vertex of the trihedral angle whose plane angles are right angles, B the vertex of the other angle. On the line se~ent AB take a point M such that 2 1 AM 1 = 1 MB I. Through the point M pass a plane perpendicular to AB. This plane will cut each of the two trihedral angles in a regular triangle with side b = a V~ . In Fig. 52, a, the triangle PQR corresponds to the section of the trihedral angle with the vertex A. The face BCD cuts off the pyramid QF KL from the pyra-mid APQR (the position of the point F is clear from Fig. 52, b). The volume of this pyramid is proportional to the product 1 QK I-I QL 1·1 QF I. Tb.e quantity I QF I, obviously, reaches the greatest value for a = n/3, /. where a = CMQ. Let us prove that I KQ I-I QL I reache, Answers, Hints, Solutions the greatest value .also for a = n/3. Since KL is tangent to the circle inscribed in PQR, the perimeter of the tri-angle KQL is constant and is equal to b. We set I KQ I = Q Q ------~----~-HA~--~~--~~----~T (b) Fig. 52 x, I QL I = y,l'ZJ.then KL = b -x -y. Write the theorem of cosines for the triangle KQL: (b -x -y)2 = x2 + y2 -xy =} b2 -2b (x + 1/) + 3xy = 0 =} b2 -4b V xy + 3xy ~ o. C -w'-b -w'-onsequently, either J' xy ~ 3' .or J' xy > b. But b b -w'-b 1 ~ x < 2 and 0 ~ y ~ ""2 · Hence, J' xy ~ 3" · Equality is obtained if x = 1/ = ~ • Thus, the volume of the pyramid QKLF is the great-b 4 .. /2 est for a=n/3. Here, I KQ I = I QL I =3=3 V 3-Further, for a=n/3, N is the midpoint of QM (Fig. 52, b). Drawing QT parallel to FB, we get I BT 1 = I MB I-Thus, IAFI IABI 3 2 IFQI=IBTI 2' IQFI=TIAQI. 1' Problems in Solid Coometry The volume of the pyramid APQR is found readily, it is equal to a3 5ra . Three pyramids equal to the pyra-mid QFKL are cut off the pyramid APQR. 1 1 2 2 The volume of each of them amounts to 3"' "3. "5 =45 the volume of the pyramid A PQ R. Thus, for a = n/3 we get the "remainder" of the pyramid APQR, that is, a polyhedron having the volume as va (1-..!..) .13a3 va 54 15 -. 810 • Reasoning exac tly in the same manner, we get that for a=n/3 from the pyramid BCDE there will remain a polyhedron of the smallest volume~ and the volume of this polyhedron will be 11a;ta Adding the obtained volumes, we get the answer: a3 va 20-256. Setting I BD I = 2x, it is easy to find x I 1-2x' I V 3 '-4,x2 V= V ABeD = 6 (i-x') . Making the substitution u = 1 -x2, and then w = 4u + 1/u, we get (6V)2= x' (1-2xl )' (3-4x2l) (1-x2)' (1-u) (2u-1)21 (4u-1) u' =(5-~-4u) (4u++-4) =(5-w) (w-4)= -w2l+9w-20. Answers, Hints, Solutions t 97 The greatest value is attained for w = 9/2, wheJh'c -.,-.. / 9±Y17 x = ... 1-u = V 1 -16 . 1 Answer; the greatest value of V ABeD equals 12 • 257. Let x denote the radius of the ball, V (x) the sum of the volume of the part of the ball situated outside the tetrahedron and the part of the tetrahedron outside the ball. It is easy to see that V' (x) = Sl (x) -S (x), where 81 (x) is the surface area of the part of the balf out-side the tetrahedron, S2 (x) is the surface area of the part of the ball enclosed inside the tetrahedron. Minimum is 1 .. /2 reached for Sl (x) = S2 (x), whence x = a"3 V 3' 258. Let a, b, c be the sides of the base, p = a+~+ c , r the radius of the inscribed circle, x, y, z the distances from the foot of the altitude of the pyramid to the sides a, b, c, and h the altitude of the pyramid. Then Slat=..!..a Yh2+X2 +...!.. b Yh2+y2 -+1- c Yh2+:2 • . , 2 2 2 Note that the function I (x) = Yh2 + x2 is concave (convex downward). And for such functions the following inequality is valid: all (Xl) + a21 (X2) + . . . + anl(xn) > I (a1x1 + a2x2 + ... + anxn), at > 0, i = 1, 2, ... , n, a1 + a2 + ... + an = 1 Let us take advantage of this inequality. We get Slat= P (..!:.. Yh2+X2 +.!!.. Yh2+ y2 + ~ Yh2+:2) 2p 2p 2p ,. ;-h2+ (..!:.. x+~ y+.!:.. :)2 JI 2p 2p 2p -----::-::---.. /h2 ' S~ase _ -"h2+ 2 V --r 4p2 -p ... r , =p which was required to be proved. {98 Problems in Solid Geometry 259. If 0 is the centre of the circle, L is the projection of N on the plane of the base, then the point M must lie on the line se~ent LO since M is a point of the circle nearest to N. On the other hand, since N is a point of the diagonal of the face nearest to M, M N is perpendicular to this diagonal, and, hence, KN is also perpendicular to this diagonal, where K is the projection of M on the face containing this diagonal (Fig. 53). Fig. 53 Let I AL I = ax, AN K is an isosceles right triangle, consequently, I LK I == I AL I == ax, I LK I ax I MK I = I OD I I LD 1= f-2x ' I KD I =T (f-4x). Writing the Pythagorean theorem for 6:. MOE (ME is parallel to AD), we get the following equations for x: (f-4x)2+ (~ x )' _ 25 4 2 f-2x -f44 ¢:> [6 (f-4z) (f-2z}1'+ [6 (f-4x)]'+ [5 (f-2x)1'. Answers, Hints, Solutions f99 Making the substitution 5' = 3' + 4' in the right-hand side and transposing it to the left, we get [6 (f -4x) (f -2x)]' -[3 (f -2X)]2 + [6 (f -4X))2 -[4 (f -2X)]2 = 0 ¢:> 9 (f -2x)' (f -Sx) (3 -Sx) + 4 (5-f6x) (f -Sx) = 0 <::=> (f -Sx) [9 (f -2X)2 (3 -Sx) + 4 (5 -f6x)) = O. It is easy to see that the point K must lie to the left of the point D, that is, 0 < x < f/4, hence, the expres-sion in the square brackets is not equal to zero, x = f/S. Answer. a V 34 . 24· 260. (a) Let I SC I = d; a, b, and c the sides of the triangle ABC, ha' hb' he the altitudes of the triangle ABC, and s its area. Then Sl" n ,., = ha "R. hb " he v..,i ' slnp=,i ' slny=V . r~+~ r~+~ ~+~ Thus, we get for d the equation V~+b2 Vd'+a2 Vd'+h~ ha + hb = f + he . Multiplying this equation by 2s, we get a V ~+b'+b V ~+a2=2s+ V c2 d2 +4s2 • (f) Multiplying and dividing both sides of (f) by the differ-ences of the corresponding quantities (assuming that A A A =1= B), we get a2_b2 c' a V d2+b2-b V d2+a2 -Vc2~+4s'-2s ' whence ac' V ~+bi-bc' V d2+a2=(a2-b') (V c2 d2+4s'-2s) (2) 200 Problems in Solid Geometry Multiplying (t) by bl -al! and adding the result to (2), we obtain a (b2+c2-al ) Va2+b2+b (b2-al -cl ) V d2+a2 =4a (b2-al ). With the aid of the theorems of cosines and sines, the last equation is transformed as follows (3) Transform the right-hand member of Equation (3) as follows: b2-al 2R = 2R (sinl! B -sin2 A) = 2R sin (A+B) sin (B-A), now, multiplying both sides of (3) by cos A • V d' + b l + cos B . V d' + a2 , we get the equation (cos2 A -cos2 B) d' + b2 cos2 A -al! cos2 B = 2R sin (A + B) sin (B -A) X (cos A • V d' + b 2 + cos B • V d' + a2). (4) In Equation (4) we see cosl! A - cos2 B = sin (A + B) sin (B -A), b2 cosl! A -a2 cosl! B = 4R2 sin (B + A) sin (B -A). Consequently, after reduc-tion, Equation (4) is transformed to Adding (3) and (4'), we get ,i . d2 2 cos A· r d'+b2 = 2R + 2R (sinl! B + cos2 A), Answers, Hints, Solutions whence (v' d2+bl -2R cos A)2=0, d' = 4R2 (cos2 A -sin2 B) = 4R2 cos (A + B) cos (A -B). Thus, I SC I = 2R V cos (A + B) cos (A- B). 20f The problem has a solution if A + B < 90°, that Is, in the triangle ABC the angle C is obtuse. (b) Let us take advantage of the notation used in I tem (a). Then our inequality is rewritten in the form If the angle C is acute, then the right-hand Side, as it follows from Item (a), is never equal to f, consequently, the inequality takes place, since it is fulfilled for d = O. And if C is an obtuse angle (or it is equal to 90°), then the right-hand side is equal to f for the unique value of d (if C is a right angle, then d = 0). But for d = 0 and sufficiently large values of d the inequality is obvioue (for large d's it follows from the triangle inequality), conseguently, if for some value of d the left-hand side were less than unity, then the left-hand side would take on the value equal to unity for two different values of d. 261. Let ABCD be the given tetrahedron. On the edges BC and BD take points M and N and solve the following ~roblem: for what position of the points M and N does the radius of the smallest circle enclosing the triangle AM N (we consider the circles lying in the plane AM N) reach the least value? (Obviously, the radius of the smallest hole cannot be less than this radius. For this purpose, it suffices to consider the instant of passing of the tetrahedron through the hole when two vertices of the tetrahedron are found on one side of the plane of the hole, the third vertex on the other Side, and tne fourth in the plane of the hole.) Suppose that the points M and N correspond to the desired triangle. Suppose that this triangfe is acute. 202 Problems in Solid Geometry Then the smallest circle containing this triangle coincides with the circumscribed circle. Circumscribe a circle about the triangle AM N and consider the solid obtained by revolving the arc AMN of this circle about the chord AN. The straight line Be must be tangent to the surface of this solid. Otherwise, on Be we could take a point Ml such that the radius of the circle circumscribed about the trian~le AM IN would be less than the radius of the circle cIrcumscribed about the triangle AMN. The more so, Be must be tangent to the surface of the sphere passi g through A, M, and N having the centre in the plane AMN. The straight line BD must also touch this sphere exactly in the same manner. Consequently, BM 1 = 1 BN I. Set 1 BM 1 = 1 BN 1 = x. Let K denote the midpoint of MN, L the projection of B on the plane A MN (L lies on the extension of A K). The fore-going implies that LM and LN are tangents to the circle circumscribed about the triangle AM N. This triangle is isosceles, 1 AMI = 1 AN 1 = V x2 -X + 1, 1 M N 1 = A x. If MAN = ct., then xl -2z+2 . x V3xt -4x-t-4 cosa=2 (xl -x+1) , SIn a= 2 (x2-x+1) , Xl V3x2-4x+4 I LK I = I MK I tan a= 2 (x2 -2.1:+2) • Consider the triangle AKB, AK'B = ~ > 180°; &-2 cos ~= , I LK I = -1 KB 1 cos ~ = V3 (3x8-4x+4) ; (2-3x) • Equating two expressions for I LK 1, 2 3.1:2-4x+4 we get for x, after simplifications, the equation 3x2 -6x2 + 7x -2 = o. (i) The radius of the circle circumscribed about the triangle AMN, will be x2-x+1 R= V3z2-4z+4 · Answers, Hints, Solutions 203 (It is possible to show that if AM N is a right triangle, then its hypotenuse is not less than V f5 -fa -y2 > 0.9.) Let us show that our tetrahedron can go through the hole of the found radius. On the edges CB and CA mark points Land P such that 1 CL 1 = 1 CP 1 = 1 BM 1 = 1 BN 1 = x, where x satisfies the equation (f). Place the tetrahedron on the plane containing the given hole so that M and N are found on the boundary of the hole. We will rotate the tetrahedron about the line MN until the edge AB, passing the hole, becomes parallel to our plane. Then, retaining AB parallel to this plane, we displace the tetrahedron ABCD so that the points P and L get on the boundary of the hole. And, finally, we shall rotate the tetrahedron about PL until the edge DC goes out from the hole. (The tetrahedron will turn out to be situated on the other side of our plane, the face ABC lying in this plane.) A nswer: the radius of the smallest hole R = x2 -X + f h . h f h . -y , w ere x IS t e root 0 t e equation 3xS-4x+4 3x3 -6x2 + 7x -2 = O. The relevant computations yield the following approximate values: x ~ 0.39f3, R ~ 0.4478 with an error not exceeding 0.00005. Section 4 262. Let S denote the vertex of the angle. Take points A, B, and C on the edges such that 1 SA 1 = 1 SB i = 1 SC I. The bisectors of the angles ASB and BSC pass through the midpoints of the line segments AB and BC, while the bisector of the angle adjacent to the angle CSA is rarallel to CA. -f -f 264. , if --- is not a whole number, 2 • a 2' a SIn~ SIn ~ 1 =f, if -f is a whole number, where [%1 2' a 2. tx 1m 2 SlOT is an integral part of z. 204 Problems in Solid Geometry 265. We shall regard the given lines as the coordinate axes. Let the straight line make angles a, ~, and y with --+ ~ these axes. Then the projections of the vectors OA 1, OBI' and Oc1 on the axes OA, OB, and OC will be respectively equal to a cos 2a, a cos 2~, and a cos 2y, a = 1 OA I. Con-sequently, the point M of intersection of the planes pass-ing through Al.' B I , and CI respectively perpendicular to OA, OB, and OC will have the coordinates (a cos 2a, a cos 2~, and a cos 2y). The set of points with the coor-dinates (cos2 a, cos'~, and cos2 y) is a triangle with vertices at the end points of the unit vectors of the axes. Consequently, the sought-for locus of points is also a triangle whose vertices have the coordinates (-a, -a, a); (-a, a, -a); (a, -a, -a). 266. Denote the gJven lines by It and l,. Through II pass a ~lane PI parallel to la, and through l2 a plane P2 parallel to '1' I t is obvious that the midpoints of the line segments with the end points on II and l2 belong to the pI.a:ne P parallel to PI and P2 and equidistant from E and P2' (It is possible to show that if we consider ad kinds of such line segments, then their midpoints will entirely fill up the plane p.) Project now these line seg-ments on the plane p parallel to the given plane. Now, their end points will be on two straight lines which are the projections of the lines II and l2.1 and the projections themselves will turn out to be parallel to the given line of the plane p representing the line of intersection of the plane P and the given plane. Hence it follows that the required locus of I'oints is a straight line. 267. (a) The whole space. (b) Proceeding exactly in the same way as in Prob-lem 266, we can prove that the locus of points dividing in a given ratio all possible line segments parallel to the given plane with the end points on the given skew lines is a straight line. ApplylOg this statement twice (first, find the locus of midpoints of sides AB, and then the locus of centres of gravity of triangles ABC), prove that in this case the locus of centres of gravity of triangles ABC is a straight line. 268. Through the common perpendicular to the straight lines, pass a plane p perpendIcular to '3' Let the line N M intersect l3 at point L; N1 , Mu Ll be the re-spective points of intersection of the lines ll' l2' l3 with Answers, Hints, Solutions 205 the common perpendicular, N 2' M 2 the projections 0 r N and M on the passed plane, ct and ~ the angles made by the lines 11 and 12 wlth this plane, K the midpoint of Nt Fig. 54 N M, K],. and K2 the projections of K on the common per-pendicular and on the plane p (Fig. 54). We have 1 KK2 I 1 NN2 1 + 1 MM2 1 1 KlK21 1 N2Nl 1 +1 M2Mli . 1 N2Nl 1 tan ct+ 1 M2Ml 1 tan ~ 1 NlNl 1 + 1 M2Ml ! __ 1 NlLl 1 tanct+ 1 MILl 1 tan~ -t -1 MILl 1 +1 MILl 1 -cons, hence, the point K describes a straight line. 269. Let us introduce a rectangular coordinate system, choOSing the origin at the point A. Let el (tit, bl , Cl)' e~ (a2, b2, C2)' •.. , en (an' bn, cn) be unit vectors J)ar-allel to the given lines, e (x, y, z) a unit vector parallel to the line satisfying the conditions of the problem. Thus, we get for e the following equation I alx + bly + ClZ I + I azx + b2y + CZZ I -t- ••• + I anx + bny + cnz I = const. 206 Prob1ems in Solid Geometry I t is now easily seen tha t the locus of termini of the vector e will be the set of circles or parts thereof situated on the surface of the unit sphere with centre at A. 270. Place equal loads at the points A, B, C, AI' BI , and C. Then the centre of gravity of the obtained system of loads will coincide with the centre of gravity of the triangle with vertices at the midpoints of the line segments AA I , BBI , CCI -On the other hand, the centre of gravity of this system coincides with the midpoint of the line segment GH, where G is the centre of gravity of the triangle ABC, H the centre of gravity of the three loads found at AH BI , and CI • With a change in AI' B I , and CI the point H mov~s in the straight line I, and the point G remains fixed. Hence, the point M, which is the midpoint of GH, will describe a straight line parallel to I. 271.. Through A draw a straight line t parallel to I. The sought-for locus of points represents a cylindrical surface, except for I and t, in which I and t are diametri-cally opposite elements. 272. Let us first prove that if the line M K is tangent to the sphere ~, then it is also tangent to the sphere ct. cv (<<, (0) Fig. 55 Consider the section of the given spheres by the plane passing throu~h points M, K, A, B, and N (Fig. 55). The ana-Ie M KB II measured by half the arc KB enclosed Answers, Hints, Solutions / '. /'. inside this angle, consequently, M KB = BAN, since the angle measures of the arcs KB and BN are equal (we take the arcs situated on different sides of the line KN if the tangency is external (Fig. 55, a) and situated on one side if the tangency is internal (Fig. 55, b)). Hence / '. /'. /'. /'. it follows that AMK = ABN or AMK = 1800 -ABN, /'. -and this means that AM K is measured by half AM, since the corresponding arcs AM and AN have the same angle measure, that is, M K touches the circle along which the considered section cuts the sphere ct. It is now possible to prove that the locus or points M is a circle. 273. Let A and B denote the given points, C the point of intersection of the line AlJ with the given plane, M the point of tangency of a baH with the plane. Since I CM 12 = I CA 1·1 ClJ I, M lies on the circle with centre at the point C and radius Y I CA 1'1 ClJ I. Con-sequently, the centre of the sphere belongs to the lateral surface of the right cylinder whose base is this circle. On the other hand, the centre of the sphere belongs to the plane passing through the midpoint of A B perpendicular to AB. Thus, the sought-for locus of points IS the line of intersection of the lateral surface of a cylinder and a plane (this line is called the eUipse). 274. ))enote by Olt O~ and Rlt R, the centres and radii of the given spheres, respectively; M is the midpoint of a common tangent. Then, it is easy to see that 101M 12 -I O~M 12 = R~ - Ri, and, consequently, M lies in the plane perpendicular to the line segment 0 10. and cutting this segment at a po in t N such that lOIN 12 -I o.,.N 12 = Rf - Ri. Let us see what is the range of variation of the quan-tity I N M I. Let I 010~ I = a and Rl > R.,., then lOIN 1= ~ (Rr:R~ +a) . 208 Problems in Solid Geometry If 2x is the length of the common tan~nt, whose mid-point is M, then 1 MN 12= 101M 12_1 OIN 1 2=x2+Rf ! (R~ -: Ri + a ) • Now, if a ~ Rl + R 2, then the quantity 4x2 changes within the interval from a2 -(R1 + R2)2 to a2 -(R 1 -R9.)2, and, hence, in this case the locus of points M wul be an annulus whose plane is perpendicular to 0 10 2 , and the centro is found at the point N, the inner radius is equal to 1- (R -R) .. ;-1- (R1 +R2)2 2 1 2 V a2 ' and the outer to 1-(R +R) 1,/' 1- (R1-R2 )" 2 1 2 Y a2· And if a < RI + R , that is, the spheres intersect, then the inner radius of t~e annulus will be equal to the radius of the circle of their intersection, that is, it will be 1 2a V(a+R1+R2) (a+R1-R2) (a+R2-R1) (R1+R2-a). 275. Denote by A and B the points of tangency of the lines 11 and '2 with the sphere, and by K the point of tan-gency of the line MN with the sphere. We will have I AM I = I MK I, I BN I = I NK I. Project II and ls on the plane perpendicular to AB. Let AIt MIt NIt and K1 denote the respective projections of tlie points A (and also B), M, N, and K. ObViously, i A1M1 I I A1N1 1 1 AM 1 =p, 1 BN 1 =q, where p and q are constants. Let now d and h be the dis-tances from K1 to the straight lines A1M1 and A1N1. Answers, Hints, Solutions We have i d """'2' i AIMII d T= i """2 1 AINI 1 h I A1N I ! S A1M1Kl \ AIMI \ -- S A1N1Kl 1 MIKI 1 1 AINI I 1 MK 1 -• ~-=---=-=--7'"" -1 NIKI 1 I AIMI liNK 1 I AM 1 1 Al NIl q -. ---I AIMI t 1 BN 1 -p • 209 Thus, the ratio of the distances from the point KI to two given straight lines in the plane is constant. This means that the point KI belongs to one of the two straight lines passing through the point Al: And the sought-for locus of points represents two circles on the surface of the given sphere. These circles are obtained when the sphere is cut by two planes passing through the lines de-scribed by the point KJ. and the straight line AB. The points A and B themselves are excluded. 279. Let BK denote the altitude of the triangle ABC, H the point of intersection of the altitudes of the tri-angle ABC, BM the altitude of the triangle DBC, N the point of intersection of the altitudes in the triangle DBC. Prove that N is the projection of the point H on the plane DBC. Indeed, KM is perpendicular to DC, since BM is perpendicular to DC, and KM is the projection of BM on the plane ADC. Thus, the plane KMB is pe~endicular to the ed~ DC, consequently, UN is perpendicular to DC. Exactly In the same way, HN is perpendicular to the edge DB. Hence, HN is perpendicular to the plane DBC. I t is not difficult to prove now, that N lies in the plane passing through AD perpendicular to BC. The required locus of points refresents a circle with diameter H L, where L is the foot 0 the altitude dropped from A on BC whose plane is perpendicular to the plane ABC. 283. Denote by P and Q the points of intersection of the opposite sides of the quadrilateral ABCD. If the section by the plane of the lateral surface of the pyra-mid ABCDM is a parallelogram, then the plane of the 1,-0"9 210 Problems in Solid Geometry section must be parallel to the plane PQM, the sides of the parallelogram being parallel to the straight lines PM and QM. Hence, in order for a section to 00 a rectangle, the angle P MQ must be equal to 90°, that is, M lies on the surface of the sphere with diameter PQ. (Thus, Item (a) has been solved.) (b) Denote by K and L the points of intersection of the diagonals of the quadrilateral ABCD and the straight line PQ. Since the diagonals of the parallelogram obtained by cutting the lateral surface of the pyramid ABCD M by a plane will be parallel to the lines M K and ML, /"'" this parallelogram will be a rhombus if KM L = 90°, that is, M lies on the surface of a sphere with diameter KL. (c) Items (a) and (b) imply that the locus of points M will be a circle which is the intersection of two spheres of diameters PQ and KL. (d) The locus of points is a conical surface with vertex at the point of intersection of the diagonals of the quadri-lateral ABCD whose directing curve is a circle from the preceding item. 284. If K and L are the midpoints of BC and AM, ° the centre of the sphere circumscribed about ABCM, then, since G is the midpoint of LK and OG is perpen-dicular to LK, I OL 1 = I OK I. Hence it follows that I A M I = I BC I, that is, M lies on the surface of the sphere of radius BC centred at A. Let, further, N be the centre of gravity of the tri-angle ABC, 0 1 the centre of the circle circumscribed about the triangle ABC, G 1 the projection of G on the j)lane ABC. Since, by the hypothesis, OG is perpendicular to AK,01G1 isalsoperpendiculartoAK.Hence, G lies in the plane passing through 0 1 and perpendicular to A K. Hence, since INGI =i-INMI, it follows that the point M also lies in the plane per .. pendicular to A K. Thus, the sought-for locus of points represents the line of intersection of a sphere and a plane, that is, generally speaking, is a circle. 285. Introduce a rectangular Cartesian coordinate system taking for ° the vertex of the trihedral an~le and Answers, Hints, Solutions 211 directing the axes along the edges of this angle. Let the plane of the circle make angles a, ~, and y with the coor-dinate planes XOY, YOZ, and ZOX, resrectively. Then the point 0 1 (the centre of the circle) wi! have the coor-dinates (R sin ~, R sin y, R sin a), where R is the radius of the circle. From the origin draw a straight line per-pendicular to the plane of the circle. This line will make angles ~, y, and a with the coordinate axes. Consequently, cosl! a + cos2 ~ + cos2 Y = 1 and, hence t Thus, the point 0 1 lies on the surface of the sphere with centre at 0 and radius R Y2. On the other hand, the distance from 0 1 to the coordinate planes does not ex-ceed R. Consequently, the sought-for set represents a spherical triangle tiounded by the planes x = R, y = R, , = R on the surface of the sphere I 001 I = R Y2, situated in the first octant. 286. Let the spider be found in the vertex A of the cube ABCDAlBlClDl . Consider the triangle DCCI • It is rather easy to prove that the shortest path joining A to any point inside the triangle DCCI intersects the edge DC. In this case, if the faces ABCD and DCClDl are "developed" so as to get a rectangle made from two squares ABCD and DCClDu then the shortest path will repre-sent a segment of a straight line. Consequently, the arc of a circle with radius of 2 cm whose centre is found at the point A of the development si tua ted inside the tri-angle DCCI will be part of the boundary of the sought-for locus of points. The entire boundary consists of six such arcs and separates the surface of the cube into two parts. The part which contains the vertex A together with the boundary is just the required locus of points. 287. We take the edges of the trihedral angle for the coordinate axes. Let (x, y, ,) be the coordinates of the ~ vector OA, (Xi' Yi' 'i) the coordinates of the ith section U· 212 Problems in Solid Geometry of the polygonal line. Each section of the polygonal line is regarded as a vector. Then z = ~ Xi' H = ~ Hi' % = ~ zit here, the conditions of thelroblem imply that all the Xi are different from zero an have a si~ coinciding witli that of X (the same is true for Yi and Ii). Obviously, 1 OA 1 < a. On the other hand, 1 X 1 + 1 H 1 + 1 % 1 = ~ (I Xi 1 + 1 Yi 1 + 1 %i I) :;;:, ~ Ii = a (Ii is the length of the ith section of the polygonallin&). It can be readily shown that any point A satisfying the conditions 1 OA 1 ~ a, 1 z 1 + 1 H 1 + 1 % 1 > at where :IJ, H, % are the coord ina tes of the point A, can be the end point of a polygonal line consisting of not more than three sections and satisfying the conditions of the problem. Let, for instance, Ml and M, be two points lying on one straight line emanating from the point 0 such that 1 Xl 1 + 1 111 1 + 1 -"1 1 = a, XlYl%l =1= 0 (Xl' HIt %1 the coordinates of the point M l ), 10M. l = a. Con-sider the polygonal line with vertices (0, 0, 0), (Xl' 0, O)t (Xl' Yu 0), (Xl' y)t st). The length of this polygonal line is equal to a. '~tretching' this line, we get all points of the line segment MlM, (excluding MIl. Thus t the desired locus of points consists of all points ying outside the octahedron 1 z 1 + 1 11 1 + 1 % 1 = a and inside or on the surface of the sphere with centre 0 and radius a. In this case, the points situated in the coordinate planes are excluded. 288. First of all note that if r is the radius of the ball inscribed in ABCD, then, firstly, all the edges of the tetrahedron ABCD are longer than 2r and, secondly, the radius of the circle inscribed in any face of the tetra-hedron is greater than r. The first assertion is obvious. To prove the second assertion t through the centre of the inscribed ball, passa plane parallel, say, to the face ABC. The section cut is a triangle AlBICl similar to the tri-angle ABC with the ratio of simi itude less than unity and containing inside itself a circle of radius r. (t) The condition determining the set of points A will be expressed by the inequality lOA 1 > 3r, the Answers, Hints, Solutions 213 equality I OA I = 3r being true for a regular tetrahedron. If for some point A the inequality I OA I < 3r were fulfilled, then the radius of the smallest ball containing the tetrahedron ABCD would be less than 3r, which is impossible (see Problem 246). (2) The condition determining the set of points B will be expressed by the inequality I OB I > r Y5. In-deed, if for some point B the inequality I OB I < r VS were fulfilled, then for the triangle DBC the radius of the circle containing this triangle would be not greater than Y5r2 -r2 = 2r, that is, the radius of the circle inscribed in the triangle DBC would not exceed r, which is impossible. (3) The condition determining the set of points C is expressed by the inequality I OC I > r vi. Indeed, if I oc I ~ r Y2, then I CD I ~ 2r. (4) The condition determining the set of points D will be expressed by the inequality I OD I > r. Let us show that I OD I can be arbitrarily large. To this effect, for the tetrahedron ABCD take a tetrahedron all faces of which are congruent isosceles triangles having sufficiently small vertex angles. Then the centres of the inscribed and circumscribed balls will coincide, and the ratio !!:.... , where R is the radius of tho circumscribed ball, r can be arbitrarily large. 289. If MC is the hypotenuse of the appropriate tri-angle, then the equality I MC 12 = I M A 12 + I MB 12 must be fulfilled. Introducing a rectangular Cartesian coordinate system, it is easy to make sure that the point M must describe the surface of a sphere. Find the centre and radius of this sphere. Let Cl be the midpoint of AB, C2 lie on the extension of CClt I ClC~ I = I CCI I (A CBC2 is a parallelogram). Denote the sides of the triangle ABC, as usualt by a, b, and c, the median to the side AB by me' We snaIl have I MA 1 1+ I MB 1 21=21 MCl 12+ I A~ II - 2IMClr2+~I. Since I MA 12 + I MB 12 = I Me 12, 214 Problems in Solid Geometry we get 2 1 Me 12_21 MCl 12= -To (1) ./'-.. Let MC2C = cp, write for the triangles MC2C and MC2Cl the theorem of cosines: I MC 12 = I MC2 I' + 4m~ -4 I MC2 I me cos cp, (2) I MCl 12 = I MC2 12 + ~ - 2 I MC. I me cos cp. (3) Multiplying (3) by 2 and subtracting the result from (2), we get (tai{ing into account (1)) c2 I MC 2 12 = 2m~ - 2" = a2 + b2 -c2 • Thus, for this case the set of l'0ints M will be non-empty if a2 + b2 -c2 > 0, that IS, the angle C in the trian~le ABC is not obtuse. Consequently, the whole set of pOInts M for an acute-angled triangle consists of three spJieres whose centres are found at the points C , A2 and B2 such that CACJ], ABA 2C, BCBiA are para~lelo-grams, the radii being respectively equal to Vai + b2 _c2 , Vb2 + c2 -a2, and Va2 + c2-b2• For the right-angled triangle ABC the sought-for set consists of two spheres and a point, and for an obtuse-angled triangle of two spheres. 290. Let 0 denote the centre of the Earth, A the point on the equator corresponding to zero meridian, M the poin t on the surface of the Earth with longitude and la ti-tude equal to cp, N the projection of M on the plane of the equator. IntrodUcing a rectangular Cartesian coordi-nate system in the plane of the equator, taking the line OA for the x-axis, and the origin at the point 0, we get tha t N has the following coordinates: x = R cos2 g>, Y = R cos cp sin cp, where R is the radius of the Earth. It is easy to check that the coordinates of the point N satisfy the equation ( R )2 R2 X- 2 +y2=T' I.e. the sought-for set is a circle with centre ( ~ , 0) and radius R /2. Answers, Hints, Solutions 215 291. Introduce the following notation: S is the vertex of the cone, N the projection of the point M on the plane passing through the points S and A parallel to the base of the cone, P a point on the straight line SN such that /" S M P = 900 (Fig. 56), M P is a normal to the surface -----9~------ A Fig. 56 of the cone. It follows from the hypothesis that A P is /" /" parallel to the renected ray. Hence AMP = MPA, 1 A M I = 1 A P I. Let CG be the angle between the altitude and generatrix of the cone 1 SA I = a. The plane passing through M parallel to the plane SPA cuts the axis of the cone at the point SIt Al is the projection of A on this plane, ~ I SSI 1 = X, MSIA I = cP, 1 MAli = y. By the theorem of cosines for the triangle SIMA It we have y2 = x2 tan2 CG + a2 -2ax tan CG cos cp. Besides, 1 PA 12 = 1 M A 12 = y2 + x2, 1 sP 1 = 1 ~M 1 x sIn CG cos CG sin CG sin 2a • (1) (2) (i) 216 Problems in Solid Geometry Writing the theorem of cosines for the triangle SPA and using the above relationships, we have 4z' 4ax x' tan' a-2ax tan a cos cp+x2= sin2 2a sin 2a cos cp, whence x = a sin 2a cos cp. If now we erect a perpendicular to SN at the point N in the plane SPA and denote by L the point of its inter-section with SA, then I SL 1= I SN 1= xtana 2asin'a. cos cp cos cp Thus, I SL 1 is constant, consequently, the point N describes a circle with diameter SL. 292. When solving this problem, we shall need the following statements from plane geometry. If in a circle of radius R through a point P found at a distance d from its centre two mutually perpendicular chords AD and BE are drawn, then (a) I AD 12 + I BE 12 = 8R2 -4cl', (b) the perpendicular dropped from P on AB bisects the chord DE. For a three-dimensional case, these two statements are generalized in the following way. If through a point P found inside a ball of radius R centred at 0 three mutually perpendicular chords AD, BE, and CF are drawn at a distance d from its centre, then (a) I AD 12 + I BE 12 + I CF 12 = 12R2 -8ell, (b) a straight line passing through P perpendicular to the plane ABC passes through the median point of the triangle DEF. Let usprove Item (a). Let Ru R 2, R3 denote the radii of the circles circumscribed respectively about the quadri-laterals ABDE, ACDF, and BCEF, du d2, dt the distances in these quadrilaterals from the centres 0 the cireumw scribed circles to the point P, and x, y, :I the respective distances from the point 0 to the planes of these quadri-laterals. Then ,xl + y2 + Z2 = rP, d¥ + ~ + d§ = 2 (xli + y2 + r) = 2rP, Ri + R~ + R~ = 3R2 -rP. Answers, Hints, Solutions 217 Thus, taking advantage of the statement of Item (a), we get I AD !2+ I BE 12+ I CF 12=+ [( I AD 12+ I BE Ii) +( I BE 1 2+ I CF 12 )+( I CF li+ I AD li)1 = ; (8Ri-4di+8Rf-4d~+8R!-4d:) = 12RI-8d2• To prove I tern (b), project the drawn line on the planes of the quadrilaterals ABDE, ACDF, and BCEF, and then take advantage of Item (b). Now, let us pass to the statement of our problem. On the line segments PA, PB, and PC construct a parallel-epiped and denote by M the vertex of this parallelepiped opposite the f,0int P. Analogous y, determine the point N for the line seg-ments PD, PE, and PF. K is the point of intersection of PM with the plane ABC, 0 the midpoint of PM, T the midpoint of PN, 0 1 the centre of tlie circle circum-scribed about the triangle ABC, and H the foot of the perpendicular dropped from P on ABC. It follows from Item (b) that H lies on the straight line N P. Further, K is the point of intersection of the medians of the triangle ABC, 1 P K I = ~ I PM I. The straight line 00 is perpendicular to the plane ABC and passes through the point 0 1 , since 0 and 0 are the centres of two spheres passing through the points A, B, and C. (Note that we have proved simultaneously that the points 01, K, andHarecollinearand 1 KH 1 = 21 01K I. As is known, this straight line is called the Euler line.) Thus, 00 is parallel to NP, the same as TO is parallel to MP. Hence, 0 is the midpoint of NM. On the line segment OP take a point S such that 1 PS I = ! I PO I. The perpendicular dropped from S on KH passes through the midpoint of KH. Consequently, 1 SK 1 = 1 SH I. But SK 110M, 1 1 I SK I ="31 OM 1=6 1 NMI. 218 Problems in Solid Geometry It follows from Item (a) that ! N AI 12 == 12R2- 8d2 (N M is the diagonal of the parallelepiped whose edges are equal to I AD I, I BE I, I CF I), that is I SK I = ~ V3R2 U2 is a quantity independent of the way in which the line segments PA, PB, PC were drawn. 293. Denote by a, b, and c the unit vectors directed • --+-along the edges of the trIhedral angle, let, further, ON = --+--+-e, P the centre of the sphere, OP = u, OA = xa, --+---+-OB = yb, OC = %c. The points 0, N, A, B, and C belong to one and the same sphere with centre at P. Thi.s means that (u -e)2 = u2, (xa -U)2 = u2, (yb -U)2 = u 2, (zc -U)2 = u 2, whence e2-2eu = 0, x-2au=O. y-2bu=O, %-2cu=0. (1) Let e = aa + ~b + yc. Multiplying the second, third, and fourth equations of System (1) respectively by a, ~, and 'V and subtracting from the first, we obtain e2 -ax -~y -'V% = O. (2) If M is the centre of gravity of the triangle ABC, then ~ 1 -+ --}to---+-1 OM=s(OA+OB+OC)=3 (xa+yb+zc). Taking into consideration Equation (2), we may conclude that the locus of points M is a plane. 294. Prove that each of these planes passes through the point symmetric to the point N with respect to the centre of gravity of the tetrahedron. 295. Prove that all these flanes pass through the point 'ijymmetric to the centre 0 the sphere circumscribed Answers, Hints, Solutions 219 about the tetrahedron with respect to its centre of gravity. 296. When solving Problem 295, we proved that Monge's point is symmetric to the centre of the sphere circumscribed about the tetrahedron with respect to the centre of gravity of the tetrahedron. Consequently, if Monge's point belongs to the plane of some face of the tetrahedron, then the centre of the circumscribed sphere is situated from this face at a distance equal to half the length of the corresponding altitude and is located on the same side of the face on which the tetrahedron itself lies. This readily leads to the statement of our problem. 297. Take advantage of the equality 1 M A 12 + 1 M B 12 = 4 I M D 1 ~ + 1 AB 12 , where D is the midpoint of AB, and also by the fact that in an arbitrary tetrahedron the Sum of the squares of its opposite edges is equal to twice the sum of the squares of the distances between the midpoints of two pairs of its remaining edges (see Problem 21). 298. Denote the areas of the faces of the tetrahedron by S1' S'4' S3, S4 and the volume of the tetrahedron by V. If r is the radius of the sphere touching all the planes forming the tetrahedron, then, with the signs of 8\, = +1, i = 1, 2, 3, 4, properly chosen, the equa ity r (81S 1 + 82S 2 + 83S 3 + 84S 4)3 = V must be fulfilled. I n this caSe if for a given set 8i the value of r determined by the last equality is positive, then the corresponding ball exists. Thus, in an arbitrary tetrahedron there always exists one inscribed ball (8i = +1) and four externally inscribed balls (one 8i = -1, the remaining ones +1), that is, four such balls each of which has the centre outside the tetrahedron and touches one of its faces at an interior point of this face. Further, obviously, if for some choice of 8i there exists a ball, then for an opposite set ei there exists no ball. This means that there are at most eight balls. There will be exactly eight balls if the sum of the areas of any two faces is not equal to the sum of the areas of two others. 220 Problems in Solid Geometry 299. For any two neighbouring sides of the quadri-lateral there are two planes equidistant from them (the bisector planes of the angle of the quadrilateral itself and the angle adjacent to it). In this case, if three such planes corresponding to three vertices of the quadrilateral inter-sect at a certain point, then through this point there passes one of the two bisector planes of the fourth veI1ex. Thus, when finding the points equidistant from the lines forming the quadrilateral, it suffices to consider the bi-sector planes of three angles of this quadrilateral. Since two planes correspond to each vertex, there will be, generally speaking, eiEht points of intersection. It remains to find out under what conditions some three such planes do not intersect. Since our quadrilateral is three-dimensional, no two bisector flanes are parallel. Hence, there remains the possibility 0 one bisector plane to be parallel to the line of intersection of two others. And this means that if, through some point in space, three planes are passed \larallel to the given ones, then these three planes will Intersect along a straight line. Let, for the sake of definiteness, the bisector planes of the three interior angles of the quadrilateral ABCD C Fig. 57 not intersect. Through the vertex C, draw straight lines parallel to the sides AB and AD (Fig. 57) and on these lines layoff line segments CP and CQ, I C P I = I CQ I. Layoff equal line segments CM and CN on the sides CB and CD. The aforegoing reasoning imply that the bisector planes of the angles MCP, PCQ, QCN, and NCM inter-sect along a straight line and, hence, all the points of this line are equidistant from the straight lines CP, CQ, Answers, Hints, Solutions 221 CN, CM, that is, the lines CP, CQ, CN, and CM lie on the surface of the cone, and PQNM is an inscribed quadrilateral. Let the plane of the quadrilateral PQNM intersect AB and AD at points Land K. The line LK is paralle to QP, and this means that NMLK is also an inscribed quadrilateral. Besides, it is easily seen that 1 LB 1 = I MB I, 1 KD 1 = I DN I, 1 KA I = I AL I. Hence, in particular, it follows that I AB I + I DC 1 = I AD 1 + 1 BC I· Let now 0 denote the centre of the circle circumscribed abollt the quadrilateral KLMN. The congruence of the triangles LOB and MOB implies that 0 is equidistant from the lines AB and BC. Proceeding in the same way, we will show that 0 is equidistant from all the lines form-ing the quadrilateral ABCD, that is, 0 is the centre of the ball touching the straight lines AB, BC, CD, and DA. Other cases are considered exactly in the same manner to obtain analogous relationships among the sides of ABCD: I AB I + I AD I = I CD I + I CB I, I AB 1 + I BC I = 1 AD I + I DC I. It is not difficult to show that the indicated relationships among the sides of the quadrilateral ABCD are the necessary and sufficient con-ditions for the existence of infinitely many balls touching the sides of the quadrilateral. In all remaining cases there are exactly eight such balls. 300. Using the formula of Problem 11 for the volume of the tetrahedron, prove that each of the relationships under consideration is equal to 4S~i;,S4, where SI' S2' Sa, S4 are the areas of the faces of the tetrahedron, V its volume. 301. If hi (i = 1, 2, 3, 4) is the altitude of the cor-responding face of the tetrahedron, then 1 V 1 4 1 Vi 4 l~-R~ _ _ '" S2 (l2_R2) __ _ '" S2h2 t 1. 3 2 ~ iii -- 3 2 ~ i i h~ . 1 • 1 1. 1= 1.= l~--R~ 1. t M t • 222 Problems in Solid Geometry If now di is the distance from the centre of the circum scribed ball to the ith face (R is the radius of this ball), then l1- R1= (l~-hV-(R2-dV+hi = (R2- (hi-di)2] -(R2_d1)-+ hi= 2hidi' Thus, we get the following radicand: '" ~=1 L.J hi (see Problem 182), which was required to be proved. (We assumed that the centre of the circumscrined ball lies inside the tetrahedron. If the centre is found out-side it, proceed in the same way regarding one of the quantities di as being negative.) 302. Denote the lengths of the edges of the tetrahed-ron ABCD as is shown in Fig. 58, a. Through the vertex A pass a plane tangent to the ball circumscribed about the tetrahedron ABCD. The tetrahedron ABCIDI in this figure is formed by this tangent plane, the planes ABC. ABD, and also by the plane passing through B xarallel to the face ADC. Analogously, the tetrahedron B 2C2D is formed by the same tangent plane, the planes ABD. ADC, and the plane passing through D parallel to ABC. From the similarity of the triangles ABC and ABCI (Fig. 58, b, ACI is a tangent line to the circle circum-/'\ scribed about the triangle ABC, consequently, BACI = / '\ /"-., /'\ BCA, besides, BCI II AC, hence, CIBA= BAC) find ac nc j ACI j = b. Analogously, find I ADt I = in' I AC2 1= mp JAB I-~ B h t' I AC D d AB C b' 2 c' ut t e nang es I I an 2 2 are similar, hence I CIDIl = j ADI I I C D I = pc2 I AC2 I I AB2 j' I I bm • Note that if the lengths of the sides of the triangle are multiplied by bm , then these lengths will turn out c Answers, Hints, Solutions 223 to be numerically equal to the quantities am, bn, and cp, thus c2 S s. AD1Cl = b2m2 Let, further, A M denote the diameter of the circumscribed ball and BK the altitude of the pyramid c, A (a) __ 'n"'I:;::------, C t c A ___ ... o (I) (e) Fig. 58 ABCID} dropped from B on ACIDI (Fig. 58, c). From the similarity of the triangles ABK and 0 LA (0 L is c2 perpendicular to AB) we find I BK 1= 2R' Hence, 224 Problems in Solid Geometry And, finally t V ADICIB S ABCIS ABDI c2 c2 c4 V = S S =-b 2 • -2 ' V AD 1C1B=b2m2 V. ABC ABD m Comparing two expressions for V AD1 C 1 B, we get the truth of the statement in question. Remark. I t follows from our reasoning that the angles of the trianfle the lengths of the sides of which are numer-ically equa to the products of the lengths of the oppo-site edges of the tetrahedron are equal to the angles be-tween the tangents to the circles circumscribed about three faces of the tetrahedron. The tangents are drawn through the vertex common for these faces and are situated in the plane of the appropriate face. It is readily seen, that the same will also be true for a degenerate tetrahedron, that is, for a plane quadrilateral. Hence, in particular, it is possible to obtain the theorem of cosines (Bretschneider's theorem, see p. 171) for a plane quadrilateral. 303. Let SI and S2 denote the areas of the faces having a common edge a, Sa, and S4 the areas of the two remaining faces. Let, further, a, m, and n denote the lengths of the edges forming the face SI, and a, y, and 6 the dihedral angles adjacent to them, V the volume of the tetrahedron. Then it is readily verified that the following equality is true: 3V 3V 3V a -S cota+m-S coty+n scot6=2S1, 1 1 1 or 2S2 a cot a-i- m cot y+ n,~ot 6 = 3V . Writing such equalities for all the faces -of the tetra-hedron, adding together the equalities corresponding to the faces SI and S2, and subtracting the two others, we get 1 acot a -b cot ~ = 3V (Sf+S~-S~-S:). Squaring this equali ty, replacing cot2 a and cot2 ~ by . \ -1 and . \ ~ -1, and taking advantage of the SIn a sIn Answers, Hints, Solutions following equalities: a2 4SiS~ b2 4S~Si sin2 a -9V2 'sin2 ~ 9V2 (see Problem 11), we fmally get 1 a2+b2+2ab cot a cot ~= 9V2 (2Q-T), 225 with Q the sum of the squares of the pairwise products of the areas of the faces, and T the sum of the fourth powers of the areas of the faces. 304. The necessity of all conditions is obvious. We are going to prove their sufficiency. (a) The statement of the problem is readily proved by making the development of the tetrahedron (to this end, the surface of the tetrahedron should be cut along three edges emanating from one vertex). (b) Make the development of the tetrahedron ABCD following Fig. 59, a in the supposition that the sums of Dz Fig. 59 the plane angles at the vertices Band C are equal to 180°. The points DI , D 2 , and Ds correspond to the vertex D. Two cases are possible: (1) 1 AD 1 = 1 BC I. In this case 1 DsA 1 + 1 D2A 1 = 2 1 BC 1 = 1 DsD2 I, that is, the triangle D2ADa degenerates, the point A must coincide with the point K which is the midpoint of D 2Da• 1/. 15-0449 226 Problems in Solid Geometry (2) 1 AB 1 = 1 CD 1 (or 1 AC 1 = 1 BD I). In this case 1 KB 1 = 1 AB I, the point A being found on the middle perpendicular to the side D 2D s. If DID2DS is an acute-angled triangle, then 1 AB 1 < 1 KB 1 for points A situated inside the triangle KBC, and 1 AB 1 > 1 KB 1 for the {loints situated outside the triangle KBC. And If the triangle DIDsDa is obtuse-angled (an obtuse angle being either at the vertex Dz or at the vertex D s), then at one of the two vertices of the tetrahedron (either B or C) one plane angle will be greater than the sum of two other angles. (c) Let 1 AB 1 = 1 CD I, 1 A C 1 = 1 DB I, and the sum of the angles at the vertex D is equal to 1800. We have: the triangle ACD is congruent to the triangle ABD, /, ~ consequently, ADB = DAC. /'-... ~ ./'-..... ~ /'.. Thus ADB + ADC + CDB = DAC + ADC + ./"... /"'--... ./"'-.. CDB = 1800 • Hence, it follows that CDB = A CD and ~ACD = ~CDB, 1 AD 1 = 1 CB I. (d) Cut the tetrahedron along the edges, and super-impose the four triangles thus obtained one over another so as to bring to coincidence their equal angles. In Fig. 59, b, identical letters corresJ>ond to one and the same vertex of the tetrahedron, and identical subscripts to one and the same face. Identical letters correspondlDg to one point show that at this point the corresponding vertices of the appropriate triangles coincide. Conse-quently, I C3A 3 1 = 1 CA I, 1 B 2D2 1 = 1 B IDI 1 and this means that AC3 is parallel to B~l which is impossible. (e) Project the tetrahedron ABCD on the plane par-allel to the edges AB and CD. Then it is possible to prove that the projections of the triangles ABC and ABD will be equiva.lent. Exactly in the same manner, the projections of the triangles A CD and BCD will also be equivalent. And this means that the parallelogram with diagonals AB and CD will be the projection of ABCD. Hence follow the equalities 1 A C 1 = 1 BD I, 1 AD 1 = 1 BC I. The equality 1 AB 1 = 1 CD 1 is proved exactly in the same way. Answers, Hints, Solutions 227 (f) Let 0 1 denote the point of tangency of the inscribed sphere with the face ABC, and O2 with the face BCD. The hypothesis implies that 0 1 and O2 are the centres of the circles circumscribed about ABC and BCD. Be-sides, the triangle BC01 is congruent to the triangle BCO! This implies that /"-... 1 /"-... 1 /........ /"-.., BAC=:- B01C=- B02C=BDC. 2 2 Reasoning in the same way, we shall obtain that all the plane angles adjacent to the vertex D are equal to the corresponding angles of the triangle ABC, that is, their sum is equal to 1800 • The same may be asserted about the remaining vertices of the tetrahedron ABCD. Further, take advantage of Item (a). (g) Complete the given tetrahedron to get a parallele-ptped in a usual way, that is, by passing through each edge of the tetrahedron a plane parallel to the opposite edge. Then the necessary and sufficient condition of the equality of the faces of the tetrahedron will be expressed by the condition that the obtained parallelepiped be rectangular. And from the fact that the edges of this parallelepiped are equal and parallel to the corresponding line segments joining the midpoints of opposite edges of the tetrahedron will follow our statement. (h) If 0 is the centre of the sphere circumscribed about the tetrahedron ABCD, then the hypothesis will imply that the triangle AOB is congruent to the triangle COD, since both triangles are isosceles with equal lateral sides, equal medians emanating from the vertex 0 (0 coincides with the midpoint of the line segment joining the mid-points of AB and CD). Consequently, I AB I = I CD I. The equality of other pairs of opposite edges is proved exactly in the same manner. (i) From the fact that the distances from the centres of gravity to all the faces are equal follows the equality of the altitudes of the tetrahedron and then also the equality of its faces (see Item (e)). 305. Let a, b, c, and d denote vectors perpendicular to the faces of the tetrahedron, directed outside and having the length numerically equal to the area of the corre-sponding face, and let ea' eb' ee' and ed denote the unit 15-228 Problems in Solid Geometry vectors having the same directions as a, b, c, and d. Let, further, B denote the sum of the cosines of the dihedral angles, and k = ea + eb + ee + ed. It is obvious tliat k2 = 4 -28. Thus, indeed, B ~ 2 and $ = 2 if and only U k = ell + e" + ee + ed = O. But since a + b + c + d = 0 (see Problem 214), we obtain that for B = 2 the lengths of the vectors a, b, c, and d are equal to one another, i.e. all the faces are equiv-alent, and from the equivalency of the faces there follows their congruence (see Problem 304 (e». To complete the proof, it remains to show that $ > 0 or that I k I < 2. For conveniency, we shall regard that I a I = 1, I b I ~ 1, I c I ~ 1, I d I ~ 1. Then ep = a, I k I = I a + b + c + d + (eb -b) + (ee -c) 1- (ed -d) I ~ 1 eb -b 1 + 1 ec -c I + 1 ed -d I = 3 -(I b 1 + I c I + I d I) ~ 3 -I b + c + d I = 3 -I a I = 2. Equality may be the case only if all the vectors a, b, c, and d are collinear; since it is not so, I k I < 2, B> O. 306. Consider the tetrahedron all faces of which are congruent triangles whose an~les are respectively equal to tne plane angles of our trIhedral angle. (Prove that such tetrahedron exists.) All the trihedral angles of this tetrahedron are equal to the given trihedral angle. The sum of the cosines of the dihedral angles of such tetra-hedron is equal to 2 (see Problem 304). Consequently, the sum of the cosines of the dihedral angles of the given trihedral angle is equal to 1. 307. Constructing a parallelepiped from the given tetrahedron, and passing through each edge a plane par-allel to the opposite edge. we shall get for the equifaced tetrahedron, as is known, a rectangular parallelepiped. The centre of the inscribed ball coincides with the centre of the parallelepiped, and the centres of the exter-nally inscribed balls are found at the vertices of the parallelepiped different from the vertices of the tetra-hedron. This implies both statements of the problem. 308. Let ABCD be the given tetrahedron, DR its altitude, DA I , DB I , and DCI the altitudes of the faces dropped from the vertex D on the sides BC, CA, and AB. Cut the surface of the tetrahedron along the edges DA, DB, and DC and make the development (Fig. 60). It is obvious that R is the point of intersection of the altitudes of the triangle DJD 2D s• Let F denote the point of inter-section of the altitudes of the triangle ABC, A K the Answers, Hints, Solutions 229 altitude of this triangle, I AF I = hI' I F K I = h2. Then I DIH I = 2hl , I DIAl I = hI + h2' I HAIl = I hI -hI I. Hence, since h is the altitune of our tetrahenron, h2 = 1 DH 12 = 1 DAI 12 -I HAl 12 = (hI + h2)2 -(hI -h2)2 = 4h1h2• Now, let M denote the centre of gravity of the triangle ABC (it also serves as the centre of gravity of the tri-angle DID2Da), 0 the centre of the circle circumscribed D3 ~~------~--------~~ Fig. 60 about this triangle. I t is known that F, M, and 0 lie on one and the same straight line (E uler' s line), M lying between F and 0, 1 F M 1 = 2 1 MO I. On the other hand, the triangle DID'J,Da is homothetic to the triangle ABC with centre at M and ratio of simil-itude equal to (-2), hence, 1 MN 1 = 2 I F M I. Hence it follows that I OH 1 = I FO I· 309. When solving the preceding problem, we proved that the centre of the sphere circumscribed about the tetrahedron is projected on each edge into the midpoint of the line segment whose end points are the foot of the altitude dropped on this face and the point of intersection of the altitudes of this face. And since the distance from the centre of the sphere circumscribed about the tetra-hedron to the face is equal to i-h, where h is the altitude of the tetrahedron, the centre of the circumscribed sphere 230 Problems in Solid Geometry is found at a distance of Y Is h2 + a2 from the given points, where a is the distance between the point of inter-section of the altitudes and the centre of the circle cir-cumscribed about the face. 310. First of all, let us note that all the triangles ABC are acute. Indeed, if H is the point of intersection of the altitudes of the triangle ABC, 0 the centre of the given circle, then 1 OH 1 = 3 10M I, M lying between 0 Fig. 61 and H, that is, H is found inside the circle circumscribed about the triangle ABC, and this means that the triangle ABC is acute, consequently, there is a pOint D such that ABCD is an equifaced tetrahedron. Let us develop this tetrahedron (Fig. 61). Obviously, Hl..' which is the point of intersection of the altitudes of the triangle D1 D 2D 3 , is the foot of the altitude dropped from D on ABC. But the triangles ABC and D1D,Da..-have a common centre of gravity M with respect to which they are homothetic with the ratio of similitude (-2), hence 1 HIM 1 = 21M H I, M lying between HI and H, HI is a fixed point. It remains to prove that the altitude of the tetra-hedron ABCD is also constant. In the triangle ABC draw the altitude AK and extend it to intersect the cir-cumscribed circle at point L. I t is known (and is readily proved) that 1 LK 1 = 1 KH I· Let 1 AH 1 = hl1 Answers, Hints, Solutions 231 1 H K 1 = h2' the altitude of the tetrahedron is h. We know (see Problem 307) that h2 = 4hlh2 = 2 1 AH 1 X 1 HL 1 = 2 (R2 -9a2), where a = 1 OM I, which was required to be proved. 311. Consider too cube AEFGAlElFlGl with edge equal to the side of the square ABCD. On the edges AIEl and A lGl take the points P and 0 such that 1 AlP 1 = 1 BP 1 = 1 CO I, 1 Al 0 1 = 1 OD 1 = 1 PC 1 (Fig. 62, a). Con-e ..;r----- Cf 'D E-------~ (a) £ A Fig. 62 sider the rectangle AlPMlO. In view of the condition 1 AlP 1 + 1 AlO 1 = 1 AIEl I, the point Ml lies on the diagonal ElGl. Consequently, if M is the projection of Ml on EG, then the tetrahedron APOM has all the faces equal to the triangle A PO. The square A BCD whose plane contains the triangle A PO is obtained from the square AEElAl by rotating about the diagonal AFI through some angle a (Fig. 62, b). Since the plane EGA I is perpendicular to the diagonal A F 1, BD belongs to this plane. But the planes AEElA u ABCD, as well as the straight lines EG, EAu AlG, and BD are tangent to the ball inscribed in the cutie. Hence it follows that the angle between the planes ABCD and A lEG has a constant size, it is equal to the angle q> between the planes AEElAl 232 Problems in Solid Geometry and A,EC for which cos cp = ~3. But the planes A,EC and ABCD intersect along the diagonal BD. Hence, the point M lies in the plane passing through BD and making an angle q> with the plane A BCD , and the locus of pro-jections of points M will be represented by two line seg-ments emanating from the midpoint of A C at an angle q> 1 }'-to AC so that cos q> = va' and having the length a 2 2 (Fig. 62, c). 312. (a) Let ABeD denote the given tetrahedron. If its altitudes intersect at the point U, then DU is perpen-dicular to the plane ABC and, hence, DH is perpendic-ular to BC. Exactly in the same way, AU is perpendic-ular to BC. Consequently, the plane DAB is perpen-dicular to BC, that is, the edges DA and BC are mutually perpendicular. Conversely, let the opposite edges of the tetrahedron ABCD be pairwise perpendicular. Through DA pass a plane perpendicular to BC. Let us show that the alti ... tudes of the tetrahedron drawn from the vertices A and D lie in this plane. Denote by K the point of intersection of the passed plane and the edge BC. The altitude DDl of the triangle ADK will be perpendicular to the lines AK and BC, hence, it is an altitude of the tetrahedron. Thus, any two altitudes of the tetrahedron intersect, hence, all the four intersect at one point. (b) It is easy to prove that if one altitude of the tetra-hedron passes through the point of intersection of the altitudes of the appropriate face, then the opposite edges of the tetrahedron are pairwise perpendicular. This follows from the theorem on three perpendiculars. Hence, Items (a) and (b) are equivalent. (c) The equality of the sums of the squares of opposit& ed~s of the tetrahedron is equivalent to the conaition of the perpendicularity of opposite edges (see Item (a»). (d) Complete the tetrahedron to a parallelepiped, as usual, by passing through each of its edges a plane parallel to the opposite edge. The edges of the obtained parallele-piped are equal to the distance between the midpoints of the skew edges of the tetrahedron. On the other hand. the condition of perpendicularity of opposite edges of Answers, Hints, Solutions 233 the tetrahedron which is, according to Item (a), equiva-lent to the condition of the orthocentricity of the given tetrahedron, is, in turn, equivalent to the condition of the equality of the edges of the obtained parallelepiped (the diagonals of each face are equal and parallel to two opposite edges of the tetrahedron, that is, each face must be a rhombus). (e) From Problems 300 and 303 it follows that this condition is equivalent to the condition of Item (c). (f) Let a and aI' band bu C and CI be the lengtbs of three pairs of opposite edges of the tetrahedron, a the angle between them. From Problem 185 it follows that of the three numbers aal cos a, bbl cos a, and CCI cos a one is equal to the sum of two others. If cos a 9= 0, then of the three numbers aal' bbu and CCI one number is equal to the sum. of two others. But this is impossible, since there is a triangle the lengths of the sides of which are numerically equal to the quantities aau bbl , and CCI (see Problem 302). 313. Let ABeD denote the/iven tetrahedron. Com-plete it to get a parallelepipe in a usual way. Since , ___ .C A Fig. 63 ABeD is an orthocentric tetrahedron, all the edges of the parallelepiped will be equal in length. Let A BI be the diagonal of a face of the parallelep~ed parallel to AB, 0 the centre of the ... \ball. circumscribed about ABeD, H the point of intersection of the altitudes, M the centre of gravity (Fig. 63). Then the triangles ABH and AJ.BIO are symmetric with respect to the point M. This follows from the fact that ABBIAI is a parallelogram and, be-£6-0449 234 Problems in Solid Geometry sides, AIO is perpendicular to the plane ACD (the points o and Al are equidistant from the points A, C, and D), and, hence, })arallel to BU. Exactly in the same manner, OBI is parallel to AU. 314. Let us introduce the notation used in the pre-ceding problem. Let K and L be the midpoints of AB and AIBI • Then KOLU is a parallelogram. Consequently, IOU [2=2[ OK [2+2[ OL [2-1 KL [2 = 2 ( R2 _ [ ~ 12 ) + 2 ( R2 _ I C~ [2 ) -12 = 4R'-~ ( 1 AB 12+ 1 CD 12) -(Io=4R2 -3I2 • 2 315. If ABCD is an orthocentric tetrahedron, then (see Problem 312 (d» I AB 12 + I CD 12 = I AD 12 + I BC I't whence I AB 12 + I AC 12 -I BC 12 = I AD 12 + I AC 12 -I CD I', ./"'-... /"... that is, the angles BAC and DAC are both acute or ob-tuse. 316. The section of an orthocentric tetrahedron by any Clane parallel to opposite ed~s and passing at an e~ua distance from these edges IS a rectangle whose dIagonals are equal to the distance between the midpoints of opposite edRE's of the tetrahedron (all these distances are equal in length, see Problem 312 (d)). Hence it follows that the midpoints of all the edges of' an orthocentric tetrahedron lie on the surface of the sphere whose centre coincides with the centre of gravity of the given tetrahedron and the diameter is equal to the distance between the opposite edges of the tetrahed-ron. Hence, all the four 9-point circles lie on the surface or this sphere. 317. Let 0, M, and U respectively denote the centre of the circumscribed ball, centre of gravity and ortho-centro (the point of intersection of altitudes) of the ortho-Answers, Hints, Solutions 235 centric tetrahedron, M the midpoint of the line segment OH (see Problem 3f3). The centres of gravity of the faces of the tetrahedron serve as the vertices of the tetrahedron, homothetic to the given one, with the centre of similitude at th~ point M an.d the ratio of ~imilitude ~qual to -(f/3). In thIS homothetic transformatIOn the pOInt 0 will move into the point 0 1 situated on the line segment MH so that I MOl I = 1.13 I OM I, 0 1 will be the centre of the sphere passing through the centres of gravity of the faces. On the other hand, the points dividing the line seg-ments of the altitudes of the tetrahedron from the vertices to the orthocentre in the ratio 2 = f serve as the vertices 0' 0; H' Fig. 64 of the tetrahedron homothetic to the given with the centre of similitude at H and the ratio of similitude equal to f/3. In this homothetic transformation the point 0, as is readily seen, will go to the same point 01" Thus, eight of twelve points lie on the surface of the sphere with centre at 0 1 and radius equal to one-third the radius of the sphere circumscribed about the tetrahedron. Prove that the points of intersection of altitudes of each face lie on the surface of the same sphere. Let 0', HI, and MI denote, respectively, the centre of the cir-cumscribed circle, the point of intersection of altitudes, and the centre of gravity of some face. 0' and H' are the respective projections of 0 and H on the plane of this face, and the point M' divides the line segment 0' H' in the ratio f : 2 as measured from the point 0' (a well-known fact from plane geometry). Now, we easily make sure (see Fig. 64) that the projection of 0 1 on the plane of this face (point OD coincides with the midpoint of the line segment M' HI, that is, 0 1 is equidistant from M' and H' which was required to tie proved. f 6-236 Problems in Solid Geometry 318. The centres of gravity of the faces of the ortho~ centric tetrahedron lie on the surface of the sphere homo~ thetic to the sphere circumscribed about the tetrahedron with the centre of similitude at the point M and the ratio of similitude equal to 1/3 (see the solution of Problem 317). Hence follows the statement of the problem. 319. The feet of the altitudes of the orthocentric tetrahedron lie on the surface of the sphere homothetic to the sphere circumscribed about the tetrahedron with the centre of similitude at the point G and ratio of simil-itude equal to -(1/3) (see the solution of Problem 317). Hence follows the statement of the problem. 320. Suppose the contrary. Let the planes containing the arcs intersect pairwise on the surface of the ball at points A and A H Band Bh C and C 1 (Fig. 65). Since each c B Fig. 65 arc measures more than 180°, it must contain at least one of any two opposite points of the circle on which it is situated. Let us enumerate these arcs and, respe~ tively, the planes they lie in: I, II, III. A and A:.., are the points of intersection of planes I and II, Band BI the points of intersection of planes II and III, C and C.\ the points of intersection of planes III and I. Each of the points A, A.\, B, B1, C, C1 must belong to one arc. Let At and C1 belong to arc I, Bl to arc II. Then Band C must belong to arc III, A to arc II. Denote by a, ~, y the plane angles of the trihedral angles, as is shown in the figure, 0 the centre' of the sphere. Since arc I does not contain the points A and C, the inequality 3600 -~ > 3000 must De fulfilled. Similarly, since arc I I does not contain the points B and AH it must be 1800 + a> 3000 and, finally, for Answers, Hints, Solutions 237 arc III we wUl have 3600 -Y > 300°. Thus, ~ < 600 , ex > 1200 , Y < 600 , hence, ex > ~ + y, which is impos-sible. 321. Let A and B denote two points on the surface of the sphere, C a point on the smaller arc of the great circle palsing through A and B. Prove that the shortest path from A to B must pass through C. Consider two circles ex and ~ on the surface of the sphere passing through C with centres on the radii OA and OB (0 the centre of the sphere). Let the line join-ing A to B does not pass through C and intersect the circle ex at point M and the circle p at N. Rotating the circle ex together with the part of the Hne enclosed inside it so that M coincides with C and the circle ~ so as to bring N in coincidence with C, we get a line joining A and B whose length, obviously, is less than the length of the line under consideration. 322. The circumscribed sphere may not exist. It can be exemplified by the polyhedron constructed in the following way. Take a cube and on its faces as on bases construct outwards regular quadrangular pyramids with dihedral angles at the base equal to 450 • As a result, we get a dodecahedron (the edges of the cube do not serve as the edges of this polyhedron), having fourteen vertices, eight of which are the vertices of the cube, and six are the vertices of the constructed pyramids not coinciding with the vertices of the cube. I t is easy to see that all the edges of this polyhedron are equal in length and equidistant from the centre of the cube, while the vertices cannot belong to one sphere. 323. Let us note, first of all, that the area of the spher-ical lune formed by the intersection of the surface of the sphere with the faces of the dihedral angle of size ex, whose edge passes through the centre of the sphere, is equal to 2exR2. This follows from the fact that this area is proportional to the magnitude of ex, and for ex = n it is equal to 2nR2. To each pair of planes forming the two faces of the given trihedral there correspond two lunes on the surface of the sphere. Adding their areas, we get the surface of the sphere enlarged by 4Sh" where S/). is the area of the desired triangle. Thus, S /). = R2 (ex + ~ + y -n). 238 Problems in Solid Geometry The quantity c£ + ~ + y -n is called the spheric excess of the spheric triangle. 324. Consider the sphere with centre inside the poly-hedron and project the edges of. the polyhedron from the centre of the sphere on its sphere. The surface of the sphere will be broken into poly-gons. If nk is the number of sides of the kth polygon, Ak the sum of its angles, Sk the area, then Sk = R' (Alt -n (nit -2)). Adding together these equalities for all K, we get 4nRS = R2 (2nN -2nk + 2nM). Hence, N -K + M = 2. 325. Let c£ denote the central angle corresponding to the spheric radius of the circle <the angle between the radii of the sphere drawn from the centre of the sphere to the centre of the circle and a point on the circle). Consider the spheric triangle corresponding to the trihedral angle with vertex at the centre of the sphere one edge of which (OL) passes through the centre of the circle, another (OA), through the point on the circle, and a third (OB) is arranged so that the plane OAB touches the circle, the dihedral angle at the edge OL being equal to <p, /'... LOA = c£. Applying the second theorem of cosines (see Prob-lem 166), find the dihedral angle at the edge OB, it is equal to arccos (cos c£ sin <Pl. Any circumscribed polygon (our polygon can be regarded as circumscribed, since otherwise its area could be reduced) can be divided into triangles of the described type. Adding their areas, we shall see that the area of the polygon reaches the smallest value together with the sum arccos (cos c£ sin <PI) + arccos (cos c£ sin <P2) + ... + arccos (cos c£ sin <PN)' where <PI' ••• , <PlY are the corresponding dihedral angles, <PI + <P2 + ... --t- <PN = 2n. Then we can take advan-tage of the fact that the function arccos (k sin <p) is a concave (or convex downward) function for 0 < k < 1. Hence it follows that the minimum of our sum is reached for <PI = <P2 = ... = <PN' Answers, Hints, Solutions 239 326. Denote, as in Problem 324, by N the number of faces, by K the number of edges, and by M the number of vertices of our polyhedron, N -K + M = 2. (d Since from each vertex there emanate at least three edges and each edge is counted twice, M < ~ K. Substituting ~ 3 M into (f), we get f N -3K~2, whence 2K::::; 6N -12, 2; < 6. The latter means that tbere is a facp having less than 6 sides. Indeed, let us number thp faces and denote by nl, n2, .•• , nN the number of sides in each face. Then n t + n2 + ... + nN _ 2K < 6 N -N . 327. If each face has more than three sides and from each vertex there emanate more than three edges, then (the same notation as in Problem 324) K > 2M, K > 2N and N -K + M < 0, which is impossible. 328. If all the faces are triangles, then the number of edges is multiple of 3. If there is at least one face with the number of sides exceeding three, then the num-ber of edges is not less than eight. An n-gon pyramid has 2n edges (n > 3); (2n + 3) ed~s (n > 3) will be found in the polyhedron which will be obtained if an n-gon pyramid is cut by a triangular plane passing sufficiently close to one of the vertices of the base. 329. If the given polyhedron has n faces, then each face can have from three to (n -f) sides. Hence it follows that there are two faces with the same number of sides. 330. Consider the so-called d-neighbourhood of our polyhedron, that is, the set of points each of which is found at a distance not greater than d from at least one point of the polyhedron. The surface of the obtained solid 240 Problems in Solid Geometry consists of plane parts equal to the corresponding faces of the polyhedron, cylindrical parts corresponding to the edges of the polyhedron (here, if I, is the length of some edge and a, is the dihedral angle at this edge, then the surface area of the part of the corresponding cylinder is equal to (n ~ a,) IJd), and spherical parts corresponding to the vertices 01 the polyhedron the total area of which is equal to the surface aNa of the sj)here of radius d. On the other hand, the surface area of the d-neighbourhood of the J>olyhedron is less than the surface area of the sphere of radius d + 1, that is, S+d ~ (n-ailli+4nd2 < 4n (d+1)2. And since ai E;;; ~ t we get 2J li < 24, which was required to be proved. 331. In Fig. 66, 0 denotes the centre of the sphere, A and B are the points of intersection of the edge of the o Fig. 66 dihedral angle with the surface of the sphere, D and C '-'" -are the midpoints of the arcs ADB and A CB, respectively, the plane ADB passes through 0, and E is the vertex of the spherical segment cut off by the plane A CB. The area of the curvilinear triangle ADC amounts to half the desired area. On the other hand (assUming a ~~ ) , SADC = SAEC ~ SAED· (f) Answers, Hints, Solutions 24i Find S AEC. If <p is tbe angle between the planes AEO and OEC, I EK I =h, then obviously, S.AEC=2~ 2nRh= <pRh; h and <p are readily found: h= I EK 1= R-I OK 1= R-a sin ex, sin <p= sin {xi = I AL I = V R2-a2 , I AK ! V R2-a2 sin2 ex . VR2-a2 <p= arCSIn ~i 2 0 2 0 r R -a2 sm ex ThU!~, VR~-a2 S AEC = R (R-a sin ex) arcsin V 2 2' 2 0 (2) R -a 8m ex Now find SAEDO As iI known (see Problem 323), SA~D =-= R2 (<p + '¢ + y ~ n), wh8re ., ,¢, al'ld y are the dihedral an2'les of the trihedral angle with vertex a.t 0 and edges OE, OA, and OD. Thy aJlgle ; N, since from each vertex there emanate at least three sides, and each side is counted twice. Now, it is easy to obtain that K < 15, N < 10. In Problem 325, we have proved that among all spherical n-gons containing the given circle a regular n-gon has the smallest area. Besides, it is possible to show that the sum of areas of regular n- and (n + 2)-gons is greater than the doubled area of a regular n-gon. (The polygons circumscribed about one circle are conSidered.) It is also obvious that the area of a re~lar circumscribed n-gon is decreased with an increase In n. Hence it follows that the sum of areas of the seven obtained polygons cannot be less than the sum of areas of five regular quadrilaterals Problems in Solid Geometry and two regular pentagons circumscribed about the circle with the SJ~herical radius to which there corresponds the central angle a= arcsin 0.7. The area of the correspond-ing regular pentagon will be S5 = 9 [10 arccos (cos a sin ~) -3n ] ' the area of the regular quadrilateral 84 =9 [8 arccos (~2 cos a ) -2nJ. We can readily prove that 286+ 584.,.> 36n. Thus, seven balls with radius 7 cannot simultaneously touch the ball with radius 3 without intersecting one another. At the same time we can easily show that it is possible in the case of six balls. 336. Consider the cube ABCDAIBIC)D~. On the edges AlB and AID take points K and L such that I AIK I = I CM I, I AlL I = I CN I. Let P and Q denote the points of intersection of the lines AK and BA I , AL and DAIf respectively. As is eaiily seen, the sides of the triangle AIPQ are equal to the corresponding line segments of the diagonal BD. And since the triangle BAlD is regular, our statement has been proved. 337. If the point P did not lie in the plane of the triangle ABC, the statement of the problem would be obvious, since in that case the points P, A 2, B" and C2 would belong to the section of the surface of the sphere circumscribed about the tetrahedron ABCP by the plane passing through P and l. The statement of our problem can now be obtained with the aid of the passage to the limit. 338. Let ABCDEF denote the plane hexagon circum-scribed about the circle. Take an arbitrary space hexagon AIBIC}DIEIFI (Fig. 67), different from ABCDEF, whose projection on our plane is the hexagon ABCDEF and whose corresponding sides pass through the points of contact of the hexagon ABCDEF and the circle. To prove the existence of such hexagon AIBICID}EIFh it suffices to take one vertex, say A If arbitrarily on the perpendic-ular to the plane erected at the point A, then the remain-Answrfs, Hints, Solutions 245 ing vertices will be determined identically. Indeed, let; a, b, e, d, e, and f be the lengths of the tangents to the circle drawn through the respective points A, B, C, D, E, F, and h the distance from A to the plane. Then BI lies on the other side of the plane as compared with A At Fig. 67 at a distance of hb, CIon the same side as Al at a distance a of hb. ~ = he from the plane, and so on. Finally, we a b a find that FI lies on the other side of the plane as compared with Al at a distance of hi and, hence, Al and FI lie on a the straight line passing through the point of tangency of AF with the circle. Any two opj>osite sides of the hexagon AIBICIDIEIFI lie in one and the same plane. This follows from the fact that all the angles formed by the sides of the hexa-gon with the given plane are congruent. Consequently, any two diagonals connecting the opposite vertices of the hexagon AIBICIDIEIFI intersect, and, hence, all the three diagonals of this hexagon (they do not lie in one plane) intersect at one point. Since the hexagon ABCDEF is the projection of the hexagon AIBICIDIEIF1' the theorem has been proved. 339. The plane configuration indicated in the problem can be regarded as three-dimensional projection~ a tri-246 Problems in Solid Geometry hedral angle cut by two planes, for which our statement is obvious. 340. This problem represents one of the possible three-dimensional analogues of Desargues' theorem (see Problem 339). For its solution, it is convenient to go out to a four-dimensional space. Let us first consider some properties of this space. The simplest fig!J-res of the four-dimensional space are: a point, a straight line, a plane, and a three-dimensional variety which will be called the hyperplane. The first three figures are our old friends from the three-dimen-sional space. Of course, some statements concerning these fi~s must be refined. For instance, the following axiom of the three-dimensional space: if two distinct planes have a common point, then they intersect along a straight line, must be replaced by the axiom: if two distinct planes belonging to one hyperplane have a com-mon point, then they intersect along a straight line. The introduction of a new geometric image, a hyperplane, prompts the necessity to introduce a group of relevant axioms, just as the passage from plane geometry to solid ~ometry requires a group of new axioms (refresh them, please) expressin~ the basic pr9perties of planes in space. This group consIsts of the following three axioms: f. Whatever a hyperplane is, there are po in ts belong-ing to it and j>oints not belonging to it. 2. If two distinct hyperplanes have a common point, then they intersect over a plane, that is, there is a plane belonging to each of the hyperplanes. 3. If a straight line not belonging to a plane has a common point with this plane, then there is a unique hyperplane containing thIs line and this plane. From these axioms it follows directly that four points not belonging to one plane determine a hyperplane; exact-ly in the same way, three straight lines not belonging to one plane, but having a common point, or two distinct planes having a common straight line determine a hyper-plane. We are not going to prove these statements, try to do it independently. For our further reasoning we need the following fact existing in the four-dimensional space: three distinct hyperplanes having a common point also have a common straight line. Indeed, by Axiom 2, any two of three hyper-planes have a common plane. Let us take two planes Answers, Hints, Solutions 247 over which one of the three hyperplanes intersects with two others. These two planes belonging to one hyper-plane have a common point and, hence, intersect along a straight line or coincide. Let us now pass to the proof of our statement. If the three planes under consideration were arranged in a four-dimensional space, then the statement would be obvious. Indeed, every trihedral angle determines a hyperplane. Two hyperplilnes intersect over a plane. This plane does not belong to a third hyperplane (by the hypothesis, these hyperplanes intersect one of the given planes along three straight lines not passing through one point) and, consequently, intersects with them along a straight line. Any three corresponding faces of trihedral angles lie in one hyperplane determined by two planes on whicb the corresponding edges lie, and therefore each triple of the corresponding faces has a common point. These three points belong to the three hyperplanes determined by the trihedral angles, and, as it was proved, lie on one straight line. Now, to complete the proof, it is sufficient to "see" in the given hypothesis the projection of the cor-responding four-dimensional configuration of planes and trihedral angles. TO THE READER Mir Publishers would be grateful for your comments on the content, translation and design of this book. We would also be pleased to receive any other suggestions you may wish to make. Our address is: Mir Publishers 2 Pervy Rizhsky Pereulok 1-110, GSP, Moscow, 129820 USSR Printed In the Union 0/ Soviet Socialist Republics |
11550 | https://www.indeed.com/career-advice/career-development/how-to-calculate-discount | How To Calculate a Discount Using 2 Methods (With Examples) | Indeed.com
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How To Calculate a Discount Using 2 Methods (With Examples)
How To Calculate a Discount Using 2 Methods (With Examples)
Written by
Indeed Editorial Team
Updated June 6, 2025
Companies offer discounts to customers, sometimes in an attempt to attract a new customer demographic or to increase demand for a certain product or service. As a business owner or sales manager, it's important to be able to calculate the discount you might offer to make sure you can still break even. In this article, we examine the two primary methods of calculating a discount and provide a step-by-step guide for each, along with examples, to help you understand the concepts.Key takeaways:
There are two main ways to calculate a discount. The first calculates the discount price exactly, while the other provides a rough estimate.
The first method can help with activities such as sales forecasts, providing an accurate picture of a discounted item's earning potential.
The second method may be useful for establishing an approximate idea of an item's profitability even with discounted prices.
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How to calculate a discount as a percentage of the original price
Calculating a discount as a percentage of the original price is a straightforward process. The resulting figure is precise down to the penny, so you can determine yourrevenue potentialand make accurate sales forecasts. Follow these steps:
1. Convert the percentage to a decimal
The first step is to represent the discount percentage in decimal form. Calculators have a function for this, but you can also do it manually. The conversion is as simple as moving the decimal point two places to the left. Consider, for instance, a 5% discount. The number 5 has an implied decimal point after it—5.0. Moving the decimal point two places leftward would give you 0.05, which is the percentage in decimal form.
Example:Shoe Mart wants to offer a 25% discount on winter boots. Moving the decimal point of 25 two spaces to the left results in the figure 0.25.
Related:How To Calculate a Percentage (With Examples)
2. Multiply the original price by the decimal
The next step determines how much the discount is in terms of dollar figures. To make the calculation, multiply the original price of the item by the decimal determined in step one.
Example:Winter boots at Shoe Mart originally sold for $147. To determine the discount value, a sales associate multiplies that price by a discount of 0.25. The product of $147 times 0.25 is $36.75. Therefore, the boots have been discounted by $36.75.
Related:Rules of Multiplication: Definition and Examples
3. Subtract the discount from the original price
The final step is a matter of subtraction. Subtract the discounted dollar amount you calculated in step two from the original price. The resulting amount is the final discounted price of the product.
Example:Since the discount value for Shoe Mart's winter boots is $36.75, the company can subtract that figure from the original price of $147. The resulting amount is $110.25, which is how much Shoe Mart will charge for the discounted boots.
Related:Basic Math Skills: Definitions, Examples and Improving Them
How to calculate a discount through estimation
Sometimes, it's not practical or efficient to calculate a discount to the exact penny. Perhaps you'd just like to get a rough sense of whether you can discount prices while remaining profitable. In such instances, it's useful to know how to calculate a discount through estimation. Follow these steps to learn how:
1. Round the original price
Rounding a number is the act of raising or lowering it to the nearest specified numeric place. In this case, round it to the nearest ten. Keep in mind that the general rule for rounding is to round up if the number is at or above the halfway point and to round down if it's below. Thus, a figure of 35 would round up to 40, while 34 would round down to 30.
Example:The original price of winter boots at Shoe Mart was $147. Here, the company can round up to the nearest tens place, which is $150.
Related:How To Round Decimals: A Complete Guide With Examples
2. Find 10% of the rounded number
Determining 10% of the rounded number further simplifies it, so you have a figure that's easy to calculate in your head. To do so, move the decimal point one place to the left.
Example:A sales associate finds that the rounded price of the winter boots is $150.00. If they move the decimal point one place to the left, they can determine that 10% of $150 is $15.
Related:Math Skills: Definition, Examples and How To Develop Them
3. Determine "10s"
Figure out how many 10s are in the discount. Don't worry about any 5s at this point.
Example:There are two 10s in 25%. They know this because 10 times 2 equals 20. A third 10 would equal 30, which would exceed the value of 25.
4. Estimate the discount
To estimate the discount, multiply the number of 10s determined in step three by 10% of the dollar amount as determined in step two.
Example:The rounded price of the winter boots is $150. Of that, 10% is $15. The discount is 25%, and there are two 10s in the number 25. Multiplying the 10% amount by 2 results in an amount of $30.
5. Account for 5%
Some discounts may have a 5 in the hundreds place. If this is the case, find the dollar amount for the remaining 5% by dividing the dollar amount for 10% by 2.
Example:The Shoe Mart discount on winter boots is 25%, and 10% of the original rounded price of $150 is $15. To account for the remaining 5%, divide $15 by 2 for an amount of $7.50. Therefore, 5% of $150 is $7.50.
Related:12 Helpful Rules of Division (With Examples)
6. Add the 5%
To determine the final discount value, add the 5% dollar amount to the initial estimated discount. That is, combine the figures you've calculated in steps four and five.
Example:In step four, they found that 20% of $150 is $30. In step five, they found that 5% of $150 is $7.50. Adding the percentage values together gives us a total discount of $37.50.
Related:How To Calculate Average Percentage: Formula and Examples
7. Calculate the sale price
Finally, subtract the final discount you calculated in step six from the rounded original price. The result is your estimated discounted price.
Example:The price of winter boots at Shoe Mart was rounded from $147 to $150. The total estimated discount amounted to $37.50. Subtracting $37.50 from $150 results in a difference of $112.50. The estimated sale price of the winter boots is $112.50.
Related: What Is Retail Margin? (With Steps and Tips to Calculate It)
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2 more examples of how to calculate a discount
Here are two additional examples showing you how to find the discounted price of a product:
Example 1
The following example shows you how to calculate an exact discount as a percentage of the original price, which is method one from above:Abbott's Clothiers intends to have a season's end sale on winter sweaters. The original price of the sweaters is $80. The store would like to discount the sweaters by 15%.To calculate the discount, the sales manager converts 15% into the decimal 0.15. Then, they multiply 0.15 by the original price of $80, resulting in a figure of $12. The result is the discount amount. Finally, the company subtracts $12 from the original sweater price of $80. With the 15% discount, the sweaters cost $68.Related:12 Common Types of Discounts
Example 2
This example demonstrates how to calculate a discount through estimation, which is method two from above:Forever Jewelers wants to run a promotion for its customers. It's considering a 15% discount on wedding bands with the purchase of a full-price engagement ring. The wedding bands vary in cost, so the store uses the price of its most popular wedding band—$1,245—as a baseline to estimate the discounted price. To calculate the estimated discount, a sales manager follows these steps:
1. Round the original baseline price to the nearest 10. In this case, $1,245 rounds up to $1,250.
2. Find 10% of $1,250, which is $125.
3. Consider that only one 10 goes into the discount percentage of 15. Multiplying 1 by $125 results in an initial estimated discount of $125.
4. To account for the 5% in the discount value, divide $125 by 2 for a total of $62.50.
5. Add $62.50 to the initial estimated discount of $125. The sum is $187.50.
6. Subtract $187.50 from the original baseline price of $1,250. The difference is $1,062.50. That's the estimated discount price.
Video: How to Calculate Percentages
Calculating percentages is key to so many parts of our lives! Whether it’s your car loan payment or your salary increase, this video will show you how to use percentages. Watch and learn tips, examples and more.
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11551 | https://problemasteoremas.files.wordpress.com/2020/07/msebooklet.pdf | Some contributions to Mathematics Stack Exchange S(L, 0) Q(0, ut) R(0, uT) y = f (x) s = vt P(x, y) x y O C ε R × × x y O γε γR z1 z2 Γ Américo Tavares July 2020 Queluz Contents I Answers 1 1 Elementary algebra 3 1.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.1.1 Inverse of y = x3 + x? . . . . . . . . . . . . . . . . . . . . .
3 1.1.2 Polynomials - Solutions . . . . . . . . . . . . . . . . . . . .
4 1.1.3 Is there a general formula for solving 4th degree equations?
5 1.2 Elementary Finite Sequences and Sums . . . . . . . . . . . . . . .
9 1.2.1 Calculating the value of Annuities . . . . . . . . . . . . . .
9 1.2.2 Proof that n ∑ k=1 k2 = n(n+1)(2n+1) 6 ?
. . . . . . . . . . . . . . .
10 1.2.3 How to compute this sum . . . . . . . . . . . . . . . . . . .
11 1.2.4 Adding powers of i . . . . . . . . . . . . . . . . . . . . . . .
12 1.3 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12 1.3.1 a √ a+b + b √ b+c + c √c+a > √ a + b + c is true for positive a,b,c 12 2 Geometry and Trigonometry 15 2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.1.1 Orthogonal projection of a point onto a line . . . . . . . . .
15 2.1.2 How to test any 2 line segments (3D) are collinear or not?
16 2.1.3 Rotation by 180◦angle . . . . . . . . . . . . . . . . . . . . .
17 2.2 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 2.2.1 Solve trigonometric equation: 1 = m cos(α) + sin(α) . . .
17 2.2.2 Is there a more efficient method of trig mastery than rote memorization?
. . . . . . . . . . . . . . . . . . . . . . . . .
20 2.2.3 Why is it that when proving trig identities, one must work both sides independently? . . . . . . . . . . . . . . . . . . .
23 2.2.4 Is it necessary to have θ in radians to obtain sin θ θ →1 as θ →0? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 i ii Contents 3 Calculus 27 3.1 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 3.1.1 Integral of 1 (1 + x2)2 . . . . . . . . . . . . . . . . . . . . . .
27 3.1.2 Calculation of R √ tan x + 2dx . . . . . . . . . . . . . . . . .
28 3.1.3 Orthogonality of sine and cosine integrals. . . . . . . . . .
30 3.1.4 Help finding integral: R dx x √ 1+x+x2 . . . . . . . . . . . . . .
31 3.1.5 Integration trig substitution R dx x √ x2+16 . . . . . . . . . . . .
32 3.1.6 Prove: Z 1 0 ln x x −1 = ∞ ∑ n=1 1 n2 . . . . . . . . . . . . . . . . . . .
34 3.1.7 Cat Dog problem using integration . . . . . . . . . . . . . .
35 3.2 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 3.2.1 Does this equality always hold?
. . . . . . . . . . . . . . .
39 3.3 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 3.3.1 Nice proofs of ζ(4) = π4 90 ? . . . . . . . . . . . . . . . . . . .
40 3.4 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 3.4.1 Proving the inequality e−2x ≤1 −x . . . . . . . . . . . . .
40 4 Fourier Series 43 4.1 Trigonometric series . . . . . . . . . . . . . . . . . . . . . . . . . .
43 4.1.1 Different methods to compute ∞ ∑ n=1 1 n2 . . . . . . . . . . . .
43 4.1.2 Fourier Series for | cos(x)| . . . . . . . . . . . . . . . . . . .
46 5 Complex Analysis 51 5.1 Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 5.1.1 Finding the Laurent series of f (z) = 1/((z −1)(z −2)) . .
51 5.2 Evaluating definite integrals by the residue theorem . . . . . . . .
53 5.2.1 Verify integrals with residue theorem . . . . . . . . . . . .
53 5.2.2 Evaluate an improper integral using complex analysis . .
55 II Questions 61 6 A few questions 63 6.1 Real Analysis and Calculus . . . . . . . . . . . . . . . . . . . . . .
63 6.1.1 Asymptotic (divergent) series . . . . . . . . . . . . . . . . .
63 6.1.2 What is the importance of Calculus in today’s Mathematics? 65 6.1.3 What does closed form solution usually mean? . . . . . . .
68 Contents iii 6.2 Number Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69 6.2.1 Binomial coefficients: how to prove an inequality on the p-adic valuation? . . . . . . . . . . . . . . . . . . . . . . . .
69 This is a small compilation of some answers and questions I posted to the site
User contributions licensed under cc by-sa 3.0 ( licenses/by-sa/3.0/) with attribution required (
com/2009/06/attribution-required/) My user profile page is
Américo Tavares (Last Updated on July 10, 2020) v Part I Answers 1 Chap. 1 Elementary algebra 1.1 Equations 1.1.1 Inverse of y = x3 + x?
Question by badatmath ( Inverse of y = x3 + x?, URL (version: 2011-10-06):
com/q/60907 Can you help me find the inverse function for y = x3 + x? (...) Answer. URL (version: 2011-09-01): Write x3 + x −y = 0 and make x = u + v. You get u3 + v3 −y + (u + v) (3uv + 1) = 0.
Then solve the system ( u3 + v3 = y u3v3 = −1 27, which is equivalent to solving a quadratic equation in u3 or v3, because you know the sum of the two numbers u3, v3 and their product, e.g. the equation u32 −yu3 −1 27 = 0.
This technique is known as Cardano’s method1.
Reference: , pp. 215-216 1 3 4 Chap. 1. Elementary algebra 1.1.2 Polynomials - Solutions Question by Dek ( Poly-nomials - Solutions, URL (version: 2013-06-27):
com/q/428508 How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial ... what methods occupy for this "problem"?
x3 + 3x2 −7x + 1 = 0 (...) Answer. URL (version: 2013-06-25): How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial You have a real polynomial of the third degree with real coefficients. There are exact formulas to find the roots of any polynomial of this kind.
A general cubic equation of the form ax3 + bx2 + cx + d = 0, (1) can be transformed by the substitution x = t −b a (2) into the reduced cubic equation t3 + pt + q = 0.
(3) In the present case, we have x3 + 3x2 −7x + 1 = 0, a = 1, b = 3, c = −7, d = 1.
(A) 1.1. Equations 5 For x = t −1, we get the reduced equation t3 −10t + 10 = 0, p = −10, q = 10.
(B) It is known from the classical theory of the cubic equation that when the dis-criminant ∆= q2 + 4p3 27 = 102 + 4 (−10)3 27 < 0, (C) the three roots tk of (3), with k ∈{1, 2, 3}, are real and can be written in the following trigonometric form2 tk = 2 p −p/3 cos 1 3 arccos −q 2 q −27/p3 + (k −1)2π 3 .
(4) The roots of (1) are thus xk = tk −b a.
(5) Consequently, x1 = 2 √ 10/3 cos 1 3 arccos −5 p 27/103 −1 ≈1.4236, x2 = 2 √ 10/3 cos 1 3 arccos −5 p 27/103 + 2π 3 −1 ≈−4.5771, x3 = 2 √ 10/3 cos 1 3 arccos −5 p 27/103 + 4π 3 −1 ≈0.15347.
(D) 1.1.3 Is there a general formula for solving 4th degree equa-tions?
Question by John Gietzen ( Is there a general formula for solving 4th degree equations?, URL (version: 2012-07-30): 2A deduction can be found in this post of mine in Portuguese.
6 Chap. 1. Elementary algebra There is a general formula for solving quadratic equations, namely the Quadratic Formula.
For third degree equations of the form ax3 + bx2 + cx + d = 0, there is a set of thee equations: one for each root.
Is there a general formula for solving equations of the form ax4 + bx3 + cx2 + dx + e = 0 ?
How about for higher degrees? If not, why not?
Answer. URL (version: 2011-08-16): Yes. As an answer I will use a shorter version of this3 Portuguese post of mine, where I deduce all the formulae. Suppose you have the general quartic equation (I changed the notation of the coefficients to Greek letters, for my con-venience): αx4 + βx3 + γx2 + δx + ε = 0.
(1) If you make the substitution x = y −β 4α, you get a reduced equation of the form y4 + Ay2 + By + C = 0, (2) with A = γ α −3β2 8α2 , B = δ α −βγ 2α2 + β 8α, C = ε α −βδ 4α2 + β2c 16α3 −3β4 256α4.
After adding and subtracting 2sy2 + s2 to the LHS of (2) and rearranging terms, we obtain the equation y4 + 2sy2 + s2 | {z } (y2+s)2 − h (2s −A) y2 −By + s2 −C i = 0.
(2a) 3Resolução da equação do 4.o grau (ou quártica), 1.1. Equations 7 Then we factor the quadratic polynomial (2s −A) y2 −By + s2 −C = (2s −A) (y −y+)(y −y−) and make y+ = y−, which will impose a constraint on s (equation (4)). We will get: y2 + s + √ 2s −Ay − B 2 √ 2s −A y2 + s + √ 2s −Ay + B 2 √ 2s −A = 0, (3) where s satisfies the resolvent cubic equation 8s3 −4As2 −8Cs + 4Ac −B2 = 0.
(4) The four solutions of (2) are the solutions of (3): y1 = −1 2 √ 2s −A + 1 2 s −2s −A + 2B √ 2s −A , (5) y2 = −1 2 √ 2s −A −1 2 s −2s −A + 2B √ 2s −A , (6) y3 = −1 2 √ 2s −A + 1 2 s −2s −A − 2B √ 2s −A , (7) y4 = −1 2 √ 2s −A −1 2 s −2s −A − 2B √ 2s −A .
(8) Thus, the original equation (1) has the solutions xk = yk −β 4α.
k = 1, 2, 3, 4 (9) Example: x4 + 2x3 + 3x2 −2x −1 = 0 8 Chap. 1. Elementary algebra y4 + 3 2y2 −4y + 9 16 = 0.
The resolvent cubic is 8s3 −6s2 −9 2s −101 8 = 0.
Making the substitution s = t + 1 4, we get t3 −3 4t −7 4 = 0.
One solution of the cubic is s1 = −q 2 + 1 2 r q2 + 4p3 27 !1/3 + −q 2 −1 2 r q2 + 4p3 27 !1/3 −b 3a, where a = 8, b = −6, c = −9 2, d = −101 8 are the coefficients of the resolvent cubic and p = −3 4, q = −7 4 are the coefficients of the reduced cubic. Numerically, we have s1 ≈1.6608.
The four solutions are : x1 = −1 2 p 2s1 −A + 1 2 s −2s1 −A + 2B √ 2s −A −β 4α, x2 = −1 2 p 2s1 −A −1 2 s −2s1 −A + 2B √ 2s −A −β 4α, x3 = −1 2 p 2s1 −A + 1 2 s −2s1 −A − 2B √ 2s −A −β 4α, x4 = −1 2 p 2s1 −A −1 2 s −2s1 −A − 2B √ 2s −A −β 4α, with A = 3 2, B = −4, C = 9 16, D = 9 16. Numerically we have x1 ≈−1.1748 + 1.6393i, x2 ≈−1.1748 −1.6393i, x3 ≈0.70062, x4 ≈−0.35095.
Another method is to expand the LHS of the quartic into two quadratic polyno-mials, and find the zeroes of each polynomial. However, this method some-times fails.
Example: x4 −x −1 = 0.
If we factor x4 −x −1 as x4 −x − 1 = x2 + bx + c x2 + Bx + C expand and equate coefficients we will get two 1.2. Elementary Finite Sequences and Sums 9 equations, one of which is −1/c −c2 1 + c22 + c = 0. This is studied in Galois theory4 The general quintic is not solvable in terms of radicals, as well as equations of higher degrees.
1.2 Elementary Finite Sequences and Sums 1.2.1 Calculating the value of Annuities Question by user78886 ( Calculating the value of Annuities, URL (version: 2014-01-13):
stackexchange.com/q/636242 Instead of investing 3000 at the end of 5 years, and $4000 at the end of 10 years, Steve wishes to make regular monthly payments that will amount to the same total after 10 years. Determine the monthly payment if interest is com-pounded monthly at an annual rate of 4% Could anyone set up the annuity formula with the numbers from the ques-tion. Im not sure if what i did was correct: 4000 = R1 −1.003120 0.004/12 The answer when you isolate for R should be $52.04.
Answer. URL (version: 2014-01-17): You have some numerical errors and have considered the contribution of $4000 only. Mathematically we can establish an equivalence between the in-vestments (at the end of 5 years and at the end of 10 years) and a series of 120 monthly constant payments.
Since there are m = 12 compounding periods per year, the (nominal) annual interest rate r = 4% = 0.04 indicates a monthly interest rate i = r m = 4 12% = 0.04 12 .
The hypothetical investment of 3000 at the end of 5 years (60 months) accumu-lates to the future value F′ = 3000 (1 + i)60 = 3000 (1 + 0.04/12)60 ≈3663.0.
Adding the second hypothetical investment F′′ = 4000 yields the total future value F = F′ + F′′ = 7663.0 at the end of 10 years. Let A (the annuity) denote each monthly payment. The payment at the end of month k increases to a future 4
10 Chap. 1. Elementary algebra value of Fk = A(1 + i)n−k at the end of n = 120 months. Summing all these Fk the resulting geometric series ( of n payments, whose ratio is c = 1 + i, should be equal to F, as a conse-quence of the equivalence mentioned above. Applying the formula(http:// math.stackexchange.com/a/29035/752) for such a sum, we get F = n ∑ k=1 Fk = n ∑ k=1 A(1 + i)n−k = n ∑ j=1 Acj−1 = Acn −1 c −1 = A(1 + i)n −1 i .
Numerically we obtain A = F i (1 + i)n −1 = 7663.0 0.04 12 (1 + 0.04 12 )120 −1 ≈52.04, which agrees with the answer you indicate.
1.2.2 Proof that n ∑ k=1 k2 = n(n+1)(2n+1) 6 ?
Question by Nathan Osman ( nathan-osman), Proof that n ∑ k=1 k2 = n(n+1)(2n+1) 6 ?, URL (version: 2013-08-14): Answer. URL (version: 2011-07-10): Proof 1. (Exercise 2.5.1 in Dias Agudo, Cândido da Silva, Matemáticas Gerais III).
Let S := ∑n k=1 k2. Consider (1 + a)3 = 1 + 3a + 3a2 + a3 and sum (1 + a)3 for a = 1, 2, . . . , n: (1 + 1)3 = 1 + 3 · 1 + 3 · 12 + 13 (1 + 2)3 = 1 + 3 · 2 + 3 · 22 + 23 (1 + 3)3 = 1 + 3 · 3 + 3 · 32 + 33 · · · (1 + n)3 = 1 + 3 · n + 3 · n2 + n3 The term (1 + 1)3 on the LHs of the 1st sum cancels the term 23 on the RHS of the 2nd, (1 + 2)3, the 33, (1 + 3)4, the 43, ..., and (1 + n −1)3 cancels n3. Hence 1.2. Elementary Finite Sequences and Sums 11 (1 + n)3 = n + 3 (1 + 2 + . . . + n) + 3S + 1 and S = n(n + 1)(2n + 1) 6 , because 1 + 2 + . . . + n = n (n + 1) 2 .
Proof 2. (Exercise 1.42 in Balakrishnan, Combinatorics, Schaum’s Outline of Com-binatorics). From k 1 + 2 k 2 = k + 2k (k −1) 2 = k2, we get S := n ∑ k=1 k2 = n ∑ k=0 k 1 + 2 k 2 = n ∑ i=1 k 1 + 2 n ∑ k=1 k 2 = n + 1 2 + 2 n + 1 3 = n (n + 1) (2n + 1) 6 .
1.2.3 How to compute this sum Question by Timo Willemsen ( timo-willemsen), How to compute this sum, URL (version: 2011-10-14): http: //math.stackexchange.com/q/72618 Answer. URL (version: 2011-10-14): Hints: expand 4(2k + 1)2 −12k and use n ∑ k=1 k = n(n + 1) 2 n ∑ k=1 k2 = n(n + 1)(2n + 1) 6 .
You can find a proof of the second formula in the post of mine (in Portuguese) Soma dos quadrados dos primeiros n números inteiros positivos5.
5 12 Chap. 1. Elementary algebra 1.2.4 Adding powers of i Question by jkottnauer ( Adding powers of i, URL (version: 2013-04-09):
com/q/4062 (...) But how do I get the result of i3 + i4 + ... + i50? (...) Answer. URL (version: 2011-10-14): Observing that i3 + i4 + . . . + i50 is a geometric progression with ratio i, first term i3 and 50 −3 + 1 = 48 terms, we have i3 + i4 + . . . + i50 = i3 × 1 −i50−3+1 1 −i = i3 × 1 −i48 1 −i = i2i × 1 −(i2)24 1 −i = −i1 −(−1)24 1 −i = −i1 −1 1 −i = 0 1.3 Inequalities 1.3.1 a √ a+b + b √ b+c + c √c+a > √ a + b + c is true for positive a,b,c Question by mgamer ( a √ a+b + b √ b+c + c √c+a > √ a + b + c is true for positive a,b,c, URL (version: 2010-10-16): How to prove that for all positive a,b,c this formula holds true: a √ a + b + b √ b + c + c √c + a > √ a + b + c (...) Answer. URL (version: 2010-10-16): Since a + b + c = q (a + b + c)2 = √ a + b + c √ a + b + c ⇐ ⇒ a + b + c √ a + b + c = √ a + b + c and 1.3. Inequalities 13 a √ a + b > a √ a + b + c b √ b + c > b √ a + b + c c √c + a > c √ a + b + c we have a √ a + b + b √ b + c + c √c + a > a + b + c √ a + b + c = √ a + b + c.
Chap. 2 Geometry and Trigonometry 2.1 Geometry 2.1.1 Orthogonal projection of a point onto a line Question by jbennet ( Orthogonal projection of a point onto a line, URL (version: 2011-09-08): http: //math.stackexchange.com/q/62633 (...) find an orthogonal projection of a point T(−4, 5) onto a line x 3 + y −5 = 1 Answer. URL (version: 2011-09-08): The slope of the line r, with equation x 3 + y −5 = 1, is mr = 5 3 (because x 3 + y −5 = 1 is equivalent to y = 5 3x −5). The slope of the line s orthogonal to r is ms = −3 5 (because mrms = −1). Hence the equation of s is of the form y = −3 5x + bs.
Since T(−4, 5) is a point of s, we have 5 = −3 5 (−4) + bs, which means that bs = 13 5 . So the equation of s is y = −3 5x + 13 5 .
15 16 Chap. 2. Geometry and Trigonometry The coordinates of the orthogonal projection of T onto r are the solutions of the system y = 5 3x −5 y = −3 5x + 13 5 , which are (x, y) = 57 17, 10 17 .
2.1.2 How to test any 2 line segments (3D) are collinear or not?
Question by niro ( How to test any 2 line segments (3D) are collinear or not?, URL (version: 2012-01-27): Answer. URL (version: 2012-01-27): An alternative method. Assume PQ and RS are the line segments. Let the 2.2. Trigonometry 17 direction cosines1 of the vectors u = − → PQ and v = − → RS be, respectively, αu, βu, γu and αv, βv, γv. The angle φ between the line segments is such that2 cos φ = αuαv + βuβv + γuγv.
Hence the line segments are collinear if cos φ = ±1.
2.1.3 Rotation by 180◦angle Question by dato datuashvili ( dato-datuashvili), rotation by 180 angle, URL (version: 2011-07-23): http: //math.stackexchange.com/q/53279 In general I know that if we rotate (x, y) about origin by the 180 degree we will get new image (−x, −y), but suppose that we make rotation not about origin but some other point (a, b) does your result be rotation around origin + or - (a, b)? for suppose we have point A(3, 27) and we want turn it by 180 around the point (2, −1), if we rotate (3, 27) about origin by 180 we get (−3, −27) but how to connect (2, −1) to this result?
Answer. URL (version: 2011-07-23): You can make a translation of axes so that (2, −1) becomes the new origin.
The new axes are X = x −2, Y = y + 1. Then you compute the new coordinates of A(3 −2, 27 + 1) = (1, 28). The symmetric point A′ with respect to (X, Y) = (0, 0) is thus A′(−1, −28) in the XY−coordinate system or A′(−1 + 2, −28 − 1) = (1, −29) in the original xy−coordinate system.
2.2 Trigonometry 2.2.1 Solve trigonometric equation: 1 = m cos(α) + sin(α) Question by ftiaronsem ( Solve trigonometric equation: 1 = m cos(α) + sin(α), URL (version: 2011-06-21): Dealing with a physics Problem I get the following equation to solve for α 1 2Formula 10.7 of 18 Chap. 2. Geometry and Trigonometry 1 = m cos(α) + sin(α) Putting this in Mathematica gives the result: a == 2ArcTan 1 −m 1 + m However I am unable to get this result myself. (...) Answer. URL (version: 2011-07-09): Added: As explained in the comments, certain trigonometric equations such as the linear equations in sin x and cos x can be solved by a resolvent quadratic equation. One method is to write the sin x and cos x functions in terms of the same trigonometric function. Since all [direct] trigonometric functions of the simple angle can be expressed rationally as a function of the tan of the half-angle, such a conversion is adequate for these equations 3.
Since cos α = 1 −tan2 α 2 1 + tan2 α 2 and sin α = 2 tan α 2 1 + tan2 α 2 your equation m cos α + sin α = 1 is equivalent to m −m tan2 α 2 + 2 tan α 2 = 1 + tan2 α 2.
One may set x = tan α 2 (α = 2 arctan x), and thus get the quadratic equation 3 states and proves: “All [direct] trigonometric functions of the double angle can be ex-pressed rationally as a function of the simple angle tangent." (my translation). For instance cos α = 1−tan2 α 2 1+tan2 α 2 can be derived as follows cos α = cos2 α 2 −sin2 α 2 = cos2 α 2 −sin2 α 2 cos2 α 2 + sin2 α 2 = 1 −tan2 α 2 1 + tan2 α 2 .
2.2. Trigonometry 19 (1 + m) x2 −2x + 1 −m = 0.
Its solutions are: x = 1 m+1 (−m + 1) (if m ̸= −1) or x = 1 (if m = −1), which gives i) If m ̸= −1, α = 2 arctan x = 2 arctan 1 −m m + 1, ii) If m = −1, α = 2 arctan 1 = π 2 .
A different technique to solve a linear equation in sin α and cos α is to use an auxiliary angle ϕ. If you set m = tan ϕ, your equation takes the form sin α + tan ϕ · cos α = 1 or sin(α + ϕ) = cos ϕ = 1 ± q 1 + tan2 ϕ = ± r 1 1 + m2, and obtain α = ± arcsin r 1 1 + m2 −arctan m.
Detailed derivation: from m cos α + sin α = 1 and m = tan ϕ, we get sin α + tan ϕ · cos α = 1 ⇐ ⇒sin α + sin ϕ cos ϕ · cos α = 1 ⇐ ⇒sin α · cos ϕ + sin ϕ · cos α cos ϕ = 1 ⇐ ⇒sin (α + ϕ) cos ϕ = 1 ⇐ ⇒sin (α + ϕ) = cos ϕ.
The identity cos ϕ = ± s 1 1 + tan2 ϕ 20 Chap. 2. Geometry and Trigonometry can be obtained as follows sin2 ϕ + cos2 ϕ = 1 ⇐ ⇒sin2 ϕ cos2 ϕ + 1 = 1 cos2 ϕ ⇐ ⇒tan2 ϕ + 1 = 1 cos2 ϕ ⇐ ⇒cos2 ϕ = 1 1 + tan2 ϕ.
Therefore sin (α + ϕ) = ± s 1 1 + tan2 ϕ ⇐ ⇒α + ϕ = arcsin ± s 1 1 + tan2 ϕ !
⇐ ⇒α + arctan m = arcsin ± r 1 1 + m2 !
(m = tan ϕ, ϕ = arctan m) and finally α = arcsin ± r 1 1 + m2 !
−arctan m.
2.2.2 Is there a more efficient method of trig mastery than rote memorization?
Question by Joe Stavitsky ( Is there a more efficient method of trig mastery than rote memorization?, URL (version: 2011-09-18): Answer. URL (version: 2011-09-19): I would emphasize how to derive trigonometric identities from a few ones. Learn: • how to derive the relations between the direct functions of the same angle from the definition of the trigonometric functions and the Pythagorean formula; • maxima, minima, zeroes and period of each function; • if a function is odd or even; • the trigonometric functions of 0◦, 30◦, 45◦, 60◦and 90◦; 2.2. Trigonometry 21 • the relations between functions of symmetric, complementary and sup-plementary angles; • the relations between functions of angles whose difference is 180◦; • the relations between functions of angles whose sum is 360◦; • the inverse trigonometric functions; • the addition formulas of sin and cos; • how to derive the subtraction formulas of sin and cos; • how to derive the addition and subtraction formulas of tan and cot; • how to derive the double and half angle formulas; • how to derive the sum to product formulas; • how to solve some elementary trigonometric equations; • (triangle) sin and cos laws; • Heron’s formula; • derivatives of direct and inverse trigonometric functions.
Added. Examples. From sin(α + β) = sin α · cos β + cos α · sin β, (A) if we set α = β = a, we get sin 2a = 2 sin a · cos a.
(1) And from cos(α + β) = cos α · cos β −sin α · sin β (B) for α = β = a, we have cos 2a = cos2 a −sin2 a.
(2) 22 Chap. 2. Geometry and Trigonometry Using the Pythagorean identity cos2 a + sin2 a = 1, (C) if cos a ̸= 0, then sin 2a = 2 sin a · cos a = 2 sin a · cos a cos2 a + sin2 a = 2sin a · cos a cos2 a cos2 a + sin2 a cos2 a = 2 sin a cos a 1 + sin2 a cos2 a = 2 tan a 1 + tan2 a.
(3) Similarly cos 2a = cos2 a −sin2 a = cos2 a −sin2 a cos2 a + sin2 a = cos2 a −sin2 a cos2 a cos2 a + sin2 a cos2 a = 1 −sin2 a cos2 a 1 + sin2 a cos2 a = 1 −tan2 a 1 + tan2 a.
(4) Then tan 2a = sin 2a cos 2a = 2 tan a 1 + tan2 a 1 −tan2 a 1 + tan2 a = 2 tan a 1 −tan2 a.
(5) Added 2. The linear equation in sin x and cos x A sin x + B cos x = C (6) can be solved by a resolvent quadratic equation in tan x 2, by writting the sin x and the cos x functions in terms of tan x 2 (set x = 2a in (3) and (4)): 2.2. Trigonometry 23 sin x = 2 tan x 2 1 + tan2 x 2 , (7) cos x = 1 −tan2 x 2 1 + tan2 x 2 .
(9) The equation (6) is equivalent to A 2 tan x 2 1 + tan2 x 2 + B 1 −tan2 x 2 1 + tan2 x 2 = C, 2A tan x 2 + B −B tan2 x 2 = C + C tan2 x 2, (B + C) tan2 x 2 −2A tan x 2 + C −B = 0.
(10) 2.2.3 Why is it that when proving trig identities, one must work both sides independently?
Question by Ord ( Why is it that when proving trig identities, one must work both sides independently?, URL (version: 2012-01-21): Suppose that you have to prove the trig identity: sin θ −sin3 θ cos2 θ = sin θ I have always been told that I should manipulate the left and right sides of the equa-tion separately, until I have transformed them each into something identical. So I would do: sin θ −sin3 θ cos2 θ = sin θ(1 −sin2 θ) cos2 θ = sin θ(cos2 θ) cos2 θ = sin θ 24 Chap. 2. Geometry and Trigonometry And then, since the left side equals the right side, I have proved the identity. My problem is: why can’t I manipulate the entire equation? In this situation it probably won’t make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated.(...) Answer. URL (version: 2014-05-03): Why can’t I manipulate the entire equation?
You can. The analytical method for proving an identity consists of starting with the identity you want to prove, in the present case sin θ −sin3 θ cos2 θ = sin θ, cos θ ̸= 0 (1) and establish a sequence of identities so that each one is a consequence of the next one. For the identity (1) to be true is enough that the following holds sin θ −sin3 θ = sin θ cos2 θ (2) or this equivalent one sin θ 1 −sin2 θ = sin θ cos2 θ (3) or finally this last one sin θ cos2 θ = sin θ cos2 θ (4) Since (4) is true so is (1).
The book indicated below illustrates this method with the following identity 1 + sin a cos a = cos a 1 −sin a a ̸= (2k + 1)π 2 It is enough that the following holds (1 + sin a)(1 −sin a) = cos a cos a 2.2. Trigonometry 25 or 1 −sin2 a = cos2 a, which is true if 1 = cos2 a + sin2 a is true. Since this was proven to be true, all the previous indentities hold, and so does the first identity.
References: 2.2.4 Is it necessary to have θ in radians to obtain sin θ θ →1 as θ →0?
Question by z_z ( Is it nec-essary to have θ in radians to obtain sin θ θ →1 as θ →0?, URL (version: 2012-08-21): Answer. URL (version: 2012-08-20): The length L of an arc of a circle with radius r and subentending angle θ is L = rθ if and only if θ is measured in radians4. Under this assumption and for 0 < θ < π/2 radians sin θ < L r = θ < tan θ and 4If θ is measured in degrees, then L = πr 180θ.
26 Chap. 2. Geometry and Trigonometry 1 < θ sin θ < 1 cos θ, which implies that lim θ→0 θ sin θ = 1 ⇔lim θ→0 sin θ θ = 1, because sin θ θ is an even function.
Chap. 3 Calculus 3.1 Integration 3.1.1 Integral of 1 (1 + x2)2 Question by Virtuoso ( Integral of 1 (1+x2)2, URL (version: 2011-04-30):
com/q/35924 Answer. URL (version: 2011-04-30): If we add and subtract x2 in the numerator, we can integrate the first integral immediately Z 1 (1 + x2)2 dx = Z 1 1 + x2 dx − Z x2 (1 + x2)2 dx = arctan x − Z x x (1 + x2)2 dx and the second integral by parts: Z x x (1 + x2)2 dx = x − 1 2 (1 + x2) + Z 1 2 (1 + x2)dx = − x 2 (1 + x2) + 1 2 arctan x.
27 28 Chap. 3. Calculus By applying this method n −1 times, we can reduce the integration of the func-tion f (x) = 1 (1 + x2)n to the integration of 1 1 + x2.
3.1.2 Calculation of R √ tan x + 2dx Question by juantheron ( Calculation of R √ tan x + 2dx, URL (version: 2014-01-04):
com/q/626942 Answer. URL (version: 2014-01-05): Using your substitution t = √ tan x + 2 we need to integrate I 2 = 1 2 Z √ tan x + 2 dx = Z t2 (t2 −2)2 + 1 dt + C, as you show in your edited question. We can reduce it to a table integral if we factorize the denominator t4 −4t2 + 5 = t2 + q 4 + √ 20t + √ 5 t2 − q 4 + √ 20t + √ 5 and expand the integrand into partial fractions t2 t4 −4t2 + 5 = At t2 + p 4 + √ 20t + √ 5 − At t2 − p 4 + √ 20t + √ 5 , where A = −B = −1 4 q 4 + √ 20 −2 + √ 5 .
The standard integral we need is the following one 3.1. Integration 29 Z t t2 + bt + c dt = 1 2 ln t2 + bt + c − b √ 4c −b2 arctan 2t + b √ 4c −b2 + C, 4c −b2 > 0.
In the case at hand 4c −b2 = 4 √ 5 − 4 + √ 20 = 2 √ 5 −4 > 0. So Z t t2 + p 4 + √ 20t + √ 5 dt = 1 2 ln t2 + q 4 + √ 20t + √ 5 − p 4 + √ 20 p 2 √ 5 −4 arctan 2t + p 4 + √ 20 p 2 √ 5 −4 + C, and Z t t2 − p 4 + √ 20t + √ 5 dt = 1 2 ln t2 − q 4 + √ 20t + √ 5 + p 4 + √ 20 p 2 √ 5 −4 arctan 2t − p 4 + √ 20 p 2 √ 5 −4 + C.
We thus get I 2 = A 1 2 ln t2 + q 4 + √ 20t + √ 5 − p 4 + √ 20 p 2 √ 5 −4 arctan 2t + p 4 + √ 20 p 2 √ 5 −4 !
−A 1 2 ln t2 − q 4 + √ 20t + √ 5 + p 4 + √ 20 p 2 √ 5 −4 arctan 2t − p 4 + √ 20 p 2 √ 5 −4 !
+ C.
Substituting back t = √ tan x + 2 we get the given integral I = 2I 2 = R √ tan x + 2 dx.
ADDED. After simplifying I’ve obtained I = 2 − √ 5 p 4 + √ 20 4 ln tan x + 2 + p 4 + √ 20√ tan x + 2 + √ 5 tan x + 2 − p 4 + √ 20√ tan x + 2 + √ 5 + p 4 + √ 20 2 × × arctan 2√ tan x + 2 − p 4 + √ 20 p√ 20 −4 + arctan 2√ tan x + 2 + p 4 + √ 20 p√ 20 −4 !
+ C.
30 Chap. 3. Calculus 3.1.3 Orthogonality of sine and cosine integrals.
Question by Frank_W ( Orthogonality of sine and cosine integrals., URL (version: 2014-01-14): http: //math.stackexchange.com/q/638308 How to prove that Z t0+T t0 sin(mωt) sin(nωt) dt will equal to 0 when m ̸= n and T 2 when m = n ̸= 0? Besides Z t0+T t0 cos(mωt) cos(nωt) dt will equal to 0 when m ̸= n and T 2 when m = n ̸= 0 and T when m = n = 0?
Answer. URL (version: 2014-01-14): I will assume that T is the period and ω is the angular frequency1 of the wave sin(ωt). In such a case, which is important to obtain the final results, the following relation holds ω = 2π T .
(1) Let x = ωt, x0 = ωt0. Then I(m, n) = Z t0+T t0 sin(mωt) sin(nωt) dt (2) = 1 ω Z x0+2π x0 sin(mx) sin(nx) dx = 1 2ω Z x0+2π x0 cos((m −n) x) −cos((m + n) x) dx, (3) because in general cos(α −β) −cos(α + β) = 2 sin α sin β, (4) as can be seen by subtracting 1 3.1. Integration 31 cos(α + β) = cos α cos β −sin α sin β from cos(α −β) = cos α cos β + sin α sin β.
• For m ̸= n, since Z x0+2π x0 cos((m −n) x) dx = sin((m −n) x) m −n x0+2π x0 = 0 Z x0+2π x0 cos((m + n) x) dx = sin((m + n) x) m + n x0+2π x0 = 0 (5) the integral I(m, n) = 0.
• For m = n ̸= 0, by (1) I(m, n) = I(m, m) = 1 2ω Z x0+2π x0 1 −cos(2mx) dx = 1 2ω x −sin(2mx) 2m x0+2π x0 = 1 2ω (2π) = π ω = T 2 .
(6) The evaluation of the second integral is similar.
3.1.4 Help finding integral: R dx x √ 1+x+x2 Question by Dave ( Help finding integral: R dx x √ 1+x+x2, URL (version: 2013-05-18): URL (version: 2013-05-18): Could someone help me with finding this integral Z dx x √ 1 + x + x2 32 Chap. 3. Calculus or give a hint on how to solve it.
Answer. URL (version: 2012-04-09): Since the integrand is a quadratic irrational function of the type R(x, √ 1 + x + x2), you may use the Euler substitution2 √ 1 + x + x2 = x + t. You get Z dx x √ 1 + x + x2 = Z 2 t2 −1 dt = −2 arctanh t + C = −2 arctanh p 1 + x + x2 −x + C.
3.1.5 Integration trig substitution R dx x √ x2+16 Question by Dantheman ( Integration trig substitution R dx x √ x2+16, URL (version: 2013-07-03):
stackexchange.com/q/434837 Z dx x √ x2 + 16 With some magic I get down to 1 4 Z 1 sin θ dθ Now is where I am lost. How do I do this? I tried integration by parts but it doesn’t work.
Answer. URL (version: 2013-07-03): UPDATE 2 (...) The first integral (...) I = Z dx x √ x2 + 16 , x = tan θ, dx = sec2 θdθ = Z sec2 θ (tan θ) 4 sec θ dθ = Z sec θ 4 tan θ dθ = Z sec θ 4 tan θ dθ = Z 1 4 sin θ dθ.
2 3.1. Integration 33 Use the Weierstrass substitution3 t = tan θ 2.
Then Z 1 sin θ dθ = Z 2 2t 1+t2 (1 + t2) dt = Z 1 t dt = ln |t| + C = ln tan θ 2 + C.
Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in sin θ, cos θ, i.e. a rational fraction of the form R(sin θ, cos θ) = P(sin θ, cos θ) Q(sin θ, cos θ), where P, Q are polynomials in sin θ, cos θ tan θ 2 = t, θ = 2 arctan t, dθ = 2 1 + t2 dt, which converts the integrand into a rational function in t. We know from trigonom-etry that cos θ = 1 −tan2 θ 2 1 + tan2 θ 2 = 1 −t2 1 + t2, sin θ = 2 tan θ 2 1 + tan2 θ 2 = 2t 1 + t2.
Proof. A possible proof is the following one, which uses the double-angle for-mulas and the identity cos2 θ 2 + sin2 θ 2 = 1: cos θ = cos2 θ 2 −sin2 θ 2 = cos2 θ 2−sin2 θ 2 cos2 θ 2 cos2 θ 2+sin2 θ 2 cos2 θ 2 = 1 −tan2 θ 2 1 + tan2 θ 2 , sin θ = 2 sin θ 2 cos θ 2 = 2 sin θ 2 cos θ 2 cos2 θ 2 cos2 θ 2+sin2 θ 2 cos2 θ 2 = 2 tan θ 2 1 + tan2 θ 2 .
3 34 Chap. 3. Calculus Another possible substitution is the Euler substitution4 p x2 + 16 = t + x.
Then I = Z dx x √ x2 + 16 = Z 2 t2 −16 dt = Z 1 4 (t −4) − 1 4 (t + 4)dt = 1 4 ln t −4 t + 4 + C = 1 4 ln √ x2 + 16 −x −4 √ x2 + 16 −x + 4 + C.
3.1.6 Prove: Z 1 0 ln x x −1 = ∞ ∑ n=1 1 n2 Question by Jozef ( Prove: R 1 0 ln x x−1dx = ∑∞ 1 1 n2, URL (version: 2014-01-01):
com/q/108248 Answer. URL (version: 2012-02-11): Hint: use the substitution u = 1 −x to obtain I := Z 1 0 ln x x −1dx = − Z 1 0 ln (1 −u) u du and the following Maclaurin series ln (1 −u) = −u −1 2u2 −1 3u3 −. . . −un+1 n + 1 −. . .
(|u| < 1).
4 3.1. Integration 35 3.1.7 Cat Dog problem using integration Question by pokrate ( Cat Dog problem using integration, URL (version: 2012-11-28):
stackexchange.com/q/244333 (...) A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed u. Without any delay, the dog starts with running with constant speed v > u to catch the cat. Initially, v is perpendicular to u and L is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of v, u and L.
(...) Answer. URL (version: 2014-05-02): (...) I assume that: (a) the dog starts at the point S = (L, 0) and the cat at the origin O = (0, 0); (b) the cat moves in the positive direction along the y-axis, and the dog describes a curve of pursuit (WolframMathWorld link http:// mathworld.wolfram.com/PursuitCurve.html) C in the xy-plane. I call y = f (x) the equation of C.
36 Chap. 3. Calculus 1. At time t the tangent line to C at the point P(x, y) passes through the point Q = (0, ut), which means that the derivative y′ = f ′(x) = dy/dx is y′ = y −ut x .
(A) Solving for t we get t = y −xy′ u .
(A’) 2. Let s be the distance traveled by the dog from S to P, i.e. the length of the arc SP measured along C. Since the arc length formula 5 is the integral s = Z L x q 1 + ( f ′(ξ))2dξ = − Z x L q 1 + ( f ′(ξ))2dξ, (B) and s = vt, we have t = s v = −1 v Z x L q 1 + ( f ′(ξ))2dξ = y −xy′ u .
(B’) Hence, equating (A′) to (B′), we get −u v Z x L q 1 + ( f ′(ξ))2dξ = y −xy′ (C) 3. Differentiate both sides and simplify −u v q 1 + (y′)2 = d dx y −xy′ −u v q 1 + (y′)2 = y′ − y′ + xy′′ = −xy′′.
to get the following differential equation q 1 + (y′)2 = kxy′′, k = v u > 1.
(D) 5 3.1. Integration 37 4. Set w = y′ and solve (D) for w applying the method of separation of variables 6. Then p 1 + w2 = kxw′ = kxdw dx ⇔ dw √ 1 + w2 = dx kx.
(E) So Z dw √ 1 + w2 = Z dx kx + C arcsinh w = 1 k ln x + ln C1.
(F) The initial condition x = L, w = y′(L) = 0 determines the constant C1 0 = 1 k ln L + ln C1 ⇒C1 = e−1 k ln L.
Consequently, arcsinh w = 1 k ln x −1 k ln L = 1 k ln x L.
(G) Solve (G) for w and rewrite in terms of exponentials using the definition of sinh z = 1 2 (ez −e−z) dy dx = w = sinh 1 k ln x L = 1 2 x L 1/k − x L −1/k (H) This last differential equation is easily integrable y = 1 2 Z x L 1/k − x L −1/k dx = 1 2 L 1/k + 1 x L 1/k+1 − L 1 −1/k x L 1−1/k + C (I) Find C making use of the initial condition x = L, y = 0 6 38 Chap. 3. Calculus 0 = 1 2 L 1/k + 1 L L 1/k+1 − L 1 −1/k L L 1−1/k!
+ C ⇒C = Lk k2 −1.
The equation of the trajectory is thus y = L 2 1 1 k + 1 x L 1 k +1 − 1 1 −1 k x L 1−1 k + Lk k2 −1.
(J) 5. To obtain the time T the dog takes to catch the cat, make x = 0 in the last equation and observe that the cat travels the distance y = f (0) = uT (point (R)): y = f (0) = Lk k2 −1 = Lv/u (v/u)2 −1 = uv v2 −u2 L = uT.
(K) Therefore T = L v v2 −u2.
(L) ADDED. Let M be the point (L/2, 0). We can easily verify that the total length of C is equal to SM + MR.
3.2. Derivatives 39 References: [1, 2, 3, 4] 3.2 Derivatives 3.2.1 Does this equality always hold?
Question by ghshtalt ( Does this equality always hold?, URL (version: 2011-06-04):
com/q/43157 Is it true in general that d dx Z x 0 f (u, x)du = Z x 0 d dx f (u, x) du + f (x, x) ?
Answer. URL (version: 2013-06-20): Yes, it is, under the conditions indicated below. Let I(x) = Z x 0 f (u, x) du.
(∗) If f (u, x) is a continuous function and ∂f /∂x exists and is continuous, then I′(x) = Z x 0 ∂f (u, x) ∂x du + f (x, x) (∗∗) follows from the Leibniz rule and chain rule.
Note: the integrand of (∗∗) is a partial derivative. It generalizes to the integral I(x) = Z v(x) u(x) f (t, x) dt.
Under suitable conditions (u(x), v(x) are differentiable functions, f (t, x) is a continuous function and ∂f /∂x exists and is continuous), we have I′(x) = Z v(x) u(x) ∂f (t, x) ∂x dt + f (v(x), x)v′(x) −f (u(x), x)u′(x).
3.3 Series 40 Chap. 3. Calculus 3.3.1 Nice proofs of ζ(4) = π4 90?
Question by Mike Spivey ( Nice proofs of ζ(4) = π4/90?, URL (version: 2012-04-04):
com/q/28329 Answer. URL (version: 2011-03-21): In the same spirit of the 1st proof of the answer in 4.1.1 to the question “Different methods to compute ∞ ∑ n=1 1 n2”. If we substitute x for π in the Fourier trigonometric series expansion of f (x) = x4, with −π ≤x ≤π, x4 = 1 5π4 + ∞ ∑ n=1 8n2π2 −48 n4 cos nπ · cos nx, we obtain π4 = 1 5π4 + ∞ ∑ n=1 8n2π2 −48 n4 cos2 nπ = 1 5π4 + 8π2 ∞ ∑ n=1 1 n2 −48 ∞ ∑ n=1 1 n4.
Hence ∞ ∑ n=1 1 n4 = π4 48 −1 + 1 5 + 8 6 = π4 48 · 8 15 = 1 90π4.
3.4 Inequalities 3.4.1 Proving the inequality e−2x ≤1 −x Question by sssuuuccc ( Proving the inequality e−2x ≤1 −x, URL (version: 2012-01-10):
stackexchange.com/q/97909 How do I prove the inequality e−2x ≤1 −x for 0 ≤x ≤1/2?
Answer. URL (version: 2014-05-24): 3.4. Inequalities 41 By Taylor expansion, we have ln 1 1 −x = ∞ ∑ n=1 xn n , whose convergence radius is R = 1.
The equation above can also be achieved by integrating both sides of 1 1 −x = ∞ ∑ n=0 xn.
For 0 ≤x ≤1/2 we have the following upper bound ∞ ∑ n=1 xn n ≤ ∞ ∑ n=1 1 n (2n) = ln 2 ≤2.
Therefore −ln (1 −x) = ∞ ∑ n=1 xn n ≤2x.
The given inequality follows.
Chap. 4 Fourier Series 4.1 Trigonometric series 4.1.1 Different methods to compute ∞ ∑ n=1 1 n2 Question by AD ( Different methods to compute ∞ ∑ n=1 1 n2, URL (version: 2012-06-19):
com/q/8337 Answer. URL (version: 2013-12-11): We can use the function f (x) = x2 with −π ≤x ≤π and find its expansion into a trigonometric Fourier series a0 2 + ∞ ∑ n=1 (an cos nx + bn sin x), which is periodic and converges to f (x) in −π ≤x ≤π. Observing that f (x) is even, it is enough to determine the coefficients an = 1 π Z π −π f (x) cos nx dx n = 0, 1, 2, 3, ..., because bn = 1 π Z π −π f (x) sin nx dx = 0 n = 1, 2, 3, ....
For n = 0 we have 43 44 Chap. 4. Fourier Series a0 = 1 π Z π −π x2dx = 2 π Z π 0 x2dx = 2π2 3 .
And for n = 1, 2, 3, ... we get an = 1 π Z π −π x2 cos nx dx = 2 π Z π 0 x2 cos nx dx = 2 π × 2π n2 (−1)n = (−1)n 4 n2, because Z x2 cos nx dx = 2x n2 cos nx + x2 n −2 n3 sin nx.
Thus f (x) = π2 3 + ∞ ∑ n=1 (−1)n 1 n2 cos nx .
Since f (π) = π2, we obtain π2 = π2 3 + ∞ ∑ n=1 (−1)n 4 n2 cos (nπ) π2 = π2 3 + 4 ∞ ∑ n=1 (−1)n(−1)n 1 n2 π2 = π2 3 + 4 ∞ ∑ n=1 1 n2.
Therefore ∞ ∑ n=1 1 n2 = π2 4 −π2 12 = π2 6 Second method (available on-line a few years ago) by Eric Rowland. From log(1 −t) = − ∞ ∑ n=1 tn n and making the substitution t = eix one gets the series expansion 4.1. Trigonometric series 45 w = Log(1 −eix) = − ∞ ∑ n=1 einx n = − ∞ ∑ n=1 1 n cos nx −i ∞ ∑ n=1 1 n sin nx, whose radius of convergence is 1. Now if we take the imaginary part of both sides, the RHS becomes ℑw = − ∞ ∑ n=1 1 n sin nx, and the LHS ℑw = arg (1 −cos x −i sin x) = arctan −sin x 1 −cos x.
Since arctan −sin x 1 −cos x = −arctan 2 sin x 2 · cos x 2 2 sin2 x 2 = −arctan cot x 2 = −arctan tan π 2 −x 2 = x 2 −π 2 , the following expansion holds π 2 −x 2 = ∞ ∑ n=1 1 n sin nx.
(∗) Integrating the identity (∗), we obtain π 2 x −x2 4 + C = − ∞ ∑ n=1 1 n2 cos nx.
(∗∗) Setting x = 0, we get the relation between C and ζ(2) C = − ∞ ∑ n=1 1 n2 = −ζ(2).
And for x = π, since 46 Chap. 4. Fourier Series ζ(2) = 2 ∞ ∑ n=1 (−1)n−1 n2 , we deduce π2 4 + C = − ∞ ∑ n=1 1 n2 cos nπ = ∞ ∑ n=1 (−1)n−1 n2 = 1 2ζ(2) = −1 2C.
Solving for C C = −π2 6 , we thus prove ζ(2) = π2 6 .
Note: this 2nd method can generate all the zeta values ζ(2n) by integrating repeatedly (∗∗). This is the reason why I appreciate it. Unfortunately it does not work for ζ(2n + 1).
4.1.2 Fourier Series for | cos(x)| Question by Cloud15 ( Fourier Series for | cos(x)|, URL (version: 2014-03-04):
com/q/697843 I’m having trouble figuring out the Fourier series of | cos(x)| from −π to π.
I understand its an even function, so all the bns are 0 a0 = 2 π Z π 0 | cos(x)| dx = 0 an = 2 π Z π 0 | cos(x)| cos(nx) dx = 2 π Z π 0 cos2(x) dx.
Since for all j, k not equal the integral is zero.
So only a1 remains. is this correct?
How would I evaluate ∑∞ n=1(−1)n−1/(4n2 −1) ?
4.1. Trigonometric series 47 Answer. URL (version: 2014-03-04): Although R π 0 cos(x) dx = 0, a0 ̸= 0 because Z π/2 0 | cos(x)| dx = Z π π/2 | cos(x)| dx.
We can evaluate it as follows, as can be seen in the plot below a0 = 1 π Z π −π | cos(x)| dx = 2 π Z π 0 | cos(x)| dx = 4 π Z π/2 0 | cos(x)| dx = 4 π Z π/2 0 cos(x) dx = 4 π.
(1) Plot of cos x (doted line) and | cos x| (solid line) in the interval [−π, π].
The coefficients bn = 0 as you concluded. As for the an coefficients only the odd ones are equal to 0 (see below). The functions cos(x) and cos(nx) are orthogonal in the interval [−π, π], but | cos(x)| and cos(nx) are not. Since |cos(x)| = cos(x) −cos(x) if if 0 ≤x ≤π/2 π/2 ≤x ≤π, (2) we have that 48 Chap. 4. Fourier Series an = 1 π Z π −π |cos(x)| cos(nx) dx = 2 π Z π 0 |cos(x)| cos(nx) dx = 2 π Z π/2 0 |cos(x)| cos(nx) dx + 2 π Z π π/2 |cos(x)| cos(nx) dx = 2 π Z π/2 0 cos(x) cos(nx) dx −2 π Z π π/2 cos(x) cos(nx) dx.
a1 = 2 π Z π/2 0 cos2(x) dx −2 π Z π π/2 cos2(x) dx = 0.
Using the following trigonometric identity, with a = x, b = nx, cos(a) cos(b) = cos(a + b) + cos(a −b) 2 , (3) we find a2m = 4 π (1 −4m2) cos(2mπ 2 ) = 4 π (1 −4m2)(−1)m a2m+1 = 4 π(1 −4(2m + 1)2) cos((2m + 1)π 2 ) = 0, m = 1, 2, 3, . . . .
(4) The expansion of |cos(x)| into a trigonometric Fourier series in the interval [−π, π] is thus |cos x| = a0 2 + ∞ ∑ n=1 (an cos(nx) + bn sin(nx)) = 2 π + 4 π ∞ ∑ m=1 (−1)m 1 −4m2 cos(2mx) (5) | sin(x)| (blue) and the partial sum 2 π + 4 π 5 ∑ m=1 (−1)m 1 −4m2 cos(2mx) (red) in [−π, π] 4.1. Trigonometric series 49 Setting x = 0 in (5), we obtain 1 = 2 π + 4 π ∞ ∑ m=1 (−1)m 1 −4m2 = 2 π −4 π ∞ ∑ n=1 (−1)n−1 1 −4n2 .
(6) Hence ∞ ∑ n=1 (−1)n−1 1 −4n2 = 1 2 −π 4 .
(7) Chap. 5 Complex Analysis 5.1 Laurent Series 5.1.1 Finding the Laurent series of f (z) = 1/((z −1)(z −2)) Question by Freeman ( Finding the Laurent series of f (z) = 1/((z −1)(z −2)), URL (version: 2011-11-04): Let f (z) = 1 (z −1)(z −2) and let R1 = n z 1 < |z| < 2 o and R2 = n z |z| > 2 o .
How do you find the Laurent series convergent on R1? Also how do you do it for R2?
I’m having serious trouble with this as I can’t see how to expand things into se-ries with n as any integer, not just natural number. Also how to apply Cauchy’s integral formula to an annulus. If anyone can explain this to me I will be ex-tremely grateful.
Answer. URL (version: 2011-11-04): The function f (z) can be expanded into two partial fractions 51 52 Chap. 5. Complex Analysis f (z) := 1 (z −1) (z −2) = 1 z −2 − 1 z −1.
We now expand each partial fraction into a geometric series. On R2 these series are 1 z −2 = 1 z (1 −2/z) = 1 z ∞ ∑ n=0 2 z n |z| > 2 = 1 z ∞ ∑ n=0 2n 1 zn = ∞ ∑ n=0 2n 1 zn+1 and 1 z −1 = 1 z (1 −1/z) = 1 z ∞ ∑ n=0 1 z n = ∞ ∑ n=0 1 zn+1 |z| > 1.
And so, the Laurent series is 1 (z −1) (z −2) = ∞ ∑ n=0 1 zn+1(2n −1) |z| > 2 > 1.
On R1, the two geometric series are 1 z −2 = −1/2 1 −z/2 = ∞ ∑ n=0 −1 2 z 2 n |z| < 2 = ∞ ∑ n=0 − 1 2n+1 zn and 1 z −1 = 1/z 1 −1/z = ∞ ∑ n=0 1 z 1 z n |z| > 1 = ∞ ∑ n=0 1 zn+1.
We thus get the following Laurent series 1 (z −1) (z −2) = ∞ ∑ n=0 − 1 2n+1 zn − 1 zn+1 1 < |z| < 2.
5.2. Evaluating definite integrals by the residue theorem 53 5.2 Evaluating definite integrals by the residue the-orem 5.2.1 Verify integrals with residue theorem Question by emka ( Ver-ify integrals with residue theorem, URL (version: 2011-10-13):
stackexchange.com/q/71932 This is another problem that I got stuck on during my self-study of Complex Variables.
Z ∞ 0 cos(ax) (1 + x2)2 dx with a > 0 This is equivalent to 1 2 Z ∞ −∞ cos(ax) (1 + x2)2 dx I know the following: (1) There is a pole at x = i (2) We work with the half-circle with radius R.
Answer. URL (version: 2011-10-13): Let f (z) = eiaz (1 + z2)2 = eiaz (z −i)2(z + i)2.
(1) The residue of f (z) at z = i is1 res z=i f (z) = 1 (2 −1)! lim z→i d dz (z −i)2 f (z) = lim z→i d dz eiaz (z + i)2 = lim z→i iaeiaz(z + i)2 −eiaz2 (z + i) (z + i)4 = −1 4i (a + 1) e−a.
(2) 1 54 Chap. 5. Complex Analysis r × x y O Cr −r i Let CR denote the boundary of the upper half of the disk |z| = R, described counterclockwise (see picture). By the residue theorem2 Z R −R eiax (1 + x2)2 dx + Z CR eiaz (1 + z2)2 dz = 2πi res z=i f (z)eiaz = 1 2π (a + 1) e−a.
(3) Then Re Z R −R eiax (1 + x2)2 dx + Re Z CR eiaz (1 + z2)2 dz = Re1 2π (a + 1) e−a, Z R −R cos ax (1 + x2)2 dx + Re Z CR eiaz (1 + z2)2 dz = 1 2π (a + 1) e−a.
(4) When |z| = R, we have 1 (1 + z2)2 = 1 |z + i|2 |z −i|2 ≤ 1 ||z| −|i|2 ||z| −|i||2 = 1 (R −1)4 =: MR, (5) which means that MR > 0 and lim R→∞MR = lim R→∞ 1 (R −1)4 = 0.
(6) Then we can apply the Jordan’s lemma for every positive constant a and con-clude that 2 5.2. Evaluating definite integrals by the residue theorem 55 lim R→∞ Z CR eiaz (1 + z2)2 dz = 0.
(7) Consequently, Z ∞ −∞ cos ax (1 + x2)2 dx = 1 2π (a + 1) e−a (8) and Z ∞ 0 cos ax (1 + x2)2 dx = 1 4π (a + 1) e−a.
(9) Note: Exercise 3 on page 265 of generalizes this integral to Z ∞ 0 cos ax (b2 + x2)2 dx = 1 4b3 π (ab + 1) e−ab (a > 0, b > 0).
5.2.2 Evaluate an improper integral using complex analysis Question by Simplyorange ( simplyorange), VEvaluate an improper integral using complex analysis, URL (version: 2020-06-21): I got stuck trying to find Z ∞ 0 log x (x + a)2 + b2 dx using complex analysis.
My attempt is to evaluate the contour integral of Z C log z (z + a)2 + b2 dz for some nicely chosen contour C. [...] Answer. URL (version: 2020-07-09): 56 Chap. 5. Complex Analysis Based on this3 answer to the question4 "How to evaluate R ∞ 0 log x (x2+a2)2 dx" and on this5 answer to the question6 "Evaluate R ∞ 0 (log x)4dx (1+x)(1+x2)". Instead of a function with log z on the numerator, we consider a function with log2 z. This is the very same method as that pointed to in the comments.
For a, b > 0, this method gives the closed formula Z ∞ 0 log x (x + a)2 + b2 dx = 1 2b arctan b a log a2 + b2 , a, b > 0.
(∗) We chose the multiple-valued function f (z) with branch cut arg z = 0 de-fined as f (z) = log2 z (z + a)2 + b2, with 0 < arg z < 2π, z = reiθ = log2 z (z −z1) (z −z2) z1 = −a + ib, z2 = −a −ib, and integrate it counterclockwise around the closed contour Γ shown in the figure. This contour is indented around the branch point O and consists of the circles γR (|z| = R) and γε (|z| = ε), 0 < ε < 1 < R, and the segment [ε, R] described in the positive sense above the x-axis and in the negative sense below the x-axis.
ε R × × x y O γε γR z1 z2 Γ 3 4 5 6 5.2. Evaluating definite integrals by the residue theorem 57 On the upper edge, θ = 0 (r ∈[ε, R]) and f (z) = (log r)2 (r + a)2 + b2.
On the lower edge, θ = 2π (r ∈[ε, R]) and f (z) = log rei2π2 (r + a)2 + b2 = (log r + i2π)2 (r + a)2 + b2 .
As such, I = lim ε→0,R→∞ I Γ (log z)2 (z + a)2 + b2 dz, = Z ∞ 0 (log r)2 (r + a)2 + b2 dr − Z ∞ 0 log rei2π2 (rei2π + a)2 + b2 dr + lim R→∞ Z γR (log z)2 (z + a)2 + b2 dz −lim ε→0 Z γε (log z)2 (z + a)2 + b2 dz = Z ∞ 0 (log r)2 −(log r + i2π)2 (r + a)2 + b2 dx = 4π2 Z ∞ 0 1 (r + a)2 + b2 dr −i4π Z ∞ 0 log r (r + a)2 + b2 dr provided that lim R→∞ Z γR (log z)2 (z + a)2 + b2 dz = lim ε→0 Z γε (log z)2 (z + a)2 + b2 dz = 0, (see below).
By the residue theorem, 58 Chap. 5. Complex Analysis I = 2πi (Resz=z1 f (z) + Resz=z2 f (z)) = 2πi " Resz=z1 (log z)2 (z −z1) (z −z2) + Resz=z2 (log z)2 (z −z1) (z −z2) # = 2πi " (log z1)2 z1 −z2 + (log z2)2 z2 −z1 # = 2πi " (log (−a + ib))2 i2b −(log (−a −ib))2 i2b # = π b [log (−a + ib)]2 −π b [log (−a −ib)]2 We now assume that a, b > 0. Then I = π b log (|−a + ib|) + i π −arctan b a 2 −π b log (|−a −ib|) + i π + arctan b a 2 = π b 1 2 log a2 + b2 + i π −arctan b a 2 −π b 1 2 log a2 + b2 + i π + arctan b a 2 = 4π2 b arctan b a −i2π b arctan b a log a2 + b2 because log (|−a + ib|) = log (|−a −ib|) = 1 2 log a2 + b2 .
Taking the imaginary part of I we obtain (∗) in the form Im (I) = −4π Z ∞ 0 log r (r + a)2 + b2 dr = −2π b arctan b a log a2 + b2 Proof that R γR f, R γε f →0. If z is any point on γR, 5.2. Evaluating definite integrals by the residue theorem 59 | f (z)| = |log z|2 (z + a)2 + b2 , z = R eiθ, R > 1, 0 < θ < 2π ≤ (log R + 2π)2 |z + (−z1)| |z + (−z2)|, ≤ (log R + 2π)2 R − √ a2 + b2 2 ≤MR where MR = 4π log R + 4π2 + log2 R R2 + (a2 + b2) −2R √ a2 + b2 because |z + (−z1)| ≥|R −|z1|| , |z + (−z2)| ≥|R −|z2|| , |z1| = |z2| = p a2 + b2.
This means that Z γR f (z) dz ≤MR × 2πR = 4π log R + 4π2 + log2 R R2 + (a2 + b2) −2R √ a2 + b2 × 2πR − →0 (R →∞) .
Similarly, if z is any point on γε | f (z)| = |log z|2 (z + a)2 + b2 , z = ε eiθ, 0 < ε < 1, 0 < θ < 2π ≤ (log ε + 2π)2 |z + (−z1)| |z + (−z2)| ≤ (log ε + 2π)2 ε − √ a2 + b2 2 ≤Mε, where 60 Chap. 5. Complex Analysis Mε = 4π log ε + 4π2 + log2 ε ε2 + (a2 + b2) −2ε √ a2 + b2 and Z γε f (z) dz ≤Mε × 2πε z = ρ eiθ, ρ < 1 ≤ 4π log ε + 4π2 + log2 ε ε2 + (a2 + b2) −2ε √ a2 + b2 × 2πε − →0 (ε →0) .
Part II Questions 61 Chap. 6 A few questions 6.1 Real Analysis and Calculus 6.1.1 Asymptotic (divergent) series Question Asymptotic (divergent) series, URL (version: 2013-06-21): http:// math.stackexchange.com/q/123632 MOTIVATION. After having read in detail an article by Alf van der Poorten I read a very short paper by Roger Apéry. I am interested in finding a proof of a series expansion in the latter, which is in not given in it. So I assumed it should be stated or derived from a theorem on the subject.
In Apéry, R., Irrationalité de ζ2 et ζ3, Société Mathématique de France, Astérisque 61 (1979) there is a divergent series expansion for a function I would like to un-derstand. Here is my translation of the relevant part for this question (...) given a real sequence a1, a2, . . . , ak, an analytic function f (x) with respect to the variable 1 x tending to 0 with 1 x admits a (unique) expansion in the form f (x) ≡∑ k≥1 ck (x + a1) (x + a2) . . . (x + ak).
(A) (...) and the translation by Generic Human ( users/26855/generic-human) of the text after the formula: (We write ≡instead of = to take into account the aversions of mathemati-cians who, following Abel, Cauchy and d’Alembert, hold divergent series to 63 64 Chap. 6. A few questions be an invention of the devil; in fact, we only ever use a finite sum of terms, but the number of terms is an unbounded function of x.) Remark. As far as I understand, based on this last text, the expansion of f (x) in (A) is in general a divergent series and not a convergent one, but the existing answer [by WimC] seems to indicate the opposite.
The corresponding finite sum appears and is proved in section 3 of Alfred van der Poorten’s article A proof that Euler missed ... Apéry’s proof of the irrational-ity of ζ(3) as For all a1, a2, . . .
K ∑ k=1 a1a2 · · · ak−1 (x + a1)(x + a2) · · · (x + ak) = 1 x − a1a2 · · · aK x(x + a1)(x + a2) · · · (x + aK), (A’) Questions: 1. Is series (A) indeed divergent?
2. Which is the theorem stating or from which expansion (A) can be derived?
3. Could you please indicate a reference?
Answer by robjohn ( Asymptotic (divergent) series, URL (version: 2012-06-13):
com/q/157890 Writing g1(x) = f (1/x) gives g1(x) ≡∑ k≥1 ckxk (1 + a1x)(1 + a2x) . . . (1 + akx) (1) which vanishes at x = 0.
Recursively define gn+1(x) = (1 + anx)gn(x) x −cn (2) where 6.1. Real Analysis and Calculus 65 cn = lim x→0 gn(x) x (3) Then gn(x) ≡∑ k≥n ckxk−n+1 (1 + anx)(1 + an+1x) . . . (1 + akx) (4) is another series like (1) (which vanishes at x = 0).
The series in (1) may or may not converge, as with the Euler-Maclaurin Sum Series. As with most asymptotic series, we are only interested in the first several terms; the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms. Therefore, convergence is not an issue.
6.1.2 What is the importance of Calculus in today’s Mathemat-ics?
Question What is the importance of Calculus in today’s Mathematics?, URL (version: 2013-04-29): For engineering (e. g. electrical engineering) and physics, Calculus is im-portant. But for a future mathematician, is the classical approach to Calculus still important? What is normally taught, as a minimum, in most Universities worldwide?
(...) Edit: Hoping it is useful, I transcribe three comments of mine (to this ques-tion): 1. I had [have] in mind for instance Tom Apostol’s books, although learning differentiation before integration. (in response to Qiaochu Yuan’s "What is the classical approach to calculus?") 2. Elementary Calculus, continuous functions, functions of several variables, partial differentiation, implicit-functions, vectors and vector fields, multi-ple integrals, infinite series, uniform convergence, power series, Fourier series and integrals, etc. (in response to a comment by Geoff Robinson).
66 Chap. 6. A few questions 3. I had [have] in mind calculus for math students, although I am a retired electrical engineer.
(in response to Andy’s comment "Are you talking about what is usually taught to engineers and physicists, or also about a calculus curriculum for math majors? ") (...) Answer by Pete L. Clark ( What is the importance of Calculus in today’s Mathematics?, URL (version: 2012-12-20): In a comment to his question, Américo has clarified that by "classical calcu-lus" he means something relatively rigorous and theoretical, as for instance in Apostol’s book (or Spivak’s). I think the answer to the question was probably yes no matter what, but when restricted in this way it becomes a big booming YES.
The methods of rigorous calculus – may I say elementary real analysis? it seems more specific – are an indispensable part of the cultural knowledge of all mathe-maticians, pure and applied. Not all mathematicians will directly use this mate-rial in their work: I for one am a mathematician with relatively broad interests almost to a fault, but I have never written "by the Fundamental Theorem of Calculus" or "by the Mean Value Theorem" in any of my research papers. But nevertheless familiarity and even deep understanding of these basic ideas and themes permeates all of modern mathematics. For instance, as an arithmetic geometer the functions I differentiate are usually polynomials or rational func-tions, but the idea of differentiation is still there, in fact abstracted in the notion of derivations and modules of differentials. One of the most important con-cepts in algebraic / arithmetic geometry is smoothness, and although you could in principle try to swallow this as a piece of pure algebra, I say good luck with that if you have never taken multivariable calculus and understood the inverse and implicit function theorems.
Eschewing "classical" mathematics in favor of more modern, abstract or spe-cialized topics is one of the biggest traps a bright young student of mathematics can fall into. (If you spend any time at a place like Harvard, as I did as a grad-uate student, you see undergraduates falling for this with distressing regular-ity, almost as if the floor outside your office was carpeted with banana peels.) The people who created the fancy modern machinery did so by virtue of their knowledge of classical stuff, and are responding to it in ways that are profound even if they are unfortunately not made explicit. Although I am very far from 6.1. Real Analysis and Calculus 67 really knowing what I’m talking about here, my feeling is that the analogy to the fine arts is rather apt: abstract modern art is very much a response to clas-sical, figurative, realistic (I was tempted to say "mimetic", so I had better end this digression soon!) art: if you decide to forego learning about perspective in favor of arranging black squares on a white canvas, you’re severely missing the point.
The material of elementary real analysis – and even freshman calculus – is re-markably rich. I have taught more or less the same freshman calculus courses about a dozen times, and each time I find something new to think about, some-times in resonance with my other mathematical thoughts of the moment but sometimes I just find that I have the chance to stop and think about something that never occurred to me before. Once for instance I was talking about com-puting volumes of solids of revolution and it occurred to me that I had never thought about proving in general that the method of shells will give the same answer as the method of washers. It was pretty good fun to do it, and I men-tioned it to a couple of my colleagues and they had a similar reaction: "No, I never thought of that before, but it sounds like fun." There are thousands of little projects and discoveries like this in freshman calculus.
I confess though that it would be interesting to hear mathematicians talk about parts of calculus that they never liked and never had any use for. As for me, I really dread the part of the course where we do related rates problems and min / max problems. The former seems like an exercise whose only point is to exploit – sometimes to the point of cruelty – the shakiness of students’ under-standing of implicit differentiation, and the latter was sort of fun for me for the first ten problems but twenty years and thousands of min / max problems later I could hardly imagine something more tedious. (Moreover I am not that good at these problems. I had a couple of embarrassing failures as a graduate student, and ever since I look to make sure I can do the problems before I assign them, something I have stopped needing to do in most other undergraduate courses.) Added: Let me be explicit that I am not answering the second part of the ques-tion, i.e., what is a minimum that is or should be taught. It goes hand in hand with the richness of these topics that if you tried to make a list of everything that it would be valuable for students to know, your (surely severely incomplete!) list would contain vastly more material than could be reasonably covered in the allotted courses. This is one subject where books which aim to be "comprehen-sive" come off as pretty daunting. For instance I own the first of Courant and John’s two volumes on advanced calculus: it’s more than six hundred pages!
Is there anything in there which I am willing to point to as "dispensable"? Not 68 Chap. 6. A few questions much. (Not to mention that the second volume of their work comes in two parts, the second part of which is itself 954 pages long!) The challenge of teaching these courses lies in the fact that the potential landscape is almost infinite and virtually none of it manifestly unimportant, so you have to make hard choices about what not to do.
6.1.3 What does closed form solution usually mean?
Question What does closed form solution usually mean?, URL (version: 2011-11-25): This is motivated by this question1 and the fact that I have no access to Tim-othy Chow’s paper What Is a Closed-Form Number?2 indicated there by Qiaochu Yuan.
If an equation f (x) = 0 has no closed form solution, what does it normally mean? Added: f may depend (and normally does) on parameters.
To me this is equivalent to say that one cannot solve it for x in the sense that there is no elementary expression g(c1, c2, . . . , cp) consisting only of a finite number of polynomials, rational functions, roots, exponentials, logarithmic and trigonometric functions, absolute values, integer and fractional parts, such that f (g(c1, c2, . . . , cp)) = 0.
Answer by J. M. ( What does closed form solution usually mean?, URL (version: 2010-11-07):
stackexchange.com/q/9203 I would say it very much depends on the context, and what tools are at your disposal. For instance, telling a student who’s just mastered the usual tricks of integrating elementary functions that Z exp u −1 u du and Z q (u + 1)(u2 + 1)du 1 2 6.2. Number Theory 69 have no closed form solutions is just the fancy way of saying "no, you can’t do these integrals yet; you don’t have the tools". To a working scientist who uses exponential and elliptic integrals, however, they do have closed forms.
In a similar vein, when we say that nonlinear equations, whether algebraic ones like x5 −x + 1 = 0 or transcendental ones like π 4 = v −sin v 2 have no closed form solutions, what we’re really saying is that we can’t represent solutions to these in terms of functions that we know (and love?). (For the first one, though, if you know hypergeometric or theta functions, then yes, it has a closed form.) I believe it is fair to say that for as long as we haven’t seen the solution to an integral, sum, product, continued fraction, differential equation, or nonlinear equation frequently enough in applications to give it a standard name and nota-tion, we just cop out and say "nope, it doesn’t have a closed form".
6.2 Number Theory 6.2.1 Binomial coefficients: how to prove an inequality on the p-adic valuation?
Question Binomial coefficients: how to prove an inequality on the p-adic valu-ation?, URL (version: 2014-05-25): In section 4 of the article by Afred van der Poorten’s A Proof That Euler Missed ... the following inequality is used: νp n m ≤ ln n ln p −νp(m) (∗) (In the original denoted ord p(·) instead of νp(·)), where νp(k) is the p-adic valuation of k ∈Q, i. e. the exponent of the prime p in the prime factorization of k. I know some properties of the foor function and that νp(a/b) = νp(a) −νp(b), νp(a · b) = νp(a) · νp(b) and 70 Chap. 6. A few questions νp(n!) = ∑ i≥1 n pi but I didn’t convince myself on the correct argument I should use to prove (∗).
Question: How can this inequality be proven?
If an equation f (x) = 0 has no closed form solution, what does it normally mean? Added: f may depend (and normally does) on parameters.
Answer by Grigory M ( Binomial coefficients: how to prove an inequality on the p-adic valuation?, URL (version: 2010-08-19): Since νp(n!) = ∑i⌊n/pi⌋, νp(n m) = ∑i(⌊n/pi⌋−⌊m/pi⌋−⌊(n −m)/pi⌋).
Observe that each summand is ≤1, and for i > j ln n ln p k it’s clearly 0. That gives vp(n m) ≤ j ln n ln p k .
Finally observe that for i ≤νp(m) ⌊m/pi⌋= m/pi, so (⌊m/pi⌋+ ⌊x/pi⌋= ⌊(m + x)/pi⌋and) the corresponding summand is also 0. That gives the in-equality in question.
The appendix fragment is used only once. Subsequent appendices can be created using the Chapter Section/Body Tag.
Bibliography Lloyd, Michael “Pursuit Curves”, Faculty/Academic_Forum/2006-7/2006-7AFPursuit.pdf Wikipedia Entry “Pursuit Curves”, Pursuit_curve German Wikipedia Entry “Radiodrome”, Radiodrome Knaust, Helmut “The Curve of Pursuit”, Math 3226 Laboratory 2B , http: //www.math.utep.edu/classes/3226/lab2b/lab2b.html Calado, J., “Compêndio de Trigonometria”, Empresa Literária Fluminense Lda., Lisboa, 1967 Silva, S. and Paulo, S. “Compêndio de Álgebra”, Vol II, Livraria Polular de Francisco Franco, Lisboa, 1963 Brown, J., Churchill, R. “Complex Variables and Applications”, McGraw Hill, International edition, 2003 Spiegel, M. and Liu, J. Manual de Fórmulas e Tabelas Matemáticas, Coleção Schaum, Portuguese translation of Schaum’s Outline Series Mathematical Handbook of Formulas and tables, 2/e Agudo, D. and Silva, C., Matemáticas Gerais III Balakrishnan Combinatorics, Schaum’s Outline of Combinatorics 71 Mathematics Stack Exchange swags |
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Differentiation of Trigonometric Functions
The process of finding the derivatives of trigonometric functions is known as the differentiation of trigonometric functions. In other words, the differentiation of trigonometric functions is finding the rate of change of the function with respect to the variable. The six trigonometric functions have differentiation formulas that can be used in various application problems of the derivative.
The six basic trigonometric functions include the following: sine (sin x), cosine (cos x), tangent (tan x), cotangent (cot x), secant (sec x) and cosecant (cosec x). In this article, we will find the derivatives of the trigonometric functions and their proofs. Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
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| 1. | What is the Differentiation of Trigonometric Functions? |
| 2. | Proofs of Trig Derivatives |
| 3. | Applications of Differentiation of Trigonometric Functions |
| 4. | Differentiation of Inverse Trigonometric Functions |
| 5. | Anti-Differentiation of Trigonometric Functions |
| 6. | FAQs on Differentiation of Trigonometric Functions |
What is the Differentiation of Trigonometric Functions?
In trigonometry, differentiation of trigonometric functions is a mathematical process of determining the rate of change of the trigonometric functions with respect to the variable angle. The differentiation of trigonometric functions can be done using the derivatives of sin x and cos x by applying the quotient rule. The differentiation formulas of the six trigonometric functions are listed below:
Derivation of sin x: (sin x)' = cos x
Derivative of cos x: (cos x)' = -sin x
Derivative of tan x: (tan x)' = sec2 x
Derivative of cot x: (cot x)' = -cosec2 x
Derivative of sec x: (sec x)' = sec x.tan x
Derivative of cosec x: (cosec x)' = -cosec x.cot x
We use d/dx to write the derivatives. Here are the tri derivatives using this notation.
Proofs of Trig Derivatives
Now, that we have the differentiation of trigonometric functions (sin x, cos x, tan x, cot x, sec x, cosec x), we will prove and derive the trig derivatives using various methods such as the quotient rule, the first principle of differentiation, and chain rule along with some limit formulas.
Derivative of sin x
We will derive the derivative of sin x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function sin x, we will use the following limit and trigonometric formulas:
sin (A+B) = sin A cos B + sin B cos A
(\lim_{x\rightarrow 0} \dfrac{\cos x -1}{x} = 0)
(\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1)
Now, we will calculate the differentiation of trigonometric function sin x:
(\begin{align}\frac{\mathrm{d} (\sin x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\sin (x + h)-\sin x}{(x+h)-x} \&= \lim_{h\rightarrow 0} \dfrac{\sin x \cos h +\cos x \sin h-\sin x}{h}\&=\lim_{h\rightarrow 0} \dfrac{\cos h -1 }{h}\sin x + \dfrac{\sin h}{h}\cos x\&=(0)\sin x + (1)\cos x\&=\cos x\end{align})
Therefore, d(sin x)/dx = cos x
Derivative of cos x
We will derive the derivative of cos x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function cos x, we will use the following limit and trigonometric formulas:
cos (A + B) = cos A cos B - sin A sin B
(\lim_{x\rightarrow 0} \dfrac{\cos x -1}{x} = 0)
(\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1)
Thus, we have
(\begin{align}\frac{\mathrm{d} (\cos x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\cos (x + h)-\cos x}{(x+h)-x} \&= \lim_{h\rightarrow 0} \dfrac{\cos x \cos h -\sin x \sin h-\cos x}{h}\&=\lim_{h\rightarrow 0} \dfrac{\cos h -1 }{h}\cos x - \dfrac{\sin h}{h}\sin x\&=(0)\cos x - (1)\sin x\&=-\sin x\end{align})
Therefore, d(cos x)/dx = -sin x
Derivative of tan x
We will determine the derivative of tan x using the quotient rule. We will use the following formulas and identities to calculate the derivative:
(sin x)' = cos x
(cos x)' = -sin x
tan x = sin x/ cos x
cos2x + sin2x = 1
sec x = 1/cos x
(tan x)' = (sin x/cos x)'
= [(sin x)' cos x - (cos x)' sin x]/cos2x
= [cos x. cos x - (-sin x). sin x]/cos2x
= (cos2x + sin2x)/cos2x
= 1/cos2x
= sec2x
Therefore, d(tan x)/dx = sec2x
Derivative of cot x
We will determine the derivative of cot x using the quotient rule. We will use the following formulas and identities to calculate the derivative:
(sin x)' = cos x
(cos x)' = -sin x
cot x = cos x/ sin x
cos2x + sin2x = 1
cosec x = 1/sin x
(cot x)' = (cos x/sin x)'
= [(cos x)' sin x - (sin x)' cos x]/sin2x
= [-sin x. sin x - cos x. cos x]/sin2x
= (-sin2x - cos2x)/sin2x
= -1/sin2x
= -cosec2x
Therefore, d(cot x)/dx = -cosec2x
Derivative of sec x
We will determine the derivative of sec x using the chain rule. We will use the following formulas and identities to calculate the derivative:
sec x = 1/cos x
tan x = sin x/cos x
(cos x)' = -sin x
(sec x)' = (1/cos x)' = (-1/cos2x).(cos x)'
= (-1/cos2x).(-sin x)
= sin x/cos2x
= (sin x/cos x).(1/cos x)
= tan x sec x
Therefore, d(sec x)/dx = tan x sec x
Derivative of cosec x
We will determine the derivative of cosec x using the chain rule. We will use the following formulas and identities to calculate the derivative:
cosec x = 1/sin x
cot x = cos x/sin x
(sin x)' = cos x
(cosec x)' = (1/sin x)' = (-1/sin2x).(sin x)'
= (-1/sin2x).(cos x)
= -cos x/sin2x
= -(cos x/sin x).(1/sin x)
= -cot x cosec x
Therefore, d(cosec x)/dx = -cot x cosec x
Applications of Differentiation of Trigonometric Functions
The differentiation of trigonometric functions has various applications in the field of mathematics and real life. Some of them are listed below:
It is used to determine the slope of the tangent line to a trigonometric curve y = f(x).
It is used to determine the slope of the normal line to a trigonometric curve y = f(x).
It helps to determine the equation of the tangent line or normal line of a curve.
Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
We use the derivatives of trigonometric functions to determine the maximum and minimum values of particular functions.
Differentiation of Inverse Trigonometric Functions
The differentiation of inverse trigonometric functions is done by setting the function equal to y and applying implicit differentiation. Let us list the derivatives of the inverse trigonometric functions along with their domains (arcsin x, arccos x, arctan x, arccot x, arcsec x, arccosec x):
(arcsin x)' = 1/√(1 - x2) , -1 < x < 1
(arccos x)' = -1/√(1 - x2) , -1 < x < 1
(arctan x)' = 1/(1 + x2) , -∞ < x < ∞
(arccot x)' = -1/(1 + x2) , -∞ < x < ∞
(arcsec x)' = 1/|x|√(x2 - 1) , x ∈ (-∞, -1) ∪ (1, ∞)
(arccosec x)' = -1/|x|√(x2 - 1) , x ∈ (-∞, -1) ∪ (1, ∞)
Anti-Differentiation of Trigonometric Functions
Anti differentiation of trigonometric functions is the reverse process of differentiation of trigonometric functions. This process is also called the integration of trigonometric functions. The list of anti-derivatives of the trigonometric functions is given below as:
∫ sinx dx = -cos x + C
∫ cosx dx = sin x + C
∫ tanx dx = ln |sec x| + C
∫ cotx dx = ln |sin x| + C
∫ secx dx = ln |sec x + tan x| + C
∫ cosecx dx = -ln |cosec x + cot x| + C
Here C is the constant of integration.
☛ Related Topics:
Inverse Trigonometric Formulas
sin cos tan
Trigonometric Identities
Sine Function
Important Notes on Differentiation of Trigonometric Functions:
Derivation of sin x: (sin x)' = cos x
Derivative of cos x: (cos x)' = -sin x
Derivative of tan x: (tan x)' = sec2 x
Derivative of cot x: (cot x)' = -cosec2 x
Derivative of sec x: (sec x)' = sec x.tan x
Derivative of cosec x: (cosec x)' = -cosec x.cot x
Trig Derivatives Examples
Example 1: Show the differentiation of trigonometric function cos x using the chain rule.
Solution: The chain rule for differentiation is: (f(g(x)))’ = f’(g(x)) . g’(x). Now, to evaluate the derivative of cos x using the chain rule, we will use certain trigonometric properties and identities such as:
(\cos (\dfrac{\pi}{2} - \theta) = \sin \theta)
(\sin (\dfrac{\pi}{2} - \theta) = \cos \theta)
d(sin x)/dx = cos x
Using the above three trigonometric properties, we can write the derivative of cos x as the derivative of sin (π/2 - x), that is, d(cos x)/dx = d (sin (π/2 - x))/dx . Using chain rule, we have,
(\begin{align} \frac{\mathrm{d} \cos x}{\mathrm{d} x} &=\frac{\mathrm{d} \sin(\dfrac{\pi}{2}-x)}{\mathrm{d} x}\&=\cos(\dfrac{\pi}{2}-x).(-1)\&=-\cos(\dfrac{\pi}{2}-x)\&= -\sin x\end{align})
Answer: Hence, we have derived the derivative of cos x as -sin x using chain rule.
2. Example 2: Show the differentiation of trigonometric function tan x by the first principle of differentiation.
Solution: We will derive the derivative of tan x using the first principle of differentiation, that is, using the definition of limits. To derive the differentiation of the trigonometric function tan x, we will use the following limit and trigonometric formulas:
tan (A + B) = (tan A + tan B)/(1 - tan A tan B)
(\lim_{x\rightarrow 0} \tan x = 0)
(\lim_{x\rightarrow 0} \dfrac{\tan x}{x} = 1)
tan x = sin x/ cos x
1 + tan2x = sec2x
(\begin{align}\frac{\mathrm{d} (\tan x)}{\mathrm{d} x} &= \lim_{h\rightarrow 0} \dfrac{\tan (x + h)-\tan x}{(x+h)-x} \&= \lim_{h\rightarrow 0} \dfrac{\frac{\tan x +\tan h}{1-\tan x \tan h}-\tan x}{h}\&=\lim_{h\rightarrow 0} \dfrac{\tan x +\tan h -\tan x+\tan^2 x \tan h }{h(1-\tan x \tan h)}\&=\lim_{h\rightarrow 0} \dfrac{\tan h}{h}\times \lim_{h\rightarrow 0} \dfrac{1+\tan^2 x}{(1-\tan x \tan h)} \&=1 \times \dfrac{\sec^2x}{1}\&=\sec^2x\end{align})
Answer: Hence, we have derived the derivative of tan x as sec2x using the first principle of differentiation.
3. Example 3: Show the differentiation of trigonometric function sec x by the quotient rule of differentiation.
Solution:
We know that sec and cos are reciprocals of each other.
d/dx (sec x) = d/dx (1/cos x)
Now, by applying the quotient rule,
= [ cos x d/dx (1) - 1 d/dx (cos x) ] / (cos2x)
Derivative of 1 is 0 and derivative of cos x is - sin x.
= [ 0 - (-sin x)] / (cos2x)
= (sin x) / (cos x) × 1 / (cos x)
= tan x sec x
Answer: The derivative of sec x is sec x tan x.
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FAQs on Differentiation of Trigonometric Functions
What is the Differentiation of Trigonometric Functions in Trigonometry?
In trigonometry, differentiation of trigonometric functions is a mathematical process of determining the rate of change of the trigonometric functions with respect to the angle variable. The process of finding the derivatives of circular trigonometric functions is known as the differentiation of trigonometric functions.
What are the Derivatives of the 6 Trig Functions?
The differentiation formulas of the six trigonometric functions are listed below:
Derivation of sin x: (sin x)' = cos x
Derivative of cos x: (cos x)' = -sin x
Derivative of tan x: (tan x)' = sec2 x
Derivative of cot x: (cot x)' = -cosec2 x
Derivative of sec x: (sec x)' = sec x.tan x
Derivative of cosec x: (cosec x)' = -cosec x.cot x
What are the Applications of Differentiation of Trigonometric Functions?
The differentiation of trigonometric functions has various applications in the field of mathematics and real life.
It helps to determine the equation of the tangent line or normal line of a curve.
Differentiation of trigonometric functions have applications in different fields such as electronics, computer programming and modeling different cyclic functions.
We use the differentiation of a trigonometric function to determine the maximum and minimum values of particular functions.
What is the Anti-Differentiation of Trigonometric Functions in Trigonometry?
Anti differentiation of trigonometric functions is the reverse process of differentiation of trigonometric functions. This process is also called the integration of trigonometric functions.
What are the Anti-Derivatives of the Six Trigonometric Functions?
The list of anti-derivatives of the trigonometric functions is given below as:
∫ sin x dx = -cos x + C
∫ cos x dx = sin x + C
∫ tan x dx = ln |sec x| + C
∫ cot x dx = ln |sin x| + C
∫ sec x dx = ln |sec x + tan x| + C
∫ cosec x dx = -ln |cosec x + cot x| + C
What are the Trig Derivatives?
Derivative of Trigonometric Functions can be calculated using various methods such as quotient rule, the first principle of differentiation, and chain rule along with some limit formulas.The trig derivatives are
(sin x)' = cos x
(cos x)' = -sin x
(tan x)' = sec2 x
(cot x)' = -cosec2 x
(sec x)' = sec x.tan x
(cosec x)' = -cosec x.cot x
How Do You Derive the Derivatives of Trigonometric Functions?
The derivatives of trigonometric functions can be found using different methods of differentiation such as the first principle of derivatives, product rule, quotient rule, and chain rule.
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11553 | https://stats.stackexchange.com/questions/470626/why-is-using-squared-error-the-standard-when-absolute-error-is-more-relevant-to | Skip to main content
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I recognize that parts of this topic have been discussed on this forum. Some examples:
Is minimizing squared error equivalent to minimizing absolute error? Why squared error is more popular than the latter?
Why square the difference instead of taking the absolute value in standard deviation?
What I still don't understand is why OLS regression is still the default solution to the problem of linear regression. It seems to me that in the majority of practical situations, the costs associated with errors are linear or approximately linear. If I order 2 extra parts, I incur twice the unnecessary cost as compared to if I order 1 extra part. Therefore, the optimal solution that OLS produces will not correspond to an optimal solution in reality. I understand that in general, you can minimize whatever error metric makes the most sense in the situation. My question is not about whether it is possible or a good idea to use MAE in a specific case; it is about the convention. Why is MSE minimized in nearly all simple cases instead of MAE when the real cost is typically linear?
The cases I have seen made for minimizing MSE include the following:
MSE is continuously differentiable
Squaring gives a greater penalty to large errors
It makes sense because we assume errors to be normally distributed
However, if we can perform regression with absolute errors easily, and we are concerned primarily with prediction, I don't see how those points lead anyone to choose squared errors. If we are to choose a convention, isn't absolute error better?
This post has received many excellent answer which have all been useful to me. Of those answers, and the answers elsewhere which the mods believe answer my question, none of them exactly address the real source of my confusion except for the answer by @richard-hardy.
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edited Jun 7, 2020 at 14:55
Ryan Volpi
asked Jun 5, 2020 at 19:47
Ryan VolpiRyan Volpi
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"it doesn't discuss the choice to minimize one loss over the other" - I see " in what instance would the Root Mean Squared Error be a more appropriate measure of error than the Mean Absolute Error" in that question, which seems to be precisely what you are asking about. As to why MSE is ubiquitous, well, on the one hand, it's the differentiability argument, and on the other hand, it is the only error that will be minimized by unbiased estimates/predictions, which is very often what we want. See my answer in that thread.
Stephan Kolassa
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Stephan Kolassa
06/05/2020 20:27:00
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@Aksakal: well, that is the definition of bias (except for the technical use in neural networks). And no, the fact that the (R)MSE (and nothing else) is optimized in expectation precisely by an unbiased forecast is not only true for symmetric distributions. The shoe is on the other foot: for symmetric distributions, the MAE is also minimized by an unbiased forecast (because it's minimized by the median, which is the expectation in the symmetric case), but in general, it isn't, and that's a reason to look at the MSE.
Stephan Kolassa
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Stephan Kolassa
06/05/2020 20:39:14
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I think the analytical tractability of squared loss has historically been a powerful point in its favour.
Daneel Olivaw
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Daneel Olivaw
06/06/2020 10:46:12
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Here is another related question: stats.stackexchange.com/q/369589/164061 . When we are optimizing the result of a cost function that depends on the absolute error, then it might still be useful to use estimates based on MSE. The relevant issue is not only the cost function, but also the error distribution. So... we use often MSE because that is how most error distributions are (as Gauss argued based on a few simple axioms). But for more funky distributions it makes sense to use other methods, and I believe that those are ubiquitous (only sometimes disguised as least squares like GLM).
Sextus Empiricus
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Sextus Empiricus
06/07/2020 14:34:31
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Does the duplicate question really answer the question here? There are many ways how you can discuss the relative use of MAE vs MSE. In this question the way of comparison seems to be about optimizing the result relative to some cost function. I don't see that in the other duplicate question. In none of the answers do I see an explanation relating to the sample distribution of the estimate and the idea that selecting the method which optimizes the lowest expected cost, might still be optimizing MSE even when the cost function is related to the absolute error.
Sextus Empiricus
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The first 5 answers fail to distinguish between estimation loss1 and prediction loss2, something that is crucial in answering the question. A priori, there is no reason that the two should coincide. I will discuss both types of loss in the context of point prediction using linear regression. The discussion can be extended to models other than linear regression and tasks other than point prediction, but the essence remains the same.
Setup
Suppose you are facing a prediction problem where the model is
y=Xβ+ε
with ε∼D(0,σ), D being some probability distribution with location 0 and scale σ. You aim to predict y0 given x0, and your point prediction will be y^0, a function of x0, the data sample, the model and the penalty (the negative of reward) function defined on the prediction error. The penalty function you are facing is LP(y−y^). It has a minimum at zero (the value LP(0) can be set to zero without loss of generality) and is nondecreasing to both sides of zero; this is a typical characterization of a sensible prediction loss function. You can freely choose an estimation loss function LE(⋅) and a point prediction function yhat0. What are your optimal choices for each? This will depend on the error distribution D and the prediction loss function LP(⋅).
Estimation loss
Estimation loss specifies how parameter estimates of a model are obtained from sample data. In our example of linear regression, it concern the estimation of β and σ. You can estimate them by minimizing the sum of squared residuals (OLS) between the actual y and the corresponding fitted values, sum of absolute residuals (quantile regression at the median) or another function. The choice of the estimation loss can be determined by the distribution of model errors. The most accurate estimator in some technical sense will be achieved by the estimation loss that makes the parameter estimator the maximum likelihood (ML) estimator. If the model errors are distributed normally (D is normal), this will be OLS; if they are distributed according to a Laplace distribution (D is Laplace), this will be quantile regression at the mean; etc.
To simplify, given a ML estimator, you may expect more accurate parameter estimates from your model than provided by alternative estimators.
Prediction loss
Prediction loss specifies how prediction errors are penalized. You do not choose it, it is given. (Usually, it is the client that specifies it. If the client is not capable of doing that mathematically, the analyst should strive to do that by listening carefully to the client's arguments.) If the prediction error causes the client's loss (e.g. financial loss) to grow quadratically and symmetrically about zero, you are facing square prediction loss. If the client's loss grows linearly and symmetrically about zero, you are facing absolute prediction loss. There are plenty of other possibilities for types of prediction loss you may be facing, too.
Prediction
Given the parameter estimates of the model and the values of the regressors of the point of interest, x0, you should choose the point prediction y^0 based on the prediction loss. For square loss, you will choose the estimated mean of y0, as the true mean minimizes square loss on average (where the average is taken across random samples of y0 subject to x=x0). For absolute loss, you will choose the estimated median. For other loss function, you will choose other features of the distribution of y0 that you have modelled.
Back to your question
Why do people frequently choose square error rather than absolute error, or correspondingly square loss rather than absolute loss, as estimation loss? Because normal errors (D being normal) are common in applications, ~~arguably~~ more so than Laplace errors (D being Laplace). They also make the regression estimators analytically tractable. They are not much easier to compute, however. Computational complexity of OLS (corresponding to ML estimation under normal errors) vs. quantile regression at the median (corresponding to ML estimation under Laplace errors) are not vastly different. Thus there are some sound arguments for the choice of OLS over quantile regression at the median, or square error over absolute error.
Why do people choose square error, or correspondingly square loss, as prediction loss? Perhaps for simplicity. As some of the previous answers might have mentioned, you have to choose some baseline for a textbook exposition; you cannot discuss all possible cases in detail. However, the case for preferring square loss over absolute loss as prediction loss is less convincing than in the case of estimation loss. Actual prediction loss is likely to be asymmetric (as discussed in some previous answers) and not more likely to grow quadratically than linearly with prediction error. Of course, in practice you should follow the client's specification of prediction loss. Meanwhile, in casual examples and discussions where there is no concrete client around, I do not see a strong argument for preferring square error over absolute error.
1 Also known as estimation cost, fitting loss, fitting cost, training loss, training cost.
2 Also known as prediction cost, evaluation loss, evaluation cost.
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edited Dec 30, 2021 at 14:05
answered Jun 6, 2020 at 20:00
Richard HardyRichard Hardy
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this is absolutely on the nose and addresses precisely the points on which I was confused. Applying the method you describe in the "prediction" section for an arbitrary error distribution and cost function requires evaluating the expected value of the cost function. This requires that the product of the error distribution and cost function be integrable. Is that correct?
Ryan Volpi
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Ryan Volpi
06/07/2020 04:06:41
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Jun 7, 2020 at 4:06
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@RyanVolpi, yes, I think this is correct. Curiously, I have encountered some situations in which this is not the case; see "In model selection, what to do if expected prediction loss of all models is infinity?".
Richard Hardy
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Richard Hardy
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'Because normal errors are common in applications, arguably more so than Laplace errors' I don't think you need to caveat this with 'arguable' - Laplacian distributed variables only arise as the difference between two exponentially distributed variables, which is clearly a pretty rare situation compared to a variable which is itself the sum of many independent variables (i.e. ~Gaussian)
stuart10
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stuart10
06/08/2020 08:10:17
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As a corollary consider a classification problem. We typically train with a cross-entropy loss (i.e. assume the data is the result of independent Bernoulli trials), but consider metrics such a accuracy or recall when looking at validation or test data
stuart10
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stuart10
06/08/2020 08:15:02
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@stuart10, thanks for the comment, I have struck "arguably" out.
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Richard Hardy
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TLDR; when nothing is known about actual cost of error to the user of the model, MSE is a better default option compared to MAE because, in my opinion, it is easier to manipulate analytically and is more likely to match the actual cost of error.
It's a great question. I like that you start with desire to make your loss function match actual costs. This is how it's supposed to be done ideally in my opinion. However, it is impractical to derive the cost function from actual costs every time you build a model, so we tend to gravitate to using one of the loss functions available in software. Least squares is one of the most popular functions mainly due to mathematical convenience. It is easier to deal with it analytically. Also, in some cases least squares produces unbiased point forecast, that is E[y]−y^=0, which is often considered desirable for sentimental reasons.
Having said this, I must argue that it is not obvious to me that absolute value loss is more realistic. Consider, drug overdoses - they are much costlier than underdoses in some situations: not getting high enough vs dying.
Within your parts example, consider this: what if you underestimated the cost of parts to be $1, and entered into a forward agreement to deliver one million parts one month later at $1.1 knowing that you will have $1M one month from today. You are going to make 10% profit!
Then comes the day and parts are actually $1.2 a piece. So, you are not only going to incur loss of $100K, but will also lack funds to deliver 1M parts. So, you are forced to default and go into bankruptcy which is very expensive. On the other hand if you overestimated the cost of parts, then you wold forego some profit but wouldn't end up in dire situation of insolvency or liquidity crisis.
This is a very common situation in business where the losses are asymmetric and highly nonlinear with rapidly escalating costs in one direction of forecast error but not the other.
Hence, I'd argue that absolute loss, which is symmetric and has linear losses on forecasting error, is not realistic in most business situations. Also, although symmetric, the squared loss is at least non linear.
Yet the differences between absolute and squared loss functions don't end here. For instance, it can be shown that the optimal point forecast in absolute loss is the median while for the squared loss it is mean.
I think that the following loss function is more suitable to business forecasting in many cases where over forecasting error e=y−y^ can become very costly very quickly:
L(e,y^)=|ln(1+ey^)|
Here, if you are forecasting a non negative quantity y, then over forecasting is potentially devastating. Imagine you are bank forecasting the deposit volume, and the actual deposit volume turned out to be much lower than you hoped for. This can have severe consequences. This type of asymmetric loss function will lead to a biased optimal point forecast, i.e. E[y]−y^≠0, but that's exactly what you want: you want to err on the side of under forecasting in this kind of business problem.
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edited Jun 6, 2020 at 16:31
answered Jun 5, 2020 at 20:03
AksakalAksakal
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How does this apply to MSE vs MAE? This seems like cons to symmetric loss functions in general.
Dave
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Dave
06/05/2020 20:12:18
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+1. Very often, costs are asymmetric: if we have too much product on hand, we may be able to sell it tomorrow if we can't sell it today, but if we have too little, then we lose a sale - in such a situation it's better to over- than to underestimate. Conversely, if we are talking about strawberries, anything we don't sell today we have to throw away, so now underestimates are better than overestimates. This leads quickly to quantile regression and appropriate error measures, i.e., pinball losses.
Stephan Kolassa
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Stephan Kolassa
06/05/2020 20:14:00
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Jun 5, 2020 at 20:14
@Dave, the detailed discussion can be found in paper "Optimal Point Forecast for Certain Bank Deposit Series" see cer.columbian.gwu.edu/sites/g/files/zaxdzs2011/f/downloads/… , the PDF has an embedded paper in it. This is a pretty standard stuff though
Aksakal
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Aksakal
06/05/2020 20:16:50
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@Aksakal: I don't think I fully understand. It seems to me your loss will reward overforecasting. (Can we agree on that?) So why would we use it if overforecasting is more expensive than underforecasting? More generally, I am uncomfortable with using a loss function without knowing which functional of the future distribution it elicits - see that little paper of mine I link. I'd rather first figure out which functional I want, then choose an appropriate loss function.
Stephan Kolassa
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Stephan Kolassa
06/05/2020 20:20:54
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I think if nothing else is known then MSE is preferable to MAE.
Aksakal
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06/05/2020 21:51:50
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I think the reason is more sociological that statistical.
Short version: We do it this way because we always have.
Longer version:
Historically, we could not do many of the things we now take for granted. Many things are computer intensive and Ronald Fisher was born before Alan Turing.
So, people did OLS regression - a lot. And people read those regressions in all sorts of substantive fields and statistics courses in those fields taught ANOVA/regression and not more modern methods.
Additionally, editors of journals learned those methods and not others, and many will reject articles with modern methods because e.g. "they won't be understood".
Many practitioners reject modern methods too; I used to be a sort of data analysis geek at a hospital. Doctors would come to ask my advice and, if it wasn't "do OLS regression" or "do logistic regression" they would reject my advice.
I got my PhD in psychometrics and many of my professors in other branches of psychology did not know any modern methods (one said: "just report the p value, that's what matters").
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edited Jun 6, 2020 at 22:49
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I think the reason why OLS is so popular is because it started in science (Laplace etc.) where forecast error cost is symmetrical and maybe nonlinear, so it fit the most important requirements plus it's easier to manipulate analytically. If this was started in business, I bet that it wouldnt be as popular because business cost of forecast error is asymmetrical often
Aksakal
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06/05/2020 20:58:56
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Jun 5, 2020 at 20:58
With logistic regression you already have an example where we deviate from minimizing MSE.
Sextus Empiricus
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I think it's worth taking a step back and considering what the two losses imply.
Looking at it from a probabilistic point of view, the loss function is equivalent to the assumed log-likelihood function and thus should correspond to how we think our measurements are distributed around their unknown 'true' values.
As you say, in the case of OLS this is equivalent to assuming a Gaussian likelihood, where as an absolute error loss function is equivalent to a Laplacian likelihood. Gaussian likelihoods are far far more often a good match to real life as a consequence of the central limit theorem.
Our predictions are in general improved by making our assumed (and implicitly generative) model as close a match to reality as possible. In many (most?) cases this will improve the predictive accuracy by any sensible metric (including e.g. mean absolute error). It is far more often the case assuming a Gaussian likelihood will achieve this.
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edited Jun 6, 2020 at 14:13
answered Jun 6, 2020 at 14:05
stuart10stuart10
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Are you claiming that a model fit by OLS will actually have a lower expected MAE on unseen data than one fit using MAE?
Ryan Volpi
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Ryan Volpi
06/06/2020 15:08:11
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Not in absolutely all cases no, but if the underlying process producing the data is (approximately) Gaussian then assuming a squared loss in training (i.e. Gaussian likelihood) will often than not produce lower MAEs on unseen data than assuming a clearly incorrect model in training (i.e. a Laplacian likelihood).
stuart10
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If errors are independent and follow the normal distribution (of any variance but consistent), then the sum of squared errors corresponds to their joint probability/likelihood.
Πe−x2i=e−Σx2i
So under those conditions minimizing the sum of square errors is the same as maximizing the likelihood.
If a cost-minimizing prediction is needed (where the cost metric is different from MSE) the general/accurate approach would be to explicitly minimize the expected cost over the entire distribution of models weighted by their likelihoods (or probabilies if you have prior knowledge). This completely decouples the problem of minimizing expected cost from the problem of estimation in the presence of noise.
Suppose you are measuring a constant quantity in the presence of Gaussian noise. Even if your cost metric for future outcomes is MAE, you would rather predict with the mean (minimizing past MSE) than the median (minimizing past MAE), if indeed you know the quantity is constant and the measurement noise is Gaussian.
Example
Consider the following spread of hits produced by a gun that was mechanically fixed in place. You place a circle of a given size somewhere on the target. If the next shot lands entirely inside your circle, you win, else you lose. The cost function is of the form fC(x,y)=sign((x−xC)2+(y−yC)2−R2).
If you minimize ∑ifC(xi,yi), you would place the circle in the blue position, containing entirely the maximum number of past shots.
But if you knew that the gun is fixed in place and the error is Gaussian, you would place the circle in the green position, centered on the data's mean/centroid (minimizing MSE), as you are optimizing future expected payoff, not average past payoff.
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edited Jun 6, 2020 at 19:40
answered Jun 6, 2020 at 14:20
MusefulMuseful
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Is there a practical reason that maximizing likelihood would be preferred to minimizing the expectation of a realistic cost metric?
Ryan Volpi
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Ryan Volpi
06/06/2020 15:12:23
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@RyanVolpi Consider the simplest case for example: trying to measure a constant quantity in the presence of Gaussian noise. Even if your cost metric for future outcomes is absolute error, you would rather predict with the mean (minimizing past square error) than the median (minimizing past absolute error), if indeed you know the quantity is constant and the measurement noise is Gaussian.
Museful
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06/06/2020 15:50:42
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@RyanVolpi In my mind minimizing cost of prediction (future) is a separate concern from suppressing measurement noise (past). The explicit way to do that would be to minimize over a distribution of weighted models.
Museful
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I never thought of it in that way. So, in the presence of gaussian noise, the mean minimizes expected MAE better than the median. Does that extend to other circumstances? For example, for a linear model with gaussian error, are the least squares estimates better than the absolute error estimates in terms of expected MAE?
Ryan Volpi
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Ryan Volpi
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@RyanVolpi I presume yes as long as the errors are coming (for practical purposes) from random gaussian noise and not from your model being over-constrained.
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Suppose one rolls one die (numered 1-6), and wants to compute its average deviation from the average value of 3.5. Two rolls would differ by 0.5, two by 1.5, and two by 2.5, for an average deviation of 1.5. If one takes the average of the squares of the values, one would have one deviation of 0.25, one of 2.25, and one of 6.25, for an average of 2.916 (35/12).
Now suppose instead of rolling one die, one rolls two. The average deviation would be 1.94 (35/18), and the average square of the deviation would be 5.833 (70/12).
If instead of rolling two dice, one wanted to estimate the expected deviation based upon what it was with one die, doubling the linear average single-die deviation (i.e. 1.5) would yield a value of 3, which is much larger than the actual linear average deviation of 1.94. On the other hand, doubling the average square of the deviation when using a single die (2.916) would yield precisely the average square of the deviation when using two dice.
In general, the square root of the average of the squares is a more useful number than the average of the squares itself, but if one wants to compute the square root of the average of a bunch of squares, it's easier to keep the values to be added as squares, than to take the square roots whenever reporting them and then have to square them before they can be added or averaged.
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answered Jun 6, 2020 at 17:35
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In my opinion, it boils to that the squared error guarantees a unique solution, easier to work with and hence much more intuition. By only two main assumptions (and linearity of the error term), a quadratic loss function guarantees that the estimated coefficient is the unique minimized. Least-absolute deviations does not have this property. There is always a potential for an infinite number of solutions. Assuming that ∃θo∈Θ such that E(y|x)=m(x,θo) and E((m(x,θ)−m(x,θo)2)>0 for all θ≠θo, then θo is the unique minimizer for non-linear least squares.
Proof: Let y=m(x,θo)+u and E(u|x)=0. Then
Eθo((y−m(x,θ))2)=Eθo((y−m(x,θo)+m(x,θ0)−m(x,θ))2)
=Eθo(u2)+Eθo((m(x,θo)−m(x,θ))2)+2Eθo(u(m(x,θo)−m(x,θ))).
By the law of iterated expectations, the third term is zero. Therefore
Eθo((y−m(x,θ))2)=u2+Eθo((m(x,θo)−m(x,θ))2)
is uniquely minimized at θo.
Another nice property is the total law of variance
Var(Y)=VarX(EY(Y|X))+EX(VarY(Y|X)),
which can be read as the variance of the dependent variable is the variance of the fitted value plus the residual's variance.
On a more technical note, the asymptotic formulas are much easier for a quadratic loss function. Importantly, the formulas don't depend on the probability density of the error term. Unfortunately, that is not true for least-absolute deviations. Therefore most practitioners end up having to assume independence of the error term (the formula has the conditional density of the error term at 0 conditioned on x, which is impossible to estimate(fu|x(0))) to estimate fu(0).
And the least rigorous point is that people have an easy time understanding what a mean or expected value is, and the quadratic loss solves for the conditional expectation. Least-absolute deviations soles for the median, which is just harder to interpret. Another reason quantile regressions aren't very popular.
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answered Jun 6, 2020 at 23:32
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11554 | https://en.wikipedia.org/wiki/Annihilator_(ring_theory) | Annihilator (ring theory) - Wikipedia
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1 Definitions
2 PropertiesToggle Properties subsection
2.1 For commutative rings
2.1.1 Relation to support
2.1.2 Short exact sequences
3 ExamplesToggle Examples subsection
3.1 Over the integers
3.2 Over a commutative ring R
3.3 Over k[x,y]
4 Chain conditions on annihilator ideals
5 Category-theoretic description for commutative rings
6 Relations to other properties of rings
7 See also
8 Notes
9 References
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From Wikipedia, the free encyclopedia
Ideal that maps to zero a subset of a module
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In mathematics, the annihilator of a subsetS of a module over a ring is the ideal formed by the elements of the ring that give always zero when multiplied by each element of S. For example, if R{\displaystyle R} is a commutative ring and I{\displaystyle I} is an ideal of R{\displaystyle R}, we can consider the quotient ringR/I{\displaystyle R/I} to be an R{\displaystyle R}-module. Then, the annihilator of R/I{\displaystyle R/I} is the ideal I{\displaystyle I}, since all of the i∈I{\displaystyle i\in I} act via the zero map on R/I{\displaystyle R/I}.
Over an integral domain, a module that has a nonzero annihilator is a torsion module, and a finitely generated torsion module has a nonzero annihilator.
The above definition applies also in the case of noncommutative rings, a subset of a left module has a left annihilator, which is a left ideal, and a subset of a right module has a right annihilator, which is a right ideal. If S{\displaystyle S} is a module, then the annihilator is always a two-sided ideal, regardless of whether the module is a left module or a right module.
Definitions
[edit]
Let R be a ring, and let M be a left R-module. Choose a non-empty subset S of M. The annihilator of S, denoted Ann R(S), is the set of all elements r in R such that, for all s in S, rs = 0. In set notation,
A n n R(S)={r∈R∣r s=0{\displaystyle \mathrm {Ann} {R}(S)={r\in R\mid rs=0} for all s∈S}{\displaystyle s\in S}}
It is the set of all elements of _R that "annihilate" S (the elements for which S is a torsion set). Subsets of right modules may be used as well, after the modification of "sr = 0" in the definition.
The annihilator of a single element x is usually written Ann R(x) instead of Ann R({x}). If the ring R can be understood from the context, the subscript R can be omitted.
Since R is a module over itself, S may be taken to be a subset of R itself, and since R is both a right and a left R-module, the notation must be modified slightly to indicate the left or right side. Usually ℓ.A n n R(S){\displaystyle \ell .!\mathrm {Ann} {R}(S)\,} and r.A n n R(S){\displaystyle r.!\mathrm {Ann} {R}(S)\,} or some similar subscript scheme are used to distinguish the left and right annihilators, if necessary.
If M is an R-module and Ann R(M) = 0, then M is called a faithful module.
Properties
[edit]
If S is a subset of a left R-module M, then Ann(S) is a left ideal of R. If S is not just a subset but also a submodule of M, then Ann R(S) is moreover a two-sided ideal: (ac)s = a(cs) = 0, since cs is another element of S.
If S is a subset of M and N is the submodule of M generated by S, then in general Ann R(N) is a subset of Ann R(S), but they are not necessarily equal. If R is commutative, then the equality holds.
An R-module M may be also viewed as an R/Ann R(M)-module using the action r¯m:=r m{\displaystyle {\overline {r}}m:=rm\,}. It is not always possible to make an R-module into an R/I-module this way, but if the ideal I is a subset of the annihilator of M, then this action is well-defined. The module M is always faithful when considered as an R/Ann R(M)-module.
For commutative rings
[edit]
Throughout this section, let R{\displaystyle R} be a commutative ring and M{\displaystyle M} an R{\displaystyle R}-module.
Relation to support
[edit]
The support of a module is defined as
SuppM={p∈SpecR∣M p≠0}.{\displaystyle \operatorname {Supp} M={{\mathfrak {p}}\in \operatorname {Spec} R\mid M_{\mathfrak {p}}\neq 0}.}
When the module is finitely generated, the support of M{\displaystyle M} is exactly the set of prime ideals containing Ann R(M){\displaystyle \operatorname {Ann} _{R}(M)}.
Short exact sequences
[edit]
Given a short exact sequence of modules,
0→M′→M→M″→0,{\displaystyle 0\to M'\to M\to M''\to 0,}
we have the relations
Ann R(M′)∩Ann R(M″)⊇Ann R(M)⊇Ann R(M′)Ann R(M″).{\displaystyle \operatorname {Ann} {R}(M')\cap \operatorname {Ann} {R}(M'')\supseteq \operatorname {Ann} {R}(M)\supseteq \operatorname {Ann} {R}(M')\operatorname {Ann} _{R}(M'').}
If the sequence splits, so that M≅M′⊕M″{\displaystyle M\cong M'\oplus M''}, then the inequality on the left is always an equality. Indeed, this holds for arbitrary direct sums of modules:
Ann R(⨁i∈I M i)=⋂i∈I Ann R(M i).{\displaystyle \operatorname {Ann} {R}\left(\bigoplus {i\in I}M_{i}\right)=\bigcap {i\in I}\operatorname {Ann} {R}(M_{i}).}
In the special case that M′=I M{\displaystyle M'=IM} and M″=M/I M{\displaystyle M''=M/IM} for some ideal I⊂R{\displaystyle I\subset R}, we have the relation Ann R(M/I M)⊇Ann R(M)+I{\displaystyle \operatorname {Ann} {R}(M/IM)\supseteq \operatorname {Ann} {R}(M)+I}.
Examples
[edit]
Over the integers
[edit]
Over Z{\displaystyle \mathbb {Z} } any finitely generated module is completely classified as the direct sum of its free part with its torsion part from the fundamental theorem of abelian groups. Then the annihilator of a finitely generated module is non-trivial only if it is entirely torsion. This is because
Ann Z(Z⊕k)={0}=(0){\displaystyle {\text{Ann}}_{\mathbb {Z} }(\mathbb {Z} ^{\oplus k})={0}=(0)}
since the only element killing each of the Z{\displaystyle \mathbb {Z} } is 0{\displaystyle 0}. For example, the annihilator of Z/2⊕Z/3{\displaystyle \mathbb {Z} /2\oplus \mathbb {Z} /3} is
Ann Z(Z/2⊕Z/3)=(6)=(lcm(2,3)),{\displaystyle {\text{Ann}}_{\mathbb {Z} }(\mathbb {Z} /2\oplus \mathbb {Z} /3)=(6)=({\text{lcm}}(2,3)),}
the ideal generated by (6){\displaystyle (6)}. In fact the annihilator of a torsion module
M≅⨁i=1 n(Z/a i)⊕k i{\displaystyle M\cong \bigoplus {i=1}^{n}(\mathbb {Z} /a{i})^{\oplus k_{i}}}
is isomorphic to the ideal generated by their least common multiple, (lcm(a 1,…,a n)){\displaystyle (\operatorname {lcm} (a_{1},\ldots ,a_{n}))}. This shows the annihilators can be easily be classified over the integers.
Over a commutative ring R
[edit]
There is a similar computation that can be done for any finitely presented module over a commutative ring R{\displaystyle R}. The definition of finite presentedness of M{\displaystyle M} implies there exists an exact sequence, called a presentation, given by
R⊕l→ϕ R⊕k→M→0{\displaystyle R^{\oplus l}\xrightarrow {\phi } R^{\oplus k}\to M\to 0}
where ϕ{\displaystyle \phi } is in Mat k,l(R){\displaystyle {\text{Mat}}_{k,l}(R)}. Writing ϕ{\displaystyle \phi } explicitly as a matrix gives it as
ϕ=[ϕ 1,1⋯ϕ 1,l⋮⋮ϕ k,1⋯ϕ k,l];{\displaystyle \phi ={\begin{bmatrix}\phi {1,1}&\cdots &\phi {1,l}\\vdots &&\vdots \\phi {k,1}&\cdots &\phi {k,l}\end{bmatrix}};}
hence M{\displaystyle M} has the direct sum decomposition
M=⨁i=1 k R(ϕ i,1(1),…,ϕ i,l(1)){\displaystyle M=\bigoplus {i=1}^{k}{\frac {R}{(\phi {i,1}(1),\ldots ,\phi _{i,l}(1))}}}
If each of these ideals is written as
I i=(ϕ i,1(1),…,ϕ i,l(1)){\displaystyle I_{i}=(\phi {i,1}(1),\ldots ,\phi {i,l}(1))}
then the ideal I{\displaystyle I} given by
V(I)=⋃i=1 k V(I i){\displaystyle V(I)=\bigcup {i=1}^{k}V(I{i})}
presents the annihilator.
Over k[x,y]
[edit]
Over the commutative ring k[x,y]{\displaystyle k[x,y]} for a fieldk{\displaystyle k}, the annihilator of the module
M=kx,y⊕kx,y{\displaystyle M={\frac {k[x,y]}{(x^{2}-y)}}\oplus {\frac {k[x,y]}{(y-3)}}}
is given by the ideal
Ann kx,y=((x 2−y)(y−3)).{\displaystyle {\text{Ann}}_{k[x,y]}(M)=((x^{2}-y)(y-3)).}
Chain conditions on annihilator ideals
[edit]
The lattice of ideals of the form ℓ.A n n R(S){\displaystyle \ell .!\mathrm {Ann} {R}(S)} where _S is a subset of R is a complete lattice when partially ordered by inclusion. There is interest in studying rings for which this lattice (or its right counterpart) satisfies the ascending chain condition or descending chain condition.
Denote the lattice of left annihilator ideals of R as L A{\displaystyle {\mathcal {LA}}\,} and the lattice of right annihilator ideals of R as R A{\displaystyle {\mathcal {RA}}\,}. It is known that L A{\displaystyle {\mathcal {LA}}\,} satisfies the ascending chain condition if and only ifR A{\displaystyle {\mathcal {RA}}\,} satisfies the descending chain condition, and symmetrically R A{\displaystyle {\mathcal {RA}}\,} satisfies the ascending chain condition if and only if L A{\displaystyle {\mathcal {LA}}\,} satisfies the descending chain condition. If either lattice has either of these chain conditions, then R has no infinite pairwise orthogonal sets of idempotents.
If R is a ring for which L A{\displaystyle {\mathcal {LA}}\,} satisfies the A.C.C. and R R has finite uniform dimension, then R is called a left Goldie ring.
Category-theoretic description for commutative rings
[edit]
When R is commutative and M is an R-module, we may describe Ann R(M) as the kernel of the action map R → End R(M) determined by the adjunct map of the identityM → M along the Hom-tensor adjunction.
More generally, given a bilinear map of modules F:M×N→P{\displaystyle F\colon M\times N\to P}, the annihilator of a subset S⊆M{\displaystyle S\subseteq M} is the set of all elements in N{\displaystyle N} that annihilate S{\displaystyle S}:
Ann(S):={n∈N∣∀s∈S:F(s,n)=0}.{\displaystyle \operatorname {Ann} (S):={n\in N\mid \forall s\in S:F(s,n)=0}.}
Conversely, given T⊆N{\displaystyle T\subseteq N}, one can define an annihilator as a subset of M{\displaystyle M}.
The annihilator gives a Galois connection between subsets of M{\displaystyle M} and N{\displaystyle N}, and the associated closure operator is stronger than the span. In particular:
annihilators are submodules
SpanS≤Ann(Ann(S)){\displaystyle \operatorname {Span} S\leq \operatorname {Ann} (\operatorname {Ann} (S))}
Ann(Ann(Ann(S)))=Ann(S){\displaystyle \operatorname {Ann} (\operatorname {Ann} (\operatorname {Ann} (S)))=\operatorname {Ann} (S)}
An important special case is in the presence of a nondegenerate form on a vector space, particularly an inner product: then the annihilator associated to the map V×V→K{\displaystyle V\times V\to K} is called the orthogonal complement.
Relations to other properties of rings
[edit]
Given a module M over a Noetherian commutative ring R, a prime ideal of R that is an annihilator of a nonzero element of M is called an associated prime of M.
Annihilators are used to define left Rickart rings and Baer rings.
The set of (left) zero divisorsD S of S can be written as
D S=⋃x∈S∖{0}A n n R(x).{\displaystyle D_{S}=\bigcup {x\in S\setminus {0}}{\mathrm {Ann} {R}(x)}.}(Here we allow zero to be a zero divisor.)In particular D R is the set of (left) zero divisors of R taking S = R and R acting on itself as a left R-module.
When R is commutative and Noetherian, the set D R{\displaystyle D_{R}} is precisely equal to the union of the associated primes of the R-module R.
See also
[edit]
Faltings' annihilator theorem
Socle
Support of a module
Notes
[edit]
^Pierce (1982), p. 23.
^Proof: If a and b both annihilate S, then for each s in S, (a+b)s = as+bs = 0, and for any r in R, (ra)s = r(as) = r 0 = 0.
^Pierce (1982), p. 23, Lemma b, item (i).
^Eisenbud 2004, p.67.
^"Lemma 10.39.5 (00L2)—The Stacks project". stacks.math.columbia.edu. Retrieved 2020-05-13.
^Anderson & Fuller 1992, p.322.
^ abLam 1999.
References
[edit]
Anderson, Frank W.; Fuller, Kent R. (1992), Rings and categories of modules, Graduate Texts in Mathematics, vol.13 (2 ed.), New York: Springer-Verlag, pp.x+376, doi:10.1007/978-1-4612-4418-9, ISBN0-387-97845-3, MR1245487
Eisenbud, David (2004). Commutative Algebra With a View Toward Algebraic Geometry. Graduate Texts in Mathematics No. 150. Springer-Verlag.
Israel Nathan Herstein (1968) Noncommutative Rings, Carus Mathematical Monographs #15, Mathematical Association of America, page 3.
Lam, Tsit Yuen (1999), Lectures on modules and rings, Graduate Texts in Mathematics No. 189, vol.189, Berlin, New York: Springer-Verlag, pp.228–232, doi:10.1007/978-1-4612-0525-8, ISBN978-0-387-98428-5, MR1653294
Richard S. Pierce. Associative algebras. Graduate Texts in Mathematics, Vol. 88, Springer-Verlag, 1982, ISBN978-0-387-90693-5
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11555 | https://brainly.com/question/23203865 | [FREE] Write a sentence that contains "cuando" or "mientras," a verb in the imperfect tense, and a verb in the - brainly.com
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Write a sentence that contains "cuando" or "mientras," a verb in the imperfect tense, and a verb in the preterite tense.
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Mientras mi hermano estaba dormido mis primos y yo fuimos a la tienda para ver cuanto valía el juguete que el quería.
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To create a sentence with 'cuando' or 'mientras', an imperfect verb, and a preterite verb, you can say, 'Cuando mis amigos jugaban fútbol, yo miré un partido en la televisión.' This structure illustrates the relationship between two past actions. The imperfect verb shows an ongoing action and the preterite shows a completed action.
Explanation
To construct a sentence that includes 'cuando' or 'mientras', a verb in the imperfect tense, and a verb in the preterite tense, we first need to understand the meanings of these elements.
Cuando denotes 'when' and is used to indicate a specific time in the past.
The imperfect tense describes actions that were ongoing or habitual in the past.
The preterite tense conveys completed actions in the past.
Here is an example sentence that meets these criteria:
"Cuando mis amigos jugaban fútbol, yo miré un partido en la televisión."
In this sentence, 'jugaban' (were playing) is in the imperfect tense, indicating that the friends were continuously playing soccer at that time, while 'miré' (watched) is in the preterite tense, indicating that the action of watching the game is completed.
Such sentences are useful for showing how one action (watching) interrupts or occurs while another action (playing) is taking place in the past.
Examples & Evidence
In this example, 'jugaban' is an ongoing action of playing, while 'miré' indicates a completed action of watching, showcasing how both actions relate temporally.
The usage of 'cuando' introduces a timeframe in which two actions are happening, fulfilling the requirements of the question by using distinct past tenses.
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Community Answer Match each sentence with the correct verb in preterite or imperfect tense
Community Answer 1 Write one sentence in Spanish to describe something that was being done because of hurricanes. Use one progressive tense verb in the preterite tense, or imperfect tense.
Community Answer read the following sentence and choose the correct verbs in preterite or imperfect tense to complete it. yo __ (escuchar) música mientras ___ (hacer) mi tarea.
Community Answer 5.0 2 Translate the following sentence into Spanish using the Imperfect and Preterite Tenses. Include the Subjects. (One Sentence: use Formula #1 - subj + imperfect verb cuando subj + preterite verb)They used to play when they visited the park.
Community Answer 4.6 3015 Now that you have worked through a lot of material that includes these basic patterns, and you have compared grammatically correct and incorrect sentences, write down what you think is a rule that could explain what makes a sentence grammatically correct or not. For example, you might write something like: "verbs always match nouns in number, and they usually come before the noun." In other words, make your best guess for the grammar rule that makes sense out of the pattern(s) you see in the phrases you have been working with
Community Answer 13 que significa cuando te echan humo de cigarro en la cara
Community Answer 4.6 303 After looking through the sentence comparison examples, what do you think the guidelines are that tell you when to use the different forms of "you?" Write down your thoughts about this grammar principle. Specifically, write down a "rule" that you think Spanish uses to explain the difference between tú and usted. Then check out the grammar pattern in the next link to see if your hunch was correct.
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Which of the following is least likely to be served as a main course during la cena? A. la carne B. el pollo C. el pescado D. el helado
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Translate the following sentence from English into Spanish: Following his retirement, he lived in Switzerland.
Translate the following sentence from English into Spanish: We went out to the party very early.
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11556 | https://www.sciencedirect.com/science/article/abs/pii/S0165587610004283 | Simple bone cyst: Report of cases and proposal for a minimal surgical intervention - ScienceDirect
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International Journal of Pediatric Otorhinolaryngology
Volume 74, Issue 12, December 2010, Pages 1449-1451
Case report
Simple bone cyst: Report of cases and proposal for a minimal surgical intervention
Author links open overlay panel Ana Luísa Homem de Carvalho, Vinicius Coelho Carrard, Manoela Domingues Martins, Pantelis Varvaki Rados, Manoel Sant’Ana Filho
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Abstract
Simple bone cyst (SBC) is an intraosseous pseudocyst that appears as a radiolucent lesion, frequently observed among young patients. In this article we report six cases of SBC and propose a protocol for minimal surgical intervention in the management of this condition. No history of trauma was reported. All patients underwent a minimal bone intervention procedure to perforate the cortical bone and stimulate blood clot formation. Complete healing and no recurrence were observed after 1-year follow-up. This treatment shows advantages such as the establishment of a definitive diagnosis and low invasiveness, particularly in pediatric patients.
Introduction
Simple bone cyst (SBC) was first described as an intraosseous pseudocyst devoid of an epithelial lining, either empty or filled with serous or sanguineous fluid , . The condition affects both men and women similarly , being more frequently observed among young patients during the second decade of life , . SBC usually appears as an asymptomatic lesion diagnosed accidentally during routine radiographic examination , , . The radiographic features of these lesions are non-pathognomonic, which may hinder accurate diagnosis. SBC often appears as a unilocular radiolucent area, with scalloped but well-defined margins, usually located between the roots of lower molars and premolars, but rarely extending into the cortical bone. When located close to the periapical region or lateral to roots, these lesions may be misdiagnosed as inflammatory odontogenic lesions. Therefore, it is important to carry out pulp sensitivity tests during clinical evaluation to rule out this hypothesis. Additionally, other entities should be included in the differential diagnosis, especially when dental lamina is preserved and pulp tests are positive, including keratocystic odontogenic tumor, central giant cell granuloma, ameloblastoma, and myxoma. The lesions exhibit different clinical behavior and require distinct treatments. Thus, biopsy and histopathological examination are necessary to establish a final diagnosis. When an empty cavity or a cavity with scarce hemorrhagic material is observed during a surgical biopsy procedure, the hypothesis of SBC may be strongly suggested .
SBC has recently become the subject of much debate, especially with regard to surgical intervention. Most authors have used surgical approaches to establish a definitive diagnosis. On the other hand, spontaneous resolution has been reported based on clinical and radiographic criteria, with the period between diagnosis and complete lesion resolution ranging from 2 to 7 years . However, when a definitive diagnosis is not established, the lesion may have a very long course until resolution and also exhibit periods of exacerbation. Taking these considerations into account, some authors argue that surgical intervention is a crucial part of the diagnosis and treatment of SBC. The aim of the present article is to report a series of SBC cases and to propose a protocol for minimal surgical intervention in the management of this condition.
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Report of cases
Cases of SBC diagnosed from 2000 to 2007 were included in the study sample (n=6). All patients’ parents or legally responsible persons signed a consent form accepting the proposed surgical protocol. Inclusion criteria were having at least one panoramic radiograph at diagnosis and one panoramic radiograph after 1 year of postoperative follow-up. The clinical data of SBC cases are shown in Table 1.
The male:female ratio was 1:1. Patient's age ranged from 8 to 14 years (12.16±2.56 years). All
Surgical technique
Surgery was performed under local anesthesia. A Neumann's incision was made to raise a mucoperiosteal flap. The minimal intervention procedure was based on lesion access with a size 4 or 5 carbide bur mounted on a low-speed handpiece in order to promote wear of the outer cortical bone and to produce a 5-mm diameter hole, taking care not to reach the inner cortical bone or the apical region of adjacent teeth (Fig. 1B). The cortical bone was perforated and an empty cavity was found in all cases,
Discussion
SBC has been referred to by different names in the literature, such as solitary bone cyst, traumatic bone cyst, hemorrhagic bone cyst, hemorrhagic cyst, unicameral bone cyst, and idiopathic bone cavity. It usually occurs in the metaphyseal region of long bones, being uncommon in the maxillofacial region. In the jaws, 56–70% of SBC cases exhibited no gender-related differences and were present during the second decade of life, which makes it an important lesion to consider in pediatric dentistry
Acknowledgement
We thank Scientific Language for their helpful comments and expertise in the English grammar.
Recommended articles
References (13)
L.S. Hansen et al.
Traumatic bone cysts of jaws: report of sixty-six cases
Oral Surg. (1974)
J. Sapone et al.
Traumatic bone cysts of jaws: diagnosis, treatment and prognosis
Oral Surg. Oral Med. Oral Pathol. (1974)
J.P. Sapp et al.
Self-healing traumatic bone cysts
Oral Surg. Med. Oral Pathol. Oral Radiol. Endod. (1990)
M. Penarrocha-Diago et al.
Surgical treatment and follow-up of solitary bone cyst of the mandible: a report of seven cases
Br. J. Oral Maxillofac. Surg. (2001)
C.D. Lucas et al.
Do all cysts in the jaws originate from dental system?
J. Am. Dent. Assoc. (1929)
There are more references available in the full text version of this article.
Cited by (8)
A management strategy for idiopathic bone cavities of the jaws
2016, Journal of Oral and Maxillofacial Surgery Citation Excerpt :
In a series of 27 IBCs with radiographic follow-up and literature review of an additional 123 IBCs, 75% of IBCs persisted.44 Other investigators also have reported high rates of persistence,4,5,45 and some have suggested a second procedure if bone fill is not observed in the first 3 months after intervention.46 The present clinical observation that IBCs are relatively common in the pediatric population, but seldom seen in adults, coupled with multiple reports of spontaneous bone fill, suggest that IBCs might resolve on their own.24 Show abstract Idiopathic bone cavities (IBCs) are benign osseous pseudocysts of unclear etiology. Their clinical course and response to treatment are poorly understood. The purpose of this study was to present a case series of patients with IBC with long-term follow-up. A retrospective case series of patients with IBC of the jaw was implemented. Medical records were reviewed for data on presentation and imaging. All patients underwent surgical exploration to confirm the diagnosis. Follow-up radiographs were evaluated for resolution or persistence after the procedure. Descriptive data were summarized. The sample included 45 patients (60% girls) with 47 IBCs who presented at a mean age of 13.5 ± 0.35 years (range, 9 to 17 yr). All lesions were asymptomatic and discovered as incidental findings; the average size was 2.0 cm (range, 0.8 to 7 cm). Most (n= 44) were solitary lesions of the mandible presenting within the body (n= 27), symphysis (n= 15), or ramus (n= 2). One patient had synchronous bilateral mandibular and maxillary IBCs. Common radiographic features included scalloping without root resorption or displacement and cortical thinning without expansion or perforation. Intraoperative findings showed an empty bone cavity often filled with blood-tinged serous fluid or blood. The histopathology of scrapings from the bony wall showed benign mixed fibrous tissue and no epithelial lining. Average radiographic follow-up was 32.7 ± 6.7 months (range, 0 to 9 yr). After exploration and curettage, 43% of patients had complete bone fill within 2 to 5 years. The other 57% had no change in lesion size at an average of 2.25 years (range, 2 to 9 yr). No patients developed symptoms, had enlargement of the lesion, or had pathologic fracture. Many IBCs persist for years despite exploration and curettage. Nonetheless, unresolved lesions did not enlarge or cause harm. A single procedure to confirm the diagnosis is sufficient for management.
### Atypical presentations of simple bone cysts of the mandible: A case series and review of literature
2013, Journal of Cranio Maxillofacial Surgery Citation Excerpt :
In cases with COD follow-up examinations may be necessary even after healing because there is a high probability that new lesions could develop in such cases (Melrose et al., 1976). In only four of the six cases could material be obtained for histological evaluation, which reinforces the fact that the absence of soft tissue, such as epithelial tissue, is one of the most frequent features of these lesions (Suomalainen et al., 2009; Homem de Carvalho et al., 2010). Although the majority of SBCs present with similar clinical and radiographic features, some cases may be confused with other entities. Show abstract Simple bone cysts are well-defined intraosseous radiolucencies that often extend between the roots and appear clinically like empty cavities. This article aims to provide more information about this lesion with limited prominence in academic literature, to illustrate atypical cases, and to provide a review of the current literature. A series of six atypical cases of simple bone cysts is presented and their clinical, radiographic and microscopic characteristics, differential diagnosis, treatment and follow-up are discussed. Correct diagnosis of this entity is of key importance, since it presents with clinical & radiographic similarities to other bone lesions, some exhibiting more aggressive behaviour.
### Simple bone cyst in ipsilateral maxilla and mandible: A case report and literature review
2024, Journal of Jilin University Medicine Edition
### Simple bone cyst in mandibular ramus with atypical features
2022, Radwaste Solutions
### Simple bone cyst in mandibular ramus with atypical features
2022, Revista Mexicana De Pediatria
### Simple bone cyst: Description of 60 cases seen at a brazilian school of dentistry and review of international literature
2020, Medicina Oral Patologia Oral Y Cirugia Bucal
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11557 | https://thischangedmypractice.com/management-of-helicobacter-pylori/ | Management of Helicobacter pylori in 2023: who should be tested, treated, and how | This Changed My Practice (TCMP) by UBC CPD
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»This Changed My Practice»Management of Helicobacter pylori in 2023: who should be tested, treated, and how
Management of Helicobacter pylori in 2023: who should be tested, treated, and how
By Dr. Sander Veldhuyzen van Zanten on August 9, 2023
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Author
Sander Veldhuyzen van Zanten, MD, FRCPC, MSc, MPH, PhD(biography and disclosures)
Disclosures: No disclosures relevant to this publication. Received peer and non-peer-reviewed funding for clinical studies of Helicobacter infection in the past. Received speaker fees and member of advisory boards for _H. pylori_ related activities. None in the last five years. Mitigating potential bias: Only published trial data is presented. Recommendations are consistent with published Canadian and International guidelines (references 4-7). Recommendations are consistent with current practice patterns.
What gaps I have noticed
Helicobacter pylori (H. pylori) continues to be an important pathogen associated with peptic ulcer disease, dyspepsia, and gastric cancer.
Data that answers these gaps
The epidemiology of H. pylori in Canada is changing.1 It can now rarely be found in second-generation Canadians, but the prevalence continues to be high in immigrants, children of immigrants, and First Nations Canadians. In an H. pylori positive individual the lifetime risk of developing gastric or duodenal ulcers is 5 to 15% and dyspepsia up to 25%. H. pylori is an important risk factor for development of gastric cancer, but this is rare, occurring in less than 1 in 1,000 infected individuals. Most infections are acquired in childhood, during the first 5–10 years of life.2,3 In a substantial proportion of individuals, the infection remains dormant and causes no symptoms. Although most H. pylori infections are acquired in childhood, it is not recommended to routinely test family members of H. pylori infected individuals. Among adults, it is very unlikely that H. pylori is transmitted, even when there is close personal contact.
Worldwide and in Canada the guidelines for the treatment of H. pylori infection changed in 2016.4-7 Details of the medications used, the dosages, and frequencies, are listed in Table 1. Importantly, the recommended duration for most regimens is 14 days.
What I recommend (practice tips)
Who should be tested for Helicobacter and how?
Generally speaking, do not test patients for H. pylori unless you are prepared to treat if they are positive.
Patients who should be tested for H. pylori:
Duodenal or gastric ulcer diagnosis (recent or in the past)
Dyspepsia (epigastric pain dominant symptoms)
First-degree family history of gastric cancer (especially if index case is young)
High-risk groups (First Nations, immigrants)
H. pylori can be diagnosed by histology when biopsies are taken during gastroscopy. The urea breath test (UBT) is a very accurate and robust test to diagnose the infection.8 More recently, the H. pylori stool antigen test (HpSAT) has been used more frequently.9 It is an accurate test but it has a higher proportion of equivocal results than the UBT. Use of serology to detect H. pylori infection is not recommended as it cannot discriminate between a current active infection or an old (treated) infection in the past. Importantly, one has to stop PPIs (proton pump inhibitors) for 14 days prior to performing a urea breath test or HpSAT. The reason is that up to 20% of people while on PPIs may get false negative results due to suppression of the organism by the PPI.10
Although a large proportion of patients with duodenal or gastric ulcers are H. pylori positive, H. pylori negative ulcers are becoming more common. Therefore, it is recommended to document that a patient is H. pylori positive before initiating treatment. The most common cause of H. pylori negative ulcers are NSAIDs and antiplatelet agents.
Helicobacter treatment recommendations have changed
The conventional triple combination of a PPI, clarithromycin (Cl), and either amoxicillin (A) or metronidazole (Met) is no longer recommended because its success rates have decreased to less than 60%.11,12 This includes the use of Hp-PAC®, the combination of lansoprazole, amoxicillin, and clarithromycin.
Recommended treatments (summarized in Table 1 below):
The new recommended first-line therapy is called concomitant or ClAMet (using the initials of clarithromycin (Cl), amoxicillin (A), and metronidazole (Met)) given for 14 days.4-7 Its efficacy is 80–85%.
The best second-line therapy is bismuth-based quadruple therapy(PPI, bismuth, metronidazole, and tetracycline) given for 14 days. The higher dose of metronidazole 500 mg is the author’s preference as there is evidence that it can better overcome partial resistance to metronidazole.13 This is a robust treatment that can also be used as first line. The downside of bismuth-based quadruple therapy (Bi-Quad) is that it is a complex regimen consisting of up to 24 pills a day. Some of the medications need to be taken twice and others four times a day. Bi-Quad also has more side effects than concomitant therapy.
The third-line therapy is the combination of a PPI, amoxicillin, and levofloxacin (PPI-AL) given for 14 days. The author uses the higher dose of levofloxacin 500 mg bid. Sometimes PPI-AL is given as second line instead of Bi-Quad because it is an easier regimen to take, but it is less well tested. PPI-AL should not be used as first-line therapy because the background rate of levofloxacin resistance is high in Canada, up to 15%.14 This high rate is likely explained by the high use of fluoroquinolones for other indications.
There is a fourth-line therapy consisting of a PPI with amoxicillin and rifabutin (Rif). In Canada, its efficacy is 62% when given as rescue therapy.15 This is the only therapy for which the duration is 10 days instead of 14 days. However, family physicians may want to consider referring patients to a GI specialist should the three first-choice therapies have failed. Use of rifabutin often needs special authorization. In the USA, a combination tablet of omeprazole, amoxicillin, and rifabutin (Talicia), has been available since 2019. It has been tested as a first-line therapy and the success rate of 10-day therapy was 84%.16 The tablet is not available in Canada and the decision to move the PPI-amoxicillin-rifabutin combination to first line requires more studies.
If the above-listed four therapies are given in sequence to individuals who failed higher order therapies (i.e., start with concomitant, followed by B-Quad, etc.) the cumulative success rate is 96–97%.17, 18 The first choice of therapy in patients with a true penicillin allergy is Bi-Quad. If that fails, the second line of therapy is triple therapy with PPI, Clarithromycin, and Metronidazole (PPI-CM) given for 14 days.4
One thing to note, although there are pharmacologic differences among the different PPIs, resulting in differences in acid suppression 19, for day-to-day practice PPIs, when given in standard doses, are generally considered equal when treating common conditions such as Gastro Esophageal Reflux Disease (GERD) or peptic ulcer disease (PUD). For treatment of H. pylori, there is some evidence that using the more potent PPIs, such as esomeprazole or rabeprazole, slightly higher success rates are achieved compared to pantoprazole.20
What patients need to know about the treatments
It is important that patients get a clear explanation about these treatments, preferably with a handout. Examples of handouts the author uses about the importance of H. pylori infection and the different treatments are available in the patient handouts section below. Prescribing anti-Helicobacter therapy is difficult to do well over the phone. It is helpful for patients to prescribe these treatments in a blister pack, as the regimens are complex. That way patients are not confused about which medications have to be taken and when. Patients should be told that compliance is very important as cure rates drop significantly if treatments are not taken for the full duration. See handouts for patients.
Is it necessary to test whether the treatment was successful?
It is generally recommended to test whether treatment was successful. In primary care, UBT or the stool antigen test are used for this. Importantly, one has to wait at least 4 weeks after the treatment was finishedbefore re-testing is done to avoid getting false negative results. Once again, it is important that PPI therapy, if there is an ongoing need for it after the antibiotics are finished, is held for 14 days prior to the repeat testing for Helicobacter.10 Patients can be managed during those 14 days with standard doses of antacids if they have symptoms while off the PPI. If the patient is negative after appropriate testing for cure was done, it is very unlikely such a patient will get re-infected later in life. In adults, most patients who are thought to have a re-infection, in fact, have a false negative test result in the past. They usually have recrudescence of their original H. pylori strain.21
Given that none of the current H. pylori therapies are 100% successful it is my practice to always test whether cure has been received after each treatment. I generally give a patient three or four rounds of treatment if needed but I should state that I have a special interest in H. pylori. What to do with patients who fail more than four therapies should be decided on a case-by-case basis. This may in part depend on the indication for treatment. An option is to refer the patient for antibiotic resistance testing for which a gastroscopy is required. Knowledge of the resistance pattern on the strain may guide further therapy. However, testing of H. pylori strains is not widely available in Canada and data on resistance are scarce. The data show that resistance to clarithromycin and metronidazole and also levofloxacin is high. Resistance to amoxicillin and tetracycline is rare.4,14, 22-24
Another management option is the decision not to treat Hp any further, but possibly observe these patients long-term. Long-term follow-up especially is a consideration, if there is a family history of gastric cancer or if more severe gastric metaplasia was present on histology.25,26
Peptic ulcer disease
There are a few important recommendations in the management of patients diagnosed with acute peptic ulcer disease (duodenal or gastric ulcers), and especially bleeding ones, for example during a hospitalization. If they turn out to be H. pylori positive it is important to continue PPI therapy after the 14-day anti-Helicobacter treatment, for at least 8–16 weeks, to ensure adequate healing of the ulcer. Importantly, in such patients, the PPI should not be stopped until there is proof that the H. pylori infection has been successfully cured. The reason for this is that should treatment have failed to cure H. pylori, these patients are at risk of getting recurrent ulceration (with possible bleeding) if PPI maintenance therapy is not given.27 A decision whether or not to continue PPI longer term should only be taken once cure of H. pylori has been established.
There is evidence that anti-Helicobacter therapy is effective in the treatment of H. pylori positive gastric or duodenal ulcers. However, a proportion of such patients will require long-term PPI therapy because of ongoing symptoms even though H. pylori was cured. Another reason for ongoing acid-suppressive therapy may be that some of these patients, especially older ones (> 65–70 years) require, for example, anticoagulation and/or antiplatelet agents for management of cardiovascular risk factors. Given that they, therefore, have an ongoing risk of developing (bleeding) ulcers it is reasonable to maintain such patients on a once-a-day PPI for ulcer prophylaxis.
The role of Helicobacter in dyspepsia and GERD
There is definite evidence that a test and treat strategy for H. pylori is efficacious in patients who present with dyspepsia symptoms in primary care.28 Dyspepsia is defined as epigastric pain or discomfort which may be accompanied by other upper GI symptoms, such as burning sensation or excessive burping. Compared to no-antibiotic treatment, more patients will have permanent resolution of their dyspepsia symptoms following cure of the H. pylori infection. However, clinical trial data, including studies from Canada, have shown that more than 50% of dyspepsia patients will have ongoing dyspepsia symptoms, often requiring acid suppressive therapy, after successful cure of the infection.28
Another important comment on the management of dyspepsia is that there is good evidence that PPIs work in epigastric pain dominant dyspepsia (ulcer-like dyspepsia), and in those who have concomitant symptoms of heartburn (reflux-like dyspepsia). However, PPIs are generally not effective in so-called dysmotility-like dyspepsia where upper abdominal bloating rather than epigastric pain or discomfort is the dominant symptom.29 In this subgroup of patients, PPIs do not provide benefits. In such patients overlap with possible IBS or other diagnoses needs to be further explored.
In Canada nowadays, gastroesophageal reflux disease (GERD) is more common than dyspepsia. The diagnosis of GERD can be made based on the presence of dominant symptoms of heartburn and/or regurgitation.30 Importantly the clinical course of GERD (frequency of symptoms, need for acid suppressive therapy) will not be altered by successful cure of a coexisting H. pylori infection.31 It is important to tell patients about this as most of them will require ongoing maintenance therapy with PPI for control of their GERD symptoms even though H. pylori has been cured. For this reason, it is not recommended to routinely test for H. pylori at the time GERD is diagnosed in patients. This is in contrast to the patient who has dyspepsia, defined by epigastric pain or discomfort as the dominant symptom rather than heartburn, where a test for Helicobacter is recommended.30
The incidence of gastric cancer has been decreasing for decades in Canada but it is still more common than esophageal cancer. 32 Apart from smoking, presence of H. pylori is the only other main modifiable risk factor for gastric cancer. It is recommended that all 1 st degree relatives of gastric cancer patients are tested and, if positive, treated for Helicobacter, especially if the index case occurred at a younger age.25,32 The risk of gastric cancer is much higher in First Nations people compared to the average Canadian population 33,34,35 and the high prevalence of H. pylori is likely a partial explanation for this. The prevalence of H. pylori is also high in immigrants from certain areas, such as Southeast Asia and Central and South America. In those patients, one should have a low threshold to test for H. pylori infection. In general, in children, who have a 1 st degree relative with gastric cancer, the recommendation is to wait with testing and treatment until after they have gone through puberty.
Table 1. The most commonly used treatments for H. pylori infection
TreatmentMedications
Concomitant Therapy (ClAMET)1. Proton Pump Inhibitor | 1 tablet given twice a day
2. Amoxicillin | 1 gram given twice a day
3. Clarithromycin | 500 mg given twice a day
4. Metronidazole | 500 mg given twice a day
Bismuth-based Quadruple Therapy (Bi-Quad)1. Proton Pump Inhibitor | 1 tablet given twice a day
2. Bismuth-salicylate | 2 tablets given 4 times a day
3. Metronidazole | 250–500 mg given 4 times a day
4. Tetracycline | 500 mg given 4 times a day
PPI-Amoxicillin-Levofloxacin (P-AL)1. Proton Pump Inhibitor | 1 tablet given twice a day
2. Levofloxacin | 250–500 mg given twice a day
3. Amoxicillin | 1 gram given twice a day
Rifabutin-based Triple Therapy (P-A-Rif)1. Proton Pump Inhibitor | 1 tablet given twice a day
2. Amoxicillin | 1 gram given twice a day
3. Rifabutin | 150 mg given twice a day
Treatments that no longer should be used:
PPI-CA Triple Therapy 1. Proton Pump Inhibitor | 1 tablet given twice a day
2. Clarithromycin | 500 mg given twice a day
3. Amoxicillin | 1 gram given twice a day
PPI-CM Triple Therapy 1. Proton pump inhibitor | 1 tablet given twice a day
2. Clarithromycin | 500 mg given twice a day
3. Metronidazole | 500 mg given twice a day
Proton Pump Inhibitor=PPI. Options: Omeprazole 20mg, Pantoprazole 40 mg, Lansoprazole 30 mg, Rabeprazole 20mg, Esomeprazole 40mg, Dexlansoprazole 60mg
Cl=Clarithromycin
M= Metronidazole
A= Amoxicillin
B= Bismuth subsalicylate
T= Tetracycline
L=Levofloxacin
Handouts for patients
Helicobacter infection of the stomach
Concomitant Therapy (CLAMET)
Bismuth-based Quadruple Therapy (Bi-Quad)
PPI-Amoxicillin-Levofloxacin Therapy (P-AL)
Rifabutin-based Triple Therapy (P-A-Rif)
(Download all)
References
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Fallone CA. Epidemiology of the antibiotic resistance of Helicobacter pylori in Canada. Can J Gastroenterol. 2000;14(10):879-882. doi:10.1155/2000/562159 (Request with CPSBC or view with UBC)
Morse AL, Goodman KJ, Munday R, et al. A randomized controlled trial comparing sequential with triple therapy for Helicobacter pylori in an Aboriginal community in the Canadian North [published correction appears in Can J Gastroenterol. 2014 Mar;28(3):167]. Can J Gastroenterol. 2013;27(12):701-706. doi:10.1155/2013/143047 (View)
Choi IJ, Kim CG, Lee JY, et al. Family History of gastric cancer and Helicobacter pylori treatment. N Engl J Med. 2020;382(5):427-436. doi:10.1056/NEJMoa1909666 (Request with CPSBC or view with UBC)
Gupta S, Li D, El Serag HB, et al. AGA Clinical Practice Guidelines on Management of Gastric Intestinal Metaplasia. Gastroenterology. 2020;158(3):693-702. doi:10.1053/j.gastro.2019.12.003 (View)
Chan FK, Chung SC, Suen BY, et al. Preventing recurrent upper gastrointestinal bleeding in patients with Helicobacter pylori infection who are taking low-dose aspirin or naproxen. N Engl J Med. 2001;344(13):967-973. doi:10.1056/NEJM200103293441304 (View)
Chiba N, Van Zanten SJ, Sinclair P, Ferguson RA, Escobedo S, Grace E. Treating Helicobacter pylori infection in primary care patients with uninvestigated dyspepsia: the Canadian adult dyspepsia empiric treatment-Helicobacter pylori positive (CADET-Hp) randomised controlled trial. BMJ. 2002;324(7344):1012-1016. doi:10.1136/bmj.324.7344.1012 (View)
Moayyedi P, Delaney BC, Vakil N, Forman D, Talley NJ. The efficacy of proton pump inhibitors in nonulcer dyspepsia: a systematic review and economic analysis. Gastroenterology. 2004;127(5):1329-1337. doi:10.1053/j.gastro.2004.08.026 (Request with CPSBC or view with UBC)
Veldhuyzen van Zanten SJ, Flook N, Chiba N, et al. An evidence-based approach to the management of uninvestigated dyspepsia in the era of Helicobacter pylori. Canadian Dyspepsia Working Group. CMAJ. 2000;162(12 Suppl):S3-S23 (View)
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Colquhoun A, Hannah H, Corriveau A, et al. Gastric Cancer in Northern Canadian Populations: A focus on cardia and non-cardia subsites. Cancers (Basel). 2019;11(4):534. Published 2019 Apr 15. doi:10.3390/cancers11040534 (View)
PathwaysBC Resources
Helicobacter pylori associated duodenal/gastric ulcer: adult patient treatment recommendations. Bugs & Drugs. Accessed July 19, 2023.
Patient handouts for tests, emailable via PathwaysBC:
Patient instructions: Helicobacter pylori stool antigen test. LifeLabs. Accessed July 19, 2023.
Patient instructions: Helicobacter pylori urea breath test. LifeLabs. Accessed July 19, 2023.
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13 responses to “Management of Helicobacter pylori in 2023: who should be tested, treated, and how”
Joanne Perry August 9, 2023 at 6:18 pmPermalink In Ontario h.pylori breath test and fecal testing is private pay rendering it inaccessible to the majority of newcomers especially the refugee population.
We are left with serology which as pointed out will always remain positive.
The proposed strategy of retesting for cure after treatment renders this even more challenging to use this approach.
Otherwise a good overview and review.
2. Mairead Keady August 9, 2023 at 6:35 pmPermalink Thank you for this excellent TCMP! You specify that the PPI should not be stopped in patients with H Pylori confirmed PUD (and especially bleeding ulcers) until the infection is confirmed eradicated – but I presume one must stop the PPI for 14 days in order to completed a UBT or HpSAT to confirm successful treatment? i.e. continue a PPI for 8-12 weeks after the 14 day eradication period, and then stop it for 2 weeks prior to testing the patient again?
3. Evan M August 9, 2023 at 8:02 pmPermalink I agree with the change to quad therapy as first line treatment.
In Sask as well I believe it is not easy access to get the urea breath test, however, stool antigen is much more accessible.
I am curious about the need to hold a PPI prior to the stool antigen or endoscopy as to what the percentage of equivocal results would be?
For Joanne in Ontario, I do hope the province moves to cover at least stool antigen as a much less invasive way than endoscopy for getting results that are not private pay.
If someone has ulcers confirmed by endoscopy my referral centre will treat empirically and after treatment will continue minimum 3 months of PPI
I think as well telling patients the pill burden before they pick up the prescription is important to increase compliance as it is a lot of medications for two weeks to take, and I have had some people say they didn’t finish the course as it was to many pills!
Very relevant TCMP, thank you!
4. R Keith Phillips August 9, 2023 at 9:21 pmPermalink While I was practicing as a GP Anesthetist, and taking an NSAID for shoulder pain, I had a GI bleed. I was H Pylori positive on gastroscopy. I had the multi pill regime.
Since then, if I have spicy pizza and/or a beer, I get some indigestion, and will take a PPI for a few days. Likely about 7 tablets per month, on average.
Is being an anesthetist still considered a risk factor for IPylori infections?
Begs the question of whether I should be worked up again.? Gastroduodenosopy. Or, whatever.
KP born 1945.
BSc (Hons. Chem) MD CCFP FCFP LM
Clinical Associate Professor, Retired
Department of Family Practice,
Faculty of Medicine, University of BC
5. Ron Abraham's August 10, 2023 at 7:53 amPermalink Thank you- good update for current treatment approach – also should mention diet and eliminating caffeine – esp the AM caffeine – on empty stomachs
6. Mustapha Olajuwon August 11, 2023 at 4:43 amPermalink Great presentation on current approach to treatment of H.pylori ; still dispense a lot of HP-PAC combos and occasionally get the Bi-QUAD therapy dispensed.
7. Dr. Cynthia L. Clark August 25, 2023 at 9:40 amPermalink Thanks for a great review of H Pylori testing and treatment. Will use the blister packs! I anticipate it will still be a challenge for me to differentiate patients with dyspepsia vs GERD. Any further tips on this?
8. Rahim November 22, 2023 at 10:12 amPermalink Thanks for coming up with this amazing piece on the subject. Can you please advise on which foods to take and avoid while treating H. Pylori?
9. Sander Veldhuyzen van Zanten November 22, 2023 at 3:03 pmPermalink @ Mairead Keady: Yes that is correct. I would continue to treat with PPI for at least 3 months if indeed an ulcer was seen. After that one can usually hold the PPI for 2. Weeks. This assumes the patient does not have any ongoing dyspepsia symptoms. Sometimes if the patient has heartburn symptoms that are troublesome one can give antacids temporarily
@ Evan M: The issue is false negative or equivocal results. If an endoscopy is done there is no need to hold the PPI as histology generally will detect Hp if the infection is persistent. Importantly biopsies need to be taken from both the antrum AND the body part of the stomach to avoid false negative results.
For UBT and stool antigen test the recommendation is to hold PPIs for 14 days prior to the test. The stool test has more equivocal results than the UBT. Not much data have been published on the rates of equivocal or false negative results.
@ R Keith Phillips: Endoscopists and possibly anesthesiologists were reported to have a higher risk of Hp infection but there are no recent data on this. It is recommended to repeat testing to determine whether Hp was cured after Hp treatment is given. This is especially true if one had a bleeding ulcer. In this case was there indeed a bleeding ulcer seen as the cause of the GI bleed? In this case the NSAID may have triggered the GI bleed but test for cure of Hp is still recommended. I do not think it likely that the spicy food is the cause of the upper GI symptoms. More likely this is due to underlying GERD (heartburn dominant symptom) or dyspepsia (epigastric discomfort dominant symptom). In this case given there is frequent recurrence of symptoms I would consider taking a PPI every day. And this even more so since there is a history of a previous upper GI bleed.
@Dr. Cynthia L. Clark: Any further tips on to differentiate patients with dyspepsia vs GERD? Not really. Importantly PPIs are effective in both conditions. There is a lot of overlap between GERD and Dyspepsia. In GERD heartburn is often the dominant symptom. In dyspepsia epigastric pain or discomfort is the dominant symptom. If gastroscopy is done in patients with dyspepsia reflux esophagitis is more commonly seen than gastric or duodenal ulcers. So GERD is more common.
@Ron Abraham A healthy diet is very important. And so is quitting smoking as this is a risk factor for GERD, peptic ulcers and upper GI bleeding. I never take people off a good cup of coffee! (I myself love a cappuccino in the morning and would not comply with stopping having them) But excessive intake of coffee can cause indigestion.
10. Mary Obstfeld December 21, 2023 at 5:46 amPermalink Does the quad-pack come prepacked like H-pak or would it be better to get the pharmacist to bubble pack for better adherence?
Thanks
Mary Obstfeld, NP
Cold Lake First Nations, Alberta
11. Dr. Memuna Tanko August 30, 2024 at 11:09 amPermalink Brilliant brilliant paper on Pylori infection
I have 2 questions and will be most grateful if you could share some answers;
1, what do you do if you started treating a patient for infective diarrhea before getting your stool for Pylori antigen results, will you advise that the patient completes the course of the 1st antibiotic before starting the quad therapy, or discontinue and start the quad therapy?
Can l also request a stool routine examination as well when I am suspecting H Pylori infection?
Dr. Memuna Tanko August 30, 2024 at 11:21 amPermalink Thank you so much.
Does the quad therapy also apply to children managed for Pylori infection?
13. Sandra Wickham July 23, 2025 at 10:37 pmPermalink When using BiQuad Therapy is it necessary to separate doses of Pepto Bismol and Tetracycline in order to avoid decreased levels of the latter medication? Is it better to use Pepto Bismol Liquid vs Tablets because of Calcium Carbonate in the Tablets? Excellent article!
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11558 | https://www.gauthmath.com/solution/1838008205305873/8-200-mod-7 | Solved: -200 mod 7 [Math]
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-200 mod 7
Expert Verified Solution
100%(2 rated)
Answer
The answer is 3
Explanation
Find the largest multiple of 7 less than or equal to -200
We want to find an integer $$k$$k such that $$7k \le -200$$7 k≤−200. Dividing -200 by 7 gives approximately -28.57. So, we take the integer part, which is -29.
2. Calculate the multiple
$$7 \times (-29) = -203$$7×(−29)=−203
3. Calculate the remainder
The remainder is the difference between -200 and -203:
$$-200 - (-203) = -200 + 203 = 3$$−200−(−203)=−200+203=3
4. Express the result in modulo notation
$$-200 \equiv 3 \pmod{7}$$−200≡3(mod 7)
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11559 | https://www.brookings.edu/wp-content/uploads/1971/12/1971c_bpea_perry_denison_solow.pdf | GEORGE L. PERRY Brookings Institution Labor Force Structure, Potential Output, and Productivity ECONOMISTS HAVE ATTEMPTED TO ESTIMATE potential gross national product for over a decade now. Potential GNP measures the output the economy would produce if it were operating at some fixed, fairly low level of unemployment, usually defined by an aggregate unemployment rate of about 4 percent. The difference between potential and actual GNP at any point in time is known as the GNP gap. In 1962, Arthur Okun published an analysis that has been the benchmark for official measures of potential GNP ever since, and in the process enunciated what came to be known as Okun's law, which relates the unemployment rate to the percentage GNP gap.' Potential GNP and Okun's law became two of the handiest tools of analysis and presentation for economic stabilization problems. Particularly during the first half of the 1960s, when GNP was running below potential and policy was devoted to closing the gap, no sophisticated analysis of the economy failed to identify the loss in real output that was associated with an economy falling short of full employment. The careful estimation of the full employment surplus in the federal budget has been an important by-product of the estimation of potential I want to thank Nancy Hwang and Herbert F. Lowrey, Jr., who did all of the com- putations in this paper. 1. Arthur M. Okun, "Potential GNP: Its Measurement and Significance," in American Statistical Association, Proceedings of the Business and Economic Statistics Section (1962), pp. 98-104. 533 534 Brookings Papers on Economic Activity, 3:1971 GNP.2 By estimating the revenues and expenditures that would be gener- ated by the economy operating at its potential level, and comparing these with actual revenues and expenditures, one can determine with a useful degree of precision that part of a budget surplus or deficit attributable to cyclical deviations in the economy around potential. Medium-term pro- jections of the economy have also come to depend on projections of potential GNP. The last two reports of the Council of Economic Advisers have used the concept in this way, projecting the total GNP that would be available in future years and the various end uses to which it could be put under alternative economic policies. The attempts over the last decade to quantify potential output have focused attention on two important matters: the implications, in terms of lost output, incomes, and revenues, of operating the economy below its potential level; and the changes that could be expected in the growth rate of the economy's potential as a result of changes in the factors that underlie it. Corresponding to the two kinds of analytic insights just mentioned are two basic ways of approaching the measurement of potential output. One can analyze the relation between the GNP gap and the unemployment rate at any moment of time. This approach involves estimating the differ- ences in employment, average hours worked by each employee, and average productivity of each employee that would accompany the difference be- tween the actual and target unemployment rate. Or one can start from a benchmark period for which potential output is known, and estimate the changes in potential output for other years from information about the sources of these changes. Although some fairly elaborate models of eco- nomic growth have been used for this purpose, this procedure, at its sim- plest, involves estimating the changes in the labor force and hence employ- ment, in average hours worked by the employed, and in output per manhour that have prevailed or can be expected to prevail along the potential path. In his original article, Okun wedded these approaches to measuring potential. He came out with estimates, first, of how potential output grew through time, and, second, of how the gap between potential and actual GNP was related to the unemployment rate at a particular time-Okun's law. According to Okun's law, the percentage GNP gap is a little over three times the excess of the unemployment rate over 4 percent, where 2. See, for example, Arthur M. Okun and Nancy H. Teeters, "The Full Employment Surplus Revisited," Brookings Papers on Economic Activity (1: 1970), pp. 77-110. George L. Perry 535 potential is defined as the output the economy would produce with 4 per- cent unemployment. In describing the other approach to the problem-measuring the ingre- dients of growth in the economy's potential output year by year-Okun found a trend growth rate of 3.5 percent a year during the years covered by his study (1954-62) and, in simplest form, decomposed this growth into a 1.2 percent and a 2.3 percent annual rate of growth in the labor force and in output per man employed, respectively, estimates that took account of a gradual decline in average hours worked per man during the period. As economic expansion narrowed the GNP gap that had existed at the time Okun made these estimates, both these methods of estimating potential GNP stood up very well. By the middle of the decade, the gap had been closed and actual GNP corresponded to estimated potential. But since 1965, some unexpected things have happened to the determinants of potential GNP. After growing at nearly the predicted potential rate of 1.2 percent during the previous decade, the labor force grew at an average annual rate of almost 2.4 percent between 1965 and 1970. With the simplified formula that allows for a 2.3 percent trend growth rate in output per man, an expansion this rapid in the potential labor force would indicate a 4.7 percent annual growth in potential GNP over the interval. This rate is much faster than anyone had projected and much faster than that which actually developed, despite falling unemployment rates over most of the period. However, during these same years, the unexpected surge in labor force growth was accompanied by an equally unexpected dip in average hours worked per employee, and disappointing growth in actual labor produc- tivity. In Table 1, the actual changes in the labor force, average hours of Table 1. Annual Growth Rates of Labor Force, Weekly Hours, and Output per Manhour, 1948-70 Percentage Average Output Labor weekly per Period force hours manhour 1948-55 1.49 -0.47 3.45 1955-65 1.16 -0.39 2.75 1965-70 2.35 -0.78 1.63 Sources: For labor force data, see appendix; other series are calculated by the author from data supplied by Edward F. Denison from a draft of his forthcoming study. See note 6, p. 538, for amplification. 536 Brookings Papers on Economic Activity, 3:1971 work, and productivity over this period are compared with their changes in earlier years. The surprises in hours of work and productivity affected output growth in the opposite direction from the surge in the labor force. As a result, until 1969, Okun's law kept working well in relation to the official estimates of potential. Official estimates, reflected in the Economic Report of the President and in the Census Bureau publication, Business Conditions Digest, accept the methodology of adding trend productivity growth to the growth of the potential manhours of labor input in the economy in order to measure the growth of potential output. But the official estimates of these compo- nents of potential output growth are periodically adjusted to reflect devel- opments in actual output and unemployment. They indicate that potential output grew at an average annual rate of about 4 percent from 1965 to 1970. This downward adjustment from the naive 4.7 percent growth pro- jection is in the right direction, but the justification for it is not clear. Which of the changes shown in Table 1 are likely to represent new trends in the components of potential output growth, which are likely to prove transitory, and how do they add up to a quantitative estimate of current potential GNP? The performance of productivity over this interval is of particular in- terest in itself. Changes in the productivity trend have important implica- tions for the level of profits in the economy3 and for the distribution of income more generally. They are significant for understanding inflation since the productivity trend determines the translation of hourly wage costs into standard wage costs per unit of output, which constitute the prime determinant of price movements.4 Elements of GNP Growth WEIGHTED LABOR FORCE VARIABLES In a recent paper I discussed the significance of the changing age-sex composition of the pool of unemployed workers for the question of infla- tion.5 I noted that large and persistent differences exist in the effective 3. Arthur M. Okun and George L. Perry, "Notes and Numbers on the Profits Squeeze," Brookings Papers on Economic Activity (3:1970), pp. 466-72. 4. Robert J. Gordon, "Inflation in Recession and Recovery," Brookings Papers on Economic Activity (1:1971), pp. 105-58. 5. George L. Perry, "Changing Labor Markets and Inflation," Brookings Papers on Economic Activity (3:1970), pp. 411-41. George L. Perry 537 labor supply offered by individuals in the several age-sex groups. In order to take account of these differences, the average wage and average number of hours worked per week by individuals in the various groups were used to weight the number of individuals in each group. This led to the concept of a weighted unemployment rate, defined as the ratio of weighted total unemployment to the weighted total labor force. The same weighting adjustments are even more directly relevant for measuring the real labor input associated with a given employment total. So it is natural to conduct an analysis of productivity and potential output in these terms. The effect of the changing composition of employment has always been implicitly accounted for in past studies of potential output growth. Changes in the relative composition of employment that change average labor productivity or average hours worked per man become one element deter- mining the average trend in these sources of growth. But if the composi- tional change is not proceeding at a steady rate, it becomes important to take explicit account of its effect in order to isolate other changes that may be taking place. In order to do this, I have used relative wages and hours of work to weight the various labor force groups. Because employ- ment and average hours are generally treated separately in analyzing potential output growth, it seemed most natural to weight employment by the relative wages of the age-sex groups and to use differences in average hours among groups to help explain the observed movements in economy- wide average hours. Weighted employment thus refers to the sum of wage-weighted employment for each age-sex group, with the wage weights corresponding to the relative wages earned by the average worker in each age-sex group. These wage weights are proxies for the relative productivi- ties of workers. In the weighted employment total, an employee whose productivity (as measured by his wage) is only half the average productivity of all workers gets only half weight. Since the composition of the work force has been shifting continually toward relatively more women and young workers, individuals who have relatively low wage weights, the growth rate of weighted employment is lower than that of officially measured employment throughout the postwar period. But while the difference between the growth rate of the two mea- sures averaged about 0.2 percentage point until 1965, it became 0.5 per- centage point during the 1965-70 period. This change makes it especially important to use' the weighted employment variable in analyzing the trend of productivity and potential output in recent years. In order to examine potential growth and some of its characteristics 538 Brookings Papers on Economic Activity, 3:1971 more carefully, I have used the weighted labor force concepts to conduct a new analysis covering the whole postwar period. This involves estimating the trend growth of labor force participation rates, average hours per worker, and productivity along the potential path, and their departure from trend values associated with operating at output levels other than potential GNP. Taking explicit account of labor force composition turns out to provide new information on these estimates of potential and assists understanding of the recent behavior of productivity. PARTICIPATION RATES For some groups in the population, notably women and young workers, the fraction of the population that is in the labor force has varied sub- stantially over the postwar period. These labor force participation rates vary cyclically, with a larger fraction of the population in the labor force when unemployment is low than when it is high. They also vary through time for reasons that are independent of the unemployment rate. Over most of the postwar period, participation rates in these groups have risen. While both the cyclical sensitivity and the rising trend of participation rates stand out clearly in the data, the precise forces governing participation rates are complex. They can be approximated only crudely by statistical relations that try to explain participation rates in terms of unemployment rates and a time trend. However, it is impossible to take explicit account of all the things that should help explain participation rates. And estimates of the potential labor force are needed.6 As a necessary compromise, I have let actual labor force growth tell most of the story. The first step was to estimate participation rate equations of the following form: (1) (pt) = A +a(l (1-ut) + bi log Tt, 6. The labor force and employment data and the data on average hours of work that are used in this paper are based on data supplied by Edward F. Denison from a draft of a forthcoming study. They are derived primarily from the establishment-based data of the Office of Business Economics, which are published in July issues of the Survey of Current Business, with additions made for workers not covered by the establishment data in order to make the labor input total comparable with the output total represented by real GNP. The appendix describes how I matched these data to the official labor force series adjusted for definition and census benchmark changes in order to allocate the total among the age-sex groups in the population. George L. Perry 539 where L = labor force P = population i = subscript designating age-sex groups t = dating subscript A = a constant u = weighted unemployment rate as defined above, with 1 - u, the weighted unemployment rate, taken as a measure of labor mar- ket tightness log T = a time-trend term in which T equals 100 in the first quarter of 1948 and rises by 4 each year. A separate equation was estimated for each age-sex group from annual data fitted to 1948-69. The estimates are given in the appendix.7 Generally, equations of this form are used to estimate potential par- ticipation rates by setting the employment rate at its potential value and solving for the participation rate in each year.8 This simply allows the estimated time trend to determine the path of potential participation rates. Rather than doing this, I have used the employment rate coefficient, ai, to adjust the actual participation rate each year to an estimated potential level for that year. The difference between the actual and potential weighted employment rate is multiplied by a, to get the estimated marginal change in participation rates that would have come from being at the potential unemployment rate. Thus the time trend in equation (1) serves simply as the best available variable to use in arriving at an accurate estimate of the effect of unemployment on participation rates. Since factors affecting the trend of the potential labor force should be expected to affect the actual labor force as well, this procedure should yield more sensible estimates, particularly for the crucial period of the late 1960s. In that period, unem- ployment was near its potential level, thus minimizing the error from the marginal adjustment that is made. 7. Equations using lagged as well as current employment rates were tried, but the lagged terms were insignificant. I also tried the employment rate squared, both adding it to equation (1) and using it by itself, to see if the effect was nonlinear, but it was not. Equations using the conventional unemployment rate in place of u gave slightly inferior fits. 8. N. J. Simler and Alfred Tella, "Labor Reserves and the Phillips Curve," Review of Economics and Statistics, Vol. 50 (February 1968), pp. 32-49; and Gordon, "Inflation in Recession and Recovery," especially Appendix B, pp. 149-53. 540 Brookings Papers on Economic Activity, 3:1971 This way of estimating makes a big difference in recent years when actual participation rates have climbed much faster than any equation fitted to earlier periods would have predicted. Had predictions been made directly from the equation, the potential labor force would have been estimated well below the actual labor force, even after adjustment of the actual labor force for unemployment experience in the period. Perhaps a one-year surge in participation rates could have been regarded as aberrant and as irrele- vant to an estimate of the trend of potential. But after five years of such ex- perience, one must accept the verdict of the actual data. The fact that lagged employment rates and nonlinear forms of the employment rate were not useful variables in the equation further strengthens the presump- tion that a change in the trend of participation rates occurred in this period. After potential participation rates for each year for each age-sex group were estimated in the manner just described, the potential labor force was projected separately for each group by multiplying its potential participa- tion rate by its population in each year. The total potential labor force series is given in the appendix. The effect of unemployment on the overall labor force participation rate varies through time with the changing composition of the work force. For recent years, the estimates indicate that a drop of one percentage point in the official unemployment rate would induce an increase of 500,000 in the labor force. Thus, taking account of this enlarged labor force, an employment increase of 1,340,000 workers, or 1.6 percent, would be needed to reduce the unemployment rate 1 percentage point. POTENTIAL WEIGHTED EMPLOYMENT Potential employment was calculated using the estimates of each group's potential labor force together with an estimate of what each group's unem- ployment rate would be along the potential path each year. These unem- ployment rate estimates were made by adjusting each year's actual un- employment rate for each group as described in the appendix. Just as in calculating actual weighted employment, potential employment in each group was multiplied by the relative wage for the group, and these esti- mates were added over all age-sex groups to arrive at potential weighted employment for the whole economy. George L. Perry 541 AVERAGE HOURS Over the postwar period, the average hours worked per year per em- ployee show clear cyclical fluctuations as well as a persistent downward trend. The cyclical fluctuations demonstrate mainly that employers vary average hours as well as employment in response to changes in their own demand for labor. Higher unemployment rates mean less overtime and an increase in short workweeks of employment. To a smaller extent, the cyclical fluctuation in average hours may reflect some relative shift in em- ployment away from manufacturing industries, where average weekly hours are high, and toward less cyclically sensitive industries, such as retail sales and services, where the standard workweek is shorter. The downward trend in average hours that is apparent once cyclical fluctuations are accounted for has not proceeded at the same rate throughout the postwar period. Through the mid-1950s, the downtrend was relatively fast as manufacturing industries moved to a standard forty-hour week. After that time, there was little further drift in the standard manufacturing work- week, and the economy-wide downtrend in average hours slowed. Then, surprisingly, average hours declined sharply again in the 1965-68 period. Because there are persistent differences in the relative number of average hours worked by different age-sex groups in the labor force, the changing mix of employment contributes to the trend in economy-wide average hours of work. In order to isolate underlying trends in average hours from the effects of the changing employment mix, I eliminated the annual change in hours that was due purely to changes in the relative mix in employment. The resulting series was then explained by its statistical relationship to the weighted unemployment rate and time trends. The best-fitting equation for the postwar period was (2) AHI = 43.45 - 0.195u - 0.209TI, + 0.073T2, - 0.135T3t, (-6.68) (-11.17) (2.92) (-2.81) W2 = 0.986; standard error = 0.125; Durbin-Watson = 1.50. The numbers in parentheses are t-statistics. where AH is average hours worked per week, u is the weighted unemploy- ment rate, and the Ts are separate time-trend variables. In the equation, T1 and T2 begin in 1947 and 1955, respectively, and continue throughout the period; T3 runs from 1966 to 1968 and remains at its 1968 level thereafter. These time-trend variables explain the data better than any alternative I tried. 542 Brookings Papers on Economic Activity, 3:1971 The last trend variable, T3, is clearly contrived to fit the data, and I have no independent reason to justify it. That it was needed even with data cleansed of employment mix effects is evidence that the 1965-68 decline in average hours is based on much more than the sharp rise in the employ- ment of workers from age-sex categories with short workweeks. I con- ducted some other tests that convinced me that the recent decline in hours should be treated as a permanent change in the level of the trend. In particular, it did not appear that employers were reacting to the very tight labor markets of these years by hiring more employees than they normally would have in case they would be needed later, compensating by reducing average hours of work more than they normally would. If this had been the case, it should have shown up in the statistics on involuntary part-time employment. The fraction of workers on part time for economic reasons is historically closely related to the unemployment rate. Regressions showed that, during these years, the historical relationship continues to explain this category of part-time employment with no unusual errors. Equation (2) should already account for normal effects of the changing composition of employment. To check further, I ran regressions on volun- tary part-time employment as a fraction of total employment for each age-sex group. This fraction is inversely related to the unemployment rate, indicating that a greater-than-average proportion of marginal workers choose to work part time. Once that source of variation is accounted for, there remains a clear increase during the last half of the 1960s in the proportion of employment that is voluntarily part time in each age-sex group taken separately. The composition change adds to this decline. The trend in average hours indicated by equation (2) was adjusted by adding back the changes due to mix effect that would occur along the potential path. Because the separate time trends give abrupt changes in this path of potential average manhours, the final series was smoothed by forming centered three-year moving averages for use in estimating poten- tial GNP. According to these estimates, a fall of 1 percentage point in the weighted unemployment rate causes a 0.20 hour rise in average weekly hours worked per employee; or, equivalently, there is a 0.18 hour rise for a fall of 1 percentage point in the official unemployment rate. The time trends in equation (2) indicate that, with a constant weighted unemployment rate, average weekly hours fall by 0.21 hour per year from 1948 to 1955, by George L. Perry 543 0.14 hour per year from 1955 to 1965, by 0.27 hour per year from 1965 to 1968, and by 0.14 hour per year thereafter.9 OUTPUT GROWTH PATHS The estimates of potential weighted unemployment and potential average hours worked per employee just described provide the basis for calculating the economy's potential labor input. It will be measured by weighted po- tential manhours, defined as the product of weighted employment and average hours, both calculated along the potential path. The next step is to use this labor input measure to analyze the behavior of productivity in the economy, and from there to analyze potential output. An awkward problem of terminology now arises. I want to retain the accepted normative definition of potential output as the output that would be produced with the economy operating at around a 4 percent official unemployment rate. Thus the modifier "potential" will continue to refer to magnitudes defined along a path corresponding to a constant official unemployment rate. But analytically, it becomes more natural for some purposes to work with the growth path defined by a constant weighted unemployment rate. So I shall use the modifier "trend" to refer to magni- tudes defined along a growth path corresponding to a constant weighted unemployment rate. Both the trend and potential paths are defined to pass through the actual real GNP level in the third quarter of 1955, a conven- tion that defines the unemployment rate along the potential path to be 3.87 percent with the labor force data used here and the weighted unem- ployment rate along the trend path to be 3.37 percent.10 Because of gradual changes in the age-sex composition of total unem- ployment, the official unemployment rate has been rising gradually along the employment path defined by a constant weighted unemployment rate. 9. The time trends in equation (2) decline by a constant amount each year rather than a constant percentage. Thus the percentage decline they project would gradually increase, but I would have no confidence in projecting so far ahead with this equation that this effect became important. 10. These unemployment rates are based on data that include the armed forces in em- ployment and the labor force and that are adjusted for census benchmark revisions and definition changes, as discussed in the appendix. At present, the official unemployment rate corresponding to the potential path is 4.05 percent. The values of weighted employ- ment and average hours along the trend path are estimated in a manner exactly analogous to the estimates made along the potential path that are described above. 544 Brookings Papers on Economic Activity, 3:1971 Thus potential output and trend output differ by noticeable amounts. But it makes little difference whether the analytics described here are con- ducted along the trend path and then adjusted to potential, as I do, or conducted along the potential path in the first place. A Model of Productivity and Output The procedure for estimating trend GNP starts by wedding proposi- tions about the behavior of labor productivity, and applies them to wage- weighted employment rather than to employment as conventionally measured. These propositions correspond to the two ways of viewing the concept of trend GNP: as a fairly smooth growth path passing through years when GNP is known to have been at its trend level (a growth view); and as a predictable relationship between the deviation of GNP from its trend and the weighted unemployment rate (a gap view). The simplest version of the model will be explained first and then some elaborations of it will be reported. The proposition that, along the trend output path, weighted labor pro- ductivity grows at a constant rate is expressed by the formula (3) M = BerTt. Here Q and H are the hypothetical levels of output (real GNP) and labor input (weighted trend manhours) along the growth trend of the economy defined by a constant (low) weighted unemployment rate. The trend of weighted labor productivity grows at the rate r each year; T is a time trend; and B is a constant. The proposition that, in any given year, weighted labor productivity is higher the higher the level of labor input can be expressed directly by a formula such as (4) = where b is expected to be greater than 1. In this equation, Q and H are again the hypothetical levels of output and labor input at trend, while Q and H are the observed levels of output and labor input at that same point George L. Perry 545 in time. This equation is equivalent to a relation expressed directly in terms of weighted labor productivity, P, of the following form: (4a) (H)6 The combining of the growth and gap equations, (3) and (4), results in an equation from which to estimate the growth rate, r, and the propor- tionality factor, b, using only the trend labor input estimated earlier and actual, observable data on output and labor input: (5) Qt = BerTt H11bHb. For estimation purposes, this equation is expressed in logarithmic form: (6) log (Q) - log B + rTt + b log () . CHANGES IN TREND PRODUCTIVITY Although the weighted productivity trend may grow smoothly, the as- sumption that its growth rate never changed throughout the postwar period, as implied by equation (3), may be too restrictive. In particular, aggregate production functions for the economy that explicitly measure the input of capital as well as of labor result in estimates of potential labor productivity that depend on the size of the capital stock at any time.11 Such models imply that, other factors being equal, the growth rate of labor productivity will be positively related to the growth rate of the capital stock with which labor works. In practice, attempts to measure such aggregate production functions have not led to estimates of labor productivity that show sharp changes in its growth rate over the postwar period. Changes in the capital-labor ratio, the infusion of new technology, and improvements in the average education of the work force-the major influences on labor productivity-come about only gradually. Still, the assumption of a completely unchanged rate of productivity growth along the trend path may be too restrictive. Equation (6) was 11. Lester C. Thurow and L. D. Taylor, "The Interaction between the Actual and the Potential Rates of Growth," Review of Economics and Statistics, Vol. 48 (November 1966), pp. 351-60; and Robert M. Solow, "Technical Progress, Capital Formation, and Economic Growth," in American Economic Association, Papers and Proceedings of the Seventy-fourth Annual Meeting, 1961 (American Economic Review, Vol. 52, May 1962), pp. 76-86. 546 Brookings Papers on Economic Activity, 3:1971 estimated separately for various subperiods of the postwar years in order to see if noticeable differences in the estimated productivity trend could be observed. Differences did arise but, invariably, changes in the estimates of r, the productivity trend, went along with changes in the estimate of b, the cyclical productivity term. This was particularly noticeable when sub- periods were broken at extremes of cyclical peaks or troughs such as 1953, indicating that the statistical estimates were confusing the two sources of change in actual labor productivity. In order to minimize this problem, the equations were constrained to keep the cyclical response of productivity measured by b unchanged throughout the period, while permitting the trend of productivity, r, to vary by introducing additional time trends. The new estimating equation embodying these assumptions is (7) log Q) = log B + riTi, + r2T2t + r3T3t + b log ()H. The several productivity terms, ri, r2, and r3, are additive for all times when their corresponding time indexes are nonzero. To examine the pos- sibility that the productivity trend changed at various times during the postwar period, the periods for starting the time indexes were varied. This method of scanning the data for possible interruptions in the productivity trend invites the discovery of breaks that may be spurious. The estimates obtained must be viewed with this in mind. On the other hand, the failure to achieve improvement in statistical fit with the additional time trends would be fairly strong evidence that the trend growth of productivity did not vary noticeably over the period. VARIATION IN GAP RESPONSE Equation (4a) implies that, at a given time, weighted labor productivity will be greater the higher the level of labor input. This tendency has been widely observed. But it is less clear that the productivity improvement should continue at the same rate as employment levels are pushed higher and higher. Should the productivity gain available from operating at 5 percent rather than 6 percent unemployment rates be expected to be the same as the improvement available in operating at 3 percent rather than 4 percent unemployment? Many factors contribute to the increased productivity that is observed. Within a given industry, some employment is of an overhead nature, vary- ing little if at all with variations in production. For the economy as a George L. Perry 547 whole, industries whose production and employment are cyclically more sensitive experience, on the average, higher levels of labor productivity than most others; thus as these high-productivity industries increase their relative share of total employment, economy-wide average labor pro- ductivity increases. It is worth noting that the weighting of employees that has been used in this study could be expected partly to offset this last effect since the workers in industries with high labor productivity are relatively high-weight workers; this means that the effect identified as productivity change in studies with body-count measures of employment will be captured here, in part at least, as a change in weighted employment rather than in productivity. On the other hand, the overhead labor effect will probably appear more strongly in the present study, since technical and supervisory personnel, whose employment experiences cyclical varia- tions below the average, tend to be high-weight individuals. There are also forces working to reduce average labor productivity as employment levels rise, most notably the likelihood that the marginally un- employed at any time constitute a labor group with lower average skills than their employed counterparts. But most of this effect should disappear when labor input is measured by weighted employment. It is beyond the scope of my purpose in this paper to try to sort out each of these effects. But I have tried to test whether their net effect changes as employment levels move nearer to potential and as they move above potential, as they did in recent years and during the Korean war period. In order to test this possibility, equation (4) was modified to allow the proportionality factor, b, to vary with the level of weighted unemploy- ment, u, giving the following equation: (8) (Q) = (H)bl+b2u In this form of the equation, the sum b1 + b2u gives the proportionality factor applicable at any time. If productivity increments become smaller the lower the unemployment rate, then bi and b2 will both have positive signs. The estimating equation in logarithmic form, which now embodies both the possibility of variations in the long-run productivity trend and the adjustment just described, is (9) log Q = log B + riTit + r2T2t + r3T3t + bi log (H) + b2 (u log). 548 Brookings Papers on Economic Activity, 3:1971 LAGS IN CYCLICAL RESPONSE The equations used thus far make no allowance for cyclical lags in the employment of labor. Reductions in the unemployment rate are known to lag behind a cyclical upturn. This situation reflects in part an increase in average hours of work that substitutes for an increase in employment. Also, there have been periods, such as 1956 and 1969, when productivity growth was interrupted following a substantial cyclical upswing. Some part of this interruption is due to the increasing employment of secondary workers at that stage of the cycle. This part of the effect would disappear with the labor input measure used here since it weights employment to eliminate productivity differences among workers and thus permits tests for those true lags in productivity that exist. Unfortunately, the use of annual data will reveal only substantial lags, and some genuine cyclical characteristics of the relation between output and labor input may be lost. The change in the labor input gap was introduced into equation (4) to account for lag effects,12 yielding (10) (, ) =tH (10) (Q)t @)~t [H)t (H)t,_ With allowance made again for changes in the productivity trend, this addition now leads to the estimating equation in logarithmic form (11) log tO log B + riTit + r2T2, + r3T3, + b log H + c [log () - log () I Statistical Estimates ESTIMATES OF PRODUCTIVITY The estimates of the various forms of the model just discussed and shown in Table 2 permit a comparison of some alternative hypotheses about the behavior of productivity. One kind of comparison is among the three forms of equations. It permits selection, from among the gap-type specifications, of the form that best captures the relation between output 12. Lageffects could also be introduced by a change term such as Ht/Ht-1. This im- plies a slightly different response when the growth rate of H changes. Estimates using this alternative were slightly inferior to those with the form shown here. 0E iXU X N o N ~~~00 t v t 10 PIR \1 N0 ; } t S b m m c X t O D & G o o ~~~~~~~CN n r- m W W 02 N mn en m>> mmn m>n mco~~~~~~~~~~~~~~~~~~~ C4 n 3- Q Q to tz o t ~~~~~~~~~~~~n CrsN 15 C's XX V~~~~~~~~~~~~~~~~~~e T- 4 ?- C t L r o t~ >;;?oo)dX = 9 be 2 N n N m N n n N m e N n n N n e N .00@~00 0 14 c lrt O N O O~~~~c0 t o Po "94 tXX e X t k 2 ~~~~~~~~~N ON " ON 0 0 \C 0 r-- N' oo 0 00 _ tz Oe o Oe o~~~~~r C h C14 en,c CJi Ch ClE Ch o o 0 Ch CS c) 5 X ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 0 F Y n Q r Q < Q .=~~~~~~~~~n 550 Brookings Papers on Economic Activity, 3:1971 and labor input at a given point in time. A second kind of comparison is among the different time intervals chosen in fitting productivity trends. This comparison offers a test of whether productivity trends changed noticeably during the postwar period. The results of changing productivity trends over various subintervals of the postwar period are shown by the letter designations in the stub of the table. Thus, for example, equation (9b) refers to the equation (9) form of the model with separate trends allowed for the two subperiods 1948-56 and 1957-70. The estimates for equation (9b) are shown in the middle bank of columns, labeled equation (9), and the second row, labeled (b). Alternative equations. Equation (7) is the simplest form tested. Esti- mates using it are shown in the first bank of columns in Table 2. The esti- mates in row (a) correspond to equation (6), which is a special case of equation (7) in which the productivity trend is unchanged throughout the period. Equation (9) allows the relation between the output gap and the labor input gap to vary with the weighted unemployment rate, which has the coefficient b2 in the table. Estimates are shown in the second bank of columns in Table 2. The estimates of the coefficient b2 have the right sign in the equation for every combination of subperiods, suggesting that there is some of the suspected effect: The productivity gains available from each increment of gap closing diminish as the gap itself gets smaller (or grows larger negatively). But while the expected sign persists, in no case is the coefficient estimate as much as twice as large as its standard error, and in most cases it is not larger than its standard error. In all cases, the standard error of estimate of the equation is worsened slightly rather than im- proved in the shift from the equation (7) to the equation (9) specification. Equation (11), estimates for which are shown in the third bank of columns, allows the dynamics of employment change to affect the relation between output and employment. This amendment to the simple model yields better results. The positive coefficients estimated on the dynamic term are typically twice their standard error and the standard errors of estimate of all the equations are smaller than their counterparts in the estimates from equation (7). The positive coefficients for the dynamic term predict the following pat- tern between labor input and output changes during a cyclical movement of the economy. Starting from a constant input gap, a surge of employment in one year will be accompanied by an especially large increase in output, George L. Perry 551 improvement in productivity, and reduction in the output gap. If the input gap remains unchanged in the second year, with its growth just equal to its trend, it will be accompanied by a widening of the output gap and an increase in productivity slower than trend. If the input gap continues un- changed in the third year, output and productivity will grow at their trend rates and the output gap will remain unchanged. If input growth is espe- cially slow in the fourth year and input gap widens, there will be an espe- cially slow growth in output and productivity and an especially large in- crease in the output gap. Alternative productivity trends. The most striking thing about com- parisons of different rows in Table 2 is how favorably the estimates in row (a) compare with the others. These estimates constrain the productivity trend to take on a single value for the whole postwar period. Allowing the data to set different productivity trends for different subperiods, as is done in rows (b) through (g), usually increases the standard error of estimate of the equation. In the estimates in the third bank, which employ the dynamic labor input gap term, no alternative improves at all on the overall fit of the row (a) equation. In the third bank, equation (1 If) is of special interest because it gives the most strikingly different results from the other equations, and it fits the data virtually as well as equation (1 Ia). It breaks the period after 1953, a peak year of very low unemployment, and again after 1961, a recession year. This produces a trend estimate of productivity growth in the 1954-61 period substantially slower than the trend either before or after that. It also leads to an estimate of virtually no pure gap effects on productivity, which is both doubtful a priori and inconsistent with all other estimates of this effect in the table. Finally equation (1 if) makes a prediction error on pro- ductivity in 1970 twice as large as that of equation (11 a). Thus, the results reported in Table 2 lead me to accept the hypothesis that the weighted productivity trend was constant throughout the postwar period. There is no reason to reject the row (a) estimates in favor of any others. PRODUCTIVITY TRENDS Equation (1la) can be used to provide estimates of productivity and output trends. Two adjustments are necessary to produce estimates in conventional form. The first is simply to convert from the trend path, 552 Brookings Papers on Economic Activity, 3:1971 defined by a constant weighted unemployment rate, to the potential path, defined by a constant official unemployment rate. This conversion is dis- cussed below in connection with estimating potential output and is not quantitatively important for estimates of productivity trends. The second adjustment is basic to the method used here. Equation (I Ia) says that, throughout the postwar period, weighted labor productivity grew at a constant rate of about 3 percent a year along the trend path. But this means that trend labor productivity as commonly measured-using un- weighted manhours-varied as the relation between weighted and un- weighted trend manhours varied. This proposition in turn implies a growth rate of trend productivity slower than 3 percent throughout the period, since the relative importance of low-weight employees grew. It also implies that the growth rate changed in various periods since the relative composi- tion of employment did not change at the same rate throughout. Table 3 shows the effect of these shifts on conventionally measured productivity growth along the potential path. The annual rate of productivity advance slows from 2.9 percent in 1948-55 to 2.7 percent in 1955-65 and to 2.4 percent in 1965-70. Table 3. Growth in Potential Output, by Source, 1948-70 Annual growth rates in percent Output and sources of growth 1965-70 1955-65 1948-55 Potential output 4.16 3.53 3.90 Sources Potential employment 2.41 1. 17 1.53 Potential average hours, total -0.67 -0.36 -0.53 With constant employment mix -0.53 -0.29 -0.48 Effect of employment mix -0.14 -0.07 -0.05 Potential productivity, total 2.39 2.71 2.87 With constant employment mix 2.93 2.97 2.99 Effect of employment mix -0.53 -0.25 -0.11 Addendum Effect of employment mix on potential output -0.67 -0.32 -0.16 Trend output 4.13 3.45 3.90 Source: Calculated by author; see text. Figures are rounded and may not add to totals. Potential Output Equation ( lla) is used to calculate potential output for the postwar years. The steps used in these calculations are as follows: George L. Perry 553 1. The weighted unemployment rate that would prevail along the poten- tial path was estimated for each year based on disaggregated unemploy- ment totals by age-sex categories. (Along the potential path, the weighted unemployment rate declined noticeably from the mid-1950s to the late 1960s.) 2. The potential labor force by age-sex categories was then calculated using the unemployment coefficient in the participation rate equations to adjust the trend labor force. Weighted potential employment was calcu- lated from this and combined with estimates of potential average hours from equation (2) to produce an estimate of potential labor input. 3. Potential output was calculated by fixing it equal to actual real GNP in the third quarter of 1955 and projecting the years before and after 1955 by using the growth rate of potential labor input each year in equation (lla). VARIATIONS IN POTENTIAL OUTPUT GROWTH Table 3 summarizes the growth rate of potential output and the sources of this potential growth for three subperiods of the postwar years. For the 1955-65 decade, it confirms the 3.5 percent potential growth rate that Okun had estimated. And it indicates that, in the earlier postwar period, potential had grown at nearly a 4 percent rate. But, of greatest interest, it shows why potential output growth accelerated by only 0.6 percentage point in the 1965-70 period compared with the previous decade, despite the strong evi- dence in Table 2 that trend-weighted productivity did not slow down in recent years and despite the jump of 1.2 percentage points in the growth rate of the potential labor force and employment. During 1965-70, shifts in the composition of the work force reduced the average workweek by 0.07 percentage point a year more than in the previous decade. In addition, the downward trend in the length of the workweek independent of such relative employment shifts was 0.24 per- centage point faster than during the previous decade. Thus 0.31 percentage point of employment growth was offset by a faster decline in the trend of average hours. At the same time, the changing composition of the work force reduced the growth of productivity-or the growth of effective employment as measured by productivity weighting-by 0.28 percentage point more than during the previous decade. Together these hours and productivity effects offset half of the 1.2 percentage point acceleration of potential employment. 554 Brookings Papers on Economic Activity, 3:1971 The yearly estimates of potential GNP in current prices are shown for the postwar period in Table 4, along with actual GNP and the gap between potential and actual. In Figure 1, the difference between potential GNP based on the model of this paper and the official estimates of potential GNP are shown for the period from 1952, the first year for which official estimates are available, up to the present. The discrepancy between the two estimates is largest at the start and end of the period, but is not particularly large at any point. However, the present estimates indicate a Table 4. Potential and Actual GNP, 1948-70 Dollar amounts in billions of current dollars Gap Dollar Year Potenitial GNP GNP Percent amount 1948 $ 267.4 $257.6 3.8% $ 9.8 1949 276.6 256.5 7.9 20.2 1950 292.0 284.8 2.5 7.2 1951 324.6 328.4 -1.2 -3.8 1952 343.8 345.5 -0.5 -1.7 1953 360.5 364.6 -1.1 -4.1 1954 379.0 364.8 3.9 14.2 1955 399.0 398.0 0.2 1.0 1956 426.3 419.2 1.7 7.1 1957 458.1 441.1 3.8 17.0 1958 485.1 447.3 8.5 37.8 1959 510.3 483.7 5.5 26.6 1960 536.7 503.7 6.5 33.0 1961 562.7 520.1 8.2 42.6 1962 589.2 560.3 5.1 28.9 1963 617.8 590.5 4.6 27.3 1964 651.0 632.4 2.9 18.6 1965 688.9 684.9 0.6 4.0 1966 736.8 749.9 -1.7 -13.1 1967 792.3 793.9 -0.2 -1.6 1968 857.5 864.2 -0.8 -6.7 1969 937.2 929.1 0.9 8.1 1970 1,030.8 974.1 5.8 56.7 Sources: Actual GNP-U.S. Office of Business Economics, The National Inicome and Product Accounts of the United States, 1929-1965: Statistical Tables (1966), Table 1.1, and Survey of Current Business, Vol. 51 (July 1971), and July issues for 1968, 1969, and 1970; potential GNP-derived by author from equation (1 ta) discussed in text. George L. Perry 555 Figure 1. Difference between Present Estimates and Official Projections of Potential GNP, 1952-70 Difference (percent of potential) 0.6 0.4 - 0.2- 0.0 -0.2- -0.6 -0.86 1952 1954 1956 1958 1960 1962 1964 1966 1968 1970 Sources: Derived from formula where Q = potential GNP estimated in this paper and Q. Q official estimate of potential GNP. slightly smaller gap during the 1955-65 interval than the official estimates do. And they indicate a more rapid growth of potential between the early 1960s and the present. GNP Gaps and Okun's Law The GNP gap based on the potential output path can be decomposed into its employment, average hours, and productivity components using the same equations that were used to estimate potential GNP. At a point in time, a lower unemployment rate is associated with a larger labor force and employment level and longer average hours of work. These add up to a higher level of labor input, measured by weighted manhours. And this, in turn, is associated with a higher level of labor productivity. The higher levels of labor input and productivity measure the higher level of output that would be produced. Thus, specific estimates of each of these effects can be used to derive a relationship between the output gap and the unem- ployment rate, analogous to Okun's law. 556 Brookings Papers on Economic Activity, 3:1971 The increased employment associated with a reduction of one point in the unemployment rate can be estimated from the participation rate equa- tions given in the appendix. Because disaggregated labor force data were used in the model, this relationship will vary slightly from year to year. For recent years, a one point difference in the unemployment rate implied about a 1.35 percent difference in weighted employment. From equation (2), a one point difference in the unemployment rate is associated with a 0.45 percent difference in average hours of work. When the hours and employment effects are added together, weighted labor input is greater by an estimated 1.8 percent for each 1 percentage point differential in the unemployment rate at a given point in time. Equation (11) implies a relation between the productivity and labor input gaps of the following form: (12) = (b - 1)(- ) + C [(MH) - H_ With b estimated at 1.25 in equation (lla), the permanent elasticity of weighted productivity with respect to labor input is only 0.25. Together with the 1.8 percent rise in labor input, this implies that only a 2.25 percent output gap is associated with one percentage point in the unemployment rate. Equation (12) also shows that a substantial additional gain in output and productivity would initially accompany a closing of the gap, since c is equal to 0.33, making the elasticity of productivity 0.58 in the first year the employment gap is narrowed. But this additional effect would be transitory. If equation (7a) is accepted in place of (1 Ia), the permanent elasticity of weighted productivity to labor input is 0.40, although the effect in the first year is smaller since there is no lag term. Together with the 1.8 percent difference in labor input, this implies that a 2.52 percent output gap accompanies each point in the unemployment rate. With the labor input effect estimated here, an elasticity of output to labor input of 1.78 would be needed to reach the 3.2 multiplier between the output gap and the unemployment rate that was estimated by Okun. None of the estimated values of b in Table 2 approaches that magnitude. ALTERNATIVE SPECIFICATIONS The model of this paper basically asks what potential output level would accompany the potential unemployment rate. Okun really asked the George L. Perry 557 opposite question: What output gain is needed to achieve a target, low unemployment rate? The Okun's law multiplier depends on which question is asked. A model of output and productivity similar in form to equation (11) can be specified in terms of the second question. Keeping equation (3), which states that the productivity trend is constant, and changing equation (10)to (13) l where the expectation is that 0 < m < 1, and n < 0, yields an estimating equation of the form (14) ln (H) = K + mrT, + m log (Q) + n [log (-Q)- log (Q)t1] (where K is a constant), which implies a relation between the productivity and output gaps of the following form: (15) p = (1-m)(-Q) - [(Q) -t (Q ) The permanent elasticity of productivity with respect to labor input is given by (1 - m)/m, with some additional transitory change in the same direction in the first year the gap changes, just as with equation (11). Estimates from equation (14) fitted to the same period as equation (1 a) show the following parameter values: r= 2.98 m= 0.68 n = -0.18 1-m = 0.32 (1 - m)/m = 0.47. The weighted productivity trend is still estimated to grow by 2.98 percent yearly, yielding the same potential GNP path estimated with equation ( lla). The Okun's law multiplier is now 2.7, with weighted productivity rising by 0.86 percent along with the 1.8 percent rise in labor input for each drop of one point in the unemployment rate. 558 Brookings Papers on Economic Activity, 3:1971 Finally, the importance of the choice of question can be seen from esti- mates that retain the model of equation (11), but make output an inde- pendent variable. Doing this leaves the estimated productivity trend at 2.97 but raises the estimate of b from 1.25 to 1.56 and lowers the estimate of c from 0.33 to 0.18. Thus, once again, the same potential GNP path is pro- jected. But the Okun's law multiplier is now raised to 2.8 for permanent gap changes, while the additional transitory change in output and productivity is reduced. Even this estimate falls well short of Okun's original 3.2. It is important to emphasize that the estimates of potential output shown in Table 3 are unchanged by any of these alternative specifications of the relation between employment and output. The potential path was calcu- lated using the productivity trend, which is estimated to be just under 3.0 percent a year in all the variants discussed. Similarly, estimates of the size of the output gap, as well as the associated shortfall of actual em- ployment, hours, and productivity from their values on the potential path, are the same for all variants since they are calculated as the difference between potential values, which are the same in all variants, and actual values in any year. However, in any year the error in explaining actual productivity will depend on the equation used, since each predicts some- what different responses of productivity to deviations of the economy from the potential path. RECENT GNP GAPS AND PRODUCTIVITY In 1970, the estimated GNP gap, as shown in Table 4, was 5.8 percent of actual GNP, or about $57 billion. An unusually large part of this esti- mated gap was associated with the fact that productivity was below its projected potential level. Average hours were 0.44 percent below poten- tial; employment was 1.43 percent below potential; and productivity was 3.87 percent below potential. The employment shortfall represented 1.2 million jobs. Of these, 420,000 were the additional jobs associated with the larger labor force that would have been expected with the economy oper- ating at its potential, and 750,000 represented a reduction in current unemployment. The GNP gap was negative in 1966 and 1968, virtually zero in 1967, and 0.9 percent of GNP in 1969. The only other years in the postwar period with negative GNP gaps were 1951 through 1953. On the basis of present evidence that real GNP is rising by a little less George L. Perry 559 than 3.0 percent in 1971, the output gap will have grown to about 7.3 percent for the year, or to an estimated $77 billion in current prices. This is about $5 billion larger than the official estimates based on the same actual GNP. If output expands at a 5 percent annual rate in the fourth quarter of the year, the gap in that quarter will be about $78 billion. The data needed to separate this output gap into the shortfall in labor input and the shortfall in productivity that exist in 1971 are not yet available. But the evidence that productivity growth in the private nonfarm sector has been rising rapidly indicates that low productivity will account for a smaller part of the total output gap in 1971 than it did in 1970. Recent productivity. The poor performance of productivity over the 1965-70 period that has attracted so much attention can now be broken down into several components. During the previous decade productivity growth had averaged 2.75 percent per year. At that rate, it would have grown by 14.5 percent by 1970; but it grew only 8.4 percent, leaving a 5.3 percent shortfall from a naive projection of past productivity trends. Of this shortfall, 1.5 percent is attributable to the acceleration that occurred in the employment of women and young workers relative to prime-age men. According to equation (1 la), another 1.9 percent is accounted for by the predicted effect of slow growth of demand over this interval, and 1.9 percent is the unexpected shortfall in productivity (the error for 1970 from that equation). According to the estimates from equation (14), 2.7 percent is accounted for by the predicted effect of slow growth over the interval and 1.1 percent by the unexpected shortfall in productivity (the error for 1970 from that equation). This equation, which predicts a larger produc- tivity difference between actual and potential output, correspondingly pre- dicts lower productivity in 1970 and oence has a smaller error in that year when actual productivity was extraordinarily low. The substantial produc- tivity gains noted for 1971 will make up at least some of the unexplained shortfall shown by both equations for 1970. A Look Ahead PROJECTED POTENTIAL OUTPUT With equation (lla), potential output growth can be projected over the decade of the 1970s. For this projection, the estimated time trends in the participation rate equations had to be relied on to project the future labor 560 Brookings Papers on Economic Activity, 3:1971 force by age-sex groups. The projections used as benchmarks the 1970 estimates of potential participation rates, which in turn were based on actual 1970 labor force data. No change was assumed to take place in the relative unemployment rates among age-sex groups at potential. The resulting employment projections by age-sex category were weighted by relative wages as before. Changes in potential average hours were esti- mated from equation (2) and adjusted for the effect of changing employ- ment mix by age-sex categories. The age-sex adjustments were not sub- stantial in the projection period. On these assumptions, potential output grows at a rate of between 4.2 and 4.4 percent each year in the 1970s. For the decade, potential output growth averages 4.3 percent annually, an indisputable record for any decade in recent history. If the price deflator rises at an average annual rate of 2.5 percent over this period, potential GNP in current prices would reach $2 trillion by the end of the seventies. UNEMPLOYMENT PREDICTIONS The projections of potential output growth in future years can be used to predict the unemployment rate that would accompany alternative rates of economic expansion. Since the aim is to predict employment for assumed output paths, equation (14) is used in this projection. For each output path considered, labor input is predicted from the equation and the labor input gap is calculated for each year. The official unemployment rate is then calculated from (16) Ut = 1.8 ( HH) + 4.05 (where A is potential labor input), since 4.05 percent is the official unem- ployment rate at potential and 1.8 percent of labor input gap is associated with one point in the unemployment rate. For 1972 and 1973, potential output is projected to grow at its fastest rate in the postwar period, averaging approximately 4.4 percent yearly. If real GNP grows at a 5 percent annual rate in the fourth quarter of 1971, calculations based on equations (15) and (16) predict a 6.0 percent official unemployment rate for the year, indicating they are tracking quite well. If real output continues to grow at a 5.0 percent rate in the following eight quarters, the unemployment rate is predicted to continue at 6.0 percent in George L. Perry 561 1972 and to decline to only 5.8 percent in 1973. This is a surprisingly slight improvement in unemployment in face of a rate of economic expansion that is substantially faster than the economy has experienced for some time. If real GNP growth proceeds at a 6.0 percent rate after the end of 1971, unemployment is projected to average 5.8 percent for all of 1972 and 5.3 percent for 1973. On this expansion path, the unemployment rate would reach about 5.5 percent by the end of 1972. If expansion proceeds at a 7.0 percent rate starting in 1972, unemployment is predicted to average 5.7 percent in that year and 4.8 percent in 1973. By these estimates, a rapid rate of expansion will have to be sustained for several years in order to restore full employment. APPENDIX Labor Force and Employment Data Definition and Census Changes The labor force concepts used in this study cover all persons aged 16 and over, including those in the armed forces. Some adjustments have been made to the official data. First, the official population, labor force, and employment data from the Current Population Reports (CPR), subdivided by age-sex groups, were adjusted for the 19501 and 19602 census population benchmark changes. The data were assumed correct in the most recent cen- sus (for example, 1960). The estimates for 1960 based on the 1950 census measured the error in the 1950 census for 1960; and years between census dates were adjusted on the assumption that the error grew linearly between census years. Second, the data for the labor force and employment were ad- 1. See U.S. Bureau of the Census, Current Population Reports, Series P-57, No. 129 (1953), pp. 5, 12. 2. See Employment and Earnings, Vol. 8 (April 1962), p. xvii. 562 Brookings Papers on Economic Activity, 3:1971 justed for the definition changes introduced by the Bureau of Labor Statis- tics in 1967. The overlapping data for 1966 provided by the Bureau of Labor Statistics were used to make the adjustment.3 These data provide estimates of employment and the labor force on both the old and new definitions. The ratio of the two estimates in the overlap period was applied to data for pre- vious years to put all series on the basis of present definitions. Linkage to Establishment Data Employment totals from the Office of Business Economics, based on establishment reports and expanded to cover sectors excluded by those re- ports,4 were used in the measure of labor input of this study. For each year, discrepancies between this employment total and the total derived from the adjusted data in the CPR series described above were allocated among secondary workers (all but males aged 25 to 64) according to the share of employment of each age-sex group in total employment of secondary workers. The same discrepancy was allocated to the labor force in each group.5 In this way, labor force and employment subtotals for each age-sex group were derived, adding up to the "establishment-Denison basis" totals used here. Table A-1 compares series on the total labor force and unemployment rate based on these adjustments with the official series. The unemployment rates differ primarily because the armed forces are not included in the official series. The adjustments described above have only a slight effect on unemployment rates. 3. Employment and Earnings, Vol. 13 (February 1967), pp. 3-30. 4. The expanded totals were kindly provided by Edward F. Denison from a draft manuscript of a forthcoming study. The establishment data appear in U.S. Office of Business Economics, Thle National Income and Product Accounts of the United States, 1929-1965: Statistical Tables (1966), Tables 6.3, 6.4, and 6.6, and in Survey of Current Business, July issues, Tables 6.3, 6.4, and 6.6. 5. This procedure amounts to assuming that the discrepancy is a reporting error in the employment estimates of secondary workers in the CPR series. Alternatively, it could have been assumed to arise from variations in multiple job holding, resulting in a some- what different allocation of the discrepancy among age-sex groups. Since either secondary workers or second jobs would receive low weights in converting to the labor input mea- sure used in this study, and since both would be cyclically sensitive totals, it makes little difference for the end result which adjustment is made. The evidence that the discrepancy was due to systematically variable multiple job holding was not persuasive. George L. Perry 563 Table A-1. Derived Labor Force and Unemployment Rates Compared with Official Data, 1948-70 Labor force data in thousands Labor force Unenmployment rate Year Derived Official Derived Official 1948 61,737 62,080 3.5% 3.8% 1949 61,816 62,903 5.7 5.9 1950 63,125 63,858 5.1 5.3 1951 65,756 65,117 3.0 3.3 1952 66,879 65,730 2.7 3.0 1953 67,691 66,560 2.6 2.9 1954 67,653 66,993 5.1 5.5 1955 68,453 68,072 4.0 4.4 1956 69,867 69,409 3.8 4.1 1957 69,923 69,729 4.0 4.3 1958 69,957 70,275 6.4 6.8 1959 70,474 70,921 5.2 5.5 1960 71,555 72,142 5.2 5.5 1961 72,236 73,031 6.4 6.7 1962 72,889 73,442 5.2 5.5 1963 73,872 74,571 5.4 5.7 1964 75,010 75,830 4.9 5.2 1965 76,789 77,178 4.3 4.5 1966 79,529 78,893 3.5 3.8 1967 81,216 80,793 3.7 3.8 1968 83,019 82,272 3.4 3.6 1969 85,126 84,239 3.3 3.5 1970 86,253 85,903 4.7 4.9 Sources: Derived data are discussed in the text of this appendix. The official data are from Manpower Report of the President, April 1971, pp. 203, 223. Potential Labor Force and Employment Table A-2 shows the participation rate equations used to estimate the potential labor force. In order to project potential employment disaggre- gated by age-sex groups, an assumption was needed about how individual group unemployment rates would be distributed at potential. Experience suggests that proportional changes for all groups would understate the change for groups with high unemployment, while the same percentage point change for all would overstate it. I chose an adjustment lying between 564 Brookings Papers on Economic Activity, 3:1971 Table A-2. Coefficients for Labor Force Participation Rate Equations Labor Standard Serial market Time error correla- Age-sex tightness trend of tion co- group Constant I - u log T W2 estimate efficient Males 16-19 years 0.0695 1.2961 -0.3268 0.910 0.0126 0.775 (0.3243) (0.2676) (0.0950) 20-24years -0.3548 0.9373 0.1524 0.781 0.0157 0.903 (0.4787) (0.3183) (0.1740) 25-64 years 1.0232 0.0106 -0.0405 0.805 0.0024 0.872 (0.0684) (0.0488) (0.0237) 65 years and 1.4900 0.4393 -0.7254 0.985 0.0089 0.860 over (0.2522) (0.1839) (0.0855) Females 16-19 years -0.5172 0.9871 -0.0121 0.565 0.0141 0.547 (0.3296) (0.3128) (0.0670) 20-24 years -0.7945 0.4991 0.3781 0.881 0.0120 0.866 (0.3420) (0.2466) (0.1172) 25-64 years -1.2908 0.4780 0.5750 0.987 0.0055 0.759 (0.1395) (0.1171) (0.0397) 65 years and -0.0889 0.0821 0.0511 0.436 0.0056 0.739 over (0.1399) (0.1197) (0.0385) Source: Based on official labor force data, adjusted as described in the text of this appendix. The dependent variable is the fraction of the population in each group that is in the labor force. The form of the equations and the results are discussed in the section, "Participation Rates," in the article text. All regressions were fitted to annual data for 1947 to 1969. The standard errors of coefficient estimates are in parentheses. these extremes: For each age-sex group in each year, I calculated the "total unemployment rate," defined as the sum of recorded unemployment plus hidden unemployment (the difference between the trend and actual labor force for the group) as a ratio to its trend labor force. At a point in time, this total unemployment rate was assumed to change proportionately for all age-sex groups. The size of the proportionate change in each year was calculated by constraining the total weighted unemployment rate to be 3.37 percent, its constant value along the trend path. Then the same proportional changes in employment were maintained in shifting from the trend to the potential path. Since the groups that experience relatively high unemploy- ment rates also experience large changes in labor force participation as unemployment varies, this way of adjusting individual rates to their po- George L. Perry 565 tential levels falls between the extreme methods of adjustment mentioned above. Potential employment in each age-sex group for each year was multiplied by the relative wage of workers in that group. These wage-weighted employ- ment estimates for each group were added together to get weighted poten- tial employment for the whole economy. Actual weighted employment was estimated in a parallel manner by adding actual wage-weighted employ- ment over all age-sex groups. Comments and Discussion Edward F. Denison: I am a little embarrassed because to discuss Perry's paper I must discuss my own recent research. It consists of another growth study of the general type I have made before, with numerous improvements and refinements. It provides annual estimates for the United States covering the postwar period, 1947-69, and a few prewar years on a comparable basis. In any such study, the first prerequisite is to obtain output and input measures that are as statistically consistent as possible. It is a great ad- vantage that the Office of Business Economics (OBE) series on compensa- tion of employees is matched by an employment series based on the same sources and processed in the same way. OBE's estimates of proprietors are also a series as consistent with proprietors' income as one can obtain. To maximize consistency between input and output, I measured output from the income side of the accounts (if one starts from the expenditure side the deflated statistical discrepancy is subtracted) and used OBE em- ployment data as the main basis for an employment series. Conceptually, the series is on a labor force basis, to permit a tie-in with labor force char- acteristics obtained from the Current Population Survey (CPS), but the movement is based essentially on the OBE employment data. While my employment series differs from the CPS employment series, the differences are not in long-term trends, nor in cyclical movement to any great extent. The main difference is that there are periodic changes in their relative levels: One is above the other for a number of years, then they cross over for another series of years, again reverse, and so on. 566 George L. Perry 567 Hours are based on establishment data from the Bureau of Labor Sta- tistics (BLS), related to this employment series. A small adjustment, based on unpublished BLS estimates, converts hours paid for to hours worked. So much for background. Perry has picked up my employment and hours series for his analysis. He obviously must use an "actual" labor force series that is statistically con- sistent with employment, and has obtained it by adding to employment an estimate of unemployment. I am convinced that, besides having the merit of consistency with output, this is a better time series for the labor force than the CPS series. Perry also required a consistent series for the potential labor force. To obtain it he needed an age-sex distribution of the actual labor force, which he derived by scattering the difference between the actual labor force as estimated and the CPS series among the groups whose labor force participation rates fluctuate. One result of Perry's analysis particularly interested me: When the new labor force series is used, a cyclical response of the labor force to the un- employment rate remains. Until he conducted his analysis I had no idea whether this response would appear. I completely agree with Perry's pro- cedure of measuring the potential labor force by starting from the actual and adding or subtracting the gap or surplus. Perry's next problem was to obtain a cyclically adjusted average hours series. His rather arbitrary solution was to break the period into three pieces and use a trend value for each. He might do a little better up to 1965 by looking at employment components. Until then the main changes in average hours resulted from compositional shifts in employment from farm to nonfarm, from men to women, from full-time to part-time work, and so on. However, the big drop after 1965 was not compositional alone; hours dropped within homogeneous groups. I have no real criticism of Perry's procedure. Any way one handles the recent period will be arbitrary. The projection assumes the old rate of decline in hours starting from the new lower level. Because changes in hours are likely to be discrete rather than gradual, they are essentially nonprojectable over periods of five or ten years or less. Nevertheless, given the assignment, I should probably have made the same assumption. The rest of the study involves attempts to measure a trend rate of pro- ductivity change and cyclical fluctuations in productivity. I would not say Perry's results are necessarily wrong, but I must raise some questions. The broadest point is simply this: Output per weighted manhour in the whole 568 Brookings Papers on Economic Activity, 3:1971 economy, his measure of productivity, is so summary a measure that it is hard to analyze it, draw any conclusions, or appraise the results. Perry and I agree in some sense on one statement: There has been no important acceleration or deceleration in the rate of productivity increase during the postwar period. If there has been any change, it was small. The trouble is that we define productivity in such different ways that if one of us is right, the other is in all probability wrong. Also, when I consider the three periods Perry distinguishes, I obtain a lower rate of change in the middle period than in the other two, while he does not. Both series attempt to eliminate cyclical movements, but otherwise they are quite different. Perry defines productivity as GNP per hour worked, weighted by age and sex, in the economy as a whole. The series in which I find stability is confined to the nonresidential business sector of the economy, and within this sector measures output per unit of total input after the effects of shifts of resources out of labor and out of nonfarm self- employment, of cyclical movements, and of minor sources of irregularity are eliminated. My first point concerns the size of the sector covered: Unless by chance all sectors march side by side, trouble arises in dealing only with the econ- omy as a whole. The government-household-institution sector is measured differently, behaves differently, and affects all numbers differently from the business part of the economy. Output per manhour in that sector has increased about one-third of a percent per year, much lower than the rate in the nonresidential business sector, and its share in the total has shifted up and down. Aside from the difference in the rate of productivity change, shifts in the importance of the sectors themselves affect their combined output per manhour. In a large part of the economy, consisting of the services of dwellings plus income from abroad, output is produced with almost no manhours at all; it is entirely a product of capital. Output of these sectors should be elimi- nated to analyze output per manhour, and added back to obtain output in the entire economy. A rise in the proportion of output in these sectors raises output per manhour in the whole economy; a decline reduces it. The proportions have changed rather substantially over this period and not at a constant rate. A broader point concerns what one is trying to do in this type of analysis. I start with the proposition that changes in manhours are only one of many determinants of output. One should go on to take separate account of all George L. Perry 569 the determinants that affect output, or output per manhour, that can be identified and measured annually. What is left is a residual whose growth one seeks to analyze for trend. Perry has taken account of one such determinant, the age-sex composi- tion of hours worked. He says that other determinants, such as education or capital, are also important but that they generally change smoothly and can be allowed to ride in the residual productivitity trend. I have tried to measure directly the effects of a much larger number of determinants. Table 1 presents some comparisons between Perry's results and mine. His numbers, however, refer to the whole economy, mine to the nonresidential business sector, which represents about four-fifths of total national income, on the average; Perry refers to potential hours, while I deal with hours actually worked. Line 1 shows growth rates of Perry's labor input per man, which changes only because of age-sex composition. Below it are three of my series. These are total labor input per person employed; total input per person em- ployed; and total input per person employed plus the effects of shifts out of agriculture and self-employment. Table 1. Comparison of Perry and Denison Measures of Annual Growth Rates of Input per Man and per Manhour, 1948-70a Percent Chlanzge Change to niext to nzext Type of measure 1948-55 period 1955-65 period 1965-70b Input per man 1. Labor input, Perryc -0.62 0.01 -0.61 -0.57 -1.18 2. Labor input, Denison 0.51 -0.36 0.15 -0.43 -0.28 3. Total input, Denison 0.86 -0.40 0.46 -0.39 0.07 4. Total input plus resource reallocation, Denison 1.24 -0.44 0.80 -0.34 0.46 Intpuit per manhour 5. Labor input, Perryc -0.11 -0.14 -0.25 -0.28 -0.53 6. Labor input, Denisonc 0.08 -0.29 -0.21 -0.20 -0.41 7. Total input plus resource reallocation, Denison 1.66 -0.51 1.15 0.18 1.33 Source: See text. a. The Perry measures concern the potential for the whole economy and potential hours; the Denison measures, the actual for the business sector only and actual hours worked. b. Perry: 1965-70; Denison: 1965-69. c. Takes account of age-sex composition only. 570 Brookings Papers on Economic Activity, 3:1971 Line 7 can be viewed as my estimate of the effects on output per man- hour in the nonresidential business sector of all the factors that can be identified and measured annually, of which age-sex composition is only one. Now, the levels do not affect Perry's analysis; what matters is the change from one time span to the next (the second and fourth columns). It turns out that changes in the growth rate of Perry's labor input per manhour are not at all similar to changes in the growth rates of the sum of all my separately identified factors per manhour shown in line 7. While my calculations have yet to be completed, I suggest that the age- sex calculation, although appropriate as far as it goes, may give a dia- metrically erroneous indication of the effects on output per manhour of all identifiable and measurable factors in total input and resource shifts. My last general point concerns the method of adjusting the data to un- tangle trend and cyclical movements. Table 2 is designed to show how different our numbers are in Perry's first period. I first set down our growth rates of actual output per manhour in the whole economy, whose differences lie only in the measure of output. Al- though Perry measures output by GNP and I by national income, the dif- ference between our estimates, shown in lines 1 and 3, is not due mainly to the difference in definition. Inclusion of line 2 brings out the fact that most Table 2. Comparison of Perry and Denison Measures of Annual Growth Rates of Economy-wide Output per Manhour, 1948-70 Percent Type of measure 1948-55 1955-65 1965-69 1965-70 Actual output per manhouir 1. GNP (product side), Perry 3.45 2.75 ... 1.63 2. GNP (income side), Denison 3.27 2.85 ... 1.64 3. GNP, national income accounts basis (income side), Denison 3. 19 2.88 1.79 ... Potential output per manhour 4. GNP (product side), Perry 2.87 2.71 ... 2.39 5. GNP, national income accounts basis (income side), Denison 3.31 2.66 2.81 ... Actual less poten7tial 6. Perry (line 1 minus line 4) 0.58 0.04 ... -0.76 7. Denison (line 3 minus line 5) -0.12 0.22 -1.02 ... Source: See text. George L. Perry 571 of the difference results from my measuring output from the income side while he measures it from the product side. It happens that for these par- ticular periods, especially for a comparison of 1948-55 with 1955-65, the choice makes a big difference. Line 4 shows the growth rate of Perry's potential output per manhour and, for a rough comparison, line 5 gives a series constructed from my estimates. I adjusted actual output in the nonresidential business sector each year to what it would have been if the labor and capital present had been used at a constant rate of utilization, added this to actual output in the other sectors, and divided by actual total hours in the whole economy. The point of the table is to be able to compare lines 6 and 7, obtained by subtracting the growth rates of potential from those of actual output per manhour to obtain the effect of the cyclical adjustment. Our numbers are, perhaps, not very far apart for the later periods. But they are very different for the first period. I estimate that 1948 saw a mod- erately more intense use of resources than 1955. Perry's growth rate of 0.58 over seven years implies that output per manhour in 1955 was roughly 4 percent higher than it would have been with the 1948 utilization rate. To be sure, lines 7 and 6 are not estimates of precisely the same thing. Line 7 estimates what output per manhour of the hours actually worked in each year would have been if they were not affected by changes in utiliza- tion, whereas line 6 provides a measure of what the output of all the poten- tial hours would have been. Still, I don't think very big differences in the numbers can be explained this way. In seeking a means to adjust productivity for changes in intensity of utilization, it seemed to me that the series most likely to be affected by such changes in the same way and at the same time as productivity was profits, or more precisely, the nonlabor share of corporate national income when depreciation is measured in current prices and on a consistent basis. I used this series to estimate the fluctuation in productivity. My approach was to compute a logarithmic trend for the most refined series on output per unit of input I could obtain, and compute an arithmetic trend (which is slightly downward) for the nonlabor share. Deviations of productivity and the nonlabor share from their trends were correlated, and the regres- sion was used to estimate from the nonlabor share the effects of intensity of utilization on, productivity each year. The correlation is pretty good; jR2 is 0.90. I also superimposed another adjustment to eliminate the effect of irregular movements in the relationship between wages and prices on 572 Brookings Papers on Economic Activity, 3:1971 the nonlabor share itself. This refinement changed some of the numbers a little, but not very basically. What were the alternatives? In his initial study, the portion of Okun's law reflecting the adjustment for productivity was obtained by a relation- ship to unemployment, as were the rest of its components. In Why Growth Rates Differ, I picked up the productivity part of Okun's formula, although I was unhappy with it. In the process of my current study, with the advan- tage of more years of data, I tested it and there was no correlation at all between productivity and unemployment; the calculation comes out literally as close to zero as one ever gets. Another possibility was the change in hours worked from one year to the next, which performed better than unemployment but still not very well, giving improbable answers in some years. Then I tried both unem- ployment and the change in hours, and this worked no better than the change in hours alone. There are good reasons why the change in hours should not be a satisfactory indicator. For example, labor hoarding reduces productivity but raises hours worked so the causation is often backward. In his analysis here, Perry first uses the unemployment rate. I suspect this does nothing for him at all. I checked directly on the relationship between output per manhour for the whole economy and the unemployment rate. The correlation is for all intents and purposes zero, and only trivially better if his weighted unemployment rate is used. True, I used national income as an output measure but this shouldn't matter unless the statistical discrep- ancy and unemployment are highly correlated. Perry also introduces a measure of the change in hours not worked. This helps quite a bit, but only about as much as the change in hours in my analysis. The two mea- sures are close substitutes and either is helpful only in periods of rapid expansion. I doubt that Perry has as good a measure of utilization as one can get. In many years his method will yield estimates close to those ob- tained by my procedure but in occasional years-particularly those like 1953 or 1957 that embrace the beginning of a downturn-it yields a very different, and I believe worse, answer. The 1948-55 movement revealed in Table 2 suggests his results are quite different from mine, but I have not seen estimates for all the other years. I do not want at this point to say that any of Perry's conclusions or his projection is wrong, yet I am reluctant to accept them as right. The whole operation is insufficiently detailed to permit much evaluation, and the procedure for cyclical adjustment of productivity is not very good. George L. Perry 573 As a final comment about the very recent period, I think it interesting that, after allowing for variation in intensity of utilization, neither of us finds any drop in the rate of productivity increase-in my case through 1969 and in Perry's through 1970. The differences in measuring inputs discussed above are probably not very important in this short-term examination of very recent changes. Robert M. Solow: When Perry first introduced the weighted employment concept in his inflation paper last year, I felt an impulse to suspend judg- ment about it because I was worried about whether it was a concept that simply happened to meet the requirements of the moment and might not last. Now I think that the success with which the weighted employment concept comes out here also lends, in my mind, more credibility to the analysis in the inflation paper as well. Not that there is any logical connec- tion between the two, but the concept seems to me to be paying off. Next, I agree with Perry that there is no evidence whatever in his paper for moving away from the hypothesis of a constant weighted productivity trend in the period. Only one equation in his Table 2 yields any worthwhile difference, and I mean worthwhile rather than statistically significant. Some differences shown in the table are not worthwhile in the sense that they are trivial and well within the specification error of any model of this kind. Equation (llf), the one equation that does yield a worthwhile difference, is suspicious on other grounds. Finally, when it comes to accounting for the relatively slow growth of his measured potential between 1965 and 1970, Perry puts a quarter of 1 per- centage point of the rate of growth on the unusually fast reduction of aver- age hours. In the model, this comes from the dummy variable for 1966-68 which, as Perry says perfectly clearly in the paper, is strictly ad hoc. In a sense it is still unexplained. It is just that the data seemed to call for this kind of unusual reduction in hours during that period. There is probably something to it, although I do not understand why it happens. It helps ex- plain another mystery-the decline (or, at best, the failure to rise) of real spendable weekly earnings of production workers between 1965 and 1970, even while real hourly compensation has been rising. Presumably, a good chunk of the reason lies in the autonomous reduction in hours for which we have no special explanation. One question I have is why Perry's potential output grows faster than what he calls official potential output between 1965 and 1970. I was puzzled 574 Brookings Papers on Economic Activity, 3:1971 too, as Denison was, but in obviously quite a different way, about the 1948-55 comparisons because Perry's productivity figures imply a gap- closing productivity gain in that interval, yet the official unemployment rate was lower in 1948 than in 1955. The discrepancy between Perry's and Denison's numbers that I would like cleared up is that Perry's series for the rate of growth of output per unit of his labor input looks constant on his potential path; and Denison's series for the rate of growth of output per unit of total input is, he says, constant along his potential path. The difference between these is the rate of growth of Perry's labor input per unit of Denison's total input. The difference between two constants should be constant. But Denison says this difference is not. To finish up, I have two very brief comments on some of the material relating to the Okun multiplier. Perry points out that he can formulate this model in different ways, essentially interchanging dependent and inde- pendent variables, and get slightly different results for the Okun multiplier. If either weighted hours or output were treated as exogenous in the model, it ought to be the independent variable. In the present case, I presume one would take output as the independent variable on the assumption that the causality runs from output to hours, with output in turn given by product demand. But it would be better still to imbed this analysis in a somewhat more complete model of the economy in which some variables were clearly exogenous, and then estimate the complete system. George L. Perry: Let me first try to respond to the array of results and observations Denison has offered. On the question of what output concept to use, I went directly to total GNP because that was what I was interested in. Measuring it from the input side is one alternative. I doubt that it makes a difference in analyzing potential since the two concepts differ only by the statistical discrepancy, although it can make a noticeable difference in measuring actual output or productivity changes between two years. I accept Denison's comment that a productivity analysis is cleaner if it excludes some parts of GNP, particularly those for which we measure or impute output that involves no labor input. But sectoral shifts are a part of all observed changes and are important even within the nonfarm busi- ness sector. Weighting labor input, as I have done, takes account of shifts from the input rather than the output side. If the growth of an imputed part of output, such as household services, changed sharply over the George L. Perry 575 period, it could still affect the results. I have treated such changes as grad- ual in dealing with total GNP and in allowing their effects to appear in the trend term. The Perry and Denison concepts that are displayed side by side in Deni- son's Table 1 are so different that they are hard to compare. Denison uses actual data for part of the economy and compares them with my data on potential for the economy as a whole. Where we are closest to measuring the same thing, labor input per manhour, the big disparities in our changes between the first and second periods are worth noting. These probably reflect the large residuals that exist in the first few years in my analysis. Then, productivity moved erratically, even after adjustment for cyclical variations in unemployment. Thus, growth rates such as his, using actual data over a few years, will vary sharply according to which of the early years one starts with. Growth rates along my potential line will not. The differences between Denison's series on labor input and his series on total input plus resource allocation indicate that everything but labor input grew unevenly when measured per hour of labor input. If some of these other factors are treated as fixed inputs, some uneven movement would be accounted for by the fact that manhours vary cyclically along the actual path. In any case, without a careful study of Denison's work it is hard to know what to make of this as a comment on my results. The nearest that we can come to comparing our work is in the data on potential growth rates given in Table 2. Here our results on economy-wide productivity trends are summarized using conventional manhours in both cases. Yet the comparison is still elusive, because of two essential dif- ferences. First, what Denison calls potential is not what I call potential; the two are defined differently. His potential is defined by making the nonlabor share a prescribed fraction of total income-almost, but not quite, a con- stant fraction through the years. My potential is defined by holding the un- ployment rate at a fixed level each year. Second, I am purposely measuring along a smoothed trend line. Denison is measuring from the end points of a potential path calculated by adjusting actual productivity in each year for his estimate of underutilization each year (defined by the ratio of the actual nonlabor share to his trend value for this share). Our most substantial and important disagreements center on the last period. If I adjust his potential growth rate for the difference between out- 576 Brookings Papers on Economic Activity, 3:1971 put measured from the product and the income sides, our differences in the first two periods are rather small despite all the differences in concept and definition. But for the crucial last five years, our disagreements matter for we say significantly different things about what has happened recently to productivity and where the economy would be if it were at "potential." Here I find Denison's results implausible if he means them to apply to something near my concept of potential. In this period, the trend in produc- tivity per manhour in my analysis slows by 0.32 percentage point from the earlier period. In Denison's figures, the trend quickens by 0.15 percentage point. My numbers imply a constant productivity trend on my labor input basis, which produces the decline in the trend on an unweighted manhour basis since so many women and young workers entered the work force over this interval. If instead I accept Denison's numbers for unweighted manhour productivity, and take them to apply to what I mean by potential, the productivity trend on my labor input basis jumps by 0.47 percentage point in this period rather than remaining constant. I find a productivity revolution of this size implausible on the face of it and inconsistent with all the statistical evidence in my paper. While I am only speculating, I suspect Denison's 1969 end point is the main source of our difference. The non- labor share was at recession levels that year. I suspect that low profits were an even bigger surprise than low productivity-that is, the unexplained residual in profits was greater. And if this is so, Denison adjusted produc- tivity up by more than he should have. The answer to the first of Solow's questions-Why is the 1948 to 1955 actual productivity gain larger than the gain on the potential path?-has to do again with the residuals for the earliest years, which reflect erratic move- ments of measured productivity and make actual end-point calculations misleading. With my data, 1948 shows a positive output gap measured using the trend line, despite a very low unemployment rate; 1950 shows a much smaller output gap despite an unemployment rate that is 1.6 percent- age points higher. Actual productivity growth from 1948 to 1955 averages more than trend even though, without residuals, we could expect growth slower than trend because of some gap opening between those years. Ac- tual productivity growth from 1950 to 1955 averages the same as trend even though, without residuals, we would expect growth to be faster than trend because of gap closing between those years. One should not take such calcu- lations from these early years too seriously. Solow also asks about the comparison of my estimates with the official George L. Perry 577 estimates of potential output growth. The official series grows more slowly because, as near as I can tell, it is not measured accurately. The ingredients of the official estimate are unclear. I think that, as the growth in the labor force surprised the Council of Economic Advisers each year, they moved the potential path up a little bit. They did not make the full leap at any one time because to retain the rest of the trend calculation-the average hours and productivity trends-while accepting the evidence on the ac- celeration of the growth in the labor force would have moved potential up very rapidly, as I have indicated. The surprising thing is that they came out pretty close, by sheer luck perhaps, since the ingredients that would allow a calculation of the change in the productivity trend measured con- ventionally were never part of the analysis. The unemployment rate and Okun's law surely kept them compromising on how fast to move up the potential path and kept them in the ball park. Basically it was telling them that something was holding down potential output growth despite the acceleration in the labor force. Finally, there is the question of which productivity trend has really been "constant." I make no claims beyond those discussed in the paper for the constancy of productivity trend on the basis of my input measure. Statisti- cal tests designed to find significant changes could not shake the hypothesis that the trend was constant throughout the period (although what in- significant evidence there was indicated a slowdown in recent years, while Denison implies an acceleration). I have seen no statistical evidence of constancy on Denison's concept. Conceivably, we could both be right since he does not mean by potential the same thing that I mean. But with- out seeing a great deal more of his analysis, including how he measures the contribution to potential output of capital, education, and other things, I cannot say whether I would accept this coincidental result. If I may borrow his phrase, I would not say Denison's results are necessarily wrong, but I must raise some questions. General Discussion Arthur Okun noted that Perry's results on the Okun's law multiplier were not surprisingly different from his own. He noted that his original work had shown differences in the multiplier similar to those shown by Perry, according to which question was asked. His original regressions, 578 Brookings Papers on Economic Activity, 3:1971 which sought to estimate the unemployment rate if output grew 1 percent faster than potential, gave a multiplier of around 3.2; asking what output gain accompanied a one-point reduction in the unemployment rate gave a multiplier of around 2.5. The difference comes because the correlations are not perfect, and they are similar to Perry's differences. He also had suspected that if lags were allowed explicitly, as in Perry's equation (11) in contrast with (7), some of what is attributed to the level of the gap would turn out to result from changing the gap. This helps account for the lower level of both of Perry's multipliers compared with Okun's original esti- mates. R. J. Gordon suggested that it would be useful to try to introduce the possible influence of wage and price expectations into the labor force participation equations to see if they helped explain the unusual rise in participation rates of recent years. This calculation would help decide if the high participation rates were permanent or transitory. He also suggested, as a possible explanation of the drop in hours in 1966-67, that tight labor mar- kets may have induced an unusual number of women who ordinarily would not work at all to take up part-time jobs in retail trade and services, and that this effect would vanish when employers could hire full-time personnel in a slacker labor market. He believed this may have been happening in 1971. |
11560 | https://www.sciencedirect.com/science/article/pii/S0012365X17303576 | No acute tetrahedron is an 8-reptile - ScienceDirect
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Abstract
Keywords
1. Introduction
2. Results
3. The proof
4. Bold conjectures
Acknowledgement
References
Figures (1)
Discrete Mathematics
Volume 341, Issue 4, April 2018, Pages 1131-1135
Note
No acute tetrahedron is an 8-reptile
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Abstract
An r-gentiling is a dissection of a shape into r≥2 parts which are all similar to the original shape. An r-reptiling is an r-gentiling of which all parts are mutually congruent. The complete characterization of all reptile tetrahedra has been a long-standing open problem. This note concerns acute tetrahedra in particular. We find that no acute tetrahedron is an r-gentile or r-reptile for any r<10. The proof is based on showing that no acute spherical diangle can be dissected into less than ten acute spherical triangles.
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Keywords
Tessellation
Tetrahedron
Reptile
1. Introduction
Let T be a closed set of points in Euclidean space with a non-empty interior. We call T an r-gentile if T admits an r-gentiling, that is, a subdivision of T into r≥2 sets (tiles) T 1,…,T r, such that each of the sets T 1,…,T r is similar to T. In other words, T is an r-gentile if we can tile it with r smaller copies of itself. This generalizes the concept of reptiles, coined by Golomb: a set T is an r-reptile if T admits an r-reptiling, that is, a subdivision of T into r≥2 sets T 1,…,T r, such that each of the sets T 1,…,T r is similar to T and all sets T 1,…,T r are mutually congruent under translation, rotation and/or reflection. In other words, T is an r-reptile if we can tile it with r equally large, possibly reflected, smaller copies of itself. Interest in reptile tetrahedra (or triangles, for that matter) exists, among other reasons, because of their application in meshes for scientific computing. In this realm techniques such as reptile-based stack-and-stream are well-developed in two dimensions, but three-dimensional space poses great challenges.
It is known what triangles are r-reptiles and r-gentiles for what r. However, for tetrahedra the situation is much less clear; in fact the identification of reptile and gentile tetrahedra and, even more general, of tetrahedra that tile space, has been a long-standing open problem.
The regular tetrahedron does not tile space, as its dihedral angles are arccos(1∕3), which is larger than 2 π∕6 but slightly smaller than 2 π∕5, so that no number of regular tetrahedra can fill the space around a common edge. Goldberg described all known tetrahedra that do tile space. Delgado Friedrichs and Huson characterize all tetrahedra that produce tile-transitive tilings, but to the best of my knowledge, without the restriction to tile-transitive tilings the problem of identifying all space-filling tetrahedra is still open.
The reptile tetrahedra must be a subset of the tetrahedra that tile space. Matoušek and Safernová argued that r-reptilings with tetrahedra exist if and only if r is a cube number. In particular, it is known that all so-called Hill tetrahedra (attributed to Hill by Hertel and Matoušek and Safernová) are 8-reptiles. It has been conjectured that the Hill tetrahedra are the only reptile tetrahedra, but this conjecture is false: Sommerville found two non-Hill tetrahedra that tile three-dimensional space and which were recognized as 8-reptiles by Liu and Joe. To the best of my knowledge, the Hill tetrahedra and the two non-Hill tetrahedra from Liu and Joe are the only tetrahedra known to be reptiles, but there might be others. This paper provides a small contribution to the answer to the question: exactly what tetrahedra are reptiles?
In mesh construction applications one typically needs to enforce certain quality constraints on the mesh elements. This has motivated studies into acute tetrahedra:
Definition 1
A tetrahedron is acute if each pair of its facets has a dihedral angle strictly less than π∕2.
All facets of an acute tetrahedron are acute triangles themselves (Eppstein et al., Lemma 2). The Hill tetrahedra, as well as the two non-Hill tetrahedra from Liu and Joe, all have right dihedral angles.1 Thus, no acute reptile tetrahedra are known.
2. Results
In this note we will prove the following statement, which may serve as evidence that acute reptile tetrahedra are probably hard to find, if they exist at all:
Theorem 1
Let T be an acute tetrahedron subdivided into r≥2 acute tetrahedra T 1,…,T r . If the diameter (longest edge) of each tetrahedron T i is smaller than the diameter (longest edge) of T , then r≥10 .
In particular we get:
Corollary 1
No acute tetrahedron is an r -gentile for any r<10 .
With the result from Matoušek and Safernová that r-reptile tetrahedra can only exist when r is a cube number, we get:
Corollary 2
No acute tetrahedron is an r -reptile for any r<27 .
3. The proof
Note that if a tetrahedron T is subdivided into tetrahedra T 1,…,T r with smaller diameter than T, then at least one tetrahedron T i, for some i∈{1,…,r}, must have a vertex v on the longest edge of T. For the proof of Theorem 1 we analyse S v, the subdivision of an infinitesimal sphere around v that is induced by the facets of T and T 1,…,T r. In such a subdivision, we find:
•faces: each face is either a spherical triangle, corresponding to a tetrahedron T i of which v is a vertex, or a spherical diangle (also called lune), corresponding to a tetrahedron that has v on the interior of an edge;
•edges: the edges of S v are segments of great circles and correspond to facets of T 1,…,T r that contain v; the angle between two adjacent edges on a face of S v corresponds to the dihedral angle of the corresponding facets of a tetrahedron T i.
•vertices: each vertex of S v corresponds to an edge of a tetrahedron T i that contains v.
Thus, S v consists of a spherical diangle D corresponding to T, subdivided into a number of spherical triangles, and possibly some spherical diangles, that correspond to the tetrahedra from T 1,…,T r that touch v. Below we will see that S v must contain at least ten faces (not counting the outer face, that is, the complement of D), which proves Theorem 1.
In what follows, when we talk about diangles and triangles, we will mean acute, spherical diangles and acute, spherical triangles on a sphere with radius 1. Note that the faces are diangles or triangles in the geometric sense, but they may have more than two or three vertices on their boundary. More precisely, a diangle or triangle has, respectively, exactly two or three vertices, called corners, where its boundary has an acute angle, and possibly a number of other vertices where its boundary has a straight angle. A chain of edges of a diangle or triangle from one corner to the next is called a side. Note that S v contains at least one triangle, since v is a vertex of at least one tetrahedron T i. Therefore, in what follows we consider a subdivision S of a diangle D into a number of diangles and triangles, among which at least one triangle. We call such subdivisions valid. Henceforth, we will assume that S has the smallest number of faces out of all possible valid subdivisions of all possible diangles D. Our goal is now to prove that S contains at least ten faces.
Lemma 1
Each face of S is a triangle.
Proof
If S would have any diangular face F, it must have the same corners as D, because the corners of any diangle must be an antipodal pair and there is only one antipodal pair within D. The removal of F would separate S into at most two diangular components with the same corners as D. By construction, at least one of these components contains a triangle. That component would then constitute a valid subdivision that has fewer faces than S, contradicting our choice of S. □
In S, we distinguish boundary vertices (vertices on the boundary of D) and interior vertices (vertices in the interior of D). Among the boundary vertices, we distinguish poles (the corners of D) and side vertices (the remaining boundary vertices). Among the interior vertices we distinguish full vertices and hanging vertices: a vertex v is a full vertex if it is a corner of each face incident on v; a vertex v is a hanging vertex if it is a non-corner vertex of one of the faces incident on v.
We will now derive a few properties of S from the acuteness of its angles.
Lemma 2
Each side of each face of S has length strictly less than π∕2 .
Proof
Consider any face F of S. Let a be the length of a particular side of F, let α be the angle in the opposite corner of F, and let β and γ be the angles in the other two corners of F. Since F is acute, the sines and cosines of α,β and γ are all positive. By the supplementary cosine rule (, Art.47) we have cos α=−cos β cos γ+sin β sin γ cos a, so cos a=(cos α+cos β cos γ)∕(sin β sin γ)>0. It follows that a<π∕2. □
Lemma 3
There are at least four side vertices: two on each side of D .
Proof
For the sake of contradiction, suppose one side of D contains only one side vertex. Then this side would consist of two edges, at least one of which has length at least π∕2, contradicting Lemma 2. □
Lemma 4
Each pole is incident on at least two edges.
Each hanging vertex and each side vertex is incident on at least four edges.
Each full vertex is incident on at least five edges.
Proof
The poles are incident on at least two edges by definition. If a side vertex or a hanging vertex would be incident on only three edges, then two of these edges make a straight angle on one side, while the third edge divides the straight angle on the other side. Thus, at least one of the angles that results from this division would be non-acute. If a full vertex would be incident on at most four edges, then at least two of those edges must make an angle of at least 2 π∕4=π∕2 on their common face, again contradicting the assumption that all faces are acute. □
Now we can combine these properties with Euler’s formula and find:
Lemma 5
The number of faces equals 2 f+h+s , where f≥2 is the number of full vertices, h is the number of hanging vertices, and s≥4 is the number of side vertices. More precisely, f≥2+p , where p is the number of edges with one end point at a pole and the other end point at an interior vertex.
Proof
Let v=f+h+s+2 be the number of vertices, let e be the number of edges and r be the number of triangles of S. By Lemma 1 all faces are triangles, so by Euler’s formula we have v+r=f+h+s+2+r=e+1, hence 2 e=2 f+2 h+2 s+2+2 r. We say that a hanging vertex is owned by the triangle of which it is a non-corner vertex; each hanging vertex is owned by exactly one triangle. The number of edges on the boundary of a triangle F is three plus the number of hanging vertices owned by F. Thus, if we add up the numbers of edges of all triangles, we obtain a total of 3 r+h. This counts all edges double, except the s+2 edges on the boundary of D, which are counted only once. Therefore we have 2 e−s−2=3 r+h. Hence we have 2 e=3 r+h+s+2=2 f+2 h+2 s+2+2 r, which solves to r=2 f+h+s.
Thus we have 2 e=3 r+h+s+2=6 f+4 h+4 s+2. By Lemma 4 we also have 2 e≥5 f+4 h+4 s+4+p, so 6 f+4 h+4 s+2≥5 f+4 h+4 s+4+p, which solves to f≥2+p.
The condition s≥4 is given by Lemma 3. □
Lemma 6
The number of faces of S is at least ten.
Proof
Suppose, for the sake of contradiction, that S has at most nine faces. Then, by Lemma 5, we cannot have f≥3, so f=2, that is, there are exactly two full vertices, which we denote by X and Y. Let N X and N Y be the vertices adjacent to X and Y, respectively. Moreover, by Lemma 5, we have p=0 (so neither N X nor N Y contains any of the poles), f+s+h≤7 (there are at most seven vertices other than the poles), and in particular h≤1 (there is at most one interior vertex other than X and Y).
By Lemma 4, we have |N X|≥5 and |N Y|≥5. Since |N X∪N Y|≤f+s+h≤7, this implies |N X∩N Y|≥3, that is, X and Y share at least three neighbours. Since there is at most one interior vertex other than X and Y, at least two of the shared neighbours must be side vertices, let us call these V and W. Now consider the cycle X V Y W X. This cycle separates the diangle D into three regions—see Fig. 1 a. Only one of these regions, namely the interior of the cycle, has both X and Y on its boundary. Therefore, the third shared neighbour of X and Y must be an interior vertex U inside the cycle X V Y W X. Since there are no other interior vertices, U can have edges only to X, V, Y, and W. By Lemma 4, all four of these edges must exist.
Now, since U is a hanging vertex, one of the angles around U must be a straight angle, say, without loss of generality, the angle X U V. But then the cycle X U V X would constitute a geometric diangle, contradicting Lemma 1. Therefore, the assumption that S has at most nine faces must be wrong. □
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Fig. 1. (a) If S has at most nine faces, two full vertices and two side vertices must form a quadrilateral with a hanging vertex inside, but we cannot get the angles around the hanging vertex right. (b) An example of a subdivision of an acute diangle into ten acute triangles, with angles and edge lengths given in degrees rounded to integers.
This concludes the proof of Theorem 1.
Note that the crucial observation of the proof is that any dissection of an acute spherical diangle into acute spherical triangles requires at least ten triangles. This bound is, in general, tight: Fig. 1 b shows a dissection of an acute spherical diangle into ten acute spherical triangles.
4. Bold conjectures
Could we exploit the observations in this paper to improve Corollary 2 further? A first step could be the following. If T is an acute tetrahedron with diameter d, and we can identify three segments x, y and z of length d∕3 on the edges of T, then the arguments presented in this paper tell us that any 27-reptiling of T must contain at least ten tiles that intersect x, ten tiles that intersect y, and ten tiles that intersect z. If additionally, one can ensure that x, y and z lie at distance more than d∕3 from each other, these three sets of ten tiles each must be mutually disjoint, so there must be at least 30 tiles in total. This would contradict the existence of a 27-reptiling and thus improve Corollary 2 to: no acute tetrahedron is an r-reptile for any r<64.
However, given that no acute tetrahedron can be an r-reptile or r-gentile for small values of r (and given, in general, that the past hundred years did not turn up any reptile tetrahedra without right dihedral angles), we may rather restate the obvious:
Conjecture 1
There are no reptile acute tetrahedra.
A stronger conjecture would be:
Conjecture 2
There are no gentile acute tetrahedra.
We conclude with an even bolder conjecture:
Conjecture 3
There are no gentile tetrahedra that do not have a dihedral angle of exactly π∕2 .
Acknowledgement
I thankthe anonymous reviewer who improved Lemma 5 and gave me a hint on how to use this to improve Lemma 6. This raised the bound in Lemma 6 and thus, many other bounds in this paper, from nine to ten.
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1
One of the non-Hill tetrahedra can be given (modulo similarity transformations) by A=(−1,0,0), B=(0,1,0), C=(1,0,0), D=(0,0,1∕2); the second type of non-Hill tetrahedron is obtained by cutting the first type along the y z-plane. Both types have right dihedral angles along the x-axis. Any Hill tetrahedron can be described as the convex hull of four vertices A=0, B=v 1, C=v 1+v 2 and D=v 1+v 2+v 3, such that the vectors v 1, v 2 and v 3 have the same length and such that the angle between each pair of these vectors is the same, say α. For ease of notation, assume that the tetrahedron is scaled, rotated and reflected such that v 1, v 2 and v 3 have length 2, the vertex C=v 1+v 2 lies on the positive x-axis, and the vertex B=v 1 lies in the first quadrant of the x y-plane. We use t to denote cos α. Note that we have t<1, otherwise we would have α=0, all vertices would lie on a single line, and they would not be the vertices of a tetrahedron. The condition on the angles of the vectors can now be written as v 1⋅v 2=v 1⋅v 3=v 2⋅v 3=2 t. Thus we must have v 1=(a,b,0) and v 2=(a,−b,0) with a=1+t and b=1−t>0, so that indeed, ‖v 1‖=‖v 2‖=a 2+b 2=2 and v 1⋅v 2=a 2−b 2=2 t. The vector v 3=(x,y,z) must now satisfy v 1⋅v 3=v 2⋅v 3⇔a x+b y=a x−b y, which, given b≠0, solves to y=0, and we get D=v 1+v 2+v 3=(2 a+x,0,z). Thus, the face A B C lies in the x y-plane and the face A C D lies in the x z-plane, and these faces meet at a right dihedral angle along the x-axis.
© 2017 Elsevier B.V.
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11561 | https://ocw.mit.edu/courses/2-25-advanced-fluid-mechanics-fall-2013/1ff7ec3782567cdcedabdfd4b95c1792_MIT2_25F13_Unstea_Bernou.pdf | 2.25 Advanced Fluid Mechanics Fall 2013 Unsteady Bernoulli Equation In addition to understanding the effects of fluid acceleration in steady flow, we are also interested in impulsively started flows and transient flows. For such flows the Euler equation will be: ∂vs ρ ∂t + ρvs ∂vs ∂s = −∂P + ρgs ∂s We now integrate this along a streamline: 2 ∂vs ρ 1 ∂t ds + 2 1 ρvs ∂vs ds = (P2 P1) ρg(z2 z1) ∂s − − − − which can be simplified to: 2 ∂vs ρ 1 ∂t ds + (P + 1 2ρv2 s + ρgz)2 −(P + 1 2ρv2 s + ρgz)1 = 0 (1) This is not a very useful result in general since ∂vs/∂t can change dramatically from one point to another; to use this in practice we need to be able to draw streamline shapes at each instant in time. It works especially for simple cases such as impulsively started confined flows where streamlines have the same shape at each instant and we are interested in time required to start the flow.
Example: Flow out of a long pipe connected to a large reservoir (steady and transient starting stages) ≀≀ L h 1 2 a b A1 A2 Figure 1: Discharge of water from a long pipe connected to a large fluid reservoir with cross section area A1 >> A2. The problem approaches steady state when the valve has been open for a “long time” but is transient in the starting stage.
Consider the flow in the discharge of water through a long pipe connected to a big reservoir. If the area of the tank is much larger than the pipe cross section area (i.e. A1 >> A2) then the 1 Notes by B.K. and G.H.M., 2013 2.25 Advanced Fluid Mechanics Fall 2013 solution for steady state case, in which the discharge valve has been open for a while, can be easily done by writing Bernoulli between points 1 and 2 : J Pa + 1 ρ(0)2 + ρgh = Pa + 1 ρv2 2 + ρg(0) ⇒ v2 = 2gh 2 2 where v1A1 = v2A2 ⇒ v1 c 0 because A2 << A1. This result was known to Torricelli in the 1630. Now consider the analysis for a more general case which includes the starting time. Although the velocity is changing with time in the pipe during the transient stage one can easily conclude that conservation of mass says that velocity has to be constant at any instant along the length of the pipe and it just changes with time. Applying unsteady Bernoulli equation, as described in equation (1) will lead to: 2 ∂vs 1 1 ρ ds + (Pa + ρ(v2)2 + ρg(0)) − (Pa + ρ(0)2 + ρgh) = 0 (2) ∂t 2 2 1 Calculating an exact value for the first term on the left hand side is not an easy job but it is possible to break it into several terms: 2 a b 2 ∂vs ∂vs ∂vs ∂vs ρ ds = ρ ds + ρ ds + ρ ds ∂t ∂t ∂t ∂t 1 1 a b If the reservoir area is much larger than the pipe area then it the integral from 1 to a is negligible compared to the integral along the pipe length ( b to 2 ) because vs in the tank is small, also knowing that the entry region is small compared to the length of the pipe we can easily neglect the integral from a to b compared to the corresponding integral over the pipe length. Thus the following estimate is an acceptable approximation for the unsteady term in the Bernoulli equation: 2 2 2 ∂vs ∂vs ∂v2 ∂v2 ρ ds c ρ ds = ρ ds = ρ L (3) ∂t ∂t ∂t ∂t 1 b b Combining (2) and (3) will result in: ∂v2 1 ρ L + ρv2 = ρgh 2 ∂t 2 It is worthy to mention that in this equation both v2 and h are in reality changing with time but for simplifying the analysis one can assume that the pressure head (i.e. h in the tank) remains almost unchanged during the transient starting stage (physically also it is right to assume that changes in h are almost negligible compared to other terms since the area of the tank is really large (A1 >> A2) and it takes a lot of fluid flow through the pipe to see changes in h). Assuming a constant value of h the simplified equation will be: dv2 1 2L 2 L + v = gh → dv2 = dt 2 2 dt 2 2gh − v /2 2 integrating from t = 0 and knowing that v2 = 0 at t = 0 gives the following integral: 2gh)−1 v2 dv2 t dt 1 v2 1 1 t dt 0 2gh − v /2 2 2 = 0 2L → 2 2gh √ 0 √ 2gh + v2 + √ 2gh − v2 dv2 = 0 2L which can be simplified to: √ J 2gh J v2 = 2gh tanh L t ⇒ v2 = 2ghtanh(t/τ) (4) √ where the characteristic time constant is τ ≡ L( . The described relationship for the transient velocity and time is plotted in Figure 2. 2 Notes by B.K. and G.H.M., 2013 2.25 Advanced Fluid Mechanics Fall 2013 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 v 2 / √ 2gh t/τ Figure 2: Velocity in the pipe as a function of time. The characteristic time constant is √ τ ≡ L( 2gh)−1 3 Notes by B.K. and G.H.M., 2013 MIT OpenCourseWare 2.25 Advanced Fluid Mechanics Fall 2013 For information about citing these materials or our Terms of Use, visit: |
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Best Practice & Research Clinical Endocrinology & Metabolism
Volume 31, Issue 4, August 2017, Pages 367-384
Physiology of milk secretion
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Milk is a unique and complete nutritive source for the mammal neonate, also providing immune protection and developmental signals. Lactation is a complex process, proper to the mother and child dyad, and including numerous variables ranging from psychological aspects to the secretory functioning of the mammary epithelial cells, all contributing to a successful breastfeeding. This review gives an integrated overview of the physiology of lactation with a particular focus on cellular and molecular mechanisms involved in milk product secretion and their regulations.
Introduction
Milk production and secretion is a complex physiological process resulting from both the previous development of the mammary gland (MG) and tight regulations by systemic hormones and local factors. All these aspects ultimately contribute to the coordinated secretory functioning of the mammary epithelial cells (MECs) in order to provide milk of adequate composition and in sufficient quantity to the newborn.
The mammary gland (MG) is a dynamic exocrine organ that can undergo repeated cycles of growth, functional differentiation, and regression, closely intertwined with the reproductive processes. Indeed, mammary development begins during early fetal life, occurs only slightly during estrous cycles, while complete mammogenesis only takes place during pregnancy to become fully functional after parturition to provide a nutritional support to the newborn. Once the child is weaned, the mammary tissue declines during involution and can re-differentiate if a new pregnancy starts (Fig. 1). All steps of the physiological development of the MG are tightly spatio-temporally coordinated, both by systemic hormones and local factors.
From about 7 weeks of gestation, the human MG develops from a single ectodermal ridge localized along the anterior body wall which extends from the epidermis into the underlying mesenchyme. Concomitantly, a loose condensation of mesenchyme extends sub-dermally to form the fat pad precursor. The ectoderm elongates to form a mammary sprout, invades the fat-pad precursor (10th-12th weeks), branches (13th-20th weeks), and canalizes to form the primary mammary ductal system (32nd weeks), which opens onto the area that gives rise to the nipple (Fig. 1, birth) , .
After birth, the MG remains as a rudimentary network of small branching ducts ending in short ductules called terminal end buds (TEBs) lined by one to two layer of epithelial and one of myoepithelial cells. These structures regress at ¼4 weeks postpartum along with a decrease in the secretion of prolactin (PRL) from the anterior pituitary gland of the infant. Until puberty, the growth of the breast is isometric. Of note, diet and/or metabolic pathologies such as diabetes may impair mammary development and subsequent lactation performance .
After a period of quiescence during childhood, the increase in ovarian estrogen secretion at puberty (812 years of age) stimulates the allometric growth of both the epithelial network and the adipose tissue within the MG (Fig. 1, puberty). Thus, while the increase in breast size is merely due to the enhanced deposition of adipose tissue, the mammary epithelium progressively elongates and further branches resulting in an extensive ductal network. These maturational changes occur in the course of ovulatory cycles and are then regulated by both systemic and locally released factors such as estrogen, progesterone, PRL, luteinizing hormone (LH), follicle stimulating hormone (FSH), growth hormone (GH) and epidermal growth factor (EGF) . During the follicular phase of the menstrual cycle, the lobules are small, with few alveoli, and there is low mitotic activity. During the luteal phase, ovarian progesterone stimulates lobulo-alveolar development, e.g. mitotic activity of the bilayered MECs, and opening of lumens in TEBs to form small alveoli . TEBs generate new branches, twigs, and small alveolar structures which cluster around a terminal duct, forming a lobule (¼11 alveoli/duct, Fig. 2B). Lobule formation occurs within 12 years after onset of the first menstrual period. Alveolar clusters grow and increase in complexity during each luteal phase and slightly regress with the onset of the menses and the loss of hormonal support, thus leading to a gradual accretion of the epithelial tissue with each successive cycle. In women, 3 types of lobules have been identified based on the size of the composing alveolar buds and their differentiation state (see Fig. 1 in ). With increasing years, mitotic activity slightly decreases until ¼35 years of age. Then, full differentiation of the MG is a gradual process taking many years, which is achieve only if pregnancy supervenes.
While in the non-pregnant adult woman connective and adipose tissues predominate and epithelial tissue is sparse, the onset of pregnancy induces progressive changes in both cellular and functional organization in the MG. Early pregnancy is characterized by growth due to a marked proliferation of both ductal and alveolar cells, concomitantly with the reduction of the fat pad. This leads to the formation of an extensive branched ductal system with a high number of alveoli of variable size and shape, gradually derived from TEBs (mammogenesis, Fig. 1, gestation). The surrounding stromal and myoepithelial cells provide essential cues for MEC survival, proliferation and differentiation. In newly formed lobules, the alveolar MECs not only increase in number due to active cell division but also increase in size, mainly because of cytoplasm enlargement. These modifications are regulated by numerous systemic hormones, including estrogen, progesterone, PRL, GH, insulin, glucocorticoids (GCs) and parathyroid hormone-related protein, as well as local factors such as insulin-like growth factor-1 (IGF-1), EGF and fibroblast growth factor (FGF), which are likely produced by the stromal cells , . Moreover, both MECs and stromal cells produce various extracellular matrix (ECM) components (e.g. proteoglycans, hyaluronan, fibronectin, and laminin), which are important for MG growth and function . The definitive structure of the ductal tree is essentially settled by the end of the first half of pregnancy and further changes until parturition are chiefly continuation and accentuation of branching and alveoli formation. Hence, small amounts of secretion product (colostrum) can be observed in enlarged lumen of alveoli and milk ducts, signing the functional secretory differentiation of MECs (referred as lactogenesis I). In the last trimester, there is a reduced proliferation of new alveoli with a further increase in their size due to distension of their lumen (terminal differentiation of MECs) by accumulation of colostrum. In addition to progesterone, PRL and/or human placental lactogen (hPL) appeared to be involved in the final stages of secretory MECs growth and differentiation. Concomitantly with the increased metabolic activity of MECs, the mammary blood flow approximately doubles in volume during pregnancy and persists during lactation until weaning.
Secretory differentiation of the alveolar MECs or lactogenesis starts around mid-pregnancy and has been divided into two successive phases: initiation or lactogenesis I and activation or lactogenesis II. These critical stages rely on variations of gene expression, structural and functional properties of alveolar cells, all of which being hormonally regulated , .
During lactogenesis I, MECs differentiate morphologically and become competent to produce and secrete some milk components referred as colostrum , due to the activation of the expression of some milk protein genes and biosynthetic enzymes, as well as the production of lactose and accumulation of lipid droplets (LDs) . However, production and secretion of milk components appear to be restricted to a limited number of alveolar MECs with incompletely developed secretory mechanisms. As colostrum is not removed by suckling, its components are reabsorbed into the blood through the paracellular pathway. At late pregnancy, milk secretion is inhibited by high plasma concentrations of progesterone and estrogen until parturition.
After parturition, the expulsion of the placenta results in a rapid withdrawal of progesterone , estrogen and hPL during the 46 days after birth, while PRL concentrations remain high in the presence of insulin and cortisol, thus triggering lactogenesis II , .
Colostrum is produced during the first 4 days postpartum, followed by a 1015 days period of transitional milk secretion, before copious production of mature milk (after 15 days) . Milk composition is dramatically altered: sodium and chloride concentrations fall while those of lactose, immunoglobulins A (IgA), lactoferrin (LTF) and other components of mature milk increase. These changes are completed by 72 h postpartum and precede the increase of milk volume by ¼24 h, accordingly to the terminal differentiation of alveolar MECs into lactocytes . These changes result from substantial variations of milk protein genes (e.g. α-lactalbumin) and biosynthetic enzymes (e.g. acetyl-CoA carboxylase and fatty acid synthetase) expression , supported by alveolar MECs reorganization, including apico-basal polarization of organelles, expansion of mitochondria and RER, maturation of the Golgi apparatus, appearance of secretory vesicles (SVs) containing casein micelles and of numerous microvilli at the APM, increase in the number of bigger LDs and closure of TJs that blocks the paracellular pathway, to adapt to their high secretory state . Moreover, there is an increase of transport activities for all substrates for milk production such as amino-acids, glucose and fatty acids, as well as ions. Indeed, with the closure of TJs, ions such sodium and chloride can no longer pass from the interstitial space into the lumen of the alveolus and then must be secreted by the cellular route.
Milk volume produced rapidly increases in the first 24 h postpartum, accordingly to the increase of both the frequency of breastfeeding and the volume consumed by the newborn, and stabilizes after ¼1 month (¼750800 ml/day) to remain fairly constant up to 6 months postpartum .
Although not essential within the first hours after birth, milk removal by day 3 is critical for the establishment of a successful lactation. Both the time of the first breastfeeding and the breastfeeding frequency on day 2 were positively correlated with milk volume on day 5 postpartum, suggesting that milk removal soon after birth increases the efficiency of milk secretion. Once lactation is established, the volume of milk produced is merely determined by the baby's appetite . Indeed, the breast is rarely completely drained during a suckling (on average 67% of the available milk is consumed). Thus, in connection with the frequency and effectiveness of the drainage-filling cycle of the alveoli , , there is a switch from endocrine to autocrine control and milk removal becomes the primary regulatory mechanism for galactopoiesis (milk synthesis) and to adjust milk volume to the requirements of the newborn. Milk can be stored for up to 48 h before the rate of milk synthesis and secretion begin to decrease. However, incomplete/inefficient milk removal or milk stasis induce multiple local effects on milk secretion: 1) an autocrine whey protein, termed feedback inhibitor of lactation (FIL) regulates milk secretion according to frequency or completeness of milk removal in each MG ; 2) other factors such as osmolarity and mechanical stress influence milk synthesis; 3) expression of the PRL receptors in MECs decreases, thereby uncoupling the stimulatory effects of PRL on milk synthesis; and 4) prolonged milk stasis triggers MECs apoptosis. Lactation is prolonged as long as milk is regularly removed from the MG .
Some particular conditions may delay lactogenesis, including placental retention, caesarean section, diabetes or stress during parturition. Obese women are more likely to experience delayed lactogenesis II, potentially due to hormonal influences on milk production, increased difficulty attaining a successful infant latch to the breast, and/or socio-cultural factors .
Early milk removal, correct attachment of the baby to the nipple, as well as the frequency and the efficiency of suction are the main key conditions contributing to a successful breastfeeding. Therefore, irregular or incomplete removal of milk leading to breast engorgement may be due to a mother's pathology such as an impaired milk ejection, inverted nipples or mastitis, as well as poor attachment and/or positioning, ineffective suckling, infrequent feeds of the infant. The best indicator of an adequate milk supply is the infant weight gain during the early neonatal period.
Lactation has a marked effect on fertility in breastfeeding women. After parturition, the systemic levels of LH and FSH, both controlled by the pulsatile release of gonadotropin-releasing hormone (GnRH), are low due to the suppression of the hypothalamicpituitary axis by placental steroids. While fertility returns approximately 69 weeks postpartum in non-breastfeeding women , in breastfeeding women GnRH secretion is suppressed by various factors such as maternal nutrition, PRL levels and the suckling stimulus and appears to be highly correlated with the breastfeeding pattern, e.g. the frequency and duration of suckling. The inhibition of GnRH release results in the disturbance of pulsatile LH secretion, which in turn suppresses ovarian activity. In addition, the increased sensitivity of the hypothalamo-pituitary system to the negative feedback effects of ovarian estrogen after parturition also contribute to the suppression of fertility during lactation.
At the end of lactation, when regular removal of milk ceases, the MG enters a tissue-remodeling process known as involution . Early after weaning, the epithelial architecture is maintained by the recent exposure to elevated systemic hormones, e.g. PRL, GCs and IGF-1, which are also critical survival (anti-apoptotic) factors for MECs. During this first phase of involution or reversible phase, the MG can revert to a state of milk production if the suckling stimulus occurs again . However, extended milk stasis in the ducts and alveolar lumen, concomitantly with PRL and GC withdrawal due to the absence of suckling, leads the MG to enter an irreversible phase (or phase 2 of involution) and to become unable to return to lactation without being re-stimulated by pregnancy levels of hormones . Milk stasis directly inhibits milk protein synthesis and secretion through both mechanical stretch and local production of various pro-apoptotic factors such as serotonin (5-hydroxytryptamine, 5-HT), LTF, Interleukin (IL)-6 family of cytokines, transforming growth factor-β (TGFβ) and α-lactalbumin. These factors lead to the inhibition of milk production by inducing the desensitization of MECs to lactogenic hormones. The fine balance between survival factors (PRL, GC, IGF-1) and cell death factors (5-HT, ILs, TGFβ, Vitamin-D receptor, IGF-binding protein-5) regulates the coordinated, multifocal and asynchronous processes resulting in a massive epithelial tissue regression (¼80%), mainly via apoptosis and autophagy of MECs. TJs gradually breakdown and the ECM is progressively remodeled by the action of both the matrix metalloproteases (MMPs) and the plasminogen system . Loss of attachment-dependent survival through integrins signaling (e.g., anoikis) together with pro-apoptotic signals leads to the elimination of MECs, collapse of acinar structures and narrowing of the tubules, while myoepithelial cells remain relatively well-organized during involution around residual ductal buds . In addition to immune cells present in the MG at all stages of development , surviving MECs play a major role in the clearance of residual milk and cell debris as they engulf casein micelles, MFGs and apoptotic cells . They also release anti-inflammatory cytokines which limit the action of the recruited leukocytes and neutrophils during early involution. Macrophages, local acute-phase response activation, and a late B-lymphocyte response complete the clearance of cell debris. These events ultimately lead to rapid regression of the epithelial tissue resulting in a rudimentary ductal tree morphologically similar to a virgin MG with some persisting alveoli (Fig. 1, involution). Concomitantly, pre-adipocytes re-differentiate (adipogenesis) and colonize a major part of the MG, while the vascular tissue is also remodeled .
After menopause, accompanied by an almost complete cessation of ovarian estrogen and progesterone production, the breast undergoes a slight regression. Nulliparous and parous breasts appear quite identical with only minimal quantitative differences in the proportion of lobule subtypes. However, nulliparous women exhibit a higher incidence of breast cancer than parous women and differentiation is suggested to protect the MG against carcinogenesis .
During lactation, the MG consists of a highly branched tubulo-alveolar glandular epithelium (or parenchyma) (Fig. 1, lactation), embedded in a stroma of both connective- and white adipose-tissue, and supported by a loose framework of fibrous connective tissue referred as Cooper's ligaments (Fig. 2A) . There is a decrease in the amount of adipose tissue relative to glandular tissue (ratio ¼1:2), which is not correlated to milk production or storage capacity, and the size and weight of the breast increases. Human breast consists of 1520 lobes, the size of which is highly variable, subdivided into lobules containing between 10 and 100 alveoli or alveoli (¼0.12 mm in diameter), which are the basic secretory units producing milk (Fig. 2B). Alveoli are clustered around ductules connected to the interlobular duct of the lobules that coalesce to form larger ducts, which are drained towards the nipple by a lactiferous duct (1.22.5 mm in diameter) that only dilate during milk ejection (no storage, only transport) . The significant variation in lobule size observed may reflect the difference in secretory activity from lobule to lobule. Moreover, growth and differentiation of MECs can occur in the same lobule, concomitantly with milk production. Each alveolus is surrounded by contractile myoepithelial cells responsible for milk ejection and an extensive capillary network . In addition to their role in milk ejection, myoepithelial cells also regulate mammary development through secreting various growth factors , spatially restrict MECs to form ducts during puberty, and act as tumor suppressors. Alveoli are embedded in a connective-tissue stroma containing adipocytes, fibroblasts and some plasma cells, which produce the Igs found in milk, as well as non-cellular components such as collagen and proteins of the ECM (Fig. 2C). Lymph is drained by two main pathways: the axillary nodes and the internal mammary nodes, which mostly drain the medial and lateral portions and the deep portion of the breast, respectively. The lymphatic network transports lipid-soluble nutrients (e.g., vitamin K and lipids) to the lactocytes, while the lymph nodes, which contain leukocytes (mainly lymphocytes and macrophages), provide an immune defense system to the MG in response to bacteria or foreign material. The MG contains only few internal innervations. Nerve fibers associate with the major duct system and are rather sparse in the region of the smaller ducts, areola, and nipple . Sympathetic nerves are associated with the arteries but not the alveoli and there is no parasympathetic innervation of the MG. However, sensory nerves present in the nipple are critical for initiating the afferent neural pathway of the milk ejection reflex. As there is no motor innervation of the mammary epithelium nor the myoepithelial cells, milk production and ejection are independent of the neural stimulation.
In the alveoli, the secretory MECs form a sealed epithelial monolayer upon the closure of their apical adherens- and tight-junctions (TJs), which segregate the lumen from the interstitial space, thus preventing paracellular transport (Fig. 2D). TJs also delimit the apical (APM) from the basolateral (BPM) plasma membranes of MECs, thus contributing to the establishment and the maintenance of the functional asymmetry (polarity) of MECs required for the vectorial secretion of milk . The basal side of alveolar MECs contacts myoepithelial cells and the basement membrane (BM), a specialized ECM, which separates the epithelium from the stroma and the vascular system. The BM results from the secretion by both stromal cells and MECs of specific ECM components further assembled in a 100 nm thick matrix at the basal surface of the mammary epithelium . Integrins are hetero-dimeric ECM receptors localized on the BPM of MECs and mediate cellmatrix adhesion and regulate various aspects of MEC development and function through integration with other signals . Integrin/BM interaction leads the formation of focal adhesion centers integrating both the assembling of the cytoskeleton and cell survival signals. Integrins signaling is thus involved in the establishment of the apico-basal polarity (e.g. apical side speciation) of MECs and lumen formation during pregnancy, enables PRL signaling through its effectors Jak2 and Stat5 to activate milk protein genes during lactation, as well as remodeling of the mammary tissue during involution . Moreover, the composition and stiffness of the BM change during MG development according to the variations in the ratio of the mammary cell types. Therefore, there is a critical interdependency of tissue architecture and cell fate for the spatio-temporal regulation of both mammary development and function . Thus, by transducing both biochemical (survival, differentiation and functional) and biophysical signals (changes in cell shape or membrane tension through cytoskeletal changes), integrins determine the fate of MECs , .
The APM of MECs borders the lumen of the alveoli, where milk product are released. As their principal function is to produce and secrete huge amounts of milk to feed the newborn, the intracellular organization of MECs reflects their highly secretory state. Indeed, the cytoplasm of alveolar MECs is filled with an extensive rough endoplasmic reticulum (ER) network, enlarged Golgi apparatus, and contains numerous mitochondria and SVs containing casein micelles. Lactose is synthesized in the Golgi, and is transported with casein micelles into the SVs towards the APM. Secretory MECs also produce LDs emerging from the ER by accumulation of neutral lipids and which grow during their transport before being released as milk fat globules (MFGs) by budding (Fig. 2D) .
After reaching the MG through the blood stream or the lymph system, nutrients and other components used to synthesize milk constituents diffuse in the interstitial space and reach the BPM of MECs. Depending on their molecular nature, they enter MECs and are secreted in milk by several routes. Most of the transport pathways are tightly regulated and coordinated, so that sufficient milk of adequate composition is available for the newborn, even during inadequate food intake by mothers.
Molecules can enter milk through paracellular or transcellular pathways (Fig. 2D), which are affected by the functional state of the MG and regulated by hormones, growth factors and probably mechanical constrains. While the mammary epithelium is leaky before lactation, the direct bi-directional paracellular exchanges of molecules between the interstitial space and the alveolar lumen (Fig. 1, Fig. 2) is inhibited during the first days of lactation after the closure of TJs triggered by the hormonal changes . Consequently, large trans-epithelial concentration gradients are established and maintained for ions and macromolecules between blood and milk.
After TJs closure, the composition of the milk reflects the highly coordinated functioning of four main transcellular pathways in MECs, which operate to produce milk components from blood-borne and interstitial molecules , . Many transporters (Fig. 2D and 2) are involved in the transfer of ions, glucose, amino acids and water are present on both the BPM and the APM . Transcytosis (Fig. 2D and 3) allows the transport numerous components originating from the bloodstream or the stroma, such as Igs, albumin, transferrin, insulin, PRL, estrogen, cytokines and lipoprotein lipase . Endogenously produced constituents such as major milk proteins, oligosaccharides, lactose, secreted through the exocytic pathway (Fig. 2D and 4) , while lipids (mainly triglycerides) are secreted by a specialized budding process (Fig. 2D and 5) .
Water is drawn across the alveolar MECs in a transcellular manner, driven by an osmotic gradient largely created by the lactose content of the milk. Water is transported by small transmembrane proteins of the aquaporin (AQP) family. AQPs are quite ubiquitous and, in addition to water, may also facilitate entry of gases such as CO2, NO and ammonia within cells. Various AQPs have been identified in the MG of various species including human, and localized in MECs, endothelial and myoepithelial cells. For example, AQP3 is localized in the BPM of alveolar MECs and may participate in the regulation of milk isotonicity by diluting milk components. The permeability of AQPs is strongly dependent on the molecular weight of the osmolytes they are exposed to. Moreover, the activity of AQPs could be up-regulated after their rapid membrane translocation in response to hormones .
Membrane transport pathway (Fig. 2D and 2) relies on the concerted activity at both the BPM and the APM, as well as in cellular membranes, of various transporter proteins, allowing transcellular transfer of ions, trace elements, glucose and amino acids from blood to milk. Ion transporters or channels for sodium, potassium and chloride are found on the BPM and the APM of MECs, while calcium, phosphate, iodide and citrate transporters appear to be limited to the BPM , . Sodium and potassium are also actively transported by Na+/K + ATPase pumps localized in the BPM but not APM of MECs. Active transport of calcium and trace elements including iron, zinc, copper, selenium, iodide, fluoride, and manganese have also been described in MECs but the underlying mechanisms have not been fully characterized. Moreover, the activity of some of these transporters, such as Ctr1 and ATP7A for copper and Zip3 for zinc, has been shown to be up-regulated by PRL, which induces their targeting to the BPM. Adequate supply of trace elements from milk is crucial to ensure neonate survival and both their uptake from blood and release in milk are tightly regulated by MECs, so that trace element concentrations remain remarkably stable, independently of the mother's diet .
As human milk contains up to 8 mM of calcium, large quantities of calcium are transported from the blood, which contains ¼3 mM of calcium, before being concentrated in milk . The presence of calcium channels has been described in the BPM and some intracellular membranes of MECs , , while the intracellular compartmentalization of calcium depends on cytoplasmic binding proteins. In milk, calcium is found associated with casein micelles (¼20%), free ionized or non-ionized (¼32%) or complexed to inorganic anions such as phosphate and citrate (¼46%).
In addition, the expression and/or the activity of some ion transporters may be hormonally regulated. For example, potassium uptake and chloride transport may be up-regulated by PRL via phosphorylation of transporters, while the expression of the sodium/iodide symporter is regulated by PRL and OT .
Plasma-derived glucose is a substrate for several key metabolic processes in MECs, including fatty acid and amino acids synthesis, triglyceride esterification, and is the obligate precursor for lactose synthesis. Hence, several types of glucose transporters (predominantly GLUT1) are found at both the BPM and the APM of MECs, as well as on Golgi, where lactose is synthesized from UDP-galactose and glucose, and SVs membrane. Lactogenic hormones such as PRL control both the expression of glucose transporters and their activity through translocation from intracellular sites to the BPM .
As amino acids are building blocks of proteins, large amounts of these precursors are required to support milk protein synthesis in the lactating MEC. Both sodium-dependent and -independent amino acid transporters are present at the BPM of MECs, but their presence at the APM remains unclear, although milk contains some amino acids. Amino acid transport has been shown to be modulated by PRL and milk stasis .
After their receptor-mediated endocytosis at the basal side of MECs, some interstitial molecules enter milk through the transcytic pathway (Fig. 2, Fig. 3). After endosomal maturation, these molecules are transported alone or in complex with their receptor to the APM of MECs, where they are secreted by exocytosis, while their receptor are degraded or recycled back to the BPM. Transcytosis has been described for IgA, insulin, PRL, serum albumin, transferrin, IGF-1 and low-density lipoprotein. Of note, the fusion of some transcytic vesicles with SVs may occur in the apical area of MECs before the exocytosis of their content .
Milk proteins are synthesized in a classical secretory pathway (Fig. 2D and 4), beginning with the transcription of their genes into mRNA, then translated in proteins and folded in the rough ER. Major milk proteins, namely caseins (α-, β-, γ-, and κ-caseins) also undergo post-translational modifications , mostly in the Golgi, associate with calcium and phosphate to form supramolecular structures called casein micelles (¼140 nm in diameter), which are packed in SVs. SVs also contain water, LTF, oligosaccharides, and high concentration of lactose, phosphate, calcium and citrate. SVs are vectorially transported via microtubules and fuse with the APM, then releasing their content into the lumen of the alveolus by exocytosis (Fig. 2D and 4). Interestingly, this pathway may be regulated by lactogenic hormones at several levels. Indeed, independently of activating casein gene expression, PRL exerts a secretagogue effect on the last steps of apical transport and possibly the exocytosis of caseins through the production of arachidonic acid . On the other hand, after binding to its cognate receptor in MECs, OT has been shown increase the number of SVs and to accelerate their transport towards the APM .
The molecular machinery responsible for membrane fusion has been characterized in many cell types, particularly in neuronal cells, and more recently in MECs . SNARE (Soluble N-ethylmaleimide-Sensitive Factor Attachment Protein Receptor) proteins mediate specific fusion of transport vesicles with target cellular membranes. To do so, the vesicular SNARE (v-SNARE) binds to cognate SNAREs located on the target membrane (t-SNAREs), thus forming a tripartite SNARE complex that promotes the fusion of the vesicle with the target membrane (Fig. 3A and B). The whole process of exocytosis is highly regulated by numerous proteins working in close association with the SNARE complex. In MECs, specific SNAREs have been observed associated with the APM, SVs and MFGs during lactation . On the other hand, several studies have shown that SNARE proteins are the targets of arachidonic acid in different neuroendocrine cell types . Thus, it is tempting to speculate that the SNARE proteins may be the target effectors of arachidonic acid produced in response to PRL, then providing a link between signal transduction, secretagogue effect and exocytosis in MECs (Fig. 3C), . Moreover, the expression of some SNARE genes has been found to be regulated by PRL . In MECs, the expression of some genes encoding SNAREs involved in casein exocytosis (e.g. SNAP23 and VAMP8) is strongly up-regulated during lactation, and our unpublished results, potentially in response to lactogenic hormones .
MECs import, synthesize and store lipids as LDs which are mainly formed by accumulation of triglycerides between the two leaflets of the ER and coated with some specific proteins (Fig. 2D and 5) , . Precursors of triglycerides include acetate, β-hydroxybutyrate, acetoacetate, fatty acids, glycerol, and monoacylglycerides, which are taken up by MECs, as well as ketone bodies. Free cholesterol also associates with LDs . LDs are thought to grow by fusing with each other during their apical transport, and are released by budding, enwrapped by APM, as MFG . Although proteins such as butyrophilin (BTN1), adipophilin (PLIN2) and xanthine oxidase (XOR) appear to play a critical role in this unique secretory process, the molecular mechanisms of MFG release have not been fully deciphered . Moreover, it occasionally results in the inclusion of a cytoplasmic crescent in the MFG, thus virtually enabling any cellular components to reach milk. The MFG is a major energy source for the newborn and also contains numerous enzymes, immunomodulatory factors, such as lactadherin/MFG-E8 and BTN1. Lipid secretion is regulated by hormones such as PRL and OT through mechanical deformation of MECs upon myoepithelial cells contraction .
As the release of milk constituents involves at least two distinct mechanisms (e.g. exocytosis and budding, Fig. 2D, 4 and 5) synchronized at time of suckling, it is likely that common activation switch and/or molecular effectors may exist to coordinate their activities. On the other hand, because of their large size (¼4 μm in diameter) and their high number, the membrane surface needed to enwrap the MFGs could exceed that of the APM of MESCs. Thus, at time of suckling, there is both membrane supply and loss at the APM of MESCs, due to SV fusion and MFG budding, respectively. Various data also reinforce the possibility of coupling of these two processes: 1) the association of SVs with the APM and the basal part of the budding MFG (Fig. 3A) has been extensively described, 2) some SNARE proteins are localized at the interface between SVs and the budding MFG (Fig. 3B) and 3) the membrane supplied by the fusion of a high number of SVs with the APM is used to enwrap the MFG , . As depicted in Fig. 3C, a possible scenario could be that, in response to PRL (secretagogue effect), local production of arachidonic acid, potentially from neutral lipid core of the MFG, stimulate membrane fusion through interacting with SNARE proteins , . Both heterotypic (SVs with APM) and homotypic (SVs with SVs) fusion may then occur (Fig. 3C), leading to the coordinated release of milk products. Furthermore, this would also partly balance the membrane loss caused by MFGs release, concomitantly with the efficient resealing of the APM (Fig. 3D). Recently, the final expulsion of MFGs has been shown to occur after OT-mediated contraction of the myoepithelial cells . This observation is compatible with the above scenario and suggest that milk secretion processes are spatio-temporally coupled and regulated by both hormonal and mechanical factors.
As MG growth and differentiation, lactation is regulated by hormones, but also by interactions between the MG and the central nervous system. PRL signals through the JAK2/STAT5 pathway to regulate the expression of target genes, and also stimulates lipid synthesis and exocytosis. On the other hand, OT is rapidly released in response to suckling and induces the contraction of myoepithelial cells surrounding the alveoli, thus triggering milk ejection .
PRL is a pleiotropic hormone produced by the anterior pituitary which is involved in homeostasis, reproduction, and lactation. During the MG development and differentiation, PRL exerts morphogenic effects, while during lactation this hormone displays lactogenic effects by stimulating milk protein and lactose synthesis and secretion, as well as other metabolic processes in MECs. PRL is thus required to maintain milk yield, but also for alveolar MECs survival and maintenance of tight junctions (TJs) , . During pregnancy, the serum PRL level slightly increases from ¼10 ng/mL in the non-pregnant women up to ¼200 ng/mL at term . In the course of lactation, levels of circulating PRL gradually decrease to return to ¼10 ng/mL after ¼6 months postpartum. PRL is episodically released in response to suckling to reach a peak in concentration in the blood 45 min after the beginning of breastfeeding, for up to 75 min in duration . However, while the amount of PRL released is related to the intensity of nipple stimulation, plasma PRL concentration does not appear to be directly correlated with the volume of milk produced. Interestingly, in serum and milk, several molecular forms of PRL are found, which arise from PRL processing such as cleavage . Whether this molecular heterogeneity can account for the various effects of PRL remains unclear. For example, while binding of the 23-kDa PRL to its cognate receptor on the BPM of MECs stimulates milk protein genes transcription, the internalization of the PRL/PRL receptor complex enables the transcytosis of PRL to the lumen, which is required for milk protein secretion .
As soon as it begins, suckling is detected by mechanoreceptors of sensory nerve terminals in the areolus of the nipple which send afferent cholinergic impulses to the paraventricular nuclei and supraoptic nuclei in the hypothalamus, that in turn stimulate the pulsatile release of OT, a nonapeptide hormone, from the posterior pituitary . Once in the bloodstream, OT reaches the MG where it interacts with specific G-protein-coupled receptors localized on myoepithelial cells, and induces their asynchronous contraction. As OT receptors are also present in MECs , this hormone may also exert direct effects on the secretory activity of MECs (Fig. 3D) , . Milk is then expelled out of the alveoli into the ducts and lactiferous sinuses. Contraction of the myoepithelial cells also shortens and widens the ducts, thus increasing the intraductal pressure and consequently the milk flow rate, ultimately leading to milk ejection from the nipple. Thus, OT mediates the milk ejection reflex (or let-down reflex), which is essential for the efficient removal of milk from the breast. As OT is released in a pulsatile manner, there are several ejections of milk during a feeding . The number of ejections is significantly correlated to the volume of milk consumed but not to the duration of the feeding . Suckling also causes an inhibition of the release of LH-releasing hormone by the hypothalamus that results in the inhibition of ovulation and a natural form of birth control.
There is also a significant psychological component in the let-down reflex, as OT release also occurs in response to such stimuli as the sight or sound of the baby . In addition to mediate the milk ejection reflex, OT also has significant roles on the central nervous system for the psychological integration of the interactions between the mother and the suckling neonate, and in maternal behavior. Furthermore, physical and psychological stress or pain of the mother has been shown to decrease milk output through the inhibition of OT release . However, responses to stress seem to be reduced, e.g. plasma levels of adrenocorticotropic hormone (ACTH), cortisol, and epinephrine are significantly decreased in lactating women stressed with graded treadmill exercise as compared to those found in non-lactating women . OT release is likely to be involved as its pulsatile release in response to suckling is accompanied by a decrease in plasma ACTH and plasma cortisol levels in lactating women.
The composition of milk varies between and within species and is specifically and ideally adapted to the needs of the neonate mammals to properly develop. Indeed, milk composition varies according to gestation, time postpartum and even during suckling.
During pregnancy, pre-colostrum contains high concentrations of protective Igs, lysozyme, and LTF, sodium, chloride, and low concentrations of casein, lactose, potassium, citrate, calcium, and phosphate. Colostrum persists for 4 or 5 days after parturition, followed by transitional milk for a further 5 days until mature milk is produced .
Mature milk is a complex emulsion of fat and aqueous fluid containing proteins (¼3.5%), sugars (¼7%), lipids (¼4%), minerals (¼0.5%) and water, constituting a unique complete nutritive source for the newborn.
Milk protein fraction includes four major proteins , e.g. α-lactalbumin and LTF (an iron-binding immunomodulatory protein with antibacterial properties), which are the most nutritionally important, caseins, and Igs (IgA for up to 10% of human milk protein, IgM and IgG). Igs provide passive immunity to the newborn and also serve as part of the immune system of the MG .
The aqueous fraction of milk (or whey) contains serum albumin, some hormones (e.g. PRL and insulin, leptin and adiponectin), growth factors (EGF, IGF-1, Ghrelin, and TGF), cytokines, lysozyme (a heme peroxidase with antibacterial and anti-oxidant properties), more than 30 enzymes (including lactoperoxidase which oxidizes bacterial components; proteases, protease activators, nucleases, glycosidases, amino-acid oxidases), vitamins, non-protein nitrogen, nucleotides, as well as minerals (sodium, potassium, chloride, citrate, calcium, magnesium, free phosphate, trace elements), and water. Growth factors such as EGF may regulate the intestinal growth, while hormones may modulate metabolism and body composition of the newborn . Factors with antimicrobial activities play important roles in protecting both the gastrointestinal tract of the newborn and the mother's breast . Of note, the sodium concentration in breast milk within the first 3 days postpartum may be predictive of lactation success, particularly in some mothers at high risk for insufficient milk supply . Indeed, high concentrations of sodium in milk are found in some clinical situations such as mastitis, inhibition of PRL secretion, and premature birth .
Milk also contains various types of carbohydrates, mainly lactose, a disaccharide unique to milk, glucose, galactose, and oligosaccharides, which display substantial protective effect against a variety of pathogens .
In human, the fat (MFGs) accounts for ¼4% of milk volume and contributes for up to 50% of the energy content. MFGs mainly contain triglycerides as well as a variety fatty acids, cholesterol, phospholipids, and steroid hormones (GCs, progesterone and estrogen) . Fat is the most variable fraction as its fatty acid composition varies with the maternal diet, and even during suckling. Bioactive lipids such as prostaglandins (PGs, including PGE2, PGD2, PGF2, PGI2), and thromboxane A2 are, which are synthesized from arachidonic acid by cyclooxygenases, are also present in milk and may exert protective effects.
Extracellular vesicles (EVs) such as exosomes (40100 nm diameters) have been identified in milk. Exosomes are vesicles formed in the multivesicular bodies (MVBs) derived from the endocytic pathway. During MVBs biogenesis, cargos, such as proteins, lipids, non-coding RNAs including micro-RNAs (miRNAs), and mRNAs are sorted into internal vesicles (e.g. exosomes), which are released into milk after exocytosis of the MVBs. In addition to play a role in nutrition, exosomes may participate to cell-to-cell communication, regulate developmental and immune processes, intestinal microflora, as well as cellular metabolism and gene expression after ingestion by the newborn .
Human milk is particularly rich in miRNAs, which are potentially involved in infant protection and development. MiRNAs are small non-coding RNA molecules that regulate gene expression at the post-transcriptional level, modulating important cell functions such as cell cycle, metabolism, proliferation, differentiation, apoptosis, and immune response .
Some cells such as MECs, macrophages, neutrophils, lymphocytes and stem cells are also found in milk , .
Human milk has also been identified as the first probiotic food as it contains a large microbial community including more than 200 phylotypes. Although not clearly established, these bacteria may be present on the mother's skin or may come from the maternal intestine after reaching the MG via lymph and/or blood circulation . In addition to enrich the intestinal flora of the newborn, milk bacteria could influence the long-term microbiota composition and activity, thus playing a key role to prevent various diseases such as allergies, disorders, and metabolic syndrome , . Therefore, breast milk not only functions as a nutritive source but also delivers both developmental and immune modulatory factors to the newborn.
Although the gross composition of mature human milk appears fairly constant with only slight changes for major components with stage of lactation, there are declines in the total fat content of the milk between 1 and 2 months, in the concentration of protein between 1 and 6 months , and in the concentration of calcium between 4 and 6 months. In addition, subtle variations occur in some constituents, such as fatty acids, vitamins, selenium and iodide, according to the maternal diet . Indeed, although the total fat content of breast milk appears unaffected by diet, the proportions of some fatty acids, e.g. omega-3 and omega-6 polyunsaturated fatty acids (PUFAs) vary substantially with the mother's diet . These variations may have important consequences due to the positive correlation between the quantity of omega-3 PUFAs in the mother's diet and the infant brain development . The fat content of milk is also known to increase with the duration of breastfeeding in proportion to the emptying of the alveoli . Thus, even if the storage capacity influences fat concentration in milk, it does not affect the total amount of fat consumed by the child . While the concentration of lactose shows no significant change with stage of lactation, variations in milk glycans, e.g. complex oligosaccharides free or covalently bound (glycolipids, glycoproteins, glycopeptides, and glycosaminoglycans), have been observed both between lactating women and during the course of lactation, according to the newborn's needs. These complex glycostructures are important dietary factors during early life as they regulate multiple functions . However, the growth rate of breast-fed babies is related to the total amount of milk they consume, rather than the concentration of fat, protein, or lactose .
Maternal nutrition affects both the quantity and quality of milk, which vary among countries, and lactation requires adjustments of maternal metabolism to adapt to the energetic demands of breastfeeding . As the milk production is almost entirely regulated by the infant demand, the maternal metabolism can be increased up to 20% of the metabolic output of the mother. This can be achieved by an increased food intake or increased weight loss to compensate for the metabolic needs to produce milk.
Milk production works according to the use it or lose it principle and current recommendations are to feed babies on demand . Indeed, babies feed according to their appetite and the mother's milk production is regulated to match the baby's needs. Although PRL stimulates the synthesis of milk proteins, it does not control the amount of milk produced once lactation is established. In fact, the quantity of milk produced is correlated to the draining efficiency of the suckling and is accordingly up-regulated if the breast is well-drained . Moreover, the efficient draining of the breast appears to be more important than the frequency of feeding to stimulate milk production. According to its appetite, a baby drains the breast one or more times per day but on average takes only 67% of the available milk . Therefore, the feeding frequency appears significantly increased for mothers with low storage capacities. Milk contents in proteins and lactose also seem to have more influence on the frequency of feeding, which is independent of the volume of milk consumed, than the quantity of lipids or the calorie value of the meal . Furthermore, the fat content of milk is related to the degree of fullness of the breast: the more the breast is filled with milk, the more the fat content of milk is low, while conversely, the more the breast is drained, the more the fat content of milk is high.
During exclusive breastfeeding, the lactating breast has a limited capacity (from 80 to 600 ml) to store milk, which varies to adapt to the child's needs . Storage capacity also varies from one breast to the other, independently of the ability to produce enough milk, but potentially affecting the feeding frequency. This may be related to the frequency and the efficiency of milk removal and to the local negative feedback regulation of milk secretion occurring when alveoli and ducts are filled with milk. As supplementary feeds are introduced, the milk storage capacity decreases along with the reduction of milk production .
The World Health Organization recommends exclusive breastfeeding for the first 6 months of life, and partial breastfeeding into the second year . When lactation is extended beyond 6 months, a significant decrease of the mammary tissue occurs gradually accompanied with a slight decline of the volume of milk produced and changes of its composition . Breast returns to its preconception size after ¼15 months of lactation.
Human milk is an optimal food for newborns as it contains both nutrients and bioactive compounds which contribute to both the short and long-term health benefits that have been reported to be directly correlated with the duration of breastfeeding. Breast-fed infants experience fewer and shorter infections, exhibit different growth patterns, have different gut microflora, show better cognitive development and even face differences in the risk of chronic diseases, such as obesity, type 1 and type 2 diabetes and cardiovascular disease. Breastfeeding also appears to be protective against sudden infant death syndrome, the risk of diarrhea, respiratory infections, and malocclusion, but does not seem to provide a protection towards either eczema or food allergy , . Breastfeeding outcomes are also related to mother genotype, phenotype, diet, disease, and lifestyle .
Human milk is also recommended to feed preterm infants as it significantly reduces complications associated with prematurity such as necrotizing enterocolitis, retinopathy of prematurity, broncho-pulmonary dysplasia and late-onset sepsis and promotes brain development and neurocognitive outcome .
Extended breastfeeding has also beneficial effects for the mother as it leads to birth spacing due to longer periods of amenorrhea, reduces risk of developing a longer-term diabetes type 2, overweight/obesity, and leukemia . However, breastfeeding seems not to have a protective effect towards hypertension and/or hypercholesterolemia . Extended breastfeeding also reduces the incidence of ovarian and breast cancer. Numerous studies suggest that high parity is associated with a decreased risk of developing breast cancer but that lactation itself, even extended, contributes no extra protection. The incidence of breast cancer appears to be reduced among pre- and post-menopausal breastfeeders, but a direct relationship between the duration of lactation and the reduction in the risk is found only for women with premenopausal cancer. Nonetheless, the mechanisms by which lactation could protect against breast cancer are not clearly identified although they probably involve the hormonal changes associated with breastfeeding and their effects on both the breast and the inhibition of ovulation .
Section snippets
Conclusion
Numerous aspects of the lactation process still remain to explore and to understand. The emergence of more efficient approaches to decipher mammary development, secretory functioning and milk composition, together with the integration of multi-scaled data from clinical trials to cellular biology, should highlight new aspects of breastfeeding and help to improve both mother and child's health, as well as infantile formulas.
Conflict of interest
The authors declare that there are no conflicts of interest.
Practice points
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Breast development and function under physiological and pathological conditions.
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Milk composition and influence of maternal and environmental factors.
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Consider important aspects to help mothers who want to breastfeed their newborn.
Research agenda
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Meta-analysis of clinical data.
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Integration of multi-scaled data, from populations to cell biology.
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Elucidate what and how milk can transfer information to the child and study of
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11563 | http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/quadtpeak.html | Time to Peak for Quadratic Drag
Time to Peak for Quadratic Drag
A vertically launched object will be presumed to experience an air resistance force proportional to the square of its speed. For a launch speed of v 0 the object is to calculate the time to the peak. The expressions will be developed for the two forms of air drag which will be used for trajectories:
but the simpler -cv 2 form will be used initially for simplicity and the forms for terminal velocity v t and characteristic time τ will be used.
The motion equation for this vertical launch is
which can be integrated for time t in the form
but this is a non-trivial integral. Using an integral table it can be expressed as
We finally reach an expression for the velocity as a function of time:
Now to find the time at the peak of the vertical trajectory, we can set the velocity equal to zero, yielding
This expression is used in the vertical trajectory calculation.Index
Fluid friction
Reference
A. Douglas Davis
Sec. 2.6
HyperPhysics Mechanics FluidsR NaveGo Back
Peak Height for Quadratic Drag
A vertically launched object will be presumed to experience an air resistance force proportional to the square of its speed. For a launch speed of v 0 the object is to calculate the peak height. The expressions will be developed for the two forms of air drag which will be used for trajectories:
but the simpler -cv 2 form will be used initially for simplicity and the forms for terminal velocity v t and characteristic time τ will be used.
Since we have an expression for velocity as a function of time from above,
it can be integrated to find an expression for distance y
Again, a good integral table is needed to obtain an expression for y:
Using the expression for velocity above, it is found that at the peak where v=0:
The expression for the peak height is then
This expression for y peak is used in the vertical trajectory calculation.Index
Fluid friction
Reference
A. Douglas Davis
Sec. 2.6
HyperPhysics Mechanics FluidsR NaveGo Back |
11564 | https://zhuanlan.zhihu.com/p/523937668 | 运筹学笔记——运输、指派和转运问题 - 知乎
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运筹学笔记——运输、指派和转运问题
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运筹学笔记——运输、指派和转运问题
王二麻子
13 人赞同了该文章
目录
1.表示运输问题
针对运输问题，我们可以构造一个线性规划模型去进行求解
以该问题为一个例子
上面是供给约束，下面是需求约束
求解LP后得到的结果示意图，以图的方式展现
总结而言，运输问题可以通过上述的规划普遍表示
如果总供给=总需求，即 ∑i=1 i=m s i=∑j=1 j=n d j\sum_{i=1}^{i=m}s_i=\sum_{j=1}^{j=n}d_j ,那么是一个平衡运输问题
如果总供给>总需求，那么我们可以设置一个伪需求点（dummy demand point）来使得等式成立。
一个运输问题具体可由供给，需求，运输成本组成，可以被概括在一个运输表中，如果有值就在表中元素左侧
如果总供给<总需求，那么显然不存在可行解，那么此时我们设一个伪供给点（dummy supply point），且从该点运输到需求点的代价为：需求点短缺时的代价。
2.找到运输问题的基本可行解
对于一个有m个供给点和n个需求点的运输问题而言，其LP有m+n个等式，对于这个问题，我们发现用大M法或者两阶段单纯形法是有困难的。
不过，针对这类结构的问题，我们很容易找到一个基本可行解。
引理：如果在一个平衡的问题下取值满足一个以外的所有约束条件，那么就一定满足其他约束条件
这样我们就可以舍弃任意一个其中的约束条件，去求解一个m+n-1个约束条件的LP。（我们假设舍弃的是第一个供给约束）
用线性方程组的方法可能找到基本解有点困难，所以我们用loop的方法判断约束是否有基本解。
定义1: 至少有四个元素的有序序列，且满足：
任何两个相邻的序列元素要么在同一行要么在同一列
没有三个在同一行/列的连续元素
序列最后的一个元素与第一个元素有共同的行/列
称为一个循环
图5:(2,1)-(2,4)-(4,4)-(4,1) 图6:(1-1)-(1-2)-(2,2)-(2,3)-(4,3)-(4,5)-(3,5)-(3,1)
定理1:在平衡运输问题中，m+n-1个变量的约束问题有的基本解当且仅当m+n-1个变量的元素中没有循环(loop)
我们有三种方法去找到基本可行解BFS
1.西北角法
1.我们从东北角(也就是X11)开始，对该点取供给和需求的更小值
如果X11=s1,那么删去表的第一行，并让d1=d1-s1，这样第一行就不再有其他的基变量了；如果X11=d1,那么删去表的第一列，并让s1=s1-d1，这样第一列就不再有其他的基变量了
如果X11=s1=d1，那么删去表的第一行或者第一列其中的一个。如果删去的是第一行，让d1=0；如果删去的是第一列，让s1=0
2.继续将该步骤用于表中最西北的单元格（单元格不在被划掉的行或列中）
最终，只有一个单元格可以赋值，在赋值后划掉其单元格所在的行和列，那么现在得到了一个基本可行解。
缺点：由于西北角法没有考虑运输成本，所以在找到一个基本可行解(bfs)后仍需变换多次去找到最优解。
下述的最小成本法会减少这一点
2.最小成本法
使用最小成本法时，我们不是选择西北角的单元格，而是在选单元格时选择运输成本最小的单元格，然后类似前者的操作。
3.Vogel法
我们计算每一行/列中最小的成本之差，记为惩罚(penalty)然后找到最大惩罚所在的行和列，然后类似前者的操作，每次重新计算未删行/列的惩罚，找到最大值进行赋值。
3.指派问题
对于指派问题而言，运输单纯形算法的效率是非常低的，所以我们要额外设计一套方法（说的就是你匈牙利）去有效解决这个问题。
一个指派问题的例子，可以看到该问题为0-1整数规划（要么指派，要么不指派,而且最后的0-1约束换成正整数约束是一样的），此时的基本可行解是高度退化的(highly degenerate)，对于一个mm的问题而言，在基本可行解中必有m个约束为1，m-1个约束为0
匈牙利指派法：
找到每行运输成本最小的元素并且在每一行减去最小的成本。找到每列运输成本最小的元素并且在每一列减去最小的成本。
针对新矩阵中值为0的元素，画出可以覆盖所有这些元素的最小线条，如果最小线条数为m，那么就已经找到最优解；如果小于m，那么进入步骤3
找到未被线条覆盖的最小元素，记其值为k，对所有未被线条覆盖的元素减去k；对被两条线覆盖的元素上加上k，回到步骤2
匈牙利方法的原理：
如果在某一行/列一同加上一个值，那么这个问题的最优解是没有改变的
所以对于步骤一和步骤三而言，其导致的最优解都与初始矩阵相同。每次执行步骤三时，都会在代价矩阵中创造出至少一个新的0。对于构建出的新的矩阵而言，令所有等于0的元素的 X_{ij} =1的分配是最优的，所以当步骤二表示需要m条线覆盖新矩阵中的所有0时，就找到了原始问题的最优解。
4.转运问题
转运问题是指除了供给点和需求点之外运输时的中间节点，并且可以转化为一个运输问题。
我们定义：
供给点（supply point）是指可以发货但不能收货的点
需求点（demand point）是指可以收货但不能发货的点
转运点（transshipment point）是同时可以收货和发货的点
以图中情形为例，左侧是供给点，中间是转运点，右侧是需求点
转化为平衡运输问题的步骤：
当总供给>总需求时，设置一个伪需求点(其供给为0，需求为总供给-总需求)，其运输成本为0
按如下方式构建运输表:每个供应点和转运点需要在表格中占一行，每个需求点和转运点需要占一列。每个供给点的供给都等于它的原始供给，每个需求点的需求都等于它的原始需求。让s=总供应量。那么每个转运点的供给将等于(点的原始供给)+s，需求将等于(点的原始需求)+s。这确保了作为净供给者任何转运点的净流出将等于该点的原始供应，同样，净需求者的净流入将等于该点的原始需求。
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11565 | https://www.pik-potsdam.de/en/output/publications/pikreports/.files/pr86.pdf | K I P FOR POTSDAM INSTITUTE CLIMATE IMPACT RESEARCH (PIK) PIK Report Carlo C. Jaeger A NOTE ON DOMAINS OF DISCOURSE Logical Know-How for Integrated Environmental Modelling Herausgeber: Dr. F.-W. Gerstengarbe Technische Ausführung: U. Werner POTSDAM-INSTITUT FÜR KLIMAFOLGENFORSCHUNG Telegrafenberg Postfach 60 12 03, 14412 Potsdam GERMANY Tel.: +49 (331) 288-2500 Fax: +49 (331) 288-2600 E-mail-Adresse:pik@pik-potsdam.de Author: Prof. Dr. Carlo C. Jaeger Potsdam Institute for Climate Impact Research P.O. Box 60 12 03, D-14412 Potsdam, Germany Phone: +49-331-288-2601 Fax: +49-331-288-2600 E-mail: Carlo.Jaeger@pik-potsdam.de POTSDAM, OKTOBER 2003 ISSN 1436-0179 Abstract Building computer models means implementing a mathematical structure on a piece of hardware in such a way that insights about some other phenomenon can be gained, remembered and commu-nicated. For meaningful computer modelling, the phenomenon to be modelled must be described in a logically coherent way. This can be quite difficult, especially when a combination of highly heterogeneous scientific disciplines is needed, as is often the case in environmental research. The paper shows how the notion of a domain of discourse as developed by logicians can be used to map out the cognitive landscape of integrated modelling. This landscape is not a fixed universe, but a multiverse resonating with an evolving pluralism of domains of discourse. Integrated modelling involves a never-ending activity of translation between such domains, an activity that often goes hand in hand with major efforts to overcome conceptual confusions within given domains. For these purposes, a careful use of mathematics, including tools of formal logic presented in the paper, can be helpful. The concept of vulnerability as currently used in global change research is discussed as an example of the challenges to be met in integrated environmental modelling.1 1Invaluable help from Rupert Klein and Alois Rust, from P. Flondor, I.Grossmann, W. Lucht, C.
Ionescu, A. Patt, D. Schr¨ oter, and from the debates in the European Climate Forum, is gratefully acknowledged. As usual, fame is shared, blame is not.
Comments are welcome to: carlo.jaeger@pik-potsdam.de.
3 1 Introduction The present note assembles know-how concerning domains of discourse. This know-how is relevant, perhaps essential, for building computer models that combine knowledge from a variety of scientific disciplines. The relevant know-how stems mainly from that border region between philosophy and mathematics where the tradition of logic is cultivated. The note should be accessible to patient readers without much training in this field; equally patient readers trained in logic should be able to skim offthe relevant arguments from paragraphs with a more introductory character than they would need.
The notation and concepts from logic that I use, in particular the concept of a domain of discourse around which the present note is built, are standard tools of logical inquiry.
The way I use them, in particular the emphasis on exchanges of arguments and on bridges between domains of discourse, is more novel. It is oriented towards those kinds of scientific debates where scholars do not try to speak in chorus, but as participants in a conversation between professionals with different backgrounds. In this sense they are contributions to the business of managing interdisciplinary research.
I introduce domains of discourse as the contexts needed for reasonably coherent exchanges of logical arguments (2). Such arguments rely on an interplay between experience and tradition taking place in evolving domains of discourse (3). I emphasize the necessary pluralism of domains of discourse used by scientific communities and by society at large.
The links between different domains are rooted in natural langauge and often rely on metaphor (4). In view of computer modelling, I am especially interested in linkages be-tween mathematical and other domains of discourse; I I therefore give a closer look at the measurement processes that evolved with the rise of modern physics and mathematics (5). I then discuss linkages between mathematical and biological domains of discourse.
Here, probability plays a key role (6). Across different domains of discourse, one can find the same rules of inference. However, rules of inference are heterogeneous, too. I argue that a pragmatic use of a variety of rules is appropriate (7). A pragmatic approach is particularly helpful whenever one must work without a consolidated domain of discourse.
This is relevant for efforts to model social systems. I consider the example of vulnerability studies and highlight some related challenges (8). In particular, I look at the domains of discourse of economics in view of modelling vulnerability to global environmental change (9). Clarifying linkages between domains of discourse can help to overcome major concep-tual confusions within a given domain as well. Mathematics is surprisingly useful, often simply indispensable to take care of existing linkages and to design new ones (10).
4 2 Exchanging arguments If one is able to model some phenomenon or effect on a computer - e.g. the influence of greenhouse gas emissions on future sea level rise - , then this computer model can be used to argue about that topic. The model can be used to assess the pros and cons of various factual claims concerning the phenomenon as well as the pros and cons of various proposals for actions dealing with it.
Any exchange of arguments presupposes some shared context of objects, properties, rela-tions, and inferences. This context is often called the domain of discourse (or the ”universe of discourse”, sometimes the ”ontology”; for the history of the concept and various ways of using it see De Morgan 1847, Boole 1854, Peirce 1885, Mead 1934, Eco 1976, Brandom 1984, Goldblatt 1984, Gamut 1991, Heim and Kratzer 1998, Benjamins et al. 1999).
A given domain of discourse corresponds to a language game (Wittgenstein 1953; more on this in section 7). As Wittgenstein emphasised, however, not all language games are concerned with exchanges of arguments - language games of greeting, e.g., although vital for the human condition, are not. In the present note, I restrict the focus to language games built around exchanges of arguments. As I will discuss, these language games form a much more heterogeneous landscape than is sometimes assumed, and it is a landscape that is considerably larger than the territory claimed by scientific traditions.
In logic, terms referring to objects are either names of individual objects or variables rang-ing over all objects in a domain of discourse. Terms expressing properties and relations are called predicates. The arguments studied by logic deal with properties of and relations between objects. They build on inferences linking given statements to other ones.
With these and further concepts, logic makes pathways of thought explicit (Brandom 1984). It is interested in describing in general terms pathways of thought that appear repeatedly in wide ranges of arguments and are generally treated as valid. Logic is reflexive in the sense that logical concepts can be used to study the arguments of logic itself.
An important and well-reserached subject of logical inquiry are the natural numbers.
They have names, e.g. ”1”, ”2”, ”957” etc. Relevant properties are: ”... is prime” or ”... is even”; relevant relations: ”... is the successor of ... ” or ”... ≥...”. Relations may involve more than two individuals, as: ”... is the greatest common divisor of ... and ... ”, or: ”... is the sum of ... and ...”. The natural numbers, along with relations of succession, of being a sum, etc., form the domain of discourse of arithmetic.
One can refer to objects by using variables and predicates together with expressions saying that an object exists or that something holds for all objects. When dealing with natural numbers 1, 2, 3, etc., one can refer to the number 1 with the statement: ”There exists an x such that for all y : y ≥x”. In standard notation this reads: ”∃x∀y (y ≥x)”. The expressions ”∃...” (there exists ... such that) and ”∀...” (for all ...) are called quantifiers.
An elementary sentence states that some objects have certain properties or stand in certain relations to each other. Elementary sentences may be denied, as in: ¬∃x∀y (x ≥y). Here ”¬” is used as negation sign. Elementary sentences may be connected into compound sentences, using sentential connectives like ”and” (often written as ∧), ”or” (∨), ”implies” (⇒), and ”if and only if” (⇔).
5 Logical argument is based on the fact of life that some sentences imply other ones. A simple example is: ”All philosophers are wise. Some Greeks are philosophers. Therefore some Greek are wise” To describe the pathway of thought involved in such an argument, logic uses variables for possible predicates along with variables referring to objects. This enables one to ask which pathways are sound for arbitrary predicates and objects - i.e. across different domains of discourse. Take ”P”, ”Q”, ”T” to be three predicate variables and read ”Px” as: ”x has property P”. Then the following sequence can be used to express the pattern of logical inference involved in our example: ∀x(Px ⇒Qx) ∃x(Tx ⇒Px) (1) ∃x(Tx ⇒Qx) The sequence as such is not an argument, it is a form of possible arguments. It becomes an argument once the object variables are assigned to identifiable objects and the predicate variables are assigned to specific properties and relations. This happened in our initial example. The same holds for the inferences: • ”All gods are immortal. Some characters in the Ilyad are gods. Therefore some characters in the Ilyad are immortal.” • ”All numbers divisble by three have a sum of the digits divisible by three. Some numbers divisible by seven are divisible by three. Therefore some numbers divisible by seven have a sum of the digits divisible by three” • ”All mammals breast-feed their babies. Some animals in the ocean are mammals.
Therefore some animals in the ocean breast-feed their babies”.
Arguments only work if the people involved in the discourse share an ability to identify relevant objects and to assess relevant properties and relations. This is a practical skill embedded in what Wittgenstein (1953) calls a form of life - a social pattern of human prac-tice. Our first example presupposes a community whose members know how to distinguish philosophers from other beings, wise from non wise people, and Greek from non-Greeks.
In a similar way, the ensuing arguments presuppose communities sharing specific domains of discourse - the ones of greek literature, arithmetic, and biology. In the words of G.H.
Mead: ”This universe of discourse is constituted by a group of individuals carrying on and participating in a common social process of experience and behavior, within which these gestures and symbols have the same or common meanings for all members of that group” (Mead 1933/1962, p.89).
A domain of discourse then involves a list of objects and a list of properties and relations.
Both lists contain items that the members of some community are able to identify. Usually, the lists are open-ended; it is understood that further objects, properties and relations will emerge in the course of further arguments and investigations. However, not all lists will do: a domain of discourse also needs some kind of coherence. This coherence is based mainly on relations of inference. Sentences referring to a given domain of discourse hang together by patterns of inference, i.e. by possible arguments.
6 3 Experience, tradition, and debate In patterns of inference, there is a fundamental difference between finite and infinite domains of discourse. To take a simple case, consider a domain of discourse consisting of Romeo and Julia. The statement that both Romeo and Juila are mortal implies that all individuals in this domain of discourse are mortal. Writing ”M” for ”is mortal”, ”R” for ”is Romeo”, and ”J” for ”is Julia” we get: ∃x(Rx ∧Mx) ∧∃x(Jx ∧Mx)) ⇒∀x(Mx) (2) In a finite domain of discourse, a conjunction of a finite number of statements of existence can imply a universal sentence. In an infinite domain, this is not possible. To take a famous example, many even numbers are known to be the sum of two primes (as with 8 = 5 + 3) and no even number is known for which this is not the case. But from these facts it does not follow that all even numbers are the sum of two primes. The latter statement - known as Goldbach’s conjecture - still waits for proof.
This feature of infinite domains of discourse lies at the roots of Popper’s (1935) influential view of how scientific inquiries work.
As he noticed, for a statement to count as an observation it has to be a statement of existence. On the other hand, key insights of science are universal statements ranging over an infinity of possible cases. This is the typical form of what are called natural laws as well as of many other scientific findings. To say that salt is soluble in water is to make a universal statement about an infinite number of possible situations where salt and water might mix. In an infinite domain of discourse, arguments built on observations and nothing else would be utterly uninteresting: all they could do would be to rearrange these observations one way or other without ever leading to a universal sentence.
In real life, arguments are less boring than that because we can combine statements of existence with universal sentences stemming from a tradition of inquiry.
Every child learning a language learns a whole lot of universal sentences with it.
They may be challenged and displaced by other universal sentences, but the burden of proof lies with the challenger, and she will hardly be successful if she tries to talk without using universal sentences at all - language does not work that way. We have no doubt that all human beings are mortal although billions of them have not yet died. We have no doubt because it is part of a received wisdom cultivated through the criticism and experience of generations.
Perhaps some day genetical engineers will advance a claim that they can produce immortal human beings. If so, a debate will be warranted on the credibility of that claim and what to do with it. For the time being, we will continue to live by our belief in the universal sentence that all human beings are mortal.
I leave open the question how far scientific inquiry really matches Popper’s peculiar ac-count of it and how scientific inquiry relates to other traditions of inquiry like democratic debate or religious search. What matters is that the coherence of a domain of discourse is due to a dynamic interplay between tradition, experience, and critical arguments. Tradi-tion is not a preliminary source of knowledge to be supplanted by experience in the course of scientific progress: no observation is possible without concepts provided by some tra-dition. These concepts come with a fabric of sentences, including universal ones. Any 7 particular sentence can be questioned, if there are reasons to do so. But to question a sentence already presupposes a whole fabric of sentences. Inquiries driven by reasonable doubt do not lead towards a world without tradition. What the progress of any particular inquiry does - be it in the realm of science or in other fields - is to cultivate traditions of inquiry by renewing them through fresh experiences and debates.
If a tradition of inquiry enables one to deal with an infinite domain of discourse, one can still cut out finite sub-domains for particular purposes. Logical arguments can be developed and exchanged in a domain of discourse involving the Earth and the sun only, as if no other bodies existed. Or one may study the permutations of just two or three numbers, disregarding all other numbers. Out of an infinite domain, other infinite sub-domains may be cut out as well.
In arithmetic, e.g., for some purposes one may be interested in prime numbers only, leaving out the other ones. In physics, out of a class of possible oscillations of a pendulum for some purposes one may wish to consider only a certain - still infinite - sub-class.
Given a finite domain of discourse, one can always try to embed it in a larger one. Some-times, it also turns out to be possible to aggregate two or more infinte domains into a more comprehensive one. As a domain of discourse is characterized by patterns of inference linking its various sentences, such aggregation only works if new patterns of inference running across the different domains considered can be found. In the development of science, successful aggregation of this kind is often seen as major progress. Such was the case with Maxwell’s treatment of electricity and magnetism, and later on with Einstein’s development of a single geometrical framework to deal with electromagnetism and gravity.
One may wonder whether integrated environmental modelling implies the possibility, per-haps even the necessity, to expand this kind of synthesis to the point where it would include biological, psychological, and socio-cultural phenomena as well. This would mean that the interplay of experience, tradition, and debate can, perhaps even must, lead to-wards one basic domain of discourse, providing a comprehensive frame in which all other domains are to be integrated.
4 The case for pluralism Such a view of integrated environmental modelling gains some plausibility from the influ-ential image of the world as ultimately consisting of physical matter. Often, such matter is visualized as an arrangement of small particles whose complex dynamic patterns lie behind the stars, stones, bodies, faces, melodies, thoughts, emotions and whatever else we experience. According to this image, there is a single ultimate reality that can be described by a single domain of discourse, the one of physics.
Recently, this image has been renewed thanks to advances in molecular biology and brain research. Our ability to relate important aspects of human behavior, perception, emotion, and thought to biochemical processes in the brain has been greatly enhanced. Moreover, several of these biochemical processes can be linked to genetic mechanisms that play out in biological evolution.
Many see this as a confirmation of the view that there is an ultimate reality consisting of physical matter.
8 However, some arguments to the contrary deserve attention here. To discuss them, one needs a domain of discourse for elementary physical objects, call them ”a, a′, a′′, ...”, char-acterized by various physical predicates, call them ”P, P ′, P ′′, ...”. Starting with these objects and predicates, one can identify further physical objects, namely sets of elemen-tary ones. To study these, one needs a predicate of membership, which I write in the usual fashion as ”.. ∈...”. I treat sets in the standard way as collections of objects fulfilling some predicate P: a′ = {x : Px}. In such a domain of discourse, logical argument about physical objects is possible in a rigorous fashion.
For example, one may form a term for a set of all red objects - with ’R’ as the predicate for ”... is red” one would simply write a′ = {x : Rx}. If one can show that such a set exists and that it is unique - these conditions are less obvious than one may expect - to say that something is red amounts to the same thing as saying that it is an element of the set of red objects. With the membership predicate one can introduce not only sets of objects, but also sets of such sets, etc.
This leads to the venerable view of an atomistic universe. One can try to sketch such a universe by taking elementary particles as the logical atoms (the a, a′, etc.), and imagining these as arranged in the atoms of modern physics, these in molecules, etc., etc. until one reaches the mountains and oceans, the planets and the stars. In these terms one can sketch a map of our galaxy and the world beyond.
4.1 The necessity of non-physical objects In such a domain of discourse one can describe physical objects in mathematical terms as follows. Consider sets containing exactly one elementary object. Call the set of all these sets the number 1. More generally, call the set of all sets with n elementary objects as elements the number n. Now to say that some set of elementary objects has 7 members amounts to the same thing as saying that the set in question is an element of the 7-set.
Along these lines, numbers can be treated as properties of elementary physical objects on a par with colors and the like.
Showing that logic, mathematics, and physics could be combined in such a way was one of the great achievements of work on the foundations of mathematics (Russell and Whitehead, 1910-13). Perhaps the greatest merit of that work, however, has been the gradual clarification of how incoherent a physicalist world-view is bound to be.
First, consider the set-theoretic proof that ”there is no universe” (Halmos 1960, sec.2). Let a be an arbitrary set and a′ the set of all elements of a that are not elements of themselves: a′ = {x : x ∈a ∧¬(x ∈x)} (3) Then ¬(a′ ∈a). Suppose otherwise. By its definition, a′ cannot be an element of itself (because it only contains elements that are not elements of themselves). But then if it were a member of a it would fulfill the definition of members of a′ and so would bound to be an element of itself. This contradiction shows that a′ cannot be an element of a.
Therefore, a cannot contain everything, as it cannot contain a′. But a was a completely arbitrary set, and so there can be no set containing all sets.
9 What is this proof about? In a way, it is about the structure of language, more precisely: of the kind of language needed to talk about sets of objects.
If one wants to complement formal proof with intuitive associations, it may be helpful to think of the world for a while as consisting of quarks, photons, electrons, and other elementary particles. An atom then is a set of such particles and not an element of itself, while elementary particles are elements of themselves (as Quine, 1969, p.30ff, shows, this is a nice way of distinguishing elementary objects from non-elementary ones).
Next, I focus on the property called spin. Photons, electrons, hydrogen atoms all have spin (polarized sunglasses use it with photons). If one considers any of these entities in isolation under suitable experimental conditions, their spin is either up or down. If one considers a pair of such entities under analogous conditions, however, their spin can correspond not only to the four combinations of up and down, but also to an additional state that is sometimes called one of entanglement.
This kind of situation was ignored by classical physics. It is widespread in modern physics as in many other fields. One can describe it by distinguishing between a photon and a set consisting of a photon: taken in isolation, photons have spin up or down, as members of sets, they have spin up, down, or entangled.
Of course one can use other words to describe this situation, but the key point is quite clear: being an element of larger structures changes the possible properties of elementary particles. The same holds when structures involving elementary particles are elements of further structures. When two hydrogen atoms, e.g., are combined in a hydrogen molecule, their two electrons behave as a two-electron cloud. In that cloud, there are no distinguishable trajectories of single electrons. To identify any property of a single electron, one must break it away from the molecule. But by studying an electron in isolation, one would never find out what electrons can do as components of molecules.
Knowing an elementary particle means knowing what it can do. One cannot learn this without considering the infinite set of possible structures they can be involved in. A world consisting of elementary particles consists of entangled pairs, n-particle clouds and suites of more complex structures as well, leading from quarks to jaguars and beyond (Gell-Man 1994). The ”no-universe” argument then says that it would be logically incoherent to imagine some ultimate suite of structures out of which all the other ones could be explained. Whenever one can talk about some such suite of structures with the means of mathematical logic, one can find some even more comprehensive suite of suites.
The ”no-universe” argument is not restricted to elementary particles. In general terms, it implies that one cannot effectively summarize all the possibilities involved in a situation where along with any given sets one also needs to consider sets of these sets.
Technical as they are, at the beginning of the 20th century issues of this kind triggered one of the deepest crises in the study of the foundations of mathematics and of philosophy in general, because they suggested that scientific discourse is unable to refer to the world as a whole. The philosophy of Wittgenstein, one of the most influential philosophers of the 20th century, can be read as a series of variations on this theme.
Second, consider the fact that any physical domain of discourse can be described with sentences expressed by finite strings of symbols. The symbols are taken from two kinds 10 of lists. There is a finite list of symbols for quantifiers, connectives, and brackets, and there are infinite, but countable lists of symbols for objects, properties, and relations.
Objects may be referred to by names, a, a′, a′′, ... or variables, x, x′, x′′, .... Properties and relations may be expressed by predicate symbols P, P ′, P ′′, .... Expressions for names, object variables, and predicates can be iterated in order to represent arbitrarily long lists of objects and predicates. On the basis of these symbols, all strings containing just one symbol can be put in lexicographic order, then the same can be done for the strings containing two symbols, and so on for strings of arbitrary length. This yields a countable list, and the meaningful formulae are bound to be an infinite subset of this list. Therefore, the whole domain of discourse can be represented by a countable set of formulae. But this means that the objects that can be identified by these formulae will form a countable set, too. In other words, each identifiable object can be associated to a natural number.
Besides natural numbers, however, real numbers - like √ 2 and π - are basic for any meaningful description of physical realities: √ 2 matters wherever the diagonal of a square shows up, π matters wherever the diameter of a circle or the shape of a wave show up. Real numbers are remarkable among other things because they cannot be counted. Suppose otherwise. Then they could be put in a list. Imagine this list in tabular form with each row representing a real number between 0 and 1 in decimal notation. Take the first digit in the first row and add 1 modulo 9 to it: if the digit is x ̸= 9, it becomes x + 1; if the digit is 9, it becomes 0. Following Cantor’s (1891) diagonal argument, do the same for the second digit in the second row, the third in the third, etc. The result will be a real number that appears nowhere in the list. But the list was assumed to contain all real numbers. In other words, the assumption that the real numbers can be counted leads to a contradiction.
This means that while one can study I R, the set of real numbers, as an object within some domain of discourse, there will always be properties of I R that cannot be specified in that domain. After all, containing some specific number as an element is a property of I R. Moreover, any formal theory of real numbers will be amenable to a rigorously defined re-interpretation that makes it a theory - although with very different meaning -over natural numbers. Any theory that can be formulated by a countable set of formulae allows for an interpretation over the set of natural numbers (Skolem 1941).
By the first argument, a physical domain of discourse does not enable one to talk about the world as a whole. By the second argument, such a domain of discourse needs objects - real numbers - and properties - those expressed with the help of real numbers - that cannot be fully specified in any given domain. A more sophisticated argument - due to G¨ odel (1931, with Hofstadter, 1979, providing an exposition accessible to the non-specialist) - shows that this already holds for any domain of discourse comprising the natural numbers. It also shows that, given a domain of discourse comprising the natural numbers, it is always possible to form a new domain specifying some additional property of these by a suitable formula.
The third argument to be considered shows that even a physical domain of discourse leads to non-physical objects. It runs as follows. Given an arbitrary set X and an arbitrary predicate P, one can always form the subset U of all elements v fulfilling that predicate: 11 ∀x∃U∀v ((v ∈U) ⇔(v ∈X ∧Pv)) (4) This is an example of a universal sentence corroborated by a tradition of critical inquiry: the various axiomatic and non-axiomatic approaches to set theory developed in the course of time all share it one way or another. Given a domain of physical objects, then, one can take an arbitrary set of physical objects (call it a′), form the predicate: ”... is not an element of a′ ” and gets: ∃U∀v ((v ∈U) ⇔(v ∈a′ ∧¬(v ∈a′))) (5) This set is empty, as nothing can simultaneously be a member of a′ and not be a member of a′. So there is an empty set. Inconspicuous as it may look, it still is a non-physical object. And as a matter of fact, it is the ancestor of a whole domain of discourse of non-physical objects.
By received wisdom, two sets are equal if they have the same elements: ∀u∀x∀y ((u ∈X) ⇔(u ∈Y )) ⇔(X = Y )) (6) As a result, there can be only one empty set, as there are no elements by which two empty sets could be distinguished from one another.
Another piece of received wisdom says that given an arbitrary set U with no more than one member, one can always form the set V containing that set.
∀U∀x∀y(((x ∈U ∧y ∈U) ⇔x = y) ⇒ ∃V ∀Z (U ∈V ∧(Z ∈V ⇔Z = U))) (7) Therefore, there is a set containing only the empty set. This is a second non-physical object. It is different from the empty set precisely because it has a different element. By the same reasoning, one can construe an infninite sequence of sets, each one containing only its predecessor. But this sequence has all the properties known from natural numbers.
Therefore, starting with the empty set, one can construe the natural numbers. From there, one can go on to construe the other kinds of numbers, functions between them, and the whole fabric of mathematical entitities.
But now we have two kinds of natural numbers: those introduced above as sets of el-ementary physical objects, and those constructed now out of the empty set. As if this were not enough, we can construe a third kind of natural numbers by repeating the first construction with sets of sets of objects - and this proliferation cannot be stopped: the attempt to describe the world in purely physicalistic terms leads to a world populated not only by elementary particles, but also by sets of such particles, by sets of such sets, ..., and by an uncountable multitude of natural number systems as well.
Quite likely, one may feel that the word ”exists” is used in a different way when one says that a planet exists than when one says that a prime number exists. But this is precisely the point of distinguishing different domains of discourse. To say that something exists places it in some domain of discourse. To say that the word ”exists” is used in different ways means to distinguish different domains of discourse.
12 4.2 Mutually irreducible objects For the introduction of mathematical objects, it turns out to be quite irrelevant whether one starts with physical ones. The former are needed to talk about the latter, and they cannot be construed out of these. One could introduce mathematical objects starting with greek gods as well, and they cannot be construed out of those, either. Therefore, one may just as well start with mathematical objects for their own sake, as has become common practice.
If one cannot reduce mathematical domains of discourse to physical ones, does it work the other way round? This is the ancient idea that the ultimate reality is the one of mathematics. It was first developed by the Pythagoreans, later in a different version by Platonists. It is so much at odds with the fabric of contemporary science and technology that it is hard to propose as a serious view. Nevertheless, at least a short discussion is appropriate.
In the times of classical physics, typical objects of investigation were physical bodies, characterized by properties like position and momentum, and linked with each other by relations like attraction and impact. Meanwhile, physics itself has evolved so as to use other domains of discourse as well. Often, the relevant objects are regions in space-time with some given shape - cubes, spheres, whatever. These regions are then characterized by properties like the fact that in a specific region some given body will experience a specific acceleration, and relations like the fact that this very acceleration will change the analogous property of another region. Systems of space-time regions of arbitrary size, characterized by shared properties and relations, can be used to construe physical fields.
By taking the limit of sequences of space-time regions converging to a single point, for example, one can construe a field for the density of some substance in arbitrary regions of space-time.
Before this background, one might argue that the three-dimensional space of real numbers, I R3, ”is” the physical space we humans live in, that time is an additional dimension of that space, with further dimensions characterizing the densities of mass, charge, and other variables in physical space.
There are many problems with such an argument. First, there is a myriad of situations that can be described in terms of three-dimensional spaces of real numbers. How are we to distinguish the description of the position of a body in physical space from, say, the description of three bodies characterized by their respective masses? We do it by relying on non-mathematical predicates like those used to talk about physical space and time as well as mass.
Second, since the days of Einstein it is clear that a description of space-time in terms of I R4, the four dimensional space of real numbers, is quite tricky: what is simultaneous for one observer need not be so for another one. Additional subtleties come into play with quantum mechanics. Depending on circumstances, physicists use different mathematical objects to talk about time - and to say that they are actually talking about time means using a non-mathematical predicate.
”Theories pass. The frog remains”, as the French biologist Jean Rostand said. ”Frog” is a non-mathematical predicate, as is ”star” or ”gravitation”. One can use mathematical ob-13 jects in many ways to talk about non-mathematical ones, but one needs non-mathematical predicates to do so.
Recently, another way of looking at the physical world as rooted in mathematical reality has found increasing interest: as computers are built out of physical materials, one can try to describe the physical world as one huge computer (Zuse 1967, Wolfram 2002). This is the kind of image that can inspire wonderful insights as well as considerable nonsense. I will not try to explore the many questions that may be asked in such a setting, but focus on the technical point that matters here: can physical predicates be defined in terms of mathematical ones?
It is clear that a given mathematical structure - say, a specific computer program - can be implemented in two different pieces of hardware. Moreover, a given physical pattern can be looked at as representing two different mathematical structures. A striking example in the domain of computation is given by quine programs.
These are programs that make a computer print exactly their own source code. They are named in honor of the logician W.v.O. Quine, whose work is relevant to our problem (Quine 1960, 1969). A quine program for the language Python looks as follows: f = lambda : [‘f = lambda : ‘, ‘; print f(), f(), f()‘]; print f(), f(), f() (8) Read as a program, the first occurrence of the expression ”f=lambda :” is part of a function definition; read as output the same expression is an element of the value returned by that very function. If we run the quine on a computer, it will be obvious from the circumstances which one is the case. If we are faced with (8) simply as a physical pattern of ink on paper, there is no way to tell.
This kind of ambiguity carries much further, as in the case - discussed in section 4.1 - of theories of real numbers that can be re-interpreted as theories defined over natural num-bers. Because of such difficulties, a description of the physical world as a huge computer is surprisingly ambiguous. If the physical world can be described as a computer, then it can be described as other computers, too - and what makes these computers equivalent is the fact that we use them to describe the same physical world. There is no one-to-one correspondence between physical and mathematical objects. In the picture of the physi-cal world as a computer as in the example of space-time discussed above, we use physical predicates to identify the non-mathematical objects we want to talk about.
With regard to the domain of discourse established with the emergence of modern physics one thing is quite clear: one does not need any mathematical objects or predicates to observe that Newton’s apple falls down to earth, or that a stone and a brick dropped simultaneously from the oblique belltower known to Galilei hit the ground simultane-ously. Physical predicates like mass, momentum, electrical charge, etc. do not depend on mathematical concepts. One certainly does not get very far in discussing physical objects without referring to mathematical ones, but the two inhabit different domains of discourse.
Besides the domains of discourse of physics and those of mathematics, a third kind of domains is unavoidable even for the most strenuous attempts to produce a physicalist account of the world: one needs a domain in which Newton, Einstein, as well as Dr.
Jones from the lab nearby, can be identified by name. This is not just a nice add-on to 14 provide due recognition to scholarly work, it is indispensable in order to discuss empirical observations. To do so, one must be able to refer to the person that made the observation.
More generally, scientific inquiry proceeds by an on-going exchange of arguments where the participants need the ability to refer to each other. Trivial as this fact is, it implies that besides physical and mathematical domains of discourse, one also needs domains of discourse in which the relevant individuals are human beings. Of course we can describe ourselves as physical objects. But we can do so only after having identified a certain physical body as our own, or as the body of Einstein or of Dr. Jones. We build up the myriads of co-ordinate systems available to do physics starting with space-time regions that we can identify with our own bodies (Brandom, p.444f).
Once one has got acquainted to the notion of a plurality of irreducible domains of dis-course, the variety of scientific disciplines and terminologies appears in a different light.
The frog is a physical object no less than the stars, but there have been good reasons to develop domains of discourse to talk about living beings in different ways than about inanimate objects. As a matter of fact, even within physics there are rather different domains of discourse. The space-time regions of quantum mechanics, e.g., are different from the ones of relativity theory. And domains of discourse dealing with living organisms and their environments are even more heterogeneous.
Mathematical discourse has its pluralism, too. Even the natural numbers, innocuous as they may look, replicate themselves through the worlds of mathematics (see section 4.1).
One way of dealing with this situation is to focus on categories of mathematical objects that form an interesting domain of mathematical discourse. Predicates like ”...is a natural number system...” or ”...is a vector space...” cannot be used to form sets. A set of all natural number systems, e.g., would be no less self-defeating than a set of all sets (the set of all natural number systems could be the first element of a still larger natural number system). But these predicates can be used to describe categories of mathematical objects that fit the same domain of discourse. ”A category may be thought of in the first instance as a universe for a particular kind of mathematical discourse” (Goldblatt 1984, p.1).
In the course of time, such a focus on categories could lessen the weight currently put on set theory as the foundation of mathematics - for example, by helping to understand why the metaphor of a house is somewhat misleading here. The argument of section 4.1 could then be cast in a less set-theoretic format. When talking about physical objects, one might try to substitute the membership predicate with a predicate for ”... is a part of ...”; when talking about mathematical objects, one might introduce the natural numbers directly via predicates of identity and succession. The literature on such attempts shows that they increase the range of ”ways of worldmaking” (Goodman 1978) at our disposal, and thereby the pluralism of domains of discourse. The physicalist worldview would not be restored; rather, mathematical physics would use numbers and functions to talk about stars and stones, and these would complement each other as mutually irreducible objects.
Dr. Jones and other human beings enter the picture as further irreducible objects. I will not discuss domains of discourse dealing with human beings here, as I will do so in sections 8 and 9. The key point of the present section will be reinforced there: from a logical point of view, the landscape we live in is not one world, but many, not a fixed universe, but a multiverse resonating with an evolving pluralism of domains of discourse.
15 In ordinary language, of course, one can move across these domains and happily talk about the world as a whole. If asked what one means by that expression, one can draw on the resources of poetry rather than those of logic. One can build up metaphors, perhaps invoking images of waves and the ocean, of people linked by family resemblances, of the melodies in a fugue, asking one’s interlocutor whether these metaphors are helpful to her, modifying them, introducing new ones until she feels comfortable to let the conversation flow further - or until one changes the subject. What one will not be able to do is to form a sensible expression for ”the world as a whole” in the kind of language required to talk in a logically coherent way about the objects of physics. To the extent to which it makes sense to talk about the world as a whole, what is called ”the universe” in physics (even if it includes all the parallel universes advocated by authors like Tegmark, 2003) is but one aspect of the world we live in.
This situation sheds new light on the relation between scientific knowledge and other forms of knowledge. Scientific knowledge as we know it usually tries to avoid moral or aesthetic judgement. But of course there are moral and aesthetic domains of discourse, and the scientific enterprise is by no means the only arena for exchanging arguments.
As soon as one starts talking about human beings as well as about quarks, one is quite likely to talk about human values as well as about the flavors of quarks. And once you recognize the plurality of domains of discourse, ”you get rid of the notion that quarks and human rights differ in ’ontological status’ ” (Rorty 1998). The plurality of domains of discourse includes a wide array of scientific ones as rich and fascinating examples, but it is not at all restricted to the horizon of the scientific enterprise.
5 Numbers, bricks, and coins Sentences about greek gods imply other sentences about greek gods and sentences about human beings imply other sentences about human beings. What is less clear, and perhaps more interesting, is the question whether, and if so, how, sentences about gods imply sentences about humans as well as the other way around. From a logical point of view, a similar relation holds between sentences about numbers and sentences about stars. The sentences of arithmetic hang together by logical inference, and so do the sentences of astronomy. It is clear that the two are related, but it is not so clear how.
Today, few people feel comfortable with seeing the world we live in as inhabited by greek gods. More feel comfortable with seeing it as inhabited by protons and electrons. Most, however, are quite comfortable with seeing our world inhabited by flowers and birds and to treat these as objects of discourse in their own right, without bothering about how they may relate to the world of quantum fields. To some extent, we have simply learnt to live in a plurality of domains of discourse. This raises the question of how we link these domains with each other in our thoughts and actions.
A particularly instructive example for studying this problem is given by the traditions of mathematics and physics. Clearly, over the last centuries, they have developed hand in hand. And the most elementary connection between mathematical and physical domains of discourse was and is provided by the practice of counting.
16 Counting things requires the ability to identify one of them, to identify a next one, and so on, without counting anything twice, and to identify a last one. When dealing with mathematical sets, one can try to establish a one-to-one correspondence between the members of a set and a finite sequence of natural numbers that starts with 1 and moves stepwise from one natural number to its successor. If one succeeds, the set has as many members as indicated by the last number in the sequence. To take a simple example, consider the set comprising the prime numbers smaller than 10, i.e. {1, 2, 3, 5, 7} (counting 1 as prime). A one-to-one correspondence can be established with the sequence [1,2,3,4,5], so the set has five members. The diagonal argument considered above shows that no such correspondence can be established between the set of real numbers and any finite sequence of natural numbers (it even shows that no one-to-one correspondence can be established between the set of real numbers and any infinite sequence of natural numbers).
When dealing with non-mathematical objects, there are no clear rules as to when they are a set and when they are not, so things are slightly different. For an emotion, a cloud, a design for an airplane to become objects of logical reasoning, there must be some domain of discourse in which they appear. It is easy to produce such a domain involving a particular emotion, cloud, and design - e.g. by telling the story of how an engineer experiencing a particular emotion was inspired by a particular cloud to design a particular airplane. It is certainly instructive to try to produce a domain of discourse for emotions, clouds, and airplane designs in general. It is instructive especially because it is difficult, perhaps non-sensical to produce such a domain. So far, psychologists, meteorologists, and engineers live happily without any agreed domains of discourse for emotions, clouds, and airplane designs. Things do not get easier when the relevant objects are further delimited by some additional criterion and one tries to count them. A symphony, e.g., may trigger emotions, but it is not clear whether the emotions triggered by a symphony can be treated as a set on a par with mathematical sets, nor is it obvious how they could be put in a sequence.
One may look at the sky and identify a particular cloud, but it is not so clear what it means to talk about the set of clouds in the sky, and perhaps even less what it would mean to count them. One may think about possible designs of airplanes for freight traffic - again it is not at all trivial to treat these as a set, let alone a countable one.
The paradoxes of set theory have taught logicians to be careful about how to introduce sets. By now there are rules for introducing mathematical sets in ways that are known to avoid logical paradoxes (although they leave more than enough other puzzles to think about). There are no similar rules for introducing non-mathematical sets. What one can say, however, is that many things - the fingers on our hands, the apples we collect from a tree, the people we meet at a party - can be put in finite sequences by a successor relationship, and then one can safely treat them as members of sets.
Statements like: ”There are twelve Apostles”, or: ”This is the fifth mistake” can be rendered with a predicate involving a relation between two nodes: a natural number specified one way or another (”twelve”, ”fifth”) and a group of objects specified by another predicate (”the Apostles”, ”the sequence of mistakes up to this one”). Writing a predicate for counting as ”...C...”, an arbitrary natural number as ”n”, and a predicate specifying the list of objects counted as ”L...”, one can express a count of things as: ∃x(xCn ∧Lx) (9) 17 The ability to handle such predicates is trained woldwide in grammar school.
It has become part of the global culture shared by humankind.
Counting is one way of associating things with numbers, namely with the natural numbers.
One may say that by counting one measures the size of a countable set in terms of its members. What is usually called measurement provides a further link between things and numbers, this time the real numbers.
Consider a pile of bricks and a rod such that one brick is much shorter than the rod. One may be able to put several bricks in a line and count what is the maximum number of bricks yielding a line that is still not longer than the rod. Suppose the number is 3, with the rod still longer than three bricks. Then one may perform a second count with smaller bricks (or marks on the original bricks). Suppose one can use a second pile of bricks such that a line of ten short bricks is of equal length as one original brick, and that now the maximum number of bricks in a line not longer than the rod is 31, with still a difference left. One may go on with such counts, and find out that they can be expressed in terms of the original bricks by a number starting with the following digits: 3.14159...
Statements like: ”The rod is 3.14159... bricks long”, or: ”The car drives at 70 kilometers per hour” can be rendered with a predicate involving a relation between three nodes: an object, a real number (including integers), and a unit of measurement whose dimension (length, weight, etc.) is specified by the predicate. Writing ”Q” for the predicate rendering the measurement result, ”r” for an arbitrary real number, ”u” for the unit, and ”P” for the predicate identifying the object to be measured, one gets: ∃x(Px ∧Q(x, u, r)) (10) Along with counting, the language of practical measurement has become a vital part of global culture - without it, we could not share the technologies that increasingly shape the world we live in. The language and practice of measurement are so pervasive in the contemporary world, and so important for purposes of integrated modelling, that they deserve careful analysis.
Practical measurement starts with the ability to make comparisons, e.g.
to say that something is longer than someting else. Among the objects that are amenable to such comparisons, they provide a strict ordering relation, i.e. a relation that is both asymmetric and transitive: ∀x∀y(x ≻y ⇒¬(y ≻x) ∧ ∀z(x ≻y ∧y ≻z) ⇒x ≻z) (11) If an object is neither longer nor shorter than another one, they stand in a tolerance relation to each other (Chajda 1991, Ungv´ ary 1983). A measure of length, however, is more than a tolerance relation.
Understanding tolerance relations is essential when looking at the relation between math-ematical and non-mathematical domains of discourse. To talk about tolerance relations in general terms, one needs a suitable domain of discourse. Such a domain can be introduced by taking the relations one is interested in as objects and characterizing them in terms 18 of further properties and relations. For example, one can say that a tolerance relation does not necessarily have the property of transitivitiy: One may not see any difference in length between a sequence of rods presented pairwise, but clearly perceive that the first rod is longer than the last one.
When engaging in practical measurement, we refine the judgements leading to tolerance relations by relying on a series of practical abilities, skills that have been developed over generations. They include the ability to produce measurable objects by combining other measurable objects - as when we put bricks in a line -, or by considering the difference between two such objects - as when we consider that part of a rod that extends beyond an adjacent brick. I write ⊕for the combination of objects, and ⊖for the forming of differences; I define ω as the empty difference, resulting when we compare an object with itself (as usual, I define a name by using the identity predicate preceded by a colon): ∀x∀y∃u∃v (x ⊕y = y ⊕x = u ∧ x ⊖y = v ∧(x ⊕y) ⊖y = x x ⊖x = y ⊖y) ω := x ⊖x (12) Of course, the ability to manipulate measurable objects is useful only because it relates to the results of the measurement activities. The key link between the two is provided by the fact of life that if one combines a given object one after the other with two objects that differ in length, the difference shows up again in the resulting combinations: ∀x∀y∀z(x ≻y ⇒x ⊕z ≻y ⊕z ∧ z ⊖y ≻z ⊖x ∧x ⊖z ≻y ⊖z) (13) In mathematical terms, the resulting structure can be described as a lattice ordered group (comparing objects provides the lattice order, combining objects into new ones provides the group structure). But keep in mind that so far I am not referring to mathematical, but rather to physical objects (this is also the reason why I am not focussing on mathematical measure theory). To discuss the link to the real numbers, I need a few more steps.
In practical measurement, we also rely on our ability to subdivide measurable objects into smaller ones - as when we put a mark on a brick to obtain a finer scale of measurement.
One can easily describe this ability as follows: ∀x∃y(x = y ⊕y) (14) Moreover, we treat something as a single measurable property only if we can trust that the relevant objects are commensurable in the following sense: by taking an arbitrary measurable object and combining it with itself again and again, one can surpass the measure of any other given object after a finite number of steps. To express this so-called Archimedean property one needs a predicate ”...A...” (for xAy read: x is a multiple of y) saying that with regard to the comparison under consideration an object x is equal to a a smaller object y combined with itself a finite numbers of times. As usual, I define a 19 predicate using the ”if and only if” connective preceded by a colon (I use three dots to indicate that the binary operation ⊕is repeated a finite numbers of times): xAy :⇔x = y ⊕y... ⊕y (15) With this predicate one can then express the Archimedean property: ∀x∀y∃z(z ≻x ∧zAy) (16) Finally, in practical measurement we trust that at least in principle we could make our measurements as fine-grained as we wish. I express this principle with the help of an operation for best approximation: x ⋆y (read: the best approximation of x by y) is the sequence of iterations of y that is the longest sequence still not longer than x: z = x ⋆y :⇔(zAy ∧∀u((uAy ∧u ̸≻x) ⇒u ̸≻z)) ∀u∀v∃x(v ≻u ⊖(u ⋆x)) (17) Subdividing and combining things so as to get further commensurable things, securing the Archimedian property, being able to make finer and finer comparisons - it is this whole combination of practical skills and orienting principles that enables us to make the step from tolerance relations between physical objects to measurements associating these objects with real numbers expressing their length.
To do so, we pick a unit of length - perhaps an iron rod placed in a building in Paris, or a more subtle device described in widely accepted documents - and express the length of objects in terms of this unit.
If we choose another unit, we can then convert the measurements performed with the first one. All we need to do ist to divide these mea-surements by the length of the second unit as measured in terms of the first one. This simple transformation is due to the rich structure described above.
I have discussed practical measurement taking the example of length. A similar account can be made for the measurement of time, although here one needs a particular concept of repetition: putting two rods in a line is not the same as repeating a periodic process like the swing of a pendulum. Other measurements deal with mass, electrical charge, and many more physical properties. Sometimes the idea of combining units is straightforward, as with mass, sometimes it is more subtle, as with mass per volume or with temperature.
I give a brief look at these cases because - due to their subtleties - they are useful to assess the role of practical measurement in other domains of discourse, in particular with regard to the role of measurement units.
As an example for comparing masss per volume, consider the case where a piece of solid matter - say the crown whose gold content Archimedes had to assess - is put in a recipient containing some liquid in which that piece of matter sinks to the ground. If this operation is repeated with a piece of some other substance, with the same mass as the original one, one can easily assess whether it displaces a larger quantity of liquid or not, thereby comparing mass per volume. (By taking a piece of pure gold of equal weight as the crown, Archimedes could prove that the crown was by no means made of pure gold.) To produce a practical measurement, one would need a seqence of comparisons where at each step the 20 mass per volume used for reference is split in smaller components. Of course, this cannot be achieved by splitting an object such as a piece of gold, as one might consider in order to measure mass. Weight per volume is not an extensive variable. One way of splitting it would be to look for a piece of solid matter such that it would be equal in mass to the original object, while its volume would be twice as large.
Temperature is similar to mass per volume in so far as it can be expressed as energy per particle - where both ”energy” and ”particle” may need to be further specified depending on the problem at hand. To split the temperature of a gas contained in a given volume, one can compress the gas to half the volume, cooling it down until its pressure is equal to the one in the original volume. To split the temperature of solids or liquids, one can use a volume of gas with the same temperature as the solid or liquid.
Temperature is particularly instructive because for historical reasons it is often expressed in physically weird units - weird because they do not allow for the simple transformation we know from meters and yards.
The reason, however, is not some odd property of temperature, but a peculiar system of introducing the measurement units. One would get the same situation with length if a distance of zero meters were defined as equal to three yards and a distance of one meter as equal to seven yards. If on the other hand one sticks to the absolute zero of physical temperature as the reference point, the same conversion as with units of length results.
I have looked at counting and practical measurement as two ways of associating physical objects to numbers - as special cases of measurement, if you like. They are not the only ones, although historically they have been the most influential. A third such link stems from the notion of probability.
It relies on a double sequence of counts, as practical measuring does. But in this case, at each iteration one counts an event that may happen or not happen - as when we ask whether tossing a coin will yield heads up or not. The event count, therefore, will never exceed the iteration count. If one labels the event count as v and the iteration count as n, then v n is the relative frequency of the event under consideration. When this relative frequency converges to an identifiable limit, this limit is usually called the probability of the event.
Based on probabilities, additional measures can be defined, like the one for entropy in statistical mechanics or the similar one for information in communication theory. Un-fortunately, much confusion has been generated by the fact that the word probability is also used in very different, but equally fruitful ways. They run through heterogeneous domains of discourse, ranging from mathematics to decision theory and other disciplines, and often they have little to do with relative frequencies. But as far as bridges between mathematics and physics are concerned, the frequentist account is the one that matters.
So far, I have discussed different kinds of measures - based on counting, practical mea-suring, and probability estimates - as linking mathematics and physics. They provide bridges between mathematics and chemistry as well, mainly on the basis of distinguishing different chemical substances and then measuring their physical properties.
In close relation with chemistry (Primas 1981), it would be interesting to analyse mea-surement processes dealing with quantum mechanical systems. There, counting acquires new relevance, because many quantities come in discrete steps only. Moreover, proba-21 bility plays a particular role. When considering, say, an electron, it is possible to give a probability distribution for its possible positions, and another such distribution for its possible momenta. These distributions, however, cannot be simply combined to gener-ate a joint probability distribution for position and momentum, as this would contradict the uncertainty relation between these two variables. And by now it is widely understood that in quantum mechanics measurement is by no means the kind of detached observation known from astronomy, but actually modifies the system under consideration. Discussing these matters, however, would require a far-reaching study adressing a whole range of unresolved issues. A few related remarks in section 7 will have to suffice.
6 Numbers in the Garden The domains of discourse used in biology rely on practical measurements of time, distance, mass, and other physical and chemical properties. Specifically biological predicates, how-ever, involve counting and probability rather than practical measurement. Examples are the number of cheetahs living today and the probability of survival for specific genes under specific environmental conditions. Without going too far into the world of biology, it is instructive to see how a biological domain of discourse can be linked to the mathematical notion of probability. First, consider a classification of kinds of behavior as agressive vs.
non-agressive. One may also establish some comparisons between more and less agressive behavior. This will result in tolerance relations of the kind discussed in section 5.
With regard to biological predicates, however, the combination of skills that enabled us to proceed from tolerance relations to practical measurement in physics don’t seem to be of much additional help. It is not clear how one would combine the objects to be measured into suitable new objects. Bricks can be lined up to yield longer objects of the same kind and split so as to yield shorter ones. With agressive behavior, it is not clear what it would mean to line-up or split items of such behavior so as to yield suitable new items. Suitable here means: the new items must be amenable to coherent comparisons of agressivity along similar lines to those I used for refining tolerance relations of ”longer than” until I reached a function expressing length as a real number (I will look at related difficulties when discussing human vulnerability in sections 8 and 9).
What is often quite fruitful is to study the probability - in the frequentist sense introduced in section 5 - that some animals will display agressive behavior under given circumstances.
It may then be useful to develop a different way of comparison by saying that one animal is more agressive than another one if it has a higher probability of agressive behavior under given circumstances. Along these lines one can then describe at least some behavioral traits as continuous variables.
As a second example for the relevance of probability measures with regard to biological domains of discourse, consider the frequencies of different genetic sequences in a popula-tion of organisms at a given moment in time. In complex ways, different genetic sequences govern different traits of living organisms. The frequencies of sequences as well as those of traits develop through an interplay between the three fundamental principles of Darwinian evolution: reproduction, mutation, and selection (Page and Nowak 2002).
22 The frequency of a given genetic sequence in the population changes with reproduction.
Via mutation, a given sequence can be transformed into another one; via selection, the relative frequency of certain sequences may increase at the cost of other ones. The selection process depends both on environmental conditions and on the relative frequencies of different sequences with the organismic traits they govern. Along these lines, evolutionary domains of discourse involving relative frequencies of genetic sequences and organismic traits can be linked to mathematical ones involving probabilities.
Evolutionary game theory (Maynard-Smith 1982), Lotka-Volterra type population dynamics (Lotka 1920, Volterra 1926, May 2001) and further approaches to Darwinian dynamics (Michod 2000) take advantage of this possibility.
For such a link to be possible, the evolutionary domain must have certain non-trivial characteristics. For example, there must be distinct genetic sequences governing different traits of organisms, these traits must have different - possibly interdependent - reproduc-tive potentials, changes in the environment must be slow in relation to the reproduction of organisms, etc. ”Nature” then influences the population of organisms living in a given environment much like a gardener breeding flowers with specific characteristics might do: certain characteristics are fostered, other ones are gradually weakened by environmental pressure. With suitable specifications, the operation of such a gardener can be modelled in terms of dynamic processes based on mathematical probabilities.
This kind of link between biological and mathematical domains of discourse is relevant for efforts to model vulnerability in biodiversity research. Consider an environment in which favorable and unfavorable conditions for the reproduction of specific genetic sequences alternate stochastically. Then one can define a variable for the evolutionary vulnerability of a genetic sequence as the probability of its extinction over a given time-span.
Of course, this is not identical to the probability of unfavorable conditions, as these do not necessarily lead to extinction, but it is related to it. Evolutionary vulnerability depends on the probability of unfavorable environmental conditions, but also on the mechanisms by which organisms can react to unfavorable conditions, and on the interaction between individual organisms.
The bridges between biological and mathematical domains of discourse are robust enough to allow for a rigorous analysis of evolutionary vulnerability. Such analyses can also be performed for geographically well-delimited territories and for well-defined changes of en-vironmental conditions. The latter possibility is particularly relevant in view of current threats to biodiversity from human activities. However, biodiversity analysis quickly leads to two hard questions. How can one compare the prospect of extinction of corals support-ing whole reefs with the prospect of extinction of some kind of cockroaches with peculiarly colored backs? And how can one assess possibilities of extinction by unprecedented en-vironmental changes for which no reliable statistical evidence is available? In section 9, I will introduce human preferences, allowing to address the first question, and subjective probabilities, allowing to address the second one.
Having looked, albeit briefly, at the relation between organisms and numbers, what about the relation between organisms and molecules? How are physical and biological domains of discourse linked to each other? To some extent, there is a rather straightforward over-lap. Organisms occupy reasonably well-defined space-time domains, and these can be 23 described with the predicates of physics and chemistry. Along these lines one then can measure the weight, speed, etc. of a given organism. However, the same organism can be described with biological predicates like the ones specifying species, sex, behavior, habi-tat, etc. Investigating inferences linking the two kinds of predicates is one of the exciting challenges of current research. The development of the biosphere has drastically changed the chemical composition of the atmosphere as well as many physical and chemical prop-erties of the surface of the Earth. On the other hand, the development of life clearly depends on physical and chemical processes.
Since the days of Aristotle, from time to time bold thinkers have wondered whether the language needed to talk about organisms - a language of birth and death, of growth and form, of chance and resilience - was not appropriate to talk about mountains and clouds as well. Still, few people think that physical predicates like mass and momentum could be explained in terms of biological ones. On the other hand, attempts to explain biological predicates - including the one for life itself - in terms of physical and chemical ones have been remarkably fruitful in some domains.
Whether the fruits of such efforts will be a reduction of biological predicates to physical ones is another question. A plurality of domains of discourse enables one to perceive, understand, and shape a much richer environment than one would recognize if one were bound to a single domain. If one were able to translate every word and expression from one domain into another one, there would be much less to be gained from their co-existence.
Since the origins of modern science, researchers have made amazing progress in unifying widely differing topics into comprehensive domains of discourse. By now, we may have reached a point where similar progress is warranted in taking care of the linkages - based on overlap, metaphor, and other means - between heterogeneous domains.
7 Changing logics So far, I have discussed heterogeneous domains of discourse as differing in the objects, predicates, and inferences involved, leaving the rules of inference in the background. Some logicians (e.g. Quine 1960) like to think that a unique pattern of inference forms the backbone of human knowledge, and that this pattern is made explicit by classical logic.
Others (e.g. Specker 1960, Putnam 1968) are less sure. As an illustration of the problems involved, consider the pattern of inference implicit in the following reasoning.
”The Greeks and the Trojans hope to win tomorrow’s battle, and the Greeks will win or the Trojans will win.” Therefore: ”The Greeks and the Trojans hope to win tomorrow’s battle and the Greeks will win, or the Greeks and the Trojans hope to win tomorrow’s battle and the Trojans will win.” The relevant pattern of inference is represented by the transition from the first to the second formula in the following scheme (S, S′, S′′ stand for arbitrary sentences): 24 S ∧(S′ ∨S′′) (18) (S ∧S′) ∨(S ∧S′′) The scheme is quite similar to the one implied by the equation 3·(2+6) = (3·2)+(3·6).
Akin to the arithmetic operators for multiplication and addition, the logical operators for conjunction (∧) and alternation (∨) are assumed to follow a distributive pattern. In fact, such is the use of these operators in standard logic. In the domain of discourse of quantum mechanics, however, it is quite standard to use rules of inference that do not allow this distributive inference without additional justification. Otherwise, false predictions for the behavior of photons, electrons, etc. follow.
As Putnam (1968) emphasizes, this is not an arcane problem whose relevance would be restricted to the study of sub-atomic particles. The example of the battle used above is instructive here. It is a fact of life that even if one has no doubt that there will be a battle tomorrow and that the battle can only end with either the Greeks or the Trojans winning, the outcome of the battle can still be open. In particular, it can depend on a myriad of human decisions not yet taken, decisions which will in turn create the context for new decisions.
In terms of logic, the sentence: ”The Greek will win or the Trojans will win” can be true in one of the following ways: the Greek will win, the Trojans wll win, or both will win (to keep things simple, I exclude the third possibility). If one knows the sentence ”The Greek will win or the Trojans will win” to be true, one needs not and cannot imply that one of its components - say: ”the Greek will win” - is true. If by a meaningful sentence one means a sentence that is either true or false, then ”the Greek will win” is not a meaningful sentence in this context. Somebody saying such a thing claims a kind of knowledge that is not available. It is not available because it presupposes a fully predetermined future, and this presupposition is not warranted (Feyerabend 1999, suggests ways to relate the -clearly distinct - universes of discourse of Homer and those of quantum mechanics).
Of course, many people claim that the future is fully predetermined even if we cannot know it in advance. In their lives, this claim may be quite helpful, it may enable them to keep their temper where others might lose it and to achieve things they would never have achieved otherwise. Still, it is a claim that is hard to justify once it has been challenged within some reasonably comprehensive tradition of inquiry. Clearly, there are religious traditions in which the claim of a predermined future plays an important role. More surprisingly, unrestricted use of standard logic implies the same claim. However, this claim is at odds with democratic culture - democratic institutions become meaningless if there is no freedom to take decisions about an open future. It is also at odds with the culture of scientific research - scientific research would be a sequence of superstititious rituals if experiments, as well as other research settings, would not depend on deliberate human choices (see Habermas 2002 for a thoughtful reflection on the interplay between religious, political, and scientific traditions of inquiry).
Of course, one can choose to simply ignore this problem. Conceptual inconsistency does not always hinder scientific progress.
But sooner or later much greater progress will become possible by addressing the problem than by ignoring it.
25 One way to proceed might be to use variants of quantum logic - i.e. a logic without the generalized distributive law - in domains of discourse dealing with human beings.
Quantum logic as such does not imply that the future is not predetermined, but it differs from standard logic by not implying the contrary either. Therefore, it may be interesting to investigate relations between quantum logic and another kind of non-standard logic, namely temporal logic (von Weizs¨ acker 1985). There, sentences are treated differently depending on how they are embedded in temporal sequences. The meaning of the phrase ”temporal sequences” is determined in part by common sense, in part by the logical patterns brought into play.
An important domain of discourse for temporal logic are computer programs, as these organize temporal sequences of connected events.
There are other possibilities for non-standard logics, and there are other reasons to look at standard logic in new ways. I will not explore these in depth here (see Agazzi 1981 for a survey), it suffices to mention a few examples.
Just as in quantum logic one relies on non-standard ways of combining conjunction and alternation, one can also use negation in non-standard ways. In computer science, e.g., there is often an important difference between two ways in which a program may fail to compute a certain result. On one hand, the program may compute some other result, on the other, it may engage in an endless flow of operations. One way of expressing this difference is by distinguishing between strong and weak negation. This can then be further elaborated by distinguishing between two ways in which a sentence can fail to be true, leading to logical patterns where sentences can be not only true and false, but also indeterminate.
The last option can be used to expand on the problem of the distributive law. The sentence ”The Greeks will win tomorrows battle” can be treated as meaningful, but indeterminate, and the compound sentence ”S ∨S′ ” can be treated as true in case its components are both indeterminate. Another reason for studying non-standard logics is the importance of intentions and norms for human life - both are hard to deal with in standard logic.
In general, non-standard logics have been developed by exploring both non-standard uses for standard logical symbols - including the quantifiers - and by introducing non-standard symbols, like symbols to say: ”it is possible that”, ”for all actual objects”, ”x believes that”, etc. The fate of Euclidean geometry can help one thinking about these innovations in logic. Today, Euclidean geometry is still considered a milestone in the history of human thought, it still invites careful study and provides fascinating research problems, but it is one variant among several. Very few mathematicians would treat it as the one and only appropriate way to describe the physical spaces human beings live in. In the case of logic, over the past decades many versions of logical reasoning have been designed and investigated. A single domain of discourse may be investigated with different logical frameworks, and none of them is likely to be the single best choice in all cases.
Arguing about logic is a subtle affair, however. What rules of inference shall one use when arguing about rules of inference? Once more, this question points to the relation between tradition and critical debate. After all, doubts about rules of inference can only arise in a given historical situation, where some rules of inference are already in use. This clearly holds when some irritating result casts doubt on seemingly well-established rules of inference.
26 As an example, consider Russel’s famous paradox (see the ”no-universe” proof in section 4.1 for further illustration). Cantor had developed set theory on the basis of a rule of inference according to which one could always form a set of those objects satisfying a given predicate. If ”P(...)” is an arbitrary predicate, then, in the following scheme one can infer the second formula from the first: ∃x(P(x)) (19) ∃y∀x (x ∈y ⇔P(x)) Russell showed that this leads into a contradiction when P(x) is taken to be ¬(x ∈x).
In a nutshell: ∃y∀x (x ∈y ⇔¬(x ∈x)) y := {x : ¬(x ∈x)} y ∈y ∨¬(y ∈y) y ∈y ⇒¬(y ∈y) ¬(y ∈y) ⇒y ∈y (20) The last two lines, both of which follow from the definition of y, are clearly contradictions.
The problem has been solved by distinguishing predicates that correspond to sets from predicates that do not. For arbitrary predicates, then, Cantor’s rule of inference has been modified so as to gear it to some predicate conferring sethood (P ′ in the scheme below): ∃z∀x (x ∈z ⇔P ′(x))∃x(P(x)) (21) ∃y∀x (x ∈y ⇔(P ′(x) ∧P(x)) Another kind of doubts about rules of inference arises when one is faced with different groups of people subscribing to different rules of inference. As an example, consider the relation between standard and fuzzy logic. In standard logic, sentences are treated as being either true or false. In fuzzy logic, sentences are assigned degrees of plausibility represented by real numbers between zero and one. A proponent of fuzzy logic might argue that standard logic misses a key feature of human thought, namely the fact that our concepts fit reality to varying degrees under different circumstances, and that one needs ways to deal with the resulting differences in plausibility of whole sentences. She might add that standard logic is a special case of fuzzy logic, namely the one where all relevant sentences are either totally plausible or totally implausible. And she might highlight the practical use of fuzzy logic for a whole range of engineering purposes. To this a proponent of standard logic might reply that the vagueness of everyday concepts is a problematic state of affairs and that rather than reinforcing it one should try hard to reach the conceptual clarity of standard logic. As for practical use, he could mention specific technologies relying on standard logic as a key design tool, and he might argue that standard logic greatly fostered scientific progress.
When trying to make up one’s mind about such debates, one interesting question is whether rules of inference proposed for some domain of discourse can also be used for domains of discourse consisting of sentences about the former one.
Are the rules of 27 inference proposed for the study of electric circuits also appropriate for the study of sentences describing electric circuits? It is one of the great strengths of standard logic that it meets this test.
To take an extremely simple example, consider the classical rule deducing the sentence ”B” from ”If A, then B” and ”A”. This rule of inference is equally appropriate when discussing electric currents as when discussing sentences about electric currents. ”If the switch is on, the current flows; the switch is on; therefore, the current flows” is an example referring to currents. ”If the sentence ’the switch controls the current’ is true, then its negation is false; the sentence is true; therefore, its negation is false” is an example referring to sentences about currents.
Now take the case of fuzzy logic applied to the same electric currents. Suppose that the sentence ”The voltage along this circuit is 7V” is assigned a plausibility of 0.7. Then according to fuzzy logic the sentence ”The voltage along this circuit is not 7V” is as-signed a plausibility of 1 −0.7 = 0.3. Next, turn to the domain of discourse consisting of sentences about electric currents. For fuzzy logic to work in the domain of electric currents, one cannot assign plausibility 0.7 - or any other plausibility except 1 or 0 - to the sentence expressing the arithmetic of plausibilities used before. The same holds for all other sentences expressing rules of fuzzy logic. In this respect, therefore, one must revert to standard logic. This argument is no objection against the use of fuzzy logic when studying electric currents, but it is an argument against the claim that fuzzy logic is a viable alternative to standard logic when studying sentences.
Debates about rules of inference, then, are necessary and possible. Over two millennia, they have led to the mathematical formulation of standard logic in the 19th century.
Subsequent debates have produced a situation where standard logic, while still treated as the main point of reference, is complemented by other varieties in a somewhat pragmatic fashion. This pluralism blurs the distinction between general rules of inference and rules for the use of specific predicates. For example, the history of the membership predicate shows how debates about rules of inference can be tied to specific predicates, as opposed to an emphasis on patterns of arguments that are independent from any particular predicate.
Debates about rules of inference do not follow fixed rules of inference, but explore tentative modifications of rules given by some tradition or other. To do so, one must be able to describe the relevant traditions as dynamic entitities, projecting them into the future, showing how the proposed modifications take care of important strengths of given rules while amending critical weaknesses. This kind of debates cannot be framed in terms of a given domain of discourse, and therefore it cannot be carried out in a formalized language.
If one could rely on a single set of formalized rules of inference, all one would want to say about these rules might be said in the same formal language used to express the rules themselves. To some extent, such an elegant form of self-reference is possible with standard logic. This is the modern version of the classical image - shaped by Plotinus, Spinoza, and others - of the world as expressing a single logical structure (Carnap, 1928).
As far as this logical machinery works, one can then describe domains of discourse simply as sets of objects, taking the rules of inference for granted, identifying names with individ-ual objects, and predicates with subsets of a universal set that can be taken for granted 28 as well. Therefore, the universal set cannot contain all the sets of modern mathematics’, because, as discussed in section 4, these don’t fit any universal set. Moreover, the names and predicates to be used must be given in a clear-cut way. The ambiguity and openness of ordinary language must somehow have evaporated before logical argument begins.
As a matter of fact, this ambiguity and openness is put aside whenever one can rely on what Brandom (1994, p.440ff) calls canonical designators. These are names used by some community to refer to objects whose existence and identity can be taken for granted.
”The sun” and ”the Earth” are such names in astronomy, ”hydrogen” and ”oxygen” in chemistry, ”the U.S.” and ”China” in political science, ”1” and ”2” in arithmetics, ”Plato” and ”Aristotle” in philosophy, etc. Without canonical designators, there would be no domains of discourse amenable to logical argument.
Canonical designators, however, come with whole lists of predicates and inferences at-tached.
Talking about domains of discourse as given sets of objects makes sense for particular purposes, but it already presupposes the availability of canonical designators.
This is the reason why I talk about domains of discourse as linked not just to some range of objects, but to a language game and the life form that comes with it. The ability to refer to objects by canonical designators is a key resource shared by those exchanging arguments within a given domain of discourse. This resource already draws on a range of rules of inference, and these rules are neither cast in stone nor completely uniform across domains. From time to time, canonical designators may change, and so may the rules of inference used in some domains.
It takes the resources of ordinary language, in particular its ability to operate as its own meta-language, and its potential to generate new meaning through metaphor, to develop rules of inference in a reasoned debate. In this sense, logical argument is embedded in broader forms of reasoning, where as participants in a debate we ”make up the rules as we go along” (Wittgenstein 1953, para. 83) - not in an arbitrary way, but in a creative one.
There is no way of establishing a universal domain of discourse in which all exchanges of arguments could be embedded. But there are ways of story-telling that work as ”ways of world-making” (Goodman 1978), of creating logical spaces for exchanges of arguments -and for their own future transformation.
8 Modelling social systems Rules of inference can be amended, and so can the properties and relations used to identify objects of discourse. For purposes of computer modelling, this is often essential.
Without a reasonably well specified domain of discourse, computer modelling is pointless.
Often, modelling begins with some vaguely circumscribed problem and without a suitable domain of discourse already at hand. As long as the relevant predicates are not clarified and specified so as to yield such a domain of discourse, computer models are bound to be about as meaningful and about as problematic as older forms of superstition.
This problem is particularly relevant for the task of modelling social systems. There are two different challenges to be met here. First, the words and concepts used to talk about social systems are much closer to ordinary language than those used to talk about the 29 systems studied by astronomers, engineers, etc. - basically because language is itself a key ingredient of social systems. Second, the words and concepts used to talk about social systems in contemporary culture are fraught with conceptual confusions of a kind that seems particularly hard to deal with. To take but one major example, after centuries of scholarly debate, story-telling by poets and playwrights, and explorations and suffering by ordinary people, the relation between passion and rationality still provides an incredibly difficult beast to wrap one’s head around (Frank 1989, provides an entry point into the literature on this topic). Before this background, I will now discuss main linkages between social and mathematical domains of discourse.
There is a huge variety of predicates expressing all sorts of social properties and relations.
They range from concepts of gender and kinship relations (like daughter, uncle, wedding) to concepts of broad collectives and their organization (like nation, power, parliament), from personal characteristics (like introverted, team-player, leader) to historical categories (like traditional, urban, post-modern). All these predicates can be used to identify count-able objects of social research. And as I have discussed with regard to physical domains of discourse, frequency counts open the doors to probabilistic analysis. The possibility of counting then provides valuable bridges between social domains of discourse and mathe-matical ones. These bridges can be greatly enlarged because objects - individuals, actions, institutions, events, etc. - identified by means of social predicates can be further described with biological and physical predicates. Distinctions between male and female as well as differences of age come to mind. Nearly as important are physical distances: population density, e.g., is a variable of huge importance for the dynamics of social systems.
With regards to modelling social systems, then, a wide variety of predicates can be used to build domains of discourse suitable for modelling in a more or less ad hoc way. Often these domains will combine specifically social predicates affording possibilities of counting with physical predicates affording possibilities of measuring. Countless mathematical models of specific phenomena have been built along these lines, in particular for purposes of statistical analysis. Broader social theories, however, are usually not amenable to formal modelling because they do not provide sufficiently well-defined domains of discourse.
As an example, consider Max Weber’s (1920) theory of the protestant ethic. A key claim of this theory is that the protestant ethic was a necessary condition for the development of capitalism. What are the objects constituting the relevant domain of discourse? They may be certain - all? - human actions, physical persons, institutions, systems of beliefs, and more. What are key predicates? Perhaps a predicate stating that somebody believes wealth to be a sign of divine grace, or a predicate stating that a certain sequence of actions is motivated by the belief that every human being has a calling to fulfill in her or his life, or a predicate describing the prevalent denomination in specific economic regions. What are rules of inference? Perhaps one can infer from the fact that somebody believes an action to be appropriate under specific circumstances that she or he will take that action under these circumstancen - and perhaps not.
What these instances of vagueness and ambiguity show is not that the theory is false, or meaningless, or uninteresting. What they show is that the theory can only be understood and used by somebody who is willing and able to settle all sorts of indeterminacy by exerting reasonable and sometimes creative judgement on a case to case basis, informed 30 by a wealth of background knowledge that is even harder to express in terms of a domain of discourse. Developing that kind of judgement is essential in all scientific disciplines, but particularly so in disciplines like the social sciences, where mechanistic procedures rarely provide reliable guidance when dealing with the problems that these disciplines tackle.
Building a mathematical model in line with a theory like the one of the protestant ethics is not impossible, but it requires one of two things. The first possibility is to drastically simplify the theory to the point of preserving only some rudimental features. For example, one may use a domain of discourse where the relevant objects are nations between the years 1500 and 1900, and where these objects are described by two predicates. First, nations may be protestant or not protestant, and second, they may experience a transition from non-capitalist ways of life to a persistent process of capital accumulation.
The theory might then be expressed by two inferences: If the first predicate is false, then the probability for the second predicate to be true is close to zero; if the first predicate is true, then the probability for the second predicate to be true is larger than half. Interesting as such a statement may be, it is clear that it misses most of the claims and insights conveyed by the theory under consideration.
The second possibility is to elaborate the theory further to the point where a domain of discourse that can be linked to a mathematical one becomes available. This may be quite rewarding, but it certainly is very demanding. It may well require substantial advances in both the social science theory under consideration and the mathematics to be used for modelling purposes. Before this background, it is understandable that up to now attempts to formalize broad social theories have been blessed with rather limited success.
As a result, in the social sciences one often finds a mixture of theories that do not come with a domain of discourse amenable to mathematical modelling, and models - usually statistical in character - that can be used to test specific empirical hypotheses, but that do not convey the main thrust of the theories on which they rely. I call the resulting style of analysis ”semi-formal theorizing” - without implying that this is only a preliminary stage on some supposed route towards the holy grail of ”fully formal” theorizing.
Still, there are two instances of highly elaborated and extremely influential formal theories dealing with social domains of discourse: the theory of formal languages on the one hand, economics on the other. I consider them in that order.
The theory of formal language has been remarkably successful in linking social and math-ematical domains of discourse by refining linguistic predicates like ”word”, ”sentence”, ”verb”, ”noun”, etc. Building on Chomsky’s (1957) pioneering idea of a generative gram-mar, researchers in linguistics, computer science, and mathematics have engaged in this endeavour with great enthusiasm. Surprisingly, perhaps, their main impact has been on software engineering. To what extent this theory is appropriate for the study of natural languages is more controversial (Baker and Hacker 1984), but it certainly has inspired a remarkable body of linguistic literature. At least as important for our present purpose is the fact that the various versions of formal logic currently available can be studied as formal languages. In principle, this opens the door to models of human knowledge.
Such models are currently explored in the scientific communities dealing with Artificial Intelligence and with Multi-Agent-Simulations.
31 A formal language is defined as a subset of all possible strings written in a finite alphabet.
To see the relevance of this definition for natural languages, compare the strings ”ac-quainted”, ”vertraut”, ”acvarsti”. How does somebody who is acquainted with English and German recognize the first word as English, the second as German, and the third as neither English nor German? Clearly, different languages somehow separate admissible from inadmissible strings in different ways even when they can be expressed by similar or identical alphabets.
In view of the theory of formal languages, it is a remarkable feature of finite alphabets that any string written in any finite alphabet can be unambiguously translated into an alphabet with a single sign. As an example, I translate the string ”she” - based on a standard alphabet with 26 characters plus a blank sign - into an alphabet with a single sign plus a blank. First, I associate to each letter of the original alphabet the number of its position in this alphabet. Then I associate the sequence of prime numbers (2,3,5,...) to the first, second, third, etc. position in the string under consideration. Now I set each prime number to the power of the number corresponding to the sign at that position. The product of these numbers can then be used to encode the text in a single number that can be unambiguously decoded again. (Numbers unambiguously representing arbitrary finite texts written in a given finite alphabet are called G¨ odel Numbers - after the math-ematician who used them to crack some hard problems in mathematical logic.) Finally I write the resulting number as a simple string of as many marks of the one-sign alphabet as correspond to that number. An example is given below.
Text: ”she” Encoding: ”s” 7→19, ”h” 7→8, ”e” 7→5 Signs to numbers: 19, 8, 5 Positions to prime-numbers: 2, 3, 5 Product of prime-numbers to the powers of numbers: 219 · 38 · 55 G¨ odel number of text: 10’749’542’400’000 Representation by one-sign alphabet: IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII...
(10’749’542’400’000 marks) Table 1: Encoding words with numbers By this procedure, a formal language - defined as a subset of possible strings from a finite alphabet - can be analysed as a subset in the set of natural numbers. When studying natural languages, finite alphabets come into play with written texts. In spoken language, an analogous role is sometimes played by phonemes, i.e. specific sounds that can carry meaning in the language under consideration.
Once strings of signs have been associated with natural numbers, two closely related questions arise. First, how can strings belonging to a given language be distinguished from other strings, and second, how can they be generated?
32 As for the first question, it is possible to define generative grammars, i.e. functions that transform a set of acceptable strings (i.e. G¨ odel numbers) into a larger such set (i.e.
further G¨ odel numbers). These functions are computable in a strict sense. To highlight the importance of this feature, consider the seemingly simple function that maps any rational number into 1, any irrational number into 0. There is no way to implement this function on a computer, because there are plenty of numbers for which it is not known whether they are rational or irrational, and there is no computational procedure to settle that question. For the functions defining generative grammars, on the other hand, it is possible to indicate procedures that compute their value for all acceptable inputs.
Starting with a finite set of initial strings, one can then produce an infinity of acceptable strings by repeatedly applying such a function to its own result. With sufficiently complex grammars, each string so generated can be characterized by some identifiable syntactic and phonetic structure. Acceptable strings can include propositions like: ”Yesterday I went to the beach” and non-propositional utterances like: ”Ladies and Gentlemen, all the way from Michigan: Slim Shady!” As to the second question, for a given generative grammar it is possible to define functions with the following property: they take a given string as argument and either return 0 (or any predefined number) if the string is not acceptable according to that grammar, or a list of numbers that label the syntactic and phonetic structure of the given string. Until now, only small patches of natural languages have been successfully described in such terms.
On the other hand, the design and analysis of formal languages has made tremendous progress over the last decades. Many of them are sufficiently similar to natural languages to be extremely useful for structuring the interface between the mathematical structures implemented in computers and the human beings designing and using these computers.
The world of computers is particularly instructive when it comes to heterogeneity of domains of discourse. Computer scientists operate on a wide spectrum of issues, some of which are closer to the operations of the machine, some to the activities of the user, some in between. In principle, it would be possible to deal with issues across the whole spectrum in terms of a unique domain of discourse. This is so because the correspondence between formal languages and arithmetics works both ways: not only is it possible to transform any formal language into a computable function, it is also possible to describe computable functions by formal languages. One way to do so is to take standard logic and use the natural numbers as the domain of discourse.
The predicates and relations needed to design and analyze computable functions can then be introduced in this setting.
In practice, however, professionals working on different aspects of computing use quite heterogeneous domains of discourse. The tasks that some user must or wants to tackle by means of computers are described in very different languages than the steps performed by a computer shuffling single bits around in its memory. In computing as elsewhere, heterogeneous domains of discourse are an essential feature of the social division of labor.
After having looked at the theory of formal languages, I will turn to the second major formal theory of social systems, the one of economics. I take that term in a broad sense, so as to encompass not only the study of mechanisms closely tied to money, but also the impressive theoretical toolkit provided by decision and game theory.
To prepare that discussion, I give a closer look at the concept of vulnerability as currently used in 33 research on sustainable development (Adger and Kelly 1999, McCarthy et al. 2001, Smith and Klein 2003). It seems fair to say that this research is performed largely in the style that earlier in this section I have characterized as ”semi-formal theorizing”.
Consider anthropogenic climate change. Given the prospect of sea-level rise induced by future climate change, there are reasons to say that The Netherlands are more vulnerable to such change than Switzerland. And given differences in well-being and in population density, it is plausible that Bangladesh is more vulnerable than The Netherlands (Nicholls et al. 1999). Here, the word ”vulnerability” is not used in the sense of evolutionary vulnerability discussed in section 6, but rather as a concept of human vulnerability.
As Wittgenstein (1953) argued, the meaning of concepts is usually defined by examples.
Only from time to time is it possible and sensible to define concepts in terms of other concepts. Most concepts are best explained by showing how to use them under specific practical circumstances. Their further use is then shaped by these examples, but so as to leave room for creative interpretation later on. Borderline cases may arise where the people using the concept will have to decide how to apply it to these cases - a decision that is usually not taken by explicit deliberation, but by the shared development of a social practice.
In this sense, I introduce the concept of human vulnerability by examples. The examples show that vulnerability is used as a comparative concept, that can be expressed with a relation ”x is more vulnerable than y” (I will also write it as: ”x ⊢y”). I take this relation to be an ordering relation (see (11) in section 5). One can then define a relation for being neither more nor less vulnerable: x ∼y :⇔¬(x ⊢y ∨y ⊢x) (22) This is a tolerance relation like those discussed in section 5. As usual with tolerance relations, x ∼y need not be transitive: Perhaps there are no good reasons to say that Saudi Arabia is more or less vulnerable to climate change than either South Africa or Spain, but there may still be strong reasons to say that South Africa is more vulnerable than Spain.
Given the importance of physical measurements in contemporary culture, one is tempted to imagine that wehrever there is a comparative predicate - like: ”... is more vulnerable than ...” - there is bound to be some property that can be expressed by real numbers.
Tolerance relations are then understood to work as follows. The relevant objects have some quantitative property Q(x) that can be expressed by real numbers and there is some number ∆that defines a tolerance. And a tolerance relation T is based on that quantitative property: xTy :⇔Q(x) −Q(y) ≤∆ (23) But tolerance relations can be well-defined without any real numbers being involved. It may be perfectly reasonable to say that I am as grateful to Peter as to Paul while it would at best be a joke to say that my gratitude to them is 3.7 with a tolerance of 0.5.
In section 5, I have discussed tolerance relations - those of physical measurement - that can be refined up to the point where they refer to quantitative properties expressed by real numbers. In section 9, I will discuss under what conditions a tolerance relation for human vulnerability can be refined in a similar, but not identical way. The relevant conditons, 34 however, cannot be taken for granted in general, they only apply under rather limited circumstances. Therefore, tolerance relations are indispensable for vulnerability analysis.
Fortunately, in the case of human vulnerability the relevant tolerance relation has some additional structure. Small island states in the pacific are clearly more vulnerable to climate change than most other countries in the world, even if they might survive as states one way or another. And states like India or Brazil, while being less vulnerable than small island states, may still be more vulnerable than, say, Norway or New Zealand.
The key features are the fact that we are talking about a finite universe of discourse (of states or regions) ordered by a tolerance relation and that no entity is incomparable with all other ones. We can describe the relevant structure as follows: ∀x∃y(x ⊢y ∨y ⊢x) (24) On this basis, we can identify countries (or whatever entities we are dealing with) with maximal vulnerability: they are more vulnerable than at least some other one without being less vulnerable than any other one. In the same vein, we can identify countries with minimal vulnerability, and treat the remaining ones as countries with intermediate vulnerability.
If some country were completely incomparable with any other country in terms of vulnerability, that country would be simultaneously maximal and minimal, making that classification pretty meaningless - but this possibility is excluded by the structure just described.
Therefore, a rather robust kind of vulnerability analysis consists of defining three broad categories of vulnerability to climate change: maximal, intermediate, and minimal. Three, rather than two, four, or twelve categories because we are dealing with a tolerance re-lation with the structure just introduced.
No metric is required to identify maximal, intermediate, and minimal elements in a set ordered by such a relation. In practice, such identification may well be possible in a way that is not very controversial amongst re-searchers and stakeholders interested in the issue. It will require a structured dialogue sorting out the relevant arguments as far as possible and accepting that a few ambiguous cases be resolved in an arbitrary way. Along these lines a predicate of human vulnerability can be introduced in the spirit of semi-formalized theorizing discussed above.
Just as we have looked at countries, we can look at regions, at social groups and eventually at single individuals. Paying special attention to individuals is characteristic for economic theory. It is particularly appropriate here, because it can help to investigate issues of social solidarity that are intimately related to human vulnerability.
9 Human Vulnerability and Global Change Consider an individual - call her Bianca - operating in a reasonably stable environment with some stochastic features. By luck, reasoning, imitation or whatever means, she has found a way of action she cannot improve upon with the means at her disposal. Now suppose something changes in Bianca’s environment.
To stick to the climate change example, two severe floods may happen in her surroundings within a surprisingly short timespan.
35 How shall we - researchers interested in human vulnerability - assess Bianca’s vulnerability after these events? As she is a human being just like ourselves, and perhaps knows more about her life than we do, we could start by asking her. She might answer that she feels scared because people have died in these floods. We could observe her, and we might notice that she starts following the daily weather reports much more carefully than she did before. We could ask ourselves whether we consider these reactions of Bianca’s to be appropriate. Given our understanding of what it means to die in a flood as well as our knowledge about those two floods, we might conclude that her reactions are quite appropriate. On the basis of these facts, we may well say that Bianca is more vulnerable than she was before.
So far, the description of Bianca’s vulnerability still looks quite similar to the tolerance relation for human vulnerability discussed in section 8. However, now I have explicitly introduced the ”subjective” view of a human individual, and balanced it with the ”ob-jective” view of an external observer. This provides the ground for assessing under what conditions and how far one can move from a tolerance relation for vulnerability towards a measurement of vulnerability in terms of real numbers (Downing et al.
2001).
In global change research, vulnerability is usually seen as a function of impact and adaptive capacity, with impact in turn a function of exposure and sensitivity (McCarthy et al.
2001, Schr¨ oter et al 2003). Accordingly, I will proceed in four steps, addressing exposure, sensitivity, adaptive capacity, and vulnerability.
9.1 Exposure When studying human vulnerability, exposure to the risk of dying certainly matters. But of course we are all bound to die some day: what really matters for vulnerability analysis is untimely, cruel, unjust death. Now these are all predicates that different people will use in very different ways - one may say that they express a strong subjective component.
I will come back to this point, for the moment I simply take it for granted that Bianca would prefer not to die in a flood and that most people have no difficulty in understanding this preference of her.
Untimely death is by no means the only event to be considered when analysing human vulnerability, but it certainly is an important one. And once such a critical event has been identified, human vulnerability can be analysed in terms of the probability of that event (Parry et al. 2001).
This enables one to move beyond the tolerance relation introduced in the last section, and also beyond the distinction between maximal, minimal, and intermediate vulnerability.
To some extent, one can now start working with a quantitative concept of vulnerability, as the probability of a critical event can cover the whole range between zero and one.
Writing ”c” for the critical event and ”¯ π(c)” for the probability of that event, one gets a simple definition for a preliminary concept of exposure that can be expressed in terms of real numbers between 0 and 1, namely E := ¯ π(c).
As in the study of other living beings, probabilities provide a link between a domain of discourse dealing with human beings and a mathematical domain. In the mathematical domain of probability theory, probability is a function from a set of sets into the interval 36 [0,1] of real numbers. The function must satisfy several requirements, in particular: the function of the union of two sets must be equal to the sum of the function of these two sets. The set of sets must satisfy a few requirements, too, in particular: it must contain the union of two sets whenever it contains these two sets.
In domains of discourse dealing with human beings, I will use the word risk to refer to situations where a human being is faced with more than one possibility, at least one of which has a consequence that will make that person suffer (in decision theory, such situations are often labelled ”lotteries”). Notice that to talk about risks in this sense probabilities are not needed, nor does one need to quantify damages; however, some kind of preference between the different possible consequences is required.
As for probabilities, in domains dealing with human beings they come in two kinds: one relies on the relative frequencies discussed in section 5 (and to be further discussed in the present section), the other relies on the subjectivity that distinguishes humans from other beings. Subjective probabilities are not an easy topic, but an extremely important one. I approach it by considering the difficulty of measuring exposure to various stresses in terms of probabilities of critical events.
Non-probabilistic data can be vital for describing actual suffering, but they are of little use for describing potential suffering. Vulnerability does not refer to actual wounds - taking wounds as metaphors for suffering - but to potential ones. Finding the data required for a probabilistic description is no trivial task. Only this kind of data, however, provides a reasonable basis for moving from a tolerance relation for vulnerability towards a variable expressing vulnerability in terms of real numbers.
Some steps in that direction are possible even where the probability of the critical event is not well known. Imagine that Bianca must choose one of two roads for some journey by car. There is an important difference between the two roads, however: one of them leads through a deep canyon prone to flash flooding, while the other one does not. She can die on both roads, be it by car accident or by drowning in the floods, and she knows that the relative frequency of people dying on the first road is much higher than on the second. The idea of a relative frequency that is somehow characteristic of a sequence is rather elementary for human beings - we all distinguish between fast and slow rythms, even when they are quite irregular. And even if Bianca cannot tell a number for the relative frequency of deaths along the two routes, she may still choose one road rather than the other. The argument can be refined with more complex cases, as when the same road is travelled more than once, and when various combinations of the two roads are feasible. This is needed to assess subjective probabilities in detail, but I will not discuss this here.
In order to describe Bianca’s behavior under these circumstances I introduce the following terminology. If in a sequence of events one can identify the relative frequency of some event well enough for practical purposes, I call that frequency a characteristic frequency.
This is not a mathematical concept, it sits in a domain of discourse dealing with human beings, not with the functions from sets to real numbers that are the stuffmathemati-cal probability is made of. In domains of discourse dealing with humans, characteristic frequencies express a frequentist view of probability.
37 Next, I consider different risks with the same possible outcomes and look at the pref-erences of some decision-maker - e.g.
Bianca - when choosing among them.
If these preferences lead to a ranking of the risks, I call the resulting ranks the preferability of the different risks. As the different risks under consideration all have the same possible outcomes, risk preferability is closely tied to the decision-maker’s knowledge of or guesses about characteristic frequencies of different risks. The nature of this tie, however, is far from simplel. Sometimes, the decision-maker knows a characteristic frequency and this knowledge shapes her preferences between risks. Sometimes the decision-maker has very strong preferences between risks - perhaps because of strong emotional experiences in the past - and these shape the characteristic frequencies the decision-maker takes into consid-eration. Sometimes, there is some mixture between these two extremes. In any case, risk preferability expresses a subjectivist view of probability. This is not to say that there is no objectivity here: risk preferability captures a relation between a decision-maker - in our case Bianca - and a range of risks. If Bianca prefers the risk of the first road to that of the second, then saying this is as objective as saying that she is more fluent in English than in French.
Mathematical probabilities can be used to model the interplay between characteristic fre-quencies and risk preferability. Characteristic frequencies can be represented by limiting values of relative frequencies, and these limits may be called first order probabilities. Risk preferability is more subtle, because here one must deal with the whole range of possible characteristic frequencies. It is often useful to represent risk preferability by using mathe-matical probability distributions defined over all possible first order probabilities, i.e. over the interval [0, 1]. The resulting probabilities may be called second order probabilities.
The distinction between first and second order probabilities is closely related to the influ-ential distinction between risk and uncertainty introduced by Knight (1921). According to that distinction, risk is characterized by well-defined probabilities of outcomes, uncer-tainty by unknown probabilities and only partial knowledge of outcomes. In this paper, I use the word risk to characterize situations with different possible outcomes. Risks can then be further characterized by first and second order probabilities. Knightian risk is the extreme case where second order probabilities select a single first order probability as the one that matters. The case where second order probabilities are uniform is as far as one can get in the direction of Knightian uncertainty in the present framework. Fascinating unresolved questions of decision theory and economics arise in this area. In particular, in some cases it can be quite misleading to assume that an agent acts on the basis of a single well-defined function of second order probabilities (Jaeger 1988). This is not always a big issue, however, as sometimes only rough features of second order probability distributions really matter Second order probabilities do not represent relative frequencies, they represent risk prefer-ability, and thereby propensities for action - different distributions of second order prob-abilities mean that Bianca will make different decisions when she has to select the roads she will travel. Preferences and probabilities are closely related: it is an important piece of received wisdom that under the circumstances Bianca is faced with in our example, any reasonable person will and should take the second road. This holds even if the choice is about a single action that will never be repeated, and where therefore it is hard to show 38 how and why relative frequencies matter at all (Putnam 1968 calls this ”Peirce’s puzzle” - after the founder of American pragmatism, who thought about these issues long ago).
By writing ˜ π(¯ π(c)) for: ”the density of second order probabilities of first order probabilities of a critical event” - remember: it’s Bianca drowning in a flood! -, one can combine first and second order probabilities and define exposure to a critical event as follows: Ec := Z 1 0 ¯ π(c) · ˜ π(¯ π(c)) d¯ π(c) (25) The resulting number is the expected value for a first order probability given a distribution of second order probabilities. Exposure to different events can then be described by a list of such expected values (I call it ¯ E). In order to aggregate them to a single magnitude, one needs to take into account human sensitivity to risk exposure. Before doing so, however, I will look into a key advantage of combining first and second order probabilities (for more background, see Kreps 1988, p.145ff).
This combination enables one to compute a meaningful measure of exposure in cases where little or nothing is known about relative frequencies.
When nothing is known, ˜ π(¯ π(c)) = 1 everywhere, because the decision-maker has no reason to put greater weight on any particular first order probability. The integral indicated above then yields: Ec = 1/2.
Incidentally, this suggests that with unknown first order probabilities, the probability of one out of two events happening twice in a row should be assessed not as 1/4 but as 1/3, because R 1 0 ¯ π(c)2d¯ π(c) = 1/3 - a point with some relevance for the assessment and management of environmental risks.
With this way of studying exposure, one can also take advantage of a mathematical result known as de Finetti’s theorem. For the present purposes, the main insight can be illustrated as follows. Consider sequences of events such that at each step one out of two possibilities is realized. At each step, one can count the relative frequency of each of these possibilities. Suppose a sequence looks like the following one: 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, ...
(26) The relative frequencies of ”1”s then look like this: 0, 1 2, 2 3, 1 2, 2 5, 2 6, 2 7, 1 4, 1 3, 2 5, 5 11, 1 2, 7 13, 4 7, 5 3, 5 8, 11 17, 2 3, 12 19, ...
(27) Relative frequencies may oscillate without ever converging to some limit (in the present sequence, characters in boldface mark turning points of an oscillation). Under such cir-cumstances, the frequentist concept of probabilities is hard to apply. Still, risk preferabil-ity works with non-converging sequences of relative frequencies, too: Bianca, like anybody else, is free to prefer one risk to another one, much as she is free to prefer apples to oranges.
Both for converging and non-converging sequences, her preferences can be expressed by means of suitably distributed second order probabilities.
De Finetti’s theorem turns out to be useful whenever Bianca is faced with a sequence that does have a limiting relative frequency. The theorem shows that she can systematically improve her second order probabilities in these cases by applying Bayes rule (see Jaynes 2003 for a Bayesian vision of logic, probability, and science in general). If Bianca improves 39 her second order probabilities each time she learns about a new event in the sequence, these probabilities will get more and more focussed on the single first order probability towards which the relative frequencies are bound to converge under the given assumption.
To take but one example, suppose that the critical event to be considered is an unusually high flash flood and that the last flash flood has been unusually high. Then second order probabilities can be improved by the following formula. (I write ¯ π′(c) for an arbitrary first order probability of the critical event, ˜ πt(¯ π′(c)) for the second order probability assigned to ¯ π′(c) at time t, and ˜ πt+1(¯ π′(c)) for the second order probability to be assigned one tide later.) ˜ πt+1(¯ π′(c)) = ˜ πt(¯ π′(c)) · ¯ π′(c) R 1 0 (˜ πt(¯ π(c)) · ¯ π(c))d¯ π(c) (28) Therefore, researchers can help Bianca even if their data are insufficient to justify an estimate of a first order probability. To take an extreme case: nobody would want to estimate a first order probability on the basis of a single event.
If however, Bianca entertains second order probabilities of a kind that give very small weights to large first order probabilities for large floods, a single large flood allows her to improve her second order probabilities by increasing those weights by a small, but well-defined amount.
For the analysis of exposure, this means that one first needs data on second order prob-abilities for the relevant agent. By observing her behavior and asking her directly, one can usually get a rough idea of how the agent assesses different possible characteristic fre-quencies. Even if the data will hardly be sufficient to fully specify a distribution of second order probabilities, they can provide a sufficient basis for showing how additional data, now relating to first order probabilities, can be used to get a more focussed distribution of second order probabilities. This interplay of data on second and first order probabilities makes a probabilistic description of exposure feasible in the rather frequent cases where a purely frequentist approach would be hopeless.
So far, I have argued that probabilistic data of exposure can be used to make a step beyond a simple tolerance relation for vulnerability by assessing the probability of critical events. That step becomes easier when one combines first and second order probabilities.
When studying exposure in these terms, however, one must check whether the people under consideration can really be characterized by single distributions of second order probabilities. If not, it may be appropriate to stick to tolerance relations for vulnerability as described in section 8.
Of course, different critical events engender different probabilities. Exposure then is a list of probabilities. It is easy to establish a tolerance relation between such lists by using the Pareto-criterion: exposure is greater in situation A than in situation B if and only if both situations involve the same critical events and all probabilities in situation A are greater than the corresponding probabilities in situation B. The attempt to aggregate the description of exposure by a list of probabilities into a single number leads to a next link between socio-economic and mathematical domains, and thereby to the notion of sensitivity.
40 9.2 Sensitivity Nearly always, people find the risk of death worse than the risk of loosing their property.
Of course, such a comparison is based on the assumption that the relevant probabilities are the same. When running over a street to stop a thief from running away with her bag, Bianca prefers a small risk of dying in a car accident to a high risk of loosing her bag to the thief. There is some sort of balance between her preferences between the two critical events and her assessment of the uncertainties she is faced with.
An additional point should be made in this context, however. There is a fundamental difference between death and the possibility of death; if somebody is willing to trade a small risk of dying for a big opportunity to get her bag back from a thief, that does not mean that she would be willing to accept death for any amounts of bags. The suffering from actual death is a different matter than the suffering from the possibility of death.
Now consider the sentence: ”Bianca prefers the risk of c2 happening with probability ¯ π(c2) to the risk of c1 happening with probability ¯ π(c1)”. I write it as: [c2, ¯ π(c2)] ▷b [c1, ¯ π(c1)] (29) It is often sensible to treat ”... ▷b ...” as an ordering relation. Writing x, y, z for different risks with probabilities attached, one gets: ∀x∀y∀z(((x ▷b y ∧y ▷b z) ⇒x ▷b z) ∧(x ▷b y ⇒¬y ▷b x)) (30) As discussed in sections 5 and 8, this kind of comparisons leads to a tolerance relation. But when dealing with critical events, one can do more. Suppose Bianca is faced with a choice between a road with car accidents, but no flash floods, and one with flash floods but no car accidents, and suppose that her second order probabilities are the same for both roads.
She may still prefer the road with car accidents. One may then say that she finds dying in a flood even worse than dying in a car accident (for the sake of the argument, I assume that in both cases the risk is one of death, and only one of death). With heterogeneous critical events, risk preferability may not only lead to subjective probabilities, but also to a further ranking, this time according to the agent’s sensitivity to the danger of different critical events.
Consider probabilities for which Bianca is indifferent between the risk of death and the risk of loosing her bag (indifference will show by her neither running after the thief without hesitation despite the traffic, nor clearly desisting from trying to catch the thief because of the traffic). Pairs of such probabilities may well be characterized by the same proportion (I come back to this point below). If so, one can use that proportion to describe a ranking of Bianca’s suffering from these two risks. Call two such probabilities ¯ π(c1)∗and ¯ π(c2)∗, take the first risk as reference point by setting s(c1) = 1 and define: s(c2) := ¯ π(c1)∗ ¯ π(c2)∗ (31) If one can identify a fixed proportion between probabilities that make Bianca indifferent between two kinds of risks, one can use that proportion to express her sensitivity to each 41 one of these risks by a single number, Bianca’s sensitivity to the critical event in question.
Subjective probabilities rank risks across different possible characteristic frequencies, sen-sitivity values rank them across different critical events. Bianca’s sensitivity values are real numbers because they are related to characteristic frequencies: they indicate the pro-portion between characteristic frequencies that would make Bianca indifferent between risks involving different critical events.
Sensitivities can be identified for the risk of loosing one’s life as well as for the risk of loosing one’s harvest, and in the latter case they can be identified for the risks of loosing various fractions of that harvest as well. It is plausible that loosing two thirds of the harvest is worse than loosing one third; learning at what probabilities Bianca will be indifferent between these two risks can enable one to find a number expressing her sensitivitiy to different losses. One can then introduce sensitivity as a function S defined over a set C of mutually exclusive critical events and yielding a real number S(c) for each critical event c ∈C: S : C ⇒I R, c 7→S(c) (32) A sensitivity function represents the way a decision maker assesses the suffering that would be experienced when a risk materializes, but also the anxiety experienced by a decision-maker living with a frightening risk. Therefore, an impact assessment by means of a sensitivity function takes the important, if often neglected, problem of anxiety into account.
The use of a sensitivity function does not imply a claim that suffering in general could be quantified, only that a specific human decision-maker will prefer some risks to other ones depending on the characteristic frequencies involved. In principle, a sensitivity function can have positive as well as negative values, where the latter may be read as expressions of happiness. This is the usual way of framing utility functions, where people can be more or less happy, but happiness as such is unattainable. Vulnerability analysis pays attention to suffering, and when doing so one may wish to pay attention to happiness as well. One way of doing so would be to identify happiness with the absence of suffering.
This would imply that suffering can be so transformed as to lead not only to further pursuit of elusive happiness, but to actual happiness (a claim sometimes associated with Buddhist traditions). In modelling terms, this would result in a function S(c) having positive values bounded below by an absolute zero, like temperature and other physical measures. In his seminal paper on optimal growth theory, Ramsey (1928) advocated such an approach long ago, arguing that economic growth might enable humankind to reach a state of bliss, as he put it. Vulnerability analysis based on sensitivity functions may or may not follow this route.
Human sensitivity to risk is different from the susceptibility of a piece of iron to rust under the influence of humidity. Human beings care about meaning, and this enables them to do wonderful as well as terrible things. We are animals suspended in webs of meaning that we ourselves have spun, to take up Geertz’s (1973, p.5) beautiful phrase.
Sensitivity is related to humans acting in the face of an uncertain future, it reflects the priorities human beings set in the face of various risks. These priorities are of course shaped by biological properties of the human body, but they are equally shaped by the webs of meaning that human beings keep creating and transforming.
42 For this reason, caution is required when working with sensitivity functions. The reality they are meant to represent is usually highly contingent on social context. As a rule, one should expect human decision-makers to draw on a whole set of sensitivity functions, where context as well as personal autonomy can determine which one of them will shape a given decision. Moreover, one should always be prepared to the situation where suffering can be described by a tolerance relation, but this relation cannot be translated into a meaningful sensitivitiy function.
9.3 Impact If and only if a sensitivity function can be identified, one can quantify the impact expe-rienced by Bianca or any other agent by combining exposure and sensitivity to different risks. The risks to be considered can then be represented by a subset of relevant critical events out of the total set of conceivable critical events. I write C∗⊂C for the subset of relevant events, and ”...⊗...” for the operation of combining exposure ¯ E and sensitivity S into a function indicating the impact I of a possible combination of risks as a real number: I := ¯ E ⊗S = X c∈C∗ Z 1 0 ¯ π(c) · ˜ π(¯ π(c)) · S(c) d¯ π(c) (33) When discussing global change, environmental research is required to identify the relevant subset of critical events, e.g., by assessing whether there is a risk of a one-meter sea-level rise from climate change in the next fifty years. Assessing the probability of that risk requires the combination of first order and second order probabilities discussed above.
Assessing the sensitivity of some agent to that risk requires data on the preferences of the agent in question when faced with choices between risks of sea-level rise and other risks.
An impact function of this type is a special case of the broader class of cardinal utility functions for decision-making under uncertainty, as developed in different variants by von Neumann and Morgenstern (1947), Savage (1954) and many others. The numerical value for the impact of different risks still has a clear meaning. It is rooted in the possibility that risks can be mixed by concatenating them. Faced with a choice between two roads, Bianca may roll a dice and take the first road if she gets a 1 or 2, the second otherwise (risks can be concatenated by destiny as well, without Bianca having to roll a dice).
Clearly, risk mixtures modify the probabilities she has to deal with. Sensitivity values indicate what characteristic frequencies would make Bianca indifferent between different risks. Impact values indicate what mixture ratios would make Bianca indifferent between mixtures of different risks.
The caveats expressed above first for exposure, then for sensitivity, mean that any impact function must be handled with considerable care. In impact assessments, monetary dam-ages are often taken as a proxy for sensitivity. That may or may not be a useful thing to do. The key point is to ask whose suffering is being assessed - it will then often be obvious whether monetary damage is a good proxy or not. When used in the wrong circumstances or in inappropriate ways, impact functions may well do more harm than good. In these cases, it is much better to work with the tolerance relation for vulnerability discussed in section 8. Sometimes, it may also be possible to work with tolerance relations for exposure 43 and for sensitivity. But these can be combined into a tolerance relation for impact only in the most obvious cases. Suppose that we wish to compare Bianca’s vulnerability in two situations characterised by two critical events, namely floods and droughts. Let the probability for both kinds of extreme events be higher in situation A than in situation B. Now suppose that Bianca’s sensitivity for the relevant risks can be characterized by a tolerance relation only. If the relation is such that she always prefers lower probabilities to higher ones, then the impact in situation A is clearly higher than in situation B. This may well be useful statement to make, but it is a long shot from a quantifiable impact.
On the other hand, in a given social context given human decision-makers often do show the sort of preferences over risks implied by a specific impact function, and making such a function explicit can help to predict, analyse and sometimes improve their decisions.
9.4 Adaptive Capacity A key issue in vulnerability analysis is the role of adaptive capacity (Burton et al. 2002, Mendelsohn and Neumann 1999). To assess the vulnerability induced by some form of global - or other - change, therefore, one must consider how Bianca - or any other agent -can shift from a pattern of behavior that was appropriate under earlier circumstances to one that is more appropriate under new ones. For this purpose, one needs some idea of Bianca’s opportunities for action, her adaptive capacity. It will depend on various kinds of resources, ranging from money to knowledge, from friends to social security. Of course, the critical events under consideration may well impair these resources, and therefore there may be considerable difference in adaptive capacity before or after some such event.
Hence the importance of environmental research providing - as far as possible - reasonable forecasts of critical events in the future.
To a considerable extent, Bianca’s resources will depend on the actions of other people.
Whether she has a job, how much she earns, how reliable her friends are, how much attention the state pays to people like her when they are in difficulties, etc. - the answer to all these questions depends critically on actions by others. When she is well endowed with such resources, a flood will be a very different event than when these resources are lacking. This raises the issue of social interaction and social solidarity, to which I will come back in section 9.5.
Adaptive capacity will rarely, if ever, have a meaningful representation as a single number.
In general, it is best represented as a function selecting feasible actions from a set of conceivable actions depending on a list of available resources.
I call the list of these resources r, the set of all possible lists of resources R, conceivable actions xi, and the set of all conceivable sets of actions P. Then one can represent Bianca’s adaptive capacity by the following function: A : R →P = {{x1}, {x1, x2}, ...}, r 7→A(r) (34) 44 9.5 Vulnerability Until now, I have looked at the consequences for Bianca of a given course of action in the face of various risks. If environmental conditions - including the social environment - change, Bianca may need to reassess her past choice of action. To analyze this choice, one must specify the probabilities for critical events as conditional probabilities. They are conditional on Bianca’s actions, but also on those of other agents. Remember that the critical event is not a hurricane, it is Bianca drowning in the ensuing flood, and the prob-ability of her drowning depends on her actions, too. With these conditional probabilities, one can then combine impacts and adaptive capacity to get Bianca’s vulnerability.
I write x for Bianca’s actions, ¯ π(c|x)) for the first order conditional probabilities of critical events, and ArgMin for the minimal value of the impact function. The minimum is taken over those actions that are feasible with Bianca’s resources, x ∈A(r). I write ”... ⊘...” for this combination of the impact function with the function for adaptive capacity and get: V := I ⊘A = ArgMin[x∈A(r)] X c∈C∗ Z 1 0 ¯ π(c|x) · ˜ π(¯ π(c|x)) · s(c) d¯ π(c|x) (35) Bianca’s vulnerability is the minimal value of the impact function over the set of her feasible actions. Whether her vulnerability increases as a result of some environmental change depends on the comparison between this value before and after the change.
Clearly, the feasibility of actions in the contemporary world often depends on the avail-ability of money - therefore, money is a key resource for adaptive capacity, too. Money provides a remarkable link between socio-economic and mathematical domains of dis-course. When we measure the length of an object like a table or a stone, the object is there and we ask how long it is. When we assess the relative frequency of heads in a sequence of throws of a coin, the sequence is there and we ask whether it can be char-acterized by a probability measure. Money as we know it, however, consists in numbers stored in peculiar ways. Money is not based on measuring some given thing one way or other, but on using numbers in a way defined by institutions like credit, accounting rules, and the like.
Of course, there are coins, and there were the times of the gold standard, when nuggets found in the mountains could be as good as coins of money. But today we live in a world where numbers stored in computers can bring governments down, without any relation with gold reserves. And numbers stored on computers determine to a considerable extent what options people all over the world do and do not have when dealing with their vulnerabilities. Whether these numbers could function even without the circulation of coins and bank notes remains to be seen.
Economic theory relies on two major links betweens socio-economic and mathematical domains of discourse. On one hand, there are utility functions, more or less similar to the index of suffering that we have used to describe human sensitivity to global change, and sometimes geared to subjective probabilities. On the other hand, there are production functions. They are used to represent physical production structures, like the amount of steel needed to produce a car. As far as they rely on physical properties of the goods 45 under consideration, they can use the broad and solid bridges between physical and math-ematical domains of discourse. Where they are formulated in more generic terms, as when capital is treated as a single magnitude, things are less clear. One way or other, monetary magnitudes must enter the picture, and a whole array of difficult questions is usually swept under the carpet in order to use those magnitudes without further delay.
This is a pity, as those questions provide opportunities for fruitful and important research.
A good example is provided by the long traditions of studies showing that human prefer-ences sometimes systematically violate the conditions required for a neat index of human suffering - or more generally, a utility function - to exist under conditions of uncertainty (milestones of this research are Allais 1953, Ellsberg 1961, Kahneman and Tversky 1979).
The problem is not that one cannot use a utility function, the problem is that one needs a plurality of such functions even to describe a single agent. In particular, this will be the case when indifference between two risks involving the same critical events results with more than one proportion between probabilities. And much as one needs more than one utility function to describe an agent, one may need more than one distribution of subjec-tive probabilities. Which one will be activated then depends on the social context that agent finds herself in. The well-known research on anchoring, framing, and similar effects shows that it is possible to identify rules linking social contexts to different probability and utility functions.
An important example of that plurality of utility functions is given by lexical preferences.
Many people feel that it is never appropriate to consider a trade-offbetween a human life and a standard consumer product: No amount of chewing gum is sufficient to justify a decision to let a human being die. The same people may feel that it is necessary and appropriate for a hospital to decide which patients to keep alive with great effort and which ones to let die as peacefully as possible.
Ethical reasoning usually has a lexical structure: ethical arguments often imply that cer-tain trade-offs are appropriate only as long as they don’t affect specific ethical norms. At the same time, ethical choices often concern trade-offs between compliance with differ-ent ethical norms. With regard to vulnerability to global environmental change, lexical preferences are particularly relevant for the difficult, but necessary debate about equity (Kasperson et al. 2001).
When possible actions can be represented by real numbers, no single utlitity function can represent such lexical preferences. Such preferences can be easily represented, however, by a series of utility functions combined according to suitable rules. A similar argument holds for subjective probability: many difficulties with designing an empirically sound distribution of subjecitve probabilities are due to an unnecessary attempt to work with a single such function. A whole set of functions together with context-dependent selection rules seems more appropriate.
So far I have discussed vulnerability in view of a single individual, in our case named Bianca. But clearly Bianca’s opportunities for action are linked to those of other individ-uals in many ways. In the face of suffering, human beings may or may not establish and maintain bonds of solidarity. Such bonds can be very effective when they are based on face-to-face relations between people who know each other personally, but they can also 46 be highly effective in other ways when they are based on institutional mechanisms like insurance schemes, provision of public goods by political action, etc. Bonds of solidarity enable people to think and act as a team, to develop shared and interlocking preferences as well as shared and interlocking resources. Ultimately, any vulnerability analysis of human communities like nations, regions, etc., must carefully check whether large parts of these communities are being instrumentalized for the purposes of other individuals and groups, or whether those communities really have developed some common preferences and a capability for joint action in the light of these.
A critical thread in the literature on vulnerability deals with social solidarity as a cor-nerstone of adaptive capacity (Sen and Dr´ eze 1999, Ribot et al. 1996). However, it is remarkably difficult, often strictly impossible, to aggregate given individual preferences into social preferences in a meaningful way (Arrow 1951, Sen 1970, Kirman 1992). Such aggregation is often thought to be performed automatically by the market. But of course, the market aggregates preferences by weighting them with individual wealth - and the distribution of wealth in today’s world is not exactly a hallmark of social solidarity.
Aggregating individual preferences into social ones is not always desirable. Where it is, it requires changes of preferences induced by exchanges of argument - something one might claim lies at the core of human freedom. Debates about preferences and solidarity are impossible, however, as long as one maintains a simple dichotomy between scientific facts - that are supposed to be amenable to rational argument - and human values - that are supposed to be either self-evident or arbitrary. One reason to pay attention to the reality of heterogeneous domains of discourse is the fact that it helps overcoming that dichotomy without falling back into a Medieval quest for a single compelling truth (Putnam A.R.
1985, Putnam H. 2001, Jaeger et al. 2001).
Debates about human vulnerability and solidarity in the face of global change may well lead to new kinds of institutions. In the lifetime of our grandchildren, they may support forms of global solidarity that humankind has not known in the past. Before rushing into the playground of institution design, however, it may be appropriate to acknowledge the need for a patient effort to enlarge our understanding of human vulnerability in the face of current global change. This will hardly be possible without major advances in our capability for integrated environmental modelling; integrated, that is, across highly heterogeneous domains of discourse.
10 Modelling in the Multiverse The word ”New York” refers to a city in America, the word ”and” does not refer at all, it fulfills another role in human conversations. Within a given domain of discourse, we can refer to its objects, without such domains, we could not talk about anything. We can refer to objects by names, but also by identifying them with suitable predicates - as when we talk about the largest city on the East coast. Domains of discourse often have some regions of overlap. In these cases, we may talk about the same object within different domains. In one domain, the horizon does not move as the sun rises in all its glory, in another one the earth rotates so that changing parts of its surface are exposed to the 47 rays of the sun (see Goodman 1978 and Putnam 1992 for further discussions of the rather interesting philosophical issues involved in this example).
Sometimes, we use an object to represent another one, as when one uses an icecube in a glass of water to show how the tip of an iceberg is much smaller than its invisible part. I use the words ”refer” and ”represent” to highlight the difference between on the one hand the relation linking a name or an expression to the object it refers to and on the other hand the relation between two objects where the first is used to represent the second.
Representing one object of discourse with another is useful in order to learn, remember, communicate all sorts of things about the object represented. Usually this is based on the fact that the representing object is either more familiar or easier to handle than the represented one.
The word ”model” is used in many ways. When we talk about a computer model, we usually refer to a piece of software running on some hardware equipment so as to represent a specific object. The hardware is a physical object made of sylicon, metals, plastics, etc., and put in a sequence of states according to the software. The software is an algorithm, a mathematical object. The relation between hardware and software raises some deep issues, sometimes triggering speculations about the relation between body and mind. I will not explore these here. What matters for the present purpose is the fact that in computer modelling we use mathematical objects to represent some other object - an organization, a conflict, a car, a stock market bubble, a prime number, an ocean, etc.
Computer modelling is a sophisticated thread in the fabric of human culture. The phys-ical equipment is linked to the represented object via mathematical objects, namely al-gorithms, but usually the link between these and the represented object is established by further mathematical objects. When modelling an insurance market, e.g., one may use a fixed point of some continuous mapping to represent market equilibrium and then an algorithm operating over a finite range of natural numbers to represent that fixed point.
The key interface between the represented object and the computer model is not the hardware, not even the software, it is the mathematical object captured by that software.
And the key question in computer modelling is how to represent the intended object by such a mathematical object. This requires a suitable domain of discourse for the object to be represented. On the one hand, it must enable one to talk about interesting features of the object one wants to represent. On the other hand, it must correspond to a logical structure that fits the relevant domains of mathematics. By and large, such domains of discourse are available for physical objects - but things tend to become murkier when one is dealing with living beings. In the social sciences of our times such domains of discourse are not to be found ready made.
In the past decades, several powerful ways of using mathematical objects to represent socio-economic ones have been developed, in particular in the domains of statistics, lin-guistics, and economics. But serious issues remain unresolved. They concern the arbitrary character of many predicates used when construing ”measures” of social phenomena, the questionable use of formal languages as models of natural languages, the restrictive as-sumptions needed to represent preference structures by utility functions, and more. Such problems cannot be resolved by simply construing more models, but only by a patient clarification of socio-economic domains of discourse and their relationships with other 48 domains. Such clarification, in turn, can be greatly facilitated by the thoughtful design and implementation of computer models.
Clarifying relations between different domains of discourse, especially when they are so heterogeneous as to involve different logical patterns of inference, is rewarding way beyond the realm of computer modelling. Coming to terms with the pluralism of domains of discourse may well be one of the most important challenges humankind is faced with in the present global culture. This holds both for the plurality of scientific and non-scientific domains and for the plurality of cultural traditions flourishing in different parts of the world. And addressing some of the most pressing issues of our times - like the ones of global environmental change and the related ones of overcoming poverty at a planetary scale - clearly require the ability to combine knowledge rooted in highly heterogeneous domains of discourse.
The business of establishing, restoring, improving mutual understanding between re-searchers specializing in widely differing domains of discourse can take advantage of some of the most highly specialized domains of discourse developed so far: those of mathemat-ics. The standards of reliability cultivated over centuries by mathematicians have led to a wealth of techniques and insights that can be used to improve the concepts of other domains of discourse. This implies an interative movement between a mathematical and a non-mathematical domain, patiently asking questions about both as well as about their relation, and boldly exploring new inventions that hold promise of fresh insights. There is no general recipe here, only the pleasure of cracking one nut after the other. The remarks on the concept of vulnerability in the present section offer a basket of nuts to be cracked in this way.
That said, let us keep in mind that links between different domains of discourse are ultimately forged in natural languages. The great dream of a unified language for scientific discourse, a language that would then provide the scaffolding for human knowledge in general, has a long and impressive history. In a variety of forms, it has been shared by Frege, Poincar´ e, Whitehead, Wittgenstein, and many others. The publication of Kuhn’s essay on ’The Structure of Scientific Revolutions’ - ironically in the first volume of the Intemational Encyclopedia of Unified Science (1962) - marks the awakening from that dream.
The linkages between different domains of discourse are due to the fact that these domains are embedded in natural languages, not to the illusion that somewhen in the future they might all be integrated with the tools of logic, or mathematics, or whatever. One cannot model the multiverse on a computer, one can only model things in the multiverse by using other things, in particular computers. If as modellers we want to talk about the world as a whole, it is probably a good idea to pause for a moment first, perhaps listen to non-scientists like various stakeholders next (Kasemir et al 2003), and then talk in natural language. And there is not one natural language lying at the root of all others (although of course languages like Latin or English play a pivotal role in specific historical periods): the unity of the human mind resides in a plurality of languages that offers opportunities for translation.
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61 National and Regional Climate Change Impact Assessments in the Forestry Sector - Workshop Summary and Abstracts of Oral and Poster Presentations M. Lindner (ed.) (Juli 2000) No.
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63 Eine Methode zur Validierung von Klimamodellen für die Klimawirkungsforschung hinsichtlich der Wiedergabe extremer Ereignisse U. Böhm (September 2000) No.
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66 The Rising Tide of Green Unilateralism in World Trade Law - Options for Reconciling the Emerging North-South Conflict F. Biermann (Dezember 2000) No.
67 Coupling Distributed Fortran Applications Using C++ Wrappers and the CORBA Sequence Type Th. Slawig (Dezember 2000) No.
68 A Parallel Algorithm for the Discrete Orthogonal Wavelet Transform M. Uhlmann (Dezember 2000) No.
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70 Stakeholder Successes in Global Environmental Management, Report of Workshop, Potsdam, 8 December 2000 M. Welp (ed.) (April 2001) No.
71 GIS-gestützte Analyse globaler Muster anthropogener Waldschädigung - Eine sektorale Anwendung des Syndromkonzepts M. Cassel-Gintz (Juni 2001) No.
72 Wavelets Based on Legendre Polynomials J. Fröhlich, M. Uhlmann (Juli 2001) No.
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11566 | https://ieee.li/pdf/viewgraphs/ltspice_piece_wise_linear_voltage_sources.pdf | Copyright © 2017 Analog Devices. All rights reserved. Piece-Wise Linear (PWL) Voltage Sources © 2017 Analog Devices 2 Creating a PWL Voltage Source Open up the simulation file titled “RCFilterTimeDomain.asc” Run the simulation and probe the IN and OUT nodes Right-Click on the voltage source and select the PWL function Configure the PWL source to manually recreate the pulse waveform as shown in the voltage source window on the right Rerun the simulation. Notice a single pulse is now present. RCFilterTimeDomain.asc © 2017 Analog Devices 3 Repeating PWL Source Open up the simulation file titled “RCFilterTimeDomainRepeat.asc” Run the simulation and probe the IN and OUT nodes Right-Click on the PWL text string and use the repeat command to create three cycles of the input square wave. RCFilterTimeDomainRepeat.asc © 2017 Analog Devices 4 Importing Externally Generated PWL Sources To import a PWL waveform from a file, Right-Click on a voltage source, select “Advanced”, and select “PWL File” The file format must contain pairs of numbers separated by white space (carriage return, spaces, tabs). The first number is time (in seconds) and the second number is voltage. Like the standard PWL, you can also repeat, as shown in these examples Example 1 (datapoints in a line): 0 0 0.1 1 0.2 0.5 0.5 0 0.7 0.3 1 0 Example 2 (datapoints in columns): 0 0 0.1 1 0.2 0.5 0.5 0 0.7 0.3 1 0 PWLExamplesLine.asc PWLExamplesColumn.asc © 2017 Analog Devices 5 PWL waveform data can be imported into LTspice from an oscilloscope waveform capture Data can be comma separated or space separated data pairs (Time Value). Ex. .csv file. Many Oscilloscopes will export in this format (ex. Save Format Spreadsheet option shown below) Importing a PWL Waveform From an Oscilloscope © 2017 Analog Devices 6 Importing a PWL Waveform From a File Open up the simulation file titled “TekImport.asc” and follow the instructions in the simulation file. TekImport.asc © 2017 Analog Devices 7 One thing to notice is that data with a negative time is ignored. Your scope may consider the trigger time to be t=0 in the output file. If you want to use this data, edit the time column in the spreadsheet as needed. Importing a PWL Waveform From a File TEK00000.CSV © 2017 Analog Devices 8 Exporting Waveforms to ASCII With a waveform window active, go to File -> Export. Select the node voltage or current to export and click OK. The file location is shown at the top of the dialog box. TekImport.asc |
11567 | https://math.stackexchange.com/questions/1865381/how-does-this-technique-for-solving-simultaneous-congruences-work | elementary number theory - How does this technique for solving simultaneous congruences work? - Mathematics Stack Exchange
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How does this technique for solving simultaneous congruences work? [duplicate]
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Why does the CRT formula yield a solution of a congruence system? (3 answers)
Closed 7 months ago.
Find x∈Z x∈Z with
x≡3 mod 7 x≡3 mod 7
x≡9 mod 11 x≡9 mod 11
x≡1 mod 5 x≡1 mod 5
So here's what I do: I first find r 1∈Z r 1∈Z with r 1≡1 mod 7 r 1≡1 mod 7 and r 1≡0 mod(11⋅5)r 1≡0 mod(11⋅5), e.g. r 1=330 r 1=330.
Then I find r 2∈Z r 2∈Z with r 2≡1 mod 11 r 2≡1 mod 11 and r 2≡0 mod(7⋅5)r 2≡0 mod(7⋅5), e.g. r 2=175 r 2=175.
Then I find r 3∈Z r 3∈Z with r 3≡1 mod 5 r 3≡1 mod 5 and r 3≡0 mod(7⋅11)r 3≡0 mod(7⋅11), e.g. r 3=154 r 3=154.
Now if we set x=3⋅r 1+9⋅r 2+1⋅r 3 x=3⋅r 1+9⋅r 2+1⋅r 3 this does the job, right?
It kind of makes sense that this would work but can someone give me a better reason than 'kind of makes sense' ?
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4 This is actually a proof of the chinese remainder theorem en.wikipedia.org/wiki/Chinese_remainder_theorem –lisyarus Commented Jul 20, 2016 at 13:54
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Sure: What's the remainder of x x on division by 7 7? Well, it's the sum of the remainders of the three terms, mod 7. The remainder of the first, mod 7, is 3⋅1 3⋅1. The remainder of the second is 0 0; so is the remainder of the third. So the total remainder, mod 7, is just 3. The same argument applies to the other two remainders.
There's one subtle point:
Why, knowing that r 2 r 2 is 0 0 mod 7⋅5 7⋅5, do I know that it's zero mod 7 mod 7? Straight from the definitions: Because if 7⋅5 7⋅5 divides evenly into r 2 r 2 (say r 2=(7⋅5)k r 2=(7⋅5)k), then 7 7 also does, since r 2=7⋅(5⋅k)r 2=7⋅(5⋅k). (Thanks to @Bill Dubuque for straightening me out on this!)
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edited Jul 21, 2016 at 18:06
answered Jul 20, 2016 at 13:58
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thank you for the kind explanation –MyNameIs Commented Jul 20, 2016 at 17:30
@MyNameIs The final paragraph above is a bit misleading. It is true for _all_ integers p,q p,q that p q∣n⇒p,q∣n.p q∣n⇒p,q∣n. Rather, it is the _reverse_ implication that requires p,q p,q coprime (and that implication is not needed here). The explanation in the first paragraph is conceptually clearer when expressed in vector notation (as in my answer), esp. if you are familiar with linear algebra. I'll be happy to elaborate on my answer if you tell me what is not clear. –Bill Dubuque Commented Jul 20, 2016 at 18:35
@BillDubuque's remarks are spot-on. I'll edit to try to clarify. I also agree about the vector notation, but didn't want to assume anything about OP's (or other future readers') knowledge --- students often mess with number theory in high school, long before encountering linear algebra. I also find that students are often able to generalize from examples quite easily, so that having them do some computations like those in the first paragraph can lead them to an "aha!" moment. But we all have our different explanation styles, and I'm glad your answer is here as well as mine. –John Hughes Commented Jul 21, 2016 at 17:00
@John The problem is still there. You claim that r 2≡0(mod 7⋅5)⇒r 2≡0(mod 7)r 2≡0(mod 7⋅5)⇒r 2≡0(mod 7) "because 7 7 and 5 5 are prime". However the justification of that inference has nothing at all to do with primality (and asserting that it does may greatly confuse students). Eliminating the modular language, the claim is equivalent to 5⋅7∣r 2⇒7∣r 2,5⋅7∣r 2⇒7∣r 2, This is true because the divisibility relation is _transitive_, hence p∣p q,p∣p q,p q∣n p q∣n⇒⇒p∣n.p∣n. That's true for all integers p,q,p,q, not only primes –Bill Dubuque Commented Jul 21, 2016 at 17:56
Fixed (I hope). –John Hughes Commented Jul 21, 2016 at 18:06
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Hint Those elements _span_ the set of all values (a 1,a 2,a 3)(mod 7,11,5),(a 1,a 2,a 3)(mod 7,11,5), i.e.
r 1≡(1,0,0)(mod 7,11,5)r 2≡(0,1,0)(mod 7,11,5)r 3≡(0,0,1)(mod 7,11,5)r 1≡(1,0,0)(mod 7,11,5)r 2≡(0,1,0)(mod 7,11,5)r 3≡(0,0,1)(mod 7,11,5)
⟹≡≡a 1 r 1+a 2 r 2+a 3 r 3(a 1,0,0)+(0,a 2,0)+(0,0,a 3)(a 1,a 2,a 3)⟹a 1 r 1+a 2 r 2+a 3 r 3≡(a 1,0,0)+(0,a 2,0)+(0,0,a 3)≡(a 1,a 2,a 3)
This will become clearer when you learn the ring-theoretic view of CRT, which here says that Z/(7⋅11⋅5)≅Z/7×Z/11×Z/5,Z/(7⋅11⋅5)≅Z/7×Z/11×Z/5, where Z/n=Z/n= integers mod n.n. Informally this means these triples (a 1,a 2,a 3)(a 1,a 2,a 3) with the natural induced componentwise addition and multiplication have the same arithmetical (ring-theoretic) structure as the integers mod 385=7⋅11⋅5.385=7⋅11⋅5.
For example r 1+r 2+r 3=(1,1,1)r 1+r 2+r 3=(1,1,1) which is the identity element in the product ring, which maps to 330−175−154=1 330−175−154=1 in the integers mod 385 385 (I corrected the signs on your r 2,r 3)r 2,r 3)
This means that we can solve arithmetical problems in the integers mod 385 385 by mapping them into a triple of corresponding problems in the _smaller_ rings of integers mod 7,11,5,7,11,5, and then, by CRT, lift that triple of solutions to the corresponding solution in the integers mod 385.385. For example, we can solve polynomial equations in this manner. For example see this answer which computes square roots in that manner.
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edited Jan 13 at 20:43
answered Jul 20, 2016 at 14:22
Bill DubuqueBill Dubuque
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For your implication, you need more than spanning (as you know): you need that the three reduced triples that you've written above the implication arrow look like the standard basis. Nice job correcting the signs. :) –John Hughes Commented Jul 20, 2016 at 17:54
@john No, the implication is correct as it stands. I suspect that you are misunderstanding something, but I'm not sure just what. Note that r≡(a,b,c)(mod 7,11,5)r≡(a,b,c)(mod 7,11,5) means r≡a(mod 7),r≡a(mod 7),r≡b(mod 11),r≡b(mod 11),r≡c(mod 5)r≡c(mod 5) –Bill Dubuque Commented Jul 20, 2016 at 18:15
The implication (in the informal, not logical, sense that I dislike is that _spanning_ implies that a 1 r 1+a 2 r 2+a 3 r 3=(a 1,a 2,a 3)a 1 r 1+a 2 r 2+a 3 r 3=(a 1,a 2,a 3), i.e., that once you have a spanning set, all you need to do is combine it using your desired outputs as coefficients. For _that_, you need a set that reduces (mod 7, 11, 5) to the standard basis. What you wrote is, of course, literally correct. I should have said not that the implication was wrong, but that I found it easy to read in a misleading way. –John Hughes Commented Jul 21, 2016 at 11:00
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Parasitism is a type of symbiotic relationship between two organisms where one organism (the parasite) benefits at the expense of the other (the host). The parasite derives nutrients from the host, often causing harm in the process.
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Host:In biology, a host is an organism that harbors another organism inside or near their body in a symbiotic relationship. It's like the house in our analogy.
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Symbiosis is any type of close and long-term biological interaction between two different biological organisms. It's like having roommates - some are beneficial (mutualistic), some are neutral (commensal), and some are harmful (parasitic).
Ectoparasite:An ectoparasite is a parasite that lives on the surface of its host. They're similar to that freeloader roommate who always hangs out in your living room but never contributes to rent or utilities.
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Advanced Inequality Concepts
This document summarizes key concepts in majorization theory and provides proofs of majorization properties and inequalities, including: - The definition of majorization between two sequenc…
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Majorization and Karamata Inequality
MathLinks - www.mathlinks.ro
Pham Kim Hung, Stanford University, US
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2
Note
Thi s is an exc erp t fro m the secon d vo lum e of ”Se cre ts In Inequ alit ies”, by Pha m Kim Hung. The aut hor thank s sin cer ely Dar ij Gri nbe rg for some of his materi als about Symmetric Majorization Theorem, posted on Mathlinks Forum. Please don’t use this excerpt for any commercial purpose.The Autho r alwa ys appriciates every Cont ributi on to this content- Majorizat ion and Karamata Inequality.Best Regard,Pham Kim Hung
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Chapter 1
Theory of Majorization
The theory of majorization and convex functions is an important and difficult part of inequalities, with many nice and powerful applications. will discuss in this article is
Karamata
inequalit y; howev er, it’s necess ary to review first some basic properti es of majorizati on.
Defini tion 1.
Given two se quenc es
(
a
) = (
a
1
,a
2
,...,a
n
)
and
(
b
) = (
b
1
,b
2
,...,b
n
)
(where
a
i
,b
i
∈
R
∀
i
∈ {
1
,
2
,...,n
}
). We say tha t the sequ enc e
(
a
)
majoriz es the sequence
(
b
)
, and write
(
a
)
(
b
)
, if the following conditions are fulfilled
a
1
≥
a
2
≥
...
≥
a
n
;
b
1
≥
b
2
≥
...
≥
b
n
;
a
1
+
a
2
+
...
+
a
n
=
b
1
+
b
2
+
...
+
b
n
;
a
1
+
a
2
+
...
+
a
k
≥
b
1
+
b
2
+
...
+
b
k
∀
k
∈{
1
,
2
,...n
−
1
}
.
Defini tion 2.
For an arbit rar y se quenc e
(
a
) = (
a
1
,a
2
,...,a
n
)
, we de no te
(
a
∗
)
,a p erm uta tio n of ele men ts of
(
a
)
which are arrange d in incr ea sing or der:
(
a
∗
) =(
a
i
1
,a
i
2
,...,a
i
n
)
with
a
i
1
≥
a
i
2
≥
...
≥
a
i
n
and
{
i
1
,i
2
,...,i
n
}
=
{
1
,
2
,...,n
}
.
Here are some basic properties of sequences.
Proposition 1.
Let
a
1
,a
2
,...,a
n
be real numbers and
a
=1
n
(
a
1
+
a
2
+
...
+
a
n
)
,
then
(
a
1
,a
2
,...,a
n
)
∗
(
a,a,...,a
)
.
Propos ition 2.
Suppo se that
a
1
≥
a
2
≥
...
≥
a
n
and
π
= (
π
1
,π
2
,...π
n
)
is an arbitrary permutation of
(1
,
2
,...,n
)
, then we have
(
a
1
,a
2
,...,a
n
)
(
a
π
(1)
,a
π
(2)
,...,a
π
(
n
)
)
.
3
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4
Proposition 3.
Let
(
a
) = (
a
1
,a
2
,...,a
n
)
and
(
b
) = (
b
1
,b
2
,...,b
n
)
be two sequences of real numbers. We have that
(
a
∗
)
majorizes
(
b
)
if the following conditions are fulfilled
b
1
≥
b
2
≥
...
≥
b
n
;
a
1
+
a
2
+
...
+
a
n
=
b
1
+
b
2
+
...
+
b
n
;
a
1
+
a
2
+
...
+
a
k
≥
b
1
+
b
2
+
...
+
b
k
∀
k
∈{
1
,
2
,...,n
−
1
}
;These properties are quite obvious: they can be proved directly from the definition of Majorization.The followi ng result s, especially the Symmetric Mjori zation Criteri on,will be most important in what follows.
Propos ition 4.
If
x
1
≥
x
2
≥
...
≥
x
n
and
y
1
≥
y
2
≥
...
≥
y
n
are positive real numbers such that
x
1
+
x
2
+
...
+
x
n
=
y
1
+
y
2
+
...
+
y
n
and
x
i
x
j
≥
y
i
y
j
∀
i <j
, then
(
x
1
,x
2
,...,x
n
)
(
y
1
,y
2
,...,y
n
)
.
Proof.
To prove this assertion, we will use induction. Because
x
i
x
1
≤
y
i
y
1
for all
i
∈{
1
,
2
,...,n
}
, we get that
x
1
+
x
2
+
...
+
x
n
x
1
≤
y
1
+
y
2
+
...
+
y
n
y
1
⇒
x
1
≥
y
1
.
Consider two sequences (
x
1
+
x
2
,x
3
,...,x
n
) and (
y
1
+
y
2
,y
3
,...,y
n
). By the inductive hypothesis, we get(
x
1
+
x
2
,x
3
,...,x
n
)
(
y
1
+
y
2
,y
3
,...,y
n
)
.
Combining this with the result that
x
1
≥
y
1
, we have the conclusion immediately.
∇
The ore m 1 (Sy mme tri c Major iza tio n Crit erio n).
Sup po se tha t
(
a
) =(
a
1
,a
2
,...,a
n
)
and
(
b
) = (
b
1
,b
2
,...,b
n
)
ar e two se que nc es of r e al num b ers; the n
(
a
∗
)
(
b
∗
)
if and only if for all real numbers
x
we have
|
a
1
−
x
|
+
|
a
2
−
x
|
+
...
+
|
a
n
−
x
|≥|
b
1
−
x
|
+
|
b
2
−
x
|
+
...
+
|
b
n
−
x
|
.
Proof.
To prove this theorem, we need to prove the following.
(i). Necessary condition.
Suppose that (
a
∗
)
(
b
∗
), then we need to prove that for all real numbers
x
|
a
1
−
x
|
+
|
a
2
−
x
|
+
...
+
|
a
n
−
x
|≥|
b
1
−
x
|
+
|
b
2
−
x
|
+
...
+
|
b
n
−
x
|
(
)Notice that (
) is just a direct application of
Karamata
inequality to the convex function
f
(
x
) =
|
x
−
a
|
; however, we will prove algebraically.
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11570 | https://www.pipeflow.com/public/documents/wizard/PipeFlowWizardNonCompressibleResultsVerification.pdf | www.pipeflow.com Pipe Flow Wizard Software for Fluid Flow and Pressure Loss Calculations Liquids Verification of Calculation Results For Non-Compressible Flow Pipe Flow Wizard Software – Liquid Results Verification 2 Table of Contents – Results Data: Systems Solved by Pipe Flow Wizard Introduction ............................................................................................................................................... 3 Find Pressure ........................................................................................................................................... 4 Case 01: Petroleum - Oil Pipeline Pressure Loss .................................................................................... 5 Case 02: Gasoline - Transport over 15 km .............................................................................................. 6 Case 03: SAE 10 Oil - Pressure Loss per Mile ........................................................................................ 7 Case 04: Water - Asbestos Cement Pipe Friction Loss ........................................................................... 8 Case 05: Lubrication Oil - Laminar Flow Example 1 ................................................................................ 9 Case 06: Lubrication Oil - Laminar Flow Example 2 .............................................................................. 10 Case 07: Water - Reynolds Number for Smooth Wall Pipe ................................................................... 11 Case 08: SAE 70 Lube Oil - Laminar Flow in Valves ............................................................................. 12 Case 09: Water and Oil – Uncoated Cast Iron Pipe ............................................................................... 13 Case 10: Water – Pressure Loss due to Friction ................................................................................... 14 Case 11: Oil – Laminar Flow in Pipeline ................................................................................................ 15 Case 12: Oil – Head loss in Cast Iron Pipeline ...................................................................................... 16 Find Flow ................................................................................................................................................ 17 Case 01: Water - Large Diameter Cast Iron Pipe .................................................................................. 18 Case 02: Ethanol - Laminar Flow ........................................................................................................... 19 Case 03: Water – Flow Between Two Reservoirs .................................................................................. 20 Case 04: Water – Elevated Pipeline with Fittings .................................................................................. 21 Case 05: Water – Flow Through Reduced Port Ball Valve .................................................................... 22 Case 06: SAE 10 Lube Oil - Laminar Flow in Valves ............................................................................. 23 Find Diameter ......................................................................................................................................... 24 Case 01: Design of a Uniform Pipeline .................................................................................................. 25 Case 02: Pump – Storage Power Scheme – Pipeline design ................................................................ 26 Case 03: Water - Galvanized Steel Pipe ................................................................................................ 27 Case 04: Heavy Fuel Oil – Sizing a Horizontal Pipe .............................................................................. 28 Find Length ............................................................................................................................................. 29 Case 01: Length of Steel Water Pipe ..................................................................................................... 30 Case 02: Water Pipeline length .............................................................................................................. 31 Case 03: Pipeline Between Two Reservoirs with Fittings ...................................................................... 32 References ............................................................................................................................................. 33 Pipe Flow Wizard Software – Liquid Results Verification 3 Introduction Pipe Flow Wizard is a software application that performs flow rate and pressure drop calculations for fluid flow in a pipe. The Pipe Flow Wizard software can ‘Find Pressure Drop', 'Find Flow Rate', 'Find Diameter Size', and 'Find Length of Pipe' depending on the information available. Each of the Find Pressure, Find Flow, Find Diameter and Find Length calculations produced by the Pipe Flow Wizard software can be verified by comparison against published results from a number of well-known sources. The information in this document provides a general description of a published problem, the Reference Source, the Published Results Data, the Pipe Flow Wizard Results Data and a commentary on the results obtained. For each of the calculation problems detailed in this document, the results data produced by the Pipe Flow Wizard software compares well with the published results data. Calculations Friction Factors are calculated using the Colebrook-White equation. Friction Loss for non-compressible fluids is calculated using the Darcy-Weisbach method, which provides accurate results for Newtonian fluids, including general process fluids. The Pipe Flow Wizard software also contains a separate compressible calculation engine that allows for the solution of compressible systems (gases) with equations such as the General Fundamental Isothermal Flow equation, and others. There is a separate verification document that compares the results produced by the compressible calculation engine against published data. Software Releases The latest release of the Pipe Flow Wizard software has been completely rewritten to support use on Microsoft Windows (PCs), Mac OS (Apple Computers), and iOS (Apple Mobile Devices), including iPhone and iPad. The original Pipe Flow Wizard software for Windows was released over 15 years ago and today Pipe Flow Wizard software is used by engineers in over 100 countries worldwide. We have clients in a variety of industries including aerospace, chemical processing, education, food and beverage, general engineering, mining, petrochemical, pharmaceutical, power generation, water distribution, and wastewater processing. Pipe Flow Wizard Software – Liquid Results Verification 4 Find Pressure Pipe Flow Wizard Software – Liquid Results Verification 5 Case 01: Petroleum - Oil Pipeline Pressure Loss Reference: Piping Calculations Manual, 2005, McGraw-Hill, E. Shashi Menon, P.E., Page 335, Example 6.16 Pipe Flow Wizard Software: Find_Pressure_Case_01_Petroleum_Oil_Pipeline_Pressure_Loss.pfwp Calculation Problem: Find the head loss in one mile of NPS16 pipeline (0.250 inch wall thickness) at a flow rate of 4000 barrel/h. Fluid Data: Petroleum oil with 0.85 specific gravity and 10 cSt viscosity. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The published data rounds the fluid velocity to 2 decimal places and the friction factor to 4 decimal places. The Pipe Flow Wizard software uses a velocity and a friction factor that are calculated to more decimal places, which accounts for the slight differences in calculated head loss. Results Comparison: Data Item Published Data Pipe Flow Wizard Head Loss (ft. hd) 29.908 29.9265 Reynolds Number 57129 57130 Fluid Velocity (ft/s) 4.76 4.7608 Friction Factor 0.0208 0.020784 Pipe Flow Wizard Software – Liquid Results Verification 6 Case 02: Gasoline - Transport over 15 km Reference: Piping Calculations Manual, 2005, McGraw-Hill, E. Shashi Menon, P.E., Page 337, Example 6.17 Pipe Flow Wizard Software: Find_Pressure_Case_02_Gasoline_Transport_Over_15km.pfwp Calculation Problem: A DN500 (10mm wall thickness) steel pipe, with an internal roughness of 0.05 mm, is used to transport gasoline over a 15 km distance. The delivery point is 200 m above the start of the pipeline. A delivery pressure of 4 kPa must be maintained at the delivery point. Calculate the pump pressure needed to deliver a flow rate of 990 m³/h. Fluid Data: Gasoline Specific Gravity = 0.736 Viscosity = 0.6 Centistokes (0.4416 Centipoise) Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The published text uses a friction factor value of 0.013 read from the Moody diagram. The Pipe Flow Wizard software uses a friction factor calculated to more decimal places which accounts for the slight difference in the pump pressure required. Results Comparison: Data Item Published Data Pipe Flow Wizard Pump Pressure Required (kPa) 1792 1796.49 Reynolds Number 1215768 1215767 Fluid Velocity (m/s) Not stated 1.52 Friction Factor 0.013 0.013289 Pipe Flow Wizard Software – Liquid Results Verification 7 Case 03: SAE 10 Oil - Pressure Loss per Mile Reference: 2500 Solved Problems in Fluid Mechanics and Hydraulics 1989, McGraw-Hill, Jack B. Evett, Ph. D., Cheng Liu, M.S., Page 211, Example problem 9.68 Pipe Flow Wizard Software: Find_Pressure_Case_03_SAE_10_Oil_Pressure_Loss_Per_Mile.pfwp Calculation Problem: A 6” wrought iron pipe carries SAE 10 oil at 68°F. Calculate the pressure loss per mile of pipe. Fluid Data: SAE 10 at 68°F. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The Pipe Flow Wizard software calculates the friction factor to a greater number of decimal places. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Loss per mile. (psi) 244 241.33 Reynolds Number 5035 5047 Friction Factor 0.038 0.037657 Pipe Flow Wizard Software – Liquid Results Verification 8 Case 04: Water - Asbestos Cement Pipe Friction Loss Reference: Basic Principles for the Design of Centrifugal Pump Installations SIHI Group, 1998, SIHI-HALBERG. Page 134, Example of Head Loss Calculation Pipe Flow Wizard Software: Find_Pressure_Case_04_Water_Asbestos_Cement_Pipe_Friction_Loss.pfwp Calculation Problem: Water flows along a 400 m long asbestos cement pipe at the rate of 360 m³/h. The pipe designation is DN200. Find the head loss in the pipe. Fluid Data: Water at 10°C. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Reynolds Number 4.9 x 105 487458 Fluid Velocity (m/s) 3.2 3.18 Total Head Loss in pipe (m. hd) 16.4 16.42 Pipe Flow Wizard Software – Liquid Results Verification 9 Case 05: Lubrication Oil - Laminar Flow Example 1 Reference: Flow of Fluids – Technical Paper No 410M, 1999, Crane Co. Page 3-12, Example 1 Pipe Flow Wizard Software: Find_Pressure_Case_05_Lubricating_Oil_Laminar_Flow_Example_1.pfwp Calculation Problem: A 6” diameter schedule 40 steel pipe carries lubricating oil of density 897 kg/m³ and viscosity 450 Centipoise. Find the pressure drop per 100 meters. Fluid Data: Lubricating Oil Viscosity = 450 Centipoise, Density = 897 kg/m3 Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Drop per 100 meters (bar) 1.63 1.628 Reynolds Number 825 824 Pipe Flow Wizard Software – Liquid Results Verification 10 Case 06: Lubrication Oil - Laminar Flow Example 2 Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page 3-12, Example 2 Pipe Flow Wizard Software: Find_Pressure_Case_06_Lubricating_Oil_Laminar_Flow_Example_2.pfwp Calculation Problem: A 3” diameter schedule 40 carries SAE 10 lube oil at a velocity of 5.0 ft/s Find the flow rate and the pressure drop per 100 feet. Fluid Data: Oil, viscosity = 95 Centipoise, density = 54.64 lb/ft3 Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Flow Rate (US gpm) 115 115 Fluid Velocity (ft/s) 5.00 4.99 Reynolds Number 1100 1092 Pressure Drop per 100 feet (psi) 3.40 3.367 Pipe Flow Wizard Software – Liquid Results Verification 11 Case 07: Water - Reynolds Number for Smooth Wall Pipe Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page 4-1, Example 4-1 Pipe Flow Wizard Software: Find_Pressure_Case_07_Water_Reynolds_Number_For_Smooth_Wall_Pipe.pfwp Calculation Problem: 70 feet of 2” diameter plastic pipe (smooth wall) carries water at 80°F. The flow rate is 50 gpm (US). Find the Reynolds number and the friction factor. Fluid Data: Water at 80°F Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Pipe Flow Wizard uses the same fluid density and viscosity as the published text to calculate the Reynolds number. The published text friction factor has been read from a chart for water at 60°F. Results Comparison: Data Item Published Data Pipe Flow Wizard Reynolds Number 89600 89702 Friction Factor 0.0182 0.018883 Pipe Flow Wizard Software – Liquid Results Verification 12 Case 08: SAE 70 Lube Oil - Laminar Flow in Valves Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page 4-4, Example 4-8 Pipe Flow Wizard Software: Find_Pressure_Case_08_SAE_70_Lube_Oil_Laminar_Flow_In_Valves.pfwp Calculation Problem: 200 feet of 8” diameter steel pipe (schedule 40) carries SAE 70 Lube Oil at 100°F. The flow rate is 600 barrels per hour. The piping includes an 8” globe valve. Find the pressure loss in the pipe and the valve. Fluid Data: SAE 70 Lube Oil at 100°F Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Loss (psi) 2.85 2.8675 Reynolds Number 318 318 Friction Factor 0.20 0.201124 Pipe Flow Wizard Software – Liquid Results Verification 13 Case 09: Water and Oil – Uncoated Cast Iron Pipe Reference: Fluid Mechanics and Hydraulics – Third Edition 1994 Ranald V. Giles, Jack B. Evett, Ph.D., Cheng Liu, Page 149, Example problem 8.15 Pipe Flow Wizard Software: Find_Pressure_Case_09_Water_Oil_Cast_Iron_Pipe.pfwp Calculation Problem: 1000 ft of new uncoated 12” internal diameter cast iron pipe carries: (a) Water 60ﹾF at 5.00 ft/sec, and (b) Medium fuel oil 60ﹾF at the same velocity. Determine the pressure loss (head loss) in the pipe. Fluid Data: (a) Water at 60ﹾF Kinematic Viscosity = 1.217 x 10-5 ft2/sec. (b) Medium Fuel Oil at 60ﹾF Kinematic Viscosity= 0.858 ft2/sec. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The Example used a relative roughness of 0.0008, and this value was also used in the Pipe Flow Wizard calculation. The Friction factor in the published data was read from Diagram A-1, given in Appendix A (page 346). The Pipe Flow Wizard software used the Colebrook- White equation to calculate the accurate friction factor. Results Comparison: Data Item Published Data Pipe Flow Wizard Published Data Pipe Flow Wizard Fluid Water Water Medium Fuel Oil Medium Fuel Oil Pressure Loss (ft head) 7.5 7.552 8.3 8.297 Reynolds Number 411000 410868 105000 105267 Friction Factor 0.0194 0.019438 0.0213 0.021356 Pipe Flow Wizard Software – Liquid Results Verification 14 Case 10: Water – Pressure Loss due to Friction Reference: Piping Calculations Manual, 2005, McGraw-Hill, E. Shashi Menon, P.E., Page 16, Example 1.9 Pipe Flow Wizard Software: Find_Pressure_Case_10_Water_Pressure_Loss.pfwp Calculation Problem: Water flows through a 16 inch pipeline (0.375 inch wall thickness) at 3000 gal/min. Assume a pipe roughness of 0.002 inches. Calculate the friction factor and head loss due to friction in 1000 ft of pipe length. Fluid Data: Water Kinematic Viscosity of 1.0 cSt Commentary: See the Results Comparison Table below. The problem does not give the temperature of water however it specifies a kinematic viscosity of 1.0 cSt. The Pipe Flow Wizard calculation used water at 20ﹾC which has a kinematic viscosity 0f 1.004008 cSt. The problem description did not specify a pipe material. The Pipe Flow Wizard software calculation used Steel Schedule 40 with a pipe roughness of 0.002 inches. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Drop due to Friction (psi) 2.12 2.122 Reynolds Number 622131 619659 Friction Factor 0.0144 0.014442 Pipe Flow Wizard Software – Liquid Results Verification 15 Case 11: Oil – Laminar Flow in Pipeline Reference: Analysis of Flow in Pipe Networks, 1976, Roland W. Jeppson Page 32, Examples 1 and 2 Pipe Flow Wizard Software: Find_Pressure_Case_11_Oil_Laminar_Flow.pfwp Calculation Problem: A flow rate of 150 gpm (0.00947 m3/s) of oil occurs in a 4-inch (0.1016 m) pipe line. Determine the Reynolds number and head loss per 1000ft (304.8m). Fluid Data: Oil Viscosity µ = 1.5 x 10-3 lb-sec/ft2 (0.0718 N·sec/m2) Density ρ = 1.7 slug/ft3 (876 kg/m3). Commentary: See the Results Comparison Table below. The problem does not specify the pipe material data. We used Steel Schedule 40 with an internal roughness of 0.001811. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Loss (ft head) 30.2 30.244 Reynolds Number 1450 1447 Pipe Flow Wizard Software – Liquid Results Verification 16 Case 12: Oil – Head loss in Cast Iron Pipeline Reference: Fluid Mechanics and Hydraulics – Third Edition, 1994, Ranald V. Giles, Jack B. Evett, Ph.D., Cheng Liu, Page 149, Example problem 8.11 Pipe Flow Wizard Software: Find_Pressure_Case_12_Oil_Head_Loss_Cast_Iron_Pipeline.pfwp Calculation Problem: Oil Flows through 3000 m of 300 mm cast iron pipe at the rate of 0.0444 m3/s. What is the lost head in the pipe? Fluid Data: Oil Absolute Viscosity = 0.101 N·s/m2 Specific Gravity = 0.850. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Pressure Loss (m head) 8.14 8.118 Reynolds Number 1582 1586 Friction Factor 0.0405 0.040356 Pipe Flow Wizard Software – Liquid Results Verification 17 Find Flow Pipe Flow Wizard Software – Liquid Results Verification 18 Case 01: Water - Large Diameter Cast Iron Pipe Reference: 2500 Solved Problems in Fluid Mechanics and Hydraulics, 1989, McGraw-Hill, Jack B. Evett, Ph. D., Cheng Liu, M.S. , Page 209, Example problem 9.64 Pipe Flow Wizard Software: Find_Flow_Case_01_Water_Large_Diameter_Cast_Iron_Pipe.pfwf Calculation Problem: A 96” diameter, new cast iron pipe, has a frictional pressure loss of 1.5 ft. hd per 1000 ft of length, when carrying water at 60°F. Calculate the discharge capacity of the pipe. Fluid Data: Water at 60°F (v = 1.21 x 10-5 ft2 /s). Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Flow Capacity (ft3/s) 397 395.58 Pressure Loss per 1000 ft. (ft. hd) 1.5 1.5 Friction Factor 0.0124 0.012467 Pipe Flow Wizard Software – Liquid Results Verification 19 Case 02: Ethanol - Laminar Flow Reference: 2500 Solved Problems in Fluid Mechanics and Hydraulics, 1989, McGraw-Hill, Jack B. Evett, Ph. D., Cheng Liu, M.S., Page 207, Example problem 9.54 Pipe Flow Wizard Software: Find_Flow_Case_02_Ethanol_Laminar_Flow.pfwf Calculation Problem: Ethanol at 20°C is transferred from an upper tank to a lower tank via a 2 mm pipe. The upper tank has 0.6 m of fluid above the exit pipe which itself is 1.2 m long, with 0.8 m of this pipe dipping into the fluid in the lower tank. Calculate the flow rate between the tanks. Fluid Data: Ethanol at 20°C (µ= 1.20 x 10-3 Pa · s) Density = 788 kg/m3 Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The published text does not list an internal roughness for the pipe. The flow in this problem is laminar, so the friction factor is independent of the inner roughness of the pipe. The calculated Reynolds number of 883 indicates that the flow type is well within the laminar flow range. Two pipes with different internal roughness values (0.046000 mm and 0.000001 mm) were used in several Pipe Flow Wizard calculations to confirm that the variation in the internal roughness of the pipe did not affect the flow rate calculation. Results Comparison: Data Item Published data Pipe Flow Wizard Flow from upper tank (l/hr) – Pipe 1 7.59 7.596 Flow from upper tank (l/hr) – Pipe 2 7.59 7.596 Pipe Flow Wizard Software – Liquid Results Verification 20 Case 03: Water – Flow Between Two Reservoirs Reference: Analysis of Flow in Pipe Networks, 1976, Roland W. Jeppson Page 35, Example 4 Pipe Flow Wizard Software: Find_Flow_Case_03_Water_Flow_Between_Two_Reservoirs.pfwf Calculation Problem: A 4-inch PVC pipe 6000 ft long is used to convey water at 68°F between two reservoirs whose surface elevations differ by 150 ft. What is the flow rate? Fluid Data: Water at 68°F. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The published problem obtains the result by first assuming a value for the flow rate and then it uses this to calculate the Reynolds number. The Reynolds number is then used to obtain a value for the friction factor, by reading it from the Moody diagram. The published solution then iterates the flow rate to achieve the final value which agrees with the head loss specified. Note: For the published solution, only a small number of iterations are performed, and it is therefore likely that the published Reynolds number and friction factor are not as accurate as those calculated by the Pipe Flow Wizard software. Results Comparison: Data Item Published Data Pipe Flow Wizard Flow Rate (m3/s) 0.0141 0.014130 Reynolds Number 157000 176365 Friction Factor 0.0165 0.016401 Pipe Flow Wizard Software – Liquid Results Verification 21 Case 04: Water – Elevated Pipeline with Fittings Reference: Chemical Engineering, 1999, J.M. Coulson, J. F. Richardson with J.R. Backhurst, J.H. Harker, Page 92, Example 3.8 Pipe Flow Wizard Software: Find_Flow_Case_04_Water_Elevated_Pipeline_With_Fittings.pfwf Calculation Problem: Water in a tank flows through an outlet 25 m below the water level into a 0.15 m diameter horizontal pipe (e/d = 0.01), 30 m long, with a 90° elbow which leads to a horizontal pipe of the same diameter, 60 m long, containing a fully open globe valve and discharging to atmosphere 10 m below the level of the water in the tank. What is the initial rate of discharge? Fluid Data: Water with viscosity of 1mN s/m2. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Note: f = fanning friction factor, f’ = Moody chart friction factor (as shown by Pipe Flow Wizard) Data Item Published Data Pipe Flow Wizard Flow Rate (m3/s) 0.043 0.042943 Velocity (m/s) 2.45 2.430079 Reynolds Number 367000 363783 Friction θ (𝜙= 𝑓 2 = 𝑓′ 8 ) 0.0045 0.004752 derived from ( 𝑓′ 8 = 0.038019 8 ) Pipe Flow Wizard Software – Liquid Results Verification 22 Case 05: Water – Flow Through Reduced Port Ball Valve Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page 4-3, Example 4-6 Pipe Flow Wizard Software: Find_Flow_Case_05_Water_Flow_Through_Reduced_Port_Ball_Valve.pfwf Calculation Problem: 200 ft of 3” diameter steel pipe (schedule 40) carries water at 60°F. The head of fluid in the supply tank is 22 ft. The piping includes 6 standard 90° elbows and a flanged ball valve with a conical seat. Find the fluid velocity in the pipe and the rate of discharge. Fluid Data: Water at 60°F Commentary: See the Results Comparison Table below. The published data and the calculated results differ by 2.3%. The published data uses an assumed friction factor of 0.018 for a 3” diameter steel pipe. As a final check, in the published data, the friction factor is read from a chart as less than 0.02, and the text concludes that the difference in the assumed friction factor and the friction factor read from the chart, is small enough so as not to require any further correction. If the chart is read accurately the real friction factor is 0.0195. The Pipe Flow Wizard software calculated a friction factor of 0.019476 A new valve fitting was created in Pipe Flow Wizard to model the flanged ball valve as this item is not included in the database of standard valves and fittings. Results Comparison: Data Item Published Data Pipe Flow Wizard Fluid Velocity in Pipe (ft/s) 8.5 8.311 Rate of Discharge (gpm US) 196 191.50 Reynolds Number Not calculated 175978 Friction Factor 0.018 (assumed) 0.019476 Pipe Flow Wizard Software – Liquid Results Verification 23 Case 06: SAE 10 Lube Oil - Laminar Flow in Valves Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page 4-4, Example 4-7 Pipe Flow Wizard Software: Find_Flow_Case_06_SAE_10_Lube_Oil_Laminar_Flow_In_Valves.pfwf Calculation Problem: 200 feet of 3” diameter steel pipe (schedule 40) carries SAE 10 Lube Oil at 60°F. The head of fluid in the supply tank is 22 ft. The piping includes 6 standard 90° elbows and a flanged ball valve with a conical seat. Find the fluid velocity in the pipe and the rate of discharge. Fluid Data: SAE 10 Lube Oil at 60°F Commentary: See the Results Comparison Table below. The published data and the calculated results differ by 3%. The published text acknowledges that the problem has two unknowns and requires a trial and error solution. The published data results are for the initial assumed velocity. The published result is therefore likely to be slightly inaccurate. The Pipe Flow Wizard software performs numerous iterations to find a solution which is accurate to within 0.0004 ft head of pressure loss. A new valve fitting was created in Pipe Flow Wizard to model the flanged ball valve as this item is not included in the database of standard valves and fittings. Results Comparison: Data Item Published Data Pipe Flow Wizard Fluid Velocity in Pipe (ft/s) 5.13 5.27 Rate of Discharge (gpm US) 118 121.46 Reynolds Number 1040 (1st Iteration) 1096 Friction Factor 0.062 (1st Iteration) 0.05840 Pipe Flow Wizard Software – Liquid Results Verification 24 Find Diameter Pipe Flow Wizard Software – Liquid Results Verification 25 Case 01: Design of a Uniform Pipeline Reference: Nalluri & Featherstone’s Civil Engineering Hydraulics sixth edition, 2016, Martin Marriott, Page 105, Example 4.7 Pipe Flow Wizard Software: Find_Diameter_Case_01_Pipeline_Diameter_Between_Two_Reservoirs.pfwd Calculation Problem: A uniform pipeline of length 20 km is to be designed to convey water at a minimum rate of 250 L/s from an impounding reservoir to a service reservoir, the minimum difference in water level between which is 160 m. Local losses, including entry loss and velocity head, total 10V2/2g. Determine the diameter of a standard commercially available lined spun iron pipeline which will provide the required flow when in new condition (k= 0.03mm). Fluid Data: Water. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The final row shows the Pipe Flow Wizard software calculated diameter at 388.78 mm for the minimum flow rate of 250 L/s, which would lead to the selection of a pipe with a 400mm diameter. Results Comparison: Data Item Published Data Pipe Flow Wizard With Local Losses At Flow Rate Inner Diameter 350 mm 350.05 mm 0 191.1 L/s Inner Diameter 400 mm 400.01 mm 0 271.5 L/s Inner Diameter 400 mm 400.02 mm 10V2/2g 269.4 L/s Inner Diameter Not Calculated 388.78 mm 10V2/2g 250 L/s Pipe Flow Wizard Software – Liquid Results Verification 26 Case 02: Pump – Storage Power Scheme – Pipeline design Reference: Nalluri & Featherstone’s Civil Engineering Hydraulics sixth edition, 2016, Martin Marriott, Page 110, Example 4.10 Pipe Flow Wizard Software: Find_Diameter_Case_02_Pipeline_Diameter_Four_Pump_Turbine.pfwd Calculation Problem: The four pump turbine units of a pumped storage hydroelectric scheme are each to be supplied by a high-pressure pipeline of length 2000 m. The minimum gross head (difference in level between upper and lower reservoirs) is 310 m and the maximum head is 340 m. The upper reservoir has a usable volume of 3.25 x106 m3 which could be released to the turbines in a minimum period of 4 hours. Maximum power output required/turbine = 110 MW Turbogenerator efficiency = 80% Effective roughness of pipeline = 0.6 mm Taking minor losses in the pipeline, power station, and draft tube to be 3.0 m, determine the minimum diameter of pipeline to enable the maximum specified power to be developed. The book calculated a flow rate of 56.42 m3/s is required to achieve a maximum power of 110 MW. Fluid Data: Water. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The book uses a slight variation on the Colebrook white equation, and this likely explains the small difference in calculated diameters. Results Comparison: Data Item Published Data Pipe Flow Wizard Inner Diameter 2.65 m 2.63 m Pipe Flow Wizard Software – Liquid Results Verification 27 Case 03: Water - Galvanized Steel Pipe Reference: Mechanics of Fluids 9th edition, 2012, Bernard S. Massey, John Ward-Smith Page 256, Example 7.3 Pipe Flow Wizard Software: Find_Diameter_Case_03_Galvanized_Steel_Pipeline_Diameter.pfwd Calculation Problem: A galvanized steel pipe carries water over a distance of 180 m at 85 L/s with head loss of 9 m. Determine the size of galvanized steel pipe needed. Fluid Data: Water at 15°C, viscosity = 1.14 mm2/s. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. The published problem reports a fanning friction factor which is one quarter of the Darcy friction factor reported by the Pipe Flow Wizard software. Results Comparison: Data Item Published Data Pipe Flow Wizard Inner Diameter 0.1867m 0.187287 m Reynolds Number 508000 506386 Friction Factor 0.0048 0.00482325 Pipe Flow Wizard Software – Liquid Results Verification 28 Case 04: Heavy Fuel Oil – Sizing a Horizontal Pipe Reference: Fluid Mechanics and Hydraulics – Third Edition, 1994, Ranald V. Giles, Jack B. Evett, Ph.D., Cheng Liu, Page 149, Example problem 8.11 Pipe Flow Wizard Software: Find_Diameter_Case_04_Horizontal_Oil_Pipe_Size.pfwd Calculation Problem: A 300 m length of horizontal pipe carries 0.0222 m3/s of heavy fuel oil with an available head loss of 6.7 m. What size pipe should be installed? Fluid Data: Heavy Fuel Oil at 16°C Viscosity = 0.000205 m2/s Specific Gravity = 0.912. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Inner Diameter 0.170 m 0.171 m Reynolds Number 812 808 Pipe Flow Wizard Software – Liquid Results Verification 29 Find Length Pipe Flow Wizard Software – Liquid Results Verification 30 Case 01: Length of Steel Water Pipe Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page B-14 Pipe Flow Wizard Software: Find_Length_Case_01_Water_Steel_Pipe.pfwl Calculation Problem: A nominal 4 inch steel sch. 40 pipe carries water with a flow rate of 1.1140 ft3/sec and velocity of 12.6 ft/sec. If the pressure drop is given to be 5.65 lbs/inch2, what is the length of pipe? Fluid Data: Water at 60°F Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Length (ft) 100 100.132 Pipe Flow Wizard Software – Liquid Results Verification 31 Case 02: Water Pipeline length Reference: Flow of Fluids – Technical Paper No 410, 1988, Crane Co. Page B-14 Pipe Flow Wizard Software: Find_Length_Case_02_Water_Steel_Pipe.pfwl Calculation Problem: A nominal 14 inch steel sch. 40 pipe carries water with a flow rate of 2.005 ft3/sec and velocity of 2.13 ft/sec. If the pressure drop is given to be 0.047 lbs/inch2, what is the length of pipe? Fluid Data: Water at 60°F Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. Results Comparison: Data Item Published Data Pipe Flow Wizard Length (ft) 100 99.99 Pipe Flow Wizard Software – Liquid Results Verification 32 Case 03: Pipeline Between Two Reservoirs with Fittings Reference: Nalluri & Featherstone’s Civil Engineering Hydraulics sixth edition, 2016 Martin Marriott, Page 96, Example 4.2 Pipe Flow Wizard Software: Find_length_Case_03_Reservoir_Pipeline_Fittings.pfwl Calculation Problem: A uniform, 200 mm diameter pipeline with an internal roughness of 0.03 mm, conveys water at 15°C between two reservoirs with a flow rate of 48.41 L/s. The difference in water level between the reservoirs is 50 m. There is an entry head loss of 0.5V2/2g, a valve with a head loss of 10V2/2g and a velocity head of αV2/2g, where α = 1.0. Calculate the pipe length required. Fluid Data: Water.at 15°C. Commentary: See the Results Comparison Table below. The published data and the calculated results compare well. This published problem states the length of pipe and uses this to calculate the steady discharge between the reservoirs, and this is only given to two decimal places. The Pipe Flow Wizard software used the given discharge flow rate and the given head loss data to calculate the length of pipe that produced these conditions. Results Comparison: Data Item Published Data Pipe Flow Wizard Length 5000 m 4996.5 m Pipe Flow Wizard Software – Liquid Results Verification 33 References 1. Piping Calculations Manual, 2005, McGraw-Hill E. Shashi Menon, P.E 2. 2500 Solved Problems in Fluid Mechanics and Hydraulics, 1989, McGraw-Hill Jack B. Evett Ph. D., Cheng Liu M.S. 3. Basic Principles for the Design of Centrifugal Pump Installations, SIHI Group, 1998, SIHI-HALBERG 4. Flow of Fluids – Technical Paper No 410M, 1999, Crane Co. 5. Flow of Fluids – Technical Paper No 410, 1988, Crane Co. 6. Analysis of Flow in Pipe Networks, 1976, Ann Arbor Science Publishers Inc. Rowland W. Jeppson. 7. Chemical Engineering, Sixth Edition, 1999, Elsevier Butterworth Heinemann J.M. Coulson, J. F. Richardson with J.R. Backhurst, J.H. Harker. 8. Nalluri & Featherstone’s Civil Engineering Hydraulics sixth edition, 2016, Martin Marriott 9. Mechanics of Fluids 9th edition, 2012 Massey, John Ward-Smith 10. Fluid Mechanics and Hydraulics – Third Edition, 1994 Ranald V. Giles, Jack B. Evett, Ph.D., Cheng Liu |
11571 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Balanced_Equations_and_Equilibrium_Constants | Skip to main content
Balanced Equations and Equilibrium Constants
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: Jan 30, 2023
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Chemical Equilibria
Calculating an Equilibrium Concentration
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: 1363
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In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the equilibrium constants because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants.
Introduction
There are several important rules for balancing an equation:
An equation can be balanced only by adjusting the coefficients.
The equation must include only the reactants and products that participate in the reaction.
Never change the equation in order to balance it.
If an element occurs in only one compound on each side of the equation, try balancing this element first.
When one element exists as a free element, balance this element last.
Example 1:
H2(g)+O2(g)⇌H2O(l)
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| Because both reactants are in their elemental forms, they can be balanced in either order. Consider oxygen first. There are two atoms on the left and one on the right. Multiply the right by 2 | H2(g)+O2(g)⇌2H2O(l) |
| Next, balance hydrogen. There are 4 atoms on the right, and only 2 atoms on the left. Multiply the hydrogen on left by 2 | 2H2(g)+O2(g)⇌2H2O(l) |
| Check the stoichiometry. Hydrogen: on the left, 2 x 2 = 4; on right 2 x 2= 4. Oxygen: on the left: 1 x 2 = 2; on the right 2 x 1 = 2 . All atoms balance, so the equation is properly balanced. | 2H2(g)+O2(g)⇌2H2O(l) |
Example 2:
Al(s)+MnSO4(aq)⇌Al2(SO4)3+Mn;(s)
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| First, consider the SO42- ions. There is one on the left side of the equation, and three on the right side. Add a coefficient of three to the left side. | Al(s)+3MnSO4(aq)⇌Al2(SO4)3+Mn(s) |
| Next, check the Mn atoms. There is one on the right side, but now there are three on the left side from the previous adjustment. Add a coefficient of three on the right side. | Al(s)+3MnSO4(aq)⇌Al2(SO4)3+3Mn(s) |
| Consider Al. There is one atom on the left side and two on the right side. Add a coefficient of two on the left side. Make sure there are equal numbers of each atom on each side. | 2Al(s)+3MnSO4(aq)⇌Al2(SO4)3+3Mn(s) |
Example 3:
P4S3+KClO3⇌P2O5+KCl+SO2
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| This problem is more difficult. First, look at the P atoms. There are four on the reactant side and two on the product side. Add a coefficient of two to the product side. | P4S3+KClO3⇌2P2O5+KCl+SO2 |
| Next, consider the sulfur atoms. There are three on the left and one on the right. Add a coefficient of three to the right side. | P4S3+KClO3⇌2P2O5+KCl+3SO2 |
| Now look at the oxygen atoms. There are three on the left and 16 on the right. Adding a coefficient of 16 to the KClO3 on the left and the KCl on the right preserves equal numbers of K and Cl atoms, but increases the oxygen. | P4S3+16KClO3⇌2P2O5+16KCl+3SO2 |
| Tripling the other three species (P4S3, P2O5, and SO2) balances the rest of the atoms. | 3P4S3+16KClO3⇌2(3)P2O5+16KCl+3(3)SO2 |
| Simplify and check. | 3P4S3+16KClO3⇌6P2O5+16KCl+9SO2 |
Chemical Equilibrium
Balanced chemical equations can now be applied to the concept of chemical equilibrium, the state in which the reactants and products experience no net change over time. This occurs when the forward and reverse reactions occur at equal rates. The equilibrium constant is used to determine the amount of each compound that present at equilibrium. Consider a chemical reaction of the following form:
aA+bB⇌cC+dD
For this equation, the equilibrium constant is defined as:
Kc=[C]c[D]d[A]a[B]b
The activities of the products are in the numerator, and those of the reactants are in the denominator. For Kc, the activities are defined as the molar concentrations of the reactants and products ([A], [B] etc.). The lower case letters are the stoichiometric coefficients that balance the equation.
An important aspect of this equation is that pure liquids and solids are not included. This is because their activities are defined as one, so plugging them into the equation has no impact. This is due to the fact that pure liquids and solids have no effect on the physical equilibrium; no matter how much is added, the system can only dissolve as much as the solubility allows. For example, if more sugar is added to a solution after the equilibrium has been reached, the extra sugar will not dissolve (assuming the solution is not heated, which would increase the solubility). Because adding more does not change the equilibrium, it is not accounted for in the expression.
K is related to to the Balanced Chemical Reaction
The following are concepts that apply when adjusting K in response to changes to the corresponding balanced equation:
When the equation is reversed, the value of K is inverted.
When the coefficients in a balanced equation are multiplied by a common factor, the equilibrium constant is raised to the power of the corresponding factor.
When the coefficients in a balanced equation are divided by a common factor, the corresponding root of the equilibrium constant is taken.
When individual equations are combined, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.
A balanced equation is very important in using the constant because the coefficients become the powers of the concentrations of products and reactants. If the equation is not balanced, then the constant is incorrect.
K IS ALSO RELATED TO THE BALANCED CHEMICAL EQUATION OF GASES
For gas-phase equilibria, the equation is a function of the reactants' and products' partial pressures. The equilibrium constant is expressed as follows:
Kp=PcCPdDPaAPbB
P represents partial pressure, usually in atmospheres. As before, pure solids and liquids are not accounted for in the equation. Kc and Kp are related by the following equation:
Kp=Kc(RT)Δn
where
Δn=(c+d)−(a+b)
This represents the change in gas molecules. a,b,c and d are the stoichiometric coefficients of the gas molecules found in the balanced equation.
Neither Kc nor Kp have units. This is due to their formal definitions in terms of activities. Their units cancel in the calculation, preventing problems with units in further calculations.
c
PbI2⇌Pb(aq)+I(aq)
First, balance the equation.
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| Check the Pb atoms. There is one on each side, so lead can be left alone for now. Next check the I atoms. There are two on the left side and one on the right side. To fix this, add a coefficient of two to the right side. | PbI2⇌Pb2+(aq)+2I−(aq) |
| Check to make ensure the numbers are equal. | PbI2⇌Pb2+(aq)+2I−(aq) |
Next, calculate find Kc. Use these concentrations: Pb- 0.3 mol/L, I- 0.2 mol/L, PbI2- 0.5 mol/L
Kc=(0.3)∗(0.2)2(0.5)
Kc=0.024
Note: If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I- would not have been squared. This would have given an incorrect answer.
utomatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Example 5
SO2(g)+O2(g)⇌SO3(g)
First, make sure the equation is balanced.
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| Check to make sure S is equal on both sides. There is one on each side. Next look at the O. There are four on the left side and three on the right. Adding a coefficient to the O2 on the left is ineffective, as the S on right must also be increased. Instead, add a coefficient to the SO2 on the left and the SO3 on the right. | 2SO2+O2⇌2SO3 |
| The equation is now balanced. | 2SO2+O2⇌2SO3 |
Calculate Kp. The partial pressures are as follows: SO2- 0.25 atm, O2- 0.45 atm, SO3- 0.3 atm
Kp=(0.3)2(0.25)2×(0.45)
Kp=3.2
Contributors and Attributions
Charlotte Hutton, Sarah Reno, Curtis Kortemeier
Chemical Equilibria
Calculating an Equilibrium Concentration |
11572 | https://www.slideshare.net/slideshow/interhalogen-compounds/250180793 | Interhalogen compounds | PPTX | Chemistry | Science
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Interhalogen compounds
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1) Interhalogen compounds are formed by the direct combination of two different halogens. They are represented as XYn, where X has a lower electronegativity and larger size than Y. 2) Interhalogen compounds can be synthesized by the direct combination of halogen molecules or by reacting a halogen with a lower interhalogen compound. Examples of synthesis reactions are given for ClF, ICl, BrF3, ClF3, BrF5, IF5, and IF7. 3) The structure and bonding of interhalogen compounds can be explained by hybridization. Diatomic, tetra-atomic, hexa-atomic, and octa-atomic interhal
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Slide 1 Introduction to Interhalogen Compounds
Overview of Interhalogen Compounds studied in the presentation, including definition, types, and structure. View
Slide 2 Definition of Interhalogen Compounds
Definition of interhalogen compounds defined as XYn where X and Y are different halogens based on electronegativity. View
Slide 3 Classification of Interhalogen Compounds
Types of interhalogen compounds categorized based on the number of halogen atoms: Diatomic (AX), Tetra-atomic (AX3), Hexa-atomic (AX5), and Octa-atomic (AX7). View
Slide 4 Synthesis Methods for Interhalogen Compounds
Two methods of synthesizing interhalogen compounds: direct combination of halogens and reaction with lower interhalogen to form a higher one. View
Slide 5 Synthesis of Chlorine Monofluoride
Chlorine monofluoride (ClF) synthesis details, properties, and reactivity. View
Slide 6 Synthesis of Iodine Monochloride
Formation of iodine monochloride (ICl) via chlorine and iodine reaction, highlighting its properties. View
Slide 7 Synthesis of Bromine Trifluoride
Details on the synthesis and properties of bromine trifluoride (BrF3), emphasizing its role as a fluorinating agent. View
Slide 8 Synthesis of Chlorine Trifluoride
Processes for synthesizing chlorine trifluoride (ClF3) and its unique characteristics. View
Slide 9 Synthesis of Bromine Pentafluoride
Production and properties of bromine pentafluoride (BrF5) as a fluorinating agent. View
Slide 10 Synthesis of Iodine Pentafluoride
Synthesis details and characteristics of iodine pentafluoride (IF5), including shape and conductivity. View
Slide 11 Synthesis of Iodine Heptafluoride
Formation of iodine heptafluoride (IF7), highlighting its properties as an oxidizing agent. View
Slide 12 Applications of Interhalogen Compounds
Various applications of interhalogen compounds as solvents, catalysts, and fluorinating agents. View
Slide 13 Structure and Bonding in Diatomic IC
Structure and bonding explained using hybridization for diatomic interhalogen compounds like ClF. View
Slide 14 Structure in Tetra-atomic IC
Hybridization principles used to explain the geometry of tetra-atomic interhalogen compounds. View
Slide 15 Structure in Hexa-atomic IC
Bonding in hexa-atomic interhalogen compounds explained via hybridization and geometry. View
Slide 16 Structure in Octa-atomic IC
Structure of octa-atomic interhalogen compounds detailed, focusing on geometry and hybridization. View
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Interhalogen compounds
MAHATMA GANDHI,ARTS, SCIENCEAND LATE N.P.COMMERCE COLLEGE ARMORI DIST. GADCHIROLI B. SC.II SEM. III - PRESENTATION UNIT-I INTERHALOGEN COMPOUNDS CONTENTS 1)DEFINITION 2)TYPES OF I. C. 3)SYNTHESIS 4)PROPERTIES 5)STRUCTURE S.M.SONTAKKE Assistant professor CHEMISTRY
INTERHALOGEN COMPOUNDS DEFINITION Compounds ProducedBy the Direct Combination of two halogen atom of different electronegativity are called as inter halogen compounds. In General they are represented as XYn Where n= 1,3,5 or 7 X= Halogen atom with lower electronegativity and bigger size Y= halogen atom with higher electronegativity and smaller size Examples ClF,BrF,BrF3,IF5,IF7, Halogen F Cl Br I EN Decreases Size Increases
Types of Interhalogen Compounds AX AX3 AX5 AX7 Diatomic IC Tetra-atomic IC Hexa-atomic IC Octa-atomic IC e.g. ClF, BrF, IF e.g. ClF3 BrF3 IF5 BrF5 IF7 Inter halogen Compounds are named as name of first halogen followed by name of second halide with number of atoms i.e. ending –ine of halogen is replaced by –de e.g. i) ClF is named as Chlorine monofluoride. ii) ClF3 is named as Chlorine trifluoride. iii) BrF5 is named as Bromine pentafluoride.
Methods of Synthesisof Inter halogen Compounds. There are two general methods of synthesis of Interhalogen Compound. 1) By direct Combination of two halogen molecules at particular temperature condition. e.g. when equal volume of chlorine and fluorine react together at 473K , chlorine monofluoride formed. 2) Halogen molecule react with lower IC to give higher IC at particular temperature condition. e.g. molecular fluorine react with iodine pentafluoride at 543K to give higher IC iodine hepta fluoride
Synthesis ofChlorine monofluoride:- (ClF) Chlorine monofluoride is prepared by the direct combinition of molecular chlorine with fluorine at 473K. Reaction:- F2 + Cl2 2ClF Chlorine monofluoride is a colorless gas with melting point117K. It is covalent molecule and diamagnetic in nature. It is more reactive than normal halogen as the molecule is polar in nature.
Synthesis ofIodine mono chloride (ICl) Iodine mono chloride is formed by passing chlorine gas over solid iodine at temperature 273K. Reaction:- It is red-brown chemical compound melt near room temperature. ICl is highly polar molecule due to difference in electronegativity. In organic synthesis, estimation of Iodine number of oils as a source of I+.
Synthesis of Brominetrifluoride (BrF3 ) Bromine trifluoride is produced by the mixing of bromine vapor and fluorine in a stream of nitrogen at 293K Reaction:- It is straw colored liquid with pungent odor It is powerful fluorinating agents It is used to produce uranium hexafluoride UF6 in the processing and reprocessing of nuclear fuel.
Synthesis of Chlorinetrifluoride (ClF3 ) Chlorine trifluoride is synthesized by the mixing of molecular chlorine and fluorine at 473-573K in copper vessel. Reaction:- Cl2 + 3F2 2ClF3 Chlorine trifluoride is a colorless gas which condenses to give pale green liquid. It is hydrolyzed by water to give ClOF. When chlorine trifluoride is reacted with molecular fluorine it gives higher inter halogen compound ClF5. Shape of Chlorine trifluoride is bent T type due to two lp of electrons.
Synthesis ofBromine pentafluoride (BrF5 ) Bromine pentafluoride is prepared by the direct reaction of molecular bromine with excess of fluorine at temperature 423K. Reaction:- Bromine pentafluoride is pale yellow liquid. It is an extremely effective fluorinating agent, converting most uranium compound to the hexafluoride at room temperature.
Synthesis ofIodine pentafluoride ( IF5 ) Iodine pentafluoride is prepared by the direct combination of Iodine with excess of fluorine. Reaction:- I2 + 5F2 2IF5 Iodine pentafluoride is a colorless liquid with melting point 283K. It is good conductor of electricity. It is hydrolyzed by water to give halogen acid and oxyacid of larger halogen atom. Iodine pentafluoride has square pyramidal shape with one lp of electron.
Synthesis ofIodine heptafluoride (IF7) Iodine heptafluoride is synthesized by passing a mixture of iodine pentafluoride vapors and fluorine through a platinum tube at 573K. Reaction:- Iodine heptafluoride is a colorless gas. It is used as strong oxidizing agent. It is used to prepare periodic acid.
Applications ofInter halogen Compounds Inter halogen compounds are used as non-aqueous solvents in many chemical reaction. They are used as Catalyst in chemical reaction. Inter halogen compounds are used as fluorinating agents in several chemical reactions.
Structure and Bondingin Interhalogen Compounds A) Diatomic IC (AX type)ClF, ICl, IBr etc. Structure and Bonding in Interhalogen Compounds can be explain on the basis of Hybridization. Valence shell EC of Central atom Chlorine is 3s2 3p5 Diatomic IC 7+7=14 e/8 i.e. 1bp and 3lp Central atom Chlorine undergo sp3 Hybridization According to VSEPR theory, Shape of ClF is linear as it contain two atoms only Out of four hybridized orbital, three contain lp of electron and one is bp having linear structure
B) Structure andBonding in Tetra-atomic IC (AX3 type) e.g. ClF3, BrF3, IF3, ICl3 Structure of Chlorine tri-fluoride can be explain on the basis of hybridization. In, ClF3 central atom Chlorine has valence shell EC 3s23p5 In tetra-atomic IC 7X4=28e/8 i.e. 3bp and 2lp Central atom Chlorine undergo sp3d hybridization resulting Trigonal bipyramidal geometry According to VSEPR theory, shape of ClF3 is bent T shape as it contain three bp and two lp Out of five hybridized orbital three contain bp and two contain lp
C) Structure andBonding in Hexa-atomic IC (AX5 type) e.g. BrF5 IF5 etc. Structure and bonding in IF5 can be explain on the basis of Hybridization In IF5 Central atom Iodine has valence shell EC 5S25P5 In hexa-atomic IC , IF5 7x6=42/8=5bp and one lp In IF5 ,Central atom Bromine undergo sp3d2 hybridization resulting Octahedral geometry According to VSEPR theory, Shape of IF5 is Square pyramidal due to 5 bp and one lp Out of six hybridized orbital five are bp and one is lp decreasing bond angle to 81.9
D) Structure andBonding in Octa-atomic IC (AX7 type) e.g. IF7 Structure and bonding in IC can be explain on the basis of hybridization In IF7, Central atom Iodine has valence shell EC 5s25p2 In octa-atomic IC ,like IF7 8x7=56/8 i.e. 7bp and zero lp. In IF7 central atom iodine undergo sp3d3 hybridization resulting in pentagonal bipyramidal geometry. According to VSEPR theory lp is zero therefore IF7 has regular shape i.e. pentagonal bipyramidal shape. In IF7 all the seven hybridized orbital forms seven covalent bond with fluorine atom having regular shape.
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World ModelingFrontpage
5
[ Question ]
About probabilities
by mikbp
27th Nov 2022
1 min read
A 49
5
There is one thing I don't understand about probabilities:
If we toss a coin, there is a 50% chance that it shows heads or tails. If we do it 20 times and all of them showed heads, there is still a 50% chance that the next one shows heads, since the tosses are independent. However, we also know that series of X tosses showing heads are increasingly improbable when X grows. So, although there is a 50% chance that the toss shows heads again, at the same time the probability that it shows heads again are lower.
Why do we have to take into account one piece of information and not the other one when finding the probability that the next toss will show heads or tails? Are there 2 (or more) types of probabilities and I am just mixing them up (I'm thinking on things like the reported "probabilities" that polls show about one party or another getting elected in an election, for example)? Is the difference related to ergodicity (time vs ensemble averages)?
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About probabilities
7 Chloe Thompson
1 mikbp
3 noggin-scratcher
2 Chloe Thompson
2 rur
1 mikbp
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4 Dagon
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Chloe Thompson
Nov 27, 2022
7 3
Ah yes this was confusing to me for a while too, glad to be able to help someone else out with it!
The key thing to realise for me, is that the probability of 21 heads in a row changes as you toss each of those 21 coins.
The sequence of 21 heads in a row does indeed have much less than 0.5 chance, to be precise 0.5 21, which is 0.000000476837158. But it only has such a tiny probability before any of those 21 coins have been tossed. However as soon as the first coin is tossed, the probability of those 21 coins all being heads changes. If first coin is tails, the probability of all 21 coins being heads goes down to 0, if first coin is heads the probability of all 21 coins being heads goes up to 0.5 20. Say you by unlikely luck keep tossing heads. Then with each additional heads in a row you toss, the probability of all 21 coins being heads goes steadily up and up, til by the time you've tossed 20 heads in a row, the probability of all 21 being heads is now.... 0.5, i.e. the same as a the probability of a single coin toss being heads! And our apparent contradition is gone :)
The more 'mathematical' way to express this would be: The unconditional probability of tossing 21 heads in a row is 0.5 21, i.e. 0.000000476837158 but the probability of tossing 21 heads in a row conditional on having already tossed 20 heads in a row is 0.5.
P(21 heads)=0.5 21=0.000000476837158
P(21 heads|20 heads)=0.5
Let me know if any of that is still confusing.
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[-]mikbp3y 1 0
I think you explain it very well!
So the thing is something like the following, right?: "Looking at it from the outside, a world where 21 heads showed in a row is incredibly unlikely: (if the coin is fair) I would happily bet against this world happening. However, I am already in an incredibly weird world where 20 heads have shown in a row, and another heads only makes it a bit more weird, so I don't know what to bet, heads or tails."
Reply
3 noggin-scratcher 3y
Yes, essentially. While 21 heads in a row is very unlikely (when you consider it ahead of flipping any coins), by the time you get to 20 heads in a row most of the unlikely-ness of it has already happened, with the odds of one more head remaining the same as ever.
2 Chloe Thompson 3y
Yep that's it! Glad my explanation helped. (Though if we want to be a bit pedantic about it, we'd say that actually a world where 21 heads in a row ever happens is not unlikely (If heaps and heaps of coin tosses happen across the world over time, like in our world), but a world where any particular given sequence of 21 coin flips is all heads is yes very unlikely (before any of them have been flipped)).)
2 rur 3y
The advice is: do not bet. Suppose you download a gambling app that bets on games where the outcome is similar to a coin flip. You start receiving emails from someone associated with the app (so they bypass your spam filters). Each day for 20 days you receive an email predicting the outcome of the game. Each of the 20 predictions is correct. What do you do? Nothing. What you are unaware of (but should suspect) is that on the first email, the sender has sent out 8 million emails making a prediction (it is a popular gambling app). 4 million of those predicted the home team wins and the other 4 million predicted the visiting team wins. The next day the emails only goes out to those that received the correct prediction. Rinse. Repeat. And you happen to be an (un)lucky recipient of the 21st email distribution. The world you live in is no weirder than the world a Powerball Lottery winner lives in.
1 mikbp 3y
That's a nice example. I heard about it long ago with investments instead of games. It is really something important to keep in mind!
[+][comment deleted]3y 1 0
Deleted by Chloe Thompson, 11/27/2022
Viliam
Nov 28, 2022
5 0
A sequence of 100 heads is only half as likely as a sequence of 99 heads. Which is why the probability of the 100th coinflip being head is exactly one half.
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Dagon
Nov 28, 2022
4 0
One way to think of this: Uncertainty (at least on this level) is in the observer, not the coin. It comes up heads or comes up tails, with 100% chance of the thing that actually happens.
Before the flip, you assign 50% to each outcome, but that’s your uncertainty, not the coin’s, and the result may as well be secretly predetermined by the universe. After you’ve seen 20 heads, that part is now probability 100% (it’s knowledge, not uncertainty, on your part), and the next flip is still 50/50 (to your knowledge, presuming you have reason to trust the coin and not update toward an unfair flipper).
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JBlack
Nov 27, 2022
3 0
Are there 2 (or more) types of probabilities and I am just mixing them up
Yes, there are conditional probabilities and unconditional probabilities.
The unconditional probability of 21 heads in a row is 0.5^21.
The conditional probability of 21 heads in a row given that the first 20 were all heads is 0.5.
Conditional probability is just a division: the conditional probability of some event A given that B happened is just the unconditional probability of both A and B divided by the probability of B. In symbols: P(A | B) = P(A & B) / P(B).
Bayes' Law comes from simple algebra on this.
^As is common, this assumes that the coin flips are independent of one another. An alternative might be that the coin was flipped "lazily" such that it more often shows the same face as the previous flip, but over the long run still flips 50% heads. A "properly" flipped coin should not depend upon the results of any or all previous flips.
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About probabilities — LessWrong
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11574 | https://www.quora.com/Ten-lights-of-different-colors-are-to-be-arranged-in-a-circle-What-is-the-number-of-arrangements-if-the-two-colors-red-and-blue-must-always-be-together | Something went wrong. Wait a moment and try again.
Special Arrangements
Blue (color)
Permutation and Combinato...
Different Colors
Colors (general)
Permeuatuons and Combinat...
Circular Arrangements
Red (color)
5
Ten lights of different colors are to be arranged in a circle. What is the number of arrangements if the two colors red and blue must always be together?
E.M. Salvador
Former Senior Technical Editor and Writer (1976–1987)
·
Author has 7.5K answers and 2.6M answer views
·
2y
In how many ways can the colours of rainbow be arranged so that the red and the blue colours are always together?
colours of rainbow be arranged so that the red and the blue colours are always together
—
Impossible!
FYI,
the colors of the rainbow displays the natural arrangement of the various color wavelengths making up white light—i.e., from the longest wavelength, RED with the lowest vibratory frequency at the outer end, to the shortest, VIOLET with the highest vibratory frequency at the inner circle.
colours of rainbow be arranged so that the red and the blue colours are always together
—
Impossible!
FYI,
the colors of the rainbow displays the natural arrangement of the various color wavelengths making up white light—i.e., from the longest wavelength, RED with the lowest vibratory frequency at the outer end, to the shortest, VIOLET with the highest vibratory frequency at the inner circle.
Håkon Søreide
B.A. in English (language) & Comparative Literature, University of Bergen (Graduated 1996)
·
Author has 1K answers and 2M answer views
·
3y
The easiest way to start thinking about it to see what is happening is you have three lights, red (R), blue (B) and white (W). Red and blue must stay together, and so that makes two possible combinations of relative positions, and in a circle, the third light would only have one possible location relative to R and B:
R-B-W
B-R-W.
The number of arrangements is the calculation 2 x 1.
If you add anothe
The easiest way to start thinking about it to see what is happening is you have three lights, red (R), blue (B) and white (W). Red and blue must stay together, and so that makes two possible combinations of relative positions, and in a circle, the third light would only have one possible location relative to R and B:
R-B-W
B-R-W.
The number of arrangements is the calculation 2 x 1.
If you add another light, green (G), red and blue still only have two possible relative positions to each other, and the four options are:
R-B-W-G
R-B-G-W
B-R-W-G
B-R-G-W.
The calculation is 2 x 1 x 2.
When you add one more, you multiply the possibilities by 3. Let’s add yellow (Y):
R-B-W-G-Y
R-B-G-W-Y
B-R-W-Y-G
B-R-G-Y-W
R-B-Y-W-G
R-B-Y-G-W
B-R-...
Joseph Newcomer
Lives in Pitttsburgh PA (1967–present)
·
Author has 81.2K answers and 33.6M answer views
·
2y
Related
In how many ways can the colours of rainbow be arranged so that the red and the blue colours are always together?
None. In a rainbow, red and blue are always on opposite sides. Now, if you had asked, “how many ways can seven colors be arranged so that two of them are always adjacent?” which is the same as asking “How many ways can you arrange six bands of colors?” and that is easy. But it’s
your
homework, so I’ll let you look up how to compute the permutations (without repetition) of
n
elements taken
n
at a time.
Related questions
What colors mix together to make red and blue?
Which two colors make teal blue?
What color does mixing red, blue, and purple make?
Can blue and another color make red?
I have a bag full of 10 balls. 2 of them are red. 3 of them are green. How many of them are blue?
Assistant
Bot
·
1y
When arranging items in a circle, we typically fix one item to account for rotational symmetry. For this problem, we treat the two colors red and blue as a single unit or block since they must always be together.
Steps to Solve:
Treat Red and Blue as One Block:
This means instead of 10 individual lights, we now have 9 units to arrange (the red-blue block plus the other 8 lights).
Arranging the Blocks in a Circle:
The number of ways to arrange n items in a circle is given by (n−1)!. Here, n=9 (the 8 individual lights plus the red-blue block).
Thus, the arrangements in a circle
When arranging items in a circle, we typically fix one item to account for rotational symmetry. For this problem, we treat the two colors red and blue as a single unit or block since they must always be together.
Steps to Solve:
Treat Red and Blue as One Block:
This means instead of 10 individual lights, we now have 9 units to arrange (the red-blue block plus the other 8 lights).
Arranging the Blocks in a Circle:
The number of ways to arrange n items in a circle is given by (n−1)!. Here, n=9 (the 8 individual lights plus the red-blue block).
Thus, the arrangements in a circle are:
(9−1)!=8!=40320
Arranging Red and Blue Within Their Block:
Within the red-blue block, red and blue can be arranged in 2 ways (Red-Blue or Blue-Red).
Total Arrangements:
Therefore, the total number of arrangements is:
8!×2=40320×2=80640
Conclusion:
The total number of arrangements of the ten lights, with red and blue always together, is 80,640.
Saswata Banerjee
Author has 1.7K answers and 2.2M answer views
·
Feb 2
Related
In how many ways can two white, two green, and two red balls be arranged in a row without any consecutive pairs of the same color?
I am assuming that balls of similar colors are distinguishable, i.e., given the white balls, it is possible to tell them apart ( for e.g., they may have different sizes or different textures or different material, etc)
Ans:240
Let us denote by W, the set of arrangements where both the white balls are next to each other, by G, the set of arrangements where both the green balls are next to each other and by R, the set of arrangements where both the red balls are next to each other
We need to determine the value of ∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣
We denote the set
I am assuming that balls of similar colors are distinguishable, i.e., given the white balls, it is possible to tell them apart ( for e.g., they may have different sizes or different textures or different material, etc)
Ans:240
Let us denote by W, the set of arrangements where both the white balls are next to each other, by G, the set of arrangements where both the green balls are next to each other and by R, the set of arrangements where both the red balls are next to each other
We need to determine the value of ∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣
We denote the set of all possible arrangements of the balls by ξ
Clearly we have ∣ξ∣=6!=720
Let us calculate the value of ∣W∣. For this purpose, we’ll consider both the white balls as one ball temporarily. We have 5 balls that can be arranged in 5! ways. Now, the two white balls can be arranged among themselves in 2! ways
∴∣W∣=5!2!=240
Similarly, we have ∣G∣=∣R∣=240
Next let’s calculate ∣W∩G∣
For this purpose let’s consider both the white balls as one ball and both the green balls as one ball. Now, we have a total of 4 balls that can be arranged in 4! ways. The white balls can be arranged among themselves in 2! ways and the green balls can be arranged among themselves in 2! ways.
∴∣W∩G∣=4!2!2!=96
Similarly, ∣G∩R∣=∣R∩W∣=96
Finally, let’s calculate ∣W∩G∩R∣
For this purpose, we temporarily consider both the white balls as one ball, both the green balls as one ball and both the red balls as one ball. So, we have a total of 3 balls that can be arranged in 3! ways. Now, we can arrange the white balls among themselves in 2! ways, the green balls among themselves in 2! ways and and the red balls among themselves in 2! ways
∴∣W∩G∩R∣=3!2!2!2!=48
By the principle of inclusion and exclusion, we have
∣W∪G∪R∣=∣W∣+∣G∣+∣R∣−∣W∩G∣−∣G∩R∣−∣R∩W∣+∣W∩G∩R∣
Substituting the values that we have calculated previously, into the R.H.S of the identity above, we have
∣W∪G∪R∣=240+240+240−96−96−96+48
=480
By de Morgan’s law, we have
¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯W∪G∪R
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯W∪G∪R∣
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=∣ξ∣−∣W∪G∪R∣
Substituting the appropriate values in the R.H.S, we have
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=720−480
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=240
In this section, I’ll assume that balls of the same color are indistinguishable. The method remains the same
Ans:30
We will have to determine the value of ∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣
In this case though, we have
∣ξ∣=6!2!2!2!=90
Let’s calculate the value of ∣W∣. As before, we will consider them as one ball. So, we now have a total of 5 balls.
The number of ways of arrangement
∣W∣=5!2!2!=30
Similarly, ∣G∣=∣R∣=30
Next, we’ll calculate ∣W∩G∣
For this purpose, we will consider both the white balls as one ball and both the green balls as one ball. We now have a total of 4 balls
The number of ways of arrangement
∣W∩G∣=4!2!=12
Similarly, ∣G∩R∣=∣R∩W∣=12
Finally, in order to calculate the value of ∣W∩G∩R∣, we consider both the white balls as one, both the green balls as one and both the red balls as one. So, we have 3 balls
The number of ways of arrangement
∣W∩G∩R∣=3!=6
By the principle of inclusion and exclusion, we have
∣W∪G∪R∣=∣W∣+∣G∣+∣R∣−∣W∩G∣−∣G∩R∣−∣R∩W∣+∣W∩G∩R∣
Substituting the values that we have calculated previously, into the R.H.S of the identity above, we have
∣W∪G∪R∣=30+30+30−12−12−12+6
=60
By de Morgan’s law, we have
¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯W∪G∪R
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯W∪G∪R∣
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=∣ξ∣−∣W∪G∪R∣
Substituting the appropriate values in the R.H.S, we have
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=90−60
⇒∣¯¯¯¯¯¯W∩¯¯¯¯G∩¯¯¯¯R∣=30
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Ten couples are being seated in a circle. How many different seating arrangements are there if each couple must sit together?
If ten couples are being seated in circle, each being treated as unit, then we can just say that 10 units are being arranged in circle.
(We will address element of “couple” towards end of answer.)
Formula for circular permutation is: Circular Permutation OR Mathematics Formula
For generic formula, (10–1)! = 9! = 362880
Since there are 10 couples, each of which have two (2) different possible arrangements, 2^10 = 1024
Multiply both of these numbers:
362880 1024 = 371589120
If ten couples are being seated in circle, each being treated as unit, then we can just say that 10 units are being arranged in circle.
(We will address element of “couple” towards end of answer.)
Formula for circular permutation is: Circular Permutation OR Mathematics Formula
For generic formula, (10–1)! = 9! = 362880
Since there are 10 couples, each of which have two (2) different possible arrangements, 2^10 = 1024
Multiply both of these numbers:
362880 1024 = 371589120
Related questions
Are there any other two colors that make blue?
Can you combine two other colors to make red? If so, what two colors do you use?
Which color is red + blue?
What two colors make red?
Which color is involved to make a light red color?
Jeff Leach
Former Civilian Computer Programmer US Dept of Defense (1977–2003)
·
Author has 814 answers and 40M answer views
·
4y
Related
In how many ways can 3 red, 4 yellow, and 2 blue bulbs be arranged in a string of Christmas tree lights with 9 sockets?
There is an excellent website that describes permutations with repetitions which is the same as our Christmas tree lights problem:
Permutations with Repetition
So in our case there are 1,260 ways that we can arrange the light bulbs:
9!3! 4! 2!=362,880(6)(24)(2)=362,880288=1,260
There is an excellent website that describes permutations with repetitions which is the same as our Christmas tree lights problem:
Permutations with Repetition
So in our case there are 1,260 ways that we can arrange the light bulbs:
9!3! 4! 2!=362,880(6)(24)(2)=362,880288=1,260
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Tim Farage
Professor, Mathematics and Computer Science, Retired
·
Author has 4.8K answers and 17.6M answer views
·
4y
Related
How many ways can three, red, four, yellow, and two blue bulbs be arranged in a string of Christmas lights with nine sockets?
If all the colors were different, there would be 9! ways to arrange them.
But because of the duplicate colors we must divide out by the factorial of the number of each color.
So we get:
9!3!4!2!
If all the colors were different, there would be 9! ways to arrange them.
But because of the duplicate colors we must divide out by the factorial of the number of each color.
So we get:
9!3!4!2!
John Gerig
Lives in Tampa, Florida, USA
·
Author has 5.7K answers and 7.2M answer views
·
7y
Related
In how many ways can 3 red, 4 yellow and 2 blue lights be arranged in a string of lights with 9 sockets?
If the lights were >>all<< unique, then there would be 9! ways of those 9 lights being arranged. But there are 3 interchangeable reds, and 4 interchangeable yellows, etc., involved, so we have to divide 9! by 3!, AND THEN by 4!, and FINALLY, by 2! (=2). (Someone else can do the arithmetic!)
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Vagabond Guy
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Updated May 1
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Airline Mistake Fare Alerts are my favorite travel hack for getting cheap flights all over the world for a fraction of the cost!
(Check out the screenshots below to show you the kinds of deals you can get)
👉 Instead of looking for cheap flights, let them find you!
Sign up for airline mistake fare and flash sales alerts and quit planning trips like it’s 1960—choosing dates and a destination first, then spending days (or weeks) trying to make those locked-in plans fit your budget. That’s like walking onto a car lot, announcing the exact model and purchase date, and hoping the price magically works out.
This makes zero sense! You would never agree to buy a certain model of car on a specific date, and then just be stuck with either paying whatever they wanted on that date, or worse you not have a car at all. There is no other thing in your life where you plan with such insanity as one does vacations. Why would you not just choose from a list of top destinations that are currently offering a huge discount on flights? You actually would, you just didn’t know there was options to be notified of every amazing deal and that’s the purpose of this post.
Mistake fare and flash sales alerts can save you 50–90% on every single flight you book! This one trick has allowed me to see more of the world than I thought I would ever be able to. Prior to planning my travel experiences in this manner, we always went on cruises because it was the only affordable way to do travel for me. Not anymore!
Examples of some of the deals I have booked, I wish I had screenshots of all of them because some of these are even better than these.
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There are a handful of websites that do this, here’s my favorite. They don’t sell airfare; they only find the best deals and alert you of those deals via email.
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If you have 10+ destinations on your list, you will receive alerts for more than half of them at some point throughout the year, it’s just when will the alert come? Regardless of when it comes, that’s when you’re going to book it!
When An Alert Comes Book Without Fear: Most are unaware that you have 24 hours to cancel any flight you book for a full refund no questions asked. This is a law passed by congress and enforced by the depart of transportation.
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Sergei Michailov
Retired System Analyst (2009–present)
·
Author has 5.5K answers and 3.5M answer views
·
3y
29!
Consider red and blue as a single object. Then you have 9 objects to arrange which can by done 9! ways. We then multiply this by 2 because red and blue can swap places.
Blake Johnson
Engineer (2007–present)
·
7y
Related
Ten couples are being seated in a circle. How many different seating arrangements are there if each couple must sit together?
This has been answered correctly but I’ll try and explain it another way—
Assumption: The implication of the words CIRCLE and ARRANGEMENT in the question mean that only the relative position of the couples is important. So let’s look at it from the perspective of the first couple— no matter where they sit, they’re at the “front of the table.”
The second couple has 9 positions to choose relative to the first couple (not 17 because there can’t be an odd number of spaces between couples or one couple will not be able to sit together, which would violate the terms of the question), the third couple
This has been answered correctly but I’ll try and explain it another way—
Assumption: The implication of the words CIRCLE and ARRANGEMENT in the question mean that only the relative position of the couples is important. So let’s look at it from the perspective of the first couple— no matter where they sit, they’re at the “front of the table.”
The second couple has 9 positions to choose relative to the first couple (not 17 because there can’t be an odd number of spaces between couples or one couple will not be able to sit together, which would violate the terms of the question), the third couple has 8 positions, the fourth has 7 and so on. Because each choice is independent and is a piece of the final arrangement, the number of choices each couple has must be multiplied together— 987654321 = (9!) = 362880
Now each couple can sit in one of two positions (think partner on your right versus partner on your left.) Each of the ten couples independently chooses who will be on the left or right and each of these decisions is a piece of the final arrangement. Therefore this choice must be multiplied together for each couple (including the first, because even at the front of the table, someone’s got to be on the left.)— 2222222222 = (2^10)= 1024
The arrangement of each couple relative to the first couple and the arrangement of each person relative to their partner are independent and a piece of the final arrangement, so multiply together— 3628801024= 371589120
With every choice accounted for, the total number of possible arrangements is 371,589,120
Mario Skrtic
IT at Public Libraries (2003–present)
·
Author has 4.4K answers and 1.1M answer views
·
Feb 3
Related
In how many ways can two white, two green, and two red balls be arranged in a row without any consecutive pairs of the same color?
Brute force in PariGP:
for (w1=1,5,
for (w2=w1+2,6,
for (g1=1,5,
if (g1!=w1 && g1!=w2,
for (g2=g1+2,6,
if (g2!=w1 && g2!=w2,
for (r1=1,5,
if (r1!=w1 && r1!=w2 && r1!=g1 && r1!=g2,
for (r2=r1+2,6,
if (r2!=w1 && r2!=w2 && r2!=g1 && r2!=g2,
print(s," ",w1,w2,g1,g2,r1,r2);
);
);
);
);
);
);
);
);
);
1 132546
2 134625
3 142536
4 142635
5 143526
6 143625
7 152436
8 153624
9 162435
10 163524
11 241536
12 241635
13 243516
14 243615
15 251346
16 251436
17 253614
18 254613
19 261435
20 263514
21 351426
22 351624
23 352416
24 352614
25 361425
26 361524
27 362415
28 362514
29 461325
30 462513
Tapas Bera
Author has 705 answers and 59.8K answer views
·
1y
Related
In how many ways can 10 candles of different colors be arranged in a circular candle holder?
We know that n different things can be arrange in circular table in (n-1)! Ways.
So, 10 different colors candle can be arranged in a circular candle holder in (10-1)!=9!=362880 ways.
Related questions
What colors mix together to make red and blue?
Which two colors make teal blue?
What color does mixing red, blue, and purple make?
Can blue and another color make red?
I have a bag full of 10 balls. 2 of them are red. 3 of them are green. How many of them are blue?
Are there any other two colors that make blue?
Can you combine two other colors to make red? If so, what two colors do you use?
Which color is red + blue?
What two colors make red?
Which color is involved to make a light red color?
What color does red, blue, and green make mixed together?
What color does blue and red make?
What TWP colors make the color blue?
What two colors make navy blue?
Are red and blue complementary colors?
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Chapter Contents
Definition
Adjacent Sides of a quadrilateral
Opposite Sides of a quadrilateral
Diagonals of a quadrilateral
Which are types of quadrilateral?
Parallelogram
Rectangle
Square
Rhombus
Trapezium
Isosceles Trapezium
Kite
Solved Examples
Practice Worksheets
Worksheet 1
Worksheet 2
Worksheet 3
Worksheet 4
Quadrilateral and Types of Quadrilateral
Found in topics:
2D Shapes
Introduction
Quadrilateral is one of the type of polygons, which we already studied in the chapter
Curve, Polygon and Types of Polygon. From this chapter, we can
recall that a quadrilateral is a polygon which has four sides and four interior angles.
In this chapter, we will learn in detail about the
quadrilateral
about its sides, diagonals. angles and its different types.
Definition
Quadrilateral is a plane closed figure formed by joining four line segments. It is a polygon with four sides.
Quadrilateral has four sides, four vertices and four angles.
In the above figure, ABCD is quadrilateral with four line segments AB, BC, CD and DA. AB, BC, CD and DA are
called sides of Quadrilateral ABCD. Also, A, B, C and D are called vertices of
quadrilateral.
Quadrilateral ABCD has four angles viz. ∠DAB, ∠ABC, ∠BCD and ∠CDA. We
can write these angles as ∠A, ∠B, ∠C and ∠D respectively.
Adjacent Sides of a quadrilateral
Two sides of a quadrilateral are said to be adjacent sides if they meet at a common end point. In above
figure, AB and BC are adjacent sides of the quadrilateral ABCD because they have a common end point B.
Also, BC and CD are adjacent sides of the quadrilateral ABCD because they have common end point C. CD and DA are
adjacent sides of the quadrilateral ABCD because they have common end point D.
DA and AB are also the adjacent sides of the quadrilateral ABCD because they have common end point A.
Opposite Sides of a quadrilateral
The sides of a quadrilateral are called as opposite sides if they do not have a common end point. Again,
referring to the above quadrilateral ABCD, AB and CD are the opposite sides because they have no common end
point.
Also, AD and BC are also the opposite sides of the quadrilateral ABCD because they also have no common end
point.
Diagonals of a quadrilateral
The line segment which joins the opposite vertices of a quadrilateral is called its diagonal.
Diagonals of quadrilateral
In the above quadrilateral ABCD, AC and BD are two diagonals of quadrilateral.
Why?
Because the AC line segment joins two opposite vertices A and C and hence AC is a diagonal of quadrilateral
ABCD.
Similarly, BD line segment joins two opposite vertices B and D, that makes BD the diagonal of quadrilateral
ABCD.
Which are types of quadrilateral?
As the number of sides of a quadrilateral are always fixed i.e. four, so, the types of Quadrilateral are
classified based on how the length of four sides vary and how inclined the sides are.
Let’s have a look at very common types of a quadrilateral.
Parallelogram
A quadrilateral, in which both pairs of opposite sides are parallel, is called parallelogram. Also,
opposite sides of a parallelogram are equal.
Parallelogram
In above figure, ABCD is a parallelogram because opposite sides AB and CD are parallel to each other and have
equal length. Also, AD and BC are parallel to each other and have equal length. Or, we can write them as:
AB = CD and AB || CD. Also, AD = BC and AD || BC
Hence, the above ABCD quadrilateral is a parallelogram.
A diagonal of parallelogram divides it into two congruent triangles and the two diagonals of a parallelogram
bisect each other.
Parallelogram
In above parallelogram ABCD, the two diagonals AC and BD bisect each other at O. Therefore, OA=OC and OB=OD.
Rectangle
A quadrilateral in which both pairs of opposite sides are parallel, equal in length and each of four angles is
of 90°, is called rectangle.
Rectangle
In given figure, ABCD is a rectangle. Because, the opposite sides are parallel, i.e. AB||CD and BC||AD.
Secondly, the length of opposite sides are equal, i.e. AB = CD and BC = AD. And the last, all of four angles are
of 90°, i.e.
∠A = 90°, ∠B = 90°, ∠C = 90°, ∠D = 90°
Diagonals of rectangle are always equal in length and bisect each other.
Rectangle
In the above figure, AC = BD, because the diagonals of a rectangle are equal in length. And OA = OC and OB = OD,
because
diagonals of a rectangle bisects each other.
Square
A quadrilateral in which all sides are parallel, equal and each of its angle is 90 degree, is called
square.
Square
In above figure, ABCD is a square because AB || CD, BC || AD and AB = BC = CD = DA. ∠A = 90°,
∠B = 90°, ∠C = 90° and ∠D = 90°.
The diagonals of a square are equal in length and bisect each other at right angle.
Diagonals of square
In above figure, AC and BD are diagonals of the square ABCD and intersect each other at O. Both diagonals are
equal in length. i.e. AC = BD.
Both diagonals bisect each other. i.e. OA = OC and OB = OD
Both diagonals intersect each other at right angle. i.e. ∠AOB = ∠BOC = ∠COD =
∠AOD = 90°
Rhombus
A quadrilateral in which both pairs of opposite sides are parallel and all four sides are equal in length, is
called rhombus. Rhombus is also called as equilateral quadrilateral.
Rhombus
In above figure, ABCD is a rhombus because AB || CD, BC || AD and AB = BC = CD = DA. The diagonals of a
rhombus bisect each other at right angle.
Diagonals of rhombus
In the above figure, the diagonals AC and BD bisect each other at right angle. i.e. OA = OC and OB = OD.
∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
Diagonals of rhombus bisect angles
Also, in above figure, diagonals of a rhombus bisect the angles of rhombus at vertices. i.e. ∠1
= ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8
Trapezium
A quadrilateral in which only one pair of opposite sides are parallel, is called trapezium.
Trapezium
In above figure, ABCD is a trapezium because AB || CD and AD and BC are non parallel sides.
Isosceles Trapezium
If two non parallel sides of trapezium are equal, it is called isosceles trapezium.
Isosceles Trapezium
In above isosceles trapezium ABCD, AB || CD and non parallel sides BC and AD are equal in length, i.e. BC = AD.
In isosceles trapezium, base angles are of equal magnitude. i.e. ∠A = ∠B and
∠C = ∠D
In isosceles trapezium, diagonals are equal in length. i.e. AC = BD.
Kite
A quadrilateral in which two pairs of adjacent sides are equal, is called kite.
Kite
In the above figure, ABCD is a kite because AB = BC and CD = DA.
Frequently Asked Questions
1) What is quadrilateral?
Quadrilateral is a plane figure which is formed by joining four line segments.
2) How many angles does a quadrilateral have?
A quadrilateral has four angles.
3) How many vertices does a quadrilateral have?
A quadrilateral has four vertices.
4) What is the sum of four angles of a quadrilateral?
Sum of four angles of a quadrilateral is 3600
5) How many diagonals does a quadrilateral have?
There are two diagonals in a quadrilateral.
Solved Examples
1) In a parallelogram, ABCD if ∠A = 600, find the other three angles.
Solution
given ∠A = 600
∴ ∠C = 600
∵ ∠A and ∠C are opposite angles of parallelogram.
∠A + ∠B = 1800
∵ adjacent angles in parallelogram are supplementary.
600 + ∠B = 1800
∠B = 1800 - 600
∠B = 1200
∴ ∠D = 1200
∵ ∠B and ∠D are opposite angles of parallelogram.
2) Angles of quadrilateral are 900, 800 and 1100. Find the fourth angle
Solution
Let fourth angle of quadrilateral be x.
x + 900 + 800 + 1100 = 3600 (∵ Sum of angles of
quadrilateral is 3600.)
x + 2800 = 3600
x = 3600 - 2800
x = 800
3) In rhombus PQRS, if ∠P = 700. Find the ∠Q, ∠R and ∠S.
Solution
∠P = 700
Also, ∠R = ∠P (∵ opposite angles of rhombus are equal)
∴ ∠R = 700
∠P + ∠Q = 1800 (∵ Sum of adjacent angles is 1800.)
700 + ∠Q = 1800
∠Q = 1800 - 700
∠Q = 1100
Also, ∠S = ∠Q (∵ opposite angles of rhombus are equal)
∴ ∠S = 1100
4) If two adjacent angles ∠A and ∠B of parallelogram ABCD are equal. Find all four angles of the parallelogram.
Solution
In parallelogram ABCD, ∠A and ∠B are adjacent angles.
Let ∠A = ∠B = x°
∴ ∠A + ∠B = 180° (∵ sum of adjacent angles of ||gm is 180°)
∴ x + x = 180°
2x = 180°
x = 90°
∴ ∠A = ∠B = 90°
Also, ∠A = ∠C (∵ diagonally opposite angles of ||gm are equal)
∠C = 90°
Also, ∠B = ∠D (∵ diagonally opposite angles of ||gm are equal)
∠D = 90°
5) Two adjacent angles ∠P and ∠Q of a parallelogram PQRS are in the ratio of 4 : 5. Find the ∠P and ∠Q.
Solution
∠P : ∠Q = 4 : 5
or it can be written as ∠P = 4x, ∠Q = 5x
Since adjacent interior angles are supplementary
∴ ∠P + ∠Q = 180°
4x + 5x = 180°
9x = 180°
x = 20°
∴ ∠P = 4x = 4 × 20 = 80°
∠Q = 5x = 5 × 20 = 100°
6) In a parallelogram ABCD ∠A = 110°. Find the values of x, y and z.
Solution
∠A = ∠BCD (∵ diagonally opposite angles of ||gm are equal)
∴ ∠BCD = 110° (∵ ∠A = 110°)
∴ y = ∠BCD = 110°
Also. ∠A + ∠B = 180° (∵ adjacent angles of ||gm are supplementary)
∴ ∠A + x = 180°
110° + = 180°
x = 180° - 110°
x = 70°
Also. ∠DCB + ∠DCE = 180° (∵ ∠DCB and ∠DCE form linear pair of angles)
∴ y + z = 180°
110° + z = 180°
z = 180° - 110°
z = 70°
7) In the parallelogram ABCD, find the values of x, y and z.
Solution
∠ADC = ∠ABC (∵ diagonally opposite angles of ||gm are equal)
∴ z = 110°
Also. ∠A and ∠B are adjacent angles of ||gm ABCD
∴ ∠A + ∠B = 180° (∵ adjacent angles of ||gm are supplementary)
(x + 30°) + 110° = 180°
x + 140° = 180°
x = 180° - 140°
z = 40°
In ΔABC
∠BCE = ∠BAC + ∠ABC (∵ exterior angle of a triangle is equal to sum of its
opposite interior angles)
y = 110° + 30°
y = 140°
8) In quadrilateral PQRS, the four angles are in the ratio of 1 : 2 : 3 : 4. Find the value of all the four angles.
Solution
Since ∠P, ∠Q, ∠R and ∠S are in the ratio of 1 : 2 : 3 : 4
Let ∠P = x, ∠Q = 2x, ∠R = 3x and ∠S = 4x
∴ ∠P + ∠Q + ∠R + ∠S = 360°
or x + 2x + 3x + 4x = 360°
10x = 360°
x = 36°
∴ ∠P = x = 36°
∠Q = 2x = 2 × 36° = 72°
∠R = 3x = 3 × 36° = 108°
∠S = 4x = 4 × 36° = 144°
Worksheet 1
Fill in the blanks
A quadrilateral has ___________ vertices.
A trapezium is a quadrilateral in which one pair of opposite sides is ___________.
Sum of interior angles of a quadrilateral is ___________.
A rectangle is a quadrilateral in which each of its angles is ___________.
In a rhombus opposite ___________ are equal.
Rhombus is a ___________ in which all its sides are equal.
In a square all sides are ___________.
A line segment which joins opposite vertices of a quadrilateral is called its
___________.
Diagonals of a ___________ are equal in length.
In a parallelogram, if one of its angles is 60° then its diagonally opposite angle must
be ___________.
Help
box
square
parallelogram
90°
four
360°
diagonal
angles
equal
60°
parallel
Download Worksheet 1
Worksheet 2
Write True or False in the boxes.
| S.N. | Statement | Write ✓ or ✕ |
---
| 1) | A quadrilateral has two diagonals. | ▭ |
| 2) | In isosceles trapezium, its base angles are equal. | ▭ |
| 3) | The diagonals of a parallelogram bisect each other at 90°. | ▭ |
| 4) | The diagonals of a rectangle bisect each other at 90°. | ▭ |
| 5) | The diagonals of a square bisect each other at 90°. | ▭ |
| 6) | The diagonals of a rhombus bisect each other at 90°. | ▭ |
| 7) | The diagonally opposite angles of a parallelogram are equal. | ▭ |
| 8) | The diagonally opposite angles of a rhombus are equal. | ▭ |
| 9) | The diagonals of a square are unequal in length. | ▭ |
| 10) | A parallelogram is a rhombus. | ▭ |
Download Worksheet 2
Worksheet 3
Match the following.
| | | | |
--- --- |
| 1) | Each angle of square is of measure | a) | parallel |
| 2) | Diagonals of rectangle are | b) | diagonals |
| 3) | Opposite sides of rhombus are | c) | congruent triangles |
| 4) | A quadrilateral has two | d) | equal |
| 5) | Sum of all angles of a rectangle is | e) | 90° |
| 6) | Diagonal divides a parallelogram into two | f) | 360° |
Download Worksheet 3
Worksheet 4
Multiple choice questions
1) Quadrilateral is a polygon with
four sides
five sides
curve
three sides
2) Quadrilateral has
four angles
six angles
three angles
five angles
3) Which of the following is a type of quadrilateral?
Triangle
Pentagon
Rhombus
Hexagon
4) The sum of all angles of a quadrilateral is
1 right angle
2 right angle
3 right angle
4 right angle
5) A quadrilateral in which one pair of opposites are parallel and one pair of opposite
sides are non parallel but equal in length is called as
Square
Rectangle
Rhombus
Isosceles trapezium
6) A parallelogram with all equal sides is called as
Rectangle
Square
Rhombus
Kite
7) A parallelogram with all sides as equal and having each angle of 900 is
called as
Rectangle
Square
Trapezium
Rhombus
8) A rectangle with any pair of adjacent sides as equal is called as a
Kite
Rhombus
Square
Trapezium
9) Consecutive angles of a parallelogram are supplementary.
True
False
Maybe
None of the above
10) A quadrilateral with both pair of opposite sides are parallel is called as
Square
Kite
Rhombus
Parallelogram
MCQs Answer Key
Download Worksheet 4
Last updated on: 24-07-2025
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Andy needs to determine the freezing point depression of 30.0 mL of 0.50 m AlCl3 aqueous solution. When dissolved, AlCl3 dissociates into Al3+ and Cl−. Andy realizes that he may need to reduce the effect of ion-pairing in his 0.50 m AlCl3 aqueous solution. Which of the following would reduce the ion-pairing effect in his AlCl3 solution?Increase the
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Question: Andy needs to determine the freezing point depression of 30.0 mL of 0.50 m AlCl3 aqueous solution. When dissolved, AlCl3 dissociates into Al3+ and Cl−. Andy realizes that he may need to reduce the effect of ion-pairing in his 0.50 m AlCl3 aqueous solution. Which of the following would reduce the ion-pairing effect in his AlCl3 solution?Increase the
Andy needs to determine the freezing point depression of 3 0.0 mL of 0.5 0 m AlCl 3 aqueous solution. When dissolved, AlCl 3 dissociates into Al 3+ and Cl−. Andy realizes that he may need to reduce the effect of ion-pairing in his 0.5 0 m AlCl 3 aqueous solution. Which of the following would reduce the ion-pairing effect in his AlCl 3 solution? Increase the concentration of the solution by adding AlCl 3 solid. Dilute the solution by adding some water. Use 4 5 mL of the 0.5 0 m AlCl 3 solution. Use only half of his initial volume
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11577 | https://www.math.lsu.edu/~mahlburg/teaching/08-18781.html | Jump to: Problem Sets Lecture Schedule Handouts Resources Links Past Updates
| | | |
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| Current Update: | 12/11/08 | The solutions to Problem Set 11 are now available. I will have office hours as usual on Thursday, Dec. 11. During finals week, Sawyer will have an office hour on Tues, Dec. 16 from 1-2, and I will hold a session on Wed., Dec. 17 from 2 - 3:30. You may pick up your graded PSets during office hours (or we can arrange another time if needed). You may now download review problems for the Final Exam. |
MATH 18.781 - Fall 2008
Introduction to the Theory of Numbers
This is an introductory course in number theory at the undergraduate level. Topics will include divisibility, greatest common divisors, the Euclidean algorithm, the Fundamental Theorem of Arithmetic, the Chinese Remainder Theorem, Hensel's Lemma, Legendre symbols, quadratic reciprocity, simple continued fractions, infinite continued fractions, and Farey fractions. There are no official prerequisites, but abstract algebra (18.700) is very helpful.
| | | |
---
| Course Information | Scheduled Time | Room |
| Lectures | TR 1:00 | 2-102 |
| Office Hours (Karl Mahlburg) | R 3:00 - 4:00 | 2-173 |
| Office Hours (Sawyer Tabony) | M 2:00 - 3:00 | 2-094 |
|
| | |
--- |
| Textbooks (recommended) | I. Niven, H. Zuckerman, H. Montgomery, An Introduction to the Theory of Numbers G. H. Hardy, E. Wright, An Introduction to the Theory of Numbers |
| | | |
---
| Instructor: | Karl | Mahlburg |
| | Phone: | (617) 253-2685 |
| | e-mail: | mahlburg (at) math (dot) mit (dot) edu. |
| | Office: | 2-173 (First floor of Math building, south wing) |
| Teaching Assistant: | Sawyer | Tabony |
| | Phone: | (617) 452-1199 |
| | e-mail: | sawyer (at) math (dot) mit (dot) edu. |
| | Office: | 2-094 (Basement) |
Problem Sets
There will be approximately 10 problem sets, due on Tuesdays in class. The majority of the problems will be taken from Niven, Zuckerman, and Montgomery. Please start working on them early, since important results and/or proofs will often appear!
Problem Set 1 - Due Sep. 16 (Solutions)
Problem Set 2 - Due Sep. 23 (Solutions)
Problem Set 3 - Due Sep. 23 (Solutions)
Problem Set 4 - Due Oct. 7 (Solutions)
Problem Set 5 - Due Oct. 14 (Solutions)
Problem Set 6 - Due Oct. 21 (Solutions)
Problem Set 7 - Due Oct. 28 (Solutions)
Problem Set 8 - Due Nov. 4 (Solutions)
Problem Set 9 - Due Nov. 13 (Solutions)
Problem Set 10 - Due Nov. 25 (Solutions)
Problem Set 11 - Due Dec. 4 (Solutions)
Lecture Schedule
Watch for continued updates throughout the semester.
| | | |
---
| Date | Topics | Reading |
| Sep. 4 (R) | Intro; Pythagorean Numbers I | Niven 5.3, 1.2 |
| Sep. 9 (T) | Pythagorean Numbers II; Division algorithm; gcd's & Euclidean algorithm | Niven 1.2, 1.3 |
| Sep. 11 (R) | More gcd's; lcm's; linear equations; primes | Niven 1.3 |
| Sep. 16 (T) | Fundamental theorem of arithmetic; factorization; congruences | Niven 1.3, 2.1, 2.2 |
| Sep. 18 (R) | Solving linear congruence equations; residue systems; Fermat's little theorem | Niven 2.1, 2.2 |
| Sep. 23 (T) | Chinese remainder theorem; Euler's totient function | Niven 2.3 |
| Sep. 25 (R) | Arithmetic functions | Niven 4.2 |
| Sep. 30 (T) | Mobius inversion; Computational techniques; primality testing | Niven 4.3, 2.4 |
| Oct. 2 (R) | Primality testing | Niven 2.4 |
| Oct. 7 (T) | RSA cryptography; Multiplicative structure modulo primes | Niven 2.5, 2.8 |
| Oct. 9 (R) | Midterm Exam I | |
| Oct. 14 (T) | Primitive roots modulo primes and composites; binomial coefficients | Niven 2.8, 1.4 |
| Oct. 16 (R) | Binomial coefficients; primitive roots modulo odd prime powers | Niven 1.4, 2.8 |
| Oct. 21 (T) | Primitive roots modulo powers of 2; solving polynomial congruences modulo prime powers | Niven 2.8, 2.6 |
| Oct. 23 (R) | Quadratic equations modulo primes; quadratic residues and Legendre symbol | Niven 3.1 |
| Oct. 28 (T) | Quadratic reciprocity | Niven 3.2 |
| Oct. 30 (R) | Jacobi symbol; generalized quadratic reciprocity | Niven 3.3 |
| Nov. 4 (T) | Pell's equation; rational continued fractions | Niven 7.1, 7.2 |
| Nov. 6 (R) | Periodic continued fractions; Pell's equation | Niven 7.3, 7.7, 7.8 |
| Nov. 11 (T) | No class - Veteran's Day | |
| Nov. 13 (R) | Midterm Exam II | |
| Nov. 18 (T) | Rational approximations of irrationals | Niven 7.4, 7.5 |
| Nov. 20 (R) | Periodic continued fractions and quadratic irrationals; Pell's equation | Niven 7.6, 7.7, 7.8 |
| Nov. 25 (T) | Farey fractions; rational approximations revisited | Niven 6.1, 6.2 |
| Nov. 27 (R) | No class - Thanksgiving | |
| Dec. 2 (T) | Rational approximations; partitions; algebraic numbers | Niven 6.2, 10.1, 10.2, 10.3, 9.2 |
| Dec. 4 (R) | Algebraic integers; unique factorization; Riemann hypothesis | Niven 9.2 - 9.8 |
| Dec. 9 (T) | Ramsey Theory; Van der Waerden's theorem | |
| | | |
---
| Final Exam | Thursday, Dec. 18 1:30PM - 4:30PM | Room 56-154 |
Handouts
Course syllabus
Exam 1 solutions
Exam 2 review problems
Exam 2 solutions
Final Exam review problems
Resources
Number Theory is one of the oldest subjects in mathematics, and there are a large number of introductory texts. Please supplement your reading with any of these:
K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory - covers a wide swathe of modern research areas from an elementary perspective. Highly recommended.
J. Silverman, A Friendly Introduction to Number Theory
Links
Homepage for SAGE - free, open-source mathematics computation package
Homepage for PARI/GP (and documentation) - another free, open-source computation package
Continued fraction calculator
Past Updates
12/10/08: Check back on Thursday to find practice problems for the final exam.
11/25/08: Problem Set 11 is now available, and is due on Thursday, Dec. 4. This is the final problem set of the semester. You may also download the solutions to Problem Set 10. My office hours are switched to Wednesday, Dec. 3 from 2:00 - 3:00 next week.
11/18/08: Problem Set 10 is now available. You can also review the solutions to Exam 2.
11/17/08: The solutions to Problem Set 9 are available. Problem Set 10 will be posted on Tues., Nov. 18 and is due on Tues, Nov. 25.
11/07/08: Review problems for Exam 2 are now available. It's a long list, so you shouldn't necessarily solve all of them in your studying! No class on Tuesday, Nov. 11th. Midterm II is in class on Thursday, Nov. 13th. My office hours this week are on Wed., Nov. 12th from 2:00 - 3:00. Sawyer's are on Monday as usual.
11/04/08: Problem Set 9 is now available, as are the solutions for Problem Set 8.
10/28/08: Problem Set 8 is now available, as are the solutions for Problem Set 7.
10/27/08: Problem Set 6 solutions are now available (sorry for the delay!).
10/21/08: Problem Set 7 is now available. I will be gone the rest of the week, so Thursday office hours are canceled. Class will be taught by Sawyer.
10/15/08: Problem Set 6 is now available, as are the solutions for Problem Set 5.
10/10/08: You may now review the solutions to Exam 1. A typo on the bonus problem of PSet 5 has been fixed.
10/3/08: Problem Set 5 is now available, as are the solutions for Problem Set 4. Midterm I will be in-class on Thursday, Oct. 9. It is open-note and open-book, but no laptops are permitted. Calculators are allowed, but are not necessary. Topics up through Niven 2.5 may be included. My office hour is changed to Wednesday, Oct. 9, 2:00 - 3:00 next week.
9/30/08: Problem Set 4 is now available, as are the solutions for Problem Set 3. Sawyer's office hours are now Monday afternoons.
9/23/08: Problem Set 3 is now available, as are the solutions for Problem Set 2.
9/16/08: Problem Set 2 is now available, as are the solutions for Problem Set 1.
9/10/08: The Lecture Schedule can be found further down on this page. Sawyer's office hours will be on Fridays, 11:00 - 12:00. The course textbook (Niven et al) has finally arrived in the bookstore.
9/04/08: PSet 1 and the course syllabus are now available. Due to a clerical error (mainly mine), Niven's book was not ordered in sufficient quantities and is apparently out at the COOP. I am working to resolve this as soon as possible, and will update when I have more information. In the meantime, almost any book in the library with a title of the sort "Introduction to Number Theory" will do - look on the shelves around QA241.
Jump to: Top Course Info Handouts Resources Links
Back to Karl Mahlburg's homepage |
11578 | https://en.wikipedia.org/wiki/Quadrature_of_the_Parabola | Quadrature of the Parabola - Wikipedia
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(Top)
1 Main theorem
2 Structure of the text
3 Geometric proofToggle Geometric proof subsection
3.1 Dissection of the parabolic segment
3.2 Areas of the triangles
3.3 Sum of the series
4 See also
5 Notes
6 Further reading
7 External links
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From Wikipedia, the free encyclopedia
Geometric treatise by Archimedes
A parabolic segment.
Quadrature of the Parabola (Greek: Τετραγωνισμὸς παραβολῆς) is a treatise on geometry, written by Archimedes in the 3rd century BC and addressed to his Alexandrian acquaintance Dositheus. It contains 24 propositions regarding parabolas, culminating in two proofs showing that the area of a parabolic segment (the region enclosed by a parabola and a line) is 4 3{\displaystyle {\tfrac {4}{3}}} that of a certain inscribed triangle.
It is one of the best-known works of Archimedes, in particular for its ingenious use of the method of exhaustion and in the second part of a geometric series. Archimedes dissects the area into infinitely many triangles whose areas form a geometric progression. He then computes the sum of the resulting geometric series, and proves that this is the area of the parabolic segment. This represents the most sophisticated use of a reductio ad absurdum argument in ancient Greek mathematics, and Archimedes' solution remained unsurpassed until the development of integral calculus in the 17th century, being succeeded by Cavalieri's quadrature formula.
Main theorem
[edit]
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord. Proposition 1 of the work states that a line from the third vertex drawn parallel to the axis divides the chord into equal segments. The main theorem claims that the area of the parabolic segment is 4 3{\displaystyle {\tfrac {4}{3}}} that of the inscribed triangle.
Structure of the text
[edit]
Archimedes' first proof of the area of a parabolic segment.
Conic sections such as the parabola were already well known in Archimedes' time thanks to Menaechmus a century earlier. However, before the advent of the differential and integral calculus, there were no easy means to find the area of a conic section. Archimedes provides the first attested solution to this problem by focusing specifically on the area bounded by a parabola and a chord.
Archimedes gives two proofs of the main theorem: one using abstract mechanics and the other one by pure geometry. In the first proof, Archimedes considers a lever in equilibrium under the action of gravity, with weighted segments of a parabola and a triangle suspended along the arms of a lever at specific distances from the fulcrum. When the center of gravity of the triangle is known, the equilibrium of the lever yields the area of the parabola in terms of the area of the triangle which has the same base and equal height. Archimedes here deviates from the procedure found in On the Equilibrium of Planes in that he has the centers of gravity at a level below that of the balance. The second and more famous proof uses pure geometry, particularly the sum of a geometric series.
Of the twenty-four propositions, the first three are quoted without proof from Euclid's Elements of Conics (a lost work by Euclid on conic sections). Propositions 4 and 5 establish elementary properties of the parabola. Propositions 6–17 give the mechanical proof of the main theorem; propositions 18–24 present the geometric proof.
Geometric proof
[edit]
Archimedes' second proof dissects the area using an arbitrary number of triangles.
Dissection of the parabolic segment
[edit]
The main idea of the proof is the dissection of the parabolic segment into infinitely many triangles, as shown in the figure to the right. Each of these triangles is inscribed in its own parabolic segment in the same way that the blue triangle is inscribed in the large segment.
Areas of the triangles
[edit]
In propositions eighteen through twenty-one, Archimedes proves that the area of each green triangle is 1 8{\displaystyle {\tfrac {1}{8}}} the area of the blue triangle, so that both green triangles together sum to 1 4{\displaystyle {\tfrac {1}{4}}} the area of the blue triangle. From a modern point of view, this is because the green triangle has 1 2{\displaystyle {\tfrac {1}{2}}} the width and 1 4{\displaystyle {\tfrac {1}{4}}} the height of the blue triangle:
Following the same argument, each of the 4{\displaystyle 4} yellow triangles has 1 8{\displaystyle {\tfrac {1}{8}}} the area of a green triangle or 1 64{\displaystyle {\tfrac {1}{64}}} the area of the blue triangle, summing to 4 64=1 16{\displaystyle {\tfrac {4}{64}}={\tfrac {1}{16}}} the area of the blue triangle; each of the 2 3=8{\displaystyle 2^{3}=8} red triangles has 1 8{\displaystyle {\tfrac {1}{8}}} the area of a yellow triangle, summing to 2 3 8 3=1 64{\displaystyle {\tfrac {2^{3}}{8^{3}}}={\tfrac {1}{64}}} the area of the blue triangle; etc. Using the method of exhaustion, it follows that the total area of the parabolic segment is given by
Area=T+1 4 T+1 4 2 T+1 4 3 T+⋯.{\displaystyle {\text{Area}}\;=\;T\,+\,{\frac {1}{4}}T\,+\,{\frac {1}{4^{2}}}T\,+\,{\frac {1}{4^{3}}}T\,+\,\cdots .}
Here T represents the area of the large blue triangle, the second term represents the total area of the two green triangles, the third term represents the total area of the four yellow triangles, and so forth. This simplifies to give
Area=(1+1 4+1 16+1 64+⋯)T.{\displaystyle {\text{Area}}\;=\;\left(1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \right)T.}
Sum of the series
[edit]
Archimedes' proof that 1/4 + 1/16 + 1/64 + ⋯ = 1/3
To complete the proof, Archimedes shows that
1+1 4+1 16+1 64+⋯=4 3.{\displaystyle 1\,+\,{\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots \;=\;{\frac {4}{3}}.}
The formula above is a geometric series—each successive term is one fourth of the previous term. In modern mathematics, that formula is a special case of the sum formula for a geometric series.
Archimedes evaluates the sum using an entirely geometric method, illustrated in the adjacent picture. This picture shows a unit square which has been dissected into an infinity of smaller squares. Each successive purple square has one fourth the area of the previous square, with the total purple area being the sum
1 4+1 16+1 64+⋯.{\displaystyle {\frac {1}{4}}\,+\,{\frac {1}{16}}\,+\,{\frac {1}{64}}\,+\,\cdots .}
However, the purple squares are congruent to either set of yellow squares, and so cover 1 3{\displaystyle {\tfrac {1}{3}}} of the area of the unit square. It follows that the series above sums to 4 3{\displaystyle {\tfrac {4}{3}}} (since 1+1 3=4 3{\displaystyle 1+{\tfrac {1}{3}}={\tfrac {4}{3}}}).
See also
[edit]
Wikimedia Commons has media related to Quadrature of the Parabola.
Quadrature (geometry)
History of calculus
Notes
[edit]
^Swain, Gordon; Dence, Thomas (1998). "Archimedes' Quadrature of the Parabola Revisited". Mathematics Magazine. 71 (2): 123–130. doi:10.2307/2691014. ISSN0025-570X. JSTOR2691014.
^Cusick, Larry W. (2008). "Archimedean Quadrature Redux". Mathematics Magazine. 81 (2): 83–95. doi:10.1080/0025570X.2008.11953535. ISSN0025-570X. JSTOR27643090. S2CID126360876.
^Towne, R. (2018). "Archimedes in the Classroom". Master's Thesis. John Carroll University.
^"Quadrature of the parabola, Introduction". web.calstatela.edu. Retrieved 2021-07-03.
^"The Illustrated Method of Archimedes". Scribd. Retrieved 2021-07-03.
^Dijksterhuis, E. J. (1987). "Quadrature of the Parabola". Archimedes. pp.336–345.
^The green triangle has 1 2{\displaystyle {\tfrac {1}{2}}} the width of blue triangle by construction. The statement about the height follows from the geometric properties of a parabola, and is easy to prove using modern analytic geometry.
^Strictly speaking, Archimedes evaluates the partial sums of this series, and uses the Archimedean property to argue that the partial sums become arbitrarily close to 4 3{\displaystyle {\tfrac {4}{3}}}. This is logically equivalent to the modern idea of summing an infinite series.
Further reading
[edit]
Ajose, Sunday and Roger Nelsen (June 1994). "Proof without Words: Geometric Series". Mathematics Magazine. 67 (3): 230. doi:10.2307/2690617. JSTOR2690617.
Ancora, Luciano (2014). "Quadrature of the parabola with the square pyramidal number". Archimede. 66 (3). Archived from the original on 2018-07-25. Retrieved 2017-01-06.
Bressoud, David M. (2006). A Radical Approach to Real Analysis (2nd ed.). Mathematical Association of America. ISBN0-88385-747-2..
Dijksterhuis, E.J. (1987) "Archimedes", Princeton U. Press ISBN0-691-08421-1
Edwards Jr., C. H. (1994). The Historical Development of the Calculus (3rd ed.). Springer. ISBN0-387-94313-7..
Heath, Thomas L. (2011). The Works of Archimedes (2nd ed.). CreateSpace. ISBN978-1-4637-4473-1.
Simmons, George F. (2007). Calculus Gems. Mathematical Association of America. ISBN978-0-88385-561-4..
Stein, Sherman K. (1999). Archimedes: What Did He Do Besides Cry Eureka?. Mathematical Association of America. ISBN0-88385-718-9.
Stillwell, John (2004). Mathematics and its History (2nd ed.). Springer. ISBN0-387-95336-1..
Swain, Gordon and Thomas Dence (April 1998). "Archimedes' Quadrature of the Parabola Revisited". Mathematics Magazine. 71 (2): 123–30. doi:10.2307/2691014. JSTOR2691014.
Wilson, Alistair Macintosh (1995). The Infinite in the Finite. Oxford University Press. ISBN0-19-853950-9..
External links
[edit]
Look up quadrature in Wiktionary, the free dictionary.
Casselman, Bill. "Archimedes' quadrature of the parabola". Archived from the original on 2012-02-04. Full text, as translated by T.L. Heath.
Xavier University Department of Mathematics and Computer Science. "Archimedes of Syracuse". Archived from the original on 2016-01-13.. Text of propositions 1–3 and 20–24, with commentary.
| v t e Archimedes |
| Written works | Measurement of a Circle The Sand Reckoner On the Equilibrium of Planes Quadrature of the Parabola On the Sphere and Cylinder On Spirals On Conoids and Spheroids On Floating Bodies Ostomachion The Method of Mechanical Theorems Book of Lemmas (apocryphal) | |
| Discoveries and inventions | Archimedean solid Archimedes's cattle problem Archimedes' principle Archimedes's screw Claw of Archimedes |
| Miscellaneous | Archimedes' heat ray Archimedes Palimpsest List of things named after Archimedes Pseudo-Archimedes |
| Related people | Euclid Eudoxus of Cnidus Apollonius of Perga Hero of Alexandria Eutocius of Ascalon |
| Category |
| v t e Ancient Greek mathematics |
| Mathematicians (timeline) | Anaxagoras Anthemius Apollonius Archimedes Archytas Aristaeus the Elder Aristarchus Autolycus Bion Bryson Callippus Carpus Chrysippus Cleomedes Conon Ctesibius Democritus Dicaearchus Dinostratus Diocles Dionysodorus of Caunus Dionysodorus of Amisene Diophantus Domninus Eratosthenes Euclid Eudemus Eudoxus Eutocius Geminus Heliodorus Heron Hipparchus Hippasus Hippias Hippocrates Hypatia Hypsicles Isidore of Miletus Leon Marinus Menaechmus Menelaus Metrodorus Nicomachus Nicomedes Nicoteles Oenopides Pandrosion Pappus Perseus Philolaus Philon Philonides Porphyry of Tyre Posidonius Proclus Ptolemy Pythagoras Serenus Sosigenes Sporus Thales Theaetetus Theodorus Theodosius Theon of Alexandria Theon of Smyrna Thymaridas Xenocrates Zeno of Elea Zeno of Sidon Zenodorus |
| Treatises | Almagest Arithmetica Conics(Apollonius) Catoptrics Data(Euclid) Elements(Euclid) Little Astronomy Measurement of a Circle On Conoids and Spheroids On the Sizes and Distances(Aristarchus) On Sizes and Distances(Hipparchus) On the Moving Sphere(Autolycus) Optics(Euclid) On Spirals On the Sphere and Cylinder Ostomachion Phaenomena(Euclid) Planisphaerium Spherics(Theodosius) Spherics(Menelaus) The Quadrature of the Parabola The Sand Reckoner |
| Concepts and definitions | Chord Circles of Apollonius Apollonian circles Apollonian gasket Problem of Apollonius Commensurability Diophantine equation Euclidean geometry Golden ratio Lune of Hippocrates Method of exhaustion Parallel postulate Platonic solid Regular polygon Straightedge and compass construction Angle trisection Doubling the cube Squaring the circle Quadratrix of Hippias Neusis construction |
| Results | | In Elements | Angle bisector theorem Exterior angle theorem Euclidean algorithm Euclid's theorem Geometric mean theorem Hinge theorem Inscribed angle theorem Intercept theorem Intersecting chords theorem Intersecting secants theorem Law of cosines Pons asinorum Pythagorean theorem Tangent-secant theorem Thales's theorem Theorem of the gnomon | Apollonius's theorem Aristarchus's inequality Heron's formula Law of sines Menelaus's theorem Pappus's area theorem Problem II.8 of Arithmetica Ptolemy's inequality Ptolemy's table of chords Ptolemy's theorem Spiral of Theodorus |
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11579 | https://ftyracing.com/tech/afr-of-any-hydrocarbon/ | Skip to content
Custom parts for performance applications
Determining Stoichiometry of pretty much any fuel
When it comes down to it, you can burn so many different things to use as a fuel for your car. What do you need? Simple – Hydrogen and Carbon, generally, in a compound that work in conjunction with Oxygen. This mixture (called a hydrocarbon because it contains both hydrogen and carbon) combusts, which means that you’re burning it!
You may be a bit like me – when I started playing with cars, I didn’t know jack about fuels. Just that if I wanted to have fun on the weekend, I could find leaded race fuel at a few select gas stations in the area. Boy how times have changed! A few years ago, I started looking at chemical compounds with industrial availability to see if there was something generally available that would give me an upper hand as far as making power. Ultimately, I came up with a list of hydrocarbons I wanted to check out.
Ethanol – C2H6O
Methanol – CH4O
Acetone – C3H6O
MTBE – C5H12O
ETBE – C6H14O
Toluene – C7H8
Xylene – C8H1O
Nitromethane – CH3NO2
I pretty quickly realized that I needed to figure out how much of each of these compounds I would need to burn, which is where stoichiometry comes into play… Stoichiometry of a fuel is the ratio that you have to mix that fuel with oxygen in order to completely combust the fuel. That ratio for Gasoline is 14.7:1. It varies a lot depending on the fuel you’re using. The good news is that you can calculate the stoichiometric ratio, or AFR, for anything that’s a known compound. And the good news about these compounds are that they are known compounds. So how does that work?
Chemistry
Yep, chemistry is how you figure this stuff out. Unfortunately, I never took any chemistry in school, so I had to figure it out through the power of the Internet. Let’s start with just one of these compounds, Ethanol, and we’ll work through that.
Here’s a quick science lesson: the complete combustion of a hydrocarbon produces Carbon Dioxide (CO2) and Water (H2O).
Balancing the Equation
In order for combustion to happen, it must happen in a balanced manner. This means that you need to take a certain number of molecules of fuel and mix it with a certain amount of oxygen to create complete molecules of carbon dioxide and water. A typical equation for that looks like this:
C2H6O + 3 O2 -> 2 CO2 + 3 H2O
This is a balanced equation. If you’re really interested in how it all breaks down, you can check it out on Wolfram Alpha. I’ll summarize it here: C2H6O is Ethanol. O2 is Oxygen. One molecule of Ethanol mixes with three Oxygen molecules and through combustion, two CO2 molecules and three H2O molecules are created, along with a lot of heat, which you use to make horsepower!
Weighing Mass
We’re not quite done yet, though. We also need to know weights for the various elements in our hydrocarbons, and the weight of Oxygen. This allows us to get the total weight of a single molecule of our compound and we’re going to use the weight of Oxygen to determine our total ratio. For this we can turn to the periodic table!
Carbon (C on the periodic table) weighs 12.011 gHydrogen (H on the periodic table) weighs 1.001 gOxygen (O on the periodic table) weighs 15.99 g
Our Ethanol contains:
2 Carbon atoms, weighing 24.022 g total
6 Hydrogen atoms, weighing 6.048 g total
1 Oxygen atom, weighing 15.99 g total
This makes for a total weight of 46.069g per Ethanol molecule. Next up, we can figure out our oxygen:compound ratio.
Finding Ratios
We’re almost done, so bear with me! In this step, we need to figure out the oxygen:compound ratio. This allows us to then find the AFR of the compound. Going back to our balanced equation, we know that we need 1 parts of Ethanol to 3 parts of Oxygen. That equation looks like this:
oxygen:fuel = ([O2 ratio] ([O2 weight] 2) / ([Compound weight] [compound ratio])
So, that means:
(3(15.992)) / (46.069 1) = 2.084
our Oxygen:Fuel ratio is 2.084:1.
Finding the AFR, Finally!
Last step here, I promise. I’m not going to go into too many details because there are lots of places where you can find this information, but normal air we all breathe is about 23.2% Oxygen as a percentage of mass. Using our oxygen:air ratio above, we can now figure out our total AFR given the following formula:
AFR = ([o:a ratio] 100) / [percentage of oxygen in air]
So, that means:
(2.084 100) / 23.2 = 8.98275
The AFR for Ethanol is 8.98:1. Close enough!
Wrapping it Up
Yes, this was a lot of jumping through hoops to find AFR for one particular fuel – Ethanol, but it demonstrates that you can, in fact, find AFR for any hydrocarbon.
So what can you do with this information? Personally, I use it to find an AFR for a fuel that I’m mixing with other fuels to come up with AFR equivalents. I’ll write up something about this in the future, because it’s fascinating to be able to make a new fuel that is a drop-in replacement for another fuel. I have a fuel that consists of a majority of Methanol, yet runs the exact same AFR as E85. So, I can tune for E85 and reap some of the benefits of Methanol. I’ll tell you about it later! 🙂
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11580 | https://people.math.wisc.edu/~aseeger/papers/gsu-tl-endpt-arxiv.pdf | THE HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS GUSTAVO GARRIG´ OS ANDREAS SEEGER TINO ULLRICH Abstract. We determine for which parameters natural enumerations of the Haar system in Rd form a Schauder basis or basic sequence on Triebel-Lizorkin spaces. The new results concern the endpoint cases.
1. Introduction and statements of main results In this paper we essentially complete the study of the basis properties for the (inhomogeneous) Haar system in the scale of Triebel-Lizorkin spaces F s p,q(Rd).
In particular, we describe the behavior at the endpoint cases which was left open in our earlier work . Similar endpoint questions for the family of Besov spaces have been presented in the companion paper .
We note that markedly different outcomes occur for each family, in both the non-endpoint situations ([12, 15, 10, 11, 4]) and the endpoint (, ) situations.
We now set the basic notation required to state the results. Given the one variable functions h(0) = 1[0,1) and h(1) = 1[0,1/2) −1[1/2,1), for each ϵ = (ϵ1, . . . , ϵd) ∈{0, 1}d, k ∈N0 and ν = (ν1, . . . , νd) ∈Zd, we define hϵ k,ν(x) := d Y i=1 h(ϵi)(2kxi −νi), x = (x1, . . . , xd) ∈Rd.
Then, the Haar system is the collection of functions Hd = n h⃗ 0 0,ν o ν∈Zd ∪ n hϵ k,ν : k ∈N0, ν ∈Zd, ϵ ∈Υ o , where we denote Υ = {0, 1}d \ {⃗ 0}.
Consider F s p,q(Rd) with the usual definition in [13, §2.3.1] or [3, §12]. To investigate the Schauder basis properties of Hd, we initially assume that 0 < p, q < ∞(so that S is dense in F s p,q, and the latter is separable), and that (1) hϵ k,ν ∈F s p,q and hϵ k,ν ∈(F s p,q)∗, ∀ϵ, k, ν.
2010 Mathematics Subject Classification. 46E35, 46B15, 42C40.
Key words and phrases. Schauder basis, basic sequence, unconditional basis, dyadic averaging operators, Haar system, Sobolev and Besov spaces, Triebel-Lizorkin spaces.
1 2 G. GARRIG´ OS A. SEEGER T. ULLRICH Given an enumeration U = un = hϵ(n) k(n),ν(n) ∞ n=1 of Hd, we consider the corresponding partial sum operators (2) SRf = SU Rf = R X n=1 u∗ n(f)un , R ∈N, where the linear functionals u∗ n are defined by (3) u∗ n(f) = 2k(n)d⟨f, hϵ(n) k(n),ν(n)⟩, f ∈S.
The condition in (1) ensures that these operators are well-defined and indi-vidually bounded in F s p,q(Rd). Also, u∗ n(um) = δn,m, n, m ≥1.
The basis properties of U are related to the validity of the bound (4) sup R∈N ∥SU R∥F s p,q→F s p,q < ∞.
Indeed, if span Hd is dense in F s p,q, then (4) is equivalent to U being a Schauder basis of F s p,q, that is (5) lim R→∞∥SU Rf −f∥F s p,q = 0 for every f ∈F s p,q. Moreover, the basis is unconditional if and only if the bound in (4) is uniform in all enumerations U. Finally, if span Hd is not assumed to be dense, then (4) still implies that U is a basic sequence of F s p,q, meaning that (5) holds for all f in the F s p,q-closure of span Hd.
unconditional 1 p s 1 d+1 d 1 1 q 1 q −1 −1 Schauder 1 p s 1 d+1 d 1 −1 0 < q < ∞ Figure 1. Parameter domain P for Hd in F s p,q(Rd). The left region corresponds to unconditionality, and right region to the Schauder basis property.
The pentagon P depicted in Figure 1 shows the natural index region for these problems; outside its closure either (1) or the density of span Hd fail.
The open pentagon corresponds to the range d d+1 < p < ∞, 0 < q < ∞, and (6) max n d( 1 p −1), 1 p −1 o < s < min n 1, 1 p o .
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 3 Triebel showed in [15, Theorem 2.21] that Hd is an unconditional basis of F s p,q(Rd) in the green shaded region, where the additional restriction (7) max d( 1 q −1), 1 q −1 < s < 1 q is imposed. The necessity of condition (7) for unconditionality was estab-lished in [10, 11] (for d = 1). On the other hand, we recently showed in that natural enumerations of Hd form a Schauder basis of F s p,q(Rd) in the full open pentagon P. Except for a few trivial cases, the behavior at the points (1/p, s) lying in the boundary of P was left unexplored.
In this paper we attempt to fill this gap by giving an answer, positive or negative, depending on the secondary index q. Moreover, when possible, the negative answer is replaced by a suitable basic sequence property.
We first state the complete range for unconditionality, which contains new negative cases and a multivariate extension of the examples in .
Theorem 1.1. Let 0 < p, q < ∞and s ∈R. Then, Hd is an unconditional basis of F s p,q(Rd) if and only if the conditions (6) and (7) are both satisfied.
In the next results we drop unconditionality, and consider the Schauder basis property for the following natural orderings of the Haar system Hd; see [4, 5].
Definition 1.2. (i) An enumeration U is said to be admissible if for some constant b ∈N the following holds: for each cube Iν = ν + [0, 1]d, ν ∈Zd, if un and un′ are both supported in Iν and | supp(un)| ≥2bd| supp(un′)|, then necessarily n < n′ .
(ii) An enumeration U is strongly admissible if for some constant b ∈N the following holds: for each cube Iν, ν ∈Zd, if I∗∗ ν denotes the five-fold dilated cube with respect to its center, and if un and un′ are supported in I∗∗ ν with | supp(un)| ≥2bd| supp(un′)| then necessarily n < n′.
Our next theorem characterizes the Schauder basis property in F s p,q for the class of strongly admissible enumerations of Hd. A new positive result is obtained in the line s = d/p −d, when d d+1 < p ≤1; see Figure 1.
The special case F 0 1,2 = h1 is classical, and was established in [1, 17]. The negative results for s = 1 are also new.
Theorem 1.3. Let 0 < p, q ≤∞and s ∈R. Then, the following statements are equivalent, i.e. (a) ⇐ ⇒(b): (a) Every strongly admissible enumeration U of Hd is a Schauder basis of F s p,q(Rd).
(b) One of the following three conditions is satisfied: (i) 1 < p < ∞, 1 p −1 < s < 1 p, 0 < q < ∞, (ii) d d+1 < p ≤1, d p −d < s < 1, 0 < q < ∞, (iii) d d+1 < p ≤1, s = d p −d, 0 < q < ∞.
4 G. GARRIG´ OS A. SEEGER T. ULLRICH As in [4, 5], a crucial tool in our analysis will be played by the dyadic averaging operators EN. That is, if DN is the set of all dyadic cubes of length 2−N, IN,ν = 2−N(ν + [0, 1)d), ν ∈Zd, then we define (8) ENf(x) = X ν∈Zd 1IN,ν(x) 2Nd Z IN,ν f(y)dy , at least for f ∈S. We shall also need the following companion operators involving Haar functions of a fixed frequency level. Namely, for N ∈N and any a = (aν,ε)ν,ε ∈ℓ∞(Zd × Υ) we set (9) TN[f, a] = X ϵ∈Υ X ν∈Zd aν,ϵ2Nd⟨f, hϵ N,ν⟩hϵ N,ν.
For these operators one looks for estimates that are uniform in ∥a∥∞≤1.
The relation between the partial sums SU R and the operators EN and TN[·, a] is explained in §10 below; see also [4, 5]. In particular, their uniform boundedness in F s p,q implies that (4) holds for all strongly admissible enu-merations U. The optimal region for the uniform boundedness for EN and TN[·, a] in F s p,q is given in the next theorem, and depicted in Figure 2 below.
Theorem 1.4. Let 0 < p ≤∞, 0 < q ≤∞and s ∈R.
(a) The operators EN admit an extension from S into F s p,q(Rd) such that sup N≥0 ∥EN∥F s p,q→F s p,q < ∞ if and only if one of the following five conditions is satisfied: (i) 1 < p ≤∞, −1 + 1 p < s < 1 p, 0 < q ≤∞, (ii) d d+1 ≤p < 1, s = 1, 0 < q ≤2, (iii) d d+1 < p ≤1, d( 1 p −1) < s < 1, 0 < q ≤∞, (iv) d d+1 < p ≤1, s = d( 1 p −1), 0 < q ≤∞, (v) p = ∞, s = 0, 0 < q ≤∞.
(b) If one of the conditions (i)-(v) is satisfied then the operators TN[·, a] are uniformly bounded on F s p,q(Rd) when N ≥0 and ∥a∥ℓ∞(Zd×Υ) ≤1.
Regarding positive results, the cases (i) and (iii) were established in .
The novel cases appear at the end-point lines in (ii) and (iv), and the special point (v); see Figure 2.
The proof of (ii) will follow from a slightly stronger result which we state next. Let η0 ∈C∞ c (Rd) be supported in {|ξ| < 3/4} with η0(ξ) = 1 for |ξ| ≤1/4, and let ΠN be defined by (10) [ ΠNf(ξ) = η0(2−Nξ) b f(ξ).
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 5 1 p s 1 d+1 d 1 0 < q ≤∞ 0 < q ≤2 −1 0 < q ≤∞ Figure 2. Region for uniform boundedness of EN (hence for the basic sequence property) in the spaces F s p,q(Rd).
Then we shall actually prove the following.
Theorem 1.5. Let d/(d + 1) ≤p < 1 and 0 < r ≤∞. Then (11) sup N ∥ENf −ΠNf∥B1 p,r ≲∥f∥F 1 p,2.
Moreover, for ∥a∥ℓ∞≤1, (12) sup N ∥TN[f, a]∥B1 p,r ≲∥f∥F 1 p,2.
Using the embeddings F s p,q ⊂F s p,2 for q ≤2, and Bs p,r ⊂F s p,r ⊂F s p,q for r ≤min{p, q}, one deduces the uniform bounds in (ii) of Theorem 1.4.
Likewise, for the end-point cases in (iv) and (v) we shall establish the following stronger results.
Theorem 1.6. Let d/(d + 1) < p ≤1, 0 < r ≤∞, and s = d p −d. Then (13) sup N ∥ENf −ΠNf∥Bs p,r ≲∥f∥F s p,∞, and likewise for the operators TN[·, a], uniformly in ∥a∥ℓ∞≤1.
Theorem 1.7. For every r > 0, it holds (14) sup N ∥ENf −ΠNf∥F 0 ∞,r ≲∥f∥B0 ∞,∞, and likewise for the operators TN[·, a], uniformly in ∥a∥ℓ∞≤1.
Finally, concerning the negative results in Theorem 1.4, the only non-trivial case appears when s = 1, for which we shall establish the following.
Theorem 1.8. Let d d+1 ≤p < 1 and 2 < q ≤∞. Then, ∥EN∥F 1 p,q→F 1 p,q ≈N 1 2 −1 q .
6 G. GARRIG´ OS A. SEEGER T. ULLRICH This paper. In §2 we set the basic notation. In §3 and §4 we prove, respec-tively, Theorems 1.5 and 1.6, except for the special case p = d/(d+1) which is treated in §5. Theorem 1.7 is shown in §6, and Theorem 1.8 in §7. In §8 we gather all these results and complete the proof of Theorem 1.4, explaining as well the meaning of the extensions of the operators EN to the full spaces F s p,q. In §9 we study the failure of density for span Hd in the case s = 1.
In §10 and 11 we pass to the operators SU R, showing their relation with EN for admissible enumerations, and establishing Theorem 1.3. Finally, §12 is devoted to unconditionality, and the proof of Theorem 1.1.
Acknowledgements The authors would like to thank the Isaac Newton Institute for Mathe-matical Sciences, Cambridge, for support and hospitality during the program Approximation, Sampling and Compression in Data Science where some work on this paper was undertaken. This work was supported by EPSRC grant no EP/K032208/1. G.G. was supported in part by grants MTM2016-76566-P, MTM2017-83262-C2-2-P and Programa Salvador de Madariaga PRX18/451 from Micinn (Spain), and grant 20906/PI/18 from Fundaci´ on S´ eneca (Regi´ on de Murcia, Spain). A.S. was supported in part by National Science Foundation grants 1500162 and 1764295. T.U. was supported in part by Deutsche Forschungsgemeinschaft (DFG), grant 403/2-1.
2. Preliminaries 2.1. Besov and Triebel-Lizorkin quasi-norms. Let s ∈R and 0 < p, q ≤∞ be given. Throughout the paper we fix a number A > d/p and an integer (15) M > A + |s| + 2.
Consider two functions β0, β ∈C∞ c (Rd), supported in (−1/2, 1/2)d, with the properties |b β0(ξ)| > 0 if |ξ| ≤1, |b β(ξ)| > 0 if 1/8 ≤|ξ| ≤1 and β has vanishing moments up to order M, that is (16) Z Rd β(x) xm1 1 · · · xmd d dx = 0, ∀mi ∈N0 with m1 + . . . + md ≤M.
The optimal value of M is irrelevant for the purposes of this paper, and (15) suffices for our results. We let βk := 2kdβ(2k·) for each k ≥1, and denote Lkf = βk ∗f whenever f ∈S′(Rd). These convolution operators, sometimes called local means, can be used to define equivalent quasi-norms in the Bs p,q and F s p,q spaces. Namely, (17) g Bs p,q ≈ 2ksLkg ∞ k=0 ℓq(Lp) HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 7 and if 0 < p < ∞, (18) g F s p,q ≈ 2ksLkg ∞ k=0 Lp(ℓq) see e.g. [14, 2.5.3 and 2.4.6]. For the latter spaces, when p = ∞(and q < ∞) one defines instead (19) g F s ∞,q ≈sup n≥0 sup I∈Dn 1 |I| Z I X k≥n 2ksq|Lkg(x)|q dx 1/q , see [3, (12.8)], . Finally, one lets F s ∞,∞= Bs ∞,∞.
Next, let η0 ∈C∞ c (Rd) be supported on {ξ : |ξ| < 3/8} and such that η0(ξ) = 1 if |ξ| ≤1/4. We define the operators d Λ0f(ξ) = η0(ξ) b β0(ξ) b f(ξ) , (20a) d Λkf(ξ) = η0(2−kξ) −η0(2−k+1ξ) b β(2−kξ) b f(ξ), k ≥1, (20b) so that (21) f = ∞ X j=0 LjΛjf with convergence in S′. Of course, one obtains (the usual) equivalent norms if in (17), (18) and (19) the operators Lk are replaced by Λk. In particular, if we let ΠN = PN j=0 LjΛj, then (22) sup N ∥ΠNf∥F s p,q ≲∥f∥F s p,q.
Below we shall be interested in uniformly bounded extensions of the dyadic averaging operators EN defined in (8). We shall denote E⊥ N = I −EN and Π⊥ N = I −ΠN, and write (23) EN −ΠN = EN Π⊥ N −E⊥ N ΠN.
Then, using (17), we have ENf −ΠNf Bs p,r ≲ 2ksLkENΠ⊥ Nf ∞ k=0 ℓr(Lp) + (24) + 2ksLkE⊥ NΠNf ∞ k=0 ℓr(Lp).
Following [4, 5], we shall prove Theorems 1.5, 1.6 and 1.7 using suitable estimates for the functions LkENLjg and LkE⊥ NLjg, for each j, k ≥0, some of which will be new in this paper.
8 G. GARRIG´ OS A. SEEGER T. ULLRICH 3. Proof of Theorem 1.5: The case p > d d+1 Let s = 1 and d/(d+1) < p < 1. For these indices, Theorem 1.5 will be a consequence of the following two results. The first result is contained in (Propositions 3.1 and 3.4), and was also implicit in (proof of inequality (19)).
Proposition 3.1. For d d+1 < p < 1 and r > 0, it holds (25) sup N ∞ X k=0 2kr Lk EN Π⊥ Nf r p 1/r ≲∥f∥B1 p,∞.
The same holds if EN is replaced by TN[·, a] with ∥a∥ℓ∞≤1.
The second result is new, and it will require a few additional arguments compared to [4, 5].
The conditions on p are also less demanding.
Here hp = F 0 p,2 is the local Hardy space; see e.g. [13, 2.5.8].
Proposition 3.2. For d d+2 < p < 1 and r > 0, it holds (26) sup N ∞ X k=0 2kr Lk E⊥ N ΠNf r p 1/r ≲∥∇f∥hp.
The same holds if E⊥ N is replaced by TN[·, a] with ∥a∥ℓ∞≤1.
We shall prove Proposition 3.2 in the next subsections, but we indicate now how (25) and (26) imply (11).
Just use the Littlewood-Paley type inequality ∥∇f∥hp ≲∥f∥F 1 p,2; (see e.g. [13, 2.3.8/3]) and the embedding F 1 p,2 , →B1 p,∞.
3.1. A pointwise estimate. As in we shall use the Peetre maximal func-tions (27) M∗∗ A,jg(x) = sup h∈Rd |g(x + h)| (1 + 2j|h|)A , typically applied to scalar or Hilbert space valued g ∈S′(Rd) satisfying (28) supp b g ⊂{ξ : |ξ| ≤2j+1}.
In it was shown that for g satisfying (28), (29) ∥M∗∗ A,jg∥p ≤Cp,A∥g∥p, 0 < p ≤∞, A > d/p.
In what follows it will be convenient to use the notation |x|∞= max 1≤i≤d |xi|, x = (x1, . . . , xd) ∈Rd.
The following lemma is a variation of [4, (35)]. The novelty here is that the operator E⊥ N is acting on ΠNf = P j≤N LjΛjf, rather than in each LjΛjf separately.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 9 Lemma 3.3. Let f ∈S′(Rd). Then (30) E⊥ NΠNf ≲ inf |y′−y|∞≤21−N M∗∗ A,N(2−NΠN∇f)(y′), y ∈Rd.
In particular, if |y − µ 2N |∞≤21−N, then, for every p > 0, (31) |E⊥ NΠNf(y)| ≲ − Z |h|∞≤22−N M∗∗ A,N(2−NΠN∇f)( µ 2N + h) p dh 1/p .
These bounds also hold if we replace E⊥ N with TN(·, a) with ∥a∥∞≤1.
Proof. Recall that [ ΠNf(ξ) = η0(2−Nξ) b f(ξ). Let Φ ∈S, with b Φ(ξ) = 1 when |ξ| ≤1, and let ΦN(z) = 2NdΦ(2Nz). Then ΠNf = ΦN ∗ΠNf.
If I ∈DN is such that y ∈I, we have E⊥ N(ΠNf)(y) = ENΦN ∗(ΠNf) −ΦN ∗ΠNf(y) = − Z I Z Rd ΦN(z) ΠNf(v −z) −ΠNf(y −z) dz dv = − Z I Z Rd ΦN(z) Z 1 0 ⟨v −y, ∇ΠNf(y + s(v −y) −z)⟩dsdz dv ≲ Z Rd |ΦN(z)|(1 + 2N|z|)A dz sup ˜ z∈Rd |2−N∇ΠNf(y′ + ˜ z)| (1 + 2N|˜ z|)A ≤CA M∗∗ A,N2−N∇ΠNf, for any y′ such that |y −y′|∞≤21−N. This shows (30). The last assertion in (31) follows easily from here.
Finally, if we replace E⊥ N with TN[·, a], the cancellation of R I hI = 0 implies that, for w ∈I, |TNΠNf, a| ≤ 1 |I| Z I hI(v) ΦN ∗ΠNf(v) −ΦN ∗ΠNf(w) dv .
The rest of the proof is then carried out as above.
□ 3.2. Norm estimates. As in , we use the notation (32) UN,k = n y ∈Rd : min 1≤i≤d dist(yi, 2−NZ) ≤2−k−1o , k > N.
Roughly speaking, this is the set of points at distance O(2−k) from [ I∈DN ∂I.
Note (or recall from [4, Lemma 2.3.i]) that if k > N then (33) Lk(ENg)(x) = 0, ∀x ∈U∁ N,k = Rd \ UN,k.
The next two results will be obtained using Lemma 3.3.
10 G. GARRIG´ OS A. SEEGER T. ULLRICH Lemma 3.4. Let 0 < p ≤1. Then for every k > N and ∥a∥∞≤1, 2k∥LkE⊥ NΠNf∥p + 2k∥LkTN[ΠNf, a]∥p ≲2−(k−N)( 1 p −1)∥∇ΠNf∥p.
Proof. The observation in (33) implies that LkE⊥ NΠNf p ≲ LkE⊥ NΠNf Lp(U∁ N,k) + LkE⊥ NΠNf Lp(UN,k) ≲ ∥LkΠNf∥Lp(U∁ N,k) + h X µ∈Zd ∥LkE⊥ NΠNf p Lp(UN,k∩IN,µ) i 1 p.
(34) Using (31) and the fact that supp βk(x −·) ⊂µ2−N + O(2−N) for x ∈IN,µ, the last term is controlled by h X µ∈Zd |IN,µ ∩UN,k| − Z |h|∞≤22−N M∗∗ N,A(2−N∇ΠNf)(2−Nµ + h) p dh i 1 p ≲ 2−k2−N(d−1) 1 p 2 Nd p M∗∗ N,A(2−N∇ΠNf) p ≲2−N 2 N−k p ∥∇ΠNf∥p.
To estimate the first term on the right hand side of (34), observe that we can write βk ∗ΠNf = 2−k ˜ βk ∗(∇ΠNf), where ˜ β = (˜ β1, . . . , ˜ βd) and each ˜ βi is a primitive of β in the xi-variable (hence with vanishing moments up to order M −1). Moreover, ∥LkΠNf∥Lp(U∁ N,k) ≤2−k ˜ βk ∗[ΦN ∗∇ΠNf] p ≲2−k2−(k−N)(M−A)∥∇ΠNf∥p, using in the last step the cancellation of ˜ βk; see [4, Lemma 2.2]. Analogous arguments apply for TN[ΠNf, a] in place of E⊥ NΠNf.
□ Lemma 3.5. Let 0 < p ≤1. Then, for every k ≤N, ∥a∥∞≤1, (35) 2k∥LkE⊥ NΠNf∥p + 2k∥LkTN[ΠNf, a]∥p ≲2(N−k)( d p −d−2)∥∇ΠNf∥p.
Proof. Since R I E⊥ NΠNf dy = 0 for I ∈DN, we may write Lk E⊥ N[ΠNf] (x) = X µ∈Zk,N(x) Z IN,µ βk(x −y) −βk(x −2−Nµ) E⊥ NΠNf dy where (36) Zk,N(x) = {µ ∈Zd : |x −2−Nµ|∞≤2−N + 2−k−1}.
Note that card Zk,N(x) ≈2(N−k)d. Now use |βk(x −y) −βk(x −2−Nµ)| ≲2kd 2k−N, if y ∈IN,µ, HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 11 in combination with Lemma 3.3 to obtain |Lk E⊥ N[ΠNf] (x)| ≲ ≲2(k−N)(d+1) X µ∈Zk,N(x) − Z |h|∞≤22−N M∗∗ A,N(2−N∇ΠNf) p( µ 2N + h) dh 1 p ≲2(k−N)(d+1) X µ∈Zk,N(x) − Z |h|∞≤22−N M∗∗ A,N(2−N∇ΠNf) p( µ 2N + h) dh 1 p , the last step using the embedding ℓ1 , →ℓ1/p, since p ≤1. From this ∥Lk E⊥ N[ΠNf] ∥p ≲2(k−N)(d+1)h Z Rd X µ∈Zk,N(x) − Z |h|∞≤22−N M∗∗ A,N(2−N∇ΠNf) p( µ 2N + h) dh dx i 1 p ≲2(k−N)(d+1)h X µ∈Zd 2−kd − Z |h|∞≤22−N M∗∗ A,N(2−N∇ΠNf) p( µ 2N + h) dh i 1 p ≲2(k−N)(d+1)2(N−k)d/p M∗∗ A,N(2−N∇ΠNf) p, and the assertion follows by the Peetre inequality for M∗∗ A,N.
Analogous arguments apply for TN[ΠNf, a] in place of E⊥ NΠNf.
□ Proof of Proposition 3.2. Using the Lemmas 3.4 and 3.5, and noticing that we may sum in k since d d+2 < p < 1, one easily obtains ∞ X k=0 2kr Lk E⊥ N ΠNf r p 1/r ≲∥ΠN∇f∥p.
The last quantity can be estimated further, applying to g = ∇f the inequal-ity (37) ∥ΠNg∥p ≤ sup N≥0 |ΠNg| p ≲∥g∥hp ≈∥g∥F 0 p,2, which follows for example using the standard maximal function characteri-zation of the hp norm. This proves (26). The proof for the operators TN[·, a] is exactly analogous.
□ 4. Proof of Theorem 1.6: the case s = d/p −d Let s = d/p −d and d/(d + 1) < p ≤1 (we will take up the endpoint case p = d/(d + 1), when s = 1 in §5). For these indices, Theorem 1.6 will be a consequence of the following two results.
The first result was already established in (Propositions 3.2 and 3.3), using the same type of analysis as in . The inequality is slightly stronger than needed due to F s p,∞, →Bs p,∞.
12 G. GARRIG´ OS A. SEEGER T. ULLRICH Proposition 4.1. Let d d+1 < p ≤1 and r > 0. Then (38) ∞ X k=0 2k(d/p−d)r LkE⊥ NΠNf r p 1/r ≲∥f∥ B d p −d p,∞ .
The same holds if E⊥ N is replaced by TN[·, a] with ∥a∥ℓ∞≤1.
The second proposition is new, and its proof will require several additional refinements compared to the arguments given in .
Proposition 4.2. Let d−1 d < p ≤1 and r > 0. Then, ∞ X k=N+1 2k(d/p−d)r X j>N LkENLjΛjf r p 1/r ≲∥f∥ F d p −d p,∞ (39) N X k=0 2k(d/p−d)r X j>N LkENLjΛjf r p 1/r ≲∥f∥ F d p −d p,∞ .
(40) The same holds if EN is replaced by TN[·, a] with ∥a∥ℓ∞≤1.
4.1. Notation and observations on dyadic cubes. Recall that every dyadic cube I is contained in a unique parent cube of double side length. Also each dyadic cube has 2d children cubes of half side length. It will be useful to single out one of the children cubes according to the following definition.
Definition 4.3. Let I be a dyadic cube. We denote by ω(I) the unique child of I with the property that its closure contains the center of the parent cube of I.
We need some further notation (taken from ). For each dyadic cube I ∈DN, we denote by DN(I) the set of all its neighboring 2−N-cubes, that is, I′ ∈DN with ¯ I ∩¯ I′ ̸= ∅. Likewise, if ℓ> N we denote by Dℓ[∂I] the set of all J ∈Dℓsuch that ¯ J ∩∂I ̸= ∅.
Lemma 4.4. (i) Let J ∈Dℓ[∂I]. Then 2−ℓ−1 ≤dist(x, ∂I)∞≤2−ℓfor all x ∈ω(J).
(ii) Let I ∈DN, let ℓ1, ℓ2 > N and consider two distinct cubes J1 ∈ Dℓ1[∂I], J2 ∈Dℓ2[∂I]. Then ω(J1), ω(J2) have disjoint interiors.
Proof. The upper bound in (i) is true for all x ∈J, by definition of Dℓ[∂I] and the lower bound follows from the definition of ω(J) since the parent cube of J is contained in I or one of its neighbors of equal side length. To see (ii) first observe that J1, J2 are disjoint if ℓ1 = ℓ2 (and hence ω(J1) and ω(J2) are disjoint). If ℓ1 ̸= ℓ2 then (ii) follows from (i).
□ HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 13 4.2. Proof of Proposition 4.2. We make a preliminary observation about maximal functions. If g is continuous, for each j ≥0 we let M∗ jg(x) := sup |h|∞≤2−j |g(x + h)|.
Then, if a cube J ∈Dj+1 has center cJ, we have (41) sup x∈J |g(x)| ≤ inf |h|∞≤2−j−1 M∗ jg(cJ + h) ≤ h − Z ω(J) |M∗ jg|p i 1 p .
Proof of (39). Let j, k > N. By (33), LkENLjΛjf = 0 if x ∈U∁ N,k.
Moreover, by [4, Lemma 2.3 (ii)] we have (42) |EN(LjΛjf)(y)| ≲2(N−j)d X I∈DN 1I(y) X J∈Dj+1[∂I] ∥Λjf∥L∞(J).
Let x ∈UN,k ∩I, for some I ∈DN. Then supp βk(x −·) ⊂S I′∈DN(I) I′, and therefore (42) implies |LkENLjΛjf| ≤ Z |βk(x −y)| EN(LjΛjf)(y) dy ≲2(N−j)d X I′∈DN(I) X J∈Dj+1[∂I′] ∥Λjf∥L∞(J) ∥βk∥1.
Using the inequality in (41), this in turn implies (since p ≤1) Ak(x)p := h X j>N |LkENLjΛjf(x)| ip (43) ≲ X j>N 2(N−j)dp X I′∈DN(I) X J∈Dj+1[∂I′] − Z ω(J) |M∗ j(Λjf)|p ≲2Ndp X I′∈DN(I) X j>N X J∈Dj+1[∂I′] Z ω(J) sup ℓ>N |2ℓ( d p −d)M∗ ℓ(Λℓf)|p.
By Lemma 4.4 the sets ω(J) for J ∈D[∂I′] are disjoint. Also since #DN[I] = 2d we obtain Ak(x)p ≲2Ndp Z I∗∗sup ℓ>N 2ℓ( d p −d)M∗ ℓ(Λℓf) p, with I∗∗the five-fold dilation of I with respect to cI. Thus, if we write (44) G = sup ℓ>N |2ℓ( d p −d)M∗ ℓ(Λℓf)|, 14 G. GARRIG´ OS A. SEEGER T. ULLRICH we obtain ∥Ak∥p p ≲ 2Ndp X I∈DN |I ∩UN,k| Z I∗∗|G|p ≲ 2N(d−d−1 p )p 2−k ∥G∥p Lp(Rd), and therefore (45) 2k( d p −d)∥Ak∥p ≲2−(k−N)(d−d−1 p ) ∥G∥p.
When p > (d −1)/d we can sum in k > N, and hence the left hand side of (40) is controlled by ∥G∥p. Now, Peetre’s inequalities imply that M∗ ℓΛℓf ≲M∗∗ A,ℓΛℓf ≲MσΛℓf, for σ = d/A, where Mσg(x) = sup R>0 h − Z BR(x) |g|σi1/σ .
Thus, ∥G∥p ≲ sup ℓ>N 2ℓ( d p −d)Mσ[Λℓf] p ≤ Mσ h sup ℓ>N 2ℓ( d p −d)|Λℓf| i p ≲ sup ℓ>N 2ℓ( d p −d)|Λℓf| p ≲∥f∥ F d p −d p,∞ (46) using the boundedness of the Hardy-Littlewood maximal operator, since σ = A/d < p. This finishes the proof of (39), for the stated version involving EN. The analogous version for TN[·, a] follows similarly, by replacing (42) with the corresponding version for the TN, as in .
□ Proof of (40). Let j > N and k ≤N. Again, we shall follow the proof of [4, (26)], applying the same changes as in the previous subsection. Namely, let e fj = (Λjf)1UN,j and note that EN[LjΛjf] = EN[Lj e fj]; see [4, Lemma 2.3]. Then we write (47) LkEN[Ljf] = Lk(EN[Lj e fj] −Lj e fj) + LkLj e fj.
As in , the last term is harmless since ∥LkLj e fj∥p ≲2−(M−A)|j−k| M∗∗ A,j(Λjf)(x) p ≲2−(M−A)|j−k| ∥Λjf∥p, by [4, Lemma 2.2].
Thus, assuming r ≤p (as we may), and using the r-triangle inequality, N X k=0 2k( d p −d)r∥ X j>N LkLj e fj∥r p 1/r ≲ N X k=0 X j>N 2(k−j)( d p −d+(M−A))r2j( d p −d)r∥Λjf∥r p 1/r ≲∥f∥ B d p −d p,∞ .
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 15 Hence it remains to prove (48) N X k=0 2k(d/p−d)r X j>N LkE⊥ NLj e fj r p 1/r ≲∥f∥ F d p −d p,∞ .
Following , and letting Zk,N(x) be as in (36), we write Aj,k(x) := |Lk E⊥ N[Lj e fj] (x)| ≤ X µ∈Zk,N(x) Z IN,µ βk(x −y) −βk(x −2−Nµ) E⊥ NLj e fj dy ≲ 2kd2k−N X µ∈Zk,N(x) Z IN,µ |ENLj e fj| + |Lj e fj(y)| dy.
In [4, p. 1332], the terms corresponding to the two summands in the integral are estimated differently, but produce essentially the same outcome, namely (49) Aj,k(x) ≲2k−N2(k−j)d X µ∈Zk,N(x) X J∈Dj+1[∂IN,µ] ∥Λjf∥p L∞(J) 1 p , see [4, (41)]. At this point we argue as in the previous subsection. That is, we use (41) to have (50) ∥Λjf∥L∞(J) ≤ h − Z ω(J) M∗ jΛjfp dy i 1 p , and conclude that Ak(x)p := h X j>N |LkE⊥ NLj e fj(x)| ip (51) ≲ X j>N 2(k−N)p2(k−j)dp X µ∈Zk,N(x) X J∈Dj+1[∂IN,µ] − Z ω(J) |M∗ j(Λjf)|p ≲2(k−N)p2kdp X µ∈Zk,N(x) Z I∗∗ N,µ |G|p with G as in (44), and using the disjointness of the sets ω(J) as before.
Thus, integrating the above expression ∥Ak∥p p ≲ 2(k−N)p2kdp X µ∈Zd 2−kd Z I∗∗ N,µ |G|p ≲ 2(k−N)p2k(d−d p )p Z Rd |G|p and therefore (52) 2k( d p −d)∥Ak∥p ≲2k−N ∥G∥p.
Therefore, one can sum in k ≤N, and obtain the desired expression in (48) using the estimate for ∥G∥p in (46). This finishes the proof of (40). The 16 G. GARRIG´ OS A. SEEGER T. ULLRICH corresponding version for TN is proved similarly (notice that in (47) the analysis of the last summand becomes unnecessary, due to the additional cancellation of TN).
□ 5. Proof of Theorem 1.5: The case p = d d+1 The end-point case p = d/(d + 1) and s = 1, was excluded from the previous proofs because of the restrictions imposed in Propositions 3.1 and 4.1. However, one can use instead Propositions 4.2 and 3.2, which are valid at this endpoint. Namely, they imply the inequalities sup N ∞ X k=0 2kr X j>N LkENLjΛjf r p 1/r ≲∥f∥F 1 p,∞, p = d d + 1, (53) and (54) sup N ∞ X k=0 2kr LkE⊥ NΠNf r p 1/r ≲∥f∥F 1 p,2 p = d d + 1.
Then, the result stated in Theorem 1.5 follows using additionally the em-bedding F 1 p,2 , →F 1 p,∞in (53). The same argument applies to TN[·, a] with ∥a∥∞≤1 if we use the corresponding versions of Propositions 4.2 and 3.2.
□ 6. Proof of Theorem 1.7: the case s = 0 and p = ∞ In view of (23), it suffices to prove the following.
Proposition 6.1. Let r > 0. Then (55) E⊥ NΠNf F 0 ∞,r + ENΠ⊥ Nf F 0 ∞,r ≲∥f∥B0 ∞,∞.
One part of the estimates will be derived from the following inequalities, proved in [5, (36a), (37a)]: X k≤N LkE⊥ NΠNf r ∞ 1/r ≲ ∥f∥B0 ∞,∞ (56) X k≤N LkENΠ⊥ Nf r ∞ 1/r ≲ ∥f∥B0 ∞,∞.
(57) We remark that these same inequalities with P k≤N replaced by P k>N are only true if r = ∞. This necessitates the use of F 0 ∞,r-norms on the left hand side of (55) To establish the proposition, let f ∈B0 ∞,∞be such that ∥f∥B0 ∞,∞= 1.
We shall prove separately each of the two inequalities.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 17 6.1. Proof E⊥ NΠNf F 0 ∞,r ≲1. By (19) we can write E⊥ NΠNf r F 0 ∞,r = sup ℓ≥0 sup I∈Dℓ A(ℓ) I where A(ℓ) I := − Z I X k≥ℓ |LkE⊥ NΠNf|r.
If 0 ≤ℓ≤N, then, for each I ∈Dℓ, (58) A(ℓ) I ≤ N X k=ℓ LkE⊥ NΠNf∥r ∞+ − Z I X k>N LkE⊥ NΠNf r =: A(ℓ) I,1 + A(ℓ) I,2.
By (56) we have A(ℓ) I,1 ≲1. For the second term, one can split I into 2(ℓ−N)d disjoint cubes J ∈DN, so that A(ℓ) I,2 ≤ sup J∈DN J⊂I A(N) J .
Thus, it suffices to show that (59) sup ℓ≥N A(ℓ) I ≲1.
Let ℓ≥N and I ∈Dℓ. Then (33) gives Lk(ENg) ≡0 in U∁ N,k, if k ≥ℓ, so A(ℓ) I = 1 |I| X k≥ℓ Z I∩UN,k LkE⊥ NΠNf r.
We shall show that (60) LkE⊥ NΠNf(x) ≲∥f∥B0 ∞,∞= 1, for x ∈I.
This inequality combined with |I ∩UN,k| ≈2−(d−1)ℓ2−k will establish the result, since A(ℓ) I ≲1 |I| X k≥ℓ |I ∩UN,k| ≲ X k≥ℓ 2ℓ−k ≲1.
It remains to show (60). Let Q ∈DN be such that I ⊂Q. By (31) (61) E⊥ NΠNf(y) ≲− Z Q∗∗M∗∗ A,N 2−NΠN∇f , for y ∈I∗.
Now, M∗∗ A,N 2−NΠN∇f ∞ ≤ 2−NΠN∇f ∞≲ X j≤N 2j−N∥Λjf∥∞ ≲ sup m≥0 ∥Λmf∥∞≲∥f∥B0 ∞,∞= 1.
(62) So, if x ∈I, then supp βk(x −·) ⊂x + O(2−k) ⊂I∗, and using (61) and (62) one deduces (60). This completes the proof of (59).
18 G. GARRIG´ OS A. SEEGER T. ULLRICH 6.2. Proof ENΠ⊥ Nf F 0 ∞,r ≲1. We now must bound ENΠ⊥ Nf r F 0 ∞,r = sup ℓ≥0 sup I∈Dℓ B(ℓ) I where B(ℓ) I = − Z I X k≥ℓ |LkENΠ⊥ Nf|r.
The cases 0 ≤ℓ≤N are handled with the same argument as in (58), this time using the inequality (57). If ℓ≥N and I ∈Dℓwe shall use B(ℓ) I ≤1 |I| X k≥ℓ |I ∩UN,k| LkENΠ⊥ Nf r ∞, so that it will suffice to show (63) LkENΠ⊥ Nf ∞≲∥f∥B0 ∞,∞= 1.
If Q ∈DN, then (42) and the argument in (43) (with ω(J) as in §4.1) implies ENLjΛjf(y) ≲ 2(N−j)d X J∈Dj+1(∂Q) ∥Λjf∥L∞(J) (64) ≲ 2(N−j)d X J∈Dj+1(∂Q) − Z ω(J) M∗ j Λjf , y ∈Q, using in the last step (41). So, summing up in j > N and using the disjoint-ness properties of the sets ω(J) we obtain X j>N ENLjΛjf(y) ≲ 2Nd Z Q∗∗sup m≥0 M∗ m Λmf ≲ sup m≥0 ∥Λmf∥∞≲∥f∥B0 ∞,∞= 1.
Finally, taking the convolution with βk one easily deduces (63). This com-pletes the proof of (55). The corresponding version for the TN[·, a] is proved similarly. Thus the proof of Proposition 6.1 is complete, and so is the proof of Theorem 1.7.
Remark 6.2. The above proof also shows that, if f ∈B0 ∞,∞, and N is fixed, then the series P∞ j=0 EN(LjΛjf) converges in the norm of F 0 ∞,r, for all r > 0.
This is a consequence of the crude bound ∥PJ2 j=J1 LjΛjf∥F 0 ∞,r ≲N 2−J1 ∥f∥F 0 ∞,∞, which can be obtained from (64); see also [5, Remark 4.5].
7. Proof of Theorem 1.8: Necessary conditions for s = 1 Here we show the assertion in Theorem 1.8, which corresponds to the optimality of the range of q stated in (ii) of Theorem 1.4. More precisely, we establish the following.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 19 Theorem 7.1. Let d/(d + 1) ≤p < 1 and 2 < q ≤∞. Then (65) ∥EN∥F 1 p,q→F 1 p,q ≈N 1 2 −1 q .
Moreover, for every N ≥1 there exists gN ∈C∞ c ((0, 1)d) such that (66) ∥gN∥F 1 p,q ≤1 and ∥EN(gN)∥F 1 p,∞≳N 1 2 −1 q .
7.1. Proof of Theorem 7.1: upper bounds. Let s = 1. Using Proposition 3.1 when d/(d + 1) < p < 1, or Proposition 4.2 when p = d/(d + 1), one has the inequality 2kq X j>N LkENLjΛjf ∞ k=0 Lp(ℓq) ≲∥f∥F 1 p,∞.
On the other hand, the proof of Proposition 3.2 gives 2kqLkE⊥ NΠNf ∞ k=0 Lp(ℓq) ≲∥ΠNf∥F 1 p,2, and by H¨ older’s inequality one has ∥ΠNf∥F 1 p,2 ≲ N X j=0 |2jΛjf|2 1 2 p ≲N 1 2 −1 q ∥f∥F 1 p,q.
Combining the above inequalities one obtains ∥EN∥F 1 p,q→F 1 p,q ≲N 1 2 −1 q .
□ Remark 7.2. If 1 < s < 1/p, the upper bound becomes exponential: EN F s p,q→F s p,q ≲2(s−1)N, for (d −1)/d < p < 1. This is a consequence of the simpler estimates for EN −ΠN : Bs p,∞→Bs p,r shown in [5, Propositions 3.1 through 3.4]. From we have also corresponding matching lower bounds, see the discussion in §8 below.
7.2. Proof of Theorem 7.1: lower bounds. To make the notation simpler, the counterexample is first presented in the 1-dimensional case, and later extended to Rd with a tensor product argument.
7.2.1. The case d = 1. Consider, for s > 0 and Λ ⊂N, a Weierstrass-type function (67) f(x) = X j∈Λ aj 2sj e2πi2jx ψ(x), x ∈R, with ψ ∈C∞ c (0, 1), and say ψ = 1 on [1/4, 3/4]. These functions satisfy (68) ∥f∥F s p,q(R) ≈ ( ∞ X k=0 |2ksβk ∗f|q) 1 q p ≲ X j∈Λ |aj|q1/q.
This can for instance be proved from Hardy’s inequalities and the following lemma 20 G. GARRIG´ OS A. SEEGER T. ULLRICH Lemma 7.3. Let βk be as in §2, and ψj(x) = e2πi2jxψ(x). Then |βk ∗ψj(x)| ≲2−|j−k|M, x ∈R, j, k = 0, 1, 2, . . .
Proof. If k > j, using that β has M-vanishing moments, |βk ∗ψj(x)| = Z R β(y) ψj(x −2−ky) − M−1 X m=0 ψ(m) j (x)(−2−ky)m dy ≲ 2(j−k)M, (69) since ∥ψ(M) j ∥∞≲2jM. If k ≤j, then Fourier inversion gives |βk ∗ψj(x)| = Z R ˆ β(ξ/2k) ˆ ψ(ξ −2j) e2πixξ dξ ≲ Z R dξ (1 + |ξ| 2k )M (1 + |ξ −2j|)M ≲ Z |ξ−2j|>2j−1 2−jM dξ (1 + |ξ| 2k )M + Z |ξ−2j|≤2j−1 2(k−j)M dξ (1 + |ξ −2j|)M ≲ 2k2−jM + 2(k−j)M ≲2−|j−k|M.
□ We now let s = 1 and ZN = {j ∈N : N/4 ≤j ≤N/2}, and consider a randomized version of (67), namely (70) fN(x, t) = X j∈ZN rj(t) 2j e2πi2jx ψ(x), where rj : [0, 1] →{−1, 1} is the sequence of Rademacher functions. Then, by (68), sup t∈[0,1] ∥fN(·, t)∥F 1 p,q ≲N1/q.
Below we shall show that (71) Z 1 0 EN[fN(·, t)] p F 1 p,∞dt 1/p ≥cN1/2.
The above inequality will be a consequence of the estimate (72) Z 1 0 2N βN ∗(EN[fN(·, t)]) p p dt 1/p ≥cN1/2, where βN = 2Ndβ(2N·) and β is a suitable test function satisfying the con-ditions in §2.1. Thus for some t0 ∈[0, 1] the function (73) gN = N−1/q fN(·, t0) will satisfy (74) ∥ENgN∥F 1 p,q(R) ≳ 2N βN ∗(ENgN) Lp(R) ≳N 1 2 −1 q , HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 21 and in particular (75) EN F 1 p,q→F 1 p,q ≳N 1 2 −1 q .
By Fubini’s theorem and Khintchine’s inequality the expression in (72) is equivalent to (76) 2N X j∈ZN 2−j βN ∗(ENψj) 2 1 2 Lp(R) ≥cN1/2.
If the operator EN is omitted in the left hand side, then this quantity be-comes uniformly bounded by Lemma 7.3, so (76) is also equivalent to (77) 2N X j∈ZN 2−j βN ∗(E⊥ Nψj) 2 1 2 Lp(R) ≥cN1/2.
Below we fix β such that supp β = (−1/8, 1/8), and denote its primitive by B(x) = R x −∞β(u)du, which also belongs to C∞ c (−1/8, 1/8). The following lemma is similar to [5, Lemma 6.4], but we include its proof below for completeness.
Lemma 7.4. Let µ ∈Z and let e IN,µ = [ µ 2N , µ+1/8 2N ]. Then (78) βN ∗(E⊥ Nψj)(x) = −2−Nψ′ j( µ 2N ) B(2Nx−µ) + O 22(j−N) , x ∈e IN,µ.
Moreover, if µ 2N ∈[ 1 4, 3 4], then ψ′ j( µ 2N ) = 2πi2je2πi2jx.
Assuming the lemma, the pth power of the left hand side of (77) can be bounded from below by X µ∈Z 1 4 ≤µ 2N ≤3 4 Z e IN,µ X j∈Λ 2N−j βN ∗(E⊥ Nψj) 2 p 2 dx ≥(2π)p X µ∈Z 1 4 ≤µ 2N ≤3 4 Z e IN,µ X j∈Λ B(2Nx −µ) 2 p 2 −c X j∈Λ |2j−N|2 p 2 dx ≳N p 2 X µ∈Z 1 4 ≤µ 2N ≤3 4 Z e IN,µ B(2Nx −µ) p dx −c′ 2−Np/2 ≳N p 2 Z 1/8 0 B(u) p du −c′ 2−Np/2 ≳N p 2 , using in the last step that β (hence B) is not identically null in (0, 1/8).
This finishes the proof modulo Lemma 7.4.
Proof of Lemma 7.4. For simplicity we write I+ = IN,µ and I−= IN,µ−1.
If x ∈e IN,µ, then supp βN(x −·) ⊂I+ ∪I−, and thus (79) βN ∗(E⊥ Nψj)(x) = X ± Z I± βN(x −y) h ψj(y) −− Z I± ψj i dy.
22 G. GARRIG´ OS A. SEEGER T. ULLRICH Now, if y ∈I±, using the linear Taylor’s expansion of ψj around y and the bound ∥ψ ′′ j ∥∞≲22j, the inner bracketed expression becomes − Z I±(ψj(y) −ψj(w))dw = − Z I± ψ′ j(y)(y −w) dw + O(22(j−N)) = ψ′ j( µ 2N ) − Z I±(y −w) dw + O(22(j−N)) = ψ′ j( µ 2N ) (y −cI±) + O(22(j−N)) = ψ′ j( µ 2N ) (y − µ 2N ) −ψ′ j( µ 2N ) ±1 2N+1 + O(22(j−N)).
Putting these quantities into (79), and using the support and the moment condition of β, we are left with (80) βN ∗(E⊥ Nψj)(x) = − X ± ψ′ j( µ 2N ) ±1 2N+1 Z I± βN(x −y) dy + O(22(j−N)).
Now, an elementary computation using the primitive, B(u), of β(u) shows that the two integrals substructed above can be written as Z µ+1 2N µ 2N − Z µ 2N µ−1 2N βN(x−y)dy = Z µ+1 µ − Z µ µ−1 β(2Nx−u) du = 2B(2Nx−µ), since B(2Nx −µ ± 1) = 0 by the support condition. Thus, placing this expression into (80) implies the asserted identity (78).
□ 7.2.2. The d-dimensional case. Consider GN(x1, x′) = gN(x1)χ(x′), where gN is the 1-dimensional function in (73), and χ ∈C∞ c (0, 1)d−1 with χ ≡1 in [ 1 8, 7 8]d−1. We shall show that (81) ∥GN∥F 1 p,q(Rd) ≲1 and (82) ENGN F 1 p,q(Rd) ≳N 1 2 −1 q .
To do so, in the definition of the F s p,q-quasinorms we shall use suitable test functions of tensor product type; see also [5, §5.1] for a similar argument.
Namely, for a fixed M ∈N we consider a non-negative even function φ0 ∈ C∞ c (−1 8, 1 8) such that φ(2M) 0 (t) > 0 for all t in some interval [−2ε, 2ε]. Since b φ0(0) = R φ0 > 0, dilating if necessary we may also assume that b φ0 ̸= 0 on [−1, 1]. Let ϕ0 ∈C∞ c ((−1 8, 1 8)d−1) be such that b ϕ0 ̸= 0 on [−1, 1]d−1 and b ϕ0(0) = 1. For M ≥1, let φ(t) := ( d dt)2Mφ0(t), ϕ(x2, . . . , xd) := ∂2 ∂x2 2 + · · · + ∂2 ∂x2 d Mϕ0(x′).
Then, we define (83) Ψ(x) := ∆Mφ0 ⊗ϕ0 = φ(x1)ϕ0(x′) + φ0(x1)ϕ(x′).
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 23 Clearly, Z Rd Ψ(y)ym1 1 · · · ymd d dy = 0, when m1 + . . . + md < 2M.
Finally, let Ψ0 = φ0 ⊗ϕ0, and Ψk(x) = 2kdΨ(2kx), k ≥1. Then, if we choose M sufficiently large we have (84) ∥f∥F s p,q ≈ 2ksΨk ∗f k≥0 Lp(ℓq), ∀f ∈F s p,q(Rd); see e.g. [14, 2.4.6]. Observe that, for k ≥1 we can write (85) Ψk = φk ⊗ϕ0,k + φ0,k ⊗ϕk, where we denote φk(x1) = 2kφ(2kx1), ϕk(x′) = 2(d−1)kϕ(2kx′), and likewise for φ0,k and ϕ0,k.
With this notation the inequality in (81) is easily proved as follows. From (85) and ∥gN∥∞, ∥χ∥∞≲1 one obtains |Ψk ∗GN|(x1, x′) ≲|φk ∗gN|(x1) + |ϕk ∗χ|(x′), k ≥1, and a similar (simpler) expression when k = 0.
Therefore (84) and the compact support of the involved functions imply ∥GN∥F 1 p,q(Rd) ≲∥gN∥F 1 p,q(R) + ∥χ∥F 1 p,q(Rd−1) ≲1.
In order to prove (82), we let E(1) N and E(d−1) N be the dyadic averaging oper-ators on R and Rd−1, respectively. For N ≥1, we observe that (86) ΨN ∗(ENGN)(x1, x′) = φN ∗(E(1) N gN)(x1), for x′ ∈( 1 4, 3 4)d−1.
Indeed, for such x′ one has ϕ0,N ∗(E(d−1) N χ)(x′) = Z ϕ0(y′)dy′ = 1, ϕN ∗(E(d−1) N χ)(x′) = Z ϕ(y′)dy′ = 0, due to the support properties of ϕ0,N(x′ −·) and ϕN(x′ −·). Therefore, (84) and (86) imply that ∥ENGN∥F 1 p,q(Rd) ≳ 2NφN ∗E(1) N gN Lp(R) ≳N 1 2 −1 q , the last inequality due to (74). This proves (82), and concludes the proof of EN F 1 p,q(Rd)→F 1 p,q(Rd) ≳N 1 2 −1 q , 2 < q ≤∞.
□ 24 G. GARRIG´ OS A. SEEGER T. ULLRICH 8. Boundedness of the dyadic averaging operators and the proof of Theorem 1.4 We now gather the results from the previous sections to complete the proof of Theorem 1.4. In §8.1 we first explain how the extension of EN, from S into F s p,q, should be defined (which is not obvious for all cases). We discuss sufficient conditions for uniform boundedness in §8.2 using theorems in previous chapters, and necessary conditions in §8.3. The proofs of some of the more tedious details about definability are given in §8.4. The proof of necessary conditions for the individual boundedness of the EN is given in §8.5. In §8.6 we include a discussion when the characteristic function of a bounded interval can be defined as a linear functional on F s p,q.
8.1. Extension of the operators EN to the space F s p,q. Let (s, p, q) be as in (i)-(v). For a distribution f ∈F s p,q we define (87) ENf := ∞ X j=0 EN(LjΛjf).
We claim that this series always converges in the F s p,q-norm (actually, in all the Bs p,r-norms, for r > 0). When max{d/p −d, 1/p −1} < s < 1/p this fact was already justified in [5, Remarks 3.5 and 4.5]. When s = d/p −d, one can reach the same conclusion with a slight modification in the proof of Proposition 4.2. We present the details in Lemma 8.4 below. When s = 0 and p = ∞, the convergence holds in all F 0 ∞,r-norms, for r > 0, by Remark 6.2.
We also remark that when f ∈F s p,q is locally integrable with polynomial growth then the above extension coincides with the usual operator, that is X j≥0 EN(LjΛjf) = X I∈DN − Z I f 1I; see Lemma 8.5 below.
8.2. Sufficient conditions in Theorem 1.4. The uniform boundedness of EN in F s p,q in the cases (i) and (iii) was established in . In the cases (ii), (iv) and (v) it is a consequence of the Theorems 1.5, 1.6 and 1.7, and elementary embeddings; see the discussion following (10) in section 1.
8.3. Necessary conditions in Theorem 1.4. We first identify the range of exponents for the continuity of an individual operator EN.
Definition 8.1. Let A be the set of all (s, p, q) for which one of the following three conditions (i), (ii) or (iii) hold: (88) (i) max{d p −d, 1 p −1} < s < 1/p, 0 < p, q ≤∞ (ii) s = d p −d, 0 < p ≤1, 0 < q ≤∞ (iii) s = 0, p = ∞, 0 < q ≤∞.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 25 1 p s 1 d+1 d d d−1 1 −1 Figure 3. Range of exponents for the continuity of the in-dividual operators EN in F s p,q(Rd).
Proposition 8.2. Let N ∈N0 be fixed. Suppose that (89) ENψ F s p,q ≤cN∥ψ∥F s p,q, ∀ψ ∈S(Rd).
Then necessarily (s, p, q) ∈A.
The proof of the proposition is based on the fact that 1[0,1)d must belong to both F s p,q and its dual space. We present the details in §8.5 below.
We now turn to the existence of uniformly bounded extensions of the oper-ators EN in the above region. This is equivalent to the uniform boundedness of the numbers OpS(EN, F s p,q) := sup n ENf∥F s p,q : f ∈S, ∥f∥F s p,q ≤1 o , when N = 0, 1, 2, . . . An example given in [4, Proposition 4.2] shows that, OpS(EN, F s p,q) ≳OpS(EN, Bs p,∞) ≳2(s−1)N, when 1 < s < 1/p and (d −1)/d < p < 1.
So the condition s ≤1 is necessary. When s = 1, Theorem 7.1 shows that 0 < q ≤2 is also necessary.
This comprises all the cases considered in Theorem 1.4, and completes the proof of all the assertions. Moreover, it also gives the following.
Corollary 8.3. Let N ∈N0 be fixed, and (s, p, q) ∈A. Then the (extended) operator EN, as in (87), satisfies OpS(EN, F s p,q) ≈ EN F s p,q→F s p,q ≈ 2(s−1)N if 1 < s < 1/p N1/2−1/q if s = 1, q ≥2 1 otherwise.
26 G. GARRIG´ OS A. SEEGER T. ULLRICH 8.4. Convergence of P j EN(LjΛjf) when f ∈F d/p−d p,∞ .
Lemma 8.4. Let d−1 d < p ≤1, s = d/p −d and r > 0. If N ≥0 and g ∈F s p,∞(Rd), then lim J1→∞sup J2≥J1 EN J2 X j=J1 LjΛjg Bs p,r = 0.
In particular, the series ENg := ∞ X j=0 EN(LjΛjg) converges in F s p,q for all 0 < q ≤∞.
Proof. Pick a non-negative function ζ ∈S(Rd) such that ζ ≥1 on [−5, 5]d, and supp b ζ ⊂{|ξ| ≤1/8}.
For each cube Q = {x : |x −x0|∞≤δ}, let ζQ(x) := ζ x −x0 δ , so that ζQ ≥1 in Q∗∗(the 5-fold dilate of Q), and b ζQ has support in {|ξ| ≤(8δ)−1}. Finally, for j ≥N + 1, we define (with UN,j as in (32)) ζj,N(x) := X Q∈Dj+1 Q⊂UN,j ζQ(x).
This function satisfies the properties (90) |ζj,N(x)| ≤ cM 1 + 2j min 1≤i≤d dist(xi, 2−NZ) M , for each M > 0, and (91) supp b ζj,N ⊂ |ξ| ≤2j−3 .
Let f = PJ2 j=J1 LjΛjg, and assume that J1 ≥L + 3 for some fixed L > N.
Then, f ∈F s p,∞and Λjf ≡0 unless j ≥L. We now follow the proof of Proposition 4.2, with the following modification. If J ∈Dj+1 is such that J ⊂UN,j, then for all y ∈J, M∗ j(Λjf)(y) = sup |h|∞≤2−j |Λjf(y + h)| ≤M∗ j ζj,NΛjf (y).
One can use this estimate in (43) (or in (50)), so the same arguments which lead to (45) (or to (52)) can be applied with the function Λℓf replaced by ζℓ,NΛℓf. That is there exists γ > 0 such that for Ak, Ak as in (43), (51), resp., 2k( d p −d)(∥Ak∥p + ∥Ak∥p) ≲2−|k−N|γ ∥GL∥p, HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 27 with GL(x) = sup ℓ≥L 2ℓ( d p −d)M∗ ℓ ζℓ,NΛℓf .
Since the spectrum of ζℓ,NΛℓf is contained in {|ξ| ≤2ℓ}, one can use Peetre’s inequality and deduce as in (46) that ∥GL∥p ≲ sup ℓ≥L 2ℓ( d p −d)ζℓ,N |Λℓf| p ≲ ζ∗ L,N sup ℓ≥0 2ℓ( d p −d) |LℓΛℓg| p, where ζ∗ L,N(x) = sup ℓ≥L ζℓ,N(x) ≲ 1 + 2L min 1≤i≤d(xi, 2−NZ) −M .
Observe that lim L→∞ζ∗ L,N(x) = 0, ∀x ∈ ∞ [ n=1 U∁ N,n, and therefore at almost every x ∈Rd. So, the assumption g ∈F s p,∞and the Dominated Convergence Theorem imply that lim L→∞∥GL∥p = 0.
□ Lemma 8.5. Let (s, p, q) be as in (i)-(v) in Theorem 1.4. Let f ∈F s p,q be locally integrable with polynomial growth, that is, f(x)/(1 + |x|)M ∈L1(Rd) for some M ≥0. Then (92) X j≥0 EN(LjΛjf) = X I∈DN − Z I f 1I, in the sense of tempered distributions.
Proof. In this lemma we restrict the notation ENg(x) := X I∈DN − Z I g 1I(x), x ∈Rd, only to locally integrable functions g with polynomial growth. In particular, ENg is another such function, hence a tempered distribution. We write e ENf for the distribution on the left hand side of (92). If ψ ∈S(Rd), then (93) e ENf, ψ = ∞ X j=0 EN(LjΛjf), ψ = ∞ X j=0 Z Rd(LjΛjf) ENψ.
The family of operators {Πn = Pn j=0 LjΛj}n≥0 is a smooth approximation of the identity, and therefore Πnf →f in L1(Rd, (1 + |x|)−Mdx), 28 G. GARRIG´ OS A. SEEGER T. ULLRICH by the condition on f. Therefore, using that |ENψ(x)| ≲(1+|x|)−M, we can pass the sum inside the integral in the last expression of (93), and continuing with Fubini’s theorem obtain e ENf, ψ = Z Rd( ∞ X j=0 LjΛjf) ENψ = Z Rd f ENψ = Z Rd ENf ψ.
Hence e ENf coincides with ENf as distributions.
□ Remark 8.6. One can extend the domain of EN further, dropping the poly-nomial growth assumption in Lemma 8.5 if in the resolution of the identity (21) we replace the operator LN by suitable compactly supported convolu-tion kernels. Indeed there are, for ε > 0, M < ∞, C∞functions φ, e φ, ψ, e ψ supported in {|x| ≤ε} such that R φ = 1, R e φ = 1 and 1 −b φ, 1 −b e φ, b ψ, b e ψ all vanish of order M at 0, and such that for distributions f (94) L0e L0f + ∞ X k=1 Lk e Lkf = f in the sense of distributions; here L0, e L0 are the convolution operators with convolution kernels φ, e φ, resp., and for k ≥1, Lk and e Lk are the convolution operators with convolution kernels 2(k−1)dψ(2k−1·), 2(k−1)d e ψ(2k−1·), resp.
The resolution in the form (94) is perhaps not widely known; a proof can be found in [9, Lemma 2.1], together with some extensions. For us the use of the nonlocal operators ΛN has the advantage that we may apply the Peetre maximal inequalities in a straightforward way.
8.5. Proof of Proposition 8.2. Since ENψ = E0ψ(2−N·), we may assume that N = 0. Then (89) takes the form (95) E0ψ F s p,q ≤cN∥ψ∥F s p,q, ∀f ∈S(Rd).
Let ψ ∈C∞ c ((0, 1)d) such that R ψ = 1. Then E0(ψ) = 1[0,1)d, and (95) implies (96) 1[0,1)d ∈F s p,q(Rd).
The validity of this property is well-known. If 0 < p < ∞, then (96) holds iffs < 1/p. If p = ∞, then (96) holds iffs ≤0. See e.g. [16, Proposition 2.50]. This gives the required upper bounds on s.
We turn to the lower bounds for the exponent s. Consider the classes of test functions F1 = span n η(x1) · · · η(xd) : η ∈C∞ c (−1, 1) odd o F2 = n η(x1)χ(x′) : η ∈C∞ c (−1, 1) odd, χ ∈C∞ c (0, 1)d−1 with R χ = 1 o .
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 29 We first show that, if f ∈Fi, i = 1, 2, then (97) E0(f) = Z [0,1]d f · hi, for some fixed functions h1, h2 ∈span{1I : I ∈D0}.
Let f ∈F1. It suffices to show (102) for f(x) = η(x1) · · · η(xd), with η ∈C∞ c (−1, 1) odd. Given ε = (ε1, . . . , εd) ∈{0, 1}d we denote Qε = [0, 1)d −ε and sign(ε) = d Y i=1 (−1)εi.
We claim that (102) holds with (98) h1 = X ε∈{0,1}d sign(ε) 1Qε.
Indeed, for such f we have E0(f) = X ε∈{0,1}d − Z Qε f · 1Qε, and since η is odd − Z Qε f = d Y i=1 Z [0,1)−εi η(xi) dxi = d Y i=1 (−1)εi Z 1 0 η(xi) dxi = sign(ε) Z [0,1)d f.
Thus, (102) follows.
Similarly, let f ∈F2, and denote Q0 = [0, 1)d and Q1 = Q0 −(1, 0, . . . , 0).
Then E0(f) = − Z Q0 f · 1Q0 + − Z Q1 f · 1Q1 = Z Q0 f 1Q0 −1Q1 , and hence (102) holds with h2 = 1Q0 −1Q1. This completes the proof of (102), and reduces the proof of Proposition 8.2 to the following result.
Lemma 8.7. Let I = (0, 1)d. Suppose that (99) Z I f ≤C ∥f∥F s p,q, ∀f ∈F1 ∪F2.
Then one of the following two conditions must hold (a) s > max{d p −d, 1 p −1}, 0 < p, q ≤∞ (b) s = d p −d, 0 < p ≤1, 0 < q ≤∞.
Proof. Let η ∈C∞ c (−1/2, 1/2), odd and such that R 1 0 η = 1. Define gj(x1, . . . , xd) := 2jdη(2jx1) · · · η(2jxd), j ≥1.
30 G. GARRIG´ OS A. SEEGER T. ULLRICH Observe that gj ∈F1 and R I gj(x) dx = 1. On the other hand, a standard computation shows that (100) ∥gj∥F s p,q(Rd) ≲2−j( d p −d−s).
So, (99) cannot hold unless s ≥d/p −d. Suppose now that s = d/p −d, and consider g = PN j=1 gj ∈F1. In this case we have R I g(x) dx = N and ∥g∥F s p,q(Rd) ≲N1/p. Thus, (99) cannot hold unless 0 < p ≤1. Observe that when d = 1 this completes the proof of the lemma.
Suppose now that d ≥2. Define now the functions Gj(x1, x′) := gj(x1) χ(x′), j ≥1, where χ ∈C∞ c (0, 1)d−1 has R χ = 1. Then, Gj ∈F2 and R I Gj = 1. On the other hand ∥Gj∥F s p,q(Rd) ≲∥gj∥F s p,q(R) ≲2−j( 1 p −1−s).
So (99) can only hold if s ≥1/p −1. In the case s = 1/p −1, consider the function G = PN j=1 Gj ∈F2. Then we have Z I G = N and ∥G∥F s p,q(Rd) ≲∥g∥F s p,q(R) ≲N1/p.
Thus, (99) can only hold if 0 < p ≤1. This does not give a new region if d ≥2.
□ 8.6. Extension of 1I as a bounded functional in F s p,q. The next result is a converse of Lemma 8.7, which in addition gives a continuous extension of the functional 1I to the whole space F s p,q.
Lemma 8.8. Let (s, p, q) be numbers satisfying (a) or (b) in Lemma 8.7, and let I ∈DN. Then, for each f ∈F s p,q the series (101) 1∗ I(f) := ∞ X j=0 Z I LjΛjf is absolutely convergent, and there exists a constant CN = CN(s, p, q) > 0 such that 1∗ I(f) ≤CN ∥f∥F s p,q, ∀f ∈F s p,q.
Moreover, (1) (101) does not depend on the specific decomposition I = P∞ j=0 LjΛj, (2) If f ∈F s p,q(Rd) is locally integrable with polynomial growth then 1∗ I(f) = Z I f(x) dx.
(3) For every ζ ∈C∞ c such that ζ ≡1 in a neighborhood of ¯ I, it holds (102) 1∗ I(ζ f) = 1∗ I(f), for all f ∈F s p,q.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 31 Proof. Since F s p,q , →F s p,∞, it suffices to prove the result for q = ∞. First notice that N X j=0 Z I LjΛjf ≤ Z I ΠNf ≤|I| ΠNf∥L∞(I).
Now, using the Peetre maximal functions, ΠNf∥L∞(I) ≲ − Z I∗∗ M∗ N(ΠNf) p1/p ≲2 Nd p ΠNf p, and letting r = min{p, 1}, the last factor is bounded by ΠNf p ≤ N X j=0 ∥LjΛjf∥r p 1 r ≲ N X j=0 2−jsr 1 r ∥f∥Bs p,∞.
So, we are left with proving that X j>N Z I LjΛjf ≲CN ∥f∥F s p,∞.
Since j > N, we can use [4, Lemma 2.3.ii] and inequality (50) to obtain Z I LjΛjf ≲ 2−jd X J∈Dj+1(∂I) ∥Λjf∥L∞(J) ≲ 2−jd X J∈Dj+1(∂I) − Z w(J) M∗ j(Λjf) p1/p .
In the case (b), i.e. s = d/p −d and 0 < p ≤1, we argue as in (43) and obtain X j>N Z I LjΛjf p ≲ Z I∗∗sup ℓ>N 2ℓ( d p −d)M∗ ℓ(Λℓf) p ≲∥f∥p F d/p−d p,∞ .
In the case (a), i.e. s > max{d p −d, 1 p −1}, one can prove in a similar fashion the stronger estimate X j>N Z I LjΛjf r ≲CN(s, p) ∥f∥r Bs p,∞ with r = min{p, 1}.
It is immediate to verify that (101) does not depend on the specific reso-lution of the identity I = P∞ j=0 LjΛj. The assertion (2) in the statement is a consequence of the convergence of the approximate identity Πnf →f in L1 loc(Rd) when f is locally integrable with polynomial growth.
We finally verify the third assertion. Let Πn = Pn j=0 LjΛj and χ = 1−ζ.
Then, 1∗ I(f) −1∗ I(ζf) = lim n→∞ Z I Πn(χf) .
32 G. GARRIG´ OS A. SEEGER T. ULLRICH Using distribution theory we can write Z I Πn(χf)(x) dx = Z I χf, Πn(x −·) dx = D f, χ R I Πn(x −·)dx E .
The result follows after checking that for Φn(y) := χ(y) Z I Πn(x −y)dx we have limn→∞Φn = 0 in the topology of the Schwartz class.
□ Let h ∈Hd. The previous result can be applied to define h∗as a contin-uous linear functional in F s p,q. Namely, h∗(f) = ∞ X j=0 Z h(x) LjΛjf(x) dx, f ∈F s p,q.
Then, Lemmas 8.7 and 8.8 imply the following.
Corollary 8.9. The Haar functions, regarded as linear functionals, can be continuously extended from S into F s p,q if and only if the indices (s, p, q) satisfy (a) or (b) in Lemma 8.7.
9. Failure of density for s = 1 Proposition 9.1. There exists a Schwartz function f supported in ( 1 16, 15 16)d such that, for all 0 < p ≤1 and 0 < q ≤∞, (103) lim inf N→∞∥ENf −f∥F 1 p,q > 0.
Moreover, if d/(d + 1) ≤p < 1 and 0 < q ≤2 then the span of Hd is not a dense set in F 1 p,q(Rd).
Proof. The proof of (103) uses the same function f as in [5, Proposition 8.3]. Namely, pick η ∈C∞ c (Rd) such that supp η ⊂( 1 16, 15 16)d and η(x) = 1 on (1/8, 7/8)d. Then consider f(x) = x1 η(x). In [5, Proposition 8.3] it was shown that this function satisfies (104) lim inf N→∞∥ENf −f∥B1 p,∞> 0.
Therefore, (103) follows from here and the embeddings F 1 p,q , →F 1 p,∞, →B1 p,∞.
We next show that (103) implies the failure of the density of span Hd in F 1 p,q, for all 0 < q ≤2 and d/(d + 1) ≤p < 1. Indeed, assume for contradiction that such density holds, and given f ∈S as in (103) and ε > 0, find g ∈span Hd such that ∥f −g∥F 1 p,q < ε. Let N0 be large enough so that HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 33 EN(g) = g for all N ≥N0. Then, the (quasi-)triangle inequality and the uniform boundedness of EN in Theorem 1.4 gives ∥f −ENf∥F 1 p,q ≲∥f −g∥F 1 p,q + ∥ENg −ENf∥F 1 p,q ≲ε, N ≥N0, which contradicts (103).
□ Remark 9.2. It would be interesting to settle the question whether span Hd is dense in the spaces F 1 p,q, when d/(d + 1) ≤p < 1 and 2 < q < ∞. As the operators EN are not uniformly bounded in this range our current argument is not sufficient to give an answer (cf. also [5, §8.1] for a similar discussion about the Besov space analogue of this question).
10. Localization and partial sums of admissible enumerations Let U = {un}∞ n=1 be a strongly admissible enumeration of Hd, as in Definition 1.2 above. Explicit examples of such enumerations are not hard to construct; see e.g. [5, §11].
Here we quote a localization lemma for such enumerations, which relates the partial operators SU R and the dyadic averages EN and TN[·, a]. We let ς ∈C∞ c be supported in a 10−2 neighborhood of [0, 1)d and so that (105) X ν∈Zd ς(· −ν) ≡1, and denote ςν = ς(· −ν), ν ∈Zd. The following identity has been proved in [5, Lemma 9.1].
Lemma 10.1. Let U be a strongly admissible enumeration of Hd. Then, for every R ∈N and ν ∈Zd there is an integer Nν = Nν(R) ≥−1 and {0, 1}-sequences aκ,ν, 0 ≤κ ≤b, such that for all g ∈L1 loc(Rd) we have (106) SU R[gςν] = ENν[gςν] + b X κ=0 TNν+κ[gςν, aκ,ν].
We next recall a localization property of the F s p,q-quasinorms; see [14, 2.4.7] (and [16, 2.4.2] for p = ∞).
Lemma 10.2. Let 0 < p, q ≤∞and s ∈R. Then it holds (107) X ν∈Zd ςνg F s p,q ≈ X ν∈Zd ςν g p F s p,q 1/p .
We are now ready to prove the uniform boundedness of the operators SU R.
We assume that (s, p, q) ∈A, as defined in Definition 8.1, so that these operators can be continuously extended to the whole space F s p,q. More precisely, if p, q < ∞, condition (1) holds and SR is well-defined as in section 1 (that is, extended from S to F s p,q by density). In order to include as well the 34 G. GARRIG´ OS A. SEEGER T. ULLRICH cases p = ∞or q = ∞, one first considers extensions of the dual functionals u∗ n to the full space F s p,q as follows (108) u∗ n(f) := ∞ X j=1 2k(n)d Z hϵ(n) k(n),ν(n) LjΛjf, f ∈F s p,q; see the details in §8.6. In this way, the identity in (106) remains valid for all g ∈F s p,q.
Proposition 10.3. Let (s, p, q) ∈A. Suppose that (109) sup N≥0 ∥EN∥F s p,q→F s p,q + sup N≥0 sup ∥a∥ℓ∞≤1 ∥TN[·, a]∥F s p,q→F s p,q < ∞.
Then, for every strongly admissible enumeration U it holds sup R≥1 S U R ∥F s p,q→F s p,q < ∞.
Proof. Consider SR = SU R as a continuous operator in F s p,q (as described in §8.6). Then, the support properties of the extension, see (102), imply that ςν′SR(fςν) = 0, whenever |ν −ν′|∞≥3.
Then, using (105) and (107), SRf F s p,q ≈ X ν′ ςν′SR X ν ςνf p F s p,q 1 p ≲ X ν′ X ν : |ν−ν′|∞≤2 ςν′SR(fςν) p F s p,q 1/p ≲ X ν SR(fςν) p F s p,q 1/p , using in the last step that ςν′ is a uniform multiplier in F s p,q; see [14, 4.2.2].
Then Lemma 10.1 and (109) give SRf F s p,q ≲ X ν ENν(fςν) p F s p,q + b X κ=0 TNν+κ[fςν, aκ,ν] p F s p,q 1/p ≲b X ν fςν p F s p,q 1/p ≈∥f∥F s p,q.
□ Remark 10.4. The equivalence in (107) is also true with ς replaced by 1[0,1]d when max nd p −1, 1 p −1 o < s < 1 p, as in that case characteristic functions of cubes are multipliers in F s p,q. In particular, for those indices the assertion in Proposition 10.3 holds as well with the weaker notion of admissible enumeration; see [4, §3]. This is in particular the case when s = 1 and d/(d + 1) < p < 1.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 35 Finally, we conclude with the following observation, which we shall use to transfer negative results between the operators EN and SR. The explicit construction is given in [5, §11].
Lemma 10.5. There exists a strongly admissible enumeration U with the following property: for every m ≥0 there exists an integer R(m) ≥1 such that (110) SU R(m)f = Emf, f ∈C∞ c ((−5, 5)d).
11. The Schauder basis property: proof of Theorem 1.3 11.1. Necessary conditions. Suppose that every strongly admissible enumer-ation U of Hd is a Schauder basis of F s p,q. This implies that span Hd must be dense (hence p, q < ∞), and (111) CU := sup R≥1 SU R F s p,q→F s p,q < ∞.
Moreover, if we select U as in Lemma 10.5, then we must have sup m≥0 OpS Em, F s p,q ≤CU < ∞.
In view of Proposition 9.1 and Corollary 8.3 this is only possible if (i), (ii) or (iii) in Theorem 1.3 hold.
11.2. Sufficient conditions. Under the assumptions in (i), (ii), and (iii) of Theorem 1.3, the operators EN and TN[·, a] are uniformly bounded in F s p,q, by Theorem (1.4). So we can use Proposition 10.3 and conclude that (111) must hold. The density of span Hd is also true in this range, so we conclude that U is a Schauder basis of F s p,q.
11.3. Consequences for the basic sequence property. Theorem 1.4 addition-ally implies convergence of basic sequences in the cases when span Hd is not dense. Namely, when p = ∞or q = ∞, let fs p,q denote the closure of the S in F s p,q. When s < 1/p the subset span Hd is dense in fs p,q, so we deduce the following.
Corollary 11.1. Let (s, p, q) be as in (i), (iii) or (iv) in Theorem 1.4. Then, every admissible enumeration U is a Schauder basis of fs p,q. That is, f = ∞ X n=1 u∗ n(f)un, for all f ∈fs p,q, with convergence in the norm of F s p,q.
Remark 11.2. Observe that we have excluded the cases (ii) and (v) in The-orem 1.4. In these cases we can only say that U is a Schauder basis of the subspace span Hd F s p,q.
36 G. GARRIG´ OS A. SEEGER T. ULLRICH A precise description of this subspace in those cases, however, is not clear.
In the range (ii), ie s = 1 (and q ≤2) this subspace cannot contain the Schwartz class S, as shown by Proposition 9.1. On the other hand, in the case (v), i.e. s = 0 and p = ∞, this subspace strictly contains f0 ∞,q. Indeed, first of all one has f0 ∞,∞∩span Hd = {0}; see [5, Proposition 5.1]. Next, for all q ≤∞, the inclusion C∞ c (Rd) ⊂span Hd F 0 ∞,q follows, when d = 1, from the elementary embedding B1/p p,∞(R) , →F 0 ∞,q(R) and the corresponding result for B1/p p,∞(R) in [5, Proposition 8.6].
When d ≥2, one can approximate each f ∈C∞ c (Rd) by a linear combination of functions g1(x1) · · · gd(xd) with gi ∈C∞ c (R), and then use the previous result.
12. The unconditional basis property: proof of Theorem 1.1 The fact that Hd is an unconditional basis of F s p,q when (6) and (7) hold was shown by Triebel in [15, Theorem 2.21]. We now indicate references for the negative end-point results, corresponding to the dotted or dashed lines around the green region in Figure 4.
The trivial cases correspond to the lines p = ∞, s = 1/p, and to the line s = 1/p −1 with p > 1. In all of them not even the Schauder basis property may hold. Namely, if p = ∞then F s ∞,q is not separable, and hence span Hd is not dense (see however Remark 12.1 below for the validity of unconditionality in the subspace fs ∞,q). The other two cases are excluded because (s, p, q) ̸∈A, and hence (1) fails.
Concerning the horizontal line s = 1, this is a borderline of the uncon-ditionality region when d/(d + 1) ≤p, q ≤1.
This case is excluded by Proposition 9.1, since span Hd is not dense in F 1 p,q, so the Schauder basis property cannot hold here.
At the line s = d−d/p, for d/(d+1) < p ≤1, we have a positive Schauder basis result for strongly admissible U, by Theorem 1.3. So we must prove that such a basis cannot be unconditional in F s p,q. This was already shown in [5, Theorem 13.1], based on an explicit example which works well in both the Besov and the Triebel-Lizorkin setting.
Finally, we consider the horizontal lines of the green region which lie inside the open pentagon P. In , the failure of unconditionality in these lines was shown in the case q > 1 and d = 1, indeed for all exponents p ≥d/(d+1) (by [10, Remark 7.1]). Here we show how to modify the arguments in that paper to cover as well the cases q ≤1, and extend the construction to all d ≥1.
HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 37 case q > 1 1/p s 1 d+1 d 1 1 q 1 q −1 −1 case q = 1 1/p s 1 1 d+1 d −1 case d d+1 < q < 1 1/p s 1 d+1 d 1 d q −d −1 Figure 4. Parameter domain for unconditionality in the cases q > 1, q = 1 and d/(d + 1) < q < 1, respectively.
We recall some notation from . To each finite set E ⊂Hd we associate the projection operator PE(f) = X h∈E ⟨f, h∗⟩h, where h∗= 2kdh is the dual functional of a Haar function h ∈Hd of fre-quency 2k. We also write HF(E) for the set of all Haar frequencies 2k of elements h ∈E.
We first remark that the results in [10, §6] remain valid when q ≤1.
Namely, for each N ≥2, an explicit construction is given of a function f = fN ∈F 1/q−1 p,q (R) and a set E = EN ⊂H1 with #HF(E) ≤N4N such that1 (112) ∥fN∥F 1/q−1 p,q (R) ≲N1/q and ∥PEN (fN)∥F 1/q−1 p,q (R) ≥N1+ 1 q , 1In the notation of [10, §6], one should consider sets A of consecutive Haar frequencies, so that the associated “density” number in [10, (43)] takes the value Z = N.
38 G. GARRIG´ OS A. SEEGER T. ULLRICH if 0 < q ≤p < ∞. In particular, for d = 1, ∥PEN ∥F 1/q−1 p,q →F 1/q−1 p,q ≳N, and hence H1 is not unconditional at the lower segment of the green region in Figure 4.
When d ≥2, the above example can be adapted in two different ways. If 1 < q < ∞, one considers the tensorized functions FN(x1, x′) := fN(x1) ⊗χ(x′), where fN is as in (112) and χ ∈C∞ c ((−1, 2)d−1) with χ ≡1 in [0, 1]d−1, and defines the sets EN := n h ⊗1[0,1]d−1 : h ∈EN o .
Then, a standard computation (as in §7.2.2 above) gives ∥FN∥F 1/q−1 p,q (Rd) ≲N1/q and ∥PEN (FN)∥F 1/q−1 p,q (Rd) ≳N1+1/q.
When q ≤1, one considers instead the natural generalization to Rd of the construction in [10, §6], namely using the test function Qd i=1 η(xi), in place of the one dimensional function η in [10, (46)].
More precisely, if 1 ≤κ ≤4Nd, bκ = κN and 1 ≤σ ≤N, one defines the functions Yκ,σ(x1, . . . , xd) = X (ν1,...,νd)∈Zd 0≤νi<2bκ−N−2 2−σd d Y i=1 η 2bκ+N−σ(xi −2N+2−bκνi) and, for t ∈[0, 1], ft = 4Nd X κ=1 rκ(t) 2−(bκ+N)( d q −d) N X σ=1 Yκ,σ, where rk(t) is a Rademacher function. Then, arguing as in [10, Lemma 6.2] and [10, Proposition 6.3], if 0 < q ≤p one verifies that ∥ft∥F d/q−d p,q (Rd) ≲N1/q, and that for some t0 and some E ⊂Hd with HF(E) ⊂{2k}1≤k≤N4dN , PE(ft0) F d/q−d p,q (Rd) ≳N1+ 1 q .
This completes the proof of Theorem 1.1.
□ Remark 12.1. If p = ∞one can ask whether the Schauder basis property in the subspace fs ∞,q, for −1 < s < 0, can be upgraded to unconditional basis. This is certainly true when 1/q −1 < s < 0, by the uniform bound-edness of the projection operators PE (which follows by duality from the corresponding result for F −s 1,q′), and by the density of span Hd in fs ∞,q.
We now show that at the endpoint s = 1/q −1 unconditionality must fail. If not, the operators PE would be uniformly bounded in f−1/q′ ∞,q , for all HAAR SYSTEM IN TRIEBEL-LIZORKIN SPACES: ENDPOINT RESULTS 39 finite E ⊂Hd. Fix p0 ∈(q, ∞), and for each N ≥1, pick a set A = AN ⊂ {2n}n≥0 of cardinality 2N and such that log2 A is N-separated. Then, by [10, Theorem 1.4], for every E ⊂Hd with #HF (E) ⊂A it holds (113) ∥PE∥F −1/q′ p0,q ≲N1/q′.
By interpolation one then has, for θ ∈(0, 1), ∥PE∥F −1/q′ p0/θ,q ≲∥PE∥θ F −1/q′ p0,q ≲N θ q′ .
But this contradicts the lower bound N1/q′ (for the supremum of all such sets E) asserted in [10, Theorem 1.4.ii]. Similar arguments also disprove the unconditionality for s below the critical 1/q −1.
Finally consider the space f0 ∞,q for 1 ≤q < ∞, on which unconditionality fails (since otherwise it would hold on its dual F 0 1,q′, see [8, §2.1.5], on which unconditionality fails by [5, Prop. 13.3]).
Remark 12.2. We now consider the spaces fs p,∞, when 1 < p < ∞. The Schauder basis property holds for 1/p−1 < s < 1/p while the unconditional basis property holds only for 1/p −1 < s < 0, already by the estimates in . The unconditional basis property does not hold on f0 p,∞since by duality ([8, §2.1.5]) it would imply it on F 0 p′,1 where it fails by . Finally when p = q = ∞then F s ∞,∞= Bs ∞,∞hence fs ∞,∞= bs ∞,∞, and the unconditional basis property holds for −1 < s < 0 (for the dual statement see ).
Remark 12.3. It would be interesting to investigate the question of uncon-ditionality of the Haar system as a basic sequence in B1 p,q and F 1 p,q when d/(d + 1) < p < 1.
References Pierre Billard. Bases dans H1 et bases de sous-espaces de dimension finie dans A.
Proc. Conf. Oberwolfach (August 14-22, 7971), ISNM vol. 20, Birkh¨ auser, Basel and Stuttgart 1972.
Hui-Qui Bui, Mitchell Taibleson. The characterization of the Triebel-Lizorkin spaces for p = ∞. J. Fourier Anal. Appl. 6 (5) (2000), 537–550.
Michael Frazier, Bj¨ orn Jawerth.A discrete transform and decompositions of distribu-tion spaces. J. Funct. Anal., 93(1):34–170, 1990.
Gustavo Garrig´ os, Andreas Seeger, T. Ullrich. The Haar system as a Schauder basis in spaces of Hardy-Sobolev type. Jour. Fourier Anal. Appl., 24 (5) (2018), 1319–1339.
, Basis properties of the Haar system in limiting Besov spaces. Preprint 2019.
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Peter Oswald. Haar system as Schauder basis in Besov spaces: The limiting cases for 0 < p ≤1. arXiv:1808.08156.
Jaak Peetre. On spaces of Triebel-Lizorkin type. Ark. Mat. 13 (1975),123–130.
Thomas Runst, Winfried Sickel. Sobolev spaces of fractional order, Nemytskij oper-ators, and nonlinear partial differential equations, volume 3 of de Gruyter Series in Nonlinear Analysis and Applications. Walter de Gruyter & Co., Berlin, 1996.
Andreas Seeger, Terence Tao.
Sharp Lorentz space estimates for rough operators.
Math. Ann. 320 (2001), no. 2, 381–415.
40 G. GARRIG´ OS A. SEEGER T. ULLRICH Andreas Seeger, Tino Ullrich. Haar projection numbers and failure of unconditional convergence in Sobolev spaces. Math. Z. 285 (2017), 91 – 119.
. Lower bounds for Haar projections: Deterministic Examples. Constr. Appr.
46 (2017), 227–242.
Hans Triebel. On Haar bases in Besov spaces. Serdica 4 (1978), no. 4, 330–343.
. Theory of function spaces. Birkh¨ auser Verlag, Basel, 1983.
. Theory of function spaces II. Monographs in Mathematics, 84. Birkh¨ auser Verlag, Basel, 1992.
. Bases in function spaces, sampling,discrepancy, numerical integration. EMS Tracts in Mathematics, 11. European Mathematical Society (EMS), Z¨ urich, 2010.
. Theory of function spaces IV. Book manuscript, 2019.
Przemys law Wojtaszczyk. The Banach space H1. Functional Analysis: Surveys and recent results III. K.-D. Bierstedt and B. Fuchssteiner (eds.) Elsevier (North Holland), 1984.
Gustavo Garrig´ os, Department of Mathematics, University of Murcia, 30100 Espinardo, Murcia, Spain Email address: gustavo.garrigos@um.es Andreas Seeger, Department of Mathematics, University of Wisconsin, 480 Lincoln Drive, Madison, WI,53706, USA Email address: seeger@math.wisc.edu Tino Ullrich, Fakult¨ at f¨ ur Mathematik, Technische Universit¨ at Chemnitz, 09107 Chemnitz, Germany Email address: tino.ullrich@mathematik.tu-chemnitz.de |
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Pulsars-Explained
Before we can understand the “secret” of the pulsars, it is necessary to understand what a pulsar actually “is”.
There are 2 steps to this understanding:
Being able to understand what a neutron star “is”.
Understanding when a neutron star becomes a “pulsar”.
What is a neutron star?
In order to conceptualize a neutron star, we can start by looking at a normal star, such as the Sun, and compare that to a neutron star.
The following are a couple of values describing the size and weight of the Sun:
The Sun is approximately 1 million miles in diameter (~864,500 miles).
The mass (weight) of the Sun is 2×1030 kilograms
(where 2×1030 = 2 times 10 to the 30th power (1 with 30 zeroes)
( which is 330,000 times the mass of the Earth )
( 1 kilogram = 2 1/2 pounds)
The following are the corresponding values for aneutron star:
Aneutron staris approximately20 miles in diameter.
Aneutron starweighs approximately2.8×1030 kilograms
(i.e. 1.4 times the mass of the Sun)
Therefore, to imagine a neutron star (ns), one can imagine all the material that comprises the Sun in a 432,000 mile radius sphere, compressed into a 10 mile radius sphere.
Since the same mass is compressed into a sphere that is 40,000 times smaller radius than the Sun, the “density” of the material goes as the cube (power of 3), which implies the density of a neutron star is 64,000,000,000,000 = 64 trillion times as dense as the Sun.
And, actually since the ns mass is 1.4 times the Sun mass, one needs to stuff an extra 1/2 of the Sun’s mass into that 10 mile radius sphere as well.To bring things closer to reality let’s calculate what one cubic foot of neutron star material weighs:
Volume of 10 mile radius sphere = 4/3 x 3.14 x 10 3 cubic miles = 4,000 cubic miles.
1 cubic mile = 5,0003 cubic ft = 125,000,000,000 = 125 billion cubic feet.
so, vol ns = 4,000 x 125 billion cu-ft = 500,000 billion cu-ft.
so density of ns = 3×1030 kilograms / 5×1014 cu-ft = 0.6 x 1016 kg/cu-ft
= 1×1016 pounds/cu-ft
i.e.1 cubic foot of ns material weighs1016 pounds =100 quadrillion pounds
Since the Earth is 1/330,000 times the weight of the Sun, the whole planet Earth would approximately be stuffed into a 1 billion cubic foot container. This can be visualized as a cube, 1,000 feet on a side.
If we assume a football stadium is 100 feet high, and roughly 1,000 feet long and 500 feet wide (including the stands plus the field), then a football stadium has a volume of about 500,000×100 cubic feet = 50 million cubic feet.
Therefore, if the Earth were compressed to the density of a neutron star, then the whole Earth would fit in about 20 football stadiums.
Since the Moon is about 1/100 mass of the Earth, it would fit in 500,000 cu-ft, or a cube about 80 feet on a side. i.e. it would fit in the middle of the football field inside the football stadium.
Pretty heavy stuff.
Therefore, a neutron star “is” a sphere containing about 1 and a half times the mass of the Sun, where the radius of the sphere is about 10 miles.
In order to compress material to the density required for a neutron star, the atoms themselves will be crushed, so that the electrons that surround the atom’s nucleus are actually forced into the nucleus, itself, which means all the protons, each with a “+” charge will be combined with all the electrons, each with a “-” charge, which will make each electron-proton pair an uncharged neutron.
The compression effectively turns all the mass of the neutron star into uncharged neutrons, which actually means thata neutron star is one giant atomic nucleuscomprised of an unfathomable number of neutrons.
Such an object seems like a concept out of science fiction. On Earth, lone neutrons are unstable and immediately decompose into an electron and a proton, which are the components of a hydrogen atom. A hydrogen bomb explodes by combining pairs of hydrogen atoms into deuterium which initially contains 2 electrons and 2 protons:
Fusion in stars
So, even if we were able to produce a tiny amount of neutron star material on Earth, it would instantly explode with a force comparable to that of an H-bomb.
Why doesn’t a neutron star explode?
Because the gravity of all the neutrons together is so strong that the neutrons are unable to decay into proton-electron pairs, as they would instantly be fused back together.So, …
What is a pulsar?
A pulsar “is” a neutron star.
But, it is a neutron star that is behaving in a particular way.
Note that if there was a neutron star somewhere nearby, say the distance to the closest visible star, which is approximately 4 light years away from us here on Earth, it is just a 1.4 solar mass ball that is 10 miles in radius and does not shine or anything like that. i.e. we would not “see” it. The neutron star would just be a dark ball way out in space that we would have no chance of seeing even with today’s most powerful telescopes.
On the other hand we can and do see “pulsars” that are thousands of light years away from the Earth.
So, how are we able to see pulsars, which actually are neutron stars?
i.e. what is it about a neutron star that makes it a “pulsar” that we can “see”?
“Pulses” are bursts of energy that a pulsar regularly emits, at a specific periodic rate. The fact that the time between pulses is slowing down (getting longer) from an isolated pulsar, means that it will eventually use up all its pulsation energy and no longer emit pulses.
So, the question becomes: “Where does the neutron star get the energy that it uses to emit pulses?”.
The first thing that distinguishes pulsars from anything else is that they “pulse”.
i.e. when we “see” a pulsar, we “see” the pulses that it emits.
It also turns out that because pulsars are capable of pulsing at 30 times per second or more, that they must be neutron stars, as was discovered with the Crab Pulsar in 1968:
Note: while the above reference gives a good description of what a pulsar “is”,
this site is in complete disagreement with the above reference’s claim that:
apulsar is created by a supernova explosion. (i.e. this stmt isfalse)
In fact, this site will show that quite the opposite is true:
i.e. thatthe pulsar causes the supernova explosion(i.e. this stmt istrue)
Based on the pulse frequency of the Crab Pulsar and the intensity of the pulses it emits, it is necessary that the Crab Pulsar must be a neutron star, and from this we conclude that all pulsars must be neutron stars:
Pulsars
(Note: there is speculation that some pulsars may be white dwarfs, however, as will be shown on this site, it is very unlikely that white dwarfs are connected in any way to the actual objects that are currently recognized to be pulsars, except by being a companion to the actual pulsar. i.e. it will be shown that pulsars can turn regular stars into white dwarfs, as is the case with milli-second pulsars, but the white dwarfs are not, themselves, pulsars.)
Continuing with what a pulsar “is”, we need to next understand what it is that is actually “pulsing”. As described earlier, it is only by observation of these pulses, that we know that neutron stars exist, and we also know that all by itself a neutron star does not pulse, it simply exists as a dark sphere in space completely unobservable.
It is not currently understood exactly what the origin of the magnetic field within a neutron star is, however, it must be related to some sort of alignment of the neutrons, themselves, such that their magnetic fields are parallel, at least within some section of the neutron star:
Neutron Star Magnetic Field Origins
Neutron Star Magnetic Fields
Therefore, a neutron star, all by itself has a very large intrinsic magnetic field. If this field is aligned with the rotation axis, then no radiation will be emitted. However, if the magnetic field is aligned at some angle, possibly even perpendicular, to the rotation axis, then the neutron star will emit electromagnetic waves, of relatively low energy in the radio spectrum.
Therefore, a “pulsar” is a rotating neutron star that has a magnetic field that is aligned at some non-zero angle with respect to the neutron star’s rotation axis. This mis-alignment causes radio pulses to be emitted by this rotating object that can be observed with radio telescopes here on Earth.
i.e. in other words: a neutron star “becomes” a pulsar when its magnetic field is misaligned with its rotation axis.
However, as we know from conservation of energy, the energy emitted by the radio pulses must take away energy from the rotating object. That energy taken away will be in the form of a “slowing down” of the rotation rate. Eventually, on the order of millions of years, the rotation rate will become so slow that pulses will no longer be observable, and the pulsar will effectively disappear.
So, now, from the above discussion, we “know” what pulsars actually are (rotating neutron stars, with magnetic fields aligned at non-zero angle to rotation axis), and that in the course of time, that they will disappear, by losing energy emitted by radio pulses from the changing magnetic field resulting from the mis-alignment.
So, the question now arises, where do these pulsars come from in the first place?
This is where the fun really begins, and this site will take the reader on a journey to a completely new understanding of supernovae, pulsars, neutron stars, and dark matter. From there the journey will continue to a completely new understanding of the galaxy and possibly the universe itself. Also, this journey will be found to be inescapable, in the sense that no other explanation currently exists to account for the observations that have been made of pulsars.
Another conceptual pulsar model
Consider a big ball with a long pipe extending out of each pole (the north and south poles).
Now consider that the big ball is a giant vacuum cleaner that can suck matter in through the two pipes.
If this ball is floating through space and encounters a star, when it comes close to the star one of the poles will hit the atmosphere of the star and start to suck in matter from the star. This is like a “whack” to the pole and will cause the big ball to start to spin.
The second pole will then shortly also get whacked by hitting the star and cause the ball to spin faster, causing the first pole to get hit by the star again and so on.
From this model we can see roughly how a spinning pulsar is created by encountering a star. It would never have started spinning without the encounter.
Finally, to complete the model, when the pulsar is really spinning fast, let us consider that the matter being sucked in along the poles is then ejected from the pulsar along the spin axis.
The result of this latter scheme is that the matter falling in adds to the angular momentum of the pulsar, while the matter being ejected does not cause a loss of angular momentum because it is being ejected along the spin axis. i.e. this process adds angular momentum to the pulsar without adding matter to the pulsar since everything that is sucked in is also ejected out.
Introduction & Overview
The Key Concepts
Supernova Explosions (SNE) Explained
What is X-ray Astronomy?
What is a pulsar?
Pulsars-Explained
Pulsars (index)
Giant Stars and NS-Capture Theory
SuperGiant -> SuperNova
Binary Pulsars: What they really are
Binary Pulsars Don’t Slow Down
What are the Be Stars?
Be X-Ray Binaries
The Data
Classification of the Pulsars
How NS-Capture Theory explains the P/P-dot diagram
P/P-dot Evidence for NS-Capture
Cen X-3: The Pulsar Spin Contradiction
Her X-1 (Hercules X-1)
Millisecond Pulsars
The Be X-ray Binary General Data
List of References for Be X-ray Binary Data Collections
The Be X-ray Binary Spin Orbit Data
What about the Double Pulsar?
The Theories
Theory Overview
NS-Capture Basic Theory
NS-Capture Theory – Explained
NS Capture Theory (index)
Simple explanation of theory
NS-Capture vs NS-Creation
Population problems with NS-Creation
The Critical Theoretical Choice
The Proofs
Deductive Reasoning
Analogies: Automobiles, propellers, muffins
Properties of the NS-Capture Theory
Proof of the Neutron Star Capture Theory
Logical Proof of the NS-Capture Theory
Definitive Empirical Proof of NS-Capture
Empirical Proof: Population Analysis
Analytical Proof of the NS-Capture Theory
Pre-Diagrammatic Proof of NS-Capture
Diagrammatic Proof of NS-Capture
NS-Capture and the Empirical Spin-up Process
How to Disprove the NS-Capture Theory
Predictions of NS-Capture
What does an isolated neutron star look like?
What does an active galactic nucleus look like along with its surrounding galaxy (AGN)?
Could an isolated neutron star look like an AGN?
What predictions did NS-Capture Theory make about the X-ray Background in 1974?
What predictions did NS-Capture make about the existence of the class of Be X-Ray Binaries?
What about PSR B1259 -63?
What about black holes?
NS-Capture Theory predicted the existence of dark matter in 1974
NS-Capture Theory predicted that a supermassive black hole exists in the center of the Milky Way in 1974
NS-Capture Theory predicted the constant galactic rotation curve in 1974.
What is the model of galactic evolution predicted by NS-Capture Theory?
Calculations
Calculate Number of NS’s in Galaxy
NS-Capture Explanation of SNEs
Calculating an Impact Parameter
Simulations
Simulation of NS-Capture Theory
NS-Capture Simulation: Part 2
Misc Diagrams
Diagram of NS-Creation vs NS-Capture
Full Diagram
Navigation
NS Capture vs NS Creation Diagrams
Diagram of NS Creation & NS Capture
Full Creation Process
NS-Capture vs NS-Creation Diagrams 2
The Necessity of NS Capture
Alternate Diagrams: MSP’s
Misc Pages
Origin of “Rock ‘n Roll”
Simulation of the NS-Capture Theory
Under Construction
How binary x-ray pulsars introduce a whole new way of thinking about neutron stars, supernova explosions, all other pulsars and the dark matter in the Milky Way galaxy.
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11582 | https://real-statistics.com/binomial-and-related-distributions/multinomial-distribution/ | Real Statistics Using Excel
Multinomial Distribution
Overview
The multinomial distribution is a generalization of the binomial distribution to two or more events.
Definition 1: For an experiment with the following characteristics:
let x1…, xk be discrete random variables whose values are the number of times outcome Ei occurs in n trials. Then x1…, xk has a multinomial distribution. The (joint) probability distribution function (pdf) is defined as follows:
Here
The case where k = 2 is equivalent to the binomial distribution.
Key properties of the multinomial distribution are
Example
Example 1: Suppose that a bag contains 8 balls: 3 red, 1 green, and 4 blue. You reach in the bag pull out a ball at random and then put the ball back in the bag and pull out another ball. This experiment is repeated a total of 10 times. What is the probability that the outcome will result in exactly 4 reds and 6 blues?
The possible outcomes for each trial in this experiment are E1 = a red ball is drawn, E2 = a green ball is drawn and E3= a blue ball is drawn. Thus p1 = 3/8, p2 = 1/8, p3 = 4/8, x1 = 4, x2 = 0 and x3 = 6, and so
Worksheet Functions
Excel Function: While Excel does not provide a function for the multinomial distribution, it does provide the following function:
MULTINOMIAL(x1…, xk) = n! / (x1!∙…∙xk!)
Thus we could also calculate the answer to Example 9.10 by using the formula
MULTINOMIAL(4,0,6)(3/8)^4(1/8)^0(4/8)^6 = .064888
We can also use a range as the argument of MULTINOMIAL as shown in Figure 1.
Figure 1 – Multinomial distribution
We can use the following Excel array formula to calculate the same result
=PRODUCT(MULTINOMIAL(B3:B5),B6:B8^B3:B5)
Alternatively, we can use the following more complicated non-array formula
=MULTINOMIAL(B3:B5)EXP(SUMPRODUCT(B3:B5,LN(B6:B8)))
Real Statistics Function: The following function in the Real Statistics Resource Pack can be used to calculate the multinomial distribution.
MULTINOMDIST(R1, R2) = the value of the pdf of the multinomial distribution where R1 is a row or column array containing the values x1,…, xk and R2 is a row or column array containing the values p1,…, pk, where R1 and R2 have the same shape.
Referring to Figure 1, we have MULTINOMDIST(B3:B5,B6:B8) = 0.064888.
Examples Workbook
Click here to download the Excel workbook with the examples described on this webpage.
Reference
Wikipedia (2021) Multinomial distribution
Chen, Y-C. (2020) Multinomial distribution
22 thoughts on “Multinomial Distribution”
HI Charles.
Is there a way to calculate the inverse of multinomial distribution?
I’m thinking about a set of success values (s), which is a vector of different states of successes (si), each one with a probability of success, (pi). Need to calculate the total of failures before getting the number of successes desired.
An example (taken from Vose software) will provide clearer understanding:
You want to make sure that at the end of the week you have called 50 people who never heard of your product, 50 people who don’t have internet at home and 200 people who use internet almost daily.
With probabilities of success pi, and states, you can tell the number of failures you’ll have before you’ve called all the people you wanted, so you know the total number of phone calls.
In brief, Vose get the total number with a function:
The total number of phone calls = the total number of successes (300) + the total number of failures (NegMultinomial({50,50,200},{p1,p2,p3})).
How can the total number be calculated in Excel without Vose function?
Hello Johann,
Are you looking for the negative multinomial distribution as described on the following web page?
Are you looking for what Vose calls type 1 or type 2?
Charles
Hello Charles, the wikipedia reference is correct, I want to know how to calculate total number of calls, result that Vose Type 1 would provide. I have not found a worked example.
Would be great to get the complimentary result of Vose Type 2 too, I guess need first the result of type 1, and then to get the number of failures in each state is straightforward.
Hello Johann,
Seems like an interesting concept. I will add it to my list of potential future enhancements.
Charles
Hi Charles, think I found an example of calculation for expectation of negative multinomial distribution. A fair 3 sided dice, and the number of expected rolls to get a specific result; however the question was answered using Markov chains.
Hi Johann,
Thanks for sharing this. I will look into it.
Charles
Thank you for the tools very much!
Hello, thank you for the tools very much.
I use this function MULTINOMDIST(R1, R2) to get the pdf, then get the cdf by adding it up.
but the cdf does not approach 1 (in fact it approach 1.17).
I’m confused for this situation.
The parameters I used are:
(x1,x2,x3,x4,x5)=(1,1,1,1,n-4)
(p1,p2,p3,p4,p5)=(0.02,0.02,0.02,0.02,0.92)
for example:
when n=4, MULTINOMDIST(Rx, Rp)=0.00000384 ; cdf=0.00000384
when n=5, MULTINOMDIST(Rx, Rp)=0.000017664 ; cdf=0.000021504
….
when n=88, MULTINOMDIST(Rx, Rp)=0.007833858 ; cdf=0.994673755
when n=89, MULTINOMDIST(Rx, Rp)=0.007833858 ; cdf=1.002507613
…
when n=170, MULTINOMDIST(Rx, Rp)=0.000125678 ; cdf=1.169895198
Excuse me, could you help me with this question.
Thank you very very much, again.
There was a bug in the MULTINOMDIST function which was corrected yesterday.
I suggest that you download and install the new version.
Charles
sorry, i type wrong.
The parameters I used are:
(p1,p2,p3,p4,p5)=(0.02,0.02,0.02,0.02,0.92) <-correct (=in fact to use.)
So. i get the probability from the stated calculations… how do i get the confidence intervals for each proportion?Is it somewhere on the real statistics site?
The site contains lots of examples of how to calculate confidence interval. Please be more specific about the confidence interval that you are looking for. Is it for a proportion distribution? If so, see the webpage Proportion Distribution.
Charles
Hi Charles,
I found the answer to my question already. No need to spend time on it.
Thanks
Hi,
I’m interested in the answer.
What would be the formula for Confidence Interval in this case ?
Thanks
Hi,
If you have a proportional multinomial distribution with probabilities p1, p2, …, pk for mutually exclusive events E1, E2, …, Ek. Then for any pi you can look at this as a binomial distribution with p = pi. 1-p is therefore the sum of the pj’s excluding pi. The confidence interval for the population version of pi is therefore as calculated for the proportional binomial distribution. This is described on the following webpage using the normal distribution approximation.
Later today I plan to add a slightly more accurate version of the confidence interval due to Edwin Wilson, which can be used for both the binomial and multionomial distributions.
Charles
Charles,
I need to sort the weights of a large sample of people within different intervals (110kg), from which I calculate proportions.
Then I compute the confidence intervals.
So I think that I can use the CI of the proportional binomial distribution. Right?
Thanks
(110kg)
less than 50;50-69;70-89;90-109; more than 110kg
Sorry, but I don’t have enough information to provide a response.
Charles
Hi Charles,
I have a question that relates to a multinomial distribution (not even 100% sure about this) that I hope you can help me with.
If I take a sample (lets assume n=400) on a categorical variable that has more than two possible outcomes (e.g. blue, black, green, yellow) and plot the frequencies so that I can get the probabilities. E.g.:
black 10%
blue 25%
green 35%
yellow 30%
How could I compute the 95% confidence interval for those probabilities?
Any help is highly appreciated.
Thanks a lot again for creating such a great resource.
Kind regards,
Dirk
Hi,
an alternative method to calculate the probability would be an array formula:
=PRODUCT(B9,B6:B8^B3:B5) + CTRL+SHIFT+ENTER
Hi Dirk,
Yes. I like your approach, especially since it is simpler than the one I wrote on the webpage. I have just added your suggestion to the referenced webpage. Thanks for your very useful comment.
Charles
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11583 | http://matlab.izmiran.ru/help/techdoc/ref/reshape.html | reshape (MATLAB Functions)
MATLAB Function Reference
B = reshape(A,m,n,p,...)
B = reshape(A,[m n p ...])
B = reshape(A,...,[],...)
B = reshape(A,siz)
Description
B = reshape(A,m,n)
returns the m-by-n matrix B whose elements are taken column-wise from A. An error results if A does not have mn elements.
B = reshape(A,m,n,p,...) or B = reshape(A,[m n p ...])
returns an n-dimensional array with the same elements as A but reshaped to have the size m-by-n-by-p-by-.... The product of the specified dimensions, mnp..., must be the same as prod(size(A)).
B = reshape(A,...,[],...)
calculates the length of the dimension represented by the placeholder [], such that the product of the dimensions equals prod(size(A)). The value of prod(size(A)) must be evenly divisible by the product of the specified dimensions. You can use only one occurrence of [].
B = reshape(A,siz)
returns an n-dimensional array with the same elements as A, but reshaped to siz, a vector representing the dimensions of the reshaped array. The quantity prod(siz) must be the same as prod(size(A)).
Examples
Reshape a 3-by-4 matrix into a2-by-6 matrix.
A =
1 4 7 10
2 5 8 11
3 6 9 12
B = reshape(A,2,6)
B =
1 3 5 7 9 11
2 4 6 8 10 12
B = reshape(A,2,[])
B =
1 3 5 7 9 11
2 4 6 8 10 12
See Also
shiftdim, squeeze
The colon operator:
reset residue
© 1994-2005 The MathWorks, Inc. |
11584 | https://www.meracalculator.com/math/percentage.php | Percentage Calculator
To figure out the percentages of numbers put values in the given box by using this percentage difference calculator.
Result
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Table of Contents:
Is This Tool Helpful?
Percentage formula
Wondering How to find percentages? Don’t worry. We have got you covered in this space.
There are several formulas of the percentage that can be used to get a percentage of an amount. The use of these formulas depends on the type of percentage we are calculating.
Percentage = (Part × 100) / Whole
By using the above percentage equation, we can derive other quantities too.
Part = (Percentage × Whole) / 100
Whole = (Part / Percentage) × 100
Percentage calculator is an online tool that calculates percentageof an amount. You can determine what percent of X is Y, how to find a percentage of two numbers, and much more using this percent calculator.
This calculator can make several types of percentage calculations. It can be used to calculate:
We have covered all types of percentages in this tool so that you can calculate the percentage of anything you want.
In this content, we will discuss percentage, how percentage calculator works, how to figure out percentages, the formula for percentage, how to calculate a percentage of a number, and much more.
How to use percentage calculator?
Calculating percentages is not a complicated task if you are calculating it using our percentage of calculator.This tools makes the calculations of percentage too easy that you will wonder why I was asking how to get a percentagefrom my teacher when this tool is available.
To use this tool, follow the steps below:
This tool has the capability of calculating 12 types of percentages as listed above. You can use its calculations to revise your concepts as well as prepare the topic of percentage for exams.
We also use percentage in income tax calculations and discounts.
What is a percentage?
We are used to hearing the word percentage since our childhood but do we really know what a percentage is?
According to Wikipedia,
“In mathematics, a percentage is a number or ratio expressed as a fraction of 100. It is often denoted using the percent sign %, or the abbreviations pct. Sometimes the abbreviation "pc" is also used to denote percentage.”
In the opinion of Merriam-Webster,
“A percentage is a part of a whole expressed in hundredths.”
To understand the percentage, suppose you have pizza in front of you. You first cut the pizza in four slices and then further cut it to make a total of eight slices. Now, if you pick up one slice, you are holding some percentage of pizza in your hand.
It means that percentage is actually a part of a whole amount of something.
How to calculate percentages?
A discussed above, there are numerous formulas to calculate the percentage. We can use any of them depending on the type of percentage we want to calculate. Let’s find out step by step method to get a percentage.
If you need to calculate the increase in percentage, use our percentage increase calculator or percentage difference calculator.
Let’s go through an example to properly understand the percent calculations.
Example 1 – Find the percentage
Calculate the percentage if a college student has obtained 700 marks out of 1200 in exams.
Solution:
Step 1: Identify and write down the values.
Part = 700, Whole = 1200
Step 2: Write down the percentage formula.
Percentage = (Part × 100) / Whole
Step 3: Substitute the values in the percentage equation.
Percentage = (700 × 100) / 1200
Step 4: Calculate the percentage by solving the equation.
Percentage = 70000 / 1200
Percentage = 58.33%
So, a student with 700 marks out of 1200 got 58.33% in exams.
Example 2 – Find the part
What is 20% of $800?
Solution:
Step 1: Identify and write down the values.
Percentage = 20%, Whole = $800
Step 2: Write down the percentage formula.
Part = (Percentage × Whole) / 100
Step 3: Substitute the values in the percentage equation.
Part = (20 × 800) / 100
Step 4: Calculate the percentage by solving the equation.
Part = 16000 / 100
Part = $160
So, 20% of $800 is $160.
Example 3 – Find whole
$10 is 30% of what?
Solution:
Step 1: Identify and write down the values.
Percentage = 30%, Part = $10
Step 2: Write down the percentage formula.
Whole = (Part / Percentage) × 100
Step 3: Substitute the values in the percentage equation.
Whole = (10 / 30) × 100
Step 4: Calculate the percentage by solving the equation.
Whole = 0.33 × 100
Whole = $33
So, $10 is 30% of $33.
FAQs
How do I calculate percentage usage?
Percentage usage or usage rate is a calculation of the percentage of team plays used by a player while he was on the floor. It can be calculated using the following formula.
Usage Rate Formula
100 × ((Player’s Field Goal Attempts) + 0.44 × (Player’s Free Throw Attempts) + (Player’s Turnovers)) × (Team’s Total Minutes) / ((Team’s Total Field Goal Attempts) + 0.44 × (Team’s Total Free Throw Attempts) + Team’s Total Turnovers)) × 5 × (Player’s Minutes)
How do I calculate 5% of a total?
Suppose our total amount is 400. To calculate 5% of the total amount:
400 × 5%
400 × 5/100 = 20
How do I calculate the percentage of a total?
To calculate the percentage of a total, multiple the total by percentage and then divide it by 100. i.e. 200 × 4/100 = 8
How do I calculate a normal percentage?
Let’s suppose we have to calculate what percentage is $200 of $800. To calculate the normal percentage:
200 × 100 = 20000
20000 / 800 = 25%
What is a 5th of 100?
A 5th of 100 can be calculated by multiplying 1/5 with 100. A 5th of 100 is 20.
1/5 × 100 = 20
What is 20% off?
20% off means that there is a discount of $20 per $100 on the total amount of the product. If the total price of a product is $600, then the 20% off will be:
$600 × 20 / 100 = $120
The discount on a $600 product will be $120 at 20%. The new price of the discounted product will be $600 - $120 = $480
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11585 | https://onlinelibrary.wiley.com/doi/10.1111/j.1365-2672.2006.03127.x | Error - Cookies Turned Off
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11586 | https://brainly.com/question/59416372 | [FREE] Determine, by substitution, whether or not the given point lies on the line. a. (3,1) \quad y = x - 2 b. - brainly.com
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Determine, by substitution, whether or not the given point lies on the line.
a. (3,1)y=x−2
b. (7,4)y=20−2 x
c. (1,−2)y=−3 x+1
d. (−5,3)2 x+3 y+1=0
e. (−1,−4)3 x−2 y−5=0
f. (−6,−4)4 x−5 y−4=0
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Sure! Let's check each point to see if it lies on its respective line by substituting the x and y values of the point into the equation of the line.
a. Point (3, 1) and line y=x−2:
Substitute x=3 and y=1 into the equation y=x−2.
Check if 1=3−2.
Simplifies to 1=1, which is true.
Therefore, the point (3, 1) lies on the line.
b. Point (7, 4) and line y=20−2 x:
Substitute x=7 and y=4 into the equation y=20−2 x.
Check if 4=20−2×7.
Simplifies to 4=6, which is false.
Therefore, the point (7, 4) does not lie on the line.
c. Point (1, -2) and line y=−3 x+1:
Substitute x=1 and y=−2 into the equation y=−3 x+1.
Check if −2=−3×1+1.
Simplifies to −2=−2, which is true.
Therefore, the point (1, -2) lies on the line.
d. Point (-5, 3) and line 2 x+3 y+1=0:
Substitute x=−5 and y=3 into the equation 2 x+3 y+1=0.
Check if 2(−5)+3(3)+1=0.
Simplifies to 0=0, which is true.
Therefore, the point (-5, 3) lies on the line.
e. Point (-1, -4) and line 3 x−2 y−5=0:
Substitute x=−1 and y=−4 into the equation 3 x−2 y−5=0.
Check if 3(−1)−2(−4)−5=0.
Simplifies to 0=0, which is true.
Therefore, the point (-1, -4) lies on the line.
f. Point (-6, -4) and line 4 x−5 y−4=0:
Substitute x=−6 and y=−4 into the equation 4 x−5 y−4=0.
Check if 4(−6)−5(−4)−4=0.
Simplifies to −8=0, which is false.
Therefore, the point (-6, -4) does not lie on the line.
These are the detailed solutions for each scenario.
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The points (3, 1), (1, -2), (-5, 3), and (-1, -4) lie on their respective lines, while the points (7, 4) and (-6, -4) do not. Each point was verified by substituting the coordinates into the given line equations. This method provides a clear way to determine if a point is part of a line.
Explanation
To determine if a point lies on a line, we will substitute the point's coordinates into the equation of the line and see if the equation holds true. Let's go through each point and equation step by step.
a. Point (3, 1) and liney=x−2:
Substitute x=3 and y=1:
1=3−2
2. Simplifying gives:
1=1, which is true.
Therefore, the point (3, 1) lies on the line.
b. Point (7, 4) and liney=20−2 x:
Substitute x=7 and y=4:
4=20−2×7
2. Simplifying gives:
4=6, which is false.
Therefore, the point (7, 4) does not lie on the line.
c. Point (1, -2) and liney=−3 x+1:
Substitute x=1 and y=−2:
−2=−3×1+1
2. Simplifying gives:
−2=−2, which is true.
Therefore, the point (1, -2) lies on the line.
d. Point (-5, 3) and line2 x+3 y+1=0:
Substitute x=−5 and y=3:
2(−5)+3(3)+1=0
2. Simplifying gives:
0=0, which is true.
Therefore, the point (-5, 3) lies on the line.
e. Point (-1, -4) and line3 x−2 y−5=0:
Substitute x=−1 and y=−4:
3(−1)−2(−4)−5=0
2. Simplifying gives:
0=0, which is true.
Therefore, the point (-1, -4) lies on the line.
f. Point (-6, -4) and line4 x−5 y−4=0:
Substitute x=−6 and y=−4:
4(−6)−5(−4)−4=0
2. Simplifying gives:
−8=0, which is false.
Therefore, the point (-6, -4) does not lie on the line.
Examples & Evidence
For example, if we have the line y=2 x+1 and point (0, 1), substituting gives 1=2(0)+1 which is true, confirming the point lies on the line. Conversely, for point (1, 3), substituting gives 3=2(1)+1 which does not hold, showing it doesn't lie on the line.
The method used is standard practice in algebra for verifying points on lines, relying on the substitution of coordinates into linear equations to check for equality.
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Community Answer REVISION EXERCISE 5 Solve the following pairs of simultaneous equations: 1. [tex][ 4x - y = 11, \quad 5x + 2y = 4 ][/tex] 2. [tex][ x + y = 12, \quad 3x - y = 20 ][/tex] 3. [tex][ 3x - 2y = 21, \quad 4x + 5y = 5 ][/tex] 4. [tex][ 2x - 3y = 10, \quad 10x - 6y = 5 ][/tex] 5. [tex][ 2x + 3y = 17, \quad x + 5y = 0 ][/tex] 6. [tex][ 3x + y = 8, \quad x^2 + xy = 6 ][/tex] 7. [tex][ x^2 - y - 1 = 0, \quad y - 2x + 2 = 0 ][/tex]
Community Answer Which systems of equations have no solutions? (Check all that apply.) 1. [tex][ x - y = 1, \quad -3x + 3y = 3 ][/tex] 2. [tex][ 2y = 3 - 4x, \quad x + y = 0 ][/tex] 3. [tex][ 12x + 2 = 4y - 10, \quad 2x + y = -11 ][/tex] 4. [tex][ -3x - y = 5, \quad 15x = 10 - 5y ][/tex] 5. [tex][ x + y = 1, \quad -4x + 2y = 7 ][/tex]
Community Answer 5.0 The solution to which system of equations has an y-value of 0? (A) x + 2y=-1; y = 3x + 3 (B) 3x – y=2; x + 2y=-1 y= 3x + 3; 2x + 3y=01 (D) 2x + 3y=0; x + 2y=-1
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11587 | https://greatbrook.com/survey-glossary/ | Survey Glossary & Survey Article Index
Survey Glossary
This glossary of survey research terminology — a survey dictionary or survey wiki — contains definitions of commonly used survey research terms. Links are also provided to our articles that discuss that survey glossary terms.
The definitions given here are in the context of a survey project. However, many of terms are used in other research fields.
For a more detailed discussion, please refer to our Survey Guidebook or consider attending one of our Survey Workshops where we discuss in detail many of these concepts as they apply to designing and executing a successful survey project.
Click to a survey term that starts with:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Survey
The process of studying some phenomenon. A questionnaire is designed that will generate data to measure various attributes of the phenomenon, the questionnaire will be administered to a target audience, and then the results will be analyzed statistically.
A — Survey Glossary
Abandonment: Occurs when a survey respondent quits a survey part way through. Abandonment raises the question whether even the submitted data are valid.
Accuracy: The extent to which a survey result represents the attribute being measured in the population. Accuracy or “Margin of Error” is usually expressed as a plus or minus percentage, e.g., “+/- 5%”, which indicates that the survey mean score likely deviates from the population mean for that attribute by less than 5%.
Articles: Survey Statistical Accuracy Defined; The Hidden Danger of Survey Bias, Statistical Confidence in a Survey: How Many is Enough?
Acquiescence Bias: One type of response bias where the respondent is predisposed to agree with statements presented to him or her.
Actionable data: An objective of a survey program to generate data that provides insights to organizational change initiatives. Proper questionnaire construction and question writing is needed to generate actionable data.
Articles: Generating Actionable Survey Data, What’s the Point? An Unactionable Transaction Survey
Adjective Checklist: A survey question type where the respondent is asked to select among a set of adjectives that describe something. The question type generates categorical data.
Administration Biases: Biases introduced into the data set through the survey administration process, resulting in data that do not properly reflect the views of the target population for the research. These biases include mode bias, selection bias, non-response (participation) bias, and response bias.
Articles: Sampling Error — And Other Reasons Polls Differ, Creating a Survey Program for a University Help Desk, What Survey Mode is Best?, Want Higher NPS & Survey Scores? Change to Telephone Survey Mode
Administrative Burden: The amount of work required to administer a survey. This work may include the effort to enter the survey in some survey tool, generating a sample, getting the invitations to the respondent, collecting data from the respondent, and transcribing the data including open-ended comments preparing it for analysis.
Article: Survey Question Types: When to Use Each
Administration Mode: The communication medium or media used to invite people to participate in the survey, present questions to them, and collect responses from them.
Articles: Tips for Selecting an Online Survey Software Tool, Impact of Mobile Surveys
Administration, Survey: The process of managing the survey process for getting responses from the selected group. It includes selecting the audience to receive the invitations, extending the invitations, collecting responses, and loading the responses into a data set. The administration process will vary according to the administration mode.
Ambiguity: Vague, confusing, or unclear question wording that could lead respondents to have multiple interpretations of the question. Leads to measurement error and loss of validity. Perhaps the most common and destructive form of instrumentation bias.
Articles:Ambiguous Questions – The Biggest Threat to Survey Findings;The importance of Good Survey Question Wording: Even Pros Make Mistakes;
Ambiguity in Data Collection Forms
Analytical Burden: The amount of work required to analyze the data generated by a survey or survey question. Textual responses, for example, have high analytical burden.
Article: Survey Question Types: When to Use Each
Anchor: A word or phrase that describes a position on a response scale, thus “anchoring” the respondent to that point. Also called “label”. Interval rating questions generally have a set of anchors that describe ranges or levels of feelings for some dimension of measurement, such as satisfaction, agreement, or likelihood.
Articles: Misleading (or Lying) With Survey Statistics, Scale Design in Surveys,
Anonymity: A concern for the survey administration process if the lack of respondent anonymity could lead to non-participation in the survey or to responses that don’t reflect the respondents true views. Related to but different from confidentiality.
Articles: An Honest Survey Invitation?
ANOVA (ANalysis Of VAriance): A statistical technique that examines the variance structures of responses among different groups to determine if those differences are statistically significant.
Attitudinal Outcome: Questions on a survey that seek to summarize a respondent’s attitudes or views of the focus of the survey. Such questions include overall satisfaction, likelihood of recommendation, likelihood of repurchase, etc. These questions may be used as dependent variables in statistical tests.
Articles: (There Is More Than) The One Number You Need to Grow
Attribute: The characteristics of the phenomenon under study that are measured through data generated by survey questions.
Attribute Identification: The critical stage in a survey project where the attributes to be measured are identified. If poorly done, the survey will not be comprehensive and may miss key points.
Auspices: One type of response bias where the respondent is predisposed to provide responses they feel will please the sponsor of the research.
Average (Arithmetic Mean): A summary statistic to describe the typical value for a set of data from a survey question. It is calculated for the survey question by adding the the numerical scores from each survey response and then dividing by the count of responses.
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Acquiescence Bias: One type of response bias where the respondent is predisposed to agree with statements presented to him or her.
Actionable data: An objective of a survey program to generate data that provides insights to organizational change initiatives. Proper questionnaire construction and question writing is needed to generate actionable data.
Articles: Generating Actionable Survey Data, What’s the Point? An Unactionable Transaction Survey
Adjective Checklist: A survey question type where the respondent is asked to select among a set of adjectives that describe something. The question type generates categorical data.
Administration Biases: Biases introduced into the data set through the survey administration process, resulting in data that do not properly reflect the views of the target population for the research. These biases include mode bias, selection bias, non-response (participation) bias, and response bias.
Articles: Sampling Error — And Other Reasons Polls Differ, Creating a Survey Program for a University Help Desk, What Survey Mode is Best?, Want Higher NPS & Survey Scores? Change to Telephone Survey Mode
Administrative Burden: The amount of work required to administer a survey. This work may include the effort to enter the survey in some survey tool, generating a sample, getting the invitations to the respondent, collecting data from the respondent, and transcribing the data including open-ended comments preparing it for analysis.
Article: Survey Question Types: When to Use Each
Administration Mode: The communication medium or media used to invite people to participate in the survey, present questions to them, and collect responses from them.
Articles: Tips for Selecting an Online Survey Software Tool, Impact of Mobile Surveys
Administration, Survey: The process of managing the survey process for getting responses from the selected group. It includes selecting the audience to receive the invitations, extending the invitations, collecting responses, and loading the responses into a data set. The administration process will vary according to the administration mode.
Ambiguity: Vague, confusing, or unclear question wording that could lead respondents to have multiple interpretations of the question. Leads to measurement error and loss of validity. Perhaps the most common and destructive form of instrumentation bias.
Articles:Ambiguous Questions – The Biggest Threat to Survey Findings;The importance of Good Survey Question Wording: Even Pros Make Mistakes;
Ambiguity in Data Collection Forms
Analytical Burden: The amount of work required to analyze the data generated by a survey or survey question. Textual responses, for example, have high analytical burden.
Article: Survey Question Types: When to Use Each
Anchor: A word or phrase that describes a position on a response scale, thus “anchoring” the respondent to that point. Also called “label”. Interval rating questions generally have a set of anchors that describe ranges or levels of feelings for some dimension of measurement, such as satisfaction, agreement, or likelihood.
Articles: Misleading (or Lying) With Survey Statistics, Scale Design in Surveys,
Anonymity: A concern for the survey administration process if the lack of respondent anonymity could lead to non-participation in the survey or to responses that don’t reflect the respondents true views. Related to but different from confidentiality.
Articles: An Honest Survey Invitation?
ANOVA (ANalysis Of VAriance): A statistical technique that examines the variance structures of responses among different groups to determine if those differences are statistically significant.
Attitudinal Outcome: Questions on a survey that seek to summarize a respondent’s attitudes or views of the focus of the survey. Such questions include overall satisfaction, likelihood of recommendation, likelihood of repurchase, etc. These questions may be used as dependent variables in statistical tests.
Articles: (There Is More Than) The One Number You Need to Grow
Attribute: The characteristics of the phenomenon under study that are measured through data generated by survey questions.
Attribute Identification: The critical stage in a survey project where the attributes to be measured are identified. If poorly done, the survey will not be comprehensive and may miss key points.
Auspices: One type of response bias where the respondent is predisposed to provide responses they feel will please the sponsor of the research.
Average (Arithmetic Mean): A summary statistic to describe the typical value for a set of data from a survey question. It is calculated for the survey question by adding the the numerical scores from each survey response and then dividing by the count of responses.
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B — Survey Glossary
Balanced Scale: A response scale that has an even number of positive and negative response options.
Balanced Scorecard: A management approach that emphasizes having a comprehensive and “balanced” set of measurements to manage an organization. While measures of the resource use efficiency are easy to find, measures of effectiveness in the marketplace are more challenging. Survey data are one source of effectiveness measures.
Articles: Customer Insight Metrics: The Issue of Validity, Hilton Hotel Customer Survey Program, Measuring Service Effectiveness
Bar Charts: A commonly used chart to present frequency distributions from survey data, contrasting the distinction among response options. The bars may be presented horizontally or vertically (column bar or histogram).
Batch Surveying: A survey administration process where survey invitations are sent out out large groups or batches. Commonly used for transactional surveys where invitations may be sent out in daily or weekly batches.
Benchmarking: A management technique to compare the performance of an organization against comparable organizations. Very difficult to do for performance metrics generated by survey data Since each organization will likely have different questionnaires and survey procedures all of which will affect survey responses.
Bias: Any phenomenon that results in survey responses not reflecting the true feelings of the respondent, making the survey findings less valid and meaningful. Bias can never be eliminated in full, but surveyors should always try to minimize it. See: Instrumentation Bias, Administration Bias, Response Bias, Non-Response (Participation) Bias, Sample Bias, Administration Mode Bias.
Articles: Bolton Local Historic District Survey, What Survey Mode is Best?
Bimodal: A distribution of data where the values cluster around two values in the data set as opposed to clustering around one value. In a survey this would indicate two different opinions held by the respondent group.
Binary Choice: A specific type of forced choice question type where the respondent is given just two response options, such as true or false, yes or no. This question type generates categorical (nominal) data. Should be used where the responses can only legitimately be an “either or,” that is, a neutral position is not viable. Pollsters sometimes use a variation where they present two statements from which the respondent must choose. Statements must be polar opposites with non-loaded phrasing, otherwise bias is introduced.
Articles: Survey Question Design: Headlines or Meaningful Information?
Bivariate Statistics: Statistical analysis that examines the relationship between two variables (survey questions). Correlation is the most common type of bivariate analysis.
Bottom Box Scoring: For a scalar question, the “bottom box” is a cumulative frequency distribution for those responses at the bottom end of the scale up to some arbitrary point on the scale. For example, in a 1-to-10 scale, the “bottom box” may be the percent of respondent scoring 1 through 6. See also Top Box Scoring and Net Scoring.
Branching (also known as Conditional Branching and “Skip and Hit”): A technique where the flow of the questionnaire for a respondent depends upon the responses to a question. Useful to get more detailed feedback or to skip respondents over questions that are not applicable.
Articles: Creating a Survey Program for a University Help Desk
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Bivariate Statistics: Statistical analysis that examines the relationship between two variables (survey questions). Correlation is the most common type of bivariate analysis.
Bottom Box Scoring: For a scalar question, the “bottom box” is a cumulative frequency distribution for those responses at the bottom end of the scale up to some arbitrary point on the scale. For example, in a 1-to-10 scale, the “bottom box” may be the percent of respondent scoring 1 through 6. See also Top Box Scoring and Net Scoring.
Branching (also known as Conditional Branching and “Skip and Hit”): A technique where the flow of the questionnaire for a respondent depends upon the responses to a question. Useful to get more detailed feedback or to skip respondents over questions that are not applicable.
Articles: Creating a Survey Program for a University Help Desk
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C — Survey Glossary
Categorical Data (also known as Nominal Data): One of four data types. Checklist questions generate these data since respondents are selecting responses that belong to distinct categories. Analysis limited to frequency distributions.
Article: Survey Question Choice: How Question Format Affects Survey Data Analysis
CATI (Computer Aided Telephone Interviewing software): Software that manages the survey delivery process for telephone interviewers.
Articles: Tips for Selecting an Online Survey Software Tool
Census: As opposed to surveying a sample from the population, census surveying is where we invite everyone in our target population to take the survey. If everyone takes the survey, then we have conducted a census. Otherwise, we have data from a sample and need to use sampling statistics.
Articles: Survey Sample Selection: The Need to Consider Your Whole Research Program
Central Tendency: A fancy term for the “typical value” in a data set. Averages (mean, mode, and median) are common measures of central tendency.
Checklist Question Type (also known as Multiple Choice Question Type): The respondent is presented a set of items from which to select the most appropriate item (single response checklist) or all that apply (multiple response checklist). Generates categorical data.
Cherry picking. A sampling technique where the researcher picks those who will provide the desired response.
Article: Biased Survey Samples
Chi Squared: A statistical test used to determine if the difference in frequency distributions between two survey data sets of categorical data is statistically significant. Can be applied to rating scale data by treating each scale point as a category, eliminating concerns about the interval properties of the data needed for a t-test.
Closed-Ended Question: As opposed to an open-ended question that generates a textual response, a closed-ended question generates a limited set of responses that can readily be coded in a data base with some number or symbol that represents a response. Multiple-choice, ordinal, interval, and ratio questions generate closed-ended responses.
Cluster Sampling: A two-stage sampling approach used most appropriately where an interviewer is traveling to respondents and constraining travel costs is a concern. In the first stage, several clusters (for example, cities or towns in a country) are randomly selected. In the second stage, members from the selected clusters are randomly selected for participation.
Cognitive Burden: The work or effort asked of the respondent to compete a survey question. Higher cognitive burden may lead to invalid responses or abandonment. Complex questions lead to higher burden.
Comparative Scales: Question types where multiple items are presented to the respondent who is asked to evaluate the items against each other, for example, forced ranking or fixed-sum question types.
Composition Effect: A non-measurement error caused by the survey administration mode. It results from the likelihood that certain invitees will more likely respond to one administration mode than another. This is an argument for mixed-mode surveying.
Concern: A type of response bias introduced by the phrasing of the invitation, introduction or survey questions that create a concern for privacy or some other effect on the part of the respondent. Likely leads to non-participation or item non-response.
Articles: An Honest Survey Invitation?
Conclusion Validity: The extent to which the findings of research study seem reasonable. If the conclusions don’t seem reasonable, that may indicate some error in the research process, such as an instrumentation bias, administration bias, or statistical error.
Conditional Branching (also known as Branching and Skip & Hit): A process where the flow of a questionnaire is determined by the respondent’s answer to a question. Useful for branching beyond not applicable questions.
Articles: Creating a Survey Program for a University Help Desk
Confidence Interval: The range or interval within which the population mean likely lies. It indicates how well the sample mean represents the population mean. By convention, a 95% confidence interval is used, indicating that if the survey were done 20 times, 19 of the 20 times (95%) the survey mean would fall within the confidence interval, as defined by the confidence statistic. Can only be applied to survey questions that generate quantitative data (interval and ratio data).
Articles: Statistical Confidence in a Survey: How Many is Enough?
Confidence Statistic: A statistic that defines the confidence interval. It can be generated for data from a survey question that generates quantitative data — interval and ratio data. The statistic incorporates the count of values and variance in the data set.
Confidentiality: A promise typically conveyed in a survey invitation that data collected from a respondent will be handled appropriately and not shared with those who should not see it. Lack of a confidential guarantee may lead to non-response or a response bias in how questions are answered. Related to but different from anonymity.
Article: HHS Hospital Quality Survey
Conformity Bias: A type of response bias where the respondent provides answers that conform to societal norms.
Continuous Scale: A scale presented as a continuous line with endpoints only. That is, the scale does not contain discrete response options, such as a 1-to-5 scale. The respondent is asked where they fall along the continuum with the response coded typically as a millimeter measurement. See Visual Analog Scale.
Convenience sample. A method for choosing who receives a survey invitation. It uses people who are readily available and thus “convenient.” However, the convenient group may also have some bias toward the topic of the research. While response rates may be higher, this is a non-probability sampling process, meaning that the results cannot be said to represent a broader population.
Article: Biased Survey Samples
Correlation: A statistical process that examines the strength of association between two data sets, as expressed by the correlation coefficient. It ranges from +1 (perfect positive correlation) to 0 (no correlation) to -1 (perfect negative correlation). Assumes ratio data properties, and with truncated, integer-only data sets in interval rating scale survey data, strong correlations may be an artifact of the data.
Critical Incident Study: A process for determining the attributes to be measured through a survey questionnaire. Interviewees are asked to describe “critical” incidents that formed their opinions. The narratives hopefully identify salient attributes to be measured for the broader group through a survey instrument.
Cumulative Frequency Distribution: For questions with ordinal properties (which includes interval and ratio questions), the percentage of respondents who chose response options up to (or down to) a specific point in the ordered scale. In other words the frequency distribution accumulates to the chosen scale point. For example, the percentage that Agrees or Strongly Agrees with some statement. Top Box and Bottom Box scores are cumulative frequency distributions, and Net Scoring is calculated from those scores.
Customer Experience Design: The practice of designing a product or service with a primary concern to optimize the customers’ experience.
Articles: Customer Experience Design – Do our designs bring out the best or the worst in our customers?, Customer Experience Management — By Design
Customer Experience Management (CEM or CX): A management approach that emphasizes the experience of the customer in use of products and related services. Strong focus on process design and quality execution as it affects the customer. Surveys provide feedback to help identify processes in need of improvement.
Articles: A Very Short Customer Journey Map, “The Effortless Experience” Book Review
Customer Feedback Management (CFM): One element of a CEM program that focuses on capturing information from customers as a basis for improving product experiences as well as service processes that involve the customer. Surveys are a common feedback method.
Article: Capturing the Value of Customer Complaints
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D — Survey Glossary
Data Cleansing: The necessary process of examining data returned from survey respondents that are incorrect or invalid. Those data are either deleted or corrected if it can be legitimately corrected.
Data Collection Form: A form that collects typically objective data from the respondent. E., a website might collect data to complete a sale and a doctor’s office collects patient information. Closely related to a survey questionnaire in that validity issues are a concern.
Data Types: Four data types exist — categorical (or m=nominal), ordinal, interval, and ratio — and the analysis capabilities are greatest for ratio and lowest for categorical. A survey question type will generate one type of data, thus defining the analysis that can be done with data from that question. Most advanced statistical tests assume ratio properties.
Articles: Survey Question Choice: How Question Format Affects Survey Data Analysis
Demographic Questions: Survey questions that identify characteristics of the respondent, such as age, gender, income level, years of education, or years as an employee or customer. Use to analyze differences across groups.
Articles: Data Collection Form Design Issues, An Honest Survey Invitation?
Descriptive Research: The second of three stages in a research program, following Exploratory and preceding Prescriptive Research. As it sounds, this research seeks to describe the phenomenon under study. Surveys are a useful tool for such research.
Descriptive Statistics: Basic statistics that are run for each survey question individually, including, mean, median, mode, standard deviation, and confidence statistic.
Discrete Scale: A response scale a series of discrete options are presented for selection by the respondent, for example, a 1-to-10 numerical and/or verbal scale.
Dispersion: Variance and Standard Deviation are statistics that measure the level of dispersion, which is how tightly clustered or data are around the mean of those data. For a survey question dispersion indicates whether the respondent group share similar feelings or whether there is difference of views.
Double Barrelled Question: A common type of instrumentation bias where the survey question is actually composed of two separate questions. Makes interpretation of survey data generated highly problematic.
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E — Survey Glossary
Endpoint Anchoring (also known as Polar Anchoring): Rating scales where only the endpoints of the scale are anchored with verbal descriptors. Can be used with discrete numerical scales or continuous scales.
Article: Scale Design in Surveys
Event Surveys (also known as Transactional or Incident Surveys): A survey program where those invited to participate are people who have just completed some transaction, for example, a training class or a hotel stay. Useful for quality control purposes. Complementary to periodic surveys.
Articles: Practical Points for an Event Survey, Lost in Translation: A Charming Hotel Stay with a Not-So-Charming Survey, Home Depot Customer Satisfaction Survey, Misleading (or Lying) with Survey Statistics, Complaint Identification: A Key Outcome of World Class CRM Surveys
Exploratory Research: The first of three stages of a research program, followed by Descriptive and Prescriptive Research. This stage seeks to develop a broad contextual understanding of the phenomenon of interest that can set the stage for later research. Personal and small-group interviews (focus groups) are useful tools here. Surveys are not a proper tool for exploratory research since the survey instrument would rely on mostly open-ended questions, which have a high respondent burden.
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F — Survey Glossary
Face Saving Bias (also known as Prestige Bias): A type of response bias where the respondent provides answers so as to avoid embarrassment.
Fatigue: Where the cognitive demands of the survey question, especially a checklist with many items, are such that the respondent’s choice will display primacy effect, recency effect, “satisficing” behavior, or an item non-response. Also used to describe the “survey fatigue” from too many survey invitations.
Field Research: Typically qualitative research processes that provides a broad understanding of a subject area. Such exploratory research provides knowledge for writing better survey questions for descriptive and prescriptive research purposes.
Fixed Sum Question Type (also known as Fixed Allocation or Constant Sum): A question type where the respondent is asked to allocation points across a number of items, for example, allocating 100 points across 5 items. Generates interval data. Useful to measure relative importance.
Follow-Up Notices (also known as Reminders): Notices sent after the initial survey invitation to remind the sample members to take the survey. Useful to increase accuracy and reduce participation bias.
Focus Groups (also known as Small Group Interviews): An exploratory research technique where a group of people are interviewed collectively. Typically done in person, these may now be done using internet-enabled communications, such as online discussion groups.
Forced Choice Scale: A design option for discrete scales where the respondent is not presented with a neutral option. Example: a 6-point rating scale with three positive and three negative options but no neutral.
Forced-Ranking Question Type (also known as Rank Order): A question type where the respondent is presented a number of items and is asked to place the items in rank order along some decision criterion. Generates ordinal data. Can be problematic for respondents to complete.
Fractionation Question Type: A question type where the respondent is presented with a visual line with numerical designations from zero to infinity along with some numeric point that represents “average” for the attribute being measured. Respondent is asked to indicate where on that numeric scale his feelings lie. Useful for measuring ongoing improvement. Generates ratio data.
Free-Form Question Type (also known as Comments, Open-Ended, and Verbatims): Questions where the respondent is asked to provide a textual response.
Frequency Distribution: When examining the results from a survey question, the frequency distribution indicates the percentage of respondents who chose each response option.
Frequency Scale: A commonly used scale to measure the frequency of some event. Commonly used labels or anchors are: Always – Frequently – Sometimes – Rarely – Never
Fully Anchored Verbal Scale: A discrete scale where every position on the scale is anchored with a verbal description or label. Contrast with Endpoint Anchored.
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G — Survey Glossary
Generalizability: The ability to make inferences (to generalize) from the sample data to the population.
Guttman Scale: An ordinal scale type where the respondent is presented an ordered series of statements such that the respondent will agree with all the statements from one end of the scale up to some point. The last agreed-to statement is an index of the respondent’s overall feelings. Best used for hierarchical constructs.
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H — Survey Glossary
Hardcopy Survey Administration Mode: Use of paper for survey invitations and data collection, such as used in postal surveys.
Headings: Short title for a topical section of the questionnaire perhaps with accompanying brief text. These help set the respondent’s mental frame for those questions.
Histograms: A graphing technique to show frequency distributions for a survey question. Also known as vertical or column bar charts.
Horizontal Numerical Scale: A scale presentation technique for discrete, numerical scales. A horizontal line with verbal endpoint anchors and numbers for each scale point is presented to the respondent.
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I, J — Survey Glossary
Ideographical Scale: Where the response scale is presented to the respondent using graphical symbols to represent points on the scale. Smiley faces or other emoticons are examples of ideographic scales. Useful where language issues may exist, such as an instrument for pain measurement in an emergency room likely administered to those without command of the local language.
Importance Measurement: A survey program whose objective is Prescriptive research, requires that we learn the importance of various causes that drive the respondents’ positive or negative attitudes, such as employee or customer satisfaction. Various question types and analytical procedures can serve this purpose.
Articles:The Importance of Measuring Importance — Correctly — on Customer Surveys,
Survey Question Type Choice: More Than One Way to Skin That Importance Cat
Incentives: A means to increase response rates, thus increasing statistical accuracy and reducing participation bias. Incentives may be inducements — providing the gift with the invitation — or rewards — providing the gift only after completion of the survey. The latter may introduce measurement error into the survey if people are “taking” the survey but not providing actual answers to the questions.
Articles: Bribes, Incentives, and Video Tape (er, Response Bias), Home Depot Customer Satisfaction Survey,
Instructions: Guidance provided to the respondent on how to complete the survey.Tendency for survey designers is to over-instruct.
Instrument, Survey (also known as Questionnaire): An instrument measures something, and a survey instrument measures respondents views on the phenomenon of interest for the research study.
Instrumentation Bias: Improper phasing of survey questions that creates a measurement error. Examples of instrumentation bias: ambiguous phrasing, double-barreled questions, loaded language, leading language, unrealistic recall expectations.
Articles: Sampling Error — And Other Reasons Polls Differ, Have You Met Mayor Menino? Lots Have, HHS Hospital Quality Survey, An Example of the Impact of Question Sequencing, Bolton Local Historic District Survey, The importance of Good Survey Question Wording: Even Pros Make Mistakes
Interactive Voice Response Administration Mode: Use of an IVR (or VRU) where a recorded script is used to deliver the survey questions and the telephone keypad is used collect the numerical responses.
Articles: Automated Phone Surveys, Tips for Selecting an Online Survey Software Tool
Interval Data: One of four data types where the response items are in order (ordinal) but also where a consistent unit of measurement is applied to the scalar items. That is, the difference between adjacent response options is equal throughout the scale. Improper scale design will compromise the interval properties. Researchers argue that the cognitive requirements make a truly interval scale on surveys highly unlikely. Interval data properties are needed for many statistical procedures. True interval data allow for addition and subtraction and arithmetic mean to be calculated; however, multiplication and division require ratio properties.
Articles: Survey Question Choice: How Question Format Affects Survey Data Analysis
Interval Rating Scale (also known as Rating Scale or Interval Scale): A response scale where a consistent unit of measurement has been used so that the data generated possess interval properties. Common examples are 1-to-5 or 1-to-10 scales. Note: badly designed scales are not interval, but likely ordinal.
Interviewer Bias: A measurement error introduced by the interviewer by an inconsistent presentation of survey questions to the respondent. The effect could be unintentional or intentional.
Introduction: Brief text at the beginning of the survey instrument meant to provide necessary information for the respondent to then take the survey, including any instructions. Should work in conjunction with text in the invitation and not be duplicative.
Invitation: A request for a member of the invitation sample to participate in the survey process. The invitation mode may be different from the data collection mode, for example, an email invitation with a link to a webform survey or they may be the same.
Articles: An Honest Survey Invitation?, What’s the Point? An Unactionable Transaction, Communicate the Survey Findings — and Your Actions, HHS Hospital Quality Survey
Invitation Sample: Those members of the population who are selected to be asked to participate in the survey. The invitation sample is distinct from the response sample, who are those people who actually do participate. Both groups may be referred to as the “sample.” Survey sample statistics are derived from the later.
Irrelevancy: One type of response bias where the respondent just provides any random answer because the survey is irrelevant to them. Offering an incentive provided only with completion of the survey may introduce this bias.
Articles: Home Depot Customer Satisfaction Survey
Item Non-Response: When the respondent does not answer certain questions — other than those that they have been branched beyond — this is considered an item non-response. Should the respondent not complete a good number of questions, then the entire survey response should probably be removed from the data set since a strong inference exists that the respondent did not take the survey seriously.
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K, L — Survey Glossary
Kurtosis: A statistic that describes how peaked or flat the distribution of values is in a data set.
Leading Language: A type of instrumentation bias where the phrasing of a question leads the respondent towards selecting a particular response option. The leading language may be intentional or unintentional, but the bias compromises the survey data validity.
Article: Bolton Local Historic District Survey — a Critique
Loaded Language: Use of highly emotive language in a survey question that is meant to drive a desired response from the respondent. Similar to leading language in effect, it creates an instrumentation bias.
Likert or Likert-Type Scale: The most widely used survey scaling approach, named for Rensis Likert, an industrial psychologist. Generally, the respondent is presented with a statement and is asked his/her level of agreement with the statement by selecting a point on the discrete scale anchored with verbal statements and frequently with numbers. The scale should be balanced between positive and negative agreement options. Widely used in part because of its flexibility. Properly constructed, the data can be assumed to possess interval properties though it is known that the acquiescence bias leads to respondents selecting positive agreement options.
Articles: Scale Design in Surveys
Looping: A special form of branching where the same block of questions is repeated for various items, for example, for each course the respondent took or for each department a customer visited. Frequently makes use of “piped text.”
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M — Survey Glossary
Mean: One of three types of averages. How to calculate: sum all the values in a data set and divide by the count of values. A blank cell in a spreadsheet is not included in that count.
Measurement Error: The difference between the measured value of an attribute and its in fact actual value. Various biases in instrument design and survey administration contribute to measurement error.
Median (also known as the 50th Percentile Value): One of three types of averages. Found by sorting in order all the values in a data set and then selecting the value in the middle. If the count of values is an even number, then the median is the average of the two values on either side of the middle.
Mental Frame: Survey invitations and introductions establish a contextual framework for the respondent in which they should (or will) consider the survey questions that follow. That is the “mental frame.” In other words, what should they be thinking about. For example, transactional surveys will set the mental frame to be a specific event that just completed typically by naming the event. Having a consistent mental frame for all respondents increases data validity. Section headings also set — and reset — the mental frame.
Mixed-Mode Administration: Use of different modes for one survey project to collect responses from different respondents, for example, offering respondents the option to take a survey by paper or webform. Goal is to increase response; however, the mode affects the nature of the responses, introducing measurement error that complicates data interpretation.
Article: Want Higher NPS & Survey Scores? Change to Telephone Survey Mode
Mode: Regarding survey administration, the medium or method of interaction. Regarding statistics, one of three types of averages. In a data set, the most frequently occurring value.
Multiple Choice Question Type (also known as Checklist Questions): A question type where the respondent is asked to choose one or more items from a list of items. Generates categorical data.
Multiple Response Checklist: A checklist or multiple choice question type where the respondent is instructed to “check all that apply” from the options presented. Instructions and edit procedures may allow the respondent to check up to a certain number or force the selection of a certain number. This variation is useful to identify key drivers of feelings.
Article: The Importance of Measuring Importance — Correctly — on Customer Surveys
Multivariate Statistics: Analytical procedures that examine the relationship among many variables (survey questions). Typically used to determine the causal factors that drive some outcome, such as satisfaction.
Article: Effortless Experience: Statistical Errors
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N — Survey Glossary
Nominal Data Type (also known as Categorical): Data that results from questions with a group of response options that are not related in any order. Only frequency distributions can be generated from such data.
Article: Survey Question Choice: How Question Format Affects Survey Data Analysis
Net Promoter Score® (NPS): A metric that has gained great prominence in customer loyalty circles, though highly controversial. Derived from a survey question that asks likelihood of recommendation posed on a 0-to-10 scale. Net Scoring is then applied with the percentage of scores from 0 to 6 (Detractors) subtracted from the percentage of scores from 9 to 10 (Promoters).
Articles: (There Is More Than) The One Number You Need to Grow, Want Higher NPS & Survey Scores? Change to Telephone Survey Mode, Net Promoter Score — Summary & Controversy, Net Promoter Score® Discussion Notes, Survey Programs Negative Impact on Customer Satisfaction
Net Scoring: A statistic to summarize the data set for a survey question, calculated by subtracting the percentage of respondents who provided low scores (so called, “bottom box scores”) from the percentage of respondents who provided high scores (so called, “top box scores”). Net scoring does not require the assumption of interval data properties and it provides a focus to the low end of the distribution.
Articles: (There Is More Than) The One Number You Need to Grow
Non-Probability Sampling: Sampling processes where every member of the population does not have an equal chance of being invited to participate.
Non-Response Bias (also known as Participation or Self-Selection Bias): A bias introduced into the data set by those who chose to not participate, the assumption being that those who do participate likely hold different views from those who do not.
Articles: Bolton Local Historic District Survey, The Hidden Danger of Survey Bias, Why the Polls Were Wrong — Response Bias Combined with Non-Response
Numeric Scales: Scalar questions where the respondent is presented with an ordered set of integers from which to chose, for example, 1 to 5 or 1 to 10, coupled with verbal anchors for at minimum the endpoints but also perhaps for each discrete point on the scale. The numeric presentation is used to hopefully provide the data interval properties beyond just ordinal properties.
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O — Survey Glossary
Open-Ended Questions (also known as Comments or Verbatims): Questions that ask for a free-form textual response from the respondent.
Ordinal Data Type: One of four data types, these data come from questions where the respondent answer indicates some order. Example: rank ordering of factors that drove a purchase decision. Only order is captured, not relative distance between items in that order.
Article: Why the Polls Were Wrong — Response Bias Combined with Non-Response
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P — Survey Glossary
Paired Comparison Question Type: A question type where numerous pairs of items are presented to respondents from which they are instructed to select the preferred. An ordinal ranking can be statistically derived.
Participation Bias (also known as Non-Response or Self-Selection Bias): A non-measurement error introduced into the data set that results from an uneven participation in the survey by various groups or by people with different intensity of feelings.
Articles: Bolton Local Historic District Survey, The Hidden Danger of Survey Bias, Why the Polls Were Wrong — Response Bias Combined with Non-Response
Periodic Surveys: A survey that is administered on a periodic basis, such as annually. Sometimes called relationship surveys since the focus of the instrument is to measure the overall relationship.
Article: Home Depot Customer Satisfaction Survey, Customer Satisfaction Surveys: The Heart of a Great Loyalty Program
Pie Charts: A common charting technique useful for checklist questions, especially demographic. However, they are especially poor for comparative analysis because the eye cannot readily compare the size of the pie slices.
Pilot Tests (also known as a Pretest): Final testing of a survey instrument. Members of the population are asked to take the survey while being observed and interviewed. Goal is to find ambiguous language and other minor imperfections that would affect instrument validity.
Article: Effortless Experience: Questionnaire Design & Survey Administration Issues, The importance of Good Survey Question Wording: Even Pros Make Mistakes
Piped Text: A branching technique where the text from a response selection is “piped” to a subsequent question.
Pivot Tables: The Excel implementation of “cross tabs.” Pivot tables allow for slicing the data set along various variables and manipulating data.
Polar Anchoring (also known as Endpoint Anchoring): Rating scales where only the endpoints of the scale are anchored with verbal descriptors. Can be used with discrete numerical scales or continuous scales.
Article: Scale Design in Surveys
Population: The group of interest for our survey research.
Population Parameters: If a census is successfully completed, meaning everyone completes the survey — an unlikely situation — then the calculated data from the responses are population parameters, not survey statistics.
Positional Checklist Question Type: An ordinal scale question type where the respondent is presented an ordered list of statements and is asked to pick the position that best represents their views. Ones below the choice would be too negative, ones above too positive.
Postal Survey Administration Mode: An administration mode where the invitation is extended via postal letter and data capture is done via a paper survey return mailed.
Article: An Old Dog’s New Tricks: Postal Mail Surveys
Precision: If some research can be repeated numerous times with near the same result, then it is said to be precise. Not to be confused with accuracy, though frequently used interchangeably in normal usage.
Article: Statistical Confidence in a Survey: How Many is Enough?
Prescriptive Research: After exploratory and descriptive research, a researcher can then engage in research that prescribes a course of action. Usually involves cause-and-effect analysis.
Pretest (also known as a Pilot Test): A final testing of a survey instrument before launching the survey. Members of the population are asked to take the survey while being observed and questionnaire. Goal is to find ambiguous language and other minor imperfections that would affect instrument validity.
Article: Effortless Experience: Questionnaire Design & Survey Administration Issues
Primacy Effect: The tendency for respondents to choose the first item presented for them to consider. Especially prominent in telephone surveys due to the need for the respondent to remember the list. Frequently seen along with recency effect.
Probability Sampling: Sampling processes where every member of the population has an equal chance of being invited to take the survey.
Progress Indicators: Some visual indicator, typically a graph, that shows the respondent how far along they are in taking the survey. Purpose is to reduce survey abandonment.
Purposive or Purposeful Sampling: A non-probability sampling process where individuals with specific traits or viewpoints are selected to participate in research to provide a range of opinion. Frequently used for focus groups.
Push Polling: A survey or poll where the objective is to push information to the respondent as opposed to gathering information from the respondent regarding their views or attitudes. Commonly done in politics to create negativity for an opponent, but may also be used in other surveys to create awareness of some product. Use of the term “aware” in questions is a good indicator.
Article: Bolton Local Historic District Survey — a Critique
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Q — Survey Glossary
Question: A measuring tool to generate data that will measure some attribute (or characteristic) of interest in the research study. The wording in the question operationalizes the attribute so that the respondent can present their views through some measurement scheme, such as using a scale or checkbox. Various question phrasing and question types could be used to measure the same underlying attribute.
Question Types: Broadly speaking, four question types exist corresponding to the type of data each generates: categorical (or nominal), ordinal, interval, and ratio. Each question type has multiple question formats.
Articles: Survey Question Types: When to Use Each, Survey Question Choice: How Question Format Affects Survey Data Analysis
Questionnaire (also known as Survey Instrument): An instrument to measure how some group of interest feels on some subject.
Quota Sampling: A non-probability sampling process where individuals from different demographic groups are invited to participate until a quota is reached, typically to reflect the group’s relative percentage in the general population.
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R — Survey Glossary
Random Error: In any measurement system, including a survey, some variation in scoring is expected and normal. This normal variation is considered random error in contrast to systematic (or abnormal) error.
Random Sampling: A probability sampling process where every member of the population has an equal chance of being invited to take the survey.
Randomization: To avoid sequencing effects, the order of a series of questions may be randomized. Also, the order of response options in a checklist question may be randomized to avoid primacy effects.
Rank, Ranking, Rank Order: (also known as “forced ranking”) When we ask the respondent to specify the order of some items, that is a ranking process. The question generates ordinal data. Note difference from “rate” — and they are frequently confused.
Articles:
Rate, Rating: The result when we ask the respondent to specify how they feel about something using a rating scale. Note difference from “rank” — and they are frequently confused.
Rating Scale (also known as Interval Rating Scale): A response scale where the response options have interval properties. That is, a consistent unit of measurement is present so that the “distance” between response options is the same.
Ratio Data: The highest level data type found typically in physical measures such as length, weight, headcount. A ratio scale has a true zero (as opposed to an arbitrary zero on some scale), meaning zero quantity of that item, such as income. Most, but not all, demographic questions are ratio data though the data may be solicited in an ordered checklist. All mathematical operations can be performed on ratio data including multiplication and division. Many multivariate analysis assume ratio data properties.
Article:
Recall Bias: A type of instrumentation bias regarding unjustifiable memory expectations on the part of the respondent. Validity of responses is in doubt.
Recency Effect: The tendency for a respondent to provide as a response the last response option encountered. Especially prominent in telephone surveys.
Reliability or Reproduceability: A critical concept for any research effort. If the research can be reproduced by other researchers, then the findings are said to be “reliable” and thus have credibility. Lacking reproduceability, the findings should be viewed skeptically. Note difference from “validity.”
Articles: The Effortless Experience Book Review, Effortless Experience: Questionnaire Design & Survey Administration Issues, Customer Insight Metrics: The Issue of Validity
Reminders (also known as Follow-Ups): A notice communicated to a potential respondent to remind them of the invitation to take the survey. Useful to increase statistical accuracy and to reduce participation bias.
Representative: A goal of the survey administration process is for the respondent group to properly reflect or “represent” the entire population of interest. Various administration biases threaten representativeness.
Respondent Burden: The amount of work or “burden” required of the respondent to complete the survey. Higher burden will lower response rates with associated non-response bias — and perhaps create a response bias.
Article: The Poetry of Surveys: A Respondent’s Survey Design Lessons
Response Bias: A bias that the respondent brings to the surveying process that may be activated by the administration process or by the survey instrument wording. The bias can lead to non-participation — the respondent not taking or quitting the survey –, item non-response, or measurement error as the respondent provides non-accurate responses. Examples are: acquiescence (yes saying), auspices, face saving (prestige), conformity, concern, and irrelevance.
Articles: Bribes, Incentives, and Video Tape (er, Response Bias), What’s the Point? An Unactionable Transaction Survey, Have You Met Mayor Menino? Lots Have, An Honest Survey Invitation?, Why the Polls Were Wrong — Response Bias Combined with Non-Response
Response Rate: The number of people who complete the survey divided into the number who actually receive the invitation.
Response Sample: The sample data from those who complete the survey. A subset of the invitation sample, though “sample” is applied to both groups.
Response Set: The collection of items from which the respondent is asked to select his/her response.
Reverse Coded Questions: To avoid respondents falling into a routine, a few rating scale questions are posed in the “reverse,” meaning the positive statement will be the opposite end of the scale from previous questions. These should be used early in the questionnaire before routine has been established. In the data set reverse coded questions should be “unreversed” to each interpretation of results.
Article: Bolton Local Historic District Survey — a Critique
Routine: A response effect where respondents start scoring a series of rating scale questions with the same response without really reading the question and considering their true feelings. Introduces measurement error into the data.
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Reminders (also known as Follow-Ups): A notice communicated to a potential respondent to remind them of the invitation to take the survey. Useful to increase statistical accuracy and to reduce participation bias.
Representative: A goal of the survey administration process is for the respondent group to properly reflect or “represent” the entire population of interest. Various administration biases threaten representativeness.
Respondent Burden: The amount of work or “burden” required of the respondent to complete the survey. Higher burden will lower response rates with associated non-response bias — and perhaps create a response bias.
Article: The Poetry of Surveys: A Respondent’s Survey Design Lessons
Response Bias: A bias that the respondent brings to the surveying process that may be activated by the administration process or by the survey instrument wording. The bias can lead to non-participation — the respondent not taking or quitting the survey –, item non-response, or measurement error as the respondent provides non-accurate responses. Examples are: acquiescence (yes saying), auspices, face saving (prestige), conformity, concern, and irrelevance.
Articles: Bribes, Incentives, and Video Tape (er, Response Bias), What’s the Point? An Unactionable Transaction Survey, Have You Met Mayor Menino? Lots Have, An Honest Survey Invitation?, Why the Polls Were Wrong — Response Bias Combined with Non-Response
Response Rate: The number of people who complete the survey divided into the number who actually receive the invitation.
Response Sample: The sample data from those who complete the survey. A subset of the invitation sample, though “sample” is applied to both groups.
Response Set: The collection of items from which the respondent is asked to select his/her response.
Reverse Coded Questions: To avoid respondents falling into a routine, a few rating scale questions are posed in the “reverse,” meaning the positive statement will be the opposite end of the scale from previous questions. These should be used early in the questionnaire before routine has been established. In the data set reverse coded questions should be “unreversed” to each interpretation of results.
Article: Bolton Local Historic District Survey — a Critique
Routine: A response effect where respondents start scoring a series of rating scale questions with the same response without really reading the question and considering their true feelings. Introduces measurement error into the data.
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S — Survey Glossary
Sample, Sampling: A subset of the population that is selected to receive invitations to participate in the survey. Sampling may be either probabilistic (every member having an equal chance of selection) or non-probabilistic.
Article: Survey Statistical Accuracy Defined
Sample Bias: Results when creation of the sampling frame excludes a certain type of person, leading to a bias in the sample selected and thus to the respondent sample. For example, tasking contact center agents to invite customers to participate in a survey of their experience likely will mean that interactions that did not go well would not receive the invitation. Very similar to a selection bias.
Article: Creating a Survey Program for a University Help Desk, What Survey Mode is Best?, Impact of Mobile Surveys — Tips for Best Practice
Sample Size Equation: The number of people to include in an Invitation Sample divided by the likely Response Rate. The resulting sample size should lead to the desired statistical accuracy.
Articles: Survey Sample Size Calculator, Statistical Confidence in a Survey: How Many is Enough?
Sample Statistics: Calculated statistics from the Respondent Sample which are presented as indications of the Population Parameters. In other words, the sample statistics are indications of how the entire group of interest would have responded, with some degree of sampling error — and probably some degree of sample and selection bias.
Sampling Error: The error introduced into our survey results by the fact that the data are from a sample rather than from the entire population. The difference between the sample mean and the population mean results from sampling error. Statistical accuracy provides an indication of the level of sampling error.
Sampling Frame: A subset of the population from which the sample is drawn. Sampling frames are used where it is not possible or impractical to draw the sample from the entire population. For example, if we do not have contact information for some people in the population, then the sampling frame would exclude those names lacking contact information.
Article: Caveat Survey Dolor: “Show Me the Questionnaire”
Satisficing: A concept from information theory where people select a “satifactory” answer, but not the optimal answer. When presented with a long list of options from which to select, something that causes “fatigue,” the respondent may select the first, good-enough answer to save time and move on.
Scales: In common survey usage, a scale is an ordered series of response options, presented verbally, numerically, or ideographically, from which the respondents select to indicate their level of feeling about the attribute being measured. More properly a scale is a composite score of a number of survey questions that each measure the same attribute. For example, a final exam for a class is a scaled score from multiple questions that measure the students knowledge of the subject matter.
Articles: Survey Question Choice: How Question Format Affects Survey Data Analysis, Scale Design in Surveys: The Impact on Performance Measurement Systems — and the Value of Dispersion
Scatter Plots: A visual depiction of the correlation between two variables found by plotting one variable on the X and another on the Y axis.
School-Grade Scale: A response scale that uses school grades — A, B, C, D, and F — as the response options. The scale is culturally dependent.
Section Headings: Short headings perhaps with accompanying brief text that lead into a section of a survey. These help set the respondent’s mental frame for those questions.
Articles: Proper Use of Section Headings
Segmentation Analysis: In addition to analyzing the survey data set as a whole, analysis is typically desired for specific segments of the population found by slicing the data set along demographic variables. Common segments are: by income level, by gender, by length of relationship (customer, employee, etc.), and frequency of experience some phenomenon.
Selection Bias: Results when some members of the sampling frame are less likely to participate in the survey due to the way the sample is generated. For example, a webform survey designed for use on a laptop only — not on a mobile device (smartphone) — may result in people who primarily use a mobile device not participating. Very similar to a sample bias.
Articles: Impact of Mobile Surveys — Tips for Best Practice, What Survey Mode is Best?, Creating a Survey Program for a University Help Desk
Self-Selection Bias (also known as Non-Response or Participation Bias): Occurs when members of the Invitation Sample who share some common characteristics choose not to participate in the survey, resulting in a biased Response Sample.
Articles: Bolton Local Historic District Survey, The Hidden Danger of Survey Bias, Why the Polls Were Wrong — Response Bias Combined with Non-Response
Semantic Differentiation Question: A question type where the respondent is presented a multi-point scale in a horizontal numerical structure where each endpoint is verbal anchored with antonyms. Few webform tools support this question type. Very useful in handling reverse coded questions.
Semi-Structured Questionnaire: For focus groups and interviews a fluid questionnaire used to guide the discussion. Good moderator skills needed to apply such a questionnaire.
Sequencing: A response effect where the answer to one question affects the respondent’s interpretation of subsequent questions. Also seen for the sequencing of items in a checklist question. Question randomization can mitigate the effect.
Articles: Survey Question Design: Headlines or Meaningful Information?, Money Grows on Trees — If You Believe the Polls, An Example of the Impact of Question Sequencing
Service Recovery: An action program to address issues that customers have experienced to recoup them as customers or to mitigate negative word of mouth. Transactional surveys serve to identify customers in need of a service recovery act.
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Skip and Hit (also known as Branching and Conditional Branching): Where the question path a respondent follows is based upon the response to a particular question. Useful for avoiding not applicable questions.
Articles: Creating a Survey Program for a University Help Desk
Spam Trigger Words: Words that email filters use to flag possible spam emails. Avoid using these words in an email invitation, lest the invitation never get to the potential respondent’s inbox.
Stacked Bar Charts: Vertical or horizontal bar charts that display frequency distributions from a survey question. Best used when there’s a limited number of response options for the question.
Standard Deviation: Square root of variance. A measure of the dispersion of values in the data set around the mean.
Statistic: A number meant to represent some characteristic of a larger data set.
Statistical Confidence: Commonly used interchangeably with statistical accuracy, it tells us how well a statistic from a sampling process represents the population mean for the data set.
Stratified Random Sampling: A variation of random sampling to generate the invitation sample used to help ensure a consistent statistical accuracy across demographic segments of comparative interest, for example, product line, business region, school district. Sampling here is a two stage process. First, stratify (group) the members of the sample frame by the demographic variable. Second, random sample within each stratum.
Survey: The process of conducting research using survey methodology, though in common parlance “survey” is used to mean the survey instrument or questionnaire.
Survey Administration: The process of inviting people to take the survey and collect data from them. Care must be taken to avoid introducing administration biases into the data set. See “Administration” for other related items.
Survey Instrument (also known as Questionnaire): The measuring instrument used to gauge how some group feels on the topic of the research study.
Survey Fatigue: When members of the population become so weary of repeated survey invitations that they stop taking the survey, except perhaps when they have an issue. Can introduce a participation bias (a.k.a. non-response or self-selection bias). With more organizations surveying, fatigue has increased.
Article: Survey Sample Selection: The Need to Consider Your Whole Research Program, Battling Survey Fatigue, The Hidden Danger of Survey Bias
Survey Manipulation: The practice of trying to influence the scores provided by respondents by those affected by the survey findings. The manipulation may be overt or subtle, but it distorts the true views of the respondents to achieve other goals.
Survey Question: A measuring tool to generate data that will measure some attribute (or characteristic) of interest in the research study. The wording in the question operationalizes the attribute so that the respondent can present their views through some measurement scheme, such as a scale or checkbox. Various question phrasing and question types could be used to measure the same underlying attribute.
Articles: Survey Question Choice: How Question Format Affects Survey Data Analysis
Systematic Sampling: One of the four probabilistic sampling approaches, generally used when the sampling frame is in the form of a list or maybe people queued in a line. Names would be selected at an interval (say, every 8th name) going through the entire list.
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T — Survey Glossary
Telephone Surveying: Use of telephone medium to invite and administer a survey.
Tests of Independence: Statistical tests to determine if two groups of categorical data are from the same underlying population or whether the differences are statistically significant.
Textual Data: Free-form textual responses provided in open-ended questions.
Threaded Branch: A branching technique in a survey where a series of follow-up questions are posed, for example to identify the rank order of a series of items.
Top Box Scoring: For a scalar question, the “top box” is a cumulative frequency distribution for those responses at the top end of the scale down to some arbitrary point on the scale. For example, in a 1-to-10 scale, the “top box” may be the percent of respondent scoring 9 through 10. See also Bottom Box Scoring and Net Scoring.
Trade-Off Analysis: An approach used in surveys to identify key drivers of respondents’ actions or feelings by asking them to examine multiple items simultaneously and score the items to indicate relative importance. Forced Ranking and Fixed Sum Questions Types are commonly used.
Transactional Surveys (also known as Event or Incident Surveys): A survey program where those invited to participate are people who have just completed some transaction, for example, a training class or a hotel stay. Useful for quality control purposes. Complementary to periodic surveys.
Articles: Practical Points for an Event Survey, Lost in Translation: A Charming Hotel Stay with a Not-So-Charming Survey, Home Depot Customer Satisfaction Survey, Misleading (or Lying) with Survey Statistics, Complaint Identification: A Key Outcome of World Class CRM Surveys
Truncated Scales: Where the scale presentation causes the respondent to not consider the entire breadth of the response options, the scale is considered truncated. For example, checklist questions with too many items lead to primacy and recency truncation effects. Numerical scales presented in oral telephone administration mode are observed to result in a truncated scale since the respondent typically hears “on a scale from 1 to 10” leading them to chose the endpoints 1 or 10 more frequently than a scale presented visually on paper or webform.
Article: Want Higher NPS & Survey Scores? Change to Telephone Survey Mode
t Test: A statistical test that determines whether two data sets are from the same underlying population or not. Used to determine whether the difference between two survey scores is a statistically significant difference or whether the difference is simply due to sampling error. Requires interval data properties.
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U, V — Survey Glossary
Ulterior Motive: A type of response bias where the respondent does not answer honestly but to serve some other purpose, for example, to purposely corrupt the survey findings.
Unit of Analysis: In a research study, the level at which the analysis is focused. In a business-to-business survey where we might survey multiple contact people in a client company, we might be concerned about individual responses or a client company’s responses. In a municipal survey, our focus might be on the individual or on households.
Validity: A valid survey question is one that measures what it purports to measure. Various types of instrumentation bias can compromise validity. Note: various types of validity exist in the research world: construct, content, conclusion, face, internal, external, criterion.
Articles: The Effortless Experience Book Review
Effortless Experience: Questionnaire Design & Survey Administration Issues
Customer Insight Metrics: The Issue of Validity
Variance: A measure of the spread of data in a data set, that is, how tightly or loosely clustered the responses for a question are around the mean score.
Verbal Scale: Where the response options are presented to the respondent using words, whether spoken or written.
Verbatims (also known as Comments or Open-Ended Questions): Free-form textual responses to open-ended questions. The term derives from interviewers capturing a respondent’s comments “verbatim,” but it is now a term used even for comments the respondent types into a webform.
Visual Analog Scale (VAS): A response scale where the respondent is presented a continuous (analog) line with only the endpoints anchored. It is considered to generate data with better interval properties. The pain scale used in hospitals is termed a VAS though in practice it is a discrete scale with numerical, verbal, and ideographic presentation of response options.
Articles: Visual Analog Scale – A Pain-ful Misnamed Scale
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W, X, Y, Z — Survey Glossary
Webform Surveying: use of webforms to administer a survey to respondents and collect data. Invitations are commonly extended via email, but other media could be used to extend the invitation.
Yes Saying: A type of response bias where the respondent is predisposed to answer “yes” to questions. Mostly a concern for personal interviews, but also a concern for rating questions using an Agreement scale.
We have not identified any survey terms for letters X or Z.
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11588 | https://fiveable.me/key-terms/introduction-probability/discrete-uniform-distribution | printables
🎲intro to probability review
key term - Discrete Uniform Distribution
Citation:
MLA
Definition
A discrete uniform distribution is a probability distribution where a finite number of outcomes are equally likely to occur. Each outcome in the sample space has the same probability, making it simple and intuitive to understand. This type of distribution is often used in scenarios where each possible event is equally likely, providing a foundation for exploring more complex probability concepts.
5 Must Know Facts For Your Next Test
In a discrete uniform distribution, if there are 'n' outcomes, the probability of each outcome is given by $$P(X = x) = \frac{1}{n}$$.
The mean or expected value of a discrete uniform distribution can be calculated using the formula $$E(X) = \frac{a + b}{2}$$, where 'a' is the minimum value and 'b' is the maximum value.
The variance of a discrete uniform distribution is given by $$Var(X) = \frac{(b - a + 1)^{2} - 1}{12}$$, which measures the spread of the distribution.
Discrete uniform distributions are commonly found in games of chance, such as rolling dice or drawing cards, where each outcome has an equal chance of happening.
The discrete uniform distribution is a special case of the more general uniform distribution, which can also be continuous.
Review Questions
How would you explain the concept of equal likelihood in a discrete uniform distribution and its importance in understanding probability?
In a discrete uniform distribution, every possible outcome has the same likelihood of occurring, which simplifies calculations and understanding of probabilities. This equal likelihood is crucial because it allows us to predict outcomes without bias or preference toward any specific event. Understanding this concept sets a foundation for exploring more complex distributions and enhances our ability to assess situations involving chance.
Discuss how the properties of a discrete uniform distribution apply when analyzing joint probability distributions for multiple random variables.
When analyzing joint probability distributions involving discrete uniform distributions, the independence and equal likelihood properties play an important role. If two or more random variables are independently and uniformly distributed, their joint probabilities can be calculated by multiplying their individual probabilities. This leads to a straightforward analysis where outcomes remain equally likely, allowing for easier computation of probabilities across combined events.
Evaluate the implications of using a discrete uniform distribution in practical applications like games or random sampling. How does this affect decision-making processes?
Using a discrete uniform distribution in practical applications like games or random sampling has significant implications for decision-making. It ensures fairness and transparency since every outcome has an equal chance, minimizing biases. In games, this creates an engaging experience for players, while in sampling, it allows researchers to make valid inferences about populations. The reliance on equal probabilities fosters trust and reliability in outcomes, influencing strategies in both gaming and research contexts.
Related terms
Probability Mass Function:
A function that gives the probability of each possible value for a discrete random variable.
Sample Space:
The set of all possible outcomes of a random experiment.
Random Variable:
A variable whose possible values are numerical outcomes of a random phenomenon.
"Discrete Uniform Distribution" also found in:
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11589 | https://allen.in/dn/qna/233093376 | The length of the sides of a triangle are 13 , 14 and 15 if R and r respectively denote circumradius nad inradius of that triangle then 8R +r=
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HomeClass 12MATHSThe length of the sides of a triang...
The length of the sides of a triangle are 13 , 14 and 15 if R and r respectively denote circumradius nad inradius of that triangle then 8 R+r=
A
84
B
65 8
C
4
D
69
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The lengths of the sides of a triangle are 13, 14 and 15. If R and r respectively denote the circum radius and inradius of that triangle, then 8R + r =
A 84 B 65/8 C 4 D 69 Submit
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11590 | https://www.reddit.com/r/askmath/comments/1j68s1p/is_there_anything_mathematically_special_about/ | Is there anything mathematically special about this? Transition patterns in PI. I looked at PI two times. The first 19 digits, then the first 1001 digits. I looked at them in base 2. Here are some findings. The first is verifiable by hand using the Windows calculator to get the binary. : r/askmath
Skip to main contentIs there anything mathematically special about this? Transition patterns in PI. I looked at PI two times. The first 19 digits, then the first 1001 digits. I looked at them in base 2. Here are some findings. The first is verifiable by hand using the Windows calculator to get the binary. : r/askmath
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Is there anything mathematically special about this? Transition patterns in PI. I looked at PI two times. The first 19 digits, then the first 1001 digits. I looked at them in base 2. Here are some findings. The first is verifiable by hand using the Windows calculator to get the binary.
Resolved
Results when looking at PI and how the number transitions state (in binary). It is very symmetric. The transition counts add up to the counts where they stay the same and don't repeat, both for 0 and 1. The results again at 1001 digits produce the same kinds of results. The thing that fluctuates slightly are the times it stays at 0 or stays at 1 (staying the same). There seems to be a pattern, or it is approaching some state of fluctuation. Here are the results. C++ code available to run and verify.
Here is PI as an integer representation. 3141592653589793238 The second results for this number
Transitions from 0 to 1: 838
Transitions from 1 to 0: 837
Stay at 0: 831
Stay at 1: 831
Transitions from 0 to 1: 838
Transitions from 1 to 0: 837
Runs of same bits: 1676
Number of bits: 3338
Transitions from 0 to 1: 52
Transitions from 1 to 0: 52
Stay at 0: 58
Stay at 1: 42
Transitions from 0 to 1: 52
Transitions from 1 to 0: 52
Runs of same bits: 105
Number of bits: 205
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11591 | https://www.weareteachers.com/vocabulary-activities/ | Skip to main content
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36 Fun and Easy Vocabulary Activities and Games for Every Grade
These activities are the definition of fun!
By Jill Staake, B.S., Secondary ELA Education
May 6, 2024
Learning new words is like adding to your writing toolbox. The more tools available, the more interesting and engaging your writing becomes. Check out these fun and engaging vocabulary activities for kids in grades K-12, and supply your students with the tools they need to build their wordsmith skills.
FREE WORKSHEETS
Vocabulary Activities Worksheets
Grab a set of free printable vocab worksheets to go along with eight of the activities below!
Download My Free Vocabulary Activities Worksheets
Jump to:
Vocabulary Games for Kids
Vocabulary Writing Activities
More Fun Vocabulary Activities
General Vocabulary-Building Activities
Vocabulary Games for Kids
1. Riddle Me This
Use our free downloadable worksheet for this fun vocabulary game. First, kids work individually or in teams to create a list of four clues for each vocab term. Then, they use these clues to try to get other students to guess the right word.
2. Explore new words with Khanmigo
Here’s a cool way to give kidsguided exploration of new words. Go into the Lesson Hooks tool in Khanmigo and fill in your grade, topic (vocabulary), and context (give kids guided exploration of new words). Khanmigo will offer up a few options like: Start a “word detective” challenge, organize a “vocabulary auction” or maybe the real-world context generator, with info from the opening paragraph of the article. This is such a great way to get new ideas for exactly what you need to teach.
3. Hot Seat
Here’s one fun way to use Riddle Me This clues. Divide the class into two teams. The first team chooses one student to sit in the “hot seat” at the front of the classroom. Then, the teacher shows their teammates the target word for that round and sets a timer for 1 minute. The team gives clues to the player, who must guess the correct term before the time is up. Switch teams, and go for round 2! Optional: The player must spell the word correctly to earn the point.
4. Taboo
This is similar to Hot Seat but with a twist. You’ll need to create the cards for this one in advance: Write each term on your list on a separate card. On each card, also include several words students are NOT allowed to use in the game. For instance, if the word is “alligator,” you might list the taboo words “teeth,” “Florida,” and “swamp.” If players use any of these words when trying to get their teammate to say the target term, they automatically lose the point!
5. Vocabulary Matchup
Create a pair of cards for each of your vocab terms. One should show the word itself, the other the definition. Mix up the cards and pass them out, one to each student. Their job is to talk to other students until they find the person who has their match. Once everyone is finished, verify their answers, then collect the cards, mix them up, and pass them out again for another round.
6. Word Pop!
Place cards with the definitions, synonyms, and antonyms of your vocabulary words in a bag or bin (bonus points if it’s popcorn-themed!). Mix in one or two cards that just say Pop! Kids take turns drawing a card and naming the vocabulary word that it describes. If they get it right, they keep the card—if not, it goes back into the bag. Be careful, though: If you get the Pop! card, you forfeit all your cards!
7. Vocab Pictionary
Just like the beloved classic, but the words are vocabulary words! Teams send one player up to draw the word on the card they draw (prepare these in advance), and their teammates have 60 seconds to guess it. This one is always a classroom hit!
8. Charades
Here’s another classic game that’s perfect for vocabulary word practice. Your students will have a blast trying to act out the words for their teammates to guess. You can also have several students prepare brief skits that demonstrate the terms, then see if their classmates can figure them out.
9. A to Z Vocabulary
This vocabulary game is fun and challenging, and it can be played by kids of any age. Have kids work in teams. One player on each team—the scribe—writes the letters A to Z down the left side of a page. Then, show the first vocabulary word and set a timer for three minutes. Teams work together to come up with words that relate to the vocab term (synonyms, antonyms, examples, etc.). When time is up, teams share their words and tally up their points. (For a fun twist, assign points to each letter, as in Scrabble, with trickier letters earning more points.)
10. Vocabulary Jeopardy
Create a Jeopardy-style board, with categories including Definition, Synonym, Antonym, and Sentence. Then, make a set of cards for your vocab word list, and lay them face down in each category. Kids choose a category and dollar amount, then flip the card to see the vocab word. They must then define the word, offer a synonym or antonym, or use it in a sentence, whichever fits the category.
Vocabulary Writing Activities
11. Write It Out
Our free downloadable worksheet makes this activity oh-so simple. Students list their vocab words, then write a sentence that uses each correctly.
12. Write the Room
Add a fun twist to Write It Out by posting the vocabulary words on sticky notes scattered around your classroom. Number them, and have kids number their worksheet too. Then, groups circulate the room, looking for the sticky notes. When they find one, they write the word by the correct number, then write their sentences.
13. Story Time
Take the next natural step by challenging kids to use their vocab terms in a short story. We’ve got a worksheet in our free bundle for this classic vocabulary practice activity.
14. Vocab RAFT
Writing a story using vocab words is a perennial favorite, but the RAFT method gives it a new twist. Give each student a Role (the point of view from which they’ll tell the story), an Audience, a Format, and a Topic. For instance, they might be an astronaut (Role) writing a postcard (Format) to their friends back home (Audience) about what they’ve seen on Mars (Topic). RAFTs are especially great for kids who claim they don’t know what to write about.
15. Vocabulary Journal Prompts
Provide kids with a journal prompt, and require them to use each of their vocab terms at least once in their response. Need writing prompt ideas? We’ve got you covered (see them all here)!
40 Gratitude Journal Prompts for Kids
101 Best Writing Prompts for Middle School
100 Argumentative Writing Prompts
150 Picture Writing Prompts
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Fun First Day of School Worksheets (Free Printables)
16. Social Vocabulary
This is one of those vocabulary activities kids will want to do over and over again. Assign each student a word and have them create a faux Facebook, Instagram, or other social media page for it. They can draw them by hand or use an online tool like Fakebook. Post the images to a shared Google Slideshow so other students can use them for review.
17. Vocab Roll
For each word on their list, students roll a die and complete the activity indicated for that number. For example, if they roll a 2, they must use the word in a sentence. If it’s a 5, they draw a picture. Get the free worksheet in our bundle.
18. Vocabulary Acrostics
Write an acrostic poem for each vocab term, using the letters to determine the first word in each line. This can get really challenging when words are longer! Find tips on writing acrostics here.
More Fun Vocabulary Activities
19. Definition Detectives
Instead of just giving kids word definitions, have them use our free downloadable worksheet to seek them out instead. First, they write down their best guess as to the definition. Then, they use the dictionary to find the correct meaning.
20. Word Maps
Creating word maps from vocabulary words encourages students to find the relationships between the vocabulary word and other words. Have them include words, pictures, examples, real-world connections, definitions, descriptive words, etc.
21. Vocabulary Gallery Walk
Hang six to eight large sheets of chart paper in various places around the room. On each sheet, write one vocabulary word. Have students work in small groups, rotating between stations. At each station, ask students to come up with a different, original way to use each word, adding it to the chart. Continue the activity until all students have visited every station, then review them all together.
22. Four Square Words
Also known as a Frayer Model, this activity requires kids to provide the following for each term on their list: definition, synonym, antonym, and a drawing. You’ll find a free worksheet for it in our collection.
23. Graffiti Wall
Think of a vocabulary graffiti wall like a collaborative word wall. In the classroom, post the words on the wall and have kids add sticky notes to illustrate the term (they can use words or pictures). Online, try a tool like Padlet or Google Slides.
24. Vocabulary Flash Cards
Simple but effective! Give kids a stack of note cards, and have them write one vocab word on each. Then, flip the cards over and ask them to write the definition, draw a picture, or make any other helpful notes they can use when they review the word independently or with friends.
25. Sketch It Out
Rather than writing out definitions, have students draw a sketch that sums up each word instead. It’s a lot more fun and gives kids an image for visual association to help them remember the word meanings.
26. Graphic Organizers
Take illustrations one step further with graphic organizers. Here’s how they work. (Tip: Want to go digital? Have kids make a slideshow, one slide per word. They can include the same information, but instead of drawing a picture, have them find one online that illustrates the concept.)
27. Vocabulary Character Match
This is a terrific way to practice vocab words pulled from books you’re reading. Ask students to use various words to describe the different characters in the book and their feelings, thoughts, and actions.
28. Parts of Speech Sort
Knowing which part of speech each vocab term is can help kids use them correctly. Our free printable worksheet makes this minimal-prep activity easy for teachers and meaningful for kids.
29. Word of the Day
Give really important terms the attention they deserve. Choose a new vocab word each day, then explore it in depth and make an effort to use it in class the whole day long.
30. Golden Word Club
Post a list of target vocab words. If a student uses one of the words in class (outside of vocabulary activities), they become a member of the Golden Word Club! You can have them sign their name on a wall in the classroom or award a badge online. You could even develop this into a reward system for homework passes or extra credit.
31. Shades of Meaning
This is a cool idea for exploring synonyms and the slight differences that make words unique. Ask for paint sample strips at your local hardware store. Write a vocab word at the top of each. Then, have students write similar words (synonyms) on each strip. Make notes about the subtle shades of difference between the various words. (Get more paint sample teaching ideas here.)
General Vocabulary-Building Activities
32. The Word Collector
This is one of those picture books that older kids will enjoy as much as little ones. Use it to remind your kids that they don’t need a vocabulary list to learn new words—new words are all around them!
Buy it: The Word Collector at Amazon
33. New Word Journal
Give kids a small notebook they can carry around with them. Whenever they encounter a word they don’t know, they write it down in their journal on a new page. They’ll find new words on signs and menus, in conversations or on TV shows, or while surfing the web. When they have time, they can look up the definitions and write them in.
34. Wordsmithery
Brush up on words people usually just pretend they know the meanings of. Listen to the word and three possible definitions. Choose the right one to score the point! (Best for high school students.)
Buy it: Wordsmithery at Amazon
35. Sticky Words
When students read a new book from your classroom library, have them keep a pad of sticky notes handy. As they encounter words they don’t know, they write each one on a sticky note, then look up the definition and write that too. Stick each note to the inside covers of the book so others who read the book can use it as a glossary.
36. Blurt!
Listen to the clues and shout out the right word first to score points in this vocab game. The clues aren’t necessarily hard (“the nut of an oak tree”), but when you’re on the spot, it makes things much harder!
Buy it: Blurt! at Amazon
37. Vocab Show-and-Tell
Think of this like everyone’s favorite preschool activity but with words! Choose a few students each week to “bring” a new word to show the class. Have them choose the word in advance and share it with you to make sure it’s appropriate and they truly understand it. Then, each student brings and presents something that “shows” and “tells” the word and its meaning. Example: If the word is “fuzzy,” a student might bring their favorite stuffed animal. For “delicious,” they could bring a favorite treat to share with the class.
Don’t forget to download your free printable vocabulary practice activity worksheets!
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Reading poetry also helps students expand their vocabulary. Check out these must-share poems for elementary school and middle and high school.
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11592 | https://orca.cardiff.ac.uk/id/eprint/115340/ | Systemic antibiotics for symptomatic apical periodontitis and acute apical abscess in adults
| |
| Cope, Anwen ORCID: Francis, Nicholas ORCID: Wood, Fiona ORCID: and Chestnutt, Ivor ORCID: 2018. Systemic antibiotics for symptomatic apical periodontitis and acute apical abscess in adults. Cochrane Database of Systematic Reviews 9 , CD010136. 10.1002/14651858.CD010136.pub3 |
| |
| Cope, Anwen ORCID: Francis, Nicholas ORCID: Wood, Fiona ORCID: and Chestnutt, Ivor ORCID: 2018. Systemic antibiotics for symptomatic apical periodontitis and acute apical abscess in adults. Cochrane Database of Systematic Reviews 9 , CD010136. 10.1002/14651858.CD010136.pub3 |
| | |
--- |
| [thumbnail of Cope_et_al-2018-Cochrane_Database_of_Systematic_Reviews.pdf] Preview | PDF - Published Version Download (506kB) | Preview |
Abstract
Background Dental pain can have a detrimental effect on quality of life. Symptomatic apical periodontitis and acute apical abscess are common causes of dental pain and arise from an inflamed or necrotic dental pulp, or infection of the pulpless root canal system. Clinical guidelines recommend that the first-line treatment for teeth with these conditions should be removal of the source of inflammation or infection by local, operative measures, and that systemic antibiotics are currently only recommended for situations where there is evidence of spreading infection (cellulitis, lymph node involvement, diffuse swelling) or systemic involvement (fever, malaise). Despite this, there is evidence that dentists frequently prescribe antibiotics in the absence of these signs. There is concern that this could contribute to the development of antibiotic-resistant bacterial colonies within both the individual and the community. This review is an update of the original version that was published in 2014. Objectives To evaluate the effects of systemic antibiotics provided with or without surgical intervention (such as extraction, incision and drainage of a swelling, or endodontic treatment), with or without analgesics, for symptomatic apical periodontitis and acute apical abscess in adults. Search methods Cochrane Oral Health’s Information Specialist searched the following databases: Cochrane Oral Health’s Trials Register (to 26 February 2018), the Cochrane Central Register of Controlled Trials (CENTRAL; 2018, Issue 1) in the Cochrane Library (searched 26 February 2018), MEDLINE Ovid (1946 to 26 February 2018), Embase Ovid (1980 to 26 February 2018), and CINAHL EBSCO (1937 to 26 February 2018). The US National Institutes of Health Ongoing Trials Register (ClinicalTrials.gov) and the World Health Organization International Clinical Trials Registry Platform were searched for ongoing trials. A grey literature search was conducted using OpenGrey (to 26 February 2018) and ZETOC Conference Proceedings (1993 to 26 February 2018). No restrictions were placed on the language or date of publication when searching the electronic databases. Selection criteria Randomised controlled trials of systemic antibiotics in adults with a clinical diagnosis of symptomatic apical periodontitis or acute apical abscess, with or without surgical intervention (considered in this situation to be extraction, incision and drainage or endodontic treatment) and with or without analgesics. Data collection and analysis Two authors screened the results of the searches against inclusion criteria, extracted data and assessed risk of bias independently and in duplicate. We calculated mean differences (MD) (standardised mean difference (SMD) when different scales were reported) and 95% confidence intervals (CI) for continuous data. A fixed-effect model was used in the meta-analysis as there were fewer than four studies. We contacted study authors to obtain missing information. Main results We included two trials in this review, with 62 participants included in the analyses. Both trials were conducted in university dental schools in the USA and compared the effects of oral penicillin V potassium (penicillin VK) versus a matched placebo when provided in conjunction with a surgical intervention (total or partial pulpectomy) and analgesics to adults with acute apical abscess or symptomatic necrotic tooth. The patients included in these trials had no signs of spreading infection or systemic involvement (fever, malaise). We assessed one study as having a high risk of bias and the other study as having unclear risk of bias. The primary outcome variables reported in both studies were participant-reported pain and swelling (one trial also reported participant-reported percussion pain). One study reported the type and number of analgesics taken by participants. One study recorded the incidence of postoperative endodontic flare-ups (people who returned with symptoms that necessitated further treatment). Adverse effects, as reported in one study, were diarrhoea (one participant, placebo group) and fatigue and reduced energy postoperatively (one participant, antibiotic group). Neither study reported quality of life measurements. Objective 1: systemic antibiotics versus placebo with surgical intervention and analgesics for symptomatic apical periodontitis or acute apical abscess Two studies provided data for the comparison between systemic antibiotics (penicillin VK) and a matched placebo for adults with acute apical abscess or a symptomatic necrotic tooth when provided in conjunction with a surgical intervention. Participants in one study all underwent a total pulpectomy of the affected tooth, while participants in the other study had their tooth treated by either partial or total pulpectomy. Participants in both trials received oral analgesics. There were no statistically significant differences in participant-reported measures of pain or swelling at any of the time points assessed within the review. The MD for pain (short ordinal numerical scale 0 to 3) was -0.03 (95% CI -0.53 to 0.47) at 24 hours; 0.32 (95% CI -0.22 to 0.86) at 48 hours; and 0.08 (95% CI -0.38 to 0.54) at 72 hours. The SMD for swelling was 0.27 (95% CI -0.23 to 0.78) at 24 hours; 0.04 (95% CI -0.47 to 0.55) at 48 hours; and 0.02 (95% CI -0.49 to 0.52) at 72 hours. The body of evidence was assessed as at very low quality. Objective 2: systemic antibiotics without surgical intervention for adults with symptomatic apical periodontitis or acute apical abscess. We found no studies that compared the effects of systemic antibiotics with a matched placebo delivered without a surgical intervention for symptomatic apical periodontitis or acute apical abscess in adults. Authors’ conclusions There is very low-quality evidence that is insufficient to determine the effects of systemic antibiotics on adults with symptomatic apical periodontitis or acute apical abscess.
| | |
--- |
| Item Type: | Article |
| Date Type: | Publication |
| Status: | Published |
| Schools: | Schools > Dentistry Schools > Medicine Research Institutes & Centres > Prime Centre Wales (PRIME) |
| Subjects: | R Medicine > RK Dentistry |
| Publisher: | Cochrane Collaboration |
| ISSN: | 1469-493X |
| Date of First Compliant Deposit: | 3 October 2018 |
| Date of Acceptance: | 25 September 2018 |
| Last Modified: | 21 Sep 2024 12:48 |
| URI: | |
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11593 | https://math.stackexchange.com/questions/1280400/prove-that-n3-n-text-mod-6-for-every-integer-n | elementary number theory - Prove that $n^3=n \text{ mod }6$ for every integer $n$. - Mathematics Stack Exchange
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Prove that n 3=n mod 6 n 3=n mod 6 for every integer n n. [duplicate]
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Prove that x 3≡x mod 6 x 3≡x mod 6 for all integers x x (2 answers)
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Prove that for every integer n n , n 3=n mod 6 n 3=n mod 6
I was having no clue how to do this, then I thought of case-by-case analysis and obviously it worked. The problem is that there were six case and together they are long.
Is there any shorter method of proving this. Kindly provide some hints.
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asked May 13, 2015 at 14:59
SinghSingh
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2 Fermat's little theorem? Or n 3−n=n(n+1)(n−1)n 3−n=n(n+1)(n−1). Or simply, Euler's theorem if you know that.Sawarnik –Sawarnik 2015-05-13 15:00:05 +00:00 Commented May 13, 2015 at 15:00
1 Hmm, wait, probably Euler's theorem won't work. But the factorization is the best :)Sawarnik –Sawarnik 2015-05-13 15:09:29 +00:00 Commented May 13, 2015 at 15:09
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Factorize n 3−n n 3−n. Is any of these factors even? Is any of these factors a multiple of three?
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edited May 13, 2015 at 15:18
answered May 13, 2015 at 15:02
ajotatxeajotatxe
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n 3−n=n(n+1)(n−1)n 3−n=n(n+1)(n−1) is always divisible by 2 and 3, so it is 0 mod 6.
Note: I marked this answer community wiki because I got the answer form the comment.
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answered May 13, 2015 at 15:03
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wythagoras
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The direct approach:
0 3≡0 mod 6 0 3≡0 mod 6
1 3≡1 mod 6 1 3≡1 mod 6
2 3≡8≡2 mod 6 2 3≡8≡2 mod 6
3 3≡27≡3 mod 6 3 3≡27≡3 mod 6
4 3≡(−2)3≡−8≡−2≡4 mod 6 4 3≡(−2)3≡−8≡−2≡4 mod 6
5 3≡(−1)3≡−1≡5 mod 6 5 3≡(−1)3≡−1≡5 mod 6
Thus
n 3≡n mod 6 n 3≡n mod 6
by exhaustive checking.
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answered May 13, 2015 at 15:20
JoelJoel
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If you want to use purely modular arithmetic, then you must prove that both 3 and 2 divide your function. So, we use Fermat's little theorem.
Clearly, n 3≡n(mod 3)⟹n 3−n≡0(mod 3)n 3≡n(mod 3)⟹n 3−n≡0(mod 3)
So, 3 divides this.
Now we move on to 2, with Fermat's little theorem again,
n 2≡n(mod 2)⟹n 3≡n 2≡n(mod 2)n 2≡n(mod 2)⟹n 3≡n 2≡n(mod 2), or, through transitivity, n 3−n≡0(mod 2)n 3−n≡0(mod 2). Thus, 2 and 3 divide this function, which immediately implies that 6 must also divide it, ie n 3−n≡0(mod 6)n 3−n≡0(mod 6).
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answered May 13, 2015 at 15:15
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Carmichael's function λ(6)=lcm(ϕ(3),ϕ(2))=2 λ(6)=lcm(ϕ(3),ϕ(2))=2, and 6 6 is square-free, so n k λ(6)+c≡n c mod 6(c>0)n k λ(6)+c≡n c mod 6(c>0) . And in particular, n 3≡n mod 6 n 3≡n mod 6.
In the same way, n 13≡n mod 910 n 13≡n mod 910.
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edited May 13, 2015 at 20:22
answered May 13, 2015 at 19:46
JoffanJoffan
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5Prove that x 3≡x mod 6 x 3≡x mod 6 for all integers x x
0Prove that for any integer n, we have n 3=n in Z 6 n 3=n in Z 6
3(∀n∈Z):n 3≡n(∀n∈Z):n 3≡n (mod 6 6)
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0Is there a solution for i to the general equation (i+x)mod(n)(i+x)mod(n) = (k i+y)mod(n)(k i+y)mod(n) if k, x and y are given?
2We have a n−1≡1(m o d n)a n−1≡1(m o d n) but a m≢1(m o d n)a m≢1(m o d n) for every divisor m m of n−1 n−1, other than itself. Prove that n n is prime.
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11594 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.08%3A_Second-Order_Reactions/2.8.01%3A_Pseudo-1st-order_reactions | Skip to main content
2.8.1: Pseudo-1st-order Reactions
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: Jun 5, 2023
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2.8: Second-Order Reactions
2.9: Third Order Reactions
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( \newcommand{\kernel}{\mathrm{null}\,})
Under certain conditions, the 2nd order kinetics can be approximated as first-order kinetics. These pseudo-1st-order reactions greatly simplify quantifying the reaction dynamics.
Introduction
A 2nd-order reaction can be challenging to follow mostly because the two reactants involved must be measured simultaneously. There can be additional complications because certain amounts of each reactant are required to determine the reaction rate, for example, which can make the cost of one's experiment high if one or both of the needed reactants are expensive. To avoid more complicated, expensive experiments and calculations, we can use the pseudo-1st-order reaction, which involves treating a 2nd order reaction like a 1st order reaction.
In second-order reactions with two reactant species,
A+B⟶products
the rate of disappearance of A is
d[A]dt=−k[A][B]
as discussed previously (Case 2a), the integrated rate equation under the condition that [A] and [B] are not equal is
1[B]o−[A]oln[B][A]o[A][B]o=kt(2.8.1.1)
However, when [B]0≫[A]0, then [B]0≈[B] and Equation 2.8.1.1 becomes
1[B]0−[A]0ln[B][A]0[B]0[A]≈1[B]ln[A]0[A]=kt
or
[A]=[A]0e−[B]kt
This functional form of the decay kinetics is similar ot the first order kinetics and the system is said to operate under pseudo-first order kinetics. To reach a pseudo-1st-order reaction, we can manipulate the initial concentrations of the reactants. One of the reactants, B, for example, would have a significantly higher concentration, while the other reactant, A, would have a significantly lower concentration. We can then assume that the concentration of reactant B effectively remains constant during the reaction because its consumption is so small that the change in concentration becomes negligible. Because of this assumption, we can multiply the reaction rate, k, with the reactant with assumed constant concentration, B, to create a new rate constant (k′=k[B]) that will be used in the new rate equation,
Rate=k′[A]
as the new rate constant so we can treat the 2nd order reaction as a 1st order reaction.
Overloading
One can also do the same by overloading the initial concentration of A so that it effectively remains constant during the course of the reaction For example, if one were to dump a liter of 5 M HCl into a 55 M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physically compared to the HCl. Even if the amount of water was one liter this would still be the case because 55 M is relatively large compared to 5 M.
If we have an instance where there are more than two reactants involved in a reaction, all we would have to do to make the reaction pseudo-1st-order is to make the concentrations of all but one of the reactants very large. If there were three reactants, for example, we would make two of the three reactants be in excess (whether in amount or in concentration) and then monitor the dependency of the third reactant.
We can write the pseudost-order reaction equation as:
[A]=[A]0e−[B]0kt(2.8.1.2)
or
[A]=[A]0e−k′t
where
[A]o is the initial concentration of A,
[B]o is the initial concentration of B,
k′ is the pseudo-1st-order reaction rate constant,
k is the 2nd order reaction rate constant, and
[A] is the concentration of A at time t.
By using natural log to both sides of the pseudo-1st-order equation we get:
ln(AA0)=−k[B]0t
or
ln(AA0)=−k′t
Example 2.8.1.1
If a 2nd order reaction has the rate equation Rate=k[A][B], and the rate constant, k, is 3.67M−1s−1, [A] is 4.5M and [B] is 99M, what is the rate constant of its pseudo-1st-order reaction?
Solution
Because [B] is in excess we multiply 99M with 3.67M−1s−1
(99M)(3.67M−1s−1)=363.33s−1
Example 2.8.1.2
If [A] = 55 M at 39 s, [A]o = 99 M, and [B]o = 1000 M, what is the 2nd order reaction rate constant?
Solution
Use the Equation 2.8.1.2
(55 M) = (99 M)e-k'(1000M)(39s)
k' = 1.507 x 10-5 M-1s-1
Example 2.8.1.3
What is the concentration of A at time 45 s if [A]o = 1M, [B]o = 45 M, and 2nd order rate constant is 0.6 M-1s-1?
Solution
Use the Equation 2.8.1.2
[A] = (1)e-(0.6 M-1s-1)(45)
[A] = 1.88 x 10-12 M
Example 2.8.1.4
What is the rate of a reaction if [A]o = 560 M, [B]o = 0.2M, and 2nd order rate constant is 0.1 M-1s-1?
Solution
Use the equation R = k'[A][B]
R = (0.1M-1s-1)(560M)(0.2M)
Rate = 11.2Ms-1
Half-Life in a Pseudo-1st Order reaction
Half-life refers to the time required to decrease the concentration of a reactant by half, so we must solve for t. Here, [B] will be the reactant in excess, and its concentration will stay constant. [A]o is the initial concentration of A; thus the half-life concentration of A is 0.5[A]o.
The pseudo-1st-order reaction equation can be written as:
[A]=[A]oe−[B]ktor[A][A]o=e−k′t
By taking natural logs on both sides of the pseudo-1st-order equation, we get:
ln([A][A]o)=k′t
Because the concentration of A for a half-life t1/2 is 1/2[A]o :
ln(1/2[A]o[A]o)=ln(12)=−k′t1/2
Recalling that k′=k[B], [B]≈[B]0 and that −ln(1/2)=ln2:
ln(2)=k[B]0t1/2
or
t1/2=ln2k[B]0
Example 2.8.1.5
What is the half-life of a reaction with [A]o = 109 M, [B]o = 1 M, k' = 45 M-1s-1?
Solution
Because [A] is in excess we can multiply the k' with [A]o to find k
(109M)(45M-1s-1) = 4905s-1
t1/2 = (ln0.5 / -k)
t1/2 = 1.41 X 10-4s
References
Chang, Raymond. Physical Chemistry for the Biosciences. Sausalito, California: University Science Books, 2005. Print.
Garrett R., Grisham C. Biochemistry 3rd Edition. California: Thomson Learning, Inc., 2005. Print.
Hikata, H., Asai, S., Ishikawa, H., Honda, M. "The Kinetics of Reactions of Carbon Dioxide with Monoethanolamine, Diethanolamine, and Triethanolamine by a Rapid Mixing Method." The Chemical Engineering Journal 13 (1977): 7-12. Web 13 Mar. 2011.
Ho, Yuh-Shan. "Citation Review of Lagergren Kinetic Rate Equation on Adsorption Reactions." Scientometrics 59.1 (2004): 171 - 177. Web. 12 Mar. 2011.
Naumann, Wolfgang, Shokhirev, Nikolai V., Szabo, Attila. "Exact Asymptotic Relaxation of Pseudo-First-Order Reversible Reactions." Physical Review Letters 79.16 (1997): 3074 - 3077. Web. 12 Mar. 2011.
Schnell, S., Mendoza, C. "The Condition for Pseudo-First-Order Kinetics in Enzymatic Reactions is Independent of the Initial Enzyme Concentration." Biophysical Chemistry 107 (2004): 165 - 174. Web. 14 Mar. 2011.
Segal, Irwin. Enzyme Kinetics. John Wiley & Sons, Inc., 1975. Print.
2.8: Second-Order Reactions
2.9: Third Order Reactions |
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Insulin is the major hormone controlling critical energy functions such as glucose and lipid metabolism. Insulin activates the insulin receptor tyrosine kinase (IR), which phosphorylates and recruits different substrate adaptors such as the IRS family of proteins. Tyrosine phosphorylated IRS then displays binding sites for numerous signaling partners. Among them, PI3K has a major role in insulin function, mainly via the activation of the Akt/PKB and the PKCζ cascades. Activated Akt induces glycogen synthesis through inhibition of GSK-3; protein synthesis via mTOR and downstream elements; and cell survival through inhibition of several pro-apoptotic agents (Bad, FoxO transcription factors, GSK-3, and MST1). Akt phosphorylates and directly inhibits FoxO transcription factors, which also regulate metabolism and autophagy. Inversely, AMPK is known to directly regulate FoxO3 and activate transcriptional activity. Insulin signaling also has growth and mitogenic effects, which are mostly mediated by the Akt cascade as well as by activation of the Ras/MAPK pathway. The insulin signaling pathway inhibits autophagy via the ULK1 kinase, which is inhibited by Akt and mTORC1, and activated by AMPK. Insulin stimulates glucose uptake in muscle and adipocytes via translocation of GLUT4 vesicles to the plasma membrane. GLUT4 translocation involves the PI3K/Akt pathway and IR-mediated phosphorylation of CAP, and formation of the CAP:CBL:CRKII complex. In addition, insulin signaling inhibits gluconeogenesis in the liver, through disruption of CREB/CBP/mTORC2 binding. Insulin signaling induces fatty acid and cholesterol synthesis via the regulation of SREBP transcription factors. Insulin signaling also promotes fatty acid synthesis through activation of USF1 and LXR. A negative feedback signal emanating from Akt/PKB, PKCζ, p70 S6K, and the MAPK cascades results in serine phosphorylation and inactivation of IRS signaling.
Selected Reviews:
Altarejos JY, Montminy M (2011)CREB and the CRTC co-activators: sensors for hormonal and metabolic signals.Nat. Rev. Mol. Cell Biol.12(3), 141–51.
Cheng Z, Tseng Y, White MF (2010)Insulin signaling meets mitochondria in metabolism.Trends Endocrinol. Metab.21(10), 589–98.
Fritsche L, Weigert C, Häring HU, Lehmann R (2008)How insulin receptor substrate proteins regulate the metabolic capacity of the liver--implications for health and disease.Curr. Med. Chem.15(13), 1316–29.
Guo S (2014)Insulin signaling, resistance, and the metabolic syndrome: insights from mouse models into disease mechanisms.J. Endocrinol.220(2), T1–T23.
Rowland AF, Fazakerley DJ, James DE (2011)Mapping insulin/GLUT4 circuitry.Traffic 12(6), 672–81.
Siddle K (2011)Signalling by insulin and IGF receptors: supporting acts and new players.J. Mol. Endocrinol.47(1), R1–10.
Shao W, Espenshade PJ (2012)Expanding roles for SREBP in metabolism.Cell Metab.16(4), 414–9.
Wong RH, Sul HS (2010)Insulin signaling in fatty acid and fat synthesis: a transcriptional perspective.Curr Opin Pharmacol 10(6), 684–91.
We would like to thank Ashley Webb and Prof. Anne Brunet Stanford University, Sanford, CA for reviewing this diagram.
created June 2003
revised September 2016
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3.4: Double and Half Angle Identities
Unit 3: Trigonometric Identities
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3.4.2: Double Angle Identities
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3.4.1: Double and Half Angle Formulas
3.4.3: Simplifying Trigonometric Expressions with Double-Angle Identities
Page ID 4222
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Table of contents
1. Double Angle, Half Angle, and Power Reducing Identities
1. Double Angle Identities
Half Angle Identities
Power Reducing Identities
Example (\PageIndex{1})
Example (\PageIndex{2})
Example (\PageIndex{3})
Example (\PageIndex{4})
Example (\PageIndex{5})
Review
Review (Answers)
Vocabulary
Additional Resources
Interactive Element
Simplifying trigonometric functions with twice a given angle.
These identities are significantly more involved and less intuitive than previous identities. By practicing and working with these advanced identities, your toolbox and fluency substituting and proving on your own will increase. Each identity in this concept is named aptly. Double angles work on finding (\sin 80^{\circ} ) if you already know (\sin 40^{\circ} ). Half angles allow you to find (\sin 15^{\circ} ) if you already know (\sin 30^{\circ} ). Power reducing identities allow you to find (\sin ^2 15^{\circ} ) if you know the sine and cosine of (30^{\circ} ).
What is (\sin ^2 15^{\circ} )?
Double Angle, Half Angle, and Power Reducing Identities
Double Angle Identities
The double angle identities are proved by applying the sum and difference identities. They are left as review problems. These are the double angle identities.
(\sin 2x=2\sin x\cos x)
(\cos 2x=\cos ^2x−\sin ^2x)
(\tan 2x=\dfrac{2\tan x}{1−\tan ^2x})
Half Angle Identities
The half angle identities are a rewritten version of the power reducing identities. The proofs are left as review problems.
(\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1−\cos x}{2}})
(\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1+\cos x}{2}})
(\tan \dfrac{x}{2}=\pm \sqrt{\dfrac{1−\cos x}{1+\cos x}})
Power Reducing Identities
The power reducing identities allow you to write a trigonometric function that is squared in terms of smaller powers. The proofs are left as examples and review problems.
(\sin ^2x=\dfrac{1−\cos 2x}{2})
(\cos ^2x=\dfrac{1+\cos 2x}{2})
(\tan ^2x=\dfrac{1−\cos 2x}{1+\cos 2x})
Power reducing identities are most useful when you are asked to rewrite expressions such as \sin 4x as an expression without powers greater than one. While (\sin x\cdot \sin x\cdot \sin x\cdot \sin x) does technically simplify this expression as necessary, you should try to get the terms to sum together not multiply together.
(\begin{aligned} \sin ^4x=(\sin ^2x)^2 &=\left(\dfrac{1−\cos 2x}{2} \right)^2 \&=\dfrac{1−2\cos 2x+\cos ^2 2x}{4} \&= \dfrac{1}{4}\left(1−2\cos 2x+\dfrac{1+\cos 4x}{2}\right)\end{aligned})
Example (\PageIndex{1})
Earlier, you were asked to find (\sin ^2 15^{\circ} ). In order to fully identify (\sin ^2 15^{\circ} ) you need to use the power reducing formula.
Solution
(\begin{aligned} \sin ^2x&=\dfrac{1−\cos 2x}{2}=\dfrac{1}{2}−\dfrac{\sqrt{3}}{4} \ \sin ^2 15^{\circ} &=\dfrac{1−\cos 30^{\circ}}{2} \ &=\dfrac{2−\sqrt{3}}{4} \end{aligned})
Example (\PageIndex{2})
Write the following expression with only (\sin x) and (\cos x ): (\sin 2x+\cos 3x).
Solution
(\begin{aligned} \sin 2x+\cos 3x &=2\sin x\cos x+\cos (2x+x) \&=2\sin x\cos x+\cos 2x\cos x−\sin 2x\sin x \&=2\sin x\cos x+(\cos ^2x−\sin ^2x)\cos x−(2\sin x\cos x)\sin x \ &=2\sin x\cos x+\cos ^3x−\sin ^2x\cos x−2\sin ^2x\cos x \ &=2\sin x\cos x+\cos ^3x−3\sin ^2x\cos x \end{aligned})
Example (\PageIndex{3})
Use half angles to find an exact value of \tan 22.5^{\circ} without using a calculator.
Solution
(\tan \dfrac{x}{2}=\pm \sqrt{\dfrac{1−\cos x}{1+\cos x}})
(\begin{aligned}
\tan 22.5^{\circ} &=\tan \dfrac{45^{\circ}}{2}=\pm \sqrt{\dfrac{1-\cos 45^{\circ}}{1+\cos 45^{\circ}}}=\pm \sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{1+\dfrac{\sqrt{2}}{2}}}=\pm \sqrt{\dfrac{\dfrac{2}{2}-\dfrac{\sqrt{2}}{2}}{\dfrac{2}{2}+\dfrac{\sqrt{2}}{2}}}=\pm \sqrt{\dfrac{2-\sqrt{2}}{2+\sqrt{2}}} \
&=\pm \sqrt{\dfrac{(2-\sqrt{2})^{2}}{2}}
\end{aligned})
Example (\PageIndex{4})
Prove the power reducing identity for sine.
(\sin ^2x=\dfrac{1−\cos 2x}{2})
Solution
Using the double angle identity for cosine:
(\begin{aligned} \cos 2x&=\cos ^2x−\sin ^2x \ \cos 2x&=(1−\sin ^2x)−\sin ^2x \ \cos 2x &=1−2\sin ^2x \end{aligned})
This expression is an equivalent expression to the double angle identity and is often considered an alternate form.
Example (\PageIndex{5})
Simplify the following identity: (\sin ^4x−\cos ^4x).
Solution
Here are the steps:
(\begin{aligned}\sin ^4x−\cos ^4x&=(\sin ^2x−\cos ^2x)(\sin ^2x+\cos ^2x)\&=−(\cos ^2x−\sin ^2x)\&=−\cos ^2x \end{aligned})
Review
Prove the following identities.
(\sin 2x=2\sin x\cos x)
(\cos 2x=\cos^2 x−\sin^2 x)
(\tan 2x=\dfrac{2\tan x}{1−\tan^2 x})
(\cos^2 x=\dfrac{1+\cos 2x}{2})
(\tan^2 x=\dfrac{1−\cos 2x}{1+\cos 2x})
(\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1−\cos x}{2}})
(\cos \dfrac{x}{2}=\pm \sqrt{\dfrac{1+\cos x}{2}})
(\tan \dfrac{x}{2}=\pm \sqrt{\dfrac{1−\cos x}{1+\cos x}})
(\csc 2x=\dfrac{1}{2} \csc x\sec x)
(\cot 2x=\dfrac{\cot ^2 x−1}{2\cot x})
Find the value of each expression using half angle identities.
(\tan 15^{\circ})
(\tan 22.5^{\circ})
(\sec 22.5^{\circ})
Show that (\tan \dfrac{x}{2}=\dfrac{1−\cos x}{\sin x}).
Using your knowledge from the answer to question 14, show that (\tan \dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}).
Review (Answers)
To see the Review answers, open this PDF file and look for section 6.4.
Vocabulary
| Term | Definition |
--- |
| Half Angle Identity | A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument. |
| identity | An identity is a mathematical sentence involving the symbol “=” that is always true for variables within the domains of the expressions on either side. |
| power reducing identity | A power reducing identity relates the power of a trigonometric function containing a given argument to a set of trigonometric functions, each containing the original argument. |
Additional Resources
Interactive Element
Video: Multiple Angle Formulas - Overview
This page titled 3.4.2: Double Angle Identities is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform.
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3.4.4: Solving Equations with Double-Angle Identities
3.4.5: Half Angle FormulasDerivation of sine and cosine formulas for half a given angle
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11598 | https://www.nature.com/articles/s41467-019-11433-0 | Multiple origins of prokaryotic and eukaryotic single-stranded DNA viruses from bacterial and archaeal plasmids | Nature Communications
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article
Multiple origins of prokaryotic and eukaryotic single-stranded DNA viruses from bacterial and archaeal plasmids
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Published: 31 July 2019
Multiple origins of prokaryotic and eukaryotic single-stranded DNA viruses from bacterial and archaeal plasmids
Darius KazlauskasORCID: orcid.org/0000-0001-6135-85491,
Arvind VarsaniORCID: orcid.org/0000-0003-4111-24152,3,
Eugene V. KooninORCID: orcid.org/0000-0003-3943-82994&
…
Mart KrupovicORCID: orcid.org/0000-0001-5486-00985
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Nature Communicationsvolume 10, Article number:3425 (2019) Cite this article
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Abstract
Single-stranded (ss) DNA viruses are a major component of the earth virome. In particular, the circular, Rep-encoding ssDNA (CRESS-DNA) viruses show high diversity and abundance in various habitats. By combining sequence similarity network and phylogenetic analyses of the replication proteins (Rep) belonging to the HUH endonuclease superfamily, we show that the replication machinery of the CRESS-DNA viruses evolved, on three independent occasions, from the Reps of bacterial rolling circle-replicating plasmids. The CRESS-DNA viruses emerged via recombination between such plasmids and cDNA copies of capsid genes of eukaryotic positive-sense RNA viruses. Similarly, the rep genes of prokaryotic DNA viruses appear to have evolved from HUH endonuclease genes of various bacterial and archaeal plasmids. Our findings also suggest that eukaryotic polyomaviruses and papillomaviruses with dsDNA genomes have evolved via parvoviruses from CRESS-DNA viruses. Collectively, our results shed light on the complex evolutionary history of a major class of viruses revealing its polyphyletic origins.
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Introduction
Viruses with single-stranded (ss)DNA genomes represent a vast, highly diverse supergroup of medically, ecologically, and economically important pathogens infecting hosts from all three domains of cellular life1,2. Although for years, ssDNA viruses have been thought to be relatively rare in the biosphere, recent metagenomics studies have increasingly revealed high abundance of these viruses in diverse environments3,4,5,6,7,8,9,10,11,12,13,14,15."). Currently, ssDNA viruses are classified into 13 families, 9 of which include (presumably) eukaryotic viruses, but many uncultivated ssDNA viruses remain unclassified. The majority of ssDNA viruses (9 families) have small circular genomes, which are known or predicted to be replicated by the rolling-circle mechanism. This mechanism of replication is initiated by the virus-encoded Rep protein of the HUH endonuclease superfamily, characterized by the signature HUH motif, in which two histidine residues are separated by a bulky hydrophobic residue2."),16."),17."),18."),19."). Informally, these viruses are often collectively referred to as circular, Rep-encoding ssDNA (CRESS-DNA) viruses1 viruses: ubiquitous viruses with small genomes and a diverse host range. Adv. Virus Res 103, 71–133 (2019)."),17."). A variation on this theme is employed by members of the _Parvoviridae_ family which have linear ssDNA genomes replicated by the rolling-hairpin mechanism initiated by a Rep protein homologous to those of the CRESS-DNA viruses18."),20."). Members of the family _Bidnaviridae_ apparently have evolved from parvoviruses by replacing the HUH endonuclease domain with the DNA polymerase from polintoviruses21.").
The Rep proteins of ssDNA viruses of prokaryotes (bacteria and archaea) and eukaryotes display distinct domain organizations2. In eukaryotic CRESS-DNA viruses, the endonuclease domain is fused to a superfamily 3 helicase (S3H) domain22, which is responsible for unwinding of the double-stranded (ds)DNA replicative intermediate and, in some viruses, packaging of the viral genome into assembled empty capsids20,23. By contrast, none of the bacterial or archaeal ssDNA viruses isolated to date encodes a Rep fused to a helicase domain2. Instead, these viruses recruit a cellular helicase for the same function24. Such dichotomy in the domain organization of the Rep proteins raises questions regarding the evolutionary relationship between ssDNA viruses infecting hosts from different cellular domains25,26. Furthermore, HUH Reps are not restricted to ssDNA viruses, but are also functional in several groups of bacterial and archaeal dsDNA viruses, including certain members of the families Sphaerolipoviridae, Rudiviridae, Corticoviridae, and Myoviridae. However, the implications of the presence of this gene for potential evolutionary links between these dsDNA viruses and ssDNA viruses remain unclear.
The HUH endonucleases are also encoded by diverse bacterial and archaeal as well as several eukaryotic plasmids and transposons, some of which have been shown experimentally to replicate and/or transpose via the rolling-circle mechanism27,28,29,30. The homology between the endonuclease domains of the viral and bacterial plasmid Reps has been initially inferred from the conservation of 3 signature motifs18,19, and subsequently validated by structural analyses16. Motif I, UUTU (U denotes hydrophobic residues), is thought to be involved in the recognition of the origin of replication. Motif II, HUH, is involved in the coordination of divalent metal ions, Mg 2+ or Mn 2+, which are essential for endonuclease activity at the origin of replication18,31. Motif III (YxxK/YxxKY, where x is any amino acid) is involved in dsDNA cleavage and subsequent covalent attachment of the Rep through the catalytic tyrosine residue to the 5’ end of the cleaved product16,18,32. The HUH endonucleases encoded by prokaryotic plasmids, viruses and transposons can have either two or one catalytic tyrosine residue in the motif III, whereas all known eukaryotic Rep-encoding viruses contain a single tyrosine residue18. Notably, whereas most prokaryotic Reps consist of stand-alone endonuclease domains, some bacterial plasmids encode Reps with the domain organization similar to that characteristic of eukaryotic ssDNA viruses, that is, a nuclease-helicase fusion. For example, it has been shown that Reps encoded by plasmids of phytoplasma, plant-pathogenic bacteria, show the highest sequence similarity to Reps of plant-infecting geminiviruses33,34. However, whether the similar domain organization is a result of convergent evolution or whether it alludes to a more recent common ancestry of the corresponding replicons remained unclear.
Here, we systematically explore the relationships among Rep-encoding DNA viruses, plasmids, and transposons from all three cellular domains. We identify 8 previously undescribed families of integrative plasmids that are widespread across different bacterial phyla and show that they have seeded the eukaryotic CRESS-DNA virosphere on at least 3 independent occasions. Similarly, the origins of bacterial and archaeal ssDNA viruses replicating by the rolling-circle mechanism can be traced to different families of prokaryotic plasmids, emphasizing tight evolutionary connections between viruses and capsid-less mobile genetic elements (MGE).
Results
Global network of the HUH replicons
To explore the evolutionary history of the HUH replicons, we collected a dataset of HUH endonucleases—the only protein encoded by all these replicons—representing each family of viruses, plasmids, and transposons associated with hosts across all three cellular domains16,27,28,29,30. In this analysis, we did not consider Mob relaxases involved in plasmid conjugation. Enzymes in this family encompass circularly permuted conserved motifs which complicate their sequence-based comparison with the HUH endonucleases involved in DNA replication or transposition16,19. The resulting dataset included 8764 sequences. These were grouped based on pairwise similarity, and clusters were identified using a convex clustering algorithm (p-value threshold of 1e−08) with CLANS35. This analysis revealed 33 clusters which varied in size from 7 to 2711 sequences (Supplementary data1). Following an inspection of the connectivity between clusters (Fig.1), we defined 2 orphan clusters and 2 superclusters, which displayed either no or very few connections to each other (Supplementary data1). Nevertheless, comparison of the available high-resolution structures for representatives of both orphan clusters and the 2 superclusters16,36 unequivocally confirm their common origin.
Fig. 1
Representative HUH superfamily Reps clustered by their pairwise sequence similarity. Lines connect sequences with P-value ≤ 1e−08. Groups were named after well-characterized plasmids, viruses or most frequent taxon
Full size image
Orphan cluster 1 includes a single family of IS 200/IS 605 transposons which are widespread in bacteria and archaea37. The HUH endonucleases of the IS 200/IS 605 insertion sequences have been extensively studied structurally and biochemically, resulting in a comprehensive understanding of their functions16,38. Although IS 200/IS 605 transposases have a structural fold common to that of other HUH endonucleases and contain all 3 signature motifs, they did not show appreciable sequence similarity to any other cluster of HUH endonucleases and thus remained disconnected from sequences in other clusters. Nevertheless, sequence diversity within the IS 200/IS 605 cluster is comparable to that within other clusters.
Orphan cluster 2 includes Rep proteins that are conserved in hyperthermophilic archaeal viruses of the family Rudiviridae39. Structural studies of the Rep protein from the rudivirus SIRV1 revealed the canonical HUH endonuclease fold and biochemical characterization of the protein confirmed the expected nicking and joining activities in vitro36. Like the IS 200/IS 605 transposases, the rudiviral Rep cluster does not connect to other HUH endonucleases, including homologs from other families of archaeal viruses and plasmids.
Conceivably, the uniqueness of the 2 orphan clusters is linked to the unusual transposition and replication mechanisms employed by the respective elements. Indeed, IS 200/IS 605 insertion sequences transpose by a unique peel-and-paste mechanism38, whereas rudiviruses, unlike most other viruses and plasmids replicating by the rolling-circle mechanism, contain relatively large (~35 kb) linear dsDNA genomes with covalently closed termini40.
Supercluster 1 is by far the largest and most diverse HUH assemblage that includes 24 clusters (Supplementary data1). Of these 24 clusters, 15 contain Reps from bona fide extrachromosomal plasmids of which 7 clusters also include Reps from diverse ssDNA (Microviridae, Inoviridae, and Pleolipoviridae) and/or dsDNA (Myoviridae and Corticoviridae) viruses of bacteria and archaea. Three clusters consist of Reps encoded by microviruses of the subfamilies Gokushovirinae and Bullavirinae, and Xanthomonas inovirus Cf1 (family Inoviridae), respectively. Notably, phiX174-like microviruses (Bullavirinae) display similarity exclusively to microviruses of the subfamily Gokushovirinae, indicative of the Rep monophyly in the two subfamilies of the Microviridae, despite high sequence divergence. The bacterial IS 91 (including IS CR subfamily) and eukaryotic Helitron family transposons, respectively, form two distinct clusters. The two groups of transposons are not directly connected to each other, but are linked to distinct groups of bacterial and, in the case of IS 91, archaeal plasmids, suggesting independent origins from bacterial extrachromosomal replicons. It has been previously suggested that helitrons might represent a missing link between eukaryotic CRESS-DNA viruses, namely, geminiviruses, and bacterial HUH replicons41 or that helitrons evolved from geminiviruses42. However, in our analysis, helitrons do not connect to any of the groups of CRESS-DNA viruses, suggesting independent evolutionary trajectories, consistent with the recent findings43.
The remaining 5 clusters do not include any recognizable plasmid, viral or transposon sequences and thus are likely to represent new families of integrated MGE. Four of these groups are predominantly found in bacteria of the taxa Clostridiales, Actinobacteria, Neisseriales, and Bacteroidetes, respectively (labeled accordingly in Fig.1), whereas the fifth group is specific to the candidate division MSBL1 (Mediterranean Sea Brine Lakes 1)44, a group of uncultured archaea found in different hypersaline environments. Most of the clusters display taxonomic uniformity at the domain level, i.e., clusters included either bacterial, or archaeal, or eukaryotic sequences (including the corresponding viruses and plasmids), suggesting that horizontal transfers of viruses or plasmids between host domains are infrequent. The two exceptions include the pUB110-like and IS 91-like bacteria-dominated clusters, which include a handful of archaeal sequences. In the case of IS 91 transposons, horizontal transfer from bacteria has been ascertained by phylogenetic analyses45. In addition, some of the clusters include sporadic sequences annotated as being eukaryotic; however, analysis of the corresponding contigs suggests that these are likely bacterial contaminants.
Of particular interest are the 7 clusters that include both viruses and plasmids. For instance, pEC316_KPC-like cluster, besides plasmids, contains evolutionarily-unrelated viruses from 3 families, Myoviridae, Corticoviridae, and Inoviridae, suggesting extensive horizontal spread of the rep genes. Notably, Reps of inoviruses are distributed among 5 clusters. Given the scarcity of inoviral sequences in the pVT736-1-like and pUB110-like clusters, which include only Pseudomonas phage Pf3 and Propionibacterium phage B5, respectively, the directionality of gene transfer, from plasmids to the corresponding viruses, appears obvious. Furthermore, many inoviruses do not encode HUH endonucleases, but rather encode replication initiators of an evolutionarily unrelated superfamily, Rep_trans (Pfam id: PF02486)15."), which also abounds in bacterial plasmids30."), whereas inoviruses of the genus _Vespertiliovirus_ lack Reps and instead replicate by transposition using IS _3_ and IS _30_ family transposases derived from the corresponding insertion sequences46."). Collectively, these observations indicate that the replication modules of inoviruses have been exchanged with distantly related and even non-homologous replication modules from various plasmid and transposon families. Similarly, archaeal pleolipoviruses are split between two clusters corresponding to different families of archaeal plasmids, pGRB1-like and pTP2-like, respectively, suggesting that exchange of replication-associated genes is common in bacterial and archaeal viruses with small, plasmid-sized genomes. ¶In some cases, it is difficult to ascertain the viral versus plasmid membership of Reps encoded in cellular chromosomes because both types of MGE can integrate into the host genomes. For example, the XacF1-like cluster includes 62 Rep sequences, 2 of which are encoded by filamentous phages, whereas the rest come from bacterial genomes. Analysis of the genomic neighborhoods suggests that only 6 of the remaining 60 Reps represent prophages. Furthermore, the pAS28-like cluster includes one plasmid, pAS28 (ref. 47.")); however, related Reps have been previously identified in prophages48."), but not in characterized viruses, giving the erroneous impression that the pAS28-like Rep is plasmid-exclusive. ¶To further characterize the evolutionary relationships between Reps encoded by different types of MGE, we constructed maximum likelihood phylogenetic trees for the 7 clusters that included Reps from both viruses and plasmids (Supplementary Fig.2a-g). The results of phylogenetic analyses suggest horizontal transfer of the _rep_ genes between plasmids and viruses, with viral sequences typically being nested among plasmid-encoded homologs.
Supercluster 2 (SC2) consists of 7 clusters (Supplementary data1) which include all known classified and unclassified eukaryotic CRESS-DNA viruses, parvoviruses, a cluster of plasmids from the red alga Pyropia pulchra49, and 4 clusters containing bacterial Rep sequences. The vast majority of the bacterial Reps in the pCPa-like and p4M-like clusters are encoded in bacterial genomes rather than in plasmids and have not been previously characterized. In our network, the CRESS-DNA viruses are connected to pCPa-like, p4M-like, pPAPh2-like and P. pulchra-like clusters, whereas the pE194/pMV158-like cluster does not form direct connections to the CRESS-DNA viruses, but joins SC2 through the pCPa-like cluster (Fig.1). Notably, geminiviruses and genomoviruses form a subcluster with plasmids of phytoplasma (pPAPh2-like cluster) and P. pulchra, which is separated from other CRESS-DNA viruses. The Parvoviridae cluster, including parvoviruses and derived endogenous viruses integrated in various eukaryotic genomes, is loosely connected directly to the CRESS-DNA viruses, suggesting that parvoviruses with linear ssDNA genomes share common ancestry with CRESS-DNA viruses which, by definition, have circular genomes. Intrigued by the seemingly close evolutionary connection between eukaryotic CRESS-DNA viruses and bacterial and algal Reps, we investigated these relationships in greater detail, as reported in the following sections.
The diversity of viral-like Reps in bacterial genomes
To investigate the extent of similarity between the Reps of eukaryotic CRESS-DNA viruses and non-viral replicons from SC2, we compared their domain organizations. With the exception of pE194/pMV158-family plasmids, which contain only the nuclease domain, bacterial and algal SC2 Reps had the same nuclease-helicase domain organization as CRESS-DNA viruses. The same two-domain organization is also characteristic of the parvovirus Reps2. Thus, domain organization analysis corroborates the results of sequence clustering and further indicates that the bacterial SC2 Reps are more closely related to the Reps of eukaryotic viruses than to those from other prokaryotic plasmids and viruses.
We then sought to obtain additional information on the diversity and taxonomic distribution of the viral-like SC2 Reps that are encoded in bacterial genomes. Maximum likelihood phylogenetic analysis revealed 9 well-supported clades (Fig.2a). Clustering and subsequent community detection analysis validated the 9 groups of bacterial Reps (Fig.2b), where groups 1–3 correspond to the p4M-like cluster shown in Fig.1, groups 4–8 to the pCPa-like cluster, and group 9 to the pPAPh2-like cluster. To emphasize their similarity to Reps of CRESS-DNA viruses, we refer to the 9 groups as pCRESS1 through pCRESS9. These groups displayed partially overlapping but distinct taxonomic distributions, covering several classes within 4 bacterial phyla (Supplementary Fig.1 and Supplementary Table1).
Fig. 2
Diversity of viral-like Rep proteins in bacteria. a Phylogenetic tree of bacterial Rep proteins and their homologs in P. pulchra. Closely related sequences are collapsed to triangles, whose side lengths are proportional to the distances between closest and farthest leaf nodes. b CLANS groups of bacterial Rep proteins and their homologs. Nodes indicate protein sequences. Lines represent sequence relationships (CLANS P-value ≤ 1e−05). The nodes belonging to the same cluster are colored with the same colors, corresponding to the clades shown in panel A. c Genome maps of integrated and extrachromosomal plasmids representing groups 1–9. Homologous genes are depicted using the same color and their functions are listed on the right side of the figure
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The majority of the Reps from pCRESS7 and pCRESS9 are encoded by extrachromosomal plasmids (Supplementary Table1). By contrast, the vast majority (97.5%) of Reps found in other groups are encoded within mobile genetic elements site-specifically integrated into bacterial chromosomes (Supplementary Table1; Fig.2c; Supplementary Fig.3; Supplementary Note1). Notably, none of the elements encoded any homologs of currently known viral structural proteins (Supplementary Note1). Collectively, these observations indicate that viral-like Reps in bacteria are encoded by diverse extrachromosomal and integrated plasmids.
Conserved features of bacterial and CRESS-DNA virus Reps
Sequence analysis showed that, despite considerable overall sequence divergence, Reps of pCRESS4 through 8 contain closely similar sequence motifs within the nuclease and helicase domains (Fig.3), consistent with the results of the clustering and phylogenetic analyses (Fig.2). In particular, these 5 pCRESS groups share a specific signature, YLxH (x, any amino acid) within motif III of the nuclease domain, which was not observed in Reps from pCRESS1–3 and 9 (Fig.3). Thus, we refer to pCRESS4–8 collectively as the YLxH supergroup (rather than the pCPa-like cluster), to emphasize this shared feature. The YLxH signature was also conserved in Reps from the pE194/pMV158-like cluster, suggesting a closer evolutionary relationship between the two clusters, despite the fact that pE194/pMV158-like Reps lack the helicase domain. Also, pCRESS9 displays motifs similar to those of the P. pulchra plasmids and thus could be unified with these plasmids into a common assemblage. By contrast, pCRESS1, -2 and -3 (p4M-like cluster) display distinctive sets of motifs (Fig.3; Supplementary Note1).
Fig. 3
Conserved sequence motifs of Rep proteins. Bacterial Rep groups are depicted in gray background. Residues are colored by their chemical properties (polar, green; basic, blue; acidic, red; hydrophobic, black; neutral, purple). The Rep groups were manually ordered according to the pairwise similarity in the aligned motifs. The HUH endonuclease and SF3 helicase domains are delineated at the top of the figure
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Origin of the SF3 helicase domain
Sequence analyses suggest that the SF3 helicase domain-containing plasmid Reps, especially those from pCRESS2, pCRESS3, and pCRESS9, and P. pulchra, are closely related to the Reps of CRESS-DNA viruses. However, the directionality of evolution, i.e., whether plasmid Reps evolved from those of CRESS-DNA viruses or vice versa, is not obvious. Although it is tempting to take the absence of the helicase domain in the pE194/pMV158-like cluster as an indication that this group is ancestral to the helicase-containing Reps, it cannot be ruled out that the helicase domain was lost by these plasmids. Thus, we set out to investigate the provenance of the SF3 helicase domain in the plasmid and viral Reps. Sensitive sequence searches with HMMER against the nr30 database showed that the helicase domains of plasmid and CRESS-DNA viral Reps are most closely related to those of eukaryotic positive-sense RNA viruses (order Picornavirales and family Caliciviridae) as well as the AAA+ ATPase superfamily50,51. In this analysis, we also included the SF3 sequences of parvoviruses, polyomaviruses, and papillomaviruses that are thought to be evolutionarily related to CRESS-DNA viruses2,25. Several groups of more distant SF3 helicases from viruses with large dsDNA genomes52 were disregarded. Due to the high sequence divergence and relatively short length, phylogenetic analyses of the SF3 helicase domains were not informative, resulting in star-shaped tree topologies, irrespective of the evolutionary models or taxonomic sampling used. However, clustering analysis based on pairwise similarities provided insights into the relationships between the different ATPase families (Fig.4a). In particular, the close relationship between the SF3 helicase domains of bacterial Reps and CRESS-DNA viruses was clearly supported. Both groups connect to the RNA viruses, but only bacterial Reps, particularly those of the YLxH supergroup, show connections to AAA+ superfamily ATPases, namely, bacterial helicase loader DnaC and, to a lesser extent, DnaA and Cdc48-like ATPases (Fig.4a). The closer similarity between the YLxH supergroup and bacterial AAA+ ATPases is supported by comparison of the catalytic motifs which revealed several shared derived characters, to the exclusion of other groups (Supplementary Fig.4). At the same clustering threshold, neither eukaryotic DNA nor RNA viruses linked to any group of ATPases other than those from bacterial plasmids. The SF3 helicases of parvoviruses linked to those of CRESS-DNA viruses, consistent with the analysis of full-length Rep sequences (Fig.1). Papillomaviruses and polyomaviruses formed 2 clusters which connected to each other and to parvoviruses.
Fig. 4
Relationships between Superfamily 3 helicases and AAA+ ATPases. a Superfamily 3 helicase and AAA+ ATPase domains clustered by their pairwise similarity using CLANS. In total, 3854 sequences were clustered with CLANS (CLANS P-value ≤ 5e−09). Groups of unclassified CRESS-DNA viruses are referred to as CRESSV1 through CRESSV6 (ref. 53). b A proposed evolutionary scenario for the origin and evolution of viral Superfamily 3 helicases. Abbreviations: SF3, superfamily 3 helicase domain; HUH, HUH superfamily nuclease domain; OBD, origin-binding domain; HGT, horizontal gene transfer; RHR, rolling-hairpin replication
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This pattern of connectivity suggests a specific vector of evolution and appears to be best compatible with the following scenario. The SF3 helicase domain of bacterial plasmids evolved from a bacterial DnaC-like ATPase; this helicase domain was appended to the nuclease domain of Reps of pE194/pMV158-like plasmids yielding the ancestor of the YLxH supergroup; bacterial plasmid Reps were passed on to the CRESS-DNA viruses; the SF3 helicase of RNA viruses was horizontally acquired either from bacterial plasmids or, more likely, from eukaryotic CRESS-DNA viruses; CRESS-DNA viruses have spawned parvoviruses which in turn gave rise to polyomaviruses and papillomaviruses (Fig.4b). The alternative scenario, under which SF3 helicases of eukaryotic RNA viruses gave rise to the universal bacterial DnaC and DnaA proteins, through bacterial plasmids, appears non-parsimonious and extremely unlikely. Indeed, DnaA is ubiquitous and essential in bacteria50,51, so the capture of the helicase from a plasmid would have to occur at the very origin of the bacterial domain of life. Notably, pCRESS9 and P. pulchra plasmids are not linked with other plasmids but are rather connected to the rest of the sequences through the CRESS-DNA viruses. The latter pattern has been also observed in the global clustering analysis of the HUH Reps (Fig.1) as well as in the clustering of the nuclease domains alone.
Origins of CRESS-DNA viruses from bacterial plasmids
Analysis of the SF3 helicase domains suggests that Reps of pE194/pMV158-like plasmids are ancestral rather than derived forms. The alternative possibility, namely, that Reps of pE194/pMV158-like plasmids have lost the helicase domain, cannot be currently ruled out. However, the fact that the helicase domain has not been lost in any of the numerous known groups of CRESS-DNA viruses or in pCRESS1 to pCRESS9 plasmids, suggests that, once acquired, the helicase domain becomes important for efficient plasmid/viral genome replication. Thus, the close similarity between the pE194/pMV158-like Reps and those of the YLxH supergroup, resulting in direct connectivity of the two groups in the global network (Fig.1), implies that the former group is an adequate outgroup for the phylogeny of Reps from bacterial plasmids and CRESS-DNA viruses. For phylogenetic analyses, we used a dataset of SC2 Reps, excluding Reps of Parvoviridae and CRESS-DNA viruses which were previously judged to be chimeric with respect to their nuclease and helicase domains53, to avoid potential artifacts resulting from conflicting phylogenetic signals. The dataset included representatives of all classified families of CRESS-DNA viruses as well as 6 groups of unclassified CRESS-DNA viruses provisionally labeled CRESSV1–6 (ref. 53) as well as a small group of GasCSV-like viruses, which have been previously noticed to encode Reps with significant similarity to bacterial Reps54. In the well-supported maximum likelihood phylogenetic tree constructed with PhyML and rooted with pE194/pMV158-like Reps, the YLxH supergroup (pCRESS4–8) is at the base of an assemblage that includes all CRESS-DNA viruses, pCRESS1–3 and pCRESS9 as well as P. pulchra plasmids. This assemblage splits into two clades (Fig.5).Clade 1 includes two subclades, one of which consists of geminiviruses and genomoviruses joining pCRESS9 plasmids of phytoplasma, and the other one includes CRESSV6 and P. pulchra plasmids. Notably, P. pulchra plasmids appear to emerge directly from within the CRESSV6 diversity, with the closest relationship to the CRESSV6 subclade of viruses sequenced from wastewater samples. The relationship between geminiviruses/genomoviruses and pCRESS9 plasmids is not resolved in the phylogeny. However, clustering analyses strongly suggest that Reps of pCRESS9 plasmids evolved from geminiviruses-genomoviruses (Figs.1 and 4). Consistent with this scenario, phytoplasmal pCRESS7 and pCRESS9 plasmids, despite encoding phylogenetically distinct Reps, share the gene content, namely, the copy number control protein, PRK06752-like SSB protein and conserved hypothetical protein (Supplementary Fig.3g, i). Furthermore, geminiviruses and CRESSV6 encode homologous capsid proteins suggesting that they evolved from a common viral ancestor rather than converged from two groups of plasmids by capturing homologous capsid protein genes.Clade 2 includes bacterial Reps of pCRESS1–3 and, as a sister group, CRESS-DNA viruses of the families Nanoviridae/Alphasatellitidae, Smacoviridae, and Circoviridae as well as unclassified CRESSV1 through CRESSV5, whereas GasCSV-like viruses are nested within bacterial pCRESS2.
Fig. 5
Maximum likelihood phylogenetic tree of Rep proteins. GasCSV—Gastropod-associated circular ssDNA virus. The tree was constructed with PhyML78. Branches with support values below 70 are contracted
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The robustness of the PhyML tree was validated by additional analyses (Supplementary Note1), including (i) maximum likelihood phylogenetic analyses using RAxML and IQ-Tree, with alternative branch support methods (FigureS5); (ii) phylogenetic reconstruction using the 20-profile mixture model (FigureS5); (iii) statistical analysis of the unconstrained and 3 constrained tree topologies (Supplementary Table2). Collectively, these results indicate that the obtained tree topology is highly robust and is likely to accurately reflect the evolutionary history of Reps encoded by CRESS-DNA viruses and plasmids.
Notably, analysis of the conserved motifs (Fig.3) suggests a specific association between the virus Reps in clade 1 and bacterial pCRESS3 (rather than pCRESS1–3 collectively), implying that the phylogenetic placement might be affected by ancient recombination events. Furthermore, bacilladnaviruses were omitted from the global phylogenetic tree because their Reps displayed unstable position in the phylogeny depending on the taxon sampling (Supplementary Fig.6), possibly, due to the small number of available sequences, their high divergence and potential chimerism. Regardless, phylogenetic analysis strongly suggests that the majority of CRESS-DNA viruses, including circoviruses, smacoviruses, nanoviruses, and CRESSV1–5, evolved from a common ancestor with bacterial Reps of pCRESS1–3, whereas the uncultivated GasCSV-like viruses emerge directly from the bacterial pCRESS2 Reps (Fig.5). The provenance of the assemblage including geminiviruses, genomoviruses and CRESSV6 is less clear but might predate the emergence of the other CRESS-DNA virus groups and possibly involved a common ancestor with the YLxH supergroup. The Reps of bacterial pCRESS9 and P. pulchra plasmids have been likely horizontally acquired more recently from the corresponding CRESS-DNA viruses.
Discussion
Here, we explored the evolutionary relationships among different classes of bacterial, archaeal, and eukaryotic replicons encoding HUH endonucleases (Reps). Our analysis revealed widespread exchange of rep genes among bacterial and archaeal viruses and plasmids, with some of the Rep clusters being particularly promiscuous, as in the case of pEC316_KPC-like Reps which are encoded not only in the corresponding plasmid but also in evolutionarily unrelated bacteriophages from 3 different families. Conversely, Reps of filamentous bacteriophages (family Inoviridae) fall into 5 distinct HUH clusters, indicating that, in this virus family, replication modules are readily exchangeable, presumably, for those better suited in particular hosts. Thus, the genome replication module of inoviruses shows extreme promiscuity and cannot serve as a phylogenetic marker, consistent with a recent analysis of 10,000 inovirus genomes15."), so that the family is held together by the shared morphogenetic module. By contrast, the _rep_ is the only gene conserved in all CRESS-DNA viruses and can serve as a vertically transmitted character against which various evolutionary events associated with the diversification of this virus class are mapped. Ultimately, however, no single gene or even functional module can fully represent the evolution of a given virus group. Instead, a more “holistic” approach is needed, where the provenance of all or most virus genes is deciphered.
Our present analysis pinpoints the origins of the replication modules of CRESS-DNA viruses. We identified 9 groups of bacterial Reps which share the nuclease-helicase domain organization with CRESS-DNA viruses. These bacterial Reps are encoded by previously unknown plasmids integrated into the genomes of diverse bacteria. By tracing the evolution of the helicase domain, we inferred the likely vector of evolution, namely, from plasmid Reps to the Reps of CRESS-DNA viruses. Although the Reps of CRESS-DNA viruses are generally considered to be monophyletic17, our analysis shows that this might not be the case sensu stricto. Instead, the CRESS-DNA virus diversity has likely been seeded on 3 independent occasions from different groups of bacterial plasmids at different stages of evolution (Fig.6). Conversely and contrary to the previous conclusion33, our results also indicate that CRESS-DNA viruses have given rise to (at least) 2 groups of plasmids in red alga and phytopathogenic phytoplasma, respectively. Thus, transitions between the virus and plasmid states appear to be bidirectional.
Fig. 6
A proposed evolutionary scenario for the origin of CRESS-DNA viruses from bacterial plasmids. The three hypothetical events of CRESS-DNA virus emergence are indicated with numbered circles. JRC, jelly-roll capsid protein-encoding genes. Non-orthologous JRC genes are indicated with different colors
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Obviously, transformation of a plasmid into a virus involves acquisition of the morphogenetic module, i.e., minimally, a gene for the capsid protein. We and others have previously shown that capsid proteins of different groups of CRESS-DNA viruses display specific relationships to single jelly-roll capsid proteins of RNA viruses of different families33,55,56,57,58. Thus, we propose that CRESS-DNA viruses evolved from plasmids through acquisition of reverse-transcribed capsid protein genes from different groups of RNA viruses (Fig.6). It seems likely that the capture of the capsid protein genes from RNA viruses has occurred in eukaryotic cells, possibly involving symbiotic or parasitic bacterial donors of the corresponding plasmids. Some of these events could have occurred at the early stages of eukaryotic evolution, as in the case of the viral assemblage including circoviruses, smacoviruses, nanoviruses and CRESSV1–5. By contrast, GasCSV-like viruses probably emerged relatively recently. Given the close relationship between pCRESS2 and GasCSV-like viruses, viruses of the latter group might infect bacteria rather than eukaryotes. Alternatively, the transition from a plasmid to a CRESS-DNA virus ancestor could have occurred once and was followed by replacement of the rep genes with counterparts from other plasmids, resulting in the 3 contemporary lineages of CRESS-DNA viruses. However, given that neither Rep nor capsid proteins appear to be orthologous in the 3 virus groups, this alternative scenario cannot be substantiated at this point. Regardless, it is clear that Rep and capsid protein genes have been repeatedly exchanged with distantly related homologs from other viruses, even in the more recent history of CRESS-DNA viruses55,58,59. Notably, it has been recently suggested based on the presence of matching CRISPR spacers that smacoviruses infect methanogenic archaea60. In our phylogeny (Fig.5), smacoviruses are deeply nested among circoviruses and nanoviruses, for which eukaryotic hosts have been confirmed experimentally1. Thus, if smacoviruses are shown to infect archaea as recently suggested60, the phylogeny is best compatible with a eukaryote to prokaryote transfer. The hosts of CRESSV1–6 are currently unknown and might include organisms from any of the 3 cellular domains of life. Furthermore, given that the SF3 helicase domain is now found in Reps of diverse bacterial replicons, this signature should be considered with caution when attributing viral genomes discovered by metagenomics to particular hosts.
Our findings further suggest that parvoviruses that have linear ssDNA genomes evolved directly from CRESS-DNA viruses. Indeed, both Rep and capsid proteins of parvoviruses are homologous to those of the CRESS-DNA viruses2. Unlike the Reps involved in rolling-circle replication, the Rep of parvoviruses lacks the joining activity used by CRESS-DNA viruses to circularize progeny genomes. Instead, the parvovirus Rep remains covalently attached to the 5′ ends of all viral DNA molecules20. The eukaryotic viruses with small, circular dsDNA genomes that comprise the families Polyomaviridae and Papillomaviridae encode major replication proteins that consist of an SF3 helicase domain and an inactivated HUH nuclease domain, lacking all 3 signature motifs. Nevertheless, structural studies have unequivocally demonstrated that the N-terminal origin-binding domains of both polyomaviruses and papillomaviruses are homologous to the HUH endonuclease domains of CRESS-DNA viruses and parvoviruses61. Thus, these viruses, most likely, evolved from ssDNA viruses but their evolution involved a drastic change in both the genome DNA structure and the replication mechanism such that the HUH domain switched from an enzymatic to a structural role. Clustering analysis of the SF3 helicase domains suggests that both polyomaviruses and papillomaviruses evolved from parvoviruses, although the driving forces behind this transition remains obscure.
The current classification of CRESS-DNA viruses largely relies on the phylogeny of the Rep proteins17,62,63. The ever-growing diversity of sequenced CRESS-DNA virus genomes calls for revision of the taxonomy of this class of viruses. Our analysis reveals two larger groupings of CRESS-DNA viruses, each including several families/clades, which could be equivalent to new orders, whereas all CRESS-DNA viruses could be unified at a yet higher taxonomic level. Finally, the membership of parvoviruses, polyomaviruses, and papillomaviruses in this assemblage could be also considered, at the highest taxonomic level. Indeed, it is not unprecedented that the same taxon contains viruses with different nucleic acids types. For instance, the order Ortervirales includes reverse-transcribing viruses with RNA and DNA genomes64, whereas members of the family Pleolipoviridae have either ssDNA or dsDNA genomes65. Notably, the ICTV has recently announced that taxonomic ranks above the order level are now officially accepted66, opening the door for the formal unification of the whole spectrum of evolutionarily related CRESS-DNA viruses.
Although the Reps of prokaryotic viruses of the families Microviridae, Inoviridae, and Sphaerolipoviridae lack the helicase domain, their HUH nucleases show clear affinities with those from different groups of plasmids, suggesting routes of evolution parallel to those of the CRESS-DNA viruses. Furthermore, the evolution of inoviruses appears to have involved multiple replacements of the rep gene with those from various plasmids.
The results presented here shed light on the origin of a major part of the virosphere, the ssDNA viruses replicating via the rolling-circle mechanism, and in particular, CRESS-DNA viruses. Arguably, evolution of the ssDNA viruses is the most compelling manifestation of the previously noted general trend in virus evolution, namely, tight evolutionary connections between viruses and capsid-less MGE67,68.
Methods
Databases
Homologs of the HUH endonucleases were retrieved by running searches against protein sequence databases filtered to 50 and 90% sequence identity (UniRef50 and UniRef90, respectively) which were downloaded from Search for bacterial homologs of CRESS-DNA virus Reps was performed against nr90 (NCBI’s nr database (ftp://ftp.ncbi.nlm.nih.gov/blast/db/) filtered to 90% identity). To detect remote sequence similarity, we used sequence profile databases which included profiles from PDB (www.pdb.org), SCOP69, Pfam70, and CDD71. For query profile generation nr70 database was used.
Sequence searches and clustering
Homologs of the HUH superfamily endonuclease domains for each representative Rep sequence were obtained by performing three jackhmmer72 iterations against the UniRef50 database. Representative Reps were selected as queries for homology searches based on exhaustive review of literature on the HUH superfamily16,17,28,53. In addition, for HUH groups with less than 10 homologs in UniRef50, we repeated searches against UniRef90 database. For homology searches only the HUH endonuclease domain was used to avoid attracting unrelated proteins, for example, containing superfamily 1 or 3 helicase domains. However, clustering was performed using full-length sequences to better reflect their evolutionary history. Dataset obtained by searches against the UniRef databases was supplemented with CRESS-DNA virus Reps devoid of obvious recombinant sequences from our previous study53. Sequences were clustered using CLANS with BLAST option35. CLANS is an implementation of the Fruchterman-Reingold force-directed layout algorithm, which treats protein sequences as point masses in a virtual multidimensional space, in which they attract or repel each other based on the strength of their pairwise similarities (CLANS p-values)35. Thus, evolutionarily more closely related sequences gravitate to the same parts of the map, forming distinct clusters. Rep clusters were identified by CLANS convex algorithm at P-value = 1e−08. To collect bacterial homologs of CRESS-DNA virus Reps, we used representative sequences as queries and performed two jackhmmer iterations against nr90 database. The resultant set of sequences was grouped using a convex clustering algorithm (at P-value = 1e−05) in CLANS. To ensure that we gathered all bacterial homologs, HMM profiles were constructed for each identified cluster and used as queries for searches against nr90 with hmmsearch72. Accessions of proteins for each group, shown in Fig.1, are available for download (Supplementary Data1). For collection of the SF3 helicase dataset, the helicase domain of a YLxH supergroup member from Streptococcus canis (WP_003048523) was used as a query for hmmer search against nr30 database available at the Bioinformatics Toolkit server73. The resulting dataset was supplemented with SF3 helicase sequences from CRESS-DNA viruses53, polyomaviruses, papillomaviruses, parvoviruses and P. pulchra-like plasmids (Supplementary Data1). Extracted helicase domains were filtered to 70% identity with CD-HIT (parameter “-c 0.7”)74.
Remote homology detection
Sequence searches based on profile-profile comparisons were used to detect remote homology. For profile generation, two iterations of jackhmmer72 were run against nr70 sequence database using E-value = 1e−03 inclusion threshold. The resulting profiles were used to search against profile databases with HHsearch75. Search results for proteins from representative bacterial plasmids and integrative elements are available in Supplementary Data1.
Multiple sequence alignments and phylogenetic analysis
To construct multiple sequence alignments for phylogenetic analysis we used MAFFT76 and TrimAl77. MAFFT options G-INS-i and L-INS-i and TrimAl gap thresholds 0.05 and 0.15 were used to generate alignments for Figs.2 and 5, respectively. The resulting alignments covered both HUH and SF3 (where available) domains and contained 743 and 508 positions, respectively. Both alignments can be found in the Supplementary Data2 and 3. Phylogenetic trees were calculated with PhyML78 using automatic model selection and aBayes branch support. Substitution models VT + G + I + F (VT, amino acid replacement matrix; G, gamma shape parameter: estimated (1.864); I, proportion of invariable sites: estimated (0.005); F, equilibrium frequencies: empirical) and LG + G (LG, amino acid replacement matrix; G: estimated (1.807)) substitution models were selected for phylogenetic analyses shown in Figs.2 and 5, respectively. Additional trees were constructed using IQ-Tree v1.6.8 (ref. 79) with Ultrafast Bootstrap Approximation branch support80, and RAxML with non-parametric bootstrapping81. Mixture model tree was constructed with IQ-Tree79 using model parameters (LG + C20 + F + G) and ultrafast bootstrap (with 1000 replicates). Alignment and guide tree (parameters “-s” and “-ft”, respectively) were the same as in Fig.5. Highly diverged sequences forming long branches were removed before constructing final trees. Bacilladnaviridae viruses were also removed, because their position was not stable in trees with different sequence sampling (Supplementary figure6). Phylogenetic trees are available from the authors upon request. The trees shown in Figs.2, 5, S5 and S6 can be found in the Supplementary Data4 to 10.
Statistical tests
Alternative topologies for the Rep tree were tested using the IQ-Tree software version 1.6.8 with the following parameters: -m LG+G -n 0 -zb 100000 -zw -au (ref. 79). As an unconstrained tree, we used the original PhyML tree (Fig.5), which was tested against each of the constrained trees. The following tests were performed: Approximately Unbiased (AU) test82, logL difference from the maximal logl in the set, RELL test83, one sided and weighted Kishino–Hasegawa (KH) tests84, Shimodaira–Hasegawa (SH) test85, weighted SH test, Expected Likelihood Weight (ELW) test86.
Sequence logos
Sequence logos for the Reps of CRESS-DNA virus families were taken from ref. 57. Alignments for other groups were obtained from an alignment used to build the tree shown in Fig.5. Sequence logos were created using WebLogo server87.
Genomic context analysis
The integrated plasmids were identified by thorough analysis of genomic neighborhoods of the Rep-encoding genes. The precise borders of integration were defined based on the presence of direct repeats corresponding to attachment sites. The repeats were searched for using Unipro UGENE88. Genes of integrated plasmids were annotated based on the HHsearch searches75. Genome maps were compared and visualized using Easyfig with tBLASTx option89.
Reporting summary
Further information on research design is available in theNature Research Reporting Summary linked to this article.
Data availability
The authors declare that the data supporting the findings of this study are available within the paper and its supplementary information files. Accession numbers for all proteins analyzed in this study as well as alignments used to generate the trees shown in Figs.2 and 5 are included in the Supplementary Data1, 2, and 3.
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Acknowledgements
The authors are grateful to Valerian V. Dolja for insightful comments on the manuscript. M.K. was supported by l’Agence Nationale de la Recherche (project ENVIRA, #ANR-17-CE15–0005-01). D.K. was partly supported by a Short Term Fellowship from the Federation of European Biochemical Societies (FEBS). E.V.K. is supported through the intramural program of the U.S. National Institutes of Health.
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Authors and Affiliations
Institute of Biotechnology, Life Sciences Center, Vilnius University, Saulėtekio av. 7, Vilnius, 10257, Lithuania
Darius Kazlauskas
The Biodesign Center for Fundamental and Applied Microbiomics, School of Life Sciences, Center for Evolution and Medicine, Arizona State University, Tempe, AZ, 85287, USA
Arvind Varsani
Structural Biology Research Unit, Department of Integrative Biomedical Sciences, University of Cape Town, Rondebosch, 7700, Cape Town, South Africa
Arvind Varsani
National Center for Biotechnology Information, National Library of Medicine. National Institutes of Health, Bethesda, MD, 20894, USA
Eugene V. Koonin
Department of Microbiology, Institut Pasteur, 25 rue du Docteur Roux, Paris, 75015, France
Mart Krupovic
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1. Darius KazlauskasView author publications Search author on:PubMedGoogle Scholar
2. Arvind VarsaniView author publications Search author on:PubMedGoogle Scholar
3. Eugene V. KooninView author publications Search author on:PubMedGoogle Scholar
4. Mart KrupovicView author publications Search author on:PubMedGoogle Scholar
Contributions
M.K. conceived the study. D.K. and M.K. performed sequence analyses. D.K., A.V., E.V.K., and M.K. interpreted the results. D.K. and M.K. wrote the first draft of the manuscript. All authors edited and approved the final version of the manuscript.
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Correspondence to Mart Krupovic.
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Kazlauskas, D., Varsani, A., Koonin, E.V. et al. Multiple origins of prokaryotic and eukaryotic single-stranded DNA viruses from bacterial and archaeal plasmids. Nat Commun10, 3425 (2019).
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Zhao, L., Rosario, K., Breitbart, M. & Duffy, S. Eukaryotic circular rep-encoding single-stranded DNA (CRESS DNA) viruses: ubiquitous viruses with small genomes and a diverse host range. Adv. Virus Res103, 71–133 (2019).
ArticlePubMedGoogle Scholar
Krupovic, M. Networks of evolutionary interactions underlying the polyphyletic origin of ssDNA viruses. Curr. Opin. Virol.3, 578–586 (2013).
ArticleCASPubMedGoogle Scholar
Wang, H. et al. Plasma virome of cattle from forest region revealed diverse small circular ssDNA viral genomes. Virol. J.15, 11 (2018).
ArticlePubMedPubMed CentralCASGoogle Scholar
Chow, C. E. & Suttle, C. A. Biogeography of viruses in the sea. Annu Rev. Virol.2, 41–66 (2015).
ArticleCASPubMedGoogle Scholar
Labonte, J. M. & Suttle, C. A. Previously unknown and highly divergent ssDNA viruses populate the oceans. ISME J.7, 2169–2177 (2013).
ArticleCASPubMedPubMed CentralGoogle Scholar
Roux, S., Krupovic, M., Poulet, A., Debroas, D. & Enault, F. Evolution and diversity of the Microviridae viral family through a collection of 81 new complete genomes assembled from virome reads. PLoS ONE7, e40418 (2012).
ArticleADSPubMedPubMed CentralCASGoogle Scholar
Wang, Y. et al. The fecal virome of red-crowned cranes. Arch. Virol.164, 3–16 (2019).
ArticleCASPubMedGoogle Scholar
Sadeghi, M. et al. Virome of >12 thousand Culex mosquitoes from throughout California. Virology523, 74–88 (2018).
ArticleCASPubMedGoogle Scholar
Rosario, K. et al. Virus discovery in all three major lineages of terrestrial arthropods highlights the diversity of single-stranded DNA viruses associated with invertebrates. PeerJ6, e5761 (2018).
ArticlePubMedPubMed CentralGoogle Scholar
Richet, C. et al. Novel circular DNA viruses associated with Apiaceae and Poaceae from South Africa and New Zealand. Arch. Virol.164, 237–242 (2019).
ArticleCASPubMedGoogle Scholar
Creasy, A., Rosario, K., Leigh, B. A., Dishaw, L. J. & Breitbart, M. Unprecedented diversity of ssDNA phages from the Family Microviridae detected within the gut of a protochordate model organism (Ciona robusta). Viruses10, E404 (2018).
Kraberger, S. et al. Identification of circular single-stranded DNA viruses in faecal samples of Canada lynx (Lynx canadensis), moose (Alces alces) and snowshoe hare (Lepus americanus) inhabiting the Colorado San Juan Mountains. Infect. Genet Evol.64, 1–8 (2018).
ArticleCASPubMedGoogle Scholar
Steel, O. et al. Circular replication-associated protein encoding DNA viruses identified in the faecal matter of various animals in New Zealand. Infect. Genet Evol.43, 151–164 (2016).
ArticleCASPubMedGoogle Scholar
Ng, T. F. et al. High variety of known and new RNA and DNA viruses of diverse origins in untreated sewage. J. Virol.86, 12161–12175 (2012).
ArticleCASPubMedPubMed CentralGoogle Scholar
Roux, S. et al. Cryptic inoviruses revealed as pervasive in bacteria and archaea across Earth’s biomes. Nat. Microbiol. (2019).
Chandler, M. et al. Breaking and joining single-stranded DNA: the HUH endonuclease superfamily. Nat. Rev. Microbiol11, 525–538 (2013).
ArticleCASPubMedPubMed CentralGoogle Scholar
Rosario, K., Duffy, S. & Breitbart, M. A field guide to eukaryotic circular single-stranded DNA viruses: insights gained from metagenomics. Arch. Virol.157, 1851–1871 (2012).
ArticleCASPubMedGoogle Scholar
Ilyina, T. V. & Koonin, E. V. Conserved sequence motifs in the initiator proteins for rolling circle DNA replication encoded by diverse replicons from eubacteria, eucaryotes and archaebacteria. Nucleic Acids Res20, 3279–3285 (1992).
ArticleCASPubMedPubMed CentralGoogle Scholar
Koonin, E. V. & Ilyina, T. V. Computer-assisted dissection of rolling circle DNA replication. Biosystems30, 241–268 (1993).
ArticleCASPubMedGoogle Scholar
Cotmore, S. F. & Tattersall, P. Parvoviruses: Small Does Not Mean Simple. Annu Rev. Virol.1, 517–537 (2014).
ArticlePubMedCASGoogle Scholar
Krupovic, M. & Koonin, E. V. Evolution of eukaryotic single-stranded DNA viruses of the Bidnaviridae family from genes of four other groups of widely different viruses. Sci. Rep.4, 5347 (2014).
ArticleADSPubMedPubMed CentralCASGoogle Scholar
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11599 | https://books.google.com/books/about/Environmental_Engineering.html?id=fBlSAAAAMAAJ | Environmental Engineering - Howard S. Peavy, Donald R. Rowe, George Tchobanoglous - Google Books
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Environmental Engineering
Howard S. Peavy, Donald R. Rowe, George Tchobanoglous
McGraw-Hill, 1985 - Technology & Engineering - 719 pages
This book brings together, and integrates the three principal areas of environmental engineering water, air, and solid waste management. It introduces a unique approach by emphasizing the relationship between the principles observed in natural purification processes and those employed in engineered systems. First, the physical, chemical, mathematical, and biological principles that define, measure and quantify environmental quality are described. Next, the processes by which nature assimilates waste material are discussed and the natural purification processes that form the basis of engineered systems are detailed. Finally, the engineering principles and practices involved in the design and operation of environmental engineering works are covered at length. Written in a lucid style and offering abundant illustrations and problems, the book provides a treatment of environmental engineering that can be understood by a wide range of readers.
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Contents
Water Purification Processes in Natural Systems 63
CHEMICAL PROCESSES 73
RESPONSE OF STREAMS TO BIODEGRADABLE 83
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Environmental Engineering
Howard S. Peavy,Donald R. Rowe,George Tchobanoglous
No preview available - 1985
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acidactivated carbonadiabatic lapse rateadsorbentadsorptionaerationair pollutionammoniaanaerobicanaerobic digestionatmospherebiologicalbiomassCaCO3carboncarbon monoxidechemicalchlorinecoagulationcollectioncombustioncompoundsconcentrationcontainersDeterminediameterdigestiondioxidedischargedisposaldissolvedeffectsefficiencyeffluentemissionsenergyEngineeringEnvironmentalequationExamplefactorsFigurefilterfiltrationflocculationflow rategaseshead losshydraulichydrocarbonshydrogenindustrialinfluentkg/m³landfillliquidm³/dm³/smassmaterialmaximummechanicalmetermethanemg/Lmunicipalnitrogennitrogen oxidesoperationorganicoxidesoxygenparticlespercentpipeplantpotableprecipitationpressureprimaryprocessespumpquantitiesreactorreducedremovalsamplesecondary clarifiersettling basinsettling velocitysewershown in Figsludgesolid wastessolublesourcesspecificSpray towersstandardsstreamsulfursulfur oxidessurfacesuspended solidsTabletanktemperatureTypicalunitsvolumewastewater treatmentwastewater-treatmentweirzoneµg/m³
References to this book
Wastewater Biosolids to Compost
Frank R. Spellman
Limited preview - 1996
Distribution System Performance Evaluation
Arun K. Deb,Yakir J. Hasit,Frank M. Grablutz
Limited preview - 1995
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Bibliographic information
Title Environmental Engineering
McGraw-Hill civil engineering series
McGraw-Hill series in water resources and environmental engineering
AuthorsHoward S. Peavy, Donald R. Rowe, George Tchobanoglous
Contributors Donald R. Rowe, George Tchobanoglous
Edition illustrated
Publisher McGraw-Hill, 1985
Original from the University of Michigan
Digitized Dec 4, 2007
ISBN 0070491348, 9780070491342
Length 719 pages
SubjectsTechnology & Engineering
›
Environmental
›
General
Technology & Engineering / Chemical & Biochemical
Technology & Engineering / Environmental / General
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