source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-35897 | 1. ## reivew for final
Three marbles are drawn at random from an urn containing 8 black, 7 white, and 5 red marbles.
What is the probability that none of them are white. Answer: $\frac{_{13}C_3}{_{20}C_3}$.
What is the probability that at least one of them is red? Answer: $1-\frac{_{15}C_3}{_{20}C_3}$
A drawer contains 8 blue socks, 4 green socks, and 6 brown socks. If you choose 2 socks at random, what is the probability that they are both brown? Answer: $\frac{_{6}C_2}{_{18}C_2}$
On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$
Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?
2. Originally Posted by dori1123
On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$
Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?
Hi!
For the 2nd last question, I think it should be:
Let X be the no. of correct answers:
$P(X \geq 8)= {10 \choose 8}(0.5)^8(0.5)^2+ {10 \choose 9}(0.5)^9(0.5)+(0.5)^{10}$
The following is multiple choice question (with options) to answer.
In a box, there are 2 yellow, 4 grey and 4 white pencils. One pencil is picked up randomly. What is the probability that it is neither yellow nor grey? | [
"2/5",
"1/5",
"3/5",
"4/5"
] | A | Explanation :
Neither yellow nor grey means the pencil drawn is white.
Total number of outcomes = (2+4+4) = 10.
Number of Favourable outcomes = 4 = Number of white pencils.
Hence, Probability of the event = 4/10 = 2/5.
Answer : A |
AQUA-RAT | AQUA-RAT-35898 | A number is divisible by 11 if and only if the difference of the sum of the odd numbered digits (the first digit, the third digit, ...) and the sum of its even numbered digits is divisible by 11. The sum of the odd numbered digits of $y$ is $250\cdot(2+7)$ and the sum of the even numbered digits of $y$ $250\cdot(7+2)$. The difference between these two quantities is $0$; so $y$ is divisible by 11.
-
This explains it very well! Thank you so much. – SNS Feb 27 '12 at 23:42
$2772=99 \times 28$
so
$277227722772\ldots277227722772 = 99 \times 28 \times 100010001\ldots000100010001$
and so is divisible by both $9$ and $11$ (and $4$ and $7$ and other numbers).
The following is multiple choice question (with options) to answer.
Which is the following is divisible by 11 | [
"A)4305",
"B)4825",
"C)6594",
"D)4905"
] | C | Explanation:
Sum of first 'n' natural numbers = n(n + 1)/2
Sum of first 9 natural numbers = 11(11 + 1)//2 = 11 x 6 = 66
Sum of first 99 natural numbers = 111(111 + 1)//2 = 111 x 60 = 6660
6660 - 66 = 6594
Answer: Option C |
AQUA-RAT | AQUA-RAT-35899 | # Probability of a certain ball drawn from one box given that other balls were drawn
Box 1 contains 2 green and 3 red balls, 2 has 4 green and 2 red, and 3 has 3 green and 3 red. Only one ball is drawn from each of the 3 boxes. What is the probability that a green ball was drawn from box 1 given that two green balls were drawn?
So in total there were exactly 2 green balls and 1 red ball drawn, from a different combinations of the 3 boxes. We could have selected the 2 greens from the first 2 boxes and a red from the last box, 2 greens from the last 2 boxes, or 2 greens from box 1 and 3. I get $\frac{2}{5} \frac{4}{6} \frac{3}{6} + \frac{2}{5} \frac{2}{6} \frac{3}{6} + \frac{3}{5} \frac{4}{6} \frac{3}{6} = \frac{2}{5}$. Now this is the probability of drawing 2 green balls. What do I do from here?
Hint: Let $G_1$ be the event a green was drawn from the first box, and let $T$ be the event two green were drawn. We want the conditional probability $\Pr(G_1|T)$, which is $\frac{\Pr(G_1\cap T)}{\Pr(T)}$.
Alternately, if the notation above is unfamiliar, you can use a "tree" argument.
The following is multiple choice question (with options) to answer.
A bag contains 2 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green. | [
"19/20",
"17/20",
"21/25",
"8/10"
] | C | Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
P(A) = (2/5) x (6/10) = 6/25.
P(B) = (3/5) x (4/10) = 6/25.
Hence, the required probability is P(A) + P(B) which is nothing but 21/25.
ANSWER:C |
AQUA-RAT | AQUA-RAT-35900 | (ii) 1/(√7 1/(√5 + √2) = 1/(√5 + √2) × (√5 − √2)/(√5 − √2) Solution: We rationalize the denominator of x: \begin{align} x &= \frac{{11}}{{4 - \sqrt 5 }} \times \frac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}\\ &= \frac{{11\left( {4 + \sqrt 5 } \right)}}{{16 - 5}}\\ &= 4 + \sqrt 5 \\ \Rightarrow x - 4 &= \sqrt 5 \end{align}. By using this website, you agree to our Cookie Policy. Oh No! That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. Ex 1.5, 5 Ask questions, doubts, problems and we will help you. &= 2 - \sqrt 3 \hfill \\ The denominator contains a radical expression, the square root of 2.Eliminate the radical at the bottom by multiplying by itself which is \sqrt 2 since \sqrt 2 \cdot \sqrt 2 = \sqrt 4 = 2.. the smallest positive integer which is divisible by each denominators of these numbers. $\begin{array}{l} 4\sqrt {12} = 4\sqrt {4 \times 3} = 8\sqrt 3 \\ 6\sqrt {32} = 6\sqrt {16 \times 2} = 24\sqrt 2 \\ 3\sqrt {48} = 3\sqrt {16 \times 3} =12\sqrt 3 \end{array}$, $\boxed{\begin{array}{*{20}{l}} To get the "right" answer, I must "rationalize" the denominator. &= \frac{{3 + 2\sqrt 3 }}{{5 - 2\sqrt 3 }} \times \frac{{5 + 2\sqrt 3 }}{{5 +
The following is multiple choice question (with options) to answer.
1/(4-√15)=? | [
"4+√15",
"9-4√5",
"9+2√5",
"9-2√5"
] | A | This question requires us to rationalize (fix) the denominator.
For more on this technique, seehttps://www.gmatprepnow.com/module/gmat ... video/1044
Given: 1/(4-√15)
Multiply top and bottom by the CONJUGATE of 4-√15, which is 4+√15
So, 1/(4-√15) = (1)(4+√15)/(4-√15)(4+√15)
= (4+√15)/(1)
= 4+√15
=
A |
AQUA-RAT | AQUA-RAT-35901 | bruce
the jet will fly 9.68 hours to cover either distance
bruce
Riley is planning to plant a lawn in his yard. He will need 9 pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs$3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound? Vonna Reply 33.336 Robinson Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have? mikayla Reply 2q=n ((2q).05) + ((q).25) = 8.75 .1q + .25q = 8.75 .35q = 8.75 q = 25 quarters 2(q) 2 (25) = 50 nickles Answer check 25 x .25 = 6.25 50 x .05 = 2.50 6.25 + 2.50 = 8.75 Melissa John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?
7-$10 21-$5
Robert
Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.)
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hour longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
8mph
michele
16mph
Robert
3.8 mph
Ped
16 goes into 80 5times while 20 goes into 80 4times and is 4mph faster
Robert
The following is multiple choice question (with options) to answer.
9 neighbors are to split the cost of topsoil. The cost of the 5 truck loads of dirt was $690.47. What is the least amount of money (in whole number of dollars) that they must add to bill if they wants to split this money evenly among his nine neighbors? | [
"690.48",
"690.49",
"690.5",
"690.51"
] | A | If the bill was $690.47 dollars, how much money should be removed with 1 cent as the smallest unit?
This is equivalent to finding the first number that is divisible by 9 that occurs after 690.47.
In order to divide the sum in 9 parts, the amount must be divisible by 9
Divisibility rule of 9: The sum of the digits must be divisible by 9
Sum of digits of 6+9+0+4+7 = 26. If you add 1 , the number is divisible by 9 (26 +1). 27 is divisible by 9.
Hence, we need to add 1 cent from this number for it to be divisible by 9.
Correct Option : A |
AQUA-RAT | AQUA-RAT-35902 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A hiker walked for 3 days. She walked 18 miles on the first day, walking 3 miles per hour. On the second day she walked for one less hour but she walked one mile per hour, faster than on the first day. On the third day she walked at 7 miles per hour for 2 hours. How many miles in total did she walk? | [
"24",
"52",
"58",
"60"
] | B | First day - 18 miles with 3 miles per hours then total - 6 hours for that day
Second day - 4 miles per hour and 5 hours - 20 miles
Third day - 7 miles per hour and 2 hours - 14 miles
Total 18+20+14 = 52
Answer: option B. |
AQUA-RAT | AQUA-RAT-35903 | ### Exercise 20
Mr. Halsey has a choice of three investments: Investment A, Investment B, and Investment C. If the economy booms, then Investment A yields 14% return, Investment B returns 8%, and Investment C 11%. If the economy grows moderately, then Investment A yields 12% return, Investment B returns 11%, and Investment C 11%. If the economy experiences a recession, then Investment A yields a 6% return, Investment B returns 9%, and Investment C 10%.
1. Write a payoff matrix for Mr. Halsey.
2. What would you advise him?
#### Solution
1. .14.08.11.12.11.11.06.09.10.14.08.11.12.11.11.06.09.10 size 12{ left [ matrix { "." "14" {} # "." "08" {} # "." "11" {} ## "." "12" {} # "." "11" {} # "." "11" {} ## "." "06" {} # "." "09" {} # "." "10"{} } right ]} {}
2. 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 010010 size 12{ left [ matrix { 0 {} ## 1 {} ## 0 } right ]} {} or 010010 size 12{ left [ matrix { 0 {} # 1 {} # 0{} } right ]} {}, 001001 size 12{ left [ matrix { 0 {} ## 0 {} ## 1 } right ]} {}, value=.11value=.11 size 12{"value"= "." "11"} {}
### Exercise 21
Mr. Thaggert is trying to decide whether to invest in stocks or in CD's(Certificate of deposit). If he invests in stocks and the interest rates go up, his stock investments go down by 2%, but he gains 1% in his CD's. On the other hand if the interest rates go down, he gains 3% in his stock investments, but he loses 1% in his CD's.
The following is multiple choice question (with options) to answer.
At the end of the first quarter, the share price of a certain mutual fund was 20 percent higher than it was at the beginning of the year. At the end of the second quarter, the share price was 60 percent higher than it was at the beginning of the year. What was the percent increase in the share price from the end of the first quarter to the end of the second quarter? | [
"20%",
"25%",
"30%",
"33.33%"
] | D | Another method is to use the formula for 2 successive percentage changes:
Total = a + b + ab/100
60 = 20 + b + 20b/100
b = 33.33
Answer (D) |
AQUA-RAT | AQUA-RAT-35904 | homework-and-exercises, kinematics, velocity, vectors, relative-motion
I've drawn the man swimming at some arbitrary angle $\theta$ at a speed $v$. The river is flowing at a speed $V$, and the time the man takes to cross is $t$. The distance swum by the man is $d_m$ and the distance the water moves is $d_r$.
The key point is that the speed the river flows affects where the man emerges on the other side of the river, but it doesn't affect the time to cross. The time to cross is simply the distance swum, $d_m$, divided by the swimming speed, $v$:
$$ t = \frac{d_m}{v} $$
and by trigonometry the distance the man swims is related to the angle $\theta$ by:
$$ d_m = \frac{W}{\sin\theta} $$
so:
$$ t = \frac{W}{v \sin\theta} $$
Both $W$ and $v$ are constants, so to minimise the time you need to maximise $\sin\theta$, and the maximum value of $\sin\theta$ is 1 when $\theta$ = 90º i.e. perpendicular to the bank.
Response to response to comment:
If we take $x$ to be the direction along the river and $y$ the direction across it, the the time taken to cross is just:
$$ t = \frac{w}{U_y} $$
where $U$ is the total velocity and $U_y$ is its $y$ component. Because $U$ is the vector sum of $v$ and $V$, its $y$ component is simply:
$$ U_y = v_y + V_y $$
But the river is flowing in the $x$ direction i.e. $V_y$ is zero, and therefore $U_y$ = $v_y$ i.e. the $y$ component of the total velocity depends only only the man's swimming speed and not on the river speed. This is why the river speed doesn't affect the time to cross.
The following is multiple choice question (with options) to answer.
A man can row upstream at 60 kmph and downstream at 90 kmph, and then find the speed of the man in still water? | [
"75 kmph",
"65 kmph",
"30 kmph",
"73 kmph"
] | A | US = 60
DS = 90
M = (60 + 90)/2 = 75
Answer:A |
AQUA-RAT | AQUA-RAT-35905 | You are asked whether |x|=|y| or in other words, is the distance of x from 0 = distance of y from 0 ? You would get a yes if x= y or x=-y. Lets evaluate the choices.
Per statement 1:$$x=-y$$ . Clearly this is one of the conditions we are looking for. Assume 2-3 different values. x=2, y = -2 --> 2 and -2 are both 2 units from 0 . Similar for 3/-3 or 10/-10 etc. Thus this statement is sufficient.
Per statement 2: $$x^2=y^2$$ ---> $$x= \pm y$$. Again both the cases, x=y and x=-y give you a "yes" for the question asked as is hence sufficient.
Both statements are sufficient ---> D is the correct answer.
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
17 Dec 2015, 06:54
statement 1 is sufficient as absolute value for -ve should give +ve
statement 2 is sufficient as
it will give either x and y both are +ve or both are -ve
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Re: Is |x| = |y|? (1) x = -y (2) x^2 = y^2 [#permalink]
### Show Tags
09 Mar 2018, 03:18
HarveyKlaus wrote:
Is |x| = |y|?
(1) x = -y
(2) x^2 = y^2
I found this question in GMAT Prep and selected ST2 as the answer but that's wrong.
Normally, x^2 would equal to lxl since it will contain both signs "positive as well as negative". How come st 1 is true?
Does it assume that since +x = -y, it must be true that -x is also equal to -y or?
What am I missing?
Question:
The following is multiple choice question (with options) to answer.
If x-y=8, which of the following must be true?
I. Both x and y are positive
II. If x is negative, y must be negative
III. If x is positive, y must be positive | [
"I only",
"II only",
"III only",
"I and II"
] | B | If x-y=8, which of the following must be true?
I. Both x and y are positive
II. If x is positive, y must be positive
III. If x is negative, y must be negative
1. x and y can be negative, for instance, -2 - (-10) = 8 =>Eliminate A and D
2. x can be positive and y can be negative, for instance, 2 - (-6) = 8 => Eliminate B and E
A) I only
B) II only
C) III only
D) I and II
E) II and III
Answer B |
AQUA-RAT | AQUA-RAT-35906 | homework-and-exercises, special-relativity, inertial-frames, observers
But how much has John aged, according to Mike?
During t1, John was stationary according to Mike and therefore aged t1. During t2 + t2, John was moving at 0.5c, so aged
$$(2\sqrt{\frac{3}{4}} + 2\sqrt{\frac{3}{4}}) * \sqrt{\frac{3}{4}} = 4 * \frac{3}{4} = 3\;years$$
So in total, John aged
$$0.8 + 3 = 3.8\;years$$
from Mike's perspective?! That can't be correct. The thing is, no matter how I do the calculations, I come out at 3.8 years. Where are those 0.2 years I am missing?
At the same time, Mike gets in his spaceship and also moves to the star at 0.5c, such that Mark and Mike come at the star simultaneously.
The following is multiple choice question (with options) to answer.
Six years back, a father was 42 years older than his son. At present the father is 4 times as old as the son. How old will the son be three years from now? | [
"20 years",
"19 years",
"18 years",
"17 years"
] | D | Using linear equations to convert the word problem:
** F = fathers current ageS = sons current age
#1 Six years back, a father was 42 years older than his son: F-6 = 42 + (S - 6)
#2 At present the father is 4 times as old as the son: F = 4*S
How old will the son be three years from now?: S + 3 = ?
With two variables and two linear equations we are able to solve the problem:
(S*4) - 6 = 42 +S -6
3S=42
S = 14
S + 3 = 14 + 3 = 17
Answer: D |
AQUA-RAT | AQUA-RAT-35907 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1904? | [
"1700",
"2777",
"2889",
"27670"
] | A | 1904 = P [1 + (5*12/5)/100]
P= 1700.Answer:A |
AQUA-RAT | AQUA-RAT-35908 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 340 m long? | [
"56 sec",
"42 sec",
"45 sec",
"48 sec"
] | A | Speed = 45 * 5/18 = 25/2 m/sec
Total distance covered = 360 + 340 = 700 m
Required time = 700 * 2/25 = 56 sec
ANSWER:A |
AQUA-RAT | AQUA-RAT-35909 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he | [
" broke even",
" lost 4 cents",
" gained 4 cents",
" lost 10 cents"
] | A | 20 % profit on $ 1.20
= $ 20/100 × 1.20
= $ 0.20 × 1.20
= $ 0.24
Similarly, 20 % loss on $ 1.20
= $ 20/100 × 1.20
= $ 0.20 × 1.20
= $ 0.24
Therefore, in one pipe his profit is $ 0.24 and in the other pipe his loss is $ 0.24.
Since both profit and loss amount is same so, it’s broke even.
The answer is (A) |
AQUA-RAT | AQUA-RAT-35910 | When we add $10$ red, we end up with $4k+10$ red. The blues remain unchanged at $7k$.
So the new proportion is $(4k+10): 7k$. We are told that the proportion $(4k+10): 7k$ is $6:7$. So $$\frac{4k+10}{7k}=\dfrac{6}{7}.$$ If we multiply through by $7k$, we get $4k+10=6k$, and therefore $k=5$. It follows that there are $35$ blues in the bag.
-
Let $x$ be the number of red cubes and $y$ be the number of blue cubes.
To start, the ratio of red cubes to blue cubes is 4:7, or for every 4 red cubes, there are 7 blue cubes. Hence, we have:
$7x = 4y$.
When 10 more red cubes are added to the bag, the ratio of red cubes to blue cubes shifts to 6:7, or:
$7(x+10) = 6y$.
Expanding, we get a system:
$7x = 4y$
$7x + 70 = 6y$
Can you solve the system of equations from here?
The following is multiple choice question (with options) to answer.
If 5 (A's capital) = 7 (B's capital) = 9 (C's capital). Then the ratio of their capitals is? | [
"63:55:35",
"63:45:38",
"62:45:35",
"63:45:35"
] | D | 5A = 7B = 9C
A:B:C = 1/5:1/7:1/9
= 63:45:35
Answer: D |
AQUA-RAT | AQUA-RAT-35911 | # For dinner, $n$ people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so ...
Problem:
For dinner, $$n$$ ($$n \geq 4$$) people came and sat at a round table at random. If Ana, Ivan and Mark were among them, how many ways could they sit so that Ana and Ivan do not sit next to each other and at least one of them sits next to Mark? (Note: the round table implies seating arrangements that differ only in rotation.)
My attempt:
If I have $$n$$ people sitting around circular table, the number of different arrangements are $$(n-1)!$$.
If I have $$2$$ people Mark and Ana number of arrangements that they can sit next to each other is $$2 \cdot (n-2)!$$. So the number of arrangements that Mark sit next to Ivan is also $$2 \cdot (n-2)!$$, and sitting next to Ana also $$2 \cdot (n-2)!$$.
• yes.....it is fixs
– josf
Nov 2 '19 at 14:49
• "... seating arrangements that differ only in rotation... " are what? Considered to be the same, or different? From the context, I assume they should be considered to be the same, but the language isn't clear. Nov 11 '19 at 16:28
Method 1: Seat Mark. We will use him as our reference point.
Only Ana sits next to Mark: She can be seated in two ways, to his left or to his right. That leaves $$n - 2$$ seats. Since Ivan cannot sit next to Ana or Mark, he may be seated in $$n - 4$$ ways. The remaining $$n - 3$$ people can be seated in the remaining $$n - 3$$ seats in $$(n - 3)!$$ ways as we proceed clockwise around the table relative to Mark. Hence, there are $$2(n - 4)(n - 3)!$$ such arrangements.
The following is multiple choice question (with options) to answer.
At a dinner party, 7 people are to be seated around a circular table. Two seating arrangement are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group? | [
"5",
"10",
"720",
"32"
] | C | Has to be C.
Simple formula: There are (n - 1)! ways to arrange n distinct objects in a circle
hence (7-1)! = 6!= 720 |
AQUA-RAT | AQUA-RAT-35912 | \$7,382.94 3.2% 01.09.08 \$19.6879
\$7,402.63 3.2% 01.10.08 \$19.7404
\$7,422.37 3.2% 01.11.08 \$19.7930
\$7,442.17 3.2% 01.12.08 \$19.8458
The following is multiple choice question (with options) to answer.
Find: S.l on Rs 68000 at 16 2/3%per annum for 9 months | [
"6500",
"7500",
"8500",
"9500"
] | C | Explanation:
P = 68000, R =50/3% & T=9 months (3/4 years)
S.I=(68000 *50/3 * 3/4)/100=Rs.8500
ANSWER IS C |
AQUA-RAT | AQUA-RAT-35913 | Q 4.Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:
If same pair of coins is tossed at random, find the probability of getting:
SOLUTION:
Number of trials = 500
Number of outcomes of two heads (HH) = 105
Number of outcomes of one head (HT or TH) = 275
Number of outcomes of no head (TT) = 120
(i) Probability of getting two heads = $\frac{Frequency\;of\;getting\;2\;heads}{Total\;No.\;of\;trails}$ = $\frac{105}{500}$ = $\frac{21}{100}$
(ii) Probability of getting one head = $\frac{Frequency\;of\;getting\;1\;heads}{Total\;No.\;of\;trails}$ = $\frac{275}{500}$ = $\frac{11}{20}$
(iii) Probability of getting no head = $\frac{Frequency\;of\;getting\;no\;heads}{Total\;No.\;of\;trails}$ = $\frac{120}{500}$ = $\frac{6}{25}$
#### Practise This Question
The lines shown below never meet at any point. So, these lines are not parallel. Say true or false.
The following is multiple choice question (with options) to answer.
If four coins are tossed, the probability of getting two heads and two tails is ? | [
"3/8",
"3/5",
"3/2",
"3/1"
] | A | Since four coins are tossed, sample space = 24
Getting two heads and two tails can happen in six ways.
n(E) = six ways
p(E) = 6/24 = 3/8
Answer:A |
AQUA-RAT | AQUA-RAT-35914 | • Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
The following is multiple choice question (with options) to answer.
A consignment of 20 picture tubes contains 5 defectives. Two tubes are selected one after the other at random. The probability that both are defective assuming that the first tube is not replaced before drawing
the second, would be: | [
"1/16",
"1/19",
"1/4",
"1/3"
] | B | Probability of drawing a defective tube in first draw is 5/20
Probability of drawing a defective tube in second draw (without replacing first tube) is 4/19.
Therefore probability of getting both defective is = (5/20)*(4/19) = 1/19
ANSWER:B |
AQUA-RAT | AQUA-RAT-35915 | # Analyzing a mixture issue.
I am having a problem with this question:
Coffee A costs $75$ cents per pound and coffee B costs $80$ cents per pound to form a mixture that costs $78$ cents per pound.IF there are $10$ pounds of Mixture how many pounds of Coffee A are used?
According to the text the answer is 4. I don't know how they came up with this answer since they haven't given the clue or percentage of the how much of each coffee is used in the mixture. Maybe I am missing something here. Any suggestions?
Here is what I could think of:
$\$0.75 + \$0.80 = 155$ so $75$ is $\left(\frac{7500}{155}\right)$% of $155$ cents
Now taking $\left(\frac{7500}{155}\right)$% of $78$ cents (Price of the mixture) we get $\frac{1170}{31}$ cents.
Now if $75$ cents is $1$ pound $\frac{1170}{31}$ would be $\frac{78}{155} pound. So for ten pounds it would be$\frac{780}{155}$which is still not the answer. - Others will tell you how to do it, but first let's just check that the answer in the text is right: 10 pounds of mixture, of which 4 pounds is A, means 6 pounds are B. 4 pounds at 75 cents per, plus 6 pounds at 80 cents per, is$4\times75+6\times80=300+480=780$cents, or \$7.80, for 10 pounds, so 78 cents per pound. – Gerry Myerson Jun 15 '12 at 1:53
I would model it with a system of equations which are relatively simple to solve.
$$A + B = 10$$ $$75A + 80B = 78 \cdot 10 \implies 75A + 80B= 780$$
Multiply the top equation through by $80$ to get
$$80A + 80B = 800$$
We also have $$75A + 80B= 780$$
Simply subtract them to get
$$5A = 20 \implies A = 4$$
The following is multiple choice question (with options) to answer.
A grocer has 400 pounds of coffee in stock, 20 percent of which is decaffeinated. If the grocer buys another 100 pounds of coffee of which 50 percent is decaffeinated, what percent, by weight, of the grocer’s stock of coffee is decaffeinated? | [
"26%",
"30%",
"32%",
"34%"
] | A | 1. 20% of 400=80 pounds of decaffeinated coffee
2. 50% of 100=50 pounds of decaffeinated coffee
3. Wt have 130 pounds of decaffeinated out of 500 pounds, that means 130/500*100%=26%. The correct answer is A. |
AQUA-RAT | AQUA-RAT-35916 | Option A, your future value will be$10,000 plus any interest acquired over the three years. Let's take a look. Discounting or Present Value Concept Compound Value Concept Vn=Vo*(1+k) ^n. Let’s take a look at a couple of examples. (Also, with future money, there is the additional risk that the … The amount of interest depends on whether there is simple interest or compound interest. There are many reasons why money loses over time. The decision is now more difficult. This concept states that the value of money changes over time. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times. However, we don't need to keep on calculating the future value after the first year, then the second year, then the third year, and so on. Aside from being known as TVM, the theory is sometimes referred to the present discount value. Future value is amount that is obtained by enhancing the value of a present payment or a series of payments at the given interest rate to reflect the time value of money. Interest is charge against use of money paid by the borrower to the lender in addition to the actual money lent. Time value of money is the concept that the value of a dollar to be received in future is less than the value of a dollar on hand today. The time value of money recognizes that receiving cash today is more valuable than receiving cash in the future. This video explains the concept of the time value of money, as it pertains to finance and accounting. Though a little crude, an established rule is the “Rule of 72” which states that the doubling period can be obtained by dividing 72 by the interest rate. The time value of money is the widely accepted conjecture that there is greater benefit to receiving a sum of money now rather than an identical sum later. We arrive at this sum by multiplying the principal amount of $10,000 by the interest rate of 4.5% and then adding the interest gained to the principal amount: $10,000×0.045=450\begin{aligned} &\10,000 \times 0.045 = \450 \\ \end{aligned}10,000×0.045=$450, $450+$10,000=$10,450\begin{aligned}
The following is multiple choice question (with options) to answer.
A sum of money becomes 7/6 of itself in 3 years at a certain rate of simple interest. The rate per annum is? | [
"5 5/9 %",
"5 5/6 %",
"5 8/9 %",
"5 1/9 %"
] | A | Let sum = x. Then, amount = 7x/6
S.I. = 7x/6 - x = x/6; Time = 3 years.
Rate = (100 * x) / (x * 6 * 3) = 5 5/9 %.
Answer: A |
AQUA-RAT | AQUA-RAT-35917 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
By selling 12 pencils for a rupee a man loses 12%. How many for a rupee should he sell in order to gain 18%? | [
"12",
"9",
"10",
"89"
] | A | 88% --- 12
112% --- ?
88/112 * 12 = 9
Answer: A |
AQUA-RAT | AQUA-RAT-35918 | Change the chapter
Question
Jogging on hard surfaces with insufficiently padded shoes produces large forces in the feet and legs. (a) Calculate the magnitude of the force needed to stop the downward motion of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of 6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to include the weight of the 75.0-kg jogger’s body.) (b) Compare this force with the weight of the jogger.
a) $16300 \textrm{ N}$
b) $22.2$
Solution Video
OpenStax College Physics Solution, Chapter 7, Problem 53 (Problems & Exercises) (3:42)
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The following is multiple choice question (with options) to answer.
On a certain island, 5% of the 10000 inhabitants are one-legged and half of the others go barefooted. What is the least number of Shoes needed in the island? | [
"9000",
"10000",
"9500",
"9700"
] | B | one-legged =5% of 10000=500
remaining=10000-500=9500
barefooted=9500/2=4750
remaining people= 9500-4750=4750
hence required number of shoes= 4750*2+500*1=10000
ANSWER:B |
AQUA-RAT | AQUA-RAT-35919 | Let us take another mixture problem:
Question 2:
Two types of rice costing $60 per kg and$40 per kg are mixed in a ratio 2: 3. What will be the cost per kg of mixed rice?
Solution:
• Can we apply alligation to this question?
• Can we assume the price of the mixed rice to be $x per kg and make this diagram? • Now does it mean that $$x – 40 = 2$$ and $$60 – x = 3$$? • We can see that we are getting 2 different values of x above • Thus, we cannot use alligation in this way here • Well, we can still use these numbers by using proportions here as follows, • $$\frac{x-40}{60-x}=\frac{2}{3}$$ $$\Rightarrow 3x-120=120-2x$$ $$\Rightarrow 5x=240$$ $$\Rightarrow x=48$$ However, as you can see that we still need to do some calculations, and hence alligation does not help us a lot here. • Hence, we always recommend, for questions such as this where we are asked to find the resultant concentration upon mixing\combining direct parameters of two entities, it is always preferred to use a weighted average. So, let us apply that. • Since the two rice are mixed in the ratio 2 : 3, let us assume the quantities mixed be 2a and 3a • Thus, after which we can write $$x=\frac{60\times 2a+40\times 3a}{2a+3a}$$ $$\Rightarrow x=\frac{120a+120a}{5a}$$ $$\Rightarrow x=\frac{240a}{5a}$$ $$\Rightarrow x=48$$ • So, the final concentration or the price of the mix will be$48 per kg
• The point to be noted here is that all mixture questions need not be tackled with the alligation method
Alligation in other topics?
Alligation is generally associated with mixtures of questions
The following is multiple choice question (with options) to answer.
In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg? | [
"5/6",
"5/9",
"5/0",
"5/3"
] | A | Explanation:
Let us say the ratio of the quantities of cheaper and dearer varieties = x : y
By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
Answer: Option A |
AQUA-RAT | AQUA-RAT-35920 | 3 , 10 , 12 , 5 , 18 , 6 = { x1 , x2 , x3 , x4 , x5 , x6 }
n = 6 , μ = 9
σ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{3^{2}+10^{2}+12^{2}+5^{2}+18^{2}+6^{2}}{6}\:&space;-\:&space;9^{2}}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{\frac{638}{6}\:&space;-\:&space;81}$ = $\dpi{80}&space;\fn_jvn&space;\sqrt{23}$ = 5.03
----------------------
With the Standard Deviation being an indicator of how far away the values are from the mean/average. Let's use it with the list of values from example (1.1).
For 5 , 7 , 3 , 5 , 6 , 4, the mean was 5.
The Standard Deviation was 1.29.
We can work out how far way one Standard Deviation is from the mean in both a positive and negative direction.
51.29 = 3.71 , 5 + 1.29 = 6.29
This means we would expect that the majority of the values will be between 3.71 and 6.29.
Looking at the list, this does turn out to be the case.
With only 3 and 7 lying out side this range.
Now looking at the list from example (1.2).
3 , 10 , 12 , 5 , 18 , 6, the mean was 9.
The Standard Deviation was 5.03.
95.03 = 3.97 , 9 + 5.03 = 14.03
This means we would expect that the majority of the values will be between 3.97 and 14.03.
Similar to the list from (1.1), this again is the case.
The following is multiple choice question (with options) to answer.
If the average (arithmetic mean) of a list of numbers is 12 and the standard deviation of that list of numbers is 1.3, then which of the following numbers is more than two standard deviations from the mean?
I. 14.7
II. 9.2
III. 9.7 | [
"I only",
"I and II only",
"II only",
"III only"
] | B | Mean = 12
SD = 1.3
2 SD above mean = 12+2*1.3 = 14.6
2 SD below mean = 12-2*1.3 = 9.4
only 9.7 lies in the range of 2SD from mean
Answer: Option B |
AQUA-RAT | AQUA-RAT-35921 | Applying this to each of the pairs $$m = \alpha_i, n = \beta _i$$, we get that
$a \times b = \prod_{i=1}^k p_i ^{\alpha_i + \beta_i } = \prod_{i=1}^k p_i ^ { \min(\alpha_i, \beta_i) + \max(\alpha_i , \beta_i)} = \gcd(a,b) \times \mbox{lcm}(a,b).$
### 3. Given that $$a$$ and $$b$$ are 2 integers such that $$13 \gcd(a,b) = \mbox{lcm}(a,b)$$ and $$a + b = 2016$$, what are the values of $$a$$ and $$b$$?
Let $$G = \gcd(a,b)$$ and $$L = \mbox{lcm}(a,b)$$.
Let $$a = a^* G$$ and $$b = b^* G$$ where $$\gcd(a^*, b^*) = 1$$ by construction. Since $$(a^*G)\times (b^*G) = ab = GL = G \times 13 G$$, we get that $$a^* b^* = 13$$. Hence, we have $$\{ a^*, b^* \} = \{ 1, 13\}$$. WLOG, we may assume that $$a\leq b$$, and thus $$a = G, b = 13 G$$.
Since $$2016 = a + b = G + 13G = 14 G$$, thus $$G = \frac{2016} { 14} = 144$$. Thus, $$\{a, b\} = \{ 144, 1872 \}$$.
Note by Arron Kau
2 years, 12 months ago
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The following is multiple choice question (with options) to answer.
Find the product of the localvalue and absolutevalue of 1 in 20168? | [
"100",
"110",
"151",
"120"
] | A | Local value of 1 = 1 x 100 = 100
Place value of 1 = 1
There fore = 1 x 100 = 100
A |
AQUA-RAT | AQUA-RAT-35922 | yields b -> 27.52553345330246, so we will use b = 28.
To check,
s = Sum[(-1)^n/(3 n + 1), {n, 0, \[Infinity]}];
s-(2Sum[(-1)^n/(3n+1),{n,0,b}]+(-1)^(b+1)/(3(b+1)+1))/2/.b->28.//Chop
(*-0.00009679302515119836 *)
We see we do indeed have an absolute value error of less than .0001 and have used just 29 terms to get it.
• You can use Chop to get rid ofthe spurious imaginary part of the final result. – corey979 Oct 12 '16 at 7:33
• @corey979 I have added that, thanks! – bobbym Oct 12 '16 at 10:17
The following is multiple choice question (with options) to answer.
On multiplying a number B by 153, the result obtained was 102325. However, it is found that both the 2's are wrong. Find the correct result. | [
"104345",
"107375",
"108385",
"109395"
] | D | The only thing you actually know about the correct number B is that it is divisible by 153 and has 5 as a factor.
You should immediately try to find the factors of 153 and look for them in the options.
153 = 9*17
Divisibility by 9 is easy to check. Only (D) satisfies. |
AQUA-RAT | AQUA-RAT-35923 | P = principal = $50,000 here. r = rate of interest = 0.055 here. n = number of years In compounded annually, A = P(1+r)^n After year 1, A = P(1+r) Atfer withdrawal of X, A = P(1+r) -X After year 2, A = [P(1+r) -X](1+r) = P(1+r)^2 -X(1+r) After withdrawal of X, A = P(1+r)^2 -X(1+r) -X After year 3, A = [P(1+r)^2 -X(1+r) -X](1+r) A = P(1+r)^3 -X[(1+r)^2 +(1+r)] After withdrawal of X, A = P(1+r)^3 -X[(1+r)^2 +(1+r)] -X . . After year 25, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] After withdrawal of X, A = P(1+r)^25 -X[(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] -X ------(i) And that is now equal to zero. The [(1+r)^24 +(1+r)^23 +(1+r)^22 +....+(1+r)] can be rewritten as [(1+r) +(1+r)^2 +(1+r)^3 +...+(1+r)^24]. It is a geometric series where common ratio = (1+r) a1 = (1+r) also n = 24 So, since (1+r) = (1+0.055) = 1.055, then, Sn = (a1)[(1 -r^n)/(1-r)]
The following is multiple choice question (with options) to answer.
P , Q, R enter into a partnership & their share are in the ratio 1/2 : 1/3 : 1/4 , after two months , P withdraws half of the capitals & after 10 months , a profit of Rs 336 is divided among them . What is R's share? | [
"96",
"120",
"134",
"144"
] | A | Explanation :
The ratio of their initial investment = 1/2 : 1/3 : 1/4
= 6 : 4: 3
Let's take the initial investment of P, Q and R as 6x, 4x and 3x respectively
A:B:C = (6x * 2 + 3x * 10) : 4x*12 : 3x*12
= (12+30) : 4*12 : 3*12
=(4+10) : 4*4 : 12
= 14 : 16 : 12
= 7 : 8 : 6
R's share = 336 * (6/21) = 96. Answer : Option A |
AQUA-RAT | AQUA-RAT-35924 | r
#DECILE 5
t_prag_top_total_income_50<-decili_total_income_neto\[5,1\]
t_prag_top_total_income_filter_50<-filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50)
t_prag_top_total_income_filter_50_tax<-sum(t_prag_top_total_income_filter_50$TAX)
t_tax_share_50<-((t_prag_top_total_income_filter_50_tax)/ZBIR_TOTAL_TAX)*100
t_prag_top_total_income_filter_50<-sum(t_prag_top_total_income_filter_50$NET_INCOME)
t_prag_top_total_income_filter_50a<-nrow(filter(data2, NET_INCOME> t_prag_top_total_income_40, NET_INCOME<=t_prag_top_total_income_50))
t_prag_top_total_income_50b<-((t_prag_top_total_income_filter_50)/ZBIR_TOTAL_NET_INCOME)*100
FINAL_CENTILE_TABLE50<-data.frame(cbind(t_prag_top_total_income_50,t_prag_top_total_income_filter_50,t_prag_top_total_income_filter_50a,t_prag_top_total_income_50b,t_prag_top_total_income_filter_50_tax , t_tax_share_50))
colnames(FINAL_CENTILE_TABLE50)<-c("Decile threshold","Total income in the decile","Number of persons in the centile","Share of the decile in total income (%)","Tax","Share tax(%)")
FINAL_DECILE_TABLE <- rbind(FINAL_DECILE_TABLE, FINAL_CENTILE_TABLE50)
The following is multiple choice question (with options) to answer.
The income derived from a Rs. 100, 13% stock at Rs. 105,is : | [
"Rs. 5",
"Rs. 8",
"Rs. 13",
"Rs. 18"
] | C | Solution
Income on Rs. 100 stock = Rs. 13
Answer C |
AQUA-RAT | AQUA-RAT-35925 | Now
$$y+5 = 2k+5$$
Now we just need to show that $$2k+5$$ is 2 times an integer plus 1
$$2k+5 = 2(k+2)+1$$
So $$2k+5$$ is odd because it can be written in the form 2*integer +1 where the integer here is $$k+2$$. So $$y+5$$ is odd since $$y+5 = 2k+5$$
So if any number is even. Then that number plus 5 is odd. It doesn't matter if the original number is 3x or 8z or 3x^2-5x+x^3 etc...
The following is multiple choice question (with options) to answer.
If y = 2 + 2K and y≠0y≠0, then 5/y + 3/y + 1/y + 1/y = ? | [
"1/(8+8k)",
"5/(1+k)",
"1/(8+k)",
"4/(8+k)"
] | B | 5/Y + 3/Y + 1/Y + 1/Y
= 10/Y
= 10/(2 + 2K)
= 5/(1 + K)
ANSWER:B |
AQUA-RAT | AQUA-RAT-35926 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If there are 6 men and 55 tasks , how many tasks will be left undone if they perform equal number of tasks? | [
"5",
"2",
"3",
"1"
] | D | There are 6 men
There are 55 tasks
If they perform equal number of tasks, divide 55 by 6 and the remainder will be the number of tasks left undone
55 = (6 x 9) + 1
So 1 task is left undone
ANSWER D |
AQUA-RAT | AQUA-RAT-35927 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 900 m long? | [
"266 Sec",
"210 Sec",
"776 Sec",
"166 Sec"
] | B | L = S*T
S= 1200/120
S= 10 m/Sec.
Total length (D)= 2100 m
T = D/S
T = 2100/10
T = 210 Sec
Answer: B |
AQUA-RAT | AQUA-RAT-35928 | #### HallsofIvy
Science Advisor
When tiny-tim says "expand the brackets first" I believe he is referring to the generalized binomial expansion. And you should only need the first few terms.
#### Mark44
Mentor
Another approach is to multiply the original expression by 1 like so:
$$(x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}$$
This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.
After using this technique, the numerator can be simplified to a single term, and the limit of the whole expression can be found without much trouble.
#### djeitnstine
Gold Member
Another approach is to multiply the original expression by 1 like so:
$$(x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}$$
This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.
The following is multiple choice question (with options) to answer.
The expression (a+b+c)10 is expanded and simplified. How many terms are in the resulting
expression? | [
"30",
"44",
"55",
"66"
] | D | (a+b+c)10 = [(a+b)+c]10. Let x = a+b. We see that (x+c)10 = x10+10x9c+
+ c10 has 11 distinct terms. Also, the ith term of this expansion has 11 - i distinct terms
(since x10-i = (a + b)10-i has 11 - i distinct terms). Hence there are 11 + 10 +...+ 1 = 66
terms in this expansion.
correct answer D |
AQUA-RAT | AQUA-RAT-35929 | 6. Hello, Migotek84!
Here's a very primitive method . . .
Compute: . $17^{-1}\text{ (mod 23)}$
Let: . $x \:\equiv\:17^{-1}\text{ (mod 23)}$
Multiply by 17: . $17x \:\equiv\:1\text{ (mod 23)}$
Then: . $17x \:=\:23a + 1\:\text{ for some integer }a.$
$\text{Solve for }x\!:\;\;x \:=\:\dfrac{23a+1}{17} \:=\:a + \dfrac{6a+1}{17}$ .[1]
Since $x$ is an integer, $6a+1$ is a multiple of 17.
. . $6a+1 \:=\:17b\:\text{ for some integer }b.$
$\text{Solve for }a\!:\;\;a \:=\:\dfrac{17b-1}{6} \:=\:2b + \dfrac{5b-1}{6}$ .[2]
Since $a$ is an integer, $5b-1$ is a multiple of 6.
. . $5b-1 \:=\:6c\:\text{ for some integer }c.$
$\text{Solve for }b\!:\;\;b \:=\:\dfrac{6c+1}{5} \:=\:c + \dfrac{c+1}{5}$ .[3]
Since $b$ is an integer, $c+1$ is a multiple of 5.
. . The first time this happens is when . $c = 4$
Substitute into [3]: . $b \:=\:4 + \frac{4+1}{5} \quad\Rightarrow\quad b \:=\:5$
The following is multiple choice question (with options) to answer.
How many integer values Q are there for x such that 1 < 3x + 5 < 17? | [
"Two",
"Three",
"Four",
"Five"
] | D | 1 < 3x + 5 < 17
=> -4 < 3x < 12
=> -4/3 < x < 4
x can take integer values Q=-1,0 , 1 , 2 , 3
Answer D |
AQUA-RAT | AQUA-RAT-35930 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
In a 500 m race, the ratio of speeds of two runners P and Q is 3:5. P has a start of 200 m then the distance between P and Q at the finish of the race is: | [
"Both reach at the same time",
"P wins by 100 m",
"Q wins by 50 m",
"Q wins by 100 m"
] | A | Explanation :
When P starts running for 300 m and Q for 500 m then, the ratio of speeds of P and Q is 3 : 5.
According to question P has already covered 200 m when Q is starting at starting point and Q needs to cove 300 m more mean while P covers this 300 m when Q can cover 500 m. So, both P and Q will reach the finishing point at the same time.
Thus, Both reach at the same time.
Answer : A |
AQUA-RAT | AQUA-RAT-35931 | # Difference between revisions of "2019 AMC 10A Problems/Problem 23"
## Problem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
## Solution 1
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$
The following is multiple choice question (with options) to answer.
Andy solves problems 75 to 125 inclusive in a Math exercise. How many problems does he solve? | [
"53",
"52",
"51",
"50"
] | C | 125-75+1= 51
'C' is the answer |
AQUA-RAT | AQUA-RAT-35932 | 10.21, 9.82, 10.81, 10.3, 9.81, 11.48, 8.51, 9.55, 10.41, 12.17, 9.9, 9.07, 10.51, 10.26, 10.62, 10.84, 9.67, 9.75, 8.84, 9.85, 10.41, 9.18, 10.93, 11.41, 9.52]
The following is multiple choice question (with options) to answer.
0,1,2.9,5.61,9.05,? | [
"11.15",
"12.15",
"13.15",
"14.15"
] | C | The series is 0, 1, 2.9, 5.61, 9.05, ...
Fisrt oder differences are 1, 1.9, 2.71, 3.44, ...
Second order differences are 0.9, 0.81, 0.73, ...
Note that the second order differences are powers of 0.9 (0.9, 0.81, 0.729, ...), rounded off to two decimal digits. So next term in second order difference is 0.9^4 = 0.6561 = 0.66. Then, the next term in first order difference is 3.44+0.66 = 4.1. So, the term in the series is 9.05+4.1 = 13.15
ANSWER:C |
AQUA-RAT | AQUA-RAT-35933 | # In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty?
5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.
My attempt:-
First choose 3 balls to be placed in 3 boxes so that none of them remain empty in ${{5}\choose{3}}\cdot3! = 60$ ways.
Now remaining 2 balls can go into any of the 3 boxes in $3\cdot3 = 9$ ways.
Total number of ways $= 60\cdot9 = 540$.
Where am I going wrong ?
• When you place the last two balls, you are over counting. Your answer better be less than $243$. Oct 1 '17 at 10:51
There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are $$\binom{3}{1}2^5$$ ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is $$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$ by the Inclusion-Exclusion Principle.
Where am I going wrong?
The following is multiple choice question (with options) to answer.
There are 6 boxes numbered 1, 2,…,.6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is | [
"21",
"52",
"45",
"63"
] | A | List down possibilities: From only 1 box all the way to all 6
If only one of the boxes has a green ball, it can be any of the 6 boxes. So, we have 6 possibilities.
If two of the boxes have green balls and then there are 5 consecutive sets of 2 boxes. 12, 23, 34, 45, 56.
If 3 of the boxes have green balls, there are 4 possibilities: 123, 234, 345, 456.
If 4 boxes have green balls, there are 3 possibilities: 1234, 2345, 3456.
If 5 boxes have green balls, there are 2 possibilities: 12345, 23456.
If all 6 boxes have green balls, there is just 1 possibility.
Total number of possibilities = 6 + 5 + 4 + 3 + 2 + 1 = 21.
Ans: A |
AQUA-RAT | AQUA-RAT-35934 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How long does a train 110 m long traveling at 90 kmph takes to cross a bridge of 170 m in length? | [
"sec",
"sec",
"sec",
"sec"
] | D | D = 110 + 170 = 280 m
S = 90 * 5/18 = 50/2
T = 280 * 2/50= 11.2 sec
Answer: D |
AQUA-RAT | AQUA-RAT-35935 | python, python-3.x, calculator
Title: Taxi fare calculator, trip recorder, and fuel calculator I solved this practice problem in preparation for school exams.
The owner of a taxi company wants a system that calculates how much money his taxis take in one day.
Write and test a program for the owner.
Your program must include appropriate prompts for the entry of data.
Error messages and other output need to be set out clearly and understandably.
All variables, constants and other identifiers must have meaningful names.
You will need to complete these three tasks. Each task must be fully tested.
TASK 1 – calculate the money owed for one trip
The cost of a trip in a taxi is calculated based on the numbers of KMs traveled and the type of taxi that you are traveling in. The taxi company has three different types of taxi available:
a saloon car, that seats up to 4,
a people carrier, that seats up 8,
a mini-van, that seats up to 12.
For a trip in a saloon car the base amount is RM2.50 and then a charge of RM1.00 per KM. For a trip in a people carrier the base amount is RM4.00 and a charge of RM1.25 per KM. For a trip in a mini- van the base amount is RM5.00 and a charge of RM1.50 per KM. The minimum trip length is 3KM and the maximum trip length is 50KM. Once the trip is complete a 6% service tax is added.
TASK 2 – record what happens in a day
The owner of the taxi company wants to keep a record of the journeys done by each one of his taxis in a day. For each one of his three taxis record the length of each trip and the number of people carried. Your program should store a maximum of 24 trips or 350km worth of trips, whichever comes first. Your program should be able to output a list of the jobs done by each one of the three taxis.
TASK 3 – calculate the money taken for all the taxis at the end of the day.
At the end of the day use the data stored for each taxi to calculate the total amount of money taken and the total number of people carried by each one of the three taxis. Using the average price of RM2.79 per litre use the information in the table below to calculate the fuel cost for each taxi:
The following is multiple choice question (with options) to answer.
A certain NYC taxi driver has decided to start charging a rate of r cents per person per mile. How much, in dollars, would it cost 6 people to travel x miles if he decides to give them a 50% discount? | [
"3xr/100",
"3x/200r",
"3r/200x",
"3xr/200"
] | A | 6xr/2 is in cents - 6xr/200 dollars = 3xr/100 dollars
Answer : A |
AQUA-RAT | AQUA-RAT-35936 | I see from your profile that you’re a programmer. Extreme cases like these are analogous to testing boundary conditions in your code (things like loops that execute 0 times).
Let x to be your gross salary and y = 1600 to be your net salary. Because tax is 20% of gross salary then we have
x = y + tax = 1600 + 0.2x
=> 0.8x = 1600
=> x = 2000 USD
The following is multiple choice question (with options) to answer.
In Township K each property is taxed at 8 percent of its assessed value. If the assessed value of a property in Township K is increased from $20,000 to $24,000, by how much will the property tax increase? | [
"$32",
"$50",
"$320",
"$400"
] | C | Property tax as a percentage of its assessed value = 8%
Increase in assessed value of the property = 24000 - 20000 = 4,000 $
Increase in property tax = (8/100)* 4000 = 320 $
Answer C |
AQUA-RAT | AQUA-RAT-35937 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train covers a distance of 12 km in 10 min. If it takes 6 sec to pass a telegraph post, then the length of the train is? | [
"100",
"110",
"120",
"130"
] | C | Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec = 20 m/sec.
Length of the train = 20 * 6 = 120 m.
Answer: Option C |
AQUA-RAT | AQUA-RAT-35938 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A fruit vendor purchased 20 dozens of apples at $15 per dozen. But one-fourth of the apples were rotten and had to be thrown away. He sold two-third of the remaining apples at a$22.50 per dozen. At what price per dozen should he sell the remaining apples to make neither a profit nor a loss? | [
"$17",
"$14",
"$15",
"$18"
] | C | C
$15
CP of 20 dozen of apples = 15 * 20 = $300
Number of apples which are rotten = 1/4 * 20 = 5 dozen.
SP of two-third of remaining apples = (2/3 * 15) * 22.5 = $225
SP of remaining 5 dozens of apples to make no profit and no loss =(300 - 225) = $75.
SP of 1 dozen apples = 75/5 = $15. |
AQUA-RAT | AQUA-RAT-35939 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
A person crosses a 1080 m long street in 10 minutes. What is his speed in km per hour? | [
"4.1",
"6.4",
"4.8",
"5.4"
] | B | Speed = 1080/(10x60) m/sec
= 1.8 m/sec.
Converting m/sec to km/hr =1.8 x(18/5) km/hr
= 6.4 km/hr.
ANSWER :B |
AQUA-RAT | AQUA-RAT-35940 | 4. (ODD)(ODD) = ODD
5. (ODD)(EVEN) = EVEN
6. (EVEN)(EVEN) = EVEN
Target question: Is integer n EVEN?
Statement 1: n² - 1 is an odd integer
n² - 1 = (n + 1)(n - 1)
So, statement 1 is telling us that (n + 1)(n - 1) = ODD
From rule #4 (above), we can conclude that BOTH (n + 1) and (n - 1) are ODD
If (n + 1) is ODD, then n must be EVEN (since 1 is ODD, we can apply rule #2 to conclude that n is EVEN)
If (n - 1) is ODD, then n must be EVEN (by rule #2 )
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: 3n + 4 is an even integer
In other words, (3n + EVEN) is EVEN
From rule #3, we can conclude that 3n is EVEN
Since 3 is odd, we can write: (ODD)(n) = EVEN
From rule #5, we can conclude that n is EVEN
So, the answer to the target question is YES, n is even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Cheers,
Brent
_________________
Test confidently with gmatprepnow.com
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Posts: 13278
Re: If n is an integer, is n even? [#permalink]
### Show Tags
20 Jun 2019, 03:27
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
The following is multiple choice question (with options) to answer.
If nn is a positive integer and (n+1)(n+3)(n+1)(n+3) is odd, then (n+2)(n+4)(n+2)(n+4) must be a multiple of which one of the following? | [
"2",
"4",
"5",
"8"
] | D | (n+1)(n+3)(n+1)(n+3) is odd only when both (n+1)(n+1) and (n+3)(n+3) are odd. This is possible only when nn is even.
Hence, n=2mn=2m, where mm is a positive integer. Then,
(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)
=4 * (product of two consecutive positive integers, one which must be even)=4 * (product of two consecutive positive integers, one which must be even) =4 * (an even number), and this equals a number that is at least a multiple of 8=4 * (an even number), and this equals a number that is at least a multiple of 8
Hence, the answer is (D). |
AQUA-RAT | AQUA-RAT-35941 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
The ratio between the length and the breadth of a rectangular park is 1 : 2. If a man cycling along the boundary of the park at the speed of 6 km/hr completes one round in 6 min, then the area of the park (in sq. m) is? | [
"600m",
"6000 m",
"20000 m",
"100 m"
] | C | Perimeter = Distance covered in 6 min. = (6000 x 6)/60 m = 600 m.
Let length = 1x metres and breadth = 2x metres.
Then, 2(1x + 2x) = 600 or x = 100.
Length = 100 m and Breadth = 200 m.
Area = (100 x 200) m2 = 20000 m
C |
AQUA-RAT | AQUA-RAT-35942 | Seven draws
The first six draws are Others: . ${9\choose6}\text{ ways } \hdots\:P(\text{6 Others}) \:=\:\frac{{9\choose6}}{{15\choose6}}$
And the 7th is a 75w: . $\frac{6}{9}$
. . Hence: . $P(\text{7 draws}) \;=\;\frac{{9\choose6}}{{15\choose6}}\left(\frac{6 }{9}\right)$
Eight draws
The first seven draws are Others: . ${9\choose7}\text{ ways }\hdots\:P(\text{7 Others}) \:=\:\frac{{9\choose7}}{{15\choose7}}$
And the 8th is a 75w: . $\frac{6}{8}$
. . Hence: . $P(\text{8 draws}) \;=\;\frac{{9\choose7}}{{15\choose7}}\left(\frac{6 }{8}\right)$
Nine draws
The first eight draws are Others: . ${9\choose8}\text{ ways }\hdots\:P(\text{8 Others}) \:=\:\frac{{9\choose8}}{{15\choose8}}$
And the 9th is a 75w: . $\frac{6}{7}$
. . Hence: . $P(\text{9 draws}) \;=\;\frac{{9\choose8}}{{15\choose8}}\left(\frac{6 }{7}\right)$
The following is multiple choice question (with options) to answer.
A standard-sized cigarette can be rolled out of 6 standard-sized cigarette butts. How many cigarettes can be made and smoked from 36 butts? | [
"2",
"3",
"5",
"7"
] | D | D
No, 6 is not the answer! It is not correct, because after the 6 cigarettes have been smoked there will again be 6 butts which can be made into another cigarette.
The answer is 7 |
AQUA-RAT | AQUA-RAT-35943 | mass, astrophysics, astronomy, sun
Title: How massive is the Sun compared to other stars? I was just reading "Rare Earth" by Peter Ward and Donald Brownlee and they claim that $95\%$ of stars are less massive than the Sun. But, I was always told the Sun is an average star. Which is it? The notion that the sun is an 'average' star, is a popular notion, not a scientific one. It is to my knowledge based on the spectral colour-classification (OBAFGKM..) of stars, where the sun (G2-class) hits about the center of this classification, depending how many classes one adds at the end.
However by number, the sun is not average, it is on the rarer end of the mass distribution function. See also this graphic, provided by the University of Colorado:
This has to do with two effects:
1. Heavier stars burn through their fuel much faster than low-mass stars.
The lifetime $t_{\rm life}$ of a star on the main sequence goes like $$t_{\rm life} \sim 10^{10}yrs \left(\frac{M}{M_{\odot}}\right)^{-2.5}$$ so you can do the math yourself how short more massive stars live.
2. Lower mass stars are formed more frequently.
In star-forming environments it is the 3D-turbulence that clumps gas together at smaller and smaller scales (as opposed to larger and larger scales). It is because of this, that there are more small clumps that become gravitationally unstable than large clumps, which then form stars of a given clump mass.
The following is multiple choice question (with options) to answer.
A certain galaxy is known to comprise approximately 2x 10^11 stars. Of every 50 million of these stars, one is larger in mass than our sun. Approximately how many stars in this galaxy are larger than the sun? | [
"800",
"1,250",
"8,000",
"4,000"
] | D | 2*10^11
50 mln = 5*10^7
we divide 10^11 by 10^7 and we get (10^4)*2 = 20,000 and divide by 5. the result is 4,000 (c)
D |
AQUA-RAT | AQUA-RAT-35944 | Split it into disjoint events, and then add up their probabilities:
• Event #$1$: choosing Jason, then Kim, then Ellie
• Event #$2$: choosing Jason, then Ellie, then Kim
• Event #$3$: choosing Kim, then Jason, then Ellie
• Event #$4$: choosing Kim, then Ellie, then Jason
• Event #$5$: choosing Ellie, then Jason, then Kim
• Event #$6$: choosing Ellie, then Kim, then Jason
As you already calculated, the probability of each event is $\frac{1}{18\cdot17\cdot16}$
Therefore, the probability of either one of these events is $\frac{6}{18\cdot17\cdot16}$
The following is multiple choice question (with options) to answer.
3 singers will pick a song among a list of 3 songs. What is the probability that exactly one singer will pick a different song? | [
"3!/3^3",
"1/3^3",
"3/3!",
"3/3^3"
] | A | Each singer out of 3 has 3 options, hence total # of outcomes is 3^3;
Favorable outcomes will be 3!, which is # of ways to assign 3 different song to 3 singers.
P=favorable/total=3!/3^3
Answer: A. |
AQUA-RAT | AQUA-RAT-35945 | python
#Replace random letters in word - word = ward, wore, wond, etc
word_section = sorted(random.sample(word_range, 2))
if word_section[0] or word_section[1] != word_len:
new_word = word[random.randint(0, word_section[0]):random.randint(word_section[1], word_len)]
new_word_len = len(new_word)
for replacement in xrange(random.randint(0, new_word_len/2)):
replacement_index = random.randint(0, new_word_len-1)
new_letter = random.choice(all_letters)
new_word = new_word[:replacement_index]+new_letter+new_word[replacement_index+1:]
#Only add to list if
if new_word != original_word:
word_list.append(new_word)
return word_list
def debug_grid(self):
"""Output a grid showing the related ID of each cell.
>>> WordSearch(4, 4).debug_grid()
00 01 02 03
04 05 06 07
08 09 10 11
12 13 14 15
>>> WordSearch(12, 3).debug_grid()
00 01 02 03 04 05 06 07 08 09 10 11
12 13 14 15 16 17 18 19 20 21 22 23
24 25 26 27 28 29 30 31 32 33 34 35
"""
count = 0
max_len = len(str(self.x*self.y-1))
for i in range(self.y):
print ' '.join(str(i+count).zfill(max_len) for i in range(self.x))
count += self.x
def get_coordinate(self, id=0, **kwargs):
"""Convert an ID into its coordinate.
Only needs the X grid value to calculate, the Y value checks it is within range.
Parameters:
id:
Coordinate ID of the cell.
The following is multiple choice question (with options) to answer.
If in a certain code "RANGE" is coded as 12345 and "RANDOM" is coded as 123678. Then the code for the word
"ROD" would be | [
"156",
"174",
"176",
"180"
] | C | R-1
O-7
D-6
so for mango the code is 176
ANSWER:C |
AQUA-RAT | AQUA-RAT-35946 | # Math Help - Integers
1. ## Integers
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
2. ## Re: Integers
Originally Posted by fsiwaju
Hello,
I am trying to solve this problem:
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
1. Set $j = k + 2m, m\in \mathbb{N}$. Then (j - k) is even.
2. Check all 5 terms, replacing j by k + 2m. For instance:
$j + 2k = k + 2m + 2k = 3k + 2m~\implies~$ ..... j+2k is only even if k is even because then j is even too.
...
$jk + j = (k+2m)k + (k + 2m) = (k + 2m)(k + 1)~\implies~$..... (jk + j) is even if k is even because then the 1st bracket would be even too; (jk + j) is even if k is odd because then the 2nd bracket would be even too. A product is even if one factor is even.
3. ## Re: Integers
If j and k are integers and j - k is even, which of the following must be even?
a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
if j - k is even, then j and k are both even or both odd.
4. ## Re: Integers
if j - k is even either j and k are both odd, or j and k are both even.
The following is multiple choice question (with options) to answer.
x and y are positive integers of J. If 1/x + 1/y < 2, which of the following must be true? | [
" x + y > 4",
" xy>1",
" x/y + y/x < 1",
" (x - y)^2 > 0"
] | B | Answer is B:
1/X + 1/Y < 2
The maximum value of 1/X is 1 because if X equals any other number greater than one it will be a fraction. The same is true with 1/Y.
So 1/X and 1/Y will always be less than 2 as long as both X and Y are not both equal to one at the same time.
Another way of putting it is:
X*Y>1.B |
AQUA-RAT | AQUA-RAT-35947 | You've got what it takes, but it will take everything you've got
Intern
Joined: 30 Nov 2017
Posts: 42
Re: Working at constant rate, pump X pumped out half of the water in a flo [#permalink]
### Show Tags
15 Feb 2018, 09:56
Suppose X extracts x liters/hour, while Y extracts y liters/hour
In 4 hours, X extracts 4 x liters
This is half of the basement capacity. So, total volume of water in the basement = 8x liters
After 4 hours, Y was started as well. They flush the remaining 4 x in 3 hours.
In 3 hours, X and Y combined would flush 3*(x + y)
It is given that 3*(x + y) = 4x
This gives x = 3y
Operating alone how much would Y take?
Total water = 8x
Y's capacity = y liters/hour
So, time taken by Y = 8x/y
We know x = 3y
So, x/y = 3
So, time taken by Y = 8x/y = 8*3 = 24
Re: Working at constant rate, pump X pumped out half of the water in a flo &nbs [#permalink] 15 Feb 2018, 09:56
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Working alone at its constant rate, pump U pumped out ¼ of the water in a tank in 2 hours. Then pumps V and W started working and the three pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 3 hours. If pump V, working alone at its constant rate, would have taken 18 hours to pump out the rest of the water, how many hours would it have taken pump W, working alone at its constant rate, to pump out all of the water that was pumped out of the tank? | [
"6",
"12",
"15",
"18"
] | B | Rate of pump U = 1/8
3 hours are required to pump out the remaining (3/4)ths of tank --> 1 hr to pump out 1/4
Rate of U + Rate of V + Rate of W = 1/4
Rate of V + Rate of W = 1/4 - 1/8 = 1/8
V takes 18 hours to pump out the remaining (3/4)ths of tank --> 6 hrs per (1/4)ths --> 24 hrs to pump out fully.
Rate of V = 1/24
1/24 + Rate of W = 1/8
Rate of W = 1/8 - 1/24 = 1/12
Time required to pump out all the water by W = 12 hrs
Answer: B |
AQUA-RAT | AQUA-RAT-35948 | c#, interview-questions, finance
Title: Sales Tax Problem, rejected for not being up to their standards I applied for a Junior .NET Developer Position recently, and I was asked to solve a problem (Sales Tax Program) and I was rejected for some reason. The question goes like this:
Basic sales tax is applicable at a rate of 10% on all goods, except food, medicines and books. For all imported goods, a tax rate of additional 5% is applied irrespective of their type. Generate a receipt which shows names of all the items purchased and their prices (including taxes), finishing with total cost of the items, and total amount sales tax paid. The rounding rules for sales tax are that for a tax rate of n%, a shelf price of P contains(np/100 rounded up to nearest 0.05) amount of sales tax. Write an application to generate receipt with all these details.
They gave me the input to take and output should come:
Input 3:
1 imported bottle of perfume at 27.99
1 bottle of perfume at 18.99
1 packet of headache pills at 9.75
1 box of imported chocolates at 11.25
Output 3:
1 imported bottle of perfume: 32.19
1 bottle of perfume: 20.89
1 packet of headache pills: 9.75
1 imported box of chocolates: 11.85
Sales Taxes: 6.70
Total: 74.68
They gave two more sets of input, but I just didn't put that to save some time.
Please review my code and let me know what's wrong with that and what are the concepts I need to focus on to solve a problem like this.
namespace Products
{
class Program
{
static void Main(string[] args)
{
try
{
// calling the Generatereciept class and invoking Generate method to display all the product details, price, taxes and total after taxes
GenerateReciept generate = new GenerateReciept();
generate.Generate();
}
catch (Exception ex)
{
// writing the exception to console.. we can log it to a database or eventviewer based on the requirements
Console.WriteLine(ex.Message);
}
The following is multiple choice question (with options) to answer.
A tourist does not have to pay tax on the first $600 of goods he purchases in Country B, but does have to pay a 7 percent tax on the portion of the total value that is in excess of $600. What tax must be paid by a tourist if he buys goods with a total value of $1720? | [
"$54.00",
"$78.40",
"$90.00",
"$100.80"
] | B | Correct Answer: B
The tourist must pay tax on $1720 - $600 = $1120. Thus, the amount of tax he has to pay is 0.07 ($1120) = $78.40. The correct answer is B. |
AQUA-RAT | AQUA-RAT-35949 | ### Show Tags
19 Aug 2015, 01:34
2
KUDOS
1
This post was
BOOKMARKED
Let's assume that the total no. of people is n and the initial average is x.
So we can assume that the total sum of ages would be nx initially.
When the guy aged 39 is added to the total, the new sum of ages would be (nx+39). The new average now would be (nx+39)/n+1. We already know that now the average(x) is increased by 2. so we can equate the two above saying:
(nx+39)/(n+1) = x+2
solving this equation:
nx+39=(x+2)(n+1)
nx+39=nx +x +2n+2
we get,
2n+x=37
When the guy aged 15 is added to the total, the new sum of ages would be (nx+15). The new average now would be (nx+15)/n+1. We already know that now the average(x) is decreased by 1. so we can equate the two above saying:
(nx+15)/(n+1) = x-1
we get
x-n=16
solving the two equation simultaneously, we get n's value as 7.
Kudos [?]: 24 [2], given: 0
Manager
Joined: 21 Jan 2015
Posts: 149
Kudos [?]: 121 [0], given: 24
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Other)
Re: When a person aged 39 is added to a group of n people, the average age [#permalink]
### Show Tags
19 Aug 2015, 01:55
1
This post was
BOOKMARKED
Bunuel wrote:
When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Ans: A
The following is multiple choice question (with options) to answer.
The average age of an adult class is 40years. 12new students with an avg age of 32years join the class. Therefore decreasing the average by 4year. Find what was theoriginal strength of class? | [
"8",
"12",
"15",
"17"
] | B | Let original strength = y
Then , 40y + 12 x 32 = ( y + 12) x 36
⇒ 40y + 384 = 36y + 432
⇒ 4y = 48
∴ y = 12
B |
AQUA-RAT | AQUA-RAT-35950 | It takes 6 days for 3 women and 2 men working together to complete a work --> sum the rates: $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
3 men would do the same work 5 days sooner than 9 women --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Solving: $$m=15$$ and $$w=90$$. $$\frac{w}{m}=6$$.
bunuel, pls help
if i equate time i am not getting it pls tell me where i am going wrong
lets take 1 woman can complete the work in 'w' days and 1 man can complete in 'm' days
so, it becomes w/3+m/2=6
and m/3+5=w/9
but i am getting the answer wrong.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Time n Work Problem [#permalink]
### Show Tags
18 Jan 2014, 02:22
saggii27 wrote:
Bunuel wrote:
nonameee wrote:
Guys, even if you know the solution right away, it takes several minutes (definitely more than 3) to just write it down to find the answer. Is it a real GMAT question? Can something like that be expected on the real test?
Below is another solution which is a little bit faster.
It takes 6 days for 3 women and 2 men working together to complete a work.3 men would do the same work 5 days sooner than 9 women.How many times does the output of a man exceed that of a woman?
A. 3 times
B. 4 times
C. 5 times
D. 6 times
E. 7 times
Let one woman complete the job in $$w$$ days and one man in $$m$$ days. So the rate of 1 woman is $$\frac{1}{w}$$ job/day and the rate of 1 man is $$\frac{1}{m}$$ job/day.
The following is multiple choice question (with options) to answer.
X is three times as fast as Y and working together, they can complete a work in 20 days. In how many days can Y alone complete the work? | [
"11 days",
"80 days",
"13 days",
"15 days"
] | B | X = 3Y
X + Y = 3Y + Y = 4Y
These 4Y people together can do the work in 20 days, which means Y can alone do the work in 4*20 = 80 days.
Answer : B |
AQUA-RAT | AQUA-RAT-35951 | # math
How is the circumference of a circle with radius 9cm related to the circumference of a circle with diameter 9cm?
i don't understand this.Help?? Thanks:)
1. 👍
2. 👎
3. 👁
c = pi x d.
The diamter is 2 x radius. So, the diameter of the first cirlce is 18cm. You know the diameter of the second circle. So, how do the circumferences compare?
1. 👍
2. 👎
2. is the second circle twice as big? like is that how they compare?
1. 👍
2. 👎
3. the circumference of the larger circle is indeed twice as long as that of the smaller one.
but...
the area of the larger circle is 4 times that of the area of the smaller.
Proof:
area of smaller = pi(4.5)^2 = 20.25pi
area of larger = pi(9^2) = 81pi
20.25pi/(81pi) = 1/4
1. 👍
2. 👎
4. alright thx!
1. 👍
2. 👎
## Similar Questions
1. ### Math ~ Check Answers ~
Find the circumference of the circle (use 3.14 for pi). Show your work. Round to the nearest tenth. the radius is 23 yd My Answer: ???? •Circumference = (2) (3.14) (23) •3.14 x 2 x 23 = 144.44
2. ### Mathematics
1. Find the circumference of the given circle. Round to the nearest tenth. (Circle with radius of 3.5 cm) -22.0 cm -38.5 cm -11.0 cm -42.8 cm 2. Find the area of the given circle. Round to the nearest tenth. (Circle with a
3. ### Math
The formula C = 2 x Pi x r is used to calculate the circumference of a circle when given the radius. What is the formula for calculating the radius when given the circumference? C/2Pi = R*** C/2R = Pi 2 x Pi x C = R 2 x Pi/ C = R
4. ### MATH
The following is multiple choice question (with options) to answer.
Find the area of circle whose radius is 9m? | [
"138",
"154",
"254.6",
"280"
] | C | 22/7 * 9 * 9
= 254.6
Answer:C |
AQUA-RAT | AQUA-RAT-35952 | 3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16
6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)
7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100
The following is multiple choice question (with options) to answer.
A sun is divided among X, Y and Z in such a way that for each rupee X gets, Y gets 45 paisa and Z gets 50 paisa. If the share of Y is RS. 63, what is the total amount? | [
"115",
"273",
"117",
"118"
] | B | x:y:z = 100:45:50
20:9:10
9 --- 63
39 --- ? => 273
ANSWER:B |
AQUA-RAT | AQUA-RAT-35953 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
The salaries of A, B and C are of ratio 2:3:5. If the increments of 15%, 10% and 20% are done to their respective salaries, then find the new ratio of their salaries | [
"20:33:60",
"21:33:60",
"22:33:60",
"23:33:60"
] | D | Explanation:
Let A salary be 2k
B salary be 3k and C salary be 5k
A's new salary = 115/100∗2k=23/10k
B's new salary = 110/100∗3k=33/10k
C's new salary = 120/100∗5k=6k
New ratio = 23k/10: 33k/10: 6k=23:33:60
Option D |
AQUA-RAT | AQUA-RAT-35954 | Q.8 Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check if each has used up an equal fraction of her/his pencils?
Total number of pencils Ramesh had = 20
Number of pencils used by Ramesh = 10
$$\therefore$$ Fraction = $$\frac{10}{20} = \frac{1}{2}$$
Total number of pencils Sheelu had = 50
Number of pencils used by Sheelu = 25
$$\therefore$$ Fraction = $$\frac{25}{50} = \frac{1}{2}$$
Total number of pencils Jamaal had = 80
Number of pencils used by Jamaal = 40
$$\therefore$$ Fraction = $$\frac{40}{80} = \frac{1}{2}$$
Yes, each has used up an equal fraction of pencils i.e $$\frac{1}{2}$$
The following is multiple choice question (with options) to answer.
When an amount was distributed among 14 boys, each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount? | [
"5040",
"5020",
"5010",
"5030"
] | A | Let the total amount be Rs. X the,
x/14 - x/18 = 80<=> 2x/126 = 80 <=> x/63 =63 x 80 = 5040.
Hence the total amount is 5040.
Answer is A. |
AQUA-RAT | AQUA-RAT-35955 | earth
Well, lets look at the values returned compared to days in a non-leap year:
Month N1 Day Diff
1. 30 31 -1
2. 61 59 +2
3. 91 90 +1
4. 122 120 +2
5. 152 151 +1
6. 183 181 +2
7. 213 212 +1
8. 244 243 +1
9. 275 273 +2
10. 305 304 +1
11. 336 334 +2
12. 366 365 +1
A bit messy, right? Now, remember that you're subtracting 30 from the total at the end to get N, and we're adding in the current date. This means that although we multiply our current month to get N1, we're actually using this to calculate the dates from the months prior to our current month! Thus if we take the value of N1, subtract it by 30, and compare it to the preceding month, the chart will come out like this:
Month N1 Day Diff
1. 31 31 0
2. 61 59 +2
3. 92 90 +2
4. 122 120 +2
5. 153 151 +2
6. 183 181 +2
7. 214 212 +2
8. 245 243 +2
9. 275 273 +2
10. 306 304 +2
11. 336 334 +2
12. --- 365 ---
From this, you can see that the value of N1 will equal 2 greater than the actual date for any day in which it is March or later. This is perfect, as N2 is already a formula determining this for catching leap days. Note, these would all equal +1 on a leap year, as in another day would have been added in February. Thus coming back to the final calculation:
N = N1 - (N2 * N3) + day - 30
The following is multiple choice question (with options) to answer.
The average salary of a person for the months of January, February, March and April is Rs.8000 and that for the months February, March, April and May is Rs.8500. If his salary for the month of May is Rs.6500, find his salary for the month of January? | [
"3999",
"7867",
"4500",
"2789"
] | C | Sum of the salaries of the person for the months of January, February, March and April = 4 * 8000 = 32000 ----(1)
Sum of the salaries of the person for the months of February, March, April and May = 4 * 8500 = 34000 ----(2)
(2)-(1) i.e. May - Jan = 2000
Salary of May is Rs.6500
Salary of January
= Rs.4500
Answer: C |
AQUA-RAT | AQUA-RAT-35956 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two trains 121 meters and 165 meters in length respectively are running in opposite directions, one at the rate of 80 km and the other at the rate of 55 kmph. In what time will they be completely clear of each other from the moment they meet? | [
"7.19",
"7.17",
"7.62",
"7.15"
] | C | T = (121 + 165)/ (80 + 55) * 18/5
T = 7.62
Answer:C |
AQUA-RAT | AQUA-RAT-35957 | A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:
(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:
99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)
Therefore, the probability is: 24,000 / 90,000 ==4 / 15
Feb 4, 2023
The following is multiple choice question (with options) to answer.
How many of the integers between 25 and 95 are even ? | [
"35",
"90",
"11",
"10"
] | A | Number start between 25 to 95 is 70 numbers
half of them is even..which is 35
ANSWER:A |
AQUA-RAT | AQUA-RAT-35958 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 6 hours. If both the taps are opened simultaneously, then after how much time will the cistern get filled? | [
"5.2 hrs",
"2.9 hrs",
"1.9 hrs",
"6 hrs"
] | D | Net part filled in 1 hour = (1/3 - 1/6) = 1/6
The cistern will be filled in 6/1 hrs i.e., 6 hrs.
Answer:D |
AQUA-RAT | AQUA-RAT-35959 | # Probability based on a percentage
We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: $${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35$$ and $${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28$$,
so the probability should be $$\frac 1{980}$$.
But I'm stuck on the second question, how should I proceed?
• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. – sdds Apr 15 at 14:47
• Your calculation is right. I have a different rounding. $0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. – callculus Apr 15 at 14:52
The following is multiple choice question (with options) to answer.
In a certain corporation, there are 300 male employees and 150 female employees. It is known that 20% of the male employees have advanced degrees and 40% of the females have advanced degrees. If one of the 450 employees is chosen at random, what is the probability this employee has an advanced degree or is female? | [
"7/15",
"3/10",
"3/5",
"2/5"
] | A | P(female) = 150/450 = 1/3
P(male with advanced degree) = 0.2*300/450 = 60/450 = 2/15
The sum of the probabilities is 7/15
The answer is A. |
AQUA-RAT | AQUA-RAT-35960 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
A car finishes a journey in nine hours at the speed of 80 km/hr. If the same distance is to be covered in six hours how much more speed does the car have to gain? | [
"8 km/hr",
"40 km/hr",
"12 km/hr",
"16 km/hr"
] | B | Distance covered by the car = 80 × 9 = 720 km
\ Speed = 720⁄6 = 120 km/hr
\ Speed gain = 120 – 80 = 40 km/hr
Answer B |
AQUA-RAT | AQUA-RAT-35961 | Thus the hands have been coincident 118 times after the starting position.
Counting the starting position, the answer is 119. But let's discount this, which is what I'm guessing is intended. Thus the answer is 118.
Assuming you want to include both 2pm and 4pm as times when the two hands line up, the answer is $119$. Here's a way to see it with a minimum of arithmetic. (The most you need to calculate is that the second hand goes around the dial $120$ times in $2$ hours.)
To make sure we count both 2pm and 4pm, let's start at one second before 2 and end at one second after 4. So with that small buffer, we can think of what we're counting as the number of times the second hand passes the minute hand as both go round in "clockwise" fashion.
Or, more exactly, the number of times the second hand passes the minute hand minus the number of times the minute hand passes the second hand.
That may seem like a silly way to put it, but consider this: It doesn't matter at what rate the two hands move, it merely matters how many times each goes around the dial.
In particular, if we "freeze" the minute hand (pointing straight up) for the moment, and let the second hand do its thing, it will go fully around $120$ times, passing the minute hand each time, and then the final two seconds, passing the minute hand once more, for a total of $121$ passes, ending at one second after the hour. If we now let the minute hand move, it goes around the dial twice, passing the (now frozen) second hand twice. Subtraction gives $121-2=119$.
The following is multiple choice question (with options) to answer.
How many times does the minute hand and the second hand make a straight line? | [
"2844",
"2832",
"2855",
"2381"
] | B | make a straight line 59 x 2 times = 118 times (both aligned, and in opposition) and there are 24 hours in a day, so in a day minutes and second hand make a straight line 24 x 118 times = 2832 times.
so answer is B |
AQUA-RAT | AQUA-RAT-35962 | # Math Help - Function word problems
1. ## Function word problems
Hi, I need help on a couple of problems. I'll be having my math mid-terms on the 26th, which is a Friday, and we were told by the teacher that some of the questions appearing in the test would be similar to the problems below. Specifically, I need help on four of them.
1) 1000 copies of a souvenir program will be sold if the price is $50 and that the number of copies sold decreases by 10 for each$1 added to the price. Write a function that determines gross revenue as a function of x. What price yields the largest gross sales revenue and what is the largest gross revenue? How many copies must be sold to yield the maximum gross revenue?
The following is multiple choice question (with options) to answer.
A used-book dealer sells paperback books at 3 times dealer's cost and hardback books at 4 times the dealer's cost. Last week, the dealer sold a total of 140 books, each of which had cost the dealer $1. If the gross profit (sales revenue minus the dealer's cost) on the sale of all of these books was $400, how many of the books sold were paperbacks? | [
"20",
"40",
"60",
"80"
] | A | The dealer earns a profit of $2 per paperback and $3 per hardback.
If all 140 books were paperbacks, the dealer would earn a profit of $280.
Since the actual profit was $400, the dealer must have sold 120 hardbacks.
Then the number of paperbacks sold was 140-120 = 20.
The answer is A. |
AQUA-RAT | AQUA-RAT-35963 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
In how many ways can a group of 4 men and 2 women be made out of a total of 7 men and 3 women? | [
"100",
"102",
"105",
"107"
] | C | We need to select 4 men from 7 men and 2 women from 3 women.
Number of ways to do this
= 7C4 × 3C2
= 7C4 × 3C1 [∵ nCr = nC(n-r)]
= 105
Ans- C |
AQUA-RAT | AQUA-RAT-35964 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
In a certain year, the population of a certain town was 9000. If in the next year the population of males increases by 4% and that of the females by 8% and the total population increases to 9600, then what was the ratio of population of males and females in that given year? | [
"1 : 2",
"5 : 4",
"2 : 3",
"Data inadequate"
] | A | Let the population of males = x; then the population of females = 9000 – x
Now, 4% of x + 8% of (9000 – x)
= (9600 – 9000 ) = 600
or 0.04x + 720 – 0.08x = 600
or 720 – 600 = 0.08x – 0.04x
or, 120 = 0.04x
x = 3000
Reqd ratio of population of males and females
3000/9000−3000=3000/6000=1:2
Answer A |
AQUA-RAT | AQUA-RAT-35965 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Average of 7 results is 121. If the average of first three results is 115 and average of last three results is 120 Then find the fourth result? | [
"142",
"150",
"126",
"185"
] | A | Option 'A' |
AQUA-RAT | AQUA-RAT-35966 | reference-request, modelling, voting
For the election $(C, \{v\})$, i.e. with $v$ being the only vote, we have two admissible outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 1, p_\mathrm{dem} \mapsto 0\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. For $(C, \{v, v\})$, we have the same two outcomes. The same holds true for three, four, and arbitrarily many copies of $v$.
Consider a new kind of vote $w(O) \equiv \mathrm{dem}$ that always votes for democrates. The election $(C, \{v,w\})$ has two admissable outcomes: $O_1 = \{p_\mathrm{rep} \mapsto 0.5, p_\mathrm{dem} \mapsto 0.5\}$ and $O_2 = \{p_\mathrm{rep} \mapsto 0, p_\mathrm{dem} \mapsto 1\}$. In the latter case, the vote $v$ "noticed" that they will never be surrounded by a like-minded peer group of size $> 40\%$ and hence switched its candidate it voted for.
Research Questions
Concerning scriptable votes, some broad questions I'd be interested in are:
Does every election have at least one admissible outcome? If not, can we relax the conditions on admissible outcomes to just choose the best ones?
How much influence do individual votes have? Is the function that maps $(C, V)$ to a set of admissible outcomes continuous in some sense?
The following is multiple choice question (with options) to answer.
In an election between two candidates, 70% of the voters cast their votes, out of which 4% of the votes were declared invalid. A candidate got 6552 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election. | [
"13000",
"12500",
"14000",
"12000"
] | A | Explanation :
Solution: let the total number of votes enrolled be x. then, number of votes cast = 70% of x. valid votes = 96% of(70% of x). .'. 75% of(96% of (70% of of x)) = 6552.
(75/100 * 96/100 * 70/100 * x) = 6552.
=> x = (6552*100*100*100)/(75*96*70) = 13000
Answer : A |
AQUA-RAT | AQUA-RAT-35967 | So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. _________________ Intern Joined: 26 May 2012 Posts: 21 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 15 Jul 2012, 02:22 Bunuel wrote: farukqmul wrote: When the price of oranges is lowered by 40%, 4 more oranges can be purchased for$12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price?
(A) 8
(B) 12
(C) 16
(D) 20
(E) 24
Say $$x$$ is the original price of an orange, then:
$$xn=12$$;
and
$$0.6x*(n+4)=12$$ --> $$x(n+4)=20$$ --> $$xn+4x=20$$ --> $$12+4x=20$$ --> $$x=2$$.
So, for 24 dollars 24/2=12 oranges can be purchased at the original price of $2. Answer: B. are there any other ways? Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 377 Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink] ### Show Tags 22 Oct 2016, 11:32 Set up: 12/p = x 12/0.60 = x+4 Manipulate and plug the first equation into the second --> you'll find p =$2
Thus $24/$2 per orange = 12 oranges
VP
Joined: 07 Dec 2014
Posts: 1128
Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
100 oranges are bought at the rate of Rs. 350 and sold at the rate of 48 per dozen. The percentage of profit is | [
"12 2/7%",
"13 2/7%",
"14 2/7%",
"15 2/7%"
] | C | Explanation:
So before solving this question we will get the C.P. and S.P. of 1 article to get the gain percent.
C.P. of 1 orange = 350/100 = Rs 3.50
S.P. of one orange = 48/12 = Rs 4 [note: divided by 12 as 1 dozen contains 12 items]
Gain = 4 - 3.50 = Rs 0.50
Gain%=0.50/ 3.50∗100=100 /7%=14 2/7%
Option C |
AQUA-RAT | AQUA-RAT-35968 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
The Kiljaro Highway is marked with Milestones denoting the distance to the town of Kiljaro. Tommy left kiljaro and drove the highway, passing the 30 km milestone at 8:30. Some time afterwards, Tommy got a phone call asking him to return home, and he made a U-turn at the 160 km milestone. At 09:A0 Tommy passed the milestone marking 70 km to Kiljaro. The variable A represents the tens digit of the minutes in 09:A0. Assuming Tommy maintained the same constant speed during the entire drive, how many kilometers did Tommy travel in one minute? | [
"220/(30+10A)",
"220/(30+60A)",
"220/36A",
"220/(30+10A)"
] | A | Since we are dealing with the variables in the answer choices,the best possible method according to me would besubstitution.
Substitute A with 3.meaning tommy would have travelled a distance of ((160-30) + (160-70)) in 60 minutes.
220 Kms in 60 minutes ==> 3.6km/hr.
Substitute A with 3 in the answer options.
Option A |
AQUA-RAT | AQUA-RAT-35969 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The compound ratio of 5:1, 3:1 and 2:5? | [
"6:1",
"1:3",
"1:5",
"1:1"
] | A | 5/1 * 3/1 * 2/5 = 30/5
6:1
Answer: A |
AQUA-RAT | AQUA-RAT-35970 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
Mahesh can do a piece of work in 45days. He works at it for 20days and then Rajesh finished it in 30days. How long will Y take to complete the work? | [
"54",
"25",
"37",
"41"
] | A | Work done by Mahesh in 45days = 20*1/45 = 4/9
Remaining work = 1 - 4/9 = 5/9
5/9 work is done by Rajesh in 30days
Whole work will be done by Rajesh is 30*9/5 = 54days
Answer is A |
AQUA-RAT | AQUA-RAT-35971 | ### Show Tags 23 Aug 2014, 11:31 kanusha wrote: On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at$0.60 per glass on the first day, what was the price per glass on the second day?
The following is multiple choice question (with options) to answer.
A lemonade stand sold only small and large cups of lemonade on Tuesday. 5/9 of the cups sold were small and the rest were large. If the large cups were sold for 12/11 as much as the small cups, what fraction of Tuesday's total revenue was from the sale of large cups? | [
"(a) 7/16",
"(b) 7/15",
"(c) 48/103",
"(d) 17/35"
] | C | A simpler way i guess would be to think that in total 9 cups were sold. Out of which 5 are small and 4 are large. Now let the small ones cost $11. so the large ones would cost $12.
So,
5*11=55 and 4*12=48.
Total revenue was 55+48=103
and Large cup sales as found above is 48
Therefore answer is 48/103
C |
AQUA-RAT | AQUA-RAT-35972 | Show Tags
18 Jun 2014, 06:09
Thanks much Bunuel
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Posts: 1820
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
Show Tags
18 Jun 2014, 20:26
1
maggie27 wrote:
Can anybody please explain me that why are we not considering the option of a 7 * 7 square inscribed into the square S as dis would give us the area less than 50.
Also note that area of inscribed square is always half than that of the original square
As Bunuel pointed out, if it goes less than 50, it means some of the vertex is not touching side of the original square.
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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01 Jul 2015, 18:04
If $$x^{2}$$ is area of square, then find x, one side of the square. If square is inscribed, then diagonal is the length of larger square and therefore the diagonal is $$10$$. To determine the side, the formula also includes the area of the square, $$x^{2}$$. So, if $$2x^{2} = 100$$ then $$x^{2}=50$$
D.
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A
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Re: The perimeter of square S is 40. Square T is inscribed in square S. [#permalink]
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31 Jul 2016, 20:43
Fabino26 wrote:
The perimeter of square S is 40. Square T is inscribed in square S. What is the least possible value of the area of square T ?
A. 45
B. 48
C. 49
D. 50
E. 52
The following is multiple choice question (with options) to answer.
A square, with perimeter 80, is inscribed in a circle. What is the area of the circle? | [
"400π",
"200π",
"250π",
"300π"
] | B | Area of circuscribed circle = pi/2*area of square
Area of square = (80/4)^2 = 20^2= 400
Area of circle = pi/2*400=200pi
Answer : B |
AQUA-RAT | AQUA-RAT-35973 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can? | [
"73",
"44",
"72",
"62"
] | B | Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.
Answer: Option B |
AQUA-RAT | AQUA-RAT-35974 | per hour. The time taken to cover the first 60% of the distance is 10 minutes more than the time taken to cover the remaining distance. What is its speed? 4) A car goes 250 km in 4 hours. The total time is 5 seconds, so t = 5. 4. Aptitude Questions and Answers. Learn how teachers can make BrainPOP-style assessments by using the Quiz Mixer with a My BrainPOP account. Marco drove from home to work at an average speed of 50 miles per hour and returned home along the same route at an average speed of 46 miles per hour. ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km. Refer how to solve speed problems to …Feb 28, 2016 · Hi Friends This Video will helps you to Understand the concept on the Time and Distance(Quantitative Aptitude). Distance, Speed and Time Problems This Math quiz is called 'Distance, Speed and Time Problems' and it has been written by teachers to help you if you are studying the subject at middle school. Speed Distance and Time. When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. 42 minutes on DVD. In this speed, distance and time worksheet, students read statements and then mentally determine the speed, distance or time in a given problem. GMAT Time, Speed, Distance and Work, GRE Time, Speed, Distance and Work, SAT Time Speed Distance and Work, SSC - CGL Time Speed Distance and Work, Tags gmat gre cat sat act time speed distance formula time speed distance concepts time speed distance problems with solutions Rate This Lesson Velocity word problems The following velocity word problems will strengthen your knowledge of speed, velocity, In the end, the difference between speed and velocity should be clear. 5, and we get an average speed of 10 miles per hour. You will have 4 minutes to complete this challenge. The distance for the second leg is 200, and the rate is v+25, so the time of the second leg is. 5 miles traveled. A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. As for your brand-new red sports car, your friend was …Now just plug in your values for speed and time to solve for distance: d = 40 miles/hour x
The following is multiple choice question (with options) to answer.
A car takes 6 hours to cover a distance of 630 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time? | [
"60",
"50",
"40",
"70"
] | D | Time = 6
Distance = 630
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 630/9 = 70 Kmph
Answer D. |
AQUA-RAT | AQUA-RAT-35975 | Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:
0.04(total fish) = 50
This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:
Quote:
.04 (total) = tagged fish
We are missing two variables. We don't know the total fish and we don't know the tagged fish.
If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.
If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."
From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.
Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
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The following is multiple choice question (with options) to answer.
In a forest 150 deer were caught, tagged with electronic markers, then released. A week later, 50 deer were captured in the same forest. Of these 50 deer, it was found that 5 had been tagged with the electronic markers. If the percentage of tagged deer in the second sample approximates the percentage of tagged deer in the forest, and if no deer had either left or entered the forest over the preceding week, what is the approximate number of deer in the forest? | [
"150",
"750",
"1,250",
"1,500"
] | D | Given 150 deers were caught and they are tagged in the first slot.
In the second slot , 50 were caught and 5 we tagged then 10% of them are tagged.
Given this 10% approximates the percentage of tagged deer in the forest..here number is 150..
then 100% is 1500.. is the number of deers population in the forest..
Answer: D is correct answer.. |
AQUA-RAT | AQUA-RAT-35976 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
At 9 am a car (A) began a journey from a point, traveling at 40 mph. At 10 am another car (B) started traveling from the same point at 60 mph in the same direction as car (A).
At what time will car B pass car A? | [
"12 pm",
"1 pm",
"2 pm",
"3 pm"
] | A | After t hours the distances D1 traveled by car A is given by
D1 = 40 t
Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it. After (t - 1) hours, distance D2 traveled by car B is given by
D2 = 60 (t-1)
When car B passes car A, they are at the same distance from the starting point and therefore D1 = D2 which gives
40 t = 60 (t-1)
Solve the above equation for t to find
t = 3 hours
Car B passes car A at
9 + 3 = 12 pm
Answer A |
AQUA-RAT | AQUA-RAT-35977 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The balance of a trader weighs 8% less than it should. Still the trader marks up his goods to get an overall profit of 20%. What is the mark up on the cost price? | [
"40%",
"8%",
"10.4%",
"16.66%"
] | C | The most natural way to deal with 'weights' questions is by assuming values.
Say the trader's balance shows 100 gms. It is actually 92 gms because it weighs 8% less. Say, the cost price is $92 ($1/gm). Since he gets a profit of 20%, the selling price must be 92+(20/100)*92 = $110.4
Since the cost price is actually supposed to be $100 (for 100 gms) and the selling price is $110.4, the mark up is simply 10.4%.
Ans : C |
AQUA-RAT | AQUA-RAT-35978 | 28 Jan 2012, 12:09
What is the units digit of $$2222^{333}*3333^{222}$$ ?
A. 0
B. 2
C. 4
D. 6
E. 8
The units digit of $$2222^{333}$$ is the same as that of $$2^{333}$$;
The units digit of $$3333^{222}$$ is the same as that of $$3^{222}$$;
Hence, the units digit of $$2222^{333}*333^{222}$$ is the same as that of $$2^{333}*3^{222}$$;
Now, the units digits of both 2 and 3 in positive integer power repeat in patterns of 4. For 2 it's {2, 4, 8, 6} and for 3 it's {3, 9, 7, 1}.
The units digit of $$2^{333}$$ will be the same as that of $$2^1$$, so 2 (as 333 divided by cyclicity of 4 yields remainder of 1, which means that the units digit is first # from pattern);
The units digit of $$3^{222}$$ will be the same as that of $$3^2$$, so 9 (as 222 divided by cyclicity of 4 yields remainder of 2, which means that the units digit is second # from pattern);
Finally, 2*9=18 --> the units digit is 8.
For more on this check Number Theory chapter of Math Book: http://gmatclub.com/forum/math-number-theory-88376.html
Hope it helps.
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Re: What is the units digit of 2222^(333)*3333^(222) ? [#permalink]
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28 Jan 2012, 12:18
I used the method described by Bunuel.
but not to be repetitive, I will post another solution -
The following is multiple choice question (with options) to answer.
What are the unit's digits of 369, 6864, 4725 respectively ? | [
"2",
"8",
"3",
"5"
] | C | Cyclicity of 3 is 4. 369 = (34)k.31(34)k.31 = (1)k.31(1)k.31 = 3
Power of 6 always gives units digit 6 only.
If power of 4 is odd, then we will get 4 as units digit, if power of 4 is even, then we get 6 as units digit.
Hence, Option 3 is correct.
Answer:C |
AQUA-RAT | AQUA-RAT-35979 | # 3 balls are drawn from a bag contains 6 white balls and 4 red balls, what is the probability that 2 balls are white and 1 ball is red?
A bag contains 6 white balls and 4 red balls. If 3 balls are drawn one by one with replacement, then what is the probability that 2 balls are white and 1 ball is red?
$$\frac{18}{125}$$
What I did Probability of getting a white ball= $$6/10=3/5$$ Probability of getting a red ball= $$4/10=2/5$$ Probability of getting 2 balls white and 1 ball red = $$6/10*6/10*4/10=18/125$$
But the answer is $$\frac{54}{125}$$. Why are we multiplying it by $$3$$? Please someone elaborate this part
This is a gmat exam question.
• Look at it this way, if you don't multiply by three, then your answer is the probability that we pick $2$ white balls and $1$ red ball in that order. – WaveX Sep 9 '17 at 14:35
This is a binomial experiment with $P(W)=\frac{6}{10}=\frac{3}{5}$. Apply the formula : $$f(2)=C_2^3\cdot \left(\frac{3}{5}\right)^2\cdot \frac{2}{5}=\frac{54}{125}.$$
The following is multiple choice question (with options) to answer.
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws,each ball being put back after it is drawn? | [
"2/27",
"1/9",
"1/3",
"4/27"
] | D | P=2∗39∗29=427P=2∗39∗29=427
We are multiplying by 2 as there are two possible wining scenarios RW and WR.
Answer: D. |
AQUA-RAT | AQUA-RAT-35980 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
In Kaya's teacher's desk there are 10 pink highlighters, 15 yellow highlighters, and 8 blue highlighters. How many highlighters are there in all? | [
"11",
"22",
"77",
"33"
] | D | Add the numbers of highlighters.
10 + 15 + 8 =33.
Answer is D. |
AQUA-RAT | AQUA-RAT-35981 | Claim: If $m$ and $n$ are integers, and $n$ is even, then $mn$ is even.
Proof: Because $n$ is even, we can write $n=2l$, where $l$ is an integer. Then $mn=m(2l)=2(ml)$ is a multiple of $2$. In other words, $mn$ is even.
Notice that each step is justified—though if we had an extremely strict teacher, we might want to point out that we got $mn=m(2l)$ by multiplying both sides of $n=2l$ by $m$, and that $m(2k)=2(ml)$ follows from the associativity and commutativity of multiplication. But you're probably safe in omitting these details unless you've been told otherwise.
Also note that I didn't assume that $m$ is odd, because this makes the equations a little more complicated. Here is a more faithful recreation of your proof:
Claim: If $m$ is an odd integer and $n$ is an even integer, then $mn$ is even.
Proof: Because $m$ is odd and $n$ is even, we can write $m=2k+1$ and $n=2l$, where $k$ and $l$ are integers. Then $mn = (2k+1)(2l) = 2\cdot(l(2k+1))$ is a multiple of $2$, hence even.
Define $2k_1k_2+k_2=k_3$ as an integer number.
You could further formalize your text, if you want (it is always handy to write down the definitions or the theorems you use, it also makes it easier for others to read and make it look more complete): By definition: $$m \text{ even }\iff \exists k \in \mathbb Z \text{ such that }2k = m$$
The following is multiple choice question (with options) to answer.
If m is an integer such that (-2)^2m=2^(21-m) then m=? | [
"5",
"6",
"7",
"8"
] | C | 2m = 21-m
3m = 21
m = 7
The answer is C. |
AQUA-RAT | AQUA-RAT-35982 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
In 1995, the Jossy spent $800 on the family’s water bills. Anticipating that water rates would increase in 1996 by 50%, the Jossy cut back their water usage. By how much must the Jossy have reduce their 1996 water usage to pay exactly the same amount in 1996 as they paid in 1995? | [
"a)\t33 1/3 %",
"b)\t40 %",
"c) 50 %",
"d)\t66 2/3 %"
] | A | Let x be the water usage and c be the cost waterof Jossy
We have
xc =800, c=800/x
Y(1.5c)=800 where Y is the water usage in 1996
Y= 800/ (1.5*800/x)= 1/1.5x=2/3x
Thus reduction will be x-2/3x=1/3x=A |
AQUA-RAT | AQUA-RAT-35983 | A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2
"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)
"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3
"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)
I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.
Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!
Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.
"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.
"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.
The following is multiple choice question (with options) to answer.
If 6 years are subtracted from the present age of Arun and the remainder is divided by 18, then the present age of his grandson Gokul is obtained.If Gokul is 2 years younger to Madan whose age is 5 years, then what is the age of Arun? | [
"48",
"60",
"84",
"96"
] | B | arun age x.
(x-6)/18=y
y is gokul age which is 3
x=60
ANSWER:B |
AQUA-RAT | AQUA-RAT-35984 | (1) Kevin spent a total of $18.00 on beer. (2) Kevin bought 3 more cans of beer than bottles of beer. Target question: How many bottles of beer did Kevin buy? Given: Kevin pays$1.00 for each can of beer and $1.50 for each bottle of beer. Kevin buys a total of 15 bottles and cans of beer Let C = the NUMBER of Cans that Kevin bought Let B = the NUMBER of Bottles that Kevin bought So, we can write: C + B = 15 Statement 1: Kevin spent a total of$18.00 on beer
The COST of C cans = ($1.00)C = 1C The COST of B bottles = ($1.50)B = 1.5B
So, we can write: 1C + 1.5B = 18.00
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
1C + 1.5B = 18.00
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
(of course, we won't solve the system, since that would be a waste of our valuable time!)
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: Kevin bought 3 more cans of beer than bottles of beer
We can write: C = B + 3
When we combine this equation with the equation we created from the given information, we have:
C + B = 15
C = B + 3
Since we COULD solve this system for C and B, we COULD determine the number of bottles of beer that Kevin bought.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
The following is multiple choice question (with options) to answer.
Rita bought 105 items from the market, and each item is either glass, cup, or spoon. If she ownsbought 5 more cups than glasses, and twice as many cups as glasses, how many glasses Rita did buy ? | [
"20",
"25",
"30",
"40"
] | B | x = the number of glasses
y = the number of cups
z = the number of spoons
From the first sentence we have
Equation #1: x + y + z = 105
...she bought 5 more spoons than glasses...
Equation #2: z = 5 + x
...twice as many cups as glasses...
Equation #3: y = 2x
Now, we can replace y with x in Equation #1
x + 2x + z = 105
Equetion 4 : 3x + z = 105
by adding Equetion #2 and Equetion #4
4x = 100
x = 100/4
x = 25
There are 25 glasses. That's the answer. Just for check, this is 5 less than the number of spoons, so z = 30, y = 50, and 25 + 50 + 30 = 105.
Answer = 25,(B) |
AQUA-RAT | AQUA-RAT-35985 | Then, in any case $(a-b)^2=8$
• -1.This is a low grade method. – N.S.JOHN Apr 17 '16 at 10:42
$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$
Also,
$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)
The following is multiple choice question (with options) to answer.
If f(x) = -2x2 + 8x - 4, which of the follwoing is true? | [
"The maximum value of f(x) is -4.",
"The graph of f opens upward.",
"The graph of f has no x-intercept",
"f is not a one to one function."
] | D | f(x) is a quadratic function and its graph is a parabola that may be intercepted by horizontal lines at two points and therefore is not a one to one function. The answer is D
correct answer D |
AQUA-RAT | AQUA-RAT-35986 | # Coordinates of point on a line defined by two other points with a known distance from one of them
I have two points in 3D space; let's call them $A=(a_x, a_y, a_z)$ and $B=(b_x, b_y, b_z).$
Now, I need to place a third point, let's call it $C=(c_x, c_y, c_z)$, which lies on the line between $A$ and $B$ and is some known distance from $A$.
Is there some easy way I can get the $c_x, c_y, c_z$ coordinates knowing the distance from $C$ to $A$?
I would easily be able to do this in 2D, but I never worked with 3D space and I can't seem to figure it out.
Thanks.
-
It's quite similar to 2D space. Consider a vector $v = B-A$ which you can imagine is the direction from $A$ to $B$. Hence for any point $C$ between $A$ and $B$ (inclusive of $A$ and $B$).
$$C = A + tv$$
where $t = 0$ implies $C = A$ and $t = 1$ implies $C = B$ and $t \in (0,1)$ are all the points in between (one of which is the desired $C$). As you can see, this representation is independent of the dimension of your space.
So, what's the value of $t$? Well, let the known distance from $A$ to $C$ be $d_{AC}$. Now, the distance between $A$ and $B$ or $d_{AB}$ is the magnitude of $v$ or $|v|$ which is nothing but
$$d_{AB} = |v| = \sqrt{(a_x-b_x)^2+(a_y-b_y)^2+(a_z-b_z)^2}$$
(You can see that this formula for the Euclidean distance between two points is similar in 2D as well)
The following is multiple choice question (with options) to answer.
In an xy-coordinate plane, a line is defined by y = kx + 1. If (3, b), (a, 4), and (a, b+1) are three points on the line, where a and b are unknown, then k = ? | [
" 1/2",
" 1",
" 3/2",
" 2/3"
] | D | b=3k+1...(1)
b+1=ak+1...(2)
4=ak+1...(3)
Taking (2) and (3)
4=b+1
b=3
Taking (1)
3=3k+1
k=2/3
Answer : D |
AQUA-RAT | AQUA-RAT-35987 | ### Show Tags
06 Feb 2015, 13:28
1
x+y=1600 , X=1600-Y
x+100=3y, 1700=4Y
Y=425
x=1600-y=1600-425=1175
D
Director
Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
Schools: ISB '19, IIMA , IIMB, XLRI
WE: Information Technology (Consulting)
Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
### Show Tags
06 Nov 2018, 20:08
pacifist85 wrote:
There are 1600 jelly beans divided between two jars, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?
A. 375
B. 950
C. 1150
D. 1175
E. 1350
Let Jelly beans in jar $$X = x$$ and $$Y = y$$
$$x + y = 1600$$ ----- ($$i$$)
Given jar $$X$$ has $$100$$ fewer jelly beans than three times the number of beans in jar $$Y$$.
Therefore; $$x + 100 = 3y$$
$$x = 3y - 100$$
Substituting value of $$x$$ in equation ($$i$$), we get;
$$3y - 100 + y = 1600$$
$$4y = 1600 + 100 = 1700$$
$$y = \frac{1700}{4} = 425$$
Substituting value of $$y$$ in equation ($$i$$), we get;
$$x + 425 = 1600$$
$$x = 1600$$ $$-$$ $$425 = 1175$$
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Re: There are 1600 jelly beans divided between two jars, X and Y. If the [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
There are 1400 jelly beans divided between two jars, Jar X and Jar Y. If there are 300 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X? | [
"925",
"950",
"975",
"1000"
] | C | X+Y=1400 so Y=1400-X
X=3Y-300
X=3(1400-X)-300
4X=3900
X=975
The answer is C. |
AQUA-RAT | AQUA-RAT-35988 | At this point we get stuck, as there is not much left to do arithmetically. We still don’t have the $3$ or the $b$ we’re looking for.
However, there are some hints to help us. The first thing to note is that we have three terms: $a,\left(a-1\right),\left(a+1\right)$. We can further re-arrange these to get: $\left(a-1\right)\left(a+0\right)\left(a+1\right)$. These are three consecutive integers.
Given $n$ consecutive integers, we can be sure that exactly one of those integers is divisible by $n$. Thus exactly one of $\left\{\left(a-1\right),\left(a+0\right),\left(a+1\right)\right\}$ is divisible by $3$. Which means that we can write the expression in three possible cases, where $c\in \mathbb{Z}$:
$\left(3c\right)\left(a+0\right)\left(a+1\right)$ $\left(a-1\right)\left(3c\right)\left(a+1\right)$ $\left(a-1\right)\left(a+0\right)\left(3c\right)$
For each of these cases, it is trivial to extract $b\in \mathbb{Z}$ such that the expression becomes $3b$.
$\left(3c\right)\left(a+0\right)\left(a+1\right)=3b$, where $b=c\left(a+0\right)\left(a+1\right)$
$\left(a-1\right)\left(3c\right)\left(a+1\right)=3b$, where $b=\left(a-1\right)c\left(a+1\right)$
The following is multiple choice question (with options) to answer.
A, B, C are three consecutive positive integers (A>B>C). What is the value of the expression 2A +B + C? | [
"6A+7.",
"5A+1.",
"4A-3.",
"6A-5."
] | C | B = A-1
C= A-2
Putting these values in 2A +B + C we get 4A-3
C is the answer |
AQUA-RAT | AQUA-RAT-35989 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
Last year, for every 100 million vehicles that travelled on a certain highway, 90 vehicles were involved in accidents. If 3 billion vehicles travelled on the highway last year, how many of those vehicles were involved in accidents? (1 billion = 1,000,000,000) | [
"288",
"320",
"2,700",
"3,200"
] | C | To solve we will set up a proportion. We know that “100 million vehicles is to 90 accidents as 3 billion vehicles is to x accidents”. To express everything in terms of “millions”, we can use 3,000 million rather than 3 billion. Creating a proportion we have:
100/90 = 3,000/x
Cross multiplying gives us:
100x = 3,000 * 90
x = 30 * 90 = 2,700
Correct answer is C. |
AQUA-RAT | AQUA-RAT-35990 | # Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
The following is multiple choice question (with options) to answer.
A hiker walked for 3 days. She walked 18 miles on the first day, walking 3 miles per hour. On the second day she walked for one less hour but she walked two mile per hour, faster than on the first day. On the third day she walked the same number of hours as on the first day, but at the same speed as on the second day. How many miles in total did she walk? | [
"24",
"44",
"73",
"60"
] | C | She walked 18 miles on the first day, walking 3 miles per hour i.e. total time of walk on Day-1 = 18/3 = 6 Hours
Second day time of walk = 6-1 = 5 hours and Speed = 3+2=5 miles per hour i.e. Distance walked on second day = 5*5 = 25 miles
Third day time of walk = 6 hours and Speed = 5 miles per hour i.e. Distance walked on second day = 6*5 = 30 miles
Total Distance travelled on three days = 18+25+30 = 73
Answer: Option C |
AQUA-RAT | AQUA-RAT-35991 | Spoiler:
The middle number is the average of the other two.
$\frac{47+63}{2} \:=\:55,\quad \frac{85+99}{2} \:=\:92,\quad \frac{73+25}{2} \:=\:{\color{red}49}$
3. ## Re: Math puzzles questions?
Originally Posted by amrithaa
1. What number should replace the question mark?
Each row is 8 times the immediate row above. 9X8=72. 72X8=576. bottom row 576X8=4608. So there will be 0 in place of question mark in the lowest row.
4. ## Re: Math puzzles questions?
Originally Posted by amrithaa
5.What number should replace the question mark?
Each two digit number is a sum of digits of one three digit number. 2+6+8=16; 3+5+9=17; 2+6+3=11; and therefore the number in place of question mark is 5+1+6=12.
5. ## Re: Math puzzles questions?
Hello again, amrithaa!
5. What number should replace the question mark?
. . $\boxed{\begin{array}{c} 268 \qquad 11 \\ \\[-3mm] 17 \qquad\quad 259 \\ ? \\ 16 \qquad\quad 516 \\ \\[-4mm] 263 \end{array}}$
Spoiler:
Each 2-digit number is the digit-sum of a 3-digit number.
. . $\begin{array}{ccc}263 & \to & 11 \\ 268 & \to & 16 \\ 359 & \to & 17 \\ 516 & \to & {\color{red}12} \end{array}$
6. ## Re: Math puzzles questions?
Lol 4 is not as bad: the number in the middle is the average of the 2 others on the sides.
7. ## Re: Math puzzles questions?
3. 5=85/17, 4=76/19 so ?=91/13
The following is multiple choice question (with options) to answer.
What should come in place of the question mark (X) in the following equation ?
60% of X + 2/3 of 39 = 44 | [
"30",
"37",
"28",
"26"
] | A | Explanation:
60% of X + 2/3 of 39 = 44
i.e. 3/5 of X + 2/3 of 39 = 44 (60% = 3/5)
3/5 * X + 2/3 * 39 = 44
3/5 * X + 26 = 44
3/5 * X = 18
X = 18 * 5/3
X = 30
ANSWER: A |
AQUA-RAT | AQUA-RAT-35992 | Since the the percentage of tagged fish in the second catch approximates the percentage of tagged fish in the pond, the approximate number of fish in the pond is:
0.04(total fish) = 50
This equation assumes that there are 50 fish tagged in the total population. We do not know that. The only thing we know is that that the percentage of tagged fish in the second catch is 4%. The question says that 4% approximates the number of tagged fish in the pond. So this is the true equation we have:
Quote:
.04 (total) = tagged fish
We are missing two variables. We don't know the total fish and we don't know the tagged fish.
If there are 16 tagged fish, then choice A is correct.
If there are 26 tagged fish, then choice B is correct, and etc.
If we assume that the number of fish in the second catch (50) is the number of fish tagged, then yes the total fish would be 1250. However, that's not what the question provides. I think this question is written poorly.
So the question states that "In a certain pond, 50 fish were caught, tagged, and returned to the pond."
From this sentence, we can deduce that there are indeed a total of 50 tagged fish in the pond. The only way to have some other number of tagged fish in the pond is if there were already some number of tagged fish in the pond (in which case, the question would have told us so) or if either more fish were tagged afterward or some of the tagged fish were removed from the pond (again, we would have been told). Since we have no such information, we cannot assume that there might be some other number of tagged fish in the pond.
Perhaps you are missing the fact that 50 fish are caught TWICE: first all of them are tagged, and the second time, the tagged fish are counted.
_________________
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Re: In a certain pond, 50 fish were caught, tagged, and returned [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
The ratio of ducks and frogs in a pond is 35 : 39 respectively. The average number of ducks and frogs in the pond is 152. What is the number of frogs in the pond ? | [
"160",
"152",
"156",
"144"
] | A | Solution:
Ratio of Ducks and Frogs in Pond,
= 35 : 39.
Average of Ducks and Frogs in Pond,
= 152.
So, total number of Ducks and Frogs in the Pond,
= 2* 152 = 304.
Therefore,
Number of Frogs, = (304 *39)/74 = 160.
Answer: Option A |
AQUA-RAT | AQUA-RAT-35993 | Since 2m+1 is greater than P1, x1 is positive. An even number minus an odd plus an odd equals an even, and half an even number is an integer. Hence, x1 is a solution to S.
In the second case, where 2m+1 is less than P1, let’s try N = 2m+1 as a solution. This yields: $x_1 = (\frac{2S}{2^{m+1}} – 2^{m+1} + 1)/2 \\ S = 2^m\cdot P_1 \cdot S_2 \implies \\ x_1 = (\frac{2 \cdot 2^m \cdot P_1 \cdot S_2}{2^{m+1}} – 2^{m+1} + 1)/2 \\ = (P_1 \cdot S_2 – 2^{m+1} + 1)/2$
Since P1 is greater than 2m+1, x1 is again positive. P1 and S2 are both odd, hence P1S2 is odd. An odd number minus an even plus an odd is an even number, and half of an even number is an integer. Hence, x1 is a solution to S.
This proves that S has a solution R for all values that are not powers of 2; it also illustrates why 23•16 requires 23 numbers (23 < 32) while 23•8 requires only 16 (16 < 23). As we would expect, 592 (37•16) requires 32 numbers, not 37 (3+4+...+34); 1184 (37•32) requires 37 (14+15+...+50). All of this proves the original hypothesis: A positive integer S is equal to the sum of consecutive positive integers iff it has odd prime factors.
## 1 Comment
1. paulhartzer
It occurred to me later that this provides a direct proof all odd numbers have N=2 solutions: All odd numbers are of the form 2^0•P_1•S_2. Since 2^(0+1) = 2 will always be lower than P_1, N=2^(0+1)=2 will always provide a solution.
The following is multiple choice question (with options) to answer.
If x/(12p) is an even prime number, where x is a positive integer and p is a prime number, what is the least value of x? | [
" 22",
" 33",
" 44",
" 48"
] | D | x/(12p) = even prime number
x = even prime number * 11p
Least value of x = Lowest even prime number * 12 * Lowest value of p
= 2 * 12* 2 = 48
Answer D |
AQUA-RAT | AQUA-RAT-35994 | $$10^\color{blue}9+1=7\times 11\times 13\times 19\times 52579.$$
• Oh wow. This compliments Arthur's answer so well! While his answer makes sense to me, I have had trouble applying his explanation to the general situation, so that I could construct (or discover) similar patterns in other number sets (and until I can do that, I can't really say that I understand the phenomenon). Thank you! Sep 5, 2018 at 11:42
(I). A number $R$ between $0$ and $1$ is represented decimally as $0.ABCDEFABCDEF...$ iff $10^6R$ is represented as $ABCDEF. ABCDEFABCDEF...$ Subtracting, we see that $(10^6-1)R$ is represented as $ABCDEF.$
(II). The digit-sequence $ABC$ represents the number $X=10^2A+10B +C$ and the digit-sequence $DEF$ represents the number $Y=10^2D+10E+F .$ Now $Y=999-X$ iff the digit-sequence $ABCDEF$ represents $(1000 X)+(999-X)=(999)(1+X).$
(III). Combining (I) and (II) we see that $X+Y=999$ iff $(10^6-1)R=(999)(1+X)$ iff $(999)(1001)R=(999)(1+X)$ iff $$R=\frac {1+X}{1001}=\frac {1+X}{(7)(11)(13)}.$$
Now $1+X=1+(10^2A+10B+C)$ can be any integer from $1$ to $1000.$ For example if $X=285$ then $(1+X)/1001=286/1001=2/7.$ Or if $X=64$ then $(1+X)/1001=65/1001=5/77.$
The following is multiple choice question (with options) to answer.
What number has a 5:1 ratio to the number 10? | [
"88",
"50",
"66",
"2881"
] | B | 5:1 = x: 10
x = 50
Answer: B |
AQUA-RAT | AQUA-RAT-35995 | there are 60 total minutes. So if the input is like hour = 12 and min := 30, then the result will be 165°. 8. 7. Formulas for Clock 1. The distance between the 2 and the 3 on the clock is 30°. (A) 100 0 (B) 80 0 (C) 120 0 (D) 200 0. The hour hand is in #3 [which is actually 15 minutes.] Angle 2 = 60°. University of California-Berkeley, Bachelor in Arts, Cellular and Molecular Biology. I understand this working all the way up until the '2n-1' part, where n is a positive integer. Write a program to determine the angle between the hands of a clock. If Varsity Tutors takes action in response to ø = 30h – 11/2 m -----when minute hand is in first 1/2 (between 12 to 6) 54 = 30 × 7 – 11/2 m. 11/2 m = 156. m = 28.36 min = 28 4/11 min = 21.8 s. at 7 hr 28 4/11 min. $${\text{192}}{\frac{1}{2}^ \circ }$$ C. 195º. M = Minute. View Answer. This problem is know as Clock angle problem where we need to find angle between hands of an analog clock at a given time. A clock is started at noon. B. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such If the required angle is A, then A = 30*H - (11*M/2) where H= hour hand, M= Minute hand. Angle Between Hands of a Clock. The angle between the two hands must be 10. Step 2: Press the "Calculate" button. as Boston University, Bachelor of Science, Business Administration and Management. It happens due to only one such incident between 12 and 1'o clock. In one hour, they will form two right angles and in 12 hours there are only 22 right angles. In each hour the minute hand covers more distance of 55 min in comparison to the hour hand. ... Find the angle between the minute hand and hour hand of a clock when the time is 7.20? Suppose we have two numbers, hour and minutes. on or linked-to by the Website infringes your
The following is multiple choice question (with options) to answer.
A clock shows the time as 3:30 P.m. If the minute hand gains 2 minutes every hour, how many minutes will the clock gain by 1 a.m.? | [
"23 Minutes",
"19 Minutes",
"25 Minutes",
"26 Minutes"
] | B | if the minute hand gains 2 minutes in 1 hour the from 3:30 p.m-4:00 p.m it gains 1 min
then from 4:00p.m-4:00 a.m =9*2=18 total=19
ANSWER:B |
AQUA-RAT | AQUA-RAT-35996 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a job in 15days and B in 20days. If they work on it together for 4 days, then the fraction of the work that is left is? | [
"2/15",
"8/15",
"3/11",
"1/12"
] | B | A's 1 day work = 1/15
B's 1day work = 1/20
A+B 1day work = 1/15 + 1/20 = 7/60
A+B 4days work = 7/60*4 = 7/15
Remaining work = 1 - 7/15 = 8/15
Answer is B |
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