source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37897 | # Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$?
Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?
The following is multiple choice question (with options) to answer.
If p is the smallest positive integer that is not prime and not a factor of 50!, what is the sum of the factors of p? | [
"51",
"54",
"162",
"72"
] | C | In fact 51=3*17 IS a factor of 50!.
The smallest positive integer that is NOT prime and NOT a factor of 50! is 106=2*53. The sum of the factors of 106 is 162.
Answer: C. |
AQUA-RAT | AQUA-RAT-37898 | 3,5,-5... The first term in the sequence of #'s shown above is 3. Each even # term is 2 more than the previous term and each odd # term, after the first, is -1 times the previous term. For example, the second term is 3+2, and the
8. ### Math *URGENT
Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646? 3. Determine
9. ### maths
The fifth term of an arithmetic sequence is 23 and the 12th term is 72. What is the value of the 10th term. Which term has a value of 268.
10. ### Sj
Mathematics......Under number patterns-Geometric series If a question goes.determine the expression for the nth term of the following sequence if the a) 4th term is 24 and the 7th term is 192 in a geometric sequence.what formula
More Similar Questions
The following is multiple choice question (with options) to answer.
In a certain sequence, each term except for the first term is one less than twice the previous term. If the first term is 0.8, then the 3rd term is which of the following? | [
"−1.5",
"−1",
"0",
"0.2"
] | D | First = 0.8
Second = 2*0.8-1 = 0.6
Second = 2*0.6-1 = 0.2
Answer: option D |
AQUA-RAT | AQUA-RAT-37899 | # Difference between revisions of "2010 AMC 10B Problems/Problem 20"
## Problem
Two circles lie outside regular hexagon $ABCDEF$. The first is tangent to $\overline{AB}$, and the second is tangent to $\overline{DE}$. Both are tangent to lines $BC$ and $FA$. What is the ratio of the area of the second circle to that of the first circle?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108$
## Solution 1
A good diagram is very helpful.
The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle $GBA$ while the second circle is inscribed in $GKJ$. From this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas $\triangle GKJ$ to $\triangle GBA$
Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is $\left(\frac{GK}{GB}\right)^2$. From the diagram, we can see that this is $9^2=\boxed{\textbf{(D)}\ 81}$
## Solution 2
As above, we note that the first circle is inscribed in an equilateral triangle of sidelength 1 (if we assume, WLOG, that the regular hexagon has sidelength 1). The inradius of an equilateral triangle with sidelength 1 is equal to $\frac{\sqrt{3}}{6}$. Therefore, the area of the first circle is $(\frac{\sqrt{3}}{6})^2 \cdot \pi =\frac{\pi}{12}$.
The following is multiple choice question (with options) to answer.
If the side of the regular hexagon above is 2, what is the circumference of the inscribed circle? | [
"2∏√3",
"3∏",
"4∏/√3",
"2∏/√3"
] | A | Now AC will be 1 because circle is touching the midpoints of the circle.
Then we get kite ACBO.
We have AC = 1 and CB = 1, now we need to find BO which is radius
from triplets we get 30:60:90 = 1:√3:2
Now CBO is right angled triangle, then we get BO as √3.
r = √3...circumference is 2∏r is 2∏√3.
Answer :option A. |
AQUA-RAT | AQUA-RAT-37900 | Example 2
In a large population of adults, 45% have a post secondary degree.
If people are selected at random from this population,
a) what is the probability that the third person selected is the first one that has a post secondary degree?
b) what is the probability that the first person with a post secondary degree is randomly selected on or before the 4th selection?
Solution to Example 2
a)
Let "having post secondary degree" be a "success". If a person from this population is selected at random, the probability of "having post secondary degree" is $p = 45\% = 0.45$ and "not having post secondary degree" (failure) is $1 - p = 1 - 0.45 = 0.55$
Selecting a person from a large population is a trial and these trials may be assumed to be independent. This is a geometric probability problem. Hence
$P(X = 3) = (1-0.45)^2 (0.45) = 0.1361$.
b)
On or before the 4th is selected means either the first, second, third or fourth person. The probability may be written as
$P(X \le 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$
Substitute by the formula $P(X = x) = (1 - 0.45)^{x-1} 0.45$ to write
$P(X \le 4) = (1 - 0.45)^{1-1} 0.45 + (1 - 0.45)^{2-1} 0.45 + (1 - 0.45)^{3-1} 0.45 + (1 - 0.45)^{4-1} 0.45 = 0.9085$
## Sums of the terms of a Geometric sequence
The following is multiple choice question (with options) to answer.
Out of 12 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them? | [
"5/55",
"21/55",
"51/55",
"41/55"
] | D | P(at least one graduate) = 1 - P(no graduates)
= 1 - 8C₃/12C₃ = 1 - (8 * 7 * 6)/(10 * 11 *12)
= 41/55
Answer: D |
AQUA-RAT | AQUA-RAT-37901 | ## [1] 22.13
### Changing the Budget
A broader social question is whether the budget is being set at an appropriate level. One way to examine this is to look at how the QALY outcome changes as the budget changes.
1. What is the rate of change of $$f_A(x_A)$$ with respect to $$x_A$$? Of course, this depends on the value of $$x_A$$, so evaluate the derivative at the optimal expenditure.
2. What is the rate of change of $$f_B(x_B)$$ with respect to $$x_B$$? Again, evaluate this at the optimal expenditure (keeping in mind that the expenditure on B is $$50-x_A$$).
3. Using the chain rule, simplify the expression $$f_A(x_A) +f_B(50-x_A)$$ and show that, if you are at the optimal values of $$x_A$$ and $$x_B$$, it must be the case that $$\frac{\partial}{\partial x_A} f_A( x_A) = \frac{\partial}{\partial x_B} f_B (x_B)$$
This suggests another way to look at the optimum, plotting out the
difference between the derivatives to find inputs $$x_A$$ and $$x_B$$
where they are equal.
dfA = D(fA(xA) ~ xA)
dfB = D(fB(xB) ~ xB)
plotFun(dfA(xA) - dfB(xB) ~ xA & xB, xA.lim = range(0, 50), xB.lim = range(0,
50))
There are many pairs of values $$(x_A, x_B)$$ on this graph where the two derivatives are equal. Find several. Then explain how the optimal value given the budget constraint $$x_A + x_B = 50$$ corresponds to just one of these.
\begin{AnswerText}
The optimal point is at the intersection of the constraint (shown in red) and the set of points where the two derivatives are equal.
The following is multiple choice question (with options) to answer.
If cost of sugar increases by 25%. How much percent consumption of sugar should be decreased in order to keep expenditure fixed? | [
"20",
"10",
"15",
"25"
] | A | 100
125
-----
125 ----- 25
100 ------ ? => 20%
ANSWER A |
AQUA-RAT | AQUA-RAT-37902 | # 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 70 percent is decaffeinated, what percent, by weight, of the grocer’s stock of coffee is decaffeinated? | [
"28%",
"30%",
"32%",
"34%"
] | B | 1. 20% of 400=80 pounds of decaffeinated coffee
2. 70% of 100=70 pounds of decaffeinated coffee
3. Wt have 150 pounds of decaffeinated out of 500 pounds, that means 150/500*100%=30%. The correct answer is B. |
AQUA-RAT | AQUA-RAT-37903 | 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.
In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race? | [
"180 sec",
"190 sec",
"290 sec",
"490 sec"
] | B | Time taken by B run 1000 meters = (1000 * 10)/50
= 200 sec.
Time taken by A = 200 - 10
= 190 sec.
Answer:B |
AQUA-RAT | AQUA-RAT-37904 | python, python-3.x, datetime
if sun == 1:
first_sunday = np.busday_offset(start, 0, roll='forward', weekmask='Sun')
last_sunday = np.busday_offset(end, 0, roll='preceding', weekmask='Sun')
sun_count = np.busday_count(first_sunday, last_sunday, weekmask='Sun') + 1
for i in trange(sun_count, desc="Sunday"):
touch.touch(file_name + " " + datetime.strptime(str(first_sunday), '%Y-%m-%d').strftime('%m-%d-%Y') + "." + file_type)
first_sunday += np.timedelta64(7, 'D')
if mon == 1:
first_monday = np.busday_offset(start, 0, roll='forward', weekmask='Mon')
last_monday = np.busday_offset(end, 0, roll='preceding', weekmask='Mon')
mon_count = np.busday_count(first_monday, last_monday, weekmask='Mon') + 1
for i in trange(mon_count, desc="Monday"):
touch.touch(file_name + " " + datetime.strptime(str(first_monday), '%Y-%m-%d').strftime('%m-%d-%Y') + "." + file_type)
first_monday += np.timedelta64(7, 'D')
if tue == 1:
first_tuesday = np.busday_offset(start, 0, roll='forward', weekmask='Tue')
last_tuesday = np.busday_offset(end, 0, roll='preceding', weekmask='Tue')
tue_count = np.busday_count(first_tuesday, last_tuesday, weekmask='Tue') + 1
for i in trange(tue_count, desc="Tuesday"):
The following is multiple choice question (with options) to answer.
If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004? | [
"Wednesday",
"Tuesday",
"Sunday",
"Monday"
] | C | The year 2004 is a leap year. So, it has 2 odd days.
Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
6th March, 2004 is Sunday
Answer: C |
AQUA-RAT | AQUA-RAT-37905 | $7|61$ gives $61=7\cdot 8 +5$
He would have 5 cows left over. So 61 can't be an answer
5. Hello, swimalot!
A cowboy was asked how many cows were on the ranch.
He replied that he was unsure, but he knew that when he counted them
by twos, threes, fours, fives, or sixes, he always had one left over.
When counted by sevens, he had none left over.
What is the smallest number of cows on the ranch?
Let $N$ = number of cows on the ranch.
The LCM of 2, 3, 4, 5, 6 is $60$
. . Hence: . $N \:=\:60a + 1$ ... for some integer $a.$
Since $N$ is divisible by 7: . $60a + 1 \:=\:7b$ ... for some integer $b.$
Solve for $b\!:\;\;b \:=\:\frac{60a+1}{7} \;=\;8a + \frac{4a+1}{7}$
Since $b$ is an integer, $4a + 1$ must be divisible by 7.
The first time this happens is: $a = 5$
Therefore: . $N \;=\;60(5)+1 \;=\;\boxed{301}$
6. Originally Posted by TheEmptySet
we know that it needs to be a multiple of 7 from all of the other clues we know it can't be even (because two can't divide it) and the last digit need to be a one because when divided by 5 it needs one left over.
here is our list,
56,63,70,77,84,91
If you check all the other conditions you will see that they hold.
I hope this helps.
Hello Tessy
91 doesn't work for ... four
91=88+3
7. You could use the Chinese Remainder Theorem, Topsquark has a lovely example in post #5 here http://www.mathhelpforum.com/math-he...nt-modulo.html
8. I will do the Chinese Remainder Theorem for you:
The following is multiple choice question (with options) to answer.
Two persons A and B take a field on rent. A puts on it 21 horses for 3 months and 15 cows for 2 months; B puts 15 cows for 6months and 40 sheep for 7 1/2 months. If one day, 3 horses eat as much as 5 cows and 6 cows as much as 10 sheep, what part of the rent should A pay? | [
"1/3",
"1/9",
"2/8",
"6/4"
] | A | 3h = 5c
6c = 10s
A = 21h*3 + 15c*2
= 63h + 30c
= 105c + 30c = 135c
B = 15c*6 + 40s*7 1/2
= 90c + 300s
= 90c + 180c = 270c
A:B = 135:270
27:52
A = 27/79 = 1/3
Answer: A |
AQUA-RAT | AQUA-RAT-37906 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
If it is 7:19 in the evening on a certain day, what time in the morning was it exactly 2,880,705 minutes earlier? (Assume standard time in one location.) | [
"7:25",
"7:34",
"7:43",
"7:47"
] | B | 7:19 minus 2,880,705 must end with 4, the only answer choice which ends with 4 is B.
Answer : B. |
AQUA-RAT | AQUA-RAT-37907 | # 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 person walking at 4 Kmph reaches his office 8 minutes late. If he walks at 6 Kmph, he reaches there 16 minutes earlier. How far is the office from his house? | [
"4 4/5 Km",
"4 1/7 Km",
"3 1/5 Km",
"3 1/7 Km"
] | A | Formula = S1*S2/S2-S1 * T1+T2/60
= 4*6/2 * 24/6
= 24/2 * 24/60
= 12 * 2/5
= 24/5 = 4 4/5 Km
A) |
AQUA-RAT | AQUA-RAT-37908 | 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 train covers a distance in 50 min, if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40min will be | [
"45 min",
"60 min",
"55 min",
"70 min"
] | B | Time = 50/60 hr = 5/6hr
Speed = 48mph
distance = S*T = 48 * 5/6 = 40km
time = 40/60hr = 2/3hr
New speed = 40* 3/2 kmph = 60kmph
Answer : B |
AQUA-RAT | AQUA-RAT-37909 | take a 5 = 48 ⇒ a r 5-1 = 48 ⇒ a 2 4 = 48 ⇒ a = 3. . To obtain the median, let us arrange the salaries in ascending order: The question asks for a three digit number: call the digits x, y and z. Example 1:Find the arithmetic mean of first five prime numbers. . . Median: Arrange the goals in increasing order We much have b – A = A- a ; Each being equal to the common difference. If we remove one of the numbers, the mean of the remaining numbers is 15. Mean with solution. The following table shows the grouped data, in classes, for the heights of 50 people. From equation ( ii ) , ( iii ) & b = -1. a = 2 and c = -4. Solution: Using the formula: Sum = Mean × Number of numbers Sum of original 6 numbers = 20 × 6 = 120 Sum of remaining 5 numbers = 15 × 5 = 75 Question 20. So the sequence is 2, -1, -4. The last two numbers are 10. Solution:First five prime numbers are 2, 3, 5, 7 and 11. Selection of the terms in an Arithmetic Progression, If number of terms is 3 then assume them as ” a-d, a & a+d” and common difference is “d”, If number of terms is 4 then assume them as ” a-3d, a-d, a+d & a+3d ” and common difference is “2d”, If number of terms is 5 then assume them as ” a-2d, a-d, a, a+d & a+2d ” and common difference is “d”, If number of terms is 6 then assume them as ” a-5d, a-3d, a-d, a+d , a+3d & a+5d ” and common difference is “2d”. . Detailed explanations and solutions to these questions … This GMAT averages problem is a vey easy question. The “n” numbers r1, r2, r3, r4, . G n are said to be Geometric means in between ‘a’ and ‘b’. a + c = 2b . Give One Example Each On Designed Experiment, Observational Study And Retrospective Study, T Distribution And Sampling Distribution Between The Difference Of Means And Justify. Solution: From graph, median = 54. Example -11: Four geometric means are
The following is multiple choice question (with options) to answer.
Given that a is the average (arithmetic mean) of the first five positive multiples of six and b is the median of the first twelve positive multiples of six, what is the ratio of a to b? | [
"3:4",
"6:13",
"5:6",
"13:10"
] | B | The first nine positive multiples of six are {6, 12, 18, 24,30}
The first twelve positive multiples of six are {6, 12, 18, 24, 30,36,42, 48, 54, 60, 66, 72}
Both sets are evenly spaced, thus their median=mean:
a=18 and b=(36+42)/2=39 --> a/b=18/39=6/13.
Answer: B. |
AQUA-RAT | AQUA-RAT-37910 | However, it seems clear that the question is meant to be something along the lines of:
In how many different ways can we pair the boys and the girls to make six couples that will all dance at the same time?
Now, since the order in which we list the pairs doesn't matter, we just care about what girl goes with what boy. So we can arrange the girls on a line and keep them fixed, and just see in how many ways we can "shuffle" the boys to pair them up with the girls: indeed, each of these will lead to a different set of six pairings. And given any set of six pairings, we can re-list them so that the girls appear in the order we selected. So we just need to count them under this assumption (the girls are ordered, say alphabetically by name, and we just need to order the boys to decide who dances with which girl). Viewed in this light, the question is equivalent to asking in how many different ways we can order the six boys.
There are six possibilities for the boy who goes first. That leaves five for the boy who goes second, four for the boy who goes third, etc. We are computing the permutations of six objects, and the answer is simply $6\times 5\times 4\times 3\times 2\times 1 = 6!$.
-
What does the term "How many couples can perform together?" mean. If it means how many distinct ways can the boys and girl be matched up then it is 6! But if it just means how many pairing of boys and girls can there be then the answer is 6x6=36 because each boy can be matched up with any one of the 6 girls. I think the question is ambiguous.
-
The following is multiple choice question (with options) to answer.
A group of medical interns at Bohemus Medical School want to go on dates. There are 5 girls and 7 guys. Assuming girls go on dates with guys, how many possible ways can these 10 medical interns date each other? | [
" 10",
" 25",
" 60",
" 2520"
] | D | 1st girl can go with 7 guys
2nd girl can go with remaining 6
3rd girl can go with remaining 5 and so on
so the total ways are 7 * 6 * 5 * 4 * 3
D should be the answer |
AQUA-RAT | AQUA-RAT-37911 | -
No, please see what I added for c) – André Nicolas Nov 18 '12 at 3:52
Your answer of $\binom{20}5$ for (a) is correct.
(b) To get the probability that $1$ and $2$ are not in the sample, you must count the $5$-card sets that don’t contain either $1$ or $2$. These are the $5$-card sets chosen from $\{3,4,5,\dots,20\}$; how many of them are there? This is really a problem just like (a). Once you have that, what fraction must you form to get the desired probability?
(c) Now observe that the events in (b) and (c) are complementary, so their probabilities add up to a known amount; what is that amount? Use it and your answer to (b) to get the answer to (c).
-
The following is multiple choice question (with options) to answer.
If the numbers 1 to 95 are written on 95 pieces of paper, (one on each) and one piece is picked at random, then What is the probability that the number drawn is neither prime nor composite? | [
"1/50",
"1/25",
"1/95",
"1"
] | C | There are 25 primes, 69 composite numbers from 1 to 95. The number
which is neither prime nor composite is 1.
Therefore, required probability = 1/95.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37912 | star
A8 1.784E+01 2.202E+00 7832. 1.900 8.004E+06 1.62 -0.12 0.20 0.10 0.27 2.02 1.74 1.93 1.63 1.47 1.34 1.26 1.23 1.20 0 0
A7 2.226E+01 2.268E+00 8176. 2.000 7.125E+06 1.38 -0.13 0.17 0.08 0.21 1.77 1.51 1.68 1.43 1.30 1.17 1.10 1.07 1.04 0 0
A3 3.330E+01 2.395E+00 8824. 2.200 5.731E+06 0.94 -0.19 0.08 0.03 0.07 1.26 1.13 1.21 1.10 1.06 0.97 0.95 0.94 0.93 0 0
A0 5.441E+01 2.512E+00 9705. 2.500 4.298E+06 0.41 -0.38 -0.02 0.00 -0.03 0.73 0.80 0.78 0.80 0.82 0.84 0.85 0.86 0.87 0 0
A0 7.341E+01 2.618E+00 10241. 2.700 3.595E+06 0.09 -0.49 -0.04 -0.01 -0.05 0.42 0.57 0.53 0.58 0.63 0.68 0.70 0.72 0.73 0 0
B9 1.059E+02 2.720E+00 11048. 3.000 2.815E+06 -0.31 -0.63 -0.07 -0.01 -0.07 0.06 0.32 0.26 0.34 0.40 0.49 0.53 0.55 0.58 0 0
The following is multiple choice question (with options) to answer.
852.68 − 4.5 + 108.98 = ? − 2132.54 | [
"6078.58",
"5225.9",
"6258.58",
"6088.58"
] | A | Option 'A' |
AQUA-RAT | AQUA-RAT-37913 | There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
2. HINT: 3k + 42 = 5k + 2
4. Originally we must have integers in the ratio $8k:5k:3k$ clearly we can't have fractional apples.
you are adding 42 to the crate that is in the ratio 3.
And then you are told this is 2 more than the amount in the crate of ratio 5.
So that means for some integer k, we need the equation posted above to be satisfied.
$42+3k=5k+2\Rightarrow 40=2k \Rightarrow k=20$
That means 20 is this common ratio we are looking for.
Crate A $=20\cdot 8=160$
Crate B $=20\cdot 5=100$
Crate C $=20\cdot 3=60$
Add these up to get 220 apples.
You check and see that if you add 42 to 60 you get 102 which is 2 more than 100 and these crates are in the proper ratios, thus the answer is correct. Well done wilmer, hope you don't mind me jumping in here, I just saw that you were not signed in and didnt want gwen to wait for a response.
5. Thank You very much, Gamma and Wilmer.
6. Originally Posted by gwen
There are 3 crates of apples A, B and C. The ratio of the number of apples in crate A to crate B to crate C is 8 : 5 : 3. David removed 42 apples from crate A and placed them into crate C. as such, there are 2 more apples in crate C than crate B. How many apples are there altogether?
You are given the ratio, and told that there are at least 42 items in A (else how could 42 be remove, right?), so one way to start might be to list triples in the given ratio, with the first value being 42 or larger. We can safely assume that we are dealing with whole numbers, so:
The following is multiple choice question (with options) to answer.
If 10 crates have 99 apples each and how many apples more is required in order to make 100 apples in each crate? | [
"5",
"10",
"15",
"20"
] | B | Each crate requires 1 apples and totally there are 10 crates so required apples = 10 * 1 = 10
Answer: B |
AQUA-RAT | AQUA-RAT-37914 | # 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.
The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet? | [
"10 a.m",
"07 a.m",
"11 a.m",
"03 a.m"
] | C | Suppose they meet x hrs after 8 a.m. Then,
(Distance moved by first in x hrs) + [Distance moved by second in (x - 1) hrs] = 330
60x + 75(x - 1) = 330 => x = 3
So, they meet at (8 + 3) i.e., 11 a.m.
Answer: C |
AQUA-RAT | AQUA-RAT-37915 | ### 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;
$$x + 425 = 1600$$
$$x = 1600$$ $$-$$ $$425 = 1175$$
Intern
Joined: 16 May 2018
Posts: 47
Location: Hungary
Schools: Queen's MBA'20
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.
Rs.6160 were divided among B, C & D in such a way that B had Rs.320 more than C and D had Rs 240 more than B . How much was C’s share? | [
"Rs.2300",
"Rs.2310",
"Rs.2320",
"Rs.2330"
] | C | Let C gets Rs x. Then We can say B gets Rs (x + 320 ) and D gets Rs ( x + 280) .
x + 320 + x + x + 560 = 6160
3x = 5280
x = 1760 .
C’s share = Rs ( 1760+560 ) = Rs.2320
C |
AQUA-RAT | AQUA-RAT-37916 | have s6 : f (fin.last n.succ) ≤ max (f (fin.last n.succ)) (fin_max n f') := le_max_left (f (fin.last n.succ)) (fin_max n f'),
have s7 : f (fin.last n.succ) ≤ fin_max n.succ f := eq.subst (eq.refl s5) s6,
have s8 : fin_max n f' ≤ max (f (fin.last n.succ)) (fin_max n f') := le_max_right (f (fin.last n.succ)) (fin_max n f'),
have s9 : fin_max n f' ≤ fin_max n.succ f := eq.subst (eq.refl s5) s8,
The following is multiple choice question (with options) to answer.
The H.C.F of 1.75, 5.6 and 7 is? | [
"0.07",
"0.7",
"3.5",
"0.35"
] | D | Given numbers with two decimal place are: 1.75, 5.60 and 7.00. Without decimal places, these numbers are: 175, 560 and 700, whose H.C.F. is 35.
H.C.F. of given numbers = 0.35.
Correct Option : D |
AQUA-RAT | AQUA-RAT-37917 | Advertisement Remove all ads
# Answer the following : The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. - Mathematics and Statistics
Sum
Answer the following :
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and the combined S.D.
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#### Solution
Let suffix 1 denote quantities for boys and suffix 2 for girls.
Given : bar(x)_1 = 70, sigma_1 = 8, bar(x)_2 = 62, sigma_2 = 10, n1 + n2 = 200
∴ n2 = 200 – n1
Combined mean = bar(x) = 65, where
bar(x) = ("n"_1bar(x)_1 + "n"_2bar(x)_2)/("n"_1 + "n"_2)
∴ 65 = (70"n"_1 + 62(200 - "n"_1))/200
∴ 13000 = 8n1 + 12400
∴ 600 = 8n1
∴ n1 = 75
∴ n2 = 200 – 75 = 125
d1 = bar(x)_1 - bar(x) = 70 – 65 = 5
d2 = bar(x)_2 - bar(x) = 62 – 65 = – 3
If combined S.D. is sigma, then
sigma = sqrt(("n"_1("d"_1^2 + sigma_1^2) + "n"_2("d"_2^2 + sigma_2^2))/("n"_1 + "n"_2)
= sqrt((75(25 + 64) + 125(9 + 100))/200
= sqrt((6675 + 13625)/200
= sqrt(101.5)
= 10.07
Hence, the number of boys = 75 and combined S.D. = 10.07.
Concept: Standard Deviation for Combined Data
Is there an error in this question or solution?
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#### APPEARS IN
The following is multiple choice question (with options) to answer.
The average height of 50 pupils in a class is 150 cm. Five of them whose height is 146 cm, leave the class and five others whose average height is 156 cm, join. The new average height of the pupils of the class (in cm) is? | [
"288",
"151",
"177",
"282"
] | B | :
Total height = 150 * 50 = 7500 cm.
New average = [7500 - 5 * 146 + 5 * 156 ] / 50 = 151 cm.
Answer: B |
AQUA-RAT | AQUA-RAT-37918 | 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.
A work crew of 5 Men takes 6 days to complete one-half of a job. If 7 men are then added to the crew and the men continue to work at the same rate, how many days will it take the enlarged crew to do the rest of the job? | [
"2 1/2",
"3",
"3 1/3",
"4"
] | A | Suppose 1 man can do work in X days..
so 5 men will do in ..
5/X=1/6*1/2 as half job is done
X=60
now 7 more are added then
12/60=1/2*1/d for remaining half job
d=2 1/2 Number of days
A |
AQUA-RAT | AQUA-RAT-37919 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\frac{Before-After}{Before}*100=\frac{45-20}{45}*100=\frac{2500}{45}=55.55$$% or ~56%
But say you took the other way $$=\frac{45-20}{20}*100=\frac{2500}{20}=125$$% .. AND the wrong answer is there in the choice.
so be careful
I would add more examples with a slight different wordings slightly later
_________________
Percentage increase/decrease- WHAT should be the denominator?? [#permalink] 29 Jan 2019, 05:54
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
20% of a number is more than 30% of 120 by 80. Find the number? | [
"520",
"560",
"580",
"620"
] | C | (20/100) * X – (30/100) * 120 = 80
1/5 X = 116
X = 580
Answer:C |
AQUA-RAT | AQUA-RAT-37920 | Puzzle of gold coins in the bag
At the end of Probability class, our professor gave us the following puzzle:
There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?
After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):
Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.
My questions are:
The following is multiple choice question (with options) to answer.
A bag contains an equal number of one rupee, 50 paise and 25 paise coins respectively. If the total value is 105, how many coins of each type are there? | [
"20 coins",
"30 coins",
"40 coins",
"60 coins"
] | D | Let number of each type of coin = x. Then,
1 × x + .50 × x + .25x = 105
⇒ 1.75x = 105 ⇒ x = 60 coins
Answer D |
AQUA-RAT | AQUA-RAT-37921 | # probability of the 3rd pick is a boy
10 students are in a class,
4 of these students are boys and 6 are girls
the teacher wanted to randomly select 3 students to represent the class in an event
what is the probability that the 3rd student is a boy?
Here is what I have tried
the probability of that 1st student is a boy = 4/10 the probability of that 1st student is a girl = 6/10
the probability of that 3rd student is a boy means one of these scenarios
1- previously 2 boys were selected
2- previously 2 girls were selected
3- previously a boy and a girl were selected
in case (1) the probability of having the 3rd is a boy would be 2/8
in case (2) the probability of having the 3rd is a boy would be 4/8
in case (3) the probability of having the 3rd is a boy would be 3/8
So which one of these is the right answer?!
• You can do it that way but there's no need to, each choice is like any other so the answer is $\frac 4{10}$. To do it the way you started, you need to weight each case by the probability of being in that scenario. – lulu Apr 28 at 22:50
• @lulu but when we select the 1st and 2nd student, number of students change and the distribution changed as well.. how come it is the same as the 1st pick? – asmgx Apr 28 at 22:53
• If you specify the first two choices then of course the answer changes, but if you don't specify them then there is no information from them, – lulu Apr 28 at 22:56
• – JMoravitz Apr 28 at 22:59
• If you were to simplify that god-awful expression however... it very simply results in $\frac{4}{10}$... which as many of the answers and comments in the linked question as well as other answers and comments here already will tell you makes sense as being "the third" person to be selected is not different enough from being "the first" person selected, so naturally the probability the third is a boy is going to be the same as the probability the first is a boy which is obviously $\frac{4}{10}$ with no tedious calculations required. – JMoravitz Apr 28 at 23:50
The following is multiple choice question (with options) to answer.
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary. What is the probability that only girls will be elected? | [
"8/125.",
"2/5.",
"1/30.",
"1/720."
] | C | probability that only girls will be elected = 4C3/10C3 = 1/30
Answer C. |
AQUA-RAT | AQUA-RAT-37922 | 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.
70 91 120
14 13 24
5 7 ?
What number should replace the question mark? | [
"2",
"5",
"3",
"19"
] | B | B
5
70 ÷ 14 = 5;
91 ÷ 13 = 7,
120 ÷ 24 = 5; |
AQUA-RAT | AQUA-RAT-37923 | 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.
Mohan's salary was first increased by 20% and then decreased by 20%. If his present salary is Rs. 7200, then what was his original salary? | [
"188",
"7500",
"2788",
"1277"
] | B | Let Mohan's salary be Rs.100.
When increased by 20%, Mohan's salary = Rs.120
Again when decreased by 20%, Mohan's salary = 120 - 24 = Rs. 96.
But present salary is Rs. 7200
for, 96 ---> 100 ; 7200 ---> ?
Required salary is 7200/96 * 100 = Rs. 7500
Answer: B |
AQUA-RAT | AQUA-RAT-37924 | (D)
Manager
Joined: 03 Aug 2017
Posts: 64
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink]
### Show Tags
05 Oct 2019, 22:16
Bunuel wrote:
If 20 typists can type 48 letters in 20 minutes, then how many letters will 30 typists working at the same rate complete in 1 hour?
A. 63
B. 72
C. 144
D. 216
E. 400
This i how i solved...
20 Typist can type 48 letters / 20 Min therofre in 1 Min 20 Typist can type = 48/20 = 2.4 Letters per Min
In 60 Min 20 Typist can type = 2.4 *60 =144 Words per min...
We are also told now there are 30 Typist so if 20 typist can type 144 words so 30 Typist can type X words per min.....
Solve for X
20/144 = 30 / x .....
x = 72*3 = 216 Words per min
Re: If 20 typists can type 48 letters in 20 minutes, then how many letters [#permalink] 05 Oct 2019, 22:16
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
How many times the keys of a typewriter have to be pressed in order to write first 400 counting numbers? | [
"1092",
"1300",
"1500",
"1125"
] | A | 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
=====> 1092
ANSWER A |
AQUA-RAT | AQUA-RAT-37925 | The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
The following is multiple choice question (with options) to answer.
Increasing the original price of an article by 10 percent and then increasing the new price by 10 percent is equivalent to increasing the original price by | [
"20%",
"20.5%",
"20.8%",
"21%"
] | D | 1.1*1.1*x = 1.21*x
The answer is D. |
AQUA-RAT | AQUA-RAT-37926 | Thanks for any help. I am totally confused.
1. 👍 0
2. 👎 0
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there are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag. what is the theoretical probability of randomly drawing either a red marble or a blue marble?
1. ### math check probability
5. A bag contains 7 green marbles and 4 white marbles. You select a marble at random. What are the odds in favor of picking a green marble? A. 7:11**? B. 7:4 C. 4:7 D. 3:7 ------------------------------------ 6. Food Express is
2. ### Math
There are 5 red marbles, 8 blue marbles, and 12 green marbles in a bag. What is the theoretical probability of randomly drawing a red marble and then a green marble? 10%
3. ### math
A bag contains 8 red marbles, 5 blue marbles, 8 yellow marbles, and 6 green marbles. What is the probability of choosing a red marble if a single choice is made from the bag? is it 8/27 ?
4. ### statistics
The following is multiple choice question (with options) to answer.
A jar contains 3 red marbles and 7 green marbles.If a marble is drawn from a jar at random,what is the probability that this marble is green? | [
"3/10",
"7/10",
"1",
"4/10"
] | B | Total number of marbles = 7+3 = 10
no. of green marbles = 7
probability of drawn a green marble= 7/10
Answer is B |
AQUA-RAT | AQUA-RAT-37927 | Substitute 0.085 for interest rate, $3000 for the amount deposited and$6000 for the amount after t years in the formula, “A=Pert” as,
$6000=$3000e0.085t
Divide both sides by $3000 as,$6000$3000=$3000e0.085t$3000e0.085t=2 Take natural logarithm on both sides as, ln(e0.085t)=ln2 Now, apply the inverse property of lnex as, 0.085t=ln2 Divide both sides by 0.085 as, 0.085t0.085=ln20.085t0.6930.0858.15 years So, the deposited amount will be doubled in approximately 8.15 years when the interest rate is 8.5%. Now, as the amount will become quadruple of the deposited amount after t years, then the amount after t years is 4P, that is, A=4P=4($3000)=\$12000
Substitute 0
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The following is multiple choice question (with options) to answer.
The equal amounts of money are deposited in two banks each at 15% per annum for 3.5 years and 5 years respectively. If the difference between their interests is Rs.144, find the each sum? | [
"337",
"640",
"297",
"276"
] | B | (P*5*15)/100 - (P*3.5*15)/100 = 144
75P/100 – 52.5P/100 = 144
22.5P = 144 * 100
=> P = Rs.640
Answer:B |
AQUA-RAT | AQUA-RAT-37928 | Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.
So 45 runs upon x innings will give an average of 4.5
Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)
Problem 6:
The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?
Sol:
If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.
Can you please explain how you arrived at 94 and 92
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Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman in his 12th inning makes a score of 60 and their by increasing his average by 4. What is his average after the 12th inning? | [
"12",
"16",
"20",
"24"
] | B | 11x + 60 = 12(x + 4)
x = 12 + 4 = 16
Answer:B |
AQUA-RAT | AQUA-RAT-37929 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
If books bought at prices ranging from Rs. 130 to Rs. 320 are sold at prices ranging from Rs. 210 to Rs 310, what is the greatest possible profit that might be made in selling 18 books ? | [
"Rs. 2500",
"Rs. 3000",
"Rs. 3500",
"Rs. 3240"
] | D | The greatest profit is possible only if the cost price of the books are minimum and selling prices are maximum.
Let lowest cost price of the 18 books = 130*18 = Rs. 2,340
Maximum selling price of 18 books = 310 *18 = Rs. 5,580
So, maximum profit = 5580 - 2340 = Rs. 3,240
ANSWER : OPTION D |
AQUA-RAT | AQUA-RAT-37930 | # 6.8 Exponential growth and decay (Page 3/9)
Page 3 / 9
If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from $100$ to $200$ bacteria as it does to grow from $10,000$ to $20,000$ bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have
$\begin{array}{ccc}\hfill 2{y}_{0}& =\hfill & {y}_{0}{e}^{kt}\hfill \\ \hfill 2& =\hfill & {e}^{kt}\hfill \\ \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}2& =\hfill & kt\hfill \\ \hfill t& =\hfill & \frac{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{k}.\hfill \end{array}$
## Definition
If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by
$\text{Doubling time}\phantom{\rule{0.2em}{0ex}}=\frac{\text{ln}\phantom{\rule{0.2em}{0ex}}2}{k}.$
## Using the doubling time
Assume a population of fish grows exponentially. A pond is stocked initially with $500$ fish. After $6$ months, there are $1000$ fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches $10,000.$ When will the owner’s friends be allowed to fish?
The following is multiple choice question (with options) to answer.
You have found a mutant algae that doubles in size every hour. It takes 18 hours for one algae plant to take up half of a certain lake.
How long would it take for two of these plants to take up half of the same lake? | [
"7",
"17",
"27",
"37"
] | B | At 17 hours the one plant will take up 1/4 of the lake (1/2 of 1/2). At 17 hours the two plants would be double the size of the one plant and double 1/4 is one half. |
AQUA-RAT | AQUA-RAT-37931 | For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since $$15 = 6*2 + 3$$.
Hence, the positive integer k is divided by the positive integer n, the remainder is 11, could be written as k = nq + 11. Divide by n: k/n = q + 11/n.
We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55.
Similar questions to practice:
http://gmatclub.com/forum/when-positive ... 06493.html
http://gmatclub.com/forum/if-s-and-t-ar ... 35190.html
Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html
Tips on Remainders: http://gmatclub.com/forum/remainders-ti ... s#p1376126
Units digits, exponents, remainders problems: http://gmatclub.com/forum/new-units-dig ... 68569.html
All DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
All PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199
P.S. Please read carefully and follow: http://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank you.
Hi Bunuel,
"We are also given that k/n = 81.2 = 81 + 0.2. So, the quotient, q, is 81 and 11/n is 0.2: 11/n = 0.2 --> n = 55."
The following is multiple choice question (with options) to answer.
The difference of two numbers is 1385. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ? | [
"274",
"270",
"295",
"360"
] | A | Let the smaller number be x. Then larger number = (x + 1385).
x + 1385 = 6x + 15
5x = 1370
x = 274
Smaller number = 274.
ANSWER A |
AQUA-RAT | AQUA-RAT-37932 | # 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.
At what price must an article costing Rs.47.50 be marked in order that after deducting 5% from the list price. It may be sold at a profit of 25% on the cost price? | [
"62.5",
"62.0",
"62.6",
"62.1"
] | A | CP = 47.50
SP = 47.50*(125/100) = 59.375
MP*(95/100) = 59.375
MP = 62.5
Answer: A |
AQUA-RAT | AQUA-RAT-37933 | 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 and B together can do a piece of work in 10 days. B alone can finish it in 20 days. In how many days can A alone finish the work ? | [
"20",
"10",
"15",
"5"
] | A | Time taken by A to finish the work = XY/(Y-X)
= 10 x 20 / (20-10)
= 200/10
= 20 days
Answer: A |
AQUA-RAT | AQUA-RAT-37934 | 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 moves fast a telegraph post and a bridge 264 m long in 8 sec and 40 sec respectively. What is the speed of the train? | [
"89",
"29.7",
"56",
"79.2"
] | B | Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/40 = y
y = 8.25 Speed = 8.25 m/sec = 8.25 * 18/5 = 29.7 km/hr.
Answer: Option B |
AQUA-RAT | AQUA-RAT-37935 | $(120,34)\simeq S_{5}$
$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$
$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$
$(144,183)\simeq S_{3}\times S_{4}$
$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$
$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$
$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$
$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$
$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$
The following is multiple choice question (with options) to answer.
Three numbers are in the ratio 3:4:5 and their L.C.M is 2400. Their H.C.F is: | [
"40",
"80",
"120",
"200"
] | A | Let the numbers be 3x, 4x and 5x.
Then, their L.C.M = 60x. So, 60x = 2400 or x = 40.
The numbers are (3 * 40), (4 * 40) and (5 * 40).
Hence, required H.C.F = 40.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37936 | -
Hint: There are ten outcomes that give a sum of 12: (4,4,4), (2,2,8), (2,8,2), (8,2,2), (2,4,6), (4,2,6), (6,2,4), (2,6,4), (4,6,2), (6,4,2).
-
This looks like a classic Bayes' rule problem. We want to find the probability of two 2s given that the three numbers sum to 12. We can apply Bayes' rule thusly: $$P(\text{two }2s| \text{sum}=12) = \frac{P(\text{sum}=12 | \text{two } 2s)*P(\text{two } 2s)}{P(\text{sum}=12)}$$ It should be easier to find these probabilities than the original one directly (especially with the help of Byron Schmuland's answer). Note that the numerator is also $P(\text{two } 2s \bigwedge \text{sum}=12)$, which you have already found.
-
The following is multiple choice question (with options) to answer.
Three students appear at an examination of Mathematics. The probability of their success are 1/2, 1/4, 1/5 respectively. Find the probability of success of at least two. | [
"1/60",
"2/50",
"9/40",
"3/40"
] | C | The probability of success of at least two students will involve the following possibilities.
The first two students are successful, the last two students are successful, the first and third students are successful and all the three students are successful.
Therefore, the required probability = 1/2 x 1/4 x 4/5 + 1/4 x 1/5 x 1/2 + 1/2 x 1/5 x 3/4 + 1/2 x 1/4 x 1/5 = 9/40
ANSWER:C |
AQUA-RAT | AQUA-RAT-37937 | # 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 shipment of 1500 heads of cabbage, each of which was approximately the same size was purchased for $600.The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head.The following day the rest were sold at a price per head equal to 10% less than the price each head sold for the day before.what was the gross profit V on this shipment? | [
"a) $100",
"b) $115",
"c) $125",
"d) $130"
] | C | I recommend one should use fractions to solve this one rather than converting it into decimals
Sol:
1500 heads -> $600
1 head -> $600/1500
1 head -> $(2/5)
25% more of (2/5) -> 125/100 * 2/5 = $(1/2)
He sold 2/3*1500 = 1000 heads for $(1/2) per head
Total revenue by selling 1000 heads = 1000 * 1/2 = $500
Heads left: 500
Cost per head: 90% of the previous price: 90/100 * 1/2 = $(9/20)
Total revenue by selling 500 heads = 9/20 * 500 = 225
Total revenue after selling 1500 cabbage heads - 225+500 = $725
Money spent on the purchase: $600
Profit V= 725-600 = $125
Ans:C |
AQUA-RAT | AQUA-RAT-37938 | 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 running at the speed of 72 km/hr takes to cross a bridge 132 m length? | [
"11",
"12.1",
"13",
"14"
] | B | Speed = 72 * 5/18 = 20 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/20 = 12.1 sec.
Answer: Option B |
AQUA-RAT | AQUA-RAT-37939 | # 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.
John purchased a fridge & a mobile for Rs. 15000&Rs. 8000 respectively. He sold the fridgeat a loss of 4% and the mobile phone at a profit of 10%. Overall how much he make a profit. | [
"Rs. 100",
"Rs. 150",
"Rs. 200",
"Rs. 300"
] | C | Let the SP of the refrigerator and the mobile phone be Rs. r and Rs. m respectively.
r = 15000(1 - 4/100) = 15000 - 600
m = 8000(1 + 10/100) = 8000 + 800
Total SP - Total CP = r + m - (15000 + 8000) = -600 + 800 = Rs. 200
As this is positive, an overall profit of Rs. 200 was made.
C |
AQUA-RAT | AQUA-RAT-37940 | # Inequality with two absolute values
I'm new here, and I was wondering if any of you could help me out with this little problem that is already getting on my nerves since I've been trying to solve it for hours.
Studying for my next test on inequalities with absolute values, I found this one:
$$|x-3|-|x-4|<x$$ (I precisely found the above inequality on this website, here to be precise, but, the problem is that when I try to solve it, my answer won't be $(-1,+\infty)$, but $(1,7)$. I took the same inequalities that the asker's teacher had given to him and then put them on a number line, and my result was definitely not $(-1,+\infty)$
Here are the inequalities: $$x−3 < x−4 +x$$ $$x−3 < −(x−4) +x$$ $$−(x−3)<−(x−4)+x$$
And here are my answers respectively: $$x>1, \quad x>-1, \quad x<7$$
I will really appreciate if anyone could help me out, because I'm already stressed dealing with this problem, that by the way, it is not demanding that I solve it, but you know, why not?
What you have is almost correct, the last step is to restrict your solution to the corresponding region.
For example, for $$x>1$$, the answer must be in the $$x>4$$ region, so your answer for this region is $$x>4$$.
For $$x>-1$$, your answer must be in $$x<3$$ region, so your answer for this region is $$-1.
And for the last one, $$x<7$$, your answer must be in the corresponding region we had first, namely $$3, so your answer for this region is $$3.
Now draw these three answers on the number line and you'll have $$x>-1$$, the desired final answer.
The following is multiple choice question (with options) to answer.
If –1 < x < 1 and x ≠ 0, which of the following inequalities must be true?
I. x^3 < x
II. x^2 < |x|
III. x^4 – x^5 > x^3 – x | [
" I only",
" II only",
" III only",
" II and III only"
] | A | Answer is D. Let's look at the statements one by one
Stmt I. x^3 < x
if 0<x<1 then x^3< x but if -1<x<0 then x^3>x
So this statement is not always true
Stmt II. x^2 < |x|
Because we know that x is a number less than one but not equal to zero then x^2 will always be less than |x|.
Why? think of positive fractions (and you can think in terms of positive fractions because the inequality is in regards to |x|). Lets set x = 1/2, then x^2 = 1/4 and 1/4<1/2
So Stmt II is always true
Stmt III. x^4 – x^5 > x^3 – x^2
This one may seem tricky but lets break it down. x^4 – x^5 > x^3 – x^2 = x^4(1-x)>x^2(x-1).
Because lets concentrate on (1-x) and (x-1). We are given that -1<x<1 which means that (1-x)>0 and (x-1)<0. x^4 will always be positive and x^2 will always be positive so without doing any math we are looking at positive > negative... which is always true.
So Stmt III is always true
A |
AQUA-RAT | AQUA-RAT-37941 | 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.
A woman is 18 years older than her son. In 4 years, her age will be twice the age of her son. The present age of her son is | [
"12",
"14",
"16",
"18"
] | B | Explanation:
Let the son's present age be x years. Then, woman's present age = (x + 18) years
=> (x + 18) + 4 = 2(x + 4)
=> x + 22 = 2x +8
So, x = 14 Answer: B |
AQUA-RAT | AQUA-RAT-37942 | 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.
Excluding stoppages, the speed of a train is 45 kmph and including stoppages it is 33 kmph. Of how many minutes does the train stop per hour? | [
"E982",
"16",
"12",
"121"
] | B | Explanation:
T = 12/45 * 60 = 16
Answer: Option B |
AQUA-RAT | AQUA-RAT-37943 | We are given: .a
5 = 120 .and .a6 = 720
Then: .r .= .a
6/a5 .= .720/120 .= .6
. . The "rule" is multiply-by-six.
Therefore, the preceding term is: .a
4 = 20.
See? .We could have eyeballed the problem . . .
5. Originally Posted by Soroban
. . There is a simpler solution.
I always tell my students that I have a tendancy to make things harder than they have to be.
-Dan
The following is multiple choice question (with options) to answer.
Find the fourth proportional to 0.5, 1.12 and 1.3? | [
"2.914",
"2.81",
"2.192",
"2.912"
] | D | Explanation:
Formula = Fourth proportional = (b × c)/a
A = 0.5 , B = 1.12 and C = 1.3
(1.12 × 1.3)/0.5
1.456/0.5 = 2.912
Answer: Option D |
AQUA-RAT | AQUA-RAT-37944 | the classroom. What does this mean?. applying Sn formula, we know that d is 3 and 4 respectively. How can you tell if a number is divisible by 3? A. (ZX81 BASIC doesn't do this automatically: both sides of an OR are evaluated, even (c, 5); BigInteger answer15 = GetSumOfNumbersDivisibleByN(c, 15); Console. Thus we do not need to remove. A conditional statement is known as an If-then statement. 2: If the last number is even. 1050 is divisible by 5 and it is But i did not understand the divisibility rule 11 example 4. Examples Approach: To solve the problem, follow the below steps: Find the sum of numbers that are divisible by 3 upto N. It's one of the easiest methods to quickly find the sum of given number series. (a) ___ 6724. This number is divisible by 5, because its last digit is 5. Find the sum of the digits. Contains All of Contains Any of Does not contain any of Count of odd digits is Divisible by Sum of all digits is Digit sum of all digits is Consecutive digits in a sequence at most Consecutive digits in a. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory. Write a program that prints “yes” if the given integer is divisible by 2 or 3 and “no” otherwise. The sum of all numbers smaller than a divisible by 3 or 5 is the same as. Therefore sum is 288350. Consider N1, N2, N3… be numbers which when divided by a divisor D, give quotients Q1, Q2, Q3… and remainders R1, R2, R3… respectively. The sum of the digits of the number is a multiple of 3. @ShadowRanger as per the task it requires the sum of numbers divisible by 3 or 5 within a given. For large numbers this rule can be applied again to the result. For example: Numbers 90, 150, 700 are divisible by 2, because they end in
The following is multiple choice question (with options) to answer.
If number divisible by both 3 and 2 ,then the same number divisible by which number? | [
"4",
"5",
"8",
"6"
] | D | Obviously,The answer is 6
Option D |
AQUA-RAT | AQUA-RAT-37945 | ### 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.
John bought 2 shares and sold them for $75 each. If he had a profit of 20% on the sale of one of the shares but a loss of 20% on the sale of the other share, then on the sale of both shares John had | [
"a profit of $10",
"a profit of $6.25",
"a loss of $6.25",
"a loss of $10"
] | C | Loss% = (%age Profit or loss / 10)^2 = (20/10)^2 = 4% loss
Total Selling Price = 75*2 = $192
Total Cost Price = 150/(0.96) = $156.25
Loss = 156.25 - 150 = $6.25
Answer: Option C |
AQUA-RAT | AQUA-RAT-37946 | # Find Smallest Positive Integer
#### anemone
##### MHB POTW Director
Staff member
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
#### Wilmer
##### In Memoriam
Find the smallest positive integer $n$ such that for every integer $m$ with $0<m<1993$, there exists an integer $k$ for which $$\displaystyle \frac{m}{1993}<\frac{k}{n}<\frac{m+1}{1994}$$.
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Unfortunately, got no exotic formula for you
#### anemone
##### MHB POTW Director
Staff member
Terrific puzzle, Anemone!
Smallest n = 3987
Example using low m (1) and high m (1992):
(m + 1) / 1994 > k / n > m / 1993
m = 1:
2 / 1994 > 3 / 3987 > 1 / 1993
m = 1992:
1993 / 1994 > 3985 / 3987 > 1992 / 1993
Hi Wilmer,
Thanks for participating and thanks for the compliment to this problem as well.
Yes, 3987 is the answer to this problem but...
Unfortunately, got no exotic formula for you
30 minutes in the corner, please...
#### Wilmer
##### In Memoriam
30 minutes in the corner, please...
No fair! You only asked: "Find the smallest positive integer [FONT=MathJax_Math-italic]n" [/FONT]
#### anemone
The following is multiple choice question (with options) to answer.
If a is the smallest positive integer such that 5,880 multiplied by a is the square of an integer, then ya must be | [
"30",
"15",
"12",
"23"
] | A | 5880= 2*2*7*7*5*6, so we need one 5 and one 6 to make it a square of a number. so 5*6= 30
Ans: A |
AQUA-RAT | AQUA-RAT-37947 | The remainder is $\,2 \,P_6(x^2)\,$, which follows for $\,n=6\,$ from the general identity:
\begin{align} P_{2n}(x^2) = \frac{x^{4n}-1}{x^2-1} &= \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}+1}{x+1} \\[5px] &= \, \frac{x^{2n}-1}{x-1} \, \frac{x^{2n}-1+2}{x+1} \\[5px] &= - \, \frac{x^{2n}-1}{x-1} \, \frac{(-x)^{2n}-1}{(-x)-1} + 2 \, \frac{x^{2n}-1}{x^2-1} \\[5px] &= - \, P_{2n}(x) P_{2n}(-x) + 2 P_n(x^2) \end{align}
The divisor $f = (\color{#c00}{x^{\large 12}\!-1})/(x-1)$ and $\,g = (1+\color{#c00}{x^{\large 12}})(1+x^{\large 2}+\cdots+x^{\large 10})\,$ is the dividend
hence $\bmod\, f\!:\,\ \color{#c00}{x^{\large 12}\equiv 1}\$ implies that $\,\ g\equiv\, (1\:+\ \color{#c00}1\,)\:(1+x^{\large 2}+\cdots+x^{\large 10})$
The following is multiple choice question (with options) to answer.
In a division sum, the remainder is 6 and the divisor is 5 times the quotient and is obtained by adding 22 to the thrice of the remainder. The dividend is | [
"74",
"166",
"86",
"92"
] | B | Divisor = (6 * 3) + 22 = 40
5 * Quotient = 40
Quotient = 8.
Dividend = (Divisor * Quotient) + Remainder
Dividend = (20 * 8) + 6 = 166.
B) |
AQUA-RAT | AQUA-RAT-37948 | or the length of the line you see in red. After finding your height, substitute your values for base and height into the formula for area of a triangle to find the area. Area of triangle = × Base × Height . Area of a rhombus. Area of Triangle (given base and height) A triangle is a 3-sided polygon. Side of triangle without height @, tan30^ @, cos30^ @, @. Deriving the formula of half the product of the line you see red! Hence, the side “ a ” units n't use 1/2 × base height! Triangle of the triangle 0.5 area of an equilateral triangle ABC area of equilateral triangle formula when height is given as... Bc * sinB its side sides and an included angle is given as area... That sinB = sin30° = 1/2 * AB * BC * sinB if we call the side a! A perpendicular AD is drawn from a to side BC, then AD is the amount of that! A triangle is given as: area of triangle without height is of cm. Triangle = so, the formula for area of triangle without height it all! Formed by height will be a/2 units long cos30^ @, or @... Of a triangle triangle is the amount of space that it occupies in a 2-dimensional surface also substitute into! ( h ) or the length of each side of the side “ ”. Know that sinB = sin30° = 1/2 = 0.5 area of a triangle 2-dimensional surface without height = sin30° 1/2... To get the height be found using the formula for area of triangle = so, =... It means all side of the site ; Geometry since this is an equilateral can. That sinB = sin30° = 1/2 * AB * BC * sinB units.! An equilateral triangle of triangle without height units long X as its side diagram at right... Of triangle without height is 10 cm, it means all side of triangle without height the included is! ) or the length of the side across from 30 degrees will be triangles... A 2-dimensional surface ca n't use 1/2 × base × height is an equilateral triangle is 10. At the right shows when to use the formula given below
The following is multiple choice question (with options) to answer.
If the sides of a triangle are 26 cm, 22 cm and 10 cm, what is its area? | [
"120",
"110",
"130",
"140"
] | B | The triangle with sides 26 cm, 22 cm and 10 cm is right angled, where the hypotenuse is 26 cm.
Area of the triangle = 1/2 * 22 * 10 = 110 cm2
Answer: Option B |
AQUA-RAT | AQUA-RAT-37949 | ### Show Tags
16 Jun 2018, 09:16
agdimple333 wrote:
During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of$25.00 Did the store sell more sweaters than shirts during the sale?
1) The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00 2) The total of the prices of all of the shirts and sweaters that the store sold during the sale was$420.00
The average of the prices of all of the shirts and sweaters that the store sold during the sale was $21.00. Since the average price of$21 is closer to $25 than it is to$15, there must be more sweaters sold than shirts. Statement one alone is sufficient.
Statement Two Alone:
The total of the prices of all of the shirts and sweaters that the store sold during the sale was $420.00. It’s possible that 12 sweaters and 8 shirts are sold since 12 x 25 + 8 x 15 = 300 + 120 =$420. It’s also possible that 6 sweaters and 18 shirts are sold since 6 x 25 + 18 x 15 = 150 + 270 = $420. In the former example, more sweaters were sold; however, in the latter example, more shirts were sold. Statement two alone is not sufficient. Answer: A _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 11 Mar 2018 Posts: 1 Re: During a sale, a clothing store sold each shirt at a price of$15 and [#permalink]
### Show Tags
24 Jan 2020, 02:24
1
Bunuel wrote:
dchow23 wrote:
from statement 2,
shirts x
sweaters y
15x +20y = 420
Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation?
If there is a common multiple for
The following is multiple choice question (with options) to answer.
The original price of a suit is $100. The price increased 20%, and after this increase, the store published a 20% off coupon for a one-day sale. Given that the consumers who used the coupon on sale day were getting 20% off the increased price, how much did these consumers pay for the suit? | [
"$88",
"$96",
"$100",
"$106"
] | B | 0.8*(1.2*100) = $96
The answer is B. |
AQUA-RAT | AQUA-RAT-37950 | - 6 years, 9 months ago
So, here's my proposed solution.
Solution: It is trivial how to prove the 'if' part. Let us proceed to prove the 'only if' part. Suppose $x,y,z$ are such integers with $xy+1, yz+1, zx+1$ not all squares. Let us also assume that $x,y,z$ is the smallest counterexample, that is, $x+y+z$ is minimum.
WLOG, $xy+1$ is not a perfect square. As Dinesh Chavan wrote below, let $s$ be the smallest positive root of $s^2+x^2+y^2+z^2-2(xy+yz+zs+sx+zx+sy)-4xyzs-4 = 0$. This can be, as he said, written equivalently as:
$\\ (x+y-z-s)^2 = 4(xy+1)(zs+1),\\ (x+z-y-s)^2 = 4(xz+1)(ys+1),\\ (x+s-y-z)^2 = 4(xs+1)(yz+1).$
But then, our quadratic formula has root $s = x+y+z+2xyz\pm 2\sqrt{(xy+1)(yz+1)(zx+1)}$ which was given to be an integer, so we have that RHS is a square. This implies that $(xs+1)(ys+1)(zs+1)$ is a square.
Since, $xs+1\ge0,ys+1\ge0,zs+1\ge0$, we have that by verifying that $x=y=z=1$ is not a solution, $s\ge-\frac{1}{\max(x,y,z)}>-1$.
The following is multiple choice question (with options) to answer.
If x, y, and z are positive integers and x = 5y = 9z, then the least possible value of x + y - z is | [
"33",
"40",
"49",
"61"
] | C | x + y - z = x/5 + x - x/9
= (45 + 9 - 5 )x/45 = 49x/45
49 is not divisible by 45, so for least value, x = 45
C |
AQUA-RAT | AQUA-RAT-37951 | 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.
A pump can fill a tank with water in 3 hours. Because of a leak, it took 3 1/3 hours to fill the tank. The leak can drain all the water of the tank in? | [
"17 hr",
"19 hr",
"30 hr",
"14 hr"
] | C | Work done by the tank in 1 hour
= (1/3 - 3 1/3)
= 1/30 Leak will empty the tank in 30 hrs.
Answer:C |
AQUA-RAT | AQUA-RAT-37952 | This simplifies to $$t(9z-x^2)+(2z-y)=0 \tag{1}$$
Now we have two cases:
Case 1: $$9z-x^2\ne0$$:
Then we have
$$t=\frac{2z-y}{9z-x^2}$$
Because of $$x,y,z \in \{0,\ldots,9\}$$ and $$9z-x^2\ne0$$ we have $$|2z-y|\le18$$ and $$|9z-x^2|\ge1.$$ So $$|t|\le18$$, and therefore $$t=11$$.
If we substitute this in $$(1)$$ we get
$$11x^2+2y=101z$$
Taking this equation $$\pmod {11}$$ we get
$$y\equiv 2z \pmod{11}$$
and we can calulate $$y$$ for every digit $$z$$. For $$z=5$$ there is no valid $$y$$, because $$y=10$$ is not a digit. From $$z$$ and $$y$$ we can calculate $$x$$. For $$z=2$$ and $$y=4$$ there is no valid $$x$$ because the calculated value is a non integer square root.
z y x
0 0 0
1 2 3
2 4 *
3 6 *
4 8 6
5 * *
6 1 *
7 3 8
8 5 *
9 7 *
Case 2: $$9z-x^2=0$$
Then we also have
$$2z-y=0$$
The number $$9z$$ must be a perfect square and therefore $$z$$ must be a perfect square. So we can calculate
z x y
0 0 0
1 3 2
4 6 8
9 9 *
Because $$9z-x^2=2z-y=0$$ the equation holds for arbitrary $$t.$$
So we can conclude that
$$88^2+33=7777$$
The following is multiple choice question (with options) to answer.
If z is a multiple of 9402, what is the remainder when z^2 is divided by 9? | [
"0",
"2",
"4",
"6"
] | A | The sum of the digits is 9+4+2=15.
Thus 3 is a factor of 9402, so 3 is a factor of z.
Then 3^3 = 9 is a factor of z^2.
Then the remainder when z^2 is divided by 9 is 0.
The answer is A. |
AQUA-RAT | AQUA-RAT-37953 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is? | [
"2:3",
"4:3",
"6:7",
"9:16"
] | B | Explanation:
Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √16 : √9 = 4:3
ANSWER IS B |
AQUA-RAT | AQUA-RAT-37954 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
The area of a square is equal to three times the area of a rectangle of dimensions 25 cm * 27 cm. What is the perimeter of the square? | [
"180 cm",
"190 cm",
"170 cm",
"150 cm"
] | A | Area of the square = s * s = 3(25 * 27)
=> s = 3 * 5 * 3 = 45 cm
Perimeter of the square = 4 * 45 = 180 cm.
Answer: A |
AQUA-RAT | AQUA-RAT-37955 | to solve for the other. Logic To Calculate Percentage Difference Between 2 Numbers. Sale Discount Calculator - Percent Off Mortgage Loan Calculator - Finance Fraction Calculator - Simplify Reduce Engine Motor Horsepower Calculator Earned Value Project Management Present Worth Calculator - Finance Constant Acceleration Motion Physics Statistics Equations Formulas Weight Loss Diet Calculator Body Mass Index BMI Calculator Light. The commonly used way is to go from right to left, which gives us a positive number. Posted: (1 week ago) Percentage difference is usually calculated when you want to know the difference in percentage between two numbers. Just subtract the past value from the current value. Click the calculate button 3. Calculate percentage difference between two columns I have a input text file in this format: ITEM1 10. The percentage difference between the two values is calculated by dividing the value of the difference between the two numbers by the average of the two numbers. First, you need Excel to subtract the first number from the second number to find the difference between them. Listing Results about Calculate Variance Between Two Numbers Real Estate. To write an increase or decrease as a percentage, use the formula actual increase or decrease original cost × 100%. The growth rate can be listed for real or nominal GDP. This difference needs to be divided between the first number (the one that doesn't change). We then append the percent sign. Follow 52 views (last 30 days) Show older comments. Enter an old number in cell A1 and a new number in cell B1. The percent difference formula or the percent difference equation of two numbers a and b is: ((a - b) / (a+b)/2) × 100, where a > b Calculate the percentage difference between the numbers 35 and 65. The first step to the equation is simple enough. Step 1: Calculate the difference (subtract one value from the other) ignore any negative sign. With a Difference From, Percent Difference From, or Percent From calculation, there are always two values to consider: the current value, and the value from which the difference should be calculated. A percentage variance, aka percent change, describes a proportional change between two numbers, an original value and a new value. It has more of an impact when you say, "There was a 50 percent increase in attendance at the concert compared to last year," versus when you say, "There were 100 more people at the concert this year than
The following is multiple choice question (with options) to answer.
Two numbers are respectively 20% and 25% more than a third number. The percentage that is first of the second is? | [
"98%",
"56%",
"96%",
"964%"
] | C | I II III
120 125 100
125----------120
100-----------? => 96%
Answer:C |
AQUA-RAT | AQUA-RAT-37956 | Now since $$x_n\in\mathbb{R}$$ and $$x_n\neq 0\implies x_n^2>0.$$ Now since $$i\ge 0\implies i.x_n^2\ge 0 \implies 2+i-n\ge 0\implies i\ge n-2.$$ Therefore $$i=n-2,n-1$$.
Now when $$i=n-2$$, we have $$(n-2)x_n^2=0.$$ Now since it is given that $$n\ge 3\implies n-2\ge 1>0$$. Therefore $$x_n^2=0\implies x_n=0$$. But $$x_n\neq 0$$. Therefore $$i\neq n-2$$.
Now when $$i=n-1$$, we have $$(n-1)x_n^2=1\implies x_n^2=\frac{1}{n-1}\implies x_n=\pm\frac{1}{\sqrt{n-1}}.$$ Therefore we have $$x_1=x_2=\cdots=x_n=\pm\frac{1}{\sqrt{n-1}}.$$
Therefore the required set of solutions are $$(x_1,x_2,\cdots,x_n)=\left(\frac{1}{\sqrt{n-1}},\frac{1}{\sqrt{n-1}},\cdots,\frac{1}{\sqrt{n-1}}\right),\left(-\frac{1}{\sqrt{n-1}},-\frac{1}{\sqrt{n-1}},\cdots,-\frac{1}{\sqrt{n-1}}\right).$$
Can someone check if my solution is correct or not? And if correct, is there a more better and efficient solution than this?
The following is multiple choice question (with options) to answer.
If n > 0 , which of the following must be true?
I 2n - 1 > 0
II n - n^2 < 0
III n^2 > 1 | [
"I only",
"II only",
"none",
"I and II only"
] | C | I. True for all values of n except 1
When n = 1 ; n^2 = 1
II. True for all values of n except 1
When n = 1 ; n - n^2 = 1 - 1 =>0
III True for all values of n
Hence the correct answer will be (C) |
AQUA-RAT | AQUA-RAT-37957 | ### Show Tags
18 Feb 2015, 07:51
1
Using smart numbers:
Jennifer Purchases watches for $5: B=5 Markup of 100%; now selling for$10: x=10
She sells 5 watches: N=5
Her total profit will be (N*x)-(B*N)
(5*10)-(5*5)= 25
T=25
Now using the variables we must look at the answer choices that will solve for the % markup of the watches, which we chose to be 100.
B=5
T=25
N=5
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
18 Feb 2015, 21:51
Cost price per watch = b
Selling price per watch = $$\frac{t}{n}$$
Let x = percent of the markup from her buy price to her sell price
$$x = 100 *\frac{t}{n} * \frac{1}{b} = \frac{100t}{nb}$$
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Re: Jennifer can buy watches at a price of B dollars per watch, which she [#permalink]
### Show Tags
22 Feb 2015, 11:17
Bunuel wrote:
Jennifer can buy watches at a price of B dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of B and T and N, what is the percent of the markup from her buy price to her sell price?
A. 100T/(NB)
B. TB/(100N)
C. 100TN/B
D. ((T/N) – B)/(100B)
E. 100(T – NB)/N
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION
The following is multiple choice question (with options) to answer.
Ranjani can buy watches at a price of X dollars per watch, which she marks up by a certain percentage before selling. If she makes a total profit of T by selling N watches, then in terms of X and T and N, what is the percent of the markup from her buy price to her sell price? | [
"100T-X",
"100T/NX",
"100TX/N",
"100TX"
] | B | Algebraic Solution:If she makes a total profit of T for N watches, then that must be a profit of T/N for each watch. That must be the markup above cost on each watch, the amount of the increase. Well, percent increase = (amount of increase)/(starting amount) x 100% = (T/N)/X *100 = 100T/(NX)
Answer : B |
AQUA-RAT | AQUA-RAT-37958 | # Is this a simple answer to the classic problem "A certain city has 10 bus routes..."
A certain city has 10 bus routes. Is it possible to arrange the routes and the bus stops so that if one route is closed, it is still possible to get from anyone stop to any other (possibly changing along the way), but if any two routes are closed, there are at least two stops such that it is impossible to get from one to the other?
Given the solution above, why did the authors feel compelled to answer: Yes. Consider 10 straight lines in the plane, no 2 are parallel & no 3 are concurrent. Let lines be bus routes & let points of intersection be stops. We get from anyone stop to any other (if the stops lie on 1 line, w/o changing; & if not, then with just 1 change). If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once. However, if we discard 2 lines, then 1 stop-their point of intersection-will have no bus routes passing thru it, & it'll be impossible to get from this stop to any other.
Source: A. M. Yaglom and l. M. Yaglom CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY SOLUTIONS Volume II Problems From Various Branches of Mathematics Translated by James McCawley, Jr. Revised and edited by Basil Gordon DOVER PUBLICATIONS, INC. NEW YORK
• Yes, the answer is correct. Jun 9, 2019 at 6:12
• My guess, Joe, is that when they wrote "changing", they meant "changing once". Jun 9, 2019 at 13:04
• Your answer is correct and fine. It is possible the authors just didn't think of this solution and thus put their solution into the book. Everybody overlooks something every now and then. Jun 9, 2019 at 15:30
• My solution fails to meet the requirement, "If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once." Jun 10, 2019 at 0:55
• Gerry, your interpretation of the wording is what makes the problem sufficiently complex. Thanks. Jun 10, 2019 at 0:58
Let’s go back to the roots.
The following is multiple choice question (with options) to answer.
The ratio of buses to cars on River Road is 1 to 17. If there are 80 fewer buses than cars on River Road, how many cars are on River Road? | [
"40",
"85",
"60",
"30"
] | B | B/C=1/17
C-B=80.........> B=C-80
(C-80)/C=1/17
Testing answers. Clearly Eliminate ACDE
Put C=85.........> (85-80)/85=5/85=1/17
Answer: B |
AQUA-RAT | AQUA-RAT-37959 | ### 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.
David paid $7,000 for 100 shares of stock X and $5,000 for 100 shares of stock Y. He later sold the same shares of both stocks, gaining 3/4 the amount he paid for stock Y. If the prices he had paid for the stocks had been reversed and everything else remained the same, then the net result would have been: | [
"A Gain of 1.67 times as much",
"A Gain of 1.4 times as much",
"A Gain of 2 times as much",
"A Gain of 1.6 times as much"
] | B | A = 7000; B = 5000; Profit = (3/4)*5000 = 3750
After prices are reversed:
A = 5000; B =7000; Profit = (3/4)*7000 =5250
5250 --> Gain of 1.4 times 3750
Answer: B |
AQUA-RAT | AQUA-RAT-37960 | A: 9
B: 12
C: 16
D: 18
E: 24
This is a copy of the following OG question: five-machines-at-a-certain-factory-operate-at-the-same-constant-rate-219084.html
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 03:08
1
2
Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
Time taken by 4 machines to fill a certain production order = 27 hours
Time taken by 1 machine to fill that production order = 27 * 4 = 108 hours
Time taken by 6 machines to fill that production order = 108/6 = 18 hours
Number of fewer hours it takes 6 machines to fill that production order = 27 - 18 = 9 hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
In a manufacturing plant, it takes 36 machines 4 hours of continuous work to fill 6 standard orders. At this rate, how many hours of continuous work by 72 machines are required to fill 12 standard orders? | [
"4",
"6",
"8",
"9"
] | A | the choices give away the answer..
36 machines take 4 hours to fill 8 standard orders..
in next eq we aredoubling the machines from 36 to 72, but thework is not doubling(only 1 1/2 times),= 4*48/72*12/6 = 4
Ans A |
AQUA-RAT | AQUA-RAT-37961 | 4.2 5 customer reviews. And the third angle is three times of the first angle. In a triangle, the second angle is 5° more than the first angle. check your answers against the answer explanation section at the end of each chap-ter. Find the three angles of the triangle. Always let x represent the unknown number. Question 1: There are 47 boys in the class. Find the length of the sides. Click the button and find the first one on your computer. For problems that require more than one step, a thorough step-by-step explanation is provided. Given : If the rod was 2 meter shorter and each meter costs $1 more. So, the two parts of the 25 are 10 and 15. He prepared this workbook (with full solutions to every problem) to share his strategies for solving algebra word problems. Get step-by-step solutions to your math problems. Distance, Rate, and Time Word Problems These Algebra 1 Equations Worksheets will produce distance, rate, and time word problems with ten problems per worksheet. Given : The total cost of 125 units of the product is$28750. Find out below how you can print these problems. \displaystyle \frac {4} {5} of a number is less than 2 less than the same number. Learn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball. Now that you can do these difficult algebra problems, you can trick your friends by doing some fancy word problems; these are a lot of fun. Question 2 : The denominator of a fraction exceeds the numerator by 5. ... College Algebra Word Problem College Algebra Algebra Word Problem. Section 1: Examples. Area and perimeter word problems. So, the three angles of the triangle are 35°, 40° and 135°. Divide 25 in two parts so that sum of their reciprocals is 1/6. The sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right angle triangle is formed. : Step 2:: Solve the equation created in the first step. Identifying Divisibility by Looking at the Final Digits. Author: Created by dh2119. In a group of 120 people, 90 have an age of more 30 years, and the others have an age of less than 20 years. Example 1. Area = length ×
The following is multiple choice question (with options) to answer.
The second angle of a triangle is 45° more than the smallest angle. The third angle is three times
the smallest. How many degrees are there in each angle? | [
"L=6/W=18",
"L=18/W=6",
"L=7/W=19",
"L=19/W=7"
] | B | We are looking for three angles.
! 1 - x
! 2 - x + 45
! 3 - 3x
You would not be able to solve this problem unless you knew that the sum of the interior angles
of a triangle is 180°.
! 1 + ! 2 + ! 3 = 180°
x + x + 45 + 3x = 180°
5x + 45 = 180°
5x = 135°
x = 27°
That means the first angle is 27°, the second angle is 27 + 45 or 72°, and the third angle is 3 times
27 or 81°.
Try this on your own. The length of a rectangle is three times the width and its perimeter is 48 ft.
Find the length and width. The answers are length is 18 and the width is 6 ft.
correct answer B |
AQUA-RAT | AQUA-RAT-37962 | $\frac {0.367}{0.978} = 0.375.$
13. ## Re: Probability with a "known"
Another way to do this is with combinations (since you don't care about the order that you pick the fruit). You will get the same answer either way. With combinations, you can just divide the number of outcomes. So, if we figure out the number of outcomes with (at least two red AND at least one red or green) divided by the number of outcomes with at least one apple (red or green), that will give us the same probability.
Just as ebaines did, we want the number of outcomes with exactly 0 red and exactly 1 red, then take the compliment (total number of outcomes minus outcomes that give us only 0 or 1 red). For exactly 0 red, that is $\binom{24}{5}$ outcomes.
For exactly 1 red, that is $\binom{24}{4}\binom{8}{1}$ outcomes (note: since we are doing combinations, we don't care which of the apples is red). So, there are $\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}$ outcomes with at least 2 red.
For at least one apple (red or green), we want the number of outcomes with 0 apples and take the compliment: $\binom{32}{5} - \binom{16}{5}$.
So, the probability that you get at least 2 red apples given that you pick at least one apple (red or green):
$\dfrac{\binom{32}{5} - \binom{24}{5} - \binom{24}{4}\binom{8}{1}}{\binom{32}{5} - \binom{16}{5}} = \dfrac{73,864}{197,008} = \dfrac{1319}{3518} \approx 0.375$
14. ## Re: Probability with a "known"
The following is multiple choice question (with options) to answer.
A basket contains 9 apples, of which 1 is spoiled and the rest are good. If we select 2 apples from the basket simultaneously and at random, what is the probability that the 2 apples selected will include the spoiled apple? | [
"2/3",
"2/5",
"2/7",
"2/9"
] | D | The total number of ways to choose 2 apples is 9C2 = 36
The number of ways that include the spoiled apple is 8C1 = 8
P(the spoiled apple is included) = 8/36 = 2/9
The answer is D. |
AQUA-RAT | AQUA-RAT-37963 | # 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.
A jogger running at 10 km/hr along side a railway track is 340 m ahead of the engine of a 120 m long train running at 46 km/hr in the same direction. In how much time will the train pass the jogger? | [
"76 sec",
"67 sec",
"98 sec",
"46 sec"
] | D | Speed of train relative to jogger = 46 - 10 = 36 km/hr.
= 36 * 5/18 = 10 m/sec.
Distance to be covered = 340 + 120 = 460 m.
Time taken = 460/10 =46 sec.
Answer:D |
AQUA-RAT | AQUA-RAT-37964 | 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}$$
The following is multiple choice question (with options) to answer.
Manish, Rahul and Bharti have some stones with each of them. Five times the number of stones with Rahul equals seven times the number of stones with Manish while five times the number of stones with Manish equals seven times the number of stones with Bharti. What is the minimum number of stones that can be there with all three of them put together? | [
"113",
"109",
"93",
"97"
] | B | Explanation :
Let the stones with Manish, rahul and Bharti be m,r and b respectively.
Given,
=>5r=7m and 5m=7b.
=>25r=35m and 35m=49b.
=>25r=35m=49b=k.
=>r/49=m/35=b/25.
The least possible integral values for r,m,b will be r=49, m=35 and b=25.
=>Total=49+35+25=109.
Answer : B |
AQUA-RAT | AQUA-RAT-37965 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
A football matches has been organized in such a fashion, every participating team plays a match against every other team once and only once.
If 36 matches are totally played, how many teams participated? | [
"7",
"8",
"9",
"10"
] | C | For a match u need 2 teams.
Suppose there are totally 'n'teams.
Now we have to choose 2 teams out of 'n' teams.
nC2 = 36;
Solve to get n=9.
n(n-1)/2 = 36
9(9-1)/2 = 36
9(8)/2 = 36
72/2 = 36
So, answer is 9 If 36 matches are totally played.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37966 | => 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.
A sum of salary is distributed among A,B,C,D in the proportion of 2:3:4:6. If D gets $700 more than C, what is the B's share? | [
"$1050",
"$2500",
"$3000",
"$3400"
] | A | Let the shares of A,B,C,D are 2x,3x,4x,6x
6x-4x = 700
x = 350
B's share = 3x = $1050
Answer is A |
AQUA-RAT | AQUA-RAT-37967 | your kids. The volume formula for a rectangular box is height x width x length, as seen in the figure below: To calculate the volume of a box or rectangular tank you need three dimensions: width, length, and height. Find the dimensions of the box that minimize the amount of material used. The calculated volume for the measurement is a minimum value. Rectangular Box Calculate the length, width, height, or volume of a rectangular shaped object such as a box or board. This is the main file. You must have a three-dimensional object in order to find volume. Volume is the amount of space enclosed by an object. Our numerical solutions utilize a cubic solver. To determine the surface area of a cube, calculate the area of one of the square sides, then multiply by 6 because there are 6 sides. Volume of a Cuboid. The largest possible volume for a box with a square bottom and no top that is constructed out of 1200$\mathrm{cm}^2$of material is 4000$\mathrm{cm}^3$. For example, enter the side length and the volume will be calculated. 314666572222 cubic feet, or 28. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. CALCULATE VOLUME OF BOX. So: Answer. A container with square base, vertical sides, and open top is to be made from 1000ft^2 of material. A cuboid is a box-shaped object. 1) Enter a valid Start value into text box below, default is "1", 2) Select an increment value from the list below, default is 1, 3) Select an accuracy (significant digits) value from the list below, default is 5, 4) Most cases the result will appear immediately, 5) Click on the "Create Table" button. Everyone has a personal profile and you can use yours to choose colours that really suit your face. Volume of a square pyramid given base side and height. Volume of a cube - cubes, what is volume, how to find the volume of a cube, how to solve word problems about cubes, nets of a cube, rectangular solids, prisms, cylinders, spheres, cones, pyramids, nets of solids, examples and step by step solutions, worksheets. If a square has one side of 4 inches, the volume would be 4 inches times 4 inches times 4 inches, or 64 cubic inches. For
The following is multiple choice question (with options) to answer.
A rectangular cube has a volume of 6 cubic feet. If a similar cube is twice as long, twice as wide, and twice as high, then the volume, in cubic feet of such cube is?
Choices | [
"24",
"48",
"64",
"80"
] | B | Lets say the sides of the rectangular cube are x, y and z. Problem states volume is 6.
i.e,
x*y*z = 6 .
If every side is increased twice as before. then new volume is
2x*2y*2z = 8*(x*y*z) = 8*6 = 48.
Answer is B |
AQUA-RAT | AQUA-RAT-37968 | (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.
Little John had $5.10. He spent $1.05 on sweets and gave to his two friends $1.00 each. How much money was left? | [
"$2.15",
"$2.05",
"$2.45",
"$2.25"
] | B | John spent and gave to his two friends a total of
1.05 + 1.00 + 1.00 = $3.05
Money left
5.10 - 3.05 = $2.05
Answer :B |
AQUA-RAT | AQUA-RAT-37969 | # How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
The following is multiple choice question (with options) to answer.
Every person at a birthday party either wants candy or a wants cake, but not both. The ratio of people who want cake to people who want candy is 9 to 6 . If 9 people decided they wanted candy instead of cake, the ratio of people who want cake to people who want candy will be 1 to 1. How many people could be at the party? | [
"60",
"50",
"90",
"45"
] | C | By checking the answer choices:
Since we have an initial ratio of 9:6, the total count should be divisible by 15.
A. 60/15=4, Cake People = 36, Candy People = 24, Since 36-9 is not equal to 24+9 this can not be the answer.
B. Out, Since not divisible by 15.
C. 90/15=6, Cake People = 54, Candy People = 36, Since 54-9 is equal to 36+9, this is the correct answer.
D. 45/15=3, Cake People = 27, Candy People = 18, Since 27-9 is not equal to 18+9 this as well can not be the answer.
E. Since choice D is not correct, choice E is also not correct.
C is the correct answer. |
AQUA-RAT | AQUA-RAT-37970 | # Reset the equation counter
\documentclass{article}
\usepackage{amsmath, amsfonts, chngcntr}
\newcounter{problem}
\newcounter{solution}
\newcommand\Problem{%
\stepcounter{problem}%
\textbf{\theproblem.}~%
\setcounter{solution}{0}%
}
\newcommand\TheSolution{%
\textbf{Solution:}\\%
}
\newcommand\ASolution{%
\stepcounter{solution}%
\textbf{Solution \thesolution:}\\%
}
\parindent 0in
\parskip 1em
\begin{document}
\section{Kinematics}
\Problem A motorboat going going downstream overcame a raft at point $\emph{A}$; $\tau$ = 60 min later it turned back and after some time passed the raft at a distance $l$ = 6.0 km from the point $\emph{A}$. Find flow velocity assuming the duty of the engine to be constant.
\TheSolution Let u be the flow velocity and v be velocity of boat in still water,
$$\frac{l}{u}=\tau + \frac{(u+v)\tau-l}{v-u}$$
$$u=\frac{l}{2\tau}=\frac{6}{2\cdot1}=3 \ km/hr$$
\Problem A point traversed half the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time, and with velocity $v_2$ for the other half of the time. Find the mean velocity of the point averaged over the whole time of motion.
\TheSolution $$v_{av} = \frac{2\cdot v_0\cdot\frac{v_1+v_2}{2}}{v_0 + \frac{v_1+v_2}{2}}$$
\end{document}
The following is multiple choice question (with options) to answer.
A man rows his boat 65 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream? | [
"5 kmph",
"4 kmph",
"9 kmph",
"8 kmph"
] | B | Speed downstream = d/t = 65/(2 1/2) = 26 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream = (26 - 18)/2 = 4 kmph
Answer:B |
AQUA-RAT | AQUA-RAT-37971 | Manager
Joined: 07 Jul 2016
Posts: 79
GPA: 4
Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
04 Aug 2016, 22:33
AbdurRakib wrote:
Two bottles are partially filled with water. The larger bottle currently holds $$\frac{1}{3}$$ of its capacity. The smaller bottle, which has $$\frac{2}{3}$$ of the capacity of the larger bottle, currently holds $$\frac{3}{4}$$ of its capacity.If the contents of the smaller bottle are poured into the larger bottle, the larger bottle will be filled to what fraction of its
capacity?
Let $$L$$ be the capacity of the larger bottle
Let $$l$$ be the current capacity of the larger bottle. $$l = \frac{1}{3}L$$
Let $$S$$ be the capacity of the smaller bottle. $$S = \frac{2}{3}L$$
Let $$s$$ be the current capacity of the smaller bottle. $$s = \frac{3}{4}S$$
Question: What is $$l + s$$
First get $$s$$ in terms of $$L$$
$$s = \frac{3}{4} \times \frac{2}{3}L = \frac{1}{2}L$$
$$s + l = (\frac{1}{2} + \frac{1}{3})L = \frac{5}{6}L$$
A. $$\frac{5}{6}$$
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Re: Two bottles are partially filled with water. The larger bottle current [#permalink]
### Show Tags
The following is multiple choice question (with options) to answer.
A big container is 40% full with water. If 28 liters of water is added, the container becomes 3/4 full. What is the capacity of the big container in liters? | [
"68",
"72",
"76",
"80"
] | D | 28 liters is 35% of the capacity C.
28 = 0.35C
C = 28/0.35 = 80 liters.
The answer is D. |
AQUA-RAT | AQUA-RAT-37972 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$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.
Two trains are running in opposite directions with the same speed. If the length of each train is 120 m and they cross each other in 6 sec, then the speed of each train is? | [
"30",
"31",
"72",
"25"
] | C | Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/6 => x = 20
Speed of each train = 20 m/sec.
= 20 * 18/5 =72km/hr.
Answer: Option C |
AQUA-RAT | AQUA-RAT-37973 | Arithmatic & Algebra: http://gmatclub.com/forum/arithmatic-algebra-93678.html
Helpful Geometry formula sheet: http://gmatclub.com/forum/best-geometry-93676.html
I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.
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Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the [#permalink]
### Show Tags
04 May 2017, 08:45
udaymathapati wrote:
In the sequence 1, 2, 4, 8, 16, 32, …, each term after the first is twice the previous term. What is the sum of the 16th, 17th, and 18th terms in the sequence?
A. 2^18
B. 3(2^17)
C. 7(2^16)
D. 3(2^16)
E. 7(2^15)
The series is in Geometric progression.
To find nth term, Tn = a x r ^ (n-1)
So, the sum of 16th, 17th and 18th term will be
=2^15 + 2 ^16 + 2^17
=(2^15) x (1 + 2 + 2^2)
=7 x (2^15)
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Joined: 06 Dec 2016
Posts: 249
Re: In the sequence 1, 2, 4, 8, 16, 32, …, each term after the [#permalink]
### Show Tags
04 May 2017, 12:18
You could even do this approach
Look at the series carefully. You will see:
2^0, 2^1, 2^2, 2 ^3
1, 2, 3, 4
16th number is 2^15
17th number is 2^16
18th number is 2^17
2^15 + 2^16 + 2^17
2^15(1 + 2^1 + 2^2)
2^15(7)
The following is multiple choice question (with options) to answer.
In a certain series, each term is m greater than the previous term. If the 19th term is 760 and the 16th term is 700, what is the first term? | [
"200",
"300",
"400",
"450"
] | C | a + 18m = 760 (17th term)
a + 15m = 700 (16th term)
You get a = 400 and m = 20
Answer: C |
AQUA-RAT | AQUA-RAT-37974 | $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\
The following is multiple choice question (with options) to answer.
7+2+2^2+2^3+2^4+2^5=？ | [
"(2^3-1)(2^3+1)",
"2^6+5",
"2^5-1",
"2^5+1"
] | B | From 7+2+2^2+2^3+2^4+2^5=69,
2^6+5=69
the correct answer is B. |
AQUA-RAT | AQUA-RAT-37975 | # Number of ways in which they can be seated if the $2$ girls are together and the other two are also together but separate from the first two
$5$ boys and $4$ girls sit in a straight line. Find the number of ways in which they can be seated if $2$ girls are together and the other two are also together but separate from the first two.
I divided the $4$ girls in two groups in $\frac{4!}{2!2!}$ ways.
I counted $10$ ways in which there is at least one boy between the two girl pairs. Boys can be arranged in $5!$ ways. Total number of ways$=\frac{4!}{2!2!}\times 10\times 5!=7200$
But the answer in the book is $43200$. I don't know where I am wrong.
• What do you mean "the two girls are together"? That makes it sound as if the pair is specified. – lulu Apr 4 '16 at 16:10
• Sorry,"the" was not given,i edited it.@lulu – mathspuzzle Apr 4 '16 at 16:13
• no problem. I'll post something below. – lulu Apr 4 '16 at 16:13
The following is multiple choice question (with options) to answer.
Q-1) Of the 84 parents who attended a meeting at a
school, 35 volunteered to supervise children during
the school picnic and 11 volunteered both to supervise
children during the picnic and to bring refreshments to
the picnic. If the number of parents who volunteered
to bring refreshments was 3 times the number of
parents who neither volunteered to supervise children
during the picnic nor volunteered to bring
refreshments, how many of the parents volunteered to
bring refreshments? | [
"25",
"36",
"22",
"42"
] | C | i used the following formula: total = group1 + group2 + neither - both
using the information from the question stem, we have:
84 total people
35 who supervise (group1)
x who neither supervise nor bring refreshments
3x people who bring refreshments (group2)
11 who supervise AND bring refreshments (both)
therefore...
84 = 35 + 3x + x - 11
solving, we get x = 15.
since we want the value of those who bring refreshments AND supervise:
1.5(15) = 22
answer is C |
AQUA-RAT | AQUA-RAT-37976 | # Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of student is $pq$,however I am not sure how to find the minimum number of students.Any ideas?
$1) \quad pq - \binom{q}{2}$
$2) \quad pq - \binom{p}{2}$
$3) \quad (p-1)(q-1)$
-
Something is missing. Is every student supposed to be on a committee? – JavaMan Aug 31 '11 at 16:24
@DJC:Not mentioned in the question,I guess we may have to consider that to get a solution. – Quixotic Aug 31 '11 at 16:28
@DJC: For the minimum number of students this does not matter. – TMM Aug 31 '11 at 16:30
@Thijs Laarhoven:Yes you are right but as the problem also asked for maximum number I have considered it in my solution. – Quixotic Aug 31 '11 at 16:31
@Thijs, FoolForMath, I guess my question is, should the minimum answer be in terms of $p$ and $q$? – JavaMan Aug 31 '11 at 16:31
For $1\leq i\leq p$, let $C_i$ be the set of students on the $i$th committee. Then by inclusion-exclusion, or more accurately Boole's inequalities, we have $$\sum_i|C_i|-\sum_{i<j}|C_i C_j|\leq |C_1\cup C_2\cup\cdots \cup C_p|\leq \sum_i |C_i|.$$
From the constraints of the problem, this means $$pq-{p\choose 2}\leq \#\mbox{ students}\leq pq.$$
The following is multiple choice question (with options) to answer.
A student committee that must consists of 4 members is to be formed from a pool of 8 candidates. How many different committees are possible | [
"5",
"8",
"70",
"56"
] | C | Out of 8 people, any 4 can be selected for a committee. So 8c4=70 is the answer
C |
AQUA-RAT | AQUA-RAT-37977 | 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 travelling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car. | [
"55 km/hr.",
"25 km/hr.",
"65 km/hr.",
"35 km/hr."
] | D | D
35 km/hr.
Time taken = 1 hr 40 min 48 sec = 1 hr 40 4/5 min. = 126/75 hrs.
Let the actual speed be x km/hr.
Then, (5/7)x X 126/75 = 42
--> x = 35 km/hr. |
AQUA-RAT | AQUA-RAT-37978 | 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 the two candidates, the candidates who gets 70% of votes polled is winned by 280 vote’s majority. What is the total number of votes polled? | [
"750",
"700",
"800",
"850"
] | B | Explanation:
Note: majority (40 %) = difference in votes polled to win (70 %) & defeated candidates (30 %)
40 % = 70 % - 30 %
40% -----> 280 (40*7 = 280)
100% -----> 700 (100*7 = 700)
Answer: Option B |
AQUA-RAT | AQUA-RAT-37979 | ### 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.
A man bought an article and sold it at a gain of 5 %. If he had bought it at 5% less and sold it for Re 3 less, he would have made a profit of 10%. The C.P. of the article was | [
"344",
"218",
"200",
"600"
] | D | Explanation:
Let original Cost price is x
Its Selling price = (105/100) * x = 21x/20
New Cost price = (95/100) * x = 19x/20
New Selling price = (110/100 )* (19x/20 )= 209x/200
[(21x/20) - (209x/200)] = 3
=> x = 600
Answer: D) Rs 600 |
AQUA-RAT | AQUA-RAT-37980 | ## Digit Problems
1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$
## Digit Problems (Contd.)
The following is multiple choice question (with options) to answer.
Find the least number of five digits which is exactly divisible by 12, 15 and 18? | [
"10080",
"1080",
"10025",
"10050"
] | A | LCM = 180
180) 10000 (55
9900
---------
100
10000 + 180 - 100 = 10080
ANSWER A |
AQUA-RAT | AQUA-RAT-37981 | mass, newtonian-gravity, statics
Title: How do you weigh a box on a scale whose limit is too low? As you will see I know nothing about physics and after being asked to solve a physics problem in a recent interview wanted to ask it of professionals and see what the response would be:
I have a set of domestic scales (actually just one scale) that weigh up to the maximum weight of 5kg and a large package that weighs more than 5kgs but less than 10kgs. How can I tell the exact weight using the inadequate scales.
The package is a long item akin to a steal girder but not in weight obviously.
I know that stack exchange has rules about this type of question so please treat it as light entertainment from someone in awe of your intellect. You put the package horizontally across three scales and add up the weight you see. If the center scale registers more than the limit, move the box, or put some pieces of paper under the scales which are registering a small weight to redistribute the weight.
A more physics-y question would be how to determine the weight when you have only one scale. This can be done by tilting the package at a small angle, by placing one end on a scale (perhaps propped up on something light, like your shoe, so that the weight momentum flow goes through the scale to the floor). To weigh this way requires knowing where the CM of the box is, and this can be determined by balancing it precisely on the edge of the scale (or on your finger, or on something else narrow).
Then the weight registered by the scale (minus the weight of the shoe) is the ratio of the length of the box to the distance from the end which is on the floor to the center of mass. If this is still too heavy, you can tilt the box up to 45 degrees, which will cut down the weight registered by the scale further by a factor of .707, which can be undone by multiplying by 1.414.
The following is multiple choice question (with options) to answer.
You have only 1 kg weight. You have to weight 31kg. Min number of measurements you have to done ? | [
"21R",
"33R",
"35R",
"31R"
] | D | You can do it in 5 steps.
1,2,4,8,16
Asume stone as S (Left Side) weight of 1kg as T (Right Side)
Move 1T to the same side as S and, Left Side is S+1T, Now weight 2T
Move 2T to the same side as S and, against S+3T, Now weight 4T
Move 4T to the same side as S and, against S+7T, weight 8T
Move 8T to the same side as S and, against S+15T, weight on right Side is 16T
15R + 16R = 31R.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37982 | area of that circle click on the calculate button... It is also known as the longest straight line segment which passes the! Segment which passes through the center of the circle demonstration is to use this calculator, by... 3 = 150687.075 mm if we know circumference as the straight line segment which passes through material... Example tire with a 30 inch diameter turning at 300 rpm will have road... Outside circumference by 3.1415 measurement with a 30 inch diameter turning at 300 rpm will have a road speed 26.8. Of 26.8 mph x π x d 3 ) 16. d 3 ) 16. d 3 150687.075... Learn C programming, data Structures Tutorials, exercises, examples, programs, hacks, tips and tricks.! Defined as the known factor to calculate the area of a circle given circumference! Background Tutorials: Simply plug in 36 for d in the Wheel Graph calculator demonstration is to PI... Use this calculator, begin by entering the circumference is about 1/32″ ( )! Plug in 36 for d in the formula for calculating the area of a.... Do you find the diameter is given to be changed to calculate diameter circumference and area are - (. Circle 's edge from one point, meeting back at that point,,... A road speed of 26.8 mph we are using functions that find the end is accessible... Also select units of measure for both input data and results to private tutoring material best their. Equal to 5.3 meters you have the diameter consider the maximum torque can... The measurement of its diameter the formula for calculating the area of Wheel... Tire diameter always how to find diameter Hamiltonian cycle in the image, only the tire diameter speed. Use PI and the circumference of a circle given its circumference if we know circumference value into the for! Speed of 26.8 mph where it meets a 30 inch diameter turning at rpm... Yes if you have the diameter that find the radius with the longest chord of the circle using.! Helps you to find diameter, circumference, and area of the circle you... This program, we separated the logic using python functions features make Nerd! Back at that
The following is multiple choice question (with options) to answer.
If a tire rotates at 400 revolutions per minute when the car is traveling 168km/h, what is the circumference of the tire? | [
"2",
"1",
"7",
"3"
] | C | 400 rev / minute = 400 * 60 rev / 60 minutes
= 24,000 rev / hour
24,000 * C = 168,000 m : C is the circumference
C = 7 meters
correct answer C |
AQUA-RAT | AQUA-RAT-37983 | php, datetime
Title: Get Next Working Date, skip weekends and holidays The object is to be able to pass some dates (start date, holiday) and the number of days you want to skip. We only want to skip working days, not weekends and holidays.
Just let me know what you'd do different than what I have. The code works, but I was told there are issues with it, but not told what issues they are.
function getWDays($startDate,$holiday,$wDays){
$d = new DateTime( $startDate );
$t = $d->getTimestamp();
$h = strtotime($holiday);
// loop for $wDays days
for($i=0; $i<$wDays; $i++){
// 1 day = 86400 seconds
$addDay = 86400;
$nextDay = date('w', ($t+$addDay));
if($nextDay == 0 || $nextDay == 6) {
$i--;
}
$t = $t+$addDay;
if ($t == $h) {
// lets make sure the holiday isn't one of our weekends
if(!$nextDay == 0 || !$nextDay == 6) {
$t = $t+$addDay;
}
}
}
$d->setTimestamp($t);
return $d->format( 'Y-m-d' );
}
echo getWDays("2013-08-29","2013-09-02", 3) Here is another way of doing the same thing
<?php
function getWDays($startDate,$holiday,$wDays) {
// using + weekdays excludes weekends
$new_date = date('Y-m-d', strtotime("{$startDate} +{$wDays} weekdays"));
$holiday_ts = strtotime($holiday);
// if holiday falls between start date and new date, then account for it
if ($holiday_ts >= strtotime($startDate) && $holiday_ts <= strtotime($new_date)) {
The following is multiple choice question (with options) to answer.
Every second Saturday and all Sundays are holidays. How many working days will be there in a
month of 30 days beginning on a Saturday? | [
"24",
"23",
"18",
"21"
] | B | Explanation :
Mentioned month has begins on a Saturday and has 30 days
Sundays = 2nd, 9th, 16th, 23rd, 30th
=> Total Sundays = 5
Second Saturdays = 8th and 22nd
Total Second Saturdays = 2
Total Holidays = Total Sundays + Total Second Saturdays = 5 + 2 = 7
Total days in the month = 30
Total working days = 30 - 7 = 23
Answer : Option B |
AQUA-RAT | AQUA-RAT-37984 | We know a = $$frac{2}{3}\\$b from the question stem. If each decreased its speed by 10 kmph, the speeds of A and B will be$$$frac{2}{3}\\$b – 10) and$b – 10) respectively.
From statement 2, ($$frac{2}{3}\\$b – 10) = $\frac{1}{2}\\$$b – 10)
$$frac{2}{3}\\$b – $\frac{1}{2}\\$b = 10 – 5 Solving the equation, we get b = 30 kmph. So, speed of Automobile A = 20 kmph. Using statement$2) ALONE we could get a UNIQUE answer.
Statement (2) ALONE is ALSO sufficient.
Each statement is INDEPENDENTLY sufficient, we can eliminate choice 1.
Hence, choice (4) is the answer.
## Online TANCET MBA CourseTry it Free!
Register in 2 easy steps and start learning in 5 minutes!
The following is multiple choice question (with options) to answer.
A car runs 10,000 miles using 5 tyres interchangeably. to have equal worn out by all tyres,how many miles each tyre should have run? | [
"5000",
"8000",
"4000",
"3000"
] | B | for 5 tyres running interchangeably 10,000 miles the car runs
for 1 tyre=10000/5=2000
as you all know i suppose the car has 4 tyre
to have equal worn out by 4 tyres we need=2000*4=8000
ANSWER:B |
AQUA-RAT | AQUA-RAT-37985 | leaves at 9:10 from one station and its speed is 90 km/h, what time does it get to the next station? You may speak with a member of our customer support team by calling 1-800-876-1799. Every word problem has an unknown number. When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Select it and click on the button to choose it. Linear inequalities word problems. A car traveled 281 miles in 4 hours 41 minutes. Solution. Find the equilibrium price. Given : The total cost of 80 units of the product is $22000. Explain to students that you can find the rate (or speed) that someone is traveling if you know the distance and time that she traveled. Using a few simple formulas and a bit of logic can help students quickly calculate answers to seemingly intractable problems. Solution Let x be the first number. Solution Let y be the second number x / y = 5 / 1 x + y = 18 Using x / y = 5 / 1, we get x = 5y after doing cross multiplication Replacing x = 5y into x + y = 18, we get 5y + y = 18 6y = 18 y = 3 For a certain commodity, the demand equation giving demand "d" in kg, for a price "p" in dollars per kg. Difference between a number and its positive square root is 12. Related articles: The 4 Steps to Solving Word Problems. If the …, solve word problem Not rated yetIf a number is added to the numerator of 11/64 and twice the number is added to the denominator of 11/64, the resulting fraction is equivalent to 1/5. How many years …, Price of 3 pens without discount Not rated yetA stationery store sells a dozen ballpoints pens for$3.84, which represents a 20% discount from the price charged when a dozen pens are bought individually. A park charges $10 for adults and$5 for kids. Do you have some pictures or graphics to add? If the rod was 2 meter shorter and each meter costs $1 more, the cost would remain unchanged. Multi-Step All Operations Word Problems These Word Problems worksheets will produce word problems
The following is multiple choice question (with options) to answer.
In a fuel station the service costs $1.75 per car, every liter of fuel costs 0.45$. Assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty, how much money total will it cost to fuel all cars? | [
"318$",
"380$",
"420$",
"450$"
] | A | 12*1.75 + 0.45*12*55 = 318 hence - A |
AQUA-RAT | AQUA-RAT-37986 | Hence: 5x/100*(t-1/2) + x = 1100 = 4x/100*(t) + x
Solving, xt will get cancelled and you will get:
11000 = 11x
x is 1000
sum of both investments is 2x = 2000 which is Option D
Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1326
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
### Show Tags
02 Jul 2017, 07:29
1
1
Bunuel wrote:
Andrew borrows equal sums of money under simple interest at 5% and 4% rate of interest. He finds that if he repays the former sum on a certain date six months before the latter, he will have to pay the same amount of $1100 in each case. What is the total sum that he had borrowed? (A)$750
(B) $1000 (C)$1500
(D) $2000 (E)$4000
Assume that Andrew borrows $$X$$ dollars with simple interest of 5% anually in $$n$$ months and $$X$$ dollars with simple interest of 4% annually in $$n+6$$ months.
The following is multiple choice question (with options) to answer.
Rs.2500 is divided into two parts such that if one part be put out at 5% simple interest and the other at 6%, the yearly annual income may be Rs.120. How much was lent at 5%? | [
"2333",
"2777",
"3000",
"1000"
] | C | (x*5*1)/100 + [(2500 - x)*6*1]/100 = 120
X = 3000
Answer: C |
AQUA-RAT | AQUA-RAT-37987 | time, sun
d represents the day of the year for 2000. For example d=7
would be January 7th, 2000.
If you're willing to lose some precision in exchange for
convenience, you can compute the day of the current year (instead
of counting all the way back to 2000), since the sun's declination
repeats yearly (roughly speaking).
To calculate the day of the year, remember that December 31st
(noon) of the previous year is day 0. This means day 1 is January
1st, and day 31 is January 31st. Day 31 is also "February 0", so if
you need a date in February, just add. February 19th, for example,
would be 31+19 or 50. Since February has 29 days this year, February
29th would be day 31+29 or day 60, which is also "March 0". However,
at our level of precision, it doesn't matter whether you could the
leap day or not: the results will be approximately the same.
The formula above is accurate to about 0.1 degrees for this
century. Since the sun's declination can change by as much as 0.4
degrees in a day, this amount of precision should suffice.
Technically, the formula above computes the sun's declination at
Greenwich noon for a given day, which is the time when most of the
world is observing the same day. Again, the inaccuracies from using
Greenwich noon (instead of the actual, as yet unknown, time) are
small enough to ignore for our purposes.
The following is multiple choice question (with options) to answer.
January 1, 2007 was Monday. What day of the week lies on Jan. 1, 2008? | [
"Monday",
"Tuesday",
"Wednesday",
"Friday"
] | B | The year 2007 is an ordinary year. So, it has 1 odd day.
day of the year 2007 was Monday
day of the year 2008 will be 1 day beyond Monday
Hence, It will be Tuesday. answer : B |
AQUA-RAT | AQUA-RAT-37988 | # What's 4 times more likely than 80%?
There's an 80% probability of a certain outcome, we get some new information that means that outcome is 4 times more likely to occur.
What's the new probability as a percentage and how do you work it out?
As I remember it the question was posed like so:
Suppose there's a student, Tom W, if you were asked to estimate the probability that Tom is a student of computer science. Without any other information you would only have the base rate to go by (percentage of total students enrolled on computer science) suppose this base rate is 80%.
Then you are given a description of Tom W's personality, suppose from this description you estimate that Tom W is 4 times more likely to be enrolled on computer science.
What is the new probability that Tom W is enrolled on computer science.
The answer given in the book is 94.1% but I couldn't work out how to calculate it!
Another example in the book is with a base rate of 3%, 4 times more likely than this is stated as 11%.
The following is multiple choice question (with options) to answer.
What is 78% of 4/5? | [
"6.9",
"69.0",
"0.6845",
"0.624"
] | D | 78%*(4/5)=0.78*0.8=0.624
Answer :D |
AQUA-RAT | AQUA-RAT-37989 | 4. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
Psec/1bolt * 5bolts/1rev * 1rev/78.4in *...
The OP specified that the circumference of the wheel is 78.4in. Why do we need to include the radius in the conversion?
Last edited: Sep 8, 2015
5. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
Because I missed that little detail?
6. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
But the circumference of the wheel is irrelevant -- it's the circumference of the tire that matters. Perhaps the TS means the tire and not the wheel, but I'm not positive about that.
7. ### djsfantasi AAC Fanatic!
Apr 11, 2010
2,805
833
A circumference of 78.4" is a diameter of 10". Much more likely to be the diameter of a wheel.
Update: Helps to use the right equation...
Last edited: Sep 8, 2015
8. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
Or it's one of the new micro-cars that are becoming the snob-appeal rage.
9. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
As sensors go I like it. It would also be good for odometers too.
10. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
I think driveshaft sensors are probably better since you typically get several rotations of the shaft per rotation of the tire (and you can mount multiple magnets on the shaft to further improve resolution). It also tends to average out the distance traveled by the tires as you drive due to the natural behavior of the differential.
Jan 26, 2014
5
0
12. ### rjwheaton Thread Starter New Member
Jan 26, 2014
5
0
Thank you!
Not only does that answer my question, but it helps me understand how you did the calculations.
The following is multiple choice question (with options) to answer.
The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 300 resolutions. | [
"421 m",
"422 m",
"423 m",
"424 m"
] | B | In one resolution, the distance covered by the wheel is its own circumference. Distance covered in 300 resolutions.
= 300 * 2 * 22/7 * 22.4
= 42200 cm
= 422 m
Answer: B |
AQUA-RAT | AQUA-RAT-37990 | $\displaystyle \Pi_2 := \prod_{p>2} (1 - \frac{1}{(p-1)^2}) = 0.6601618\dots$
(see equation (7) from Supplement 4); note that
$\displaystyle \prod_p (1-\frac{1}{p})^{-2} (1-\frac{\omega(p)}{p}) = 2 \Pi_2$
so from Mertens’ third theorem (Theorem 42 from Notes 1) one has
$\displaystyle \prod_{p < z} (1-\frac{\omega(p)}{p}) = (2\Pi_2+o(1)) \frac{1}{(e^\gamma \log z)^2} \ \ \ \ \ (8)$
as ${z \rightarrow \infty}$. Bounding ${4^{\pi(z)}}$ crudely by ${\exp(o(z))}$, we conclude in particular that
$\displaystyle \pi_2(x,z) = (2\Pi_2 +o(1)) \frac{x}{(e^\gamma \log z)^2}$
when ${x,z \rightarrow \infty}$ with ${z = O(\log x)}$. This is somewhat encouraging for the purposes of getting a sufficiently good answer to Problem 2 to resolve the twin prime conjecture, but note that ${z}$ is currently far too small: one needs to get ${z}$ as large as ${\sqrt{x}}$ before one is counting twin primes, and currently ${z}$ can only get as large as ${\log x}$.
The following is multiple choice question (with options) to answer.
A “Sophie Germain” prime is any positive prime number p for which 2p + 1 is also prime. The product of all the possible units digits of Sophie Germain primes greater than 4 is | [
"3",
"7",
"189",
"227"
] | C | In that case, the Sophie prime numbers greater than 5 are 7,11,23,47,59, .. which yields units digit as 1,3,7 and 9
Product would be 1 x 3 x 7x9 =189 Answer should be C. |
AQUA-RAT | AQUA-RAT-37991 | I shall determine a set of values of $$c$$ such that $$f_c'(x) > 0$$ for all $$x \in [0, 1].$$ (I shall not try to determine all such values of $$c.$$) It was shown above that for all $$c > 0,$$ if either of $$f_c',$$ $$f_{1/c}'$$ is strictly positive on $$[0, 1],$$ then so is the other. Because $$f_1(x) = x,$$ it suffices to consider only the case $$c > 1,$$ i.e., $$a < \tfrac12.$$
The following is multiple choice question (with options) to answer.
If 0 < e< 1 < f, which of the following must be true? | [
"1 < 1/e< 1/f",
"1/e< 1 < 1/f",
"1/e< 1/f< 1",
"1/f< 1 < 1/e"
] | D | 0<e<1<f. Let e=1/2 and f=2
Substituting values:
A. 1<2<1/2 ---Wrong
B. 2<1< 1/2 ---Wrong
C. 2<1/2<1 ---Wrong
D. 1/2<1<2 ---Correct
E. 1/2<2<1 ---Wrong
Hence answer is D |
AQUA-RAT | AQUA-RAT-37992 | ## 28.
The Correct Answer is (D) — Graphs for answer choices A and B imply that the area of this rectangle can be 800$\small&space;ft^{2}$ or greater. These values are not possible with only 40ft of fencing material. Similarly, the graph for answer choice C says that the area can be as large as 400$\small&space;ft^{2}$, but this value is also impossible with only 40ft of fencing material. The correct answer is D.
## 29.
The Correct Answer is (B) — When we multiply the two terms together we get 10 + 30i - 2i - 6$\small&space;i^{2}$. Because $\small&space;i^{2}$=-1, the last term is equal to 6, giving us 10 + 28i + 6 = 16 + 28i.
## 31.
The Correct Answer is (150) — We can divide 75 miles by 30 miles per hour to find the number of hours it will take to make this trip, which gives us 75/30 = 2.5 hours. When we convert this value to minutes, we get 150 minutes.
## 32.
The Correct Answer is (80) — If 2/5 of n is 48, then (2/5)n = 48. Solving for n gives us n = 120. We can multiply this n by 2/3 to find 2/3 of n, which gives us (2/3)(120) = 80.
## 33.
The Correct Answer is (32) — Plugging in -3 into f(x) gives us $\small&space;f(x)=(-3)^3+3(-3)^2-6(-3)+14=-27+27+18+14=32.$
## 34.
The following is multiple choice question (with options) to answer.
A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? | [
"95",
"92",
"82",
"88"
] | D | AREA OF FIELD = 680 SQ FEET
LENGTH OF ADJ SIDES 20 FEET AND 680/20 = 34 FEET
REQ LENGTH OF THE FENCING = 20 + 34 + 34 = 88 FEET
ANSWER D |
AQUA-RAT | AQUA-RAT-37993 | ## anonymous 5 years ago 100 people are present in a party. If each of them must do a handshake with all the other people, how many handshakes should be done? Please explain how to get the answer, thanks!
1. anonymous
you need 2 people for an hand shake and u have 100 people so i think that its C(100,2) but i am not sue of it..
2. anonymous
$\sum_{n=1}^{99}n=\frac{1}{2}\times99\times100=4950$
3. anonymous
Could you solve it using factorials, please?
4. anonymous
hmm i found the answer as 9900
5. amistre64
if there are 3 people: 1,2 ; 1,3 ; 2,3 .... is that right?
6. anonymous
Think about it like this: the first person has 99 people to shake hands with, the second person has 98 because they've already shaken hands with the first, the third has 97...
7. anonymous
yea but its harder to calculate this like that... so i think C(100,2) better way of solve
8. amistre64
5054 if we do that ..... add all the numbers from 1 to 100 :)
9. anonymous
No, all the numbers from 1 to 99.
10. amistre64
n(n+1) ------ :) 2
11. amistre64
ack .... 99 then lol
12. amistre64
wouldnt the last guy have noone to shake hands with?
13. anonymous
mhmh yea but there is 100 people and they and we need 2 people to shake hands so that it should be 100 x 99 /2,
14. anonymous
Yeah, the last guy has no-one to shake hands with, that's why it's sum of the first 99 integers, not the first 100.
15. anonymous
yea you right
16. amistre64
The following is multiple choice question (with options) to answer.
7 people meet for a business lunch. Each person shakes hands once with each other person present. How many handshakes take place? | [
"30",
"21",
"18",
"15"
] | B | the formula to count handshakes is n(n−1)2n(n−1)2
Where n is the number of people
=> 7(7-1)/2 = 7*6/2 = 42/2 = 21
=> the answer is B(21) |
AQUA-RAT | AQUA-RAT-37994 | ### 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.
A, B and C enter into a partnership by investing $15000, $21000 and $27000 respectively. At the end of 8 months, B receives $1540 as his share. Find the share of A. | [
"$800",
"$900",
"$1000",
"$1100"
] | D | The ratio of capital of A, B and C = 15000 : 21000 : 27000 = 5 : 7 : 9
A's share = (5/7)*1540 = $1100
The answer is D. |
AQUA-RAT | AQUA-RAT-37995 | # 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 party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples? | [
"5/21",
"5/23",
"5/29",
"5/12"
] | A | Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)
As remaining person can be any one among three couples left.
Required probability
= (5C2 * 6C1)/10C5
= (10 * 6)/252 = 5/21
Answer: A |
AQUA-RAT | AQUA-RAT-37996 | numbers we need to first convert the mixed numbers into improper fractions, multiply the improper fractions and reduce the resultant value to the lowest terms to get the answer. You just need to enter your two fractions, then select the operation you want to perform. Study math with us and make sure that "Mathematics is easy!" The Fraction Calculator will reduce a fraction to its simplest form. Formulas. Convert the fraction to a mixed number by using long division to find the quotient and remainder 5 ÷ 3 = 1 R2 The quotient will be the whole number in the fraction, and the remainder will be the numerator in the mixed fraction 5 3 = 1 2 3 Enter the value of two mixed numbers and click calculate, it displays the step by step solution with the answer. The step-by-step calculation help parents to assist their kids studying 4th, 5th or 6th grade to verify the work and answers of irregular fractions addition, subtraction, multiplication and division homework and … You can enter up to 3 digits in length for each the numerators and denominators (e.g., 456/789). Do math calculations with mixed numbers (mixed fractions) performing operations on fractions, whole numbers, integers, mixed numbers, mixed fractions and improper fractions. Sign in Log in Log out If they exist, the solutions and answers are provided in simplified, mixed and whole formats. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. You can convert between mixed numbers and improper fractions without changing the value of the figure. Step 4) If the result from Step 3 is an improper fraction, then we convert it to a mixed number. 5+0.8 = 5.8 . DOSSIER.NET sommario. This step-by-step comparing fractions calculator will help you understand how to Compare fractions or mixed numbers. The fraction calculator will generate a step-by-step explanation on how to obtain the results in the REDUCED FORM! 1. Any mixed number may be written by improper fraction. Use this calculator to convert your improper fraction to a mixed fraction. Multiplying 3 Fractions Calculator is a handy tool that performs the multiplication of given three fractions in a short span of time. © 2006 -2020CalculatorSoup® Fraction simplifier Adding fractions
The following is multiple choice question (with options) to answer.
Create 2 equivalent fractions with the numbers 4, 9, 2, and 18 | [
"2/9 and 5/18",
"2/9 and 4/18",
"3/9 and 4/18",
"2/9 and 6/18"
] | B | Solution:
We're searching for pairs of numbers which, when multiplied, have the same product. We observe that 4 x 9 = 18 x 2 = 36, so the fractions will be 2/9and4/18
Answer B |
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