source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-37297 | 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.
The sum of the present ages of A, B, C is 39 years. Three years ago their ages were in the ratio 1:2:3. What is the present age of A? | [
"5",
"6",
"7",
"8"
] | D | Three years ago:
a:b:c = 1:2:3
Let a = 1x, b = 2x, and c = 3x.
Today:
(x+3) + (2x+3) + (3x+3) = 39
x = 5
So the present age of A is x+3 = 8
The answer is D. |
AQUA-RAT | AQUA-RAT-37298 | ### Show Tags
23 Jan 2019, 14:19
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
In a set of consecutive integers, the sum of the first two is 37
(1) First integer, $$x$$?
$$x + (x+1)=37$$
$$2x+1=37$$
$$x=18$$
(2) Last integer?
If $$x$$ is the first integer, and there are 20 integers total, then there are 19 more
Add 19 to $$x=18$$ (Consecutive = +1 for each integer in the set.)
$$18+19=37$$
If not sure, use a simpler example: There are 4 consecutive integers in the set.
$$x=18$$. List them. 18, 19, 20, 21
There are 4 integers, but we need only +3 added to $$x=18$$ to find the last integer.
20 integers in this set, but add only 19 to the first integer. $$18+19=37$$
(3) Sum of last three integers?
Last three integers: 37, 36, 35
Units digit = 8
(Or, 37 + 36 + 35 = 108)
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Posts: 1224
Re: In a series of twenty consecutive integers, the sum of the first two [#permalink]
### Show Tags
23 Jan 2019, 15:36
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
The following is multiple choice question (with options) to answer.
Three consecutive odd integers are in increasing order such that the sum of the last two integers is 17 more than the first integer. Find the three integers? | [
"11,13,15",
"7, 2, 10",
"7, 9, 10",
"7, 9, 11"
] | A | Explanation:
Let the three consecutive odd integers be x, x + 2 and x + 4 respectively.
x + 4 + x + 2 = x + 17 => x = 11
Hence three consecutive odd integers are 11, 13 and 15
Answer: A |
AQUA-RAT | AQUA-RAT-37299 | Difficult Probability Solved QuestionAptitude Discussion
Q. If the integers $m$ and $n$ are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 equals:
✔ A. $\dfrac{1}{4}$ ✖ B. $\dfrac{1}{7}$ ✖ C. $\dfrac{1}{8}$ ✖ D. $\dfrac{1}{49}$
Solution:
Option(A) is correct
Table below can be scrolled horizontally
Form of the exponent
$m$ $4x +1$ $4x+3$ $4x+2$ $4x$
$n$ $4y+3$ $4y+1$ $4y$ $4y+2$
last digit of
$7^m+7^n$
$0$ $0$ $0$ $0$
Number of
selections
$25 \times 25$ $25 \times 25$ $25 \times 25$ $25 \times 25$
If a number ends in a 0 then the number must be divisible by 5.
Hence required probability is,
$=\dfrac{625 \times 4}{100^2}$
$=\dfrac{1}{4}$
Edit: Thank you, Barry, for the very good explanation in the comments.
Edit 2: Thank you Vaibhav, corrected the typo now it's $25 \times 25$ and not $26 \times 25$.
Edit 3: For yet another approach of solving this question, check comment by Murugan.
(7) Comment(s)
Murugan
()
This sum is very simple. Power cycles of 7 are 7, 9, 3, 1
So totally 4 possibilities.
Total possibilities are $^4P_1 \times ^4P_1=16$
For selecting a number from 1 to 100 which are divisible by 5,
$m=9$, $n=1$ or $m=1$, $n=9$ or $m=7$, $n=3$ or $m=3$, $n=7$
i.e. 4 chances.
The following is multiple choice question (with options) to answer.
If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being divisible by 8? | [
"a. 25%",
"b. 50%",
"c. 62.5%",
"d. 72.5%"
] | C | There is another way I approach such questions if I am short of time (otherwise I prefer the logical approach given above) - Brute Force/Pattern Recognition/Intuition - whatever you may want to call it.
We need to find the numbers in which the product is a multiple of 8. I know I get a multiple of 8 after every 8 numbers. I will also get an 8 when I multiply 4 by an even number. In first 8 numbers, I have exactly two multiples of 4.
Basically, I figure that I should look at the first 8 cases. In all other cases, the pattern will be repeated. It helps that n can be from 1 to 96 i.e. a multiple of 8:
1*2*3 N
2*3*4 Y
3*4*5 N
4*5*6 Y
5*6*7 N
6*7*8 Y
7*8*9 Y
8*9*10 Y
5 of the first 8 products are divisible by 8 so my answer would be 5/8 = 62.5% C |
AQUA-RAT | AQUA-RAT-37300 | Capture question 1.PNG [ 2.18 KiB | Viewed 1988 times ]
Step 2.1. Find all values of $$x$$ those make each factor equals to 0.
$$x-2=0 \iff x=2$$
$$x-4=0 \iff x=4$$
In the first row, list all these values in ascending order: $$-\infty , 2, 4, +\infty$$.
Step 2.2. In next 2 rows, review the signs of each factor.
This step based simply on this rule: If $$x<a \implies x-a<0$$. If $$x>a \implies x-a>0$$.
For example, $$x-2=0 \implies x=2$$. Hence, any value of $$x<2$$, then $$x-2$$ will be negative (-); any value of $$x>2$$, then $$x-2$$ will be positive (+).
We dont need to care about what is the value of $$x-2$$ if $$x=4$$ or what is the value of $$x-4$$ if $$x=2$$. These values are expressed as sign "|".
Step 2.3. In final row, review the sign of the expression.
The sign of the expression simply based on this rule:
$$(+) \times (+) = (+)$$
$$(-) \times (-) = (+)$$
$$(+) \times (-) = (-)$$
$$(-) \times (+) = (-)$$
Step 3. Come to solution.
We need to find sign (-) of the expression. Based on the table, we simply have $$2<x<4$$.
That's it.
Now let's solve Question 2.
$$x=0$$
$$x+1=0 \implies x=-1$$
$$x-5=0 \implies x=5$$
The factor table with sign:
Attachment:
Capture question 2.PNG [ 2.97 KiB | Viewed 1982 times ]
The following is multiple choice question (with options) to answer.
If 2 : 9 :: x : 18, then find the value of x | [
"2",
"3",
"4",
"6"
] | C | Explanation:
Treat 2:9 as 2/9 and x:18 as x/18, treat :: as =
So we get 2/9 = x/18
=> 9x = 36
=> x = 4
ANSWER IS C |
AQUA-RAT | AQUA-RAT-37301 | that equation, depending on what we need it for. succeed. Write the standard form of the equation of the circle with center that also contains the point . 4. This formula reads, “Area equals pi are squared.” Find the radius, circumference, and area of a circle if its diameter is equal to 10 feet in length. 1. The radius of the circle that passes through the three non-collinear points P 1, P 2, and P 3 is given by = | → − → | , where θ is the angle ∠P 1 P 2 P 3.This formula uses the law of sines.If the three points are given by their coordinates (x 1,y 1), (x 2,y 2), and (x 3,y 3), the radius can be expressed as The area of the clock is 50 in2. Drag either orange dot at the ends of the diameter line. lessons in math, English, science, history, and more. {{courseNav.course.topics.length}} chapters | Also find the Area of a circle. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. What is the standard form equaton of a circle? Circles. If the area of a circular orange slice is 12 in2 , what is its radius? A circle can have many radii (the plural form of radius) and they measure the same. Radius is 1 2 the diameter. This website uses cookies to ensure you get the best experience. Example: Find the diameter of the circle whose radius is 8 cm ? Calculate the area of a segment of a circle with a central angle of 1 6 5 degrees and a radius of 4. All formulas for radius of a circumscribed circle. Dividing both sides of this equation by π gives. Uncheck the "fixed size" box. There are basically two forms of representation: MEMORY METER. Remember that $$\pi$$ is about 3.14. Add to Library ; Share with Classes; … Similarly, if you enter the area, the radius needed to get that area will be calculated, along with the diameter and circumference. What is the radius of the wheel? How Do I Use Study.com's Assign Lesson Feature? All radii in
The following is multiple choice question (with options) to answer.
If the area of a circle is 81pi square feet, find its circumference. | [
"18 pi feet",
"28 pi feet",
"38 pi feet",
"48 pi feet"
] | A | The area is given by pi *r* r. Hence
pi * r *r = 81 pi
r * r = 81 ; hence r = 81 feet
The circumference is given by
2 * pi * r = 2 * pi * 9 = 18 pi feet
correct answer A |
AQUA-RAT | AQUA-RAT-37302 | 1. ## More Probability
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
Hope I did these right! Part d is especially tricky so I wouldn't be surprised if I tripped over on that one.
2. Originally Posted by hansel13
I never took Stats before, so I'm just trying to get a grasp of statistics and make sure I understand all of it.
A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs are randomly selected.
(c) What is the probability that one bulb of each type is selected?
$
P = {{\binom{4}{1}} {\binom{5}{1}} {\binom{6}{1}}}/{\binom{15}{3}} = .264
$
(d) Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the probability that it is necessary to examine at least six bulbs?
$
P = {{\binom{6}{0}}{\binom{9}{6}}}/{\binom{15}{6}} = .017
$
The following is multiple choice question (with options) to answer.
One out of every 500 light bulbs are defected. If 2 out of every 10 defected light bulbs have a broken glass and there are 16 broken glass light bulbs in the shipment, how many light bulbs total are there in the shipment? | [
"2,000",
"5,000",
"10,000",
"40,000"
] | D | out of 500 only 1 bulb is defective.So for 10 defective bulbs, we must have 5000 bulbs
Now out of these 10 bulbs, 2 bulbs, have broken glass
i.e 2 broken glass out of 5000 bulbs
16 broken glass will be from 40000 bulbs
Answer is D |
AQUA-RAT | AQUA-RAT-37303 | (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.
Kanul spent $80000 in buying raw materials, $30000 in buying machinery and 20% of the total amount he had as cash with him. What was the total amount? | [
"$135656",
"$137500",
"$134446",
"$123265"
] | B | Let the total amount be x
then, (100-20)% of x = 80000+30000
80% of x = 110000
80x/100 = 110000
x = $1100000/8
x = $137500
Answer is B |
AQUA-RAT | AQUA-RAT-37304 | Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Originally posted by bbear on 17 Jun 2016, 14:31.
Last edited by bbear on 17 Jun 2016, 16:04, edited 1 time in total.
##### General Discussion
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Joined: 07 Dec 2014
Posts: 1157
Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 15:50
1
6t=4*27
t=18 hours
27-18=9 fewer hours
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
17 Jun 2016, 23:37
2
1
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
Total work = 27 * 4
Time taken when 6 machines work = $$\frac{(27*4)}{6}$$ => 18 hours
So, working together 6 machines take 9 hours less ( 27 - 18 )
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Re: Six machines at a certain factory operate at the same constant rate. [#permalink]
### Show Tags
18 Jun 2016, 01:43
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?
The following is multiple choice question (with options) to answer.
A TaTa company has two types of machines, type R and type S. Operating at a constant rate, a machine of type R does a certain job in 14 hrs and a machine of type S does the same job in 7 hours. If the company used the same number of each type of machine to do the job in 2 hours, how many machines of type R were used? | [
"3",
"4",
"6",
"9"
] | A | Rate of machine R =1/14
Rate of machine S =1/7
since same no of machines used for R and S to do the same work in 2 hrs
So collective rate needed to finish the work in 2 hrs= 1/2
Let the no of machine be x
So, x/14 +x/7 =1/2
3x/14=1/2
x=14/6=2.3=3
So no of machine R is 3
Answer A |
AQUA-RAT | AQUA-RAT-37305 | # 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.
In one hour, a boat goes 6 km/hr along the stream and 2 km/hr against the stream. The speed of the boat in still water (in km/hr) is: | [
"3 km/hr",
"4 km/hr",
"5 km/hr",
"6 km/hr"
] | B | Upstream relative speed is u + v=6km/hr
Downstream speed is u-v = 2
Where u = speed of boat in still water and v is speed of stream
Then adding two equations u+v + u-v =6+2
2u=8
Finally, u=4.
ANSWER : B |
AQUA-RAT | AQUA-RAT-37306 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Albert borrowed a total of $ 6000 from john and Charlie. He paid simple interest at the rate of 7 % Per yr and 9 % per yr respectively.At the end of three years he paid $ 1494 as total interest. What is the interest paid to Brian at the end of three years? | [
"571",
"581",
"441",
"445"
] | C | Let x be the amount borrowed form john. So amount borrowed form Charlie = 6000-x.
1494 = 21/100x + 27/100(6000-x)
=>x = 2100.
Interest paid = 3*7/100*2100 = 441.
C |
AQUA-RAT | AQUA-RAT-37307 | homework-and-exercises, pressure, fluid-statics
Title: Which tank fills up first? Which tank would fill first. My first guess was 3 and 4 simultaneously due to Pascal's Law of pressure distribution. Then tank 2 and then 1. Could you please help? This is my first question ever on Stack Exchange. Tank 1 will to the level of the pipe. Then water will flow into 2.
If the pipe is blocked, 2 will fill. When the water in 2 reaches the level of the upper pipe, tanks 1 and 2 will stay even with each other. When tank 2 reaches the top, water will spill out. It ends there.
If the pipe to 2 is open, tank 2 will fill to the level of the lower pipe. Then water will flow into 3. Water in tank 3 will stay even with the level in the pipe to 4.
It looks like the level of the upper part of both pipes from 3 are the same. When the level in 3 rises to the pipes, water will begin to spill into 4. When 4 is full up to the pipe, the level will rise in 2, 3, and 4 until it spills over the top of 3 and 4.
The following is multiple choice question (with options) to answer.
A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 5 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely? | [
"4 min. to empty",
"6 min. to full",
"9 min. to empty",
"9 min. to full"
] | A | Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 2/5
Part emptied by (A + B) in 1 minute = (1/5 - 1/10) = 1/10
1/10 : 2/5 :: 1 : x
x = (2/5 * 1 * 10) = 4 min.
So, the tank will be emptied in 4 min.
ANSWER:A |
AQUA-RAT | AQUA-RAT-37308 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can do a piece of work in 5 days and B can do it in 7 days how long will they both work together to complete the work? | [
"6/11",
"8/11",
"7/12",
"9/12"
] | D | Explanation:
A’s one day work = 1/5
B’s one day work = 1/7
(A+B)’s one day work = 1/5 + 1/7 = 12/35 => time = 35/12 = 2 9/12 days
Answer: Option D |
AQUA-RAT | AQUA-RAT-37309 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
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## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
Fourteen people are planning to share equally the cost of a rental van. If one person withdraws from the arrangement and the others share equally the cost of the rental van, then the share of each of the remaining people will increase by...? | [
"1/15",
"1/14",
"1/13",
"13/14"
] | C | Let P = total cost of the rental van.
The original share per person is P/14.
The new share per person is P/13.
P/13 = P/14 * 14/13 = (1 + 1/13)*original share
The answer is C. |
AQUA-RAT | AQUA-RAT-37310 | 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.
A $72.95 lawn chair was sold for $59.95 at a special sale. By approximately what percent was the price decreased? | [
"18%",
"20%",
"25%",
"60%"
] | A | Listed selling price of chair = 72.95 $
Discounted selling price of chair = 59.95 $
Discount = 72.95 - 59.95 = 13 $
% decrease in price of chair = (13/72.95) * 100% = 18 % approx
Answer A |
AQUA-RAT | AQUA-RAT-37311 | # Thread: Help with probability question/
1. ## Help with probability question/
I'm struggling with probability questions, even those that should be easy! We didn't really go over techniques of solving these types of questions, and the textbook doesn't really address these types of problems (it's more of a stats book than a probability book).
The question is:
Suppose that the last 3 men out of a restaurant all lose their hatchecks, so that the hostess hands back their 3 hats in random order. What is the probabability...
a) That no man will get the right hat?
b) That exactly 1 man will?
c) That exactly 2 men will?
d) That all 3 will?
My reasoning is that that there are six combinations of returning the hats. Let's say the men are A, B, and C. There are six combinations:
1) ABC
2) ACB
3) BAC
4) BCA
5) CAB
6) CBA
My reasoning for part a) So I assume that, let's say ABC is the correct order. The probability that no man will get the right hat is any order in which there are no A's in position one, no B's in position 2, and no C's in position 3. So these are 3, 4, 5, 6. This is 4 out of the 6, so is the probability 2/3? This answer just doesn't seem right to me. How do I solve this? What is the reasoning behind this?
Reasoning for part b) Again, I assume that ABC is the right order. 2, 3, 6 are the positions in which A, B, or C are the only ones in the right position. So I think it is 1/2, but is this right? Is there a correct way of thinking about this and getting the right answer?
reasoning for part c) Again, I assume ABC is the right order. But there is no position in which only two letters are in that place, since there are three letters?! So I'm assuming my answers above are wrong too.
d) I reason that there is only one combination out of 6 in which all 3 men their hats, so 1/6?
Please help! Thanks!
The following is multiple choice question (with options) to answer.
6% of a towns eligible voters did not vote in a recent election. If 30 of the towns eligible voters did not vote, how many did vote? | [
"470",
"500",
"440",
"400"
] | A | 30 is 6% of the total number of voters.
If the total number of voters in the town is X, then 6/100 = 30/X
Therefore 100/6 * 30 = X
3000/6 = X
There are 500 total voters in the town.
500 - 30 = 470
A) |
AQUA-RAT | AQUA-RAT-37312 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Tom, working alone, can paint a room in 6 hours. Peter and John, working independently, can paint the same room in 3 hours and 3 hours, respectively. Tom starts painting the room and works on his own for one hour. He is then joined by Peter and they work together for an hour. Finally, John joins them and the three of them work together to finish the room, each one working at his respective rate. What fraction of the whole job was done by Peter? | [
"1/3",
"2/5",
"3/10",
"7/15"
] | D | Tom paints 1/6 of the room in the first hour.
Tom and Peter paint 1/6+1/3 = 1/2 of the room in the next hour for a total of 4/6.
The three people then paint the remaining 2/6 in a time of (2/6) / (5/6) = 2/5 hours
Peter worked for 7/5 hours so he painted 7/5 * 1/3 = 7/15 of the room.
The answer is D. |
AQUA-RAT | AQUA-RAT-37313 | ### 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, C and D enter into partnership. A subscribes 1/3 of the capital B 1/4, C 1/5 and D the rest. How much share did A get in a profit of Rs.2460? | [
"14029",
"14027",
"11867",
"14000"
] | D | 25*12: 30*12: 35*8
15:18:14
14/47 * 47000 = 14000
Answer: D |
AQUA-RAT | AQUA-RAT-37314 | 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 1515. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder. What is the smaller number ? | [
"A)270",
"B)280",
"C)290",
"D)300"
] | D | Let the smaller number be x. Then larger number = (x + 1515).
x + 1515 = 6x + 15
5x = 1500
x = 300
Smaller number = 300.
D) |
AQUA-RAT | AQUA-RAT-37315 | -
You say the answer is 1/2 and seem to argue that the answer is 2/3. – Aryabhata Aug 18 '10 at 20:10
He says the answer is 2/3 for the cards example, which is correct, not for the boys/girls paradox, which should be 1/2 and he so states in the opening paragraph. – Abel Aug 18 '10 at 20:24
Ok, I clarified. – Nate Eldredge Aug 18 '10 at 20:32
Let A be the event "Family has at least one girl". Let B be the event "Boy opens door"
Pr(A and B)=1/4 *1/2 + 1/4*1/2 = 1/4
AND
Pr(B)=1*1/4 + 1/4*1/2 +1/4*1/2 = 1/2
So Pr(A|B) = 1/4 / ( 1/2) = 1/2.
-
We will assume all the obvious implicit assumptions (eg. random child being boy of girl is 50/50, boys and girls open the door uniformly, etc.).
If you had a slightly different question, i.e. if you asked the couple if they have at least one boy, and the answer is yes, then the chance of the other one being a girl is 2/3. Intuitively, the probability is not 1/2 because in this case the answer depends on both the children, i.e. it is a function of both of them considered together.
However, if you asked the couple to pick a child at random, then she/he bears no information about the other child, and consequently his/her gender does not give you any information about the sibling.
Your case is the second case, where the child opening the door is selected at random, and she happens to be female. This does not bear any information regarding the other child.
Mathematically,
The following is multiple choice question (with options) to answer.
Two brother X and Y appeared for an exam. The probability of selection of X is 1/3 and that of B is 2/7. Find the probability that both of them are selected. | [
"2/69",
"2/21",
"2/23",
"2/29"
] | B | Explanation:
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/3, P(B) = 2/7.
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/3) × (2/7) = 2/21
Answer: B) 2/21 |
AQUA-RAT | AQUA-RAT-37316 | They settle and use the first price that comes to mind, copy competitors, or (even worse) guess. Prove that the product of three consecutive positive integer is divisible by 6. gl/9WZjCW prove that the product of three consecutive positive integers is divisible by 6. Therefore, n = 3p or 3p + 1 or 3p + 2 , where p is some integer. 1 3 + 2 3 + 3 3 +. (the alphanumeric value of MANIC SAGES) + (the sum of all three-digit numbers you can get by permuting digits 1, 2, and 3) + (the number of two-digit integers divisible by 9) - (the number of rectangles whose sides are composed of edges of squares of a chess board) 91 + 1332 (12*111) + 10 - 1296 = 137. By the laws of divisibility, anything divisible by 2 and 3 is divisible by 6. Whenever a number is divided by 3 , the remainder obtained is either 0,1 or 2. Find the smallest number that, when. Essentially, it says that we can divide by a number that is relatively prime to. Let the three consecutive positive integers be n , n + 1 and n + 2. Let n be a positive integer. 1 Consecutive integers with 2p divisors. If A and B are set of multiples of 2 and 3 respectively, then show that A = B and A∪B. Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. The array contains integers in the range [1. Prove that one of any three consecutive positive integers must be divisible by 3. ← Prev Question Next Question →. Btw jayshay - if you said 7n, 7n+1 and 7n+2 then your 'proof' would effectively be proving that the product of 3 consecutive integers is a multiple of 7. We have to prove this for any arbitrary k ∈Z, so fix such a k. (Examples: Prove the sum of 3 consecutive odd integers is divisible by 3. Prove: The product of any three consecutive integers is divisible by 6; the product of any four consecutive integers is divisible by 24; the product of any five consecutive integers is divisible by 120. If A =40, B =60 and
The following is multiple choice question (with options) to answer.
The product of three consecutive numbers is 720. Then the sum of the smallest two numbers is ? | [
"11",
"15",
"17",
"38"
] | C | Product of three numbers = 720
720 = 8*9*10
So, the three numbers are 8, 9 and 10.
And Sum of smallest of these two = 8 + 9 = 17.
ANSWER : OPTION C |
AQUA-RAT | AQUA-RAT-37317 | # How many numbers made out of 5 distinct digits contain 4 and are even
This is an exercise from a book on combinatorics and I can't seem to wrap my head around it: How many numbers are there, made out of 5 distinct digits, which contain '4' and are even. The answer according to the book is 7686.
I distinguished some cases (not sure if I distinguished in a 'smart' way)
• the last digit is 4. In that case, there are 8 possibilities for the first digit (not 0, not 4). The second digit has 8 possibilities (not the first, not 4), the third has 7, the fourth has 6. The total amount is $$8\cdot 8\cdot 7\cdot 6 = 2688$$.
• the last digit isn't 4. Therefore, the last digit must be 0,2,6 or 8. We must pick 3 digits from the remaining 8 digits. This can be done in $$8 \cdot 7 = 56$$ ways. These 3 digits along with '4' have to be distributed over the first 4 places, giving a total of $$56 \cdot 4 \cdot 4! = 5376$$ ways. However, these include some invalid numbers: those starting with $$0$$. These have to be substracted. The first digit is fixed, the last digit can be $$2,6,8$$, so 3 possibilities. We have to pick 2 digits from the remaining 7 digits. This can be done in $$7 \cdot 6/2 = 21$$ ways. We then need to distribute these two digits and '4' over 3 spaces, so $$3!$$ possibilities, giving a total of $$3 \cdot 21 \cdot 3! = 378$$ invalid numbers.
The total therefore equals $$2688 + 5376 - 378 = 7686$$ ways.
This seems like a brute force solution. Anyone who can 'smoothen' this, for example by making a smarter choice of distinguished cases?
I would break the problem into three main cases instead of two. There are $$5$$ positions for the $$4$$: the first position, the middle three, or the last position.
The following is multiple choice question (with options) to answer.
find the number of 5 digits numbers can be formed using the digits 1,2,3,4,5 if no digit is reapeted and two adjoint digits are not together | [
"11",
"12",
"13",
"14"
] | B | 13524
14253
24153
24135
25314
31425
31524
+ 6 more numbers replicas of above numbers (total 12)
ANSWER:B |
AQUA-RAT | AQUA-RAT-37318 | 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.
Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days, six men left and six women joined, then in hoe many more days will the work be completed? | [
"2 (1/8) days",
"2 (1/6) days",
"3 (1/2) days",
"2 (1/2) days"
] | D | Work done by a women in one day = 1/2 (work done by a man/day)
One women's capacity = 1/2(one man's capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women - 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.
Answer: D |
AQUA-RAT | AQUA-RAT-37319 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
At the faculty of Aerospace Engineering, 312 students study Random-processing methods, 232 students study Scramjet rocket engines and 114 students study them both. If every student in the faculty has to study one of the two subjects, how many students are there in the faculty of Aerospace Engineering? | [
"438",
"436",
"430",
"442"
] | C | 312 + 232 - 114 (since 112 is counted twice )= 430
C is the answer |
AQUA-RAT | AQUA-RAT-37320 | Capture question 1.PNG [ 2.18 KiB | Viewed 1988 times ]
Step 2.1. Find all values of $$x$$ those make each factor equals to 0.
$$x-2=0 \iff x=2$$
$$x-4=0 \iff x=4$$
In the first row, list all these values in ascending order: $$-\infty , 2, 4, +\infty$$.
Step 2.2. In next 2 rows, review the signs of each factor.
This step based simply on this rule: If $$x<a \implies x-a<0$$. If $$x>a \implies x-a>0$$.
For example, $$x-2=0 \implies x=2$$. Hence, any value of $$x<2$$, then $$x-2$$ will be negative (-); any value of $$x>2$$, then $$x-2$$ will be positive (+).
We dont need to care about what is the value of $$x-2$$ if $$x=4$$ or what is the value of $$x-4$$ if $$x=2$$. These values are expressed as sign "|".
Step 2.3. In final row, review the sign of the expression.
The sign of the expression simply based on this rule:
$$(+) \times (+) = (+)$$
$$(-) \times (-) = (+)$$
$$(+) \times (-) = (-)$$
$$(-) \times (+) = (-)$$
Step 3. Come to solution.
We need to find sign (-) of the expression. Based on the table, we simply have $$2<x<4$$.
That's it.
Now let's solve Question 2.
$$x=0$$
$$x+1=0 \implies x=-1$$
$$x-5=0 \implies x=5$$
The factor table with sign:
Attachment:
Capture question 2.PNG [ 2.97 KiB | Viewed 1982 times ]
The following is multiple choice question (with options) to answer.
If 2 : 9 :: x : 18, then find the value of x | [
"2",
"3",
"4",
"5"
] | C | Explanation:
Treat 2:9 as 2/9 and x:18 as x/18, treat :: as =
So we get 2/9 = x/18
=> 9x = 36
=> x = 4
Option C |
AQUA-RAT | AQUA-RAT-37321 | # Permutations of the word $\text{TRIANGLE}$ with no vowels together.
First of all, $$\text{TRIANGLE}$$ has $$8$$ distinct letters, $$3$$ of which are vowels($$\text{I, A, E}$$) and rest are consonants($$\text{T, R, N, G, L}$$).
While attempting this, I came up with the idea of putting alternate vowels and consonants not to group same types together.
So, I decided to form two 'batteries'. [$$\text{V}$$ stands for Vowels and $$\text{C}$$ stands for consonants.]
$$\text{V} \text{ C}\text{ V} \text{ C}\text{ V}$$
And,
$$\text{C} \text{ V}\text{ C} \text{ V}\text{ C}$$
If we count all the permutations and then add them up (Mutually Exclusive Events), we can get total number of permutations.
Now, For the first case,
$$3$$ vowels can be arranged in the $$3$$ spaces required in $$3! = 6$$ ways
From $$5$$ consonants, $$2$$ spaces can be filled with consonants in $$^5P_2 = 20$$ ways
One battery, $$(8 - 3- 2) = 3$$ letters to arrange.
Total number of permutations : $$6 * 20 * 4! = 2880$$.
In Second case,
From $$3$$ vowels, $$2$$ spaces can be filled with vowels in $$^3P_2 = 6$$ ways
From $$5$$ consonants, $$3$$ spaces can be filled with consonants in $$^5P_3 = 60$$ ways.
One battery, $$(8 - 2- 3) = 3$$ letters to arrange.
Total number of permutations : $$6 * 60 * 4! = 8640$$
So, Total number of permutations for the word $$\text{TRIANGLE} = 2880 + 8640 = 11520$$
The following is multiple choice question (with options) to answer.
How many words can be formed from the letters of the word ‘EXTRA’ so that the vowels are never together? | [
"56",
"72",
"48",
"60"
] | B | The given word contains 5 different letters.
Taking the vowels EA together, we treat them as one letter.
Then, the letters to be arranged are XTR (EA).
These letters can be arranged in 4! = 24 ways.
The vowels EA may be arranged amongst themselves in 2! = 2 ways.
Number of words, each having vowels together = (24x2) = 48 ways.
Total number of words formed by using all the letters of the given words
= 5! = (5x4x3x2x1) = 120.
Number of words, each having vowels never together = (120-48) = 72.
Ans: B |
AQUA-RAT | AQUA-RAT-37322 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work? | [
"6days",
"5days",
"4days",
"3days"
] | A | B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
ANSWER A 6days |
AQUA-RAT | AQUA-RAT-37323 | $$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games
The following is multiple choice question (with options) to answer.
Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability q that the World Series will consist of fewer than 7 games? | [
" 12.5%",
" 25%",
" 31.25%",
" 68.75%"
] | D | In order to determine the probability that the World Series will last less than 7 games, we can first determine the probability q that the World Series WILL last exactly 7 games and then subtract this value from 1.
In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.
Let's analyze one way this could happen for Team 1:
Game 1 Game 2 Game 3 Game 4 Game 5 Game 6
T1 Wins T1 Wins T1 Wins T1 Loses T1 Loses T1 Loses
There are many other ways this could happen for Team 1. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for Team 1 to win 3 games and lose 3 games in the first 6 games.
Logically, there are also 20 ways for Team 2 to win 3 games and lose 3 games in the first 6 games.
Thus, there are a total of 40 ways for the World Series to last the full 7 games.
The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.
Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is.
Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:
Thus the probability that the World Series will last less than 7 games is 100% - 31.25% = 68.75%.
The correct answer is D. |
AQUA-RAT | AQUA-RAT-37324 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
A rectangular field has area equal to 150 sq m and perimeter 50 m. Its length and breadth must be? | [
"15 m,80 m",
"15 m,18 m",
"15 m,60 m",
"15 m,10 m"
] | D | lb = 150
2(l + b) = 50 => l + b = 25
l – b = 5
l = 15 b = 10
Answer:D |
AQUA-RAT | AQUA-RAT-37325 | 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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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 man whose bowling average is 22.2, takes 4 wickets for 36 runs and thereby decreases his average by 1.2. The number of wickets, taken by him before his last match is : | [
"14",
"22",
"38",
"40"
] | D | Explanation :
Let the number of wickets taken before the last match is x.
Then, (22.2x + 36)/(x + 4) = 21
=> 22.2x+36 = 21x+84
=> 1.2x = 48
=> x = 48 / 1.2
= 40
Answer : D |
AQUA-RAT | AQUA-RAT-37326 | Question
# Three natural numbers are taken at random from a set of numbers $$\left \{ 1, 2, .... 50 \right \}$$.The probability that their average value taken as $$30$$ is equals
A
30C289C2
B
89C250C47
C
89C8750C3
D
None of these
Solution
The following is multiple choice question (with options) to answer.
Find the average of first 25 natural numbers? | [
"12",
"16",
"13",
"18"
] | C | Sum of first 100 natural numbers = 25*26/2 = 325
Required average = 325/25 = 13
Answer is C |
AQUA-RAT | AQUA-RAT-37327 | A question on the same concept
The number of television sets sold by Store R last month was approximately what percent less than the number of television sets sold by Store T last month? ( The number of television sets sold by Store R was 20 and number of television sets sold by Store T was 45 as per the attached figure)
A) 40%
B) 56%
C) 86%
D) 95%
E) 125%
so simplify it -
R is what % less than T
so T is after THAN and becomes BEFORE and R becomes AFTER.
Now we are looking for % less = $$\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.
Seventy five percent of a number is 21 less than four fifth of that number. What is the number ? | [
"240",
"340",
"420",
"130"
] | C | Let the number be x.
Then, 4*x/5 –(75% of x) = 21
4x/5 – 75x/100 = 21
x = 420.
Answer C. |
AQUA-RAT | AQUA-RAT-37328 | ### 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.
A retailer sells 5 shirts. He sells first 2 shirts for $30 and $20. If the retailer wishes to sell the 5 shirts for an overall average price of over $30, what must be the minimum average price of the remaining 3 shirts? | [
"$28.50",
"$30.50",
"$33.33",
"$40.50"
] | C | First 2 shirts are sold for $30 and $20=$50.
To get average price of $30,total sale should be 5*$30=$150
So remaining 3 shirts to be sold for $150-$50=$100
Answer should be 100/3=$33.33 that is C |
AQUA-RAT | AQUA-RAT-37329 | # When will train B catch up with train A?
Printable View
• January 27th 2010, 08:19 PM
bball20
When will train B catch up with train A?
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
• January 27th 2010, 08:31 PM
VonNemo19
Quote:
Originally Posted by bball20
Trains A & B are traveling in the same direction on a parallel tracks. Train A is traveling at 60 mph and train B is traveling at 80 mph. Train A passes a station at 5:15 PM. If train B passes the same station at 5:30 PM, at what time will train B catch up to train A?
When will train B catch up with train A?
When B passes the station, A is a quarter hour ahead of B. Or, A is, $\frac{1}{4}hr\cdot60\frac{mi}{hr}=15mi$ ahead of B.
B is traveling $20\frac{mi}{hr}$ faster than A, so how long will it take B to go 15 miles at 20mph?
• January 27th 2010, 08:49 PM
bball20
Ok, so I am still lost? (Headbang)
• January 27th 2010, 08:49 PM
fishcake
At 5.30 pm, train A is 15m away from the station, while train B is 0m away from the station. The distance of these two trains from the station in respect to the time (in hour) can be respresented as functions:
$f(t) = 60t + 15$ (for train A)
$g(t) = 80t$ (for train B)
The moment when train B catches up with train A is when both of the trains are at the same distance away from the station. That is, $f(t) = g(t)$. You'll end up with an equation:
$60t + 15 = 80t$
The following is multiple choice question (with options) to answer.
Two good train each 750 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km / hr and 30 km /hr respectively. Find the time taken by the slower train to pass the driver of the faster one. | [
"12 sec",
"24 sec",
"48 sec",
"72 sec"
] | D | Sol.
Relative Speed = (45 + 30) km / hr
= (75 x 5 / 18) m/sec
= (125/6) m/sec.
Distance covered = (750 + 750) m = 1500m.
Required time = (1500 x 6 / 125) sec = 72 sec.
Answer D |
AQUA-RAT | AQUA-RAT-37330 | concentration, mole, density
Title: Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution I’m trying to solve a question about a saturated salt solution but am unsure of how to answer one of the parts or what it’s asking for.
The Problem
The maximum solubility of table salt (NaCl) in water at 25 degrees Celsius is approximately 359 g/L. Assume the volume of the solution does not change with the addition of salt.
a) What is the molar concentration of the mixture in mol/L?
b) What is the molal concentration of the mixture in mol/kg? Assume that the density of water is 980 kg/m3.
c) How many moles of Cl2 gas would need to be present in order to make up an equivalent amount of saturated salt solution?
d) What is the density of the mixture?
My Attempt
a) molar concentration
359 g NaCl * 1 mol NaCl / 58.44 g NaCl = 6.14 mol => 6.14 mol/L
b) molal concentration
(1 L H2O * 1 m3 H2O * 980 kg) / (1000 L H2O * 1 m3) = 6.14 mol / 0.980 kg = 6.26 mol/kg
c) moles of Cl2 gas
I thought it was asking how many moles of chlorine gas is equivalent to the amount of salt in the 1 L solution (359 g NaCl). There are 70.9 g in 1 mole of Cl2, so that gives me an answer of 5.06 mol Cl2 gas (359 g NaCl / 70.9 g Cl2).
But the answer key states the correct solution as 3.07 mol Cl2 gas. How should I go about this part?
d) density of the mixture
density of water = 980 kg/m3
density of salt = 359 g/L => 359 kg/m3
density of mixture = (980 + 359) kg/m3 = 1339 kg/m3 a) molar concentration
There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.
The following is multiple choice question (with options) to answer.
One litre of water is evaporated from 6 litre of a solution containing 5% salt. The percentage of salt ? | [
"4 4/19%",
"5 5/7%",
"5%",
"6%"
] | D | Answer
Salt in 6 liter = 5% of 6 = 0.30 liter
Salt in new solution = (0.30/5) x 100% = 6%
Correct Option: D |
AQUA-RAT | AQUA-RAT-37331 | ## anonymous one year ago Water is draining from a small cylindrical tank into a larger one below it. The small cylindrical tank has a radius of 4 feet and a height of 6 feet; the large cylindrical tank has a radius of 8 feet and a height of 16 feet. The small tank is initially full of water, and the water drains out at a rate of 12 cubic feet per second. Note: The volume of a cylinder is V = π(r^2)h. A. Find the volume Vs of the water remaining in the small tank as a function of time. B. How long does it take for the small tank to completely empty? C. Let z be the depth of the water in the large tank, which...
1. anonymous
...empty. Compute dz/dt. D. What fraction of the total amount of water is in the large tank at time t = 6?
2. anonymous
@dan815 look at the first comment, it continues the rest of the question :)
3. Michele_Laino
I think another way to solve the problem is to start from this differential equation: $\Large \frac{{d{V_1}}}{{dt}} + \frac{{d{V_2}}}{{dt}} = 0$ and substituting the formula for each volume V, we get: $\Large \frac{{d{h_2}}}{{dt}} = - {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\frac{{d{h_1}}}{{dt}}$
4. dan815
am i wrong to assume that water is just draining out of the small one at 12 cubic feet/sec
5. TrojanPoem
You were right, volume as function of time is V(t) = pi (r)^2h - 12 t
6. dan815
oh dang
7. TrojanPoem
V = 0 to get the time when it's empty .
8. anonymous
So because it is draining out of the little one at a rate of twelve, it is draining into the larger one at a rate of 12? and this rate is dv/dt or v'(t)?
9. Michele_Laino
The following is multiple choice question (with options) to answer.
A leak in the bottom of a tank can empty the full tank in 4 hours. An inlet pipe fills water at the rate of 6 litres a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litres does the cistern hold? | [
"7580",
"7960",
"8290",
"2160"
] | D | Solution
Work done by the inlet in 1 hour = (1/4 - 1/12)
= 1/6.
Work done by the inlet in 1 min. = (1/6 × 1/60)
= 0.002778
Volume of 0.002778 part = 6 litres.
Therefore, Volume of whole = (.002778×6)
‹=› 2160 litres.
Answer D |
AQUA-RAT | AQUA-RAT-37332 | # 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.
One train leaves Los Angeles at 15mph heading for New York. Another train leaves from New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide, how far will the bird have traveled? | [
"5L/7 miles",
"4L/7 miles",
"3L/7 miles",
"2L/7 miles"
] | A | This is an infinite round problem means the bird will complete infinite rounds in this manner now for finding the total distance traveled by the bird let us consider the total distance between the two given stations as L. Here since the distance between these two stations is not given so answer will be in terms of L.
speed of the two trains moving in opposite directions are 15 mph and 20 mph.
so time taken for the collision between the trains will be L/(15+20) = L/35
In this much time the bird will move 25*L/35 miles = 5L/7 miles
ANSWER:A |
AQUA-RAT | AQUA-RAT-37333 | Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%.
So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best.
We will post the detailed solution tomorrow and then we will show another innovative method of solving this question.
Regards,
Saquib
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The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 07 Aug 2018, 06:11
2
1
Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post.
We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say,
The following is multiple choice question (with options) to answer.
If x and y are positive numbers and z = xy^2, a 60 percent increase in x and a 30 percent decrease in y would result in which of the following changes in z? | [
"decrease of 12.5%",
"decrease of 21.6%",
"increase of 4.8%",
"increase of 12.5%"
] | B | After the changes, z = 1.6x*(0.7y)^2 = (1.6*0.49)xy^2 = 0.784*z
The answer is B. |
AQUA-RAT | AQUA-RAT-37334 | However, as clearly stated in the text, the probabilities for components $B$ and $C$ are interdependent! This means that, before you can do a calculation as in the previous paragraph, you must first explore fully the interdependence of these probabilities.
Okay, let us do this. It is given that $C$ has a failure chance of $0.01$. Furthermore it is given that if $C$ fails, then the chance that $B$ fails is $0.55%$. This leads us to the conclusion that the probability that both $C$ and $B$ fail is equal to $0.01 * 0.55 = 0.0055$.
With a bit more effort we can derive that the probability that $C$ fails and $B$ works is equal to $0.0045$. The chance that $C$ works and $B$ fails is $0.1455$. The chance that both $C$ and $B$ work is $0.8455$. We now have the four values that determine the interdependence of $C$ and $B$.
Finally we can compute the overall probability that the system works. We get:
$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$
$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$
The following is multiple choice question (with options) to answer.
The probability that a computer company will get a computer hardware contract is 2/3 and the probability that it will not get a software contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability that it will get both the contracts ? | [
"14/45",
"4/9",
"4/5",
"4/6"
] | A | Explanation :
Let, A ≡ event of getting hardware contract
B ≡ event of getting software contract
AB ≡ event of getting both hardware and software contract.
P(A) = 2/3, P(~B) =5/9
=> P(B) = 1- (5/9) = 4/9.
A and B are not mutually exclusive events but independent events. So,
P(at least one of A and B ) = P(A) + P(B) - P(AB).
=> 4/5 = (2/3) + (4/9) -P(AB).
=> P(AB) = 14/45.
Hence, the required probability is 14/45.
Answer : A |
AQUA-RAT | AQUA-RAT-37335 | # How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$?
How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $a/c = 2/5$ but that is not a correct answer.
-
Hint: Ratio $\,a:b = 2:5\,$ is the same as $$\frac{a}{b}=\frac{2}{5}$$ – DonAntonio Aug 20 '12 at 15:48
First thing, your c:d is not clear, – Rahul Taneja Aug 20 '12 at 16:46
Thanks, I fixed it. – jbman223 Aug 20 '12 at 16:48
Maybe it helps you to simply set e.g. $a=30$ and figure out what the other numbers must be in that case. – celtschk Aug 20 '12 at 17:01
These ratios are just simple equations. For example $a:b=2:5$ is $$a= \frac{2}{5}b$$ No need for confusing tricks here. Just substitutions : $$a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$ So that $$a:c = 8:75$$
-
The following is multiple choice question (with options) to answer.
The ratio between the present ages of A and B is 7:3 respectively. The ratio between A's age 4 years ago and B's age 4 years hence is 1:1. What is the ratio between A's age 4 years hence and B's age 4 years ago? | [
"3:4",
"3:0",
"9:1",
"9:2"
] | C | Let the present ages of A and B be 7x and 3x years respectively.
Then, (7x - 4)/(3x + 4) = 1/1
4x = 8 => x = 2
Required ratio = (5x + 4):(3x - 4) = 18:2 = 9:1.Answer: C |
AQUA-RAT | AQUA-RAT-37336 | Question
# In a school, there are $$1000$$ student, out of which $$430$$ are girls. It is known that out of $$430, 10$$% of the girls study in class $$XII$$. What is the probability that a student chosen randomly studies in class $$XII$$ given that the chosen student is a girl?.
Solution
## Total number of students $$= 1000$$Total number of girls $$= 430$$Girls studying in class $$XII = 10 \% \text{ of } 430$$ $$\\ = \cfrac{10}{100} \times 430$$ $$\\ = 43$$We need to find the probability that a student chosen randomly studies in class $$XII$$, given that the chosen student is a girl.$$A$$ : Student is in class $$XII$$$$B$$ : Studenet is a girlTherefore,$$P{\left( A | B \right)}$$ $$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$$Mathematics
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The following is multiple choice question (with options) to answer.
If 47.5% of the 880 students at a certain college are enrolled in biology classes, how many students at the college are NOT enrolled in a biology class? | [
"110",
"330",
"550",
"462"
] | D | Students enrolled in biology are 47.5% and therefore not enrolled are 52.5%. so of 880 is 880*.525 = 462
Answer is D 462 |
AQUA-RAT | AQUA-RAT-37337 | => $$\frac{74}{100} \times x$$ = 1,11,000
=> $$x$$ = 1,11,000 $$\times \frac{100}{74}$$ = 1,50,000
$$\therefore$$ Total profit = 4,00,000 + 1,50,000 = Rs. 5,50,000
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The following is multiple choice question (with options) to answer.
Anil started a business with Rs.5000 and Rajesh joined afterwards with Rs.10,000. If profit at the end of the year is divided equally, then when did Rajesh join him? | [
"6 months afterwards",
"8 months afterwards",
"9 months afterwards",
"5 months afterwards"
] | A | 5000*12=10000*(12-x)
x=6
ANSWER:A |
AQUA-RAT | AQUA-RAT-37338 | Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Intern Joined: 12 Mar 2015 Posts: 4 Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink] ### Show Tags 05 Oct 2015, 23:05 Barkatis wrote: It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages? A. 5 B. 6 C. 8 D. 10 E. 12 Hello, why can't it be time taken by tom= T and time taken by Jack= T+2 ? And then the eq becomes $$\frac{20}{T}$$ + $$\frac{20}{T+2}$$ = $$\frac{25}{3}$$ Please help? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8196 Location: Pune, India Re: It takes Jack 2 more hours than Tom to type 20 pages. If [#permalink] ### Show Tags 05 Oct 2015, 23:34 2 amanlalwani wrote: Barkatis wrote: It takes Jack 2 more hours than Tom to type 20 pages. If working together, Jack and Tom can type 25 pages in 3 hours, how long will it take Jack to type 40 pages? A. 5 B. 6 C. 8 D. 10 E. 12 Hello, why can't it be time taken by tom= T and time taken by Jack= T+2 ? And then the eq becomes $$\frac{20}{T}$$ + $$\frac{20}{T+2}$$ = $$\frac{25}{3}$$ Please help? It can be. If you solve it, you will get T = 4. Then, to get time taken by Jack, you just add 2 to it to get 6 hrs. Note that you can take either variable but it is often better to assume what you have to find (time taken by Jack) since sometimes, we forget the last step. If you directly get the value of J, great. If you first
The following is multiple choice question (with options) to answer.
Two consultants can type up a report in 12.5 hours and edit it in 7.5 hours. If Mary needs 30 hours to type the report and Jim needs 12 hours to edit it alone, how many T hours will it take if Jim types the report and Mary edits it immediately after he is done? | [
"41.4",
"34.1",
"13.4",
"12.4"
] | A | Break down the problem into two pieces: typing and editing.
Mary needs 30 hours to type the report--> Mary's typing rate = 1/30 (rate reciprocal of time)(point 1 in theory below);
Mary and Jim can type up a report in 12.5and --> 1/30+1/x=1/12.5=2/25 (where x is the time needed for Jim to type the report alone)(point 23 in theory below)--> x=150/7;
Jim needs 12 hours to edit the report--> Jim's editing rate = 1/12;
Mary and Jim can edit a report in 7.5and --> 1/y+1/12=1/7.5=2/15 (where y is the time needed for Mary to edit the report alone) --> y=20;
How many T hours will it take if Jim types the report and Mary edits it immediately after he is done--> x+y=150/7+20=~41.4
Answer: A. |
AQUA-RAT | AQUA-RAT-37339 | harpazo
#### harpazo
##### Pure Mathematics
Sure you do...you know that the number $$54$$ has a value of $$5\cdot10+4$$, right?
Yes but???
#### MarkFL
##### La Villa Strangiato
Staff member
Moderator
Math Helper
Yes but???
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
harpazo
#### harpazo
##### Pure Mathematics
But what? If a number has the two digits from left to right as x and y, then the value of the number (assuming base 10) is 10x + y. If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x.
Does that make sense?
You said:
"If y is twice x, then the number's value is 10x + 2x. Switching the digits gives the number a value of 20x + x."
How does switching the digits yield
20x + x?
Staff member
The following is multiple choice question (with options) to answer.
The difference of 2 digit number & the number obtained by interchanging the digits is 36. What is the difference the sum and the number if the ratio between the digits of the number is 1:2 ? | [
"5",
"8",
"10",
"14"
] | B | Let the number be xy.
Given xy – yx = 36.
This means the number is greater is than the number got on reversing the digits.
This shows that the ten’s digit x > unit digit y.
Also given ratio between digits is 1 : 2 => x = 2y
(10x + y) – (10y +x) = 36 => x – y = 4 => 2y – y =4.
Hence, (x + y) – (x – y) = 3y – y = 2y = 8
B |
AQUA-RAT | AQUA-RAT-37340 | # 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 Ram's firm's annual revenue grows twice as fast as its costs. In 2007 it operated at a $1000 loss, it broke even in 2008, and in 2009 its revenues were 44% higher than in 2007. If the firm's revenues and costs grew at a constant rate over this period, what was its profit in 2009? | [
"700",
"1000",
"1300",
"1600"
] | C | At Ram's firm's annual revenue grows twice as fast as its costs.First solve for how much revenue is increasing each year:
R2008= R2007*(1+x)
R2009= R2007*(1+x)^2=1.44*R2007
(1+x)^2 = 1.44
1+x=1.2
x=0.2 aka revenue increases 20% each year and cost increases 10% annually
Next solve for R2007and C2007:
R2007= C2007- 1000
1.2*R2007- 1.1*C2007= 0
1.2*[C2007- 1000] - 1.1*C2007= 0
0.1*C2007= 1,200
C2007= 12,000
R2007= 11,000
Finally find 2009 profits:
Profit2009= 1.44*11,000 - 1.21*12,000
Profit2009= 15,840 - 14,520
Profit2009= 1320
Answer: C |
AQUA-RAT | AQUA-RAT-37341 | # 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 boat moves upstream at the rate of 1 km in 30 minutes and down stream 1 km in 12 minutes. Then the speed of the current is : | [
"1.5 kmph",
"2 kmph",
"3 kmph",
"2.5 kmph"
] | A | Rate upstream = (1/30 *60) = 2 kmph
Rate down stream = 1/12 * 60 = 5 kmph
Rate of the current = ½ (5-2) = 1.5 kmph
ANSWER:A |
AQUA-RAT | AQUA-RAT-37342 | 51. Trisaba
thanks for the help
52. dan815
|dw:1441062793507:dw|
53. dan815
this is the full break down
54. dan815
they will probably ask u a final question like the chance it will rain, when there is a prediction of rain
55. Trisaba
60%
56. Trisaba
bayes' theorem
57. Trisaba
thanks for helping dan
58. dan815
wait i got 36% lemme see
59. dan815
okay think about it like this out of every 97 days, 5% of those days say it will rain, and it wont rain out of every 3 days, 90% of those days say it will rain, and it will rain! what is the total number of days it will rain to not raining then?
60. Trisaba
61. dan815
0.03*0.9 : 0.05*0.97 this means that (0.03*0.9)/((0.03*0.9)+(0.05*0.97)) = chance it rains when it say it rains
62. Trisaba
60% is what i got
63. dan815
what did u say p(b) was
64. Trisaba
P(B) is .0755
65. Trisaba
look at my sheet posted pic
66. dan815
hard to read, what did u say P(B|A) was then
67. Trisaba
.9
68. Trisaba
$\frac{ 90 }{ 151 }$
69. Trisaba
it rounds to 60%
70. dan815
hmm i am not seeing anything wrong with the way im finding the answer, ill think about this more later but for now for every 100 days 3 days really rain 90% of 3 days is 2.7 so every 2.7 days out of the 3 days it rains, it will predict right now an additional bad predictions of 5% of 97 days is thrown in there 0.05*97=4.85 right preiction and rain = 2.7 total prediction of rainy days = 4.85+2.7 therefore 2.7/(4.85+2.7)= ~36%
71. Trisaba
|dw:1441064489549:dw|
72. Trisaba
The following is multiple choice question (with options) to answer.
It rained as much as on Wednesday as on all the other days of the week combined. If the average rainfall for the whole week was 3.5 cms, How much did it rain on Wednesday? | [
"10.9",
"10.5",
"12.25",
"10.1"
] | C | Explanation:
Let the rainfall on wednesday = 6x.
∴ Rainfall on the remaining days = 6x
Given,
(6x + 6x )/7 = 3.5
⇒12x = 24.5
⇒6x = 12.25
Answer: C |
AQUA-RAT | AQUA-RAT-37343 | 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 moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train? | [
"270",
"288",
"266",
"299"
] | A | Relative speed = (72 - 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.
Answer: A |
AQUA-RAT | AQUA-RAT-37344 | # How many $2$'s are needed?
There is a positive integer $N$. $N$ is made up of only two distinct digits- $2$ and $3$. $N+18$ is divisible by $37$. What is the minunum amount of times the number $2$ can appear in $N$? I'm pretty sure the answer is only one time. But how can I prove this?
• are there any better tags I could use? – Joao Dec 3 '14 at 22:22
• I'm afraid I don't share your optimism; the smallest count of twos I managed to come up was three. – Peter Košinár Dec 3 '14 at 22:33
• @PeterKošinár but is it possible to prove that the smallest amount of two's is 3? – Joao Dec 3 '14 at 22:39
• Note that $1000\equiv 1 \mod 37$ so you can reduce the problem to analysing the sum of digits (with some cases to check). – Mark Bennet Dec 3 '14 at 23:01
• @MarkBennet Thanks. That should be an answer. – Joao Dec 3 '14 at 23:08
## 3 Answers
Let's start with a simple observation:
The number $37$ plays very nicely with powers of $10$... The value $10^m$ can only have three possible remainders modulo $37$: $1$, $10$ and $26$. If $N+18$ is divisible by $37$, $N$ must be congruent to $19$ modulo $37$. This observation will be quite useful in subsequent thoughts.
The number $N=2323323$ satisfies the given conditions and contains three twos. In order to show that we cannot satisfy the conditions given with fewer twos, we need to consider three cases:
The following is multiple choice question (with options) to answer.
How many positive integers less than 22 can be expressed as the sum of a positive multiple of 2 and a positive multiple of 3? | [
"14",
"17",
"12",
"11"
] | B | The number = 2a + 3b < 20
When a = 1, b = 1, 2, 3, 4, 5, 6 -> 2a = 2; 3b = 3, 6, 9, 12, 15, 18 -> the number = 5, 8, 11, 14, 17, 20 --> 6 numbers
when a =2, b = 1,2,3,4,5, 6 -> ....--> 6 numbers
when a =3, b = 1,2,3,4, 5 --> ....--> 5 numbers
Total number is already 17. Look at the answer there is no number greater than 17 --> we dont need to try any more
Answer must be B |
AQUA-RAT | AQUA-RAT-37345 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
a school has 11 maths 8 physics and 5 chemistry teachers each teacher can teach 3 subjects max what is he minimum number of teachers required | [
"2",
"4",
"6",
"8"
] | D | Total subjects = 11+8+5=24
Max subjects by 1 teacher = 3
So, min of teachers required = 24/3 = 8
ANSWER:D |
AQUA-RAT | AQUA-RAT-37346 | The only possible answer is E i.e 12 hrs.
Yes, your approach is very good. The only thing I have an issue with is the approximation used.
Their combined time is 4.8 hrs and hence we know that Jack will take more than 9.6 hrs. 10 hrs is a possible candidate for the correct option in that case. Though I would say that if Jack took just a wee bit more than 9.6 hrs, then Tom would have taken a tiny bit less than 9.6 hrs and then the difference in their individual time taken could not be 2 hrs. So yes, (E) must be the answer.
_________________
Karishma
Veritas Prep GMAT Instructor
The following is multiple choice question (with options) to answer.
Harish can paint a wall in just 3 hours. However it takes 6 hours for Ganpat to complete the same job of painting the wall.
How long would it take for both of them to paint the wall , if both work together ? | [
"3 Hours",
"1 Hours",
"3.20 Hours",
"2 Hours"
] | D | D
Two hours.
Harish who could paint the wall in 3 hours could paint 2/3 of the wall in just two hours. Ganpat who paint the wall in 6 hours could paint 1/3 of the wall in two hours.
2/3 + 1/3 = 1. |
AQUA-RAT | AQUA-RAT-37347 | # If $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$ = $$K(10)^9$$, then k is equal to
## Solution :
$$K(10)^9$$ = $$(10)^9$$ + $$2(11)^1(10)^8$$ + $$3(11)^2(10)^7$$ + …… + $$10(11)^9$$
K = 1 + 2$$({11\over 10})$$ + 3$$({11\over 10})^2$$ + ….. + 10$$({11\over 10})^9$$ ……(i)
$$({11\over 10})$$K = 1$$({11\over 10})$$ + 2$$({11\over 10})^2$$ + 3$$({11\over 10})^3$$ + ….. + 10$$({11\over 10})^{10}$$ …..(ii)
On subtracting equation (ii) from (i), we get
K$$(1 – {11\over 10})$$ = 1 + $$({11\over 10})$$ + $$({11\over 10})^2$$ + …. + $$({11\over 10})^9$$ – 10$$({11\over 10})^{10}$$
$$\implies$$ K$$({10 – 11\over 10})$$ = $$1[({11\over 10})^{10} – 1]\over ({11\over 10} – 1)$$ – 10$$({11\over 10})^{10}$$
The following is multiple choice question (with options) to answer.
If 2 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution? | [
" -7",
" -4",
" -5",
" 1"
] | C | The phrase “2 is one solution of the equation” means that one value of x is 2. Thus, we first must plug 2 for x into the given equation to determine the value of k. So we have
2^2 + (3)(2) + k = 10
4 + 6 + k = 10
10 + k = 10
k =0
Next we plug 0 into the given equation for k and then solve for x.
x^2 + 3x = 10
x^2 + 3x – 10 = 0
(x+5)(x-2) = 0
x = -5 or x = 2
Thus, -5 is the other solution. Answer C. |
AQUA-RAT | AQUA-RAT-37348 | ### 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.
Suzie’s Discount Footwear sells all pairs of shoes for one price and all pairs of boots for another price. On Monday the store sold 22 pairs of shoes and 16 pairs of boots for $650. On Tuesday the store sold 8 pairs of shoes and 32 pairs of boots for $760. How much more do pairs of boots cost than pairs of shoes at Suzie’s Discount Footwear? | [
"$2.50",
"$5.00",
"$5.50",
"$7.50"
] | B | Let X be pair of shoes and Y be pair of boots.
22x+16y = 650 ... eq1
8x +32y = 760....eq 2.
Now multiply eq 1 by 2 and sub eq 2.
44x = 1300
8x = 760.
36x = 540 => x = 15.
Sub x in eq 2.... we get 120 + 32y = 760... then we get 32y = 640 then y = 20
Difference between X and Y is 5.
Answer: option B is correct answer. |
AQUA-RAT | AQUA-RAT-37349 | A triangle with two sides 6 and 8 units long is inscribed in a circle. If the third side is a diameter, find the length of the diameter.
Solution
A triangle with one side the diameter is a right triangle.
The lengths of the sides this triangle are 6 : 8 : h. This is a multiple of a
3 : 4 : 5 right triangle, so the length of the hypotenuse that is a diameter is 10.
Find the values of x, y and z.
Solution
The circle is the sum of the arcs, so start by finding 3z°.
360° = 110° + 130° + 54° + 3z° so 3z° = 66° and z = 22.
There are two methods to find the other two angles.
Method 1
The measures of inscribed angles are equal to half the sum of the opposite arcs.
Using the opposite arcs:
2x° = (\dfrac{1}{2})(66° + 110°)
so 2x° = (\dfrac{1}{2})176° and x = 44.
6y° = (\dfrac{1}{2})(66° + 54°)
so 6y° = 60° and y = 10.
Method 2
Since the inscribed polygon is a quadrilateral, opposite angles of the quadrilateral add to 180°.
The measure of inscribed
C = (\dfrac{1}{2})(130° + 54°) = 92°.
Add the opposite angles.
2x° + 92° = 180° so 2x° = 88° and x = 44.
The measure of inscribed
B = (\dfrac{1}{2})(130° + 110°) = 120°.
6y° + 120° = 180°
so 6y° = 60° and y = 10.
A square is inscribed in a circle of diameter 20. Determine the ratio of the area of the circle to the area of the square.
Solution
Sketch the figure.
You need to find the area of the circle and the square.
Be careful! Since diameter = 20,
r = 10, and the area of the circle is A = πr2 = 100π.
The following is multiple choice question (with options) to answer.
A triangle has a perimeter 16. The two shorter sides have integer lengths equal to x and x + 1. Which of the following could be the length of the other side? | [
"2",
"4",
"6",
"7"
] | C | The SHORTER sides have integral lengths equal to x and x + 1
Let the longest side be 'a'
So, a + x + (x +1) = 16
a + 2x = 15.......eqn (1)
We know that the sum of the lengths of the shorter sides has to be more than the length of the longer one. i.e 2x+1> a
a =6
C |
AQUA-RAT | AQUA-RAT-37350 | Re: If a, b, c, and d, are positive numbers, is a/b < c/d? [#permalink] 21 Apr 2017, 03:36
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16 A, B, C and D are positive integers such that A/B = C/D. Is C divisibl 7 16 Mar 2016, 06:12
5 For non–zero integers a, b, c and d, is ab/cd positive? 8 05 Sep 2016, 15:05
If a, b, c, and d are positive, is a/b > c/d? 1 04 May 2017, 19:50
If a, b, c, and d are positive numbers, and a/b = c/d, what is the val 2 28 Jul 2016, 14:34
8 If a, b, c, and d are positive numbers and a/b < c/d , which of the fo 7 27 Mar 2017, 22:14
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The following is multiple choice question (with options) to answer.
If a·b·c·d=360, where a, b, c and d are positive integers, and a<b<c<d, which of the following could be the value of d−a? | [
"9",
"10",
"11",
"13"
] | B | prime factorize : 360 = 2*3*5*12
Hence a = 2 and d = 12 .
Hence answer is 12 - 2 = 10
Answer B |
AQUA-RAT | AQUA-RAT-37351 | It helps to try out the strategy with a smaller number of students. With only five students you teacher's method gives $$4^5-\binom{4}{1}\cdot3^5+\binom{4}{2}\cdot2^5-\binom{4}{3}\cdot1^5=1024-972+192-4=240$$ ways. You can check that this is right by observing that some book must be distributed to two students and each of the other books to only one. There are four choices for which book goes to two students, and $$\binom{5}{2}$$ choices of the two students, and $$3!$$ ways to distribute the other three books to the remaining students: $$4\cdot10\cdot6=240$$.
Now try your method, which gives $$\binom{5}{4}\cdot4!\cdot4^1=480$$ ways. Why is your number too big? It's because distributing different books to students $$1$$, $$2$$, $$3$$, $$4$$ and any book to student $$5$$ includes some of the same configurations as distributing different books to students $$1$$, $$3$$, $$4$$, $$5$$ and any book to student $$2$$. For example, if students $$1$$, $$2$$, $$3$$, $$4$$ get $$H$$, $$R$$, $$L$$, $$E$$ in that order and student $$5$$ gets $$R$$, that's the same as students $$1$$, $$3$$, $$4$$, $$5$$ getting $$H$$, $$L$$, $$E$$, $$R$$ in that order and student $$2$$ getting $$R$$.
The following is multiple choice question (with options) to answer.
The maximum number of students among them 1008 pens and 928 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils? | [
"16",
"18",
"36",
"19"
] | A | number of pens = 1008
number of pencils = 928
Required number of students = H.C.F. of 1008 and 928 = 16
Answer is A |
AQUA-RAT | AQUA-RAT-37352 | 5. Hello, James!
Another approach . . .
12 Students are in a class.
Five can go to room A, Four to room B, and Three to room C.
How many ways can this happen?
Assign 5 students to room A.
. . There are: . $_{12}C_5 \:=\:\frac{12!}{5!7!} \:=\:792$ ways.
From the remaining 7 students, assign 4 students to room B.
. . There are: . $_7C_4 \:=\:\frac{7!}{4!3!} \:=\:35$ ways.
From the remaining 3 students, assign 3 students to room C.
. . Of course, there is: . $_3C_3 \:=\:1$ way.
Therefore, there are: . $792 \times 35 \times 1 \:=\:27,\!720$ ways.
The following is multiple choice question (with options) to answer.
Every student in a room is either a junior or a senior. There is at least one junior and at least one senior in the room. If 1/2 of the juniors is equal to 2/3 of the seniors, what fraction of the students in the room are juniors? | [
"4/7",
"1/3",
"5/3",
"12/7"
] | A | Let total number of juniors= J
total number of seniors =S
(1/2) J = (2/3) S
=> S = 3/4 J
Total number of students = J+S = (7/4) J
Fraction of the students in the room are juniors = J/(J+S) = J/[(7/4) J]
=4/7
Answer A |
AQUA-RAT | AQUA-RAT-37353 | ### Show Tags
30 Jun 2018, 18:53
12/(.6x)=12/x+4
20=12+4x
x=2
24/2
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Re: When the price of oranges is lowered by 40%, 4 more oranges [#permalink]
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04 Jul 2018, 18:25
1
farukqmul wrote:
When the price of oranges is lowered by 40%, 4 more oranges can be purchased for $12 than can be purchased for the original price. How many oranges can be purchased for 24 dollars at the original price? (A) 8 (B) 12 (C) 16 (D) 20 (E) 24 We use the equation: price per item x no. of items = total cost. Here, we let p = the original price of an orange and q = the original number of oranges purchased. We can create the equation for the original total cost: pq = 12 q = 12/p After the orange’s price is lowered, we have that 0.6p = the new (reduced) price of an orange and (q + 4) = the new number of oranges that can be purchased at the reduced price. Our new equation for total cost is: (0.6p)(q + 4) = 12 0.6pq + 2.4p = 12 Substituting for q, we have: 0.6p(12/p) + 2.4p = 12 7.2 + 2.4p = 12 2.4p = 4.8 p = 2 That is, each orange is$2. So for \$24, we can buy 24/2 = 12 oranges.
_________________
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Re: When the price of oranges is lowered by 40%, 4 more oranges &nbs [#permalink] 04 Jul 2018, 18:25
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The following is multiple choice question (with options) to answer.
Rani bought more apples than oranges. She sells apples at ₹23 apiece and makes 15% profit. She sells oranges at ₹10 apiece and makes 25% profit. If she gets ₹653 after selling all the apples and oranges, find her profit percentage. | [
"16.8%",
"17.4%",
"17.9%",
"18.5%"
] | B | Given: Selling price of an Apple = 23 --> Cost price = 23/1.15 = 20
Selling price of an orange = 10 --> Cost price = 10/1.25 = 8
A > O
23*(A) + 10*(O) = 653
653 - 23*(A) has to be divisible by 10 --> Units digit has to be 0
Values of A can be 1, 11, 21, 31, .... --> 1 cannot be the value
Between 11 and 21, If A = 11, O = 30 --> Not possible
If A = 21, O = 17 --> Possible
Cost price = 20*21 + 8*17 = 420 + 136 = 556
Profit = 653 - 556 = 97
Profit% = (97/556)*100 = 17.4%
Answer: B |
AQUA-RAT | AQUA-RAT-37354 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
The C.P of 10 pens is equal to the S.P of 12 pens. Find his gain % or loss%? | [
"16 2/7%",
"16 6/3%",
"16 2/3%",
"16 1/3%"
] | C | 10 CP = 12 SP
12 --- 2 CP loss
100 --- ? => 16 2/3%
Answer: C |
AQUA-RAT | AQUA-RAT-37355 | # Math Help - help
1. ## help
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
name a fraction that is between 1/2 and 1/3....?
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
2. Originally Posted by BeBeMala
mr.lopez needs 3/4 kg of apples to make a pie.how many pies can he make if he uses 20 kg of apples?
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
Originally Posted by BeBeMala
name a fraction that is between 1/2 and 1/3....?
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Originally Posted by BeBeMala
which of the following fractions is closest to one?
a)2/3 b)3/4 c)4/5 d)5/6
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
Originally Posted by BeBeMala
three pizzas were evenly shared among some friends. if each of them go 3/5 of the pizza. how many were there altogether?
This one works just like the first one, above.
3. Originally Posted by stapel
How many 3/4 kilograms can fit inside 20 kilograms? (Hint: Divide.)
There are infinitely-many choices. One quick way of finding such a fraction is to convert the two given fractions to a common denominator, and then find some other fraction of the same denominator that is between the two numerators.
Convert them all to the same denominator, and subract each from 1. Whichever one has the smallest answer was the biggest (and thus closest to 1).
This one works just like the first one, above.
sorry....
The following is multiple choice question (with options) to answer.
Jen has a jar of jam, and she ate 1/3 of the jam for lunch. If Jen ate 1/7 of the remaining jam for dinner, what fraction of the jam was left after dinner? | [
"2/7",
"3/7",
"4/7",
"9/14"
] | C | Let x be the amount of jam at the beginning.
After lunch, the remaining jam was (2/3)x.
After dinner, the remaining jam was (6/7)(2/3)x = (4/7)x.
The answer is C. |
AQUA-RAT | AQUA-RAT-37356 | # 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",
"62.6",
"62.1",
"62.7"
] | 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-37357 | $$41=2\sum_{k=1}^9n_k$$
There is no integer $\sum_{k=1}^9n_k$ that satisfies this.
-
But my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive – 403 Forbidden May 25 '13 at 8:33
@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. – Ataraxia May 25 '13 at 8:34
As it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) – 403 Forbidden May 25 '13 at 9:09
The following is multiple choice question (with options) to answer.
Two whole numbers whose sum is 40 cannot be in the ratio | [
"4:10",
"2:3",
"1:3",
"2:4"
] | D | D) 2:4 |
AQUA-RAT | AQUA-RAT-37358 | 1. Four $1$'s. Every column and row contains exactly one $1$, so the number of possible arrangements equals: $$4! = 24$$
2. Six $1$'s. One column contains three $1$'s, the three others contain one $1$. Once the former $1$'s have been placed in the grid, either all three remaining $1$'s are placed in the empty row and columns, or one $1$ is placed in the empty row and columns, and two $1$'s are placed in one of the non-empty rows. The number of possible arrangements thus equals: $${4 \choose 1}{4 \choose 3}\bigg(1+{3 \choose 1}{3 \choose 2}\bigg) = 4 \cdot 4 \cdot (1 + 3 \cdot 3) = 160$$
3. Eight $1$'s. Two columns contain three $1$'s and two columns contain one $1$. The six $1$'s must be distributed over all rows, otherwise it is impossible to have the three rows contain an odd number of $1$'s. Once these $1$'s have been placed in the grid, the remaining two $1$'s must be placed in the rows which contain two $1$'s. As such, the number of possible arrangements equals: $${4 \choose 2}{4 \choose 3}{3 \choose 2}{2 \choose 1} = 6 \cdot 4 \cdot 3 \cdot 2 = 144$$
The total number of arrangements thus equals:
$$24 + 160 + 144 + 160 + 24 = 512$$
The following is multiple choice question (with options) to answer.
The integers from 1 to 1851 are entered sequentially in 12 columns (numbered 1 to 12); i.e. Row 1 contains the numbers 1 to 12, Row 2 contains the next 12 numbers and so on. If the number 1369 occurs in row p and column q, then the value of p + q is | [
"114",
"115",
"116",
"117"
] | C | 1369=114*12+1
1368=114*12 mean 1368 occurs in 114 row and 12 column.
so 1369 occur in 115 row and 1st column.as elements are entered sequentially.
so p=115 q=1.p+q=116
ANSWER:C |
AQUA-RAT | AQUA-RAT-37359 | # STRNO - Editorial
Setter: Kritagya Agarwal
Tester: Felipe Mota
Editorialist: Taranpreet Singh
Easy
Number theory
# PROBLEM:
Given two integers X and K, you need to determine whether there exists an integer A with exactly X factors and exactly K of them are prime numbers.
# QUICK EXPLANATION
A valid choice for A exists if the prime factorization of X has the number of terms at least K. So we just need to compute prime factorization of X.
# EXPLANATION
Let’s assume a valid A exists, with exactly K prime divisors.
The prime factorization of A would be like \prod_{i = 1}^K p_i^{a_i} where p_i are prime factors of A and a_i are the exponents. Then it is well known that the number of factors of A are \prod_{i = 1}^K (a_i+1). Hence we have X = \prod_{i = 1}^K (a_i+1) for a_i \geq 1, hence (a_i+1) \geq 2
So our problem is reduced to determining whether is it possible to write X as a product of K values greater than 1 or not.
For that, let us find the maximum number of values we can split X into such that each value is greater than 1. It is easy to prove that all values shall be prime (or we can further split that value). If the prime factorization of X is \prod r_i^{b_i}, then \sum b_i is the number of terms we can split X into.
For example, Consider X = 12 = 2^2*3 = 2*2*3. Hence we can decompose 12 into at most 3 values such that their product is X. However, if K < \sum b_i, we can merge the values till we get exactly K values. Suppose we have X = 12 and K = 2, so after writing 2, 2, 3, we can merge any two values, resulting in exactly two values.
So a valid A exists when the number of prime factors of X with repetition is at least K.
The following is multiple choice question (with options) to answer.
For any integer p, *p is equal to the product of all the integers between 1 and p, inclusive. How many prime numbers are there between *6 + 3 and *6 + 6, inclusive? | [
" One",
" None",
" Two",
" Three"
] | B | Generally *p or p! will be divisible by ALL numbers from 1 to p. Therefore, *6 would be divisible by all numbers from 1 to 6.
=> *6+3 would give me a number which is a multiple of 3 and therefore divisible (since *6 is divisible by 3)
In fact adding anyprimenumber between 1 to 6 to *6 will definitely be divisible.
So the answer is none (A)!
Supposing if the question had asked for prime numbers between *6 + 3 and *6 + 11 then the answer would be 1. For *6 +3 and *6+ 13, it is 2 and so on...
B |
AQUA-RAT | AQUA-RAT-37360 | 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
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Andrew borrows equal sums of money under simple interest at 5% and 4% [#permalink]
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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.
At 5% per annum simple interest, Rahul borrowed Rs. 500. What amount will he pay to clear the debt after 4 years | [
"750",
"700",
"650",
"600"
] | D | Explanation:
We need to calculate the total amount to be paid by him after 4 years, So it will be Principal + simple interest.
So,
=>500+500∗5∗4 /100=>Rs.600
Option D |
AQUA-RAT | AQUA-RAT-37361 | # 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 can row a boat at 20 kmph in still water. If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream? | [
"3:7",
"3:6",
"3:1",
"3:2"
] | C | The ratio of the times taken is 2:1.
The ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3/1 = 3:1
Speed of the stream = 42/3 = 14 kmph.Answer:C |
AQUA-RAT | AQUA-RAT-37362 | AAB,
ABA,
BAA,
which matches your question. If you want it like this, then say your string contains for example three equal sizes groups "AABBCC". One of the requirements for a viable increment should be that the first letters of group should appear in increasing order. So e.g. $$\mathtt{\underline A\underline BB\underline CAC}$$ is allowed, but $$\mathtt{\underline AA\underline C\underline BBC}$$ is not.
As for counting, you are on the right track. You need to count the number of groups of each size. For example, let's say that "ABCCDDDEEE" has type $$1^2 2^1 3^2$$, meaning $$2$$ groups of $$1$$ letter, $$1$$ group of $$2$$ letters, and $$2$$ groups of $$3$$ letters. Then a string of length $$n$$ with type $$k_1^{e_1}\ldots k_m^{e_m}$$ will have $$\frac{n!}{(k_1!)^{e_1}\cdots (k_m!)^{e_m}\cdot(e_1!)\cdots (e_m!)}$$ non-equivalent permutations in your sense. The factors $$(e_i)!$$ are the only thing new, and they take care of the ways the groups can be permuted among each other.
So for example, take "AABB". It's got type $$2^2$$, so we get $$\frac{4!}{(2!)^2 2!} = 3.$$ These are represented by "AABB", "ABAB" and "ABBA".
Your numbers can be found at http://oeis.org/A178867.
The following is multiple choice question (with options) to answer.
How many different 6-letter sequences are there that consist of 1A, 20B's and 15C's ? | [
"6",
"60",
"120",
"360"
] | D | How many different 6-letter sequences are there that consist of 1 A, 20B's, and 15 C's?
(A) 6
(B) 60
(C) 120
(D) 360
(E) 720 (ANS D) |
AQUA-RAT | AQUA-RAT-37363 | reads 760 mm of mercury, how much work is done by the system comprising the helium initially in the bottle, if the balloon is light and requires no stretching. In order to find the surface area or volume of a sphere, we will need to use a formula, and as with any formula that involves circles, the number 'π ' is included in both. Do not forget the units. 11" round latex balloons at 5280' will become 11 1/4" balloons at 7500' (assuming spherical balloons, this is more than a 2% increase in diameter, since diameter scales as the cube root of volume for a sphere. Write the formula for volume of the balloon as a function of time. Give your answer to 2 decimal places. Problem A meteorologist is inflating a spherical balloon with a helium gas. So, You have to figure out how fast the radius is changing, so. A balloon is not a straight edged polygon shape, usually, so the mathematical equations get that much harder, on the flip side, it may be a spherical or ovalish shape, but measurements with math alone are detrimental due to the uneven sizes of the balloon. (b) Using the chain rule, or otherwise, find an expression in terms of r and t for t r d d. Hint: Use composite function relationship V sphere = 4/3 π r 3 as a function of x (radius), and x (radius) as a function of t (time). Write a formula for the volume Mt (in cubic meters) of the balloon after t seconds. This calculator can be used to calculate the lifting force of a volume with lower density than surrounding air. Because the balloon is in the shape of sphere, we can use the formula of volume of a sphere to find volume of air in the balloon. The volume formula for a sphere is 4/3 x π x (diameter / 2)3, where (diameter / 2) is the radius of the sphere (d = 2 x r), so another way to write it is 4/3 x π x radius3. How fast does the volume of a balloon change with respect to time? D. A spherical cap is a portion of a sphere that is separated from the rest of the sphere by a plane. Two concentric spheres have radii of 5" and 6". 8 The volume is about 38. , ball,spherical balloon. What is the volume formula for a
The following is multiple choice question (with options) to answer.
A volume of 10976 l water is in a container of sphere. How many hemisphere of volume 4l each will be required to transfer all the water into the small hemispheres? | [
"2812",
"8231",
"2734",
"2744"
] | D | a volume of 4l can be kept in 1 hemisphere
therefore, a volume of 10976l can be kept in (10976/4) hemispheres
ans.2744
ANSWER:D |
AQUA-RAT | AQUA-RAT-37364 | 27&305893372041&12,5,5,7,5,8\\ 28&801042337577&12,5,5,7,7,8\\ 29&2097687354880&12,5,7,7,7,8\\ 30&5493183075966&12,7,7,7,7,8\\ 31&14383060457018&12,7,7,7,7,10\\ 32&37658422859324&14,7,7,7,7,10\\ 33&98594676094434&14,7,7,9,7,10\\ 34&258133753770289&14,7,7,9,9,10\\ 35&675827901330148&14,7,9,9,9,10\\ 36&1769404155218244&14,9,9,9,9,10\\ 37&4632452165313827&16,9,9,9,9,10\\ \end{array}
The following is multiple choice question (with options) to answer.
576, 529, 484, 441, 400, ? | [
"324",
"361",
"289",
"256"
] | B | Explanation :
The pattern is 24^2, 23^2, 22^2, 21^2, 20^2, ...
So next number is 19^2 = 361
Answer : Option B |
AQUA-RAT | AQUA-RAT-37365 | ### Show Tags
23 Dec 2016, 09:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink]
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09 Nov 2017, 02:28
4
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Responding to a pm:
Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315
w1/w2 = (A2 - Aavg)/(Aavg - A1)
w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36
So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons
The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink]
### Show Tags
09 Nov 2017, 11:23
1
1
bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
The following is multiple choice question (with options) to answer.
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained? | [
"9:1",
"9:8",
"9:0",
"9:5"
] | A | Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Answer: Option A |
AQUA-RAT | AQUA-RAT-37366 | permutations around a round table with labelled seats
Two men, Adam and Charles, and two women, Beth and Diana, sit at a table where there are seven places for them to sit down. Two people are sitting next to each other if they occupy consecutive chairs. A non-trivial rotation defines a different seating arrangement, meaning that if all four people rotate their positions by moving k chairs to the right, it is the same way for them to be seated if and only if k divides 7.
Determine the number of ways that these four people can be seated so that every man is next to a woman and every woman is next to a man.
Answer is 224 . Here is the link with (source with explanation).
But my answer is 252. My calculations are given below.
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man any of the 3 far away seats(3)
seat remaining woman in adjacent seat(2)
so 7*2*2*3*2= 168
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the woman(1)
seat remaining woman in adjacent seat(2)
so 7*2*2*1*2= 56
place adam in any seat(7)
select any woman and place any of the adjacent seat (2*2)
place a man in adjacent seat to the man(1)
seat remaining woman in adjacent seat(1)
so 7*2*2*1*1= 28
So total = 168+ 56+28 = 252
Can someone help me to figure out whether any of this answer is right? If my answer is wrong, please help to understand why it has gone wrong and how correct answer can be reached.
The explanation given in the site derives the answer as 224 but it is in a different approach than mine. Is that correct?
The correct answer is $224$. Your calculations are almost fine
The following is multiple choice question (with options) to answer.
In how many ways can 5 Indians and 5 Englishmen be seated along a circle so that they are alternate? | [
"6!8!",
"4!5!",
"3!9!",
"8!5!"
] | B | 5 Indians can be seated along a circle in 4! ways
Now there are 5 places for 5 Englishmen
5 Englishmen can be seated in 5! ways
Required number = 4!5!
Answer is B |
AQUA-RAT | AQUA-RAT-37367 | This is an A.P. The below workout with step by step calculation shows how to find what is the sum of natural numbers or positive integers from 1 to 250 by applying arithmetic progression. is 56. If n is an integer, then n, n+1, and n+2 would be consecutive integers. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are . OPtion 1) 312750 2) 2502 62500 is a sum of number series by applying the values of input parameters in the formula. On each iteration, we add the number num to sum, which gives the total sum in the end. Write a program in java to print the sum of all numbers from 50 to 250(inclusive of 50 and 250) that are multiples of 3 and not divisible by 9. i.e. asked Jan 14 in Binomial theorem by Ritik01 (48.1k points) The sum of numbers from 250 to 1000 which are divisible by 3 is (A) 135657 (B) 136557 (C) 161575 (D) 156375. binomial theorem; jee; jee mains; Share It On Facebook Twitter Email. Here, we will not only tell you what the sum of integers from 1 to 300 is, but also show you how to calculate it fast. , 249, 250.The first term a = 1The common difference d = 1Total number of terms n = 250 Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … Then, ⇒ a n = 999 ⇒ a + (n – 1)d = 999 ⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250 ∴ Required sum … The number series 1, 2, 3, 4, . The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. + 249 + 250 = 31375 Therefore, 31375 is the sum of positive integers upto 250. I. DIVISIBILITY OF NUMBERS 1. (1). Sum of first n integers = n/2(n+1), in this case 25 x 51 = 1275. Next, it’s going to add those numbers
The following is multiple choice question (with options) to answer.
The sum of the first 50 positive even integers is 2,000. What is the sum of the odd integers from 101 to 200, inclusive? | [
" 5,050",
" 7,500",
" 8,050",
" 15,000"
] | C | 101+103+.......199
If we remove 100 from each of these it will be sum of 1st 100 odd numbers.
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......)
sum of 1st 100 natural numbers = (100 * 101) / 2 = 5050
Sum of 1st 50 positive even integers = 2000
sum of 1st 100 odd numbers = 5050 - 2000 = 3050
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......) = 5000 + 3050 =8050
C is the answer. |
AQUA-RAT | AQUA-RAT-37368 | 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.
The tax on a commodity is diminished by 20% but its consumption is increased by 10%. Find the decrease percent in the revenue derived from it? | [
"20%",
"18%",
"15%",
"12%"
] | D | Answer: Option D
100 * 100 = 10000
80 * 110 = 8800
10000------- 1200
100 ------- ? = 12% |
AQUA-RAT | AQUA-RAT-37369 | # 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.
Two kinds of Vodka are mixed in the ratio 1:2 and 2:1 and they are sold fetching the profit 40% and 20% respectively. If the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4/3 and 5/3 times respectively, then the mixture will fetch the profit of | [
"18%",
"20%",
"21%",
"33%"
] | D | Answer: D. |
AQUA-RAT | AQUA-RAT-37370 | A short computer code shows there are: 24,000 5-digit integers that are NOT divisible by all the elements in the set {1, 2, 3, 4, 5, 6}. They begin like this:
(10001 , 10003 , 10007 , 10009 , 10013 , 10019 , 10021 , 10027 , 10031 , 10033 , 10037 , 10039.......and end like this:
99953 , 99959 , 99961 , 99967 , 99971 , 99973 , 99977 , 99979 , 99983 , 99989 , 99991 , 99997)
Therefore, the probability is: 24,000 / 90,000 ==4 / 15
Feb 4, 2023
The following is multiple choice question (with options) to answer.
What is the total number of integers between 100 and 500(exclusive) that are divisible by 5? | [
"51",
"63",
"79",
"66"
] | C | 105, 110, 115, ..., 490,495
This is an equally spaced list; you can use the formula:
n = (largest - smallest) / ('space') + 1 = (495 - 105) / (5) + 1 = 78 + 1 = 79
Answer is C |
AQUA-RAT | AQUA-RAT-37371 | ## A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
##### This topic has expert replies
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### A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners
by Vincen » Sat Nov 27, 2021 4:38 am
00:00
A
B
C
D
E
## Global Stats
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
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### Re: A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the dine
by [email protected] » Sat Nov 27, 2021 7:31 am
00:00
A
B
C
D
E
## Global Stats
Vincen wrote:
Sat Nov 27, 2021 4:38 am
A number of people shared a meal, intending to divide the cost evenly among themselves. However, several of the diners left without paying. When the cost was divided evenly among the remaining diners, each remaining person paid $$\12$$ more than he or she would have if all diners had contributed equally. Was the total cost of the meal, in dollars, an integer?
(1) Four people left without paying.
(2) Ten people in total shared the meal.
Source: Veritas Prep
Target question: Was the total cost of the meal, in dollars, an integer?
This is a great candidate for rephrasing the target question
The following is multiple choice question (with options) to answer.
Rs.770 were divided among A,B,C in such a way that A had Rs.40 more than B and C had Rs 30 more than A . How much was B’s share? | [
"Rs.260",
"Rs.270",
"Rs.280",
"Rs.290"
] | D | Let B gets Rs x. Then We can say A gets Rs (x + 40 ) and C gets Rs ( x + 70) .
x + 40 + x + x + 70 = 770
3x = 660
x = 220 .
B’s share = Rs ( 220 + 70 ) = Rs.290
D |
AQUA-RAT | AQUA-RAT-37372 | Examveda
# In an institute, 60% of the students are boys and the rest are girls. Further 15% of the boys and 7.5% of the girls are getting a fee waiver. If the number of those getting a fee waiver is 90, find the total number of students getting 50% concessions if it is given that 50% of those not getting a fee waiver are eligible to get half fee concession?
A. 360
B. 280
C. 320
D. 330
E. 350
### Solution(By Examveda Team)
Let us assume there are 100 students in the institute.
Then, number of boys = 60
And, number of girls = 40
Further, 15% of boys get fee waiver = 9 boys
7.5% of girls get fee waiver = 3 girls
Total = 12 students who gets fee waiver
But, here given 90 students are getting fee waiver. So we compare
12 = 90
So, 1 = $$\frac{{90}}{{12}}$$ = 7.5
Now number of students who are not getting fee waiver = 51 boys and 37 girls
50% concession = 25.5 boys and 18.5 girls (i.e. total 44)
Hence, required students = 44 × 7.5 = 330
1. 60%*15%+40%*7.5%=12%
12%=90
1=750
750-90=660
50%= 330
2. let total students = x
then
(15/100*60/100*x)+(7.5/100*40/100*x)=90
900x+300x=90,0000
x=750
number of students who are not getting fee waiver=750-90=660
50% of those not getting a fee waiver are eligible=660/2=330
required students=330
Related Questions on Percentage
The following is multiple choice question (with options) to answer.
A student gets 60% in one subject, 70% in the other. To get an overall of 70% how much should get in third subject. | [
"75%",
"80%",
"45%",
"55%"
] | B | Let the 3rd subject % = x
60+70+x = 3*70
130+x =210
x = 210-130 = 80
Answer :B |
AQUA-RAT | AQUA-RAT-37373 | # How do i find Ways such that $1$ T in STATISTICS will be alone while other $2$ $Ts$ are together
Find the number of distinguishable ways the word "STATISTICS" can be arranged if only $$1$$ T will be alone while the other $$2$$ T will be together.
How do I solve this? Or does it need complex workings?
I Have done many practices on permutation and Combination. But all I see all the time is questions such as "Number of ways letter T come together" or Number of ways no $$2$$ Ts are together which also means T cannot be together.
Therefore, the question above is come out by my own self. How do we attempt it?
We set aside the $$T$$s for now and arrange the seven letters $$S, S, S, I, I, A, C$$. We can fill three of the seven positions with an $$S$$ in $$\binom{7}{3}$$ ways, fill two of the remaining four positions with an $$I$$ in $$\binom{4}{2}$$ ways, and fill the remaining two positions with an $$A$$ and a $$C$$ in $$2!$$ ways. Hence, there are $$\binom{7}{3}\binom{4}{2}2! = \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} = \frac{7!}{3!2!}$$ distinguishable arrangements of the letters $$S, S, S, I, I, A, C$$.
The following is multiple choice question (with options) to answer.
In how many ways can the letters L, I, M, I, T, S be arranged so that the two I's are not next to each other? | [
"36",
"48",
"240",
"296"
] | C | 1 L
2 I
1 M
1 T
1 S
Number of ways these letters can be arranged = 6!/2! (2! to account 2 same Is) = 360
Consider 2 Is as 1 entity and thus the number of arrangements for (II)LMTS = 5! = 120
Total allowed cases = 360-120 =240
C is the correct answer. |
AQUA-RAT | AQUA-RAT-37374 | java, optimization, finance
eloan=Integer.parseInt(loan.getString());
einterest=Float.parseFloat(interest.getString());
eyears=Integer.parseInt(years.getString());
months=0;
months=eyears*12;
double temp=(1.0d+(einterest/(12.0*100.0d)));
double original_temp=temp;
for(i=0;i<months;i++)
{
temp*=original_temp;
}
emi=(eloan*(einterest/(12.0*100.0d)))*(temp);
The following is multiple choice question (with options) to answer.
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest? | [
"3.46%",
"3.96%",
"3.49%",
"4.46%"
] | A | Explanation:
Let the original rate be R%. Then, new rate = (2R)%.
Note: Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.1/3 year(s).
(2175 + 725) R = 33.50 x 100 x 3
(2175 + 725) R = 10050
(2900)R = 10050
Original rate = 3.46%
Answer: A) 3.46% |
AQUA-RAT | AQUA-RAT-37375 | eq11 =d^2*b^2*a1+c^2*b^2*a2+a^2*c^2*a3+d^2*a^2*a4;
sol1=solve(eq1,eq2,eq3,eq4,eq5,eq6,eq9,eq10,'a1','a2','a3','a4','a5','a6','a7','a8') % one of the possible solutions
sol1=solve(eq1,eq2,eq3,eq4,eq5,eq8,eq9,eq10,'a1','a2','a3','a4','a5','a6','a7','a8') % another possible solution
sol1=solve(eq1,eq2,eq3,eq4,eq5,eq7,eq9,eq10,'a1','a2','a3','a4','a5','a6','a7','a8') % another possible solution
a9=1; % free variable
w1=sol1.a1;
w2=sol1.a2;
w3=sol1.a3;
w4=sol1.a4;
w5=sol1.a5;
w6=sol1.a6;
w7=sol1.a7;
w8=sol1.a8;
w9=a9;
% Test problem
syms x y
f=@(x,y) 1 +1*x +1*y +2*x*y +1*x^2 +1*y^2 +1*x^2*y+1*x*y^2+1*x^3+1*y^3+x^4*y;
f_x=diff(f,x);
f_xy=diff(f_x,y);
f_xy_00=subs(f_xy,[x y], [0 0]);
The following is multiple choice question (with options) to answer.
Solve the given two equations and find the inequality of X and Y. (a) x^2 = 729 and (b) Y= (729)^(1/2) | [
"If x>y",
"If x>=y",
"If x<y",
"If x<=y"
] | D | Solution:-
X=+27, -27 and Y= +27
Comparing we get X<=Y
D |
AQUA-RAT | AQUA-RAT-37376 | • Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
The following is multiple choice question (with options) to answer.
A box contains 25 mangoes out of which 5 are spoilt. If five mangoes are chosen at random, find the probability that A number is selected at random from first fifty natural numbers. What is the chance that it is a multiple of either 7 or 9? | [
"17/25",
"1/25",
"6/25",
"7/25"
] | C | The probability that the number is a multiple of 7 is 7/50. (Since 7*7= 49).
Similarly the probability that the number is a multiple of 9 is 5/50. {Since 9*5 =
45).
Neither 7 nor 9 has common multiple from 1 to 50. Hence these events are mutually exclusive events. Therefore chance that the selected number is a multiple of 7 or 9 is (7+5)/50 = 6/25.
ANSWER:C |
AQUA-RAT | AQUA-RAT-37377 | Lets assume like finding a pair for each man,
for the 1st guy -- can choose 1 from 5 women
for the 2nd guy -- can choose 1 from 4 women
.
.
for the 5th guy -- can choose 1 from 1 woman
so 5!
The following is multiple choice question (with options) to answer.
7 members have to be selected from 12 men and 3 women ,Such that no two women can come together .In how many ways we can select them ? | [
"12c6*3c1 + 12c7",
"12c6*3c1 + 2c7",
"1c6*3c1 + 12c7",
"11c6*3c1 + 12c7"
] | A | since no two women can come together,therfore women can be selected in 3c1 ways ...and men will be chosen in 12c6 so to complete the 7 group members...
=12c6*3c1
and only selection of men also possible = 12c7
so final ans : 12c6*3c1 + 12c7
ANSWER:A |
AQUA-RAT | AQUA-RAT-37378 | Since the sum of the ages of all 48 people must be equal to the sum of the ages of the 22 men plus the sum of the ages of the 26 women, we have
48(35) = 22(38) + 26x
1680 = 836 + 26x
26x = 844
x = 844/26
x = 32 12/26 ≈ 32.5
_________________
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A total of 22 men and 26 women were at a party, and the average [#permalink]
### Show Tags
04 May 2016, 09:23
Bunuel wrote:
A total of 22 men and 26 women were at a party, and the average (arithmetic mean) age of all of the adults at the party was exactly 35 years. If the average age of the men was exactly 38 years, which of the following was closest to the average age, in years, of the women?
(A) 31
(B) 31.5
(C) 32
(D) 32.5
(E) 33
Kudos for a correct solution.
Total age of men and women = 48*35 => 1,680
Total age of men is = 22*38 => 836
So, total age of women in = 1680 - 836 => 844
Average age of women is 844/26 => 32.46
Hence answer will be (D) 32.5
_________________
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Abhishek....
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Re: A total of 22 men and 26 women were at a party, and the average [#permalink]
The following is multiple choice question (with options) to answer.
In a class, the average age of 30 boys is 13 years and the average of 20 girls is 12 years. what is the average age of the whole class? | [
"12 Years",
"12.6 Years",
"13.6 Years",
"14 Years"
] | B | Total age of 50 students
(30X 13+20 X 12) = 630
Average = 630/50 = 12.6 Years \
B |
AQUA-RAT | AQUA-RAT-37379 | 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.
In what time will a railway train 60 m long moving at the rate of 36 kmph pass a telegraph post on its way? | [
"6 sec",
"9 sec",
"7 sec",
"11 sec"
] | A | Explanation:
T = 60/36 * 18/5 = 6 sec
A) |
AQUA-RAT | AQUA-RAT-37380 | # Math Help - Annual Compounding Interest
1. ## Annual Compounding Interest
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
2. Originally Posted by magentarita
Jane has $6 and Sarah has$8. Over the next few years, Jane invests her money at 11%. Sarah invests her money at 8%. When they have the same amount of money, how much will they have? Assume annual compounding interest, and round to the nearest cent.
a-10.50
b-They will never have the same amount of money
c-17.23
d-17.95
Thanks
It will take 10.5 years for them both to have the same amount. See soroban's conclusion for finding that amount. Here is my work.
$6(1+.11)^t=8(1+.08)^t$
$\log 6(1.11)^t=\log 8(1.08)^t$
Etc. and so forth to find that $t \approx 10.4997388$
3. Hello, magentarita!
I got a different result . . .
Jane has $6 and Sarah has$8. Over the next few years,
Jane invests her money at 11%. Sarah invests her money at 8%.
When they have the same amount of money, how much will they have?
Assume annual compounding interest, and round to the nearest cent.
$a)\;\10.50 \qquad b)\;\text{never equal}\qquad c)\;\17.23 \qquad d)\;\17.95$
At the end of $n$ years, Jane will have: . $6\left(1.11^n\right)$ dollars.
At the end of $n$ years, Sarah will have: . $8\left(1.08^n\right)$ dollars.
The following is multiple choice question (with options) to answer.
Lucy bought a 1-year,10000 certificate of deposit that paid interest at an annual rate of 8 percent compunded semiannually. What was the total amount of interest paid on this certificate at maturity ?
. | [
"10464",
"864",
"816",
"800"
] | D | Solution:
It says, 8 percent per annum per compounded semi annually.
Therefore r = 8/2 = 4% .
So, P = 10000
r = 4
n = 2
Net Amount A = 10816, I = 800.
ANSWER:D |
AQUA-RAT | AQUA-RAT-37381 | metres (m) and kilometres (km). The angle subtended at the center of the circle by the arc is called the central angle. The sector to the right is a fraction of the circle to the left so the the area of the sector is. - 1133971 2:15 75.0k LIKES. Dec 15,2020 - If the perimeter of a sector of a circle of radius 5.2 cm. Let the angle subtended by the arc of the sector at the centre of the circle (radius assumed to be $r$) be $\theta$. What multiple of the radius is the area of the sector?a)5thb)3rdc)4thd)2ndCorrect answer is option 'B'. (a) Write down, in terms of r and θ, expressions for P and A. A sector of a circle is like a slice of pizza or pie. Units Metric units. In the Given Figure, Radius of Circle is 3.4 Cm and Perimeter of Sector P-abc is 12.8 Cm . Thus the perimeter of the sector is r+θr+r. Problem 6 : Find the length of arc, if the perimeter of sector is 45 cm and radius is 10 cm. Or P = d + L ( arc) => P = 2r + {(β*2 pi r)/360°} , where β is sector angle. = π* 25 * 42 /180+2* 25. thus the perimeter of the sector is L+2r units. Sector of a circle ↺ Theta is an angle that can be defined as the figure formed by two rays meeting at a common endpoint. The most common sector of a circle is a semi-circle which represents half of a circle. The curved part of the sector is of the circumference of the circle but to find the perimeter of the sector we must add (the radius of the circle is) so the perimeter of the sector is. Perimeter of a Circle Sector A region bounded by two radii and an arc is called a sector of a circle. 15.8k SHARES. Click hereto get an answer to your question ️ The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Working with ratios (one question) Angles in quadrilateral (one question) We are learning about: The area and perimeter of a sector. If and the area of the
The following is multiple choice question (with options) to answer.
The 8 spokes of a custom circular bicycle wheel radiate from the central axle of the wheel and are arranged such that the sectors formed by adjacent spokes all have different central angles, which constitute an arithmetic series of numbers (that is, the difference between any angle and the next largest angle is constant). If the largest sector so formed has a central angle of 40°, what fraction of the wheel’s area is represented by the smallest sector? | [
"5/36",
"1/36",
"1/18",
"1/12"
] | A | Its an AP question .... it is given clearly in the question .
Let the smallest angle be a
and the circle has 8 sectors and hence 8 angle with a common difference d
hence all the angles can be written in AP form with Cd as d ,
a, a+d, a+2d, a+3d ,a+4d, a+5d, a+6d ,a+7d,
given that a+7d = 40 --------1
also
a + a+d + a+2d + a+3d +
a+4d + a+5d + a+6d + a+7d = 360 ( as sum of all the angle is 360)
which is 8a + 28d = 360 --------2
solving 1 and 2
we get a=50
We are almost done ,
now the question ask what fraction of the wheel’s area is represented by the smallest sector ?
(50/360)( pie r*r)/ (pie r*r) = 50/360= 5/36
A ans .... |
AQUA-RAT | AQUA-RAT-37382 | Author Message
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$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink] ### Show Tags 26 Nov 2010, 13:52 3 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 67% (02:54) correct 33% (02:25) wrong based on 199 sessions ### HideShow timer Statistics Source: Knewton$686,000 in bonus money is to be divided among 6 employees. No employee is to receive a bonus more than 20% greater than the bonus received by any other employee. What is the minimum possible bonus that an employee can receive?
(A) $96,000 (B)$97,000
(C) $98,000 (D)$99,000
(E) $100,000 [Reveal] Spoiler: OA _________________ I appreciate the kudos if you find this post helpful! +1 Manager Joined: 02 Apr 2010 Posts: 103 Followers: 5 Kudos [?]: 120 [0], given: 18 Re:$686,000 in bonus money is to be divided among 6 employees. No employe [#permalink]
### Show Tags
26 Nov 2010, 14:03
The question stem states that the difference between the minimum and maximum bonus may not exceed 20%. To determine the minimum possible bonus for an employee you have to assume that the other 5 employees obtain the maximum possible bonus.
If x denotes the minimum possible bonus and 1.2x denotes the maximum possible bonus you can set up the equation as follows:
The following is multiple choice question (with options) to answer.
In a company of 12 employees, 5 employees earn $36,000, 4 employees earn $45,000, and the 3 highest-paid employees earn the same amount. If the average annual salary for the 12 employees is $47,500, what is the annual salary for each of the highest-paid employees? | [
"$60,000",
"$65,000",
"$70,000",
"$75,000"
] | C | 5*36,000+4*45,000+3x=12*47,500
3x=570,000-180,000-180,000
3x=210,000
x=70,000
The answer is C. |
AQUA-RAT | AQUA-RAT-37383 | # 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.
If an article is sold at 20% profit instead of 14% profit, then the profit would be Rs. 180 more. What is the cost price? | [
"Rs.1000",
"Rs. 2000",
"Rs. 3000",
"Rs. 4000"
] | C | Explanation:
Let the cost price of an article be Rs. x.
(20% of x) - (14% of x) = 180
20x/100 - 14x/100 = 105 => 6x = 180 * 100
=> x = 3000
Cost price = Rs. 3000
Answer:C |
AQUA-RAT | AQUA-RAT-37384 | Difficulty:
65% (hard)
Question Stats:
53% (03:18) correct 47% (03:14) wrong based on 97 sessions
### HideShow timer Statistics
Question of the Week #7
Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours respectively, while R can empty the tank in 5 hours. P is opened at 10 am and Q is opened at 11 am, while R is kept open throughout. If the tank was initially empty, approximately at what earliest time it will be full if P or Q cannot be opened together and each of them needs to be kept closed for at least 15 minutes after they have been opened for 1 hour?
A. $$4:30 PM$$
B. $$6:00 PM$$
C. $$6: 30 PM$$
D. $$8:30 PM$$
E. $$9:30 PM$$
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The following is multiple choice question (with options) to answer.
Pipe P can fill a tank in 16 minutes and pipe Q cam empty it in 24 minutes. If both the pipes are opened together after how many minutes should pipe Q be closed, so that the tank is filled in 30 minutes? | [
"21",
"20",
"25",
"22"
] | A | 30/16-X/24==>1
X/24=30/16-1==14/16
X=14/16*24 = 21
ANSWER A |
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Re: At a certain university, the ratio of the number of teaching [#permalink]
### Show Tags
26 Jan 2016, 10:18
budablasta wrote:
At a certain restaurant, the ratio of the number of chefs to the number of burgers on any day must always be greater than 3:80. At this restaurant, what is the maximum number of burgers possible on a day that has 5 chefs.
A) 130
B) 131
C) 132
D) 133
E) 134
[EDIT] The same problem has been solved elsewhere:
problem-solving-question-on-ratios-79240.html
Sorry, I couldn't delete this post!
same for me
Please help. The phrase "must always be greater than" states it has to be 134 & not 133
what is the catch here?
I thought I was good at ratios!
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Re: At a certain university, the ratio of the number of teaching [#permalink]
### Show Tags
26 Jan 2016, 14:53
1
KUDOS
The question states that the ratio must always be greater than 3:80, not the number of students (or burgers). So when you calculate the ratio $$\frac{5}{x}>\frac{3}{80}$$, increasing the value of $$x$$ will decrease the ratio $$\frac{5}{x}$$, and decreasing the value of $$x$$ will increase the ratio $$\frac{5}{x}$$.
The following is multiple choice question (with options) to answer.
At a certain restaurant, the ratio of the number of cooks to the number of waiters is 3 to 10. When 12 more waiters are hired, the ratio of the number of cooks to the number of waiters changes to 3 to 14. How many cooks does the restaurant have? | [
"4",
"6",
"9",
"12"
] | C | Originally there were 3k cooks and 10k waiters.
14k = 10k+12
k=3
There are 9 cooks.
The answer is C. |
AQUA-RAT | AQUA-RAT-37386 | \;=\;\frac{9}{128}$ 6. Originally Posted by alexandrabel90 for part (b),Archie, what do you mean by adding the probability of the 2 players? does it mean (9 choose 7) x ( 0.5)^10 x 2? Yes, that was it, because you calculated the probability of only one out of the two winning on the tenth throw. Soroban has shown that really nicely. Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even on one's own throws while the other has more wins than losses or vice versa on their throws. In this situation, one could break even on one's own throws but make a profit if the other player does badly, or make a loss if the other player does well, "if" losing a toss loses a dollar and winning a toss wins a dollar. Or, if the two players lost every time, "player A" wins "player B"'s 6 dollars and "player B" wins "player A"'s 6 dollars, so they break even. They could also both win 1 and lose 5 each, win 2 and lose 4 each, win 4 and lose 2 each, win 5 and lose 1 each, win 6 and lose 0 each. That's the thing with probability questions, quite often open to alternative interpretations. The answer book will distinguish given it's answers. We've been interpreting this as..... If one player loses on a throw, then the second player wins the first player's dollar. In this case then, both would have to win 3 times and lose 3 times to end up with the 6 dollars they began with. It is probably safest to assume there is just 6 tosses rather than 6 each. 7. Originally Posted by alexandrabel90 2 gamblers bet$1 each on the successive tosses of a coin. each as $6. what is the probability that b. one player wins all the money on the tenth toss? Originally Posted by Archie Meade Yes, that was it, Plato has made an interesting point also. If we interpret "break-even" as not winning or not losing on one's own throws, then that is 3 wins and 3 losses, but one can break even
The following is multiple choice question (with options) to answer.
A gambler bought $3,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back? | [
"$2,040",
"$2,120",
"$1,960",
"$1,920"
] | B | In order to maximize the amount of money that the gambler kept, we should maximize # of $20 chips lost and minimize # of $100 chips lost, which means that # of $20 chips lost must be 2 more than # of $100 chips lost.
So, if # of $20 chips lost is x then # of $100 chips lost should be x-2. Now, given that total # of chips lost is 16: x+x-2=16 --> x=9: 9 $20 chips were lost and 9-2=7 $100 chips were lost.
Total worth of chips lost is 9*20+7*100=$880, so the gambler kept $3,000-$880=$2,120.
Answer: B. |
AQUA-RAT | AQUA-RAT-37387 | 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, B and C, each working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn $ 2340, what will be B's share of the earnings? | [
"$1100",
"$520",
"$780",
"$1170"
] | C | Explanatory Answer
A, B and C will share the amount of $2340 in the ratio of the amounts of work done by them.
As A takes 6 days to complete the job, if A works alone, A will be able to complete 1/6 th of the work in a day.
Similarly, B will complete 1/8th and C will complete 1/12th of the work.
So, the ratio of the work done by A : B : C when they work together will be equal to 1/6 : 1/8 : 1/12
Multiplying the numerator of all 3 fractions by 24, the LCM of 6, 8 and 12 will not change the relative values of the three values.
We get 24/6 : 24/8 : 24/12 = 4 : 3 : 2.
i.e., the ratio in which A : B : C will share $2340 will be 4 : 3 : 2.
Hence, B's share will be 3*2340/9 = 780
correct choice is (C) |
AQUA-RAT | AQUA-RAT-37388 | python, python-3.x, datetime
return next_lecture
def launch_timer(time):
call(["termdown", time.isoformat()])
if __name__ == "__main__":
next_lecture = get_next_lecture()
print(lecture)
# launch_timer(next_lecture) Your algorithm seems good, but the while loop, and lecture_hour and lecture_minute variables make your code a lot more complicated.
If we KISS then a simple algorithm is to just remove () from Lectures and iterate through it, since it is sorted.
The first lecture that is after the current time is the lecture we want.
This is nice and simple:
import datetime
LECTURES = [(0, 10, 15), (2, 12, 15), (3, 8, 15)]
def _get_next_lecture(now):
today = (now.weekday(), now.hour, now.minute)
for lecture in LECTURES:
if today < lecture:
return lecture
def get_next_lecture(now=None):
if now is None:
now = datetime.datetime.now()
day, hour, minute = _get_next_lecture(now)
return (
now.replace(hour=hour, minute=minute, second=0, microsecond=0)
+ datetime.timedelta(day - now.weekday())
)
From here we can see if the weekday is 4-6 then _get_next_lecture will return nothing and so will error.
This is easy to solve, we just return the first lecture with +7 days.
def _get_next_lecture(now):
today = (now.weekday(), now.hour, now.minute)
for lecture in LECTURES:
if today < lecture:
return lecture
day, hour, minute = LECTURES[0]
return day + 7, hour, minute
With only 3 lectures there's not much point in optimizing further. However if you have more, here is some food for thought:
You can use bisect to find where to insert into in \$O(\log n)\$ time.
The following is multiple choice question (with options) to answer.
The timing of a college is from 11 p.m to 4.20 p.m. Five lectures are held in the given duration and a break of 5 minutes after each lecture is given to the students. Find the duration of each lecture. | [
"56 minutes",
"52 minutes",
"30 minutes",
"48 minutes"
] | B | Explanation:
Total time a student spends in college = 5 hours 20 minutes = 280 minutes
As there are 5 lectures, the number of breaks between lectures is 4. Total time of the break = 20 minutes
Hence, the duration of each lecture is = (280 – 20)/5
= 52 minutes
ANSWER B |
AQUA-RAT | AQUA-RAT-37389 | ## Dipping cubes into paint
A little break from the complexity of dice sums to focus on painting cubes:
You submerge an unpainted 3x3x3 cube, made up of 27 1x1x1 cubes, into red paint and break it apart into the small individual cubes. Then you pick a small cube at random and roll it. What is the probability that you’ll end up with a red painted side on top?
Solution
This is a classic example of a problem that really only requires 30 seconds to solve, but at first sight seems much more complex. Most systematically-minded people will take the longer route:
The small 1x1x1 cubes will either have no sides painted (the one in the middle of the 3×3 cube, aka interior piece), one side painted (the center of each face, aka face piece), two sides painted (the middle cube of each edge, aka edge piece), or three sides painted (the corners). There are 27 cubes total, and there is 1 center piece, 6 faces, 12 edges, and 8 vertices. The probability of selecting an interior piece is thus $\frac{1}{27}$, and $\frac{6}{27}$ for selecting a face piece, and so on. With $n$ sides painted, the probability of rolling a painted side up is $n/6$ because a cube has 6 sides. Therefore our total probability of rolling a painted face up is:
$\displaystyle \underbrace{\frac{1}{27}\left(\frac{0}{6}\right)}_\text{interior piece}+\underbrace{\frac{6}{27}\left(\frac{1}{6}\right)}_\text{face piece}+\underbrace{\frac{12}{27}\left(\frac{2}{6}\right)}_\text{edge piece}+\underbrace{\frac{8}{27}\left(\frac{3}{6}\right)}_\text{corner}=\frac{6+24+24}{162}=\boxed{\frac{1}{3}}$.
The following is multiple choice question (with options) to answer.
Two identical cubes if one of them is painted pink on its 4 side and blue on the remaining two side then how many faces painted pink to other cube so that probability of getting cube is 1/3 when we roll both the cube. | [
"0",
"1",
"2",
"3"
] | A | 0 faces should be painted pink i.e all faces blue
ANSWER:A |
AQUA-RAT | AQUA-RAT-37390 | Sort by:
Solution for Q-3:
Using binomial theorem, one can note that,
$P(x)=(x-3)^4-10x+15$
We have, using Remainder-Factor Theorem and the fact that $a,b,c,d$ are roots of $P(x)$ that,
$P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}$
Using the last result, we have the sum as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)$
Using Vieta's formulas, we have,
$\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120$
Hence, the required sum evaluates as,
$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}$
- 5 years, 11 months ago
Nicely done.
Staff - 5 years, 11 months ago
Your hint did most of the work. So, the credit actually goes to you. :)
- 5 years, 11 months ago
$Q-1$
The following is multiple choice question (with options) to answer.
if p/q=3/4
then3p+9q=? | [
"25/4",
"25/2",
"20/4",
"45/4"
] | D | 3p+9q=?
Divided by q,
3(p/q)+9=x
3*(3/4)+9=45/4
ANSWER:D |
AQUA-RAT | AQUA-RAT-37391 | 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.
In what time will a railway train 70 m long moving at the rate of 36 kmph pass a telegraph post on its way? | [
"8 sec",
"1 sec",
"9 sec",
"7 sec"
] | D | T = 70/36 * 18/5
= 7 sec
Answer:D |
AQUA-RAT | AQUA-RAT-37392 | (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.
Sonika bought a V.C.R. at the list price of 18,600. If the rate of sales tax was 8%, find the amount she had to pay for purchasing the V.C.R. | [
"20,088",
"19,780",
"19,680",
"19,380"
] | A | Sol. List price of V.C.R. =18,600
Rate of sales tax = 8%
∴ Sales tax = 8% of 18,600
= 8⁄100 × 18600 = 1488
So, total amount which Sonika had to pay for purchasing the V.C.R. = 18,600 + 1488
= 20,088. Answer A |
AQUA-RAT | AQUA-RAT-37393 | 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.
A number when divided by 44, gives 432 as quotient and 0 as remainder. What will be the remainder when dividing the same number by 31 | [
"5",
"3",
"4",
"6"
] | A | Explanation:
P ÷ 44 = 432
=> P = 432 * 44 = 19008
P / 31 = 19008 / 31 = 613, remainder = 5
Option A |
AQUA-RAT | AQUA-RAT-37394 | ## A committee of 2 people is to be selected out of
##### This topic has expert replies
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### A committee of 2 people is to be selected out of
by VJesus12 » Thu Mar 15, 2018 4:23 am
A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
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by [email protected] » Thu Mar 15, 2018 5:37 am
VJesus12 wrote:A committee of 2 people is to be selected out of 3 teachers and 4 preachers. If the committee is selected at random, what is the probability that the committee will be composed of at least 1 preacher?
A. 1/4
B. 1/3
C. 2/3
D. 6/7
E. 8/9
The OA is D.
Should I use combinations here? Or probability? I am confused. <i class="em em-confused"></i>
Yes! (You can think of probability as a ratio of combinations or permutations.)
Useful equation P(x) = 1 - P(not x)
P( at least 1 preacher) = 1 - P(no preachers)
The following is multiple choice question (with options) to answer.
In a certain group of 13 members, 4 members teach only French and the rest teach only Spanish or German. If the group is to choose 3-member committee, which must have at least 1 member who teaches French, how many different committee can be chosen ? | [
"198",
"154",
"202",
"201"
] | C | Case 1
1french 2either German or Spanish: 4C1 * 9C2 = 144
Case 2
2french 1either German or Spanish: 4C2 * 9C1 = 54
Case 3
3french 4C3 = 4
Answer is C |
AQUA-RAT | AQUA-RAT-37395 | 1. Let total distance=X.
1st distance =30x/100.Speed=20kmph
2nd distance=60x/100 Speed=40kmph
3rd distance=10x/100Speed=10kmph
T1=1/20*30x/100 = 3x/200
T2=1/40*60x/100 =3x/200
T3=1/10*10x/100 =X/100
Applying S=D/T Formula
30x/100+60x/100+10x/100
______________________________
3x/200+ 3x/200+ X/100.
=100x*200/8x*100
=25
2. Thank you so much
3. excellent
4. Palal,post your calculation so we can rectify. It may consist a calculation error. Otherwise, answer would be same.
5. if we have taken x instead of 100, why answer doesnt comes the same way
Related Questions on Speed Time and Distance
The following is multiple choice question (with options) to answer.
Car A runs at the speed of 58km/hr & reaches its destination in 8hr. Car B runs at the speed of 70 km/h & reaches its destination in 4h. What is the respective ratio of distances covered by Car A & Car B? | [
"11 : 6",
"12 : 7",
"29 : 18",
"15 : 6"
] | C | Sol. Distance travelled by Car A = 58 × 8 = 464 km
Distance travelled by Car B = 70 × 4 = 280 km
Ratio = 464/280 = 29 : 18
C |
AQUA-RAT | AQUA-RAT-37396 | (A) 1
(B) 2
(C) 4
(D) 6
(E) 8
11. What is the area of the shaded region of the given 8 X 5 rectangle?
The following is multiple choice question (with options) to answer.
A rectangle has a perimeter of 176 inches. The length of the rectangle is 8 inches more than its width. What is the area of the rectangle? | [
"1,408 square inches",
"1,920 square inches",
"1,936 square inches",
"2,304 square inches"
] | B | Explanation:
We have: (l - b) = 8 and 2(l + b) = 176 or (l + b) = 88.
Solving the two equations, we get: l = 48 and b = 40.
Area = (l x b) = (48 x 40) in2 = 1920 in2.
Correct answer: B |
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