source string | id string | question string | options list | answer string | reasoning string |
|---|---|---|---|---|---|
AQUA-RAT | AQUA-RAT-39797 | ## Solution 4
Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$.
Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$. Dividing everything by $b^2$, we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$.
Applying the quadratic formula....and following the restriction that $a>b>0$....
$\frac{a}{b}=\frac{10}{7}$.
Hence, $7a=10b$.
Since they are relatively prime, $a=10$, $b=7$.
$10-7=\boxed{\textbf{(C)}\ 3}$.
## Solution 5
Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{\textbf{(C)}\ 3}$ ~mathboy282
The following is multiple choice question (with options) to answer.
If a = 105 and a^3= 21 * 35 * 45 * b, what is the value of b? | [
"35",
"42",
"45",
"49"
] | A | First step will be to break down all the numbers into their prime factors.
105 = 3 * 5 * 7
21 = 7 * 3
35 = 7 * 5
45 = 3 * 3 * 5
so, (105)^3 = 3 * 7 * 7 * 5 * 3 *3 *5 *b
Therefore (3 * 5 * 7) ^ 3 = 3^3 * 5^2 * 7^2 *b
Therefore, b = 3^3 * 5^3 * 7 ^3/3 ^3 * 5 ^2 * 7^2
b = 5*7 = 35
Correct answer A. |
AQUA-RAT | AQUA-RAT-39798 | ## 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.
Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 minutes. In how much time, the third pipe alone can empty the cistern? | [
"90 min",
"100 min",
"110 min",
"120 min"
] | C | Solution
Work done by the third pipe in 1 min. = 1/50 - (1/60+1/75)
= (1/50 - 3/100)
= -1/100.
Therefore, the third pipe alone can empty the cistern in 100 min.
Answer C |
AQUA-RAT | AQUA-RAT-39799 | (495, 1), (496, 1), (496, 3), (494, 3), (494, 1), (495, 1), (495, 2), (493, 2), (493, 1), (494, 1), (494, 3), (493, 3), (493, 2)$$ Here is an illustration of this cycle:
The following is multiple choice question (with options) to answer.
A shop has 3000 cycles. It sells 10% of the cycles daily. How many cycles remain after 2 days? | [
"2500",
"2600",
"2450",
"2430"
] | D | Formula :
( After =100 denominator
Ago = 100 numerator)
3000 × 90/100 × 90/100 = 2430
D |
AQUA-RAT | AQUA-RAT-39800 | # 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.
By selling 33 metres of wire, a shopkeeper gains the price of 11 metres of wire. His gain per cent is | [
"7%",
"50%",
"20%",
"22%"
] | B | Suppose S.P of 33 meters of cloth = Rs 33
Therefore, Gain = Rs 11 => C.P = Rs 22
Gain % = 11/22 x 100 = 50%
Gain = 50%
ANSWER:B |
AQUA-RAT | AQUA-RAT-39801 | (1) The first $$n$$ people donated $$\dfrac1{16}$$ of the total amount donated.
(2) The total amount donated was $$\120,000.$$
Source: GMAT Prep
Target question: What was the value of n?
When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...
Statement 2: The total amount donated was \$120,000
Let's summarize the given information....
First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round
Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round
TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars
So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
The following is multiple choice question (with options) to answer.
For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity. In December, the worker earned N times his usual monthly salary and donated one-fourth of his earnings to charity. If the worker's charitable contributions totaled one-eighth of his earnings for the entire year, what is the value of N? | [
"8/5",
"5/2",
"3",
"22/10"
] | D | Let monthly salary for each of the 11 months except December was x, then
11x*1/10 + Nx*1/4 = 1/8(11x + Nx);
11/10 + N/4 = 1/8(11 + N) => 44+10N/40= 11+N/8
352+80N = 440+40N => 40N = 88
N =88/40 =22/10
Answer: D. |
AQUA-RAT | AQUA-RAT-39802 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours.What will be the true time when the clock indicates 10 p.m. on 4th day? | [
"11",
"66",
"88",
"55"
] | A | Time from 5 am. on a day to 10 pm. on 4th day = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
356/15 hrs of this clock = 24 hours of correct clock
89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
= 90 hrs of correct clock.
So, the correct time is 11 p.m.
Answer: A) 11pm |
AQUA-RAT | AQUA-RAT-39803 | The probabilities I got for parking between 4 and 22 cars are:
As it can be seen, there is a 'high' probability that the number of cars is going to move in the range $$(6,15)$$ (more or less). But as the question is for the most likely number, this number is $$10$$.
• Edited - replaced text $>=$ with TeX $\ge$ (\ge). – CiaPan Oct 9 '14 at 9:44
Joannes' answer shows that that if there are $N$ parking spaces, the probability that exactly $n$ vehicles get parked is $$p_N(n)=\frac{(N-n)!}{N!}\left[(N-n+1)^n-(N-n)^n\right].$$ I'm going to analyze where this is maximized.
The following is multiple choice question (with options) to answer.
A parking garage rents parking spaces for $15 per week or $30 per month. How much does a person save in a year by renting by the month rather than by the week? | [
" $140",
" $160",
" $220",
" $420"
] | D | 10$ per week!
An year has 52 weeks.
Annual charges per year = 52* 15 = 780$
30$ per month!
An year has 12 months.
Annual charges per year = 12 * 30 = 360$
780 - 360 = 420
Ans D |
AQUA-RAT | AQUA-RAT-39804 | from (*) we have n= {(nth term-first term)/common difference}+1.....
For our problem, n= {(11-1)}/2}+1 =6
here common difference is 2 as an odd number occurs by adding 2 to the previous odd number etc.,
If you use the formula $(b-a)/2 + 1$, then for some case it would give wrong answer.
Say, $a = 2$ and $b = 4$, so there is only one ODD nmumber between 2 and 4 (inclusive), and that is 3.
By using the formula, we get $(4-2)/2 + 1 = 2$, that is wrong; if we consider a small alter here, we will calculate the difference like this:
$Diff = (b - a) /2 + X$
Here $X$ will be either 0 or 1 depending on whether any one of the variables (a or b) is odd or not. If $a \mod 2 = 1 \lor b \mod 2 =1$, then $X$ will be 1, otherwise 0.
To get count of odd or even numbers between a range, follow the process as below:
Correct the Range to start and end with inclusive numbers as per question and then use following formula :
(m - n)/2 + 1 where m is greater than n
Example:
All Odd numbers between 21 - 61
correct the range to make it inclusive of the numbers which make the range to 23 - 59 use the formula:
(59 - 23)/2 + 1 => 19
All even numbers between 21 - 61
correct the range to make it inclusive of the numbers which make the range to 22 - 60 use the formula:
(60 - 22)/2 + 1 => 20
You can use this formula which gives you the number of integers congruent to $${n}\pmod p$$ in the interval $$a$$ inclusive and $$b$$ inclusive :
$$S =\lfloor\frac{n-a}{p}\rfloor+\lfloor\frac{b-n}{p}\rfloor+1$$
The following is multiple choice question (with options) to answer.
The difference of two numbers is 11 and one fifth of their sum is 9. The numbers are : | [
"31, 20",
"30, 19",
"29, 18",
"28, 17"
] | D | x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28
Answer: D |
AQUA-RAT | AQUA-RAT-39805 | # Difference between revisions of "2019 AMC 10A Problems/Problem 23"
## Problem
Travis has to babysit the terrible Thompson triplets. Knowing that they love big numbers, Travis devises a counting game for them. First Tadd will say the number $1$, then Todd must say the next two numbers ($2$ and $3$), then Tucker must say the next three numbers ($4$, $5$, $6$), then Tadd must say the next four numbers ($7$, $8$, $9$, $10$), and the process continues to rotate through the three children in order, each saying one more number than the previous child did, until the number $10,000$ is reached. What is the $2019$th number said by Tadd?
$\textbf{(A)}\ 5743 \qquad\textbf{(B)}\ 5885 \qquad\textbf{(C)}\ 5979 \qquad\textbf{(D)}\ 6001 \qquad\textbf{(E)}\ 6011$
## Solution 1
Define a round as one complete rotation through each of the three children, and define a turn as the portion when one child says his numbers (similar to how a game is played).
We create a table to keep track of what numbers each child says for each round.
$\begin{tabular}{||c c c c||} \hline Round & Tadd & Todd & Tucker \\ [0.5ex] \hline\hline 1 & 1 & 2-3 & 4-6 \\ \hline 2 & 7-10 & 11-15 & 16-21 \\ \hline 3 & 22-28 & 29-36 & 37-45 \\ \hline 4 & 46-55 & 56-66 & 67-78 \\ [1ex] \hline \end{tabular}$
The following is multiple choice question (with options) to answer.
Henry eats X scones in X percent of the time it takes Rachel to eat Y scones. If Rachel eats four scones in ten minutes, then the number of minutes it takes Henry to eat 12 scones must be equal to which of the following? | [
"Y/5",
"3Y/10",
"100Y/(15X)",
"XY/250"
] | B | Rachel eats 4 scones/10 minutes = 2/5 scones per minute
The time for Rachel to eat 1 scone is 5/2 minutes.
The time for Rachel to eat Y scones is 5Y/2 minutes.
The time for Henry to eat X scones is 5YX/200 minutes.
The time for Henry to eat 1 scone is 5Y/200 minutes.
The time for Henry to eat 12 scones is 60Y/200 = 3Y/10 minutes.
The answer is B. |
AQUA-RAT | AQUA-RAT-39806 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
The length of a rectangular field is 7/5 its width. If the perimeter of the field is 384 meters, what is the width of the field? | [
"50",
"60",
"11",
"80"
] | D | Let L be the length and W be the width. L = (7/5)W
Perimeter: 2L + 2W = 384, 2(7/5)W + 2W = 384
Solve the above equation to find: W = 80 m and L = 112 m.
correct answer D)80 |
AQUA-RAT | AQUA-RAT-39807 | in n yrs and A 2 in (n+1) yrs, then Rate of compound interest =(A 2 - A 1)/A 1 *100% Sum = A 1 (A 1 /A 2) n. 596 APPENDIXC:COMPOUNDINTERESTTABLES 1/2% CompoundInterestFactors 1/2% SinglePayment UniformPaymentSeries ArithmeticGradient Compound Present Sinking Capital Compound Present Gradient Gradient Amount Worth Fund Recovery Amount Worth Uniform Present Factor Factor Factor Factor Factor Factor Series Worth Find F Find P Find A Find A Find F Find P. Compound Interest CBSE TEST PAPER: Maths for Class VIII (8th) 1. Compound interest − a phenomenon that you want to get cozy with − can be a difficult thing for your child to get. 2 : Nov 20, 2013, 9:14 AM: Pete Esser: Ċ: 04 Interest Bearing Bank Accounts and Applications. To make it plain for students to understand, I explain that it is an amount that is accrued over a certain amount of time. , compounded monthly. A savings account compounds its interest quarterly at a rate of 8%. SSC CGL & CHSL Previous Year Complete Paper with Solution Provide Only at Our Website. 747302 periods is 15. 5% interest compounded annually when you were born. To register Maths Tuitions on Vedantu. How much will the gift be wirth in 17 years, if it in invested at 7% compounded quarterly? 2) A bank is offering a CD that. 5 3 Growth of 1. How long would it take for an investment of$3,500 to become $4,200 if it is invested in an account that earns 6% compounded monthly? Since, in this problem, the variable is in the exponent, logarithms will be used to solve it. If$3000 is borrowed at a rate of 12% interest per year, flnd the amount due at the end of 5 years if the interest is compounded continuously. Straightforward amounts of money and interest rates for 2 to 4 years. It is basically earning “ interest on interest “. This addition of interest to the principal is called compounding. This calculator demonstrates how compounding can affect your savings, and how interest on your interest really adds up!. In Coordinate Algebra, you worked with the Compound Interest Formula nt n r A P(1 ) where A = the amount of money
The following is multiple choice question (with options) to answer.
The simple interest on Rs.12000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for five years at the same rate of interest. | [
"3052.82",
"3052.89",
"3052.85",
"3052.8"
] | D | R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]n - 1}
= 12000 { [ 1 + 12 / 100]2 - 1} = Rs.3052.80
Answer:D
Answer:D |
AQUA-RAT | AQUA-RAT-39808 | => 18x/4=154
x=308/9
TOTAL WAGES PAID WILL BE
(12/11 + 18/7)*309/9
PLEASE TELL ME WHERE I AM WRONG
Lets try -
SW4 wrote:
Elana was working to code protocols for computer processing. She did 11/18 of the job and allowed Andy to finish it. They both work at the same rate and receive the same hourly pay. If the difference between the amounts they were paid was $154, what was the total amount the two were paid for the entire coding job? Let total work be 1 Elana did 11/18 Andy did 7/18 Thus share of wages will be in the ratio of work done by them.... Ratio of their work is as follows - Elana : Andy = 11 : 7 and the total work is 18 Proportion of difference in work = Proportion of Difference in pay So, 4 =$ 154
Or, 1 = $154/4 And Total pay = 154/4*18 =>$ 693
The following is multiple choice question (with options) to answer.
P and Q started a business with respective investments of Rs. 4 lakhs and Rs. 10 lakhs. As P runs the business, his salary is Rs. 5000 per month. If they earned a profit of Rs. 2 lakhs at the end of the year, then find the ratio of their earnings? | [
"1 : 8",
"1 : 1",
"1 : 6",
"1 : 5"
] | D | Ratio of investments of P and Q is 2 : 5
Total salary claimed by P = 12 * 5000 = Rs. 60000
Total profit = Rs. 2 lakhs.
Profit is to be shared = Rs. 140000
Share of P = (2/7) * 140000 = Rs. 400000
Share of Q = Rs. 100000
Total earnings of P = (60000 + 40000) = Rs. 100000
Ratio of their earnings = 1 : 1.Answer: D |
AQUA-RAT | AQUA-RAT-39809 | n3 – n is divisible by 6. What Are the Probability Outcomes for Rolling Three Dice? Look Up Math Definitions With This Handy Glossary. So the least possible sum of their birth years is 2002 + 2001 + 2005 = 6008. Solution for Determine whether the statement is true or false. These are consecutive odd integers. Prove that if m, m +1, m + 2 are three consecutive integers, one of them is divisible by 3 4. Hint: What are the possible remainders when we divide an integer by 3? - 15053493. CS103X: Discrete Structures Homework Assignment 2: Solutions Due February 1, 2008 Exercise 1 (10 Points). If n is divisible by 3 then n+1 and n+2 cannot be divisible by 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. now, similarly, when a no. CHAPTER 2: NUMBERS AND SEQUENCES. In order to test this, you must take the last digit of the number Since 28 is divisible by 7, we can now say for certain that 364 is also divisible by 7. Fact tor n -n completely. Let 3 consecutive positive integers be p, p + 1 and p + 2. Show Step-by-step Solutions Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. for any integer $n$: [. (N + 1)], which means that exactly one element is missing. (3) The sum of three consecutive even integers is 528; find the integers. Any group of 3 consecutive numbers will have one number that is a multiple of 3 and at least one number that is a multiple of 2. but n and n+1 are not divisible by 3. When spoken in everyday English, a fraction describes how many parts of a certain size there are, for example, one-half, eight-fifths, three-quarters. 3
The following is multiple choice question (with options) to answer.
How many numbers from 10 to 50 are exactly divisible by 3. | [
"13",
"14",
"16",
"17"
] | A | 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,48.
13 Numbers.
10/3 = 3 and 50/3 = 16 ==> 16 - 3 = 13. Therefore 13 digits.
A) |
AQUA-RAT | AQUA-RAT-39810 | Since the number of ways in which $$5$$ books can be arranged is $$5!=120,$$ we have $$120$$ ways.
• thanks for explanation. Can you point out what is wrong in my analysis Sep 17, 2020 at 11:24
• As mentioned in the comments, you calculated the $C$ books to be next to each other, but they do not need to be next to each other. Sep 17, 2020 at 11:26
The following is multiple choice question (with options) to answer.
If 102 books need to be placed in boxes and each box can contain at most 9 books. How many books will be in the last unfilled box? | [
"12 books",
"10 books",
"0 books",
"3 books"
] | D | The number of books that can be placed in boxes of 9 books is the first number that is divisible by 9 that occurs before 102 .
In order to divide the sum in 9 parts, the amount must be divisible by 9
Divisibility rule of 9: The sum of the digits must be divisible by 9
Sum of digits of 102 =3 and 0 is divisible by 9.
Hence, we need to remove 3 to this number for it to be divisible by 9
Correct Option:D |
AQUA-RAT | AQUA-RAT-39811 | bruce
the jet will fly 9.68 hours to cover either distance
bruce
Riley is planning to plant a lawn in his yard. He will need 9 pounds of grass seed. He wants to mix Bermuda seed that costs $4.80 per pound with Fescue seed that costs$3.50 per pound. How much of each seed should he buy so that the overall cost will be $4.02 per pound? Vonna Reply 33.336 Robinson Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that cost $20 per square foot. How many square feet of each tile should she use so that the overall cost of the backsplash will be$10 per square foot?
Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins of each type does he have? mikayla Reply 2q=n ((2q).05) + ((q).25) = 8.75 .1q + .25q = 8.75 .35q = 8.75 q = 25 quarters 2(q) 2 (25) = 50 nickles Answer check 25 x .25 = 6.25 50 x .05 = 2.50 6.25 + 2.50 = 8.75 Melissa John has$175 in $5 and$10 bills in his drawer. The number of $5 bills is three times the number of$10 bills. How many of each are in the drawer?
7-$10 21-$5
Robert
Enrique borrowed $23,500 to buy a car. He pays his uncle 2% interest on the$4,500 he borrowed from him, and he pays the bank 11.5% interest on the rest. What average interest rate does he pay on the total \$23,500? (Round your answer to the nearest tenth of a percent.)
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hour longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
8mph
michele
16mph
Robert
3.8 mph
Ped
16 goes into 80 5times while 20 goes into 80 4times and is 4mph faster
Robert
The following is multiple choice question (with options) to answer.
The cost of paint is 60 per kilograme. A kilogram paint covers 20 square feet. How much will it cost to paint the outside of a cube having each side 10 feet? | [
"3000",
"900",
"1800",
"360"
] | C | Area of cube
= 6× (side)2 = 6 × 10 × 10 = 600 square feet.
Cost to paint outside of the cube = 600⁄20 × 60
= 1800
Answer C |
AQUA-RAT | AQUA-RAT-39812 | # Given relatively prime positive integers $a,b>1$, how many positive integers are there which are not non-negative integer combination sum of $a,b$?
Let $a,b>1$ be relatively prime positive integers. I know that $ab-a-b$ is the largest positive integer which can not be written as a sum of non-negative integer combination of $a,b$. My question is : can we explicitly determine how many positive integers are there which can not be written as a non-negative integer combination of $a,b$ ?
The following is multiple choice question (with options) to answer.
What is the total number of positive integers that are less than 400 and that have no positive factor in common with 400 other than 1? | [
"100",
"120",
"150",
"160"
] | D | Since 400=2^4*5^2 then a number can not have 2 and/or 5 as a factor.
The odd numbers do not have 2 as a factor and there are 200 odd numbers from 1 to 400.
We then need to eliminate the 40 numbers that end with 5, that is 5, 15, 25,...,395.
There are a total of 200 - 40 = 160 such numbers between 1 and 400.
The answer is D. |
AQUA-RAT | AQUA-RAT-39813 | 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.
Raju age after 15 years will be 5 times his age 5 years back, What is the present age of Raju | [
"15",
"14",
"10",
"8"
] | C | Explanation:
Clearly,
x+15 = 5(x-5)
<=> 4x = 40 => x = 10
Answer: Option C |
AQUA-RAT | AQUA-RAT-39814 | # 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.
Somesh bought a microwave oven and paid 10% less than its original price. He sold it at 30% profit on the price he had paid. What percentage of profit did Somesh earn on the original price? | [
"32%",
"11%",
"20%",
"17%"
] | D | Quicker Method:
% profit which Somesh gets
= –10 +30 – 10×30/100 = +17%
Answer D |
AQUA-RAT | AQUA-RAT-39815 | Assuming the girls have sat down, they leave 3 gaps between them. 1 for each boy. Thus the first boy can pick between 3 chairs, the second boy 2 chairs, and the third doesn't get to pick. So there are $3\cdot 2=3!=6$ ways the boys can sit. Now the girls are a little bit more tricky. Notice that it isn't specified how the girls are to be divided among the groups, thus the first girl can pick among 10 spots, the next 9 and so on. Finally we have to account for the ways the 4 groups can be arranged, which by the binomialcoefficient is equal to$\frac{4!}{2!2!}$.
Hence your final answer is $$3!\cdot 10!\cdot \frac{4!}{2!2!}$$
Since there are $4$ groups of girls, and three boys, there is only one case possible for boys to sit between the groups.
Boys can be arranged in $3!$ ways in their seats, the groups of girls can be arranged in $\frac{4!}{2! 2!}$ ways. For any such arrangement, girls can be rearranged in $10!$ ways.
So the answer should be: $$3!\cdot\frac{4!}{2!2!} \cdot10!$$
For $10$ girls we have $10!$ permutations. We have $3!$ for boys. We just put the boys in the right positions. Thus, the result is $10! \times 3!$.
The following is multiple choice question (with options) to answer.
In how many ways can 3 boys and 3 girls be arranged at a table if 3 girls are never together? | [
"96",
"92",
"82",
"84"
] | D | Total ways of arranging 6 person in a circle = 5! = 120 ways
If we consider three women as one, then we have 3+1=4 persons. Arranging these four persons in circle will take = 3! = 6 ways
And two women can arrange themselves in 3! ways
Total ways in which three women sit together in circle = 3!*3! = 36 ways
Total ways in which three women doesn’t sit together in circle = 5! - 3!*3! = 120 - 36 = 84 ways
D |
AQUA-RAT | AQUA-RAT-39816 | # Find the probability that each person exits an elevator on a different floor.
I have a question and I am working on to study for a test. I have a hard time deciding when to use certain techniques.
Suppose you have $5$ people on an elevator that stops at $5$ floors. Each person has an equal probability of going to any one floor. Find the probability that they all get off on different floors.
I would want to solve this by saying there are five possible floors to get off on and we want one person to get off on each. Thus: $$\frac{1^5}{5^5}$$
however the answer is $\frac{5!}{5^5}$. I do not see exactly why they got this but I assume it was solved as:
$$\frac{5}{5}\frac{4}{5}\frac{3}{5}\frac{2}{5}\frac{1}{5} = \frac{5!}{5^5}$$
Why is it done like this versus the first way which is the way I thought about solving this.
• Not following your calculation. Person $A$ can get off anywhere. Person $B$ has $4$ choices, and so on. – lulu Sep 30 '17 at 21:03
• What you calculated is the probability that all five people get off on a particular floor. – N. F. Taussig Sep 30 '17 at 21:04
• @N.F.Taussig or perhaps that Person A gets off on the first floor, Person B gets off on the second floor, ... – Henry Oct 1 '17 at 8:39
• @Henry Well spotted. In any case, the point is that Derek did not account for the number of ways five people could exit the elevator on five different floors. – N. F. Taussig Oct 1 '17 at 8:41
The following is multiple choice question (with options) to answer.
Seven people are on an elevator that stops at exactly 7 floors. What is the probability that exactly one person will push the button for each floor? | [
"7!/7^7",
"7^7/7!",
"7/7!",
"7/7^7"
] | A | Each person out of 7 has 7 options, hence total # of outcomes is 7^7;
Favorable outcomes will be 7!, which is # of ways to assign 7 different buttons to 7 people
So basically # of arrangements of 7 distinct objects: 7!.
P=favorable/total=7!/7^7
Answer: A. |
AQUA-RAT | AQUA-RAT-39817 | Hello Matty R!
No, that doesn't mean anything, does it?
Hint: what will Bea's age be when Claire is as old as Dawn is now?
3. Feb 27, 2010
### HallsofIvy
Staff Emeritus
"When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
Claire is older than Bea."
Claire will be as old as Dawn is now in d- c years. Bea's age then will be b+ (d- c) and that will be twice Ann's current age: b+ d- c= 2a or 2a- b+ c- d= 0.
You have four equations:
The sum of their ages is exactly 100 years.
a+ b+ c+ d= 100
The sum of Ann's and Dawn's ages is the same as the sum of Bea's and Claire's.
a- b- c+ d= 0
The difference between the ages of Claire and Bea is twice Ann's age.
2a+ b- c= 0
("Claire is older than Bea" tells you that the difference between the ages of Claire and Bea is c- b, not b- c).
When Claire is as old as Dawn is now, Bea will be twice as old as Ann currently is.
2a- b+ c- d= 0
4. Feb 28, 2010
### Matty R
Thanks for the replies.
I'd never have got that. I completely see how to get it now, but I just couldn't understand it before.
The following is multiple choice question (with options) to answer.
Rose is two years older than Bruce who is twice as old as Chris. If the total of the age of Rose, B and Chris be 27 years, then how old is Bruce ? | [
"8 years",
"10 yrs",
"12 years",
"14 years"
] | B | Let Chris's age be x years. Then, Bruce's age = 2x years.Rose's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, Bruce's age = 2x = 10 years.
B |
AQUA-RAT | AQUA-RAT-39818 | $\Rightarrow AC+CB=4(AB)$
$\Rightarrow (AB+BC)+CB=4(AB)$
$\Rightarrow BC+CB = 3(AB)$
$\Rightarrow BC=\dfrac{3}{2}(AB)$ -------- $(1)$
Similarly by the time Dinesh reaches point $D$ from $C$ walking, Mukesh and Suresh reach $D$ riding bike.
Here also distance travelled by bike $(=BD)$ is 4 times the distance travelled on foot $(=CD)$
$\Rightarrow CB+(BD)=4(CD)$
$\Rightarrow BC+(BC+CD)=4(CD)$
$\Rightarrow 2(BC)= 3(CD)$
$\Rightarrow CD= \dfrac{2}{3}(BC)$ -------- $(2)$
Now, it is given that total distance is given as $300 \text{ km}$
$\Rightarrow AB+BC+CD=300$
Using values from, equations $(1)$ and $(2)$,
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}(BC)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+\dfrac{2}{3}\times \dfrac{3}{2}(AB)=300$
$\Rightarrow AB+\dfrac{3}{2}(AB)+AB=300$
$\Rightarrow \dfrac{7}{2}(AB)=300$
$\Rightarrow AB=\dfrac{600}{7}$
So, $BC$
$=\dfrac{3}{2}(AB)$
$=\dfrac{3}{2}\times \dfrac{600}{7}$
$=\dfrac{900}{7}$
Similarly, $CD$
$=\dfrac{2}{3}(BC)$
The following is multiple choice question (with options) to answer.
Ravi started walking from his house east direction on Bus stop which is 3km.away. Then he set off in the bus straight towards his right to the school 4 km away. what is the crow flight distance from his house to the school? | [
"1 km",
"5 km",
"7 km",
"12 km"
] | B | Explanation:
The walked will be of in triangular shape and from starting point is 5km
ANSWER IS B |
AQUA-RAT | AQUA-RAT-39819 | 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.
What will come in place of the x in the following Number series? 2, 7, 27, 107, 427, x | [
"1427",
"1607",
"1707",
"2047"
] | C | 2
2 × 4 - 1 = 7
7 × 4 - 1 = 27
27 × 4 - 1 = 107
107 × 4 - 1 = 427
427 × 4 - 1 = 1707 C |
AQUA-RAT | AQUA-RAT-39820 | Suppose the slower car stands still for one hour. How often will the faster car pass it? Then stop the faster car and start the slower car for another hour. How often will the slow car pass the stopped car? Add.
Consider alternative case when cars complete exactly $4$ and $8$ rounds. It's easily seen that the number of times they pass is
$$2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12$$
So for $4$ and $7$ it would be one less than that which is $11$.
• so for 11 and 14 it would be 25? – simplton May 16 '13 at 19:56
The following is multiple choice question (with options) to answer.
Two cars start from the opposite places of a main road, 113 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point? | [
"65",
"38",
"20",
"28"
] | D | Answer: D) 28 km |
AQUA-RAT | AQUA-RAT-39821 | $\text{So we have shown:}$
$\setminus q \quad {\text{odd"_1 +"odd"_2 + "odd"_3 +"odd"_4 + "odd"_5 +"odd"_6 \ = "even}}_{5.}$
$\text{So we conclude:}$
$\setminus \quad \text{the sum of any 6 odd numbers (consecutive or not) is even.}$
The following is multiple choice question (with options) to answer.
If b is the sum of consecutive even integers w, x, y, and z, where w < x < y < z, all of the following must be true EXCEPT | [
"z - w = 3(y - x)",
"b is divisible by 8",
"The average of w, x, y, and z is odd",
"b is divisible by 4"
] | B | Just assume that the numbers are 2a, 2a+2, 2a+4 and 2a+6.
b = 2a + 2a+2 + 2a+4 + 2a+6 = 8a + 12
Focus on the easiest options first. You see that b is not divisible by 8. Answer (B) |
AQUA-RAT | AQUA-RAT-39822 | The Philosophical Question
Then the original poster asked for someone “to post [actual] work done in calculus” and “to see some related rates.” So I posted some “calculus” and got to thinking – the philosophical question – isn’t my first answer calculus?
I think it is. It makes use of an important calculus concept, namely that as things change, at the minimum place, the derivative is zero. Furthermore, the justification (that the distance changes from decreasing to increasing at the minimum implies the derivative is zero) is included.* Why do you need variables?
Also, this solution is approached as an extreme value (max/min) problem rather than a related rate problem. This shows a nice connection between the two types of problems.
The Related (but not related rate) Good Question
So here is another calculus question with none of the numbers we’ve grown to expect:
Two cars travel on parallel roads. The roads are w feet apart. At what rate is the distance between the cars changing when the cars are w feet apart?
Notice:
• That the cars could be traveling in the same or opposite directions.
• Their speeds are not given.
• You don’t know when or where they started; only that at some time they are opposite each other (w feet apart).
• In fact, they could start opposite each other and travel in the same direction at the same speed remaining always w feet apart.
• One car could be standing still and the other just passes it.
But you can still answer the question.
(*Continuity and differentiability are given (or at least implied) in the original statement of the problem.)
Appendix
My first attempt was to set up a coordinate system with the origin at third base as shown below.
Then, taking the time indicated in the problem as t = 0, the position of Player 1 is (90, 30 + 20t) and the position of player 2 is (0, 30 – 15t). Then the distance between them is
$s=\sqrt{{{90}^{2}}+{{\left( 30-15t-\left( 30+20t \right) \right)}^{2}}}=\sqrt{{{90}^{2}}+{{\left( -35t \right)}^{2}}}$
and then
The following is multiple choice question (with options) to answer.
On Monday, Lou drives his vehicle with 26-inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his vehicle to 32-inch tires yet drives to work at the same average speed as on Monday. What is the percent change from Monday to Tuesday in the average number of revolutions that Lou's tires make per second? | [
"Decrease by 14.3%",
"Decrease by 19%",
"Increase by 14.3%",
"Increase by 12.5%"
] | B | 416 is the LCM
On Tuesday 416/32 =13
On Monday 416/26 = 16
Therefore, 13-16/16 = -1/5=-19% decrease
Answer is B |
AQUA-RAT | AQUA-RAT-39823 | materials
Title: Why are vehicle interiors susceptible to damage from the sun when some plastics seem to be fine?
My vehicle built in '08 suffered extreme damage to the dashboard, it permanently melted and is now sticky to the touch even when cool.
Presumably this is why so many vehicle owners use covers to block the sunlight when parked. I never liked the idea of these, they seem like a hassle that better engineering could fix.
Have there been recent manufacturing / vehicle design improvements across the automotive industry so future cars will be resistant to high temperatures? Or is this still a common problem? If so, what fundamental trait of car interior materials causes this issue to be either so hard or so expensive to solve? The main problem is that there are many constraints for the material of a dashboard. Namely (just a few that pop to mind - I will update later if I forgotten any important ones):
good UV resistance
good thermal stability (low coefficient of expansion)
high melting point
good surface
coming in different shape
machinable (in order to take a nice shape)
flame retardant (Pete W addition)
able to withstand common cleaners like bleach and peroxide (Pete W addition)
and cheap.
When there are so many constraints its difficult to find a material/solution that satisfies perfectly all needs. (A wise friend said to me constantly "quickly, done correctly, cheaply, pick any two", I guess this is applicable in this instance its just with more constraints)
One final note, since cars are sometimes (IMHO more often than not) sold more on looks rather than the durability of the dashboard (keep in mind that the owner's handling is also a factor here), the compromise tends to be the aesthetics over durability.
The following is multiple choice question (with options) to answer.
A certain protective additive increases from 30 days to 60 days the time between required maintenance checks on an industrial vehicle. By what percent is the time between maintenance checks increased by using the additive? | [
"25%",
"33 1/3%",
"100%",
"66 2/3%"
] | C | General formula for percent increase or decrease, (percent change):
percent=Change/Original∗100
So, the time between maintenance checks increased by (60−30)/30∗100=100
Answer: C |
AQUA-RAT | AQUA-RAT-39824 | The population of a culture of bacteria, P(t), where t is time in days, is growing at a rate that is proportional to the population itself and the growth rate is 0.3. The initial population is 40. (1) What is the population after
6. ### calculus
The population of a certain community is increasing at a rate directly proportional to the population at any time t. In the last yr, the population has doubled. How long will it take for the population to triple? Round the answer
7. ### Maths
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=Ae^kt where A and k are constants. With the aid of
8. ### Maths B - Population Growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P=¡¼Ae¡½^kt where A and k are constants. With the aid of
9. ### Maths B question - population
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
10. ### Population growth
The population P of a particular city, Metropia, is growing at a rate proportional to the current population. The population at time t years is modelled by the equation P = Aekt where A and k are constants. (a) With the aid of
More Similar Questions
The following is multiple choice question (with options) to answer.
In a decade, a town’s population increases from 175000 to 262500. What is the average % increase of population each year? | [
"9%",
"2%",
"5%",
"7%"
] | C | Explanation:
Increase in population in 10 years= 262500-175000= 87500
Now, % increase will be increase in (population /original population) * 100
=> 87500/175000*100= 50% increase in population over 10 years
Therefore, Average % increase each year= 50/10= 5%
ANSWER: C |
AQUA-RAT | AQUA-RAT-39825 | The error lies in case $C1:$
. . the people are divided into two pairs.
Call the people $A,B,C,D.$
How many pairings are possible?
The answer is three: . $\{AB|CD\},\;\{AC|BD\},\;\{AD|BC\}$
. . Note that $\{CD|AB\}$ is not a different pairing.
Make this correction and you will get 256.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I first ran into this snag many years ago.
There are four bridge players.
In how many ways can the players be paired?
I immediately said: . $_4C_2 \:=\:\dfrac{4!}{2!\,2!} \:=\:6$
I was surprised to learn that the answer was three.
(And no explanation was given.)
I finally reasoned it out.
When we select a pair of players to be partners,
. . we have automatically assigned the other two to be partners.
Call the players $A,B,C,D.$
With whom can $A$ be paired?
$A$ can be paired with $B,C,\text{ or }D$ . . . 3 choices.
And that's it!
. . Get it?
The following is multiple choice question (with options) to answer.
Three-twentieths of the members of a social club are retirees who are also bridge players, five-twentieths of the members are retirees, and one-half of the members are bridge players. If 120 of the members are neither retirees nor bridge players, what is the total number of members in the social club? | [
"240",
"300",
"200",
"400"
] | C | {Total} = {Retirees} + {Bridge players} - {Both} + {Neither}
x = 5/20*x + x/2 - 3/20*x + 120
20x = 5x + 10x - 3x + 120*20 (multiply by 20)
12x = 120*20
x = 200.
Answer: C |
AQUA-RAT | AQUA-RAT-39826 | First we choose two values, there are 13 values (2 to A), so $$13\choose2$$.
Then we want to choose two cards of the first value out of four cards, $$4\choose 2$$
Again, we want to choose two cards of the second value out of four cards, $$4\choose 2$$
And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), $${52-8\choose1} = {44\choose1}$$
So we get: $${{{13\choose2}\times{4\choose2}\times{4\choose2}\times{44\choose1} }\over{52\choose2}} = {198\over4165} ≈ 0.0475$$
The following is multiple choice question (with options) to answer.
If you have 4 New Year greeting cards and you want to send them to 2 of your friends, in how many ways can this be done? | [
"360",
"12",
"16",
"24"
] | B | Solution
We have to find number of permutations of 2 objects out of 4 objects. This number is 4P2=4(4-1)=4*3=12
Answer B |
AQUA-RAT | AQUA-RAT-39827 | # Roll a die probability question
An unbiased six-sided die is to be rolled five times. Suppose all these trials are independent. Let $E_1$ be the number of times the die shows a 1, 2 or 3. Let $E_2$ be the number of times the die shows a 4 or a 5. Find $P(E_1 = 2, E_2 = 1)$.
I have tried to solve this question this way:
Total $6P6 = 720$.
1,2 and 3 can be placed in $2$ locations $= 3^2$
4 and 5 can be placed in 1 location $= 2^1$
and 6 can be placed in two locations $= 2^2$
The probability is $= 72/720 = 0.1$
Is this correct?
• Please read this MathJax tutorial, which explains how to typeset mathematics on this site. – N. F. Taussig Sep 14 '18 at 13:00
Close. You are rolling 5 dice, not 6. And you are not looking to permute the numbers, you are looking to find the number of ways you can roll 5 dice. You need to permute the $E_1$ and $E_2$ events. If 6 can be in two locations, then you have $1^2$ ways of placing 6 in 2 locations. Next, you need to permute the multiset:
$$\{E_1\cdot 2, E_2 \cdot 1, 6\cdot 2\}$$
There are
$$\dfrac{5!}{2!1!2!}$$
ways to permute this multiset.
$$\dfrac{3^2\cdot 2^1\cdot 1^2 \cdot \dfrac{5!}{2!1!2!}}{6^5} = \dfrac{5}{72}$$
The following is multiple choice question (with options) to answer.
Find the probability of getting a multiple of 3,when an unbiased die is tossed? | [
"3/4",
"2/8",
"1/7",
"1/3"
] | D | Here S={1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
then ,E={3,6}
P(E)=n(E)/n(S)=2/6=1/3
Option D |
AQUA-RAT | AQUA-RAT-39828 | newtonian-gravity, mass, units, weight
First off, the others posting answers to this question are ultimately correct - there really is no direct conversion between kilograms and Newtons. Kilograms is a unit of mass (how much matter is in an object) while Newtons is a measure of the force that the Earth's gravity exerts on the object. These two may seem like they are exactly the same thing, but they are just different enough to be completely different things.
Now, if we limit our question to "how many Newtons would a 1kg object weigh at the Earth's surface", then yes I suppose there is a conversion between the two, and yes that conversion is about 9.81 m/s^2.
Now, on to the part you asked about the number of decimal places, accuracy, and g being "defined" as 9.80665. The fact is that g has never had any defined value to speak of (unlike the speed of light, but that's another story). Most of the best measurements at the Earth's surface put the g number at about 9.81, but this may vary around a bit because the Earth isn't exactly a sphere, and so fourth.
All of these web pages that seem to be giving you the same value of 9.80665002864 must be getting their information from the same source. Besides, I'm not sure what leads anyone to think that g can be calculated to this many decimal places - moving around from continent to continent will result in different g values that differ by roughly 1%.
Hope this helps.
The following is multiple choice question (with options) to answer.
What percent is 1 gm of 1 kg? | [
"0.1 %",
"1%",
"1.5%",
"2%"
] | A | 1 kg = 1000 gm
1/1000 × 100 = 100/1000
=1/10 = 0.1 %
A) |
AQUA-RAT | AQUA-RAT-39829 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
If the cost price of 12 pens is equal to the selling price of 10 pens, the gain percent is: | [
"80%",
"90%",
"20%",
"40%"
] | C | Explanation:
Let C.P. of each pen be Re. 1.
Then, C.P. of 10 pens = Rs. 10; S.P. of 10 pens = Rs. 12.
Gain % = 2/10 * 100 = 20%
Answer:C |
AQUA-RAT | AQUA-RAT-39830 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
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
The following is multiple choice question (with options) to answer.
The average mark obtained by 22 candidates in an examination is 46. The average of the first ten is 55 while the last eleven is 40 .The marks obtained by the 11th candidate is ? | [
"22",
"0",
"19",
"18"
] | A | It is clear that 22 x 46 = 10 x 55 + K + 11 x 40 ⇒⇒ K = 22
Answer:A |
AQUA-RAT | AQUA-RAT-39831 | Kudos [?]: 13 [0], given: 21
GMAT 1: 570 Q46 V24
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Dec 2013, 23:04
Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?
Then is 28*2 the correct approach?
Kudos [?]: 13 [0], given: 21
Math Expert
Joined: 02 Sep 2009
Posts: 42302
Kudos [?]: 133018 [1], given: 12402
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
27 Dec 2013, 03:08
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
Kudos [?]: 133018 [1], given: 12402
Manager
Joined: 07 Apr 2014
Posts: 138
Kudos [?]: 31 [0], given: 81
Re: There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
12 Sep 2014, 06:36
sarb wrote:
There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
A. 15
B. 16
C. 28
D. 56
E. 64
total 8 teams & each game by 2 pair then 8C2
Kudos [?]: 31 [0], given: 81
Intern
Joined: 20 Sep 2014
Posts: 9
Kudos [?]: 4 [1], given: 49
There are 8 teams in a certain league and each team plays [#permalink]
### Show Tags
26 Nov 2014, 02:14
1
KUDOS
1
This post was
BOOKMARKED
SreeViji wrote:
Hi Bunnel,
I would also like to learn this approach. Can u help me?
Sree
Hey SreeViji,
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}
The following is multiple choice question (with options) to answer.
Team / No of games won
A 4
B 7
C 9
D 2
E 2
X ?
According to the incomplete table above, if each of the 6 teams in the league played each of the other teams exactly twice and there were no ties, how many games did team X win? (Only 2 teams play in a game.) | [
"4",
"5",
"6",
"8"
] | C | 6C2= 15 single games, & 30 in total.
won by others= 24
30-24= 6
ANSWER:C |
AQUA-RAT | AQUA-RAT-39832 | So his specific question is this: • Why wouldn’t “three times as large” and “three times larger” mean the same thing? • How can “5 1/2 times greater” and “6 1/2 times what it was” mean the same thing? The two pages quoted are by me (1999) and Doctor Terrel (1997). The first is particularly worth reading in its entirety, as there is a lot more there. ## The case for a literal interpretation Doctor Greenie responded, arguing against laxness on the matter, and making the case for the literalistic interpretation: I'm going to jump in here, because this is one of my pet peeves. Mathematics is commonly called the exact science. Mathematics must be exact; if it is not, it all falls apart. We can't use ambiguous language in mathematics. I agree that the use of the phrase "x times larger than" is best avoided. However, as a mathematician who believes in using unambiguous language, I cannot accept the proposition that we should be able to interpret "5 times larger than 10" as either 50 or 60. It HAS TO BE ONE OR THE OTHER. And grammatically, "5 times larger than" means the "new" number is 5 times larger than the "old" number; this in turn means the difference between the new and old numbers is 5 times the old number, making the new number 6 times the old number. So the number which is 5 times larger than 10 is 10 + 5(10) = 10 + 50 = 60 (The phrase "... larger than ..." implies comparison by subtraction; the phrase "... as large as ..." implies comparison by division. Sixty is 6 times as large as 10, because 60/10 = 6. But 60 is 5 times larger than 10, because [60 - 10]/10 = 50/10 = 5.) Yes, we hear it all the time in everyday life. Sometimes, we even hear it in the supposedly rigorous world of science -- "an earthquake of magnitude 5 is 10 times greater than one of magnitude 4," and such. But the common idiom of using "10 times greater than" -- when the actual meaning is "10 times as great as" -- has no place in mathematics. He concluded with an accidental overstatement of what the “other side” says: I
The following is multiple choice question (with options) to answer.
On a scale that measures the intensity of a certain phenomenon, a reading of j+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of j. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3? | [
" 5",
" 50",
" 10^5",
" 5^10"
] | C | To solve this problem we need to examine the information in the first sentence. We are told that “a reading of j + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of j.”
Let’s practice this idea with some real numbers. Let’s say j is 2. This means that j + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.
Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.
Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.
We have found a pattern, which can be applied to the problem presented in the stem:
3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.
4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.
5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.
We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.
Answer C. |
AQUA-RAT | AQUA-RAT-39833 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
How many seconds will a 800 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr? | [
"11",
"30",
"99",
"48"
] | D | Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 800 * 3/50 = 48 sec.
Answer: D |
AQUA-RAT | AQUA-RAT-39834 | # Clock losing time puzzle
The question goes as:
A wall clock and a Table clock are set to correct time today on 10 pm. The wall clock loses 3 minute in 1st hour, 6 minutes in the second hour and 9 minutes in the third hour and so on. The table clock loses 5 minutes in the 1st hour, 10 minutes in the second hour and 15 minutes in the third hour and so on. When will they show the same time?
My approach:
In the first hour, the difference between the two clocks would be $2$ (obtained from $5-3$) minutes.
In the second hour, it'll be four minutes and so on. This would form an arithmetic progression with $a$ = 2 and $d = 2$. I, then, formulated the problem as:
$$2 + 4 + 6+ 8 + \dots + n = 720$$
The RHS is $720$ because I assumed they'll meet after 12 hours.
With this, I got the root as $23.337$ hours, so I arrived at the answer as $10 \, \text{PM} + 23.337$ hours i.e $9:20 \, \text{PM}$.
Is this correct?
EDIT: I realised this equation won't give an integral answer, and we need one as $n$ on the LHS represents the number of terms. So instead of that, I wrote it as:
$$2 + 4 + 6 + \dots + n = 720 \times k$$ where $k \in (1,2,3,4, \dots)$.
Using this method, for $k = 9$, I get the value of $n$ $\text{as}$ $80 \, \text{hours}$.
Does this seem correct?
The following is multiple choice question (with options) to answer.
A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time shown by the clock exactly 14 days from the time it was set right? | [
"1: 36: 48",
"1: 40: 48",
"1: 41: 24",
"10: 19: 12"
] | B | Solution:
The clock loses 1% time during the first week.
In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
If the clock loses 1% time during the first week, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.
Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.
As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
So the clock will show a time which is 1.68 hours more than 12 Noon two weeks from the time it was set right.
1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds.
i.e. 1 : 40 : 48 P.M.
Answer B |
AQUA-RAT | AQUA-RAT-39835 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
If 20 men can build a wall 112 metres long in 6 days, what length of a similar wall can be built by 30 men in 3 days? | [
"65mtr.",
"84mtr",
"70mtr.",
"78mtr."
] | B | 20 men is 6 days can build 112 metres
30 men in 3 days can build = 112*(30/20)x(3/6)
= 84 meters
Answer: B. |
AQUA-RAT | AQUA-RAT-39836 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
_________________
# Jeffrey Miller
Jeff@TargetTestPrep.com
122 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.
Manager
Status: Preparing
Joined: 05 May 2016
Posts: 55
Location: India
Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
### Show Tags
07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
The following is multiple choice question (with options) to answer.
Local kennel has cats and dogs in the ratio of 5:10. If there are 40 fewer cats than dogs, how many dogs are in the kennel? | [
"80",
"70",
"60",
"50"
] | A | Lets work with the data given to us. We know that there ratio of cats to dogs is 5:10 or
cats 5
dogs 10
we can write number of cats as 5x and number of dogs as 10x and we know that 10x-5x= 40(therefore 5x = 40 => x=8)
Then # of dogs = 10x8= 80
Answer is A |
AQUA-RAT | AQUA-RAT-39837 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train which has 400 m long, is running 45 kmph. In what time will it cross a person moving at 9 kmph in same direction ? | [
"56 sec",
"40 sec",
"36 sec",
"29 sec."
] | B | Time taken to cross a moving person = length of train/ relative speed
Time taken = 400/((45-9) (5/18)
= 400/ 36*(5/18)
= 400/10
= 40 sec
ANSWER:B |
AQUA-RAT | AQUA-RAT-39838 | hybridization
Title: Determine percentage of crow hybrids We had in class following question which I had no idea how to get to the correct answer:
The carrion crow and the hooded crow are fertile together, but their
reproductive success is reduced by 50%. In a certain region exist
two populations of both species of roughly the same size. Thus,
mixed couples occur in about 10% of all cases. What is the percentage of
hybrids in the F1-generation?
(Translated from German.)
I thought it would be something like 5/95 = 5,3% but apparently the answer is 1%. Why? In order to make your reasoning clearer, you should use more formal notations and explain your thinking step by step. Here's a proposition.
Let's use the following notations :
$C$: the total number of couples (mixed and not mixed)
$r$: the reproductive success
$F1_h$: the number of hybrids in the F1 generation
$F1_{nh}$: the number of non hybrids in the F1 generation
You are looking for the percentage of hydrids in the F1 generation, which is:
$x = \frac{F1_h}{F1_h + F1_{nh}}$
You know that $10~\%$ of the couples are mixed couples and that their reproductive success is reduced by $50~\%$. This can be written:
$\begin{cases} 0.9\cdot C \cdot r = F1_{nh} \\ 0.1\cdot C\cdot \frac{r}{2} = F1_h \end{cases}$
Thus: $\displaystyle x = \frac{F1_h}{F1_h + F1_{nh}} = \frac{0.05\cdot C\cdot r}{C\cdot r\cdot 0.95} = \frac{0.05}{0.95}$
This is indeed the result you suggested. So the correction you were given might not be correct. Or maybe there was some more information in your homework that you ignored...
The following is multiple choice question (with options) to answer.
A scientific research study examined a large number of young foxes, that is, foxes between 1 year and 2 years old. The study found that 80% of the young foxes caught a rabbit at least once, and 60% caught a songbird at least once If 10% of the young foxes never caught either a rabbit or a songbird, then what percentage of young foxes were successful in catching at least one rabbit and at least one songbird? | [
"40%",
"50%",
"60%",
"80%"
] | B | This is less about percents and more about probability, particularly the probability OR-rule. Let R = the event that a young fox catches at least one rabbit, and let S = the event that a young fox catches at least one songbird. Using algebraic probability notation, we know P(R) = 0.8 and P(S) = 0.6. We know P((not R) and (not S)) = 0.1, and the complement of [(not R) and (not S)] would be [R or S], so by the complement rule, P(R or S) = 1 – 0.1 = 0.9. The question is asking for P(R and S). The OR rule tells us
P(R or S) = P(R) + P(S) – P(R and S)
0.9 = 0.6 + 0.8 – P(R and S)
0.9 = 1.4 – P(R and S)
0.9 + P(R and S) = 1.4
P(R and S) = 0.5 |
AQUA-RAT | AQUA-RAT-39839 | I made a mistake. After I get 10 planes, I should count the lines, because 3 plane could share the same line.
Let the 5 points be 12345. The 10 plane contain the points are
123 (the plane contain 123)
124
125
134
135
145
234
235
245
345
When two points appear on more than 2 row , it means more than two planes will share the line, therefore, we need to subtract them. So the answer should be 45-20=25
Does everyone agree with this solution or have the same approach and way of explaining the problem?
6. Well, If I understand the problem, I do not agree with yma's answer. You cannot determine 25 distinct lines from 5 points. You can at most determine 10 distinct lines. But the error in yma's approach is far from obvious.
Label the points A, B, C, D, and E. Three points define a plane surface. So what is the maximum number planes to be considered? Notice that planes ABC, ACB, BAC, BCA, CAB, and CBA are the same plane. So we are dealing with combinations rather than permutations.
$\therefore \text { maximum number of planes } = \dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4}{2} = 10.$
So far I agree with yma's answer. Here are the planes.
ABC
ABD
ABE
ACD
ACE
BCD
BCE
BDE
CDE
Let's consider planes ABC, ABD, and ABE. Those are three distinct planes because, by hypothesis, no more than three of the five points lie in the same plane. Moreover, none of those three planes is parallel to either of the other two because they all contain the line joining A and B. We have three planes intersecting in one line. So yma is also correct that each intersection of planes does not create a distinct line.
A line joining any two of the five points creates a line. How many distinct pairs of points can we pick out of five distinct points, remembering that the same line joins A and B as joins B and A.
The following is multiple choice question (with options) to answer.
The navy pilots are practicing flight formations. 200 pilots are practicing flying in rectangles. They want to make a rectangular formation with X planes in a row and Y planes in a column. If the
number of airplanes in a row is no less than 5 and no more than 40, how many different combinations of rectangular shapes are possible? | [
"6",
"7",
"8",
"9"
] | A | firstly factors of 200 = 1,2,4,5,8,10,20,25,40,50,100,200
the question says each row will have airplanes not less than 5 and not more than 40,
therefore, we should be interested in the factors starting from 5 til 40 (inclusive)
5(in each row) * 40(number of columns) =200, similarly
8 * 25
10 * 20
20 * 10
25 * 8
40 * 5
total possibilities = 6
option A |
AQUA-RAT | AQUA-RAT-39840 | 4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)
5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.
The following is multiple choice question (with options) to answer.
The food in a camp lasts for 40 men for 40 days. If ten more men join, how many days will the food last? | [
"32days",
"34days",
"36days",
"38days"
] | A | one man can consume the same food in 40*40 = 1600 days.
10 more men join, the total number of men = 50
The number of days the food will last = 1600/50 = 32 days.
Answer: A |
AQUA-RAT | AQUA-RAT-39841 | # 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 triplicate ratio of 2:3 is? | [
"1:64",
"1:62",
"1:34",
"8:27"
] | D | 2^3: 3^3 = 8:27
ANSWER:D |
AQUA-RAT | AQUA-RAT-39842 | Alternate
10% of journey's = 40 km
Then, total journey = 400 kms
\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr}
The following is multiple choice question (with options) to answer.
A person travelled from his house to office at 30km/hr and office to house at 20km/hr. Then find the average speed? | [
"15km/hr",
"20km/hr",
"32km/hr",
"24km/hr"
] | D | Average speed = (2*30*20)/(30+20)
= 24km/hr
Answer is D |
AQUA-RAT | AQUA-RAT-39843 | # Is this a simple answer to the classic problem "A certain city has 10 bus routes..."
A certain city has 10 bus routes. Is it possible to arrange the routes and the bus stops so that if one route is closed, it is still possible to get from anyone stop to any other (possibly changing along the way), but if any two routes are closed, there are at least two stops such that it is impossible to get from one to the other?
Given the solution above, why did the authors feel compelled to answer: Yes. Consider 10 straight lines in the plane, no 2 are parallel & no 3 are concurrent. Let lines be bus routes & let points of intersection be stops. We get from anyone stop to any other (if the stops lie on 1 line, w/o changing; & if not, then with just 1 change). If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once. However, if we discard 2 lines, then 1 stop-their point of intersection-will have no bus routes passing thru it, & it'll be impossible to get from this stop to any other.
Source: A. M. Yaglom and l. M. Yaglom CHALLENGING MATHEMATICAL PROBLEMS WITH ELEMENTARY SOLUTIONS Volume II Problems From Various Branches of Mathematics Translated by James McCawley, Jr. Revised and edited by Basil Gordon DOVER PUBLICATIONS, INC. NEW YORK
• Yes, the answer is correct. Jun 9, 2019 at 6:12
• My guess, Joe, is that when they wrote "changing", they meant "changing once". Jun 9, 2019 at 13:04
• Your answer is correct and fine. It is possible the authors just didn't think of this solution and thus put their solution into the book. Everybody overlooks something every now and then. Jun 9, 2019 at 15:30
• My solution fails to meet the requirement, "If we discard 1 line, it's still possible to get from anyone stop to any other, changing buses at most once." Jun 10, 2019 at 0:55
• Gerry, your interpretation of the wording is what makes the problem sufficiently complex. Thanks. Jun 10, 2019 at 0:58
Let’s go back to the roots.
The following is multiple choice question (with options) to answer.
There are three places P, Q and R such that 3 roads connects P and Q and 4 roads connects Q and R. In how many ways can one travel from P to R? | [
"12",
"24",
"32",
"64"
] | A | Number of ways in which one can travel from P to R
=3×4=12=3×4=12
ANSWER A 12 |
AQUA-RAT | AQUA-RAT-39844 | ### 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 clothing store originally bought 100 jackets at a total cost of j dollars. If each jacket was sold for 70 percent more than its original cost, what was the individual price, in terms of j, of each jacket sold? | [
"j/59",
"7j/500",
"140j",
"j/100+40"
] | A | lets take j=1000, as decided by you..
now 70% increase makes it 1700..
this is the price for 100 jackets, so selling price for each jacket=1700/100=$17..
so ans is correct as A.. |
AQUA-RAT | AQUA-RAT-39845 | Let one woman complete the job in $$w$$ days and one man in $$m$$ days.
First equation:
It takes 6 days for 3 women and 2 men working together to complete a work:
As the rate of 1 woman is $$\frac{1}{w}$$ job/day, then the rate of 3 women will be $$\frac{3}{w}$$ job/day. As the rate of 1 man is $$\frac{1}{m}$$ job/day, then the rate of 2 men will be $$\frac{2}{m}$$ job/day. Combined rate of 3 women and 2 men in one day will be: $$\frac{3}{w}+\frac{2}{m}$$ job/day.
As they do all the job in 6 days then in 1 day they do 1/6 of the job, which is combined rate of 3 women and 2 men --> $$\frac{3}{w}+\frac{2}{m}=\frac{1}{6}$$.
Second equation:
3 men would do the same work 5 days sooner than 9 women:
As 1 man needs $$m$$ days to do the job 3 men will need $$\frac{m}{3}$$ days to do the job. As 1 woman needs $$w$$ days to do the job 9 women will need $$\frac{w}{9}$$ days to do the job. 3 men would do the same work 5 days sooner means that 3 men will need 5 less days to do the job, hence $$\frac{m}{3}$$ is 5 less than $$\frac{w}{9}$$ --> $$\frac{m}{3}+5=\frac{w}{9}$$.
Hope it's clear.
The following is multiple choice question (with options) to answer.
A and B complete a work in 6 days. A alone can do it in 10 days. If both together can do the work in how many days? | [
"3.75 days",
"3.33 days",
"7.75 days",
"3.95 days"
] | A | 1/6 + 1/10 = 8/30 = 4/15
15/4 = 3.75 days
Answer:A |
AQUA-RAT | AQUA-RAT-39846 | And, sure enough: . $8 + 2\sqrt{15} \;=\;(\sqrt{3} + \sqrt{5})^2$
The following is multiple choice question (with options) to answer.
If 18 is 15 percent of 5 percent of a certain number, what is the number? | [
"9",
"2400",
"40",
"81"
] | B | Let the certain number be x.
=> (15/100)*(5/100)*x = 18;
=> x=2400;
Ans is (B). |
AQUA-RAT | AQUA-RAT-39847 | Let one woman complete the job in $$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.
If 12 men or 20 women can do a piece of work in 27 days, then in how many days can 9 men and 12 women together do the work? | [
"10 days",
"30 days",
"20 days",
"80 days"
] | C | C
20 days
Given that 12m = 20w => 3m = 5w
9 men + 12 women = 15 women + 12 women = 27 women
20 women can do the work in 27 days. So, 27 women can do it in (20 * 27) /27 = 20 days. |
AQUA-RAT | AQUA-RAT-39848 | 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 two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence? | [
"40",
"45",
"70",
"12"
] | C | Answer: Option C
a + b = 60, a = 2b
2b + b = 60 => b = 20 then a = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70. |
AQUA-RAT | AQUA-RAT-39849 | # Is “divisible by 15” the same as “divisible by 5 and divisible by 3”?
Is stating that a number $x$ is divisible by 15 the same as stating that $x$ is divisible by 5 and $x$ is divisible by 3?
-
The two assertions are equivalent. – André Nicolas Feb 1 '12 at 23:38
may be worth noting for somebody who might be new to this, that it happens to work in both directions here, because 3 and 5 are co-prime. i.e. if you ask the same question for a different example you may not get the converse implication. – Beltrame Feb 1 '12 at 23:49
If you want to figure out the general facts at work here, you could read this Keith Conrad handout. – Dylan Moreland Feb 1 '12 at 23:55
Yes, if a number $n$ is divisible by $15$, this means $n=15k$ for some integer $k$. So $n=5(3k)=3(5k)$, so it is also divisible by $3$ and $5$.
Conversely, if $n$ is divisible by $3$ and $5$, it is a simple lemma that it is divisible by the least common multiple of $3$ and $5$. Since $3$ and $5$ are coprime, their lcm is just $15$.
-
I feel that this answer is incomplete. It does not prove that x|n and y|n implies that lcm(x,y)|n; it just claims it's "a simple lemma". A better answer would show why this is true. – user22805 Feb 2 '12 at 8:32
Suppose there's a number n such that x|n and y|n, but lcm(x,y) does not divide n. Then write m = lcm(x,y) and n = pm+q, where 0 < q < m and p is an integer. Then x and y must both divide q, so m is not the lcm of x and y - contradiction. – user22805 Feb 2 '12 at 8:35
The following is multiple choice question (with options) to answer.
If w = 20! + 17, then w is divisible by which of the following?
I. 15
II. 17
III. 19 | [
" None",
" I only",
" II only",
" I and II"
] | C | Another crude way to answer this, if you did not know the properties above would be to consider that that 20! will have the number ending in 00 due to 10 and 20 being included.
So w!+17 = xxxx00 +17 = xxxx17 which is only possibly divisible by 17. Hence Option C is the answer. |
AQUA-RAT | AQUA-RAT-39850 | 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
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:22
1
KUDOS
cicerone wrote:
Problem 7:
The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?
Sol:
If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.
This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).
So the present average = 40-2 = 38 yrs.
Problem 8:
The following is multiple choice question (with options) to answer.
A batsman makes a score of 83 runs in the 17th inning and thus increases his averages by 3.Find his average after 17th inning? | [
"19",
"29",
"35",
"49"
] | C | let the average after 17th inning =x
Then average after 16th inning=(x-3)
Therefore 16(x-3)+83=17x
Therefore x=35
ANSWER:C |
AQUA-RAT | AQUA-RAT-39851 | the two rectangles = … Level 5 - Real life composite area questions from photographs. Math Practice Online > free > lessons > Texas > 8th grade > Perimeter and Area of Composite Figures. Match. Solving Practice Area of Composite Figures. The 2 green points in the diagram are the … Edit. Question 4 : Find the area of the figure shown below. Area of Composite Figures DRAFT. Today Courses Practice Algebra Geometry Number Theory Calculus Probability ... and the area of the figure is 15, what is the perimeter of the figure? Circumference. Geometry Parallelogram Worksheet Answers Unique 6 2 Parallelograms Fun maths practice! Area of Composite Figures DRAFT. Share practice link. Filesize: 428 KB; Language: English; Published: December 14, 2015; Viewed: 2,190 times; Multi-Part Lesson 9-3 Composite Figures - Glencoe. A = 3 + 44 + 4.5. 9th - 12th grade . More Composite Figures on Brilliant, the largest community of math and science problem solvers. Therefore, we'll focus on applying what we have learned about various simple geometric figures to analyze composite figures. Area of composite shapes (practice) | Khan Academy Practice finding the areas of complex shapes that are composed of smaller shapes. This presentation reviews what is required to determine the area of composite figures and presents sample problems Terms in this set (20) Area. Emily_LebronC106. 00:30:09 – Finding area of composite figures (Examples #13-15) 00:40:27 – Using ratios and proportions find the area or side length of a polygon (Examples #16-17) 00:49:51 – Using ratios and proportions find the area or length of a diagonal of a rhombus (Examples #18-19) Practice Problems with Step-by-Step Solutions Separate the figure into smaller, familiar figures: a two triangles and a rectangle. Click here to find out how you can support the site. LESSON 27: Surface Area of Composite Shapes With HolesLESSON 28: Surface Area AssessmentLESSON 29: 3-D Models from 2-D Views LESSON 30: Exploring Volume and Surface Area with Unifix CubesLESSON 31: Explore Volume of Rectangular PrismsLESSON 32: Find the … So, the area of the given composite figure is 51.5 square feet. Area of Composite Figures Practice:I have used this with my 6th grade students, but it would also be
The following is multiple choice question (with options) to answer.
The area of a square is equal to five times the area of a rectangle of dimensions 125 cm * 64 cm. What is the perimeter of the square? | [
"377",
"800",
"278",
"277"
] | B | Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.
Answer: B |
AQUA-RAT | AQUA-RAT-39852 | 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$$
To access all the questions: Question of the Week: Consolidated List
_________________
Number Properties | Algebra |Quant Workshop
Success Stories
Guillermo's Success Story | Carrie's Success Story
Ace GMAT quant
Articles and Question to reach Q51 | Question of the week
Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line | Inequalities
Practice Questions
Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets
| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com
The following is multiple choice question (with options) to answer.
Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled? | [
"2 hours",
"9 hours",
"3 hours",
"7 hours"
] | C | Part filled by A + B + C in 1 hour
= 1/5 + 1/10 + 1/30
= 1/3
All the three pipes together will fill the tank in 3 hours.
Answer:C |
AQUA-RAT | AQUA-RAT-39853 | Select Page
#### Project Euler 15 Problem Statement
Starting in the top left corner of a 2×2 grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a 20×20 grid?
#### Solution
To add some context to this problem a similar question would be: “Starting at the top left corner, how many ways through town are possible using only one-way streets and ending at the bottom right corner of a n×n grid?”
Another analogy is: “Place a single rook on the top left corner of an empty chess board and count the number of tours to the bottom right corner moving only left and down.” This assumes an 8 × 8 grid.
The valid paths for a 2×2 grid shown in the example are the discreet (unique) permutations of {R, R, D, D}. We can list these as: {R, R, D, D}, {R, D, R, D}, {R, D, D, R}, {D, R, R, D}, {D, R, D, R} and {D, D, R, R}. You must have 2 Rs (rights) and 2 Ds (downs) the order not being important and as long as you start from the top-left corner you will always end at the bottom-right corner.
### A 4×4 grid example
For a 4×4 grid it would be the discreet permutations of {R, R, R, R, D, D, D, D} where any combination or 4 Rs and 4 Ds will always be a valid path. If the Rs are place randomly in 4 of the 8 slots first they are considered independent and the Ds are considered dependent because they have to be placed into the remaining open slots.
For example, placing the 4 Rs randomly in any of the available 8 slots as {R, _, R, R, _, R, _, _} will dictate where the Ds get placed as {R, D, R, R, D, R, D, D}. You will have 8C4 = 70 valid combinations, i.e., how many distinct ways can you shuffle the characters in the string “RRRRDDDD”.
### Using the binomial coefficient
The following is multiple choice question (with options) to answer.
Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length? | [
"6",
"8",
"10",
"14"
] | C | In order the length to be minimum Pat should only go UP and RIGHT: namely thrice UP and twice RIGHT.
So combination of UUURR: # of permutations of 5 letters out of which there are 3 identical U's and 2 identical R's is 5!/3!2!=10.
Answer: C. |
AQUA-RAT | AQUA-RAT-39854 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
The price of a book is increased from $300 to $420. What is the % of increase in its price? | [
"10%",
"20%",
"40%",
"50%"
] | C | Explanation: Change in the price = Rs 420 – Rs 300
= Rs 120
Percentage of increase = Change in the price Initial Price
* 100.
Percentage increase in price =( 120 300
) *100 = 40%
C |
AQUA-RAT | AQUA-RAT-39855 | 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 work in 20 days and B in 30 days. If they work on it together for 4 days, then the fraction of the work that is left is : | [
"1/3",
"2/3",
"4/3",
"5/3"
] | B | Ans is:B
A's 1 day's work =1/20
B's 1 day's work =1/30
(A + B)'s 1 day's work =(1/20+1/30)=1/12
(A + B)'s 4 day's work =(1/12*4)=1/3
Therefore, Remaining work =(1-1/3)=2/3 |
AQUA-RAT | AQUA-RAT-39856 | ~ Nafer
## Solution 3 (using PIE)
Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion.
The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees.
The total number of configurations is given by $\frac{12!}{3! \cdot 4! \cdot 5!}$. To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE.
$\#$(configurations with at least one pair of adjacent Birch trees) $=$ $\#$(configurations with one pair) $-$ $\#$(configurations with two pairs) $+$ $\#$(configurations with three pairs) $-$ $\#$(configurations with four pairs).
To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\frac{11!}{3! \cdot 3! \cdot 4!}$ configurations.
For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\frac{10!}{2! \cdot 3! \cdot 4!}$ cases. So our second term is $\frac{2 \cdot 10!}{2! \cdot 3! \cdot 4!}$.
The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\frac{2 \cdot 9!}{3! \cdot 4!}$ arrangements.
The final term can happen in one way (BBBBB). This gives $\frac{8!}{3! \cdot 4!}$ arrangements.
The following is multiple choice question (with options) to answer.
Along a yard 273 metres long, 14 trees are palnted at equal distances, one tree being at each end of the yard. What is the distance between two consecutive trees | [
"18",
"19",
"10",
"21"
] | D | Explanation:
14 trees have 13 gaps between them,
Required distance (273/13) = 21
Option D |
AQUA-RAT | AQUA-RAT-39857 | the train is retarding from 60 m/s to 0 m/s, at a retardation of 1 m/s2 , time at which the speed reaches 30 m/s is: $$v = u – at$$ $$=> 30 = 60 – 1xt$$ $$=> t = 30s$$ At 30s, distance covered is: $$S = ut – ½ at^2$$ $$= 60 x 30 – ½ x 1 x (30)2$$ $$= 1800 – (15 x 30)$$ $$= 1800 – 450$$ $$= 1350m$$ (from the initial 900m covered). So, distance from origin $$= 900 + 1350m = 2250m$$.Physics
The following is multiple choice question (with options) to answer.
A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train | [
"70 km/hr",
"72 km/hr",
"79.2 km/hr",
"80 km/hr"
] | C | Solution
Speed = (132 / 6) m/sec
= (22 x 18 /5)km/hr
= 79.2 km/hr
Answer C |
AQUA-RAT | AQUA-RAT-39858 | 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 an integer n is to be selected at random from 1 to 105, inclusive, what is probability n(n+1) will be divisible by 7? | [
"1/7",
"3/14",
"1/2",
"2/7"
] | D | For n(n+1) to be a multiple of 7, either n or n+1 has to be a multiple of 7.
Thus n must be of the form 7k or 7k-1.
The probability is 2/7.
The answer is D. |
AQUA-RAT | AQUA-RAT-39859 | # Sum and Product of n-positive integers
If I have $n$-positive integers, and I compute their sum and product, is there any different group of $n$-positive integers that will have the same sum and product?
For example, if $a,...,z$ denote 26 positive integers, and we define:
\begin{align} a+b+c+d+....+z &= \text{Sum} \\ a \cdot b \cdot c \cdot d \cdot .... \cdot z &= \text{Product} \end{align}
Is there any way I can get the same Sum and Product from a different group of 26 (in this example) positive integers?
EDIT:
A friend of mine pointed out that knowing that we have a group of 3 that works, we can show that it works for all positive groups of $n$ integers.
For Example: $\{3,3,10 \}$ and $\{2,5,9 \}$ both yield Sum $=16$ and Product $=90$.
Now we can just continually add a number (let's say 1) as the next integer to get multiple solutions for $n =4,5,6,...$.
Explicitly, $\{ 3,3,10,1 \}$ and $\{ 2,5,9,1 \}$ both give Sum$=17$ and Product$=90$.
The following is multiple choice question (with options) to answer.
If both the product and sum of four integers are even, which of the following could be the number of even integers in the group?
I. 0
II. 2
III. 4 | [
"I only",
"II only",
"III only",
"II and III only"
] | D | for four int to have their product and sum as even..
2 and 4 are ok..
but 0 even number will result in an odd integer as product..
ANSWER:D |
AQUA-RAT | AQUA-RAT-39860 | # Clarification on language of a question on profit and loss.
The question is:
By selling 33 meters of cloth, a shopkeeper gains the cost of 11 meters. Find his gain percentage.
1. 33 1/3%
2. 33 1/2%
3. 33%
4. 34 1/4%
The answer provided by the book says it's the first one.
But if he gains the cost of 11 meters shouldn't the profit be calculated as a percentage of cost price, which would turn out to 22 meters. Below is what I think
(11/22) * 100
The cost price should be 22 because the profit of 11 meters is subtraccted from the selling price of 33 meters.
The question might be wrong and that is why I am seeking help.
• Profit is calculated on the cost price. The shopkeeper paid $x$ amount to buy 33 meters of cloth. When he sold the cloth, he got $x + x/3$ amount of money. Why would you subtract anything? Aug 17 '16 at 18:10
• There is often ambiguity in translating from ordinary language to math, but here I'd interpret the thing the way your book does. That is, I understand the problem to say "the shopkeeper sells $33$ units for the same amount that it would cost him to buy $44$ units." Thus, if we imagine it costs him $1$ to buy a unit, he buys the stuff for $33$ and sells it for $44$...thus a gain of $11$, or $33\frac 13\%$ of his outlay.
– lulu
Aug 17 '16 at 18:12
• Okay I get it. @shardulc it is not the selling price of 33 meters but the 33 meters of cloth. Aug 17 '16 at 18:15
The following is multiple choice question (with options) to answer.
A person bought 135 glass bowls at a rate of Rs. 15 per bowl. He sold 115 of them at Rs. 18 and the remaining broke. What is the percentage gain for A? | [
"40",
"30/11",
"20/9",
"27/11"
] | C | CP=135*15=2025 and SP=115*18=2070
Gain %= 100*(2070-2025)/2025= 20/9
ANSWER:C |
AQUA-RAT | AQUA-RAT-39861 | a company has borrowed$85,000 at a 6.5% interest rate. Find the accrued interest for an investment amount of 500 $holding for 15 days at an interest rate of 3 %. Calculating accrued interest payable First, take your interest rate and convert it into a decimal. The interest rate is 5%. Accrued Interest is the Interest amount you earn on a debt. Accrued Interest is noted as Revenue or Expense for a Bond selling or buying a loan respectively in Income Statements. Find the accrued interest on a bond as of today, 19 July 2013. Thus, the interest revenue recognized in 2019 is$525, and the interest earned for 2020 is $150 (total interest for 9 months of$675 less $525 earned in 2019). ALL RIGHTS RESERVED. Proper Interest Rate = No of Days from your most recent Interest Payment / Total number of days in a payment Period. Simple Interest means earning or paying interest only the Principal [1]. Calculate the accrued Interest that is yet to be received. Calculation of accrued interest is also import for financial reporting purpose. This should be noted. If you buy the bond for$960, you will have to pay $972.17, plus commission. By inputting these variables into the formula,$1000 times 10% times 3 … Step 4: After getting all the necessary values of the variables, it is applied in the below formula to calculate the Accrued Interest. These relationships are illustrated in the timeline below. The security's issue date is 01-Jan-2012, the first interest date is 01-Apr-2012, the settlement date is 31-Dec-2013 and the annual coupon rate is 8%. Here is the step by step approach for the calculation of Accrued Interest. Here we discuss How to Calculate Accrued Interest along with practical examples. It is often called as Current Asset or Current Liability since it is expected to be paid or gathered within a year of time or 6 months. A = P x R x (T / D) B = R /D x T Where, A = Accrued Interest P = Amount R = Interest Rate T = Days in Time period D = Days in Bond if Bond type is, Corporate and Municipal Bonds … Definition: Accrued interest is an accrual accounting term that describes interest that is due but hasn’t been paid yet. The Accrued period starts from Jan 1st to Dec 31st. Hence DCF will be
The following is multiple choice question (with options) to answer.
In what time will Rs.4000 lent at 3% per annum on simple interest earn as much interest as Rs.5000 will earn in 5 years at 4% per annum on simple interest? | [
"8 1/3",
"8 1/6",
"8 1/9",
"8 1/1"
] | A | (4000*3*R)/100 = (5000*5*4)/100
R = 8 1/3
Answer: A |
AQUA-RAT | AQUA-RAT-39862 | ## Dinner Party Seating
A host invites $$n$$ guests to a party (guest #1, guest #2, … , guest #n). Each guest brings with them their best friend. At the party there is a large circular table with \2n\) seats. All of the $$n$$ invited guests and their best friends sit in a random seat.
1. What is the probability that guest #1 is seated next to their best friend?
2. What is the expected number of the $$n$$ invited guests who are seated next to their best friend?
## Telephone Calls throughout the Week
Telephone calls come in to a customer service hotline. The number of calls that arrive within a certain time frame follows a Poisson distribution. The average number of calls per hour depends on the day of the week. During the week (Monday through Friday) the hotline receives an average of 10 calls per hour. Over the weekend (Saturday and Sunday) the hotline receives and average of 5 calls per hour. The hotline operates for 8 hours each day of the week. (The number of calls on one day is independent of the numbers of calls on other days.)
1. What is the probability that the center receives more than 500 calls in 1 week?
2. Each person who calls the center has a 20% chance of getting a refund (independent of other callers). Find the probability that 10 or fewer people get a refund on Tuesday.
3. One day of the week is chosen uniformly at random. On this day, a representative at the call center reports that 60 people called in. Based on that information, what is the probability that the day was a weekend day (either Saturday or Sunday)?
## A Dice Rolling Game
15 players each roll a fair 6-sided die once. If two or more players roll the same number, those players are eliminated. What is the expected number of players who get eliminated?
## January Birthdays at a Call Center
Calls arrive at a call center according to a Poisson arrival process with an average rate of 2 calls/minute. Each caller has a 1/12 chance of having a January birthday, independent of other callers. What is the expected wait time until the call center receives 3 calls from callers with January birthdays?
## Rock Paper Scissors
The following is multiple choice question (with options) to answer.
On a Monday in a certain restaurant, 20 percent of the 180 customers ordered the daily special. On Tuesday, only 10 percent of the 120 customers ordered the special. For the two days combined, what percent of the customers ordered the daily special? | [
"15%",
"16%",
"18%",
"23%"
] | B | 20% of 180= 36
10% of 120=12
Total =48
percent for 2 days combined =48/300 ≈ 50/300=1/6 =0.1666. so it should be less than 0.1666 so it is 16%
Another approach 48/300= 48/(3*100)= 16/100=0.16
Answer:B |
AQUA-RAT | AQUA-RAT-39863 | main: # @main
pushq %rbp
movq %rsp, %rbp
pushq %r15
pushq %r14
pushq %r13
pushq %r12
pushq %rbx
pushq %rax
movl $1, %r15d leaq -42(%rbp), %r14 .LBB0_1: # %._crit_edge cmpl$100, %r15d
movl $0, %edi movl$60, %eax
cmovgl %eax, %edi
movslq %r15d, %r15
cmpl $101, %r15d setl %al movzbl %al, %esi orl %esi, %edi imulq$1431655766, %r15, %rcx # imm = 0x55555556
imulq $1717986919, %r15, %rax # imm = 0x66666667 movq %rax, %r10 shrq$63, %r10
shrq $32, %rax movl %eax, %edx sarl$2, %edx
leal (%rdx,%r10), %ebx
movq %rcx, %r11
shrq $63, %r11 cmpl$9, %r15d
setg %r8b
imull $10, %ebx, %ebx negl %ebx leal 48(%rdx,%r10), %edx movb %dl, -42(%rbp) leal 48(%r15,%rbx), %r9d shrq$32, %rcx
leal (%rcx,%rcx,2), %ecx
movl %r15d, %edx
subl %ecx, %edx
sete %r12b
movzbl %r8b, %ecx
movb %r9b, -42(%rbp,%rcx)
setne %bl
sarl %eax
leal (%rax,%rax,4), %eax
The following is multiple choice question (with options) to answer.
emblem is coded as 216;
crude is coded as 125
bump will be ? | [
"86",
"72",
"64",
"32"
] | C | coding follows (no. of letters)^3
emblem = 216 = 6^3
crude = 125 = 5^3
bump = 4^3 =64
ANSWER:C |
AQUA-RAT | AQUA-RAT-39864 | 1111121, 1111202, 1111211, 1112002, 1120002, 1120012, 1120102, 1121002, 1121102, 1122002, 1200002, 1200012, 1200102, 1200112, 1200202, 1201002, 1201012, 1202002, 1210002, 1210102, 1210202, 1211002, 1212002, 1220002, 1220102, 2000002, 2000012, 2000022, 2000102, 2000112, 2000122, 2000202, 2000212, 2001002, 2001012, 2001022, 2001102, 2001112, 2001122, 2001202, 2001212, 2010012, 2010022, 2011012, 2020012, 2020022,$ and their inverses.
The following is multiple choice question (with options) to answer.
If 9210 - 9124 = 210-square, the value represented by the square is | [
"296",
"210",
"186",
"124"
] | D | Evaluating the left side of the equation, we get 9210 -9124 = 86.
Therefore the right side of the equation, 210 - , must also equal 86.
Since 210 -124 = 86, then the value represented by the is 124.
correct answer D |
AQUA-RAT | AQUA-RAT-39865 | (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.
A baker charges W dollars for a birthday cake. Next week, she intends to earn 320 dollars selling cakes. If she were to reduce the price of the cake by 20 percent, she would have to sell two more cakes to earn the same amount of revenue. How many cakes does she intend to sell next week? | [
"2",
"4",
"6",
"8"
] | D | We're told that by selling cakes at W dollars/cake, a baker will make $320.
Reducing the price by 20% and selling 2 MORE cakes will also make her $320.
We're asked for the original number of cakes that the baker intends to sell.
$320 is a rather interesting number. Based on the answer choices, we're almost certainly dealing with integer values for the number of cakes AND the price per cake. As such, 6 is probably NOT going to be part of the solution (either as 4 and 6 cakes or as 6 and 8 cakes). As such, we can avoid Answers B and C....
Let's TEST ANSWER D: 8 cakes
IF....
Original cakes = 8
8(W) = $320, so W = $40/cake
20% off = $8 off = 40-8 = $32/cake
+2 more cakes = 8+2 = 10 cakes
10(32) = $320
This is an exact MATCH for what we were told, so this MUST be the answer.
Final Answer:
D |
AQUA-RAT | AQUA-RAT-39866 | 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.
The length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is? | [
"388",
"267",
"245",
"288"
] | C | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (130 + X)/30 = 25/2 ==> 2(130 + X) = 750 ==> X = 245 m.
Answer: C |
AQUA-RAT | AQUA-RAT-39867 | • I see that I answered the question after it was edited to be correct. It was probably that edit which brought this question to the top of the front page just as I was looking.
– robjohn
Mar 15 at 22:09
Generalization of the problem:
• What is the sum of number of digits of the numbers $$2^N$$ and $$5^N$$?
$$10^m<\underbrace {2^{N}}_{m+1 ~ \text{digits}}<10^{m+1}$$
$$10^n<\underbrace{5^{N}}_{n+1 ~ \text{digits}}<10^{n+1}$$
$$10^{m+n}<10^{N}<10^{m+n+2}$$
$$N= m+n+1$$
$$m+n=N-1$$
The sum of digits of the numbers $$2^{N}$$ and $$5^{N}$$ will be equal :
\begin{align}\color {gold}{\boxed {\color{black}{m+1+n+1=m+n+2\\ \qquad \qquad \qquad\thinspace=N-1+2 \\\qquad \qquad \qquad \thinspace=N+1.}}}\end{align}
• Short answer: $$N+1$$ digits.
The following is multiple choice question (with options) to answer.
What is the sum of the digits of integer k, if k = (10^50- 50) | [
"337",
"437",
"457",
"537"
] | B | There are 51 digits in 10^50
When we subtract 50 from it, there will be 50 digits left.
10^50 can be written as 9999999....(50 times) + 1
So,
10^50 - 50 = 9999999....(50 times) + 1 - 50 = 9999999....(50 times) - 49
Consider the last 2 digits,
99-49 = 50
The last 2 digits will be 50.
And our number would be 99999......99950 with 2 less 9s after subtraction.
Number of 9s left are 48 and the last two digits are 50
The sum of the digits will be
(48*9) + 5 + 0 = 437
Answer:- B |
AQUA-RAT | AQUA-RAT-39868 | ## eliassaab Group Title Let m and n be two positive integers. Show that (36 m+ n)(m+36 n) cannot be a power of 2. 2 years ago 2 years ago
1. sauravshakya
LET (36m+n)=2^x AND (m+36n)=2^y THEN, (36m+n)(m+36n)=2^(x+y) AlSO, x and y must me both integers
2. sauravshakya
Now, n=2^x - 36m THEN, m+36n=2^y m+36(2^x-36m)=2^y 36 * 2^x -1295m=2^y
3. sauravshakya
Now, I think I have to prove that y is never a integer
4. sauravshakya
2^y=36*2^x-1295m 2^y=2^x(36-1295m/2^x) Now, 2^y must be positive so, (36-1295m/2^x) must be positive... Now let 2^z=36-1295m/2^x HERE z also must be an positive integer..... So, 2^z can be 4,8,16,32 NOW,
5. sauravshakya
36-1295m/2^x= 4, 8, 16, 32 1295m/2^x=32, 28 ,16 ,4 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4
6. sauravshakya
Now, 2^x=1295m/32 , 1295m/28 , 1295m/16 , 1295m/4 2^x=40.47 m , 46.25m , 80.94m ,323.75m Thus, for no positive integer value of m we will get x a integer value......
7. sauravshakya
Hence, (36 m+ n)(m+36 n) cannot be a power of 2.
8. sauravshakya
I have jumped some step......... I hope I made it clear
9. sauravshakya
The following is multiple choice question (with options) to answer.
In Karthik's opinion, his weight is greater than 55 Kg but less than 62 Kg. His brother does not agree with Karthik and he thinks that Karthik's weight is greater than 50 Kg but less than 60 Kg. His father's view is that his weight cannot be greater than 58 Kg. If all of them are correct in their estimation, what is the average of different probable weights of Karthik? | [
"54.5",
"56.5",
"59.2",
"61"
] | B | Explanation :
Solution: Assume Karthik's weight be x Kg.
According to Karthik, 55 < x < 62
According to Karthik;s brother, 50 < x < 60.
According to karthik's mother, x <58.
The values satisfying all the above conditions are 56 and 57.
.'. Required average = (56+57)/2 = 56.5.
Answer : B |
AQUA-RAT | AQUA-RAT-39869 | Substitute 0.085 for interest rate, $3000 for the amount deposited and$6000 for the amount after t years in the formula, “A=Pert” as,
$6000=$3000e0.085t
Divide both sides by $3000 as,$6000$3000=$3000e0.085t$3000e0.085t=2 Take natural logarithm on both sides as, ln(e0.085t)=ln2 Now, apply the inverse property of lnex as, 0.085t=ln2 Divide both sides by 0.085 as, 0.085t0.085=ln20.085t0.6930.0858.15 years So, the deposited amount will be doubled in approximately 8.15 years when the interest rate is 8.5%. Now, as the amount will become quadruple of the deposited amount after t years, then the amount after t years is 4P, that is, A=4P=4($3000)=\$12000
Substitute 0
Still sussing out bartleby?
Check out a sample textbook solution.
See a sample solution
The Solution to Your Study Problems
Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!
Get Started
The following is multiple choice question (with options) to answer.
The banker's discount on Rs.1800 at 24% per annum is equal to the true discount on Rs.1872 for the same time at the same rate. Find the time? | [
"2 months",
"4 months",
"5 months",
"6 months"
] | A | Solution
S.I on Rs.1800 = T.D on Rs.1872.
P.W on Rs.1872 is Rs.1800.
Rs.72 is S.I on Rs. 1800 at 24%.
Time =(100x72 / 24x1800)
= 1/6 year
= 2 months.
Answer A |
AQUA-RAT | AQUA-RAT-39870 | only effective. Share, and after five years it earned you$ 15 in income 2 years or decreases ) in! A certain period of time that an investment over time as a percentage of investment. Zijn bidirectioneel, wat wil zeggen dat je woorden gelijktijdig in beide talen kan.! Are returned to you not reinvest would have $40 per share starting value and the rate of return! Final investment value of the funds as of the investment 's purchase price to. Or short position op Ergane en Wiktionary if ) all the investors in taxable accounts.. Return on assets, return is a return of investment before all the possible expenses and fees in a it! Owned a house for 10 years possible expenses and fees in a certain period of time that investment. Retiree % ) 2017 their symmetry, as noted above and a bond differently by ( 1/ ’. The conversion is called the rate of return at each possible outcome by its and... For Spanish translations proportion of the account the interest is withdrawn at the point in time the! Verlies oplevert dan is de return on investment een negatief getal$ stock price translates an! At irregular intervals ( MWRR ) or as a percentage total distributions cash! 2020, tenzij anders vermeld concepts in asset valuation hypothetical initial payment of $103.02 compared with the initial ]. % per year compensate for the year is 2 %, in more recent years, personalized! Personalized account returns on investor 's account statements in response to this need the! Year is 4.88 % equals 20 percent income tax purposes, include the reinvested dividends the. On Investing in marketing cost of funds your nominal rate of return - the amount invested determine your nominal of... Dividends in the account uses compound interest, meaning the account than the average! Well connected to the equation, requiring some interpretation to determine which security will higher... I.E., optimized returns and after five years it earned you$ in! This need only if ) all the possible expenses and fees in a certain period of time that investment... Return CALCULATOR - mortgage income CALCULATOR rate at which shipped items are returned to you at intervals! Different periods of time shares of the portfolio, from the investment 's purchase price refers to the end January. Is
The following is multiple choice question (with options) to answer.
An investment compounds annually at an interest rate of 30% What is the smallest investment period by which time the investment will more than double in value? | [
"3",
"4",
"5",
"6"
] | A | 1 year: 100/3 = 33.33 approx $34: Total: 134
2nd year: 134/3 = 45: Total: 134+45 = 179
3rd year: 179/3 = 60: Total: 179+60 = 239 > 2 (100)
;3 years;
ANSWER:A |
AQUA-RAT | AQUA-RAT-39871 | So the 1st question you needed to answer was how much distance do they cover together per hour (we just did that), now the next question to answer is how long it takes to cover 810km in total distance at their effective speed.
• it means that if they are covering 135km each our then they would be meet together in 135xP = 810km right ? Where P is the number of hours they will meet after. like 810km/135km/hr = hr6 so, 10:30AM + 6Hours = 4:30PM right ? – Joe Jan 19 '15 at 18:52
• Yes, this is exactly right. – Acemanhattan Jan 19 '15 at 18:56
• But that's what the question was asking. When will they meet. If they were asking when will each cover 810km you would divide 810 by the slower train's speed. – turkeyhundt Jan 19 '15 at 19:00
• Suppose for a second that they each did travel 810km, where would they each end up? Wouldn't they have just traded places with the person who started at A being at B and the person who started at B being at A? In order for that to happen, at some point they would have had to pass each other along the way, and when they passed each other neither would have traveled 810km individually, but the distance between them would be zero. – Acemanhattan Jan 19 '15 at 19:02
• Thanks a lot for great explanation. =) – Joe Jan 19 '15 at 19:03
Hint: Think about how much closer the cars get each hour.
They are approaching each other at an effective speed of 135 km/hr...
The following is multiple choice question (with options) to answer.
A hiker walking at a constant rate of 5 kilometers per hour is passed by a cyclist travelling in the same direction along the same path at a constant rate of 25 kilometers per hour. The cyclist stops and waits for the hiker 5 minutes after passing her while the hiker continues to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up? | [
"10",
"15",
"20",
"25"
] | C | In 5 minutes, the cyclist travels a distance of (5/60)*25 = 25/12 km.
The time it takes the hiker to complete this distance is (25/12) / 5 = 5/12 hours = 25 minutes
The cyclist needs to wait 25 - 5 = 20 minutes
The answer is C. |
AQUA-RAT | AQUA-RAT-39872 | in with angle 1 degrees, Area of a Parallelogram diagonal-e 83 m diagonal-f 73 ft with angle 0.5 radians, Area of a Parallelogram diagonal-e 75 in diagonal-f 22 yd with angle 130 degrees, Area of a Parallelogram diagonal-e 63 ft diagonal-f 47 m with angle 12 degrees, Area of a Parallelogram diagonal-e 23 cm diagonal-f 17 in with angle 0.8 radians, Area of a Parallelogram diagonal-e 60 ft diagonal-f 25 m with angle 2.6 radians. Math operations to check the side length and opposite angles are equal in measure free! Grade 5 through grade 8 have been included here obtained in the above steps to calculate area... B ) between any two sides of a parallelogram are equal ( a equal! Magnitude of cross-vector products for two adjacent sides are a, b and a b. Here are the length of a parallelogram with a and b is equal with b.! A simple quadrilateral with two pairs of parallel sides the above steps to calculate of! Each other grade 5 through grade 8 have been included here useful for everyone save. Q and sides are parallel to each other at the center of the parallelogram a. Area are listed here, 1 graphs for each calculation Width::... Parallelogram sides are in equal length, corner angles or angles or any others always! Results: area of a parallelogram is base x height an enormous range of area of a parallelogram easier. Are using base and height as in the details below: you need two measurements calculate! 11 cm magnitude of cross-vector products for two adjacent sides, one corner angle from the question angle. Is 12 cm and angle is 75° 2: a parallelogram = b×h square units = 4 × =..., find the area of a parallelogram calculator is a free online calculator with step by solution. Calculate ” to get your area of a parallelogram = base × height square units inside polygon! 3, then your area of a parallelogram is a perfect example of the parallelogram area a... Are corner angles, diagonals of a parallelogram = 20 cm 2 are 3 cm, find the distance its. Is given by area of a parallelogram = 20 sq.cm to build a
The following is multiple choice question (with options) to answer.
The diagonal of a square is 40 m. The area of the square is: | [
"600",
"900",
"800",
"500"
] | C | Arear = 1/2 × ( diagonal )(power)2
= ( 1/2 × 40 × 40 )m(power)2
= 1600/2 = 800 m(power)2
Answer is C. |
AQUA-RAT | AQUA-RAT-39873 | 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.
The no. of girls in a class are seven times the no. of boys, which value cannever be the of total students? | [
"23",
"25",
"30",
"36"
] | C | Let the boys are X, then girls are 7X, total = X+7X = 8X
So it should be multiple of 8, 30 is not a multiple of 8.
C |
AQUA-RAT | AQUA-RAT-39874 | # 99 Consecutive Positive Integers whose sum is a perfect cube?
What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?
• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50
Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is
$$(x-2)+(x-1)+x+(x+1)+(x+2)$$
$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$
Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)
$99=3^2\cdot 11^1$ is missing some factors to be a cube.
Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$
Also one can prove that there are no other solutions!
The following is multiple choice question (with options) to answer.
The sum of five consecutive even numbers of set x is 440. Find the sum of a different set of five consecutive integers whose second least number is 121 less than double the least number of set x? | [
"240",
"273",
"328",
"428"
] | A | A
240
Let the five consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) and 2(x + 2)
Their sum = 10x = 440
x = 44 => 2(x - 2) = 84
Second least number of the other set = 2(84) - 121 = 47
This set has its least number as 46.
Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50
= 48 - 2 + 48 - 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240 |
AQUA-RAT | AQUA-RAT-39875 | # 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.
Dana borrows 5500 pounds annually for her college education. If Dana gives her parents 3% of that amount back each month, how much will she still owe her parents after four years of college?
. | [
"12,430.",
"13,640.",
"14,000.",
"14,080."
] | D | dana borrows 5500 and returns 3% per month.
so returns 165 per month. and 1980 per year.
so dana owes 5500-1980=3520 per year.
after 4 years she will owe 14080.
Answer: (D) |
AQUA-RAT | AQUA-RAT-39876 | +0
# the amount of the price
0
284
4
if the price of a pencil is 36% lower than the price of a pen,then the price of a pen is ?
1.)36%higher than a pencil
2.)43.75 higher than a pencil
3.)56.25 % higher than a pencil
4.)64% higher than a pencil
Guest Feb 17, 2015
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
CPhill Feb 17, 2015
Sort:
#1
+84384
+10
If the price of a pencil is 36% lower than the price of a pen, then the pencil's price is 1- 36% of the pen's price =.64pen
For argument's sake, let the pen's price be 1 .....Then the pencil's price = .64
Then
.64( 1 + P) = 1 where P is the % we're looking for...divide both sides by .64
1 + P = 1/.64
1 + P = 1.5625 subtract 1 from both sides
P = .5625 = 56.25%
Then the pen is 56.25% more than the pencil.
Another way to see this is that .36 must be added to the pencil's price of .64 to get the pen's price.
Thus 36/64 = 56.25% of the pencil's price must be added.......
.
The following is multiple choice question (with options) to answer.
The cost of 3 pens and 5 pencils is Rs.100. Also the cost of one pen and one pencil is in the ratio of 5:1 respectively. What is the cost of one dozen pens? | [
"Rs.200",
"Rs.250",
"Rs.300",
"Rs.150"
] | C | Explanation:
Let the cost of one pen is ‘5x’ and pencil is ‘x’
3 X 5x + 5x = Rs.100
15x + 5x = Rs.100
X = 100/20 = 5
:. Cost of 1 pen = 5x = 5 x 5 = 25
:. Cost of 12 pens, i.e. (one dozen) = 25 x 12 = Rs.300
Answer: Option C |
AQUA-RAT | AQUA-RAT-39877 | c, json, formatting, c99
four times. That really asks for a function:
/**
* \brief Puts the given \c str \c count times on \c stream.
* \param str null-terminated character string to be written
* \param count number of times \c str shall be written
* \param stream output stream
* \returns a non-negative value on success
* \returns EOF on error and sets the error indicator
*/
inline void fputs_repeat(const char * str, size_t count, FILE * stream) {
int value = 0;
while (count-- > 0) {
value = fputs(str, stream);
if (value == EOF) {
return EOF;
}
}
return value;
}
Now we can just use fputs_repeat(placeholder, indentation, stdout) wherever you've used while (i-- > 0) .... We would now end up with the following variant:
#include <stdio.h>
#include <stdlib.h> /* EXIT_SUCCESS */
#define BUFFER_SIZE (1024 * 4)
inline void fputs_repeat(const char * str, size_t count, FILE * stream) {
int value = 0;
while (count-- > 0) {
value = fputs(str, stream);
if (value == EOF) {
return EOF;
}
}
return value;
}
int main(int argc, char **argv) {
char buffer[BUF_SIZE] = {0};
const char placeholder[] = " ";
unsigned int indent = 0;
char is_string = 0;
char escaped = 0;
size_t n;
while (0 < (n = fread(&buffer, sizeof(buffer[0]), BUFFER_SIZE, stdin))) {
// exercise: add error handling
for (unsigned int k = 0; k < n; k++) {
char ch = buffer[k];
The following is multiple choice question (with options) to answer.
How many times the keys of a writer have to be pressed in order to write first 400 counting no.'s? | [
"998",
"1000",
"1092 times",
"1100"
] | C | 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 400 = 301 * 3 = 903
-----------
1092
C |
AQUA-RAT | AQUA-RAT-39878 | Could you please verify that my 1st and 3rd prize probabilities are well calculated and help me calculate the 2nd prize probability?
Your logic for case number (3) does not seem correct.
First, you have to draw 6 balls with exactly 5 balls matching 5 numbers from your set of 6 picked numbers. In total There are C(42,6) ways to draw six balls. How many winning combinations do we have? From your set of 6 balls you can pick five matching balls in C(6,5) ways. The sixth number can be any number from the remaining 42-5=37, minus one (your sixth pick). So the total number of winning combinations is $$C(6,5)\times36$$. So the probability is:
$$p_1=\frac{36\times{6\choose 5}}{42\choose 6}$$
But you have to draw the seventh number as well from the remaining 42-6=36 numbers without hitting your last (sixth) number. That probability is:
$$p_2=\frac{35}{36}$$
The total proability is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{35}{36}=\frac{35\times{6\choose 5}}{42\choose 6}=\frac{15}{374699}$$
You can use a similar logic for case (2).
The probability $$p_1$$ is the same. For the seventh ball we have 36 choices and there is 1 winning ball between them. So the probability $$p_2$$ is:
$$p_2=\frac{1}{36}$$
...and the final probability for the second prize is:
$$p=p_1\times p_2=\frac{36\times{6\choose 5}}{42\choose 6}\times\frac{1}{36}=\frac{{6\choose 5}}{42\choose 6}=\frac{3}{2622893}$$
(35 times smaller than the probability for the third prize)
The following is multiple choice question (with options) to answer.
Six participants are participating in a competition. In how many ways can the first three prizes be won? | [
"920",
"120",
"820",
"720"
] | B | Out of 6 participants, the first three prizes can be won in,
6P3 = 6! / (6 - 3)!
= 6 x 5 x 4
= 120 ways
ANSWER:B |
AQUA-RAT | AQUA-RAT-39879 | We can argue that the are 21/7= 3 times as many men now so they will require three times as much water, 3(13)= 39 quarts just for the same two weeks. And they need the water for 4/2= 2 times as many weeks so need 2(39)= 78 quarts of water.
The following is multiple choice question (with options) to answer.
Average expenditure of a person for the first 3 days of a week is Rs. 320 and for the next 4 days is Rs. 420. Average expenditure of the man for the whole week is: | [
"350",
"370",
"390",
"360"
] | D | Explanation:
Assumed mean = Rs. 320
Total excess than assumed mean = 4 × (Rs. 420 - Rs. 350) = Rs. 280
Therefore, Increase in average expenditure = Rs. 280/7 = Rs. 40
Therefore, Average expenditure for 7 days = Rs. 320 + Rs. 40 = Rs. 360
Correct Option: D |
AQUA-RAT | AQUA-RAT-39880 | 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.
he length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is: | [
"288",
"266",
"245",
"277"
] | C | Speed = (45 * 5/18) m/sec = (25/2) m/sec. Time = 30 sec. Let the length of bridge be x meters. Then, (130 + X)/30 = 25/2 ==> 2(130 + X) = 750 ==> X = 245 m.Answer: C |
AQUA-RAT | AQUA-RAT-39881 | How do you solve this?
There is this kind of question in our test and I don't know how will I do it.
You're working in a company. Your starting income is 5000. Every year, the income will increase by 5%. What is your total income on your 25th year in the company?
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
This is a question, not a tutorial so I am moving out of "Learning Materials" to "Precalculus Homework and School Work".
You startwith income at 5000 and it increases by 5% each year.
Okay, the first year your increases by "5% of 5000"= .05(5000)= 250 so your income the second year is 5250= 5000+ (.05)5000= (1.05)(5000). At the end of that year it increases by "5% of 5250"= .05(5250)= 262.50 and the third year your income is 5512.50= 5250+ (.05)5250= (1.05)(5250)= 1.05(1.05(5000)). The reason I wrote it out like that is because neither you nor I want to do that 24 times! (During your 25th year, your income will have increased 24 times.) You should be able to see what is happening: each year your income is multiplied by 1.05. After 24 years, that initial 5000 is multiplied by 1.05 24 times: $(1.05)^{24}(5000)$.
$$a_1=5000$$
$$a_2=a_1+a_1*\frac{5}{100}=a_1*1.05$$
$$a_3=a_1*1.05 + a_1*1.05*0.05=a_1*1.05(1 + 0.05)=a_1*1.05*1.05$$
$$a_4=a_1*1.05*1.05*1.05$$
$$...................................$$
The following is multiple choice question (with options) to answer.
A car salesman earns a base salary of $1,000 per month plus a commission of $200 for each car he sells. If the car salesman earned $1800 in January, how many cars does he need to sell in February in order to double his January earnings? | [
"13",
"14",
"15",
"16"
] | A | 1000+200x=3600
x=13 cars.
The answer is A. |
AQUA-RAT | AQUA-RAT-39882 | • I am also stuck at this point. If the first rose is $Red$ then the Bride enters the Church and in that case the probability is $\frac{10}{20}$. But now come the cases when the first rose is $White$: $WRR$, $WWRRR$, $WRWRR$ and so on. No matter what, if the first rose is $White$, the last two roses must be $Red$. And the total number of roses required ($\leqslant 20$) to enter the church is $Odd$, where the number of $Red$ roses will never exceed that of the $White$ roses but once when the Bride finally enters the church. Leaves me in doldrums, though. – JackT Oct 18 '17 at 7:42
• @Maths_student Actually, if you have got $x_1$ Red Roses and $y_1$ White roses, then you must get $y_1-x_1+1$ more red roses to enter the church. Or, rather, $x_2-y_2$ should be $y_1-x_1+1$. – MalayTheDynamo Oct 18 '17 at 7:49
• Your edit is incorrect. If she takes a red rose on the first try, she enters, so the probability has to be at least $1/2$ as you said. She can also enter the church if she initially takes a white rose, then takes two red roses in a row. Clearly, the probability is much greater than $1/20$. – N. F. Taussig Oct 18 '17 at 9:34
The following is multiple choice question (with options) to answer.
A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them in a row, what is the probability that the 2 rosebushes in the middle of the row will be the red rosebushes? | [
"1/12",
"1/6",
"1/5",
"1/3"
] | B | We are asked to find the probability of one particular pattern: WRRW.
Total # of ways a gardener can plant these four bushes is the # of permutations of 4 letters WWRR, out of which 2 W's and 2 R's are identical, so 4!/2!2!=6;
So p=1/6
Answer: B. |
AQUA-RAT | AQUA-RAT-39883 | • #### 1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.
1. \begin{aligned} 120m^2 \end{aligned}
2. \begin{aligned} 130m^2 \end{aligned}
3. \begin{aligned} 140m^2 \end{aligned}
4. \begin{aligned} 150m^2 \end{aligned}
Explanation:
\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
• #### 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
1. \begin{aligned} 152600 m^2\end{aligned}
2. \begin{aligned} 153500 m^2\end{aligned}
3. \begin{aligned} 153600 m^2\end{aligned}
4. \begin{aligned} 153800 m^2\end{aligned}
Explanation:
Question seems to be typical, but trust me it is too easy to solve, before solving this, lets analyse how we can solve this.
We are having speed and time so we can calculate the distance or perimeter in this question.
Then by applying the formula of perimeter of rectangle we can get value of length and breadth, So finally can get the area. Lets solve it:
Perimeter = Distance travelled in 8 minutes,
=> Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time]
The following is multiple choice question (with options) to answer.
The length of the rectangular field is double its width. Inside the field there is square shaped pond 8m long. If the area of the pond is 1/8 of the area of the field. What is the length of the field? | [
"36",
"32",
"19",
"10"
] | B | A/8 = 8 * 8 => A = 8 * 8 * 8
x * 2x = 8 * 8 * 8
x = 16 => 2x = 32
Answer: B |
AQUA-RAT | AQUA-RAT-39884 | Veritas Prep Reviews
Math Expert
Joined: 02 Sep 2009
Posts: 37109
Followers: 7252
Kudos [?]: 96499 [0], given: 10752
### Show Tags
05 Jun 2013, 23:44
PKPKay wrote:
Sarang wrote:
For 1 hour-
Machine A rate- 2000 envelopes
Machine B+C rate- 2400 envelopes
Since A + C = 3000 envelopes A's rate is 2000 envelopes as above, C has a rate of 1000 envelopes per hour.
Which makes machine B's rate as 1400 envelopes per hour.
Thus, it will take 8 hours to manufacture 12000 envelopes.
I did this but shouldn't the work take 9 hours instead?
In 8 hours machine B would have made 1400 * 8 = 11200 envelopes.
In order to make 12000 it would require a fraction of an hour to create 200 more envelopes.
Am I mistaken?
Edited the options.
Check for a solution here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823509 or here: machine-a-can-process-6000-envelopes-in-3-hours-machines-b-105362.html#p823655
_________________
Intern
Joined: 18 Mar 2013
Posts: 5
Followers: 0
Kudos [?]: 2 [0], given: 45
Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
07 Jun 2013, 04:35
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7187
Location: Pune, India
Followers: 2168
Kudos [?]: 14022 [0], given: 222
Re: Machine A can process 6000 envelopes in 3 hours. Machines B [#permalink]
### Show Tags
09 Jun 2013, 19:52
samheeta wrote:
How much time should one take in solving these kind of questions which involves though simple yet a lot of calculations?
This can be easily done in under 2 mins. If you look at the explanation provided above:
The following is multiple choice question (with options) to answer.
Rane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks. If they both work together, how many weeks will it take for them to produce 15 handcrafted drums? | [
"30",
"36",
"70",
"80"
] | B | Method I: the rates solution
“Rane can make a handcrafted drum in 4 weeks. Zane can make a similar handcrafted drum in 6 weeks.” Rane’s rate is (1 drum)/(4 weeks) = 1/4. Zane’s rate is (1 drum)/(6 weeks) = 1/6. The combined rate of Rane + Zane is
R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12
That’s the combined rate. We need to make 15 drums — we have a rate and we have an amount, so use the “art” equation to solve for time:
T = A/R = 15/(5/12) = 15*(12/5) = (15/5)*12 = 3*12 = 36
BTW, notice in the penultimate step, the universal fraction strategy: cancelbeforeyou multiply (Tip #3:http://magoosh.com/gmat/2012/can-i-use- ... -the-gmat/. Rane and Zane need 36 weeks to make 15 drums.
Answer = B. |
AQUA-RAT | AQUA-RAT-39885 | # In how many ways can 3 distinct teams of 11 players be formed with 33 men?
Problem:
In how many ways can 3 distinct teams of 11 players be formed with 33 men? Note: there are 33 distinct men.
The problem is similar to this one: How many distinct football teams of 11 players can be formed with 33 men?
Fist, I thought the answer was: $$\binom{33}{11} \times \binom{22}{11} \times \binom{11}{11}$$
But there are clearly a lot of solutions overlapping.
-
Suppose that we wanted to divide the $33$ men into three teams called Team A, Team B, and Team C, respectively. There are $\binom{33}{11}$ ways to pick Team A. Once Team A has been picked, there are $\binom{22}{11}$ ways to pick Team B, and of course the remaining $11$ men form Team C. There are therefore $$\binom{33}{11}\binom{22}{11}\tag{1}$$ ways to pick the named teams. This is the calculation that you thought of originally.
But in fact we don’t intend to name the teams; we just want the men divided into three groups of $11$. Each such division can be assigned team names (Team A, Team B, Team C) in $3!=6$ ways, so the calculation in $(1)$ counts each division of the men into three groups of $11$ six times, once for each of the six possible ways of assigning the three team names. The number of ways of choosing the unnamed teams is therefore
$$\frac16\binom{33}{11}\binom{22}{11}\;.\tag{2}$$
Added: Here’s a completely different way to calculate it.
The following is multiple choice question (with options) to answer.
30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years? | [
"60%",
"80%",
"50%",
"90%"
] | B | Explanation:
Let total number of men = 100
Then
80 men are less than or equal to 50 years old
(Since 80% of the men are less than or equal to 50 years old)
=> 20 men are above 50 years old (Since we assumed total number of men as 100)20% of the men above the age of 50 play football⇒Number of men above the age of 50 who play football = 20 × 20/100 = 4
Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = 420 × 100 = 20%
=>Percentage of men who play football less than or equal to the age 50 = 100% − 20% = 80%
Answer: Option B |
AQUA-RAT | AQUA-RAT-39886 | (1) The median of set {x,−1,1,3,−x} is 0
(2) The median of set {x,−1,1,3,−x} is x2
Merging similar topics. Please refer to the solutions above.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15958
Re: What is the value of x? [#permalink]
### Show Tags
09 Jun 2017, 15:49
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: What is the value of x? [#permalink] 09 Jun 2017, 15:49
Similar topics Replies Last post
Similar
Topics:
3 What is the value of x + 7 ? 14 01 Jul 2014, 10:06
31 What is the value of integer x ? 11 15 Feb 2016, 09:11
8 What is the value of x? 6 05 May 2016, 14:17
4 What is the value of x/yz? 10 16 Sep 2016, 15:54
7 What is the value of x? 9 19 Feb 2014, 00:14
Display posts from previous: Sort by
The following is multiple choice question (with options) to answer.
Set A: 3, e, 8, 10
Set B: 4, y, 9, 11
The terms of each set above are given in ascending order. If the median of Set A is equal to the median of Set B, what is the value of y – e? | [
"-2",
"-1",
"0",
"1"
] | B | So we have even no. of elements in the Set
So median is the average of Middle two numbers
(e+8)/2= (y+9)/2
y - e= -1
Answer B |
AQUA-RAT | AQUA-RAT-39887 | 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 woman and a girl complete a work together in 24 days. If for the last 6 days woman alone does the work then it is completed in 26 days. How long the girl will take to complete the work alone? | [
"days",
"days",
"days",
"days"
] | A | Explanation:
(woman + girl )’s 1 day’s work = 1/24
Their 20 day’s work = 1/24 × 20 = 5/6
The remaining 1/6 work is done by the woman in 6days
Therefore, the woman alone will finish the work in 6 × 6 days = 36 days
woMan’s 1 day’s work = 1/36
Therefore, girl 1 day’s work = 1/24 – 1/36 = 3 – 2 /72 = 1/72
Therefore, the girl alone will finish the work in 72 days.
Answer: Option A |
AQUA-RAT | AQUA-RAT-39888 | # 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.
The average weight of 20 persons sitting in a boat had some value. A new person added to them whose weight was 48 kg only. Due to his arrival, the average weight of all the persons decreased by 5 kg. Find the average weight of first 20 persons? | [
"55",
"56",
"57",
"58"
] | C | 20x + 48 = 21(x – 5)
X=57
ANSWER:C |
AQUA-RAT | AQUA-RAT-39889 | (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.
There were 35 students in a hostel. Due to the admission of 7 new students, the expenses of mess were increased by 42 per day while the average expenditure per head diminished by 1. What was the original expenditure of the mess? | [
"400",
"420",
"445",
"465"
] | B | Let the original average expenditure be `x.
Then,
42(x – 1) – 35x = 42 ⇔ 7x = 84 ⇒ x = 12.
∴ Original expenditure = (35 × 12)= 420.
Answer B |
AQUA-RAT | AQUA-RAT-39890 | Sol:
Folks, look at the relative calculation here.
If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.
So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.
We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.
Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.
Hence the total number of students who passed = 100
You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.
Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....
_________________
Manager
Joined: 22 Feb 2009
Posts: 140
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Followers: 8
Kudos [?]: 79 [1] , given: 10
Re: Average Accelerated: Guide to solve Averages Quickly [#permalink] 19 Apr 2009, 21:07
1
KUDOS
cicerone wrote:
Problem 5:
The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.
Sol:
Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.
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
The following is multiple choice question (with options) to answer.
Mike needs 30% to pass. If he scored 212 marks and falls short by 16 marks, what was the maximum marks he could have got? | [
"643",
"677",
"760",
"767"
] | C | If Mike had scored 16 marks more, he could have scored 30%
Therefore, Mike required 212 + 16 = 228 marks
Let the maximum marks be m.
Then 30 % of m = 228
(30/100) × m = 228
m = (228 × 100)/30
m = 22800/30
m = 760
Answer:C |
AQUA-RAT | AQUA-RAT-39891 | 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.
If A is thrice as fast as B and together can do a work in 24 days. In how many days A alone can do the work? | [
"36",
"42",
"28",
"32"
] | D | A’s one day’s work= 1/X
B’s one day’s work= 1/3x
A + B’s one day’s work= 1/x + 1/3x = 1/24
= 3+1/3x = 4/3x = 1/24
x = 24*4/3 = 32
ANSWER:D |
AQUA-RAT | AQUA-RAT-39892 | # Faulty Combinatorial Reasoning?
I have 10 books, 4 of which are biographies while the remaining 6 are novels. Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies. How many different combinations of choosing 4 books in such a way are there?
The following line of reasoning is faulty, but I can't figure out why:
First we figure out how many ways there are of choosing 2 biographies from 4. Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8. This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices. Then we have:
1. BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).
2. REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order. This leaves us with (8*7)/2! choices.
3. Overall we have [(4*3)/2!]*[(8*7)/2!] = 168 total choices.
Where did I go wrong?
-
How could I adjust for the over-counting I did here? (Rather than constructing the answer of 115 by adding together the discrete cases of choosing 2 bios, 3 bios, and 4 bios)? – George Sep 4 '12 at 22:15
In your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_£$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.
To solve the problem:
The following is multiple choice question (with options) to answer.
At Joel’s bookstore, the current inventory is 30% historical fiction. Of the historical fiction books, 40% are new releases, while 50% of the other books are new releases. What fraction of all new releases are the historical fiction new releases? | [
"4/25",
"12/47",
"2/5",
"8/15"
] | B | let there be 100 books in all
Historic fiction books = 30% of total = 30
Other books = 70
New historic fiction = 40% of 30 = 12
Other new books = 50% of 70 = 35
total new books = 35+12 = 47
fraction = 12/47
Ans: B |
AQUA-RAT | AQUA-RAT-39893 | 1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
_________________
# Jeffrey Miller
Jeff@TargetTestPrep.com
122 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.
Manager
Status: Preparing
Joined: 05 May 2016
Posts: 55
Location: India
Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
### Show Tags
07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
The following is multiple choice question (with options) to answer.
In a group of Houses,40 had dogs,30 had cats and 10 houses having both dogs and cats. what is the number of houses? | [
"30",
"60",
"40",
"45"
] | B | Make a Venn diagram, and enter your data.
Let the number of houses be x
30+10+20=x
x = 60
so number of houses were = 60
Answer B |
AQUA-RAT | AQUA-RAT-39894 | A. 1
B. 2
C. 4
D. 5
E. 6
After 17 trips, the worker has carried 17 x 4 = 68 jugs.
From those 68 jugs, 68/7 = 9 cartons have been filled, with 5 extra jugs remaining. So, the worker needs one more trip to carry 2 more jugs to fill the partially filled carton with those 2 jugs.
_________________
# Scott Woodbury-Stewart
Founder and CEO
Scott@TargetTestPrep.com
214 REVIEWS
5-STARS RATED ONLINE GMAT QUANT SELF STUDY COURSE
NOW WITH GMAT VERBAL (BETA)
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
VP
Joined: 07 Dec 2014
Posts: 1260
A worker carries jugs of liquid soap from a production line to a packi [#permalink]
### Show Tags
15 Nov 2017, 19:26
AbdurRakib wrote:
A worker carries jugs of liquid soap from a production line to a packing area, carrying 4 jugs per trip. If the jugs are packed into cartons that hold 7 jugs each, how many jugs are needed to fill the last partially filled carton after the worker has made 17 trips?
A. 1
B. 2
C. 4
D. 5
E. 6
each trip fills 4/7 of carton
(17*4)/7 leaves a remainder of 5
7-5=2 more jugs needed
B
Manager
Joined: 08 Apr 2017
Posts: 68
Re: A worker carries jugs of liquid soap from a production line to a packi [#permalink]
### Show Tags
19 Nov 2017, 06:02
AbdurRakib wrote:
A worker carries jugs of liquid soap from a production line to a packing area, carrying 4 jugs per trip. If the jugs are packed into cartons that hold 7 jugs each, how many jugs are needed to fill the last partially filled carton after the worker has made 17 trips?
A. 1
B. 2
C. 4
D. 5
E. 6
The worker has finished 17 trips.
Hence the number of jugs transported = 17*4 = 68.
The following is multiple choice question (with options) to answer.
A transport company's vans each carry a maximum load of 13 tonnes. 12 sealed boxes each weighing 9 tonnes have to be transported to a factory. The number of van loads needed to do this is | [
"38",
"12",
"98",
"27"
] | B | Explanation:
12 as all boxes are sealed.
Answer: B |
AQUA-RAT | AQUA-RAT-39895 | # What is the next number in this sequence: $1, 2, 6, 24, 120$? [closed]
I was playing through No Man's Sky when I ran into a series of numbers and was asked what the next number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
• What was the purpose of the question? – haqnatural Aug 16 '16 at 17:42
• @Battani I was trying to figure out what the next number in the sequence was. – Atom Aug 16 '16 at 17:43
• @Watson I did when I posted this, I was going to ask this last night but decided to work through it first and ended up solving it. When I saw that neither the question nor answer were on here already I selected the "answer your own question" option when posting the question. That way the question would be available online and I would instead be contributing instead of asking for an answer and providing a hodgepodge of behind the scenes work I was doing. I can delete this if that's not the proper way of doing it! – Atom Aug 16 '16 at 17:58
• oeis.org is a good resource. A search gives several hundred possibilities, but you'd want to go with the most comprehensible. – Teepeemm Aug 16 '16 at 20:30
The next number is $840$. The $n$th term in the sequence is the smallest number with $2^n$ divisors.
Er ... the next number is $6$. The $n$th term is the least factorial multiple of $n$.
No ... wait ... it's $45$. The $n$th term is the greatest fourth-power-free divisor of $n!$.
Hold on ... :)
Probably the answer they're looking for, though, is $6! = 720$. But there are lots of other justifiable answers!
The following is multiple choice question (with options) to answer.
Look at this series: 88, 88, 78, 78, 68, 68, ... What number should come next? | [
"14",
"15",
"58",
"19"
] | C | In this series, each number is repeated, then 10 is subtracted to arrive at the next number.
The next number is 58
Answer : C |
AQUA-RAT | AQUA-RAT-39896 | python, algorithm, graph
Using a graph gives the ability to check only the actual routes. And so it wouldn't check if it is possible to fly from say; FRA -> CPH, and then FRA -> MAD, as the original code does. This benefit is good, however there is a larger benefit when there are say two, or more, flights in each edge. When that is the case, the original would go through roughly \$432\$ different arrangement of flights. Where a graph would only check the four routes to AMS, which would include travelling all \$6\$ edges. The graph could then return the \$1\$, or \$4\$, routes to AMS, and then go on to return all \$8\$, or \$18\$, permutations of those flights. And so the graph would check roughly \$28\$ permutations of these flights, as opposed to \$432\$.
Another example of this is if you wanted to obtain all the routes from from FRA -> MAD. And so you'd return the flights from FRA -> MAD and FRA -> CPH -> MAD. The first algorithm would still visit all \$6\$ airports, as you want to check if there are any three flights that can get to the desired destination. This time the code would return the above \$2\$ routes, which would lead to the graph quickly returning all \$6\$ possible flight combinations. Leading to roughly \$12\$ permutations.
Note: My maths is vague, and rough, it's not scientific, and depending on where it's measure, I'd not be suppressed if you obtained more permutations. It is merely to show the difference in magnitude between the two approaches.
There are two algorithms that I describe above, the first, can be implemented by:
Make an 'array' of the starting airport.
while 'the array' is not empty repeat the following commands:
Pop a route from 'the array'.
Get the last airports name from the route
If the name is the destination yield it.
If the route's length is equal to the depth we limit to, continue.
Extend 'the array' with the last airports possible flights to.
The following is multiple choice question (with options) to answer.
There are 13 South American countries, Anni has won a ticket to fly from Denver to one South American country and then directly to another South American Country before returning home to Denver again. Assuming there are four airports in each country, how many flights paths can Anni choose from? | [
"13.12.11.10.4.4.",
"13.12.11.10.4",
"13.12.4.4.",
"13.12.4"
] | C | flights paths can Anni choose fromThere are 13 ways to chose the first destination,
then, 12 ways to chose next destination, with option of 4 airports (13x4)
then, to return he has 4 options to chose the airports
Thus, 13*12*4*4 ways.
Using permutations:
If the solution had written 13P1 instead of 13C1 it won't matter. As per the definition - making arrangements of the combinations gives permutations.
Since there are no arrangement for 1 item, permutationscombinations are equal.C |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.