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Working alone, pump A can empty a pool in 3 hours. Working alone, pump

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Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 23 Mar 2015, 06:34
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A
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Question Stats:

95% (01:12) correct 5% (01:52) wrong based on 155 sessions

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Working alone, pump A can empty a pool in 3 hours. Working alone, pump B can empty the same pool in 2 hours. Working together, how many minutes will it take pump A and pump B to empty the pool?

A. 72
B. 75
C. 84
D. 96
E. 108

Kudos for a correct solution.

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Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 23 Mar 2015, 08:02
1
Pump A can empty the pool in 3 hours therefore the rate at which it empties is 1/3 pool/hour
Pump b can empty the pool in 2 hours therefore the rate at which it empties is 1/2 pool/hour.

If they work together, the resulting rate is the addition of both rates (1/3 +1/2)pool/hour = 5/6 pool/hour

Now we have the following:

(5/6pool)/60min = 1pool/x

x = 72minutes

Answer is A
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Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 23 Mar 2015, 08:16
1
Bunuel wrote:
Working alone, pump A can empty a pool in 3 hours. Working alone, pump B can empty the same pool in 2 hours. Working together, how many minutes will it take pump A and pump B to empty the pool?

A. 72
B. 75
C. 84
D. 96
E. 108

Kudos for a correct solution.



Speed of B is 1.5 times of A .
combined speed B+A = 1.5A + A = \(\frac{5}{2}\) A . so if ideally A takes 3 hrs , with \(\frac{5}{2}\) *A speed it will take --> \(\frac{3*2}{5}* 60\) = 72 minutes.

Answer A.
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 23 Mar 2015, 08:58
1
Rate of A: \(\frac{1}{3h}\)
Rate of B: \(\frac{1}{2h}\)
Combined rate: \(\frac{1}{3h}+\frac{1}{2h}=\frac{5}{6h}=\frac{5}{360 minutes}=\frac{1}{72 minutes}\)

Answer A is correct
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 24 Mar 2015, 02:50
Answer = A = 72

Rate of pump A \(= \frac{1}{180}\)

Rate of pump B = \(\frac{1}{120}\)

Combined rate \(= \frac{5}{360}\)

Time required for combined work \(= \frac{360}{5} = 72\)
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Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 24 Mar 2015, 10:55
1
Bunuel wrote:
Working alone, pump A can empty a pool in 3 hours. Working alone, pump B can empty the same pool in 2 hours. Working together, how many minutes will it take pump A and pump B to empty the pool?

A. 72
B. 75
C. 84
D. 96
E. 108

Kudos for a correct solution.


Let the total work be 1
Rate of work A = Total work done/ Total time taken = 1/3
Rate of work B = Total work done/ Total time taken = 1/2

Total Rate \(= 1/2 + 1/3 = 5/6\)
In 1 hour work done is = 5/6
Time taken to complete the entire work (1) = 6/5 hours

\((6/5)*60 = 72\) minutes.

Answer A
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 25 Mar 2015, 21:21
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1
Hi All,

In this prompt, we have two 'entities' sharing a task (with no *twists*), so the Work Formula will be perfect for this question.

Work = (A)(B)/(A+B) where A and B are the rates of the two entities.

We're given the respective rates for two pumps to empty a pool:

Pump A can empty the pool in 3 hours.
Pump B can empty the pool in 2 hours.

We're asked how long it takes the two pumps, working together, to empty the pool.

Plugging in the respective numbers (the 3 and the 2), we have...

(3)(2)/(3+2) = 6/5 = 1.2 hours = 1 hour 12 minutes

The question asks for an answer in MINUTES. 1 hour 12 minutes = 72 minutes.

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Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 25 Mar 2015, 21:36
1
Combined rate is 1/3+1/2 is 5/6.

time taken to empty pool is 6/5*60 = 72 Mins
Answer is A
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 25 Mar 2015, 23:46
1
Total Work Done in 1 hr = (1/2) +(1/3)

Total time in minutes = 6/5 *60 = 72 minutes

Ans : A
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 25 Mar 2015, 23:59
1
Bunuel wrote:
Working alone, pump A can empty a pool in 3 hours. Working alone, pump B can empty the same pool in 2 hours. Working together, how many minutes will it take pump A and pump B to empty the pool?

A. 72
B. 75
C. 84
D. 96
E. 108

Kudos for a correct solution.


+1 for A
IN 1 hour, Pump A can empty=1/3 of the pool
In 1 hour, Pump B can empty=1/2 of the pool
Working together, in 1 hour, they can empty=1/3+1/2=5/6 of the pool
Therefore, to empty the pool. both will take=6/5 Hours=6/5*60 minutes=72
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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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New post 30 Mar 2015, 03:31
Bunuel wrote:
Working alone, pump A can empty a pool in 3 hours. Working alone, pump B can empty the same pool in 2 hours. Working together, how many minutes will it take pump A and pump B to empty the pool?

A. 72
B. 75
C. 84
D. 96
E. 108

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
pumpsaandb_text.PNG
pumpsaandb_text.PNG [ 17.61 KiB | Viewed 10926 times ]

_________________

New to the Math Forum?
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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