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

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Math Expert
Joined: 02 Sep 2009
Posts: 52344
Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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23 Mar 2015, 05:34
00:00

Difficulty:

5% (low)

Question Stats:

91% (00:59) correct 9% (01:32) wrong based on 194 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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23 Mar 2015, 07:02
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

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

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23 Mar 2015, 07:16
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.

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

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23 Mar 2015, 07:58
3
Rate of A: $$\frac{1}{3h}$$
Rate of B: $$\frac{1}{2h}$$
Combined rate: $$\frac{1}{3h}+\frac{1}{2h}=\frac{5}{6h}=\frac{5}{360 minutes}=\frac{1}{72 minutes}$$

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

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24 Mar 2015, 01:50

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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24 Mar 2015, 09:55
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.

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

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25 Mar 2015, 20:21
1
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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Intern
Joined: 02 Jan 2014
Posts: 38
Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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25 Mar 2015, 20:36
1
Combined rate is 1/3+1/2 is 5/6.

time taken to empty pool is 6/5*60 = 72 Mins
Manager
Joined: 06 Jan 2014
Posts: 58
Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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25 Mar 2015, 22:46
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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Director
Joined: 21 May 2013
Posts: 660
Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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25 Mar 2015, 22:59
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
Math Expert
Joined: 02 Sep 2009
Posts: 52344
Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump  [#permalink]

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30 Mar 2015, 02:31
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 [ 17.61 KiB | Viewed 12490 times ]

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

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13 Oct 2018, 17:25
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

The combined rate of pumps A and B is:

1/3 + 1/2 = 2/6 + 3/6 = 5/6, so the time is 1/(5/6) = 6/5 hours, which is 6/5 x 60 = 72 minutes.

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Re: Working alone, pump A can empty a pool in 3 hours. Working alone, pump &nbs [#permalink] 13 Oct 2018, 17:25
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