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Two water pumps, working simultaneously at their respective [#permalink]
12 Jul 2013, 06:26

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Difficulty:

25% (medium)

Question Stats:

77% (02:10) correct
23% (02:03) wrong based on 105 sessions

Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

Re: Two water pumps, working simultaneously at their respective [#permalink]
12 Jul 2013, 07:25

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This post received KUDOS

Expert's post

kmasonbx wrote:

Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.

Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.

The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.

Re: Two water pumps, working simultaneously at their respective [#permalink]
12 Jul 2013, 09:01

i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong...

Re: Two water pumps, working simultaneously at their respective [#permalink]
12 Jul 2013, 09:07

Expert's post

AMITAGARWAL2 wrote:

i am little confused. If i take slower pump as X then 1/X + 1/1.5X = 1/4 which would result X = 20/3 and so faster pump as 10..... where i am going wrong...

In my solution x is the rate in your solution x is the time.

In your solution x is the time of the faster pump and 1.5x is the time of the slower pump (faster pump needs less time).

Re: Two water pumps, working simultaneously at their respective [#permalink]
08 Sep 2013, 11:25

2

This post received KUDOS

kmasonbx wrote:

Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?

A. 5 B. 16/3 C. 11/2 D. 6 E. 20/3

Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!

Re: Two water pumps, working simultaneously at their respective [#permalink]
25 Sep 2013, 07:48

1

This post received KUDOS

Expert's post

Skag55 wrote:

I did, A+B = 4 A = 1.5B

4-B = 1.5B => B = 8/5

Why is this wrong?

We are told that "two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool".

If one pump needs A hours to fill the pool (rate=1/A) and another need B hours to fill the same pool (rate=1/B), then 1/A + 1/B = 1/4.

Solving 1/A + 1/B = 1/4 and A = 1.5B gives A=10 and B=20/3.