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2 pumps working simultaneously at their respective rates

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Manager
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Joined: 21 Sep 2006
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2 pumps working simultaneously at their respective rates [#permalink]

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New post 23 Dec 2007, 22:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2 pumps working simultaneously at their respective rates took exactly 4 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 take for the faster pump to fill the pool if it had worked alone at its constant rate?

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

This is what I did:

Step 1: 1/0.5x + 1/x = 1/4.

Step 2: Solve for x. I got x = 12.

Step 3: Hence the faster pump can complete the job in 6 hours (i.e. 0.5 of 12)

But the OA is E. Appreciate your help.

Thanks.
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Re: Rates Q: Pumpus [#permalink]

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New post 23 Dec 2007, 22:21
Hang Tuah wrote:
2 pumps working simultaneously at their respective rates took exactly 4 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 take for the faster pump to fill the pool if it had worked alone at its constant rate?

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

This is what I did:

Step 1: 1/0.5x + 1/x = 1/4.

Step 2: Solve for x. I got x = 12.

Step 3: Hence the faster pump can complete the job in 6 hours (i.e. 0.5 of 12)

But the OA is E. Appreciate your help.

Thanks.


1/x+1/y=1/4

--> 1/x+3/2*(1/x)=1/4

(2+3)/2x=1/4 --> 20=2x --> x=10.

so 3/2*(1/10)--> 3/20 which is 3 pools in 20 hours.

now we want the time for 1 pool so 1/(3/20) --> 20/3.

E
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Re: Rates Q: Pumpus [#permalink]

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New post 24 Dec 2007, 07:16
Hang Tuah wrote:
2 pumps working simultaneously at their respective rates took exactly 4 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 take for the faster pump to fill the pool if it had worked alone at its constant rate?

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

This is what I did:

Step 1: 1/0.5x + 1/x = 1/4.

Step 2: Solve for x. I got x = 12.

Step 3: Hence the faster pump can complete the job in 6 hours (i.e. 0.5 of 12)

But the OA is E. Appreciate your help.

Thanks.


Your first equation was wrong.

1/0.5x + 1/x = 1/4.

I dont know where did u get 0.5 from?

The basic formula is

(1/r) + (1/s) = (1/h)

r and s are number of hours person 1 and person 2 needs working alone for a job and h is the number of hours it takes when they do the same job together

(1/r) = rate of working for first person for a job

(1/s) = rate of working for the second for a job

(1/h) = combined rate of working for a job.

Hope this helps you!
Re: Rates Q: Pumpus   [#permalink] 24 Dec 2007, 07:16
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2 pumps working simultaneously at their respective rates

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