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Two water pumps, working simultaneously at their respective

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Two water pumps, working simultaneously at their respective [#permalink] New post 12 Jul 2013, 06:26
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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!
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Jul 2013, 07:02, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 12 Jul 2013, 07:25
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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.

Answer: E.

Theory on work/rate problems: work-word-problems-made-easy-87357.html

All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66

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

Hope it's clear.
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 12 Jul 2013, 10:34
yes it does. Thanks...
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 08 Sep 2013, 11:25
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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!


sol:

Rates be A and B

(A+B) * Time = Work
(A+B) * 4 = 1 ---->eq 1

A= 3B/2 ----->eq 2

substituting eq 2 in eq 1

B = 1/10 --->eq 3

substituting eq 3 in eq 2

A= 3/20

Time= work/rate
= 1/(3/20) =>20/3
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 25 Sep 2013, 04:13
I did,
A+B = 4
A = 1.5B

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

Why is this wrong?
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 25 Sep 2013, 07:48
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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.

Check the links provided here: two-water-pumps-working-simultaneously-at-their-respective-155865.html#p1245761 for more.
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 01 Sep 2014, 19:06
Quote:
I did,
A+B = 4
A = 1.5B

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

Why is this wrong?


A+B=(1/T)
NOT
A+B=T

Per formula,
(A+B)T = 1 where A is rate of pump A, B is rate of pump B, T time to complete, and 1 marking completion.
[(A+B)T]/T = 1/T
(A+B) = (1/T)

With B being the rate of the faster pump, B = (3/2)A

A + (3/2)A = (1/4)
(5/2)A = 1/4
A = 1/10
B = 3/20

B = 1/T
(3/20) = 1/T
T = (20/3)
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 15 Sep 2014, 05:40
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AMITAGARWAL2 wrote:
yes it does. Thanks...


Let me elaborate the math so that it's absolutely clear:

Let's calculate the combined rate first:

Rate x Time = Work
Rate x 4 = 1 [It takes 4 hours for both the pumps to fill up the pool]
Rate = 1/4 [So, 1/4 is the rate for the pumps working together]

Now, the let's assume the rate for the slower pump is x ; so the rate for the faster pump will be 1.5x

According to our previous calculations,
Slower pump + faster pump = 1/4
x + 1.5x = 1/4
2.5x = 1/4
x = 1/10 [slower pump's rate]

so, the faster pump's rate is 1/10 x 1.5 = 3/20

Now let's calculate the time it will take for the faster pump

Rate x Time = Work
3/20 x Time = 1
Time = 1 x 20/3 = 20/3 the answer :)
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 14 Feb 2015, 13:38
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!



fast pump takes x hour
Slow pump takes 1.5x hour

so

1/x+1/1.5x = 1/4

> (1.5+1)/1.5x = 1/4
> 2.5/1.5x = 1/4
> 1.5 x = 10
>x = 10/1.5
>x = 20/3


Somebody confirm whether this is a right approach to do this type of problem or not. Thanks
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Re: Two water pumps, working simultaneously at their respective [#permalink] New post 14 Feb 2015, 20:23
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Hi Salvetor,

Yes, your approach is correct. In 'Work' questions, there are usually several different ways to organize the given information, but they all end up involving a ratio at some point.

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Re: Two water pumps, working simultaneously at their respective   [#permalink] 14 Feb 2015, 20:23
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