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Two water pumps, working simultaneously at their respective [#permalink]

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12 Jul 2013, 07:26

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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!

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]

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12 Jul 2013, 10:01

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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...

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]

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08 Sep 2013, 12:25

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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!

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.

Re: Two water pumps, working simultaneously at their respective [#permalink]

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15 Sep 2014, 06:40

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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

Re: Two water pumps, working simultaneously at their respective [#permalink]

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14 Feb 2015, 14:38

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!

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.

Re: Two water pumps, working simultaneously at their respective [#permalink]

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05 Feb 2016, 19:29

Salvetor wrote:

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!

Somebody confirm whether this is a right approach to do this type of problem or not. Thanks

I got all of this, and I get the whole logic, even why you have to flip etc. The only part I didn't get is why 1/x + 1/1.5x gives you a numerator (1.5+1)/1.5x rather than (1.5x+x)/1.5x.......I get the denominator, it's just the numerator part which confuses me, seems like I missed a fundamental concept in fractions.

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

We are given that the rate of 1 pump is 1.5 times faster than the rate of the other pump. Since 1 pool is being filled and rate = work/time, the rate of the faster pump is 1/x, in which x = the time it takes for the faster pump to fill the pool, and the rate of the slower pump = 1/(1.5x) = 1/(3x/2) = 2/3x.

Since when the pumps work together they take 4 hours to fill 1 pool, we can create the following equation:

work of faster pump + work of slower pump = 1

(1/x)4 + (2/3x)4 = 1

4/x + 8/3x = 1

Multiplying the entire equation by 3x, we have:

12 + 8 = 3x

20 = 3x

20/3 = x

Answer: E
_________________

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GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

Re: Two water pumps, working simultaneously at their respective [#permalink]

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15 Mar 2017, 22:30

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!

This problem asks us what is the amount of hours it would have taken the faster pump to fill the pool by itself- to solve this question you don't necessarily need an exact value- well, you might with a 700 level question but here's my process for this question:

We can pick and choose values- again, this may not be the best approach for a 700 or maybe 600 level question but for a 500 it might be faster If it takes the faster pump, say "6" hours to complete 1 job (filling the entire pool) then it takes the slower pump 4 hours to fill the pool by itself (you can do 60/15 to get 4 or you can multiply 4 by 4 and add half of 4 to get 6). Now let's set the denominator's equal and combine the rates ( I learned this from Mcgoosh) 1 job /6 hours +1 job /4 hours = 4 jobs /24 hours +6 jobs /24 hours= 10 jobs/24 hours 1 job/ 2.4 hours

This does not fit our answer for what the combined work rate should be- therefore, we can eliminate any answer less than 6. "E" by elimination must be the answer.

Salvetor 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

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.

daviddaviddavid , your approach is almost identical to Salvetor's. You just used slightly different algebra.

I did the algebra your way one time and a second time as a fusion of yours and Salvetor's.

He added. You subtracted. You used a different LCM or this shortcut: \(\frac{a}{b} + \frac{c}{d} = \frac{(da + bc)}{bd}\). So I fused by adding, as Salvetor did, and by using your LCM. Both your method and fusion method gave me the correct answer.

I think you should swagger a bit in your head. You took the way four people here asked about, which also seems to me to be the most direct way.

Here's the math done in a way that fuses Salvetor's method, which has expert EMPOWERgmatRichC 's stamp of approval, and yours:

\(\frac{1}{x} + \frac{1}{1.5x} = \frac{1}{4}\)

\(\frac{(1.5x + x)}{1.5x^2} = \frac{1}{4}\)

\(4 (1.5x + x) = 1.5 x^2\)

\(4 (2.5x) = 1.5 x^2\)

\(10 x = 1.5 x^2\), divide by x (we're allowed b/c we know x is not 0):

Decimals in denominators look strange to me, too. So I made myself a little rule: if decimal in denominator isn't in answer choices, either multiply the fraction by 10/10 or 100/100 etc., or convert the decimal to a fraction.

You just got stuck on the part where multiplying your answer by 2 would match one of the choices. Very frustrating. And common. Just ask me.

Salvetor 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

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.

daviddaviddavid , your approach is almost identical to Salvetor's. You just used slightly different algebra.

I did the algebra your way one time and a second time as a fusion of yours and Salvetor's.

He added. You subtracted. You used a different LCM or this shortcut: \(\frac{a}{b} + \frac{c}{d} = \frac{(da + bc)}{bd}\). So I fused by adding, as Salvetor did, and by using your LCM. Both your method and fusion method gave me the correct answer.

I think you should swagger a bit in your head. You took the way four people here asked about, which also seems to me to be the most direct way.

Here's the math done in a way that fuses Salvetor's method, which has expert EMPOWERgmatRichC 's stamp of approval, and yours:

\(\frac{1}{x} + \frac{1}{1.5x} = \frac{1}{4}\)

\(\frac{(1.5x + x)}{1.5x^2} = \frac{1}{4}\)

\(4 (1.5x + x) = 1.5 x^2\)

\(4 (2.5x) = 1.5 x^2\)

\(10 x = 1.5 x^2\), divide by x (we're allowed b/c we know x is not 0):

Decimals in denominators look strange to me, too. So I made myself a little rule: if decimal in denominator isn't in answer choices, either multiply the fraction by 10/10 or 100/100 etc., or convert the decimal to a fraction.

You just got stuck on the part where multiplying your answer by 2 would match one of the choices. Very frustrating. And common. Just ask me.

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