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# Hoses X and Y simultaneously fill an empty swimming pool

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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
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Hose X fills 1/28 pool in one hour.
Hose Y fills 1/36 pool in one hour.

Combine X and Y will fill 1/28+1/36 pool in our.Say this comes out to be (a/b) pool in one hour.

To fill full pool it will require 1/(a/b) hours which is b/a .

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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]

Rate-work-formula: R*T=W
W = 50,000 liters

Rx = Rate of hose x
Ry = Rate of hose y
Rt = Combined rate of both hoses = Rx + Ry

Question: Given Rt * T = 50,000 what is T?

Statement (1) allows you to calculate Rx: Rx = 50,000/26. But you have no information about Ry. Not sufficient.
Statement (2) allows you to calculate Ry: Ry = 50,000/36. But you have no information about Rx. Not sufficient.

Combined, you can calculate each rate and add them to get Ry. You then can calculate the time T = 50,000/Rt.

You actually do not need to do any of these calculations. Just imagining the requirement in your head (without even writing down anything), you can see quickly that answer is C.

Btw. I feel question is Sub-600 difficulty.
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
beatthegmat05 wrote:
Hose X fills 1/28 pool in one hour.
Hose Y fills 1/36 pool in one hour.

Combine X and Y will fill 1/28+1/36 pool in our.Say this comes out to be (a/b) pool in one hour.

To fill full pool it will require 1/(a/b) hours which is b/a .

Your answer includes "b/a hours", which is a variable. I think the question asks for a specific number of hours.
You need both statements to get that.
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
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let A be the Hours taken by hose X alone to fill the pool.
Let B be the hours taken by hose Y alone to fill the pool.

So working together the time taken to fill the pool is given by $$\frac{1}{A}+\frac{1}{B}=\frac{1}{T}$$
Where T is the time taken together to fill the pool.

(1) Hose X alone would take 28 hours to fill the pool.
We are given A no info of B insufficient.

(2) Hose Y alone would take 36 hours to fill the pool.
we are given B no info of A insufficient.

1+2

we have both A and B hence we can calculate T using
$$\frac{1}{28}+\frac{1}{36}= \frac{1}{T}$$

Sufficient
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
One point worth mentioning here is that one actually does not require the value of total amount of work (i.e. 50,000 litres) to answer this question. The way statements I and II are written, whether the total capacity is 50,000 or 10 litres, the answer will be the same.
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
Bunuel wrote:
SOLUTION

Hoses X and Y simultaneously fill an empty swimming pool that has a capacity of 50,000 liters. If the flow in each hose is independent of the flow in the other hose, how many hours will it take to fill the pool?

We have the amount of the work to be done, hence to answer the question we need the rates of both hoses X and Y.

(1) Hose X alone would take 28 hours to fill the pool. We have the rate of hose X only. Not sufficient.
(2) Hose Y alone would take 36 hours to fill the pool. We have the rate of hose Y only. Not sufficient.

(1)+(2) We have both rates. Sufficient.

Hi everyone, I don't quite understand this. Yea I understand why you took C, but how come it is C and not D? The question says: "if the flow in each hose is independent of the flow in the other hose", doesn't that mean that if we just have one hose working (with the other hose not doing anything) we can figure out how long it takes? So as long as X OR Y is turned on (independent of each other) we can fill the pool?
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
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Re: Hoses X and Y simultaneously fill an empty swimming pool [#permalink]
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