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Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone,

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Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone,  [#permalink]

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New post 21 Dec 2017, 19:46
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Question Stats:

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Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone, Hose X can fill 1/x of the pool in 15 minutes. What fraction of the pool can Hose Y fill in 15 minutes if, working together, the two hoses can fill the entire pool in one hour?

A. (x−4)/x

B. (x−4)/(4x)

C. (4−x)/x

D. x/(4−x)

E. 4x/(x−4)

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Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone,  [#permalink]

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New post 22 Dec 2017, 17:35
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4
Bunuel wrote:
Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone, Hose X can fill 1/x of the pool in 15 minutes. What fraction of the pool can Hose Y fill in 15 minutes if, working together, the two hoses can fill the entire pool in one hour?

A. (x−4)/x

B. (x−4)/(4x)

C. (4−x)/x

D. x/(4−x)

E. 4x/(x−4)

This question has a hitch: different units for work rates (minutes vs. hours)

Rate of X

In 15 minutes, Hose X can fill \(\frac{1}{x}\) of the pool
15 minutes = \(\frac{1}{4}\) hour
So in one hour, Hose X can do 4 times as much work as it can in 15 minutes

\((\frac{1}{x} * 4) = \frac{4}{x}\) = hourly work rate of Hose X

Rate of Y

X and Y can fill one pool in one hour
X's rate + Y's rate = \(\frac{1 pool}{1 hr}\) = \(1\)
\(\frac{4}{x}\) + Y's rate = \(1\)
\((1 - \frac{4}{x})\) = Y's rate

\((1 - \frac{4}{x}) = (\frac{x - 4}{x})\) = hourly work rate of Hose Y

Y fills how much?

At that rate, what fraction of the pool can Y fill in 15 minutes (= \(\frac{1}{4}\) hour)?

(Y's work rate per hour, r) * (\(\frac{1}{4}\) hour t) = (work finished W)

\((\frac{x - 4}{x}) * (\frac{1}{4}) = \frac{(x - 4)}{(4x)}\)

Answer B
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Re: Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone,  [#permalink]

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New post 21 Dec 2017, 21:43
1
Bunuel wrote:
Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone, Hose X can fill 1/x of the pool in 15 minutes. What fraction of the pool can Hose Y fill in 15 minutes if, working together, the two hoses can fill the entire pool in one hour?

A. (x−4)/x

B. (x−4)/(4x)

C. (4−x)/x

D. x/(4−x)

E. 4x/(x−4)


The answer is E

Hose X can fill 1/x in 15 minutes , then it will fill the tank in 15x minutes
Similarly let us suppose that the fill rate of hose Y is y in 15 minutes then to fill full tank it will take 15y
Setting up the equation we have =1/15x+1/15y=1/60
solving we get 4x/(x-4)
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Re: Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone,  [#permalink]

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New post 22 Dec 2017, 02:17
1
in 1 hour horse X can fill up: 4/x pool
assuming in 1 hour, horse Y can fill up: 1/y pool
working together horse X and horse Y fill up the tank: 4/x + 1/y = 1
=> 1/y = 1-4/x
in 15 minutes, horse Y will fill up:
1/4y = 1/4*(1-4/x) = (x-4)/4x
So B may be?
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Re: Two hoses, Hose X and Hose Y, are used to fill a pool. Working alone, &nbs [#permalink] 22 Dec 2017, 02:17
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