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The water from one outlet, flowing at a constant rate, can [#permalink]
23 Mar 2012, 20:18

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The water from one outlet, flowing at a constant rate, can fill a swimming pool in 9 hours. The water from a second outlet, flowing at a constant rate, can fill the same pool in 5 hours. If both outlets are used at the same time, approximately what is the number of hours required to fill the pool?

The water from one outlet, flowing at a constant rate, can fill a swimming pool in 9 hours. The water from a second outlet, flowing at a constant rate, can fill the same pool in 5 hours. If both outlets are used at the same time, approximately what is the number of hours required to fill the pool?

(A) 0.22 (B) 0.31 (C) 2.50 (D) 3.21 (E) 4.56

Always remember RT=W i.e Rate*Time = Work

Also remember that rate can be added or subtracted. For e.g if A do a work in 2 day and B do a work in 2 day. They both of them together will do a work in 1 day.

So now your question first determine both outlets rate. 1st outlet rate = 1/9 (R=W/T here W=1 work, T = 9hrs) 2nd outlet rate = 1/5 (R=W/T here W=1 work, T = 5hrs)

Both of them working together rate = 1st outlet rate + 2nd outlet rate = 1/9+1/5 = 14/45

again apply the formula RT=W T = W/R = 1/14/45 = 45/14 =3.21

Re: The water from one outlet, flowing at a constant rate, can [#permalink]
24 Mar 2012, 01:04

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calreg11 wrote:

The water from one outlet, flowing at a constant rate, can fill a swimming pool in 9 hours. The water from a second outlet, flowing at a constant rate, can fill the same pool in 5 hours. If both outlets are used at the same time, approximately what is the number of hours required to fill the pool?

A. 0.22 B. 0.31 C. 2.50 D. 3.21 E. 4.56

Remember we can add the rates of individual entities to get the combined rate.

Generally for multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously and \(t_1\), \(t_2\), ..., \(t_n\) are individual times needed for them to complete the job alone.

So for two pumps, workers, etc. we'll have \(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\) --> \(T=\frac{t_1*t_2}{t_1+t_2}\) (general formula for 2 workers, pumps, ...).

Back to the original problem: for two outlets the formula becomes: \(\frac{1}{9}+\frac{1}{5}=\frac{1}{T}\) --> \(\frac{14}{45}=\frac{1}{T}\) --> \(T=\frac{45}{14}=3.something\) (or directly \(T=\frac{t_1*t_2}{t_1+t_2}=\frac{5*9}{5+9}=\frac{45}{14}\)).

Answer: D.

Alliteratively you can do: if both outlets were as slow as the first one, so if both needed 9 hours, then together they would fill the pool in 9/2=4.5 hours, since we don't have two such slow outlets then the answer must be less than 4.5. Similarly, if both outlets were as fast as the second one, so if both needed 5 hours, then together they would fill the pool in 5/2=2.5 hours, since we don't have two such fast outlets then the answer must be more than 2.5. Only answer choice D meets these requirements.

Re: The water from one outlet, flowing at a constant rate, can [#permalink]
25 Feb 2015, 06:38

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Re: The water from one outlet, flowing at a constant rate, can [#permalink]
16 Apr 2015, 19:03

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Hi All,

This question is a standard "Work Formula" question. When you have 2 entities sharing a task, you can use the following formula to figure out how long it takes for the 2 entities to complete the task together.

Work = (A)(B)/(A+B) where A and B are the individual times required to complete the task

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