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The water from one outlet, flowing at a constant rate, can [#permalink]

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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

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]

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25 Feb 2015, 06:38

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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

Here, we're given the rates as 9 hours and 5 hours. Using the Work Formula, we have...

(9)(5)/(9+5) = 45/14

45/14 is a little more than 3.....there's only one answer that matches...

Re: The water from one outlet, flowing at a constant rate, can [#permalink]

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17 Dec 2015, 20:42

Considering the answer choices are not close enough, another quick way to answer this is to work with approximate percentages:

First outlet takes 9 hours to fill the pool, i.e. it fills approx. 11% of the pool every hour. Similarly, second outlet fills 20% of the pool in the same time. Thus, together they will fill approx. 31% of the pool in 1 hour, so to fill 100% of it they will take a little over 3 hours but definitely less than 4 hours. Only choice (D) meets this criteria.

Re: The water from one outlet, flowing at a constant rate, can [#permalink]

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09 Jun 2016, 13:07

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

This problem is called a combined work problem. In these problems we use the formula:

Work (of machine 1) + Work (of machine 2) = Total Work Done

In this particular problem we can define “machine” as “outlet”. We are given that the water from one outlet, flowing at a constant rate, can fill a swimming pool in 9 hours and that the water from a second outlet, flowing at a constant rate, can fill the same pool in 5 hours. This means the hourly rate for one outlet is 1/9 pool per hour and the rate of the other outlet is 1/5 pool per hour. We also are told that the two outlets work together to fill the pool. Thus they both work together for “T” hours. We can fill these values into a simple table.

We can plug in the two work values for outlet one and outlet two into the combined worker formula.

Work (of outlet 1) + Work (of outlet 2) = Total Work Done

T/9 + T/5 = 1

To eliminate the need for working with fractions, let's multiply the entire equation by 45.

45(T/9 + T/5 = 1)

5T + 9T = 45

14T = 45

T = 45/14 = 3 3/14 ≈ 3.21 hours

Answer D
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Jeffrey Miller Jeffrey Miller Head of GMAT Instruction

Re: The water from one outlet, flowing at a constant rate, can [#permalink]

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29 Nov 2016, 13:47

Consider tank capacity to be 45l, a smart number divisible by 9 & 5. First outlet fills the tank in 45/9 = 5 Hrs. (9 liters / Hr.) Second outlet fills the tank in 45/5 = 9 Hrs. (5 liters / Hr.) In 1 hour, first and second outlet can fill 9+5 = 14 Liters. In x hours, first and second outlet can fill 45 Liters.

X = (45*1)/14 = 3.21 Hrs.
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Re: The water from one outlet, flowing at a constant rate, can
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