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Pumping alone at their respective constant rates, one inlet

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by BTGmoderatorLU » Wed Dec 26, 2018 2:07 pm

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Source: Official Guide

Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5

The OA is B
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Source: — Problem Solving |

by GMATGuruNY » Wed Dec 26, 2018 6:36 pm
BTGmoderatorLU wrote:Source: Official Guide

Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5
Let the tank = 36 gallons.

Since the first pipe takes 3 hours to fill 1/2 the tank -- and (1/2)(36) = 18 gallons -- the rate for the first pipe = w/t =18/3 = 6 gallons per hour.
Since the second pipe takes 6 hours to fill 2/3 of the tank -- and (2/3)(36) = 24 gallons -- the rate for the second pipe = w/t = 24/6 = 4 gallons per hour.
Combined rate for the two pipes = 6+4 = 10 gallons per hour.
Since the two pipes have a combined rate of 10 gallons per hour, the time to fill the entire 36-gallon tank = w/r = 36/10 = 3.6 hours.

The correct answer is B.
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by Scott@TargetTestPrep » Mon Feb 04, 2019 5:25 pm
BTGmoderatorLU wrote:Source: Official Guide

Pumping alone at their respective constant rates, one inlet pipe fills an empty tank to 1/2 of capacity in 3 hours and a second inlet pipe fills the same empty tank to 2/3 of capacity in 6 hours. How many hours will it take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity?

A. 3.25
B. 3.6
C. 4.2
D. 4.4
E. 5.5

The OA is B

We are given that the first inlet pipe fills an empty tank to 1/2 capacity in 3 hours. Since rate = work/time, the rate of the first inlet pipe is (1/2)/3 = 1/6.

We are also given that the second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate of the second inlet pipe is (2/3)/6 = 1/9.

We need to determine how many hours it will take both pipes, pumping simultaneously at their respective constant rates, to fill the empty tank to capacity.

If we let t = the time, in hours, the two inlet pipes are working together, then the work of the first inlet pipe = (1/6)t, and the work of the second inlet pipe = (1/9)t.

Since the tank is filled, we can set the total work to 1 and create the following equation:

(1/6)t + (1/9)t = 1

Multiplying the entire equation by 18, we obtain:

3t + 2t = 18

5t = 18

t = 18/5 = 3.6

Answer: B

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