Pumping alone at their

Hi All,

This question can be solved in a couple of different ways, but it's essentially just a Work Formula question.

Work = (A)(B)/(A+B) where A and B are the respective times it takes for two entities to individually complete a task.

To start, we have to figure out how long it takes each pipe to fill the FULL tank....

First pipe = fills 1/2 the tank in 3 hours... so it fills the FULL tank in 6 hours
Second pipe = fills 2/3 the tank in 6 hours... so it fills the FULL tank in 9 hours

Working together, the two pipes will fill the tank in (6)(9)/(6+9) = 54/15 = 3 9/15 hours = 3.6 hours

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

boomgoesthegmat wrote: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

OA: B

The rate of inlet pipe 1 = 3/(1/2) = 6 hours. It can alone fill the empty tank in 6 hours.

The rate of inlet pipe 2 = 6/(2/3) = 9 hours. It can alone fill the empty tank in 9 hours.

Once both are in operation, the part of the tank filled by them is given by:

1/6 + 1/9 = 5/18

=> The empty tank would be filled by both in 18/5 = 3.60 hours.

Answer: B

-Jay
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boomgoesthegmat wrote: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

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

We are also given that a second inlet pipe fills the same empty tank to 2/3 capacity in 6 hours. Thus, the rate 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 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