chicagocubsrule wrote:

Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill the tank in 2 hours. How many hours does it take pumps A, B, and C, operating simultaneously, to fill the tank.

A. 1/3

B. 1/2

C. 1/4

D. 1

E. 5/6

We are given that pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours. Recall that rate = work/time. If we consider the work (filling the tank) as 1, then the combined rate of pumps A and B is 1/(6/5). We can express this in the following equation in which a = the time it takes pump A to fill the tank when working alone (thus 1/a is pump A’s rate) and b = the time it takes pump B to fill the tank when working alone (thus 1/b is pump B’s rate):

1/a + 1/b = 1/(6/5)

1/a + 1/b = 5/6

We are next given that pumps A and C, operating simultaneously, can fill the tank in 3/2 hours. We can create another equation in which c = the time it takes pump C to fill the tank alone.

1/a + 1/c = 1/(3/2)

1/a + 1/c = 2/3

Finally, we are given that pumps B and C, operating simultaneously, can fill the tank in 2 hours. We can create the following equation:

1/b + 1/c = ½

Next we can add all three equations together:

(1/a + 1/b = 5/6) + (1/a + 1/c = 2/3) + (1/b + 1/c = ½)

2/a + 2/b + 2/c = 5/6 + 2/3 + 1/2

2/a + 2/b + 2/c = 5/6 + 4/6 + 3/6

2/a + 2/b + 2/c = 12/6

2/a + 2/b + 2/c = 2

Since we need to determine the combined rate of all three machines when filling 1 tank, we can multiply the above equation by ½:

(2/a + 2/b + 2/c = 2) x ½

1/a + 1/b + 1/c = 1

Since the combined rate of all 3 pumps is 1 and time = work/rate, the time needed to fill the tank when all 3 pumps are operating simultaneously is 1/1 = 1 hour.

Answer: D

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