BTGmoderatorDC wrote:A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/7
(E) 4/7

OA E

Source: Veritas Prep

Given all the 5 pipes are open, to fill the tank, say each of the three narrow pipes fills y liters; thus, each of the two wode pipes would fill 2y liters. So,

"¢ Three narrow pipes fill 3y liters and
"¢ Two wide pipes fill 4y liters

So, the capacity of the tank = 3y + 4y = 7y liters

Thus, all the pipes working together did 7y/4y = 7/4 of the work done by the two wide pipes working together to fill the tank. Since time and work are inverse;y proportional, all the pipes working together, will take 1/(7/4) = 4/7 of time taken by the two wide pipes working together to fill the tank.

The correct answer: E

Hope this helps!

-Jay
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BTGmoderatorDC wrote:A tank has 5 inlet pipes. Three pipes are narrow and two are wide. Each of the three narrow pipes works at 1/2 the rate of each of the wide pipes. All the pipes working together will take what fraction of time taken by the two wide pipes working together to fill the tank?

(A) 1/2
(B) 2/3
(C) 3/4
(D) 3/7
(E) 4/7

OA E

Source: Veritas Prep

We can let x = the rate of each small inlet pipe, and thus 2x = the rate of each large inlet pipe.

Let's now assume that it will take 10 hours to fill the pool if all the pipes work together. Thus, the capacity of the pool is 10(3x + 2(2x)) = 10(7x) = 70x.

If the pool is filled only by the 2 large pipes, then it would take 70x/(2(2x)) = 70x/(4x) = 35/2 = 17.5 hours. Therefore, the ratio of the time if all 5 pipes work together to the time if only the 2 large pipes work together is 10/17.5 = 20/35 = 4/7.

Answer: E