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Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is three-fifth full, how many hours will be needed to fill the cistern?

Pipes A, B and C are attached to an empty cistern. While the first two can fill the cistern in 4 and 10 hours, respectively, the third can drain the cistern, when filled, in 6 hours. If all the three pipes are opened simultaneously when the cistern is three-fifth full, how many hours will be needed to fill the cistern? Correct Answer 24/11

Given

A and B can fill the empty cistern in 4 and 10 hours respectively and C can empty the filled cistern in 6 hours.

Calculation:

Let the total capacity of cistern = L.C.M. of (4, 6, 10) = 60

⇒ 1 hour work of A = 15

⇒ 1 hour work of B = 6

⇒ 1 hour work of C = -10

⇒ 1 hour work of (A + B + C) = 15 + 6 - 10 = 11

Given cistern is already 3/5 filled,

Remaining capacity = 1 - 3/5 = 2/5

⇒ Remaining capacity = 2/5 × 60 = 24

∴ Time taken to fill the cistern = 24/11 hours

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