Given: There are three pipes A, B and C. A and B are filling pipes whereas C is the emptying pipe. B can fill a tank in 36 hours; C can empty the same tank in 20 hours. If all the three pipes open simultaneously, they will fill the tank in 360 hours.

Formula: If the time taken by pipes P1, P2,..Pn to fill a tank is p1, p2, ...pn and the time taken by pipes Q1, Q2, ..., Qn to empty it be q1, q2,..., qn respectively and the total time taken by all the pipes to fill the tank is t.

1/t = (1/p1 + 1/p2 + ... + 1/pn) – (1/q1 + 1/q2 + ... + 1/qn)

Calculation:

Let pipe A can fill the tank in a hours

In 1 hour A fills = 1/a part

B can fill the tank in 36 hours

In 1 hour b fills = 1/36 part

C can empty the tank in 20 hours

In 1 hour C empties = 1/20 part

A + B + C together fill the tank in 360 hours

In 1 hour they fill = 1/360 part

From the question

1/a + 1/36 – 1/20 = 1/360

⇒ 1/a = 1/360 – 1/36 + 1/20

⇒ 1/a = (1 – 10 + 18)/360

⇒ 1/a = 9/360

⇒ a = 40

∴ A will take 40 hours.