Given:

Pipe A can fill a cistern in 15 minutes Pipe B can fill it in 9 minutes. With the help of pipe C, all the pipes can fill a cistern in 5 minutes. D, with the help of C filled the cistern in 15 minutes.

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)

Efficiency is inversely proportional to time taken for doing work.

Total work = Efficiency × time

Calculation:

Let the time taken by pipes C and D to fill the cistern be c and d respectively.

In 1 minute pipe A fill = 1/15 part

In 1 minute pipe B fill = 1/9 part

In 1 minute pipe C fill = 1/c part

In 1 minute pipe D fill = 1/d part

In 1 minute pipe (A + B + C) fill = 1/5 part

In 1 minute pipe (C + D) fill = 1/15 part

According to question

1/a + 1/b + 1/c = 1/5

⇒ 1/c = (1/5) – (1/15) – (1/9)

⇒ 1/c = (9 - 3 - 5)/45

⇒ 1/c = 1/45

⇒ c = 45 minutes

Thus, C can fill the tank in 45 minutes.

C and D can fill the tank in 15 minutes, then

1/c + 1/d = 1/15

⇒ 1/d = (1/15) – (1/45)

⇒ 1/d = (3 - 1)/45 = 2/45

Thus, d can fill the tank in 22.5 minutes.

Now, let the time required by B and D to fill full tank be t.

1/b + 1/d = 1/t

⇒ 1/9 + 1/(45/2) = 1/t

⇒ 1/t = 1/9 + 2/45 = 7/45

⇒ t = 45/7

B and D can fill the full tank in 45/7 minutes.

∴ They will fill half tank in 45/14 minutes.