There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opene… — Answer & Discussion

There are three taps A, B and C in a tank. They can fill the tank in 25 hrs, 20 hrs and 10 hrs respectively. At first all of them are opened simultaneously. Then after 1 hrs, tap C is closed and tap A and B are kept running. After the 4^th hour, tap B is also closed. The remaining work is done by tap A alone. Find the percentage of work done by tap A itself?

A72%
B70%
C71%
D65%

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CCommunity09 May 2019

Tap A can fill the tank in 25 hrs.

Tap B cab fill the tank in 20 hrs.

Tap C can fill the tank in 10 hrs.

Total volume of the tank filled by 3 pipes = LCM of (25, 20, 10) = 100 units

⇒ Pipe A can fill 4 units of water in 1 hr.

⇒ Pipe B can fill 5 units of water in 1 hr.

⇒ Pipe C can fill 10 units of water in 1 hr.

⇒ Pipe (A + B + C) can fill 19 units of water in 1 hr.

According to question,

Pipe (A + B + C) are opened for 1 hr, then tap C is closed.

⇒ 19 units of water are filled in the tank.

After 4th hr, tap B is also closed

⇒ for 3 hrs, pipe A and B are open

⇒ (4 + 5) × 3 = 27 units of water are filled.

⇒ 100 - (19 + 27) = 54 units of water are filled by tap A alone.

Total units of water filled by tap A is 4 × 1 + 4 × 3 + 54 = 4 + 12 + 54 = 70 units

∴ Percentage of work done by tap A = 70%

CCommunity20 May 2019