There are 3 taps A, B and C in a tank. These can fill the tank in 10 hours, 20 hours and 25 hours, respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours from the beginning, tap B is also closed. The remaining tank is filled by tap A alone. Find the percentage of work done by tap A itself.

72%

Tap A can fill the tank in 10 hrs.

Tap B cab fill the tank in 20 hrs.

Tap C can fill the tank in 25 hrs.

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

⇒ Pipe A can fil 10 units of water in 1 hr.

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

⇒ Pipe C can fill 4 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 2 hr, then tap C is closed.

= 19 × 2 = 38  units

After 4th hour from beginning , tap B is also closed

= for 2 hrs, pipe A and B are open

⇒ (10 + 5) × 2 = 30 units

⇒ 100 - (38+ 30) = 32 units ( of water are filled by tap A alone )

Total work by A = 20 + 20 + 32 = 72

Required Percentage = \(\frac{72}{100}\) × 100 = 72%