Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?

Pipe A's work in % = 100/10 = 10%
Pipe B's work in % = 100/20 = 5%
Pipe C's work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours,
tap C is closed + After the 4th hour, tap B is also closed = 100

According to the question,
(10 + 5 + 2.5) × 2 + (10 + 5) × 2 + work by tap A alone = 100
⇒ 17.5 × 2 + 15 × 2 + work by tap A alone = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100 - 65 = 35%.