There are 3 taps, A, B and C, in a tank. These can fill the tank in 10 h, 20 h and 25 h, respectively. At first, all three taps are opened simultaneously. After 2 h, tap C is closed and tap A and B keep running. After 4 h, 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%

A+ = 10 h, B+ = 20 h, C+ = 25 h,

⇒ C opened for 2 hours = 4 x 2 = 8 units,

⇒ B opened for 4 hours = 5 x 4 = 20 units,

⇒ Remaining units = 100 - (20 + 8) = 72 units,

⇒ Remaining was filled by A, therefore, \(\frac{72}{100}\) x 100 = 72%.