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.

Answer: Option D

Solution

Answer: Option D

Solution:

2 hours total → 2 × 19 = 38
4 hours total = 38 + 2 × 15 = 68
Remaining work = 100 - 68 = 32
Total work done by A = 32 + 40 = 72
Required % = $$\frac{72}{100}$$ × 100 = 72%