Question

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.

• A. 75%
• B. 52%
• C. 72%
• D. 32%

Answer 1

Answer:

C. 72%

Total volume of tank filled by taps = LCM(10, 20 and 25) = 100 tap A in 1 hr = 100/10 = 10 tap B in 1 hr = 100/20 = 5 tap C in 1 hr = 100/25 = 4 Tap A, B and C can fill together = 10 + 5 + 4 = 19 units After 2 hrs, tap C is closed, Amount of tank filled = 2 × 19 = 38 After 4 hrs, tap B is also closed, For 2 hours, tap A and B was open. Units of water filled = (10 + 5) × 2 = 30 Amount of water filled by A alone = 100 – (30 + 38) = 32 Total units of water filled by tap A = 20 + 20 + 32= 72