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.
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 72%.
Three taps A, B, and C can fill a tank in 10 h, 20 h and 25 h respectively.
Total work (capacity of the tank) = LCM (10, 20, 25) = 100 units
The efficiency of the taps A, B, and C = (100/10), (100/20) and (100/25)
A: B: C = 10: 5: 4
According to the question
(A + B + C) * 2 + (A + B) * 2+ A * h = Total work = 100
(10 + 5 + 4) * 2 + (10 + 5) * 2 + 10 * h = 100
38 + 30 + 10*h = 100
h = 32/10 = 3.2
Hence, the percentage of work done by tap A itself = 7.2*10/100 = 72%