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.
Show answer & explanation
Correct answer: D
Explanation
The correct answer is option 4 i.e. 72%.
A can fill the tank = 10 hours
B can fill the tank = 20 hours
C can fill the tank = 25 hours
Capacity of the tank = LCM of 10, 20, and 25 = 100
Efficiency of tank A = 100/10 = 10
Efficiency of tank B = 100/20 = 5
Efficiency of tank C = 100/25= 4
tank fill in first two hours by A, B, and C = (10 + 5 + 4) × 2 = 38
For next two hours Pipe A and B are opened = (10 + 5) × 2 = 30
Remaining capacity of the tank = 100 - 68 = 32
So, total tank filled by A = (10 × 2 + 10 × 2 + 32) = 72
Required percentage = 72/100 × 100 = 72%