Question : 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.
Option 1: 32%
Option 2: 75%
Option 3: 52%
Option 4: 72%
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Correct Answer: 72%
Solution :
The capacity of the tank = LCM of (10, 20, and 25) = 100
Efficiency of A = $\frac{100}{10}$ = 10
Efficiency of B = $\frac{100}{20}$ = 5
Efficiency of C = $\frac{100}{25}$ = 4
Quantity of tank filled in 2 hours by A, B, and C = (10 + 5 + 4) × 2 = 38 units
According to the question,
After 4 hours from the beginning, tap B is also closed.
Quantity of tank filled in 2 hours by A & B = (10 + 5) × 2 = 30 units
Now, tap B is also closed.
The remaining capacity of the tank = 100 – 38 – 30 = 32 units
These 32 units are filled by A alone.
So, the total quantity of tank filled by A = Work done in (1st 2 hrs + next 2 hrs + 32 units)
= 10 × 2 + 10 × 2 + 32 = 72 units
Percentage of a tank filled by A = ($\frac{72}{100}$) × 100 = 72%.
Hence, the correct answer is 72%.
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