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 taps 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.
Option 1: 75%
Option 2: 52%
Option 3: 72%
Option 4: 32%
Correct Answer: 72%
Solution :
LCM of 10, 20 and 25 is 100
Let the capacity of the tank be 100 units
Efficiency of A = $\frac{100}{10}=10$
Efficiency of B = $\frac{100}{20}=5$
Efficiency of C = $\frac{100}{25}=4$
After 2 h, tap C is closed, and taps A and B keep running
Tank filled for 2 hours by A, B, and C = (10 + 5 + 4) × 2 = 38 units
After 4 hours from starting, B is closed.
Tank filled for 2 hours by A and B = (10 + 5) × 2 = 30 units
Remaining capacity of the tank = 100 – 38 – 30 = 32 units
$\therefore$ Total work done by A = 1st 2 hours + next 2 hours + 32 units
= 10 × 2 + 10 × 2 + 32 = 72 units
Required percentage = $\frac{72}{100}\times100=72$%
Hence, the correct answer is 72%.
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