Two taps A and B can fill a tank in 20 min and 30 min, respectively A . n outlet pipe C can empty the full tank in 15 min. A, B and C are opened alternatively, each for 1 min. How long will the tank take to be filled?
Correct Answer: Description for Correct answer:
Part of the tank filled by the taps A, B and C in 3 min.
= \( \Large \frac{1}{20}+\frac{1}{30}-\frac{1}{15}=\frac{3+2-4}{60}=\frac{1}{60} \)
Therefore, Time taken to fill \( \Large \left[ 1 - \left(\frac{1}{2}+\frac{1}{30}\right) \right] \) or
\( \Large \frac{55}{60} \)th part of the tank = \( \Large 3 \times 55 \) = 165 min.
Remaining part of the tank = \( \Large 1 - \frac{55}{60} = \frac{5}{60} = \frac{1}{12} \)
Tap A fills \( \Large \frac{1}{20} \) part in 1 min,
Then, Remaining part = \( \Large \frac{1}{12}-\frac{1}{220}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30} \)
i.e., \( \Large \frac{1}{30} \)th part is filled by B in 1 min.
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