Question : Pipe A can fill an empty tank in 30 hours and B can fill it in 10 hours. Due to a leakage in the tank, it takes pipes A and B together 1.5 hours more to fill completely than it would have otherwise taken. What is the time taken by the leakage alone to empty the same tank completely, starting when the tank is completely full?
Option 1: 42 hours
Option 2: 40 hours
Option 3: 36 hours
Option 4: 45 hours
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Correct Answer: 45 hours
Solution : Time taken by pipe A to fill a tank = 30 hours Part of tank filled by pipe A in an hour = $\frac{1}{30}$ Time taken by pipe B to fill a tank = 10 hours Part of tank filled by pipe B in an hour = $\frac{1}{10}$ Part of the tank filled by pipes A and B together in an hour $=\frac{1}{30}+\frac{1}{10}=\frac{1+3}{30} = \frac{2}{15}$ Time taken by pipes A and B together to fill the tank = $\frac{15}{2}$ = 7.5 hours Let $x$ be the time taken by the leakage to empty the tank. Part of tank emptied by the leakage in an hour = $\frac{1}{x}$ Time taken by pipes A and B together to fill the tank along with the leakage = 7.5 + 1.5 = 9 hours Part of the tank filled by pipes A and B along with the leakage = $\frac{1}{9}$ So, $\frac{1}{30}+\frac{1}{10}-\frac{1}{x} = \frac{1}{9}$ ⇒ $\frac{1}{x} = (\frac{1+3}{30}) - \frac{1}{9} $ $= \frac{12-10}{90}$ $= \frac{2}{90}$ $= \frac{1}{45}$ $\therefore x = 45$ Hence, the correct answer is 45 hours.
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