Two pipes A and B can fill a cistern in 24 hours and 32 hours respectively. If both pipes are opened simultaneously, find out when the pipe A must be turned off so that the cistern can be filled in exactly 16 hours?
Solution
Correct Answer: Option E
Suppose the first pipe was closed offer x hr.
Then, the first pipe was functional for x hours, while second pipe was functional for 16 hours.
\(\Rightarrow \frac{x}{{24}} + \frac{{16}}{{32}} = 1\)
\(\Rightarrow \frac{x}{{24}} = 1 - \frac{1}{2} = \frac{1}{2}\)
⇒ x = 12 hrs
Hence, first pipe must be closed after 12 hours to fill the tank in exactly 16 hours.
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