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A cistern can be filled by two pipes filling separately in 12 and 16 min respectively. Both pipes are opened together for a certain time but being closed, only 7/8 of the full quantity of water flows through the former and only 5/6 through the latter pipe. The obstruction however being suddenly removed the cistern is filled in 3 min from that moment. How long was it before the full flow began?1). 2.5 min2). 4.5 min3). 3.5 min4). 5.5 min |
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Answer» Let the PIPE be opened for x min ⇒ WORK DONE in 1 min by first pipe = 1/12 ⇒ Work done in 1 min by second pipe = 1/16 ⇒ Work done by both in 3 min = 3 × {(1/12) + (1/16)} = 21/48 ⇒ REMAINING work i.e. 27/48 is done by (7/8) first pipe and (5/6) second pipe ⇒ x × [{(7/8) × (1/12)} + {(5/6) × (1/16)} = 27/48 ⇒ (1/8) x = 27 / 48 ∴ x = 4.5 min |
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