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Pipe A, B and C can fill an empty tank alone in 6 hours, 8 hours and 12 hours respectively. A tank is initially 1/6 filled. Pipes A and B are simultaneously opened in the tank and closed after 2 hours. Now, if pipe C is opened in the tank, in how much time will it be filled?1). 2 hrs.2). 3 hrs.3). 4 hrs.4). 6 hrs. |
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Answer» Part filled by A in 1 hr. = 1/6 Part filled by B in 1 hr. = 1/8 Part filled by C in 1 hr. = 1/12 Now, pipes A and B are opened for 2 hrs. ⇒ Part filled by A and B together in 2 hrs. = 2(1/6 + 1/8) = 2 × 7/24 = 7/12 ? TANK is initially 1/6 filled, Remaining part of tank to be filled = 1 - 1/6 - 7/12 = 1 - 3/4 = 1/4 ∴ Pipe C will fill the remaining part in = (1/4)/(1/12) = 3 hrs |
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