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Pipes A, B and C are attached to a cistern. A and B can fill it in 20 min and 30 min respectively, while C can empty it in 15 min. If A, B and C are kept in operation successively for 1 minute each, how soon will the cistern be filled? (a) 167 min (b) 160 min (c) 166 min (d) 164 min |
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Answer» (a) 167 min Work done in 3 min. = \(\big(\frac{1}{20}+\frac{1}{30}-\frac{1}{15}\big)\) = \(\frac{1}{60}\) \(\therefore\) Work done in 3 × 55 = 165 min = \(\frac{55}{60}\) \(\therefore\) Remaining tank = \(\big(1-\frac{55}{60}\big)\) = \(\frac{5}{60}\) = \(\frac{1}{12}\) Now its A’s turn, \(\frac{1}{20}\) part of the tank is filled by A in 1 min. Since there is still \(\big(\frac{1}{12}-\frac{1}{20}\big)\) = \(\frac{1}{30}\) tank to be filled, which will be filled by B in 1 min. Therefore, required time = (165 + 2) = 167 min. |
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