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A drum can be filled with oil with a supply pipe in 40 minutes. Again the full drum can be made empty by another pipe in 60 minutes. When 2/3 part of the drum was filled, the second pipe was opened and it was stopped after 15 minutes. If the supply pipe is opened now, time required to fill the drum is(a) \(23\frac{1}{3}\) min(b) \(25\frac{2}{3}\) min(c) \(27\frac{1}{3}\) min(d) \(28\frac{2}{3}\) min |
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Answer» (a) \(23\frac{1}{3}\) min Part of the drum that can be emptied in 15 min. by the second pipe = 15 x \(\frac{1}{60}\) = \(\frac{1}{4}\) After 15 min, part of the drum filled with oil = \(\frac{2}{3}\) - \(\frac{1}{4}\) = \(\frac{8-3}{12}\) = \(\frac{5}{12}\) After 15 min, part of the drum emptied = 1 - \(\frac{5}{12}\) = \(\frac{7}{12}\) Given, the whole drum can be filled in 40 min. \(\therefore\) \(\frac{7}{12}\) part of the drum can be filled in \(\big(40\times\frac{7}{12}\big)\)min = \(\frac{70}{3}\) min = 23\(\frac{1}{3}\) min. |
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