(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.