(b) 13\(\frac{1}{3}\) min
A pipe with double the diameter can empty a tank in half the time, i.e., in 20 min.
\(\therefore\) Part of the tank emptied by the two pipes together
= \(\big(\frac{1}{40}+\frac{1}{20}\big)\) = \(\frac{1+2}{40}\) = \(\frac{3}{40}\)
\(\therefore\) The tank will be emptied by the two pipes together in \(\frac{40}{3}\) min. = \(13\frac{1}{3}\) min.