Question

Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and they B is turned off. How much time will A take to fill the remaining tank?

Answer 1

$$\begin{gathered} \mathrm{Pipe}\mathrm{A}\mathrm{can}\mathrm{fill}\mathrm{the}\mathrm{tank}\mathrm{in}10\mathrm{hours},\mathrm{and}\mathrm{pipe}\mathrm{B}\mathrm{can}\mathrm{fill}\mathrm{the}\mathrm{tank}\mathrm{in}15\mathrm{hours}. \\ \therefore \mathrm{In}1\mathrm{hour},\mathrm{A}\mathrm{can}\mathrm{fill}\frac{1}{10}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank}. \\ \mathrm{In}1\mathrm{hour},\mathrm{B}\mathrm{can}\mathrm{fill}\frac{1}{15}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank}. \\ \therefore \mathrm{In}1\mathrm{hour},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{can}\mathrm{fill}\left(\frac{1}{10}+\frac{1}{15}\right) \\ =\frac{3+2}{30}=\frac{5}{30}=\frac{1}{6}\mathrm{th}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank} \\ \therefore \mathrm{In}4\mathrm{hours},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{can}\mathrm{fill}\left(\frac{1}{6}\times 4\right)=\frac{2}{3}\text{rd}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank} \\ \mathrm{Remaining}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank}=1-\frac{2}{3}=\frac{1}{3} \\ \mathrm{Now},\mathrm{A}\mathrm{can}\mathrm{fill}\mathrm{the}\mathrm{tank}\mathrm{in}10\mathrm{hours}. \\ \therefore \frac{1}{3}\text{rd}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{tank}\mathrm{can}\mathrm{be}\mathrm{filled}\mathrm{by}\mathrm{A}\mathrm{in}(\frac{1}{3}\times 10)\mathrm{hours}\mathrm{or}\frac{10}{3}\mathrm{hours}\mathrm{or}3\frac{1}{3}\mathrm{hours}. \end{gathered}$$