Pipes and Cisterns Practice Test 2 with solutions

Before B was closed, A worked for 6 hours and B worked for 2 hours.
Part of tank filled by A in 6 hours = $\frac{1}{8}\times 6$ = $\frac{3}{4}$
Part of tank emptied by B in 2 hours = $2\times \frac{1}{3}$ = $\frac{2}{3}$
Part of tank filled in 6 hours by A and B = $\frac{3}{4}-\frac{2}{3}$
= $\frac{9-8}{12}$ = $\frac{1}{12}$
Part of the tank that is to be filled = $\frac{11}{12}$
Complete tank filled by A alone = 8 hours
So, time taken to fill $\frac{11}{12}$ th part of the tank by A alone = $8\times \frac{11}{12}$ = $\frac{22}{3}$ = $7\frac{1}{3}$ hours.