Two pipes A and B can fill cistern in $$12\frac{1}{2}$$ hours and 25 hours, respectively. The pipes were opened simultaneously, and it was found that, due to leakage in the bottom, it took one hour 40 minutes more to fill the cistern. It the cistern is full, in how much time (in hours) will the leak alone empty 70% of the cistern?

Solution (By Examveda Team)

$$\eqalign{ & A = 12\frac{1}{2} = \frac{{25}}{2} \cr & B = 25 \cr} $$

$$\eqalign{ & A + B = \frac{{25}}{{2 + 1}} = \frac{{25}}{3} \cr & = 8\,{\text{hr }}20\,\min \cr & A + B - C \cr & = 8\,{\text{hr }}20\,\min + 1\,{\text{hr 4}}0\,\min \cr & = 10{\text{ hr}} \cr} $$

C = 8 + 4 - 10 = 2
70% of tank capacity $$ = 100 \times \frac{{70}}{{100}} = 70$$
Time taken in leakage $$ = \frac{{70}}{2} = 35\,{\text{hrs}}$$