A tank can be filled by one tap in 20 minutes and by another in 25 minutes. Both the taps are kept open for 5 minutes and then the second tap is turned off. In how many minutes more the tank is completely filled ?

Answer : Option C

Explanation :

Part of the tank filled by both pipes in 5 minutes $=5\left(\frac{1}{20}+\frac{1}{25}\right)$ $=5\left(\frac{5+4}{100}\right)=\frac{9}{20}$ Remaining part $=1-\frac{9}{20}=\frac{11}{20}$ This part will be filled by first pipe. $\therefore$ Required time $=\frac{11}{20} \times 20$ $=11$ minutes Alternative:

Part of tank filled by pipe A and B in 5 minutes

= 9 × 5 = 45 units

Rest empty tank = 100 – 45

= 55 units

Time taken by pipe A to fill the rest empty tank

$=\frac{55}{5}=11 \text { minutes }$