Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is :

Part of tank filled by pipes A and B in 1 minute $=\frac{1}{30}+\frac{1}{45}=\frac{3+2}{90}=\frac{1}{18} \text { part }$ $\therefore$ Part of tank filled in 12 minutes $=\frac{12}{18}=\frac{2}{3} \text { part }$ Remaining part $=1-\frac{2}{3}=\frac{1}{3}$ part When pipe C is opened, Part of tank filled by all three pipes $=\frac{1}{30}+\frac{1}{45}-\frac{1}{36}=\frac{6+4-5}{180}=\frac{5}{180}=\frac{1}{36}$ $\therefore$ Time taken in filling $\frac{1}{3}$ part $=\frac{1}{3} \times 36=12$ minutes $\therefore$ Total time $=12+12=24$ miuntes Alternative:

According to the question,

Part of tank filled by A & B in 12 minutes

= 12 × 10 = 120 units

∴ Empty tank = 180 – 120 = 60 units

Time taken to fill the rest of tank by three pipes

$=\frac{60}{6+4-5}=\frac{60}{5}=12$ minutes

$\therefore$ Total time $=12+12$

$=24$ minutes