Question

Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tak be full?

(a) 7 hrs 15 min

(b) 7 hrs 30 min

(c) 7 hrs 45 min

(d) 8 hrs

Answer 1

First inlet pipe's 1 hour work $=\frac{1}{10}$

Second inlet pipe's 1 hour work $=\frac{1}{12}$

Third outlet pipe's one hour work $=\frac{1}{20}$

$\therefore$ Three pipes 1 hour work

$=\frac{1}{10}+\frac{1}{12}-\frac{1}{20}$

$=\frac{6+5-3}{60}=\frac{8}{60}=\frac{2}{15}$

$\therefore$ The tank will be full in $\frac{15}{2}$hours =7 hrs 30 min