A cistern has three pipes A, B, C. A and B can fill it in 4 and 5 hr respectively; and C can empty it in 2 hr. If these pipes are opened in…

A cistern has three pipes A, B, C;

A and B can fill it in 4 and 5 hr respectively; and C can empty it in 2 hr

So in 1 hr the amount of cistern will be filled = $(\frac{1}{4} + \frac{1}{5} - \frac{1}{2})$ (As c empty the cistern so minus sign)

$(= {\rm{\;}}\frac{{5 + 4 - 10}}{{20}})$

= -1/20

Minus sign, so the cistern will be emptied if the three pipes are opened at a time.

A open at 1 pm and till 3 pm C does not open so amount of cistern filled = $(\frac{1}{4} \times 2 = \frac{1}{2})$

B open at 2 pm and till 3 pm C does not open so amount of cistern filled = $(\frac{1}{5} \times 1 = \frac{1}{5})$

Amount of cisten filled till 3 pm = $(\frac{1}{2} + \frac{{1\;}}{5} = \;\frac{{5 + 2\;}}{{10}} = \;\frac{7}{{10}})$

7/10 amount of cistern will be emptied by opening all pipes in $(= {\rm{\;}}\frac{{\frac{7}{{10}}}}{{\frac{1}{{20}}}}{\rm{\;}} = {\rm{\;}}14)$ hr [As the cistern emptied 1/20 per hour]

So, the cistern will empty at (3pm + 14 hr) = 5 a.m.