Question

A cistern can be filled by two pipes $a$and $b$in $10$and $15$hours respectively and is then emptied by a tap in $8$hours.If all the taps are opened,then cistern will be fill in how many hours?

Answer 1

Pipe ‘$a$’ fills the cistern in $10$hours, so in $1$hour the part of cistern filled by pipe $a=\frac{1}{10}$

Pipe ‘$b$’ fills the csitern in $15$hours, so in $1$hour the part of cistern filled by pipe $b=\frac{1}{15}$

A tap can empty the cistern in $8$hours, so in $1$hour the part of cistern emptied by tap $=\frac{1}{8}$

Now if we open all the pipes and taps, then the cistern is filled in $1hour=\frac{1}{10}+\frac{1}{15}–\frac{1}{8}=\frac{5}{120}=\frac{1}{24}$part

Hence, in$1hour,\frac{1}{24}$part of cistern is filled.

Therefore, in $24hours$ the cistern is completely filled.