Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Two pipes can fill a tank in 12 hours and 16 hours respectively. A third pipe can empty the tank in 30 hours. If all three pipes are opened and functions simultaneously, how much time will the tank take to be full?( in hours )

a) $$10\frac{4}{9}$$
b) $$9\frac{1}{2}$$
c) $$8\frac{8}{9}$$
d) $$7\frac{2}{9}$$

Answer: c
Explanation: First pipe fill the tank in 1 hour = $$\frac{1}{{12}}$$ part of tank
Second pipe fill the tank in 1 hour = $$\frac{1}{{16}}$$ part of tank
Third pipe empty the tank in 1 hour = $$\frac{1}{{30}}$$ part of tank
When all three pipes are opened simultaneously, part of the tank filled in 1 hour
$$ = \frac{1}{{12}} + \frac{1}{{16}} – \frac{1}{{30}}$$
LCM of 12, 16 and 30 = 240
$$\eqalign{ & {\text{ = }}\frac{{20 + 15 – 8}}{{240}} \cr & = \frac{{27}}{{240}} \cr} $$
Required time taken by all the three pipes
$${\text{ = }}\frac{{240}}{{27}} = \frac{{80}}{9} = 8\frac{8}{9}\,{\text{Hours}}$$

Join The Discussion