# Easy Time and Work Solved QuestionAptitude Discussion

 Q. Two pipes can fill the cistern in 10hr and 12 hr respectively, while the third empty it in 20hr. If all pipes are opened simultaneously, then the cistern will be filled in
 ✔ A. 7.5 hr ✖ B. 8 hr ✖ C. 8.5 hr ✖ D. 10 hr

Solution:
Option(A) is correct

Work done by all the tanks working together in 1 hour.

$\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{20}=\dfrac{2}{15}$

Hence, tank will be filled in $\dfrac{15}{2}=7.5\text{ hour}$

## (1) Comment(s)

K Anudeep
()

LCM OF 10,12,20=60 units of tank

1st pipe=60/10 =6 units/hr

2nd pipe=60/12=5 units/hr

3rd pipe=60/20=3 units/hr

total units=6+5-3=8units/hr

no of hours needed to fill the tank is =60/8=7.5hrs