SSC CHSL 20 Dec 2015 Shift 2 Paper

Let the capacity of the reservoir = L.C.M. $\left(3,\frac{15}{4},1\right)=15$ units
1st inlet pipe can fill in 3 hours, its efficiency = $\frac{15}{3}=5$ units/hour
2nd inlet pipe can fill in $3\frac{3}{4}=\frac{15}{4}$ hours, its efficiency = $\frac{15}{\frac{15}{4}}=4$ units/hour
Similarly,efficiency of outlet pipe = $\frac{15}{1}=-15$ units/hour
Reservoir filled by 1st inlet pipe in 2 hours (1 pm - 3 pm) = $5\times 2=10$ units
Reservoir filled by 2nd inlet pipe in 1 hour (2 pm - 3 pm) = $4\times 1=4$ units
⇒ Reservoir filled = 10 + 4 =14 units
Now, reservoir emptied in 1 hour when all three are opened = $\left(5+4-15\right)=-6$ units/hour
⇒ Time taken to empty 14 units (from 3 pm) = $\frac{14}{6}=\frac{7}{3}$ hours
= $\frac{7}{3}\times 60=140$ minutes
∴ Reservoir will be emptied at = 3 pm + 140 minutes = 5:20 pm