A water reservoir has two inlets and one outlet. Through the inlet it can be filled in 3 hour and 3 hours 45 mins respectively. It can be emptied completely in 1 hour by the outlet. If the two inlets are opened at 1 pm and 2 pm respectively and the outlet at 3 pm, then it will be emptied at

Let the capacity of the reservoir = L.C.M. $$(3,\frac{15}{4},1)=15$$ 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) = $$5\times2=10$$ units

Reservoir filled by 2nd inlet pipe in 1 hour (2 pm - 3 pm) = $$4\times1=4$$ units

=> Reservoir filled = $$10+4=14$$ units

Now, reservoir emptied in 1 hour when all three are opened = $$(5+4-15)=-6$$ units/hour

=> Time taken to empty 14 units (from 3 pm) = $$\frac{14}{6}=\frac{7}{3}$$ hours

= $$\frac{7}{3}\times60=140$$ minutes

$$\therefore$$ Reservoir will be emptied at = 3 pm + 140 minutes = 5:20 pm

=> Ans - (C)