Let the capacity of the reservoir = L.C.M.
(3,,1)=15 units
1st inlet pipe can fill in 3 hours, its efficiency =
=5 units/hour
2nd inlet pipe can fill in
3= hours, its efficiency =
=4 units/hour
Similarly,efficiency of outlet pipe =
=−15 units/hour
Reservoir filled by 1st inlet pipe in 2 hours (1 pm - 3 pm) =
5×2=10 units
Reservoir filled by 2nd inlet pipe in 1 hour (2 pm - 3 pm) =
4×1=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) =
= hours
=
×60=140 minutes
∴ Reservoir will be emptied at = 3 pm + 140 minutes = 5:20 pm