Hello,

Can you please help with the following:

A tank has both an inlet pipe and an outlet pipe. Working alone, the inlet pipe can fill up the tank in 5 hours. Working alone, the outlet pipe can empty out the tank in 15 hours. If it is desired that the tank should be filled, from empty, exactly 6 hours after the inlet pipe is turned on at 9:30am, then at what time should the outlet pipe be opened ?

A)10:00am
B)10:45am
C)12:00pm
D)12:30pm
E)1:30pm

OA: D


Rate x Time = Work

Input Pipe: 1/5 x 5 = 1
Output Pipe: 1/15 x 15 = 1

So, rate at which the tank is getting filled is 1/5 - 1/15 = 2/15

However, I am stuck at this point. Can you please assist?

Thanks a lot,
Sri


Update:

I just tried solving as follows now:

The tank should be filled 6 hours after the inlet pipe is opened at 9:30 a.m. In these 6 hours, 1/5 x 6 = 6/5 of the tank is filled by the inlet pipe.

So we need to take the extra quantity out of the tank.

i.e. 6/5 - x = 1 => x = 1/5.

Hence, this extra 1/5 of the water needs to be removed by the outlet pipe.

So, for the outlet pipe 1/15 x t = 1/5
=> t = 3 hours.

Hence, it has to be turned 3 hours before 3:30 p.m. Hence, 12:30 p.m.

Is this approach correct? Also, is there a better way to solve this problem?

Thanks a lot,
Sri