Pipes A and B can fill a tank in 4 hours and 6 hours 40 minutes, respectively; whereas a third pipe C can empty the full tank in 1 hour 40 minutes. A and B are opened at 2:00 p.m. and 3:00 p.m., respectively and C is opened at 4:00 p.m. The tank will be completely empty at:

First, let's express all the times in hours:
Time taken by A to fill =4 hours

Time taken by B to fill =6 hours 40 minutes=6+
60
40
​
=6+
3
2
​
=
3
20
​
 hours

Time taken by C to empty =1 hour 40 minutes=1+
60
40
​
=1+
3
2
​
=
3
5
​
 hours

Let the total capacity of the tank be the LCM of the numerators (4,20,5), which is 20 units.

Now, let's find the efficiencies of the pipes:

Efficiency of A =
4
20
​
=5 units/hour (filling)

Efficiency of B =
3
20
​

20
​
=3 units/hour (filling)

Efficiency of C =
3
5
​

20
​
=12 units/hour (emptying)

Let's calculate the water filled in the tank up to 4:00 p.m.:

Pipe A is open from 2:00 p.m. to 4:00 p.m. (2 hours):

Water filled by A=5×2=10 units
Pipe B is open from 3:00 p.m. to 4:00 p.m. (1 hour):

Water filled by B=3×1=3 units
Total water in the tank at 4:00 p.m. =10+3=13 units

After 4:00 p.m., all three pipes A, B, and C are open. Their combined efficiency is:

Net Efficiency=Efficiency of A+Efficiency of B−Efficiency of C
Net Efficiency=5+3−12=−4 units/hour
The negative sign implies that the tank is being emptied at a rate of 4 units/hour.

The time required to completely empty the 13 units of water currently in the tank is:

Time=
4
13
​
 hours=3 hours 15 minutes
Adding this to 4:00 p.m.:

4:00 p.m.+3 hours 15 minutes=7:15 p.m.