Three pipes can fill a tank in 15 hours, 12 hours and 10 hours, respectively. If all the three pipes are opened simultaneously for 3 hours, then what percentage of the tank will remain unfilled?

Let the capacity of the tank be the LCM of 15, 12, and 10, which is 60 units.
Efficiency of Pipe 1 =
15
60
​
=4 units/hour

Efficiency of Pipe 2 =
12
60
​
=5 units/hour

Efficiency of Pipe 3 =
10
60
​
=6 units/hour

Total efficiency of all three pipes together = 4+5+6=15 units/hour.

In 3 hours, the work done by all three pipes = 15×3=45 units.

Remaining unfilled capacity of the tank = 60−45=15 units.

Percentage of the tank remaining unfilled:

Percentage=
60
15
​
×100=
4
1
​
×100=25%