There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. the second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. the approximate time required by each pipe to fill the tank simultaneously is

There are three pipes fitted in a tank. First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone. The second pipe fills the tank 7 hours faster than the first pipe and 3hours slower than the third pipe. The approximate time required by each pipe to fill the tank simultaneously is
Solution :
Let  the third pipe fills the tank in x hours
So, second pipe will fill the tank in x+3 hours
       first pipe will fill the tank in x+3+7=x+10 hours
The third pipe in 1hour will fill 1/x part of the tank
The second pipe in 1hour will fill 1/x+3 part of the tank
The first pipe in 1hour will fill 1/x+10 part of the tank

First two pipes when operated simultaneously, fill the tank in same time as the third pipe alone ,so
(1/x+3) + (1/x+10)=1/x
=>x=(30)^1/2=5.4...=5 and 1/2 hours=Filling hour of third pipe
:. filling hour of first pipe=x+10=15 and 1/2 hrs
    filling hour of second pipe=x+3=8 and 1/2hrs(Answer)