# A swimming pool is filled with 3 pipes with uniform flow of water.The first two pipes operating simultaneously fill the pool in the same time during which the poo; is filled by the 3rd pipe alone.The 2nd pipe fills the pool in 5 hours faster than the 1st pipe and 4 hours slower than the 3rd pipe.Find the time required by each pipe to fill the pool seperately?

Vikas TU
14149 Points
7 years ago
For volume V of the pool,
the time taken by the pipe 1,2 and 3 respectively are:
(x+5), x and (x-5) hours.
And volume they can fill on the behalf of volume V are:
=> V/(x+5), V and V/(x-5) respectively.
Now given that,
time taken by third pipe is equal to the sum ofthe first two pipes.
Therefore,
V/(x+5) + V/x = V/(x-4)
Solving we get,
x = 10 and -2
Hence,
time required by each pipe to fill the pool is:
10, 15 and 6 hours.
jvbvd
17 Points
6 years ago
A swimming pool is fitted with three pipes. The first two pipes working simultaneously, fill the pool in the same time as the third pipe alone. The second pipe alone fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. In what time will the second and third pipes together fill the pool?
rahul gupta
21 Points
5 years ago
Let volume of the pool = v
And time taken by the second pipe is 'x' hours, then
Given that time taken by the third pipe is equal to the total time taken by the first and second pipe,
v/(x+5) + v/x = v/x-4
After solving we get , x =10, -2 ( negative value is neglected because time can not be negative)
Then time taken by the second pipe is 10 hrs
Similarly, time taken by first pipe = 10+5 = hrs
Time taken for first pipe = 10-4 = 6hrs
Then total time for first and second pipes together fill the pool = (10*6)/(10+6)= 3.75hrs.