Question

A swimming pool is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is `6 h r s` b. `10 h r s` c. `15 h r s` d. `30 h r s`

Answer 1

Let V be the volume of pool.
`{:("Taps:",I,II,III),("No .of hours:",x+5,x,x-4):}`
Parts of the pool filled by first, second and third pipe in one hour are respectively
`(V)/(x+5),(V)/(x),(V)/(x-4)`
It is given that, `(V)/(x+5)+(V)/(x)=(V)/(x-4)`
`implies(1)/(5+x)+(1)/(x)=(1)/(x-4)`
`implies(x+x+5)/(x(x+5))=(1)/(x-4)`
`implies(2x+5)(x-4)=x(x+5)`
`implies2x^(2)-3x-20=x^(2)+5x`
`impliesx^(2)-8x-20=0`
`implies(x-10)(x+2)=0`
`because"either "x=10orx=-2`
But, hours cannot be in negative.
So, we reject x=-2
`because` x=10
Hence, the first, second and third pipe fill the pool in `10+5 ,10,10-4 i.e.,15,10` and 6 hrs respectively.