Question

To fill a swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. find how long it would take for each pipe to fill the pool separately, if pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.

Answer 1

Let the time taken by larger pipe alone to fill the tank= x hrs
Therefore, the time taken by the smaller pipe = x+10 hrs
Let us assume that, the quantity water to be filled is 1lt.
Water filled by larger pipe running for 4hrs=$\left(\frac{1lt}{x}\right)\times 4=\frac{4}{x}$ lt
Water filled by smaller pipe running for 9hrs= $\frac{9}{(x+10)}$
We know that
$\left(\frac{4}{x}\right)+\left(\frac{9}{(x+10)}\right)=\frac{1}{2}\frac{(4x+40+9x)}{(x^2+10x)}=\frac{1}{2}\frac{(13x+40)}{\left(x^{2}10x\right)}=\frac{1}{2}26\times 80=x^2+10x{x}^2-16x-80=0x^2-20x4x-80=0$
x(x-20) + 4(x-20)= 0
(x4)(x-20)= 0
x=-4, 20
x cannot be negative.
Thus, x=20
x+10= 30
Larger pipe would alone fill the tank in 20 hrs and smaller pipe would fill the tank alone in 30 hrs.