To fill a swimming pooltwo pipes …

According to question,two pipes are used to fill a swimming pool.Pipe with larger diameter is used for 4 hours and pipe with smaller diameter is used for 9 hours.
Let x hours be the total time taken by the larger pipe to fill the tank
so in 1 hour it would fill {tex}\frac{1}{x}{/tex} part of the tank.
Similarly, y hours are needed for the smaller pipe,
then in 1 hour it would fill {tex}\frac{1}{y}{/tex} part.
So, y=10+x  .........(1)
{tex}\frac{4}{x}{/tex} + {tex}\frac{9}{y}{/tex} = 1/2
using (1), {tex}\frac{4}{x}{/tex} + {tex}\frac{9}{10+x}{/tex} =1/2  ............(2)
solve (2), to get
x² -16x - 80=0
(x-20)(x+4)=0
since value of x cannot be -ve
therefore x=20 and y=30
Larger diameter pipe fills in 20 hours, Smaller diameter pipe fills in 30 hours. Therefore it would take 20 hours and 30 hours respectively.