Question

Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter takes $10hours$ less then the smaller one to fill the tank sepectarly. Find the time that each tap take to fill the tank seperately.

Answer 1

Given that in $x$ hours the tap can fill the tank seperately.

That is one work done in $x$ hours.

$\therefore \frac{1}{x}$ of the work done in $1$ hour.

That is $\frac{1}{x}$ part of the tank is filled in one hour with smaller tap.

Now, consider another tap with larger diameter.

The tap with larger diameter takes $10$ hours less than the smaller one to fill the tank seperately.

That is one work done in $=x-10$hours.

That is $\frac{1}{x-10}$ part of the tank is filled in one hour with larger tap.

Given that two taps together can fill a tank in $9\frac{3}{8}$ hours.

Total work done in $\frac{75}{8}$ hours.

$\therefore ,\frac{8}{75}$ of work done in $1$ hour.

That is $\frac{8}{75}$ part of the tank is filled in one hour with two taps.

$\therefore$ we have

$\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}$

$\Rightarrow \frac{x-10+x}{x^2-10x}=\frac{8}{75}$

$\Rightarrow \frac{2x-10}{x^2-10x}=\frac{8}{75}$

$\Rightarrow -150(x-5)=8x^2-80x$

$\Rightarrow 4x^2-40x+75x-375=0$

$\Rightarrow 4x^2+35x-375=0$

$\Rightarrow 4x^2-60x+25x-375=0$

$\Rightarrow 4x(x-15)+25(x-15)=0$

$\Rightarrow (x-15)(4x+25)=0$

$\Rightarrow x=15,x=\frac{-25}{4}$

So the tap with smaller diameter fills the tank seperately in $15$ hours and the tap with larger diameter fills the tank seperately in $15-10=5$hours.