Question

Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter takes $10$ hours less than the smaller one to fill the tank separately. find the time in which each tap can separately fill the tank.

Answer 1

Let the time taken by the smaller diameter tap be $x$ hours

the time taken by the larger diameter tap $=x-10$ hours

Total time taken to fill the tank $=9\frac{3}{8}=\frac{75}{8}$ hours

Portion filled in one hour by smaller diameter tap $=\frac{1}{x}$

and,Portion filled by larger diamter tap $=\frac{1}{x-10}$ hours

Portion filled by both tap in one hour $=\frac{8}{75}$

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

$\Rightarrow \frac{x-10+x}{x(x-10)}=\frac{8}{75}$

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

$\Rightarrow 8(x^2-10x)=75\times 2(x-5)$

$\Rightarrow \frac{8}{2}(x^2-10x)=75(x-5)$

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

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

$\Rightarrow 4x^2-115x+375=0$

$\Rightarrow 4x^2-100x-15x+375=0$

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

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

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

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

If $x=25$

then $x-10=25-10=15$

if $x=\frac{15}{4}$

$x-10=\frac{15}{4}-10=\frac{15-40}{4}=-\frac{25}{4}$

since, time cannot be negative therefore $x=25$

Hence, the time taken by the small dia tap is $25$ hours and the large dia tap is $15$ hours.