Question

If two pipes function simultaneously, a reservoir will be filled in $12$ hours.

One pipe fills the reservoir $10$ hours faster than the other.

Hours taken by second pipe take to fill the reservoir -------hrs.

Answer 1

$\Rightarrow$ Let the slower pipe alone can fill the reservoir in $xhrs$.

$\Rightarrow$ Then, faster pipe alone can fill the reservoir in $(x-10)hrs$

$\Rightarrow$ Part filled by slower pipe in $1hr=\frac{1}{x}$

$\Rightarrow$ Part filled by faster pipe in $1hr=\frac{1}{x-10}$

If pipes functyions simultaneously, the reservoir will be filled in $12hrs$

$\Rightarrow$ Part filled by both the pipes in $1hr=\frac{1}{12}$

$\therefore$ $\frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}$

$\Rightarrow$ $12(x-10)+12x=x(x-10)$

$\Rightarrow$ $12x-120+12x=x^2-10x$

$\Rightarrow$ $x^2-34x+120=0$

$\Rightarrow$ $x^2-30x-4x+120=0$

$\Rightarrow$ $x(x-30)-4(x-30)=0$

$\Rightarrow$ $(x-30)(x-4)=0$

$\therefore$ $x=30or4$

$x$ cannot be $4$ because $(x-10) will be negative.

$\therefore$ $x=30$

$\Rightarrow$ Hours taken by second pipe take to fill the reservoir $30hours$.