Question

Two taps running together can fill a tank in $2$ hours $40$ minutes. If one tap takes $4$ hours more then the other to fill a tank, then how much time will each tap take to fill the tank?

Answer 1

Let are tap fills tank in (x) hrs

other tap fills in (x+4) hrs

$\therefore$ Work Done by both tap in 1 hr

$\Rightarrow \frac{1}{x}+\frac{1}{x+4}=\frac{1}{2\frac{40}{60}}$

$\Rightarrow \frac{(x+4)+\left(x\right)}{x(x+4)}=\frac{1}{8/3}$

$\Rightarrow \frac{2x+4}{x(x+4)}=\frac{3}{8}$

$\Rightarrow 8(2x+4)=3\left(x\right)(x+4)$

$16x+32=3x^2+12x$

$\Rightarrow 3x^2-4x-32$

$\Rightarrow 3x^2-12x+8x-32$

$(3x+8)(x-4)\Rightarrow x=4$

$\therefore$ 1 tap fill in 4 hrs

other tap fills in : x+4

$\Rightarrow 4+4=8hrs$