Question

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 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 tap to fill the tank = x hours and

time taken by larger tap = x - 9

In 1 hour, the smaller tap will fill $\frac{1}{x}$ of tank

In 1 hour, the larger tap will fill $\frac{1}{x-9}$ of tank.

In 1 hour both the tank will fill the tank = $\frac{1}{x}$+$\frac{1}{x-9}$

But its given that in 1 hour both the tank will fill $\frac{1}{6}$ of the tank
$\therefore$ $\frac{1}{6}$ =$\frac{1}{x}$ +$\frac{1}{x-9}$

$\Rightarrow$ $\frac{1}{6}=\frac{x-9+x}{x(x-9)}$ $\Rightarrow$ $\frac{1}{6}=\frac{2x-9}{x^2-9x}$
Solving by cross multiplication,

$6(2x-9)=x^2-9x$

$12x-54=x^2-9x$

$x^2-9x-12x+54=0$

$x^2-21x+54=0$

Factorizing by splitting the middle term,

$x^2-18x-3x+54=0$

$x(x-18)-3(x-18)=0$

$(x-18)(x-3)=0$
$x=18,x=3$

If we take x= 18

Smaller tap = (x) = 18 h

Larger tap = (x-9) = 18-9 = 9h

Hence, the time taken by the smaller tap to fill the tank = 18 hrs & the time taken by the larger tap to fill the tank = 9 h