Question

Two water taps together can fill a tank is $6$ 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 'v' $m^3$ be the volume,

$(x+y)\to 6$ hours

$x\to k$ hours

$y\to (k-9)$ hours

Rate of $x\Rightarrow \frac{v}{k}{m}^3/hr$

Rate of $y\Rightarrow \frac{v}{(k-9)}{m}^3/hr$

Rate of $(x+y)\Rightarrow \frac{v}{6}{m}^3/hr$

$\therefore \frac{v}{k}+\frac{v}{k-9}=\frac{v}{6}$

$\Rightarrow \frac{1}{k}+\frac{1}{k-9}=\frac{1}{6}$

$6(k-9+k)=k(k-9)$

$\Rightarrow 6(2k-9)=k^2-9k$

$\Rightarrow 12k-54=k^2-9k$

$\Rightarrow k^2-21k+54=0$

$(k-18)(k-2)=0$

$k=18,3$

$\therefore x\Rightarrow 18$ or $3$ hours

$y\Rightarrow 9$ or $-6$ hours (Not possible)

$\therefore x=18$ hours & $y=9$ hours.