**Question:**

Two pipes running together can fill a tank in $11 \frac{1}{0}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill

the tank separately.

**Solution:**

Let the time taken by one pipe to fill the tank be *x* minutes.

∴ Time taken by the other pipe to fill the tank = (*x* + 5) min

Suppose the volume of the tank be *V*.

Volume of the tank filled by one pipe in *x* minutes = *V*

$\therefore$ Volume of the tank filled by one pipe in 1 minute $=\frac{V}{x}$

$\Rightarrow$ Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes $=\frac{V}{x} \times 11 \frac{1}{9}=\frac{V}{x} \times \frac{100}{9}$

Similarly,

Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes $=\frac{V}{(x+5)} \times 11 \frac{1}{9}=\frac{V}{(x+5)} \times \frac{100}{9}$

Now,

Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes $+$ Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes $=V$

$\therefore V\left(\frac{1}{x}+\frac{1}{x+5}\right) \times \frac{100}{9}=V$

$\Rightarrow \frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}$

$\Rightarrow \frac{x+5+x}{x(x+5)}=\frac{9}{100}$

$\Rightarrow \frac{2 x+5}{x^{2}+5 x}=\frac{9}{100}$

$\Rightarrow 200 x+500=9 x^{2}+45 x$

$\Rightarrow 9 x^{2}-155 x-500=0$

$\Rightarrow 9 x^{2}-180 x+25 x-500=0$

$\Rightarrow 9 x(x-20)+25(x-20)=0$

$\Rightarrow(x-20)(9 x+25)=0$

$\Rightarrow x-20=0$ or $9 x+25=0$

$\Rightarrow x=20$ or $x=-\frac{25}{0}$

∴ *x* = 20 (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min