Question

Two pipes running together can fill a tank in $11\frac{1}{9}$ 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. [CBSE 2010]

Answer 1

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

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

⇒ 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 = $\frac{V}{\left(x+5\right)}\times 11\frac{1}{9}=\frac{V}{\left(x+5\right)}\times \frac{100}{9}$

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
$$\begin{gathered} \therefore V\left(\frac{1}{x}+\frac{1}{x+5}\right)\times \frac{100}{9}=V \\ \Rightarrow \frac{1}{x}+\frac{1}{x+5}=\frac{9}{100} \\ \Rightarrow \frac{x+5+x}{x\left(x+5\right)}=\frac{9}{100} \\ \Rightarrow \frac{2x+5}{x^2+5x}=\frac{9}{100} \end{gathered}$$
$$\begin{gathered} \Rightarrow 200x+500=9x^2+45x \\ \Rightarrow 9x^2-155x-500=0 \\ \Rightarrow 9x^2-180x+25x-500=0 \\ \Rightarrow 9x\left(x-20\right)+25\left(x-20\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x-20\right)\left(9x+25\right)=0 \\ \Rightarrow x-20=0\mathrm{or}9x+25=0 \\ \Rightarrow x=20\mathrm{or}x=-\frac{25}{9} \end{gathered}$$
∴ 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