Question

A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely ?

Answer 1

Let the larger tap take x hours.
Smaller tap take $x+3$ hours.
Both fill the tank in 2 hours if they are open together.
Amount of water filled by larger tap in 1 hour = $\frac{1}{x}$
Amount of water filled by smaller tap in 1 hour = $\frac{1}{x+3}$
Amount of water filled in 2 hours = $\frac{1}{2}$
$$\begin{gathered} \frac{1}{x}+\frac{1}{x+3}=\frac{1}{2} \\ \Rightarrow \frac{2x+3}{x^2+3x}=\frac{{\displaystyle 1}}{{\displaystyle 2}} \\ \Rightarrow 4x+6=x^2+3x \\ \Rightarrow x^2-x-6=0 \\ \Rightarrow x^2-3x+2x-6=0 \\ \Rightarrow x\left(x-3\right)+2\left(x-3\right)=0 \\ \Rightarrow \left(x-3\right)\left(x+2\right)=0 \\ \Rightarrow x=3,-2 \end{gathered}$$
Thus, larger tap takes 3 hours while the smaller tap takes 3 + 3 = 6 hours.