Question

Two water taps together fill a tank in $1\frac{7}{8}$ hours. The tap with longer diameter takes 2 hours less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer 1

Let the smaller tap takes x hours to fill the tank.
Then, the larger one takes (x − 2) hours to fill the tank.

Tank filled in 1 hour by smaller tap = $\frac{1}{x}$
Tank filled in 1 hour by larger tap = $\frac{1}{x-2}$
Tank filled in 1 hour by both the taps = $\frac{8}{15}$

According to the question,

$$\begin{gathered} \frac{1}{x}+\frac{1}{x-2}=\frac{8}{15} \\ \Rightarrow \frac{\left(x-2\right)+\left(x\right)}{\left(x\right)\left(x-2\right)}=\frac{8}{15} \\ \Rightarrow \frac{2x-2}{x^2-2x}=\frac{8}{15} \\ \Rightarrow 15\left(2x-2\right)=8\left(x^2-2x\right) \\ \Rightarrow 30x-30=8x^2-16x \\ \Rightarrow 8x^2-46x+30=0 \\ \Rightarrow 4x^2-23x+15=0 \\ \Rightarrow x=\frac{-\left(-23\right)\pm \sqrt{{\left(-23\right)}^2-4\left(4\right)\left(15\right)}}{2\left(4\right)} \\ \Rightarrow x=\frac{23\pm \sqrt{529-240}}{8} \\ \Rightarrow x=\frac{23\pm \sqrt{289}}{8} \\ \Rightarrow x=\frac{23\pm 17}{8} \\ \Rightarrow x=\frac{23+17}{8}\mathrm{or}x=\frac{23-17}{8} \\ \Rightarrow x=5\mathrm{or}x=0.75 \\ \mathrm{But}x\ne 0.75\left(\because \left(x-2\right)\mathrm{becomes}\mathrm{negative}\right) \\ \mathrm{Thus},x=5 \end{gathered}$$

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.