Question

Two water taps together can fill a tank in 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. [CBSE 2011]

Answer 1

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = $\frac{V}{x}$

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = $\frac{V}{x}\times 6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = $\frac{V}{\left(x-9\right)}\times 6$

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
$$\begin{gathered} \therefore V\left(\frac{1}{x}+\frac{1}{x-9}\right)\times 6=V \\ \Rightarrow \frac{1}{x}+\frac{1}{x-9}=\frac{1}{6} \\ \Rightarrow \frac{x-9+x}{x\left(x-9\right)}=\frac{1}{6} \\ \Rightarrow \frac{2x-9}{x^2-9x}=\frac{1}{6} \end{gathered}$$
$$\begin{gathered} \Rightarrow 12x-54=x^2-9x \\ \Rightarrow x^2-21x+54=0 \\ \Rightarrow x^2-18x-3x+54=0 \\ \Rightarrow x\left(x-18\right)-3\left(x-18\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x-18\right)\left(x-3\right)=0 \\ \Rightarrow x-18=0\mathrm{or}x-3=0 \\ \Rightarrow x=18\mathrm{or}x=3 \end{gathered}$$

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

∴ x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.