Question

Two water taps together can fill a tank in $9\frac{3}{8}hours$. The tap of larger diameter takes $10hrs$ less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Let the tap with smaller diameter fills the tank alone in $x$ hours

Let the tap with larger diameter fills the tank alone in $(x–10)$hours.

In $1$ hour, the tap with a smaller diameter can fill $\frac{1}{x}$ part of the tank.

In $1$ hour, the tap with larger diameter can fill $\frac{1}{x-10}$ part of the tank.

The tank is filled up in $=9\frac{3}{8}=\frac{75}{8}hrs$

Thus, in $1$ hour the taps fill $\frac{8}{75}$ part of the tank.

Now, according to question,

$$\begin{gathered} \frac{1}{x}+\frac{1}{x-10}=\frac{75}{8} \\ \frac{x-10+x}{x\left(x-10\right)}=\frac{8}{75} \\ 75\left(2x-10\right)=8x\left(x-10\right) \\ 150x-750=8x^2-80x \\ 8x^2-80x-150x+750=0 \\ 8x^2-230x+750=0 \\ 4x^2-115x+375=0 \\ 4x^2-100x-15x+375=0 \\ 4x\left(x-25\right)-15\left(x-25\right)=0 \\ \left(4x-15\right)\left(x-25\right)=0 \\ 4x-15=0x-25=0 \\ x=\frac{15}{4}x=25 \end{gathered}$$

Now, there are two values of $x$.

Therefore, we will have two cases.

Case (a)- When $x=\frac{15}{4}hrs$, then

$\left(x-10\right)=\frac{15}{4}-10=\frac{-25}{4}hrs$.

Since, time cannot be in negative. therefore this case is not possible.

Case (b)- When $x=25hrs$

$\left(x-10\right)=25-10=15hrs$.

∴ Hence, the tap of smaller diameter can separately fill the tank in $25$ hours, and the time taken by the larger tap to fill the tank $15hours.$