Question

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

• A. $25,15$ hr
• B. $15,5$ hr
• C. $35,25$ hr
• D. $10,20$ hr

Answer 1

The correct option is A $25,15$ hr

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

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

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

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

According to the question, tank fills in $\frac{75}{8}$ hours.

Then in $1$ hour, both the taps together fills $\frac{8}{75}$ part of the tank

$\therefore \frac{1}{x}+\frac{1}{(x-10)}=\frac{8}{75}$

$\Rightarrow \frac{(x-10)+x}{x(x-10)}=\frac{8}{75}$

$\Rightarrow \frac{2(x-5)}{x^2-10x}=\frac{8}{75}$

$\Rightarrow \frac{x-5}{x^2-10x}=\frac{4}{75}$

$\Rightarrow 4x^2-40x=75x-375$

$\Rightarrow 4x^2-40x-75x+375=0$

$\Rightarrow 4x^2-115x+375=0$

$\Rightarrow 4x^2-100x-15x+375=0$

$\Rightarrow 4x(x-25)-15(x-25)=0$

$\Rightarrow (x-25)(4x-15)=0$

$\Rightarrow x=25$ or $x=\frac{15}{4}$

If we take the fraction value,then $x=\frac{15}{4}$ and $x-10=\frac{15}{4}-10=-\frac{25}{4}$

But time can't be negative

Then $x=25$ and $x-10=25-10=15$

Hence smaller tap fill the tank in $25$ hr.

Larger tap fill the tank in $15$ hr.