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 hrs.) in which tap of smaller diameter can separately fill the tank.

• A. $\frac{-25}{4}$
• B. $25$
• C. $24$
• D. $\frac{25}{4}$

Answer 1

Answer: (B)

Consider that the tap with smaller diameter fills the tank in $x$ hours.

Then, the tap with larger diameter fills the tank in $x-10$ hours.

This shows that the tap with a smaller diameter can fill $\frac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with larger diameter can fill $\frac{1}{x-10}$ part of the tank in 1 hour.

It is given that the tank is filled in $\frac{75}{8}$ hours that is, the taps fill $\frac{8}{75}$ part of the tank in 1 hour. Then,

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

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

$\frac{2x-10}{x^2-10x}=\frac{8}{75}$

$75\left(2x-10\right)=8\left(x^2-10x\right)$

$150x-750=8x^2-80x$

$8x^2-230x+750=0$

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

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

$4x\left(x-25\right)-15\left(x-25\right)=0$

$\left(4x-15\right)\left(x-25\right)=0$

$4x-15=0$

$x=\frac{15}{4}$

Or,

$x-25=0$

$x=25$

When $x=\frac{15}{4}$, then, $x-10=\frac{15}{4}-10$

$=\frac{15-40}{4}$

$=-\frac{25}{4}$

This cannot be possible because time can never be negative.

When $x=25$, then,

$x-10=25-10$

$x=25$

Therefore, the tap of smaller diameter can separately fill the tank in 25 hours.