# 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.

#### Complete Python Prime Pack

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack

9 Courses     2 eBooks

Given:

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

To do:

We have to find the time in which each tap can separately fill the tank.

Solution:

Time taken by both the taps to fill the tank$=9\frac{3}{8}=\frac{9\times8+3}{8}=\frac{72+3}{8}=\frac{75}{8}$ hours.

Let the time taken by the tap of the larger diameter to fill the tank be $x$ hours.

This implies,

The time taken by the tap of the smaller diameter to fill the tank$=x+10$ hours.

The portion of the tank filled by the larger tap in one hour $=\frac{1}{x}$.

The portion of the tank filled by the smaller tap in one hour $=\frac{1}{x+10}$.

The portion of the tank filled by both the taps in one hour $=\frac{1}{\frac{75}{8}}=\frac{8}{75}$.

Therefore,

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

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

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

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

$75(2x+10)=8(x^2+10x)$

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

$8x^2+80x-150x-750=0$

$8x^2-70x-750=0$

$2(4x^2-35x-375)=0$

$4x^2-35x-375=0$

Solving for $x$ by factorization method, we get,

$4x^2-60x+25x-375=0$

$4x(x-15)+25(x-15)=0$

$(x-15)(4x+25)=0$

$x-15=0$ or $4x+25=0$

$x=15$ or $4x=-25$

Therefore, the value of $x=15$.    ($x$ cannot be negative)

$x+10=15+10=25$

The time taken by the tap with larger diameter to fill the tank is $15$ hours and the time taken by the tap with smaller diameter is $25$ hours.

Updated on 10-Oct-2022 13:20:12