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# 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.

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.

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