Two pipes running together can fill a tank in $11\frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

Given:

Two pipes running together can fill a tank in $11\frac{1}{9}$ minutes.

One pipe takes 5 minutes more than the other to fill the tank separately.

To do:

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

Solution:

Time taken by both the pipes to fill the tank$=11\frac{1}{9}=\frac{11\times9+1}{9}=\frac{99+1}{9}=\frac{100}{9}$ minutes.

Let the time taken by the faster pipe to fill the tank be $x$ minutes. 

This implies,

The time taken by the slower pipe to fill the tank$=x+5$ minutes.

The portion of the tank filled by the faster pipe in one minute $=\frac{1}{x}$.

The portion of the tank filled by the slower pipe in one minute $=\frac{1}{x+5}$.

The portion of the tank filled by both the pipes in one minute $=\frac{1}{\frac{100}{9}}=\frac{9}{100}$.

Therefore,

$\frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}$

$\frac{1(x+5)+1(x)}{(x+5)x}=\frac{9}{100}$

$\frac{x+5+x}{x^2+5x}=\frac{9}{100}$

$\frac{2x+5}{x^2+5x}=\frac{9}{100}$

$100(2x+5)=9(x^2+5x)$

$200x+500=9x^2+45x$

$9x^2+45x-200x-500=0$

$9x^2-155x-500=0$

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

$9x^2-180x+25x-500=0$

$9x(x-20)+25(x-20)=0$

$(x-20)(9x+25)=0$

$x-20=0$ or $9x+25=0$

$x=20$ or $9x=-25$

Therefore, the value of $x=20$.    ($x$ cannot be negative)

$x+5=20+5=25$

The time taken by the faster pipe to fill the tank is $20$ minutes and the time taken by the slower pipe to fill the tank is $25$ minutes. 

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Updated on: 10-Oct-2022

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