If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?

Given:

If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. 

To do:

We have to find the time the second pipe takes to fill the reservoir.

Solution:

Let the time taken by the faster pipe to fill the reservoir be $x$ hours. 

This implies,

The time taken by the slower pipe to fill the reservoir$=x+10$ hours.

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

The portion of reservoir filled by the slower pipe in one hour $=\frac{1}{x+10}$.

The portion of the reservoir filled by both the pipes in one hour $=\frac{1}{12}$.

Therefore,

$\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}$

$\frac{1(x+10)+1(x)}{(x+10)x}=\frac{1}{12}$

$\frac{x+10+x}{x^2+10x}=\frac{1}{12}$

$\frac{2x+10}{x^2+10x}=\frac{1}{12}$

$12(2x+10)=1(x^2+10x)$

$24x+120=x^2+10x$

$x^2+10x-24x-120=0$

$x^2-14x-120=0$

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

$x^2-20x+6x-120=0$

$x(x-20)+6(x-20)=0$

$(x-20)(x+6)=0$

$x-20=0$ or $x+6=0$

$x=20$ or $x=-6$

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

$x+10=20+10=30$

The time taken by the second pipe to fill the reservoir is $30$ hours. 

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Updated on: 10-Oct-2022

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