To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Given:

If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled.

The pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.

To do:

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

Solution:

Let the time taken by the pipe of the larger diameter to fill the pool be $x$ hours. 

This implies,

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

The portion of the pool filled by the larger pipe in four hours $=\frac{4}{x}$.

The portion of the pool filled by the smaller pipe in nine hours $=\frac{9}{x+10}$.

Therefore,

$\frac{4}{x}+\frac{9}{x+10}=\frac{1}{2}$

$\frac{4(x+10)+9(x)}{(x+10)x}=\frac{1}{2}$

$\frac{4x+40+9x}{x^2+10x}=\frac{1}{2}$

$\frac{13x+40}{x^2+10x}=\frac{1}{2}$

$2(13x+40)=1(x^2+10x)$

$26x+80=x^2+10x$

$x^2+10x-26x-80=0$

$x^2-16x-80=0$

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

$x^2-20x+4x-80=0$

$x(x-20)+4(x-20)=0$

$(x-20)(x+4)=0$

$x-20=0$ or $x+4=0$

$x=20$ or $x=-4$

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

$x+10=20+10=30$

The time taken by the pipe with larger diameter to fill the pool is $20$ hours and the time taken by the pipe with smaller diameter is $30$ hours. 

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Updated on: 10-Oct-2022

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