# A large pipe can fill a tank in 6 hours less than it takes the small pipe. Working together, they can fill it in 4 hours. How long would it take the small pipe to fill the tank if it was working alone?

##### 1 Answer
Oct 26, 2016

Let the time it takes to fill the smaller pipe be $x$ and the time it takes the larger pipe be $x - 6$.

Then, the amount of tank that can be filled in $1$ hour is:

$\frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4}$

Solve this equation.

$\frac{4 \left(x - 6\right)}{4 x \left(x - 6\right)} + \frac{4 \left(x\right)}{4 x \left(x - 6\right)} = \frac{x \left(x - 6\right)}{4 \left(x\right) \left(x - 6\right)}$

We can now eliminate the denominators.

$4 x - 24 + 4 x = {x}^{2} - 6 x$

$0 = {x}^{2} - 14 x + 24$

$0 = \left(x - 12\right) \left(x - 2\right)$

$x = 12 \mathmr{and} 2$

Two solutions may seem non sensical, but if you determine the length of time it takes using the large pipe, you will get $6$ and $- 4$. A negative answer is not possible, so we discredit $x = 2$.

So, it takes the small pipe $12$ hours to fill the tank.

Hopefully this helps!