# 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?

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!