# Pool filling word problem

• Jan 27th 2011, 03:48 AM
Ellla
Pool filling word problem
There are 2 pipes. One is mainly used to fill up pool with water and another - to drain. It takes 2 hours more to fill up pool with water, than to drain. When we filled $1/3$ of the pool with water, all 2 pipes were opened and the pool drained in 8 hours. How long does it take to fill up pool, when only 1 pipe is open?

I've spent long time figuring this out, but never came close to answer - 8 hours.

What the equation should look like? ;\

Thanks.
• Jan 27th 2011, 04:07 AM
tonio
Quote:

Originally Posted by Ellla
There are 2 pipes. One is mainly used to fill up pool with water and another - to drain. It takes 2 hours more to fill up pool with water, than to drain. When we filled $1/3$ of the pool with water, all 2 pipes were opened and the pool drained in 8 hours. How long does it take to fill up pool, when only 1 pipe is open?

I've spent long time figuring this out, but never came close to answer - 8 hours.

What the equation should look like? ;\

Thanks.

This problem belongs to prealgebra, not to calculus. Let $x$ be the time in hours it takes pipe 2 to drain the pool, so

that $x+2$ is the time it takes pipe 1 to fill it up, and from here $\displaystyle{\frac{1}{x}\,,\,\frac{1}{x+2}}$ are the

relative work done by pipe 2 draining (resp. pipe 1 filling) in 1 hour.

When then we open up both pipes they drain the pool by a rate of $\displaystyle{\frac{1}{x}-\frac{1}{x+2}=\frac{2}{x(x+2)}}$-th of a pool every hour

and we're said it takes then 8 hours to empty the pool, so we get:

$\displaystyle{\frac{1}{3}-8\cdot \frac{2}{x(x+2)}=0\Longrightarrow (x+8)(x-6)=x^2+2x-48=0\Longrightarrow x = 6$ , as $x$ must be

positive, and thus indeed the time it takes to fill up the pool with pipe 1 alone is $x+2=6+2=8$ hours

Tonio