• February 3rd 2006, 10:37 PM
pochoa
Next-door neighbors Bob and Jim use hoses from both houses to fill Bob's swimming pool. They know it takes 18 h using both hoses. They also know that Bob's hose, used alone, takes 20% less time than Jim's hose alone. How much time is required to fill the pool by each hose alone?

• February 4th 2006, 12:56 AM
earboth
Quote:

Originally Posted by pochoa
Next-door neighbors Bob and Jim use hoses from both houses to fill Bob's swimming pool. They know it takes 18 h using both hoses. They also know that Bob's hose, used alone, takes 20% less time than Jim's hose alone. How much time is required to fill the pool by each hose alone?

Hello,

I'm very sorry, but your answer couldn't be right because one hose needs a lot more time to fill the swimmingpool than both hoses together.

Let's have another try:

(I use SP as abbreviation for "swimmingpool")

Bob's hose needs b hours to fill the SP.
Jim's hose needs j hours to fill the SP.
That means:
Bob's hose delivers $\frac{SP}{b}$ per hour.
Jim's hose delivers $\frac{SP}{j}$ per hour.
Both hoses together delivers $\frac{SP}{b}+\frac{SP}{j}$ per hour.
They need 18 hours to fill the SP:
$18 \cdot \left( \frac{SP}{b}+\frac{SP}{j} \right)=SP$
Expanded:
$\frac{18 \cdot SP}{b}+\frac{18 \cdot SP}{j} =SP$
Divide both sides of the equation by SP and you get:
$\frac{18}{b}+\frac{18}{j} =1$

As we know that $b=0.8 \cdot j$ we can substitute b by j:
$\frac{18}{0.8 \cdot j}+\frac{18}{j} =1$

I leave the next steps to you.

As the final result you'll get b=32.4 hrs (32 h; 24 min) and j = 40.5 hrs (40 h; 30 min).

Bye