# Thread: application problem

1. ## application problem

One maid can clean the house three times faster than another. working together they can clean the entire house in 3 hours. How long would it take the faster maid cleaning alone?

Help!

2. $f =$ faster maid.
$s =$ slower maid.

Originally Posted by kelsikels
One maid can clean the house three times faster than another.
$3s = f \implies s = \frac{f}{3}$

Originally Posted by kelsikels
Working together they can clean the entire house in 3 hours.
$s + f = \color{red}{\frac{1}{3}}$

...

3. Originally Posted by kelsikels
One maid can clean the house three times faster than another. working together they can clean the entire house in 3 hours. How long would it take the faster maid cleaning alone?

Help!
We can measure house-cleaning speed with the unit: houses/hour. (We consider the house in question to be the "standard" house.)

Let s be the slower speed, and f be the faster speed.

Then s+f = 1/3.

We also know that f = 3s.

So s + 3s = 1/3.

s = 1/12 houses/hour.

f = 3s = 3/12 = 1/4 houses/hour.

So it would take the faster maid 4 hours.