# Thread: Problem solving with proportions

1. ## Problem solving with proportions

Not sure how to solve this:

A high school graduate recently started a new summer job.
She worked 20% more hours in her third week on the job than
she did in the second week. She worked 30% more hours
in the second week than she did in the first week, and she
worked 10% fewer hours in the first week than her regularly
scheduled weekly hours. If the high school graduate worked
46 hours in her third week on the job, approximately what are
her regularly scheduled weekly work hours?

2. Originally Posted by Sentinel
Not sure how to solve this:

A high school graduate recently started a new summer job.
She worked 20% more hours in her third week on the job than
she did in the second week. She worked 30% more hours
in the second week than she did in the first week, and she
worked 10% fewer hours in the first week than her regularly
scheduled weekly hours. If the high school graduate worked
46 hours in her third week on the job, approximately what are
her regularly scheduled weekly work hours?
Here is the general idea:

Let $x$ be the number of regularly scheduled weekly hours.

Number of first week hours: $x - \frac{x}{10} = \frac{9x}{10}$.

Number of second week hours: $\frac{9x}{10} + \frac{3}{10} \left( \frac{9x}{10} \right) = \frac{117x}{100}$.

Number of third week hours: $\frac{117x}{100} + \frac{2}{10} \left( \frac{117x}{100} \right) = \frac{1404x}{1000}$.

Solve $46 = \frac{1404x}{1000}$ for $x$.

No doubt I have made several arithmetic mistakes so you better check all this very carefully.