Working alone at their respective constant rates, Audrey can complete a certain job in $4$ hours, while Ferris can do the same job in $3$ hours. Audrey and Ferris worked together on the job and completed it in $2$ hours, but while Audrey worked this entire time, Ferris worked for some of the time and took $3$ breaks of equal length. How many minutes long was each of Ferris's breaks?
My solution:
- Audrey's rate of work = $1/4$ per hour
- Ferris's rate of work = $1/3$ per hour
So, together they can complete $7/12$ of the job per hour.
So, if no one takes rest they can complete the job in $1/(7/12)= 12/7$ hours
But as one of them did not work full time, they needed to work $2$ hours. hence, Audrey took rest $= 2 - 12/7$ hours $= 2/7$.
So, each of Audre's break $= 2/21$ hours.
Is my approach right? If not, where is my mistake?
Your calculation would be correct if both A and F took breaks together. Then the total break time for each would indeed be $\frac{2}{7}$. However, A worked full time.
I think it may be easiest to let the number of hours that F actually worked be $h$. Then F completed $\frac{h}{3}$ of the job, while A, working for $2$ hours, completed $\frac{2}{4}$ of the job. Thus $$\frac{h}{3}+\frac{2}{4}=1.$$ It follows that $h=\frac{3}{2}$. Thus F's breaks were for a total of $\frac{1}{2}$ of an hour, $10$ minutes each.
Remark: We could argue without equations. It is clear that A, working for $2$ hours, did half the job, so F did the other half. Since F can do the job in $3$ hours, F's working time must have been an hour and a half.