James, by himself, can paint 4 rooms in 10 hours. If he hires June to help, they can do the same job together in 6 hours. If he lets June work alone, how long will it take her to paint 4 rooms?
The general formula is,
$\displaystyle \frac{ab}{a+b}$
Where $\displaystyle a,b$ is the amount of time it takes alone and the fraction is the time together they work.
Thus, the Man can do it in $\displaystyle a=10$. The Woman can do it in $\displaystyle b$ hours. Together they spend,
$\displaystyle \frac{10b}{10+b}=6$
Multiply by denominator,
$\displaystyle 10b=6(10+b)$
$\displaystyle 10b=60+6b$
$\displaystyle 10b-6b=60$
$\displaystyle 4b=60$
$\displaystyle b=15$ hours.
Another way to look at this problem:
We know that James can paint 4 rooms in 10 hours; thus, in 30 hours he can paint 12 rooms.
We know that together they can paint 4 rooms in 6 hours. Thus, in 30 hours, they can point 20 rooms.
From this, we can see that June can paint 8 rooms in 30 hours. Thus, in order to paint just 4 rooms, it would take 15 hours.
James paints 4/10=0.4 rooms per hour
suppose June paints x rooms per hour.
Together they paint x+0.4 rooms per hour.
They paint 4 rooms in 6 hours so their combined rate is 4/6=2/3 rooms per hour
So x+0.4=0.66.., so x=2/3-2/5=4/15 rooms per hour.
Time for June to paint 4 rooms = 4/x =4*15/4=15 hours.
RonL