A can do a piece of work in 10 days, B in 12 days and C in 15 days. They all start the work together, but A leaves after 2 days and B leaves 3 days before the work is completed. How many days did the work last?

In 2 days, A, B, and C together complete $2\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\right)=2\times\frac{1}{4}=\frac{1}{2}$ of the work. Since B leaves 3 days before completion, C works alone for the last 3 days, finishing $3\times\frac{1}{15}=\frac{1}{5}$. So the middle part done by B and C together is $1-\frac{1}{2}-\frac{1}{5}=\frac{3}{10}$. Their combined rate is $1/12+1/15=3/20$, so time for $3/10$ work is $(3/10)\div(3/20)=2$ days. Total time $=2+2+3=7$ days.