smc

Let us say that there are $f$ fifth graders. According to the given information, there must be $2f$ fourth graders and $4f$ third graders. The average time run by each student is equal to the total amount of time run divided by the number of students. This gives us $\frac{12\cdot 4f+15\cdot 2f+10\cdot f}{4f+2f+f}=\frac{88f}{7f}=\frac{88}{7}\Rightarrow \boxed{\frac{88}{7}}$ If you want to simplify the problem even more, just imagine/assume that only $1$ fifth grader existed. Then you can simply get rid of the variables.