Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2 s}{5}$, which simplifies to $n=\frac{6 s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2 s}{5}}{n+s}$, we can just substitute $\frac{6 s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{6 s}{5}+\frac{2 s}{5}}{\frac{6 s}{5}+s}=\frac{\frac{6 s+6 s}{15}}{\frac{11 s}{5}}=\frac{\frac{4 s}{5}}{\frac{11 s}{5}}=\left(\right.$ B) $\frac{4}{11}$. 
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy? (2015 AMC 8 Problems, Question \#16) 

1. $\frac{2}{15}$ 
2. $\frac{4}{11}$ 
3. $\frac{11}{30}$ 
4. $\frac{3}{8}$ 
5. $\frac{11}{15}$ 