# 2015 AMC 8 Problems/Problem 16

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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?

$\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}$

~savannahsolver

## Solutions

### Solution 1

Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$

### Solution 2

We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth-grader with a buddy, which would mean there are $2.5$ sixth graders, and that's impossible (of course unless you really do have half of a person). With $6$ ninth-graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth-graders total, two of whom have a buddy. Thus, the desired fraction is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.

(Easy Solution)