# Difference between revisions of "2015 AMC 8 Problems/Problem 16"

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 $\tfrac{1}{3}$ of all the ninth graders are paired with $\tfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

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

## 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 $2.5$ sixth graders with a buddy, and that's impossible. 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 $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.

## Solution 3

Let the number of sixth graders be $s$, and the number of ninth-graders be $n$. Then you get $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $5n=6s$. We can figure out that $n=6$ and $s=5$ is a solution to the equation. Then you substitute and figure out that $\frac{5\frac{2}{5}+6\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$

## See Also

 2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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