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

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 probability is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$ .

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2015 AMC 8 Problems/Problem 16

Solution 1

Solution 2

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