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 $\boxed{\textbf{(B)}~\frac{4}{11}}$

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 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.

Page

Toolbox

Search

2015 AMC 8 Problems/Problem 16

Solution 1

See Also