2015 AMC 8 Problems/Problem 16

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?

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

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 $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$ .

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/kJAJXJTKjSk

~Education, the Study of Everything

Video solution

https://www.youtube.com/watch?v=u3otXEQgsUU ~David

https://youtu.be/7md_65nXjTw

~savannahsolver

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2015 AMC 8 Problems/Problem 16

Contents

Problem

Solution 1

Video Solution (HOW TO THINK CRITICALLY!!!)

Video solution

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