2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 16 Hard

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?

  • A.

    \frac{2}{15}

  • B.

    \frac{4}{11}

  • C.

    \frac{11}{30}

  • D.

    \frac{3}{8}

  • E.

    \frac{11}{15}

Answer:B

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, which is 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}}.

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