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 of all the ninth graders are paired with
of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
Solution 1
Let the number of sixth graders be , and the number of ninth graders be
. Thus,
, which simplifies to
. Since we are trying to find the value of
, we can just substitute
for
into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$ (Error compiling LaTeX. Unknown error_msg)
Solution 2
We see that the minimum number of ninth graders is , because if there are
then there is
ninth grader with a buddy, which would mean
sixth graders with a buddy, and that's impossible. With
ninth graders,
of them are in the buddy program, so there
sixth graders total, two of whom have a buddy. Thus, the desired probability is
.
See Also
2015 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AJHSME/AMC 8 Problems and Solutions |
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