Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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==Solution 1== | ==Solution 1== | ||
− | Let the number of sixth graders be <math>s</math>, and the number of ninth graders be <math>n</math>. Thus, <math>\frac{n}{3}=\frac{2s}{5}</math>, which simplifies to <math>n=\frac{6s}{5}</math>. Since we are trying to find the value of <math>\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}</math>, we can just substitute <math>n</math> for <math>\frac{6s}{5}</math> into the equation. We then get a value of <math>\frac{\frac{\frac{ | + | Let the number of sixth graders be <math>s</math>, and the number of ninth graders be <math>n</math>. Thus, <math>\frac{n}{3}=\frac{2s}{5}</math>, which simplifies to <math>n=\frac{6s}{5}</math>. Since we are trying to find the value of <math>\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}</math>, we can just substitute <math>n</math> for <math>\frac{6s}{5}</math> into the equation. We then get a value of <math>\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 15:39, 30 November 2015
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
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 | |
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 |
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