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>\frac{6s}{5}</math> for <math>n</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}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}</math> | 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>\frac{6s}{5}</math> for <math>n</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}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}</math> | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 00:48, 11 November 2019
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 
noooooooppppeee
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 fraction 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
