Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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</math> | </math> | ||
| − | + | ==Solution 1== | |
| + | Let the number of sixth graders be <math>s</math>, and the number of ninth graders be <math>n</math>. Thus, <math>n/3=2s/5</math>, which simplifies to <math>n=6s/5</math>. Since we are trying to find the value of <math>\frac{n/3+2s/5}{n+s}</math>, we can just substitute <math>n</math> for <math>6s/5</math> into the equation. We then get a value of <math>\boxed{\textbf{(B)}~\frac{4}{11}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=15|num-a=17}} | {{AMC8 box|year=2015|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:43, 25 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 
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. 
