Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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− | We see that the minimum number of ninth graders is <math>6</math>, because if there are <math>3</math> then there is <math>1</math> ninth-grader with a buddy, which would mean there are <math>2.5</math> sixth graders, | + | We see that the minimum number of ninth graders is <math>6</math>, because if there are <math>3</math> then there is <math>1</math> ninth-grader with a buddy, which would mean there are <math>2.5</math> sixth graders, which is impossible (of course unless you really do have half of a person). With <math>6</math> ninth-graders, <math>2</math> of them are in the buddy program, so there <math>\frac{2}{\tfrac{2}{5}}=5</math> sixth-graders total, two of whom have a buddy. Thus, the desired fraction is <math>\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}</math>. |
===Video solution=== | ===Video solution=== |
Revision as of 09:03, 28 March 2023
Contents
[hide]Problem
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?
Solutions
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 (Easy)
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 there are sixth graders, which is impossible (of course unless you really do have half of a person). 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 .
Video solution
https://www.youtube.com/watch?v=u3otXEQgsUU
~savannahsolver
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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