Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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<math>\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}</math> | <math>\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}</math> | ||
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− | ===Solution 2 | + | ==Solution 1== |
− | 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 <math>2.5</math> sixth graders | + | 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 (Easy)== | ||
+ | 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>. | ||
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+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/kJAJXJTKjSk | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ===Video solution=== | ||
+ | |||
+ | https://www.youtube.com/watch?v=u3otXEQgsUU ~David | ||
+ | |||
+ | https://youtu.be/7md_65nXjTw | ||
+ | |||
+ | ~savannahsolver | ||
==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}} |
Latest revision as of 00:57, 18 January 2024
Contents
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?
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 (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video solution
https://www.youtube.com/watch?v=u3otXEQgsUU ~David
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.