Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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==Solution 3== | ==Solution 3== | ||
− | Let the number of sixth graders be <math>s</math>, and the number of ninth-graders be <math>n</math>. Then you get <math>\frac{n}{3}=\frac{2s}{5}</math>, which simplifies to <math>5n=6s</math>. We can figure out that <math>n=6</math> and <math>s=5</math> is a solution to the equation. Then you substitute and figure out that <math>\frac{5\cdot\frac{2}{5}+6\cdot\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}</math><br>-RedFireTruck | + | Let the number of sixth graders be <math>s</math>, and the number of ninth-graders be <math>n</math>. Then you get <math>\frac{n}{3}=\frac{2s}{5}</math>, which simplifies to <math>5n=6s</math>. We can figure out that <math>n=6</math> and <math>s=5</math> is a solution to the equation. Then you substitute and figure out that <math>\frac{5\cdot\frac{2}{5}+6\cdot\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}</math><br><b>-RedFireTruck</b> |
==See Also== | ==See Also== |
Revision as of 19:30, 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?
Contents
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 fraction is
.
Solution 3
Let the number of sixth graders be , and the number of ninth-graders be
. Then you get
, which simplifies to
. We can figure out that
and
is a solution to the equation. Then you substitute and figure out that
-RedFireTruck
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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