Difference between revisions of "2015 AMC 8 Problems/Problem 16"
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− | 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 <math>\ | + | ==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 <math>\frac{1}{3}</math> of all the ninth graders are paired with <math>\frac{2}{5}</math> of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy? | ||
− | <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> |
− | \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> | ||
− | ==Solution 1== | + | ==Solutions== |
− | 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> | + | ===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> | ||
− | ==Solution 2== | + | ===Solution 2=== |
− | 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 with a buddy, and that's impossible. 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 | + | 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 with a buddy, and that's impossible. 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>. |
==See Also== | ==See Also== | ||
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{{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 16:42, 16 January 2021
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
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 | |
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All AJHSME/AMC 8 Problems and Solutions |
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