Difference between revisions of "2015 AMC 8 Problems/Problem 16"

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==Solution 1==
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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 18:43, 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 $\tfrac{1}{3}$ of all the ninth graders are paired with $\tfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

$\textbf{(A) } \frac{2}{15} \qquad \textbf{(B) } \frac{4}{11} \qquad \textbf{(C) } \frac{11}{30} \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{11}{15}$

Solution 1

Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $n/3=2s/5$, which simplifies to $n=6s/5$. Since we are trying to find the value of $\frac{n/3+2s/5}{n+s}$, we can just substitute $n$ for $6s/5$ into the equation. We then get a value of $\boxed{\textbf{(B)}~\frac{4}{11}}$

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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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