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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>\tfrac{1}{3}</math> of all the ninth graders are paired with <math>\tfrac{2}{5}</math> of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
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==Problem==
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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>\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?
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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>
  
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<math>
 
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\textbf{(A) } \frac{2}{15} \qquad
 
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\textbf{(B) } \frac{4}{11} \qquad
 
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\textbf{(C) } \frac{11}{30} \qquad
 
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\textbf{(D) } \frac{3}{8} \qquad
 
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\textbf{(E) } \frac{11}{15}
 
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</math>
 
  
 
==Solution 1==
 
==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>\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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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)==
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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, 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
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==Solution 2==
 
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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 <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>.
 
  
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==Solution 3==
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===Video solution===
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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⋅\frac{2}{5}+6⋅\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}</math>
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https://www.youtube.com/watch?v=u3otXEQgsUU  ~David
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https://youtu.be/7md_65nXjTw
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 +
~savannahsolver
  
 
==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 23:57, 17 January 2024

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 $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ 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, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$.

Solution 2 (Easy)

We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth-grader with a buddy, which would mean there are $2.5$ sixth graders, which is impossible (of course unless you really do have half of a person). With $6$ ninth-graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth-graders total, two of whom have a buddy. Thus, the desired fraction is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.


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

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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