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

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

@@ Line 1: / Line 1: @@
-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?
+==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>\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>
 ==Solution 1==
-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>
+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 (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>.
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/kJAJXJTKjSk
-==Solution 2==
+~Education, the Study of Everything
-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 probability is <math>\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}</math>.
+===Video solution===
+https://www.youtube.com/watch?v=u3otXEQgsUU   ~David
+https://youtu.be/7md_65nXjTw
+~savannahsolver
 ==See Also==
 {{AMC8 box|year=2015|num-b=15|num-a=17}}
 {{MAA Notice}}

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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