Difference between revisions of "2011 AMC 8 Problems/Problem 14"
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<math> \textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45} </math> | <math> \textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45} </math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=697 | ||
==Solution== | ==Solution== | ||
At Colfax Middle School, there are <math>\frac49 \times 270 = 120</math> girls. At Winthrop Middle School, there are <math>\frac59 \times 180 = 100</math> girls. The ratio of girls to the total number of students is <math>\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}</math> | At Colfax Middle School, there are <math>\frac49 \times 270 = 120</math> girls. At Winthrop Middle School, there are <math>\frac59 \times 180 = 100</math> girls. The ratio of girls to the total number of students is <math>\frac{120+100}{270+180} = \frac{220}{450} = \boxed{\textbf{(C)}\ \frac{22}{45}}</math> | ||
+ | |||
+ | ==Solution 2 (Guess and Check)== | ||
+ | Since we know there are <math>270</math> kids at Colfax Middle School and we know the ratio of boys to girls is <math>5:4</math>, we can use guess and check to find the number of girls at the dance (and vice versa). <math>5:4=150:120</math> and <math>4:5=80:100</math>. Now that the ratios sum to <math>270</math> and <math>180</math>, we know that in Colfax there are <math>120</math> girls and in Winthrop, there are <math>100</math> girls. <math>120+100=220</math> total girls and <math>270+180=450</math> total people. <math>\frac{220}{450}=\boxed{\textbf{(C)} \frac{22}{45}}</math> | ||
+ | |||
+ | ==Solution 3 (Educated Guess)== | ||
+ | |||
+ | We know that since <math>270</math> is greater than <math>180</math> and how the ratios are opposites, we know that there will be more boys than girls, so the fraction of students at the dance that are girls should be less than <math>\frac{1}{2}</math>. We also know that the amount of girls and the amount of boys are close, so the fraction of students at the dance that are girls should be really close to <math>\frac{1}{2}</math>. The only answer choice that/best satisfies these conditions is <math>\boxed{\textbf{(C)}\ \frac{22}{45}}</math>. | ||
+ | |||
+ | ~NXC | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/zAVxg6FmLyk | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=13|num-a=15}} | {{AMC8 box|year=2011|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:25, 18 November 2024
Contents
[hide]Problem
There are students at Colfax Middle School, where the ratio of boys to girls is
. There are
students at Winthrop Middle School, where the ratio of boys to girls is
. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=697
Solution
At Colfax Middle School, there are girls. At Winthrop Middle School, there are
girls. The ratio of girls to the total number of students is
Solution 2 (Guess and Check)
Since we know there are kids at Colfax Middle School and we know the ratio of boys to girls is
, we can use guess and check to find the number of girls at the dance (and vice versa).
and
. Now that the ratios sum to
and
, we know that in Colfax there are
girls and in Winthrop, there are
girls.
total girls and
total people.
Solution 3 (Educated Guess)
We know that since is greater than
and how the ratios are opposites, we know that there will be more boys than girls, so the fraction of students at the dance that are girls should be less than
. We also know that the amount of girls and the amount of boys are close, so the fraction of students at the dance that are girls should be really close to
. The only answer choice that/best satisfies these conditions is
.
~NXC
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
Video Solution by WhyMath
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.