Difference between revisions of "2016 AMC 8 Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | + | Let there be <math>b</math> boys and <math>g</math> girls in the school. We see <math>g=b</math>, which means <math>\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b</math> kids went on the trip and <math>\frac{3}{4}b</math> kids are girls. So, the answer is <math>\frac{\frac{3}{4}b}{\frac{17}{12}b}=\frac{9}{17}</math>, which is <math>\boxed{\textbf{(B)} \frac{9}{17}}</math>. | |
− | <math> | + | |
− | <math> | + | ~CHECKMATE2021 |
− | + | ||
+ | ==Solution 2== | ||
+ | |||
+ | Using WLOG (Without loss of generativity), Let there be <math>12</math> boys and <math>12</math> girls in the school. Now we can do <math>\frac{3}{4}\times{12}</math> + <math>\frac{2}{3}\times{12}</math> to get the total number of students going to the field trip to be <math>17</math>. Since we already know the number of girls to be <math>9</math>. We have our answer to be <math>\frac{9}{17}</math>. So, the answer is <math>\boxed{\textbf{(B)} \frac{9}{17}}</math>. | ||
+ | |||
+ | ~algebraic_algorithmic | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/Y4N4L_HcnKY | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
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{{AMC8 box|year=2016|num-b=11|num-a=13}} | {{AMC8 box|year=2016|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 18:06, 25 June 2025
Contents
Problem
Jefferson Middle School has the same number of boys and girls. of the girls and
of the boys went on a field trip. What fraction of the students on the field trip were girls?
Solution 1
Let there be boys and
girls in the school. We see
, which means
kids went on the trip and
kids are girls. So, the answer is
, which is
.
~CHECKMATE2021
Solution 2
Using WLOG (Without loss of generativity), Let there be boys and
girls in the school. Now we can do
+
to get the total number of students going to the field trip to be
. Since we already know the number of girls to be
. We have our answer to be
. So, the answer is
.
~algebraic_algorithmic
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.