Difference between revisions of "2008 AMC 8 Problems/Problem 20"
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For <math>g</math> and <math>b</math> to be integers, <math>g</math> must cancel out with the numerator, and the smallest possible value is <math>8</math>. This yields <math>9</math> boys. The minimum number of students is <math>8+9=\boxed{\textbf{(B)}\ 17}</math>. | For <math>g</math> and <math>b</math> to be integers, <math>g</math> must cancel out with the numerator, and the smallest possible value is <math>8</math>. This yields <math>9</math> boys. The minimum number of students is <math>8+9=\boxed{\textbf{(B)}\ 17}</math>. | ||
| − | ==Video Solution== | + | ==Video Solution by WhyMath== |
https://youtu.be/HseEFQDuh_c | https://youtu.be/HseEFQDuh_c | ||
Revision as of 15:28, 21 April 2021
Problem
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and 
of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
Solution
Let 
be the number of boys and 
be the number of girls.
![\[\frac23 b = \frac34 g \Rightarrow b = \frac98 g\]](http://latex.artofproblemsolving.com/6/c/2/6c23a00f6e01044ecb7518140f37df28977b236d.png)
For and to be integers, must cancel out with the numerator, and the smallest possible value is 
. This yields 
boys. The minimum number of students is 
.
Video Solution by WhyMath
~savannahsolver
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
