Difference between revisions of "1985 AJHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | Let the number of boys be <math>2x</math> | + | Let the number of boys be <math>2x</math>. It follows that the number of girls is <math>3x</math>. These two values add up to <math>30</math> students, so <cmath>2x+3x=5x=30\Rightarrow x=6</cmath> |
− | The [[subtraction|difference]] between the number of girls and the number of boys is <math>3x-2x=x</math>, which is <math>6</math>, so <math>\boxed{\text{D}}</math> | + | The [[subtraction|difference]] between the number of girls and the number of boys is <math>3x-2x=x</math>, which is <math>6</math>, so the answer is <math>\boxed{\text{D}}</math>. |
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+ | ==Video Solution== | ||
+ | https://youtu.be/UkTqNBFHJaA | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1985|num-b=15|num-a=17}} | {{AJHSME box|year=1985|num-b=15|num-a=17}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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+ | {{MAA Notice}} |
Latest revision as of 07:14, 13 January 2023
Contents
Problem
The ratio of boys to girls in Mr. Brown's math class is . If there are students in the class, how many more girls than boys are in the class?
Solution
Let the number of boys be . It follows that the number of girls is . These two values add up to students, so
The difference between the number of girls and the number of boys is , which is , so the answer is .
Video Solution
~savannahsolver
See Also
1985 AJHSME (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.