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>. |
==See Also== | ==See Also== | ||
Revision as of 09:49, 13 August 2009
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 ![\[2x+3x=5x=30\Rightarrow x=6\]](http://latex.artofproblemsolving.com/7/c/5/7c5cba1c4d3b8b83efd4057cc728b08d7e2630b1.png)
The difference between the number of girls and the number of boys is 
, which is 
, so the answer is 
.
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 | ||
