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Difference between revisions of "1985 AJHSME Problems/Problem 16"

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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>.
 
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==
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https://youtu.be/UkTqNBFHJaA
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~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

Problem

The ratio of boys to girls in Mr. Brown's math class is $2:3$. If there are $30$ students in the class, how many more girls than boys are in the class?

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$

Solution

Let the number of boys be $2x$. It follows that the number of girls is $3x$. These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]

The difference between the number of girls and the number of boys is $3x-2x=x$, which is $6$, so the answer is $\boxed{\text{D}}$.

Video Solution

https://youtu.be/UkTqNBFHJaA

~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


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