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Difference between revisions of "1989 AJHSME Problems/Problem 9"

(New page: ==Problem== There are <math>2</math> boys for every <math>3</math> girls in Ms. Johnson's math class. If there are <math>30</math> students in her class, what percent of them are boy...)
 
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From the first statement, we can deduce that <math>2</math> of every <math>2+3=5</math> students are boys.  Thus, <math>2/5=40\% \rightarrow \boxed{\text{C}}</math> of the students are boys.
 
From the first statement, we can deduce that <math>2</math> of every <math>2+3=5</math> students are boys.  Thus, <math>2/5=40\% \rightarrow \boxed{\text{C}}</math> of the students are boys.
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==Solution 2==
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<cmath>\frac{30}{2+3} = \frac{30}{5} = 6,</cmath> and that is the base unit. Since there are 2 boys for every 5 people, <math>6\cdot2=12.</math> Applying this into a fraction gives <math>\frac{12}{30} = \frac{2}{5} = 40\% = \boxed{\text{C}}.</math>
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~DuoDuoling0
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1989|num-b=8|num-a=10}}
 
{{AJHSME box|year=1989|num-b=8|num-a=10}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 13:29, 28 December 2021

Problem

There are $2$ boys for every $3$ girls in Ms. Johnson's math class. If there are $30$ students in her class, what percent of them are boys?

$\text{(A)}\ 12\% \qquad \text{(B)}\ 20\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 60\% \qquad \text{(E)}\ 66\frac{2}{3}\%$

Solution

Besides ensuring the situation is possible, the $30$ students information is irrelevant.

From the first statement, we can deduce that $2$ of every $2+3=5$ students are boys. Thus, $2/5=40\% \rightarrow \boxed{\text{C}}$ of the students are boys.

Solution 2

\[\frac{30}{2+3} = \frac{30}{5} = 6,\] and that is the base unit. Since there are 2 boys for every 5 people, $6\cdot2=12.$ Applying this into a fraction gives $\frac{12}{30} = \frac{2}{5} = 40\% = \boxed{\text{C}}.$

~DuoDuoling0

See Also

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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