Difference between revisions of "1989 AJHSME Problems/Problem 9"
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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== | ||
| + | <math></math>\frac{30}{2+3} = \frac{30}{5} = 6,<math> 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}}.$ | ||
==See Also== | ==See Also== | ||
Revision as of 12:29, 28 December 2021
Contents
Problem
There are 
boys for every 
girls in Ms. Johnson's math class. If there are 
students in her class, what percent of them are boys?
Solution
Besides ensuring the situation is possible, the students information is irrelevant.
From the first statement, we can deduce that of every 
students are boys. Thus, 
of the students are boys.
Solution 2
$$ (Error compiling LaTeX. Unknown error_msg)\frac{30}{2+3} = \frac{30}{5} = 6,
6\cdot2=12.
\frac{12}{30} = \frac{2}{5} = 40\% = \boxed{\text{C}}.$
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
