Difference between revisions of "1997 PMWC Problems/Problem I7"

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== Problem ==
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<math>40\%</math> of girls and <math>50\%</math> of boys in a class got an <math>\mathrm{'A'}</math>. If there are only <math>12</math> students in the class who got <math>\mathrm{'A'}</math>s and the ratio of boys and girls in the class is <math>4:5</math>, how many students are there in the class?
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== Solution ==
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<cmath>\frac{2}{5}g + \frac{1}{2}b = 12</cmath>
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<cmath>5b = 4g \Longrightarrow b = \frac{4}{5}g</cmath>
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Substituting, <math>\frac{2}{5}g + \frac{1}{2}\left(\frac 45g\right) = 12 \Longrightarrow \frac{4}{5}g = 12 \Longrightarrow g = 15</math>. So there are 12 boys, and <math>12 + 15 = 27</math> students in the class.
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== See Also ==
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{{PMWC box|year=1997|num-b=I6|num-a=I8}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 14:23, 20 April 2014

Problem

$40\%$ of girls and $50\%$ of boys in a class got an $\mathrm{'A'}$. If there are only $12$ students in the class who got $\mathrm{'A'}$s and the ratio of boys and girls in the class is $4:5$, how many students are there in the class?

Solution

\[\frac{2}{5}g + \frac{1}{2}b = 12\] \[5b = 4g \Longrightarrow b = \frac{4}{5}g\]

Substituting, $\frac{2}{5}g + \frac{1}{2}\left(\frac 45g\right) = 12 \Longrightarrow \frac{4}{5}g = 12 \Longrightarrow g = 15$. So there are 12 boys, and $12 + 15 = 27$ students in the class.

See Also

1997 PMWC (Problems)
Preceded by
Problem I6
Followed by
Problem I8
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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