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

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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.
 
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.
  
== See also ==
+
== See Also ==
 
{{PMWC box|year=1997|num-b=I6|num-a=I8}}
 
{{PMWC box|year=1997|num-b=I6|num-a=I8}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 16:04, 15 May 2012

Problem

40% of girls and 50% of boys in a class got an 'A'. If there are only 12 students in the class who got '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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