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 | + | == 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 15:04, 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
![\[\frac{2}{5}g + \frac{1}{2}b = 12\]](http://latex.artofproblemsolving.com/d/9/5/d95d42d65cb204933a631de26774b6fa115417c8.png)
![\[5b = 4g \Longrightarrow b = \frac{4}{5}g\]](http://latex.artofproblemsolving.com/7/8/7/78796c063ee13ba00f45399fa6ba802d2e37400b.png)
Substituting, 
. So there are 12 boys, and 
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 | ||
