Difference between revisions of "2013 AMC 8 Problems/Problem 16"
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| + | The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40. Then there are <math>40*\frac{3}{5}=24</math> 6th graders and <math>40*\frac{5}{8}=25</math> 7th graders. The numbers of students is <math>40+24+25=\boxed{\textbf{(E)}\ 89}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=15|num-a=17}} | {{AMC8 box|year=2013|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:28, 27 November 2013
Problem
Solution
The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40. Then there are 
6th graders and 
7th graders. The numbers of students is 
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
