Difference between revisions of "2013 AMC 8 Problems/Problem 16"

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==Solution==
 
==Solution==
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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 09:28, 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 $40*\frac{3}{5}=24$ 6th graders and $40*\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\boxed{\textbf{(E)}\ 89}$

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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