Difference between revisions of "2013 AMC 8 Problems/Problem 16"
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==Problem== | ==Problem== | ||
| + | A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of <math>8^\text{th}</math>-graders to <math>6^\text{th}</math>-graders is <math>5:3</math>, and the the ratio of <math>8^\text{th}</math>-graders to <math>7^\text{th}</math>-graders is <math>8:5</math>. What is the smallest number of students that could be participating in the project? | ||
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| + | <math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89</math> | ||
==Solution== | ==Solution== | ||
Revision as of 08:28, 27 November 2013
Problem
A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 
-graders to 
-graders is 
, and the the ratio of -graders to 
-graders is 
. What is the smallest number of students that could be participating in the project?
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. 
