Difference between revisions of "2013 AMC 8 Problems/Problem 16"
(→Solution) |
(→Problem) |
||
Line 1: | Line 1: | ||
==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? | ||
+ | |||
+ | <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 09:28, 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.