Difference between revisions of "2013 AMC 8 Problems/Problem 16"
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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> | <math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89</math> | ||
| − | ==Solution== | + | ==Solutions== |
| − | 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> | + | ===Solution 1: Algebra=== |
| + | We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together: | ||
| + | |||
| + | <math>5:3 = 5(8):3(8) = 40:24</math> | ||
| + | |||
| + | <math>8:5 = 8(5):5(5) = 40:25</math> | ||
| + | |||
| + | Therefore, the ratio of 8th graders to 7th graders to 6th graders is <math>40:25:24</math>. Since the ratio is in lowest terms, the smallest number of students participating in the project is <math>40+25+24 = \boxed{\textbf{(E)}\ 89}</math>. | ||
| + | |||
| + | ===Solution 2: Fakesolving=== | ||
| + | The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). 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> | ||
| + | |||
| + | ==Video Solution by OmegaLearn == | ||
| + | https://youtu.be/rQUwNC0gqdg?t=949 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | |||
| + | https://youtu.be/s7dIYGdXYPU ~savannahsolver | ||
==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}} | ||
Latest revision as of 15:05, 10 November 2023
Contents
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?
Solutions
Solution 1: Algebra
We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:
Therefore, the ratio of 8th graders to 7th graders to 6th graders is 
. Since the ratio is in lowest terms, the smallest number of students participating in the project is 
.
Solution 2: Fakesolving
The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are 
6th graders and 
7th graders. The numbers of students is 
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=949
~ pi_is_3.14
Video Solution 2
https://youtu.be/s7dIYGdXYPU ~savannahsolver
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. 
