Difference between revisions of "1985 AJHSME Problems/Problem 23"
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If each student has <math>5</math> classes, and there are <math>1200</math> students, then they have a total of <math>5\times 1200=6000</math> classes among them. | If each student has <math>5</math> classes, and there are <math>1200</math> students, then they have a total of <math>5\times 1200=6000</math> classes among them. | ||
| − | Each class has <math>30</math> students, so there must be <math>\frac{6000}{30}=200</math> classes. Each class has <math>1</math> | + | Each class has <math>30</math> students, so there must be <math>\frac{6000}{30}=200</math> classes. Each class has <math>1</math> teacher, so the teachers have a total of <math>200</math> classes among them. |
Each teacher teaches <math>4</math> classes, so if there are <math>t</math> teachers, they have <math>4t</math> classes among them. This was found to be <math>200</math>, so <cmath>4t=200\Rightarrow t=50</cmath> | Each teacher teaches <math>4</math> classes, so if there are <math>t</math> teachers, they have <math>4t</math> classes among them. This was found to be <math>200</math>, so <cmath>4t=200\Rightarrow t=50</cmath> | ||
Revision as of 17:07, 17 May 2021
Problem
King Middle School has 
students. Each student takes 
classes a day. Each teacher teaches 
classes. Each class has 
students and 
teacher. How many teachers are there at King Middle School?
Solution
If each student has classes, and there are students, then they have a total of 
classes among them.
Each class has students, so there must be 
classes. Each class has teacher, so the teachers have a total of 
classes among them.
Each teacher teaches classes, so if there are 
teachers, they have 
classes among them. This was found to be , so ![\[4t=200\Rightarrow t=50\]](http://latex.artofproblemsolving.com/4/5/f/45fcdd0d7945c7294b7071e3d42baca6ff32773e.png)
This is answer choice 
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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. 
