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> |
This is answer choice <math>\boxed{\text{E}}</math> | This is answer choice <math>\boxed{\text{E}}</math> | ||
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{{AJHSME box|year=1985|num-b=22|num-a=24}} | {{AJHSME box|year=1985|num-b=22|num-a=24}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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+ | {{MAA Notice}} |
Latest revision as of 10:04, 10 June 2024
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
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.