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Difference between revisions of "1985 AJHSME Problems/Problem 23"

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m (→‎Solution: change the 1200 to 600 or might be there is mistake in the question ., please correct it)
 
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==Problem==
 
==Problem==
  
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King Middle School has <math>1200</math> students.  Each student takes <math>5</math> classes a day.  Each teacher teaches <math>4</math> classes.  Each class has <math>30</math> students and <math>1</math> teacher.  How many teachers are there at King Middle School?   
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King Middle School has <math>600</math> students.  Each student takes <math>5</math> classes a day.  Each teacher teaches <math>4</math> classes.  Each class has <math>30</math> students and <math>1</math> teacher.  How many teachers are there at King Middle School?   
  
 
<math>\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50</math>
 
<math>\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50</math>
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==Solution==
 
==Solution==
  
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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.  
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If each student has <math>5</math> classes, and there are <math>600</math> students, then they have a total of <math>5\times 1200=6000</math> classes among them.  
  
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Each class has <math>30</math> students, so there must be <math>\frac{6000}{30}=200</math> classes. Each class has <math>1</math> teachers, so the teachers have a total of <math>200</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> teacher, so the teachers have a total of <math>200</math> classes among them.
  
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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>  
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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>
  
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This is answer choice <math>\boxed{\text{D}}</math>
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This is answer choice <math>\boxed{\text{E}}</math>
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==Solution 2 (Similar to above solution) ==
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Each teacher teaches <math>4</math> classes and each class has <math>30</math> students, so each teacher teaches <math>120</math> students in one class.
 +
 
 +
So, to teach five classes, there has to be <math>5</math> teachers, but there is <math>1200</math> students, so multiply by <math>\frac{1200}{120}</math> which is <math>50</math>, or <math>\boxed{\textbf{(E)}\ 50}</math>
  
 
==See Also==
 
==See Also==
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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 15:50, 18 November 2024

Problem

King Middle School has $600$ students. Each student takes $5$ classes a day. Each teacher teaches $4$ classes. Each class has $30$ students and $1$ teacher. How many teachers are there at King Middle School?

$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$

Solution

If each student has $5$ classes, and there are $600$ students, then they have a total of $5\times 1200=6000$ classes among them.

Each class has $30$ students, so there must be $\frac{6000}{30}=200$ classes. Each class has $1$ teacher, so the teachers have a total of $200$ classes among them.

Each teacher teaches $4$ classes, so if there are $t$ teachers, they have $4t$ classes among them. This was found to be $200$, so \[4t=200\Rightarrow t=50\]

This is answer choice $\boxed{\text{E}}$

Solution 2 (Similar to above solution)

Each teacher teaches $4$ classes and each class has $30$ students, so each teacher teaches $120$ students in one class.

So, to teach five classes, there has to be $5$ teachers, but there is $1200$ students, so multiply by $\frac{1200}{120}$ which is $50$, or $\boxed{\textbf{(E)}\ 50}$

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


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