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

m (→‎Problem)
m (→‎Solution: change the 1200 to 600 or might be there is mistake in the question ., please correct it)
 
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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.  
  
 
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 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.

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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