Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 10"

(Solution)
(Solution)
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==Solution==
 
==Solution==
 
Let the number of total students by <math>s</math>, and number of classes by <math>c</math>. Thus, our problem implies that  
 
Let the number of total students by <math>s</math>, and number of classes by <math>c</math>. Thus, our problem implies that  
<cmath>\frac{s}{c+1} = 20 \rightarrow s = 20c + 20</cmath>
+
<cmath>\frac{s}{c+1} = 20 \Longrightarrow s = 20c + 20</cmath>
<cmath>\frac{s}{c-1} = 30 \rightarrow s = 30c - 30</cmath>
+
<cmath>\frac{s}{c-1} = 30 \Longrightarrow s = 30c - 30</cmath>
  
 
We solve this system of linear equations to get <math>c = 5</math> and <math>s = 120</math>. Thus, our answer is <math>\frac{s}{c} = \boxed{24}</math>.
 
We solve this system of linear equations to get <math>c = 5</math> and <math>s = 120</math>. Thus, our answer is <math>\frac{s}{c} = \boxed{24}</math>.
  
 
~Bradygho
 
~Bradygho

Revision as of 22:30, 10 July 2021

Problem

In a certain school, each class has an equal number of students. If the number of classes was to increase by $1$, then each class would have $20$ students. If the number of classes was to decrease by $1$, then each class would have $30$ students. How many students are in each class?

Solution

Let the number of total students by $s$, and number of classes by $c$. Thus, our problem implies that \[\frac{s}{c+1} = 20 \Longrightarrow s = 20c + 20\] \[\frac{s}{c-1} = 30 \Longrightarrow s = 30c - 30\]

We solve this system of linear equations to get $c = 5$ and $s = 120$. Thus, our answer is $\frac{s}{c} = \boxed{24}$.

~Bradygho