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

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==Solution 2==
 
==Solution 2==
 
We have the number of classes is <math>x</math>, so <math>20(x+1)=30(x-1)</math>, or <math>x=5</math>. Plugging back the total number of students is <math>120</math>, so the answer is <math>\frac{120}{5} = \boxed{24}</math>
 
We have the number of classes is <math>x</math>, so <math>20(x+1)=30(x-1)</math>, or <math>x=5</math>. Plugging back the total number of students is <math>120</math>, so the answer is <math>\frac{120}{5} = \boxed{24}</math>
 +
<math>\linebreak</math>
 +
~Geometry285

Revision as of 11:28, 11 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

Solution 2

We have the number of classes is $x$, so $20(x+1)=30(x-1)$, or $x=5$. Plugging back the total number of students is $120$, so the answer is $\frac{120}{5} = \boxed{24}$ $\linebreak$ ~Geometry285

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