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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>
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<math>\linebreak</math>
 +
~Geometry285
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 +
 +
 +
==See also==
 +
#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
 +
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 16:24, 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


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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