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

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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Revision as of 16:20, 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 Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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