Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 10"
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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 | |
+ | <cmath>\frac{s}{c+1} = 20 \Longrightarrow s = 20c + 20</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>. | ||
+ | |||
+ | ~Bradygho | ||
+ | |||
+ | ==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> | ||
+ | <math>\linebreak</math> | ||
+ | ~Geometry285 | ||
+ | |||
+ | |||
+ | |||
+ | ==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
Contents
Problem
In a certain school, each class has an equal number of students. If the number of classes was to increase by , then each class would have students. If the number of classes was to decrease by , then each class would have students. How many students are in each class?
Solution
Let the number of total students by , and number of classes by . Thus, our problem implies that
We solve this system of linear equations to get and . Thus, our answer is .
~Bradygho
Solution 2
We have the number of classes is , so , or . Plugging back the total number of students is , so the answer is ~Geometry285
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.