Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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Revision as of 05:37, 2 March 2021
Contents
Problem
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the students in the program,
of the juniors and
of the seniors are on the debate team. How many juniors are in the program?
Solution 1
Say there are juniors and
seniors in the program. Converting percentages to fractions,
and
are on the debate team, and since an equal number of juniors and seniors are on the debate team,
Cross-multiplying and simplifying we get Additionally, since there are
students in the program,
It is now a matter of solving the system of equations
and the solution is
Since we want the number of juniors, the answer is
Solution 2 (Fast and not rigorous)
We immediately see that is the only possible amount of seniors, as
can only correspond with an answer choice ending with
. Thus the number of seniors is
and the number of juniors is
. ~samrocksnature
Solution 3
Since there are an equal number of juniors and seniors on the debate team, suppose there are juniors and
seniors. This number represents
of the juniors and
of the seniors, which tells us that there are
juniors and
seniors. There are
juniors and seniors in the program altogether, so we get
Which means there are
juniors on the debate team,
.
Video Solution by OmegaLearn (System of Equations)
Video Solution by TheBeautyofMath
https://youtu.be/gLahuINjRzU?t=319
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=182
~Interstigation
Video Solution by WhyMath
~savannahsolver
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.