Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
We immediately see that <math>E</math> is the only possible amount of seniors, as <math>10\%</math> can only correspond with an answer choice ending with <math>0</math>. Thus the number of seniors is <math>20</math> and the number of juniors is <math>28-20=8\rightarrow \boxed{C}</math>. ~samrocksnature | We immediately see that <math>E</math> is the only possible amount of seniors, as <math>10\%</math> can only correspond with an answer choice ending with <math>0</math>. Thus the number of seniors is <math>20</math> and the number of juniors is <math>28-20=8\rightarrow \boxed{C}</math>. ~samrocksnature | ||
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+ | == Video Solution by OmegaLearn (System of Equations) == | ||
+ | https://youtu.be/BtEF-hJBGV8 | ||
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{{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}} |
Revision as of 14:59, 12 February 2021
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
-PureSwag
Solution 2
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
Video Solution by OmegaLearn (System of Equations)
2021 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 4 | |
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