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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+ | {{AMC10 box|year=2021|ab=B|before=[[2021 AMC 10A]]|after=[[2022 AMC 10A]]}} |
Revision as of 00:31, 12 February 2021
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
2021 AMC 10B (Problems • Answer Key • Resources) | ||
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