2021 AMC 10B Problems/Problem 3: Difference between revisions
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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 | ||
{{AMC10 box|year=2021|ab=B|before=[[2021 AMC 10A]]|after=[[2022 AMC 10A]]}} | |||
Revision as of 23:31, 11 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
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