Difference between revisions of "2021 AMC 10B Problems/Problem 3"

(This is an overlapping problem for AMC 10/12 B.)
(Undo revision 148154 by MRENTHUSIASM (talk) Oops. This is not a overlapping problem. I am undoing my edit.)
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==See Also==
{{AMC12 box|year=2021|ab=B|num-b=1|num-a=3}}
{{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}}
{{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}}
{{MAA Notice}}

Revision as of 05:40, 2 March 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 $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?

$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution 1

Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$

Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations \[5j=2s\]\[j+s=28,\] and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is \[\boxed{(C) \text{ } 8}.\]


Solution 2 (Fast and not rigorous)

We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$. Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{C}$. ~samrocksnature

Solution 3

Since there are an equal number of juniors and seniors on the debate team, suppose there are $x$ juniors and $x$ seniors. This number represents $25\% =\frac{1}{4}$ of the juniors and $10\%= \frac{1}{10}$ of the seniors, which tells us that there are $4x$ juniors and $10x$ seniors. There are $28$ juniors and seniors in the program altogether, so we get \[10x+4x=28,\] \[14x=28,\] \[x=2.\] Which means there are $4x=8$ juniors on the debate team, $\boxed{\text{(C)} \, 8}$.

Video Solution by OmegaLearn (System of Equations)


Video Solution by TheBeautyofMath



Video Solution by Interstigation



Video Solution by WhyMath



2021 AMC 10B (ProblemsAnswer KeyResources)
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
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