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

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==Problem==
 
==Problem==
  
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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 <math>28</math> students in the program, <math>25\%</math> of the juniors and <math>10\%</math> of the seniors are on the debate team. How many juniors are in the program?
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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 <math>28</math> students in the program, <math>25\%</math> of the juniors as a class and <math>10\%</math> of the seniors as a class are on the debate team. How many juniors are in the program?
  
 
<math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20</math>
 
<math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20</math>
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Say there are <math>j</math> juniors and <math>s</math> seniors in the program. Converting percentages to fractions, <math>\frac{j}{4}</math> and <math>\frac{s}{10}</math> are on the debate team, and since an equal number of juniors and seniors are on the debate team, <math>\frac{j}{4} = \frac{s}{10}.</math>
 
Say there are <math>j</math> juniors and <math>s</math> seniors in the program. Converting percentages to fractions, <math>\frac{j}{4}</math> and <math>\frac{s}{10}</math> are on the debate team, and since an equal number of juniors and seniors are on the debate team, <math>\frac{j}{4} = \frac{s}{10}.</math>
  
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Cross-multiplying and simplifying we get <math>5j=2s.</math> Additionally, since there are <math>28</math> students in the program, <math>j+s = 28.</math> It is now a matter of solving the system of equations <cmath>5j=2s</cmath><cmath>j+s=28,</cmath> and the solution is <math>j = 8, s = 20.</math> Since we want the number of juniors, the answer is <cmath>\boxed{(C) \text{ } 9}.</cmath>
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Cross-multiplying and simplifying we get <math>5j=2s.</math> Additionally, since there are <math>28</math> students in the program, <math>j+s = 28.</math> It is now a matter of solving the system of equations <cmath>5j=2s</cmath><cmath>j+s=28,</cmath> and the solution is <math>j = 8, s = 20.</math> Since we want the number of juniors, the answer is <cmath>\boxed{(C) \text{ } 8}.</cmath>
  
 
-[[User:PureSwag|PureSwag]]
 
-[[User:PureSwag|PureSwag]]
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~savannahsolver
 
~savannahsolver
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==Video Solution==
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https://youtu.be/F47TKAJT-XE
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 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
 
{{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}}
 
{{MAA Notice}}

Latest revision as of 18:56, 11 October 2024

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 $28$ students in the program, $25\%$ of the juniors as a class and $10\%$ of the seniors as a class 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 \[\boxed{(C) \text{ } 8}.\]

-PureSwag

Solution 2 (Fast but 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, $\boxed{\text{(C)} \, 8}$.

Solution 4 (Elimination)

The amount of juniors must be a multiple of $4$, since exactly $\frac{1}{4}$ of the students are on the debate team. Thus, we can immediately see that $\boxed{C}$ and $\boxed{E}$ are the only possibilities for the number of juniors. However, if there are $20$ juniors, then there are $8$ seniors, so it is not true that $1/10$ of the seniors are on the debate team, since $\frac{1}{10} \cdot 8 = \frac{4}{5}$, which is not an integer. Thus, we conclude that there are $8$ juniors, so the answer is $\boxed{C}$.

~mathboy100

Video Solution by OmegaLearn (System of Equations)

https://youtu.be/BtEF-hJBGV8

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

https://youtu.be/owSnyec69FM

~savannahsolver

Video Solution

https://youtu.be/F47TKAJT-XE

~Education, the Study of Everything

See Also

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AMC 10 Problems and Solutions

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