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

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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>
 
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>
  
-PureSwag
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-[[User:PureSwag|PureSwag]]
  
==Solution 2==
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==Solution 2 (Fast and not rigorous)==
 
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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==Solution 3==
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Since there are an equal number of juniors and seniors on the debate team, suppose there are <math>x</math> juniors and <math>x</math> seniors. This number represents <math>25\% =\frac{1}{4}</math> of the juniors and <math>10\%= \frac{1}{10}</math> of the seniors, which tells us that there are <math>4x</math> juniors and <math>10x</math> seniors. There are <math>28</math> juniors and seniors in the program altogether, so we get
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<cmath>10x+4x=28,</cmath>
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<cmath>14x=28,</cmath>
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<cmath>x=2. </cmath>
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Which means there are <math>4x=8</math> juniors on the debate team, <math>\boxed{\text{(C)} \, 8}</math>.
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== Solution 4 (Elimination) ==
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The amount of juniors must be a multiple of <math>4</math>, since exactly <math>\frac{1}{4}</math> of the students are on the debate team. Thus, we can immediately see that <math>\boxed{C}</math> and <math>\boxed{E}</math> are the only possibilities for the number of juniors. However, if there are <math>20</math> juniors, then there are <math>8</math> seniors, so it is not true that <math>1/10</math> of the seniors are on the debate team, since <math>\frac{1}{10} \cdot 8 = \frac{4}{5}</math>, which is not an integer. Thus, we conclude that there are <math>8</math> juniors, so the answer is <math>\boxed{C}</math>.
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~mathboy100
  
 
== Video Solution by OmegaLearn (System of Equations) ==
 
== Video Solution by OmegaLearn (System of Equations) ==
 
https://youtu.be/BtEF-hJBGV8
 
https://youtu.be/BtEF-hJBGV8
  
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==Video Solution by TheBeautyofMath==
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https://youtu.be/gLahuINjRzU?t=319
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 +
~IceMatrix
 +
 +
==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw?t=182
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 +
~Interstigation
 +
 +
==Video Solution by WhyMath==
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https://youtu.be/owSnyec69FM
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~savannahsolver
  
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==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}}
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{{MAA Notice}}

Latest revision as of 12:20, 5 March 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 $28$ 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 \[\boxed{(C) \text{ } 8}.\]

-PureSwag

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, $\boxed{\text{(C)} \, 8}$.

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

See Also

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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