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

Revision as of 21:12, 19 October 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 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 ur mom lol

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

@@ Line 13: / Line 13: @@
 ==Solution 2 (Fast but 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 ur mom lol
 ==Solution 3==

2021 AMC 10B (Problems • Answer Key • Resources)
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