Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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==Problem== | ==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 <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? | + | 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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~mathboy100 | ~mathboy100 | ||
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== Video Solution by OmegaLearn (System of Equations) == | == Video Solution by OmegaLearn (System of Equations) == | ||
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~savannahsolver | ~savannahsolver | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/F47TKAJT-XE | ||
| + | |||
| + | ~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
Contents
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 
students in the program, 
of the juniors as a class and 
of the seniors as a class 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 ![\[5j=2s\]](http://latex.artofproblemsolving.com/e/9/0/e90eca73861ccf271ca585eb49b39608553b1018.png)
![\[j+s=28,\]](http://latex.artofproblemsolving.com/f/2/d/f2d9da87d46e3ec7ee67241f9f6c951877716ebb.png)
and the solution is 
Since we want the number of juniors, the answer is ![\[\boxed{(C) \text{ } 8}.\]](http://latex.artofproblemsolving.com/5/7/2/57218434e7e1414f2a6eff3803ced2168d20ead8.png)
Solution 2 (Fast but Not Rigorous)
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
Solution 3
Since there are an equal number of juniors and seniors on the debate team, suppose there are 
juniors and seniors. This number represents 
of the juniors and 
of the seniors, which tells us that there are 
juniors and 
seniors. There are juniors and seniors in the program altogether, so we get
![\[10x+4x=28,\]](http://latex.artofproblemsolving.com/e/b/9/eb94e454a148b39c6eadf9bbb4ee96091f4a8935.png)
![\[14x=28,\]](http://latex.artofproblemsolving.com/3/a/7/3a79f8e171bd340ca1010a549dd4024a272b91a6.png)
![\[x=2.\]](http://latex.artofproblemsolving.com/9/1/7/9172ebebf917720816b359a892e83b16aed9c07e.png)
Which means there are 
juniors on the debate team, 
.
Solution 4 (Elimination)
The amount of juniors must be a multiple of 
, since exactly 
of the students are on the debate team. Thus, we can immediately see that 
and 
are the only possibilities for the number of juniors. However, if there are juniors, then there are 
seniors, so it is not true that 
of the seniors are on the debate team, since 
, which is not an integer. Thus, we conclude that there are juniors, so the answer is .
~mathboy100
Video Solution by OmegaLearn (System of Equations)
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
~savannahsolver
Video Solution
~Education, the Study of Everything
See Also
| 2021 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
