Difference between revisions of "2021 AMC 10B Problems/Problem 3"
Avi.agarwal (talk | contribs) (→Solution 2 (Logic)) |
Avi.agarwal (talk | contribs) (→Solution 2 (Logic)) |
||
| Line 14: | Line 14: | ||
==Solution 2 (Logic)== | ==Solution 2 (Logic)== | ||
| − | Clearly, the number of seniors must be a number divisible by <math>10</math>, since <math>10\%</math> points to a whole number, so the number of seniors must be either <math>10</math> or <math>20</math>. If there are ten seniors, there is <math>1</math> senior on the debate team, which would mean that there are <math>4</math> juniors, since <math>10\%s</math> = <math>25\%j</math>. | + | Clearly, the number of seniors must be a number divisible by <math>10</math>, since <math>10\%</math> points to a whole number, so the number of seniors must be either <math>10</math> or <math>20</math>. If there are ten seniors, there is <math>1</math> senior on the debate team, which would mean that there are <math>4</math> juniors, since <math>10\%s</math> = <math>25\%j</math>. This is impossible, so the number of seniors is 20, and the number of juniors is <math>28-20=8\rightarrow \boxed{C}</math>. |
Note that by using the answer choices, we could automatically rule out 10. | Note that by using the answer choices, we could automatically rule out 10. | ||
Latest revision as of 18:26, 25 October 2025
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 (Logic)
Clearly, the number of seniors must be a number divisible by 
, since points to a whole number, so the number of seniors must be either or 
. If there are ten seniors, there is 
senior on the debate team, which would mean that there are 
juniors, since 
= 
. This is impossible, so the number of seniors is 20, and the number of juniors is 
.
Note that by using the answer choices, we could automatically rule out 10.
~sigmacuber632
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
