Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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==Solution 2 (Fast and not rigorous)== | ==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 | ||
| + | |||
| + | ==Solution 3== | ||
| + | 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 | ||
| + | <cmath>10x+4x=28,</cmath> | ||
| + | <cmath>14x=28,</cmath> | ||
| + | <cmath>x=2. </cmath> | ||
| + | Which means there are <math>4x=8</math> juniors on the debate team, <math>\boxed{\text{(C)} \, 8}</math>. | ||
== Video Solution by OmegaLearn (System of Equations) == | == Video Solution by OmegaLearn (System of Equations) == | ||
Revision as of 05:42, 14 February 2021
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 and 
of the seniors 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)
-PureSwag
Solution 2 (Fast and 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, 
.
Video Solution by OmegaLearn (System of Equations)
| 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 | ||
