Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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+ | ==Problem== | ||
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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 and <math>10\%</math> of the seniors are on the debate team. How many juniors are in the program? | ||
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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 | + | -[[User:PureSwag|PureSwag]] |
− | ==Solution 2== | + | ==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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | == Solution 4 (Elimination) == | ||
+ | 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>. | ||
+ | |||
+ | ~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== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:42, 12 June 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 and the solution is Since we want the number of juniors, the answer is
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 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
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.