Difference between revisions of "2021 AMC 10B Problems/Problem 3"
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+ | ==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 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 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/F47TKAJT-XE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==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}} |
Revision as of 18:14, 16 August 2022
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
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.