Difference between revisions of "2025 AMC 8 Problems/Problem 7"
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− | ==Problem== | + | == Problem == |
+ | |||
On the most recent exam on Prof. Xochi's class, | On the most recent exam on Prof. Xochi's class, | ||
− | <math>5</math> students earned a score of at least <math>95</math> | + | <math>5</math> students earned a score of at least <math>95\%</math>, |
− | <math>13</math> students earned a score of at least <math>90</math> | + | <math>13</math> students earned a score of at least <math>90\%</math>, |
− | <math>27</math> students earned a score of at least <math>85</math> | + | <math>27</math> students earned a score of at least <math>85\%</math>, |
− | <math>50</math> students earned a score of at least <math>80</math> | + | <math>50</math> students earned a score of at least <math>80\%</math>, |
− | How many students earned a score of at least 80% and less than 90%? | + | How many students earned a score of at least <math>80\%</math> and less than <math>90\%</math>? |
<math>\textbf{(A)}\ 8\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 22\qquad \textbf{(D)}\ 37\qquad \textbf{(E)}\ 45 </math> | <math>\textbf{(A)}\ 8\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 22\qquad \textbf{(D)}\ 37\qquad \textbf{(E)}\ 45 </math> | ||
− | ==Solution== | + | == Solution 1 == |
− | <math>50</math> people scored at least <math>80\%</math>, and out of these <math>50</math> people, <math>13</math> of them earned | + | <math>50</math> people scored at least <math>80\%</math>, and out of these <math>50</math> people, <math>13</math> of them earned a score that was not less than <math>90\%</math>, so the number of people that scored in between at least <math>80\%</math> and less than <math>90\%</math> is <math>50-13 = \boxed{\text{(D)\ 37}}</math>. |
~Soupboy0 | ~Soupboy0 | ||
− | ==Solution 2== | + | == Solution 2 == |
− | Let <math> | + | Let <math>a</math> denote the number of people who had a score of at least <math>85</math>, but less than <math>90</math>, and let <math>b</math> denote the number of people who had a score of at least <math>80</math> but less than <math>85</math>. Our answer is equal to <math>a+b</math>. We find <math>a = 27 - 13 = 14</math>, while <math>b = 50 - 27 = 23</math>. Thus, the answer is <math>23 + 14 = \boxed{\text{(D)\ 37}}</math>. |
-vockey | -vockey | ||
+ | |||
+ | == Video Solution 1 by Cool Math Problems == | ||
+ | |||
+ | https://youtu.be/BRnILzqVwHk?si=kOvMjPSxqVZR8Mmt&t=89 | ||
− | == | + | ==Video Solution 2 by SpreadTheMathLove== |
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
− | ==Video Solution | + | == Video Solution 3 == |
− | + | [//youtu.be/VP7g-s8akMY?si=P0Nar6jhTGl1yKZb&t=427 ~hsnacademy] | |
− | ~hsnacademy | ||
− | ==Video Solution by Thinking Feet== | + | == Video Solution 4 by Thinking Feet == |
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
− | ==Video Solution by Daily Dose of Math== | + | == Video Solution 5 by Daily Dose of Math == |
− | + | [//youtu.be/nkpdskFVgdM ~Thesmartgreekmathdude] | |
− | + | == See Also == | |
− | |||
{{AMC8 box|year=2025|num-b=6|num-a=8}} | {{AMC8 box|year=2025|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 13:05, 4 February 2025
Contents
Problem
On the most recent exam on Prof. Xochi's class,
students earned a score of at least
,
students earned a score of at least
,
students earned a score of at least
,
students earned a score of at least
,
How many students earned a score of at least and less than
?
Solution 1
people scored at least
, and out of these
people,
of them earned a score that was not less than
, so the number of people that scored in between at least
and less than
is
.
~Soupboy0
Solution 2
Let denote the number of people who had a score of at least
, but less than
, and let
denote the number of people who had a score of at least
but less than
. Our answer is equal to
. We find
, while
. Thus, the answer is
.
-vockey
Video Solution 1 by Cool Math Problems
https://youtu.be/BRnILzqVwHk?si=kOvMjPSxqVZR8Mmt&t=89
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution 3
Video Solution 4 by Thinking Feet
Video Solution 5 by Daily Dose of Math
See Also
2025 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.