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Difference between revisions of "2025 AMC 8 Problems/Problem 7"
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==Problem 7==
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==Problem==
 
On the most recent exam on Prof. Xochi's class,  
 
On the most recent exam on Prof. Xochi's class,  
  
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<math>50</math> people scored at least <math>80\%</math>, and out of these <math>50</math> people, <math>13</math> of them earned at least <math>90\%</math>, so the people that scored in between <math>80\%</math> and <math>90\%</math> is <math>50-13 = \boxed{\text{(D)\ 37}}</math>.
 
<math>50</math> people scored at least <math>80\%</math>, and out of these <math>50</math> people, <math>13</math> of them earned at least <math>90\%</math>, so the people that scored in between <math>80\%</math> and <math>90\%</math> is <math>50-13 = \boxed{\text{(D)\ 37}}</math>.
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~Soupboy0
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==Vide Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 19:16, 30 January 2025

Problem

On the most recent exam on Prof. Xochi's class,


$5$ students earned a score of at least $95$%,

$13$ students earned a score of at least $90$%,

$27$ students earned a score of at least $85$%,

$50$ students earned a score of at least $80$%,


How many students earned a score of at least 80% and less than 90%?

$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 22\qquad \textbf{(D)}\ 37\qquad \textbf{(E)}\ 45$

Solution

$50$ people scored at least $80\%$, and out of these $50$ people, $13$ of them earned at least $90\%$, so the people that scored in between $80\%$ and $90\%$ is $50-13 = \boxed{\text{(D)\ 37}}$.

~Soupboy0

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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. AMC logo.png

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