Difference between revisions of "2025 AMC 8 Problems/Problem 7"
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+ | ==Solution 2== | ||
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+ | Let <math>b</math> denote the number of people who had a score of at least <math>85</math>, but less than <math>90</math>. Similarly, let <math>d</math> be the number of people who had a score of at least <math>80</math> but less than <math>85</math>. Now we can see the question is just asking for <math>c+d</math>, we find <math>c = 27 - 13 = 14</math>, while <math>d = 50 - 27 = 23</math>. Thus, the answer is <math>23 + 14 = \boxed{\text{(D)\ 37}}</math>. | ||
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+ | -vockey | ||
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==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI |
Latest revision as of 22:58, 31 January 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 80% and less than 90%?
Solution
people scored at least , and out of these people, of them earned at least , so the people that scored in between and is .
~Soupboy0
Solution 2
Let denote the number of people who had a score of at least , but less than . Similarly, let be the number of people who had a score of at least but less than . Now we can see the question is just asking for , we find , while . Thus, the answer is .
-vockey
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=P0Nar6jhTGl1yKZb&t=427 ~hsnacademy
Video Solution by Thinking Feet
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
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.