Difference between revisions of "2025 AMC 8 Problems/Problem 7"
Hsnacademy (talk | contribs) (→Video Solution by Thinking Feet) |
(added a new solution) |
||
Line 22: | Line 22: | ||
~Soupboy0 | ~Soupboy0 | ||
+ | ==Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | -vockey | ||
+ | |||
==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI |
Revision as of 23:58, 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.