Difference between revisions of "2023 AMC 12B Problems/Problem 18"
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Therefore, the impossible solution is | Therefore, the impossible solution is | ||
− | <math>\boxed{\textbf{(A)}}</math> | + | <math>\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}</math> |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 19:48, 15 November 2023
Problem
Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was 3 points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was 18 points higher than her average for the first semester and was again 3 points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
Yolanda's quiz average for the academic year was 22 points higher than Zelda's.
Zelda's quiz average for the academic year was higher than Yolanda's.
Yolanda's quiz average for the academic year was 3 points higher than Zelda's.
Zelda's quiz average for the academic year equaled Yolanda's.
If Zelda had scored 3 points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.
Solution
Denote by the average of person with initial in semester Thus, , , .
Denote by the average of person with initial in the full year. Thus, can be any number in and can be any number in .
Therefore, the impossible solution is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.