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Difference between revisions of "2023 AMC 12B Problems/Problem 18"

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==Solution 2==
 
==Solution 2==
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We can use process of elimination by finding possible solutions to answer choices. Let <math>y_1</math> and <math>y_2 be the number of quizzes Yolanda took in the first and second semesters, respectively. Define </math>z_1<math> and </math>z_2<math> similarly for Zelda.
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Answer choice B is satisfied by </math>(y_1,y_2,z_1,z_2) = (289,1,1,289)<math>.
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Answer choice C and E are both satisfied by </math>(y_1,y_2,z_1,z_2) = (17,17,17,17)<math>.
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Answer choice D is satisfied by </math>(y_1,y_2,z_1,z_2) = (7,5,5,7)<math>.
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Therefore the answer is </math>\boxed{(\text{A})}$.
  
 
~cantalon
 
~cantalon

Revision as of 20:05, 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?

$\textbf{(A)}$ Yolanda's quiz average for the academic year was 22 points higher than Zelda's.

$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.

$\textbf{(C)}$ Yolanda's quiz average for the academic year was 3 points higher than Zelda's.

$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.

$\textbf{(E)}$ 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 1

Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$, $Y_2 = Y_1 + 18$, $Y_2 = Z_2 + 3$.

Denote by $A_{12}$ the average of person with initial $A$ in the full year. Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$.

Therefore, the impossible solution is $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

We can use process of elimination by finding possible solutions to answer choices. Let $y_1$ and $y_2 be the number of quizzes Yolanda took in the first and second semesters, respectively. Define$z_1$and$z_2$similarly for Zelda.

Answer choice B is satisfied by$ (Error compiling LaTeX. Unknown error_msg)(y_1,y_2,z_1,z_2) = (289,1,1,289)$.

Answer choice C and E are both satisfied by$ (Error compiling LaTeX. Unknown error_msg)(y_1,y_2,z_1,z_2) = (17,17,17,17)$.

Answer choice D is satisfied by$ (Error compiling LaTeX. Unknown error_msg)(y_1,y_2,z_1,z_2) = (7,5,5,7)$.

Therefore the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{(\text{A})}$.

~cantalon

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

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