2023 AMC 12B Problems/Problem 18

Revision as of 19:03, 15 November 2023 by Gabew1234 (talk | contribs) (Solution)

Solution

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

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