AMC12 2023 B

AMC12 2023 B · Q18

AMC12 2023 B · Q18. It mainly tests Averages (mean), Weighted average.

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?

上学年，Yolanda和Zelda选修了不同的课程，每学期测验数量不一定相同。第一学期，Yolanda所有测验的平均分比Zelda第一学期所有测验平均分高3分。第二学期，Yolanda所有测验的平均分比她第一学期平均分高18分，并且再次比Zelda第二学期所有测验平均分高3分。以下哪个陈述不可能为真？

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

(B) Zelda's quiz average for the academic year was higher than Yolanda's. Zelda's quiz average for the academic year was higher than Yolanda's.

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

(D) Zelda's quiz average for the academic year equaled Yolanda's. Zelda's quiz average for the academic year equaled Yolanda's.

(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. 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.

Answer

Correct choice: (A)

正确答案：（A）

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

设$A_i$为人名首字母为$A$者在学期$i \in \left\{1, 2 \right\}$的平均分。于是，$Y_1 = Z_1 + 3$，$Y_2 = Y_1 + 18$，$Y_2 = Z_2 + 3$。设$A_{12}$为人名首字母为$A$者全年平均分。于是，$Y_{12}$可以是$(Y_1 , Y_2)$中的任意数，$Z_{12}$可以是$(Z_1 , Z_2)$中的任意数。因此，不可能的情况是 $\boxed{\textbf{(A)}~\text{Yolanda全年测验平均分比Zelda高22分。}}$

Topics