Question

上一学年，Yolanda 和 Zelda 选了不同的课程，这些课程在两个学期中不一定安排了相同数量的小测。Yolanda 第一学期所有小测的平均分比 Zelda 第一学期所有小测的平均分高 $3$ 分。 Yolanda 第二学期所有小测的平均分比她第一学期的平均分高 $18$ 分，并且又比 Zelda 第二学期所有小测的平均分高 $3$ 分。以下哪一项不可能为真？

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 的学年小测平均分比 Zelda 高 $22$ 分。

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

Zelda 的学年小测平均分高于 Yolanda。

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

Yolanda 的学年小测平均分比 Zelda 高 $3$ 分。

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

Zelda 的学年小测平均分等于 Yolanda。

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

如果 Zelda 每次小测都多得 $3$ 分，那么她的学年平均分会与 Yolanda 相同。

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.

答案：A

知识点：加权平均数不等式逻辑推理

难度评级：1800

解答：

令 Zelda 第一学期平均分为 $0,18$ ，则 Yolanda 第一学期为 $3,21$ ，她第二学期为 $p$ ，Zelda 第二学期为 $q$ 。每个人的学年平均分都是其两个学期平均分的加权平均，所以 Yolanda 的学年平均分介于 $0\lt p,q\lt 1$ 和 $21$ 之间，Zelda 的学年平均分介于和之间。 Yolanda Zelda 的最大可能差值至多为，所以不可能是 $22$ 。 $(3+18p)-18q=3+18(p-q).$

其他陈述都可以通过适当选择小测次数实现。 $(p,q)=(\tfrac14,\tfrac34)$ $p=q$ $3$ $3$ $(p,q)=(\tfrac13,\tfrac12)$

因此，正确答案是 A。

Subtracting Zelda's first-semester average from all four semester averages does not affect the comparison. We may therefore write the semester averages as $0,18$ for Zelda and $3,21$ for Yolanda. Let $p$ and $q$ be the fractions of Yolanda's and Zelda's quizzes, respectively, that occurred in the second semester. Their yearly-average difference is $(3+18p)-18q=3+18(p-q).$ Because $0\lt p,q\lt 1,$ this difference is less than $21,$ so it cannot be $22.$

The other options really can occur. Taking $(p,q)=(\tfrac14,\tfrac34)$ makes Zelda's average higher; taking $p=q$ makes Yolanda's average $3$ points higher (and also verifies the last option after adding $3$ to every Zelda score); and taking $(p,q)=(\tfrac13,\tfrac12)$ makes the yearly averages equal. Each displayed fraction can be realized by positive integer quiz counts.

Thus, the correct answer is A.

2023 AMC 12B 第 18 题

答案：A

难度评级：1800

解答：

其他年份的第 18 题