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2023 AMC 12B 第 18 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

18.

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 33 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 1818 分，并且又比 Zelda 第二 学期所有小测的平均分高 33 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 33 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 1818 points higher than her average for the first semester and was again 33 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 2222 分。

Yolanda’s quiz average for the academic year was 2222 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 33 分。

Yolanda’s quiz average for the academic year was 33 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 33 分，那么她的学年平均分会与 Yolanda 相同。

If Zelda had scored 33 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
小提示：

令 Zelda 第一学期的平均分为 00，再用给定差值写出另外三个学期平均分

Let Zelda’s first-semester average be 0,0, and write the other three semester averages using the given gaps

大提示：

若 p,qp,q 分别是 Yolanda 和 Zelda 在第二学期完成的小测所占的比例，则她们的学年平均分之差为 3+18(p−q)3+18(p-q)

If p,qp,q are the fractions of Yolanda’s and Zelda’s quizzes taken in semester two, their yearly-average difference is 3+18(p−q)3+18(p-q)

解答：

把 Zelda 第一学期的平均分从四个学期平均分中同时减去，并不影响比较。于是可以把学期平均分写成 Zelda 的 0,180,18 和 Yolanda 的 3,213,21。设 pp 和 qq 分别是 Yolanda 和 Zelda 在第二学期完成的小测所占的比例。她们的学年平均分之差为 (3+18p)−18q=3+18(p−q)。 (3+18p)-18q=3+18(p-q)\text{。} 因为 0<p,q<10\lt p,q\lt 1，这个差小于 2121，所以不可能等于 2222。

其他陈述确实可能发生。取 (p,q)=(14,34)(p,q)=(\tfrac14,\tfrac34) 会使 Zelda 的平均分更高；取 p=qp=q 会使 Yolanda 的平均分高 33 分（把 Zelda 的每次小测成绩都加上 33 分之后，这也验证了最后一个选项）；取 (p,q)=(13,12)(p,q)=(\tfrac13,\tfrac12) 会使两人的学年平均分相等。上面出现的每个分数都能用正整数的小测次数实现。

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 0,180,18 for Zelda and 3,213,21 for Yolanda. Let pp and qq 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). (3+18p)-18q=3+18(p-q). Because 0<p,q<1,0\lt p,q\lt 1, this difference is less than 21,21, so it cannot be 22.22.

The other options really can occur. Taking (p,q)=(14,34)(p,q)=(\tfrac14,\tfrac34) makes Zelda’s average higher; taking p=qp=q makes Yolanda’s average 33 points higher (and also verifies the last option after adding 33 to every Zelda score); and taking (p,q)=(13,12)(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.

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