2007 AMC 8 第 7 题

先试着解答 2007 AMC 8 第 7 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2007 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

7.

一个房间里 55 个人的平均年龄是 3030 岁。一名 1818 岁的人离开房间。剩下四个人的平均年龄是多少?
The average age of 55 people in a room is 3030 years. An 1818-year-old person leaves the room. What is the average age of the four remaining people?
2525
2626
2929
3333
3636
答案:D
知识点:平均数
小提示:
把原来的平均年龄转化为总年龄
Convert the original average into a total age.
大提示:
先减去离开者的年龄,再除以剩余人数
Subtract the age of the person who leaves before dividing by the remaining number of people.
解答:
起初,房间里所有人的总年龄为 5⋅30=1505 \cdot 30 = 150 岁。
这个人离开后,总年龄为 150−18=132150 - 18 = 132。剩下 44 人,平均年龄为 1324=33\dfrac{132}{4} = 33。
所以正确答案是 D。
Initially, the total age of everyone in the room is 5⋅30=1505 \cdot 30 = 150 years.
After the person leaves, the total age is 150−18=132.150 - 18 = 132. With 44 people remaining, the average age becomes 1324=33.\dfrac{132}{4} = 33.
Thus, D is the correct answer.
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