2007 AMC 10A 第 10 题

先试着解答 2007 AMC 10A 第 10 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 10A 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。

10.

Dunbar 家庭由母亲、父亲和若干孩子组成。全家成员的平均年龄为 2020,父亲 4848 岁,母亲和孩子们的平均年龄为 1616。这个家庭有多少个孩子?
The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is 20,20, the father is 4848 years old, and the average age of the mother and children is 16.16. How many children are in the family?
22
33
44
55
66
答案:E
知识点:平均数一次方程
小提示:
设孩子数为 NN,则全家共有 N+2N + 2 人。
Let NN be the number of children; the family has N+2N + 2 members
大提示:
全家总年龄为 20(N+2)20(N + 2);去掉父亲后,其余人的总年龄为 16(N+1)16(N + 1)。
The total age is 20(N+2);20(N + 2); removing the father leaves 16(N+1)16(N + 1)
解答:
设孩子数为 NN,全家总年龄为 TT。
则 20=TN+220 = \dfrac{T}{N + 2},且 16=T−48N+116 = \dfrac{T - 48}{N + 1}。
这给出 20N+40=T20N + 40 = T 和 16N+64=T16N + 64 = T,所以 20N+40=16N+6420N + 40 = 16N + 64。
因此 4N=244N = 24,所以 N=6N = 6。
所以正确答案是 E。
Let NN be the number of children and TT the total age of the family.
Then 20=TN+220 = \dfrac{T}{N + 2} and 16=T−48N+1.16 = \dfrac{T - 48}{N + 1}.
These give 20N+40=T20N + 40 = T and 16N+64=T,16N + 64 = T, so 20N+40=16N+64.20N + 40 = 16N + 64.
Hence 4N=244N = 24 and N=6.N = 6.
Thus, the correct answer is E.
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