2010 AMC 12B 第 10 题

先试着解答 2010 AMC 12B 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 2010 AMC 12B 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.

1,2,3,,98,991, 2, 3, \ldots, 98, 99xx 的平均数是 100x100x。求 xx

The average of the numbers 1,2,3,,98,99,1, 2, 3, \ldots, 98, 99, and xx is 100x.100x. What is x?x?

49101\dfrac{49}{101}

50101\dfrac{50}{101}

12\dfrac{1}{2}

51101\dfrac{51}{101}

5099\dfrac{50}{99}

答案:B
知识点:平均数求和一次方程
难度评级:1410
解答:

119999 的和是 991002=4950\dfrac{99\cdot100}{2}=4950

由平均数的条件可得 所以 4950+x=10000x4950+x=10000x9999x=49509999x=49504950+x100=100x, \frac{4950+x}{100}=100x,

因此 x=49509999=50101x=\dfrac{4950}{9999}=\dfrac{50}{101}

因此,正确答案是 B

The numbers 11 through 9999 sum to 991002=4950.\dfrac{99\cdot100}{2}=4950.

The average condition is 4950+x100=100x, \frac{4950+x}{100}=100x, so 4950+x=10000x4950+x=10000x and 9999x=4950.9999x=4950.

Thus x=49509999=50101.x=\dfrac{4950}{9999}=\dfrac{50}{101}.

Thus, the correct answer is B.

← 第 9 题#9
完整试卷

其他年份的第 10 题