AMC12 2010 B

AMC12 2010 B · Q10

AMC12 2010 B · Q10. It mainly tests Linear equations, Averages (mean).

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

数$1, 2, 3,\cdots, 98, 99,$和$x$的平均数是$100x$。$x$是多少？

(A) \frac{49}{101} \frac{49}{101}

(B) \frac{50}{101} \frac{50}{101}

(D) \frac{51}{101} \frac{51}{101}

(E) \frac{50}{99} \frac{50}{99}

Answer

Correct choice: (B)

正确答案：（B）

Solution

We first sum the first $99$ numbers: $\frac{99(100)}{2}=99\cdot50$. Then, we know that the sum of the series is $99\cdot50+x$. There are $100$ terms, so we can divide this sum by $100$ and set it equal to $100x$: \[\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x\] Using difference of squares: \[x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}\] Thus, the answer is $\boxed{\textbf{(B) }\frac{50}{101}}$.

先求前$99$个数的和：$\frac{99(100)}{2}=99\cdot50$。则总和为$99\cdot50+x$。共有$100$项，所以将其除以$100$并令其等于$100x$： \[\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x\] 利用平方差： \[x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}\] 因此答案是$\boxed{\textbf{(B) }\frac{50}{101}}$。

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