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AMC数学竞赛真题2016年12A 3-4

2018-08-31 重点归纳

AMC12是针对高中学生的数学测验，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 3

The remainder can be defined for all real numbers $x$ and $y$ with $y \neq 0$ by $\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor$ where $\left \lfloor \tfrac{x}{y} \right \rfloor$ denotes the greatest integer less than or equal to $\tfrac{x}{y}$ . What is the value of $\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )$ ?

$\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}$

Solution

The value, by definition, is 美国数学竞赛

Solution 2

Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is $\boxed{B}$

Problem 4

The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$ . What is the value of $x$ ?

$amc数学竞赛$

Solution

Since $x$ is the mean, amc竞赛

Therefore, $7x=540+x$ , so $x=\boxed{\textbf{(D) }90}.$

Check

Order the list: $\{40,50,60,90,100,200\}$ . $x$ must be either $60$ or $90$ because it is both the median and the mode of the set. Thus $90$ is correct.