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考题7-8 2015 AMC 10B

2018-08-06 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，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。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：

Problem 7

Consider the operation "minus the reciprocal of," defined by $a\diamond b=a-\frac{1}{b}$ . What is $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))$ ?

$\textbf{(A) } -\dfrac{7}{30} \qquad\textbf{(B) } -\dfrac{1}{6} \qquad\textbf{(C) } 0 \qquad\textbf{(D) } \dfrac{1}{6} \qquad\textbf{(E) } \dfrac{7}{30}$

Solution

$1\diamond2=1-\dfrac{1}{2}=\dfrac{1}{2}$ , so $(1\diamond2)\diamond3=\dfrac{1}{2}\diamond3=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ . Also, $2\diamond3=2-\dfrac{1}{3}=\dfrac{5}{3}$ , so $1\diamond(2\diamond3)=1-\dfrac{1}{5/3}=1-\dfrac{3}{5}=\dfrac{2}{5}$ . Thus, $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))=\dfrac{1}{6}-\dfrac{2}{5}=\boxed{\mathbf{(A)}\ -\dfrac{7}{30}}$

Problem 8

The letter F shown below is rotated $90^\circ$ clockwise around the origin, then reflected in the $y$ -axis, and then rotated a half turn around the origin. What is the final image?

考题7-8 2015 AMC 10B

Solution

The first rotation moves the base of the $F$ to the negative $y$ -axis, and the stem to the positive $x$ -axis. The reflection then moves the stem to the negative $x$ -axis, with the base unchanged. Then the half turn moves the stem to the positive $x$ axis and the base to the positive $y$ -axis, choice $\boxed{\mathbf{(E)}}$ .