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AMC数学竞赛真题2017年10B 9-10

2018-08-30 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数学竞赛真题以及官方解答吧：

Problem 9

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$amc真题$

Solution 1

There are two ways the contestant can win.

Case 1: The contestant guesses all three right. This can only happen $\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}$ of the time.

Case 2: The contestant guesses only two right. We pick one of the questions to get wrong, $3$ , and this can happen $\frac{1}{3} * \frac{1}{3} * \frac{2}{3}$ of the time. Thus, $\frac{2}{27} * 3$ = $\frac{6}{27}$ .

So, in total the two cases combined equals $\frac{1}{27} + \frac{6}{27}$ = $\boxed{\textbf{(D)}\ \frac{7}{27}}$ .

Solution 2 (complementary counting)

Complementary counting is good for solving the problem and checking work if you solved it using the method above.

There are two ways the contestant can lose.

Case 1: The contestant guesses zero questions correctly.

The probability of guessing incorrectly for each question is $\frac{2}{3}$ . Thus, the probability of guessing all questions incorrectly is $\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}$ .

Case 2: The contestant guesses one question correctly. There are 3 ways the contestant can guess one question correctly since there are 3 questions. The probability of guessing correctly is $\frac{1}{3}$ so the probability of guessing one correctly and two incorrectly is $amc数学竞赛$ .

The sum of the two cases is $\frac{8}{27} + \frac{4}{9} = \frac{20}{27}$ . This is the complement of what we want to the answer is $1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}$

Problem 10

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$ . What is $c$ ?

$美国数学竞赛$

Solution

Writing each equation in slope-intercept form, we get $y=\frac{a}{2}x-\frac{1}{2}c$ and $amc真题$ . We observe the slope of each equation is $\frac{a}{2}$ and $-\frac{2}{b}$ , respectively. Because the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$ , we see that $\frac{a}{2}=-\frac{1}{-\frac{2}{b}}$ because it is given that the two lines are perpendicular. This equation simplifies to $a=b$ .

Because $(1, -5)$ is a solution of both equations, we deduce $a \times 1-2 \times -5=c$ and $2 \times 1+b \times -5=-c$ . Because we know that $a=b$ , the equations reduce to $a+10=c$ and $2-5a=-c$ . Solving this system of equations, we get $c=\boxed{\textbf{(E)}\ 13}$