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考题9-10 2015 AMC 12A

2018-08-22 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 9

A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?

$AMC12的官方真题以及官方解答$

Solution 1

If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certian color is $\frac{4}{6} * \frac{3}{5} * \frac{2}{4} * \frac{1}{3} = \frac{1}{15}$ . Since there are three different colors, our final probability is $3 * \frac{1}{15} = \frac{1}{5} \textbf{ (C)}$ .

Solution 2

The order of the girls' drawing the balls really does not matter. Thus, we can let Cheryl draw first, so after she draws one ball, the other must be of the same color. Thus, the answer is $\frac{1}{5} \textbf{ (C)}$ .

Solution 3

The total number of ways they can draw is ${6 \choose 2}$ ${4 \choose 2}$ $AMC12的官方真题以及官方解答$ . Let Cheryl draw first and since there are three colors, there are ${3 \choose 1}$ ways she can get 2 marbles of the same color. The other two pick two each, which leads to ${4 \choose 2}$ and ${2 \choose 2}$ , respectively. $AMC12的官方真题以及官方解答$

Problem 10

Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$ . What is $x$ ?

$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$

Solution

Use SFFT to get $(x+1)(y+1)=81$ . The terms $(x+1)$ and $(y+1)$ must be factors of $81$ , which include $1, 3, 9, 27, 81$ . Because $x > y$ , $x+1$ is equal to $27$ or $81$ . But if $x+1=81$ , then $y=0$ and so $x=\boxed{\textbf{(E)}\ 26}$ .