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考题11-12 2014 AMC 8

2018-08-04 重点归纳

AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高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AMC 8数学竞赛试题及答案吧：

Problem 11

Jack wants to bike from his house to Jill's house, which is located three blocks east and two blocks north of Jack's house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill's house by biking a total of five blocks?

$考题11-12 2014 AMC 8$

Solution

We can apply complementary counting and count the paths that DO go through the blocked intersection, which is $\dbinom{2}{1}\dbinom{3}{1}=6$ . There are a total of $\dbinom{5}{2}=10$ paths, so there are $10-6=4$ paths possible. $\boxed{A}$ is the correct answer.

Solution 2

We can make a diagram of the roads available to Jack.

[asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(A--B--H--F--A); draw(D--C); draw(E--G); [/asy]

Then, we can simply list the possible routes.

[asy] size(50); defaultpen(linewidth(0.8)); pair A=(0,2),B=origin,C=(2,0),D=(2,2),E=(2,1),F=(3,2),G=(3,1),H=(3,0); draw(B--H--F); draw(D--C,dotted); draw(E--G,dotted); draw(F--A--B,dotted); [/asy]

There are 4 possible routes, so our answer is $\boxed{A}$ .

Problem 12

A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby pictures. What is the probability that a reader guessing at random will match all three correctly?

$amc数学竞赛试题及答案$

Solution

Let's call the celebrities A, B, and C. There is a $\frac{1}{3}$ chance that celebrity A's picture will be selected, and a $\frac{1}{3}$ chance that his baby picture will be selected. That means there are two celebrities left. There is now a $\frac{1}{2}$ chance that celebrity B's picture will be selected, and another $\frac{1}{2}$ chance that his baby picture will be selected. This leaves a $\frac{1}{1}$ chance for the last celebrity, so the total probability is $amc数学竞赛试题及答案$ . However, the order of the celebrities doesn't matter, so the final probability will be $3!\cdot \frac{1}{36}=\frac{1}{6}$ (B).

Solution 2

There is a $\frac{1}{3}$ chance that the reader will choose the correct baby picture for the first person. Next, the second person gives a $\frac{1}{2}$ chance, and the last person leaves only 1 choice. Thus, the probability is $amc8$

Solution 3

There are $3!$ ways assign the pictures to each of the celebrities. There is one favorable outcome where all of them are matched correctly, so the answer is $amc8$ .