2006 AMC 8 真题与在线限时测验

计时

40:00

1.

Mindy 买了三样东西，价格分别是 $\$1.98$ 、 $\$5.04$ 和 $\$9.89$ 。她的总花费四舍五入到最接近的美元是多少？

Mindy made three purchases for $\$1.98$ , $\$5.04$ , and $\$9.89$ . What was her total, to the nearest dollar?

$\$10$

$\$15$

$\$16$

$\$17$

$\$18$

答案：D

知识点：估算钱币

难度评级：370

小提示：

先把每个价格四舍五入到最接近的美元。

Round each price to the nearest dollar before adding.

大提示：

题目只问最接近的美元，所以不需要精确到美分。

The exact cents are not needed because the question asks for the nearest dollar.

解答：

这些数都接近整数美元，可以把价格四舍五入后相加。

这些价格分别约为 $\$ 2$ 、 $\$ 5$ 和 $\$ 10$ ，总和为 $\$ 17$ 。

所以正确答案是 D。

Since all the values are already close to a whole number, we can just add the sums of the rounded numbers.

These prices round to $\$ 2,$ $\$ 5,$ and $\$ 10,$ which add to $\$ 17.$

Thus, D is the correct answer.

2.

在 AMC $8$ 竞赛中，Billy 答对 $13$ 题，答错 $7$ 题，最后 $5$ 题未作答。他的分数是多少？

On the AMC $8$ contest Billy answers $13$ questions correctly, answers $7$ questions incorrectly and doesn’t answer the last $5.$ What is his score?

$1$

$6$

$13$

$19$

$26$

答案：C

知识点：基本计数

难度评级：370

小提示：

在 AMC $8$ 中，只有答对的题会加分。

On the AMC $8,$ only correct answers add to the score.

大提示：

答错和空题不会改变答对题数。

Incorrect and blank answers do not change the number of correct answers.

解答：

AMC $8$ 每答对一题得 $1$ 分，所以 Billy 得 $13$ 分。

所以正确答案是 C。

Since the AMC $8$ only awards $1$ point for each correct question, Billy will get $13$ points.

Thus, C is the correct answer.

3.

Elisa 在泳池里游圈。刚开始时，她用 $25$ 分钟游完 $10$ 圈。现在她能用 $24$ 分钟游完 $12$ 圈。她每圈用时减少了多少分钟？

Elisa swims laps in the pool. When she first started, she completed $10$ laps in $25$ minutes. Now, she can finish $12$ laps in $24$ minutes. By how many minutes has she improved her lap time?

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$1$

$2$

$3$

答案：A

知识点：速率

难度评级：560

小提示：

分别求 Elisa 改进前和改进后的每圈分钟数。

Find Elisa’s minutes per lap before and after the improvement.

大提示：

比较 $\frac{25}{10}$ 分钟每圈和 $\frac{24}{12}$ 分钟每圈。

Compare $\frac{25}{10}$ minutes per lap with $\frac{24}{12}$ minutes per lap.

解答：

起初 Elisa 每圈用 $\frac{25}{10}=2.5$ 分钟。现在她每圈用 $\frac{24}{12}=2$ 分钟。

因此每圈用时减少 $2.5-2=0.5=\dfrac{1}{2}$ 分钟。

所以正确答案是 A。

Initially, Elisa swam one lap in $\frac{25}{10}=2.5$ minutes. Now she swims one lap in $\frac{24}{12}=2$ minutes.

Therefore, she improved her lap time by $2.5-2=0.5=\dfrac{1}{2}$ minute.

Thus, A is the correct answer.

4.

起初，一个转盘指向西。Chenille 将它顺时针转 $2 \frac{1}{4}$ 圈，然后逆时针转 $3 \frac{3}{4}$ 圈。两次移动后转盘指向什么方向？

Initially, a spinner points west. Chenille moves it clockwise $2 \frac{1}{4}$ revolutions and then counterclockwise $3 \frac{3}{4}$ revolutions. In what direction does the spinner point after the two moves?

北

North

东

East

南

South

西

West

西北

Northwest

答案：B

知识点：分数变换

难度评级：660

小提示：

整圈旋转不会改变最终方向。

Whole revolutions do not change the final direction.

大提示：

去掉整圈后，比较顺时针四分之一圈和逆时针四分之三圈。

After removing whole turns, compare a clockwise quarter-turn with a counterclockwise three-quarter-turn.

解答：

转盘顺时针转 $2 \frac{1}{4}$ 圈，再逆时针转 $3 \frac{3}{4}$ 圈，等价于总共逆时针转 $1 \frac{1}{2}$ 圈。

因此转盘最终指向东。

所以正确答案是 B。

If the spinner goes $2 \frac{1}{4}$ revolutions clockwise and $3 \frac{3}{4}$ revolutions counterclockwise, it goes $1 \frac{1}{2}$ revolutions counterclockwise.

This means that the spinner will be pointing east.

Thus, B is the correct answer.

5.

点 $A$ 、 $B$ 、 $C$ 和 $D$ 是大正方形各边的中点。如果大正方形面积为 $60$ ，小正方形的面积是多少？

Points $A,$ $B,$ $C$ and $D$ are midpoints of the sides of the larger square. If the larger square has area $60,$ what is the area of the smaller square?

$15$

$20$

$24$

$30$

$40$

答案：D

知识点：正方形（几何）中点面积分割

难度评级：820

小提示：

小正方形周围的四个角三角形全等。

The four corner triangles around the smaller square are congruent.

大提示：

那四个外侧三角形可重新排列覆盖小正方形。

Those four outside triangles can be rearranged to cover the smaller square.

解答：

可以把所有三角形向内折叠，正好覆盖小正方形。这说明小正方形的面积是大正方形面积的一半。

所以小正方形面积为 $60 \div 2 = 30$ 。

所以正确答案是 D。

Note that we can fold all the triangles in to perfectly cover the smaller square. This means that area of the smaller square is half the area of the larger square.

This makes the area of the smaller square $60 \div 2 = 30.$

Thus, D is the correct answer.

6.

字母 T 由两个 $2 \times 4$ 英寸的长方形相邻放置而成，如图所示。这个 T 的周长是多少英寸？

The letter T is formed by placing two $2 \times 4$ inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?

$12$

$16$

$20$

$22$

$24$

答案：C

知识点：周长矩形

难度评级：930

小提示：

先分别计算两个长方形的周长。

Start with the perimeters of the two rectangles separately.

大提示：

共用的内部线段在分别计算周长时被计入，但不属于外周长。

The shared interior segments are counted in the separate perimeters but not in the outside perimeter.

解答：

如果分别计算两个长方形的总周长，会得到 $2(2(2 + 4)) = 2 \cdot 2 \cdot 6 = 24\text{。}$

在字母 T 中，它们的相交部分从每个长方形中去掉了一段长度 $2$ 的边。因此 T 的周长为 $24 - 2 \cdot 2 = 20\text{。}$

所以正确答案是 C。

If we found the total perimeter of the two rectangles separately, we would have gotten $2(2(2 + 4)) = 2 \cdot 2 \cdot 6 = 24.$

In the letter T, we can see that their intersection removes a piece of length $2$ from each of the rectangles. Therefore, the perimeter of the T is $24 - 2 \cdot 2 = 20.$

Thus, C is the correct answer.

7.

圆 $X$ 的半径为 $\pi$ 。圆 $Y$ 的周长为 $8 \pi$ 。圆 $Z$ 的面积为 $9 \pi$ 。按半径从小到大排列这些圆。

Circle $X$ has a radius of $\pi.$ Circle $Y$ has a circumference of $8 \pi.$ Circle $Z$ has an area of $9 \pi.$ List the circles in order from smallest to largest radius.

$X$ 、 $Y$ 、 $Z$

$X,$ $Y,$ $Z$

$Z$ 、 $X$ 、 $Y$

$Z,$ $X,$ $Y$

$Y$ 、 $X$ 、 $Z$

$Y,$ $X,$ $Z$

$Z$ 、 $Y$ 、 $X$

$Z,$ $Y,$ $X$

$X$ 、 $Z$ 、 $Y$

$X,$ $Z,$ $Y$

答案：B

知识点：圆圆周长圆面积

难度评级：960

小提示：

把三个条件都转化为半径。

Convert all three pieces of information into radii.

大提示：

对圆 $Y$ 使用 $C=2\pi r$ ，对圆 $Z$ 使用 $A=\pi r^2$ 。

Use $C=2\pi r$ for circle $Y$ and $A=\pi r^2$ for circle $Z$ .

解答：

公式为 $C = 2 \pi r$ 和 $A = \pi r^2$ 。由此圆 $Y$ 的半径是 $8 \pi \div (2 \pi) = 4\text{。}$ 圆 $Z$ 的半径是 $\sqrt{9 \pi \div \pi} = 3\text{。}$ 因为 $\pi$ 大于 $3$ 且小于 $4$ ，所以正确顺序为 $Z$ 、 $X$ 、 $Y$ 。

所以正确答案是 B。

Recall that $C = 2 \pi r$ and $A = \pi r^2.$ Using these formulas we get that the radius of $Y$ is $8 \pi \div (2 \pi) = 4.$ We also get that the radius of $Z$ is $\sqrt{9 \pi \div \pi} = 3.$ As $\pi$ is greater than $3$ and less than $4,$ the correct order is $Z,$ $X,$ $Y.$

Thus, the answer is B .

8.

表格显示了广播电台 KAMC 一项调查的部分结果。接受调查的男性中有百分之多少收听这个电台？

The table shows some of the results of a survey by radio station KAMC. What percentage of the males surveyed listen to the station?

$39$

$48$

$52$

$55$

$75$

答案：E

知识点：数据与图表解读百分数

难度评级：1000

小提示：

用总数填出接受调查的男性人数。

Use the totals to fill in the number of males surveyed.

大提示：

用男性总数减去不收听的男性数，得到男性听众数。

Find the male listeners by subtracting the male non-listeners from the male total.

解答：

接受调查的男性总数是调查总人数减去女性人数： $200-96=104$ 。

这些男性中 $26$ 人不收听，所以 $104-26=78$ 名男性收听。所求百分比为 $\dfrac{78}{104}\cdot 100\%=75\%$ 。

所以正确答案是 E。

The total number of males surveyed is the total number surveyed minus the number of females surveyed: $200-96=104$ .

Of those males, $26$ do not listen, so $104-26=78$ males listen. The desired percentage is $\dfrac{78}{104}\cdot 100\%=75\%$ .

Thus, E is the correct answer.

9.

下列乘积是多少？ $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}\text{？}$

What is the product of $\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\cdots\times\dfrac{2006}{2005}?$

$1$

$1002$

$1003$

$2005$

$2006$

答案：C

知识点：裂项相消

难度评级：1130

小提示：

大多数相邻的分子和分母会约掉。

Most adjacent numerators and denominators cancel.

大提示：

约分后只剩第一个分母和最后一个分子。

After cancellation, only the first denominator and the last numerator remain.

解答：

每个分数的分子都会与下一个分数的分母约掉。最后只剩 $\dfrac{2006}{2} = 1003\text{。}$

所以正确答案是 C。

Note that the numerator of every fraction cancels with the denominator of the following fraction. This leaves two numbers: $\dfrac{2006}{2} = 1003.$

Thus, C is the correct answer.

10.

Jorge 的老师让他画出所有有序数对 $(w, l)$ ，其中 $w$ 是面积为 $12$ 的长方形的宽， $l$ 是长，并且宽和长都是正整数。他的图应该是什么样？

Jorge’s teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area $12.$ What should his graph look like?

答案：A

知识点：因数坐标几何

难度评级：1120

小提示：

列出 $12$ 的正整数因数对。

List the positive integer factor pairs of $12$ .

大提示：

当两个因数不同时， $(w,l)$ 和 $(l,w)$ 是不同的点。

Remember that $(w,l)$ and $(l,w)$ are different plotted points when the factors differ.

解答：

满足 $wl=12$ 的正整数因数对 $(w,l)$ 为 $(1,12)$ 、 $(2,6)$ 、 $(3,4)$ 、 $(4,3)$ 、 $(6,2)$ 和 $(12,1)$ 。

这些点形成六个递减排列的点，只有图 A 与它们匹配。

所以正确答案是 A。

The positive integer factor pairs $(w,l)$ with $wl=12$ are $(1,12),$ $(2,6),$ $(3,4),$ $(4,3),$ $(6,2),$ and $(12,1)$ .

These points form a decreasing set of six points, and only graph A matches them.

Thus, A is the correct answer.

11.

有多少个两位数的各位数字和是完全平方数？

How many two-digit numbers have digits whose sum is a perfect square?

$13$

$16$

$17$

$18$

$19$

答案：C

知识点：数字完全平方数分类讨论

难度评级：1310

小提示：

最大可能数字和是 $18$ 。

The largest possible digit sum is $18$ .

大提示：

数出数位和为 $1$ 、 $4$ 、 $9$ 和 $16$ 的两位数。

Count the two-digit numbers whose digit sums are $1,$ $4,$ $9,$ and $16$ .

解答：

数位和为 $1$ 的数有 $1$ 个： $10$ 。

数位和为 $4$ 的数有 $4$ 个： $13$ 、 $22$ 、 $31$ 和 $40$ 。

数位和为 $9$ 的数有 $9$ 个： $18$ 、 $27$ 、 $36$ 、 $45$ 、 $54$ 、 $63$ 、 $72$ 、 $81$ 和 $90$ 。

数位和为 $16$ 的数有 $3$ 个： $79$ 、 $88$ 和 $97$ 。

因此共有 $17$ 个数满足题意。

所以正确答案是 C。

There is $1$ number whose digit sum is $1$ : $10.$

There are $4$ numbers whose digit sum is $4$ : $13,$ $22,$ $31,$ and $40.$

There are $9$ numbers whose digit sum is $9$ : $18,$ $27,$ $36,$ $45,$ $54,$ $63,$ $72,$ $81,$ and $90.$

There are $3$ numbers whose digit sum is $16$ : $79,$ $88,$ and $97.$

Therefore, there are $17$ numbers that satisfy the problem statement.

Thus, C is the correct answer.

12.

Antonette 在一场有 $10$ 道题的测试中得分率为 $70 \%$ ，在一场有 $20$ 道题的测试中得分率为 $80 \%$ ，在一场有 $30$ 道题的测试中得分率为 $90 \%$ 。如果三次测试合并成一场有 $60$ 道题的测试，哪个百分数最接近她的总成绩？

Antonette gets $70 \%$ on a $10$ -problem test, $80 \%$ on a $20$ -problem test and $90 \%$ on a $30$ -problem test. If the three tests are combined into one $60$ -problem test, which percent is closest to her overall score?

$40$

$77$

$80$

$83$

$87$

答案：D

知识点：平均数百分数

难度评级：1100

小提示：

把每次测试百分比转化为答对题数。

Convert each test percentage into a number of correct answers.

大提示：

这是按题目数量加权的平均，不是三个百分比的普通平均。

This is a weighted average by number of problems, not a plain average of the three percentages.

解答：

Antonette 第一次测试答对 $0.7\cdot 10=7$ 题。类似地，她第二、三次测试分别答对 $0.8\cdot 20=16$ 题和 $0.9\cdot 30=27$ 题。

总共答对 $50$ 题。合并测试成绩为 $100\cdot \dfrac{50}{60}=83\frac{1}{3}\%$ ，最接近 $83\%$ 。

所以正确答案是 D。

Antonette got $0.7\cdot 10=7$ questions right on the first test. Similarly, she got $0.8\cdot 20=16$ and $0.9\cdot 30=27$ problems right on her second and third tests.

Adding these up yields $50$ correct questions. Her score on the combined test would have been $100\cdot \dfrac{50}{60}=83\frac{1}{3}\%$ , closest to $83\%$ .

Thus, D is the correct answer.

13.

Cassie 上午 $8:30$ 从 Escanaba 骑车前往 Marquette，速度恒定为每小时 $12$ 英里。Brian 上午 $9:00$ 从 Marquette 骑车前往 Escanaba，速度恒定为每小时 $16$ 英里。他们骑在 Escanaba 和 Marquette 之间同一条 $62$ 英里的路线上。上午几点他们相遇？

Cassie leaves Escanaba at $8:30$ AM heading for Marquette on her bike. She bikes at a uniform rate of $12$ miles per hour. Brian leaves Marquette at $9:00$ AM heading for Escanaba on his bike. He bikes at a uniform rate of $16$ miles per hour. They both bike on the same $62$ -mile route between Escanaba and Marquette. At what time in the morning do they meet?

$10:00$

$10:15$

$10:30$

$11:00$

$11:30$

答案：D

知识点：路程、速度与时间

难度评级：1270

小提示：

先考虑 Cassie 的半小时提前出发。

First account for Cassie’s half-hour head start.

大提示：

Brian 出发后，两人的距离以速度之和缩小。

After Brian starts, the distance between them closes at the sum of their speeds.

解答：

Brian 开始骑车时，Cassie 已经骑了 $\dfrac{1}{2}\cdot 12=6$ 英里。因此两人相距 $62-6=56$ 英里。

他们合起来以每小时 $12+16=28$ 英里的速度缩短距离。因此他们在上午 $9{:}00$ 后 $56\div 28=2$ 小时相遇。

所以他们在上午 $11:00$ 相遇。

所以正确答案是 D。

By the time Brian starts biking, Cassie has already traveled $\dfrac{1}{2}\cdot 12=6$ miles. This means that Cassie and Brian are then $62-6=56$ miles apart.

Together, they close the distance at $12+16=28$ miles per hour. This means that they meet $56\div 28=2$ hours after $9{:}00$ AM.

This means that they meet at $11:00$ AM.

Thus, D is the correct answer.

14.

第 $14$ 、 $15$ 和 $16$ 题涉及 Reed 老师的英语作业。

小说阅读作业

Reed 老师英语课上的学生都在读同一本 $760$ 页的小说。Alice、Bob 和 Chandra 是班上的三个朋友。Alice 读一页需要 $20$ 秒，Bob 读一页需要 $45$ 秒，Chandra 读一页需要 $30$ 秒。

如果 Bob 和 Chandra 都读完整本书，Bob 会比 Chandra 多花多少秒阅读？

Problems $14,$ $15,$ and $16$ involve Mrs. Reed’s English assignment.

A Novel Assignment

The students in Mrs. Reed’s English class are reading the same $760$ -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in $20$ seconds, Bob reads a page in $45$ seconds and Chandra reads a page in $30$ seconds.

If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?

$7{,}600$

$11{,}400$

$12{,}500$

$15{,}200$

$22{,}800$

答案：B

知识点：速率分配律

难度评级：980

小提示：

按每页秒数比较 Bob 和 Chandra。

Compare Bob and Chandra by seconds per page.

大提示：

Bob 在 $760$ 页中的每一页都多花同样秒数。

Bob takes the same extra number of seconds on each of the $760$ pages.

解答：

Bob 读完整本书需要 $760 \cdot 45$ 秒，Chandra 需要 $760 \cdot 30$ 秒。

两人阅读时间相差 $760 \cdot 45 - 760 \cdot 30$ $= 760(45 - 30)$ $= 760 \cdot 15$ $= 11,400$ 秒。

所以正确答案是 B。

Bob will take $760 \cdot 45$ seconds to read the book, and Chandra will take $760 \cdot 30$ seconds.

The difference between the time they spent reading is $760 \cdot 45 - 760 \cdot 30$ $= 760(45 - 30)$ $= 760 \cdot 15$ $= 11,400$ seconds.

Thus, B is the correct answer.

15.

Chandra 和 Bob 各有一本书，他们决定用“团队阅读”来节省时间。按照这种方法，Chandra 从第 $1$ 页读到某一页，Bob 从下一页读到第 $760$ 页，读完整本书。读完后他们会互相讲述自己读的部分。为了让 Chandra 和 Bob 阅读时间相同，Chandra 应该读到最后哪一页？

Chandra and Bob, who each have a copy of the book, decide that they can save time by “team reading” the novel. In this scheme, Chandra will read from page $1$ to a certain page and Bob will read from the next page through page $760,$ finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?

$425$

$444$

$456$

$484$

$506$

答案：C

知识点：速率一次方程

难度评级：1380

小提示：

设 $x$ 为 Chandra 阅读的页数。

Let $x$ be the number of pages Chandra reads.

大提示：

令 Chandra 阅读 $x$ 页的时间等于 Bob 阅读剩余页数的时间。

Set Chandra’s time for $x$ pages equal to Bob’s time for the remaining pages.

解答：

设 Chandra 阅读 $x$ 页，那么 Bob 阅读 $760 - x$ 页。

为了让阅读时间相同， $\begin{align*} 30x &= 45(760 - x) \\ 30x &= 45 \cdot 760 - 45x \\ 75x &= 45 \cdot 760 \\ 5x &= 3 \cdot 760 \\ x &= 3 \cdot 152 \\ x &= 456 \end{align*}\text{。}$

所以正确答案是 C。

Let $x$ be the number of pages that Chandra will read. Then Bob will read $760 - x$ pages.

For them to read for the same amount of time, $\begin{align*} 30x &= 45(760 - x) \\ 30x &= 45 \cdot 760 - 45x \\ 75x &= 45 \cdot 760 \\ 5x &= 3 \cdot 760 \\ x &= 3 \cdot 152 \\ x &= 456. \end{align*}$

Thus, C is the correct answer.

16.

在 Chandra 和 Bob 开始阅读前，Alice 说她也想和他们一起团队阅读。如果他们把书分成三部分，使每个人阅读时间相同，那么每个人需要阅读多少秒？

Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?

$6400$

$6600$

$6800$

$7000$

$7200$

答案：E

知识点：速率比与比例

难度评级：1440

小提示：

比较三人在相同时间内各自读多少页。

Compare how many pages each person reads in the same amount of time.

大提示：

像 $180$ 秒这样的共同时间会让三人的页数都是整数。

A common time such as $180$ seconds makes all three page counts whole numbers.

解答：

如果 $3$ 人阅读相同时间，那么 Bob、Chandra 和 Alice 阅读页数之比为 $4:6:9$ 。

因此他们分别读 $160$ 、 $240$ 和 $360$ 页。因为三人阅读时间相同，只需计算 Bob 读他那部分所需时间。

Bob 读完这部分需要 $45 \cdot 160 = 7200$ 秒。

所以正确答案是 E。

If all $3$ individuals read the same amount of time, then the number of pages Bob, Chandra, and Alice will read will be in the ratio $4:6:9$ respectively.

This means that they will read $160,$ $240,$ and $360$ pages respectively. Since they all read for the same amount of time, we can just calculate how long it takes for Bob to read his portion.

Bob will take $45 \cdot 160 = 7200$ seconds to read his portion.

Thus, E is the correct answer.

17.

Jeff 旋转转盘 $P$ 、 $Q$ 和 $R$ ，并把得到的数字相加。他的和为奇数的概率是多少？

Jeff rotates spinners $P,$ $Q$ and $R$ and adds the resulting numbers. What is the probability that his sum is an odd number?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

答案：B

知识点：基本概率奇偶性

难度评级：1400

小提示：

只跟踪奇偶性，不需要具体和。

Track only odd and even values, not the exact sums.

大提示：

转盘 $Q$ 总是给出偶数，转盘 $R$ 总是给出奇数。

Spinner $Q$ always gives an even number, and spinner $R$ always gives an odd number.

解答：

转盘 $Q$ 总是落在偶数上，所以不改变总和的奇偶性。转盘 $R$ 总是落在奇数上。

为了让总和为奇数，转盘 $P$ 的数必须是偶数。转盘 $P$ 的三个等可能区域中只有一个是偶数。

因此概率为 $\dfrac{1}{3}$ 。

所以正确答案是 B。

Spinner $Q$ always lands on an even number, so it does not change the parity of the total. Spinner $R$ always lands on an odd number.

For the full sum to be odd, the number from spinner $P$ must be even. Only one of the three equal sectors of spinner $P$ contains an even number.

Therefore the probability is $\dfrac{1}{3}$ .

Thus, B is the correct answer.

18.

一个棱长为 $3$ 英寸的立方体由 $27$ 个棱长为 $1$ 英寸的小立方体组成。其中十九个小立方体是白色，八个是黑色。如果八个黑色小立方体放在大立方体的八个角上，那么大立方体表面积中白色部分占几分之几？

A cube with $3$ -inch edges is made using $27$ cubes with $1$ -inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?

$\dfrac{1}{9}$

$\dfrac{1}{4}$

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{19}{27}$

答案：D

知识点：正方体表面积分数

难度评级：1390

小提示：

看大立方体的一个面。

Look at one face of the large cube.

大提示：

每个面有四个来自黑色角立方体的单位正方形，还有五个其他单位正方形。

Each face has four corner unit squares from black corner cubes and five other unit squares.

解答：

因为每个面的黑白表面积比例相同，只需分析一个面的白色比例。

一个面上有 $9$ 个单位正方形，其中 $4$ 个是黑色。因此每个面有 $\dfrac{5}{9}$ 是白色。

所以正确答案是 D。

Since each face has the same black and white surface area, we can analyze what fraction of one side is white.

On one side, there are $9$ unit squares. $4$ of them are black. This means that $\dfrac{5}{9}$ of each face is white.

Thus, D is the correct answer.

19.

三角形 $ABC$ 是等腰三角形，且 $\overline{AB}=\overline{BC}$ 。点 $D$ 同时是 $\overline{BC}$ 和 $\overline{AE}$ 的中点，且 $\overline{CE}$ 长 $11$ 个单位。三角形 $ABD$ 与三角形 $ECD$ 全等。 $\overline{BD}$ 的长度是多少？

Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}.$ Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE},$ and $\overline{CE}$ is $11$ units long. Triangle $ABD$ is congruent to triangle $ECD.$ What is the length of $\overline{BD}?$

$4$

$4.5$

$5$

$5.5$

$6$

答案：D

知识点：全等（几何）等腰三角形中点

难度评级：1410

小提示：

用全等三角形关系连接 $EC$ 和 $AB$ 。

Use the congruent triangles to relate $EC$ and $AB$ .

大提示：

再用 $AB=BC$ 和 $D$ 是 $BC$ 中点。

Then use $AB=BC$ and the fact that $D$ is the midpoint of $BC$ .

解答：

由全等条件可知 $AB = EC = 11\text{。}$

由等腰条件， $BC = AB = 11\text{。}$

因为 $D$ 是 $\overline{BC}$ 的中点， $BD = BC \div 2 = 5.5\text{。}$

所以正确答案是 D。

By the congruency condition, we know that $AB = EC = 11.$

Also from the isosceles condition, we know that $BC = AB = 11.$

Since $D$ is the midpoint of $\overline{BC},$ we know that $BD = BC \div 2 = 5.5.$

Thus, D is the correct answer.

20.

一场单打锦标赛有六名选手。每名选手与其他每名选手只比赛一次，没有平局。如果 Helen 赢了 $4$ 场，Ines 赢了 $3$ 场，Janet 赢了 $2$ 场，Kendra 赢了 $2$ 场，Lara 赢了 $2$ 场，那么 Monica 赢了几场？

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won $4$ games, Ines won $3$ games, Janet won $2$ games, Kendra won $2$ games and Lara won $2$ games, how many games did Monica win?

$0$

$1$

$2$

$3$

$4$

答案：C

知识点：组合

难度评级：1260

小提示：

数出六人循环赛共有多少场比赛。

Count how many games are played in a six-player round robin.

大提示：

每场比赛恰好有一个胜者，所以总胜场数等于总比赛数。

Every game has exactly one winner, so total wins equal total games.

解答：

每场比赛恰好有一个胜者。六名选手共有 $6 \cdot 5 \div 2 = 15$ 场比赛，因此共有 $15$ 个胜场。

已经计入的胜场数为 $4 + 3 + 2 + 2 + 2 = 13$ 场，所以 Monica 赢了 $15 - 13 = 2$ 场。

所以正确答案是 C。

In every match, there was exactly one winner. There are $6 \cdot 5 \div 2 = 15$ games and therefore $15$ wins.

There are already $4 + 3 + 2 + 2 + 2 = 13$ wins accounted for, so Monica won $15 - 13 = 2$ games.

Thus, C is the correct answer.

21.

一个水族箱的长方形底面为 $100$ cm 乘 $40$ cm，高 $50$ cm。水族箱中水深 $37$ cm。随后放入一个体积为 $1000\text{ cm}^3$ 的石头并完全浸没。水位上升多少厘米？

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a depth of $37$ cm. A rock with volume $1000\text{ cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?

$0.25$

$0.5$

$1$

$1.25$

$2.5$

答案：A

知识点：体积长方体

难度评级：1310

小提示：

完全浸没的物体会排开等于自身体积的水。

A submerged object raises the water by its own volume.

大提示：

用石头体积除以水族箱底面积。

Divide the rock’s volume by the base area of the aquarium.

解答：

水族箱底面积为 $100\cdot 40=4000\text{ cm}^2$ 。

完全浸没的石头排开 $1000\text{ cm}^3$ 的水，所以水位上升 $1000\div 4000=0.25$ cm。

所以正确答案是 A。

The base area of the aquarium is $100\cdot 40=4000\text{ cm}^2$ .

The submerged rock displaces $1000\text{ cm}^3$ of water, so the water level rises by $1000\div 4000=0.25$ cm.

Thus, A is the correct answer.

22.

三个不同的一位正整数放在底行格子中。相邻格子的数相加，和放在它们上方的格子中。在第二行继续同样过程，得到顶格中的数。顶格中可能的最大数与最小数之差是多少？

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

$16$

$24$

$25$

$26$

$35$

答案：D

知识点：最优化

难度评级：1610

小提示：

将底行三个数写成 $a$ 、 $b$ 和 $c$ 。

Write the three bottom entries as $a,$ $b,$ and $c$ .

大提示：

底行中间的数在顶格中被计算两次。

The middle bottom entry is counted twice in the top number.

解答：

如果底行格子中的数依次为 $a$ 、 $b$ 和 $c$ ，那么中间一行就是 $a+b$ 和 $b+c$ 。

因此顶格为 $a+2b+c$ 。要使它最小，把 $b=1$ 放在中间，外侧放 $2$ 和 $3$ ，得到顶格数 $7$ 。

要使它最大，把 $b=9$ 放在中间，外侧放 $7$ 和 $8$ ，得到顶格数 $33$ 。所求差为 $33-7=26$ 。

所以正确答案是 D。

If the lower cells contain $a,$ $b,$ and $c,$ the middle row will have $a+b$ and $b+c$ .

This means that the top row will have $a+2b+c$ . To minimize this, put $b=1$ in the middle and $2$ and $3$ in the outer cells. This yields a top number of $7$ .

To maximize it, put $b=9$ in the middle and $7$ and $8$ in the outer cells. This yields a top number of $33$ . The desired difference is $33-7=26$ .

Thus, D is the correct answer.

23.

一个盒子里有金币。如果把金币平均分给六个人，会剩下四枚。如果平均分给五个人，会剩下三枚。如果盒子中装的是满足这两个条件的最少金币数，那么平均分给七个人时会剩下几枚？

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

$0$

$1$

$2$

$3$

$5$

答案：A

知识点：模运算系统列举

难度评级：1450

小提示：

列出除以 $6$ 余 $4$ 的较小正整数。

List small numbers that leave remainder $4$ when divided by $6$ .

大提示：

等价地，加两枚金币后应同时能被 $5$ 和 $6$ 整除。

Equivalently, adding two coins would make the number divisible by both $5$ and $6$ .

解答：

除以 $6$ 余 $4$ 的正整数为 $4, 10, 16, 22, 28, 34, 40, \cdots\text{。}$ 除以 $5$ 余 $3$ 的正整数为 $3, 8, 13, 18, 23, 28, 33, \cdots\text{。}$

可见满足条件的最少金币数是 $28$ 。它除以 $7$ 余 $0$ 。

所以正确答案是 A。

The positive integers that leave a remainder of $4$ when divided by $6$ are $4, 10, 16, 22, 28, 34, 40, \cdots.$ The positive integers that leave a remainder of $3$ when divided by $5$ are $3, 8, 13, 18, 23, 28, 33, \cdots.$

From this, we can see that the smallest number of coins that work is $28.$ This leaves a remainder of $0$ when divided by $7.$

Thus, A is the correct answer.

24.

在下面的乘法题中， $A$ 、 $B$ 、 $C$ 、 $D$ 是不同的数字。 $A+B$ 是多少？ $\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}$

In the multiplication problem below $A,$ $B,$ $C,$ $D$ are different digits. What is $A+B?$ $\begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}$

$1$

$2$

$3$

$4$

$9$

答案：A

知识点：数字谜位值

难度评级：1610

小提示：

将 $CDCD$ 写成 $100\cdot CD+CD$ 。

Write $CDCD$ as $100\cdot CD+CD$ .

大提示：

因为 $CD$ 非零，可将方程除以两位数 $CD$ 。

Since $CD$ is nonzero, divide the equation by the two-digit number $CD$ .

解答：

注意 $CDCD = 101 \cdot CD\text{。}$ 这迫使 $ABA = 101$ 。所以 $A = 1$ ， $B = 0$ ，且 $A + B = 1$ 。

所以正确答案是 A。

Note that $CDCD = 101 \cdot CD.$ This forces $ABA = 101.$ Then $A = 1,$ $B = 0,$ and $A + B = 1.$

Thus, A is the correct answer.

25.

Barry 写了 $6$ 个不同的数，分别写在 $3$ 张卡片的两面，并把卡片如图摆在桌上。每张卡片两面数字之和都相等。三个隐藏面上的数都是质数。隐藏的质数的平均数是多少？

Barry wrote $6$ different numbers, one on each side of $3$ cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?

$13$

$14$

$15$

$16$

$17$

答案：B

知识点：质数奇偶性平均数

难度评级：1650

小提示：

每张卡片的公共和对三个可见数必须有相同奇偶性。

The common card sum must have the same parity for all three visible numbers.

大提示：

使用 $2$ 是唯一偶质数这一事实。

Use the fact that $2$ is the only even prime.

解答：

公共和必须是奇数。如果公共和是偶数，那么 $44$ 和 $38$ 背后的隐藏数都必须是偶质数，但偶质数只有一个。

所以奇数 $59$ 背后的隐藏质数必须是 $2$ ，公共和为 $59+2=61$ 。

另外两个隐藏质数是 $61-44=17$ 和 $61-38=23$ 。

隐藏质数的平均数为 $\dfrac{2+17+23}{3}=14$ 。

所以正确答案是 B。

The common sum must be odd. If the common sum were even, then the hidden numbers behind $44$ and $38$ would both have to be even primes, but there is only one even prime.

So the prime hidden behind the odd visible number $59$ must be $2$ , making the common sum $59+2=61$ .

The other two hidden primes are $61-44=17$ and $61-38=23$ .

The average of the hidden primes is $\dfrac{2+17+23}{3}=14$ .

Thus, B is the correct answer.