2003 AMC 8 真题与在线限时测验

计时

40:00

1.

Jamie 数了一个立方体的棱数，Jimmy 数了顶点数，Judy 数了面数。然后他们把这三个数相加。得到的和是多少？

Jamie counted the number of edges of a cube, Jimmy counted the number of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?

$12$

$16$

$20$

$22$

$26$

答案：E

知识点：正方体

难度评级：370

小提示：

立方体的棱、顶点和面都有固定数量

A cube has a fixed number of edges, corners, and faces.

大提示：

把棱数、顶点数和面数相加

Add the counts for edges, corners, and faces.

解答：

一个立方体有 $12$ 条棱、 $8$ 个顶点和 $6$ 个面。相加得到 $12 + 8 + 6 = 26\text{。}$

所以正确答案是 E。

A cube has $12$ edges, $8$ corners, and $6$ faces. Adding these together yields $12 + 8 + 6 = 26.$

Thus, E is the correct answer.

2.

下列哪个数的最小质因数最小？

Which of the following numbers has the smallest prime factor?

$55$

$57$

$58$

$59$

$61$

答案：C

知识点：质数奇偶性

难度评级：450

小提示：

寻找质因数为 $2$ 的数

Look for a number with prime factor $2.$

大提示：

在选项中，只有一个数是偶数

Among the choices, only one is even.

解答：

最小的质数是 $2$ 。因此任何偶数都会是答案。

所以正确答案是 C。

Note that the smallest prime number is $2.$ This means that any even number would be our answer.

Thus, C is the correct answer.

3.

Ricky C’s 的一个汉堡重 $120$ 克，其中 $30$ 克是填充物。汉堡中不是填充物的部分占百分之多少？

A burger at Ricky C’s weighs $120$ grams, of which $30$ grams are filler. What percent of the burger is not filler?

$60 \%$

$65 \%$

$70 \%$

$75 \%$

$90 \%$

答案：D

知识点：百分数

难度评级：560

小提示：

先求有多少克不是填充物

First find how many grams are not filler.

大提示：

将非填充物的克数与 $120$ 比较

Compare the non-filler grams to $120.$

解答：

不是填充物的部分为 $120 - 30 = 90$ 克。因此百分比为 $100 \cdot \dfrac{90}{120} = 100 \cdot \dfrac{3}{4} = 75 \%\text{。}$

所以正确答案是 D。

We get that $120 - 30 = 90$ grams are not filler. The percentage is therefore $100 \cdot \dfrac{90}{120} = 100 \cdot \dfrac{3}{4} = 75 \%.$

Thus, D is the correct answer.

4.

一群骑自行车和三轮车的孩子经过 Billy Bob 家。Billy Bob 数到 $7$ 个孩子和 $19$ 个轮子。有多少辆三轮车？

A group of children riding on bicycles and tricycles rode past Billy Bob’s house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?

$2$

$4$

$5$

$6$

$7$

答案：C

知识点：方程组

难度评级：720

小提示：

如果 $7$ 个孩子都骑自行车，先数轮子

If all $7$ children had bicycles, count the wheels.

大提示：

每辆三轮车比自行车多一个轮子

Each tricycle adds one wheel compared with a bicycle.

解答：

设 $b$ 为自行车数量， $t$ 为三轮车数量。可列方程组： $\begin{gather*} b + t = 7, \\ 2b + 3t = 19\text{。} \end{gather*}$ 将第一个方程乘以 $2$ ，再从第二个方程中减去所得等式，得到 $t = 5$ 。

所以正确答案是 C。

Let $b$ be the number of bicycles and $t$ be the number of tricycles. Then we can set up the following system of equations: $\begin{gather*} b + t = 7, \\ 2b + 3t = 19. \end{gather*}$ Multiplying the first equation by $2$ and subtracting from the second equation, we get $t = 5.$

Thus, C is the correct answer.

5.

如果一个数的 $20 \%$ 是 $12$ ，那么同一个数的 $30\%$ 是多少？

If $20 \%$ of a number is $12,$ what is $30\%$ of the same number?

$15$

$18$

$20$

$24$

$30$

答案：B

知识点：百分数

难度评级：730

小提示：

$20\%$ 是五分之一

$20\%$ is one fifth.

大提示：

用 $10\%$ 作为从 $20\%$ 到 $30\%$ 的中间量

Use $10\%$ as a stepping stone from $20\%$ to $30\%.$

解答：

因为这个数的 $20\%$ 是 $12$ ，所以这个数的 $10\%$ 是 $6$ 。

因此同一个数的 $30\%$ 是 $3\cdot6=18$ 。

所以正确答案是 B。

Since $20\%$ of the number is $12,$ $10\%$ of the number is $6.$

Therefore $30\%$ of the same number is $3\cdot6=18.$

Thus, B is the correct answer.

6.

根据图中三个正方形的面积，内部三角形的面积是多少？

Given the areas of the three squares in the figure, what is the area of the interior triangle?

$13$

$30$

$60$

$300$

$1800$

答案：B

知识点：勾股数三角形面积

难度评级：920

小提示：

正方形面积给出了三个正方形的边长

The square areas give the side lengths of the three squares.

大提示：

这些边长形成 $5-12-13$ 直角三角形

The side lengths form a $5-12-13$ right triangle.

解答：

三个正方形的边长分别为 $\sqrt{169} = 13\text{、}$ $\sqrt{144} = 12$ 和 $\sqrt{25} = 5\text{。}$ 这些长度构成一个勾股数组。

因此内部三角形是直角三角形，面积为 $\dfrac{1}{2} \cdot 5 \cdot 12 = 30\text{。}$

所以正确答案是 B。

The side lengths of the squares are $\sqrt{169} = 13,$ $\sqrt{144} = 12,$ and $\sqrt{25} = 5.$ These lengths form a Pythagorean triple.

Therefore, the interior triangle is right. Its area is $\dfrac{1}{2} \cdot 5 \cdot 12 = 30.$

Thus, B is the correct answer.

7.

Blake 和 Jenny 各参加了四次 $100$ 分测试。Blake 四次测试平均分为 $78$ 。Jenny 第一次测试比 Blake 高 $10$ 分，第二次比他低 $10$ 分，第三次和第四次都比他高 $20$ 分。Jenny 的平均分和 Blake 的平均分相差多少？

Blake and Jenny each took four $100$ -point tests. Blake averaged $78$ on the four tests. Jenny scored $10$ points higher than Blake on the first test, $10$ points lower than him on the second test, and $20$ points higher on both the third and fourth tests. What is the difference between Jenny’s average and Blake’s average on these four tests?

$10$

$15$

$20$

$25$

$40$

答案：A

知识点：平均数

难度评级：900

小提示：

把 Jenny 四次分数相对 Blake 的差值相加

Add Jenny’s four score differences from Blake’s.

大提示：

将总差值除以 $4$ 来比较平均分

Divide the total difference by $4$ to compare averages.

解答：

两人四次测试的总分差为 $10 - 10 + 20 \cdot 2 = 40\text{，}$ 所以平均分之差为 $40 \div 4 = 10$ 。

所以正确答案是 A。

The total point difference between the two is $10 - 10 + 20 \cdot 2 = 40.$ The average of this difference is $40 \div 4 = 10.$

Thus, A is the correct answer.

8.

第 $8$ ， $9$ 和 $10$ 题使用随附段落和图形中的数据。

烘焙义卖

四个朋友 Art、Roger、Paul 和 Trisha 烤饼干，所有饼干厚度相同。饼干的形状如下所示。

• Art 的饼干是梯形：

• Roger 的饼干是长方形：

• Paul 的饼干是平行四边形：

• Trisha 的饼干是三角形：

每个朋友使用相同数量的面团，且 Art 正好做了 $12$ 块饼干。

谁用一批饼干面团做出的饼干数量最少？

Problems $8,$ $9,$ and $10$ use the data found in the accompanying paragraph and figures.

Bake Sale

Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.

• Art’s cookies are trapezoids:

• Roger’s cookies are rectangles:

• Paul’s cookies are parallelograms:

• Trisha’s cookies are triangles:

Each friend uses the same amount of dough, and Art makes exactly $12$ cookies.

Who gets the fewest cookies from one batch of cookie dough?

Art

Paul

Roger

Trisha

最少数量出现并列

There is a tie for fewest.

答案：A

知识点：面积最优化

难度评级：1000

小提示：

面团量和厚度相同，饼干面积越大，数量越少

Same dough and thickness means larger cookie area gives fewer cookies.

大提示：

比较四种饼干形状的面积

Compare the areas of the four cookie shapes.

解答：

因为所有饼干厚度相同且使用相同数量的面团，所以面积最大的饼干形状会产生最少的饼干。

Art 的梯形面积为 $\frac12(3+5)\cdot3=12$ 。Roger 的长方形面积为 $4\cdot2=8$ 。Paul 的平行四边形面积为 $3\cdot2=6$ ，Trisha 的三角形面积为 $\frac12\cdot3\cdot4=6$ 。

Art 的饼干面积最大，所以 Art 用一批面团得到的饼干最少。

所以正确答案是 A。

Since the cookies all have the same thickness and use the same amount of dough, the largest cookie shape produces the fewest cookies.

Art’s trapezoid has area $\frac12(3+5)\cdot3=12.$ Roger’s rectangle has area $4\cdot2=8.$ Paul’s parallelogram has area $3\cdot2=6,$ and Trisha’s triangle has area $\frac12\cdot3\cdot4=6.$

Art has the largest cookie area, so Art gets the fewest cookies from one batch.

Thus, A is the correct answer.

9.

Art 的饼干每块卖 $60$ ¢。为了从一批面团中赚到同样金额，Roger 的一块饼干应卖多少钱？

Art’s cookies sell for $60$ ¢ each. To earn the same amount from a single batch, how much should one of Roger’s cookies cost?

$18$ ¢

$25$ ¢

$40$ ¢

$75$ ¢

$90$ ¢

答案：C

知识点：面积比与比例

难度评级：1060

小提示：

先求 Art 一批饼干收入多少

First find the total money Art earns from a batch.

大提示：

使用同一批面团能做出的 Roger 饼干数量

Use the number of Roger’s cookies from the shared dough amount.

解答：

Art 做 $12$ 块饼干，每块卖 $60$ 分，所以一批收入 $12\cdot60=720$ 分。

这批面团面积为 $12\cdot12=144$ 平方英寸，每块 Roger 饼干面积为 $4\cdot2=8$ 。因此 Roger 能做 $144\div8=18$ 块饼干。

一批需要赚到 $720$ 分，而 Roger 有 $18$ 块饼干，因此每块应卖 $720\div18=40$ 分。

所以正确答案是 C。

Art makes $12$ cookies that sell for $60$ cents each, so one batch earns $12\cdot60=720$ cents.

The batch has $12\cdot12=144$ square inches of dough area, and each Roger cookie has area $4\cdot2=8.$ Thus Roger makes $144\div8=18$ cookies.

To earn $720$ cents from $18$ cookies, each Roger cookie should cost $720\div18=40$ cents.

Thus, C is the correct answer.

10.

一批 Trisha 的饼干会有多少块？

How many cookies will be in one batch of Trisha’s cookies?

$10$

$12$

$16$

$18$

$24$

答案：E

知识点：面积比与比例

难度评级：1030

小提示：

Trisha 的饼干面积是 Art 饼干面积的一半

Trisha’s cookie area is half of Art’s cookie area.

大提示：

面团量相同，面积减半的饼干数量会翻倍

Same dough amount means half-size cookies make twice as many.

解答：

Art 的一块饼干面积为 $12$ 平方英寸，所以整批面团面积为 $12\cdot12=144$ 平方英寸。

Trisha 的三角形饼干面积为 $\frac12\cdot3\cdot4=6$ 平方英寸。

因此 Trisha 每批可以做 $144\div6=24$ 块饼干。

所以正确答案是 E。

Art’s cookie area is $12$ square inches, so the whole batch has area $12\cdot12=144$ square inches.

Trisha’s triangular cookie has area $\frac12\cdot3\cdot4=6$ square inches.

Therefore Trisha can make $144\div6=24$ cookies per batch.

Thus, E is the correct answer.

11.

Lou’s Fine Shoes 生意有点慢，于是 Lou 决定促销。星期五，Lou 将星期四所有价格提高 $10 \%$ 。周末，Lou 做广告：“标价九折，促销星期一开始。”星期四售价 $\$40$ 的一双鞋，星期一卖多少钱？

Business is a little slow at Lou’s Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday’s prices by $10 \%.$ Over the weekend, Lou advertises the sale: “Ten percent off the listed price. Sale starts Monday.” How much does a pair of shoes cost on Monday that cost $\$40$ on Thursday?

$\$36$

$\$39.60$

$\$40$

$\$40.40$

$\$44$

答案：B

知识点：百分数

难度评级：1060

小提示：

星期一的折扣是从星期五提高后的价格上打折

The Monday discount is taken from Friday’s increased price.

大提示：

将原价先乘以 $1.10$ ，再乘以 $0.90$

Multiply the original price by $1.10,$ then by $0.90.$

解答：

星期五鞋价提高 $10\%$ ，所以标价变为 $40\cdot1.10=44$ 美元。

星期一再打 $10\%$ 的折扣，从 $\$44$ 降到 $44\cdot0.90=39.60$ 美元。

所以正确答案是 B。

On Friday, the shoes are marked up by $10\%,$ so the listed price becomes $40\cdot1.10=44$ dollars.

On Monday, the $10\%$ discount is taken from $\$44,$ giving $44\cdot0.90=39.60$ dollars.

Thus, B is the correct answer.

12.

当一个公平的六面骰子被掷到桌面上时，底面看不见。可见的五个面上的数字乘积能被 $6$ 整除的概率是多少？

When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by $6?$

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{5}{6}$

$1$

答案：E

知识点：基本概率整除性

难度评级：1120

小提示：

检查唯一可能看不见数字 $6$ 的情况

Check the only case where the face numbered $6$ is hidden.

大提示：

如果 $6$ 在底面，可见面仍包含 $2$ 和 $3$

If $6$ is hidden, the visible faces still include $2$ and $3.$

解答：

如果数字 $6$ 的面可见，那么可见数字的乘积能被 $6$ 整除。

如果数字 $6$ 在底面，那么可见面仍包含 $2$ 和 $3$ ，所以它们的乘积仍能被 $6$ 整除。

每种可能的掷骰结果都满足条件，所以概率为 $1$ 。

所以正确答案是 E。

If the face numbered $6$ is visible, then the visible product is divisible by $6.$

If the face numbered $6$ is on the bottom, then the visible faces include both $2$ and $3,$ so their product is still divisible by $6.$

Every possible toss works, so the probability is $1.$

Thus, E is the correct answer.

13.

十四个白色立方体拼成右图。这个立体的整个表面，包括底面，都被涂成红色。然后将这个立体拆成单个立方体。有多少个单个立方体恰好有四个面是红色的？

Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?

$4$

$6$

$8$

$10$

$12$

答案：B

知识点：立体几何表面积

难度评级：1240

小提示：

一个小立方体恰好有四个涂色面，当且仅当它接触两个其他立方体

A cube has four painted faces exactly when it touches two other cubes.

大提示：

分别考虑顶部立方体、底部角落立方体和剩余立方体

Separate the top cubes, bottom corner cubes, and remaining cubes.

解答：

一个小立方体恰好有四个面被涂色，当且仅当它恰好与两个其他立方体相连。

$4$ 个顶部立方体只接触一个其他立方体，所以它们有 $5$ 个涂色面。 $4$ 个底部角落立方体接触三个其他立方体，所以它们有 $3$ 个涂色面。

剩下的 $14-4-4=6$ 个立方体各恰好接触两个其他立方体，所以它们恰好有四个涂色面。

所以正确答案是 B。

A cube has exactly four painted faces exactly when it is attached to exactly two other cubes.

The $4$ top cubes touch only one other cube, so they have $5$ painted faces. The $4$ bottom corner cubes touch three other cubes, so they have $3$ painted faces.

The remaining $14-4-4=6$ cubes each touch exactly two other cubes, so they have exactly four painted faces.

Thus, B is the correct answer.

14.

在这个加法题中，每个字母代表一个不同数字。 $\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$ 如果 $T = 7$ ，且字母 $O$ 表示一个偶数，那么 $W$ 唯一可能的值是多少？

In this addition problem, each letter stands for a different digit. $\begin{array}{cccc}&T & W & O\\ +&T & W & O\\ \hline F& O & U & R\end{array}$ If $T = 7$ and the letter $O$ represents an even number, what is the only possible value for $W?$

$0$

$1$

$2$

$3$

$4$

答案：D

知识点：数字谜分类讨论

难度评级：1380

小提示：

由 $T = 7$ 考察百位相加

Use the hundreds column after $T = 7.$

大提示：

个位和 $O$ 是偶数会确定 $O$

The ones column and the fact that $O$ is even force $O.$

解答：

因为两个 $T$ 都是 $7$ ，所以 $O$ 是 $4$ 或 $5$ 。由于 $O$ 是偶数，得到 $O = 4$ 。

于是 $R = 4 + 4 = 8$ 。还知道 $W + W$ 不会进位，否则 $O$ 会是 $5$ 。

因此 $W$ 小于 $5$ ，且不能是 $4$ 或 $1$ 。如果 $W = 0$ ，则 $U = 0$ ，两个字母会代表同一个数字。如果 $W = 2$ ，则 $U = 4$ ，这也不允许。

所以 $W = 3$ 。

所以正确答案是 D。

Since both $T$ ’s are $7,$ we get that $O$ is either $4$ or $5.$ Since $O$ is even, we get that $O = 4.$

Then, we get that $R = 4 + 4 = 8.$ We also know that $W + W$ doesn’t carry over, since otherwise $O$ would be $5.$

Therefore, $W$ is less than $5$ and cannot be $4$ or $1.$ If $W = 0,$ then $U = 0,$ which gives two letters the same digit. If $W = 2,$ then $U = 4,$ which is also not allowed.

This makes $W = 3.$

Thus, D is the correct answer.

15.

一个图形由单位立方体构成。每个立方体至少与另一个立方体共享一个面。若要得到图中所示的正视图和侧视图，最少需要多少个立方体？

A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?

$3$

$4$

$5$

$6$

$7$

答案：B

知识点：立体几何最优化

难度评级：1330

小提示：

尝试只用三个立方体同时满足两个视图

Try to satisfy both views with only three cubes.

大提示：

图中给出了四个立方体的构造；只需排除三个的情况

A four-cube construction is shown; you only need to rule out three.

解答：

正视图需要一个有三个可见位置的 L 形，所以至少需要三个立方体。

假设恰好有三个立方体。正视图中上下相叠的两个立方体必须共享一个面，因此它们处于同一深度。第三个立方体在正视图中位于下方立方体的旁边。若要和前两个立方体中的一个共享面，它也必须和下方立方体处于同一深度；否则它与这两个立方体都不相连。因此三个立方体都在同一个深度列中，侧视图便只有一列，而不是题目要求的 L 形。

图中的四立方体构造具有两个所需视图，所以最小值是 $4$ 。

所以正确答案是 B。

The front view requires an L-shape with three visible positions, so at least three cubes are needed.

Suppose there were exactly three cubes. The two cubes that appear one above the other must share a face, so they have the same depth. The third cube appears beside the lower one. To share a face with either of the first two cubes, it must share the lower cube’s depth as well; otherwise it would be disconnected from both. Thus all three cubes would occupy one depth column, giving a one-column side view instead of the required L-shape.

The shown four-cube construction has both required views, so the minimum is $4.$

Thus, B is the correct answer.

16.

Ali、Bonnie、Carlo 和 Dianna 将一起开车去附近的主题公园。他们使用的车有四个座位：一个驾驶座、一个前排乘客座和两个后排座位。只有 Bonnie 和 Carlo 会开这辆车。有多少种可能的座位安排？

Ali, Bonnie, Carlo and Dianna are going to drive together to a nearby theme park. The car they are using has four seats: one driver’s seat, one front passenger seat and two back seats. Bonnie and Carlo are the only two who can drive the car. How many possible seating arrangements are there?

$2$

$4$

$6$

$12$

$24$

答案：D

知识点：有限制的排列乘法原理

难度评级：1230

小提示：

先选择驾驶员

Choose the driver first.

大提示：

选定驾驶员后，将另外三个人安排到剩余座位

After the driver is chosen, arrange the other three people in the remaining seats.

解答：

驾驶座有 $2$ 种选择。另一个前排座位有 $3$ 种选择，第一个后排座位有 $2$ 种选择。

最后一个人必须坐最后一个座位，所以共有 $2 \cdot 3 \cdot 2 = 12$ 种座位安排。

所以正确答案是 D。

There are $2$ options for who sits in the driver’s seat. There are $3$ options for the other front seat, and $2$ options for the first back seat.

The last person has to sit in the last seat, for a total of $2 \cdot 3 \cdot 2 = 12$ possible seating arrangements.

Thus, D is the correct answer.

17.

下面列出的六个孩子来自两个家庭，每个家庭有三个兄弟姐妹。每个孩子有蓝色或棕色眼睛，以及黑色或金色头发。同一个家庭的孩子至少有其中一个特征相同。哪两个孩子是 Jim 的兄弟姐妹？

姓名	眼睛颜色	头发颜色
Benjamin	蓝色	黑色
Jim	棕色	金色
Nadeen	棕色	黑色
Austin	蓝色	金色
Tevyn	蓝色	黑色
Sue	蓝色	金色

The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim’s siblings?

Child	Eye Color	Hair Color
Benjamin	Blue	Black
Jim	Brown	Blond
Nadeen	Brown	Black
Austin	Blue	Blond
Tevyn	Blue	Black
Sue	Blue	Blond

Nadeen 和 Austin

Nadeen and Austin

Benjamin 和 Sue

Benjamin and Sue

Benjamin 和 Austin

Benjamin and Austin

Nadeen 和 Tevyn

Nadeen and Tevyn

Austin 和 Sue

Austin and Sue

答案：E

知识点：逻辑推理

难度评级：1380

小提示：

Jim 的兄弟姐妹必须各自与 Jim 共享眼睛颜色或头发颜色

Jim’s siblings must each share eye color or hair color with Jim.

大提示：

这两个兄弟姐妹彼此之间也必须至少共享一个特征

The two siblings must also share at least one characteristic with each other.

解答：

Nadeen、Austin 和 Sue 是唯一与 Jim 共享某个特征的 $3$ 个人。需要找出其中哪一个与其他人完全不同。

Austin 和 Sue 都有蓝色眼睛，所以 Nadeen 是不属于这一组的人。因此 Austin 和 Sue 是 Jim 的兄弟姐妹。

所以正确答案是 E。

Note that Nadeen, Austin, and Sue are the only individuals who share a characteristic with Jim. We need to find which of the $3$ are completely different from the others.

Austin and Sue both have blue eyes, which makes Nadeen the odd one out. Therefore, Austin and Sue are Jim’s siblings.

Thus, E is the correct answer.

18.

下图中的二十个点各代表 Sarah 的一位同学。互为朋友的同学之间用线段连接。Sarah 生日聚会只邀请以下同学：她所有的朋友，以及至少和她某个朋友是朋友的同学。有多少位同学不会被邀请参加 Sarah 的聚会？

Each of the twenty dots on the graph below represents one of Sarah’s classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah’s party?

$1$

$4$

$5$

$6$

$7$

答案：D

知识点：图论补集计数

难度评级：1400

小提示：

Sarah 邀请距离她为 $1$ 或 $2$ 条线段的顶点

Sarah invites vertices at distance $1$ or $2$ from her.

大提示：

数出与 Sarah 的距离超过两条线段的点

Count the dots not within two line segments of Sarah.

解答：

Sarah 邀请她的朋友，以及至少和她某个朋友是朋友的同学。用图论语言说，她邀请距离 Sarah 为 $1$ 或 $2$ 条线段的点。

从图中可见，有 $4$ 个点与 Sarah 所在连通分量断开，另外还有 $2$ 个点在 Sarah 所在连通分量中但距离为 $3$ 条线段。

因此不会被邀请的同学有 $4+2=6$ 位。

所以正确答案是 D。

Sarah invites her friends and the classmates who are friends with at least one of her friends. In graph terms, she invites dots that are $1$ or $2$ line segments away from Sarah.

From the graph, $4$ dots are disconnected from Sarah’s component, and $2$ more dots in Sarah’s component are $3$ segments away.

Those $4+2=6$ classmates will not be invited.

Thus, D is the correct answer.

19.

$1000$ 和 $2000$ 之间有多少个整数同时以 $15$ ， $20$ 和 $25$ 为因数？

How many integers between $1000$ and $2000$ have all three of the numbers $15,$ $20,$ and $25$ as factors?

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：最小公倍数区间内整数计数

难度评级：1430

小提示：

使用 $15$ 、 $20$ 、 $25$ 的最小公倍数

Use the least common multiple of $15,$ $20,$ and $25.$

大提示：

数出 $1000$ 和 $2000$ 之间该最小公倍数的倍数

Count the multiples of that LCM between $1000$ and $2000.$

解答：

如果一个数 $x$ 以这三个数为因数，那么它们的最小公倍数也必须整除 $x$ 。

这些数的质因数分解为 $15 = 3 \cdot 5\text{，}$ $20 = 2^2 \cdot 5\text{，}$ $25 = 5^2\text{。}$

因此最小公倍数为 $2^2 \cdot 3 \cdot 5^2 = 300\text{。}$

$300$ 的倍数中，位于 $1000$ 和 $2000$ 之间的是 $1200$ 、 $1500$ 和 $1800$ 。

所以正确答案是 C。

If a number $x$ has these three numbers as factors, then their least common multiple must also divide $x.$

These numbers have the following prime factorizations: $15 = 3 \cdot 5,$ $20 = 2^2 \cdot 5,$ $25 = 5^2.$

From these values, we get that the least common multiple is $2^2 \cdot 3 \cdot 5^2 = 300.$

Therefore, the multiples of $300$ between $1000$ and $2000$ are $1200,$ $1500,$ and $1800.$

Thus, C is the correct answer.

20.

凌晨 $4{:}20$ 时，钟表两根指针形成的锐角是多少度？

What is the measure of the acute angle formed by the hands of a clock at $4{:}20$ a.m.?

$0^{\circ}$

$5^{\circ}$

$8^{\circ}$

$10^{\circ}$

$12^{\circ}$

答案：D

知识点：时钟

难度评级：1450

小提示：

$4{:}20$ 时，分针指向 $4$

At $4{:}20,$ the minute hand points at $4.$

大提示：

时针已经从 $4$ 到 $5$ 移动了三分之一

The hour hand has moved one third of the way from $4$ to $5.$

解答：

在 $4{:}20$ ，时针走过了相邻小时刻度间距的 $\frac{1}{3}$ ，因而位于 $4$ 和 $5$ 之间。

每个小时刻度代表 $360 \div 12 = 30^{\circ}$ 。所以时针又移动了 $30 \div 3 = 10^{\circ}$ ，超过 $4$ 的位置。

$20$ 分钟时，分针指向 $4$ ，所以两根指针形成的锐角为 $10^{\circ}$ 。

所以正确答案是 D。

At $4{:}20,$ the hour hand will be $\frac{1}{3}$ of the way between $4$ and $5.$

Each hour represents $360 \div 12 = 30^{\circ}.$ This means the hour hand will be $30 \div 3 = 10^{\circ}$ past $4.$

At $20$ minutes, the minute hand points to $4,$ so the acute angle between the hands is $10^{\circ}.$

Thus, D is the correct answer.

21.

梯形 $ABCD$ 的面积是 $164\text{ cm}^2$ 。高为 $8$ cm， $AB$ 为 $10$ cm， $CD$ 为 $17$ cm。 $BC$ 等于多少厘米？

The area of trapezoid $ABCD$ is $164\text{ cm}^2.$ The altitude is $8$ cm, $AB$ is $10$ cm, and $CD$ is $17$ cm. What is $BC,$ in centimeters?

$9$

$10$

$12$

$15$

$20$

答案：B

知识点：梯形勾股定理面积分割

难度评级：1590

小提示：

从 $B$ 和 $C$ 向底边作垂线

Drop perpendiculars from $B$ and $C$ to the base.

大提示：

使用 $6-8-10$ 和 $8-15-17$ 直角三角形

Use the $6-8-10$ and $8-15-17$ right triangles.

解答：

从 $B$ 和 $C$ 向 $\overline{AD}$ 作垂线，垂足分别为 $E$ 和 $F$ 。

在左右两个直角三角形中， $AE=\sqrt{10^2-8^2}=6$ ， $FD=\sqrt{17^2-8^2}=15$ 。

两侧三角形面积分别为 $\frac12\cdot6\cdot8=24$ 和 $\frac12\cdot15\cdot8=60$ 。中间长方形面积为 $8\cdot BC$ 。

因此 $164=24+60+8BC$ ，所以 $8BC=80$ ， $BC=10$ 。

所以正确答案是 B。

Drop perpendiculars from $B$ and $C$ to $\overline{AD},$ meeting it at $E$ and $F.$

In the left and right right triangles, $AE=\sqrt{10^2-8^2}=6$ and $FD=\sqrt{17^2-8^2}=15.$

The two side triangles have areas $\frac12\cdot6\cdot8=24$ and $\frac12\cdot15\cdot8=60.$ The middle rectangle has area $8\cdot BC.$

Thus $164=24+60+8BC,$ so $8BC=80$ and $BC=10.$

Thus, B is the correct answer.

22.

下列图形由正方形和圆组成。哪个图形的阴影区域面积最大？

The following figures are composed of squares and circles. Which figure has a shaded region with largest area?

只有 $A$

$A$ only

只有 $B$

$B$ only

只有 $C$

$C$ only

$A$ 和 $B$

both $A$ and $B$

全部相等

all are equal

答案：C

知识点：圆面积面积

难度评级：1580

小提示：

图形 $A$ 和 $B$ 留下相同的阴影面积

Figures $A$ and $B$ leave the same shaded area.

大提示：

对于 $C$ ，比较 $\pi-2$ 和 $4-\pi$

For $C,$ compare $\pi-2$ with $4-\pi.$

解答：

在图形 $A$ 中，阴影面积是一个 $2$ 乘 $2$ 正方形面积减去半径为 $1$ 的圆面积，即 $4-\pi$ 。

图形 $B$ 由四个边长为图形 $A$ 一半的副本组成，所以每个副本的面积是原图的四分之一，总阴影面积也为 $4-\pi$ 。

在图形 $C$ 中，圆的半径为 $1$ ，内接正方形的对角线为 $2$ ，所以其面积为 $2$ 。阴影面积为 $\pi-2$ 。

因为 $\pi>3$ ，所以 $\pi-2>4-\pi$ ，因此图形 $C$ 的阴影面积最大。

所以正确答案是 C。

In figure $A,$ the shaded area is the area of a $2$ by $2$ square minus a circle of radius $1,$ so it is $4-\pi.$

Figure $B$ is made of four copies of figure $A$ with half the side length, so each copy has one-fourth the area and the total shaded area is also $4-\pi.$

In figure $C,$ the circle has radius $1,$ and the inscribed square has diagonal $2,$ so its area is $2.$ The shaded area is $\pi-2.$

Since $\pi>3,$ we have $\pi-2>4-\pi,$ so figure $C$ has the largest shaded area.

Thus, C is the correct answer.

23.

在下面的图案中，猫在四个正方形中顺时针移动，老鼠沿四个正方形外侧的八条线段逆时针移动。

如果图案继续下去，第 $247$ 次移动后猫和老鼠会在哪里？

In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.

If the pattern is continued, where would the cat and mouse be after the $247$ th move?

答案：A

知识点：模运算找规律

难度评级：1540

小提示：

猫的位置每 $4$ 次移动重复一次

The cat repeats every $4$ moves.

大提示：

老鼠的位置每 $8$ 次移动重复一次；使用 $247$ 的余数

The mouse repeats every $8$ moves; use the remainders of $247.$

解答：

猫的位置每 $4$ 次移动重复一次，老鼠的位置每 $8$ 次移动重复一次。

因为 $247\equiv3\pmod{4}$ ，猫的位置与第 $3$ 次移动后相同：在右下方正方形。

因为 $247\equiv7\pmod{8}$ ，老鼠的位置与第 $7$ 次移动后相同：在左下方正方形的左边。

所以正确答案是 A。

The cat’s position repeats every $4$ moves, and the mouse’s position repeats every $8$ moves.

Since $247\equiv3\pmod{4},$ the cat is in the same position as after the $3$ rd move: the lower right square.

Since $247\equiv7\pmod{8},$ the mouse is in the same position as after the $7$ th move: the left side of the lower left square.

Thus, A is the correct answer.

24.

一艘船沿半圆路径从点 $A$ 到点 $B$ ，该半圆以岛 $X$ 为圆心。然后它沿直线路径从 $B$ 到 $C$ 。下列哪张图最能表示船沿路线移动时到岛 $X$ 的距离？

A ship travels from point $A$ to point $B$ along a semicircular path, centered at Island $X.$ Then it travels along a straight path from $B$ to $C.$ Which of these graphs best shows the ship’s distance from Island $X$ as it moves along its course?

答案：B

知识点：数据与图表解读圆

难度评级：1540

小提示：

沿半圆移动时，到 $X$ 的距离保持不变

Along the semicircle, the distance from $X$ is constant.

大提示：

在直线路段上，船先靠近 $X$ ，再远离它

On the straight segment, the ship first gets closer to $X,$ then farther away.

解答：

从 $A$ 到 $B$ 的半圆路径上每一点到圆心 $X$ 的距离都相同，所以图像一开始是水平的。

在从 $B$ 到 $C$ 的直线路径上，船先靠近 $X$ ，然后远离 $X$ 。

只有图 $B$ 先保持水平，然后下降再上升。

所以正确答案是 B。

Every point on the semicircular path from $A$ to $B$ is the same distance from the center $X,$ so the graph starts horizontally.

On the straight path from $B$ to $C,$ the ship first gets closer to $X$ and then farther away from $X.$

Only graph $B$ starts flat and then decreases before increasing.

Thus, B is the correct answer.

25.

图中，正方形 $WXYZ$ 的面积为 $25 \text{ cm}^2$ 。四个较小正方形边长为 $1$ cm，边要么与大正方形的边平行，要么重合。在 $\triangle ABC$ 中， $AB = AC$ ，且当 $\triangle ABC$ 沿边 $\overline{BC}$ 折叠时，点 $A$ 与 $O$ （正方形 $WXYZ$ 的中心）重合。 $\triangle ABC$ 的面积是多少平方厘米？

In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2.$ The four smaller squares have sides $1$ cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC,$ $AB = AC,$ and when $\triangle ABC$ is folded over side $\overline{BC},$ point $A$ coincides with $O,$ the center of square $WXYZ.$ What is the area of $\triangle ABC,$ in square centimeters?

$\dfrac{15}{4}$

$\dfrac{21}{4}$

$\dfrac{27}{4}$

$\dfrac{21}{2}$

$\dfrac{27}{2}$

答案：C

知识点：折纸三角形面积

难度评级：1680

小提示：

设 $M$ 为 $\overline{BC}$ 的中点

Let $M$ be the midpoint of $\overline{BC}.$

大提示：

折叠说明 $A$ 到 $\overline{BC}$ 的距离等于 $O$ 到 $\overline{BC}$ 的距离

Folding makes the distance from $A$ to $\overline{BC}$ equal the distance from $O$ to $\overline{BC}.$

解答：

正方形 $WXYZ$ 的边长为 $5$ cm，因为 $\sqrt{25} = 5$ 。

$\overline{WZ}$ 到 $\overline{BC}$ 的距离是 $2$ ，因为它等于 $2$ 个单位正方形的边长之和。

最后， $A$ 到 $\overline{BC}$ 的距离等于 $\overline{BC}$ 到 $O$ 的距离，即 $2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}\text{。}$

现在可求 $BC$ ： $WZ - 2 = 5 - 2 = 3 \text{ cm}\text{。}$

因此 $\triangle ABC$ 的面积为 $\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2\text{。}$

所以正确答案是 C。

The side length of $WXYZ$ is $5$ cm, since $\sqrt{25} = 5.$

We also know that the distance from $\overline{WZ}$ to $\overline{BC}$ is $2$ since it is the sum of the side lengths of $2$ unit squares.

Finally, the distance from $A$ to $\overline{BC}$ is the same as the distance from $\overline{BC}$ to $O,$ which is $2 + \dfrac{5}{2} = \dfrac{9}{2} \text{ cm}.$

Now, we can find $BC,$ which is $WZ - 2 = 5 - 2 = 3 \text{ cm}.$

Therefore, the area of $\triangle ABC$ is $\dfrac{1}{2} \cdot 3 \cdot \dfrac{9}{2} = \dfrac{27}{4} \text{ cm}^2.$

Thus, C is the correct answer.