2007 AMC 8 真题与在线限时测验

计时

40:00

1.

Theresa 的父母同意，如果她平均每周花 $10$ 小时帮忙做家务，连续 $6$ 周，就给她买票去看她最喜欢的乐队。前 $5$ 周她分别帮忙 $8$ 、 $11$ 、 $7$ 、 $12$ 和 $10$ 小时。为了得到门票，最后一周她必须工作多少小时？

Theresa’s parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8,$ $11,$ $7,$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?

$9$

$10$

$11$

$12$

$13$

答案：D

知识点：平均数

难度评级：560

小提示：

求六周总共需要多少小时

Find the total number of hours needed over all six weeks.

大提示：

将这个总数与 Theresa 前五周已经工作的小时数比较

Compare that total with the hours Theresa already worked in the first five weeks.

解答：

前 $5$ 周 Theresa 总共工作 $8+11+7+12+10=48$ 小时。

她承诺总共工作 $10\cdot 6=60$ 小时。

因此她最后一周必须工作 $60-48=12$ 小时才能得到门票。

所以正确答案是 D。

During the first $5$ weeks, Theresa works for a total of $8+11+7+12+10=48$ hours.

She promised to work $10\cdot 6=60$ hours in all.

This means that she has to work $60-48=12$ hours during the final week to earn the tickets.

Thus, D is the correct answer.

2.

对 $650$ 名学生进行了意大利面偏好调查。选项包括千层面、意式夹馅卷、意式馄饨和意大利面条。调查结果显示在条形图中。喜欢意大利面条的学生人数与喜欢意式夹馅卷的学生人数之比是多少？

$650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{5}{4}$

$\dfrac{5}{3}$

$\dfrac{5}{2}$

答案：E

知识点：数据与图表解读比与比例

难度评级：450

小提示：

从图中读出意大利面条和意式夹馅卷的条形高度

Read the spaghetti and manicotti bar heights from the graph.

大提示：

所求比值是意大利面条比意式夹馅卷，顺序不能反

The requested ratio is spaghetti to manicotti, in that order.

解答：

喜欢意大利面条的学生有 $250$ 人，喜欢意式夹馅卷的学生有 $100$ 人。

比值为 $\dfrac{250}{100} = \dfrac{5}{2}\text{。}$

所以正确答案是 E。

There are $250$ students who preferred spaghetti and $100$ that preferred manicotti.

The ratio is therefore $\dfrac{250}{100} = \dfrac{5}{2}.$

Thus, E is the correct answer.

3.

$250$ 的两个最小质因数之和是多少？

What is the sum of the two smallest prime factors of $250?$

$2$

$5$

$7$

$10$

$12$

答案：C

知识点：质因数分解

难度评级：560

小提示：

将 $250$ 分解成质因数

Factor $250$ into primes.

大提示：

寻找两个最小质因数时，只需要不同的质因数

Only distinct prime factors matter when looking for the two smallest prime factors.

解答：

对 $250$ 分解质因数： $250 = 2 \cdot 5^3\text{。}$ 因此 $250$ 只有两个质因数： $2$ 和 $5$ 。它们的和为 $7$ 。

所以正确答案是 C。

We can prime factorize $250$ to get $250 = 2 \cdot 5^3.$ From this, we can see that $250$ only has two prime factors: $2$ and $5.$ The sum of these is $7.$

Thus, C is the correct answer.

4.

一座鬼屋有六扇窗。幽灵 Georgie 从一扇窗进入，并从另一扇不同的窗离开，共有多少种方式？

A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

$12$

$15$

$18$

$30$

$36$

答案：D

知识点：乘法原理

难度评级：660

小提示：

先选择进入的窗户

Choose the entrance window first.

大提示：

进入窗户确定后，离开的窗户少一个选择

After the entrance is chosen, the exit has one fewer option.

解答：

Georgie 有 $6$ 种选择进入的窗户。但离开时只能选择不同的窗户，所以有 $5$ 种选择。

路径总数为 $6 \cdot 5 = 30$ 。

所以正确答案是 D。

Georgie has $6$ options for which window he enters through. He, however, only has $5$ options for the exit since it must be different from the entrance.

The total number of paths is therefore $6 \cdot 5 = 30.$

Thus, D is the correct answer.

5.

Chandler 想买一辆 $\$500$ 的山地车。生日时，他的祖父母给他 $\$50$ ，姑姑给他 $\$35$ ，表亲给他 $\$15$ 。他送报每周赚 $\$16$ 。他会使用所有生日钱和所有送报赚的钱。多少周后他可以买这辆山地车？

Chandler wants to buy a $\$500$ mountain bike. For his birthday, his grandparents send him $\$50$ , his aunt sends him $\$35$ and his cousin gives him $\$15$ . He earns $\$16$ per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?

$24$

$25$

$26$

$27$

$28$

答案：B

知识点：钱币

难度评级：770

小提示：

先加出 Chandler 已有的生日钱

First add the birthday money Chandler already has.

大提示：

用剩余价格除以每周送报收入

Divide the remaining cost by the weekly paper-route earnings.

解答：

Chandler 的生日钱总额为 $50+35+15=100$ 美元。因此他还需要 $500-100=400$ 美元。

送报需要 $400\div 16=25$ 周赚到这笔钱。

所以正确答案是 B。

The total amount of birthday money Chandler gets is $50+35+15=100$ dollars. Therefore, he still needs $500-100=400$ dollars.

This can be earned in $400\div 16=25$ weeks through his paper route.

Thus, B is the correct answer.

6.

$1985$ 年，美国长途电话平均费用为每分钟 $41$ 美分； $2005$ 年为每分钟 $7$ 美分。求长途电话每分钟费用大约下降了百分之多少。

The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance call in the USA in $2005$ was $7$ cents per minute. Find the approximate percent decrease in the cost per minute of a long-distance call.

$7$

$17$

$34$

$41$

$80$

答案：E

知识点：百分数估算

难度评级：940

小提示：

百分比下降等于下降量除以原始量

Percent decrease is decrease divided by the original amount.

大提示：

费用从 $41$ 美分降到 $7$ 美分，所以将减少的 $34$ 美分与原价 $41$ 美分比较

The decrease is from $41$ cents to $7$ cents, so compare $34$ to $41.$

解答：

费用下降了 $41-7=34$ 美分每分钟。

百分比下降为 $100\cdot\dfrac{34}{41}\%$ ，略大于 $80\%$ ，所以最接近的答案是 $80$ 。

所以正确答案是 E。

The decrease in cost is $41-7=34$ cents per minute.

The percent decrease is $100\cdot\dfrac{34}{41}\%$ , which is a little more than $80\%$ , so the closest answer is $80$ .

Thus, E is the correct answer.

7.

一个房间里 $5$ 个人的平均年龄是 $30$ 岁。一名 $18$ 岁的人离开房间。剩下四个人的平均年龄是多少？

The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves the room. What is the average age of the four remaining people?

$25$

$26$

$29$

$33$

$36$

答案：D

知识点：平均数

难度评级：900

小提示：

把原来的平均年龄转化为总年龄

Convert the original average into a total age.

大提示：

先减去离开者的年龄，再除以剩余人数

Subtract the age of the person who leaves before dividing by the remaining number of people.

解答：

起初，房间里所有人的总年龄为 $5 \cdot 30 = 150$ 岁。

这个人离开后，总年龄为 $150 - 18 = 132$ 。剩下 $4$ 人，平均年龄为 $\dfrac{132}{4} = 33$ 。

所以正确答案是 D。

Initially, the total age of everyone in the room is $5 \cdot 30 = 150$ years.

After the person leaves, the total age is $150 - 18 = 132.$ With $4$ people remaining, the average age becomes $\dfrac{132}{4} = 33.$

Thus, D is the correct answer.

8.

在梯形 $ABCD$ 中， $\overline{AD}$ 垂直于 $\overline{DC}$ ， $AD = AB = 3$ ，且 $DC = 6$ 。此外， $E$ 在 $\overline{DC}$ 上，且 $\overline{BE}$ 平行于 $\overline{AD}$ 。求 $\triangle BEC$ 的面积。

In trapezoid $ABCD,$ $\overline{AD}$ is perpendicular to $\overline{DC},$ $AD = AB = 3,$ and $DC = 6.$ In addition, $E$ is on $\overline{DC},$ and $\overline{BE}$ is parallel to $\overline{AD}.$ Find the area of $\triangle BEC.$

$3$

$4.5$

$6$

$9$

$18$

答案：B

知识点：三角形面积梯形

难度评级：960

小提示：

求出三角形 $BEC$ 的底和高

Figure out the base and height of triangle $BEC$ .

大提示：

因为 $ABED$ 是正方形，所以 $DE$ 和 $BE$ 的长度都是 $3$

Since $ABED$ is a square, $DE$ and $BE$ are both $3.$

解答：

我们有 $EC = DC - DE$ $= 6 - 3$ $= 3\text{。}$ 还知道 $BE = AD = 3\text{。}$ 因此 $\triangle BEC$ 的面积为 $\dfrac{1}{2} \cdot 3 \cdot 3 = \dfrac{9}{2}\text{。}$

所以正确答案是 B。

We know that $EC = DC - DE$ $= 6 - 3$ $= 3.$ We also know that $BE = AD = 3.$ Therefore, the area of $\triangle BEC$ is $\dfrac{1}{2} \cdot 3 \cdot 3 = \dfrac{9}{2}.$

Thus, B is the correct answer.

9.

要完成下面的方格，每一行和每一列都必须各出现一次数字 $1$ 到 $4$ 。右下角方格中会是什么数？

To complete the grid below, each of the digits $1$ through $4$ must occur once in each row and once in each column. What number will occupy the lower right-hand square?

$1$

$2$

$3$

$4$

无法确定

cannot be determined

答案：B

知识点：逻辑推理

难度评级：1120

小提示：

使用每行每列的限制逐步确定格子

Use the row and column restrictions to force entries one at a time.

大提示：

从第二行最后一个格子开始，再移到右上角格子

Start with the last square in the second row, then move to the top-right square.

解答：

第二行最后一个格子必须是 $1$ ，因为该行已有 $2$ 和 $3$ ，最后一列已有 $4$ 。

接着右上角格子必须是 $3$ ，因为第一行已有 $1$ 和 $2$ ，最后一列已有 $1$ 和 $4$ 。

这迫使 $2$ 位于右下角方格。

所以正确答案是 B。

The last square in the second row must be $1$ , since that row already contains $2$ and $3$ , and the last column already contains $4$ .

Then the top-right square must be $3$ , since the top row already contains $1$ and $2$ , and the last column already contains $1$ and $4$ .

This forces the $2$ to be in the bottom-right square.

Thus, B is the correct answer.

10.

对任意正整数 $n$ ，定义 $\boxed{n}$ 为 $n$ 的所有正因数之和。例如， $\boxed{6} = 1 + 2 + 3 + 6 = 12\text{。}$ 求 $\boxed{\boxed{11}}$ 。

For any positive integer $n,$ define $\boxed{n}$ to be the sum of the positive factors of $n.$ For example, $\boxed{6} = 1 + 2 + 3 + 6 = 12.$ Find $\boxed{\boxed{11}}.$

$13$

$20$

$24$

$28$

$30$

答案：D

知识点：因数之和自定义运算

难度评级：1150

小提示：

先对 $11$ 应用一次方框运算

Apply the boxed operation once to $11$ first.

大提示：

再对所得结果应用同样的因数和运算

Then apply the same sum-of-factors operation to the result.

解答：

首先，由于 $11$ 是质数， $\boxed{11} = 1 + 11 = 12\text{。}$ 接着求 $\boxed{12}$ 。 $12$ 的因数为 $1, 2, 3, 4, 6, 12\text{。}$ 相加得到 $\boxed{12} = 28$ 。

所以正确答案是 D。

First, we find that $\boxed{11} = 1 + 11 = 12$ since $11$ is prime. Then we need to find $\boxed{12}.$ The factors of $12$ are $1, 2, 3, 4, 6, 12.$ Adding these yields $\boxed{12} = 28.$

Thus, D is the correct answer.

11.

瓷片 $I$ 、 $II$ 、 $III$ 和 $IV$ 被平移，使每个瓷片分别与长方形 $A$ 、 $B$ 、 $C$ 和 $D$ 中的一个重合。在最终排列中，任意两个相邻瓷片公共边上的两个数字必须相同。哪个瓷片被平移到长方形 $C$ ？

Tiles $I,$ $II,$ $III$ and $IV$ are translated so one tile coincides with each of the rectangles $A,$ $B,$ $C$ and $D.$ In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle $C?$

$I$

$II$

$III$

$IV$

无法确定

cannot be determined

答案：D

知识点：逻辑推理

难度评级：1270

小提示：

寻找只出现在一个瓷片边上的数字

Look for edge numbers that appear on only one tile.

大提示：

$0$ 和 $5$ 会先确定瓷片 $III$ 的位置，再决定 $C$ 中的瓷片

The $0$ and $5$ force the placement of Tile $III$ before deciding which tile is in $C$ .

解答：

只有瓷片 $III$ 的边上有 $0$ ，所以瓷片 $III$ 必须放在 $C$ 或 $D$ ，使这个边可以在外侧。

瓷片 $III$ 的右边还有 $5$ ，且没有其他瓷片有 $5$ 。因此瓷片 $III$ 必须放在 $D$ ，让 $5$ 在外侧。

$1$ 标在瓷片 $III$ 的左边，唯一能与之匹配的是瓷片 $IV$ ，所以瓷片 $IV$ 放在 $C$ 。

所以正确答案是 D。

Only Tile $III$ has a $0$ on an edge, so Tile $III$ must be placed in $C$ or $D$ , where that edge can be on the outside.

Tile $III$ also has a $5$ on its right edge, and no other tile has a $5$ . Therefore Tile $III$ must be placed in $D$ , with its $5$ on the outside.

The only tile that can match the $1$ on the left edge of Tile $III$ is Tile $IV$ , so Tile $IV$ is placed in $C$ .

Thus, D is the correct answer.

12.

一个单位六角星由一个边长为 $1$ 的正六边形和它的 $6$ 个等边三角形延伸部分组成，如图所示。延伸部分的面积与原正六边形面积之比是多少？

A unit hexagram is composed of a regular hexagon of side length $1$ and its $6$ equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

$1:1$

$6:5$

$3:2$

$2:1$

$3:1$

答案：A

知识点：面积比等边三角形

难度评级：1150

小提示：

将正六边形分成六个全等等边三角形

Divide the regular hexagon into six congruent equilateral triangles.

大提示：

每个外侧三角形都与六个内部三角形之一共边

Each exterior triangle shares a side with one of the six interior triangles.

解答：

可以把六边形分成 $6$ 个全等等边三角形，如下图。

因为每个内部三角形都与一个外侧三角形共边，所以所有这些三角形全等。因此面积比为 $1:1$ 。

所以正确答案是 A。

Note that we can split the hexagon into $6$ congruent equilateral triangles as follows.

Since each of them share an edge with an exterior triangle, all the triangles are congruent. Therefore, the ratio of areas is $1:1.$

Thus, A is the correct answer.

13.

如维恩图所示，集合 $A$ 和 $B$ 的元素个数相同。它们的并集有 $2007$ 个元素，交集有 $1001$ 个元素。求集合 $A$ 的元素个数。

Sets $A$ and $B,$ shown in the Venn diagram, have the same number of elements. Their union has $2007$ elements and their intersection has $1001$ elements. Find the number of elements in $A.$

$503$

$1006$

$1504$

$1507$

$1510$

答案：C

知识点：韦恩图容斥原理

难度评级：1290

小提示：

设每个集合都有 $x$ 个元素

Let each set have $x$ elements.

大提示：

使用 $|A\cup B|=|A|+|B|-|A\cap B|$

Use $|A\cup B|=|A|+|B|-|A\cap B|$ .

解答：

设 $x$ 为集合 $A$ 和 $B$ 各自的元素个数，并设 $y$ 为它们交集的元素个数。

条件给出 $2x - y = 2007$ 且 $y = 1001\text{。}$ 将 $y$ 代入第一式得到 $\begin{align*} 2x - 1001 &= 2007 \\ 2x &= 3008 \\ x &= 1504 \end{align*}\text{。}$

所以正确答案是 C。

Let $x$ be the number of elements in each $A$ and $B.$ Also let $y$ be the number of elements in their intersection.

The conditions give us that $2x - y = 2007$ and $y = 1001.$ Plugging $y$ into the first equation yields $\begin{align*} 2x - 1001 &= 2007 \\ 2x &= 3008 \\ x &= 1504. \end{align*}$

Thus, C is the correct answer.

14.

等腰三角形 $\triangle ABC$ 的底边长为 $24$ ，面积为 $60$ 。其中一条全等边的长度是多少？

The base of isosceles $\triangle ABC$ is $24$ and its area is $60.$ What is the length of one of the congruent sides?

$5$

$8$

$13$

$14$

$18$

答案：C

知识点：等腰三角形三角形面积勾股定理

难度评级：1350

小提示：

从顶点向底边作高

Draw the altitude from the top vertex to the base.

大提示：

在等腰三角形中，这条高会把底边平分

In an isosceles triangle, that altitude splits the base into two equal parts.

解答：

作 $\overline{BD}$ 为从 $B$ 到 $\overline{AC}$ 的高。

则 $60 = \dfrac{1}{2} \cdot BD \cdot 24\text{，}$ 得到 $BD = 5$ 。

在 $\triangle ABD$ 中使用勾股定理： $AB^2 = 5^2 + 12^2 = 169 = 13^2\text{。}$ 因此 $AB = 13$ 。

所以正确答案是 C。

Construct $\overline{BD}$ as the altitude from $B$ to $\overline{AC}.$

Then $60 = \dfrac{1}{2} \cdot BD \cdot 24,$ which gives us that $BD = 5.$

From this, we apply the Pythagorean Theorem on $\triangle ABD:$ $AB^2 = 5^2 + 12^2 = 169 = 13^2.$ This gives us that $AB = 13.$

Thus, C is the correct answer.

15.

设 $a$ ， $b$ 和 $c$ 是满足 $0 \lt a \lt b \lt c$ 的数。下列哪一项不可能？

Let $a,$ $b$ and $c$ be numbers with $0 \lt a \lt b \lt c.$ Which of the following is impossible?

$a + c \lt b$

$a \cdot b \lt c$

$a + b \lt c$

$a \cdot c \lt b$

$\dfrac{b}{c} = a$

答案：A

知识点：不等式反例

难度评级：1380

小提示：

使用 $b<c$ 且 $a>0$ 。

Use the fact that $b<c$ and $a>0$ .

大提示：

对其他选项，只需一个例子就能说明可能

For the other choices, one counterexample is enough to show they are possible.

解答：

已知 $b \lt c$ 且 $0 \lt a$ 。把这两个不等式相加得到 $b \lt c + a\text{。}$ 这说明 A 不可能，因此就是正确答案。

为确认这一点，可以给出其他选项可行的例子。

B 和 C：取 $a = 1$ 、 $b = 2$ 、 $c = 4$ 。

D：取 $a = \dfrac{1}{3}$ 、 $b = \dfrac{1}{2}$ 、 $c = 1$ 。

E：取 $a = \dfrac{1}{2}$ 、 $b = 1$ 、 $c = 2$ 。

所以正确答案是 A。

We know that $b \lt c$ and $0 \lt a.$ Adding these two inequalities together yields $b \lt c + a.$ This shows that A is impossible, and therefore the right answer.

To ensure that this is correct, we can show that the other options are possible.

B and C : $a = 1,$ $b = 2,$ and $c = 4$

D : $a = \dfrac{1}{3},$ $b = \dfrac{1}{2},$ and $c = 1$

E : $a = \dfrac{1}{2},$ $b = 1,$ and $c = 2$

Thus, A is the correct answer.

16.

Amanda Reckonwith 画了五个半径为 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $5$ 的圆。然后对每个圆，她描点 $(C, A)$ ，其中 $C$ 是圆的周长， $A$ 是圆的面积。下列哪一个可能是她的图？

Amanda Reckonwith draws five circles with radii $1,$ $2,$ $3,$ $4$ and $5.$ Then for each circle she plots the point $(C, A),$ where $C$ is its circumference and $A$ is its area. Which of the following could be her graph?

答案：A

知识点：圆面积圆周长二次方程

难度评级：1400

小提示：

用半径表示周长和面积

Write circumference and area in terms of the radius.

大提示：

当周长随半径线性增长时，面积随半径平方增长

As the circumference grows linearly with radius, the area grows quadratically.

解答：

半径 $1$ 到 $5$ 的圆的周长分别为 $2\pi$ 、 $4\pi$ 、 $6\pi$ 、 $8\pi$ 和 $10\pi$ ，对应面积分别为 $\pi$ 、 $4\pi$ 、 $9\pi$ 、 $16\pi$ 和 $25\pi$ 。

当 $C$ 等量增加时， $A$ 增加得越来越快，所以这些点形成递增的二次形状。只有图 A 有这种形状。

所以正确答案是 A。

The circumferences of circles with radii $1$ through $5$ are $2\pi,$ $4\pi,$ $6\pi,$ $8\pi,$ and $10\pi$ . Their respective areas are $\pi,$ $4\pi,$ $9\pi,$ $16\pi,$ and $25\pi$ .

As $C$ increases by equal amounts, $A$ increases more and more quickly, so the points form an increasing quadratic pattern. Only graph A has that shape.

Thus, A is the correct answer.

17.

$30$ 升油漆混合物中有 $25\%$ 红色颜料、 $30\%$ 黄色颜料和 $45\%$ 水。向原混合物中加入五升黄色颜料。新混合物中黄色颜料占百分之多少？

A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow tint and $45\%$ water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?

$25$

$35$

$40$

$45$

$50$

答案：C

知识点：混合问题百分数

难度评级：1340

小提示：

先求原混合物中黄色颜料有多少升

Find the original number of liters of yellow tint.

大提示：

加入黄色颜料后，黄色颜料量和总混合物量都会改变

After adding yellow tint, both the yellow amount and the total mixture amount change.

解答：

原混合物中黄色颜料为 $0.3 \times 30 = 9$ 升。加入 $5$ 升后，黄色颜料共有 $14$ 升。

新混合物总量为 $35$ 升，所以黄色颜料百分比为 $100 \cdot \dfrac{14}{35} = 100 \cdot \dfrac{2}{5} = 40 \%\text{。}$

所以正确答案是 C。

The amount of yellow tint in the original mixture is $0.3 \times 30 = 9$ liters. Adding $5$ liters results in a total of $14$ liters of yellow tint.

The new mixture has a total of $35$ liters, so the percent of yellow tint is $100 \cdot \dfrac{14}{35} = 100 \cdot \dfrac{2}{5} = 40 \%.$

Thus, C is the correct answer.

18.

两个 $99$ 位数 $303,030,303,\ldots,030,303$ 和 $505,050,505,\ldots,050,505$ 的乘积，其千位数字为 $A$ ，个位数字为 $B$ 。 $A+B$ 是多少？

The product of the two $99$ -digit numbers $303,030,303,\ldots,030,303$ and $505,050,505,\ldots,050,505$ has thousands digit $A$ and units digit $B.$ What is $A+B?$

$3$

$5$

$6$

$8$

$10$

答案：D

知识点：模运算数字

难度评级：1400

小提示：

只有最后四位会影响千位和个位

Only the last four digits can affect the thousands and units digits.

大提示：

先把每个 $99$ 位数化为其最后四位再相乘

Reduce each $99$ -digit number to its last four digits before multiplying.

解答：

只有最后四位会影响乘积的千位和个位。这两个数分别以 $0303$ 和 $0505$ 结尾，所以只需计算 $303\cdot 505$ 。

因为 $303\cdot 505=153015$ ，完整乘积的最后四位是 $3015$ 。

因此 $A=3$ ， $B=5$ ，所以 $A+B=8$ 。

所以正确答案是 D。

Only the last four digits can affect the thousands digit and units digit of the product. The two numbers end in $0303$ and $0505$ , so it is enough to compute $303\cdot 505$ .

Since $303\cdot 505=153015$ , the last four digits of the full product are $3015$ .

This gives $A=3$ and $B=5$ , so $A+B=8$ .

Thus, D is the correct answer.

19.

选择两个连续正整数，它们的和小于 $100$ 。把这两个整数分别平方，然后求平方差。下列哪一个可能是这个差？

Pick two consecutive positive integers whose sum is less than $100.$ Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

$2$

$64$

$79$

$96$

$131$

答案：C

知识点：平方差奇偶性

难度评级：1430

小提示：

设连续整数为 $x$ 和 $x+1$

Let the consecutive integers be $x$ and $x+1$ .

大提示：

它们平方差可化简为这两个整数的和

The difference of their squares simplifies to the sum of the two integers.

解答：

设连续正整数为 $x$ 和 $x+1$ 。它们的和为 $2x+1$ ，小于 $100$ 。

它们的平方差为 $\begin{gathered} (x+1)^2-x^2 \\ = (x+1+x)(x+1-x) \\ = 2x+1 \end{gathered}\text{。}$

所以这个差必须是小于 $100$ 的奇数。选项中只有 $79$ 可能，它由 $39$ 和 $40$ 得到。

所以正确答案是 C。

Let the consecutive positive integers be $x$ and $x+1$ . Their sum is $2x+1$ , which is less than $100$ .

The difference of their squares is $\begin{gathered} (x+1)^2-x^2 \\ = (x+1+x)(x+1-x) \\ = 2x+1. \end{gathered}$

So the difference must be an odd number less than $100$ . Among the choices, $79$ is the only possibility, and it occurs for $39$ and $40$ .

Thus, C is the correct answer.

20.

在分区赛前，Unicorns 赢了他们篮球比赛的 $45 \%$ 。分区赛中，他们又赢六场、输两场，最终以胜率一半结束赛季。Unicorns 总共打了多少场比赛？

Before district play, the Unicorns had won $45 \%$ of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

$48$

$50$

$52$

$54$

$60$

答案：A

知识点：一次方程百分数

难度评级：1550

小提示：

设 $x$ 为分区赛前的比赛数

Let $x$ be the number of games before district play.

大提示：

写出分区赛后总胜场等于总比赛数一半的方程

Write an equation comparing total wins after district play with half of all games played.

解答：

设 Unicorns 分区赛前打了 $x$ 场比赛。那么分区赛前赢了 $0.45x$ 场，分区赛后共赢了 $0.45x+6$ 场。

他们总共打了 $x+8$ 场。题意给出 $0.45x+6=\dfrac{x+8}{2}\text{。}$

解得 $0.45x+6=0.5x+4$ ，所以 $2=0.05x$ ， $x=40$ 。因此 Unicorns 总共打了 $40+8=48$ 场。

所以正确答案是 A。

Let $x$ be the number of games the Unicorns had played before district play. Then they had won $0.45x$ games before district play and $0.45x+6$ games after district play.

In total, they played $x+8$ games. The problem statement tells us that $0.45x+6=\dfrac{x+8}{2}.$

Solving gives $0.45x+6=0.5x+4$ , so $2=0.05x$ , and $x=40$ . Therefore, the Unicorns played $40+8=48$ games in all.

Thus, A is the correct answer.

21.

从一副牌中发出两张牌。这副牌有四张红牌，分别标为 $A$ 、 $B$ 、 $C$ 、 $D$ ，还有四张绿牌，也分别标为 $A$ 、 $B$ 、 $C$ 、 $D$ 。一对获胜牌是两张同色牌或两张同字母牌。抽到一对获胜牌的概率是多少？

Two cards are dealt from a deck of four red cards labeled $A,$ $B,$ $C,$ $D$ and four green cards labeled $A,$ $B,$ $C,$ $D.$ A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

$\dfrac{2}{7}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{4}{7}$

$\dfrac{5}{8}$

答案：D

知识点：基本概率无放回抽样

难度评级：1450

小提示：

抽出第一张牌后，数有利的第二张牌

After the first card is drawn, count favorable second cards.

大提示：

第一张牌被移除后，同色牌和同字母牌不会重叠

The same-color cards and same-letter card do not overlap after the first card is removed.

解答：

抽出第一张牌后，还剩 $7$ 张。其中 $3$ 张同色， $1$ 张同字母。因此会与第一张牌形成获胜牌对的有 $4$ 张，它们都在剩余的 $7$ 张牌中。

概率为 $\dfrac{4}{7}$ 。

所以正确答案是 D。

After drawing the first card, there are $7$ left. $3$ of them have the same color, and $1$ of them has the same letter. Therefore, there are $4$ out of $7$ possibilities that result in a winning pair.

The probability of drawing a pair is then $\dfrac{4}{7}.$

Thus, D is the correct answer.

22.

一只旅鼠坐在边长为 $10$ 米的正方形的一个角上。它沿对角线朝对面的顶点跑了 $6.2$ 米。它停下，向右转 $90^{\circ}$ ，又跑了 $2$ 米。一位科学家测量旅鼠到正方形每条边的最短距离。这四个距离的平均数是多少米？

A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?

$2$

$4.5$

$5$

$6.2$

$7$

答案：C

知识点：正方形（几何）不变量

难度评级：1550

小提示：

对正方形内任一点，到两条相对边的距离和等于边长

For any point inside a square, distances to opposite sides add to the side length.

大提示：

对两组相对边分别应用这个想法

Apply that idea to both pairs of opposite sides.

解答：

根据题目给出的长度，旅鼠停下后仍在正方形内。

因为旅鼠仍在正方形内，它到两条水平边的距离和为 $10$ 米，到两条竖直边的距离和也一样。因此 $4$ 个距离之和为 $20$ 米，平均数为 $20 \div 4 = 5$ 米。

所以正确答案是 C。

Based on the lengths given in the problem, the lemming is still in the square after it stops.

Since the lemming is still in the square, the sum of the distances to the horizontal sides is $10$ meters and the same for the vertical sides. Therefore, the $4$ distances sum to $20$ meters, making the average $20 \div 4 = 5$ meters.

Thus, C is the correct answer.

23.

图中 $5 \times 5$ 方格里的阴影风车形面积是多少？

What is the area of the shaded pinwheel shown in the $5 \times 5$ grid?

$4$

$6$

$8$

$10$

$12$

答案：B

知识点：面积分割三角形面积

难度评级：1610

小提示：

从整个方格中减去未阴影面积更容易

It is easier to subtract the unshaded area from the whole grid.

大提示：

未阴影部分由四个角落正方形和四个全等三角形组成

The unshaded part consists of four corner squares and four congruent triangles.

解答：

可以用总面积减去未阴影面积来求阴影面积。

有 $4$ 个边长为 $1$ 的正方形，每个面积为 $1$ ，对未阴影面积贡献共 $4$ 。

还有 $4$ 个底为 $3$ 、高为 $\dfrac{5}{2}$ 的三角形，总面积为 $4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15\text{。}$

未阴影总面积为 $4 + 15 = 19$ 。

整个方格面积为 $5^2 = 25$ ，所以阴影面积为 $25 - 19 = 6$ 。

所以正确答案是 B。

We can find the area of the shaded part by subtracting the area of the unshaded part from the whole area.

There are $4$ squares with side length $1,$ contributing $1$ each to the unshaded area, for a total of $4.$

There are also $4$ triangles with base $3$ and height $\dfrac{5}{2}.$ This contributes a total area of $4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15.$

The total unshaded area is therefore $4 + 15 = 19.$

The total area is $5^2 = 25,$ and subtracting the unshaded area yields $25 - 19 = 6$ as the area of the shaded region.

Thus, B is the correct answer.

24.

一个袋子里有四张纸片，分别标有数字 $1$ 、 $2$ 、 $3$ 、 $4$ ，没有重复。从中不放回地一次抽出三张，用来组成一个三位数。这个三位数是 $3$ 的倍数的概率是多少？

A bag contains four pieces of paper, each labeled with one of the digits $1,$ $2,$ $3$ or $4,$ with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

答案：C

知识点：整除性基本概率数字

难度评级：1580

小提示：

一个数能被 $3$ 整除，当且仅当它的数字和能被 $3$ 整除

A number is divisible by $3$ exactly when its digit sum is divisible by $3.$

大提示：

不要先排列抽出的三个数字，先看留下了哪个数字

Instead of ordering the three drawn digits, look at which digit is left out.

解答：

一个数能被 $3$ 整除，当且仅当它的数字和能被 $3$ 整除。

从 $1,2,3,4$ 中取三个不同数字，数字和能被 $3$ 整除的只有 $(1,2,3)$ 和 $(2,3,4)$ 。

因此所组成的数是 $3$ 的倍数，当且仅当袋中剩下的是 $1$ 或 $4$ 。

每种事件概率都是 $\dfrac{1}{4}$ ，总概率为 $2\cdot\dfrac{1}{4}=\dfrac{1}{2}$ 。

所以正确答案是 C。

Recall that a number is divisible by $3$ if the sum of its digits is divisible by $3$ .

The only triples of distinct digits from $1,2,3,4$ whose sum is divisible by $3$ are $(1,2,3)$ and $(2,3,4)$ .

This means the constructed number is a multiple of $3$ exactly when either $1$ or $4$ is left in the bag.

Each of these events has probability $\dfrac{1}{4}$ , for a total probability of $2\cdot\dfrac{1}{4}=\dfrac{1}{2}$ .

Thus, C is the correct answer.

25.

下图的飞镖盘中，外圆半径为 $6$ ，内圆半径为 $3$ 。三条半径把每个圆分成三个全等区域，并标出分值。飞镖击中某区域的概率与该区域面积成正比。当两支飞镖击中这个靶时，得分为击中区域分值之和。得分为奇数的概率是多少？

On the dart board shown in the figure below, the outer circle has radius $6$ and the inner circle has radius $3.$ Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?

$\dfrac{17}{36}$

$\dfrac{35}{72}$

$\dfrac{1}{2}$

$\dfrac{37}{72}$

$\dfrac{19}{36}$

答案：B

知识点：几何概率奇偶性

难度评级：1680

小提示：

奇数得分来自一支飞镖落在 $1$ 分区域，另一支落在 $2$ 分区域

An odd score comes from one dart landing on a $1$ -region and the other on a $2$ -region.

大提示：

先按面积权重求落在 $1$ 分区域的总概率，再求落在 $2$ 分区域的总概率

Compute the total area-weighted probability of landing on $1,$ then of landing on $2.$

解答：

外圆面积为 $6^2\pi = 36\pi$ ，内圆面积为 $3^2\pi = 9\pi$ 。因此外环面积为 $36\pi - 9\pi = 27\pi$ 。

所以击中一个外环扇区的概率为 $\dfrac{9\pi}{36\pi} = \dfrac{1}{4}$ 。类似地，击中一个内圆扇区的概率为 $\dfrac{3\pi}{36\pi} = \dfrac{1}{12}$ 。

汇总所有可能，击中 $1$ 分区域的概率为 $\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}\text{。}$ 击中 $2$ 分区域的概率为 $2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}\text{。}$

得到奇数得分的唯一方式是一支击中 $1$ 分，另一支击中 $2$ 分。

两种顺序的总概率为 $2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}\text{。}$

所以正确答案是 B。

The area of the outer circle is $6^2\pi = 36\pi,$ and the area of the inner circle is $3^2\pi = 9\pi.$ Therefore, the area of the outer ring is $36\pi - 9\pi = 27\pi.$

This means that the probability of hitting an outer segment is $\dfrac{9\pi}{36\pi} = \dfrac{1}{4}.$ Similarly, the probability of hitting an inner segment is $\dfrac{3\pi}{36\pi} = \dfrac{1}{12}.$

Summing over all possibilities, the probability of hitting a $1$ is $\dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}.$ Similarly, the probability of hitting a $2$ is $2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}.$

The only way to get an odd score is to hit one $1$ and one $2.$

The probability of hitting these numbers in either order is $2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}.$

Thus, B is the correct answer.