2022 AMC 8 真题与在线限时测验

计时

40:00

1.

数学队设计了一个形状像乘号的标志，如下图所示，画在边长为 $1$ 英寸的小方格网格上。这个标志的面积是多少平方英寸？

The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of $1$ -inch squares. What is the area of the logo in square inches?

$10$

$12$

$13$

$14$

$15$

答案：A

知识点：面积分割

难度评级：770

小提示：

把标志分成几个全等的倾斜正方形

Break the logo into congruent tilted squares

大提示：

每个倾斜正方形是一个 $2$ 乘 $2$ 方格正方形的一半

Each tilted square is half of a $2$ by $2$ grid square

解答：

标志由 $5$ 个全等的倾斜正方形组成。每个倾斜正方形的两条对角线长度均为 $2$ ，所以面积为 $\frac{2\cdot2}{2}=2$ 平方英寸。

总面积为 $5\cdot2=10$ 。

正确答案是 A。

The logo is made from $5$ congruent tilted squares. Each tilted square has both diagonals of length $2$ , so its area is $\frac{2\cdot2}{2}=2$ square inches.

The total area is $5\cdot2=10$ .

Thus, the correct answer is A.

2.

考虑下面两个运算： $\begin{align*} a\,\blacklozenge\,b &= a^2 - b^2 \\ a \star b &= (a - b)^2 \end{align*}$

计算

$(5\,\blacklozenge\,3) \star 6$

Consider these two operations: $\begin{align*} a\,\blacklozenge\,b &= a^2 - b^2 \\ a \star b &= (a - b)^2 \end{align*}$

Compute the value:

$(5\,\blacklozenge\,3) \star 6$

$-20$

$4$

$16$

$100$

$220$

答案：D

知识点：自定义运算运算顺序

难度评级：560

小提示：

先计算 $5\,\blacklozenge\,3$

Evaluate $5\,\blacklozenge\,3$ first

大提示：

把第一个运算的结果代入 $\star 6$

Substitute the value from the first operation into $\star 6$

解答：

根据定义， $\begin{align*} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align*}$

正确答案是 D。

Using the definitions, $\begin{align*} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align*}$

Thus, the correct answer is D.

3.

三个正整数 $a$ 、 $b$ 、 $c$ 相乘，乘积为 $100$ 。假设 $a < b < c$ 。这些数有多少种选择方式？

When three positive integers $a,$ $b,$ and $c$ are multiplied together, their product is $100.$ Suppose $a < b < c.$ In how many ways can the numbers be chosen?

$0$

$1$

$2$

$3$

$4$

答案：E

知识点：因数分类讨论

难度评级：1190

小提示：

最小数必须是 $100$ 的因数

The smallest number must be a factor of $100$

大提示：

依次尝试 $a$ 的可能值，并利用 $a^3<100$ 缩小范围。

Try possible values of $a$ with $a^3<100$

解答：

因为 $a< b< c$ 且 $abc=100$ ，所以 $a^3<100$ ，因此 $a\le4$ 。同时 $a$ 必须整除 $100$ ，故 $a\in\{1,2,4\}$ 。

若 $a=1$ ，则 $(b,c)$ 可为 $(2,50),(4,25),(5,20)$ 。若 $a=2$ ，唯一可行的数对是 $(5,10)$ 。若 $a=4$ ，则 $bc=25$ ，不存在整数 $b$ 满足 $4< b< c$ 。

共有 $4$ 种可能的三元组。

所以正确答案是 E。

Since $a< b< c$ and $abc=100$ , we have $a^3<100$ , so $a\le4$ . Also $a$ must divide $100$ , so $a\in\{1,2,4\}$ .

If $a=1$ , the pairs $(b,c)$ are $(2,50),(4,25),(5,20)$ . If $a=2$ , the only possible pair is $(5,10)$ . If $a=4$ , then $bc=25$ , and there is no integer $b$ with $4< b< c$ .

There are $4$ possible triples.

Thus, the correct answer is E.

4.

下图中的字母 M 先关于直线 $q$ 反射，再关于直线 $p$ 反射。得到的图像是哪一个？

The letter M in the figure below is first reflected over the line $q$ and then reflected over the line $p.$ What is the resulting image?

答案：E

知识点：反射（几何）变换

难度评级：770

小提示：

关于 $q$ 反射会把字母移到对角线的另一侧

Reflecting over $q$ moves the letter across the diagonal line

大提示：

第一次反射后，再把新图像关于水平线 $p$ 反射

After the first reflection, reflect the new image over the horizontal line $p$

解答：

先把字母关于斜线 $q$ 镜像，再把得到的图像关于水平线 $p$ 镜像。最终图像位于 $p$ 下方、竖直线右侧，方向与选项 E 相同。

正确答案是 E。

Reflecting over the diagonal line $q$ changes the orientation of the letter as if the diagonal were a mirror. Reflecting that image over the horizontal line $p$ then places the final letter below $p$ , to the right of the vertical line, with the orientation shown in choice E.

Thus, the correct answer is E.

5.

Anna 和 Bella 正在一起庆祝生日。五年前，Bella 满 $6$ 岁时，她收到了一只刚出生的小猫作为生日礼物。今天两个孩子和这只小猫的年龄之和为 $30$ 岁。Anna 比 Bella 大几岁？

Anna and Bella are celebrating their birthdays together. Five years ago, when Bella turned $6$ years old, she received a newborn kitten as a birthday present. Today the sum of the ages of the two children and the kitten is $30$ years. How many years older than Bella is Anna?

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：年龄问题一次方程

难度评级：720

小提示：

现在是 $5$ 年后，而当时 Bella 刚满 $6$ 岁。

Bella is $5$ years older now than when she turned $6$

大提示：

先求 Bella 和小猫现在的年龄，再求 Anna 的年龄

Find Bella’s current age and the kitten’s current age before finding Anna’s

解答：

Bella 五年前 $6$ 岁，所以现在 $11$ 岁。

小猫五年前刚出生，所以现在 $5$ 岁。

三者共有 $3$ 个年龄，总和为 $30$ ，所以 Anna 的年龄是 $30-5-11 = 14$

Anna $14$ 岁，Bella $11$ 岁，所以 Anna 大 $3$ 岁。

正确答案是 C。

If Bella was $6$ five years ago, then she is $11$ right now.

If the kitten was a newborn five years ago, then it is $5$ right now.

Since the sum of all $3$ ages is $30,$ Anna’s age is $30-5-11 = 14$

Since Anna is $14$ and Bella is $11,$ she is $3$ years older than Bella.

Thus, the correct answer is C.

6.

三个正整数在数轴上等距排列。中间的数是 $15$ ，最大的数是最小数的 $4$ 倍。这三个数中最小的数是多少？

Three positive integers are equally spaced on a number line. The middle number is $15$ and the largest number is $4$ times the smallest number. What is the smallest of these three numbers?

$4$

$5$

$6$

$7$

$8$

答案：C

知识点：等差数列一次方程

难度评级：820

小提示：

三个等距数的中间数是最小数和最大数的平均数

The middle of three equally spaced numbers is their average

大提示：

若最小数为 $x$ ，最大数为 $4x$

If the smallest is $x$ , then the largest is $4x$

解答：

设最小数为 $x$ ，则最大数为 $4x$ 。因为中间数是最小数和最大数的平均数，所以 $\frac{x+4x}{2}=15\text{。}$ 因此 $5x=30$ ， $x=6$ 。

正确答案是 C。

Let the smallest number be $x$ . Then the largest is $4x$ . Since the three numbers are equally spaced, the middle number is the average of the smallest and largest: $\frac{x+4x}{2}=15.$ Thus $5x=30$ , so $x=6$ .

Thus, the correct answer is C.

7.

万维网在二十世纪九十年代刚流行时，下载速度最高约为每秒 $56$ 千比特。以这个速度下载一首 $4.2$ 兆字节的歌曲大约需要多少分钟？（注意一兆字节等于 $8000$ 千比特。）

When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about $56$ kilobits per second. Approximately how many minutes would the download of a $4.2$ -megabyte song have taken at that speed? (Note that there are $8000$ kilobits in a megabyte.)

$0.6$

$10$

$1800$

$7200$

$36000$

答案：B

知识点：单位换算速率

难度评级：960

小提示：

先把兆字节转换成千比特

Convert megabytes to kilobits first

大提示：

用 $4.2\cdot8000$ 千比特除以每秒 $56$ 千比特，得到秒数

$4.2\cdot8000$ kilobits divided by $56$ kilobits per second gives seconds

解答：

歌曲大小为 $4.2\cdot8000=33{,}600$ 千比特。以每秒 $56$ 千比特下载，需要 $\frac{33{,}600}{56}=600$ 秒，即 $10$ 分钟。

正确答案是 B。

The song has $4.2\cdot8000=33{,}600$ kilobits. At $56$ kilobits per second, the download takes $\frac{33{,}600}{56}=600$ seconds, which is $10$ minutes.

Thus, the correct answer is B.

8.

下列表达式的值是多少？

$\dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}$

What is the value of:

$\dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}$

$\displaystyle \dfrac{1}{462}$

$\displaystyle \dfrac{1}{231}$

$\displaystyle \dfrac{1}{132}$

$\displaystyle \dfrac{2}{213}$

$\displaystyle \dfrac{1}{22}$

答案：B

知识点：裂项相消分数

难度评级：1020

小提示：

大多数分子和分母因子会相消

Most numerator and denominator factors cancel

大提示：

相消后，分子只剩 $1\cdot2$

After cancellation, only $1\cdot2$ remains on top

解答：

从 $3$ 到 $20$ 的每个整数都既作为分母出现一次，也作为分子出现一次，所以全部相消。

相消后，分子只剩 $1$ 和 $2$ ，分母只剩 $21$ 和 $22$ 。

剩下的分数是 $\dfrac{1 \cdot 2}{21 \cdot 22}$ ，化简为 $\dfrac{2}{462} = \dfrac{1}{231}$ 。

正确答案是 B。

Since every integer from $3$ to $20$ occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only $1$ and $2$ left as numerators and $21$ and $22$ left as denominators.

The remaining fraction is $\dfrac{1 \cdot 2}{21 \cdot 22} .$ This simplifies to $\dfrac{2}{462} = \dfrac{1}{231}$

Thus, the correct answer is B.

9.

一杯沸水（ $212^\circ$ F）放在一个温度保持 $68^\circ$ F 的房间中冷却。假设水温与室温的差每 $5$ 分钟减半。 $15$ 分钟后，水温是多少华氏度？

A cup of boiling water ( $212^\circ$ F) is placed to cool in a room whose temperature remains constant at $68^\circ$ F. Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?

$77$

$86$

$92$

$98$

$104$

答案：B

知识点：等比数列

难度评级：900

小提示：

跟踪水温与室温的差

Track the difference from room temperature

大提示：

$15$ 分钟内，这个差减半 $3$ 次

In $15$ minutes, the difference is halved $3$ times

解答：

当前水温和室温的差为 $212-68 = 144$ 。

$15$ 分钟包含三个 $5$ 分钟，所以温差会减半 $3$ 次。

也就是说温差乘以 $(\dfrac{1}{2})^3 = \dfrac{1}{8}$ ，新的温差是 $\dfrac{1}{8}\cdot144=18$ 。水温为 $68+18=86$ 。

正确答案是 B。

The current difference is $212-68 = 144 .$

Since we have $15$ minutes while halving every $5$ minutes, we halve the difference $3$ times.

This means the difference is multiplied by $(\dfrac{1}{2})^3 = \dfrac{1}{8} ,$ so our new difference is $\dfrac{1}{8}\cdot144=18.$ This makes our final temperature $68+18=86.$

Thus, the correct answer is B.

10.

一个晴天，Ling 决定去山里徒步。她上午 $8$ 点离开家，以每小时 $45$ 英里的恒定速度开车，上午 $10$ 点到达步道入口。徒步 $3$ 小时后，Ling 以每小时 $60$ 英里的恒定速度开车回家。下列哪幅图最能表示整个行程中 Ling 的车与她家的距离？

One sunny day, Ling decided to take a hike in the mountains. She left her house at $8$ a.m., drove at a constant speed of $45$ miles per hour, and arrived at the hiking trail at $10$ a.m. After hiking for $3$ hours, Ling drove home at a constant speed of $60$ miles per hour. Which of the following graphs best illustrates the distance between Ling’s car and her house over the course of her trip?

答案：E

知识点：数据与图表解读路程、速度与时间

难度评级：900

小提示：

由去程开车先求家到步道入口的距离

Find the distance to the trail from the first drive

大提示：

回程速度更快，所以用时比去程少

The return drive is faster, so it takes less time than the outward drive

解答：

她以每小时 $45$ 英里的速度开 $2$ 小时，所以步道入口离家 $90$ 英里。图像应从 $0$ 英里上升到 $90$ 英里，对应上午 $8$ 点到 $10$ 点，然后在徒步的 $3$ 小时内保持水平。

她下午 $1$ 点返程。返程路程为 $90$ 英里，以每小时 $60$ 英里的速度行驶需要 $1.5$ 小时，所以下午 $2{:}30$ 到家。符合的是选项 E。

正确答案是 E。

She drives $45$ miles per hour for $2$ hours, so the trail is $90$ miles from her house. The graph must rise from $0$ to $90$ miles between $8$ AM and $10$ AM, then stay flat for the $3$ -hour hike.

She starts home at $1$ PM. Driving $90$ miles at $60$ miles per hour takes $1.5$ hours, so she gets home at $2{:}30$ PM. This matches choice E.

Thus, the correct answer is E.

11.

驴 Henry 有一根很长的意大利面。他咬了若干口，每次都从某一段意大利面的中间吃掉 $3$ 英寸。最后他有 $10$ 段意大利面，总长度为 $17$ 英寸。他最开始的那根意大利面长多少英寸？

Henry the donkey has a very long piece of pasta. He takes a number of bites of pasta, each time eating $3$ inches of pasta from the middle of one piece. In the end, he has $10$ pieces of pasta whose total length is $17$ inches. How long, in inches, was the piece of pasta he started with?

$34$

$38$

$41$

$44$

$47$

答案：D

知识点：不变量

难度评级：1020

小提示：

每咬一口，意大利面段数增加 $1$

Each bite increases the number of pasta pieces by $1$

大提示：

最后有 $10$ 段，所以 Henry 咬了 $9$ 口

To end with $10$ pieces, Henry made $9$ bites

解答：

从一整段变成 $10$ 段，需要咬 $9$ 口。咬了 $9$ 口，每口吃掉 $3$ 英寸，所以共吃掉 $27$ 英寸。

已经吃掉 $27$ 英寸，最后还剩 $17$ 英寸，因此原长为 $27+17 = 44$ 英寸。

正确答案是 D。

Since there are $10$ pieces, there were $9$ locations where a bite was made. Since we have $9$ bites and $3$ inches are removed per bite, a total of $27$ inches were removed.

With $27$ inches removed and $17$ inches remaining, we know we started with $27+17 = 44$ inches.

Thus, the correct answer is D.

12.

旋转下图中两个转盘上的指针。令 $N$ 等于转盘 A 上的数的 $10$ 倍，再加上转盘 B 上的数。 $N$ 是完全平方数的概率是多少？

The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner A, added to the number on Spinner B. What is the probability that $N$ is a perfect square number?

$\displaystyle \dfrac{1}{16}$

$\displaystyle \dfrac{1}{8}$

$\displaystyle \dfrac{1}{4}$

$\displaystyle \dfrac{3}{8}$

$\displaystyle \dfrac{1}{2}$

答案：B

知识点：基本概率完全平方数

难度评级：1100

小提示：

转盘 A 给十位数字，转盘 B 给个位数字

Spinner A gives the tens digit and spinner B gives the ones digit

大提示：

检查哪些两位平方数的十位是 $5,6,7,8$ ，个位是 $1,2,3,4$

Check which two-digit squares have tens digit $5,6,7,8$ and ones digit $1,2,3,4$

解答：

转盘 A 决定十位数字，转盘 B 决定个位数字。共有 $4\cdot4=16$ 个等可能的两位数。

在 $51$ 到 $84$ 之间，可能的完全平方数是 $64$ 和 $81$ ，且二者都可以由转盘得到。因此概率为 $\frac{2}{16}=\frac{1}{8}\text{。}$

正确答案是 B。

Spinner A gives the tens digit and Spinner B gives the ones digit. There are $4\cdot4=16$ equally likely two-digit numbers.

The possible perfect squares between $51$ and $84$ are $64$ and $81$ , and both can occur. Thus the probability is $\frac{2}{16}=\frac{1}{8}.$

Thus, the correct answer is B.

13.

有多少个正整数可以填入下面句子的空格？

“一个正整数比另一个正整数的两倍多 ___，并且这两个数的和是 $28$ 。”

How many positive integers can fill the blank in the sentence below?

“One positive integer is ___ more than twice another, and the sum of the two numbers is $28$ ”

$6$

$7$

$8$

$9$

$10$

答案：D

知识点：一次方程区间内整数计数

难度评级：1100

小提示：

设较小的正整数为 $x$

Let the smaller of the two positive integers be $x$

大提示：

若空格为 $c$ ，两个数可以写成 $x$ 和 $2x+c$

If the blank is $c$ , the two numbers can be $x$ and $2x+c$

解答：

设较小的数为 $x$ ，空格中的数为 $c$ 。那么另一个数是 $2x+c$ ，其中 $x$ 和 $c$ 都是正整数。

由两数之和得 $x+(2x+c)=28$ ，所以 $c=28-3x$ 。要使 $c$ 为正数，需要 $28-3x>0$ ，即 $x<\frac{28}{3}$ 。

因此 $x$ 可以是从 $1$ 到 $9$ 的任意整数，对应 $9$ 个可能的空格值。

正确答案是 D。

Let the smaller number be $x$ , and let the blank be $c$ . Then the other number is $2x+c$ , where both $x$ and $c$ are positive integers.

The sum condition gives $x+(2x+c)=28$ , so $c=28-3x$ . For $c$ to be positive, $28-3x>0$ , so $x<\frac{28}{3}$ .

Thus $x$ can be any integer from $1$ through $9$ , giving $9$ possible values of the blank.

Thus, the correct answer is D.

14.

单词 BEEKEEPER 的字母有多少种重新排列方式，使得不会有两个或更多个 E 连在一起？

In how many ways can the letters in BEEKEEPER be rearranged so that two or more E’s do not appear together?

$1$

$4$

$12$

$24$

$120$

答案：D

知识点：有限制的排列排列

难度评级：1190

小提示：

这个词有 $5$ 个 E 和 $4$ 个其他字母

There are $5$ E’s and $4$ other letters

大提示：

四个非 E 字母必须填入五个 E 之间的四个空隙

The four non-E letters must fill the four gaps between the E’s

解答：

有 $5$ 个 E 和 $4$ 个非 E 字母 B、K、P、R。为了不让两个 E 相邻，唯一可能的排列形式是 $E\_E\_E\_E\_E\text{，}$ 四个非 E 字母分别填入四个空隙。

B、K、P、R 可以用 $4!=24$ 种方式排列。

正确答案是 D。

The word has $5$ E’s and $4$ other letters: B, K, P, and R. To keep no two E’s adjacent, the only possible pattern is $E\_E\_E\_E\_E,$ with the four non-E letters in the four gaps.

The letters B, K, P, and R can be arranged in those gaps in $4!=24$ ways.

Thus, the correct answer is D.

15.

Laszlo 上网购买黑胡椒，找到了三十种不同的黑胡椒选项，重量和价格如下方散点图所示。按盎司计算，哪种重量的黑胡椒每盎司价格最低？

Laszlo went online to shop for black pepper and found thirty different black pepper options varying in weight and price, shown in the scatter plot below. In ounces, what is the weight of the pepper that offers the lowest price per ounce?

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：数据与图表解读斜率最优化

难度评级：1140

小提示：

比较价格除以重量

Compare price divided by weight

大提示：

对每种重量，只需考虑该重量中价格最低的点

For each weight, only the lowest plotted price can be best

解答：

对于每种重量，只有价格最低的那个点才可能给出最低的每盎司价格。

在 $1$ 盎司时，最低价格超过 $1$ 美元，所以每盎司价格超过 $1$ 美元。

在 $2$ 盎司时，最低价格为 $2$ 美元，所以每盎司价格为 $1$ 美元。

在 $3$ 盎司时，最低价格约为 $2.5$ 美元，所以每盎司价格约为 $\frac{2.5}{3}$ ，小于 $1$ 美元。

在 $4$ 盎司时，最低价格接近 $3.9$ 美元，每盎司约为 $\frac{3.9}{4}$ ，仍高于 $3$ 盎司选项。

在 $5$ 盎司时，最低价格接近 $4.5$ 美元，每盎司约为 $\frac{4.5}{5}$ ，也高于 $3$ 盎司选项。

所以每盎司价格最低的是 $3$ 盎司。

正确答案是 C。

For each weight, only the lowest-price point at that weight can give the lowest price per ounce.

At $1$ ounce, the lowest price is more than $1$ dollar, so the price per ounce is more than $1$ .

At $2$ ounces, the lowest price is $2$ dollars, so the price per ounce is $1$ .

At $3$ ounces, the lowest price is about $2.5$ dollars, so the price per ounce is about $\frac{2.5}{3}$ , less than $1$ .

At $4$ ounces, the lowest price is close to $3.9$ dollars, so the price per ounce is close to $\frac{3.9}{4}$ , still larger than the $3$ -ounce option.

At $5$ ounces, the lowest price is close to $4.5$ dollars, so the price per ounce is close to $\frac{4.5}{5}$ , also larger than the $3$ -ounce option.

The price per ounce is lowest at $3$ ounces.

Thus, the correct answer is C.

16.

四个数排成一行。前两个数的平均数是 $21$ ，中间两个数的平均数是 $26$ ，后两个数的平均数是 $30$ 。第一个数和最后一个数的平均数是多少？

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers?

$24$

$25$

$26$

$27$

$28$

答案：B

知识点：平均数方程组

难度评级：1100

小提示：

把每个平均数转换成对应的和

Convert each average into a sum

大提示：

先求 $a+b+c+d$ ，再减去 $b+c$

Find $a+b+c+d$ , then subtract $b+c$

解答：

设四个数依次为 $a,b,c,d$ 。

因为 $a$ 和 $b$ 的平均数是 $21$ ，所以它们的和是 $21\cdot2 = 42$ 。
因为 $c$ 和 $d$ 的平均数是 $30$ ，所以它们的和是 $30\cdot2 = 60$ 。
也就是 $a+b = 42$ ， $c+d = 60$ ，从而 $\begin{align*} a+b+c+d &= 42 + 60\\ &= 102 \end{align*}$

现在中间两个数 $b$ 和 $c$ 的平均数是 $26$ ，所以它们的和是 $26\cdot2=52$ 。因此 $b+c = 52$ 。

用所有数的总和减去这个结果，得 $a+d =50$ 。
所以第一个和最后一个数的平均数是 $\dfrac{a+d}{2}=25$ ，答案为 $25$ 。

正确答案是 B。

Let the numbers be $a,b,c,d$ in order.

Since the average of $a$ and $b$ is $21,$ we know their sum is $21\cdot2 = 42.$
Since the average of $c$ and $d$ is $30,$ we know their sum is $30\cdot2 = 60.$
Since $a+b = 42$ and $c+d = 60 ,$ we know $\begin{align*} a+b+c+d &= 42 + 60\\ &= 102 \end{align*}$

Now, with the average of $b$ and $c$ being $26,$ we know their sum is $26\cdot2=52.$ This means $b+c = 52.$

Subtracting this result from the sum of all the terms yields $a+d =50.$
Since $\dfrac{a+d}{2}=25,$ our answer is $25.$

Thus, the correct answer is B.

17.

若 $n$ 是正偶数，双阶乘记号 $n!!$ 表示从 $2$ 到 $n$ 的所有偶数的乘积。例如：

$8!! = 2 \times 4 \times 6 \times 8$

下列和的个位数字是多少？

$\begin{aligned} &2!! + 4!! + 6!! + \cdots \\ &\quad {}+ 2018!! + 2020!! + 2022!! \end{aligned}$

If $n$ is an even positive integer, the double factorial notation $n!!$ represents the product of all the even integers from $2$ to $n.$ For example:

$8!! = 2 \times 4 \times 6 \times 8$

What is the units digit of the following sum?

$\begin{aligned} &2!! + 4!! + 6!! + \cdots \\ &\quad {}+ 2018!! + 2020!! + 2022!! \end{aligned}$

$0$

$2$

$4$

$6$

$8$

答案：B

知识点：阶乘个位数字

难度评级：1190

小提示：

从 $10!!$ 起，任何偶数双阶乘都含有因子 $10$

Any even double factorial at least $10!!$ has a factor of $10$

大提示：

只有 $2!!,4!!,6!!$ 和 $8!!$ 会影响个位数字

Only $2!!,4!!,6!!,$ and $8!!$ can affect the units digit

解答：

对于每个满足 $n\ge10$ 的偶数，乘积 $n!!$ 都含有因子 $10$ ，所以该项的个位数字是 $0$ 。只有前四项会影响个位数字。

这四项的和为 $\begin{gathered} 2!!+4!!+6!!+8!!\\ =2+8+48+384\\ =442 \end{gathered}\text{。}$ 因此整个和的个位数字为 $2$ 。

正确答案是 B。

For every even $n\ge10,$ the product $n!!$ contains a factor of $10,$ so that term has units digit $0.$ Only the first four terms can affect the units digit.

They sum to $\begin{gathered} 2!!+4!!+6!!+8!!\\ =2+8+48+384\\ =442. \end{gathered}$ Therefore the full sum has units digit $2.$

Thus, the correct answer is B.

18.

一个矩形四条边的中点分别是 $(-3, 0)$ 、 $(2, 0)$ 、 $(5, 4)$ 、 $(0, 4)$ 。这个矩形的面积是多少？

The midpoints of the four sides of a rectangle are $(-3, 0),$ $(2, 0),$ $(5, 4),$ and $(0, 4).$ What is the area of the rectangle?

$20$

$25$

$40$

$50$

$80$

答案：C

知识点：坐标几何平行四边形中点

难度评级：1310

小提示：

这些中点形成矩形内部的一个平行四边形

The given midpoints form a parallelogram inside the rectangle

大提示：

中点平行四边形面积是原矩形面积的一半

The midpoint parallelogram has half the area of the original rectangle

解答：

四个中点形成一个平行四边形。它的一条水平底边长度为 $2-(-3)=5$ ，高度为 $4$ ，面积为 $5\cdot4=20$ 。

连接任意矩形四边中点形成的平行四边形面积等于原矩形面积的一半，所以原矩形面积为 $2\cdot20=40$ 。

正确答案是 C。

The four given midpoints form a parallelogram. Its horizontal base has length $2-(-3)=5$ , and its height is $4$ , so its area is $5\cdot4=20$ .

For any rectangle, the parallelogram formed by joining the side midpoints has half the area of the rectangle. Therefore the rectangle’s area is $2\cdot20=40$ .

Thus, the correct answer is C.

19.

Ramos 老师给 $20$ 名学生进行了一次测试。下面的点图显示了成绩分布。

后来 Ramos 老师发现某题评分有误。他重新评分，给一些学生额外加了 $5$ 分，使得测试成绩的中位数提高到 $85$ 。至少有多少名学生得到了额外分数？

（注意，中位数等于中间 $2$ 个成绩的平均数；这里 $20$ 个成绩按从小到大排列。）

Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores.

Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85.$ What is the minimum number of students who received extra points?

(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)

$2$

$3$

$4$

$5$

$6$

答案：C

知识点：中位数（数据）数据与图表解读

难度评级：1370

小提示：

中位数由第 $10$ 个和第 $11$ 个成绩决定

The median is controlled by the $10$ th and $11$ th scores

大提示：

重新评分前后，数一数至少为 $85$ 的成绩有多少个

Count how many scores are at least $85$ before and after regrading

解答：

所有成绩都是 $5$ 的倍数，加上 $5$ 分后仍然如此。如果第 $10$ 和第 $11$ 个成绩的平均数是 $85$ ，但两者不都是 $85$ ，那么第 $11$ 个成绩至少为 $90$ 。这就要求至少有 $10$ 个成绩不低于 $90$ 。

可是重新评分后，只有原来不低于 $85$ 的成绩才可能达到 $90$ 。点图中这样的成绩只有 $7$ 个，所以中间两个成绩必须都是 $85$ 。

点图中原来有 $7$ 个不低于 $85$ 的成绩。要使第 $10$ 和第 $11$ 个成绩都为 $85$ ，至少要有 $11$ 个成绩不低于 $85$ 。把四个 $80$ 分提高到 $85$ 分既是必要的，也是充分的。

所以正确答案是 C。

All scores are multiples of $5,$ and adding $5$ preserves that fact. If the $10$ th and $11$ th scores average $85$ but are not both $85,$ then the $11$ th score must be at least $90.$ That would require at least $10$ scores of $90$ or more.

However, even after regrading, only the original scores of $85$ or more can reach $90.$ The plot has only $7$ such scores, so the two middle scores must both be $85.$

The plot initially has $7$ scores of at least $85.$ To make the $10$ th and $11$ th scores both $85,$ there must be at least $11$ scores of at least $85.$ Raising four of the $80$ ’s to $85$ is both necessary and sufficient.

Thus, the correct answer is C.

20.

下方网格要填入整数，使得每一行的和与每一列的和都相同。四个数字缺失。左下角的数字 $x$ 大于另外三个缺失数字。 $x$ 的最小可能值是多少？

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x?$

$-1$

$5$

$6$

$8$

$9$

答案：D

知识点：幻方不等式

难度评级：1430

小提示：

已完整的行给出每行和每列的共同和

The completed row and column sums must match the top row

大提示：

用 $x$ 表示另外三个缺失项

Express the other missing entries in terms of $x$

解答：

由顶行可知每行、每列的和都必须为 $12$ 。

在第一列中， $x$ 上方的缺失数为 $14-x$ 。在底行中， $x$ 右边的缺失数为 $4-x$ 。中间行剩下的缺失数为 $x-1$ 。

因为 $x$ 大于另外三个缺失数，需要 $x>14-x$ 、 $x>4-x$ 、 $x> x-1$ 。最强条件是 $x>7$ ，所以最小整数为 $8$ 。

正确答案是 D。

Adding the numbers in the top row shows that every row and column must have sum $12$ .

In the first column, the missing number above $x$ is $14-x$ . In the bottom row, the missing number to the right of $x$ is $4-x$ . In the middle row, the remaining missing number is $x-1$ .

Since $x$ is larger than the other missing numbers, we need $x>14-x$ , $x>4-x$ , and $x> x-1$ . The strongest condition is $x>7$ , so the smallest possible integer value is $8$ .

Thus, the correct answer is D.

21.

Steph 在一场比赛上半场投进 $15$ 球，共出手 $20$ 次；下半场投进 $10$ 球，共出手 $10$ 次。Candace 上半场出手 $12$ 次，下半场出手 $18$ 次。在每个半场中，Steph 的命中率都高于 Candace。令人惊讶的是，她们最终总命中率相同。Candace 下半场比上半场多进了多少球？

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?

$7$

$8$

$9$

$10$

$11$

答案：C

知识点：百分数不等式

难度评级：1460

小提示：

Steph 总共投进 $25$ 球，共出手 $30$ 次。

Steph made $25$ baskets in $30$ attempts overall

大提示：

Candace 也必须总共投进 $25$ 球。

Candace must also have made $25$ baskets total

解答：

Steph 总共进了 $25$ 球，出手 $30$ 次。Candace 也出手了 $12+18=30$ 次，且两人的总命中率相同，所以 Candace 也进了 $25$ 球。

设 Candace 上半场进 $f$ 球，下半场进 $s$ 球，则 $f+s=25$ 。

Steph 上半场命中率为 $\frac{15}{20}=\frac34$ ，所以 $\frac{f}{12}<\frac34$ ，得到 $f<9$ 。Steph 下半场命中率为 $1$ ，所以 $s<18$ 。

因此 $f\le8$ 且 $s\le17$ 。又 $f+s=25$ ，唯一可能是 $f=8$ 、 $s=17$ ，所以 $s-f=9$ 。

正确答案是 C。

Steph made $25$ baskets in $30$ attempts. Candace also took $12+18=30$ attempts, and their overall percentages were the same, so Candace also made $25$ baskets.

Let $f$ be Candace’s first-half baskets and $s$ be her second-half baskets. Then $f+s=25$ .

Steph’s first-half percentage was $\frac{15}{20}=\frac34$ , so $\frac{f}{12}<\frac34$ , giving $f<9$ . Steph’s second-half percentage was $1$ , so $s<18$ .

Thus $f\le8$ and $s\le17$ . Since $f+s=25$ , the only possibility is $f=8$ and $s=17$ , so $s-f=9$ .

Thus, the correct answer is C.

22.

公交车从一个站开到下一站需要 $2$ 分钟，并在每站等 $1$ 分钟让乘客上车。Zia 从一个公交站走到下一站需要 $5$ 分钟。当 Zia 到达一个公交站时，如果公交车在前一站，或已经离开前一站，那么她会等公交车。否则她会继续走向下一站。假设公交车和 Zia 同时朝图书馆方向出发，公交车在 Zia 后面 $3$ 站。多少分钟后 Zia 会上公交车？

A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus?

$17$

$19$

$20$

$21$

$23$

答案：A

知识点：过程模拟路程、速度与时间

难度评级：1510

小提示：

只在 Zia 到达公交站时检查她的决定

Check Zia’s decision only when she reaches a stop

大提示：

比较 $5$ 、 $10$ 、 $15$ 分钟时两者的位置

Compare positions at $5$ , $10$ , and $15$ minutes

解答：

公交车需要 $2$ 分钟开到下一站，再等 $1$ 分钟，所以每 $3$ 分钟完成一个站距周期。

Zia 只在到达公交站时做决定，也就是每 $5$ 分钟一次。以公交车出发站为第零站，则 Zia 从第 $3$ 站出发。

$5$ 分钟后，Zia 在第 $4$ 站，公交车在第 $2$ 站等待，她继续走。 $10$ 分钟后，Zia 在第 $5$ 站，公交车在第 $3$ 站和第 $4$ 站之间，她继续走。 $15$ 分钟后，Zia 在第 $6$ 站，公交车在第 $5$ 站，正好是前一站，所以她停下等待。

公交车再用 $2$ 分钟到达她所在的站，所以她在 $17$ 分钟后上车。

正确答案是 A。

The bus takes $2$ minutes to drive to the next stop and then waits $1$ minute, so it moves through one-stop cycles every $3$ minutes.

Zia makes a decision only when she reaches a stop, every $5$ minutes. Measure stops from the bus’s starting stop. Zia starts at stop $3$ .

After $5$ minutes, Zia is at stop $4$ , while the bus is waiting at stop $2$ , so she keeps walking. After $10$ minutes, Zia is at stop $5$ , while the bus is between stops $3$ and $4$ , so she keeps walking. After $15$ minutes, Zia is at stop $6$ , while the bus is at stop $5$ , the previous stop, so she waits.

The bus then takes $2$ more minutes to reach her stop, so she boards after $17$ minutes.

Thus, the correct answer is A.

23.

在九个小方格中，每格放入一个 $\bigtriangleup$ 或 $\bigcirc$ ，这些方格组成一个 $3 \times 3$ 网格。下面显示了一个有三个 $\bigtriangleup$ 连成一线的示例。

有多少种配置同时有三个 $\bigtriangleup$ 连成一线，并且有三个 $\bigcirc$ 连成一线？

A $\bigtriangleup$ or $\bigcirc$ is placed in each of the nine squares in a $3 \times 3$ grid. Shown below is a sample configuration with three $\bigtriangleup$ ’s in a line.

How many configurations will have three $\bigtriangleup$ ’s in a line and three $\bigcirc$ ’s in a line?

$39$

$42$

$78$

$84$

$96$

答案：D

知识点：分类讨论对称性

难度评级：1670

小提示：

按网格中三角形的个数分类

Split by how many triangles are in the grid

大提示：

正好 $3$ 个三角形时，对角线三角形线不行

With exactly $3$ triangles, diagonal triangle lines do not work

解答：

设三角形个数为 $k$ 。要同时有三角形线和圆形线，必须有 $3\le k\le6$ 。

若 $k=3$ ，三角形必须成一条线。行或列有效，因为剩下的六个圆形中仍有完整的一行或一列；对角线无效，因为其余位置没有完整的圆形线。因此共有 $6$ 种 $k=3$ 的配置。由对称性，也有 $6$ 种 $k=6$ 的配置。

若 $k=4$ ，先选一条三角形线。若这条线是行或列，额外的三角形有 $6$ 个位置可选，每一种都可行，因为总有一条平行的行或列全是圆形，共 $6\cdot6=36$ 种；若这条线是对角线，则无论额外的三角形放在哪里，都不能留下完整的圆形线。由对称性， $k=5$ 也有 $36$ 种。

总数为 $6+36+36+6=84$ 。

正确答案是 D。

Let $k$ be the number of triangles. To have both a triangle line and a circle line, we need $3\le k\le6$ .

If $k=3$ , the three triangles must form a line. A row or column works, because the remaining six circles contain a full circle row or column. A diagonal does not work, because its complement contains no full line of circles. Thus there are $6$ configurations for $k=3$ . By symmetry, there are also $6$ configurations for $k=6$ .

If $k=4$ , choose the triangle line and then the extra triangle. If the line is a row or column, all $6$ possible extra positions work, because one parallel row or column remains all circles. This gives $6\cdot6=36$ configurations. If the triangle line is a diagonal, no extra position leaves a full circle line. By symmetry, $k=5$ also gives $36$ configurations.

The total number of configurations is $6+36+36+6=84$ .

Thus, the correct answer is D.

24.

下图显示了多边形 $ABCDEFGH$ ，由矩形和直角三角形组成。把它剪下并沿虚线折叠后，这个多边形形成一个三棱柱。已知

$AH = EF = 8$

且

$GH = 14\text{。}$

这个棱柱的体积是多少？

The figure below shows a polygon $ABCDEFGH,$ consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that:

$AH = EF = 8$

and

$GH = 14.$

What is the volume of the prism?

$112$

$128$

$192$

$240$

$288$

答案：C

知识点：展开图（立体几何）体积

难度评级：1510

小提示：

找出棱柱的直角三角形底面

Find the right-triangle base of the prism

大提示：

折叠网给出底面两条直角边为 $6$ 和 $8$ ，棱柱长度为 $8$

The folded net gives a base with legs $6$ and $8$ , and prism length $8$

解答：

折叠后匹配边长给出 $GF=EF=8$ 。因为 $GFCB$ 是矩形，所以 $BC=8$ 。折叠时 $AB$ 与 $BC$ 对应，所以 $AB=8$ 。在矩形 $HJBA$ 中， $HJ=8$ ，且 $BJ$ 等于 $AH=8$ 。

因为 $GH=14$ ，所以 $GJ=14-8=6$ 。一个三角形底面是直角三角形 $BJG$ ，面积为 $\frac{6\cdot8}{2}=24\text{。}$ 棱柱长度为 $GF=8$ ，所以体积为 $24\cdot8=192$ 。

正确答案是 C。

When the net is folded, the matching side lengths give $GF=EF=8$ . Since $GFCB$ is a rectangle, $BC=8$ . The fold identifies $AB$ with $BC$ , so $AB=8$ . In rectangle $HJBA$ , this gives $HJ=8$ , and the opposite side $BJ$ equals $AH=8$ .

Since $GH=14$ , we have $GJ=14-8=6$ . Thus one triangular base of the prism is right triangle $BJG$ , with area $\frac{6\cdot8}{2}=24.$ The prism length is $GF=8$ , so the volume is $24\cdot8=192$ .

Thus, the correct answer is C.

25.

一只蟋蟀在 $4$ 片叶子之间随机跳跃，每次跳到另外 $3$ 片叶子中的一片，概率相等。跳 $4$ 次后，蟋蟀回到起始叶子的概率是多少？

A cricket randomly hops between $4$ leaves, on each turn hopping to one of the other $3$ leaves with equal probability. After $4$ hops, what is the probability that the cricket has returned to the leaf where it started?

$\displaystyle \dfrac{2}{9}$

$\displaystyle \dfrac{19}{80}$

$\displaystyle \dfrac{20}{81}$

$\displaystyle \dfrac{1}{4}$

$\displaystyle \dfrac{7}{27}$

答案：E

知识点：递推概率对称性

难度评级：1670

小提示：

只跟踪蟋蟀是否在起始叶子上

Track only whether the cricket is on the starting leaf

大提示：

如果蟋蟀不在起点，下一跳以 $\frac13$ 的概率回到起点

If the cricket is not at the start, the next hop returns with probability $\frac13$

解答：

设 $p_n$ 为跳 $n$ 次后在起始叶子上的概率， $p_0=1$ 。

若蟋蟀在起点，下一跳一定离开；若不在起点，下一跳有 $3$ 个等可能选择，其中一个会回到起点。因此 $p_{n+1}=\frac{1-p_n}{3}\text{。}$

逐次计算得 $\begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27} \end{gathered}\text{。}$

正确答案是 E。

Let $p_n$ be the probability that the cricket is on its starting leaf after $n$ hops. We have $p_0=1$ .

If the cricket is on the starting leaf, the next hop must leave it. If the cricket is not on the starting leaf, exactly one of the $3$ possible hops returns to the start. Therefore $p_{n+1}=\frac{1-p_n}{3}.$

Thus $\begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27}. \end{gathered}$

Thus, the correct answer is E.