2008 AMC 8 详解 | LIVE by Po-Shen Loh

1.

苏珊有 $\$50$ 可以在嘉年华会上花。她花了 $\$12$ 买食物，又花了这个数的两倍去玩游乐项目。她还剩多少美元可以花？

Susan had $\$50$ to spend at the carnival. She spent $\$12$ on food and twice as much on rides. How many dollars did she have left to spend?

$12$

$14$

$26$

$38$

$50$

知识点：钱币

难度评级：370

小提示：

先求苏珊在游乐项目上花了多少钱，再从 $50$ 中减去

Find how much Susan spent on rides before subtracting from $50.$

大提示：

$\$12$ 的两倍是 $\$24$

Twice as much as $\$12$ is $\$24$ .

解答：

苏珊在游乐项目上花了 $2\cdot 12=24$ 美元，所以她一共花了 $12+24=36$ 美元。

她还剩 $50-36=14$ 美元。

所以正确答案是 B。

Susan spent $2\cdot 12=24$ dollars on rides, so she spent $12+24=36$ dollars total.

She had $50-36=14$ dollars left.

Thus, B is the correct answer.

2.

十个字母的代码 $\text{BEST OF LUCK}$ 依次表示数字 $0-9$ 。代码单词 $\text{CLUE}$ 表示哪个 $4$ 位数？

The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9,$ in order. What $4$ -digit number is represented by the code word $\text{CLUE}?$

$\ 8671$

$\ 8672$

$\ 9781$

$\ 9782$

$\ 9872$

知识点：位值

难度评级：450

小提示：

把每个数字写在 BEST OF LUCK 中对应字母的下面

Write the digit under each letter in BEST OF LUCK.

大提示：

因为第一个字母表示 $0$ ，每个字母表示的数字比它在代码中的位置少一

Because the first letter represents $0,$ each letter represents one less than its position in the code.

解答：

代码 BEST OF LUCK 依次表示数字 $0$ 到 $9$ ，所以 $C=8$ 、 $L=6$ 、 $U=7$ 、 $E=1$ 。

因此 $\text{CLUE}=8671$ 。

所以正确答案是 A。

The code BEST OF LUCK represents the digits $0$ through $9$ in order, so $C=8$ , $L=6$ , $U=7$ , and $E=1$ .

Therefore, $\text{CLUE}=8671$ .

Thus, A is the correct answer.

3.

如果某年二月的 $13$ 号是星期五，那么二月 $1$ 号是星期几？

If February is a month that contains Friday the $13$ th, what day of the week is February $1?$

星期日

Sunday

星期一

Monday

星期三

Wednesday

星期四

Thursday

星期六

Saturday

知识点：日期与时间模运算

难度评级：560

小提示：

相隔一周的日期在同一个星期几

Dates one week apart fall on the same weekday.

大提示：

先把二月 $13$ 号与二月 $6$ 号比较，再倒数到二月 $1$ 号

Compare February $13$ to February $6,$ then count back to February $1.$

解答：

因为 $13$ 号是星期五，所以 $6$ 号也是星期五。 $1$ 号比这个星期五早 $5$ 天，因此是星期日。

所以正确答案是 A。

Since the $13$ th is a Friday, we know that the $6$ th is also a Friday. The $1$ st is $5$ days before the Friday, making it a Sunday.

Thus, the answer is A .

4.

如图，外面的等边三角形面积为 $16$ ，里面的等边三角形面积为 $1$ ，三个梯形全等。一个梯形的面积是多少？

In the figure, the outer equilateral triangle has area $16,$ the inner equilateral triangle has area $1,$ and the three trapezoids are congruent. What is the area of one of the trapezoids?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

知识点：面积梯形等边三角形

难度评级：720

小提示：

从外三角形面积中减去内三角形面积

Subtract the inner triangle area from the outer triangle area.

大提示：

剩余面积被三个全等梯形平均分

The remaining area is split equally among the three congruent trapezoids.

解答：

外三角形面积为 $16$ ，内三角形面积为 $1$ 。

因此三个全等梯形的总面积为 $16-1=15$ ，每个梯形的面积为 $15\div3=5$ 。

所以正确答案是 C。

The outer triangle has area $16$ , and the inner triangle has area $1$ .

So the three congruent trapezoids have total area $16-1=15$ , and each trapezoid has area $15\div3=5$ .

Thus, C is the correct answer.

5.

巴尼·施温注意到他的自行车里程表显示 $1441$ ，这是一个回文数，因为正读反读相同。当天又骑了 $4$ 小时、第二天骑了 $6$ 小时后，他注意到里程表显示另一个回文数 $1661$ 。他的平均速度是多少英里每小时？

Barney Schwinn notices that the odometer on his bicycle reads $1441,$ a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661.$ What was his average speed in miles per hour?

$\ 15$

$\ 16$

$\ 18$

$\ 20$

$\ 22$

知识点：路程、速度与时间

难度评级：770

小提示：

用两个里程表读数相减求出骑行的英里数

Find the miles ridden by subtracting the odometer readings.

大提示：

总骑行时间是 $4$ 小时加 $6$ 小时

The total riding time is the $4$ hours plus the $6$ hours.

解答：

里程表增加了 $1661-1441=220$ 英里。

巴尼一共骑了 $4+6=10$ 小时，所以平均速度为 $220\div10=22$ 英里每小时。

所以正确答案是 E。

The odometer increased by $1661-1441=220$ miles.

Barney rode for $4+6=10$ hours, so his average speed was $220\div10=22$ miles per hour.

Thus, E is the correct answer.

6.

图中阴影正方形的面积与非阴影正方形的面积之比是多少？

In the figure, what is the ratio of the area of the shaded squares to the area of the unshaded squares?

$\ 3:10$

$\ 3:8$

$\ 3:7$

$\ 3:5$

$\ 1:1$

知识点：面积比与比例

难度评级：900

小提示：

把中央的阴影正方形划分成与其他小正方形同样大小的单位块

Subdivide the central shaded square into unit-size pieces matching the others.

大提示：

统计阴影和非阴影的小正方形个数

Count shaded and unshaded equal-area small squares.

解答：

把中央阴影正方形细分后，图中共有 $16$ 个相等的小正方形。

其中 $6$ 个是阴影， $10$ 个不是阴影，所以所求比值为 $6:10=3:5$ 。

所以正确答案是 D。

After subdividing the central shaded square, the figure has $16$ equal small squares.

Of these, $6$ are shaded and $10$ are unshaded, so the desired ratio is $6:10=3:5$ .

Thus, D is the correct answer.

7.

如果 $\dfrac{3}{5}=\dfrac{M}{45}=\dfrac{60}{N}$ ，那么 $M+N$ 是多少？

If $\dfrac{3}{5}=\dfrac{M}{45}=\dfrac{60}{N},$ what is $M+N?$

$\ 27$

$\ 29$

$\ 45$

$\ 105$

$\ 127$

知识点：分数比与比例

难度评级：820

小提示：

利用 $\frac{M}{45}=\frac{3}{5}$ 求 $M$

Solve for $M$ using $\frac{M}{45}=\frac{3}{5}$ .

大提示：

利用 $\frac{60}{N}=\frac{3}{5}$ 求 $N$ ，然后相加

Solve for $N$ using $\frac{60}{N}=\frac{3}{5}$ , then add.

解答：

由 $\dfrac{3}{5}=\dfrac{M}{45}$ ，得 $M=\dfrac{3\cdot45}{5}=27$ 。

由 $\dfrac{3}{5}=\dfrac{60}{N}$ ，得 $3N=300$ ，所以 $N=100$ 。

因此 $M+N=27+100=127$ 。

所以正确答案是 E。

From $\dfrac{3}{5}=\dfrac{M}{45}$ , we get $M=\dfrac{3\cdot45}{5}=27$ .

From $\dfrac{3}{5}=\dfrac{60}{N}$ , we get $3N=300$ , so $N=100$ .

Therefore, $M+N=27+100=127$ .

Thus, E is the correct answer.

8.

助威俱乐部从一月到四月的糖果销售额如图所示。平均每月销售额是多少美元？

Candy sales of the Boosters Club for January through April are shown. What were the average sales per month in dollars?

$\ 60$

$\ 70$

$\ 75$

$\ 80$

$\ 85$

知识点：数据与图表解读平均数

难度评级：900

小提示：

从图表中读出四个月的柱形高度

Read the four monthly bar heights from the chart.

大提示：

平均值等于总销售额除以 $4$ 个月

Average means total sales divided by $4$ months.

解答：

四个月的销售额分别为 $100$ 、 $60$ 、 $40$ 和 $120$ 美元。

它们的平均值为 $\dfrac{100+60+40+120}{4}$ $=\dfrac{320}{4}$ $=80$ 美元。

所以正确答案是 D。

The four monthly sales are $100,$ $60,$ $40,$ and $120$ dollars.

Their average is $\dfrac{100+60+40+120}{4}$ $=\dfrac{320}{4}$ $=80$ dollars.

Thus, D is the correct answer.

9.

$2005$ 年，投资大亨塔米投入了 $\$100$ ，为期两年。第一年她的投资亏损 $15\%$ ，但第二年剩余投资增长了 $20\%$ 。这两年内塔米的投资变化是多少？

In $2005$ Tycoon Tammy invested $\$100$ for two years. During the first year her investment suffered a $15\%$ loss, but during the second year the remaining investment showed a $20\%$ gain. Over the two-year period, what was the change in Tammy’s investment?

亏损 $5\%$

$5\%$ loss

亏损 $2\%$

$2\%$ loss

增长 $1\%$

$1\%$ gain

增长 $2\%$

$2\%$ gain

增长 $5\%$

$5\%$ gain

知识点：百分数

难度评级：1040

小提示：

亏损 $15\%$ 后，剩下原投资的 $85\%$

After a $15\%$ loss, $85\%$ of the original investment remains.

大提示：

把 $20\%$ 的增长作用在减少后的金额上，而不是原金额上

Apply the $20\%$ gain to the reduced amount, not to the original amount.

解答：

$15\%$ 的亏损使投资从 $\$100$ 变为 $0.85\cdot\$100=\$85$ 。

$20\%$ 的增长再使其变为 $1.2\cdot\$85=\$102$ 。

投资从 $\$100$ 变为 $\$102$ ，也就是增长 $2\%$ 。

所以正确答案是 D。

The $15\%$ loss changes the investment from $\$100$ to $0.85\cdot\$100=\$85$ .

The $20\%$ gain then changes it to $1.2\cdot\$85=\$102$ .

The investment went from $\$100$ to $\$102$ , which is a $2\%$ gain.

Thus, D is the correct answer.

10.

A 房间里 $6$ 个人的平均年龄是 $40$ 。B 房间里 $4$ 个人的平均年龄是 $25$ 。如果把两组人合在一起，所有人的平均年龄是多少？

The average age of the $6$ people in Room A is $40.$ The average age of the $4$ people in Room B is $25.$ If the two groups are combined, what is the average age of all the people?

$\ 32.5$

$\ 33$

$\ 33.5$

$\ 34$

$\ 35$

知识点：平均数

难度评级：960

小提示：

把每个房间的平均年龄转化为年龄总和

Convert each room average into a total age.

大提示：

合并后的平均年龄等于合并后的年龄总和除以全部 $10$ 个人

The combined average is the combined total age divided by all $10$ people.

解答：

A 房间的年龄总和为 $6\cdot 40 = 240\text{。}$

B 房间的年龄总和为 $4\cdot 25 = 100\text{。}$

总和为 $240+100 = 340\text{。}$

因此平均年龄为 $\dfrac{340}{10} =34\text{。}$

所以正确答案是 D。

The sum of the ages in Room A is $6\cdot 40 = 240.$

The sum of the ages in Room B is $4\cdot 25 = 100.$

The total sum is $240+100 = 340.$

The average is therefore $\dfrac{340}{10} =34.$

Thus, the answer is D .

11.

林肯中学八年级的 $39$ 名学生中，每人养一只狗、一只猫，或狗和猫都养。有二十名学生养狗， $26$ 名学生养猫。有多少名学生狗和猫都养？

Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?

$\ 7$

$\ 13$

$\ 19$

$\ 39$

$\ 46$

知识点：容斥原理

难度评级：1000

小提示：

把养狗人数和养猫人数相加时，同时养两种宠物的学生被数了两次

Adding dog owners and cat owners counts students with both pets twice.

大提示：

用容斥原理，总人数就是并集大小

Use inclusion-exclusion with the total number of students as the union.

解答：

同时养两种动物的人数等于养猫人数加养狗人数，再减去至少养一种动物的人数。

因此同时养两种动物的人数为 $20+26-39 = 7\text{。}$

所以正确答案是 A。

The number of people that have both animals is equal to the number of people that own a cat plus the number of people that own a dog minus the number of people that own either.

Therefore, the number of people who own both is $20+26-39 = 7.$

Thus, the answer is A .

12.

一个球从 $3$ 米高处落下。第一次反弹时它升到 $2$ 米高。之后每次下落并反弹到前一次反弹高度的 $\frac{2}{3}$ 。第几次反弹时它不会升到 $0.5$ 米高？

A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?

$\ 3$

$\ 4$

$\ 5$

$\ 6$

$\ 7$

知识点：等比数列

难度评级：1150

小提示：

每次乘以 $\frac{2}{3}$ ，列出反弹高度

List the bounce heights by multiplying by $\frac{2}{3}$ each time.

大提示：

将第四次和第五次反弹高度与 $\frac{1}{2}$ 比较

Compare the fourth and fifth bounce heights with $\frac{1}{2}$ .

解答：

第一次反弹升到 $2$ 米。之后每次反弹都是前一次反弹高度的 $\dfrac{2}{3}$ 。

第四次反弹升到 $2\left(\dfrac{2}{3}\right)^3=\dfrac{16}{27}$ ，大于 $\dfrac{1}{2}$ 。

第五次反弹升到 $2\left(\dfrac{2}{3}\right)^4=\dfrac{32}{81}$ ，小于 $\dfrac{1}{2}$ 。

所以正确答案是 C。

The first bounce rises to $2$ meters. Each later bounce is $\dfrac{2}{3}$ of the previous bounce.

The fourth bounce rises to $2\left(\dfrac{2}{3}\right)^3=\dfrac{16}{27}$ , which is greater than $\dfrac{1}{2}$ .

The fifth bounce rises to $2\left(\dfrac{2}{3}\right)^4=\dfrac{32}{81}$ , which is less than $\dfrac{1}{2}$ .

Thus, C is the correct answer.

13.

哈曼先生想知道他要邮寄的三个箱子的总重量，单位为磅。不过唯一可用的秤在重量小于 $100$ 磅或大于 $150$ 磅时不准确。于是他把箱子按所有可能的两两组合称重，结果是 $122$ 、 $125$ 和 $127$ 磅。这三个箱子的总重量是多少磅？

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122,$ $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?

$\ 160$

$\ 170$

$\ 187$

$\ 195$

$\ 354$

知识点：方程组

难度评级：1060

小提示：

三个两箱重量之和中，每个箱子都恰好出现两次

Each box appears in exactly two of the three pair weights.

大提示：

把三个两箱重量相加，再除以 $2$

Add the three pair weights, then divide by $2.$

解答：

设三个箱子的重量为 $a$ 、 $b$ 、 $c$ 。已知 $\begin{cases}a+b = 122\\a+c = 125\\b+c = 127\end{cases}\text{。}$ 把这些式子相加得到 $2(a+b+c) =$ $122+125+127=374\text{。}$ 所以 $a+b+c = 187\text{。}$

所以正确答案是 C。

Let the weights be $a,$ $b,$ $c.$ We know $\begin{cases}a+b = 122\\a+c = 125\\b+c = 127.\end{cases}$ Adding all of this yields $2(a+b+c) =$ $122+125+127=374.$ This makes $a+b+c = 187.$

Thus, the answer is C .

14.

三个 A、三个 B 和三个 C 被放入九个格子中，使每一行和每一列都各含一个字母。若 A 放在左上角，共有多少种排列？

Three A’s, three B’s, and three C’s are placed in the nine spaces so that each row and column contain one of each letter. If A is placed in the upper left corner, how many arrangements are possible?

$\ 2$

$\ 3$

$\ 4$

$\ 5$

$\ 6$

知识点：有限制的排列分类讨论

难度评级：1240

小提示：

左上角的 A 固定后，先决定另外两个 A 可以放在哪里

After the top-left A is fixed, decide where the other two A entries can go.

大提示：

对每一种 A 的位置模式，第一行中的 B 和 C 有两种顺序

For each A-pattern, the top row can place B and C in two orders.

解答：

另外两个 A 必须各占剩余两行和剩余两列中的一个格子。三个 A 的位置有两种可能的对角模式。

对于任意一种 A 的位置模式，第一行剩余两个位置可以填成 B,C 或 C,B，之后其余格子都被确定。

因此共有 $2\cdot2=4$ 种排列。

所以正确答案是 C。

The other two A’s must occupy one square in each of the remaining two rows and columns. There are two possible diagonal patterns for the three A’s.

For either A-pattern, the two remaining positions in the top row can be filled as B,C or C,B, and then the rest of the grid is forced.

So there are $2\cdot2=4$ arrangements.

Thus, C is the correct answer.

15.

特蕾莎前 $8$ 场篮球比赛分别得了 $7$ 、 $4$ 、 $3$ 、 $6$ 、 $8$ 、 $3$ 、 $1$ 和 $5$ 分。第九场她得分少于 $10$ 分，并且九场比赛的场均得分是整数。类似地，第十场她得分少于 $10$ 分，并且 $10$ 场比赛的场均得分也是整数。她第九场和第十场得分的乘积是多少？

In Theresa’s first $8$ basketball games, she scored $7,$ $4,$ $3,$ $6,$ $8,$ $3,$ $1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

$\ 35$

$\ 40$

$\ 48$

$\ 56$

$\ 72$

知识点：平均数整除性

难度评级：1310

小提示：

先把特蕾莎前 $8$ 场的得分相加

First add Theresa’s scores from the first $8$ games.

大提示：

$9$ 场总分必须是 $9$ 的倍数， $10$ 场总分必须是 $10$ 的倍数

The $9$ -game total must be a multiple of $9,$ and the $10$ -game total must be a multiple of $10.$

解答：

她前 $8$ 场的得分和为 $37$ 。因为前 $9$ 场的平均分是整数，所以前 $9$ 场的总分是 $9$ 的倍数。

由于第九场得分少于 $10$ ，前 $9$ 场的总分在 $37$ 到 $47$ 之间，并且是 $9$ 的倍数，因此总分为 $45$ 。所以第 $9$ 场得分为 $45-37=8$ 。

特蕾莎前 $9$ 场的总分是 $45$ 。因为前 $10$ 场的平均分是整数，所以前 $10$ 场的总分是 $10$ 的倍数。

由于第十场得分少于 $10$ ，前 $10$ 场的总分在 $45$ 到 $55$ 之间，并且是 $10$ 的倍数，因此总分为 $50$ 。所以第 $10$ 场得分为 $50-45=5$ 。

因此乘积为 $8\cdot 5=40$ 。

所以正确答案是 B。

The sum of her first $8$ scores is $37.$ Since the average of the first $9$ scores is an integer, the sum of the first $9$ scores is a multiple of $9.$

Since the score is less than $10,$ the sum of the scores after $9$ games is between $37$ and $47,$ and is a multiple of $9,$ making the sum $45.$ Thus, the score of the $9$ th game is $45-37=8.$

The sum of Theresa’s first $9$ scores is $45.$ Since the average of the first $10$ scores is an integer, the sum of the first $10$ scores is a multiple of $10.$

Since the score is less than $10,$ the sum of the scores after $10$ games is between $45$ and $55,$ and is a multiple of $10,$ making the sum $50.$ Thus, the score of the $10$ th game is $50-45=5.$

Therefore, their product is $8\cdot 5=40.$

Thus, the answer is B .

16.

如图，七个单位立方体拼成一个立体图形。体积的立方单位数与表面积的平方单位数之比是多少？

A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?

$\:1 : 6$

$\: 7 : 36$

$\: 1 : 5$

$\: 7 : 30$

$\: 6 : 25$

知识点：表面积体积正方体

难度评级：1150

小提示：

体积就是单位立方体的个数

The volume is just the number of unit cubes.

大提示：

计算表面积时，外侧的六个立方体各有 $5$ 个外露面

For surface area, each of the six outer cubes has $5$ exposed faces.

解答：

这个立体由 $7$ 个单位立方体组成，所以体积是 $7$ 立方单位。

中央立方体没有外露面。六个外侧立方体各有 $5$ 个外露面，所以表面积为 $6\cdot5=30$ 平方单位。

体积与表面积之比为 $7:30$ 。

所以正确答案是 D。

The shape uses $7$ unit cubes, so its volume is $7$ cubic units.

The center cube has no exposed faces. Each of the six outer cubes has $5$ exposed faces, for a total surface area of $6\cdot5=30$ square units.

The ratio of volume to surface area is $7:30$ .

Thus, D is the correct answer.

17.

奥斯本女士要求班上每个学生画一个边长为整数、周长为 $50$ 单位的长方形。所有学生都计算了自己所画长方形的面积。这些长方形可能面积的最大值与最小值之差是多少？

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?

$\ 76$

$\ 120$

$\ 128$

$\ 132$

$\ 136$

知识点：矩形面积最优化

难度评级：1180

小提示：

周长为 $50$ 表示两条相邻边长之和为 $25$

A perimeter of $50$ means the side lengths have sum $25.$

大提示：

正整数和固定时，最大乘积来自尽可能接近的两个数

For a fixed positive integer sum, the largest product uses numbers as close as possible.

解答：

若边长为 $l$ 和 $w$ ，则 $2l+2w=50$ ，所以 $l+w=25$ 。

最小可能面积来自边长 $1$ 和 $24$ ，面积为 $24$ 。

最大可能面积来自和为 $25$ 且最接近的整数对，即 $12$ 和 $13$ ，面积为 $156$ 。

差为 $156-24=132$ 。

所以正确答案是 D。

If the side lengths are $l$ and $w$ , then $2l+2w=50$ , so $l+w=25$ .

The smallest possible area comes from side lengths $1$ and $24$ , giving area $24$ .

The largest possible area comes from the closest integer pair with sum $25$ , namely $12$ and $13$ , giving area $156$ .

The difference is $156-24=132$ .

Thus, D is the correct answer.

18.

两个圆同心，半径分别为 $10$ 米和 $20$ 米。一只土豚沿图示路线奔跑，从 $A$ 出发，到 $K$ 结束。它跑了多少米？

Two circles that share the same center have radii $10$ meters and $20$ meters. An aardvark runs along the path shown, starting at $A$ and ending at $K.$ How many meters does the aardvark run?

$\ 10\pi+20$

$\ 10\pi+30$

$\ 10\pi+40$

$\ 20\pi+20$

$\ 20\pi+40$

知识点：弧圆周长

难度评级：1380

小提示：

把路线分成圆弧和直的径向线段

Break the route into circular arcs and straight radial segments.

大提示：

半径为 $r$ 的四分之一圆弧长度为 $\frac{\pi r}{2}$

A quarter-circle of radius $r$ has length $\frac{\pi r}{2}$ .

解答：

圆的周长是 $\pi d = 2r\pi$ ，所以走四分之一圆周的长度是 $\dfrac {\pi r }2$ 。

它先沿大圆走四分之一圈，这部分是 $10 \pi$ 米。然后从大圆到小圆，长度为 $20-10=10$ 米。

接着沿小圆走四分之一圈，这部分是 $5 \pi$ 米。然后穿过小圆的直径，长度为 $10\cdot 2=20$ 米。

再沿小圆走四分之一圈，这部分是 $5 \pi$ 米。最后从小圆到大圆，长度为 $20-10=10$ 米。

总长度为 $10\pi + 10+5\pi + 20 + 5\pi + 10$ $=20\pi + 40\text{。}$

所以正确答案是 E。

The circumference of a circle is $\pi d = 2r\pi ,$ so going a quarter of the way around is $\dfrac {\pi r }2.$

He goes a quarter of the way around the large circle, so this part is $10 \pi$ meters. He then goes from the larger circle to the smaller circle, which is $20-10=10$ meters.

He goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He then goes through the diameter of the smaller circle, which is $10\cdot 2=20$ meters.

He then goes a quarter of the way around the smaller circle, so this part is $5 \pi$ meters. He finally goes from the smaller circle to the larger circle, which is $20-10=10$ meters.

The total length is $10\pi + 10+5\pi + 20 + 5\pi + 10$ $=20\pi + 40.$

Thus, the answer is E .

19.

如图，八个点以一个单位的间隔分布在一个 $2\times2$ 正方形的周围。从这 $8$ 个点中随机选两个点。它们相距一个单位的概率是多少？

Eight points are spaced at intervals of one unit around a $2\times2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the points are one unit apart?

$\ \dfrac{1}{4}$

$\ \dfrac{2}{7}$

$\ \dfrac{4}{11}$

$\ \dfrac{1}{2}$

$\ \dfrac{4}{7}$

知识点：基本概率数对计数

难度评级：1310

小提示：

先数无序的两点组合总数

Count the total number of unordered pairs of points.

大提示：

正方形周围恰好有 $8$ 条相邻的单位线段

There are exactly $8$ adjacent unit segments around the square.

解答：

每个点都有 $2$ 个点与它相距一个单位。

因此无论第一个点怎么选，其余 $7$ 个点中有 $2$ 个与它相距一个单位，所以概率为 $\dfrac 27$ 。

所以正确答案是 B。

Each dot has $2$ dots that are one unit away from it.

Therefore, regardless of the choice of the first dot, $2$ of the other $7$ dots would be within one unit, so the probability is $\dfrac 27.$

Thus, the answer is B .

20.

尼特金先生班上的学生参加了一次书法测试。男生中的三分之二和女生中的 $\frac{3}{4}$ 通过了测试，并且通过测试的男生人数和女生人数相等。这个班最少可能有多少名学生？

The students in Mr. Neatkin’s class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?

$\ 12$

$\ 17$

$\ 24$

$\ 27$

$\ 36$

知识点：最小公倍数比与比例

难度评级：1380

小提示：

设通过测试的男生数和女生数都为 $p$

Let the common number of passing boys and passing girls be $p$ .

大提示：

则男生总数和女生总数分别是 $\frac{3p}{2}$ 和 $\frac{4p}{3}$ ，选使二者都为整数的最小 $p$

Then total boys and girls are $\frac{3p}{2}$ and $\frac{4p}{3}$ , so choose the smallest $p$ that makes both whole.

解答：

设通过测试的男生数和女生数都为 $p$ 。

因为男生中的 $\dfrac{2}{3}$ 通过了测试，男生人数为 $\dfrac{3p}{2}$ 。因为女生中的 $\dfrac{3}{4}$ 通过了测试，女生人数为 $\dfrac{4p}{3}$ 。

使两个人数都为整数的最小正数 $p$ 是 $6$ 。此时有 $9$ 名男生和 $8$ 名女生，最小总人数为 $17$ 。

所以正确答案是 B。

Let $p$ be the common number of boys and girls who passed.

Since $\dfrac{2}{3}$ of the boys passed, the number of boys is $\dfrac{3p}{2}$ . Since $\dfrac{3}{4}$ of the girls passed, the number of girls is $\dfrac{4p}{3}$ .

The smallest positive $p$ that makes both counts whole is $6$ . Then there are $9$ boys and $8$ girls, for a minimum total of $17$ students.

Thus, B is the correct answer.

21.

杰里从一个 $6$ 厘米长的圆柱形博洛尼亚香肠中切下一块楔形部分，如虚线曲线所示。哪一个选项最接近这块楔形部分的体积，单位为立方厘米？

Jerry cuts a wedge from a $6$ -cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?

$\ 48$

$\ 75$

$\ 151$

$\ 192$

$\ 603$

知识点：圆柱体积估算

难度评级：1240

小提示：

圆柱半径为 $4$ 厘米，长度为 $6$ 厘米

The cylinder has radius $4$ cm and length $6$ cm.

大提示：

虚线切割得到圆柱的一半

The dashed cut takes half the cylinder.

解答：

圆柱直径是 $8$ 厘米，所以半径是 $4$ 厘米。它的长度是 $6$ 厘米。

整个圆柱的体积是 $\pi r^2h=\pi\cdot4^2\cdot6=96\pi$ 。

虚线切割得到圆柱的一半，所以楔形部分体积为 $48\pi$ ，约为 $151$ 立方厘米。

所以正确答案是 C。

The cylinder has diameter $8$ cm, so its radius is $4$ cm. Its length is $6$ cm.

The whole cylinder has volume $\pi r^2h=\pi\cdot4^2\cdot6=96\pi$ .

The dashed cut takes half of the cylinder, so the wedge has volume $48\pi$ , which is about $151$ cubic centimeters.

Thus, C is the correct answer.

22.

有多少个正整数 $n$ ，使得 $\dfrac{n}{3}$ 和 $3n$ 都是三位整数？

For how many positive integer values of $n$ are both $\dfrac{n}{3}$ and $3n$ three-digit whole numbers?

$\ 12$

$\ 21$

$\ 27$

$\ 33$

$\ 34$

知识点：区间内整数计数不等式

难度评级：1340

小提示：

令 $x=\frac{n}{3}$ 。则 $n=3x$ ，且 $3n=9x$

Let $x=\frac{n}{3}$ . Then $n=3x$ and $3n=9x$ .

大提示：

$x$ 和 $9x$ 都必须是三位整数

Both $x$ and $9x$ must be three-digit whole numbers.

解答：

令 $x=\dfrac{n}{3}$ 。则 $n=3x$ ，所以 $3n=9x$ 。

$x$ 和 $9x$ 都必须是三位整数。因此 $100\le x$ ，并且 $9x\le999$ ，所以 $100\le x\le111$ 。

共有 $111-100+1=12$ 个整数 $x$ ，每个都对应一个正整数 $n$ 。

所以正确答案是 A。

Let $x=\dfrac{n}{3}$ . Then $n=3x$ , so $3n=9x$ .

Both $x$ and $9x$ must be three-digit whole numbers. Thus $100\le x$ and $9x\le999$ , so $100\le x\le111$ .

There are $111-100+1=12$ integer values of $x$ , and each gives one positive integer value of $n$ .

Thus, A is the correct answer.

23.

在正方形 $ABCE$ 中， $AF=2FE$ ，且 $CD=2DE$ 。 $\triangle BFD$ 的面积与正方形 $ABCE$ 的面积之比是多少？

In square $ABCE,$ $AF=2FE$ and $CD=2DE.$ What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE?$

$\ \dfrac{1}{6}$

$\ \dfrac{2}{9}$

$\ \dfrac{5}{18}$

$\ \dfrac{1}{3}$

$\ \dfrac{7}{20}$

知识点：面积分割三角形面积正方形（几何）

难度评级：1470

小提示：

因为只要求比值，可以为正方形选一个方便的边长

Because only a ratio is needed, choose a convenient side length for the square.

大提示：

取边长为 $3$ ，这样分段长度就是 $2$ 和 $1$

Take the side length as $3$ so the split lengths are $2$ and $1.$

解答：

因为答案是比值，取正方形边长为 $3$ 。则 $AF=2$ 、 $FE=1$ 、 $CD=2$ 、 $DE=1$ 。

正方形面积为 $9$ 。三角形 $ABF$ 和 $BCD$ 的面积都为 $\dfrac{1}{2}\cdot3\cdot2=3$ 。

三角形 $DEF$ 的面积为 $\dfrac{1}{2}\cdot1\cdot1=\dfrac{1}{2}$ 。

所以 $[BFD]=9-3-3-\dfrac{1}{2}=\dfrac{5}{2}$ ，所求比值为 $\dfrac{\frac{5}{2}}{9}=\dfrac{5}{18}$ 。

所以正确答案是 C。

Because the answer is a ratio, choose the side length of the square to be $3$ . Then $AF=2$ , $FE=1$ , $CD=2$ , and $DE=1$ .

The square has area $9$ . The areas of triangles $ABF$ and $BCD$ are each $\dfrac{1}{2}\cdot3\cdot2=3$ .

Triangle $DEF$ has area $\dfrac{1}{2}\cdot1\cdot1=\dfrac{1}{2}$ .

So $[BFD]=9-3-3-\dfrac{1}{2}=\dfrac{5}{2}$ , and the desired ratio is $\dfrac{\frac{5}{2}}{9}=\dfrac{5}{18}$ .

Thus, C is the correct answer.

24.

编号为 $1$ 到 $10$ 的十块牌面朝下放着。随机翻开一块牌，并掷一个骰子。牌上的数与骰子点数的乘积是平方数的概率是多少？

Ten tiles numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?

$\ \dfrac{1}{10}$

$\ \dfrac{1}{6}$

$\ \dfrac{11}{60}$

$\ \dfrac{1}{5}$

$\ \dfrac{7}{30}$

知识点：基本概率完全平方数分类讨论

难度评级：1630

小提示：

共有 $10\cdot 6$ 个等可能的牌和骰子组合

There are $10\cdot 6$ equally likely tile-die pairs.

大提示：

对每个骰子点数，数一数哪些牌号能使乘积成为平方数

For each die roll, count tile numbers that make the product a square.

解答：

如果骰子点数为 $1$ ，牌可以是 $1$ 、 $4$ 、 $9$ ，有 $3$ 种组合。

如果骰子点数为 $2$ ，牌可以是 $2$ 、 $8$ ，有 $2$ 种组合。

如果骰子点数为 $3$ ，牌可以是 $3$ ，有 $1$ 种组合。

如果骰子点数为 $4$ ，牌可以是 $1$ 、 $4$ 、 $9$ ，有 $3$ 种组合。

如果骰子点数为 $5$ ，牌可以是 $5$ ，有 $1$ 种组合。

如果骰子点数为 $6$ ，牌可以是 $6$ ，有 $1$ 种组合。

组合总数为 $3+2+1+3+1+1=11\text{。}$ 共有 $60$ 个等可能组合，所以概率是 $\dfrac{11}{60}$ 。

所以正确答案是 C。

If the rolled number was $1,$ then the tile can be $1,$ $4,$ $9,$ yielding $3$ combinations.

If the rolled number was $2,$ then the tile can be $2,$ $8,$ yielding $2$ combinations.

If the rolled number was $3,$ then the tile can be $3,$ yielding $1$ combination.

If the rolled number was $4,$ then the tile can be $1,$ $4,$ $9,$ yielding $3$ combinations.

If the rolled number was $5,$ then the tile can be $5,$ yielding $1$ combination.

If the rolled number was $6,$ then the tile can be $6,$ yielding $1$ combination.

The total number of combinations is $3+2+1+3+1+1=11.$ There are $60$ combinations each with equal likelihood, so the probability is $\dfrac{11}{60} .$

Thus, the answer is C .

25.

玛吉获奖的艺术设计如图所示。最小的圆半径为 $2$ 英寸，每个后续圆的半径都增加 $2$ 英寸。这个设计中大约百分之多少被涂上阴影？

Margie’s winning art design is shown. The smallest circle has radius $2$ inches, with each successive circle’s radius increasing by $2$ inches. Approximately what percent of the design is shaded?

$\ 42$

$\ 44$

$\ 45$

$\ 46$

$\ 48$

知识点：圆面积面积比百分数

难度评级：1560

小提示：

设计的总面积是最大圆的面积

The total design area is the area of the largest circle.

大提示：

阴影区域是半径每次增加 $2$ 的交替圆盘或圆环

The shaded regions are alternating disks or annuli with radii increasing by $2.$

解答：

最大圆的半径为 $12$ ，所以整个设计的面积为 $12^2\pi=144\pi$ 。

阴影部分的面积为 $2^2\pi=4\pi$ 、 $(6^2-4^2)\pi=20\pi$ 和 $(10^2-8^2)\pi=36\pi$ 。

阴影总面积为 $4\pi+20\pi+36\pi=60\pi$ ，所以阴影比例为 $\dfrac{60\pi}{144\pi}=\dfrac{5}{12}$ ，约为 $42\%$ 。

所以正确答案是 A。

The largest circle has radius $12$ , so the entire design has area $12^2\pi=144\pi$ .

The shaded parts have areas $2^2\pi=4\pi$ , $(6^2-4^2)\pi=20\pi$ , and $(10^2-8^2)\pi=36\pi$ .

The total shaded area is $4\pi+20\pi+36\pi=60\pi$ , so the shaded fraction is $\dfrac{60\pi}{144\pi}=\dfrac{5}{12}$ , which is about $42\%$ .

Thus, A is the correct answer.