2011 AMC 8 详解

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所有题目均经美国数学协会(MAA)官方合法授权使用。

1.

Margie 买了 33 个苹果,每个苹果 5050 美分。她用一张 55 美元纸币付款。Margie 收到多少找零?

Margie bought 33 apples at a cost of 5050 cents per apple. She paid with a 55-dollar bill. How much change did Margie receive?

$1.50\$1.50

$2.00\$2.00

$2.50\$2.50

$3.00\$3.00

$3.50\$3.50

知识点:钱币

难度评级:370

解答:

苹果总共花了 350=1503 \cdot 50 = 150 美分,也就是 1.50$1.50 \$。因此 Margie 收到 51.5=3.55 - 1.5 = 3.5 美元找零。

所以正确答案是 E

The apples cost a total of 350=1503 \cdot 50 = 150 cents, which equals 1.50$.1.50 \$. This means that Margie received 51.5=3.55 - 1.5 = 3.5 dollars in change.

Thus, E is the correct answer.

2.

Karl 的长方形菜园是 2020 英尺乘 4545 英尺,Makenna 的菜园是 2525 英尺乘 4040 英尺。谁的菜园面积更大?

Karl's rectangular vegetable garden is 2020 feet by 4545 feet, and Makenna's is 2525 feet by 4040 feet. Whose garden is larger in area?

Karl 的菜园大 100100 平方英尺。

Karl’s garden is larger by 100100 square feet.

Karl 的菜园大 2525 平方英尺。

Karl’s garden is larger by 2525 square feet.

两个菜园一样大。

The gardens are the same size.

Makenna 的菜园大 2525 平方英尺。

Makenna’s garden is larger by 2525 square feet.

Makenna 的菜园大 100100 平方英尺。

Makenna’s garden is larger by 100100 square feet.

知识点:面积矩形

难度评级:450

解答:

Karl 菜园的面积为 20ft45ft=900ft2.20 \text{ft} \cdot 45 \text{ft} = 900 \text{ft}^2. Makenna 菜园的面积为 25ft40ft=1000ft2.25 \text{ft} \cdot 40 \text{ft} = 1000 \text{ft}^2.

两者面积差为 1000ft2900ft2=100ft2.1000 \text{ft}^2 - 900 \text{ft}^2 = 100 \text{ft}^2. 因此 Makenna 的菜园比 Karl 的大 100ft2100 \text{ft}^2

所以正确答案是 E

The area of Karl's garden is 20ft45ft=900ft2.20 \text{ft} \cdot 45 \text{ft} = 900 \text{ft}^2. The area of Makenna's garden is 25ft40ft=1000ft2.25 \text{ft} \cdot 40 \text{ft} = 1000 \text{ft}^2.

The difference of these areas is 1000ft2900ft2=100ft2.1000 \text{ft}^2 - 900 \text{ft}^2 = 100 \text{ft}^2. Therefore, Makenna's garden is 100ft2100 \text{ft}^2 larger than Karl's.

Thus, E is the correct answer.

3.

一个正方形图案中有 88 块阴影方砖和 1717 块非阴影方砖。若在该正方形周围加上一圈阴影方砖来扩展图案,扩展后阴影方砖与非阴影方砖的比是多少?

Extend the square pattern of 88 shaded and 1717 unshaded square tiles by attaching a border of shaded tiles around the square. What is the ratio of shaded tiles to unshaded tiles in the extended pattern?

8:178:17

25:4925:49

36:2536:25

32:1732:17

36:1736:17

难度评级:720

解答:

在扩展后的图形中,有 3232 块阴影方砖和 1717 块非阴影方砖。因此阴影方砖与非阴影方砖的比为 32:1732:17

所以正确答案是 D

In the extended figure, there are 3232 shaded tiles and 1717 unshaded tiles. Therefore, the ratio of shaded tiles to unshaded tiles is 32:17.32:17.

Thus, D is the correct answer.

4.

下面是 Tyler 去年夏天九次外出钓到的鱼的数量: 2,0,1,3,0,3,3,1,2.2,0,1,3,0,3,3,1,2. 关于平均数、中位数和众数,哪一个说法正确?

Here is a list of the numbers of fish that Tyler caught in nine outings last summer: 2,0,1,3,0,3,3,1,2.2,0,1,3,0,3,3,1,2. Which statement about the mean, median, and mode is true?

中位数 < 平均数 < 众数

median < mean < mode

平均数 < 众数 < 中位数

mean < mode < median

平均数 < 中位数 < 众数

mean < median < mode

中位数 < 众数 < 平均数

median < mode < mean

众数 < 中位数 < 平均数

mode < median < mean

难度评级:720

解答:

为了更容易求这些值,先得到有序列表: 0,0,1,1,2,2,3,3,3. 0, 0, 1, 1, 2, 2, 3, 3, 3.

由此可知众数是 33,中位数是 22,平均数是 15/9=5/315 / 9 = 5 / 3

因为 53<2<3\dfrac{5}{3} < 2 < 3,所以有:

平均数 << 中位数 << 众数。

所以正确答案是 C

To find these values more easily, we can get the following ordered list: 0,0,1,1,2,2,3,3,3. 0, 0, 1, 1, 2, 2, 3, 3, 3.

From this, we see that the mode is 3,3, the median is 2,2, and the mean is 15/9=5/3.15 / 9 = 5 / 3.

Since 53<2<3,\dfrac{5}{3} < 2 < 3, we get that:

mean << median << mode.

Thus, C is the correct answer.

5.

20112011 年一月 11 日开始,经过 20112011 分钟后是什么时间?

What time was it 20112011 minutes after the beginning of January 1,1, 2011?2011?

11 月一日晚上 9:319:31

January 11 at 9:319:31 PM

11 月一日晚上 11:5111:51

January 11 at 11:5111:51 PM

一月 22 日凌晨 3:113:11

January 22 at 3:113:11 AM

一月 22 日上午 9:319:31

January 22 at 9:319:31 AM

一月 22 日下午 6:016:01

January 22 at 6:016:01 PM

难度评级:770

解答:

20112011 除以 6060 的余数是 3131。也就是说 2011=6033+312011 = 60 \cdot 33 + 31,所以 20112011 分钟等于 3333 小时 3131 分钟。

经过 2424 小时到达一月 22 日,所以此时是 22 日当天开始后的 99 小时 3131 分钟。

所以正确答案是 D

The remainder when 20112011 is divided by 6060 is 31.31. This means that 2011=6033+31,2011 = 60 \cdot 33 + 31, which means that 20112011 minutes is the same as 3333 hours and 3131 minutes.

2424 hours takes us to January 2,2, so we get that we are 99 hours and 3131 minutes into January 2.2.

Thus, D is the correct answer.

6.

一个镇上有 351351 名成年人,每个成年人拥有汽车、摩托车,或两者都有。若 331331 名成年人拥有汽车,4545 名成年人拥有摩托车,那么有多少名汽车拥有者不拥有摩托车?

In a town of 351351 adults, every adult owns a car, a motorcycle, or both. If 331331 adults own cars and 4545 adults own motorcycles, how many of the car owners do not own a motorcycle?

2020

2525

4545

306306

351351

知识点:容斥原理

难度评级:870

解答:

已知有 4545 人拥有摩托车,所以 35145=306351 - 45 = 306 人不拥有摩托车。

所以正确答案是 D

We know that 4545 people own motorcycles, so 35145=306351 - 45 = 306 people do not own motorcycles.

Thus, D is the correct answer.

7.

下列四个大的全等正方形分别被分成若干个全等三角形或长方形,并有部分区域被涂阴影。阴影面积占总面积的百分之多少?

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is shaded. What percent of the total area is shaded?

121212 \dfrac{1}{2}

2020

2525

331333 \dfrac{1}{3}

371237 \dfrac{1}{2}

知识点:面积分数

难度评级:960

解答:

左上和右下的阴影区域各占所在正方形的四分之一。右上占八分之一,左下占八分之三。它们的总面积为 14+14+18+38=1.\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} = 1.

因此阴影区域合起来等于一个正方形的面积,占四个正方形总面积的 25%25 \%

所以正确答案是 C

The top left and the bottom right shaded regions are both a quarter of each square. The top right is one-eighth, and the bottom left is three-eights. Their combined area is 14+14+18+38=1.\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} = 1.

Therefore, the shaded regions combined equal the area of one square, so they are 25%25 \% of the total area.

Thus, C is the correct answer.

8.

A 袋中有三枚分别标有 113355 的筹码。B 袋中有三枚分别标有 224466 的筹码。如果从每个袋中各取出一枚筹码,两枚筹码上的数字之和可能有多少个不同的值?

Bag A contains three chips labeled 1,1, 3,3, and 5.5. Bag B contains three chips labeled 2,2, 4,4, and 6.6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?

44

55

66

77

99

难度评级:900

解答:

可以列一个表来查看所有可能结果及其对应的和。

由此可知,可以得到 55 个不同的值。

所以正确答案是 B

We can create a table to look at all the possible outcomes and their respective sums.

From this, we can see that there are 55 distinct values that we can get.

Thus, B is the correct answer.

9.

Carmen 在一条多坡的公路上进行长途骑行。图像显示了她骑行过程中随时间变化的行驶英里数。Carmen 整个骑行的平均速度是多少英里每小时?

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

22

2.52.5

44

4.54.5

55

难度评级:870

解答:

Carmen 在 77 小时内行驶 3535 英里,所以平均速度为 35/7=535 / 7 = 5 英里每小时。

所以正确答案是 E

Carmen travels 3535 miles in 77 hours, so her average speed is 35/7=535 / 7 = 5 miles per hour.

Thus, E is the correct answer.

10.

Gotham City 的出租车在前 12\dfrac{1}{2} 英里收费 $2.40\$2.40,之后每额外 0.10.1 英里收费 $0.20\$0.20。你计划给司机 $2\$2 小费。用 $10\$10 最多可以乘坐多少英里?

The taxi fare in Gotham City is $2.40\$2.40 for the first 12\dfrac{1}{2} mile and additional mileage charged at the rate $0.20\$0.20 for each additional 0.10.1 mile. You plan to give the driver a $2\$2 tip. How many miles can you ride for $10\$10?

3.03.0

3.253.25

3.33.3

3.53.5

3.753.75

知识点:速率钱币

难度评级:1100

解答:

小费固定为 $2\$2,从总额中减去后剩 $8\$8。这大于 $2.40\$2.40,所以再减去前半英里的车费,并把 12\dfrac{1}{2} 英里计入总距离。

现在还有 $5.60\$5.60 可用于额外里程。每 0.10.1 英里收费 $0.20\$0.20,等价于每 11 英里收费 $2\$2。因此这笔钱还能乘坐 5.60/2=2.85.60 / 2 = 2.8 英里。总距离为 2.8+12=3.32.8 + \dfrac{1}{2} = 3.3 英里。

所以正确答案是 C

There is a guaranteed $2\$2 tip, so we can subtract that from the total, leaving $8.\$8. This is greater than $2.40,\$2.40, so we can subtract that and add 12\dfrac{1}{2} miles to the total distance.

We now have $5.60\$5.60 to use for additional miles. $0.20\$0.20 per 0.10.1 mile is the same as $2\$2 for 11 mile. That means one can ride for 5.60/2=2.85.60 / 2 = 2.8 more miles with this much money. This leaves a total of 2.8+12=3.32.8 + \dfrac{1}{2} = 3.3 miles.

Thus, C is the correct answer.

11.

图中显示一周内 Asha(左柱)和 Sasha(右柱)每天学习的分钟数。平均每天 Sasha 比 Asha 多学习多少分钟?

The graph shows the number of minutes studied by both Asha (left bar) and Sasha (right bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?

66

88

99

1010

1212

难度评级:1020

解答:

可以通过每天的差值来计算平均分钟数之差。

从星期一开始,Sasha 与 Asha 的差值分别为 101010-102020303020-20。总差为 3030 分钟,因此平均差为 30÷5=630 \div 5 = 6

所以正确答案是 A

We can calculate the difference in average minutes by looking at the differences per day.

Starting with Monday, the differences between Sasha and Asha are 10,10, 10,-10, 20,20, 30,30, and 20.-20. This is a total of 3030 minutes. Therefore, the average difference is 30÷5=6.30 \div 5 = 6.

Thus, A is the correct answer.

12.

Angie、Bridget、Carlos 和 Diego 随机坐在一张方桌四周,每边坐一个人。Angie 和 Carlos 坐在相对两边的概率是多少?

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

难度评级:960

解答:

设 Angie 的座位已经确定。Carlos 等可能坐在另外 33 个座位中的任意一个,但其中只有一个座位与 Angie 相对。因此概率为 13\dfrac{1}{3}

所以正确答案是 B

Consider that Angie's seat is chosen. Carlos has an equal probability of being in any of the other 33 seats. Only one of them is opposite Angie, however. Therefore, the probability is 13.\dfrac{1}{3}.

Thus, B is the correct answer.

13.

两个全等正方形 ABCDABCDPQRS,PQRS, 的边长均为 1515。它们重叠形成图中所示的 15152525 长方形 AQRDAQRD。长方形 AQRDAQRD 的面积中有百分之多少被涂阴影?

Two congruent squares, ABCDABCD and PQRS,PQRS, have side length 15.15. They overlap to form the 1515 by 2525 rectangle AQRDAQRD shown. What percent of the area of rectangle AQRDAQRD is shaded?

1515

1818

2020

2424

2525

知识点:面积百分数

难度评级:1140

解答:

SC=DC+SRDR=15+1525=5.\begin{align*} SC &= DC + SR - DR \\ &= 15 + 15 - 25 \\ &= 5. \end{align*}

因此 PBSCPBSC 的面积为 515=755 \cdot 15 = 75AQRDAQRD 的面积为 2515=37525 \cdot 15 = 375

75375=15\dfrac{75}{375} = \dfrac{1}{5},也就是 20%20 \%

所以正确答案是 C

We get that SC=DC+SRDR=15+1525=5.\begin{align*} SC &= DC + SR - DR \\ &= 15 + 15 - 25 \\ &= 5. \end{align*}

This means that the area of PBSCPBSC is 515=75.5 \cdot 15 = 75. The area of AQRDAQRD is 2515=375.25 \cdot 15 = 375.

75375=15,\dfrac{75}{375} = \dfrac{1}{5}, which is 20%.20 \%.

Thus, C is the correct answer.

14.

Colfax Middle School 有 270270 名学生,男生与女生的比为 5:45 : 4。Winthrop Middle School 有 180180 名学生,男生与女生的比为 4:54 : 5。两所学校举办舞会,所有学生都参加。舞会上女生占学生总数的几分之几?

There are 270270 students at Colfax Middle School, where the ratio of boys to girls is 5:4.5 : 4. There are 180180 students at Winthrop Middle School, where the ratio of boys to girls is 4:5.4 : 5. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?

718\dfrac{7}{18}

715\dfrac{7}{15}

2245\dfrac{22}{45}

12\dfrac{1}{2}

2345\dfrac{23}{45}

知识点:比与比例分数

难度评级:1140

解答:

女生总数为 49270+59180=\dfrac{4}{9} \cdot 270 + \dfrac{5}{9} \cdot 180 = 120+100=220. 120 + 100 = 220.

学生总数为 270+180=450270 + 180 = 450,所以女生所占比例为 220450=2245\dfrac{220}{450} = \dfrac{22}{45}

所以正确答案是 C

The total number of girls is 49270+59180=\dfrac{4}{9} \cdot 270 + \dfrac{5}{9} \cdot 180 = 120+100=220. 120 + 100 = 220.

There are 270+180=450270 + 180 = 450 students total, so the fraction of girls is 220450=2245.\dfrac{220}{450} = \dfrac{22}{45}.

Thus, C is the correct answer.

15.

乘积 455104^5 \cdot 5^{10} 有多少位数字?

How many digits are in the product 45510?4^5 \cdot 5^{10}?

88

99

1010

1111

1515

知识点:指数数字

难度评级:1100

解答:

为了求位数,可以尝试把这个数写成 10.10. 的幂。

45510=210510=1010.\begin{align*} 4^5 \cdot 5^{10} &= 2^{10} \cdot 5^{10} \\ &= 10^{10}. \end{align*}

这说明所求数是 11 后面跟着 1010 个零,共有 1111 位数字。

所以正确答案是 D

To find the number of digits, we can try to express this number in terms of powers of 10.10.

We get that 45510=210510=1010.\begin{align*} 4^5 \cdot 5^{10} &= 2^{10} \cdot 5^{10} \\ &= 10^{10}. \end{align*}

This shows that the desired number is 11 followed by 1010 zeros, for a total of 1111 digits.

Thus, D is the correct answer.

16.

AA 为边长 25,2525, 253030 的三角形面积。设 BB 为边长 25,2525, 254040 的三角形面积。AABB 有什么关系?

Let AA be the area of the triangle with sides of length 25,25,25, 25, and 30.30. Let BB be the area of the triangle with sides of length 25,25,25, 25, and 40.40. What is the relationship between AA and B?B?

A=916BA = \dfrac{9}{16}B

A=34BA = \dfrac{3}{4}B

A=BA = B

A=43BA = \dfrac{4}{3}B

A=169BA = \dfrac{16}{9}B

难度评级:1340

解答:

因为这些三角形是等腰三角形,可以作高,把它们分成如图所示的两个全等直角三角形。

用勾股定理,面积为 AA 的三角形的高为 252152=20.\sqrt{25^2 - 15^2} = 20. 类似地,面积为 BB 的三角形的高为 252202=15.\sqrt{25^2 - 20^2} = 15.

因此 A=122030=300. A = \dfrac{1}{2} \cdot 20 \cdot 30 = 300. 同样 B=121540=300. B = \dfrac{1}{2} \cdot 15 \cdot 40 = 300.

所以 A=BA = B

所以正确答案是 C

Since these triangles are isosceles, we can drop altitudes to create two congruent right triangles as shown in the diagram.

Using the Pythagorean theorem, we get that the altitude of the triangle with area AA equals 252152=20.\sqrt{25^2 - 15^2} = 20. Similarly, we get that the altitude of the triangle with area BB equals 252202=15.\sqrt{25^2 - 20^2} = 15.

With these altitudes, we can calculate the areas of the triangles. We get that A=122030=300. A = \dfrac{1}{2} \cdot 20 \cdot 30 = 300. Similarly, B=121540=300. B = \dfrac{1}{2} \cdot 15 \cdot 40 = 300.

Therefore, A=B.A = B.

Thus, C is the correct answer.

17.

wwxxyyzz 为整数。若 2w3x5y7z=588,2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588, 那么 2w+3x+5y+7z2w + 3x + 5y + 7z 等于多少?

Let w,w, x,x, y,y, and zz be whole numbers. If 2w3x5y7z=588,2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588, then what does 2w+3x+5y+7z2w + 3x + 5y + 7z equal?

2121

2525

2727

3535

5656

知识点:质因数分解

难度评级:1140

解答:

为了找到所需指数,注意所有底数都是质数。因此质因数分解会有帮助。

588=223172.588 = 2^2 \cdot 3^1 \cdot 7^2.

由此可知 w=2w = 2x=1x = 1y=0,y = 0,z=2z = 2,其中 y=0y = 0 使 5y5^y 这一项等于 11

代入这些指数可得 2w+3x+5y+7z=22+31+50+72=21. \begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}

所以正确答案是 A

To find the desired exponents, note that all the bases are prime numbers. This means that finding the prime factorization will be helpful.

We get that 588=223172.588 = 2^2 \cdot 3^1 \cdot 7^2.

From this, it is clear that w=2,w = 2, x=1,x = 1, y=0,y = 0, and z=2z = 2 (y=0y = 0 since that makes the 5y5^y term equal 11).

Therefore, 2w+3x+5y+7z=22+31+50+72=21. \begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}

Thus, A is the correct answer.

18.

一枚公平的六面骰子掷两次。第一次掷出的数大于或等于第二次掷出的数的概率是多少?

A fair six-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

16\dfrac{1}{6}

512\dfrac{5}{12}

12\dfrac{1}{2}

712\dfrac{7}{12}

56\dfrac{5}{6}

难度评级:1310

解答:

掷骰子两次时有 33 类结果:第一次大于第二次、两次相等、第一次小于第二次。第一类和第三类由于对称而概率相同。

第二类发生的概率是 16\dfrac{1}{6},因为第一次可以是任意数,而第二次必须等于第一次。其余两类的总概率为116=561 - \dfrac{1}{6} = \dfrac{5}{6}。因此每一类的概率为56÷2=512\dfrac{5}{6} \div 2 = \dfrac{5}{12}

所求概率是第一类加第二类,共为512+16=712\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}

所以正确答案是 D

There are 33 possible outcomes when rolling a die twice: the first number is greater than the second, both numbers are equal, or the first number is less than the second number. The first and third outcomes have the same probability since they are symmetric.

The second outcome has a 16\dfrac{1}{6} chance of happening, since the first number can be anything, and the second number must equal the first number. The other two outcomes have a combined probability of 116=56.1 - \dfrac{1}{6} = \dfrac{5}{6}. This means that each outcome has a 56÷2=512\dfrac{5}{6} \div 2 = \dfrac{5}{12} chance of happening.

The desired probability is the first outcome plus the second outcome, for a total probability of 512+16=712.\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}.

Thus, D is the correct answer.

19.

这个图形中有多少个长方形?

How many rectangles are in this figure?

88

99

1010

1111

1212

难度评级:1430

解答:

可以把图形分成这些区域,使统计长方形更容易。

图中的长方形是 bbccddababbcbccdcdcfcfdedeabcdabcdcdef,cdef,bcfgbcfg。这些共有 1111 个长方形。

所以正确答案是 D

We can split the figure into these regions to make it easier to count the rectangles.

The rectangles in this figure are b,b, c,c, d,d, ab,ab, bc,bc, cd,cd, cf,cf, de,de, abcd,abcd, cdef,cdef, and bcfg.bcfg. These form 1111 rectangles.

Thus, D is the correct answer.

20.

四边形 ABCDABCD 是梯形,AD=15AD = 15AB=50AB = 50BC=20BC = 20,高为 1212。这个梯形的面积是多少?

Quadrilateral ABCDABCD is a trapezoid, AD=15,AD = 15, AB=50,AB = 50, BC=20,BC = 20, and the altitude is 12.12. What is the area of the trapezoid?

600600

650650

700700

750750

800800

知识点:梯形勾股数

难度评级:1410

解答:

可以作如下高,更容易求面积。

用勾股定理可得 DE=152122=9 DE = \sqrt{15^2 - 12^2} = 9 FC=202122=16. FC = \sqrt{20^2 - 12^2} = 16.

还知道 EF=AB=50, EF = AB = 50, 所以 DC=DE+EF+FC=75. DC = DE + EF + FC = 75.

因此 ABCDABCD 的面积为 12(DC+50)12=6125 \dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125 =750. = 750.

所以正确答案是 D

We can drop the following altitudes to more easily find the area.

We can use the Pythagorean Theorem to get that DE=152122=9 DE = \sqrt{15^2 - 12^2} = 9 and FC=202122=16. FC = \sqrt{20^2 - 12^2} = 16.

We also know that EF=AB=50, EF = AB = 50, so DC=DE+EF+FC=75. DC = DE + EF + FC = 75.

Then the area of ABCDABCD is 12(DC+50)12=6125 \dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125 =750. = 750.

Thus, D is the correct answer.

21.

学生们猜 Norb 的年龄分别是 24,28,30,32,36,38,41,44,47,24, 28, 30, 32, 36, 38, 41, 44, 47,4949。Norb 说:“你们中至少一半猜得太低,有两个人只差一岁,而且我的年龄是质数。”Norb 多少岁?

Students guess that Norb's age is 24,28,30,32,36,38,41,44,47,24, 28, 30, 32, 36, 38, 41, 44, 47, and 49.49. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

2929

3131

3737

4343

4848

知识点:逻辑推理质数

难度评级:1470

解答:

陈述的第一部分表示 Norb 的年龄大于 3636

第二部分表示 Norb 的年龄要么在 36363838 之间,要么在 47474949 之间。

因为 3737 是质数而 4848 不是,所以 Norb 的年龄是 3737

所以正确答案是 C

The first part of the statement means that Norb's age is greater than 36.36.

The second part means that Norb's age is either between 3636 and 3838 or between 4747 and 49.49.

Since 3737 is prime and 4848 is not, Norb's age is 37.37.

Thus, C is the correct answer.

22.

720117^{2011} 的十位数字是什么?

What is the tens digit of 72011?7^{2011}?

00

11

33

44

77

难度评级:1520

解答:

要找十位数字,只需跟踪 77 的幂模 100100 的结果。因为 74=24011(mod100)7^4=2401\equiv1\pmod{100},末两位每四次方重复。

因为 2011=4502+32011=4\cdot502+3,所以 7201173=3437^{2011}\equiv7^3=34343(mod100)\equiv43\pmod{100}。因此十位数字是 44

所以正确答案是 D

To find the tens digit, it is enough to track powers of 77 modulo 100100. Since 74=24011(mod100)7^4=2401\equiv1\pmod{100}, the last two digits repeat every four powers.

Because 2011=4502+32011=4\cdot502+3, we have 7201173=3437^{2011}\equiv7^3=34343(mod100)\equiv43\pmod{100}. Thus the tens digit is 44.

Thus, D is the correct answer.

23.

有多少个 44-位正整数满足:四个数字互不相同,首位不是零,该整数是 55 的倍数,并且 55 是最大的数字?

How many 44-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5,5, and 55 is the largest digit?

2424

4848

6060

8484

108108

难度评级:1690

解答:

一个数能被 5,5, 整除,个位数字必须是 0055

如果个位数字是 00,另外三位中必须有一位是 55。剩下两位必须从 {1,2,3,4}.\{1, 2, 3, 4\}. 中选择。有 66 种方式选这两个数字,并有 66 种方式排列这三位数字,共 66=366 \cdot 6 = 36 个数。

如果个位数字是 55,千位数字有 44 种选择。选定千位后,另外 22 位共有 43=124 \cdot 3 = 12 种选择和排列方式。因此此情况共有 412=484 \cdot 12 = 48 个数。

合并两种情况,总数为 36+48=8436 + 48 = 84

所以正确答案是 D

For a number to be divisible by 5,5, the units digit must be either 00 or 5.5.

If the units digit is 0,0, one of the other three digits must be 5.5. The remaining two digits must be chosen from {1,2,3,4}.\{1, 2, 3, 4\}. There are 66 ways to choose the pair, and there are 66 ways to arrange the three digits for a total of 66=366 \cdot 6 = 36 numbers.

If the units digit is 5,5, there are 44 ways to choose the thousands digit. There are 43=124 \cdot 3 = 12 ways to choose the other 22 digits. This leaves a total of 412=484 \cdot 12 = 48 numbers for this case.

Combining both cases, we get the total number of such integers is 36+48=84.36 + 48 = 84.

Thus, D is the correct answer.

24.

1000110001 可以用多少种方式写成两个质数之和?

In how many ways can 1000110001 be written as the sum of two primes?

00

11

22

33

44

知识点:奇偶性质数

难度评级:1200

解答:

两个数相加得到奇数时,一个必须是奇数,另一个必须是偶数。唯一的偶质数是 22,所以另一个数只能是 99999999。但 99999999 不是质数,因此 1000110001 不能写成两个质数之和。

所以正确答案是 A

For two numbers to add to an odd number, one of them must be odd and the other even. Thus the only even prime is 2,2, so the other number is forced to be 9999.9999. 99999999 is not prime, however, so 1000110001 cannot be written as the sum of two primes.

Thus, A is the correct answer.

25.

一个半径为 11 的圆内切于一个正方形,并外接于另一个正方形,如图所示。圆内阴影面积与两个正方形之间阴影面积的比,最接近下列哪个分数?

A circle with radius 11 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the shaded area between the two squares?

12\dfrac{1}{2}

11

32\dfrac{3}{2}

22

52\dfrac{5}{2}

难度评级:1660

解答:

圆内阴影面积等于圆面积减去小正方形面积。内正方形的边长可由勾股定理求得: 12+12=2.\sqrt{1^2 + 1^2} = \sqrt{2}.

因此内正方形面积为 22=2\sqrt{2}^2 = 2。圆内阴影面积为 12π2=π21^2\pi - 2 = \pi - 2

外正方形面积为 22=42^2 = 4,所以两个正方形之间的阴影面积为 42=24 - 2 = 2

所求比值为 π223.142212.\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} \approx \dfrac{1}{2}.

所以正确答案是 A

The circle's shaded area is equal to the area of the circle minus the area of the smaller square. The side length of the inner square can be calculated using the Pythagorean Theorem to get 12+12=2.\sqrt{1^2 + 1^2} = \sqrt{2}.

Therefore, the area of the inner square is 22=2.\sqrt{2}^2 = 2. The circle's shaded area is then 12π2=π2.1^2\pi - 2 = \pi - 2.

The area of the outside square is 22=4,2^2 = 4, so the shaded area between the two squares is 42=2.4 - 2 = 2.

The desired fraction is π223.142212.\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} \approx \dfrac{1}{2}.

Thus, A is the correct answer.