三个数 (5)/(19) 、 (7)/(21) 和 (9)/(23) 按从小到大排列的正确顺序是什么？

(B) (5)/(19) \lt (7)/(21) \lt (9)/(23)

(A) (9)/(23) \lt (7)/(21) \lt (5)/(19)

(C) (9)/(23) \lt (5)/(19) \lt (7)/(21)

(D) (5)/(19) \lt (9)/(23) \lt (7)/(21)

(E) (7)/(21) \lt (5)/(19) \lt (9)/(23)

2012 AMC 8 真题与在线限时测验

计时

40:00

1.

Rachelle 用 $3$ 磅肉为家人做 $8$ 个汉堡。她要为社区野餐做 $24$ 个汉堡，需要多少磅肉？

Rachelle uses $3$ pounds of meat to make $8$ hamburgers for her family. How many pounds of meat does she need to make $24$ hamburgers for a neighborhood picnic?

$6$

$6\dfrac23$

$7\dfrac12$

$8$

$9$

答案：E

知识点：比与比例

难度评级：370

小提示：

从 $8$ 个汉堡按比例放大到 $24$ 个汉堡。

Scale from $8$ hamburgers to $24$ hamburgers.

大提示：

汉堡数量变为三倍，所以肉的数量也变为三倍。

The hamburger count triples, so the meat amount triples.

视频讲解：

Click to load, then click again to play

文字解答：

$8$ 个汉堡是 $24$ 个汉堡的 $\dfrac{1}{3}$ ，所以 $3$ 磅也是所需肉量的 $\dfrac{1}{3}$ 。因此总肉量为 $\dfrac{1}{\dfrac{1}{3}} \cdot 3 = 3\cdot3 = 9$ 。

所以正确答案是 E。

If we have $8$ hamburgers, we have $\dfrac{1}{3}$ of the $24$ hamburgers. This means we have $\dfrac{1}{3}$ of the meat when we have $3$ pounds. The total amount of meat is therefore $\dfrac{1}{\dfrac{1}{3}} \cdot 3 = 3\cdot3 = 9.$

Thus, the answer is E .

2.

在 East Westmore 县，统计学家估计每 $8$ 小时出生一个婴儿，每天死亡一人。四舍五入到最接近的百位，East Westmore 每年人口增加多少人？

In the county of East Westmore, statisticians estimate there is a baby born every $8$ hours and a death every day. To the nearest hundred, how many people are added to the population of East Westmore each year?

$600$

$700$

$800$

$900$

$1000$

答案：B

知识点：速率估算

难度评级：660

小提示：

先求一天内人口的净变化。

Find the net population change in one day.

大提示：

每 $8$ 小时出生一个婴儿，表示每天出生 $3$ 个婴儿。

A birth every $8$ hours means $3$ births per day.

视频讲解：

Click to load, then click again to play

文字解答：

每 $8$ 小时出生 $1$ 个婴儿，所以每 $24$ 小时出生 $3$ 个婴儿。因此每天有 $3$ 人出生、 $1$ 人死亡，平均每天人口净增加 $2$ 人。一年有 $365$ 天，每天净增加 $2$ 人，所以一年净增加约 $2\cdot 365 = 730$ 人，约为 $700$ 。

所以正确答案是 B。

Since we have $1$ birth every $8$ hours, we have $3$ births every $24$ hours. Therefore, we have $3$ births a day and $1$ death a day. The net change in population every day should be on average $2.$ Since we have $365$ days in a year and $2$ added to the population every day, the net change in population should be around $2\cdot 365 = 730 .$ This is approximately $700.$

Thus, the answer is B .

3.

二月 $13$ 日，The Oshkosh Northwester 刊登的白昼时长为 $10$ 小时 $24$ 分钟，日出时间为 $6:57\text{ AM}$ 并把日落时间标为 $8:15\text{PM}$ ，而白昼时长和日出时间是正确的，但日落时间是错误的。太阳实际什么时候落下？

On February $13$ The Oshkosh Northwester listed the length of daylight as $10$ hours and $24$ minutes, the sunrise as $6:57\text{ AM},$ and the sunset as $8:15\text{PM} .$ The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?

$5$ : $10\text{ PM}$

$5$ : $21\text{ PM}$

$5$ : $41\text{ PM}$

$5$ : $57\text{ PM}$

$6$ : $03\text{ PM}$

答案：B

知识点：日期与时间

难度评级：720

小提示：

将正确的白昼时长加到日出时间上。

Add the correct daylight length to the sunrise time.

大提示：

从上午 $6:57$ 到中午是 $5$ 小时 $3$ 分钟。

From $6:57$ AM to noon is $5$ hours $3$ minutes.

视频讲解：

Click to load, then click again to play

文字解答：

从 $6:57 \text{AM}$ 起 $10$ 小时后是 $4:57 \text{PM}$ 。也就是说，日出后 $10$ 小时 $3$ 分钟是 $5:00 \text{PM}$ 。距日落还有 $21$ 分钟，所以实际日落时间是 $5:21 \text{PM}$ 。

所以正确答案是 B。

Since $10$ hours after $6:57 \text{AM}$ is $4:57 \text{PM} ,$ we can then say $10$ hours and $3$ minutes after sunrise is $5:00 \text{PM} .$ We then have $21$ more minutes until sunset, so sunset is $5:21 \text{PM} .$

Thus, the answer is B .

4.

Peter 一家晚餐订了一个切成 $12$ 片的披萨。Peter 吃了一片，又和弟弟 Paul 平分了另一片。Peter 吃了这个披萨的几分之几？

Peter’s family ordered a $12$ -slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?

$\dfrac{1}{24}$

$\dfrac{1}{12}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

答案：C

知识点：分数

难度评级：450

小提示：

Peter 吃了一整片，又吃了另一片的一半。

Peter ate one whole slice plus half of another slice.

大提示：

将 $1+\frac{1}{2}$ 片转换成整个 $12$ 片披萨的分数。

Convert $1+\frac{1}{2}$ slices into a fraction of $12$ slices.

视频讲解：

Click to load, then click again to play

文字解答：

Peter 吃了 $1$ 整片，又吃了他分得的那片的 $\dfrac 12$ 。

因此他吃了整个披萨的 $\dfrac{\frac{3}{2}}{12}$ ，也就是 $\dfrac 18$ 。

所以正确答案是 C。

Peter ate $1$ full slice, and he ate $\dfrac 12$ of the slice that he split.

Therefore, he ate $\dfrac{\frac{3}{2}}{12}$ of the pizza, which is equivalent to $\dfrac 18.$

Thus, the answer is C .

5.

图中所有角都是直角，边长以厘米为单位给出。注意图形不是按比例绘制的。 $X$ 是多少厘米？

In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is $X$ , in centimeters?

$1$

$2$

$3$

$4$

$5$

答案：E

知识点：一次方程

难度评级：870

小提示：

分别把图形两侧的竖直长度相加。

Add the vertical lengths on each side of the shape.

大提示：

左侧和右侧的总高度必须相等。

The left and right total heights must be equal.

视频讲解：

Click to load, then click again to play

文字解答：

先从右侧高度相加求图形总高度，得 $1+2+1+6 = 10$ 。

再从左侧高度相加，得 $1+1+1+2+X = 5+X$ 。

两个高度相同，所以 $10=5+X$ ，因此 $X = 5$ 。

所以正确答案是 E。

First, we can find the height of the object by getting the sum of the heights on the right. Therefore, the height is $1+2+1+6 = 10.$

Next, we can find the height of the object by getting the sum of the heights on the left. Therefore, the height is $1+1+1+2+X = 5+X.$

Since the heights are the same, we know $10=5+X ,$ so $X = 5.$

Thus, the answer is E .

6.

一张长方形照片放入相框后，照片四周形成宽二英寸的边框。照片高 $8$ 英寸、宽 $10$ 英寸。边框的面积是多少平方英寸？

A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures $8$ inches high and $10$ inches wide. What is the area of the border, in square inches?

$36$

$40$

$64$

$72$

$88$

答案：E

知识点：面积矩形

难度评级：820

小提示：

求带相框照片的外部长方形尺寸。

Find the outside dimensions of the framed photograph.

大提示：

从外部长方形面积中减去照片面积。

Subtract the photograph area from the outside rectangle area.

视频讲解：

Click to load, then click again to play

文字解答：

每一边都增加 $2$ 英寸，所以高度和宽度各总共增加 $4$ 英寸。外部长方形尺寸为 $12 \times 14$ ，面积为 $12\cdot 14 = 168$ 。

照片本身面积为 $8 \cdot 10 =80$ 。

因此边框面积为 $168-80=88$ 。

所以正确答案是 E。

If we add $2$ inches on each side, we add $4$ inches total on both sides. This means that the dimensions of the outer part of the frame are $12 \times 14.$ The area of this is $12\cdot 14 = 168.$

However, we must take out the area of the inner part of the frame which has area $8 \cdot 10 =80.$

Therefore, the total area is $168-80=88.$

Thus, the answer is E .

7.

Isabella 的数学课要进行四次满分为 $100$ 分的测试。她的目标是四次测试的平均分至少达到 $95$ 。她前两次测试成绩是 $97$ 和 $91$ 。看到第三次测试成绩后，她意识到自己仍然可以达到目标。她第三次测试最低可能得多少分？

Isabella must take four $100$ -point tests in her math class. Her goal is to achieve an average grade of at least $95$ on the tests. Her first two test scores were $97$ and $91.$ After seeing her score on the third test, she realized that she could still reach her goal. What is the lowest possible score she could have made on the third test?

$90$

$92$

$95$

$96$

$97$

答案：B

知识点：平均数最优化

难度评级：1070

小提示：

四次测试平均至少为 $95$ 分，需要总分至少为 $4\cdot95$ 。

A $95$ average on four tests requires a total of $4\cdot95$ .

大提示：

要使第三次成绩尽量低，就让第四次成绩尽量高。

Make the fourth test as large as possible to minimize the third score.

视频讲解：

Click to load, then click again to play

文字解答：

若平均分为 $95$ ，四次测试总分为 $95\cdot 4 = 380$ 。由于已知前两次成绩，后两次测试总分是 $380$ 减去前两次成绩。

后两次测试总分必须为 $380-91-97 = 192$ 。因此后两次成绩之和为 $192$ 。

要使第三次成绩最低，就让第四次成绩最高，为 $100$ 。于是第三次成绩为 $192 -100 = 92$ 。

所以正确答案是 B。

If the average is $95,$ then the sum of the tests is $95\cdot 4 = 380 .$ Since we have the first two tests, the sum of the last two tests is $380$ minus the first two scores.

This makes the sum of the last two scores equal to $380-91-97 = 192.$ Her last two scores therefore have a sum of $192.$

Given the sum of the tests we try to minimize one score, then we must maximize the other test. Therefore, we maximize the fourth test by making it $100.$ This would make the third test equal to $192 -100 = 92.$

Thus, the answer is B .

8.

一家商店广告称“今天促销所有商品半价”。此外，一张优惠券可在促销价基础上再享受 $20\%$ 的折扣。使用优惠券后，今天的价格相对于原价的折扣率是多少？

A shop advertises that everything is “half price in today’s sale.” In addition, a coupon gives a $20\%$ discount on sale prices. Using the coupon, the price today represents what percentage discount off the original price?

$10$

$33$

$40$

$60$

$70$

答案：D

知识点：百分数

难度评级：980

小提示：

半价后，促销价是原价的 $50\%$ 。

After half price, the sale price is $50\%$ of the original.

大提示：

优惠券让价格保留促销价的 $80\%$ 。

The coupon keeps $80\%$ of the sale price.

视频讲解：

Click to load, then click again to play

文字解答：

设原价为 $p$ 。半价后价格为 $0.5p$ 。

再打 $20\%$ 折扣表示支付价格的 $80\%$ ，所以最终价格为 $0.5p\cdot 0.8 =0.4p$ 。

因此减少了 $0.6p$ ，折扣率为 $\dfrac{0.6p}{p} = 0.6$ ，即 $60\%$ 。

所以正确答案是 D。

Let $p$ be the original price. If everything is half off, we have the new price as $0.5p.$

Having a $20\%$ discount makes it such that we keep $80\%$ of the price, so the price is $0.5p\cdot 0.8 =0.4p.$

This would have $0.6p$ off, so we get a discount of $\dfrac{0.6p}{p} = 0.6,$ which is $60\%.$

Thus, the answer is D .

9.

Fort Worth Zoo 有一些两条腿的鸟和一些四条腿的哺乳动物。Margie 某次参观动物园时数到 $200$ 个头和 $522$ 条腿。Margie 数到的动物中有多少是两条腿的鸟？

The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted $200$ heads and $522$ legs. How many of the animals that Margie counted were two-legged birds?

$61$

$122$

$139$

$150$

$161$

答案：C

知识点：方程组

难度评级：1100

小提示：

先假设每只动物都有两条腿。

Start by giving every animal two legs.

大提示：

多出来的腿决定四条腿哺乳动物的数量。

The extra legs determine the number of four-legged mammals.

视频讲解：

Click to load, then click again to play

文字解答：

设 $f$ 为 $4$ 条腿动物的数量， $t$ 为 $2$ 条腿动物的数量。

数腿得到 $2t+4f = 522$ ，数头得到 $t + f = 200$ 。将第二式乘以四，得 $4t+4f = 800$ 。减去第一式，得到 $2t = 278 \implies t = 139$ 。因此有 $139$ 只两条腿的鸟。

所以正确答案是 C。

Let $f$ be the number of animals with $4$ legs and let $t$ be the number of animals with $2$ legs.

Counting the number of legs yields $2t+4f = 522$ and counting the number of heads yields $t + f = 200.$ This means $4t+4f = 800,$ and subtracting the first equation from the second yields $2t = 278 \implies t = 139.$ This means there are $139$ two-legged birds.

Thus, the answer is C .

10.

使用 $2012$ 的四个数字，可以组成多少个大于 $1000$ 的 $4$ 位数？

How many $4$ -digit numbers greater than $1000$ are there that use the four digits of $2012?$

$6$

$7$

$8$

$9$

$12$

答案：D

知识点：多重集排列

难度评级：1070

小提示：

第一位不能是 $0$ 。

The first digit cannot be $0$ .

大提示：

分别统计以 $1$ 开头和以 $2$ 开头的排列。

Count arrangements starting with $1$ , then arrangements starting with $2$ .

视频讲解：

Click to load, then click again to play

文字解答：

$0$ 不能放在千位。因此有 $3$ 个位置可放。放好零后，数字 $1$ 有 $3$ 个可用位置，两个 $2$ 自动占据剩余位置。因此共有 $3\cdot 3 = 9$ 个数。

所以正确答案是 D。

First, we can’t have the $0$ in the thousands position. Therefore, we have $3$ spots we can put it. Then, we have $3$ available positions for the $1,$ and then the two $2$ s are placed. This makes it such that we have $3\cdot 3 = 9$ combinations.

Thus, the answer is D .

11.

正整数 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $6$ 、 $7$ 和 $x$ 的平均数、中位数和唯一众数都相等。 $x$ 的值是多少？

The mean, median, and unique mode of the positive integers $3,$ $4,$ $5,$ $6,$ $6,$ $7,$ and $x$ are all equal. What is the value of $x?$

$5$

$6$

$7$

$11$

$12$

答案：D

知识点：平均数中位数（数据）众数

难度评级：1240

小提示：

唯一众数必须是列表中已经重复的数。

The unique mode must be the repeated listed number.

大提示：

如果平均数也是那个数，那么总和就被确定。

If the mean is that number, the total sum is fixed.

视频讲解：

Click to load, then click again to play

文字解答：

已列出的数中，除 $6$ 外每个数只出现一次，而 $6$ 出现两次。若 $x$ 等于其他已列出的数，就会有两个众数，不再有唯一众数。因此唯一众数必须是 $6$ ，平均数也必须是 $6$ 。共有 $7$ 个数，所以总和为 $6\cdot 7 = 42$ 。另一方面，总和还等于 $3+4+5+6+6+7+x$ $= 31+x$ $= 42\text{，}$ 所以 $x = 11$ 。

所以正确答案是 D。

Every listed value except $6$ appears once while $6$ appears twice. If $x$ equals any of the other listed values, then we have two modes, which means we do not have a unique mode. Otherwise, the only value that shows up more than once is $6$ making that the unique mode. This also means $6$ is the mean. Since there are $7$ elements, the sum of the elements is $6\cdot 7 = 42.$ The sum is also $3+4+5+6+6+7+x$ $= 31+x$ $= 42,$ so $x = 11.$

Thus, the answer is D .

12.

$13^{2012}$ 的个位数字是什么？

What is the units digit of $13^{2012}?$

$1$

$3$

$5$

$7$

$9$

答案：A

知识点：个位数字模运算

难度评级：1020

小提示：

只需要考虑底数的个位数字。

Only the units digit of the base matters.

大提示：

$3$ 的幂的个位数字循环长度为 $4$ 。

Powers of $3$ have a units-digit cycle of length $4$ .

视频讲解：

Click to load, then click again to play

文字解答：

要求个位数字，只需计算 $13^{2012} \mod 10$ 。

$\begin{align*}13^{2012} &\equiv 3^{2012} \mod 10\\&\equiv 81^{503} \mod 10\\&\equiv 1^{503} \mod 10\\&\equiv 1 \mod 10\end{align*}$

这说明 $13^{2012}$ 的个位数字与 $1$ 相同，所以 $13^{2012}$ 的个位数字是 $1$ 。

所以正确答案是 A。

We have to find $13^{2012} \mod 10.$ The following is true:

$\begin{align*}13^{2012} &\equiv 3^{2012} \mod 10\\&\equiv 81^{503} \mod 10\\&\equiv 1^{503} \mod 10\\&\equiv 1 \mod 10\end{align*}$

This means $13^{2012}$ has the same units digit as $1,$ so the units digit of $13^{2012}$ is $1.$

Thus, the answer is A .

13.

Jamar 在学校书店买了一些铅笔，每支铅笔价格超过一美分，共支付 $\$1.43$ 。Sharona 买了同样的铅笔，共支付 $\$1.87$ 。Sharona 比 Jamar 多买了多少支铅笔？

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $\$1.43.$ Sharona bought some of the same pencils and paid $\$1.87.$ How many more pencils did Sharona buy than Jamar?

$2$

$3$

$4$

$5$

$6$

答案：C

知识点：最大公约数整除性

难度评级：1310

小提示：

铅笔单价以美分计，必须整除 $143$ 和 $187$ 。

The pencil price in cents divides both $143$ and $187$ .

大提示：

使用 $143$ 和 $187$ 的最大公因数。

Use the greatest common divisor of $143$ and $187$ .

视频讲解：

Click to load, then click again to play

文字解答：

设每支铅笔价格为 $c$ 美分。则 $c$ 同时整除 $143$ 和 $187$ ，且 $c\gt1$ 。

因为 $143=11\cdot13$ ，且 $187=11\cdot17$ ，最大公因数是 $11$ 。所以每支铅笔 $11$ 美分。

Sharona 多付了 $187-143=44$ 美分，所以她多买了 $\frac{44}{11}=4$ 支铅笔。所以正确答案是 C。

Let $c$ be the price of one pencil in cents. Then $c$ divides both $143$ and $187$ , and $c\gt1$ .

Since $143=11\cdot13$ and $187=11\cdot17$ , the greatest common divisor is $11$ . Thus each pencil costs $11$ cents.

Sharona paid $187-143=44$ cents more, so she bought $\frac{44}{11}=4$ more pencils. Thus, the answer is C .

14.

在中学橄榄球联盟 BIG N 中，每支球队与其他每支球队恰好比赛一次。如果 $2012$ 赛季共进行了 $21$ 场联盟比赛，那么 BIG N 联盟有多少支球队？

In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of $21$ conference games were played during the $2012$ season, how many teams were members of the BIG N conference?

$6$

$7$

$8$

$9$

$10$

答案：B

知识点：组合三角形数

难度评级：1100

小提示：

列出每增加一支球队会新增多少场比赛。

List how many games are added when each new team joins.

大提示：

$n$ 支球队时，比赛场数为 $\frac{n(n-1)}{2}$ 。

With $n$ teams, the number of games is $\frac{n(n-1)}{2}$ .

视频讲解：

Click to load, then click again to play

文字解答：

若有 $N$ 支球队，每支球队要与 $N-1$ 支球队比赛。不过，每场比赛涉及 $2$ 支球队，所以把 $N$ 与 $N-1$ 相乘会把每场比赛数两次。因此 $N(N-1) = 42$ 。这给出 $N^2 -N +0.25 = 42.25$ 从而 $(N-0.5)^2 = 6.5^2\text{，}$ 于是 $N-0.5 = 6.5$ $N = 7$

所以正确答案是 B。

Each of the $N$ teams plays $N-1$ games. However, $2$ teams play each game, so multiplying $N$ and $N-1$ would be twice the number of games. Therefore, we know $N(N-1) = 42.$ This leads to $N^2 -N +0.25 = 42.25$ which implies $(N-0.5)^2 = 6.5^2.$ In turn, this suggests: $N-0.5 = 6.5$ $N = 7$

Thus, the answer is B .

15.

大于 $2$ 且除以 $3$ 、 $4$ 、 $5$ 或 $6$ 时余数都为 $2$ 的最小数，位于哪两个数之间？

The smallest number greater than $2$ that leaves a remainder of $2$ when divided by $3,$ $4,$ $5,$ or $6$ lies between what numbers?

$40$ 和 $50$

$40$ and $50$

$51$ 和 $55$

$51$ and $55$

$56$ 和 $60$

$56$ and $60$

$61$ 和 $65$

$61$ and $65$

$66$ 和 $99$

$66$ and $99$

答案：D

知识点：最小公倍数模运算

难度评级：1240

小提示：

从未知数中减去 $2$ 。

Subtract $2$ from the unknown number.

大提示：

结果必须是 $3$ 、 $4$ 、 $5$ 和 $6$ 的公倍数。

The result must be a common multiple of $3$ , $4$ , $5$ , and $6$ .

视频讲解：

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文字解答：

设这个数为 $x$ 。因为它除以 $3$ 、 $4$ 、 $5$ 和 $6$ 时余数都是 $2$ ，所以 $x-2$ 是 $3$ 、 $4$ 、 $5$ 和 $6$ 的公倍数。这表示 $x-2$ 是 $\operatorname{lcm}(3,4,5,6)$ 的倍数，而这个最小公倍数是 $60$ 。因此 $x-2$ 必须是 $60$ 的倍数。下一个满足条件的数出现在 $x-2 = 60 \implies x = 62$ 时。

所以正确答案是 D。

Let the number be $x.$ Since it leaves a remainder of $2$ when divided by $3$ , $4$ , $5$ , and $6,$ we know $x-2$ is a multiple of $3$ , $4$ , $5$ , and $6.$ This means $x-2$ is a multiple of $\operatorname{lcm}(3,4,5,6)$ , which is $60.$ Therefore, $x-2$ must be a multiple of $60.$ The next number such that this occurs is when $x-2 = 60 \implies x = 62 .$

Thus, the answer is D .

16.

数字 $0$ 、 $1$ 、 $2$ 、 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $7$ 、 $8$ 和 $9$ 各使用一次，组成两个五位数，使它们的和尽可能大。下列哪一个可能是其中一个数？

Each of the digits $0,$ $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $7,$ $8,$ and $9$ is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?

$76531$

$86724$

$87431$

$96240$

$97403$

答案：C

知识点：位值最优化

难度评级：1480

小提示：

把最大的数字放在最大的位值上。

Put the largest digits in the largest place values.

大提示：

万位应为 $9$ 和 $8$ ，然后继续成对分配。

The ten-thousands digits should be $9$ and $8$ , then continue in pairs.

视频讲解：

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文字解答：

要让两个五位数的和最大，从左到右的各位数字应尽可能大。因此两个数的第一位必须是 $9$ 或 $8$ ，第二位是 $7$ 或 $6$ ，第三位是 $5$ 或 $4$ ，第四位是 $3$ 或 $2$ ，最后一位是 $1$ 或 $0$ 。

给出的数中，唯一满足这种位置要求的是 $87431$ 。

所以正确答案是 C。

To construct two five-digit numbers, the digits in the largest place values must be as great as possible. Therefore, the leftmost digits must be the greatest two digits. This means the first digit must be either $9$ or $8,$ the second digit must be either $7$ or $6,$ the third digit must be either $5$ or $4,$ the fourth digit must be either $3$ or $2,$ and the last digit must be either $1$ or $0.$

The only one of the given numbers that satisfy this is $87431.$

Thus, the answer is C .

17.

一个边长为整数的正方形被切成 $10$ 个正方形，所有小正方形边长都是整数，并且至少 $8$ 个小正方形的面积为 $1$ 。原正方形边长的最小可能值是多少？

A square with an integer side length is cut into $10$ squares, all of which have integer side length and at least $8$ of which have area $1.$ What is the smallest possible value of the length of the side of the original square?

$3$

$4$

$5$

$6$

$7$

答案：B

知识点：铺砖极端原理

难度评级：1540

小提示：

边长为 $3$ 时，总面积太小，无法切成十个整数边长正方形。

A side length of $3$ gives area too small for ten integer squares.

大提示：

一个 $4$ 乘 $4$ 正方形可以按图示方式切分。

A $4$ by $4$ square can be cut as shown.

视频讲解：

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文字解答：

因为这 $10$ 个小正方形的边长都是正整数，所以每个小正方形的面积至少为 $1$ ，总面积至少为 $10$ 。边长不超过 $3$ 的整数边长正方形面积至多为 $3^2=9$ ，所以大正方形的边长不能小于 $4$ 。

下图把一个 $4\times4$ 正方形切成十个整数边长的正方形，其中八个是单位正方形。因此边长 $4$ 可以达到，而且是最小值。

所以正确答案是 B。

Since all $10$ squares have positive integer side lengths, each has area at least $1$ . Their total area is therefore at least $10$ . A square with integer side length at most $3$ has area at most $3^2=9$ , so its side length cannot be less than $4$ .

The following configuration cuts a $4\times4$ square into ten integer-sided squares, eight of which are unit squares. Therefore side length $4$ is attainable and is the minimum.

Thus, the answer is B .

18.

最小的正整数是多少，要求它既不是质数也不是平方数，并且没有小于 $50$ 的质因数？

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than $50?$

$3127$

$3133$

$3137$

$3139$

$3149$

答案：A

知识点：质因数分解质数

难度评级：1560

小提示：

这个数必须是合数，但不能是平方数。

The number must be composite but not a square.

大提示：

使用不小于 $50$ 的两个最小不同质数。

Use the smallest two distinct primes that are at least $50$ .

视频讲解：

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文字解答：

这个数是合数，不是质数，并且没有小于 $50$ 的质因数。因此最小可能的质因数是 $53$ 、 $59$ 、 $61$ ，依此类推。

像 $53^2$ 这样的平方数不允许，而 $53^3$ 远大于 $53\cdot59$ 。最小的非平方合数是 $53\cdot59=3127$ 。

所以正确答案是 A。

The number is composite, not prime, and has no prime factor below $50$ . The smallest possible prime factors are therefore $53$ , $59$ , $61$ , and so on.

A square such as $53^2$ is not allowed, and $53^3$ is much larger than $53\cdot59$ . The smallest allowed nonsquare composite is $53\cdot59=3127$ .

Thus, the answer is A .

19.

一个罐子里有红色、绿色和蓝色弹珠。除了 $6$ 个以外全是红弹珠，除了 $8$ 个以外全是绿弹珠，除了 $4$ 个以外全是蓝弹珠。罐子里有多少个弹珠？

In a jar of red, green, and blue marbles, all but $6$ are red marbles, all but $8$ are green, and all but $4$ are blue. How many marbles are in the jar?

$6$

$8$

$9$

$10$

$18$

答案：C

知识点：方程组

难度评级：1370

小提示：

将每个“除了……以外全是”语句翻译成另外两种颜色的和。

Translate each “all but” statement into a sum of two colors.

大提示：

把三个方程相加时，每个弹珠被数了两次。

Adding the three equations counts every marble twice.

视频讲解：

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文字解答：

设红、绿、蓝弹珠数分别为 $r,g,b$ 。由题意可知 $r+g+b -r=6\text{，}$ $r+g+b-g = 8\text{，}$ $r+g+b-b = 4\text{。}$ 把这些方程相加，得到 $3(r+g+b) -(r+g+b) = 18$ 。因此 $2(r+g+b) = 18$ ，所以 $r+g+b = 9$ 。也就是说，弹珠总数为 $9$ 。

所以正确答案是 C。

Let $r,g,b$ be the number of red marbles, green marbles, and blue marbles respectively. We then know $r+g+b -r=6,$ $r+g+b-g = 8,$ $r+g+b-b = 4$ by the statements given. Adding these equations yields $3(r+g+b) -(r+g+b) = 18.$ This would mean $2(r+g+b) = 18,$ so $r+g+b = 9.$ Therefore, the sum of all of the marbles is $9.$

Thus, the answer is C .

20.

三个数 $\frac{5}{19}$ 、 $\frac{7}{21}$ 和 $\frac{9}{23}$ 按从小到大排列的正确顺序是什么？

What is the correct ordering of the three numbers $\frac{5}{19} ,$ $\frac{7}{21} ,$ and $\frac{9}{23} ,$ in increasing order?

$\dfrac{9}{23} \lt \dfrac{7}{21} \lt \dfrac{5}{19}$

$\dfrac{5}{19} \lt \dfrac{7}{21} \lt \dfrac{9}{23}$

$\dfrac{9}{23} \lt \dfrac{5}{19} \lt \dfrac{7}{21}$

$\dfrac{5}{19} \lt \dfrac{9}{23} \lt \dfrac{7}{21}$

$\dfrac{7}{21} \lt \dfrac{5}{19} \lt \dfrac{9}{23}$

答案：B

知识点：分数不等式

难度评级：1420

小提示：

将这些分数与 $\frac{1}{3}$ 比较。

Compare the fractions to $\frac{1}{3}$ .

大提示：

把每个分数写成 $\frac{1}{3}$ 加上或减去一个小量。

Write each fraction as $\frac{1}{3}$ plus or minus a small amount.

视频讲解：

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文字解答：

中间的分数为 $\dfrac{7}{21}=\dfrac13$ 。又因为 $3\cdot5=15\lt19$ ，所以 $\dfrac{5}{19}\lt\dfrac13$ ；因为 $3\cdot9=27\gt23$ ，所以 $\dfrac{9}{23}\gt\dfrac13$ 。因此 $\dfrac{5}{19}\lt\dfrac{7}{21}\lt\dfrac{9}{23}\text{。}$

所以正确答案是 B。

The middle fraction is $\dfrac{7}{21}=\dfrac13$ . Also, $\dfrac{5}{19}\lt\dfrac13$ because $3\cdot5=15\lt19$ , while $\dfrac{9}{23}\gt\dfrac13$ because $3\cdot9=27\gt23$ . Therefore $\dfrac{5}{19}\lt\dfrac{7}{21}\lt\dfrac{9}{23}.$

Thus, the answer is B .

21.

Marla 有一个边长 $10$ 英尺的大白色立方体。她还有足够覆盖 $300$ 平方英尺的绿色油漆。Marla 用完所有油漆，在每个面上画出一个居中的白色正方形，周围是绿色边框。每个白色正方形的面积是多少平方英尺？

Marla has a large white cube that has an edge of $10$ feet. She also has enough green paint to cover $300$ square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?

$5\sqrt2$

$10$

$10\sqrt2$

$50$

$50\sqrt2$

答案：D

知识点：表面积正方体

难度评级：1070

小提示：

绿色油漆覆盖了立方体表面积的一半。

The green paint covers half of the cube surface area.

大提示：

剩余未涂区域平均分布在六个面上。

The remaining unpainted area is split equally among the six faces.

视频讲解：

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文字解答：

立方体总表面积为 $6\cdot 10\cdot 10 = 600$ 。因此未被绿色覆盖的面积为 $600-300 = 300$ 平方英尺，平均到每个面是 $\dfrac{300}{6} = 50$ 平方英尺。

所以正确答案是 D。

The total surface area is $6\cdot 10\cdot 10 = 600.$ Therefore, $600-300 = 300$ square feet aren’t covered. This would be $\dfrac{300}{6} = 50$ square feet per face.

Thus, the answer is D .

22.

设 $R$ 是一个由九个互不相同整数构成的集合。其中六个元素是 $2$ 、 $3$ 、 $4$ 、 $6$ 、 $9$ 和 $14$ 。 $R$ 的中位数可能有多少个不同的值？

Let $R$ be a set of nine distinct integers. Six of the elements of the set are $2,$ $3,$ $4,$ $6,$ $9,$ and $14.$ What is the number of possible values of the median of $R?$

$4$

$5$

$6$

$7$

$8$

答案：D

知识点：中位数（数据）分类讨论

难度评级：1720

小提示：

九个整数排序后，中位数是第五个数。

The median is the fifth number after sorting the nine integers.

大提示：

测试从 $3$ 到 $9$ 的哪些整数可以占据第五个位置。

Test which integers from $3$ through $9$ can occupy the fifth position.

视频讲解：

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文字解答：

九个互不相同整数排序后，中位数是第五个数。

中位数不能小于 $3$ ，否则 $3$ 、 $4$ 、 $6$ 、 $9$ 和 $14$ 已经给出至少五个更大的元素。中位数也不能大于 $9$ ，否则 $2$ 、 $3$ 、 $4$ 、 $6$ 和 $9$ 已经给出至少五个更小的元素。

从 $3$ 到 $9$ 的每个整数都可以通过适当选择另外三个整数成为中位数。因此可能的中位数有 $7$ 个。所以正确答案是 D。

In a sorted set of nine distinct integers, the median is the fifth number.

The median cannot be below $3$ , since then $3$ , $4$ , $6$ , $9$ , and $14$ would already give at least five larger elements. It cannot be above $9$ , since $2$ , $3$ , $4$ , $6$ , and $9$ would already give at least five smaller elements.

Each integer from $3$ through $9$ can be made the median by choosing the three missing integers appropriately. Therefore there are $7$ possible medians. Thus, the answer is D .

23.

一个等边三角形和一个正六边形周长相等。如果三角形面积为 $4$ ，那么六边形面积是多少？

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is $4$ , what is the area of the hexagon?

$4$

$5$

$6$

$4\sqrt3$

$6\sqrt3$

答案：C

知识点：等边三角形正多边形面积比

难度评级：1540

小提示：

周长相等使六边形边长为三角形边长的一半。

Equal perimeters make the hexagon side half the triangle side.

大提示：

正六边形由六个全等等边三角形组成。

A regular hexagon is made from six congruent equilateral triangles.

视频讲解：

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文字解答：

设三角形边长为 $s$ 。其周长为 $3s$ ，所以六边形边长为 $\frac{3s}{6} = \frac s2$ 。

如图，正六边形可分成 $6$ 个边长为 $\dfrac{s}{2}$ 的等边三角形。每个小三角形是原三角形按比例 $\dfrac{1}{2}$ 缩小得到的，所以面积缩小为 $(\dfrac{1}{2})^2 = \dfrac{1}{4}$ 。因此每个小三角形面积为 $4 \cdot \dfrac{1}{4} = 1$ 。共有 $6$ 个，所以六边形面积为 $6\cdot 1 = 6$ 。

所以正确答案是 C。

Let the side length of the triangle be $s.$ This means the perimeter is $3s.$ Therefore, the side length for the hexagon is $\frac{3s}{6} = \frac s2.$

A hexagon can be made of $6$ equilateral triangles with side length $\dfrac{s}{2}$ as shown above. Each triangle is the original triangle scaled down by $\dfrac{1}{2},$ so the area is scaled down by $(\dfrac{1}{2})^2 = \dfrac{1}{4} .$ Therefore, the area of each of these triangles is $4 \cdot \dfrac{1}{4} = 1.$ Since there are $6$ of them, the area is $6\cdot 1 = 6.$

Thus, the answer is C .

24.

一个半径为 $2$ 的圆被分成四段全等圆弧。将这四段圆弧连接成图中的星形。星形面积与原圆面积之比是多少？

A circle of radius $2$ is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?

$\dfrac{4-\pi}{\pi}$

$\dfrac{1}\pi$

$\dfrac{\sqrt2}{\pi}$

$\dfrac{\pi-1}{\pi}$

$\dfrac{3}\pi$

答案：A

知识点：圆面积面积分割

难度评级：1860

小提示：

将四段圆弧重新放回原圆中思考。

Rearrange the four circular arcs back inside the original circle.

大提示：

将星形与连接四个切分点形成的正方形比较。

Compare the star with the square formed by joining the four cut points.

视频讲解：

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文字解答：

原圆面积为 $\pi\cdot2^2=4\pi$ 。

连接四个四分之一圆弧端点形成一个正方形。这个正方形的两条对角线分别是 $4$ 和 $4$ ，所以面积为 $\frac12\cdot4\cdot4=8$ 。

圆内但正方形外的部分面积为 $4\pi-8$ 。这四块与正方形内但星形外的四块全等。

因此星形面积为 $8-(4\pi-8)=16-4\pi$ 。所求比值为 $\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}$ 。所以正确答案是 A。

The area of the original circle is $\pi\cdot2^2=4\pi$ .

Join the four quarter-circle endpoints to form a square. The square has diagonals $4$ and $4$ , so its area is $\frac12\cdot4\cdot4=8$ .

The part inside the circle but outside this square has area $4\pi-8$ . Those four pieces are congruent to the pieces inside the square but outside the star.

Thus the star area is $8-(4\pi-8)=16-4\pi$ . The desired ratio is $\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}$ . Thus, the answer is A .

25.

一个面积为 $4$ 的正方形内接于一个面积为 $5$ 的正方形，较小正方形的每个顶点分别在较大正方形的一条边上。较小正方形的一个顶点将较大正方形的一条边分成两段，长度分别为 $a$ 和 $b$ 。 $ab$ 的值是多少？

A square with area $4$ is inscribed in a square with area $5$ , with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$ and the other of length $b$ . What is the value of $ab ?$

$\dfrac{1}5$

$\dfrac{2}5$

$\dfrac{1}{2}$

$1$

$4$

答案：C

知识点：正方形（几何）全等（几何）面积分割

难度评级：1790

小提示：

两个正方形之间的面积被分成四个全等直角三角形。

The area between the squares is split into four congruent right triangles.

大提示：

每个小三角形的面积是 $\frac{ab}{2}$ 。

Each small triangle has area $\frac{ab}{2}$ .

视频讲解：

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文字解答：

四个三角形都可以通过旋转互相得到，所以它们全等。因此可以如图标出 $a$ 。两个正方形之间的总面积为 $5-4 = 1$ ，所以有 $4$ 个全等三角形的总面积为 $1$ 。每个三角形面积为 $\dfrac{1}{4}$ 。每个三角形的面积也等于 $\frac {ab}2$ ，所以 $\frac{ab}2 = \dfrac{1}{4}$ 。因此 $ab = \dfrac{1}{2}$ 。

所以正确答案是 C。

Since all the triangles can be made from each other by rotating them around, they are all congruent. Therefore, we can place the $a$ as we have. The total area of the triangles is $5-4 = 1,$ so we have $4$ congruent triangles with a combined area of $1.$ This means the area of each triangle is $\dfrac{1}{4}.$ The area of each triangle is also $\frac {ab}2,$ so $\frac{ab}2 = \dfrac{1}{4}.$ This means $ab = \dfrac{1}{2} .$

Thus, the correct answer is C .