2016 AMC 8 真题与在线限时测验

计时

40:00

1.

有史以来最长的职业网球比赛总共持续了 $11$ 小时 $5$ 分钟。这是多少分钟？

The longest professional tennis match ever played lasted a total of $11$ hours and $5$ minutes. How many minutes was this?

$605$

$655$

$665$

$1005$

$1105$

答案：C

知识点：单位换算

难度评级：370

小提示：

先把小时转换成分钟

Convert the hours to minutes first.

大提示：

将 $11$ 乘以 $60$ 后，再加上额外的 $5$ 分钟。

Add the extra $5$ minutes after multiplying $11$ by $60$ .

视频讲解：

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文字解答：

一小时有 $60$ 分钟，所以总时间为 $60 \cdot 11 + 5 = 665$ 分钟。

所以正确答案是 C。

There are $60$ minutes in an hour, so the total time is $60 \cdot 11 + 5 = 665$ minutes.

Thus, C is the correct answer.

2.

在长方形 $ABCD$ 中， $AB=6$ ， $AD=8$ 。点 $M$ 是 $\overline{AD}$ 的中点。 $\triangle AMC$ 的面积是多少？

In rectangle $ABCD,$ $AB=6$ and $AD=8.$ Point $M$ is the midpoint of $\overline{AD}.$ What is the area of $\triangle AMC?$

$12$

$15$

$18$

$20$

$24$

答案：A

知识点：三角形面积中点

难度评级：450

小提示：

因为 $M$ 是 $\overline{AD}$ 的中点，先求 $AM$

Since $M$ is the midpoint of $\overline{AD}$ , find $AM$ .

大提示：

用 $AM$ 作底，长方形的竖直边长作高

Use $AM$ as the base and the rectangle’s vertical side length as the height.

视频讲解：

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文字解答：

因为 $M$ 是 $\overline{AD}$ 的中点，所以底边 $AM$ 长 $4$ 。从 $C$ 到 $\overline{AM}$ 的高为 $AB=6$ 。因此面积为 $\dfrac{1}{2} \cdot 4 \cdot 6 = 12$ 。

所以正确答案是 A。

Since $M$ is the midpoint of $\overline{AD},$ the base $AM$ has length $4.$ The altitude from $C$ to $\overline{AM}$ has length $AB=6.$ The area is therefore $\dfrac{1}{2} \cdot 4 \cdot 6 = 12.$

Thus, A is the correct answer.

3.

四名学生参加考试。其中三人的分数是 $70$ ， $80$ 和 $90$ 。如果四人平均分为 $70$ ，那么剩下那人的分数是多少？

Four students take an exam. Three of their scores are $70,$ $80,$ and $90.$ If the average of their four scores is $70,$ then what is the remaining score?

$40$

$50$

$55$

$60$

$70$

答案：A

知识点：平均数

难度评级：450

小提示：

先求平均分为 $70$ 时所需的总分

First find the total score needed for an average of $70$ .

大提示：

从所需总分中减去三个已知分数

Subtract the three known scores from the needed total.

视频讲解：

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文字解答：

由平均分可知四人总分为 $4 \cdot 70 = 280$ 。因此剩下的分数为 $280 - 70 - 80 - 90 = 40\text{。}$

所以正确答案是 A。

From the average, we can calculate the sum of the scores to be $4 \cdot 70 = 280.$ This means that the remaining score is $280 - 70 - 80 - 90 = 40.$

Thus, A is the correct answer.

4.

奇努小时候能在 $3$ 小时 $30$ 分钟内跑 $15$ 英里。现在他年老了，能在 $4$ 小时内走 $10$ 英里。现在他走一英里比小时候跑一英里多花多少分钟？

When Cheenu was a boy he could run $15$ miles in $3$ hours and $30$ minutes. As an old man he can now walk $10$ miles in $4$ hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?

$6$

$10$

$15$

$18$

$30$

答案：B

知识点：速率单位换算

难度评级：720

小提示：

比较两种速度对应的每英里分钟数

Compare the two paces in minutes per mile.

大提示：

先将 $3$ 小时 $30$ 分钟转换成分钟，再除以 $15$

Convert $3$ hours $30$ minutes to minutes before dividing by $15$ .

视频讲解：

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文字解答：

为了更容易比较速度，将它们都换成每英里所需分钟数。

小时候他跑 $15$ 英里用了 $3 \cdot 60 + 30 = 210$ 分钟，所以速度为 $\frac{210}{15} = 14$ 分钟每英里。

现在他走 $10$ 英里用了 $4 \cdot 60 = 240$ 分钟，所以速度为 $\frac{240}{10} = 24$ 分钟每英里。

相减得，现在他走一英里多花 $10$ 分钟。

所以正确答案是 B。

To better compare the rates, we can change his speed into minutes per mile.

As a boy he ran $15$ miles in $3 \cdot 60 + 30 = 210$ minutes, which means that he ran at a pace of $\frac{210}{15} = 14$ minutes per mile.

As an adult, he can walk $10$ miles in $4 \cdot 60 = 240$ minutes, which means he walks at a pace of $\frac{240}{10} = 24$ minutes per mile.

Subtracting the two, we get that he takes $10$ more minutes to walk a mile as an adult.

Thus, B is the correct answer.

5.

数 $N$ 是一个两位数，满足以下性质：

第一个条件是 $N$ 除以 $9$ 的余数为 $1$ ；第二个条件是 $N$ 除以 $10$ 的余数为 $3$ 。

$N$ 除以 $11$ 的余数是多少？

The number $N$ is a two-digit number with the following properties:

When $N$ is divided by $9,$ the remainder is $1.$
When $N$ is divided by $10,$ the remainder is $3.$

What is the remainder when $N$ is divided by $11?$

$0$

$2$

$4$

$5$

$7$

答案：E

知识点：模运算系统列举

难度评级：940

小提示：

除以 $10$ 的余数告诉你 $N$ 的个位数字

The remainder when divided by $10$ tells you the units digit of $N$ .

大提示：

列出以 $3$ 结尾的两位数，并测试模 $9$ 的条件

List the two-digit numbers ending in $3$ and test the condition modulo $9$ .

视频讲解：

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文字解答：

第二个条件说明 $N$ 的个位数字是 $3$ ，所以可能的数为 $13$ 、 $23$ 、 $33$ 、 $43$ 、 $53$ 、 $63$ 、 $73$ 、 $83$ 、 $93$ 。其中只有 $73$ 除以 $9$ 余 $1$ ，因为它的各位数字之和是 $10$ 。最后， $73=6\cdot11+7$ ，所以所求余数是 $7$ 。

所以正确答案是 E。

The second condition says that $N$ ends in $3$ , so the possibilities are $13,$ $23,$ $33,$ $43,$ $53,$ $63,$ $73,$ $83,$ $93.$ Of these, only $73$ leaves remainder $1$ when divided by $9$ , because its digit sum is $10$ . Finally, $73=6\cdot11+7$ , so the requested remainder is $7$ .

Thus, E is the correct answer.

6.

下方条形图表示 $19$ 个人的名字长度（字母数）。这些名字长度的中位数是多少？

The following bar graph represents the length (in letters) of the names of $19$ people. What is the median length of these names?

$3$

$4$

$5$

$6$

$7$

答案：B

知识点：中位数（数据）数据与图表解读

难度评级：770

小提示：

有 $19$ 个名字，排序后中位数是第 $10$ 个值

With $19$ names, the median is the $10$ th value after sorting.

大提示：

从最短名字开始累计频数

Count cumulative frequencies from the shortest names upward.

视频讲解：

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文字解答：

因为有 $19$ 个人，每人对应一个名字长度，所以中间的长度是第十个值。从左侧开始数，数到第十个时得到 $4$ 。

所以正确答案是 B。

Since there are $19$ people, each with one corresponding name length, the middle length will be the tenth one. Counting from the left side, the tenth value that we arrive upon is $4.$

Thus, B is the correct answer.

7.

下列哪个数不是完全平方数？

Which of the following numbers is not a perfect square?

$1^{2016}$

$2^{2017}$

$3^{2018}$

$4^{2019}$

$5^{2020}$

答案：B

知识点：完全平方数指数

难度评级：870

小提示：

偶数指数一定产生完全平方数

Even exponents automatically make perfect squares.

大提示：

也要记得，完全平方数作为底数的任意正整数次幂仍是平方数

Also remember that a perfect-square base raised to any positive integer power stays a square.

视频讲解：

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文字解答：

任何指数为偶数的数都是完全平方数，所以可排除 A、C 和 E。另外，平方数的任意次幂仍然是平方数，所以也排除 D。

所以正确答案是 B。

Since any number with an even exponent is a perfect square, we can eliminate A, C, and E. Also, a square number to any power remains a square number, so that rules out D.

Thus, B is the correct answer.

8.

求下列表达式的值： $\begin{gathered} 100 - 98 + 96 - 94 + 92 - 90 \\ {}+ \cdots + 8 - 6 + 4 - 2 \end{gathered}$

Find the value of the expression $\begin{gathered} 100 - 98 + 96 - 94 + 92 - 90 \\ {}+ \cdots + 8 - 6 + 4 - 2. \end{gathered}$

$20$

$40$

$50$

$80$

$100$

答案：C

知识点：配对与分组

难度评级：900

小提示：

将表达式按相邻两项分组

Group the expression into pairs of consecutive terms.

大提示：

统计有多少对形如 $n-(n-2)$ 的项

Count how many pairs of the form $n-(n-2)$ appear.

视频讲解：

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文字解答：

可以把这个和分组为 $\begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2) \end{gathered}$ 每一对的值都是 $2$ ，共有 $25$ 对。因此总和为 $2 \cdot 25 = 50$ 。

所以正确答案是 C。

We can group the sum as follows: $\begin{gathered} (100 - 98) + (96 - 94) \\ {}+ \cdots + (4 - 2). \end{gathered}$ Note that each pair evaluates to $2$ and there are $25$ pairs. Therefore, the total sum is $2 \cdot 25 = 50.$

Thus, C is the correct answer.

9.

$2016$ 的不同质数因子之和是多少？

What is the sum of the distinct prime integer divisors of $2016?$

$9$

$12$

$16$

$49$

$63$

答案：B

知识点：质因数分解

难度评级：960

小提示：

将 $2016$ 分解为质因数

Factor $2016$ into primes.

大提示：

“不同”表示每个质因子只数一次

The word “distinct” means to count each prime divisor only once.

视频讲解：

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文字解答：

将 $2016$ 质因数分解为 $2^5 \cdot 3^2 \cdot 7$ 。因此 $2016$ 的质因子是 $2$ 、 $3$ 和 $7$ 。它们的和为 $12$ ，所以正确答案是 B。

We can prime factorize $2016$ as $2^5 \cdot 3^2 \cdot 7.$ This shows that the prime divisors of $2016$ are $2,$ $3,$ and $7.$ The sum of these is $12,$ so B is the correct answer.

10.

假设 $a \ast b$ 表示 $3a - b$ 。若下式成立，那么 $x$ 的值是多少？ $2 \ast (5 \ast x) = 1\text{？}$

Suppose that $a \ast b$ means $3a - b.$ What is the value of $x$ if $2 \ast (5 \ast x) = 1?$

$\dfrac{1}{10}$

$2$

$\dfrac{10}{3}$

$10$

$14$

答案：D

知识点：自定义运算一次方程

难度评级：1030

小提示：

先化简内部运算 $5\ast x$

Start by simplifying the inner operation $5\ast x$ .

大提示：

从内到外两次应用定义 $a\ast b=3a-b$

Apply the definition $a\ast b=3a-b$ twice, from the inside out.

视频讲解：

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文字解答：

先算内层运算，得 $5\ast x=15-x$ 。再应用一次定义： $\begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9\text{。} \end{aligned}$ 因此 $x-9=1$ ，所以 $x=10$ 。

所以正确答案是 D。

First, $5\ast x=15-x$ . Applying the operation again gives $\begin{aligned} 2\ast(15-x)&=6-(15-x)\\ &=x-9. \end{aligned}$ Thus $x-9=1$ , so $x=10$ .

Thus, D is the correct answer.

11.

有多少个两位数满足以下性质？

将这个数与它的数字反序得到的数相加，和为 $132$ 。

Determine how many two-digit numbers satisfy the following property:

When the number is added to the number obtained by reversing its digits, the sum is $132.$

$5$

$7$

$9$

$11$

$12$

答案：B

知识点：数字位值

难度评级：1100

小提示：

将两位数写成 $10a+b$

Write the two-digit number as $10a+b$ .

大提示：

加上反序数会得到 $11(a+b)$

Adding the reversed number gives $11(a+b)$ .

视频讲解：

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文字解答：

设十位数字为 $a$ ，个位数字为 $b$ 。原数与反序数相加得 $\begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132\text{，} \end{aligned}$ 所以 $a+b=12$ 。可行的数字对为 $(3,9)$ 、 $(4,8)$ 、 $(5,7)$ 、 $(6,6)$ 、 $(7,5)$ 、 $(8,4)$ 、 $(9,3)$ ，共得到 $7$ 个两位数。

所以正确答案是 B。

Let $a$ be the tens digit and $b$ the units digit. Adding the number and its reversal gives $\begin{aligned} (10a+b)+(10b+a)&=132\\ 11(a+b)&=132, \end{aligned}$ so $a+b=12$ . The valid digit pairs are $(3,9),$ $(4,8),$ $(5,7),$ $(6,6),$ $(7,5),$ $(8,4),$ $(9,3),$ giving $7$ numbers.

Thus, B is the correct answer.

12.

杰斐逊中学的男生和女生人数相同。女生中的四分之三和男生中的三分之二参加了实地考察。参加实地考察的学生中，女生占几分之几？

Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students on the field trip were girls?

$\dfrac{1}{2}$

$\dfrac{9}{17}$

$\dfrac{7}{13}$

$\dfrac{2}{3}$

$\dfrac{14}{15}$

答案：B

知识点：分数比与比例

难度评级：1170

小提示：

选一个方便且相等的男生和女生人数

Choose a convenient equal number of boys and girls.

大提示：

用 $12$ 名男生和 $12$ 名女生可以让两个分数都变成整数

Using $12$ boys and $12$ girls makes both fractions whole numbers.

视频讲解：

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文字解答：

假设有 $12$ 名女生和 $12$ 名男生，那么参加实地考察的有 $9$ 名女生和 $8$ 名男生，所以参加者中女生所占的比例是 $\dfrac{9}{9+8}=\dfrac9{17}$ 。

所以正确答案是 B。

Suppose there are $12$ girls and $12$ boys. Then $9$ girls and $8$ boys go on the field trip, so the fraction of field-trip students who are girls is $\dfrac{9}{9+8}=\dfrac9{17}.$

Thus, B is the correct answer.

13.

从集合 $\{ - 2, -1, 0, 3, 4, 5\}$ 中随机选出两个不同的数并相乘。乘积为 $0$ 的概率是多少？

Two different numbers are randomly selected from the set $\{ - 2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is $0?$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

答案：D

知识点：基本概率数对计数

难度评级：1020

小提示：

乘积为 $0$ 当且仅当选出的数中有一个是 $0$

A product is $0$ exactly when one of the selected numbers is $0$ .

大提示：

在所有选两个不同数的方式中，统计包含 $0$ 的选择

Count selections containing $0$ out of all ways to choose two different numbers.

视频讲解：

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文字解答：

从六个数中选两个不同的数共有 $\binom62=15$ 种方法。乘积为 $0$ 当且仅当选出的一个数是 $0$ ，另一个数有 $5$ 种选择。因此所求概率为 $\dfrac5{15}=\dfrac13$ 。

所以正确答案是 D。

There are $\binom62=15$ ways to select two different numbers. A product is $0$ exactly when one selected number is $0$ , and there are $5$ choices for the other number. Thus the probability is $\dfrac5{15}=\dfrac13.$

Thus, D is the correct answer.

14.

卡尔的汽车每 $35$ 英里用一加仑汽油，油箱加满时可装 $14$ 加仑。

某天，卡尔出发时油箱是满的，开了 $350$ 英里后买了 $8$ 加仑汽油，然后继续开到目的地。到达时，他的油箱还有一半满。卡尔那天一共开了多少英里？

Karl’s car uses a gallon of gas every $35$ miles, and his gas tank holds $14$ gallons when it is full.

One day, Karl started with a full tank of gas, drove $350$ miles, bought $8$ gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?

$525$

$560$

$595$

$665$

$735$

答案：A

知识点：速率

难度评级：1170

小提示：

先跟踪汽油加仑数，而不是英里数

Track gallons of gas rather than miles at first.

大提示：

开 $350$ 英里并加油后，将新的油量与半箱油比较

After the $350$ -mile drive and the refill, compare the new amount of gas to half a tank.

视频讲解：

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文字解答：

卡尔开 $350$ 英里用了 $\frac{350}{35}$ 加仑汽油。

他又加了 $8$ 加仑到原来剩下的 $14 - 10 = 4$ 加仑中，一共有 $12$ 加仑。

到达时油箱半满，所以后来又用了 $12 - 7 = 5$ 加仑，对应 $5 \cdot 35 = 175$ 英里。

因此他行驶的总路程为 $350+175=525$ 英里。

所以正确答案是 A。

If Karl drove $350$ miles, then he used $\frac{350}{35}$ gallons of gas.

When he bought more gas, he added $8$ gallons to $14 - 10 = 4$ gallons, attaining a total of $12$ gallons.

If his tank was half full when he arrived, he used $12 - 7 = 5$ gallons, which equates to $5 \cdot 35 = 175$ miles.

Therefore, he traveled a total distance of $350+175=525$ miles.

Thus, A is the correct answer.

15.

能整除 $13^4 - 11^4$ 的最大的 $2$ 的幂是多少？

What is the largest power of $2$ that is a divisor of $13^4 - 11^4?$

$8$

$16$

$32$

$64$

$128$

答案：C

知识点：平方差 2的幂

难度评级：1310

小提示：

将 $13^4-11^4$ 分解为平方差

Factor $13^4-11^4$ as a difference of squares.

大提示：

分解后只统计乘积中 $2$ 的幂

After factoring, count only the powers of $2$ in the product.

视频讲解：

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文字解答：

使用平方差分解：

$\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435 \end{aligned}\text{。}$

因为 $435$ 是奇数，所以 $32$ 是能整除该表达式的最大的 $2$ 的幂。

所以正确答案是 C。

We can factor this expression using difference of squares.

$\begin{aligned} 13^4-11^4 &=(13^2+11^2)\\ &\qquad\cdot(13^2-11^2)\\ &=290\cdot48\\ &=32\cdot435. \end{aligned}$

Because $435$ is odd, $32$ is the largest power of $2$ that divides the expression.

Thus, C is the correct answer.

16.

安妮和邦妮正在绕一个 $400$ 米椭圆形跑道跑圈。她们同时出发，但安妮已经领先，因为她比邦妮快 $25\%$ 。安妮第一次追上并超过邦妮时，她已经跑了多少圈？

Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

$1\dfrac{1}{4}$

$3\dfrac{1}{3}$

$4$

$5$

$25$

答案：D

知识点：相对速度路程、速度与时间比与比例

难度评级：1240

小提示：

快 $25\%$ 表示同样时间内安妮跑的距离是邦妮的 $\frac{5}{4}$

A $25\%$ faster pace means Annie runs $\frac{5}{4}$ as far as Bonnie in the same time.

大提示：

安妮第一次追上邦妮时，她已经多跑整整一圈

Annie first passes Bonnie when she has gained one full lap.

视频讲解：

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文字解答：

因为安妮比邦妮快 $25\%$ ，所以邦妮每完成一圈时，安妮完成 $1 \dfrac{1}{4}$ 圈。也就是说，邦妮每跑一圈，安妮领先四分之一圈。

安妮要整整领先一圈，需要邦妮完成 $4$ 圈，此时安妮完成 $5$ 圈。

所以正确答案是 D。

Since Annie is $25\%$ faster than Bonnie, for every lap Bonnie finishes, Annie completes $1 \dfrac{1}{4}$ laps. Therefore, Annie gains a quarter lap every time Bonnie finishes a lap.

With this in mind, for Annie to completely lap Bonnie, Bonnie must finish $4$ laps, which means that Annie finished $5$ laps.

Thus, D is the correct answer.

17.

弗雷德银行的 ATM 密码由 $0$ 到 $9$ 中的四个数字组成，允许数字重复。如果密码不能以序列 $9$ 、 $1$ 、 $1$ 开头，那么有多少个可能密码？

An ATM password at Fred’s Bank is composed of four digits from $0$ to $9,$ with repeated digits allowable. If no password may begin with the sequence $9,$ $1,$ $1,$ then how many passwords are possible?

$30$

$7290$

$9000$

$9990$

$9999$

答案：D

知识点：乘法原理补集计数

难度评级：1020

小提示：

先统计允许前导零时所有四位密码数

First count all four-digit passwords when leading zeroes are allowed.

大提示：

再减去前三位被固定为 $9$ 、 $1$ 、 $1$ 的密码

Then subtract the passwords whose first three digits are forced to be $9,$ $1,$ $1$ .

视频讲解：

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文字解答：

没有限制时，密码总数为 $10^4$ 。这个条件排除了 $10$ 个密码，因为前 $3$ 位已经固定，最后一位可以任意。因此可接受密码数为 $10{,}000 - 10 = 9990\text{。}$

所以正确答案是 D。

The total number of passwords with no conditions is $10^4.$ The condition removes $10$ possible passwords since the first $3$ are determined, and the last one can be anything. Therefore, the number of acceptable passwords is $10{,}000 - 10 = 9990.$

Thus, D is the correct answer.

18.

在一次全地区田径赛中， $216$ 名短跑运动员参加 $100$ 米短跑比赛。跑道有 $6$ 条道，所以一次只能有 $6$ 名运动员比赛。每场比赛结束后，五名非获胜者被淘汰，获胜者将在后面的比赛中继续参赛。

需要多少场比赛才能决出短跑冠军？

In an All-Area track meet, $216$ sprinters enter a $100$ -meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race.

How many races are needed to determine the champion sprinter?

$36$

$42$

$43$

$60$

$72$

答案：C

知识点：基本计数

难度评级：1170

小提示：

思考每场比赛淘汰多少名运动员

Think about how many sprinters are eliminated in each race.

大提示：

要从 $216$ 名运动员中留下一个冠军，必须淘汰除一人外的所有人

To leave one champion from $216$ sprinters, all but one sprinter must be eliminated.

视频讲解：

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文字解答：

每场比赛淘汰 $5$ 人。要决出冠军，需要淘汰 $215$ 人。因此需要 $\frac{215}{5} = 43$ 场比赛。

所以正确答案是 C。

Note that each race eliminates $5$ people. For there to be a winner, $215$ must be eliminated. Therefore, $\frac{215}{5} = 43$ races are required to eliminate this number of people.

Thus, C is the correct answer.

19.

$25$ 个连续偶整数的和是 $10{,}000$ 。这 $25$ 个连续偶整数中最大的是多少？

The sum of $25$ consecutive even integers is $10{,}000.$ What is the largest of these $25$ consecutive even integers?

$360$

$388$

$412$

$416$

$424$

答案：E

知识点：等差数列

难度评级：1100

小提示：

奇数个连续项的平均数就是中间项

For an odd number of consecutive terms, the average is the middle term.

大提示：

有 $12$ 个偶整数比中间项大

There are $12$ even integers larger than the middle term.

视频讲解：

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文字解答：

这些数的平均数为 $\frac{10{,}000}{25} = 400$ 。因为共有二十五个连续偶整数，平均数就是中间项。最大数比中间项大 $12$ 个偶数间隔，所以等于 $400 + 12 \cdot 2 = 424$ 。

所以正确答案是 E。

The average of these numbers is $\frac{10{,}000}{25} = 400.$ The largest number is $12$ even numbers away, which means that it equals $400 + 12 \cdot 2 = 424.$

Thus, E is the correct answer.

20.

$a$ 和 $b$ 的最小公倍数是 $12$ ， $b$ 和 $c$ 的最小公倍数是 $15$ 。 $a$ 和 $c$ 的最小公倍数的最小可能值是多少？

The least common multiple of $a$ and $b$ is $12,$ and the least common multiple of $b$ and $c$ is $15.$ What is the least possible value of the least common multiple of $a$ and $c?$

$20$

$30$

$60$

$120$

$180$

答案：A

知识点：最小公倍数

难度评级：1390

小提示：

因为 $b$ 出现在两个最小公倍数中，先考虑 $b$ 的可能值

Since $b$ appears in both least common multiples, start with possible values of $b$ .

大提示：

尽量让 $b$ 吸收 $12$ 和 $15$ 中共有的因子 $3$ 。

Try to make $b$ absorb the common factor $3$ from $12$ and $15$ .

视频讲解：

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文字解答：

$b$ 必须同时整除 $12$ 和 $15$ ，所以它只能是 $1$ 或 $3$ 。

如果 $b = 1$ ，则可取 $a = 12$ 、 $c = 15$ ，它们的最小公倍数为 $60$ 。如果 $b = 3$ ，则可取 $a = 4$ 、 $c = 5$ ，此时最小公倍数为 $20$ 。

所以正确答案是 A。

We know that $b$ has to divide both $12$ and $15,$ so it must equal either $1$ or $3.$

If $b = 1,$ then $a = 12$ and $c = 15,$ making their least common multiple $60.$ If $b = 3,$ then we may take $a = 4$ and $c = 5.$ The least common multiple in this scenario is $20.$

Thus, A is the correct answer.

21.

一个盒子里有 $3$ 枚红色筹码和 $2$ 枚绿色筹码。每次随机抽出一枚筹码且不放回，直到 $3$ 枚红色筹码都被抽出，或直到两枚绿色筹码都被抽出。 $3$ 枚红色筹码都被抽出的概率是多少？

A box contains $3$ red chips and $2$ green chips. Chips are drawn randomly, one at a time without replacement, until all $3$ of the reds are drawn or until both green chips are drawn. What is the probability that the $3$ reds are drawn?

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

答案：B

知识点：基本概率对称性

难度评级：1490

小提示：

停止时的结果由完整排列中哪种颜色最后出现决定

The stopping result is determined by which color appears last in the full ordering of the chips.

大提示：

在五个筹码位置中统计两枚绿色筹码的位置

Count equally likely positions for the two green chips among the five total chip positions.

视频讲解：

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文字解答：

$3$ 枚红色筹码先于两枚绿色筹码都被抽出，当且仅当完整抽取顺序中的最后一枚筹码是绿色。选择两枚绿色筹码位置的等可能方式共有 $10$ 种，其中最后一个位置为绿色的有 $4$ 种。

因此所求概率为 $\dfrac{4}{10} = \dfrac{2}{5}\text{。}$

所以正确答案是 B。

The $3$ reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are $10$ equally likely ways to choose the two positions of the green chips, and $4$ of them have a green chip in the last position.

Therefore, the desired probability is $\dfrac{4}{10} = \dfrac{2}{5}.$

Thus, B is the correct answer.

22.

下图中长方形 $DEFA$ 是一个 $3 \times 4$ 长方形，且 $DC=CB=BA=1$ 。“蝙蝠翅膀”（阴影部分）的面积是

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1.$ The area of the “bat wings” (shaded area) is

$2$

$2 \dfrac{1}{2}$

$3$

$3 \dfrac{1}{2}$

$4$

答案：C

知识点：相似三角形面积

难度评级：1640

小提示：

使用形成蝙蝠翅膀的两条对角线的交点

Use the intersection point of the two diagonals that form the bat wings.

大提示：

相似三角形给出顶部被切掉的小三角形的高度

Similar triangles give the height of the small triangle cut off at the top.

视频讲解：

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文字解答：

设 $I$ 为 $\overline{AD}$ 的中点， $G$ 为 $\overline{EF}$ 的中点。再设 $H$ 为 $\overline{CF}$ 与 $\overline{BE}$ 的交点。

$\triangle BCE$ 的面积是 $\dfrac12\cdot1\cdot4=2$ 。由对称性可知， $\triangle BCH$ 与 $\triangle EFH$ 相似。它们的底边之比为 $1 : 3$ ，所以高之比也相同。因此 $3IH = HG$ ，从而 $IH = 1$ 。

所以 $[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12$ 。于是 $[\triangle ECH]=2-\dfrac12=\dfrac32$ 。图形关于中线对称，因此蝙蝠双翼的总面积为 $2\cdot\dfrac32=3$ 。

所以正确答案是 C。

Define $I$ to be the midpoint of $\overline{AD}$ and $G$ to be the midpoint of $\overline{EF}.$ Also define $H$ to be the intersection of $\overline{CF}$ and $\overline{BE}.$

The area of $\triangle BCE$ is $\dfrac12\cdot1\cdot4=2.$ By symmetry, we can see that $\triangle BCH$ and $\triangle EFH$ are similar. Since their bases are in a $1 : 3$ ratio, so are their altitudes. This means that $3IH = HG,$ which implies that $IH = 1.$

Therefore, $[\triangle BCH]=\dfrac12\cdot1\cdot1=\dfrac12.$ It follows that $[\triangle ECH]=2-\dfrac12=\dfrac32.$ Since the figure is symmetric, the total area of the bat wings is $2\cdot\dfrac32=3.$

Thus, C is the correct answer.

23.

两个全等圆分别以点 $A$ 和 $B$ 为圆心，并且每个圆都经过另一个圆的圆心。经过 $A$ 和 $B$ 的直线延长后与两个圆分别交于点 $C$ 和 $D$ 。

两圆相交于两点，其中一点为 $E$ 。 $\angle CED$ 的度数是多少？

Two congruent circles centered at points $A$ and $B$ each pass through the other circle’s center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D.$

The circles intersect at two points, one of which is $E.$ What is the degree measure of $\angle CED?$

$90$

$105$

$120$

$135$

$150$

答案：C

知识点：等边三角形圆周角导角

难度评级：1510

小提示：

两个圆心和一个交点构成等边三角形

The centers and one intersection point form an equilateral triangle.

大提示：

使用直径所对圆周角是直角这一事实

Use the fact that angles subtending diameters are right angles.

视频讲解：

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文字解答：

因为 $AE = EB = AB$ ，它们都是全等圆的半径，所以三点构成等边三角形， $\angle AEB = 60^{\circ}$ 。

又因为 $\overline{DB}$ 和 $\overline{AC}$ 是直径，所以 $\angle DEB=\angle AEC=90^\circ$ 。因此 $\begin{aligned} \angle CED &= \angle DEB+\angle AEC\\ &\qquad-\angle AEB\\ &=90^\circ+90^\circ-60^\circ\\ &=120^\circ\text{。} \end{aligned}$

所以正确答案是 C。

We know that $AE = EB = AB$ since they are all radii of congruent circles, so they form an equilateral triangle, which means that $\angle AEB = 60^{\circ}.$

The segments $\overline{DB}$ and $\overline{AC}$ are diameters, so $\angle DEB=\angle AEC=90^\circ.$ Therefore, $\begin{aligned} \angle CED &= \angle DEB+\angle AEC\\ &\qquad-\angle AEB\\ &=90^\circ+90^\circ-60^\circ\\ &=120^\circ. \end{aligned}$

Thus, C is the correct answer.

24.

数字 $1$ 、 $2$ 、 $3$ 、 $4$ 、 $5$ 各使用一次，写成五位数 $PQRST$ 。三位数 $PQR$ 能被 $4$ 整除，三位数 $QRS$ 能被 $5$ 整除，三位数 $RST$ 能被 $3$ 整除。 $P$ 是多少？

The digits $1,$ $2,$ $3,$ $4,$ and $5$ are each used once to write a five-digit number $PQRST.$ The three-digit number $PQR$ is divisible by $4,$ the three-digit number $QRS$ is divisible by $5,$ and the three-digit number $RST$ is divisible by $3.$ What is $P?$

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：整除性分类讨论

难度评级：1580

小提示：

能被 $5$ 整除的条件确定了 $S$ 。

The divisibility by $5$ condition determines $S$ .

大提示：

确定 $S$ 后，列出使 $PQR$ 能被 $4$ 整除的可能两位结尾 $QR$ 。

Once $S$ is known, list the possible two-digit endings $QR$ that make $PQR$ divisible by $4$ .

视频讲解：

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文字解答：

因为 $QRS$ 能被 $5$ 整除，所以 $S = 5$ 。

因为 $PQR$ 能被 $4$ 整除，所以 $QR$ 可以是 $12$ 、 $24$ 或 $32$ 。

如果 $QR=12$ ，则 $R=2$ ，那么 $RST=25T$ ，用剩余数字无法使其被 $3$ 整除。如果 $QR=32$ ，则 $R=2$ ，也遇到同样问题。因此 $QR=24$ ，于是 $RST=45T$ 。在剩余数字中，只有 $T=3$ 使 $453$ 能被 $3$ 整除。

因此 $PQRST = 12453$ 。

所以正确答案是 A。

Since $QRS$ is divisible by $5,$ we know that $S = 5.$

Since $PQR$ is divisible by $4,$ $QR$ equals either $12,$ $24,$ or $32.$

If $QR=12,$ then $R=2,$ so $RST=25T$ cannot be divisible by $3$ using the remaining digits. If $QR=32,$ then $R=2$ again, giving the same obstacle. Thus $QR=24,$ and then $RST=45T.$ Among the remaining digits, only $T=3$ makes $453$ divisible by $3$ .

Therefore, $PQRST = 12453.$

Thus, A is the correct answer.

25.

一个半圆内切于一个底为 $16$ 、高为 $15$ 的等腰三角形中，半圆的直径位于三角形的底边上，如图所示。半圆的半径是多少？

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

$4 \sqrt{3}$

$\dfrac{120}{17}$

$10$

$\dfrac{17\sqrt{2}}{2}$

$\dfrac{17\sqrt{3}}{2}$

答案：B

知识点：三角形面积勾股定理

难度评级：1610

小提示：

画出到半圆与三角形一边相切点的半径

Draw the radius to the point where the semicircle touches a side of the triangle.

大提示：

用两种不同的底来比较三角形右半部分的面积

Compare the area of the right half of the triangle using two different bases.

视频讲解：

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文字解答：

设 $O$ 是圆心，也就是 $\overline{AB}$ 的中点。

由勾股定理可得 $BC = 17$ 。

三角形右半部分的面积为 $[\triangle BOC]=\dfrac12\cdot8\cdot15=60\text{。}$ 半径 $OE$ 垂直于相切的边 $BC$ ，所以同一面积也等于 $\dfrac12\cdot OE\cdot17$ 。因此 $OE=\dfrac{120}{17}\text{。}$

所以正确答案是 B。

Let $O$ be the center of the circle, which is the midpoint of $\overline{AB}.$

We then get that $BC = 17$ via the Pythagorean theorem.

The right half of the isosceles triangle has area $[\triangle BOC]=\dfrac12\cdot8\cdot15=60.$ The radius $OE$ is perpendicular to the tangent side $BC$ , so the same area is $\dfrac12\cdot OE\cdot17.$ Hence $OE=\dfrac{120}{17}.$

Thus, B is the correct answer.