2015 AMC 8 详解与视频讲解 | LIVE by Po-Shen Loh

LIVE

中文

2015 AMC 8 详解

向下滚动即可查看来自 LIVE by Po-Shen Loh 的专业视频讲解与文字解答，打印PDF 解答，查看答案，或参加完整限时模拟考试。

所有题目均经美国数学协会（MAA）官方合法授权使用。

1.

要铺满一个长 $12$ 英尺、宽 $9$ 英尺的长方形地板，需要多少平方码地毯？（一码等于 $3$ 英尺。）

How many square yards of carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are $3$ feet in a yard.)

$12$

$36$

$108$

$324$

$972$

知识点：单位换算面积

难度评级：370

小提示：

先把两个边长都从英尺转换成码

Convert both side lengths from feet to yards first.

大提示：

再将两个以码为单位的长度相乘，得到平方码

Then multiply the two yard measurements to get square yards.

视频讲解：

Click to load, then click again to play

文字解答：

一边长 $12$ 英尺，即 $\dfrac{12}{3} = 4$ 码。

另一边长 $9$ 英尺，即 $\dfrac{9}{3} = 3$ 码。

因为尺寸是 $4$ 码乘 $3$ 码，所以面积为 $4\cdot3=12$ 平方码。

所以正确答案是 A。

Since one side is $12$ feet, it would be $\dfrac{12}{3} = 4$ yards.

Since another side is $9$ feet, it would be $\dfrac{9}{3} = 3$ yards.

Since the dimensions are $4$ yards by $3$ yards, the area is $4\cdot3=12$ square yards.

Thus, the correct answer is A .

2.

点 $O$ 是正八边形 $ABCDEFGH$ 的中心， $X$ 是边 $\overline{AB}$ 的中点。阴影部分占八边形面积的几分之几？

Point $O$ is the center of the regular octagon $ABCDEFGH,$ and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?

$\dfrac{11}{32}$

$\dfrac{3}{8}$

$\dfrac{13}{32}$

$\dfrac{7}{16}$

$\dfrac{15}{32}$

知识点：正多边形面积分割

难度评级：1020

小提示：

从 $O$ 向八边形的每个顶点连线

Draw segments from $O$ to every vertex of the octagon.

大提示：

阴影部分包含三个完整的中心三角形和另一个的一半

The shaded region contains three full central triangles and half of another.

视频讲解：

Click to load, then click again to play

文字解答：

先注意，八边形可分成 $8$ 个面积相等的三角形，每个三角形都以 $O$ 为一个顶点。因此每个三角形占八边形总面积的 $\dfrac {1}{8}$ 。

阴影部分包含三个完整三角形和三角形 $ABO$ 的一半。因此阴影面积是八边形总面积的 $\dfrac{3.5}{8} = \dfrac{7}{16}$ 倍。

所以正确答案是 D。

First notice that there are $8$ equally sized triangles that can be created with $O$ and any two consecutive points. Therefore, they each take up $\dfrac {1}{8}$ of the total area of the octagon.

The shaded area has three complete triangles and half of the triangle $ABO.$ Therefore, the shaded area is $\dfrac{3.5}{8} = \dfrac{7}{16}$ of the total area of the octagon.

Thus, the correct answer is D .

3.

杰克和吉尔要去离家一英里的游泳池游泳。他们同时离家。吉尔以每小时 $10$ 英里的恒定速度骑自行车去游泳池。杰克以每小时 $4$ 英里的恒定速度步行去游泳池。吉尔比杰克早到多少分钟？

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$5$

$6$

$8$

$9$

$10$

知识点：路程、速度与时间

难度评级：660

小提示：

求每个人走完一英里分别需要多少分钟

Find how many minutes each person takes to travel one mile.

大提示：

用杰克的时间减去吉尔的时间

Subtract Jill’s travel time from Jack’s travel time.

视频讲解：

Click to load, then click again to play

文字解答：

杰克走 $4$ 英里需要 $60$ 分钟，所以他到游泳池需要 $\dfrac{60}{4} = 15$ 分钟。

吉尔骑行 $10$ 英里需要 $60$ 分钟，所以她到游泳池需要 $\dfrac{60}{10} = 6$ 分钟。

两人的时间差为 $15-6 = 9$ 分钟。

所以正确答案是 D。

Jack travels at a rate of $4$ miles per $60$ minutes. Therefore, it takes him $\dfrac{60}{4} = 15$ minutes to get to the pool.

Jill travels at a rate of $10$ miles per $60$ minutes. Therefore it takes her $\dfrac{60}{10} = 6$ minutes to get to the pool.

Therefore, the difference in their times is $15-6 = 9$ minutes.

Thus, the correct answer is D .

4.

森特维尔中学国际象棋队由两名男生和三名女生组成。一位摄影师想为当地报纸拍一张队伍照片。她决定让他们坐成一排，两端各坐一名男生，中间坐三名女生。这样的排列有多少种？

The Centerville Middle School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?

$2$

$4$

$5$

$6$

$12$

知识点：排列乘法原理

难度评级：770

小提示：

先把两名男生放在两端

Place the two boys at the two ends first.

大提示：

再排列中间三个座位上的女生

Then arrange the three girls in the middle seats.

视频讲解：

Click to load, then click again to play

文字解答：

两名男生坐在两端有 $2! = 2$ 种方式。三名女生坐在中间三个座位有 $3! = 6$ 种方式。

因此总排列数为 $2\cdot6=12$ 。

所以正确答案是 E。

There are $2! = 2$ ways to place the two boys at the two ends. There are $3! = 6$ ways to arrange the three girls in the middle seats.

Thus the total number of arrangements is $2\cdot6=12.$

Thus, E is the correct answer.

5.

比利的篮球队在赛季前 $11$ 场比赛中的得分如下：

$42$ 、 $47$ 、 $53$ 、 $53$ 、 $58$ 、 $58$ 、 $58$ 、 $61$ 、 $64$ 、 $65$ 、 $73$ 。

如果他的球队在第 $12$ 场比赛中得到 $40$ 分，下列哪一个统计量会增加？

Billy’s basketball team scored the following points over the course of the first $11$ games of the season:

$42,$ $47,$ $53,$ $53,$ $58,$ $58,$ $58,$ $61,$ $64,$ $65,$ $73.$

If his team scores $40$ in the $12$ th game, which of the following statistics will show an increase?

极差

range

中位数

median

平均数

mean

众数

mode

中程数

mid-range

知识点：极差

难度评级：770

小提示：

加入 $40$ 会产生一个新的最低分

Adding $40$ creates a new lowest score.

大提示：

检查哪个统计量取决于最大值与最小值之差

Check which statistic depends on the difference between the largest and smallest scores.

视频讲解：

Click to load, then click again to play

文字解答：

考虑全部 $12$ 场比赛时，得分 $40$ 来自第 $12$ 场，并会成为最低分。因此，与前 $11$ 场的极差相比，全部 $12$ 场的极差从 $73-42 = 31$ 增加到 $73-40=33$ 。

所以正确答案是 A。

When considering all $12$ games, $40,$ from the $12$ th game, will be the lowest score. Therefore, compared to the range of just the first $11$ games, the range of all $12$ games would increase from $73-42 = 31$ to $73-40=33.$

Thus, the correct answer is A .

6.

在 $\triangle ABC$ 中， $AB=BC=29$ ，且 $AC=42$ 。 $\triangle ABC$ 的面积是多少？

In $\triangle ABC,$ $AB=BC=29,$ and $AC=42.$ What is the area of $\triangle ABC?$

$100$

$420$

$500$

$609$

$701$

知识点：等腰三角形勾股定理三角形面积

难度评级：1140

小提示：

从 $B$ 向 $AC$ 作高

Drop the altitude from $B$ to $AC.$

大提示：

这条高平分 $AC$ ，得到一个斜边为 $29$ 的直角三角形

The altitude bisects $AC,$ giving a right triangle with hypotenuse $29.$

视频讲解：

Click to load, then click again to play

文字解答：

从 $B$ 向 $AC$ 作高，交 $AC$ 于 $X$ 。因为 $AB=BC$ ，点 $X$ 是 $AC$ 的中点，所以 $AX=21$ 。

在直角三角形 $ABX$ 中，高为 $\begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20 \end{aligned}\text{。}$

因此 $\triangle ABC$ 的面积为 $\dfrac12\cdot42\cdot20=420$ 。

所以正确答案是 B。

Drop the altitude from $B$ to $AC,$ meeting $AC$ at $X.$ Since $AB=BC,$ point $X$ is the midpoint of $AC,$ so $AX=21.$

In right triangle $ABX,$ the altitude is $\begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20. \end{aligned}$

The area of $\triangle ABC$ is $\dfrac12\cdot42\cdot20=420.$

Thus, B is the correct answer.

7.

两个盒子中各有三枚编号为 $1$ 、 $2$ 、 $3$ 的筹码。从每个盒子中随机取出一枚筹码，并将两枚筹码上的数字相乘。乘积为偶数的概率是多少？

Each of two boxes contains three chips numbered $1,$ $2,$ $3.$ A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

$\dfrac{1}{9}$

$\dfrac{2}{9}$

$\dfrac{4}{9}$

$\dfrac{1}{2}$

$\dfrac{5}{9}$

知识点：对立事件概率奇偶性

难度评级：960

小提示：

统计补集，也就是乘积为奇数，可能更容易

It may be easier to count the complement: an odd product.

大提示：

只有当两枚筹码都是奇数时，乘积才是奇数

The product is odd only if both drawn chips are odd.

视频讲解：

Click to load, then click again to play

文字解答：

只有当两枚筹码都是奇数时，乘积才是奇数。每个盒子中有两个奇数筹码 $1$ 和 $3$ ，共三个筹码，所以乘积为奇数的概率是 $(\dfrac23)^2=\dfrac49$ 。

乘积为偶数的概率是补集，等于 $1-\dfrac49=\dfrac59$ 。

所以正确答案是 E。

The product is odd only when both chips are odd. Each box has two odd chips, $1$ and $3,$ out of three chips, so the probability of an odd product is $(\dfrac23)^2=\dfrac49.$

The probability of an even product is the complement, $1-\dfrac49=\dfrac59.$

Thus, E is the correct answer.

8.

大于任意一个有一条边长为 $5$ 、另一条边长为 $19$ 的三角形周长的最小整数是多少？

What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19?$

$24$

$29$

$43$

$48$

$57$

知识点：三角不等式周长

难度评级：980

小提示：

使用三角不等式限制第三边

Use the triangle inequality to bound the third side.

大提示：

周长可以任意接近 $5+19$ 的两倍，但不能等于它

The perimeter can get arbitrarily close to twice $5+19,$ but not equal it.

视频讲解：

Click to load, then click again to play

文字解答：

设第三边长为 $s$ 。三角不等式给出 $s<5+19=24$ ，所以周长满足 $5+19+s<48\text{。}$

周长可以从下方任意接近 $48$ ，因此大于所有这类三角形周长的最小整数是 $48$ 。

所以正确答案是 D。

Let the third side length be $s.$ The triangle inequality gives $s<5+19=24,$ so the perimeter satisfies $5+19+s<48.$

Perimeters can be made arbitrarily close to $48$ from below, so the smallest whole number larger than the perimeter of any such triangle is $48.$

Thus, D is the correct answer.

9.

珍娜贝尔第一天上班卖出一个小部件。第二天卖出三个。第三天卖出五个，此后每天都比前一天多卖两个。工作 $20$ 天后，珍娜贝尔总共卖出多少个小部件？

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?

$39$

$40$

$210$

$400$

$401$

知识点：等差数列求和

难度评级：960

小提示：

每日销量是前 $20$ 个正奇数

The daily sales are the first $20$ positive odd numbers.

大提示：

将第一项与最后一项、第二项与倒数第二项依次配对

Pair the first and last terms, second and next-to-last terms, and so on.

视频讲解：

Click to load, then click again to play

文字解答：

我们要计算 $1+3 + \cdots + 39\text{。}$ 把两端的项依次配对，可得 $\begin{aligned} &(1+39)+(3+37)+\cdots\\ &\qquad +(19+21) \end{aligned}\text{。}$ 共有 $10$ 对，每对和为 $40$ ，所以 $10\cdot40=400\text{。}$

所以正确答案是 D。

We want to find $1+3 + \cdots + 39.$ Pairing terms from the ends gives $\begin{aligned} &(1+39)+(3+37)+\cdots\\ &\qquad +(19+21). \end{aligned}$ There are $10$ pairs, and each pair sums to $40.$ Therefore, the sum is $10\cdot40=400.$

Thus, the correct answer is D .

10.

从 $1000$ 到 $9999$ 之间，有多少个四位整数的四个数字互不相同？

How many integers between $1000$ and $9999$ have four distinct digits?

$3024$

$4536$

$5040$

$6480$

$6561$

知识点：乘法原理数字

难度评级：1070

小提示：

先选择千位数字

Choose the thousands digit first.

大提示：

之后每一位都必须避开之前已经选择的数字

After that, each later digit must avoid all previously chosen digits.

视频讲解：

Click to load, then click again to play

文字解答：

首先，千位有 $9$ 种选择，因为 $0$ 不能选。

接着，百位有 $9$ 种选择，十位有 $8$ 种选择，个位有 $7$ 种选择。因此这样的整数共有 $9\cdot9\cdot8\cdot7 = 4536$ 个。

所以正确答案是 B。

First, there are $9$ digits to choose for the thousands digit since $0$ can’t be chosen.

Then, after that, there are $9$ ways to choose the hundreds digit, $8$ ways to choose the tens digit, and $7$ ways to choose the ones digit. Therefore, we get $9\cdot9\cdot8\cdot7 = 4536$ ways to choose such an integer.

Thus, the correct answer is B .

11.

在小国“数学国”，所有汽车牌照都有四个符号。第一个必须是元音字母（A、E、I、O 或 U），第二个和第三个必须是 $21$ 个非元音字母中的两个不同字母，第四个必须是一个数字（零到 $9$ ）。如果符号在满足这些条件下随机选择，牌照读作“AMC8”的概率是多少？

In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel (A, E, I, O, or U), the second and third must be two different letters among the $21$ non-vowels, and the fourth must be a digit (0 through $9$ ). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read “AMC8”?

$\dfrac{1}{22{,}050}$

$\dfrac{1}{21{,}000}$

$\dfrac{1}{10{,}500}$

$\dfrac{1}{2{,}100}$

$\dfrac{1}{1{,}050}$

知识点：基本概率乘法原理

难度评级：1100

小提示：

统计规则允许的所有牌照数

Count all license plates allowed by the rules.

大提示：

其中只有一个牌照正好是 AMC8

Only one of those plates is exactly AMC8.

视频讲解：

Click to load, then click again to play

文字解答：

第一个符号有 $5$ 种选择，第二个有 $21$ 种选择，第三个必须是不同的非元音字母，所以有 $20$ 种选择，最后一个数字有 $10$ 种选择。

因此可能牌照数为 $5\cdot21\cdot20\cdot10=21{,}000$ 。其中正好一个是 AMC8，所以概率为 $\dfrac{1}{21{,}000}$ 。

所以正确答案是 B。

There are $5$ choices for the first symbol, $21$ choices for the second, $20$ choices for the third because it must be a different non-vowel, and $10$ choices for the final digit.

Thus there are $5\cdot21\cdot20\cdot10=21{,}000$ possible plates. Exactly one of these is AMC8, so the probability is $\dfrac{1}{21{,}000}.$

Thus, B is the correct answer.

12.

一个立方体有多少对平行棱，例如 $\overline{AB}$ 和 $\overline{GH}$ ，或 $\overline{EH}$ 和 $\overline{FG}$ ？

How many pairs of parallel edges, such as $\overline{AB}$ and $\overline{GH}$ or $\overline{EH}$ and $\overline{FG},$ does a cube have?

$6$

$12$

$18$

$24$

$36$

知识点：正方体数对计数

难度评级：1030

小提示：

按方向给立方体的棱分组

Group cube edges by direction.

大提示：

每条棱与另外三条棱平行，但按棱统计会把每对数两次

Each edge is parallel to three other edges, but pair-counting double counts.

视频讲解：

Click to load, then click again to play

文字解答：

立方体有 $12$ 条棱。对任意一条棱，都有另外 $3$ 条棱与它平行。

这样每一对平行棱都被数了两次，所以平行棱对数为 $\dfrac{12\cdot3}{2}=18$ 。

所以正确答案是 C。

A cube has $12$ edges. For any edge, there are $3$ other edges parallel to it.

This counts each pair twice, once from each edge in the pair, so the number of pairs of parallel edges is $\dfrac{12\cdot3}{2}=18.$

Thus, C is the correct answer.

13.

从集合 $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ 中可以移除多少个不同的二元素子集，使剩余九个数的平均数为 $6$ ？

How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the nine remaining numbers is $6?$

$1$

$2$

$3$

$5$

$6$

知识点：平均数系统列举

难度评级：1030

小提示：

求剩余九个数必须有的总和

Find the sum the remaining nine numbers must have.

大提示：

因此被移除的两个数必须有固定的和

So the two removed numbers must have a fixed sum.

视频讲解：

Click to load, then click again to play

文字解答：

原集合的和为 $1+2+\cdots+11=66$ 。移除两个数后剩下 $9$ 个数，平均数要为 $6$ ，所以剩余数总和必须是 $9\cdot6=54$ 。

因此被移除的两个数之和必须是 $66-54=12$ 。可能的二元素子集为 $\{1,11\}$ 、 $\{2,10\}$ 、 $\{3,9\}$ 、 $\{4,8\}$ 、 $\{5,7\}$ ，共有 $5$ 种。

所以正确答案是 D。

The original set has sum $1+2+\cdots+11=66.$ After removing two numbers, $9$ numbers remain and must have mean $6,$ so their sum must be $9\cdot6=54.$

Therefore the two removed numbers must have sum $66-54=12.$ The possible two-element subsets are $\{1,11\},$ $\{2,10\},$ $\{3,9\},$ $\{4,8\},$ $\{5,7\},$ so there are $5$ choices.

Thus, D is the correct answer.

14.

下列哪个整数不能写成四个连续奇整数之和？

Which of the following integers cannot be written as the sum of four consecutive odd integers?

$16$

$40$

$72$

$100$

$200$

知识点：整除性代数变形

难度评级：980

小提示：

用代数式表示四个连续奇整数

Write the four consecutive odd integers algebraically.

大提示：

它们的和总是 $8$ 的倍数

Their sum is always a multiple of $8.$

视频讲解：

Click to load, then click again to play

文字解答：

设四个连续奇整数为 $2k+1$ 、 $2k+3$ 、 $2k+5$ 和 $2k+7$ 。它们的和为 $8k+16=8(k+2)\text{。}$

所以这样的和必须是 $8$ 的倍数。选项中唯一不能被 $8$ 整除的是 $100$ 。

所以正确答案是 D。

Let the four consecutive odd integers be $2k+1,$ $2k+3,$ $2k+5,$ and $2k+7.$ Their sum is $8k+16=8(k+2).$

So any such sum must be a multiple of $8.$ The only answer choice that is not divisible by $8$ is $100.$

Thus, D is the correct answer.

15.

在欧拉中学， $198$ 名学生对学校公投中的两个议题进行了投票，结果如下： $149$ 人支持第一个议题， $119$ 人支持第二个议题。如果正好有 $29$ 名学生两个议题都反对，那么有多少名学生两个议题都支持？

At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results. $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues?

$49$

$70$

$79$

$99$

$149$

知识点：容斥原理

难度评级：1100

小提示：

先求有多少学生至少支持一个议题

First find how many students voted for at least one issue.

大提示：

对两个支持人数使用容斥原理

Use inclusion-exclusion on the two yes-vote counts.

视频讲解：

Click to load, then click again to play

文字解答：

因为 $29$ 名学生两个议题都反对，所以 $198-29 = 169$ 名学生至少支持一个议题。

已知 $149$ 人支持第一个议题， $119$ 人支持第二个议题，且 $169$ 人至少支持一个议题。由容斥原理，两个都支持的人数为 $149+119-169 = 99\text{。}$

所以正确答案是 D。

Since $29$ students voted against both, we know that $198-29 = 169$ people voted for at least one.

As we know that $149$ students voted for the first issue, and $119$ students voted for the second issue, and $169$ students that voted for at least one issue, we conclude that the number of students that voted for both is $149+119-169 = 99.$

Thus, the correct answer is D .

16.

在一个中学导师项目中，一些六年级学生与一名九年级学生配成伙伴。没有九年级学生被分配超过一名六年级伙伴。如果所有九年级学生的 $\dfrac{1}{3}$ 与所有六年级学生的 $\dfrac{2}{5}$ 配成伙伴，那么六年级和九年级学生总数中有伙伴的比例是多少？

In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\dfrac{1}{3}$ of all the ninth graders are paired with $\dfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

$\dfrac{2}{15}$

$\dfrac{4}{11}$

$\dfrac{11}{30}$

$\dfrac{3}{8}$

$\dfrac{11}{15}$

知识点：比与比例分数

难度评级：1300

小提示：

设六年级和九年级学生人数分别为 $s$ 和 $n$

Let $s$ and $n$ be the numbers of sixth and ninth graders.

大提示：

配对的六年级学生数与配对的九年级学生数相同

The paired sixth graders and paired ninth graders are the same number of pairs.

视频讲解：

Click to load, then click again to play

文字解答：

设六年级学生有 $s$ 人，九年级学生有 $n$ 人。配对的九年级学生数等于配对的六年级学生数，所以 $\dfrac13 n=\dfrac25 s\text{。}$

得 $5n=6s$ ，所以 $n:s=6:5$ 。学生总数对应 $11$ 份。

配对的九年级学生占全体学生的 $\dfrac13\cdot\dfrac6{11}=\dfrac2{11}$ ，配对的六年级学生占全体学生的 $\dfrac25\cdot\dfrac5{11}=\dfrac2{11}$ 。合计有 $\dfrac4{11}$ 的学生有伙伴。

所以正确答案是 B。

Let there be $s$ sixth graders and $n$ ninth graders. The number of paired ninth graders equals the number of paired sixth graders, so $\dfrac13 n=\dfrac25 s.$

This gives $5n=6s,$ so $n:s=6:5.$ The total number of students is therefore proportional to $11$ parts.

The paired ninth graders make up $\dfrac13\cdot\dfrac6{11}=\dfrac2{11}$ of all students, and the paired sixth graders make up $\dfrac25\cdot\dfrac5{11}=\dfrac2{11}$ of all students. Altogether, $\dfrac4{11}$ of the students have a buddy.

Thus, B is the correct answer.

17.

杰里米的父亲在交通高峰期开车送他上学要 $20$ 分钟。某天没有交通拥堵，所以父亲能把车速提高每小时 $18$ 英里，并在 $12$ 分钟内把他送到学校。到学校有多少英里？

Jeremy’s father drives him to school in rush hour traffic in $20$ minutes. One day there is no traffic, so his father can drive him $18$ miles per hour faster and gets him to school in $12$ minutes. How far in miles is it to school?

$4$

$6$

$8$

$9$

$12$

知识点：路程、速度与时间一次方程

难度评级：1280

小提示：

将 $20$ 分钟和 $12$ 分钟转换成小时

Convert $20$ minutes and $12$ minutes to hours.

大提示：

令两种距离表达式相等

Set the two distance expressions equal.

视频讲解：

Click to load, then click again to play

文字解答：

设高峰期速度为每小时 $s$ 英里。 $20$ 分钟是 $\dfrac13$ 小时，所以距离为 $\dfrac{s}{3}$ 。

没有交通拥堵时，速度为每小时 $s+18$ 英里，路程用时 $12$ 分钟，即 $\dfrac15$ 小时。同一距离为 $\dfrac{s+18}{5}$ 。

令两种距离相等： $\dfrac{s}{3}=\dfrac{s+18}{5}$ 得 $5s=3s+54$ ，所以 $s=27$ 。距离为 $\dfrac{27}{3}=9$ 英里。

所以正确答案是 D。

Let the rush-hour speed be $s$ miles per hour. The $20$ -minute rush-hour trip takes $\dfrac13$ hour, so the distance is $\dfrac{s}{3}.$

Without traffic, the speed is $s+18$ miles per hour and the trip takes $12$ minutes, or $\dfrac15$ hour. The same distance is $\dfrac{s+18}{5}.$

Set the distances equal: $\dfrac{s}{3}=\dfrac{s+18}{5}.$ Then $5s=3s+54,$ so $s=27.$ The distance is $\dfrac{27}{3}=9$ miles.

Thus, D is the correct answer.

18.

等差数列是指从第二项开始，每一项都由前一项加上一个常数得到的数列。例如， $2$ 、 $5$ 、 $8$ 、 $11$ 、 $14$ 是一个五项等差数列，第一项是 $2$ ，每次加上的常数是 $3$ 。这个 $5\times5$ 阵列中，每一行和每一列都是五项等差数列。 $X$ 的值是多少？

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,$ $5,$ $8,$ $11,$ $14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant $3$ is added. Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. What is the value of $X?$

$21$

$31$

$36$

$40$

$42$

知识点：等差数列

难度评级：1280

小提示：

在五项等差数列中，中间项是首项和末项的平均数

In a five-term arithmetic sequence, the middle term is the average of the first and last terms.

大提示：

将这个事实用于顶行、底行，然后用于中间列

Apply that fact to the top row, bottom row, and then the middle column.

视频讲解：

Click to load, then click again to play

文字解答：

在任意五项等差数列中，中间项是首项和末项的平均数。

顶行的中间项为 $\dfrac{1+25}{2}=13$ ，底行的中间项为 $\dfrac{17+81}{2}=49$ 。

再对中间列使用同样事实，得 $X=\dfrac{13+49}{2}=31$ 。

所以正确答案是 B。

In any five-term arithmetic sequence, the middle term is the average of the first and last terms.

The middle entry of the top row is $\dfrac{1+25}{2}=13,$ and the middle entry of the bottom row is $\dfrac{17+81}{2}=49.$

Now apply the same fact to the middle column: $X=\dfrac{13+49}{2}=31.$

Thus, B is the correct answer.

19.

顶点为 $A=(1,3)$ 、 $B=(5,1)$ 、 $C=(4,4)$ 的三角形画在一个 $6\times5$ 网格上。这个三角形覆盖了网格的几分之几？

A triangle with vertices at $A=(1,3),$ $B=(5,1),$ and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

知识点：坐标几何三角形面积

难度评级：1320

小提示：

将三角形放入一个与网格对齐的小长方形中

Put the triangle inside a small rectangle aligned to the grid.

大提示：

从长方形面积中减去三个角上的直角三角形

Subtract the three corner right triangles from the rectangle.

视频讲解：

Click to load, then click again to play

文字解答：

整个网格的面积是 $6\cdot5=30$ 。要求三角形面积，可把它放在下图所示的宽为 $4$ 、高为 $3$ 的长方形中。

长方形的面积为 $4\cdot3=12$ 。三个角上三角形的面积分别为 $\dfrac12(4)(2)=4$ 、 $\dfrac12(1)(3)=\dfrac32$ 和 $\dfrac12(3)(1)=\dfrac32$ 。因此 $\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5 \end{aligned}\text{。}$

因此覆盖比例为 $\dfrac{5}{30} = \dfrac{1}{6}$ 。

所以正确答案是 A。

The total area of the grid is $6\cdot5=30.$ To find the area of the triangle, place it inside the $4$ -by- $3$ rectangle shown below.

The rectangle has area $4\cdot3=12.$ The three corner triangles have areas $\dfrac12(4)(2)=4,$ $\dfrac12(1)(3)=\dfrac32,$ and $\dfrac12(3)(1)=\dfrac32.$ Therefore, $\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5. \end{aligned}$

Therefore, the fraction of the area is $\dfrac{5}{30} = \dfrac{1}{6} .$

Thus, the correct answer is A .

20.

拉尔夫去商店买了 $12$ 双袜子，总价 $\$24$ 。他买的袜子中有些每双 $\$1$ ，有些每双 $\$3$ ，有些每双 $\$4$ 。如果每种袜子他至少买一双，那么拉尔夫买了多少双 $\$1$ 的袜子？

Ralph went to the store and bought $12$ pairs of socks for a total of $\$24$ . Some of the socks he bought cost $\$1$ a pair, some of the socks he bought cost $\$3$ a pair, and some of the socks he bought cost $\$4$ a pair. If he bought at least one pair of each type, how many pairs of $\$1$ socks did Ralph buy?

$4$

$5$

$6$

$7$

$8$

知识点：方程组模运算

难度评级：1390

小提示：

设价格为 $\$1$ 、 $\$3$ 和 $\$4$ 的袜子双数为变量

Let the numbers of $\$1,$ $\$3,$ and $\$4$ pairs be variables.

大提示：

用总价方程减去总双数方程

Subtract the pair-count equation from the cost equation.

视频讲解：

Click to load, then click again to play

文字解答：

设 $a$ 、 $b$ 、 $c$ 分别为 $\$1$ 、 $\$3$ 、 $\$4$ 的袜子双数。则 $a+b+c=12\text{，}$ $a+3b+4c=24\text{。}$

两式相减得 $2b+3c=12$ 。因为每种至少买一双，所以 $b>0$ 、 $c>0$ 。又 $3c<12$ ，所以 $c<4$ 。从方程模 $2$ 看， $c$ 必须是偶数，所以 $c=2$ 。

于是 $2b+6=12$ ，得 $b=3$ ，所以 $a=12-3-2=7$ 。

所以正确答案是 D。

Let $a,$ $b,$ and $c$ be the numbers of $\$1,$ $\$3,$ and $\$4$ pairs, respectively. Then $a+b+c=12,$ $a+3b+4c=24.$

Subtracting gives $2b+3c=12.$ Since at least one pair of each type was bought, $b>0$ and $c>0.$ Also $3c<12,$ so $c<4.$ Modulo $2,$ the equation gives $c$ even, so $c=2.$

Then $2b+6=12,$ so $b=3,$ and $a=12-3-2=7.$

Thus, D is the correct answer.

21.

如图，六边形 $ABCDEF$ 是等角六边形， $ABJI$ 和 $FEHG$ 是面积分别为 $18$ 和 $32$ 的正方形， $\triangle JBK$ 是等边三角形，且 $FE=BC$ 。 $\triangle KBC$ 的面积是多少？

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC.$ What is the area of $\triangle KBC?$

$6\sqrt{2}$

$9$

$12$

$9\sqrt{2}$

$32$

知识点：等角多边形等边三角形三角形面积

难度评级：1510

小提示：

由两个正方形面积求出 $JB$ 和 $BC$

Find $JB$ and $BC$ from the two square areas.

大提示：

在图中， $BK$ 垂直于 $BC$

In the diagram, $BK$ is perpendicular to $BC.$

视频讲解：

Click to load, then click again to play

文字解答：

面积为 $18$ 的正方形边长为 $JB=\sqrt{18}=3\sqrt2$ 。因为 $\triangle JBK$ 是等边三角形，所以 $BK=3\sqrt2$ 。

面积为 $32$ 的正方形边长为 $FE=\sqrt{32}=4\sqrt2$ 。因为 $FE=BC$ ，所以 $BC=4\sqrt2$ 。

在 $B$ 点处，等边三角形、正方形和等角六边形的内角分别为 $60^\circ$ 、 $90^\circ$ 和 $120^\circ$ 。因此 $\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ \end{aligned}\text{。}$ 所以 $[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12\text{。}$

所以正确答案是 C。

The square with area $18$ has side length $JB=\sqrt{18}=3\sqrt2.$ Since $\triangle JBK$ is equilateral, $BK=3\sqrt2.$

The square with area $32$ has side length $FE=\sqrt{32}=4\sqrt2.$ Since $FE=BC,$ we have $BC=4\sqrt2.$

At $B,$ the equilateral triangle, square, and equiangular hexagon contribute angles of $60^\circ,$ $90^\circ,$ and $120^\circ,$ respectively. Hence $\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ. \end{aligned}$ Therefore $[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.$

Thus, C is the correct answer.

22.

六月 $1$ 日，一组学生排成若干行，每行 $15$ 人。六月 $2$ 日，同一组学生排成一长行。六月 $3$ 日，同一组学生每行只站一人。六月 $4$ 日，同一组学生每行 $6$ 人。这个过程一直持续到六月 $12$ 日，每天每行人数都不同。然而到六月 $13$ 日，他们找不到新的排列方式。这组学生最少可能有多少人？

On June $1,$ a group of students is standing in rows, with $15$ students in each row. On June $2,$ the same group is standing with all of the students in one long row. On June $3,$ the same group is standing with just one student in each row. On June $4,$ the same group is standing with $6$ students in each row. This process continues through June $12$ with a different number of students per row each day. However, on June $13,$ they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

$21$

$30$

$60$

$90$

$1080$

知识点：因数个数最小公倍数分类讨论

难度评级：1460

小提示：

不同的每行人数就是总人数的因数

The different row sizes are the divisors of the group size.

大提示：

需要最小的同时为 $6$ 和 $15$ 的倍数，并且恰好有 $12$ 个因数

You need the smallest multiple of both $6$ and $15$ with exactly $12$ divisors.

视频讲解：

Click to load, then click again to play

文字解答：

每行可站的人数正好是学生总数的正因数。因为六月 $1$ 日至六月 $12$ 日有不同排列，而六月 $13$ 日没有新的排列方式，所以总人数必须恰好有 $12$ 个正因数。

总人数必须同时能被 $15$ 和 $6$ 整除，因此能被 $\operatorname{lcm}(15,6)=30=2\cdot3\cdot5$ 整除。而这个数只有 $8$ 个因数。

最小的 $30$ 的倍数且有 $12$ 个因数的是 $60=2^2\cdot3\cdot5$ ，其因数个数为 $(2+1)(1+1)(1+1)=12$ 。

所以正确答案是 C。

The possible numbers of students per row are exactly the positive divisors of the total number of students. Since June $1$ through June $12$ give different arrangements and June $13$ gives no new one, the total number of students must have exactly $12$ positive divisors.

The number must be divisible by both $15$ and $6,$ hence by $\operatorname{lcm}(15,6)=30=2\cdot3\cdot5.$ This number has only $8$ divisors.

The smallest multiple of $30$ with $12$ divisors is $60=2^2\cdot3\cdot5,$ which has $(2+1)(1+1)(1+1)=12$ divisors.

Thus, C is the correct answer.

23.

汤姆有十二张纸条，他想把它们放入标有 $A$ 、 $B$ 、 $C$ 、 $D$ 、 $E$ 的五个杯子中。

他希望每个杯子中纸条上的数之和都是整数，而且这五个整数从 $A$ 到 $E$ 连续递增。纸条上的数是 $2$ 、 $2$ 、 $2$ 、 $2.5$ 、 $2.5$ 、 $3$ 、 $3$ 、 $3$ 、 $3$ 、 $3.5$ 、 $4$ 、 $4.5$ 。如果一张写有 $2$ 的纸条放入杯子 $E$ ，一张写有 $3$ 的纸条放入杯子 $B$ ，那么写有 $3.5$ 的纸条必须放入哪个杯子？

Tom has twelve slips of paper which he wants to put into five cups labeled $A,$ $B,$ $C,$ $D,$ $E.$

He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from $A$ to $E.$ The numbers on the papers are $2,$ $2,$ $2,$ $2.5,$ $2.5,$ $3,$ $3,$ $3,$ $3,$ $3.5,$ $4,$ $4.5.$ If a slip with $2$ goes into cup $E$ and a slip with $3$ goes into cup $B,$ then the slip with $3.5$ must go into what cup?

$A$

$B$

$C$

$D$

$E$

知识点：逻辑推理分类讨论

难度评级：1610

小提示：

五个杯子的和必须是总和为 $35$ 的连续整数

The five cup sums must be consecutive integers with total $35.$

大提示：

先填好杯子 $B$ ，再检查把 $3.5$ 放在哪里仍能使其他杯子的和满足要求

After filling cup $B,$ test where the $3.5$ slip can still leave possible sums.

视频讲解：

Click to load, then click again to play

文字解答：

所有纸条的和为 $35$ ，所以五个连续整数杯子和的平均数必须是 $7$ 。因此杯子 $A$ 、 $B$ 、 $C$ 、 $D$ 、 $E$ 的和分别为 $5$ 、 $6$ 、 $7$ 、 $8$ 、 $9$ 。

杯子 $B$ 已经有一张 $3$ ，且总和必须为 $6$ ，所以它还必须有另一张 $3$ 。杯子 $E$ 已经有一张 $2$ ，所以 $E$ 中其他纸条之和必须为 $7$ 。

$3.5$ 不能放在 $A$ ，因为杯子 $A$ 还需要 $1.5$ 。它不能放在 $B$ ，因为这个杯子已经满了。它不能放在 $C$ 或 $E$ ，因为那样还需要另一个 $3.5$ ，而剩余纸条不能凑出这个和。杯子 $D$ 可行，例如 $3.5+4.5=8$ 。

所以正确答案是 D。

The sum of all the slips is $35,$ so the five consecutive integer cup sums must average $7.$ Therefore cups $A,$ $B,$ $C,$ $D,$ and $E$ must have sums $5,$ $6,$ $7,$ $8,$ and $9,$ respectively.

Cup $B$ already contains a $3$ and must sum to $6,$ so it must contain another $3.$ Cup $E$ already contains a $2,$ so the other slips in $E$ must sum to $7.$

The $3.5$ slip cannot go in $A,$ because cup $A$ would need another $1.5.$ It cannot go in $B,$ which is already full. It cannot go in $C$ or $E,$ because either would then need another $3.5,$ and no remaining slips can make that total. Cup $D$ works, for example with $3.5+4.5=8.$

Thus, D is the correct answer.

24.

一个棒球联盟由两个四队分区组成。每支球队与同分区其他每支球队比赛 $N$ 场。每支球队与另一个分区的每支球队比赛 $M$ 场，其中 $N > 2M$ ，且 $M > 4$ 。每支球队赛程共 $76$ 场。

每支球队在本分区内打多少场比赛？

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N > 2M$ and $M > 4.$ Each team plays a $76$ game schedule.

How many games does a team play within its own division?

$36$

$48$

$54$

$60$

$72$

知识点：丢番图方程模运算

难度评级：1560

小提示：

写出赛程方程 $3N+4M=76$

Write the schedule equation $3N+4M=76.$

大提示：

使用 $N>2M$ 、 $M>4$ ，以及方程模 $3$ 的结果

Use $N>2M,$ $M>4,$ and the equation modulo $3.$

视频讲解：

Click to load, then click again to play

文字解答：

每支球队在本分区内打 $3N$ 场，对另一个分区打 $4M$ 场，所以 $3N+4M=76\text{。}$

因为 $N>2M$ ，所以 $3N>6M$ ，从而 $76=3N+4M>10M$ 。因此 $M<7.6$ 。又 $M>4$ ，所以 $M\in\{5,6,7\}$ 。

将 $3N+4M=76$ 模 $3$ ，得 $M\equiv1\pmod3$ ，所以 $M=7$ 。因此每支球队打 $4M=28$ 场非分区比赛，本分区比赛为 $76-28=48$ 场。

所以正确答案是 B。

Each team plays $3N$ games within its own division and $4M$ games against the other division, so $3N+4M=76.$

Since $N>2M,$ we have $3N>6M,$ and hence $76=3N+4M>10M.$ Thus $M<7.6.$ Together with $M>4,$ this gives $M\in\{5,6,7\}.$

Reducing $3N+4M=76$ modulo $3$ gives $M\equiv1\pmod3,$ so $M=7.$ Therefore the team plays $4M=28$ non-division games and $76-28=48$ division games.

Thus, B is the correct answer.

25.

从这个 $5$ 英寸正方形的四个角各切去一个一英寸正方形。能放入剩余空间中的最大正方形面积是多少平方英寸？

One-inch squares are cut from the corners of this $5$ inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?

$9$

$12\dfrac{1}{2}$

$15$

$15\dfrac{1}{2}$

$17$

知识点：正方形（几何）面积分割

难度评级：1590

小提示：

放入的正方形可以围绕中央 $3\times3$ 正方形倾斜

The fitted square can be tilted around the central $3\times3$ square.

大提示：

将中央正方形面积和位于四角的四个全等三角形面积相加

Add the central square area and the four congruent corner triangles.

视频讲解：

Click to load, then click again to play

文字解答：

对于能放入的最大正方形，它的每条边都必须接触一个被挖去的边长为 $1$ 英寸的小正方形的内角，否则还可以继续放大。因此它围住中央的 $3\times3$ 正方形，并在四边各增加一个全等直角三角形。

中央正方形面积为 $3\cdot3=9$ 。每个增加的三角形两条直角边为 $3$ 和 $1$ ，四个三角形总面积为 $4\left(\dfrac{3\cdot1}{2}\right)=6\text{。}$

放入的正方形面积为 $9+6=15$ 。

所以正确答案是 C。

For a largest fitted square, each side must touch an inner corner of one of the removed $1$ -inch squares; otherwise the fitted square could be enlarged. Thus it surrounds the central $3\times3$ square and adds four congruent right triangles, one along each side.

The central square has area $3\cdot3=9.$ Each added triangle has base $3$ and height $1,$ so the four triangles have total area $4\left(\dfrac{3\cdot1}{2}\right)=6.$

The fitted square has area $9+6=15.$

Thus, C is the correct answer.