1992 AMC 8 真题与在线限时测验

计时

40:00

1.

下列表达式的值是多少？

$\scriptsize \dfrac{10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1}{1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9}$

What is the value of the following expression?

$\scriptsize \dfrac{10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1}{1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9}$

$-1$

$1$

$5$

$9$

$10$

答案：B

知识点：运算顺序配对与分组

难度评级：720

小提示：

把分子分成 $(10-9)$ 、 $(8-7)$ 、 $\ldots$ 这样的组；分母也用类似方式分组。

Group the numerator into pairs $(10-9),$ $(8-7),$ $\ldots$ , and do the same for the denominator

大提示：

分子中每一组都等于 $1$ ；分母可分成 $(1-2)$ 、 $(3-4)$ 、 $\ldots$ 这样的组，每组等于 $-1$ ，最后剩下 $+9$ 。

Each numerator pair equals $1$ ; in the denominator group $(1-2),$ $(3-4),$ $\ldots$ , where each pair is $-1$ , with a $+9$ left over

解答：

分子两两分组得 $(10-9)$ $+ (8-7)$ $+ (6-5)$ $+ (4-3)$ $+ (2-1) = 5$ 。

分母分组为 $(1-2)$ $+ (3-4)$ $+ (5-6)$ $+ (7-8)$ $+ 9$ ，有四组 $-1$ ，再加 $9$ ，结果也是 $5$ 。

所以原式为 $\dfrac{5}{5} = 1$ 。

所以正确答案是 B。

Grouping the numerator in pairs gives $(10-9)$ $+ (8-7)$ $+ (6-5)$ $+ (4-3)$ $+ (2-1) = 5.$

Grouping the denominator as $(1-2)$ $+ (3-4)$ $+ (5-6)$ $+ (7-8)$ $+ 9$ gives four pairs of $-1$ plus $9$ , which is $5.$

So the expression is $\dfrac{5}{5} = 1.$

Thus, the correct answer is B .

2.

下列哪一个不等于 $\dfrac54$ ？

Which of the following is not equal to $\dfrac54?$

$\dfrac{10}{8}$

$1\dfrac14$

$1\dfrac{3}{12}$

$1\dfrac15$

$1\dfrac{10}{40}$

答案：D

知识点：分数

难度评级：560

小提示：

把 $\dfrac54$ 写成带分数 $1\dfrac14$ 。

Rewrite $\dfrac54$ as the mixed number $1\dfrac14$

大提示：

带分数 $1\dfrac{a}{b}$ 要等于 $\dfrac54$ ，必须有 $\dfrac{a}{b} = \dfrac14$ ；逐项检查。

A mixed number $1\dfrac{a}{b}$ equals $\dfrac54$ only when $\dfrac{a}{b} = \dfrac14$ ; check each choice

解答：

$\dfrac54 = 1\dfrac14$ 。逐项检查： $\dfrac{10}{8} = \dfrac54$ ， $1\dfrac14 = \dfrac54$ ， $1\dfrac{3}{12} = 1\dfrac14 = \dfrac54$ ，且 $1\dfrac{10}{40} = 1\dfrac14 = \dfrac54$ 。

但 $1\dfrac15 = \dfrac65 \neq \dfrac54$ 。

所以正确答案是 D。

Note $\dfrac54 = 1\dfrac14.$ Now check each choice: $\dfrac{10}{8} = \dfrac54,$ $1\dfrac14 = \dfrac54,$ $1\dfrac{3}{12} = 1\dfrac14 = \dfrac54,$ and $1\dfrac{10}{40} = 1\dfrac14 = \dfrac54.$

But $1\dfrac15 = \dfrac65 \neq \dfrac54.$

Thus, the correct answer is D .

3.

从集合 $\{-16, -4, 0, 2, 4, 12\}$ 中选两个数相减，能够得到的最大差是多少？

What is the largest difference that can be formed by subtracting two numbers chosen from the set $\{-16, -4, 0, 2, 4, 12\}?$

$10$

$12$

$16$

$28$

$48$

答案：D

知识点：最优化

难度评级：450

小提示：

差最大时，应当用最大的数减去最小的数。

A difference is largest when you subtract the smallest number from the largest number

大提示：

减去负数等于加上正数；计算 $12 - (-16)$ 。

Subtracting a negative adds; compute $12 - (-16)$

解答：

最大差应使用最大数 $12$ 减去最小数 $-16$ 。

这给出 $12 - (-16) = 28$ 。

所以正确答案是 D。

The largest difference uses the largest number, $12$ , minus the smallest number, $-16.$

This gives $12 - (-16) = 28.$

Thus, the correct answer is D .

4.

在垒球赛季中，Judy 有 $35$ 次安打。其中包括 $1$ 次本垒打、 $1$ 次三垒安打和 $5$ 次二垒安打。其余安打都是一垒安打。她的安打中有百分之几是一垒安打？

During the softball season, Judy had $35$ hits. Among her hits were $1$ home run, $1$ triple, and $5$ doubles. The rest of her hits were singles. What percent of her hits were singles?

$28\%$

$35\%$

$70\%$

$75\%$

$80\%$

答案：E

知识点：百分数

难度评级：720

小提示：

先求有多少次安打不是一垒安打。

First find how many hits were not singles

大提示：

一垒安打数 $= 35 - (1+1+5)$ ；再除以 $35$ 并化成百分数。

Singles $= 35 - (1+1+5)$ ; divide by $35$ and convert to a percent

解答：

非一垒安打共有 $1 + 1 + 5 = 7$ 次，所以一垒安打有 $35 - 7 = 28$ 次。

一垒安打所占比例为 $\dfrac{28}{35} = \dfrac45 = 80\%$ 。

所以正确答案是 E。

The non-single hits number $1 + 1 + 5 = 7,$ so the singles number $35 - 7 = 28.$

The fraction of singles is $\dfrac{28}{35} = \dfrac45 = 80\%.$

Thus, the correct answer is E .

5.

如图，一个直径为 $1$ 的圆被从 $2 \times 3$ 的长方形中去掉。阴影区域的面积最接近哪个整数？

A circle of diameter $1$ is removed from a $2 \times 3$ rectangle, as shown. Which whole number is closest to the area of the shaded region?

$1$

$2$

$3$

$4$

$5$

答案：E

知识点：圆面积面积分割估算

难度评级：820

小提示：

长方形面积是 $2 \times 3 = 6$ ；再减去圆的面积。

The rectangle has area $2 \times 3 = 6$ ; subtract the area of the circle

大提示：

圆半径为 $\dfrac12$ ，面积为 $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4}$ ，略小于 $1$ 。

The circle has radius $\dfrac12,$ so its area is $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4},$ a little less than $1$

解答：

长方形面积为 $6$ ，去掉的圆面积为 $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4} \approx 0.79$ 。

阴影区域面积约为 $6 - 0.79 \approx 5.2$ ，最接近的整数是 $5$ 。

所以正确答案是 E。

The rectangle has area $6,$ and the removed circle has area $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4} \approx 0.79.$

So the shaded region has area $6 - 0.79 \approx 5.2,$ whose closest whole number is $5.$

Thus, the correct answer is E .

6.

如下图所示，一个上方数为 $a$ 、左下方数为 $b$ 、右下方数为 $c$ 的三角表达式表示 $a+b-c$ 。图中所示的和是多少？

A triangular expression with top entry $a$ , lower-left entry $b$ , and lower-right entry $c$ means $a+b-c$ , as shown below. What is the indicated sum?

$-2$

$-1$

$0$

$1$

$2$

答案：D

知识点：自定义运算

难度评级：820

小提示：

对每组三个数分别应用规则：上方数加左下方数，再减右下方数。

Apply the rule to each triple separately: top plus lower-left minus lower-right

大提示：

计算 $(1 + 3 - 4)$ 和 $(2 + 5 - 6)$ ，再把两个结果相加。

Compute $(1 + 3 - 4)$ and $(2 + 5 - 6),$ then add the two results

解答：

第一组三个数给出 $1 + 3 - 4 = 0$ ，第二组三个数给出 $2 + 5 - 6 = 1$ 。

它们的和是 $0 + 1 = 1$ 。

所以正确答案是 D。

The first triple gives $1 + 3 - 4 = 0,$ and the second gives $2 + 5 - 6 = 1.$

Their sum is $0 + 1 = 1.$

Thus, the correct answer is D .

7.

$998$ 的数位和是 $9 + 9 + 8 = 26$ 。有多少个 $3$ 位整数的数位和为 $26$ ，且是偶数？

The digit-sum of $998$ is $9 + 9 + 8 = 26.$ How many $3$ -digit whole numbers, whose digit-sum is $26$ , are even?

$1$

$2$

$3$

$4$

$5$

答案：A

知识点：数字奇偶性

难度评级：900

小提示：

三位数的数位和为 $26$ ，说明三个数字都非常大。

A digit-sum of $26$ forces the three digits to be very large

大提示：

唯一可行的三个数字是 $9$ 、 $9$ 、 $8$ ；列出它们的排列并保留偶数。

The only digits that work are $9,$ $9,$ and $8$ ; list the arrangements and keep the even ones

解答：

三个数字的和为 $26$ ，只能是数字 $9$ 、 $9$ 、 $8$ 。由它们组成的 $3$ 位数是 $899$ 、 $989$ 和 $998$ 。

其中只有 $998$ 是偶数，所以这样的数恰好有 $1$ 个。

所以正确答案是 A。

A digit-sum of $26$ with three digits requires the digits $9,$ $9,$ and $8.$ The $3$ -digit numbers using them are $899,$ $989,$ and $998.$

Of these, only $998$ is even, so there is exactly $1$ such number.

Thus, the correct answer is A .

8.

一位店主买入 $1500$ 支铅笔，每支价格为 $\$0.10$ 。如果他以每支 $\$0.25$ 出售，那么他必须卖出多少支，才能恰好获利 $\$100.00$ ？

A store owner bought $1500$ pencils at $\$0.10$ each. If he sells them for $\$0.25$ each, how many of them must he sell to make a profit of exactly $\$100.00?$

$400$

$667$

$1000$

$1500$

$1900$

答案：C

知识点：钱币

难度评级：900

小提示：

先求 $1500$ 支铅笔的总成本。

First find the total cost of the $1500$ pencils

大提示：

收入必须等于成本加 $\$100$ ；再用所需收入除以每支 $\$0.25$ 的售价。

The revenue must be the cost plus $\$100;$ divide that needed revenue by the $\$0.25$ price

解答：

铅笔成本为 $1500 \times \$0.10 = \$150$ 。要获利 $\$100$ ，收入必须是 $\$150 + \$100 = \$250$ 。

每支售价 $\$0.25$ ，所以必须卖出 $\dfrac{\$250}{\$0.25} = 1000$ 支。

所以正确答案是 C。

The pencils cost $1500 \times \$0.10 = \$150.$ To make a $\$100$ profit, the revenue must be $\$150 + \$100 = \$250.$

At $\$0.25$ each, he must sell $\dfrac{\$250}{\$0.25} = 1000$ pencils.

Thus, the correct answer is C .

9.

一个小镇的人口是 $480$ 。图中表示该镇女性和男性的人数，但省略了纵轴的刻度值。标有 $F$ 和 $M$ 的柱分别表示女性和男性。这个小镇有多少名男性？

The population of a small town is $480.$ The graph indicates the numbers of females and males in the town, but the vertical scale values are omitted. The bars labeled $F$ and $M$ represent females and males, respectively. How many males live in the town?

$120$

$160$

$200$

$240$

$360$

答案：B

知识点：比与比例

难度评级：720

小提示：

设男性人数为 $M$ ，那么女性人数是 $2M$ 。

Let the number of males be $M$ ; then the number of females is $2M$

大提示：

因为 $M + 2M = 480$ ，解方程 $3M = 480$ 。

Since $M + 2M = 480,$ solve $3M = 480$

解答：

若男性人数为 $M$ ，则女性人数为 $2M$ ，总人口满足 $M + 2M = 480$ 。

所以 $3M = 480$ ，得到 $M = 160$ 。

所以正确答案是 B。

If there are $M$ males, then there are $2M$ females, and together $M + 2M = 480.$

So $3M = 480,$ giving $M = 160.$

Thus, the correct answer is B .

10.

一个直角边长均为 $8$ 的等腰直角三角形，如图被分成 $16$ 个全等三角形。阴影部分面积是

An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is

$10$

$20$

$32$

$40$

$64$

答案：B

知识点：三角形面积面积分割

难度评级：930

小提示：

整个三角形面积为 $\dfrac12 \times 8 \times 8 = 32$ ，被分成 $16$ 个等面积小三角形。

The whole triangle has area $\dfrac12 \times 8 \times 8 = 32,$ split into $16$ equal pieces

大提示：

每个小三角形面积为 $2$ ；数一数有多少个小三角形被涂色。

Each small triangle has area $2$ ; count how many are shaded

解答：

大三角形面积为 $\dfrac12 \times 8 \times 8 = 32$ ，所以 $16$ 个全等小三角形中每个面积为 $\dfrac{32}{16} = 2$ 。

图中有十个小三角形被涂色，所以阴影面积是 $10 \times 2 = 20$ 。

所以正确答案是 B。

The large triangle has area $\dfrac12 \times 8 \times 8 = 32,$ so each of the $16$ congruent small triangles has area $\dfrac{32}{16} = 2.$

Ten of the small triangles are shaded, so the shaded area is $10 \times 2 = 20.$

Thus, the correct answer is B .

11.

条形图显示了一项颜色偏好调查的结果。各柱从左到右依次表示红、蓝、棕、粉、绿。偏好蓝色的人占百分之几？

The bar graph shows the results of a survey on color preferences. From left to right, the bars represent red, blue, brown, pink, and green. What percent preferred blue?

$20\%$

$24\%$

$30\%$

$36\%$

$42\%$

答案：B

知识点：数据与图表解读百分数

难度评级：770

小提示：

先把五个数量相加，得到调查总人数。

Add the five counts to get the total number of people surveyed

大提示：

偏好蓝色的百分比等于蓝色人数除以总人数，再化成百分数。

The percent preferring blue is (blue count) divided by (total), converted to a percent

解答：

调查总人数是 $50 + 60 + 40 + 60 + 40 = 250$ 。

偏好蓝色的比例为 $\dfrac{60}{250} = 24\%$ 。

所以正确答案是 B。

The total number surveyed is $50 + 60 + 40 + 60 + 40 = 250.$

The percent preferring blue is $\dfrac{60}{250} = 24\%.$

Thus, the correct answer is B .

12.

一辆车的五个轮胎（四个行驶轮胎和一个全尺寸备胎）进行轮换，使得汽车行驶前 $30{,}000$ 英里期间每个轮胎使用的里程数相同。每个轮胎使用了多少英里？

The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first $30{,}000$ miles the car traveled. For how many miles was each tire used?

$6000$

$7500$

$24{,}000$

$30{,}000$

$37{,}500$

答案：C

知识点：速率

难度评级：980

小提示：

任意时刻有 $4$ 个轮胎在地面上，而轮胎共有 $5$ 个，所以总轮胎里程为 $4 \times 30{,}000$ 。

At any moment $4$ of the $5$ tires are on the ground, so the total tire-mileage is $4 \times 30{,}000$

大提示：

把这个总轮胎里程平均分给 $5$ 个轮胎。

Share that total tire-mileage equally among the $5$ tires

解答：

在 $30{,}000$ 英里的行驶中，始终有 $4$ 个轮胎在使用，所以总轮胎里程为 $4 \times 30{,}000 = 120{,}000$ 轮胎英里。

平均分给 $5$ 个轮胎，每个轮胎使用 $\dfrac{120{,}000}{5} = 24{,}000$ 英里。

所以正确答案是 C。

During the $30{,}000$ miles, $4$ tires are always in use, so the total tire-mileage is $4 \times 30{,}000 = 120{,}000$ tire-miles.

Split equally among $5$ tires, each tire is used $\dfrac{120{,}000}{5} = 24{,}000$ miles.

Thus, the correct answer is C .

13.

五个考试分数的平均数是 $90$ ，中位数是 $91$ ，众数是 $94$ 。最低的两个考试分数之和是

Five test scores have a mean (average score) of $90,$ a median (middle score) of $91,$ and a mode (most frequent score) of $94.$ The sum of the two lowest test scores is

$170$

$171$

$176$

$177$

由已给信息无法确定

not determined by the information given

答案：B

知识点：平均数中位数（数据）众数

难度评级：1060

小提示：

五个分数总和为 $5 \times 90 = 450$ ，中间的分数是 $91$ 。

The five scores add up to $5 \times 90 = 450,$ and the middle score is $91$

大提示：

众数 $94$ 至少出现两次，而且这两个分数都必须不低于中位数，所以三个最高分是 $91$ 、 $94$ 、 $94$ 。

The mode $94$ must appear at least twice, and both copies must be at or above the median, so the three highest scores are $91,$ $94,$ and $94$

解答：

五个分数总和为 $5 \times 90 = 450$ 。中位数是第三个分数 $91$ 。因为 $94$ 是众数，它至少出现两次，且这两次都在中位数以上，所以最高两个分数分别是 $94$ 和 $94$ 。

三个最高分是 $91$ 、 $94$ 、 $94$ ，和为 $279$ 。因此最低两个分数之和是 $450 - 279 = 171$ 。

所以正确答案是 B。

The five scores sum to $5 \times 90 = 450.$ The median is the third score, $91.$ Since $94$ is the mode, it must appear at least twice, and both copies lie above the median, so the two highest scores are $94$ and $94.$

The three highest scores are $91,$ $94,$ and $94,$ summing to $279.$ So the two lowest sum to $450 - 279 = 171.$

Thus, the correct answer is B .

14.

向一个已装满三分之一的水箱中加入四加仑水后，水箱变为装满一半。这个水箱的容量是多少加仑？

When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is

$8$

$12$

$20$

$24$

$48$

答案：D

知识点：分数一次方程

难度评级：930

小提示：

加入 $4$ 加仑使水箱从 $\dfrac13$ 满变为 $\dfrac12$ 满。

Adding $4$ gallons raises the tank from $\dfrac13$ full to $\dfrac12$ full

大提示：

所以 $4$ 加仑等于容量的 $\dfrac12 - \dfrac13 = \dfrac16$ 。

So $4$ gallons equals $\dfrac12 - \dfrac13 = \dfrac16$ of the capacity

解答：

加入的 $4$ 加仑对应从 $\dfrac13$ 满到 $\dfrac12$ 满的增加量，即容量的 $\dfrac12 - \dfrac13 = \dfrac16$ 。

如果容量的 $\dfrac16$ 是 $4$ 加仑，那么总容量是 $4 \times 6 = 24$ 加仑。

所以正确答案是 D。

The $4$ gallons account for the change from $\dfrac13$ full to $\dfrac12$ full, which is $\dfrac12 - \dfrac13 = \dfrac16$ of the capacity.

If $\dfrac16$ of the capacity is $4$ gallons, the full capacity is $4 \times 6 = 24$ gallons.

Thus, the correct answer is D .

15.

这个序列中的第 $1992$ 个字母是什么？

ABCDEDCBAABCDEDCBAABCDEDCBAABCDEDC $\cdots$

What is the $1992$ nd letter in this sequence?

ABCDEDCBAABCDEDCBAABCDEDCBAABCDEDC $\cdots$

$A$

$B$

$C$

$D$

$E$

答案：C

知识点：模运算找规律

难度评级：910

小提示：

字符串 $ABCDEDCBA$ 一直重复；先数一个循环块有多少个字母。

The string $ABCDEDCBA$ repeats over and over; count how many letters are in one block

大提示：

一个循环块有 $9$ 个字母，所以用 $1992$ 除以 $9$ ，利用余数找到循环块中的位置。

One block has $9$ letters, so divide $1992$ by $9$ and use the remainder to find the position within a block

解答：

这个序列是 $9$ 个字母的循环块 $ABCDEDCBA$ 不断重复。

因为 $1992 = 9 \times 221 + 3$ ，所以第 $1992$ 个字母是一个循环块中的第 $3$ 个字母，即 $C$ 。

所以正确答案是 C。

The sequence is the $9$ -letter block $ABCDEDCBA$ repeated again and again.

Since $1992 = 9 \times 221 + 3,$ the $1992$ nd letter is the $3$ rd letter of a block, which is $C.$

Thus, the correct answer is C .

16.

哪个圆柱体的体积是右图所示圆柱体体积的两倍？

Which cylinder has twice the volume of the cylinder shown at right?

以上都不是

None of the above

答案：B

知识点：圆柱体积

难度评级：980

小提示：

圆柱体体积为 $\pi r^2 h$ ；先求给定圆柱体的体积。

The volume of a cylinder is $\pi r^2 h$ ; first compute the volume of the given cylinder

大提示：

高加倍会使体积加倍，但半径加倍会使体积变为 $4$ 倍；检查每个选项中的 $r$ 和 $h$ 。

Doubling the height doubles the volume, but doubling the radius multiplies the volume by $4$ ; check the $r$ and $h$ labels on each choice

解答：

给定圆柱体的体积是 $\pi \times 10^2 \times 5 = 500\pi$ ，所以两倍体积是 $1000\pi$ 。

各选项体积为：A 是 $\pi \times 20^2 \times 5 = 2000\pi$ ，B 是 $\pi \times 10^2 \times 10 = 1000\pi$ ，C 是 $\pi \times 5^2 \times 20 = 500\pi$ ，D 是 $\pi \times 20^2 \times 10 = 4000\pi$ 。只有 B 等于 $1000\pi$ 。

所以正确答案是 B。

The given cylinder has volume $\pi \times 10^2 \times 5 = 500\pi,$ so a cylinder of twice the volume has volume $1000\pi.$

The choices have volumes $\pi \times 20^2 \times 5 = 2000\pi$ (A), $\pi \times 10^2 \times 10 = 1000\pi$ (B), $\pi \times 5^2 \times 20 = 500\pi$ (C), and $\pi \times 20^2 \times 10 = 4000\pi$ (D). Only (B) equals $1000\pi.$

Thus, the correct answer is B .

17.

一个三角形的边长为 $6.5$ 、 $10$ 和 $s$ ，其中 $s$ 是整数。 $s$ 的最小可能值是多少？

The sides of a triangle have lengths $6.5,$ $10,$ and $s,$ where $s$ is a whole number. What is the smallest possible value of $s?$

$3$

$4$

$5$

$6$

$7$

答案：B

知识点：三角不等式

难度评级：960

小提示：

对三角形来说，较短的两边之和必须大于最长边。

For a triangle, the two shorter sides must add to more than the longest side

大提示：

需要 $6.5 + s \gt 10$ ；找满足条件的最小整数 $s$ 。

You need $6.5 + s \gt 10$ ; find the smallest whole number $s$

解答：

由三角形不等式， $6.5 + s$ 必须大于最长边 $10$ ，所以 $s \gt 3.5$ 。

大于 $3.5$ 的最小整数是 $4$ ，而它确实能组成三角形。

所以正确答案是 B。

By the triangle inequality, $6.5 + s$ must exceed the longest side $10,$ so $s \gt 3.5.$

The smallest whole number greater than $3.5$ is $4,$ and it does form a valid triangle.

Thus, the correct answer is B .

18.

一辆车旅行时，先在一个半小时内行驶 $80$ 英里，然后在交通堵塞中停了 $30$ 分钟，接着行驶 $100$ 英里，用时 $2$ 小时。这辆车在整个 $4$ 小时旅程中的平均速度是多少英里每小时？

On a trip, a car traveled $80$ miles in an hour and a half, then was stopped in traffic for $30$ minutes, then traveled $100$ miles during the next $2$ hours. What was the car’s average speed in miles per hour for the $4$ -hour trip?

$45$

$50$

$60$

$75$

$90$

答案：A

知识点：路程、速度与时间

难度评级：930

小提示：

平均速度等于总路程除以总时间。

Average speed is total distance divided by total time

大提示：

总路程是 $80 + 100$ 英里；总时间是 $1.5 + 0.5 + 2 = 4$ 小时，因为停车时间也要算入旅程时间。

Total distance is $80 + 100$ miles; total time is $1.5 + 0.5 + 2 = 4$ hours, since the stop still counts

解答：

这辆车总共行驶 $80 + 100 = 180$ 英里，总时间为 $1.5 + 0.5 + 2 = 4$ 小时，其中堵车停止时间也计入。

平均速度为 $\dfrac{180}{4} = 45$ 英里每小时。

所以正确答案是 A。

The car covered $80 + 100 = 180$ miles in total, over $1.5 + 0.5 + 2 = 4$ hours (the traffic stop still counts as time).

The average speed is $\dfrac{180}{4} = 45$ miles per hour.

Thus, the correct answer is A .

19.

一条州际公路上第 $5$ 个出口和第 $26$ 个出口之间的距离是 $118$ 英里。如果任意两个出口之间至少相距 $5$ 英里，那么在第 $5$ 个和第 $26$ 个出口之间，相邻两个出口之间最多可以相距多少英里？

The distance between the $5$ th and $26$ th exits on an interstate highway is $118$ miles. If any two exits are at least $5$ miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the $5$ th and $26$ th exits?

$8$

$13$

$18$

$47$

$98$

答案：C

知识点：极端原理

难度评级：1110

小提示：

从第 $5$ 个出口到第 $26$ 个出口之间有 $21$ 个间隔。

Between the $5$ th and $26$ th exits there are $21$ gaps

大提示：

要让一个间隔尽可能大，就让其余 $20$ 个间隔都取允许的最小值 $5$ 英里。

To make one gap as large as possible, make the other $20$ gaps as small as allowed, $5$ miles each

解答：

从第 $5$ 个出口到第 $26$ 个出口，共有 $26 - 5 = 21$ 个相邻出口间隔，每个至少 $5$ 英里。

要最大化其中一个间隔，就让其他 $20$ 个间隔都等于 $5$ 英里，共占 $20 \times 5 = 100$ 英里。剩下的间隔是 $118 - 100 = 18$ 英里。

所以正确答案是 C。

From the $5$ th exit to the $26$ th exit there are $26 - 5 = 21$ gaps between consecutive exits, each at least $5$ miles.

To maximize one gap, make the other $20$ gaps exactly $5$ miles, using $20 \times 5 = 100$ miles. The remaining gap is $118 - 100 = 18$ miles.

Thus, the correct answer is C .

20.

下列哪一种由全等正方形组成的图案，不能沿所示线折成一个立方体？

Which pattern of identical squares could not be folded along the lines shown to form a cube?

答案：D

知识点：展开图（立体几何）正方体

难度评级：1090

小提示：

立方体有 $6$ 个面，所以每个图案都有 $6$ 个正方形；在脑中试着把它们折起来。

A cube has $6$ faces, so each pattern has $6$ squares; try folding each one up in your mind

大提示：

追踪每个正方形折叠后落在哪个面上；如果两个正方形会盖到同一个面上，图案就失败。

Track where each square lands as the pattern folds; a pattern fails when two squares would cover the same face

解答：

五个图案都由 $6$ 个正方形组成。把 A、B、C 和 E 折叠时，各正方形能分别覆盖立方体的六个面。

对图案 D，无论怎样折叠，都会迫使两个正方形落在同一个面上而重叠，所以不能形成立方体。

所以正确答案是 D。

Each of the five patterns has $6$ squares. Folding patterns (A), (B), (C), and (E) wraps the squares neatly onto the six faces of a cube.

For pattern (D), any attempt to fold forces two of the squares to land on the same face, so they overlap and no cube can be formed.

Thus, the correct answer is D .

21.

Northside 的鼓号队为旅行筹款。鼓手和号手分别记录销售额。图中每组的空心柱表示鼓手，阴影柱表示号手。根据下面的双条形图，哪个月份中一个小组的销售额超过另一个小组的百分比最大？

Northside’s Drum and Bugle Corps raised money for a trip. The drummers and bugle players kept separate sales records. In each pair on the graph, the open bar represents drums and the shaded bar represents bugles. According to the double bar graph of monthly sales below, in what month did one group’s sales exceed the other’s by the greatest percent?

一月

Jan

二月

Feb

三月

Mar

四月

Apr

五月

May

答案：B

知识点：数据与图表解读百分数

难度评级：1130

小提示：

对每个月，读出两根柱子的高度并求它们的差。

For each month, read off the two bar heights and find the gap between them

大提示：

用两者之差除以较小值，就得到较大值超过较小值的百分比；比较各月份的这个比值。

Divide the difference by the smaller value to find the percent by which the larger exceeds it; compare this ratio across the months

解答：

从图上读出（鼓手，号手）的销售额：一月 $(7, 9)$ ，二月 $(5, 3)$ ，三月 $(9, 6)$ ，四月 $(9, 12)$ ，五月 $(8, 10)$ 。

较大值超过较小值的百分比分别为：一月 $\tfrac{2}{7} \approx 29\%$ ，二月 $\tfrac{2}{3} \approx 67\%$ ，三月 $\tfrac{3}{6} = 50\%$ ，四月 $\tfrac{3}{9} \approx 33\%$ ，五月 $\tfrac{2}{8} = 25\%$ 。最大的是二月，此时 $5$ 比 $3$ 大约 $67\%$ 。

所以正确答案是 B。

Reading the graph gives (drums, bugles): January $(7, 9),$ February $(5, 3),$ March $(9, 6),$ April $(9, 12),$ and May $(8, 10).$

The percent excess of the larger over the smaller is $\tfrac{2}{7} \approx 29\%$ in January, $\tfrac{2}{3} \approx 67\%$ in February, $\tfrac{3}{6} = 50\%$ in March, $\tfrac{3}{9} \approx 33\%$ in April, and $\tfrac{2}{8} = 25\%$ in May. The greatest is February, where $5$ exceeds $3$ by about $67\%.$

Thus, the correct answer is B .

22.

如图，八块 $1 \times 1$ 的正方形瓷砖拼成一个图形，其外边界周长为 $14$ 个单位。再加入两块同样大小的瓷砖，使每块新增瓷砖至少有一条边与原图中某个正方形的一条边重合。下列哪一个可能是新图形的周长？

Eight $1 \times 1$ square tiles are arranged as shown so their outside edges form a polygon with a perimeter of $14$ units. Two additional tiles of the same size are added to the figure so that at least one side of each added tile is shared with a side of one of the squares in the original figure. Which of the following could be the perimeter of the new figure?

$15$

$17$

$18$

$19$

$20$

答案：C

知识点：周长分类讨论

难度评级：1170

小提示：

新加一块瓷砖若只共享一条边，周长增加 $+2$ ；若共享两条边，周长增加 $0$ 。

Adding a tile that shares exactly one side changes the perimeter by $+2$ ; sharing two sides changes it by $0$

大提示：

加两块瓷砖时，总变化量可能是 $0$ 、 $2$ 或 $4$ ；把它们加到 $14$ 上。

With two added tiles, the total change can be $0,$ $2,$ or $4$ ; add these to $14$

解答：

一块只共享一条边的新增瓷砖会使周长增加 $2$ ：四条新边中有两条因共享边而不在外边界上。若共享两条边，则周长增加 $0$ 。

加入两块瓷砖后，周长可能增加 $0$ 、 $2$ 或 $4$ ，新周长可能为 $14$ 、 $16$ 或 $18$ 。选项中只有 $18$ 可能。

所以正确答案是 C。

A tile that shares exactly one side adds $2$ to the perimeter (four new edges minus two hidden), while a tile sharing two sides adds $0.$

With two added tiles the perimeter can change by $0,$ $2,$ or $4,$ giving new perimeters of $14,$ $16,$ or $18.$ Of the choices, only $18$ is possible.

Thus, the correct answer is C .

23.

掷两颗骰子，朝上点数的乘积大于 $10$ 的概率是

If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than $10$ is

$\dfrac37$

$\dfrac{17}{36}$

$\dfrac12$

$\dfrac58$

$\dfrac{11}{12}$

答案：B

知识点：骰子（概率）系统列举

难度评级：1150

小提示：

共有 $6 \times 6 = 36$ 个等可能结果；数出乘积大于 $10$ 的结果。

There are $6 \times 6 = 36$ equally likely outcomes; count those whose product exceeds $10$

大提示：

做一个 $6 \times 6$ 的乘法表，并标出所有大于 $10$ 的乘积。

Make a $6 \times 6$ multiplication table and mark every product greater than $10$

解答：

共有 $36$ 个等可能结果。按第一颗骰子的点数，计算乘积大于 $10$ 的有序数对：若为 $2$ ，有 $1$ 个结果满足，即 $2 \times 6$ ；若为 $3$ ，有 $3$ 个；若为 $4$ ，有 $4$ 个；若为 $5$ ，有 $4$ 个；若为 $6$ ，有 $5$ 个。

有利结果总数为 $1 + 3 + 4 + 4 + 5 = 17$ ，所以概率是 $\dfrac{17}{36}$ 。

所以正确答案是 B。

There are $36$ equally likely outcomes. Counting the ordered pairs whose product exceeds $10$ : with a first die of $2$ there is $1$ (namely $2 \times 6$ ); of $3$ , there are $3$ ; of $4$ , there are $4$ ; of $5$ , there are $4$ ; of $6$ , there are $5.$

That is $1 + 3 + 4 + 4 + 5 = 17$ favorable outcomes, so the probability is $\dfrac{17}{36}.$

Thus, the correct answer is B .

24.

如图排列四个半径为 $3$ 的圆，它们的圆心是一个正方形的四个顶点。阴影区域的面积最接近

Four circles of radius $3$ are arranged as shown. Their centers are the vertices of a square. The area of the shaded region is closest to

$7.7$

$12.1$

$17.2$

$18$

$27$

答案：A

知识点：圆面积面积分割估算

难度评级：1170

小提示：

相邻圆相切，所以圆心组成的正方形边长等于两个半径，即 $6$ ，面积为 $36$ 。

Since neighboring circles are tangent, the square’s side equals two radii, so its side is $6$ and its area is $36$

大提示：

正方形内的四个四分之一圆合起来正好是一个整圆，面积为 $9\pi$ ；从正方形面积中减去它。

The four quarter-circles inside the square together make one full circle of area $9\pi$ ; subtract that from the square

解答：

相邻圆相切，所以圆心组成的正方形边长为 $2 \times 3 = 6$ ，面积为 $36$ 。

在正方形内，每个圆贡献一个四分之一圆，四个四分之一圆合成一个完整圆，面积为 $9\pi \approx 28.3$ 。阴影面积为 $36 - 9\pi \approx 7.7$ 。

所以正确答案是 A。

Because adjacent circles are tangent, the square through the centers has side $2 \times 3 = 6$ and area $36.$

Inside the square, each circle contributes a quarter-circle, and the four quarters make one full circle of area $9\pi \approx 28.3.$ The shaded region is $36 - 9\pi \approx 7.7.$

Thus, the correct answer is A .

25.

一个装满水的容器先倒出其中的一半。然后倒出剩余水的三分之一。继续这个过程：第三次倒出剩余水的四分之一，第四次倒出剩余水的五分之一，依此类推。倒多少次后，正好剩下原来水量的十分之一？

One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, and so on. After how many pourings does exactly one tenth of the original water remain?

$6$

$7$

$8$

$9$

$10$

答案：D

知识点：裂项相消分数

难度评级：1200

小提示：

第一次倒出后剩下 $\dfrac12$ ；第二次倒出后剩下 $\dfrac12 \times \dfrac23$ 。

After the first pouring $\dfrac12$ remains; after the second $\dfrac12 \times \dfrac23$ remains

大提示：

剩余比例的连乘积会逐项约分：倒 $n$ 次后剩下 $\dfrac{1}{n+1}$ ；令它等于 $\dfrac{1}{10}$ 。

The remaining fraction telescopes: after $n$ pourings it is $\dfrac{1}{n+1}$ ; set this equal to $\dfrac{1}{10}$

解答：

第 $n$ 次倒出后，剩余比例为 $\dfrac12 \times \dfrac23 \times \dfrac34 \times \cdots \times \dfrac{n}{n+1}$ ，逐项约分后等于 $\dfrac{1}{n+1}$ 。

令 $\dfrac{1}{n+1} = \dfrac{1}{10}$ ，得到 $n = 9$ 。

所以正确答案是 D。

After the $n$ th pouring, the fraction remaining is $\dfrac12 \times \dfrac23 \times \dfrac34 \times \cdots \times \dfrac{n}{n+1},$ which telescopes to $\dfrac{1}{n+1}.$

Setting $\dfrac{1}{n+1} = \dfrac{1}{10}$ gives $n = 9.$

Thus, the correct answer is D .