1997 AMC 8 真题与在线限时测验

计时

40:00

1.

$\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000} =$

$0.0026$

$0.0197$

$0.1997$

$0.26$

$1.997$

答案：C

知识点：小数位值

难度评级：370

小提示：

分别相加十分位、百分位、千分位和万分位上的数

Add the tenths, hundredths, thousandths, and ten-thousandths digits

大提示：

把每个分数都按它的位值写成小数

Convert each fraction to its decimal place value

解答：

把所有分数化成小数，得到 $\begin{aligned} 0.1 + 0.09 + 0.009 &+ 0.0007 \\ &= 0.1997 \end{aligned}\text{。}$

所以正确答案是 C。

Converting all the fractions to decimals, we get $\begin{aligned} 0.1 + 0.09 + 0.009 &+ 0.0007 \\ &= 0.1997. \end{aligned}$

Thus, C is the correct answer.

2.

安选择一个两位整数，从 $200$ 中减去它，再把结果加倍。安能得到的最大数是多少？

Ahn chooses a two-digit integer, subtracts it from $200,$ and doubles the result. What is the largest number Ahn can get?

$200$

$202$

$220$

$380$

$398$

答案：D

知识点：最优化

难度评级：450

小提示：

最小的两位整数是 $10$

The smallest two-digit integer is $10$

大提示：

要使结果最大，应减去尽可能小的两位数

To maximize the result, subtract the smallest possible two-digit number

解答：

要得到最大结果，应从 $200$ 中减去尽可能小的数。

最小的两位数是 $10$ 。让安选择这个数，就会得到 $2 (200 - 10) = 2 \cdot 190 = 380$ 作为最终结果。

所以正确答案是 D。

To get the largest number, we would want to subtract the smallest number possible from $200.$

The smallest two-digit number is $10.$ Having Ahn choose this number will give us $2 (200 - 10) = 2 \cdot 190 = 380$ as the final result.

Thus, D is the correct answer.

3.

下列哪个数最大？

Which of the following numbers is the largest?

$0.97$

$0.979$

$0.9709$

$0.907$

$0.9089$

答案：B

知识点：小数位值

难度评级：560

小提示：

先比较十分位，再比较百分位，然后比较千分位

Compare tenths first, then hundredths, then thousandths

大提示：

把这些小数按位值对齐

Line the decimals up by place value

解答：

所有数的十分位数字相同，所以比较百分位数字。

最大的百分位数字是 $7$ ，因此只需看百分位为七的选项。

其中最大的千分位数字是 $9$ ，对应 $0.979$ 。

所以正确答案是 B。

We have that all the tenths place digits are the same, so we then look at the hundredths digit.

The largest hundredths digit is $7,$ so we can limit the answer choices to the ones with this value.

The largest thousandths digit is $9,$ which is achieved by $0.979.$

Thus, B is the correct answer.

4.

朱莉正在为班级准备演讲。她的演讲必须持续半小时到四分之三小时之间。理想语速是每分钟 $150$ 个词。如果朱莉按理想语速演讲，下列哪个词数适合作为她演讲的长度？

Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is $150$ words per minute. If Julie speaks at the ideal rate, which of the following number of words would be an appropriate length for her speech?

$2250$

$3000$

$4200$

$4350$

$5650$

答案：E

知识点：速率单位换算

难度评级：660

小提示：

把分钟数的上下界乘以每分钟 $150$ 个词

Multiply the minute bounds by $150$ words per minute

大提示：

把这段时间范围换算成分钟

Convert the time range into minutes

解答：

半小时是 $30$ 分钟，朱莉可以说 $150 \cdot 30 = 4500$ 个词。四分之三小时是 $45$ 分钟，她可以说 $150 \cdot 45 = 6750$ 个词。选项中唯一介于这两个数之间的是 $5650$ 。

所以正确答案是 E。

One-half hour is $30$ minutes, in which Julie can speak $150 \cdot 30 = 4500$ words. Three-quarters of an hour is $45$ minutes, in which Julie can speak $150 \cdot 45 = 6750$ words. The only answer choice in between these two values is $5650.$

Thus, E is the correct answer.

5.

有许多两位数是 $7$ 的倍数，但其中只有两个倍数的数位和为 $10$ 。这两个 $7$ 的倍数之和是

There are many two-digit multiples of $7,$ but only two of the multiples have a digit sum of $10.$ The sum of these two multiples of $7$ is

$119$

$126$

$140$

$175$

$189$

答案：A

知识点：倍数数字系统列举

难度评级：730

小提示：

只保留数位和为 $10$ 的倍数

Keep only the multiples whose digits add to $10$

大提示：

列出 $7$ 的两位数倍数

List the two-digit multiples of $7$

解答：

$7$ 的两位数倍数为 $14, 21, 28, 35, 42, 49, 56, 63, 70\text{、}$ $77, 84, 91, 98\text{。}$ 其中数位和为 $10$ 的是 $28$ 和 $91$ 。

它们的和是 $28 + 91 = 119$ 。

所以正确答案是 A。

Listing out all the two-digit multiples of $7,$ we get $14, 21, 28, 35, 42, 49, 56, 63, 70,$ $77, 84, 91, 98.$ We have that $28$ and $91$ are the two multiples whose digits add to $10.$

The sum of these numbers is $28 + 91 = 119.$

Thus, A is the correct answer.

6.

在数 $74982.1035$ 中，数字 $9$ 所在位的位值是数字 $3$ 所在位的位值的多少倍？

In the number $74982.1035$ the value of the place occupied by the digit $9$ is how many times as great as the value of the place occupied by the digit $3?$

$1{,}000$

$10{,}000$

$100{,}000$

$1{,}000{,}000$

$10{,}000{,}000$

答案：C

知识点：位值指数

难度评级：820

小提示：

位值每向左移动一位就乘以 $10$

Each step left in place value multiplies by $10$

大提示：

比较 $9$ 和 $3$ 所在位置的位值

Compare the place values occupied by $9$ and $3$

解答：

每个数字所在位的位值都是它右边一位的十倍。

数字 $9$ 所在位比数字 $3$ 所在位向左 $5$ 位。

因此它的位值是 $10^5 = 100,000$ 倍。

所以正确答案是 C。

Note that each digit has ten times the value of the digit to its right.

The place occupied by $9$ is $5$ spaces to the left of the place occupied by $3.$

This means that it is $10^5 = 100,000$ times as great.

Thus, C is the correct answer.

7.

能容纳一个半径为 $4$ 的圆的最小正方形的面积是

The area of the smallest square that will contain a circle of radius $4$ is

$8$

$16$

$32$

$64$

$128$

答案：D

知识点：面积正方形（几何）

难度评级：900

小提示：

圆的直径 $8$ 就是正方形的边长

Use diameter $8$ as the side length

大提示：

最小的外接正方形边长等于圆的直径

The smallest containing square has side equal to the circle diameter

解答：

可以把圆内切在正方形中，使它与正方形四边的中点相切。

因此正方形边长是圆半径的两倍，即 $2 \cdot 4 = 8$ 。

正方形面积为 $8^2 = 64\text{。}$

所以正确答案是 D。

We can inscribe the circle inside the square so that it is tangent to the midpoints of each side of the square.

This means that the side length of the square is two times the radius of the circle, making it $2 \cdot 4 = 8.$

Then the area of the square is $8^2 = 64.$

Thus, D is the correct answer.

8.

沃尔特早上 $6:30$ 起床， $7:30$ 赶上校车，有 $6$ 节课，每节 $50$ 分钟，有 $30$ 分钟午餐时间，并且在学校还有额外 $2$ 小时时间。他乘校车回家，下午 $4:00$ 到家。他在校车上花了多少分钟？

Walter gets up at $6:30$ a.m., catches the school bus at $7:30$ a.m., has $6$ classes that last $50$ minutes each, has $30$ minutes for lunch, and has $2$ hours additional time at school. He takes the bus home and arrives at $4:00$ p.m. How many minutes has he spent on the bus?

$30$

$60$

$75$

$90$

$120$

答案：B

知识点：日期与时间单位换算

难度评级：930

小提示：

从总经过时间中减去上课、午餐和额外在校时间

Subtract the class, lunch, and extra school time

大提示：

先求从上校车到回到家之间经过了多少时间

Find the elapsed time from bus pickup to arriving home

解答：

从沃尔特上校车到回到家之间经过 $8.5$ 小时。

换算成分钟，共有 $60 \cdot 8.5 = 510$ 分钟。

他在学校的总时间是 $6 \cdot 50 + 30 + 2 \cdot 60 =$ $300 + 30 + 120 = 450$ 分钟。

因此他在校车上的时间是 $510 - 450 = 60$ 分钟。

所以正确答案是 B。

There are $8.5$ hours between the time Walter catches the school bus and arrives at home.

This is a total of $60 \cdot 8.5 = 510$ minutes.

The total time Walter spends at school is $6 \cdot 50 + 30 + 2 \cdot 60 =$ $300 + 30 + 120 = 450$ minutes.

This means that Walter spends $510 - 450 = 60$ minutes on the bus.

Thus, B is the correct answer.

9.

三个姓名不同的学生排成一列。从前到后按字母顺序排列的概率是多少？

Three students, with different names, line up single file. What is the probability that they are in alphabetical order from front-to-back?

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

答案：C

知识点：基本概率排列

难度评级：1010

小提示：

只有一种排列是按字母顺序的

Only one ordering is alphabetical

大提示：

数出三个学生所有可能的排列顺序

Count all possible orders of the three students

解答：

队首有 $3$ 种选择，中间有 $2$ 种选择。

最后剩 $1$ 种选择，所以共有 $3 \cdot 2 \cdot 1 = 6$ 种排列。

其中只有一种按字母顺序排列，所以概率为 $\dfrac{1}{6}$ 。

所以正确答案是 C。

There are $3$ options for the person in front. Then, there are $2$ options for the person in the middle.

This leaves $1$ choice for the person at the end. There are $3 \cdot 2 \cdot 1 = 6$ ways for the people to line up.

Only one of these lines is correct, and the probability it occurs is $\dfrac{1}{6}.$

Thus, C is the correct answer.

10.

这个正方形区域有多少比例被涂色？条纹宽度相等，图按比例绘制。

What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale.

$\dfrac{5}{12}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

答案：C

知识点：面积分割分数

难度评级：1030

小提示：

数出涂色小正方形占总小正方形的比例

Count shaded small squares out of the total

大提示：

按条纹宽度把正方形分成相等的小正方形

Divide the square into equal small squares matching the stripe widths

解答：

可以把大正方形分成 $6^2 = 36$ 个小正方形。涂色小正方形数为 $3 + 7 + 11 = 21\text{。}$

涂色比例为 $\dfrac{21}{36} = \dfrac{7}{12}\text{。}$

所以正确答案是 C。

We can split the square into $6^2 = 36$ unit squares. The number of shaded small squares is $3 + 7 + 11 = 21.$

The fraction of the square that is shaded is $\dfrac{21}{36} = \dfrac{7}{12}.$

Thus, C is the correct answer.

11.

令 $\boxed{N}$ 表示 $N$ 的正整数因数个数。例如， $\boxed{3}=2$ ，因为 $3$ 有两个因数 $1$ 和 $3$ 。求下式的值： $\boxed{\boxed{11}\times\boxed{20}}\text{。}$

Let $\boxed{N}$ mean the number of whole number divisors of $N.$ For example, $\boxed{3}=2$ because $3$ has two divisors, $1$ and $3.$ Find the value of $\boxed{\boxed{11}\times\boxed{20}}.$

$6$

$8$

$12$

$16$

$24$

答案：A

知识点：因数个数质因数分解

难度评级：1140

小提示：

用质因数分解来数因数个数

Use prime factorization to count divisors

大提示：

先计算里面的方框，再计算外面的方框

Evaluate the inside boxes before the outside box

解答：

$11$ 是质数，所以它只有 $2$ 个正因数。

$20$ 的质因数分解是 $2^2 \cdot 5\text{。}$ 一个数的因数个数等于其质因数分解中各指数加一后的乘积。

这里得到 $(2 + 1) (1 + 1) = 3 \cdot 2 = 6\text{。}$

因此里面的乘积是 $2 \cdot 6 = 12$ 。而 $12$ 的质因数分解是 $2^2 \cdot 3\text{，}$ 所以它也有 $(2 + 1) (1 + 1) = 3 \cdot 2 = 6$ 个因数。

所以正确答案是 A。

We know that $11$ is prime, which means that it only has $2$ divisors.

The prime factorization of $20$ is $2^2 \cdot 5.$ Recall that the number of divisors a number has is the product of all the exponents plus one in the prime factorization.

Here, that product would be $(2 + 1) (1 + 1) = 3 \cdot 2 = 6.$

Then $2 \cdot 6 = 12.$ We have the prime factorization of $12$ is $2^2 \cdot 3.$ This also has $(2 + 1) (1 + 1) = 3 \cdot 2 = 6$ divisors.

Thus, A is the correct answer.

12.

$\angle 1 + \angle 2 = 180^\circ$ $\angle 3 = \angle 4$ 求 $\angle 4$ 。

$\angle 1 + \angle 2 = 180^\circ$ $\angle 3 = \angle 4$ Find $\angle 4$

$20^\circ$

$25^\circ$

$30^\circ$

$35^\circ$

$40^\circ$

答案：D

知识点：导角角度和

难度评级：1170

小提示：

利用互补条件，然后使用角 $3$ 等于角 $4$

Use the supplementary condition and then angle $3$ equals angle $4$

大提示：

先从左边三角形求出角 $1$

First find angle $1$ from the left triangle

解答：

我们有 $\angle 1 = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}\text{，}$ 这里用到了三角形内角和为 $180^{\circ}$ 这一事实。

由此可得 $\angle 2 = 180^{\circ} - \angle 1 = 110^{\circ}\text{，}$ 因为 $\angle 1$ 和 $\angle 2$ 互补。

最后， $\angle 3 + \angle 4 + 110^{\circ} = 180^{\circ} \Rightarrow$ $2 \angle 4 = 70^{\circ} \Rightarrow \angle 4 = 35^{\circ}\text{。}$

所以正确答案是 D。

We have that $\angle 1 = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}$ using the fact that the interior angles of a triangle add to $180^{\circ}.$

This tells us that $\angle 2 = 180^{\circ} - \angle 1 = 110^{\circ}$ since $\angle 1$ and $\angle 2$ are supplementary.

Finally, $\angle 3 + \angle 4 + 110^{\circ} = 180^{\circ} \Rightarrow$ $2 \angle 4 = 70^{\circ} \Rightarrow \angle 4 = 35^{\circ}.$

Thus, D is the correct answer.

13.

三袋软糖分别有 $26$ ， $28$ 和 $30$ 颗。这三袋中黄色软糖所占比例分别为 $50\%$ 、 $25\%$ 和 $20\%$ 。把三袋糖全部倒入一个碗中，黄色软糖占碗中所有软糖的比例最接近下列哪一个？

Three bags of jelly beans contain $26,$ $28,$ and $30$ beans. The ratios of yellow beans to all beans in each of these bags are $50\%,$ $25\%,$ and $20\%,$ respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?

$31\%$

$32\%$

$33\%$

$35\%$

$95\%$

答案：A

知识点：百分数平均数

难度评级：1150

小提示：

用黄色软糖总数除以软糖总数

Divide total yellow beans by total beans

大提示：

先求每袋中黄色软糖的颗数

Find the number of yellow beans in each bag

解答：

第一袋中有 $26 \cdot 0.5 = 26 \div 2 = 13$ 颗黄色软糖，第二袋中有 $28 \cdot 0.25 = 28 \div 4 = 7$ 颗黄色软糖，第三袋中有 $30 \cdot 0.2 = 30 \div 5 = 6$ 颗黄色软糖。

黄色软糖总数为 $13 + 7 + 6 = 26\text{，}$ 软糖总数为 $26 + 28 + 30 = 84\text{。}$ 黄色软糖所占的百分比为 $\dfrac{26}{84} \cdot 100 \% = \dfrac{13}{42} \cdot 100 \%$ $\approx 30.9 \%\text{。}$

所以正确答案是 A。

There are $26 \cdot 0.5 = 26 \div 2 = 13$ yellow jelly beans in the first bag, $28 \cdot 0.25 = 28 \div 4 = 7$ yellow jelly beans in the second bag, and $30 \cdot 0.2 = 30 \div 5 = 6$ yellow jelly beans in the third bag.

The total number of yellow jelly beans is $13 + 7 + 6 = 26$ and the total number of jelly beans is $26 + 28 + 30 = 84.$ The ratio of yellow jelly beans to all the beans is $\dfrac{26}{84} \cdot 100 \% = \dfrac{13}{42} \cdot 100 \%$ $\approx 30.9 \%.$

Thus, A is the correct answer.

14.

有一组五个正整数，它们的平均数是 $5$ ，中位数是 $5$ ，且唯一的众数是 $8$ 。这组数中最大整数与最小整数的差是多少？

There is a set of five positive integers whose average (mean) is $5,$ whose median is $5,$ and whose only mode is $8.$ What is the difference between the largest and smallest integers in the set?

$3$

$5$

$6$

$7$

$8$

答案：D

知识点：平均数中位数（数据）众数

难度评级：1270

小提示：

众数 $8$ 必须恰好出现两次

The mode $8$ must appear exactly twice

大提示：

用平均数求出五个数的总和

Use the mean to get the total sum

解答：

五个数的和是 $5 \cdot 5 = 25$ 。

按从小到大排列，中间的数是 $5$ 。因为 $8$ 是唯一的众数，最后两个数都必须是 $8$ ；但不能有三个 $8$ ，否则中位数也会是 $8$ 。

这三个已知数的和是 $5 + 8 + 8 = 21$ ，所以两个较小的数之和必须是 $25 - 21 = 4$ 。它们必须是不同的正整数，否则另一个数也会与 $8$ 并列为众数。因此它们是 $1$ 和 $3$ 。

因此所求的差为 $8 - 1 = 7\text{。}$

所以正确答案是 D。

The sum of all the numbers in the list is $5 \cdot 5 = 25.$

In increasing order, the middle number is $5$ . Because $8$ is the only mode, the last two numbers must both be $8$ ; there cannot be three $8$ s, since then the median would be $8$ .

These three known numbers sum to $5 + 8 + 8 = 21,$ so the two smaller numbers must add to $25 - 21 = 4.$ They must be distinct positive integers; otherwise a second value would tie $8$ as a mode. Thus they are $1$ and $3$ .

The desired difference is then $8 - 1 = 7.$

Thus, D is the correct answer.

15.

图中大正方形的每条边都被三等分。内接正方形的顶点在这些三等分点上，如图所示。内接正方形面积与大正方形面积的比是

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is

$\dfrac{\sqrt{3}}{3}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

$\dfrac{\sqrt{5}}{3}$

$\dfrac{7}{9}$

答案：B

知识点：面积比勾股定理

难度评级：1330

小提示：

用勾股定理求内接正方形的边长

Use the Pythagorean theorem for the inscribed square side

大提示：

设大正方形边长被分成三段相等的长度

Let the large square side be split into three equal parts

解答：

设大正方形边长为 $3x$ 。于是内接正方形的边长可以由 $\sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5}$ 根据勾股定理求出。

大正方形面积为 $(3x)^2 = 9x^2\text{，}$ 内接正方形面积为 $(x\sqrt{5})^2 = 5x^2\text{。}$

因此面积比为 $\dfrac{5}{9}$ 。

所以正确答案是 B。

Let $3x$ be the side length of the large square. Then we can find the side length of the inner square via $\sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5}$ from the Pythagorean Theorem.

The area of the larger square is $(3x)^2 = 9x^2$ and that of the inner square is $(x\sqrt{5})^2 = 5x^2.$

The ratio of the areas is then $\dfrac{5}{9}.$

Thus, B is the correct answer.

16.

佩妮·普里赛斯利在三家公司各买了价值 $\$ 100$ 的股票：阿拉巴马杏仁公司、波士顿豆业公司和加利福尼亚花椰菜公司。一年后，AA 上涨 $20 \%$ ，BB 下跌 $25 \%$ ，CC 不变。第二年，AA 比前一年下跌 $20 \%$ ，BB 比前一年上涨 $25 \%$ ，CC 仍不变。如果 $A$ 、 $B$ 和 $C$ 是这些股票的最终价值，那么

Penni Precisely buys $\$ 100$ worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower. After one year, AA was up $20 \%,$ BB was down $25 \%,$ and CC was unchanged. For the second year, AA was down $20 \%$ from the previous year, BB was up $25 \%$ from the previous year, and CC was unchanged. If $A,$ $B,$ and $C$ are the final values of the stock, then

$A = B = C$

$A = B \lt C$

$C \lt B = A$

$A \lt B \lt C$

$B \lt A \lt C$

答案：E

知识点：百分数

难度评级：1210

小提示：

先上涨百分之 $20$ ，再下跌百分之 $20$ ，并不会回到原值

A $20$ percent gain followed by a $20$ percent loss does not return to the start

大提示：

百分比变化要用乘法因子逐步计算

Apply each percent change multiplicatively

解答：

第一年后，AA 的价值变为 $\$ 100 \cdot 1.2 = \$ 120\text{，}$ BB 的价值变为 $\$ 100 \cdot 0.75 = \$ 75\text{。}$

第二年后，AA 的价值为 $\$ 120 \cdot 0.8 = \$ 96\text{，}$ BB 的价值为 $\$ 75 \cdot 1.25 = \$ 93.75\text{。}$

CC 价值仍不变，所以最终顺序是 $B \lt A \lt C\text{。}$

所以正确答案是 E。

After the first year, AA’s stock’s worth goes up to $\$ 100 \cdot 1.2 = \$ 120$ and BB’s goes down to $\$ 100 \cdot 0.75 = \$ 75.$

After the second year, AA’s stock is worth $\$ 120 \cdot 0.8 = \$ 96$ and BB’s is worth $\$ 75 \cdot 1.25 = \$ 93.75.$

CC’s stock worth remains the same, so the ordering of the stock worths is now $B \lt A \lt C.$

Thus, E is the correct answer.

17.

一个立方体有八个顶点和十二条棱。连接两个不由一条棱相连的顶点的线段，如 $x$ ，称为对角线。线段 $y$ 也是对角线。一个立方体有多少条对角线？

A cube has eight vertices (corners) and twelve edges. A segment, such as $x,$ which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have?

$6$

$8$

$12$

$14$

$16$

答案：E

知识点：正方体对角线

难度评级：1270

小提示：

每个面有两条面对角线，还有连接相对顶点的空间对角线

Each face has two diagonals, and opposite vertices give space diagonals

大提示：

分别数面对角线和空间对角线

Count face diagonals and space diagonals separately

解答：

每个面有两条对角线，连接这个面上两对相对顶点。

另外，对于每个顶点，都有一个与它在立方体内部相对的顶点。

这样空间对角线共有 $8 \div 2 = 4$ 条。

总对角线数为 $6 \cdot 2 + 4 = 12 + 4 = 16\text{。}$

所以正确答案是 E。

Each face has two diagonals connecting each of the two pairs of opposite vertices.

Also, for each vertex, there is one corresponding vertex that lies opposite it on the cube.

There are then $8 \div 2 = 4$ interior space diagonals in the cube.

The total number of diagonals is then $6 \cdot 2 + 4 = 12 + 4 = 16.$

Thus, E is the correct answer.

18.

上周杂货店小盒面巾纸的价格是 $4$ 盒 $\$5$ 。本周促销价是 $5$ 盒 $\$4$ 。促销期间每盒价格下降的百分比最接近

At the grocery store last week, small boxes of facial tissue were priced at $4$ boxes for $\$5.$ This week they are on sale at $5$ boxes for $\$4.$ The percent decrease in the price per box during the sale was closest to

$30\%$

$35\%$

$40\%$

$45\%$

$65\%$

答案：B

知识点：百分数速率

难度评级：1280

小提示：

百分比下降等于价格变化除以原价

Percent decrease is change divided by original price

大提示：

分别计算原来和促销时每盒的价格

Compute the old and sale prices per box

解答：

原来每盒价格为 $\$ 5 \div 4 = \$ 1.25\text{。}$

现在每盒价格为 $\$ 4 \div 5 = \$ 0.8\text{。}$

下降百分比为 $\begin{gathered} \dfrac{1.25 - 0.8}{1.25} \cdot 100 \% \\ = \dfrac{45}{125} \cdot 100 \% \\ = \dfrac{9}{25} \cdot 100 \% \\ = 36 \% \end{gathered}\text{。}$

所以正确答案是 B。

Originally, each box is worth $\$ 5 \div 4 = \$ 1.25.$

Now, each box is worth $\$ 4 \div 5 = \$ 0.8.$

The percent decrease is then $\begin{gathered} \dfrac{1.25 - 0.8}{1.25} \cdot 100 \% \\ = \dfrac{45}{125} \cdot 100 \% \\ = \dfrac{9}{25} \cdot 100 \% \\ = 36 \%. \end{gathered}$

Thus, B is the correct answer.

19.

如果乘积 $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9\text{，}$ 那么 $a$ 与 $b$ 的和是多少？

If the product $\dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \ldots\cdot \dfrac{a}{b} = 9,$ what is the sum of $a$ and $b?$

$11$

$13$

$17$

$35$

$37$

答案：D

知识点：裂项相消分数

难度评级：1340

小提示：

约分后只剩第一个分母和最后一个分子

After cancellation, only the first denominator and last numerator remain

大提示：

这个乘积会逐项相消

The product telescopes

解答：

每个分数的分子都会与右边分数的分母约掉。

全部约分后得到 $\dfrac{a}{2} = 9 \Rightarrow a = 18 \Rightarrow b = 17\text{。}$

所求和为 $18 + 17 = 35\text{。}$

所以正确答案是 D。

Note that the numerator of each fraction cancels with the denominator of the fraction to its right.

We can then cancel out all these terms to get a final equation of $\dfrac{a}{2} = 9 \Rightarrow a = 18 \Rightarrow b = 17.$

The desired sum is then $18 + 17 = 35.$

Thus, D is the correct answer.

20.

一对 $8$ 面骰子的面分别标有 $1$ 到 $8$ 。每一面朝上的概率相同。两个朝上数字的乘积大于 $36$ 的概率是

A pair of $8$ -sided dice has sides numbered $1$ through $8.$ Each side has the same probability (chance) of landing face up. The probability that the product of the two numbers that land face-up exceeds $36$ is

$\dfrac{5}{32}$

$\dfrac{11}{64}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

答案：A

知识点：骰子（概率）分类讨论

难度评级：1450

小提示：

对第一颗骰子掷出 $5$ 到 $8$ 的情况，分别数第二颗骰子哪些结果会使乘积大于 $36$

For first rolls $5$ through $8,$ count which second rolls make the product exceed $36$

大提示：

数出两个 $8$ 面骰子的有序结果

Count ordered pairs from two $8$ -sided dice

解答：

按第一颗骰子的点数分类。如果点数为 $1$ 、 $2$ 、 $3$ 或 $4$ ，乘积不可能大于 $36$ 。

如果点数是 $5$ ，另一颗必须掷出 $8$ ，否则乘积小于 $36$ 。

如果第一颗是 $6$ ，另一颗必须是 $7$ 或 $8$ ，有 $2$ 种可能。如果第一颗是 $7$ ，另一颗必须是 $6$ 、 $7$ 或 $8$ ，有 $3$ 种可能。

最后，如果第一颗掷出 $8$ ，另一颗可以是 $5$ 、 $6$ 、 $7$ 或 $8$ ，又有 $4$ 种可能。

有利结果共有 $1 + 2 + 3 + 4 = 10$ 种，总结果数是 $8^2 = 64$ ，所以概率为 $\dfrac{10}{64} = \dfrac{5}{32}\text{。}$

所以正确答案是 A。

We can case on the value of the first die. If its value is $1,$ $2,$ $3,$ or $4,$ then it is impossible for the product to be greater than $36.$

If it is a $5,$ then the other die has to roll an $8,$ otherwise the product is less than $36.$

If the first die is a $6,$ then the other die must be $7$ or $8,$ giving $2$ possibilities. If the first die is a $7,$ then the other die must be $6,$ $7,$ or $8,$ giving $3$ possibilities.

Finally, if the first roll is an $8,$ the other die can be $5,$ $6,$ $7,$ or $8,$ giving $4$ more possibilities.

The total number of working pairs is $1 + 2 + 3 + 4 = 10,$ and the total number of pairs is $8^2 = 64.$ The desired probability is then $\dfrac{10}{64} = \dfrac{5}{32}.$

Thus, A is the correct answer.

21.

从这个 $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ 的立方体中去掉每个角上的小立方体。剩余图形的表面积是

Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube. The surface area of the remaining figure is

$19\text{ 平方厘米}$

$19\text{ sq.cm}$

$24\text{ 平方厘米}$

$24\text{ sq.cm}$

$30\text{ 平方厘米}$

$30\text{ sq.cm}$

$54\text{ 平方厘米}$

$54\text{ sq.cm}$

$72\text{ 平方厘米}$

$72\text{ sq.cm}$

答案：D

知识点：表面积正方体不变量

难度评级：1390

小提示：

总表面积保持不变

The total surface area stays the same

大提示：

去掉一个角会失去三个面，同时露出三个新面

Removing a corner loses three faces and reveals three faces

解答：

任意两个角立方体之间都有一个单位立方体隔开，所以去掉一个角不会影响其他角。

去掉一个角时，失去三个单位正方形表面，但也露出三个新的单位正方形表面。

因此去掉角立方体不改变表面积。原立方体表面积为 $6 \cdot 3^2 = 6 \cdot 9 = 54$ 平方厘米。

所以正确答案是 D。

Note that there is one unit cube between any pair of corner cubes, so the removal of each does not affect the others.

When we remove a corner, we are losing three unit squares. We, however, gain these back from the three faces that get uncovered.

This means that removing a corner cube does not change the surface area. The surface area of the original cube is $6 \cdot 3^2 = 6 \cdot 9 = 54$ square centimeters.

Thus, D is the correct answer.

22.

一个两英寸的银立方体 $(2\times 2\times 2)$ 重 $3$ 磅，价值 $\$ 200$ 。一个三英寸的银立方体价值多少？

A two-inch cube $(2\times 2\times 2)$ of silver weighs $3$ pounds and is worth $\$ 200.$ How much is a three-inch cube of silver worth?

$\$300$

$\$375$

$\$450$

$\$560$

$\$675$

答案：E

知识点：体积长度、面积与体积的缩放关系

难度评级：1380

小提示：

比较两个立方体中单位立方体的个数

Compare the numbers of unit cubes in the two cubes

大提示：

对同一种材料，价值与体积成正比

Value scales with volume for the same material

解答：

两英寸立方体由 $2^3 = 8$ 个单位立方体组成。每个单位立方体价值 $200 \div 8 = 25$ 美元。三英寸立方体需要 $3^3 = 27$ 个单位立方体，所以价值 $25 \cdot 27 = 675$ 美元。

所以正确答案是 E。

The two-inch cube consists of $2^3 = 8$ unit cubes. Each of these unit cubes is worth $200 \div 8 = 25$ dollars. To form a three-inch cube, you need $3^3 = 27$ unit cubes. This means that it is worth $25 \cdot 27 = 675$ dollars.

Thus, E is the correct answer.

23.

有一些正整数满足：

I. 它们各位数字的平方和为 $50$ ，并且

II. 每个数字都大于它左边的数字。

满足这两个条件的最大整数，其各位数字的乘积是

There are positive integers that have these properties:

I. the sum of the squares of their digits is $50,$ and

II. each digit is larger than the one to its left.

The product of the digits of the largest integer with both properties is

$7$

$25$

$36$

$48$

$60$

答案：C

知识点：数字分类讨论

难度评级：1670

小提示：

从可能的最大末位数字开始尝试，并逐步降低

Try the largest possible final digit and work downward

大提示：

如果数字太多，最小平方和也会太大

More digits make the minimum square-sum too large

解答：

如果这个数有五位，最小的递增正数字会使平方和为 $1^2+2^2+3^2+4^2+5^2=55$ ，已经太大。因此最大的有效数最多有四位。

对于四位数 $abcd$ ，其中 $0<a<b<c<d$ ，末位不能是 $7$ 或更大，因为即使是 $1^2+2^2+3^2+7^2=63>50$ 。

尝试 $d=6$ ，其余数字的平方和必须是 $50-36=14$ ，而 $1^2+2^2+3^2=14$ 。这给出有效数 $1236$ 。

任何以 $5$ 或更小数字结尾的数都小于 $1236$ ，所以最大的有效整数是 $1236$ 。

它的各位数字乘积为 $1\cdot2\cdot3\cdot6=36$ 。

所以正确答案是 C。

If the number had five digits, the smallest possible increasing positive digits would give square-sum $1^2+2^2+3^2+4^2+5^2=55,$ already too large. So the largest valid number has at most four digits.

For a four-digit number $abcd$ with $0<a<b<c<d$ , the last digit cannot be $7$ or larger, since even $1^2+2^2+3^2+7^2=63>50$ .

Trying $d=6,$ the remaining squares must sum to $50-36=14,$ and $1^2+2^2+3^2=14.$ This gives the valid number $1236.$

Any number ending with $5$ or less is smaller than $1236,$ so the largest valid integer is $1236.$

The product of its digits is $1\cdot2\cdot3\cdot6=36.$

Thus, C is the correct answer.

24.

点 $C$ 将直径 $ACE$ 按 $2:3$ 分成两段。两个半圆 $ABC$ 和 $CDE$ 把圆形区域分成上方阴影区域和下方区域。上方区域面积与下方区域面积的比是

Diameter $ACE$ is divided at $C$ in the ratio $2:3.$ The two semicircles, $ABC$ and $CDE,$ divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is

$2:3$

$1:1$

$3:2$

$9:4$

$5:2$

答案：C

知识点：圆面积面积分割

难度评级：1610

小提示：

把上方区域看成大半圆减去一个小半圆再加上另一个小半圆

Compare the upper region as a large semicircle minus one small semicircle plus the other

大提示：

选取符合 $2:3$ 比例的方便长度

Choose convenient lengths in the ratio $2:3$

解答：

取 $AE=10$ ，使 $AC=4$ 、 $CE=6$ 。大半圆半径为 $5$ ，面积为 $\frac{25\pi}{2}$ 。

以 $AC$ 为直径的半圆半径为 $2$ ，面积为 $2\pi$ ；以 $CE$ 为直径的半圆半径为 $3$ ，面积为 $\frac{9\pi}{2}$ 。

上方阴影区域面积为 $\frac{25\pi}{2}-2\pi+\frac{9\pi}{2}=15\pi\text{。}$

整个圆面积为 $25\pi$ ，所以下方区域面积为 $25\pi-15\pi=10\pi$ 。

所求比为 $15\pi:10\pi=3:2$ 。

所以正确答案是 C。

Choose $AE=10$ so that $AC=4$ and $CE=6.$ The large semicircle has radius $5,$ so its area is $\frac{25\pi}{2}.$

The semicircle on $AC$ has radius $2$ and area $2\pi,$ while the semicircle on $CE$ has radius $3$ and area $\frac{9\pi}{2}.$

The upper shaded region has area $\frac{25\pi}{2}-2\pi+\frac{9\pi}{2}=15\pi.$

The whole circle has area $25\pi,$ so the lower region has area $25\pi-15\pi=10\pi.$

The desired ratio is $15\pi:10\pi=3:2.$

Thus, C is the correct answer.

25.

把从 $2$ 到 $98$ （含端点）的所有偶数中，不以 $0$ 结尾的数全部相乘。这个乘积最右边的数字（个位数字）是什么？

All of the even numbers from $2$ to $98$ inclusive, excluding those ending in $0,$ are multiplied together. What is the rightmost digit (the units digit) of the product?

$0$

$2$

$4$

$6$

$8$

答案：D

知识点：个位数字配对与分组

难度评级：1580

小提示：

按个位数字的循环模式把这些因数分组

Group the factors by their units digit pattern

大提示：

只需要考虑个位数字

Only the units digits matter

解答：

我们只关心个位数字，所以十位数字不重要。

因数的个位可以分成 $10$ 组，每组个位乘积为 $2 \cdot 4 \cdot 6 \cdot 8 = 384\text{。}$

每组的个位数字是 $4$ 。因此需要求 $4^{10}$ 的个位数字。

$4^2 = 16$ 的个位是 $6$ ，所以只需求 $6^5$ 的个位。

$6$ 的任何正整数次幂都以 $6$ 结尾，例如 $6$ 、 $36$ 、 $216$ 、 $\ldots$ 。

所以正确答案是 D。

We only care about the units digit, which means that the tens digits don’t matter.

Then we have $10$ groups of $2 \cdot 4 \cdot 6 \cdot 8 = 384.$

The units digit of each group is $4.$ We now need to find the units digit of $4^{10}.$

The units digit of $4^2 = 16$ is $6.$ This means we only need to find the units digit of $6^5.$

Note that every power of $6$ always ends in a $6$ (e.g. $6,$ $36,$ $216,$ $\ldots$ ).

Thus, D is the correct answer.