1985 AMC 8 真题与在线限时测验

计时

40:00

1.

下列乘积的值是多少？

$\dfrac{3 \times 5}{9 \times 11} \times \dfrac{7 \times 9 \times 11}{3 \times 5 \times 7}$

What is the value of the following product?

$\dfrac{3 \times 5}{9 \times 11} \times \dfrac{7 \times 9 \times 11}{3 \times 5 \times 7}$

$1$

$0$

$49$

$\dfrac{1}{49}$

$50$

答案：A

知识点：分数

难度评级：560

小提示：

先把分子相乘、分母相乘，再化简

Multiply the numerators together and the denominators together before simplifying

大提示：

相同的因数 $3$ 、 $5$ 、 $7$ 、 $9$ 、 $11$ 同时出现在分子和分母中

The same factors $3,$ $5,$ $7,$ $9,$ $11$ appear in both the numerator and the denominator

解答：

合并两个分数后，分子是 $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ ，分母是 $9 \cdot 11 \cdot 3 \cdot 5 \cdot 7$ 。它们是同一个乘积。

因此这个值是 $1$ 。

所以正确答案是 A。

Combining the two fractions, the numerator is $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ and the denominator is $9 \cdot 11 \cdot 3 \cdot 5 \cdot 7.$ These are the same product.

So the value is $1.$

Thus, the correct answer is A .

2.

$90 + 91 + 92 + \cdots + 98 + 99$ 的值是多少？

What is the value of $90 + 91 + 92 + \cdots + 98 + 99?$

$845$

$945$

$1005$

$1025$

$1045$

答案：B

知识点：等差数列配对与分组

难度评级：450

小提示：

从 $90$ 到 $99$ 一共有 $10$ 项

There are $10$ terms, from $90$ up to $99$

大提示：

和等于首尾两项的平均数乘以项数

The sum equals the average of the first and last term times the number of terms

解答：

一共有 $10$ 项。首项和末项的平均数是 $\dfrac{90 + 99}{2} = 94.5$ 。

所以总和是 $10 \cdot 94.5 = 945$ 。

所以正确答案是 B。

There are $10$ terms. The average of the first and last is $\dfrac{90 + 99}{2} = 94.5.$

So the sum is $10 \cdot 94.5 = 945.$

Thus, the correct answer is B .

3.

下列表达式的值是多少？

$\dfrac{10^7}{5 \times 10^4}$

What is the value of the following expression?

$\dfrac{10^7}{5 \times 10^4}$

$0.002$

$0.2$

$20$

$200$

$2000$

答案：D

知识点：指数

难度评级：560

小提示：

$\dfrac{10^7}{10^4} = 10^3$

大提示：

然后把 $10^3 = 1000$ 除以 $5$

Then divide $10^3 = 1000$ by $5$

解答：

因为 $\dfrac{10^7}{10^4} = 10^3 = 1000$ ，所以原式是 $\dfrac{1000}{5} = 200$ 。

所以正确答案是 D。

Since $\dfrac{10^7}{10^4} = 10^3 = 1000,$ the expression is $\dfrac{1000}{5} = 200.$

Thus, the correct answer is D .

4.

多边形 $ABCDEF$ 的面积是多少平方单位？

The area of polygon $ABCDEF,$ in square units, is

$24$

$30$

$46$

$66$

$74$

答案：C

知识点：面积面积分割

难度评级：820

小提示：

延长边 $AF$ 和 $DC$ ，补成一个完整的 $6 \times 9$ 矩形

Extend sides $AF$ and $DC$ to complete a full $6 \times 9$ rectangle

大提示：

减去左下角被挖去的小矩形面积

Subtract the area of the small rectangular notch that was cut from the lower-left corner

解答：

把图形补成完整的 $6 \times 9$ 矩形，面积是 $54$ 。左下角被去掉的是一个 $2$ 乘 $4$ 的矩形，面积是 $8$ 。

所以多边形面积是 $54 - 8 = 46$ 。

所以正确答案是 C。

Completing the figure to the full $6 \times 9$ rectangle gives an area of $54.$ The piece removed from the lower-left corner is a rectangle measuring $2$ by $4,$ with area $8.$

So the polygon’s area is $54 - 8 = 46.$

Thus, the correct answer is C .

5.

某数学班最近一个评分阶段的成绩如下： $5$ 名学生得 A， $4$ 名得 B， $3$ 名得 C， $3$ 名得 D， $5$ 名得 F。如果 A、B、C、D 都是及格成绩，那么及格成绩占全部成绩的几分之几？

The grades in a mathematics class for the last grading period were: $5$ students earned an A, $4$ earned a B, $3$ earned a C, $3$ earned a D, and $5$ earned an F. If A, B, C, and D are satisfactory grades, what fraction of the grades are satisfactory?

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$\dfrac{9}{10}$

答案：C

知识点：数据与图表解读分数

难度评级：660

小提示：

把及格成绩 A、B、C、D 的人数加起来

Add the counts for the satisfactory grades A, B, C, and D

大提示：

用及格人数除以总人数

Divide the satisfactory count by the total number of grades

解答：

及格成绩共有 $5 + 4 + 3 + 3 = 15$ 个，总成绩数是 $15 + 5 = 20$ 。

所以及格成绩所占的分数是 $\dfrac{15}{20} = \dfrac{3}{4}$ 。

所以正确答案是 C。

The satisfactory grades number $5 + 4 + 3 + 3 = 15,$ and the total number of grades is $15 + 5 = 20.$

So the fraction that is satisfactory is $\dfrac{15}{20} = \dfrac{3}{4}.$

Thus, the correct answer is C .

6.

一令纸有 $500$ 张，厚 $5$ 厘米。大约多少张这种纸叠起来会有 $7.5$ 厘米高？

A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?

$250$

$550$

$667$

$750$

$1250$

答案：D

知识点：比与比例

难度评级：560

小提示：

纸张数量与纸叠高度成正比

The number of sheets is proportional to the height of the stack

大提示：

$7.5$ 厘米的纸叠是 $5$ 厘米纸叠的 $1.5$ 倍高

A $7.5$ cm stack is $1.5$ times as tall as a $5$ cm stack

解答：

因为 $7.5$ 厘米是 $5$ 厘米的 $1.5$ 倍，所以纸张数也是 $1.5$ 倍。

也就是 $1.5 \times 500 = 750$ 张。

所以正确答案是 D。

Since $7.5$ cm is $1.5$ times $5$ cm, the stack holds $1.5$ times as many sheets.

That is $1.5 \times 500 = 750$ sheets.

Thus, the correct answer is D .

7.

一个“阶梯形”图案由每行中交替出现的涂色和未涂色方格组成。图中显示了第 $1$ 行到第 $4$ 行。所有行都以未涂色方格开始并以未涂色方格结束。第 $37$ 行中的涂色方格数是

A “stair-step” figure is made up of alternating shaded and unshaded squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with an unshaded square. The number of shaded squares in the $37$ th row is

$34$

$35$

$36$

$37$

$38$

答案：C

知识点：找规律基本计数

难度评级：860

小提示：

第 $n$ 行有 $2n - 1$ 个方格，并且从未涂色方格开始、到未涂色方格结束

Row $n$ has $2n - 1$ squares, and they alternate starting and ending with an unshaded square

大提示：

因为两端都是未涂色方格且两种方格交替，所以每行未涂色方格比涂色方格多一个

Since the ends are unshaded and the squares alternate, there is one more unshaded square than shaded square in each row

解答：

第 $n$ 行有 $2n - 1$ 个方格。由于两端都是未涂色方格且两种方格交替，每行未涂色方格比涂色方格多一个，所以涂色方格数是 $n - 1$ 。

对第 $37$ 行，这就是 $37 - 1 = 36$ 。

所以正确答案是 C。

Row $n$ contains $2n - 1$ squares. Because both ends are unshaded and the squares alternate, each row has one more unshaded square than shaded square, so the number of shaded squares is $n - 1.$

For the $37$ th row, this is $37 - 1 = 36.$

Thus, the correct answer is C .

8.

如果 $a = -2$ ，集合

$\left\{ -3a,\ 4a,\ \dfrac{24}{a},\ a^2,\ 1 \right\}$

中最大的数是

If $a = -2,$ the largest number in the set

$\left\{ -3a,\ 4a,\ \dfrac{24}{a},\ a^2,\ 1 \right\}$

is

$-3a$

$4a$

$\dfrac{24}{a}$

$a^2$

$1$

答案：A

知识点：换元法

难度评级：730

小提示：

把 $a = -2$ 代入每个表达式

Substitute $a = -2$ into each expression

大提示：

先检查符号： $4a$ 和 $\frac{24}{a}$ 为负，而 $-3a$ 和 $a^2$ 为正

Check signs first: $4a$ and $\frac{24}{a}$ are negative, while $-3a$ and $a^2$ are positive

解答：

代入 $a = -2$ 得到 $-3a = 6$ ， $4a = -8$ ， $\dfrac{24}{a} = -12$ ， $a^2 = 4$ ，以及 $1$ 。

其中最大的是 $6$ ，也就是 $-3a$ 。

所以正确答案是 A。

Substituting $a = -2$ gives the values $-3a = 6,$ $4a = -8,$ $\dfrac{24}{a} = -12,$ $a^2 = 4,$ and $1.$

The largest of these is $6,$ which is $-3a.$

Thus, the correct answer is A .

9.

下面 $9$ 个因数的乘积是多少？

$\begin{aligned} &\left(1 - \tfrac12\right)\left(1 - \tfrac13\right)\left(1 - \tfrac14\right) \cdots \\ &\quad {}\cdot \left(1 - \tfrac{1}{10}\right) \end{aligned}$

What is the value of the product of the $9$ factors below?

$\begin{aligned} &\left(1 - \tfrac12\right)\left(1 - \tfrac13\right)\left(1 - \tfrac14\right) \cdots \\ &\quad {}\cdot \left(1 - \tfrac{1}{10}\right) \end{aligned}$

$\dfrac{1}{10}$

$\dfrac{1}{9}$

$\dfrac{1}{2}$

$\dfrac{10}{11}$

$\dfrac{11}{2}$

答案：A

知识点：裂项相消分数

难度评级：860

小提示：

改写每个因数： $1 - \dfrac1k = \dfrac{k-1}{k}$

Rewrite each factor: $1 - \dfrac1k = \dfrac{k-1}{k}$

大提示：

把乘积写成 $\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}$ ，观察相邻项相消

Write the product as $\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}$ and watch consecutive terms cancel

解答：

每个因数 $1 - \dfrac1k$ 都等于 $\dfrac{k-1}{k}$ ，所以乘积是

$\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}\text{。}$

每个分子都会和前一个分母相消，最后留下 $\dfrac{1}{10}$ 。

所以正确答案是 A。

Each factor $1 - \dfrac1k$ equals $\dfrac{k-1}{k},$ so the product is

$\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}.$

Every numerator cancels the previous denominator, leaving $\dfrac{1}{10}.$

Thus, the correct answer is A .

10.

在数轴上， $\dfrac15$ 和 $\dfrac13$ 正中间的分数是多少？

What fraction lies halfway between $\dfrac15$ and $\dfrac13$ on the number line?

$\dfrac{1}{4}$

$\dfrac{2}{15}$

$\dfrac{4}{15}$

$\dfrac{53}{200}$

$\dfrac{8}{15}$

答案：C

知识点：平均数分数

难度评级：730

小提示：

两个数正中间的点是它们的平均数

The point halfway between two numbers is their average

大提示：

$\dfrac15 + \dfrac13 = \dfrac{8}{15}$ ；然后除以 $2$

$\dfrac15 + \dfrac13 = \dfrac{8}{15};$ then divide by $2$

解答：

$\dfrac15$ 和 $\dfrac13$ 的中点是它们的平均数。

$\dfrac{\frac15 + \frac13}{2} = \dfrac{\frac{8}{15}}{2} = \dfrac{4}{15}\text{。}$

所以正确答案是 C。

The midpoint of $\dfrac15$ and $\dfrac13$ is their average.

$\dfrac{\frac15 + \frac13}{2} = \dfrac{\frac{8}{15}}{2} = \dfrac{4}{15}.$

Thus, the correct answer is C .

11.

一张纸由六个连在一起的正方形组成，标记如图所示。沿正方形的边折叠后形成一个立方体。与标有 $X$ 的面相对的面，其标记是

A piece of paper containing six joined squares labeled as shown is folded along the edges of the squares to form a cube. The label of the face opposite the face labeled $X$ is

$Z$

$U$

$V$

$W$

$Y$

答案：E

知识点：展开图（立体几何）正方体

难度评级：920

小提示：

选一个面作为正面，再沿正方形的边折叠其余部分

Pick one face to be the front and fold the rest of the net around it

大提示：

如果 $X$ 成为底面，追踪哪个正方形会成为顶面

If $X$ becomes the bottom face, track which square ends up on top

解答：

把这个展开图折成让 $X$ 成为底面。那么 $U$ 、 $V$ 、 $W$ 、 $Z$ 会围成四个侧面。

唯一剩下的正方形 $Y$ 成为顶面，与底面 $X$ 相对。

所以正确答案是 E。

Fold the net so that $X$ is the bottom face. Then $U,$ $V,$ $W,$ and $Z$ wrap around to become the four side faces.

The only remaining square, $Y,$ becomes the top face, which is opposite the bottom face $X.$

Thus, the correct answer is E .

12.

一个正方形和一个三角形的周长相等。三角形三边长分别为 $6.2$ 厘米、 $8.3$ 厘米和 $9.5$ 厘米。这个正方形的面积是多少平方厘米？

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are $6.2$ cm, $8.3$ cm, and $9.5$ cm. The area of the square, in square centimeters, is

$24 \text{ cm}^2$

$36 \text{ cm}^2$

$48 \text{ cm}^2$

$64 \text{ cm}^2$

$144 \text{ cm}^2$

答案：B

知识点：周长正方形（几何）

难度评级：730

小提示：

把三角形三条边相加，得到共同的周长

Add the triangle’s three sides to get the common perimeter

大提示：

正方形边长是它周长的四分之一

The square’s side is one fourth of its perimeter

解答：

三角形周长是 $6.2 + 8.3 + 9.5 = 24$ 厘米，所以正方形周长也是 $24$ 厘米。

正方形边长是 $\dfrac{24}{4} = 6$ 厘米，因此面积是 $6^2 = 36 \text{ cm}^2$ 。

所以正确答案是 B。

The triangle’s perimeter is $6.2 + 8.3 + 9.5 = 24$ cm, so the square also has perimeter $24$ cm.

The square’s side is $\dfrac{24}{4} = 6$ cm, so its area is $6^2 = 36 \text{ cm}^2.$

Thus, the correct answer is B .

13.

如果你先以每小时 $4$ 英里的速度步行 $45$ 分钟，再以每小时 $10$ 英里的速度跑 $30$ 分钟，那么一小时 $15$ 分钟后，你一共行进了多少英里？

If you walk for $45$ minutes at a rate of $4$ mph and then run for $30$ minutes at a rate of $10$ mph, how many miles have you gone at the end of one hour and $15$ minutes?

$3.5$ 英里

$3.5$ miles

$8$ 英里

$8$ miles

$9$ 英里

$9$ miles

$25\dfrac13$ 英里

$25\dfrac13$ miles

$480$ 英里

$480$ miles

答案：B

知识点：路程、速度与时间单位换算

难度评级：820

小提示：

把时间化成小时： $45$ 分钟 $= \dfrac34$ 小时， $30$ 分钟 $= \dfrac12$ 小时

Convert each time to hours: $45$ minutes $= \dfrac34$ hour and $30$ minutes $= \dfrac12$ hour

大提示：

每一段距离 $=$ 速度 $\times$ 时间，然后把两段距离相加

Distance $=$ rate $\times$ time for each part, then add the two distances

解答：

步行距离： $4 \times \dfrac34 = 3$ 英里。跑步距离： $10 \times \dfrac12 = 5$ 英里。

总距离是 $3 + 5 = 8$ 英里。

所以正确答案是 B。

Walking: $4 \times \dfrac34 = 3$ miles. Running: $10 \times \dfrac12 = 5$ miles.

The total distance is $3 + 5 = 8$ miles.

Thus, the correct answer is B .

14.

一件商品税前价格为 $\$20$ 。对它征收 $6.5\%$ 销售税与征收 $6\%$ 销售税之间的差额是

The difference between a $6.5\%$ sales tax and a $6\%$ sales tax on an item priced at $\$20$ before tax is

$\$0.01$

$\$0.10$

$\$0.50$

$\$1$

$\$10$

答案：B

知识点：百分数

难度评级：730

小提示：

税率差是 $6.5\% - 6\% = 0.5\%$

The difference in tax rates is $6.5\% - 6\% = 0.5\%$

大提示：

求 $\$20$ 的 $0.5\%$

Find $0.5\%$ of $\$20$

解答：

两种税相差价格的 $0.5\%$ 。

也就是 $0.005 \times \$20 = \$0.10$ 。

所以正确答案是 B。

The two taxes differ by $0.5\%$ of the price.

That is $0.005 \times \$20 = \$0.10.$

Thus, the correct answer is B .

15.

$100$ 到 $400$ 之间有多少个整数含有数字 $2$ ？

How many whole numbers between $100$ and $400$ contain the digit $2?$

$100$

$120$

$138$

$140$

$148$

答案：C

知识点：数字容斥原理分类讨论

难度评级：1050

小提示：

从 $200$ 到 $299$ 的每个数都已经含有 $2$

Every number from $200$ to $299$ already contains a $2$

大提示：

对 $100$ 至 $199$ 和 $300$ 至 $399$ ，数十位或个位含 $2$ 的数，并减去重复计算的数

For $100$ – $199$ and $300$ – $399,$ count those with a $2$ in the tens or units place, subtracting the ones counted twice

解答：

从 $200$ 到 $299$ 的 $100$ 个数全部含有 $2$ 。

在 $100$ 到 $199$ 中，十位是 $2$ 的有 $10$ 个，个位是 $2$ 的有 $10$ 个，但 $122$ 被重复计算了一次，所以有 $10 + 10 - 1 = 19$ 个。 $300$ 到 $399$ 同样有 $19$ 个。

总数是 $100 + 19 + 19 = 138$ 。

所以正确答案是 C。

All $100$ numbers from $200$ to $299$ contain a $2.$

Among $100$ to $199,$ there are $10$ with a $2$ in the tens place and $10$ with a $2$ in the units place, but $122$ is counted twice, giving $10 + 10 - 1 = 19.$ The range $300$ to $399$ similarly contributes $19.$

The total is $100 + 19 + 19 = 138.$

Thus, the correct answer is C .

16.

布朗老师的数学班中，男生与女生人数之比是 $2 : 3$ 。如果班上一共有 $30$ 名学生，那么女生比男生多多少人？

The ratio of boys to girls in Mr. Brown’s math class is $2 : 3.$ If there are $30$ students in the class, how many more girls than boys are in the class?

$1$

$3$

$5$

$6$

$10$

答案：D

知识点：比与比例

难度评级：730

小提示：

全班分成 $2 + 3 = 5$ 个相等的份

The class splits into $2 + 3 = 5$ equal parts

大提示：

女生比男生正好多一份，所以求出每份有多少人

Girls exceed boys by exactly one part, so find the size of one part

解答：

$30$ 名学生分成 $5$ 个相等的份，每份 $6$ 人。男生占 $2$ 份，即 $12$ 人；女生占 $3$ 份，即 $18$ 人。

所以女生比男生多 $18 - 12 = 6$ 人。

所以正确答案是 D。

The $30$ students split into $5$ equal parts of $6.$ Boys make up $2$ parts ( $12$ ) and girls $3$ parts ( $18$ ).

So there are $18 - 12 = 6$ more girls than boys.

Thus, the correct answer is D .

17.

如果你前六次数学测验的平均分是 $84$ ，前七次数学测验的平均分是 $85$ ，那么你第七次测验的分数是

If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85,$ then your score on the seventh test was

$86$

$88$

$90$

$91$

$92$

答案：D

知识点：平均数

难度评级：820

小提示：

七次测验的总分是 $7 \times 85$ ，六次测验的总分是 $6 \times 84$

The total for seven tests is $7 \times 85,$ and the total for six tests is $6 \times 84$

大提示：

第七次测验的分数是这两个总分的差

The seventh score is the difference between those two totals

解答：

七次测验的平均分是 $85$ ，总分为 $7 \times 85 = 595$ ；六次测验的平均分是 $84$ ，总分为 $6 \times 84 = 504$ 。

第七次测验的分数是 $595 - 504 = 91$ 。

所以正确答案是 D。

Seven tests averaging $85$ total $7 \times 85 = 595$ points; six tests averaging $84$ total $6 \times 84 = 504$ points.

The seventh score is $595 - 504 = 91.$

Thus, the correct answer is D .

18.

某种小册子九本的总价不到 $\$10.00$ ，而十本的总价超过 $\$11.00$ 。一本小册子的价格是多少？

Nine copies of a certain pamphlet cost less than $\$10.00$ while ten copies of the same pamphlet (at the same price) cost more than $\$11.00.$ How much does one copy of this pamphlet cost?

$\$1.07$

$\$1.08$

$\$1.09$

$\$1.10$

$\$1.11$

答案：E

知识点：不等式极限情形界定

难度评级：950

小提示：

设价格为 $p$ ；则 $9p \lt 10$ 且 $10p \gt 11$

Let the price be $p;$ then $9p \lt 10$ and $10p \gt 11$

大提示：

合并得 $1.10 \lt p \lt 1.111\ldots$ ，再找以美分计价的价格

Combine into $1.10 \lt p \lt 1.111\ldots$ and find the price in whole cents

解答：

由 $9p \lt 10$ 得 $p \lt 1.111\ldots$ ，由 $10p \gt 11$ 得 $p \gt 1.10$ 。

在 $\$1.10$ 和 $\$1.111\ldots$ 之间唯一的整美分价格是 $\$1.11$ 。

所以正确答案是 E。

From $9p \lt 10$ we get $p \lt 1.111\ldots,$ and from $10p \gt 11$ we get $p \gt 1.10.$

The only price in whole cents between $\$1.10$ and $\$1.111\ldots$ is $\$1.11.$

Thus, the correct answer is E .

19.

如果一个矩形的长和宽都增加 $10\%$ ，那么这个矩形的周长增加了

If the length and width of a rectangle are each increased by $10\%,$ then the perimeter of the rectangle is increased by

$1\%$

$10\%$

$20\%$

$21\%$

$40\%$

答案：B

知识点：百分数周长

难度评级：800

小提示：

周长是 $2(\ell + w)$

The perimeter is $2(\ell + w)$

大提示：

把 $\ell$ 和 $w$ 都乘以 $1.1$ ，会把整个周长乘以 $1.1$

Multiplying both $\ell$ and $w$ by $1.1$ multiplies the whole perimeter by $1.1$

解答：

新周长是 $2(1.1\ell + 1.1w) = 1.1 \cdot 2(\ell + w)$ ，也就是旧周长的 $1.1$ 倍。

这表示增加了 $10\%$ 。

所以正确答案是 B。

The new perimeter is $2(1.1\ell + 1.1w) = 1.1 \cdot 2(\ell + w),$ which is $1.1$ times the old perimeter.

That is a $10\%$ increase.

Thus, the correct answer is B .

20.

某一年，一月正好有四个星期二和四个星期六。那一年的一月 $1$ 日是星期几？

In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January $1$ fall that year?

星期一

Monday

星期二

Tuesday

星期三

Wednesday

星期五

Friday

星期六

Saturday

答案：C

知识点：日期与时间模运算

难度评级：1090

小提示：

一月有 $31$ 天，也就是 $4$ 个完整星期加 $3$ 天

January has $31$ days, which is $4$ full weeks plus $3$ extra days

大提示：

出现五次的三个星期几，就是一月 $1$ 日、 $2$ 日和 $3$ 日所对应的星期几；星期二和星期六不能在其中

The three weekdays that occur five times are the days of January $1,$ $2,$ and $3;$ Tuesday and Saturday must not be among them

解答：

因为 $31 = 4 \cdot 7 + 3$ ，一月 $1$ 日、 $2$ 日和 $3$ 日所对应的三个星期几各出现五次，其余星期几各出现四次。

要让星期二和星期六都只出现四次，它们都不能是一月 $1$ 日、 $2$ 日或 $3$ 日。唯一可行的开始日是星期三：这样星期三、星期四、星期五出现五次，而星期二和星期六各出现四次。

所以正确答案是 C。

Since $31 = 4 \cdot 7 + 3,$ the weekdays falling on January $1,$ $2,$ and $3$ each occur five times that month, and every other weekday occurs four times.

For Tuesday and Saturday to occur only four times, neither may be January $1,$ $2,$ or $3.$ The only starting day that works is Wednesday: then Wednesday, Thursday, and Friday occur five times, while Tuesday and Saturday each occur four times.

Thus, the correct answer is C .

21.

格林先生每年加薪 $10\%$ 。连续四次这样的加薪后，他的薪水总共上涨了百分之多少？

Mr. Green receives a $10\%$ raise every year. His salary after four such raises has gone up by what percent?

小于 $40\%$

less than $40\%$

$40\%$

$44\%$

$45\%$

大于 $45\%$

more than $45\%$

答案：E

知识点：百分数指数

难度评级：950

小提示：

一次 $10\%$ 的加薪会把薪水乘以 $1.1$

Applying a $10\%$ raise multiplies the salary by $1.1$

大提示：

四次加薪后薪水乘以 $1.1^4$ ；把它和 $1.45$ 比较

After four raises the salary is multiplied by $1.1^4;$ compare that to $1.45$

解答：

四次加薪后薪水被乘以 $1.1^4 = 1.4641$ 。

这表示增加了 $46.41\%$ ，大于 $45\%$ 。

所以正确答案是 E。

After four raises the salary is multiplied by $1.1^4 = 1.4641.$

That is an increase of $46.41\%,$ which is more than $45\%.$

Thus, the correct answer is E .

22.

假设每个 $7$ 位整数都可能是电话号码，但以 $0$ 或 $1$ 开头的除外。电话号码中，以 $9$ 开头并以 $0$ 结尾的占几分之几？

Assume every $7$ -digit whole number is a possible telephone number except those that begin with $0$ or $1.$ What fraction of telephone numbers begin with $9$ and end with $0?$

$\dfrac{1}{63}$

$\dfrac{1}{80}$

$\dfrac{1}{81}$

$\dfrac{1}{90}$

$\dfrac{1}{100}$

答案：B

知识点：基本概率乘法原理

难度评级：1000

小提示：

第一位可以是 $2$ 到 $9$ 中的任意一个，共 $8$ 种选择

The first digit can be any of $2$ through $9,$ which is $8$ choices

大提示：

把以 $9$ 开头的比例乘以以 $0$ 结尾的比例

Multiply the fraction of telephone numbers that begin with $9$ by the fraction that end in $0$

解答：

第一位是 $8$ 个允许数字之一（ $2$ 到 $9$ ），所以 $\dfrac18$ 的号码以 $9$ 开头。最后一位可以是 $10$ 个数字之一，所以 $\dfrac{1}{10}$ 的号码以 $0$ 结尾。

这两个条件相互独立，所以所求比例是 $\dfrac18 \cdot \dfrac{1}{10} = \dfrac{1}{80}$ 。

所以正确答案是 B。

The first digit is one of $8$ allowed digits ( $2$ through $9$ ), so $\dfrac18$ of the numbers begin with $9.$ The last digit is any of $10$ digits, so $\dfrac{1}{10}$ end in $0.$

These conditions are independent, so the fraction is $\dfrac18 \cdot \dfrac{1}{10} = \dfrac{1}{80}.$

Thus, the correct answer is B .

23.

金氏初中有 $1200$ 名学生。每名学生每天上 $5$ 节课。每位老师教 $4$ 节课。每节课有 $30$ 名学生和 $1$ 位老师。金氏初中有多少位老师？

King Middle School has $1200$ students. Each student takes $5$ classes a day. Each teacher teaches $4$ classes. Each class has $30$ students and $1$ teacher. How many teachers are there at King Middle School?

$30$

$32$

$40$

$45$

$50$

答案：E

知识点：双重计数

难度评级：950

小提示：

先数每天的学生上课总次数： $1200 \times 5$

Count the total number of student-class attendances: $1200 \times 5$

大提示：

除以 $30$ 得到每天开设的课数，再除以 $4$ 得到老师人数

Divide by $30$ to get the number of classes, then by $4$ to get the number of teachers

解答：

每天共有 $1200 \times 5 = 6000$ 次学生上课。每节课有 $30$ 名学生，所以共有 $\dfrac{6000}{30} = 200$ 节课。

每位老师教 $4$ 节课，所以有 $\dfrac{200}{4} = 50$ 位老师。

所以正确答案是 E。

Each day there are $1200 \times 5 = 6000$ student-class attendances. Since each class holds $30$ students, there are $\dfrac{6000}{30} = 200$ classes.

Each teacher teaches $4$ classes, so there are $\dfrac{200}{4} = 50$ teachers.

Thus, the correct answer is E .

24.

在一个魔术三角形中，六个整数 $10$ 至 $15$ 各被放入一个圆内，使三角形每条边上三个数的和 $S$ 都相同。 $S$ 的最大可能值是

In a magic triangle, each of the six whole numbers $10$ through $15$ is placed in one of the circles so that the sum $S$ of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is

$36$

$37$

$38$

$39$

$40$

答案：D

知识点：幻方最优化

难度评级：1140

小提示：

把三条边的和相加时，每个顶点数被计算两次，每个中点数被计算一次

Adding the three side sums counts each corner number twice and each midpoint number once

大提示：

设 $C$ 为三个顶点数之和，则 $3S = 75 + C$ ；要使它最大，应把三个最大的数放在顶点上

Let $C$ be the sum of the corner numbers. Then $3S = 75 + C;$ put the three largest numbers at the corners

解答：

设 $C$ 为三个顶点数之和。把三条边的和相加可得 $3S = 75 + C$ ，因为六个数之和为 $75$ ，且每个顶点数都被多计算了一次。要使 $S$ 最大，应把三个最大的数 $13$ 、 $14$ 、 $15$ 放在顶点，此时 $C = 42$ 。

于是 $3S = 75 + 42 = 117$ ，所以 $S = 39$ 。这个值可以达到：在顶点 $13$ 与 $14$ 之间放 $12$ ，在 $14$ 与 $15$ 之间放 $10$ ，在 $15$ 与 $13$ 之间放 $11$ 。

所以正确答案是 D。

Let $C$ be the sum of the corner numbers. Adding the three side sums gives $3S = 75 + C,$ because the six numbers sum to $75$ and every corner is counted one extra time. To maximize $S,$ place the three largest numbers $13,$ $14,$ $15$ at the corners, giving $C = 42.$

Then $3S = 75 + 42 = 117,$ so $S = 39.$ This is achievable: put $12$ between corners $13$ and $14,$ put $10$ between $14$ and $15,$ and put $11$ between $15$ and $13.$

Thus, the correct answer is D .

25.

桌上放着五张卡片。每张卡片一面有一个字母，另一面有一个整数。可见的面显示 $P$ 、 $Q$ 、 $3$ 、 $4$ 和 $6$ 。简说：“如果任意一张卡片的一面是元音字母，那么另一面就是偶数。”玛丽翻开一张卡片，证明简错了。玛丽翻开的是哪张卡片？

Five cards are lying on a table. Each card has a letter on one side and a whole number on the other side. The visible faces show $P,$ $Q,$ $3,$ $4,$ and $6.$ Jane said, “If a vowel is on one side of any card, then an even number is on the other side.” Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?

$3$

$4$

$6$

$P$

$Q$

答案：A

知识点：逻辑推理反例

难度评级：1140

小提示：

要否定这个规则，需要找到一张可能把元音字母和奇数配在一起的卡片

To disprove the rule, find a card that could hide a vowel paired with an odd number

大提示：

可见的没有元音字母（ $P$ 和 $Q$ 是辅音），所以要检查显示奇数的那张卡

No vowel is visible ( $P$ and $Q$ are consonants), so the card to check is the one showing an odd number

解答：

要证明“元音字母必对应偶数”这个规则是错的，玛丽需要找到一张一面是元音字母、另一面是奇数的卡片。没有卡片显示元音字母，因为 $P$ 和 $Q$ 是辅音。

唯一显示奇数的卡片是 $3$ 。翻开它后露出元音字母，就会反驳简的说法。

所以正确答案是 A。

To show the rule “a vowel forces an even number” is false, Mary needs a card with a vowel on one side and an odd number on the other. No card shows a vowel, since $P$ and $Q$ are consonants.

The only card showing an odd number is $3.$ Turning it over reveals a vowel, which contradicts Jane’s claim.

Thus, the correct answer is A .