2005 AMC 8 真题与在线限时测验

计时

40:00

1.

康妮把一个数乘以 $2$ ，得到答案 $60$ 。可是她本应把这个数除以 $2$ 才能得到正确答案。正确答案是多少？

Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?

$7.5$

$15$

$30$

$120$

$240$

答案：B

知识点：逆推法

难度评级：370

小提示：

先撤销康妮错误的乘法，求出原来的数

Undo Connie’s mistaken multiplication to find the original number.

大提示：

然后把原来的数除以 $2$

Then divide that original number by $2$ .

解答：

因为康妮乘以 $2$ ，所以原来的数是 $60 \div 2 = 30$ 。正确做法是除以 $2$ ，得到 $30 \div 2 = 15$ 。

所以正确答案是 B。

Since Connie multiplied by $2,$ her original number was $60 \div 2 = 30.$ To get the correct answer, she must divide by $2$ to get $30 \div 2 = 15.$

Thus, B is the correct answer.

2.

卡尔在佩阿洛特商店买了五个文件夹，每个 $\$2.50$ 。第二天佩阿洛特商店有 $20\%$ 折扣促销。如果卡尔等一天再买，他可以省多少钱？

Karl bought five folders from Pay-A-Lot at a cost of $\$2.50$ each. Pay-A-Lot had a $20\%$ -off sale the following day. How much could Karl have saved on the purchase by waiting a day?

$\$ 1.00$

$\$ 2.00$

$\$ 2.50$

$\$ 2.75$

$\$ 5.00$

答案：C

知识点：百分数钱币

难度评级：450

小提示：

先求五个文件夹的总价

Find the total price of the five folders.

大提示：

节省的钱是总价的 $20\%$

The savings would be $20\%$ of that total.

解答：

卡尔为文件夹支付了 $5\cdot\$2.50=\$12.50$ 。

$20\%$ 的 $\$12.50$ 折扣会省下 $0.20\cdot12.50=2.50$ 美元。

所以正确答案是 C。

Karl paid $5\cdot\$2.50=\$12.50$ for the folders.

A $20\%$ discount on $\$12.50$ would save $0.20\cdot12.50=2.50$ dollars.

Thus, C is the correct answer.

3.

最少还必须涂黑多少个小正方形，才能使对角线 $\overline{BD}$ 成为正方形 $ABCD$ 的一条对称轴？

What is the minimum number of small squares that must be shaded so that a line of symmetry lies on the diagonal $\overline{BD}$ of square $ABCD$ ?

$1$

$2$

$3$

$4$

$5$

答案：D

知识点：对称性变换

难度评级：660

小提示：

将已涂黑的小正方形关于对角线 $\overline{BD}$ 反射

Reflect the shaded squares across diagonal $\overline{BD}$ .

大提示：

只有没有对应反射涂黑方格的已涂黑方格需要配对

Only shaded squares without a matching reflected square need to be paired.

解答：

要使对角线 $\overline{BD}$ 成为对称轴，对角线外每个已涂黑小正方形都必须在对角线另一侧有一个对应的反射涂黑小正方形。

有 $4$ 个已涂黑小正方形的反射位置尚未涂黑，所以还必须涂黑 $4$ 个小正方形。

所以正确答案是 D。

For diagonal $\overline{BD}$ to be a line of symmetry, every shaded square off the diagonal must have a matching reflected shaded square on the other side of the diagonal.

There are $4$ shaded squares whose reflected partners are not shaded, so $4$ additional small squares must be shaded.

Thus, D is the correct answer.

4.

一个正方形和一个三角形周长相等。三角形的三边长分别是 $6.1$ cm、 $8.2$ cm 和 $9.7$ cm。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 $6.1$ cm, $8.2$ cm and $9.7$ cm. What is the area of the square in square centimeters?

$24$

$25$

$36$

$48$

$64$

答案：C

知识点：周长正方形（几何）

难度评级：660

小提示：

先求三角形的周长

First find the triangle’s perimeter.

大提示：

这个周长也是正方形的周长

That perimeter is also the square’s perimeter.

解答：

三角形的周长为 $6.1 + 8.2 + 9.7 = 24 \text{ cm。}$ 这意味着正方形的边长为 $24 \div 4 = 6 \text{ cm。}$ 因此正方形面积为 $6^2 = 36 \text{ cm}^2\text{。}$

所以正确答案是 C。

The perimeter of the triangle is $6.1 + 8.2 + 9.7 = 24 \text{ cm.}$ This means that the side length of the square is $24 \div 4 = 6 \text{ cm.}$ Therefore, the area of the square is $6^2 = 36 \text{ cm}^2.$

Thus, C is the correct answer.

5.

汽水以 $6$ 、 $12$ 和 $24$ 罐装出售。要正好买 $90$ 罐汽水，最少需要多少包？

Soda is sold in packs of $6,$ $12$ and $24$ cans. What is the minimum number of packs needed to buy exactly $90$ cans of soda?

$4$

$5$

$6$

$8$

$15$

答案：B

知识点：最优化

难度评级：730

小提示：

在总数不超过 $90$ 的情况下尽量多用 $24$ 罐装

Use as many $24$ -packs as possible without going over $90$ .

大提示：

三包 $24$ 罐装后，用最少包数补齐剩余罐数

After three $24$ -packs, finish the remaining cans with the fewest packs.

解答：

四包不可能够用。四个 $24$ 罐装共有 $96$ 罐，把其中任意一个 $24$ 罐装换成较小的包装，罐数会减少 $12$ 或 $18$ 。因此不可能恰好从 $96$ 减少 $6$ 罐而得到 $90$ 罐。

五包可以做到：三个 $24$ 罐装、一个 $12$ 罐装和一个 $6$ 罐装共有 $3\cdot24+12+6=90$ 罐。因此最少需要 $5$ 包。

所以正确答案是 B。

Four packs cannot be enough. Four $24$ -packs would contain $96$ cans, and replacing any $24$ -pack by a smaller pack decreases the total by either $12$ or $18$ cans. There is no way to decrease $96$ by exactly $6$ cans to reach $90$ .

Five packs do work: three $24$ -packs, one $12$ -pack, and one $6$ -pack contain $3\cdot24+12+6=90$ cans. Therefore, the minimum number is $5$ .

Thus, B is the correct answer.

6.

设 $d$ 是一个数字。有多少个 $d$ 的取值满足 $2.00d5 > 2.005\text{？}$

Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005?$

$0$

$4$

$5$

$6$

$10$

答案：C

知识点：小数位值

难度评级：870

小提示：

比较 $2.00d5$ 和 $2.005$ 的千分位

Compare the thousandths digit of $2.00d5$ with that of $2.005$ .

大提示：

数字 $d$ 必须至少为 $5$

The digit $d$ must be at least $5$ .

解答：

$2.00d5$ 和 $2.005$ 第一次不同出现在千分位：比较 $d$ 和 $5$ 。

因此 $2.00d5>2.005$ 当且仅当 $d\ge 5$ 。可能数字为 $5$ 、 $6$ 、 $7$ 、 $8$ 、 $9$ ，共有 $5$ 个值。

所以正确答案是 C。

The numbers $2.00d5$ and $2.005$ first differ in the thousandths place: $d$ is compared with $5$ .

Thus $2.00d5>2.005$ exactly when $d\ge 5$ . The possible digits are $5,$ $6,$ $7,$ $8,$ $9$ , for $5$ values.

Thus, C is the correct answer.

7.

比尔先向南走 $\frac{1}{2}$ 英里，再向东走 $\frac{3}{4}$ 英里，最后又向南走 $\frac{1}{2}$ 英里。沿直线计算，他离起点有多少英里？

Bill walks $\frac{1}{2}$ mile south, then $\frac{3}{4}$ mile east, and finally $\frac{1}{2}$ mile south. How many miles is he, in a direct line, from his starting point?

$1$

$1 \frac{1}{4}$

$1 \frac{1}{2}$

$1 \frac{3}{4}$

$2$

答案：B

知识点：勾股定理分数

难度评级：940

小提示：

将两段向南的路合并成一个竖直位移

Combine the two southward parts into one vertical displacement.

大提示：

用直角边 $1$ 和 $\frac{3}{4}$ 应用勾股定理

Use the Pythagorean theorem with legs $1$ and $\frac{3}{4}$ .

解答：

所求是对角线长度。使用勾股定理： $\sqrt{\left(\dfrac{1}{2} + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}^2}$ $= \sqrt{1 + \dfrac{9}{16}} = \sqrt{\dfrac{25}{16}} = \dfrac{5}{4}$

所以正确答案是 B。

Note that what we want is the length of the diagonal. We can use the Pythagorean theorem to find this. $\sqrt{\left(\dfrac{1}{2} + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}^2}$ $= \sqrt{1 + \dfrac{9}{16}} = \sqrt{\dfrac{25}{16}} = \dfrac{5}{4}$

Thus, B is the correct answer.

8.

设 $m$ 和 $n$ 是正奇数。下列哪一个一定也是奇数？

Suppose $m$ and $n$ are positive odd integers. Which of the following must also be an odd integer?

$m + 3n$

$3m - n$

$3m^2 + 3n^2$

$(nm + 3)^2$

$3mn$

答案：E

知识点：奇偶性

难度评级：1000

小提示：

用奇数的奇偶性规则检验各选项

Test the choices using parity rules for odd numbers.

大提示：

奇数的乘积是奇数

The product of odd integers is odd.

解答：

先回顾以下四条规则：

• 奇数加奇数、偶数加偶数，结果都是偶数

• 偶数加奇数，结果是奇数

• 偶数乘任何整数，结果都是偶数

• 奇数乘奇数，结果是奇数

任意奇数可写成 $2m+1$ ，任意偶数可写成 $2n$ ，其中 $m$ 、 $n$ 为整数，因此很容易验证这些规则。

下面逐一检查各选项：

A:

注意 $3$ 是奇数，因此对模 $2$ 而言，该式为 $1 + 1 \cdot 1 \equiv 0 \pmod 2\text{。}$ 根据上述规则，结果是偶数。

B: $1 \cdot 1 - 1 \equiv 0 \pmod 2\text{。}$ 结果仍是偶数。

C: $1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2\text{。}$ 结果也是偶数。

D: $(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2\text{。}$ 结果同样是偶数。

E: $1 \cdot 1 \cdot 1 \equiv 1 \pmod 2\text{。}$ 结果是奇数。

因此只有 E 一定是奇数。

所以正确答案是 E。

Recall the four following rules:

• odd plus odd and even plus even is even

• even plus odd is odd

• even times anything is even

• odd times odd is odd

These rules can be easily verified by representing arbitrary odd numbers as $2m+1$ and arbitrary even numbers as $2n$ respectively, for integers $m,$ $n.$

With this in mind, let’s examine each answer choice individually:

A:

Note that $3$ is odd. Modulo $2$ , this gives us $1 + 1 \cdot 1 \equiv 0 \pmod 2.$ From our above rules, we know that this is even.

B: $1 \cdot 1 - 1 \equiv 0 \pmod 2.$ Once again, this is even.

C: $1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2.$ This is also even.

D: $(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2.$ Unfortunately, this is also even.

E: $1 \cdot 1 \cdot 1 \equiv 1 \pmod 2.$ This is odd.

Therefore, E is the only answer choice that is an odd integer.

Thus, E is the correct answer.

9.

在四边形 $ABCD$ 中，边 $\overline{AB}$ 和 $\overline{BC}$ 长度都为 $10$ ，边 $\overline{CD}$ 和 $\overline{DA}$ 长度都为 $17$ ，且角 $ADC$ 的度数是 $60^\circ$ 。对角线 $\overline{AC}$ 的长度是多少？

In quadrilateral $ABCD,$ sides $\overline{AB}$ and $\overline{BC}$ both have length $10,$ sides $\overline{CD}$ and $\overline{DA}$ both have length $17,$ and the measure of angle $ADC$ is $60^\circ.$ What is the length of diagonal $\overline{AC}?$

$13.5$

$14$

$15.5$

$17$

$18.5$

答案：D

知识点：等腰三角形等边三角形角度和

难度评级：1020

小提示：

只看 $\triangle ADC$

Look only at $\triangle ADC$ .

大提示：

它是等腰三角形，并且有一个 $60^\circ$ 角

It is isosceles and has a $60^\circ$ angle.

解答：

$\triangle ADC$ 是等腰三角形，所以 $\angle DAC = \angle DCA$ 。

我们得到 $\begin{gather*} \angle DAC + \angle DCA \\ {}+ \angle ADC = 180^{\circ} \\ \angle DAC + \angle DCA = 120^{\circ} \\ \angle DAC = \angle DCA = 60^{\circ} \end{gather*}\text{。}$

所以 $\triangle ADC$ 是等边三角形，因此 $AC = CD = 17\text{。}$

所以正确答案是 D。

Note that $\triangle ADC$ is isosceles. This means that $\angle DAC = \angle DCA.$

We get that $\begin{gather*} \angle DAC + \angle DCA \\ {}+ \angle ADC = 180^{\circ} \\ \angle DAC + \angle DCA = 120^{\circ} \\ \angle DAC = \angle DCA = 60^{\circ}. \end{gather*}$

This shows that $\triangle ADC$ is equilateral. This gives us that $AC = CD = 17.$

Thus, D is the correct answer.

10.

乔从家到学校走到一半时发现自己迟到了，于是剩下的路跑到学校。他跑步速度是步行速度的 $3$ 倍。乔走到学校一半用了 $6$ 分钟。他从家到学校总共用了多少分钟？

Joe had walked half way from home to school when he realized he was late. He ran the rest of the way to school. He ran $3$ times as fast as he walked. Joe took $6$ minutes to walk half way to school. How many minutes did it take Joe to get from home to school?

$7$

$7.3$

$7.7$

$8$

$8.3$

答案：D

知识点：路程、速度与时间比与比例

难度评级：1030

小提示：

乔跑的是与之前步行相同的一半距离

Joe runs the same half-distance he previously walked.

大提示：

速度变为三倍，所用时间变为三分之一

Running three times as fast takes one third as much time.

解答：

乔走前半段用了 $6$ 分钟。

后半段距离相同，他以步行速度的 $3$ 倍跑，所以这一半用 $6\div3=2$ 分钟。

总时间为 $6+2=8$ 分钟。

所以正确答案是 D。

Joe took $6$ minutes to walk the first half of the distance.

He ran the second half, the same distance, at $3$ times his walking speed, so that half took $6\div3=2$ minutes.

His total time was $6+2=8$ minutes.

Thus, D is the correct answer.

11.

伯格维尔的销售税率为 $6\%$ 。伯格维尔大衣店促销期间，一件原价为 $\$90.00$ 的外套打 $20\%$ 折扣。两名店员杰克和吉尔各自计算账单。杰克先按 $\$90.00$ 加收 $6\%$ 的销售税，再从总额中减去 $20\%$ 。吉尔先从 $\$90.00$ 中减去原价的 $20\%$ ，再对折后价加收 $6\%$ 的销售税。杰克的总额减去吉尔的总额是多少？

The sales tax rate in Bergville is $6\%$ . During a sale at the Bergville Coat Closet, the price of a coat is discounted $20\%$ from its $\$90.00$ price. Two clerks, Jack and Jill, calculate the bill independently. Jack rings up $\$90.00$ and adds $6\%$ sales tax, then subtracts $20\%$ from this total. Jill rings up $\$90.00$ , subtracts $20\%$ of the price, then adds $6\%$ of the discounted price for sales tax. What is Jack’s total minus Jill’s total?

$-\$ 1.06$

$-\$ 0.53$

$\$ 0$

$\$ 0.53$

$\$ 1.06$

答案：C

知识点：百分数

难度评级：1060

小提示：

加税和打折都是乘法

Adding tax and taking the discount are both multiplications.

大提示：

乘法顺序不同结果相同

Multiplication gives the same result in either order.

解答：

杰克的总额是 $90.00\cdot1.06\cdot0.80$ 美元。

吉尔的总额是 $90.00\cdot0.80\cdot1.06$ 美元。

因为乘法满足交换律，这两个乘积相等，所以杰克的总额减去吉尔的总额为 $\$0$ 。

所以正确答案是 C。

Jack’s total is $90.00\cdot1.06\cdot0.80$ dollars.

Jill’s total is $90.00\cdot0.80\cdot1.06$ dollars.

These products are equal because multiplication is commutative, so Jack’s total minus Jill’s total is $\$0$ .

Thus, C is the correct answer.

12.

大艾尔这只猩猩从五月 $1$ 日到五月 $5$ 日一共吃了 $100$ 根香蕉。他每天都比前一天多吃六根。大艾尔在五月 $5$ 日吃了多少根香蕉？

Big Al, the ape, ate $100$ bananas from May $1$ through May $5.$ Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May $5?$

$20$

$22$

$30$

$32$

$34$

答案：D

知识点：等差数列平均数

难度评级：1100

小提示：

五天的数量形成等差数列

The five daily amounts form an arithmetic sequence.

大提示：

中间那天等于五天的平均数

The middle day equals the average of the five days.

解答：

设大艾尔在五月 $3$ 日吃了 $x$ 根香蕉。那么从五月 $1$ 日开始，五天的数量为 $x - 12, x - 6, x, x + 6, x + 12\text{。}$

它们的和为 $5x$ ，等于 $100$ 。所以 $x = 20$ 。于是五月 $5$ 日大艾尔吃了 $20 + 12 = 32$ 根香蕉。

所以正确答案是 D。

Let $x$ be the number of bananas Big Al ate on May $3.$ Then Big Al ate the following amounts starting from May $1:$ $x - 12, x - 6, x, x + 6, x + 12.$

The sum of this is $5x,$ which equals $100.$ This gives us that $x = 20.$ Then on May $5,$ Big Al ate $20 + 12 = 32$ bananas.

Thus, D is the correct answer.

13.

多边形 $ABCDEF$ 的面积为 $52$ ，且 $AB=8$ ， $BC=9$ ， $FA=5$ 。 $DE+EF$ 是多少？

The area of polygon $ABCDEF$ is $52$ with $AB=8,$ $BC=9$ and $FA=5.$ What is $DE+EF?$

$7$

$8$

$9$

$10$

$11$

答案：C

知识点：面积分割矩形

难度评级：1170

小提示：

将图形补成一个长方形

Complete the figure to a rectangle.

大提示：

从长方形面积中减去多边形面积，得到缺失长方形

Subtract the polygon area from the rectangle area to get the missing rectangle.

解答：

将图形补成长方形 $ABCG$ ，其面积为 $8\cdot9=72$ 。

缺失长方形 $FEDG$ 的面积为 $72-52=20$ 。它的高是 $DE=BC-FA=9-5=4$ 。

因此 $EF=20\div4=5$ ，所以 $DE+EF=4+5=9$ 。

所以正确答案是 C。

Complete the shape to rectangle $ABCG$ , which has area $8\cdot9=72$ .

The missing rectangle $FEDG$ has area $72-52=20$ . Its height is $DE=BC-FA=9-5=4$ .

Therefore $EF=20\div4=5$ , so $DE+EF=4+5=9$ .

Thus, C is the correct answer.

14.

小十二篮球联盟有两个分区，每个分区有六支球队。每支球队与本分区其他每支球队比赛两次，并与另一分区每支球队比赛一次。联盟共安排了多少场比赛？

The Little Twelve Basketball Conference has two divisions, with six teams in each division. Each team plays each of the other teams in its own division twice and every team in the other division once. How many conference games are scheduled?

$80$

$96$

$100$

$108$

$192$

答案：B

知识点：双重计数基本计数

难度评级：1220

小提示：

先数每支球队的比赛数，再修正重复计数

Count each team’s games, then correct for double-counting.

大提示：

每场比赛会被两支球队各数一次

Each game is counted once for each of its two teams.

解答：

关注一支球队。它与本分区另外 $5$ 支球队各比赛两次，共 $2 \cdot 5 = 10$ 场。

它还与另一分区的 $6$ 支球队各比赛一场，所以总共打 $10 + 6 = 16$ 场。

共有 $12$ 支球队，于是按球队计数为 $12 \cdot 16 = 192$ 场。但每场比赛涉及 $2$ 支球队，所以要除以 $2$ 。

实际总场数为 $192 \div 2 = 96$ 。

所以正确答案是 B。

Let us focus on one team. This team plays against $5$ other teams twice in its own division, for a total of $2 \cdot 5 = 10$ games.

This team also plays $6$ games with teams from the other division. Therefore, they play a total of $10 + 6 = 16$ games.

There are $12$ teams total, so there are $12 \cdot 16 = 192$ games conducted. Each game, however, involves $2$ teams, so we have to divide by $2.$

This gives us the actual total number of games to be $192 \div 2 = 96.$

Thus, B is the correct answer.

15.

有多少个不同的等腰三角形，其边长均为整数且周长为 $23$ ？

How many different isosceles triangles have integer side lengths and perimeter $23?$

$2$

$4$

$6$

$9$

$11$

答案：C

知识点：三角不等式奇偶性分类讨论

难度评级：1290

小提示：

设相等边长为 $a$ ，底边为 $b$

Let the equal side length be $a$ and the base be $b$ .

大提示：

使用 $2a+b=23$ 和三角形不等式

Use $2a+b=23$ and the triangle inequality.

解答：

设相等边长为 $a$ ，底边为 $b$ 。则 $2a+b=23$ ，所以 $b$ 必须是奇数。

三角形不等式要求 $b<2a$ 。由于 $b=23-2a$ ，得到 $23-2a<2a$ ，即 $a>\frac{23}{4}$ 。所以 $a\ge6$ 。

还要 $b\ge1$ ，所以 $2a\le22$ ，即 $a\le11$ 。可能的值 $a=6$ 、 $7$ 、 $8$ 、 $9$ 、 $10$ 和 $11$ 都可行，共 $6$ 个三角形。

所以正确答案是 C。

Let the equal side length be $a$ and the base be $b$ . Then $2a+b=23$ , so $b$ must be odd.

The triangle inequality requires $b<2a$ . Since $b=23-2a$ , this gives $23-2a<2a$ , so $a>\frac{23}{4}$ . Thus $a\ge6$ .

Also $b\ge1$ , so $2a\le22$ and $a\le11$ . The possible values $a=6,$ $7,$ $8,$ $9,$ $10,$ and $11$ all work, giving $6$ triangles.

Thus, C is the correct answer.

16.

一个五条腿的火星人有一抽屉袜子，每只袜子是红色、白色或蓝色，并且每种颜色至少有五只。火星人不看地一次取出一只袜子。为了保证有 $5$ 只同色袜子，火星人必须从抽屉里取出多少只袜子？

A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be $5$ socks of the same color?

$6$

$9$

$12$

$13$

$15$

答案：D

知识点：抽屉原理

难度评级：1230

小提示：

最坏情况下，火星人先拿到每种颜色各 $4$ 只

In the worst case, the Martian gets only $4$ socks of each color first.

大提示：

下一只袜子必然使某种颜色达到五只

The next sock must make five of some color.

解答：

火星人可能先取出每种颜色各 $4$ 只袜子，而还没有 $5$ 只同色。第 $13$ 只袜子必然是某种颜色的第 $5$ 只。

所以正确答案是 D。

Note that the Martian can pull out $4$ socks of each color without drawing $5$ of the same color. The $13$ th sock, however, must be the $5$ th sock of some color.

Thus, D is the correct answer.

17.

图中显示了一支越野队训练跑的结果。哪位学生的平均速度最大？

The results of a cross-country team’s training run are graphed below. Which student has the greatest average speed?

$\text{Angela}$

$\text{Briana}$

$\text{Carla}$

$\text{Debra}$

$\text{Evelyn}$

答案：E

知识点：数据与图表解读斜率

难度评级：1250

小提示：

平均速度等于距离除以时间

Average speed is distance divided by time.

大提示：

在图上，比较从原点到各点的斜率

On the graph, compare slopes from the origin.

解答：

平均速度是距离除以时间，也就是从原点到该点连线的斜率。

Evelyn 的斜率最大，因此她的平均速度最大。

所以正确答案是 E。

Average speed is distance over time, which is given by the slope of the line through the point and the origin.

Evelyn has the steepest slope, telling us that she had the greatest average speed.

Thus, E is the correct answer.

18.

有多少个三位数能被 $13$ 整除？

How many three-digit numbers are divisible by $13?$

$7$

$67$

$69$

$76$

$77$

答案：C

知识点：区间内整数计数倍数

难度评级：1260

小提示：

三位数的 $13$ 的倍数形式为 $13k$

A three-digit multiple of $13$ has form $13k$ .

大提示：

求出 $k$ 的最小和最大可能值

Find the smallest and largest possible values of $k$ .

解答：

要求有多少个 $k$ 的取值使得 $13k$ 是三位数。

最小的 $k$ 满足 $13k \gt 99$ ，因此 $k = 8$ 。

最大的 $k$ 满足 $13k \lt 1000$ ，因此 $k = 76$ 。

所以 $k$ 可以取从 $8$ 到 $76$ 的整数，也就是说共有 $69$ 个可能的 $k$ 值。

所以正确答案是 C。

We want to find how many $k$ exist such that $13k$ is a three-digit number.

The smallest possible $k$ such that $13k \gt 99$ is $k = 8.$

The largest possible $k$ such that $13k \lt 1000$ is $k = 76.$

This tells us that $k$ can range from $8$ to $76,$ which gives us $69$ values for $k.$

Thus, C is the correct answer.

19.

梯形 $ABCD$ 的周长是多少？

What is the perimeter of trapezoid $ABCD?$

$180$

$188$

$196$

$200$

$204$

答案：A

知识点：梯形勾股数周长

难度评级：1390

小提示：

作垂线，将梯形分成直角三角形和长方形

Drop perpendiculars to split the trapezoid into right triangles and a rectangle.

大提示：

使用 $18-24-30$ 和 $7-24-25$ 直角三角形

Use the $18-24-30$ and $7-24-25$ right triangles.

解答：

从 $C$ 作高，交 $\overline{AD}$ 于 $F$ 。已有的从 $B$ 作的高交 $\overline{AD}$ 于 $E$ 。

由勾股定理， $AE=\sqrt{30^2-24^2}=18$ ， $FD=\sqrt{25^2-24^2}=7$ 。同时 $EF=BC=50$ 。

因此 $AD=18+50+7=75$ ，梯形周长为 $30+50+25+75=180$ 。

所以正确答案是 A。

Drop the altitude from $C$ to meet $\overline{AD}$ at $F$ . The existing altitude from $B$ meets $\overline{AD}$ at $E$ .

By the Pythagorean theorem, $AE=\sqrt{30^2-24^2}=18$ and $FD=\sqrt{25^2-24^2}=7$ . Also $EF=BC=50$ .

Thus $AD=18+50+7=75$ , and the trapezoid perimeter is $30+50+25+75=180$ .

Thus, A is the correct answer.

20.

爱丽丝和鲍勃在一个圆上玩游戏，圆周上有 $12$ 个等距点。这些点按顺时针方向编号为 $1$ 到 $12$ 。两人都从点 $12$ 开始。爱丽丝顺时针移动，鲍勃逆时针移动。每轮游戏中，爱丽丝顺时针移动 $5$ 个点，鲍勃逆时针移动 $9$ 个点。当他们停在同一点时游戏结束。需要多少轮？

Alice and Bob play a game involving a circle whose circumference is divided by $12$ equally-spaced points. The points are numbered clockwise, from $1$ to $12.$ Both start on point $12.$ Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves $5$ points clockwise and Bob moves $9$ points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$6$

$8$

$12$

$14$

$24$

答案：A

知识点：模运算最小公倍数

难度评级：1450

小提示：

跟踪爱丽丝和鲍勃在圆上的相对运动

Track the relative motion of Alice and Bob around the circle.

大提示：

每轮两人的相对位移为 $14$ 个点，按模 $12$ 计算

Together they close the gap by $14$ points each turn, modulo $12$ .

解答：

一轮后，爱丽丝顺时针移动 $5$ 个点，鲍勃逆时针移动 $9$ 个点。他们每轮的相对位移为 $5+9=14$ 个点。

当 $14k$ 是 $12$ 的倍数时他们相遇，其中 $k$ 是轮数。

因为 $14k\equiv2k\pmod{12}$ ，使 $2k$ 成为 $12$ 的倍数的最小正整数 $k$ 是 $6$ 。

所以正确答案是 A。

After one turn, Alice has moved $5$ points clockwise and Bob has moved $9$ points counterclockwise. Their relative movement is $5+9=14$ points per turn.

They meet when $14k$ is a multiple of $12$ , where $k$ is the number of turns.

Since $14k\equiv2k\pmod{12}$ , the smallest positive $k$ making $2k$ a multiple of $12$ is $6$ .

Thus, A is the correct answer.

21.

用下图中的三个点作为顶点，可以画出多少个不同的三角形？

How many distinct triangles can be drawn using three of the dots below as vertices?

$9$

$12$

$18$

$20$

$24$

答案：C

知识点：组合补集计数

难度评级：1390

小提示：

总共有 $6$ 个点

There are $6$ dots total.

大提示：

从所有选 $3$ 个点的方式中减去三点共线的选择

Subtract the choices of three collinear dots from all $3$ -dot choices.

解答：

任取 $3$ 个点通常会形成三角形，只有 $2$ 组三点共线的情况除外。

选择三个点有 $\binom{6}{3} = 20$ 种方式，减去 $2$ 得到 $20 - 2 = 18$ 。

所以正确答案是 C。

Notice that choosing any of these $3$ dots forms a triangle, except for the $2$ triples that form a straight line.

There are $\binom{6}{3} = 20$ ways to choose the points, and then we subtract $2$ to get $20 - 2 = 18.$

Thus, C is the correct answer.

22.

一家公司销售三种不同大小的洗涤剂盒：小号（S）、中号（M）和大号（L）。中号价格比小号高 $50 \%$ ，含量比大号少 $20 \%$ 。大号含量是小号的两倍，价格比中号高 $30 \%$ 。按性价比从好到差排列这三种大小。

A company sells detergent in three different sized boxes: small (S), medium (M) and large (L). The medium size costs $50 \%$ more than the small size and contains $20 \%$ less detergent than the large size. The large size contains twice as much detergent as the small size and costs $30 \%$ more than the medium size. Rank the three sizes from best to worst buy.

$SML$

$LMS$

$MSL$

$LSM$

$MLS$

答案：E

知识点：百分数比与比例

难度评级：1550

小提示：

为小号和大号选择方便的数量

Choose convenient numbers for the small and large sizes.

大提示：

比较每种大小的每盎司价格

Compare price per ounce for each size.

解答：

选择方便数值：设小号价格为 $\$1.00$ ，含量为 $10$ 盎司。那么大号含量为 $20$ 盎司。

中号价格为 $\$1.50$ ，含量比大号少 $20\%$ ，即 $16$ 盎司。大号价格比中号高 $30\%$ ，即 $\$1.95$ 。

每盎司价格分别为 $S: \frac{1.00}{10}=0.100$ ， $M: \frac{1.50}{16}=0.09375$ ， $L: \frac{1.95}{20}=0.0975$ 。

从好到差的顺序是 $M$ 、 $L$ 、 $S$ 。

所以正确答案是 E。

Choose convenient values: let the small box cost $\$1.00$ and contain $10$ ounces. Then the large box contains $20$ ounces.

The medium box costs $\$1.50$ and contains $20\%$ less than the large box, or $16$ ounces. The large box costs $30\%$ more than the medium, or $\$1.95$ .

The costs per ounce are $S: \frac{1.00}{10}=0.100$ , $M: \frac{1.50}{16}=0.09375$ , and $L: \frac{1.95}{20}=0.0975$ .

From best to worst buy, the order is $M,$ $L,$ $S$ .

Thus, E is the correct answer.

23.

等腰直角三角形 $ABC$ 包含一个面积为 $2\pi$ 的半圆。圆心 $O$ 在斜边 $\overline{AB}$ 上，并且圆与边 $\overline{AC}$ 和 $\overline{BC}$ 相切。三角形 $ABC$ 的面积是多少？

Isosceles right triangle $ABC$ encloses a semicircle of area $2\pi.$ The circle has its center $O$ on hypotenuse $\overline{AB}$ and is tangent to sides $\overline{AC}$ and $\overline{BC}.$ What is the area of triangle $ABC?$

$6$

$8$

$3 \pi$

$10$

$4 \pi$

答案：B

知识点：圆面积对称性

难度评级：1610

小提示：

将三角形和半圆关于斜边反射

Reflect the triangle and semicircle across the hypotenuse.

大提示：

这会形成一个内切于正方形的圆

This makes a circle inscribed in a square.

解答：

将三角形和半圆关于斜边 $\overline{AB}$ 反射，会形成一个内切于正方形的整圆。

半圆面积为 $2\pi$ ，所以整圆面积为 $4\pi$ ，半径为 $2$ 。因此正方形边长为 $4$ 。

正方形面积为 $16$ ，原三角形面积是其一半，即 $8$ 。

所以正确答案是 B。

Reflect the triangle and semicircle across hypotenuse $\overline{AB}$ . This forms a full circle inscribed in a square.

The semicircle has area $2\pi$ , so the full circle has area $4\pi$ and radius $2$ . The square side length is therefore $4$ .

The square area is $16$ , so the original triangle has half that area, $8$ .

Thus, B is the correct answer.

24.

某计算器只有两个键 [ $+1$ ] 和 [ $\times 2$ ]。按下其中一个键时，计算器会自动显示结果。例如，如果计算器原来显示“ $9$ ”，按 [ $+1$ ] 后会显示“ $10$ ”。如果接着按 [ $\times 2$ ]，会显示“ $20$ ”。从显示“ $1$ ”开始，要达到“ $200$ ”最少需要按多少次键？

A certain calculator has only two keys [ $+1$ ] and [ $\times 2$ ]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “ $9$ ” and you pressed [ $+1$ ], it would display “ $10.$ ” If you then pressed [ $\times 2$ ], it would display “ $20.$ ” Starting with the display “ $1,$ ” what is the fewest number of keystrokes you would need to reach “ $200$ ”?

$8$

$9$

$10$

$11$

$12$

答案：B

知识点：逆推法最优化

难度评级：1610

小提示：

从 $200$ 反向推回 $1$

Work backward from $200$ to $1$ .

大提示：

为了给出下界，注意八次按键在 $200$ 附近能到达和不能到达什么

For a lower bound, note what eight keystrokes can and cannot reach near $200$ .

解答：

从 $200$ 反向操作。数为偶数时，用除以 $2$ 撤销 $\times2$ ；数为奇数时，用减去 $1$ 撤销 $+1$ 。

$200,100,50,25,24,12,6,3,2,1$ 使用 $9$ 个反向步骤，所以 $9$ 次按键足够。

对于目标 $200$ ， $8$ 次按键不够：小于目标的最大可达值由先 $\times2,+1$ ，再六次加倍得到，即 $3\cdot2^6=192$ 。更大的可能会至少达到 $256$ ，所以 $200$ 不能在 $8$ 次按键内到达。

所以正确答案是 B。

Work backward from $200$ . When the number is even, undo $\times2$ by dividing by $2$ ; when it is odd, undo $+1$ by subtracting $1$ .

$200,100,50,25,24,12,6,3,2,1$ uses $9$ reverse steps, so $9$ keystrokes are enough.

Eight keystrokes are not enough: the largest value below $200$ reachable in $8$ keystrokes is obtained by $\times2,+1$ followed by six doublings, giving $3\cdot2^6=192$ . The next larger possibilities overshoot to at least $256$ , so $200$ cannot be reached in $8$ keystrokes.

Thus, B is the correct answer.

25.

一个边长为 $2$ 的正方形和一个圆有相同的中心。圆内正方形外的区域总面积等于圆外正方形内的区域总面积。圆的半径是多少？

A square with side length $2$ and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

$\dfrac{2}{\sqrt{\pi}}$

$\dfrac{1+\sqrt{2}}{2}$

$\dfrac{3}{2}$

$\sqrt{3}$

$\sqrt{\pi}$

答案：A

知识点：圆面积正方形（几何）

难度评级：1560

小提示：

圆内方外与方内圆外两部分的面积相等

The two outside-area totals are equal.

大提示：

这会使正方形和圆的总面积相等

That makes the square and circle have equal total area.

解答：

设 $S$ 为同时在正方形和圆内的公共面积。题目说明圆独有区域面积等于正方形独有区域面积。

在两个相等面积上都加上 $S$ ，可知圆的总面积等于正方形的总面积。

正方形面积为 $2^2=4$ ，所以 $\pi r^2=4$ 。因此 $r=\dfrac{2}{\sqrt{\pi}}$ 。

所以正确答案是 A。

Let $S$ be the common area inside both the square and circle. The problem says the circle-only area equals the square-only area.

Adding $S$ to both equal areas shows that the total area of the circle equals the total area of the square.

The square area is $2^2=4$ , so $\pi r^2=4$ . Hence $r=\dfrac{2}{\sqrt{\pi}}$ .

Thus, A is the correct answer.