2001 AMC 8 详解 | LIVE by Po-Shen Loh

1.

凯西的手工课正在制作一个高尔夫奖杯。他需要给一个高尔夫球上的 $300$ 个凹点上色。如果给一个凹点上色需要 $2$ 秒，他需要多少分钟完成这项工作？

Casey’s shop class is making a golf trophy. He has to paint $300$ dimples on a golf ball. If it takes him $2$ seconds to paint one dimple, how many minutes will he need to do his job?

$4$

$6$

$8$

$10$

$12$

知识点：速率单位换算

难度评级：370

小提示：

先把总上色时间算成秒

Convert the total painting time to seconds first

大提示：

一分钟有 $60$ 秒

There are $60$ seconds in one minute

解答：

凯西完成工作需要 $2\cdot300=600$ 秒。因为 $60$ 秒等于一分钟，所以这是 $600\div60=10$ 分钟。

所以正确答案是 D。

It will take Casey $2\cdot300=600$ seconds to do the job. Since $60$ seconds make one minute, this is $600\div60=10$ minutes.

Thus, D is the correct answer.

2.

我想到了两个整数。它们的乘积是 $24$ ，和是 $11$ 。较大的数是多少？

I’m thinking of two whole numbers. Their product is $24$ and their sum is $11.$ What is the larger number?

$3$

$4$

$6$

$8$

$12$

知识点：因数系统列举

难度评级：450

小提示：

列出 $24$ 的因数对

List the factor pairs of $24$

大提示：

找到和为 $11$ 的那一对

Find the pair whose sum is $11$

解答：

$24$ 的因数对为 $(1,24)$ 、 $(2,12)$ 、 $(3,8)$ 和 $(4,6)$ 。

唯一和为 $11$ 的因数对是 $(3,8)$ ，所以较大的数是 $8$ 。

所以正确答案是 D。

The factor pairs of $24$ are $(1,24),$ $(2,12),$ $(3,8),$ and $(4,6)$ .

The only pair with sum $11$ is $(3,8)$ , so the larger number is $8$ .

Thus, D is the correct answer.

3.

史密斯奶奶有 $\$63$ 。埃尔伯塔比安茹多 $\$2$ ，而安茹的钱数是史密斯奶奶的三分之一。埃尔伯塔有多少美元？

Granny Smith has $\$63.$ Elberta has $\$2$ more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?

$17$

$18$

$19$

$21$

$23$

知识点：分数

难度评级：560

小提示：

先求史密斯奶奶钱数的三分之一

Find one-third of Granny Smith’s money

大提示：

埃尔伯塔比安茹多 $\$2$

Elberta has $\$2$ more than Anjou

解答：

安茹有 $\$63\div3=\$21$ 。埃尔伯塔比安茹多 $\$2$ ，所以埃尔伯塔有 $\$21+\$2=\$23$ 。

所以正确答案是 E。

Anjou has $\$63\div3=\$21$ . Elberta has $\$2$ more than Anjou, so Elberta has $\$21+\$2=\$23$ .

Thus, E is the correct answer.

4.

数字 $1$ 、 $2$ 、 $3$ 、 $4$ 和 $9$ 各使用一次，组成尽可能小的偶数五位数。十位上的数字是

The digits $1,$ $2,$ $3,$ $4,$ and $9$ are each used once to form the smallest possible even five-digit number. The digit in the tens place is

$1$

$2$

$3$

$4$

$9$

知识点：位值最优化

难度评级：720

小提示：

这个数必须以 $2$ 或 $4$ 结尾

The number must end in $2$ or $4$

大提示：

尽量让前面的数字小

Make the earlier digits as small as possible

解答：

个位数字必须是 $2$ 或 $4$ 。若个位是 $2$ ，把其余数字按尽可能小的顺序排列，得到 $13492$ 。若个位是 $4$ ，最小的排列是 $12394$ ，它更小。因此最小的偶数是 $12394$ ，其十位数字是 $9$ 。

所以正确答案是 E。

The units digit must be $2$ or $4$ . If it is $2$ , arranging the remaining digits as small as possible gives $13492$ . If it is $4$ , the smallest arrangement is $12394$ , which is smaller. Therefore the smallest even number is $12394$ , whose tens digit is $9$ .

Thus, E is the correct answer.

5.

在一个漆黑的暴风雨夜，史努比突然看见一道闪电。十秒后他听到雷声。声速是每秒 $1088$ 英尺，一英里是 $5280$ 英尺。估计史努比距离闪电有多远，精确到最近的半英里。

On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is $1088$ feet per second and one mile is $5280$ feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.

$1$

$1\frac{1}{2}$

$2$

$2\frac{1}{2}$

$3$

知识点：路程、速度与时间估算单位换算

难度评级：730

小提示：

使用距离等于速度乘以时间

Use distance equals rate times time

大提示：

将这个距离与 $2$ 英里比较

Compare the distance with $2$ miles

解答：

在 $10$ 秒内，雷声传播了 $10 \cdot 1088 = 10880$ 英尺。

$2$ 英里约为 $2 \cdot 5280 = 10560$ 英尺，接近上面的距离。

所以正确答案是 C。

In $10$ seconds, the thunder was able to travel $10 \cdot 1088 = 10880$ feet.

Note that $2$ miles is about $2 \cdot 5280 = 10560$ feet, which is close to what we found above.

Thus, C is the correct answer.

6.

六棵树沿着一条直路的一侧等间距排列。第一棵树到第四棵树的距离是 $60$ 英尺。第一棵树到最后一棵树之间的距离是多少英尺？

Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is $60$ feet. What is the distance in feet between the first and last trees?

$90$

$100$

$105$

$120$

$140$

知识点：植树问题比与比例

难度评级：820

小提示：

第一棵到第四棵之间有三个相等间隔

The first to fourth trees contain three equal gaps

大提示：

第一棵到最后一棵之间有五个相等间隔

The first to last trees contain five equal gaps

解答：

第一棵和第四棵之间有 $3$ 个间隔，因此每个间隔长 $60 \div 3 = 20$ 英尺。

第一棵和最后一棵之间有 $5$ 个间隔，所以它们相距 $5 \cdot 20 = 100$ 英尺。

所以正确答案是 B。

There are $3$ gaps between the first and fourth trees, which means that one gap is $60 \div 3 = 20$ feet long.

There are $5$ gaps between the first and last trees. This means that they are $5 \cdot 20 = 100$ feet apart.

Thus, B is the correct answer.

7.

第 $7$ 、 $8$ 和 $9$ 题与这些风筝有关。

为了宣传学校一年一度的风筝奥林匹克活动，吉纳维芙为公告板展示制作了一个小风筝和一个大风筝。风筝形状如图。对于小风筝，吉纳维芙在一英寸方格纸上画风筝。对于大风筝，她将整个方格的高度和宽度都扩大为原来的三倍。

小风筝的面积是多少平方英寸？

Problems $7,$ $8,$ and $9$ are about these kites.

To promote her school’s annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.

What is the number of square inches in the area of the small kite?

$21$

$22$

$23$

$24$

$25$

知识点：筝形面积对角线

难度评级：900

小提示：

使用方格上的风筝两条对角线

Use the kite diagonals on the grid

大提示：

面积等于两条对角线乘积的一半

The area is half the product of the diagonals

解答：

风筝面积等于两条对角线的乘积除以 $2$ 。

因此小风筝面积为 $\dfrac{6 \cdot 7}{2} = \dfrac{42}{2} = 21\text{。}$

所以正确答案是 A。

Recall that the area of a kite is the product of its diagonals divided by $2.$

Therefore, the area of the small kite is $\dfrac{6 \cdot 7}{2} = \dfrac{42}{2} = 21.$

Thus, A is the correct answer.

8.

吉纳维芙在她的大风筝上加支撑条，形式是连接风筝相对顶点的十字。她需要多少英寸支撑材料？

Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?

$30$

$32$

$35$

$38$

$39$

知识点：长度、面积与体积的缩放关系对角线

难度评级：960

小提示：

支撑材料总长是两条对角线之和

The bracing is the sum of the two diagonals

大提示：

方格扩大三倍后，两条对角线长度也都扩大三倍

Tripling the grid triples both diagonal lengths

解答：

长对角线是 $7$ 个单位，短对角线是 $6$ 个单位。

在大风筝中，一个单位为 $3$ 英寸，所以需要的支撑材料总长为 $3(7 + 6) = 3 \cdot 13 = 39\text{。}$

所以正确答案是 E。

The long diagonal is $7$ units, and the short one is $6$ units.

In the large kite, one unit is $3$ inches, so the total amount of bracing material needed is $3(7 + 6) = 3 \cdot 13 = 39.$

Thus, E is the correct answer.

9.

大风筝要覆盖金箔。金箔从一张刚好覆盖整个方格的长方形纸上裁出。从四个角裁掉的废料有多少平方英寸？

The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?

$63$

$72$

$180$

$189$

$264$

知识点：长度、面积与体积的缩放关系面积筝形

难度评级：1020

小提示：

高和宽都扩大三倍，面积扩大 $9$ 倍

Tripling both dimensions multiplies area by $9$

大提示：

废料面积等于大风筝面积

The waste area equals the large kite area

解答：

整个方格的面积是 $3 \cdot 7 \cdot 3 \cdot 6 = 378\text{。}$ 由风筝面积公式，风筝面积是这个面积的一半，所以废料面积为 $378 \div 2 = 189$ 。

所以正确答案是 D。

The area of the entire grid would be $3 \cdot 7 \cdot 3 \cdot 6 = 378.$ The area of the kite is one-half this area from the formula for the area of the kite, so the wasted material is $378 \div 2 = 189.$

Thus, D is the correct answer.

10.

一位收藏者提出按州纪念二十五分硬币面值的 $2000\%$ 购买。按这个价格，布赖登的四枚州纪念二十五分硬币能卖多少钱？

A collector offers to buy state quarters for $2000\%$ of their face value. At that rate how much will Bryden get for his four state quarters?

$\$20$

$\$50$

$\$200$

$\$500$

$\$2000$

知识点：百分数钱币

难度评级：980

小提示：

四枚二十五分硬币的面值是 $\$1$

Four quarters have face value $\$1$

大提示：

$2000\%$ 表示面值的 $20$ 倍

$2000\%$ means $20$ times face value

解答：

四枚州纪念二十五分硬币的面值为 $\$1$ 。

因为 $2000\%=20$ ，收藏者支付面值的 $20$ 倍，也就是 $\$20$ 。

所以正确答案是 A。

Four state quarters have face value $\$1$ .

Since $2000\%=20$ , the collector pays $20$ times face value, or $\$20$ .

Thus, A is the correct answer.

11.

点 $A$ 、 $B$ 、 $C$ 和 $D$ 的坐标分别为 $A(3,2)$ 、 $B(3,-2)$ 、 $C(-3,-2)$ 和 $D(-3,0)$ 。四边形 $ABCD$ 的面积是多少？

Points $A,$ $B,$ $C$ and $D$ have these coordinates: $A(3,2),$ $B(3,-2),$ $C(-3,-2)$ and $D(-3,0).$ What is the area of quadrilateral $ABCD?$

$12$

$15$

$18$

$21$

$24$

知识点：坐标几何梯形面积

难度评级：1140

小提示：

将 $ABCD$ 看作梯形

View $ABCD$ as a trapezoid

大提示：

两条平行的竖边长度分别为 $4$ 和 $2$

The parallel vertical sides have lengths $4$ and $2$

解答：

因为 $\overline{DC} \parallel \overline{AB}$ ，所以 $ABCD$ 是梯形。

梯形面积公式为 $A = \dfrac{1}{2} (b_1 + b_2) h\text{。}$

代入 $b_1 = DC = 2\text{、}$ $b_2 = AB = 4$ 和 $h = CB = 6\text{，}$ 可得 $A = \dfrac{1}{2} (2 + 4) \cdot 6$ $= \dfrac{1}{2} \cdot 6 \cdot 6 = 18\text{。}$

所以正确答案是 C。

We can see that $ABCD$ is a trapezoid since $\overline{DC} \parallel \overline{AB}.$

Recall that the formula for the area of a trapezoid is $A = \dfrac{1}{2} (b_1 + b_2) h.$

Plugging in $b_1 = DC = 2,$ $b_2 = AB = 4,$ and $h = CB = 6,$ we get that $A = \dfrac{1}{2} (2 + 4) \cdot 6$ $= \dfrac{1}{2} \cdot 6 \cdot 6 = 18.$

Thus, C is the correct answer.

12.

如果 $a \otimes b = \dfrac{a + b}{a - b}\text{，}$ 那么 $(6 \otimes 4) \otimes 3 =$

If $a \otimes b = \dfrac{a + b}{a - b},$ then $(6 \otimes 4) \otimes 3 =$

$4$

$13$

$15$

$30$

$72$

知识点：自定义运算分数

难度评级：1170

小提示：

先计算 $6\otimes4$

Evaluate $6\otimes4$ first

大提示：

再把结果作为第一个输入，与 $3$ 运算

Then use the result as the first input with $3$

解答：

计算如下： $\begin{align*} (6 \otimes 4) \otimes 3 &= \dfrac{6 + 4}{6 - 4} \otimes 3 \\ &= 5 \otimes 3 \\&= \dfrac{5 + 3}{5 - 3} \\&= 4 \end{align*}\text{。}$

所以正确答案是 A。

We can evaluate it as follows $\begin{align*} (6 \otimes 4) \otimes 3 &= \dfrac{6 + 4}{6 - 4} \otimes 3 \\ &= 5 \otimes 3 \\&= \dfrac{5 + 3}{5 - 3} \\&= 4. \end{align*}$

Thus, A is the correct answer.

13.

瑞雪尔班上有 $36$ 名学生，其中 $12$ 人喜欢巧克力派， $8$ 人喜欢苹果派， $6$ 人喜欢蓝莓派。剩下的学生中一半喜欢樱桃派，一半喜欢柠檬派。瑞雪尔用这些数据画扇形统计图时，樱桃派应占多少度？

Of the $36$ students in Richelle’s class, $12$ prefer chocolate pie, $8$ prefer apple, and $6$ prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle’s pie graph showing this data, how many degrees should she use for cherry pie?

$10$

$20$

$30$

$50$

$72$

知识点：数据与图表解读分数

难度评级：1190

小提示：

先求前三种派之后还剩多少学生

Find how many students remain after the first three pies

大提示：

樱桃派占剩余学生的一半

Cherry is half of the remaining group

解答：

喜欢樱桃派或柠檬派的学生数为 $36 - 12 - 8 - 6 = 10\text{。}$ 其中一半喜欢樱桃派，即 $10 \div 2 = 5$ 人。

樱桃派对应的角度为 $360^{\circ} \cdot \dfrac{5}{36} = 5 \cdot 10^{\circ} = 50^{\circ}\text{。}$

所以正确答案是 D。

The number of students that prefer cherry or lemon pie is $36 - 12 - 8 - 6 = 10.$ Half of these like cherry pie, which is $10 \div 2 = 5.$

The number of degrees for cherry pie would then be $360^{\circ} \cdot \dfrac{5}{36} = 5 \cdot 10^{\circ} = 50^{\circ}.$

Thus, D is the correct answer.

14.

泰勒进入一条自助餐队伍，他要选择一种肉、两种不同的蔬菜和一种甜点。如果食物选择的顺序不重要，他可能选择多少种不同的餐食？

肉类：牛肉、鸡肉、猪肉

蔬菜：烤豆、玉米、土豆、番茄

甜点：布朗尼、巧克力蛋糕、巧克力布丁、冰淇淋

Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose?

Meat: beef, chicken, pork

Vegetables: baked beans, corn, potatoes, tomatoes

Dessert: brownies, chocolate cake, chocolate pudding, ice cream

$4$

$24$

$72$

$80$

$144$

知识点：组合乘法原理

难度评级：1270

小提示：

将两种蔬菜看作无序的一对

Choose the two vegetables as an unordered pair

大提示：

将肉类选择数、蔬菜组合数和甜点选择数相乘

Multiply meat choices, vegetable pairs, and dessert choices

解答：

他有 $3$ 种肉类选择和 $4$ 种甜点选择。

他必须从 $4$ 种蔬菜中选 $2$ 种。因为顺序不重要，蔬菜选择有 $\dfrac{4\cdot3}{2}=6$ 种。

因此可能的餐食数为 $3\cdot6\cdot4=72$ 。

所以正确答案是 C。

He has $3$ choices for the meat and $4$ choices for dessert.

He must choose $2$ of the $4$ vegetables. Since order does not matter, there are $\dfrac{4\cdot3}{2}=6$ vegetable choices.

This gives $3\cdot6\cdot4=72$ possible meals.

Thus, C is the correct answer.

15.

荷马开始以每分钟削 $3$ 个土豆的速度削一堆 $44$ 个土豆。四分钟后克里斯滕加入，以每分钟削 $5$ 个土豆的速度削。完成时，克里斯滕削了多少个土豆？

Homer began peeling a pile of $44$ potatoes at the rate of $3$ potatoes per minute. Four minutes later Christen joined him and peeled at the rate of $5$ potatoes per minute. When they finished, how many potatoes had Christen peeled?

$20$

$24$

$32$

$33$

$40$

知识点：速率

难度评级：1290

小提示：

先求克里斯滕开始时还剩多少个土豆

Find how many potatoes remain when Christen starts

大提示：

之后他们合起来每分钟削 $8$ 个

After that, their combined rate is $8$ per minute

解答：

克里斯滕加入时，荷马已经削了 $3 \cdot 4 = 12$ 个土豆，还剩 $44 - 12 = 32$ 个。

荷马和克里斯滕合起来每分钟削 $5 + 3 = 8$ 个土豆，因此还需要 $32 \div 8 = 4$ 分钟削完剩下的。

$4$ 分钟内，克里斯滕削了 $4 \cdot 5 = 20$ 个土豆。

所以正确答案是 A。

Homer had peeled $3 \cdot 4 = 12$ potatoes by the time Christen joined him, leaving $44 - 12 = 32$ potatoes.

Together, Homer and Christen peel $5 + 3 = 8$ potatoes per minute, taking them $32 \div 8 = 4$ minutes to peel the rest.

In $4$ minutes, Christen peeled $4 \cdot 5 = 20$ potatoes.

Thus, A is the correct answer.

16.

一张边长 $4$ 英寸的正方形纸沿竖直方向对折。然后将两层纸沿平行于折痕的方向切成两半。形成三个新的长方形：一个大的和两个小的。一个小长方形的周长与大长方形周长之比是多少？

A square piece of paper, $4$ inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$\dfrac{5}{6}$

知识点：折纸周长

难度评级：1350

小提示：

对折后，纸变成一个 $4\times2$ 的长方形

After folding, the paper is a $4\times2$ rectangle

大提示：

小长方形是 $4\times1$

The small rectangles are $4\times1$

解答：

正方形对折后形成一个 $4\times2$ 的长方形。

将折叠后的纸沿平行于折痕的方向切成两半，得到两个 $4\times1$ 的小长方形和一个 $4\times2$ 的大长方形。

一个小长方形周长与大长方形周长之比为 $\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56\text{。}$

所以正确答案是 E。

The square is folded in half to create a $4\times2$ rectangle.

Cutting the folded paper in half parallel to the fold gives two small $4\times1$ rectangles and one large $4\times2$ rectangle.

The ratio of the perimeter of a small rectangle to the perimeter of the large rectangle is $\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56.$

Thus, E is the correct answer.

17.

游戏节目 《谁想成为百万富翁？》中每道题的奖金如下表所示（其中 $\text{K}=1000$ ）。 $\begin{array}{ccccc} \boldsymbol{\#} & 1 & 2 & 3 \\ \boldsymbol{\$} & 100 & 200 & 300 \end{array}$ $\begin{array}{cccccc} 4 & 5 & 6 & 7 & 8 \\ 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} \end{array}$ $\begin{array}{ccccc} 9 & 10 & 11 & 12 \\ 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} \end{array}$ $\begin{array}{ccc} 13 & 14 & 15 \\ 250\text{K} & 500\text{K} & 1000\text{K} \end{array}$ 哪两道题之间奖金的百分比增加最小？

For the game show Who Wants To Be a Millionaire? the dollar values of each question are shown in the following table (where $\text{K}=1000$ ). $\begin{array}{ccccc} \boldsymbol{\#} & 1 & 2 & 3 \\ \boldsymbol{\$} & 100 & 200 & 300 \end{array}$ $\begin{array}{cccccc} 4 & 5 & 6 & 7 & 8 \\ 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} \end{array}$ $\begin{array}{ccccc} 9 & 10 & 11 & 12 \\ 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} \end{array}$ $\begin{array}{ccc} 13 & 14 & 15 \\ 250\text{K} & 500\text{K} & 1000\text{K} \end{array}$ Between which two questions is the percent increase of the value the smallest?

从 $1$ 到 $2$

From $1$ to $2$

从 $2$ 到 $3$

From $2$ to $3$

从 $3$ 到 $4$

From $3$ to $4$

从 $11$ 到 $12$

From $11$ to $12$

从 $14$ 到 $15$

From $14$ to $15$

知识点：百分数

难度评级：1380

小提示：

大多数列出的增加都是翻倍

Most listed increases are doublings

大提示：

比较少数不是恰好增加 $100\%$ 的情况

Compare the few increases that are not exactly $100\%$

解答：

列出的多数增加都是翻倍，也就是增加 $100\%$ 。

在选项中，例外是从 $2$ 到 $3$ 、从 $3$ 到 $4$ 、以及从 $11$ 到 $12$ 。

这些百分比增加分别是 $\dfrac{300-200}{200}=50\%$ ， $\dfrac{500-300}{300}=66\dfrac23\%$ ，以及 $\dfrac{125000-64000}{64000}$ ，约为 $95\%$ 。

最小的是从第 $2$ 题到第 $3$ 题。

所以正确答案是 B。

Most of the listed increases are doublings, which are $100\%$ increases.

The exceptions among the answer choices are from $2$ to $3$ , from $3$ to $4$ , and from $11$ to $12$ .

These percent increases are $\dfrac{300-200}{200}=50\%$ , $\dfrac{500-300}{300}=66\dfrac23\%$ , and $\dfrac{125000-64000}{64000}$ , which is about $95\%$ .

The smallest is from question $2$ to question $3$ .

Thus, B is the correct answer.

18.

掷两个骰子。两个点数的乘积是 $5$ 的倍数的概率是多少？

Two dice are thrown. What is the probability that the product of the two numbers is a multiple of $5?$

$\dfrac{1}{36}$

$\dfrac{1}{18}$

$\dfrac{1}{6}$

$\dfrac{11}{36}$

$\dfrac{1}{3}$

知识点：骰子（概率）对立事件概率整除性

难度评级：1400

小提示：

乘积是 $5$ 的倍数，当且仅当至少一个骰子掷出 $5$

The product is a multiple of $5$ exactly when a die shows $5$

大提示：

数没有骰子掷出 $5$ 的情况更容易

It is easier to count no die showing $5$

解答：

乘积是 $5$ 的倍数的唯一方法是至少一个骰子掷出 $5$ 。

用补集计数。两个骰子都不掷出 $5$ 的概率是 $\dfrac{5}{6} \cdot \dfrac{5}{6} = \dfrac{25}{36}\text{。}$

因此至少一个骰子掷出 $5$ 的概率为 $1 - \dfrac{25}{36} = \dfrac{11}{36}\text{。}$

所以正确答案是 D。

The only way for the product to be a multiple of $5$ is if at least one of the rolls is $5.$

We can do complementary counting. The probability that neither rolls is a $5$ is $\dfrac{5}{6} \cdot \dfrac{5}{6} = \dfrac{25}{36}.$

Therefore, the probability that at least one roll is a $5$ is $1 - \dfrac{25}{36} = \dfrac{11}{36}.$

Thus, D is the correct answer.

19.

汽车 $M$ 以恒定速度行驶了一段给定时间，如虚线所示。汽车 $N$ 以两倍速度行驶相同距离。如果用实线表示汽车 $N$ 的速度和时间，哪张图表示这种情况？

Car $M$ traveled at a constant speed for a given time. This is shown by the dashed line. Car $N$ traveled at twice the speed for the same distance. If Car $N$ ’s speed and time are shown as a solid line, which graph illustrates this?

知识点：路程、速度与时间数据与图表解读

难度评级：1430

小提示：

两倍速度意味着相同距离只需一半时间

Twice the speed means half the time for the same distance

大提示：

实线应更高且更短

The solid line should be higher and shorter

解答：

汽车 $N$ 的速度是汽车 $M$ 的两倍，所以它在速度轴上的高度应是汽车 $M$ 的两倍。

因为汽车 $N$ 以更快速度行驶与汽车 $M$ 相同的距离，所以它用一半时间完成行程。

因此表示汽车 $N$ 速度的线段长度是表示汽车 $M$ 速度的线段长度的一半。

两条线都水平，因为速度恒定。查看给出的图，唯一满足条件的是图 D。

所以正确答案是 D。

Car $N$ travels at twice the speed of car $M,$ so it is represented by a point that is twice as high on the speed axis as car $M.$

Since car $N$ travels at the same distance as car $M$ but at a faster speed, it takes half the time to complete the trip.

Therefore, the line representing car $N$ ’s speed is half the length of the line representing car $M$ ’s speed.

Both lines are horizontal because the speeds are constant. Examining the given graphs, the only one that meets these conditions is graph D .

Thus, D is the correct answer.

20.

卡莱安娜把自己的考试分数给奎伊、马蒂和莎娜看，但其他人都把自己的分数藏起来。奎伊想：“我们中至少有两个人分数相同。”马蒂想：“我不是最低分。”莎娜想：“我不是最高分。”按从低到高列出马蒂（ $\text{M}$ ）、奎伊（ $\text{Q}$ ）和莎娜（ $\text{S}$ ）的分数。

Kaleana shows her test score to Quay, Marty and Shana, but the others keep theirs hidden. Quay thinks, “At least two of us have the same score.” Marty thinks, “I didn’t get the lowest score.” Shana thinks, “I didn’t get the highest score.” List the scores from lowest to highest for Marty ( $\text{M}$ ), Quay ( $\text{Q}$ ) and Shana ( $\text{S}$ ).

$\text{S,Q,M}$

$\text{Q,M,S}$

$\text{Q,S,M}$

$\text{M,S,Q}$

$\text{S,M,Q}$

知识点：逻辑推理不等式

难度评级：1520

小提示：

奎伊只看到了卡莱安娜的分数

Quay only sees Kaleana’s score

大提示：

根据每个隐藏分数的陈述，与卡莱安娜的分数比较

Use each hidden-score statement relative to Kaleana

解答：

因为奎伊只知道卡莱安娜的分数，所以奎伊和卡莱安娜的分数相同，即 $\text{K} = \text{Q}$ 。

马蒂知道他的分数高于卡莱安娜，所以 $\text{M} \gt \text{K}$ 。莎娜知道她的分数低于卡莱安娜，所以 $\text{S} \lt \text{K}$ 。

用 $\text{Q}$ 替换 $\text{K}$ ，得到 $\text{S} \lt \text{Q} \lt \text{M}$ 。

所以正确答案是 A。

Since Quay only knows Kaleana’s score, we get that Quay and Kaleana have the same score, so $\text{K} = \text{Q}.$

Marty knows that his score is higher than Kaleana’s, so $\text{M} \gt \text{K}.$ Shana knows that her score is lower than Kaleana’s, so $\text{S} \lt \text{K}.$

Substituting $\text{Q}$ for $\text{K},$ we get the following inequality: $\text{S} \lt \text{Q} \lt \text{M}.$

Thus, A is the correct answer.

21.

一组五个互不相同的正整数的平均数是 $15$ ，中位数是 $18$ 。这五个整数中最大数的最大可能值是

The mean of a set of five different positive integers is $15.$ The median is $18.$ The maximum possible value of the largest of these five integers is

$19$

$24$

$32$

$35$

$40$

知识点：平均数中位数（数据）最优化

难度评级：1550

小提示：

五个数的和是 $75$

The five numbers sum to $75$

大提示：

让另外四个数尽可能小

Make the other four numbers as small as possible

解答：

中位数是第三大的数，即 $18$ 。有两个数小于 $18$ ，两个数大于它。

平均数为 $15$ ，所以五个数的和为 $5 \cdot 15 = 75$ 。要在这个总和下使最大数尽量大，其余数必须尽量小。

小于 $18$ 的两个数必须是不同的正整数，所以最小为 $1$ 和 $2$ 。

紧接在 $18$ 后的数也应尽量小，所以是 $19$ 。

因此剩下的数，也就是最大可能值，为 $75 - 1 - 2 - 18 - 19 = 35\text{。}$

所以正确答案是 D。

The median of the set of numbers is the third largest number, which is $18.$ There are two numbers less than $18$ and two numbers greater than it.

The mean of the set is $15,$ so the sum of all the numbers is $5 \cdot 15 = 75.$ In order to maximize the largest number with this sum, the other numbers must be as small as possible.

The two numbers less than $18$ must be positive and distinct, so they must be $1$ and $2.$

The number immediately after $18$ must also be as small as possible, so it must be $19.$

Therefore, the remaining number, the maximum possible value in the set, is $75 - 1 - 2 - 18 - 19 = 35.$

Thus, D is the correct answer.

22.

在一场二十道题的测试中，每答对一题得 $5$ 分，未作答一题得 $1$ 分，答错一题得 $0$ 分。下列哪个分数不可能出现？

On a twenty-question test, each correct answer is worth $5$ points, each unanswered question is worth $1$ point and each incorrect answer is worth $0$ points. Which of the following scores is NOT possible?

$90$

$91$

$92$

$95$

$97$

知识点：分类讨论

难度评级：1550

小提示：

检查接近 $100$ 的分数

Check scores near $100$

大提示：

若答对 $19$ 题，只可能得到两个分数

With $19$ correct answers, only two scores are possible

解答：

答对 $20$ 题时，分数是 $100$ 。

答对 $19$ 题时，根据剩下一题是答错还是未作答，分数是 $95$ 或 $96$ 。

答对 $18$ 题时，剩下两题可以贡献 $0$ 、 $1$ 或 $2$ 分，所以可得 $90$ 、 $91$ 或 $92$ 。

因此列出的分数 $90,91,92,95$ 都可能，而 $97$ 不可能。

所以正确答案是 E。

With $20$ correct answers, the score is $100$ .

With $19$ correct answers, the score is either $95$ or $96$ , depending on whether the remaining question is incorrect or unanswered.

With $18$ correct answers, the remaining two questions can contribute $0$ , $1$ , or $2$ points, giving $90$ , $91$ , or $92$ .

Thus the listed scores $90,91,92,95$ are possible, while $97$ is not.

Thus, E is the correct answer.

23.

点 $R$ 、 $S$ 和 $T$ 是一个等边三角形的顶点，点 $X$ 、 $Y$ 和 $Z$ 是它各边的中点。用这六个点中的任意三个作为顶点，可以画出多少个互不全等的三角形？

Points $R,$ $S$ and $T$ are vertices of an equilateral triangle, and points $X,$ $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?

$1$

$2$

$3$

$4$

$20$

知识点：图形中的形状计数全等（几何）对称性

难度评级：1650

小提示：

按边长分类三角形

Classify triangles by side lengths

大提示：

利用对称性避免重复计数全等情况

Use symmetry to avoid recounting congruent cases

解答：

以小等边三角形的边长为一个单位，可能的三角形边长模式有：

边长为 $2,2,2$ 的等边三角形；边长为 $1,1,1$ 的等边三角形；边长为 $1,1,\sqrt3$ 的小等腰三角形；以及边长为 $1,\sqrt3,2$ 的较大等腰三角形。

这四种模式都在图中出现，而且由对称性可知，用六个点中任意三个形成的三角形都属于其中一种。

因此有 $4$ 个互不全等的三角形。

所以正确答案是 D。

Using the side length of the small equilateral triangles as one unit, the possible triangle side-length patterns are:

equilateral with sides $2,2,2$ ; equilateral with sides $1,1,1$ ; small isosceles with sides $1,1,\sqrt3$ ; and larger isosceles with sides $1,\sqrt3,2$ .

These four patterns all occur in the figure, and symmetry shows every triangle formed by three of the six points matches one of them.

Thus there are $4$ noncongruent triangles.

Thus, D is the correct answer.

24.

这个图形的每一半都由 $3$ 个红三角形、 $5$ 个蓝三角形和 $8$ 个白三角形组成。当上半部分沿中心线向下折叠时， $2$ 对红三角形重合， $3$ 对蓝三角形也重合。有 $2$ 对红白三角形重合。多少对白三角形重合？

Each half of this figure is composed of $3$ red triangles, $5$ blue triangles and $8$ white triangles. When the upper half is folded down over the centerline, $2$ pairs of red triangles coincide, as do $3$ pairs of blue triangles. There are $2$ red-white pairs. How many white pairs coincide?

$4$

$5$

$6$

$7$

$9$

知识点：折纸分类讨论

难度评级：1650

小提示：

先核算所有红三角形

Account for all red triangles first

大提示：

再跟踪未匹配的蓝三角形必须与什么配对

Then track how the unmatched blue triangles must pair

解答：

每一半都有 $3$ 个红、 $5$ 个蓝和 $8$ 个白三角形。

$2$ 对红红配对各使用每一半中的 $2$ 个红三角形，所以每一半剩下的一个红三角形用于 $2$ 对红白配对。于是所有红三角形都已计算。

$3$ 对蓝蓝配对各使用每一半中的 $3$ 个蓝三角形，每一半还剩 $2$ 个蓝三角形。由于没有更多蓝蓝配对，这 $4$ 个蓝三角形必须与白三角形配对。

所以每一半中， $1$ 个白三角形与红色配对， $2$ 个白三角形与蓝色配对，剩下 $8-3=5$ 个白三角形与白三角形配对。

所以正确答案是 B。

Each half has $3$ red, $5$ blue, and $8$ white triangles.

The $2$ red-red pairs use $2$ red triangles from each half, so the remaining red triangle from each half is used in the $2$ red-white pairs. Thus all red triangles are accounted for.

The $3$ blue-blue pairs use $3$ blue triangles from each half, leaving $2$ blue triangles from each half. Since no more blue-blue pairs occur, those $4$ blue triangles must pair with white triangles.

So on each half, $1$ white is used with red and $2$ whites are used with blue, leaving $8-3=5$ white triangles to pair with white triangles.

Thus, B is the correct answer.

25.

有 $24$ 个四位整数，每个都恰好使用数字 $2$ ， $4$ ， $5$ 和 $7$ 各一次。这些四位数中，只有一个是另一个的倍数。下列哪一个是这个倍数？

There are $24$ four-digit whole numbers that use each of the four digits $2,$ $4,$ $5,$ and $7$ exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

$5724$

$7245$

$7254$

$7425$

$7542$

知识点：排列整除性分类讨论

难度评级：1680

小提示：

一个倍数必须是较小排列的两倍或三倍

A multiple must be twice or three times a smaller permutation

大提示：

对每个可行选项尝试除以 $2$ 或 $3$

Divide each answer choice by $2$ or $3$ when possible

解答：

一个在 $4000$ 、 $5000$ 或 $7000$ 范围内的数乘以 $2$ 或更大的数后，不可能仍是这些四位排列之一。因此较小的数必须以 $2$ 开头，较大的数必须是它的两倍或三倍。

对于两倍，只有以 $4$ 结尾的选项可能。但 $\frac{5724}{2}=2862$ ， $\frac{7254}{2}=3627$ ，都没有恰好使用数字 $2,4,5,7$ 。

对于三倍，可能的选项是能被 $3$ 整除的 $7245,7425,7542$ 。分别除以三得到 $2415,2475,2514$ ，只有 $2475$ 恰好使用所需数字。

因此 $7425=3\cdot2475$ 是唯一列出的倍数。

所以正确答案是 D。

A number in the $4000$ , $5000$ , or $7000$ range cannot be multiplied by $2$ or more and remain one of the given four-digit permutations. So the smaller number must start with $2$ , and the larger number must be either double or triple it.

For doubles, only answer choices ending in $4$ can work. But $\frac{5724}{2}=2862$ and $\frac{7254}{2}=3627$ , neither of which uses exactly the digits $2,4,5,7$ .

For triples, the possible answer choices are those divisible by $3$ : $7245,7425,7542$ . Dividing gives $2415,2475,2514$ , and only $2475$ uses exactly the required digits.

Therefore $7425=3\cdot2475$ is the unique listed multiple.

Thus, D is the correct answer.