2002 AMC 8 详解 | LIVE by Po-Shen Loh

1.

在一张纸上画一个圆和两条不同的直线。这些图形最多可能有多少个交点？

A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?

$2$

$3$

$4$

$5$

$6$

知识点：交点计数最优化

难度评级：370

小提示：

分别数每一类图形对的交点。

Count each type of pair of figures separately.

大提示：

每条直线最多与圆相交两次，两条直线最多相交一次。

Each line can meet the circle at most twice, and the two lines can meet once.

解答：

一条直线和一个圆最多相交于两个点，两条不同直线最多相交于一个点。

安排两条直线都与圆相交两次，并且两条直线在另一个点相交。这样交点的总数为 $2+2+1=5\text{。}$

所以正确答案是 D。

A line and a circle can intersect at no more than two points, and two distinct lines can intersect at no more than one point.

Arrange the two lines so each meets the circle twice and the two lines meet at a different point. This gives $2+2+1=5.$

Thus, D is the correct answer.

2.

可以用多少种不同的 $\$5$ 纸币和 $\$2$ 纸币组合凑成总额 $\$17$ ？本题中顺序不重要。

How many different combinations of $\$5$ bills and $\$2$ bills can be used to make a total of $\$17?$ Order does not matter in this problem.

$2$

$3$

$4$

$5$

$6$

知识点：丢番图方程奇偶性分类讨论

难度评级：610

小提示：

因为 $\$17$ 是奇数， $\$5$ 纸币的张数必须是奇数。

Because $\$17$ is odd, the number of $\$5$ bills must be odd.

大提示：

尝试不会超过总额 $\$17$ 的奇数张 $\$5$ 纸币。

Try the possible odd numbers of $\$5$ bills before the total goes over $\$17.$

解答：

因为总额 $\$17$ 是奇数， $\$5$ 纸币的张数必须是奇数。

一张 $\$5$ 纸币还剩 $\$12$ ，也就是六张 $\$2$ 纸币。三张 $\$5$ 纸币还剩 $\$2$ ，也就是一张 $\$2$ 纸币。五张 $\$5$ 纸币太多。

因此共有 $2$ 种组合。

所以正确答案是 A。

Since the total $\$17$ is odd, the number of $\$5$ bills must be odd.

One $\$5$ bill leaves $\$12,$ which is six $\$2$ bills. Three $\$5$ bills leave $\$2,$ which is one $\$2$ bill. Five $\$5$ bills is too much.

Therefore, there are $2$ combinations.

Thus, A is the correct answer.

3.

四个互不相同的正偶数的平均数最小可能是多少？

What is the smallest possible average of four distinct positive even integers?

$3$

$4$

$5$

$6$

$7$

知识点：平均数最优化

难度评级：560

小提示：

要让平均数尽可能小，就选尽可能小的整数。

To make the average as small as possible, choose the smallest possible integers.

大提示：

使用四个最小的互不相同的正偶数。

Use the four smallest distinct positive even integers.

解答：

要得到最小可能平均数，应选 $4$ 个最小的正偶数。

这给出 $\dfrac{2 + 4 + 6 + 8}{4} = \dfrac{20}{4}$ $= 5\text{。}$

所以正确答案是 C。

To get the smallest possible average, we want to use the smallest $4$ positive even integers.

This can be achieved as follows: $\dfrac{2 + 4 + 6 + 8}{4} = \dfrac{20}{4}$ $= 5.$

Thus, C is the correct answer.

4.

年份 $2002$ 是一个回文数（从左往右读和从右往左读相同）。 $2002$ 之后下一个回文年份的各位数字乘积是多少？

The year $2002$ is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after $2002$ that is a palindrome?

$0$

$4$

$9$

$16$

$25$

知识点：回文数数字

难度评级：660

小提示：

$2002$ 之后的下一个回文数仍以 $2$ 开头和结尾。

The next palindrome after $2002$ still starts and ends with $2.$

大提示：

中间两位必须增加到最小的相同数字。

The middle two digits must increase to the smallest matching pair.

解答：

我们不想增加千位，所以千位仍可保持为 $2$ 。

因此必须把百位和十位增加到 $1$ ，得到下一个回文数 $2112$ 。它的各位数字乘积为 $4$ 。

所以正确答案是 B。

We don’t want to increase the thousands digit, so we can keep that as $2.$

This means that we have to increase the tens and hundreds digits to $1,$ to yield the next palindrome of $2112.$ The product of its digits is $4.$

Thus, B is the correct answer.

5.

Carlos Montado 出生于 $2002$ 年十一月 $9$ 日（星期六）。Carlos 满 $706$ 天时是星期几？

Carlos Montado was born on Saturday, November $9,$ $2002.$ On what day of the week will Carlos be $706$ days old?

星期一

Monday

星期三

Wednesday

星期五

Friday

星期六

Saturday

星期日

Sunday

知识点：模运算日期与时间

难度评级：730

小提示：

计算 $706$ 除以 $7$ 的余数。

Reduce $706$ modulo $7.$

大提示：

$700$ 天恰好是整数个星期。

$700$ days is an exact number of weeks.

解答：

星期每 $7$ 天循环一次。经过 $700$ 天后，星期仍然是星期六。

再过 $6$ 天后，就是星期五。

所以正确答案是 C。

The days of the week cycle every $7$ days. After $700$ days, the day of the week will still be Saturday.

After $6$ more days, the day of the week will be Friday.

Thus, C is the correct answer.

6.

一个鸟浴盆被设计成会溢流，从而可以自我清洁。水以每分钟 $20$ 毫升的速度流入，以每分钟 $18$ 毫升的速度排出。下列某个图显示了鸟浴盆在注水阶段并继续到溢流阶段的水量。是哪一个？

A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of $20$ milliliters per minute and drains at the rate of $18$ milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?

知识点：速率数据与图表解读

难度评级：820

小提示：

溢流前，水量以恒定的正速率增加。

Before overflow, the volume rises at a constant positive rate.

大提示：

鸟浴盆装满后，多余的水会溢出，所以水量保持不变。

After the birdbath is full, extra water overflows so the volume stays constant.

解答：

鸟浴盆溢流前，每分钟净增加 $20-18=2$ 毫升水，所以水量稳定上升。

鸟浴盆装满后，流入的多余水会溢出，所以水量保持不变。只有图 A 显示先上升再水平的图线。

所以正确答案是 A。

Before the birdbath overflows, it gains $20-18=2$ milliliters of water per minute, so its volume increases steadily.

Once the birdbath is full, the incoming extra water overflows, so the volume remains constant. Only graph A shows an increasing line followed by a horizontal line.

Thus, A is the correct answer.

7.

Sawyer 老师班上的学生被要求对五种糖果进行口味测试。每个学生选择一种糖果。图中显示了他们偏好的条形图。她班上选择糖果 $E$ 的学生占百分之多少？

The students in Mrs. Sawyer’s class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy $E?$

$5$

$12$

$15$

$16$

$20$

知识点：数据与图表解读百分数

难度评级：860

小提示：

将五个条形的高度相加得到班级人数。

Add the heights of all five bars to get the class size.

大提示：

比较 $E$ 的条形高度和学生总数。

Compare the height of bar $E$ with the total number of students.

解答：

班上共有 $6 + 8 + 4 + 2 + 5 = 25$ 名学生。选择 $E$ 的百分比为 $100 \cdot \dfrac{5}{25} = \dfrac{100}{5} = 20 \%\text{。}$

所以正确答案是 E。

There are a total of $6 + 8 + 4 + 2 + 5 = 25$ students in the class. The percent that chose $E$ is $100 \cdot \dfrac{5}{25} = \dfrac{100}{5} = 20 \%.$

Thus, E is the correct answer.

8.

第 $8$ 、 $9$ 和 $10$ 题使用随附段落和表格中的数据。

Juan 的旧集邮天地

Juan 按国家和邮票发行年代整理他的收藏。他在邮票店购买这些邮票的价格为：巴西和法国每张 $6$ ¢，秘鲁每张 $4$ ¢，西班牙每张 $5$ ¢。（巴西和秘鲁是南美国家，法国和西班牙在欧洲。）

各年代邮票数量

他的欧洲邮票中，有多少张是在 $80$ 年代发行的？

Problems $8,$ $9,$ and $10$ use the data found in the accompanying paragraph and table.

Juan’s Old Stamping Grounds

Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, $6$ ¢ each, Peru $4$ ¢ each, and Spain $5$ ¢ each. (Brazil and Peru are South American countries and France and Spain are in Europe.)

Number of Stamps by Decade

How many of his European stamps were issued in the ‘ $80$ s?

$9$

$15$

$18$

$24$

$42$

知识点：数据与图表解读

难度评级：890

小提示：

欧洲指表中的法国和西班牙。

European means France and Spain in the table.

大提示：

只使用 $80$ 年代那一列的条目。

Use only the entries in the ‘ $80$ s column.

解答：

法国和西班牙是欧洲国家。这两个国家在 $80$ 年代的邮票数分别是 $15$ 和 $9$ ，总计 $15 + 9 = 24$ 张。

所以正确答案是 D。

Note that France and Spain are the European countries. The number of ‘ $80$ s stamps from these countries respectively is $15$ and $9$ for a total of $15 + 9 = 24$ stamps.

Thus, D is the correct answer.

9.

他在 $70$ 年代以前发行的南美邮票花费了

His South American stamps issued before the ‘ $70$ s cost him

$\$0.40$

$\$1.06$

$\$1.80$

$\$2.38$

$\$2.64$

知识点：数据与图表解读钱币

难度评级：1010

小提示：

南美指巴西和秘鲁。

South American means Brazil and Peru.

大提示：

$70$ 年代以前指 $50$ 年代和 $60$ 年代两列。

Before the ‘ $70$ s means the ‘ $50$ s and ‘ $60$ s columns.

解答：

巴西和秘鲁是南美国家。

巴西在 $70$ 年代以前有 $4+7=11$ 张邮票，费用为 $11\cdot6=66$ 美分。秘鲁有 $6+4=10$ 张这样的邮票，费用为 $10\cdot4=40$ 美分。

总费用为 $66+40=106$ 美分，也就是 $\$1.06$ 。

所以正确答案是 B。

Brazil and Peru are the South American countries.

Brazil has $4+7=11$ stamps before the ‘ $70$ s, costing $11\cdot6=66$ cents. Peru has $6+4=10$ such stamps, costing $10\cdot4=40$ cents.

The total cost is $66+40=106$ cents, or $\$1.06.$

Thus, B is the correct answer.

10.

他 $70$ 年代邮票的平均价格最接近

The average price of his ‘ $70$ s stamps is closest to

$3.5$ ¢

$4$ ¢

$4.5$ ¢

$5$ ¢

$5.5$ ¢

知识点：数据与图表解读平均数

难度评级：1070

小提示：

计算所有 $70$ 年代邮票的总费用。

Compute the total cost of all ‘ $70$ s stamps.

大提示：

用总费用除以 $70$ 年代邮票数量，再选择最接近的选项。

Divide the total cost by the number of ‘ $70$ s stamps, then choose the closest option.

解答：

$70$ 年代邮票的费用为 $\begin{aligned} &12\cdot6+12\cdot6 \\ &\quad {}+6\cdot4+13\cdot5 \\ &\quad {}=72+72+24+65 \\ &\quad {}=233 \end{aligned}$ 美分。

邮票共有 $12+12+6+13=43$ 张，所以平均费用为 $233\div43$ ，略大于 $5$ 美分，最接近 $5.5$ 美分。

所以正确答案是 E。

The ‘ $70$ s stamps cost $\begin{aligned} &12\cdot6+12\cdot6 \\ &\quad {}+6\cdot4+13\cdot5 \\ &\quad {}=72+72+24+65 \\ &\quad {}=233 \end{aligned}$ cents.

There are $12+12+6+13=43$ stamps, so the average cost is $233\div43,$ a little more than $5$ cents and closest to $5.5$ cents.

Thus, E is the correct answer.

11.

一列正方形由相同的正方形瓷砖组成。每个正方形的边长比前一个正方形多一块瓷砖长度。图中显示了前三个正方形。第七个正方形比第六个正方形多需要多少块瓷砖？

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

$11$

$12$

$13$

$14$

$15$

知识点：完全平方数找规律

难度评级：1060

小提示：

第六个和第七个正方形边长分别为 $6$ 和 $7$ 。

The sixth and seventh squares have side lengths $6$ and $7.$

大提示：

比较 $7^2$ 和 $6^2$ 。

Compare $7^2$ and $6^2.$

解答：

第六个和第七个正方形的边长分别相当于 $6$ 块和 $7$ 块瓷砖的边长。

它们分别使用 $6^2=36$ 和 $7^2=49$ 块瓷砖，所以第七个正方形多需要 $49-36=13$ 块瓷砖。

所以正确答案是 C。

The sixth and seventh squares have side lengths $6$ and $7$ tile lengths.

They use $6^2=36$ and $7^2=49$ tiles, respectively, so the seventh square requires $49-36=13$ more tiles.

Thus, C is the correct answer.

12.

一个桌游转盘分成标有 $A$ 、 $B$ 、 $C$ 的三个区域。指针停在区域 $A$ 的概率是 $\frac{1}{3}$ ，停在区域 $B$ 的概率是 $\frac{1}{2}$ 。指针停在区域 $C$ 的概率是

A board game spinner is divided into three regions labeled $A,$ $B$ and $C.$ The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}.$ The probability of the arrow stopping on region $C$ is

$\dfrac{1}{12}$

$\dfrac{1}{6}$

$\dfrac{1}{5}$

$\dfrac{1}{3}$

$\dfrac{2}{5}$

知识点：基本概率对立事件概率

难度评级：1090

小提示：

$A$ 、 $B$ 和 $C$ 的概率和为 $1$ 。

The probabilities of $A,$ $B,$ and $C$ add to $1.$

大提示：

从 $1$ 中减去 $A$ 和 $B$ 的概率。

Subtract the probabilities for $A$ and $B$ from $1.$

解答：

总概率为 $1$ 。要得到 $C$ 的概率，需要减去落在 $A$ 和 $B$ 的概率： $1 - \dfrac{1}{3} - \dfrac{1}{2} = \dfrac{1}{6}\text{。}$

所以正确答案是 B。

The total probability is $1.$ We need to subtract the probability of the spinner landing on $B$ and $A$ to get $C,$ which is $1 - \dfrac{1}{3} - \dfrac{1}{2} = \dfrac{1}{6}.$

Thus, B is the correct answer.

13.

Bert 生日时得到一个盒子，装满时能容纳 $125$ 颗糖豆。几周后，Carrie 得到一个装满糖豆的更大盒子。她的盒子在高、宽、长三个方向上都是 Bert 盒子的两倍。Carrie 大约得到了多少颗糖豆？

For his birthday, Bert gets a box that holds $125$ jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert’s. Approximately, how many jellybeans did Carrie get?

$250$

$500$

$625$

$750$

$1000$

知识点：体积长度、面积与体积的缩放关系

难度评级：1140

小提示：

长、宽、高都加倍，体积乘以 $2\cdot2\cdot2$ 。

Doubling length, width, and height multiplies volume by $2\cdot2\cdot2.$

大提示：

按体积倍数放大糖豆数量。

Scale the number of jellybeans by the volume factor.

解答：

更大的盒子大约能装 $125 \cdot 2 \cdot 2 \cdot 2 = 1000$ 颗糖豆。

所以正确答案是 E。

The larger box will have approximately $125 \cdot 2 \cdot 2 \cdot 2 = 1000$ jellybeans.

Thus, E is the correct answer.

14.

一位商人把一大批商品降价 $30\%$ 。后来，商人又在促销价基础上降价 $20\%$ ，并声称这些商品的最终价格比原价低 $50\%$ 。总折扣实际是

A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices and claims that the final price of these items is $50\%$ off the original price. The total discount is

$35\%$

$44\%$

$50\%$

$56\%$

$60\%$

知识点：百分数

难度评级：1180

小提示：

降价 $30\%$ 后，剩下原价的 $70\%$ 。

After a $30\%$ discount, $70\%$ of the original price remains.

大提示：

第二次折扣是促销价的 $20\%$ ，不是原价的同样比例。

The second discount is $20\%$ off the sale price, not the original price.

解答：

设原价为 $x$ 。降价 $30\%$ 后，价格是 $0.70x$ 。

再从该促销价上减去 $20\%$ ，剩下 $0.80\cdot0.70x=0.56x$ 。

顾客支付原价的 $56\%$ ，所以总折扣是 $44\%$ 。

所以正确答案是 B。

Let the original price be $x.$ After the $30\%$ discount, the price is $0.70x.$

Taking another $20\%$ off that sale price leaves $0.80\cdot0.70x=0.56x.$

The customer pays $56\%$ of the original price, so the total discount is $44\%.$

Thus, B is the correct answer.

15.

下列哪个多边形面积最大？

Which of the following polygons has the largest area?

$\text{A}$

$\text{B}$

$\text{C}$

$\text{D}$

$\text{E}$

知识点：面积格点

难度评级：1290

小提示：

把每个多边形分成单位正方形和半单位三角形。

Break each polygon into unit squares and half-unit triangles.

大提示：

通过数整方格和半方格来比较五个面积。

Compare the five areas by counting full squares plus half-squares.

解答：

每个多边形包含的方格数可通过把多边形分成单位正方形和边长为 $1$ 的直角三角形，再相加得到。

单位正方形面积算作 $1$ ，三角形面积算作 $0.5$ 。

$A$ 的总面积为 $5$ ， $B$ 为 $5$ ， $C$ 为 $5$ ， $D$ 为 $4.5$ ， $E$ 为 $5.5$ 。

所以正确答案是 E。

The number of boxes enclosed by each polygon can be obtained by dividing the polygon into unit squares and right triangles with sidelength $1$ and adding up their values.

The unit squares count as $1$ and the triangles count as $0.5.$

$A$ has a total area of $5,$ $B$ has $5,$ $C$ has $5,$ $D$ has $4.5,$ and $E$ has $5.5.$

Thus, E is the correct answer.

16.

如图，在一个 $3-4-5$ 直角三角形的各边上分别构造等腰直角三角形。大写字母表示每个三角形的面积。下列哪一项正确？

Right isosceles triangles are constructed on the sides of a $3-4-5$ right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?

$X + Z = W + Y$

$W + X = Z$

$3X + 4Y = 5Z$

$X + W = \dfrac{1}{2}(Y + Z)$

$X + Y = Z$

知识点：勾股定理面积

难度评级：1310

小提示：

每个等腰直角三角形的面积是其所在边长平方的一半。

Each right isosceles triangle has area half the square of the side it is built on.

大提示：

计算建在边长 $3$ 、 $4$ 和 $5$ 上的面积。

Compute the areas on the sides $3,$ $4,$ and $5.$

解答：

对于建在边长为 $s$ 的边上的等腰直角三角形，两条直角边长为 $s$ ，所以面积为 $\frac{s^2}{2}$ 。

$\begin{aligned} &W=\frac{3\cdot4}{2}=6, \\ &\quad X=\frac{3^2}{2}=4.5, \\ &\quad Y=\frac{4^2}{2}=8, \\ &\quad Z=\frac{5^2}{2}=12.5 \end{aligned}\text{。}$

这些值满足 $X+Y=Z$ ，而其他列出的等式不成立。

所以正确答案是 E。

For a right isosceles triangle built on a side of length $s,$ the two legs have length $s,$ so its area is $\frac{s^2}{2}.$

$\begin{aligned} &W=\frac{3\cdot4}{2}=6, \\ &\quad X=\frac{3^2}{2}=4.5, \\ &\quad Y=\frac{4^2}{2}=8, \\ &\quad Z=\frac{5^2}{2}=12.5. \end{aligned}$

These values satisfy $X+Y=Z,$ and the other listed equations do not.

Thus, E is the correct answer.

17.

在一场有十道题的数学竞赛中，学生答对一题得 $5$ 分，答错一题扣 $2$ 分。Olivia 回答了所有题，得分为 $29$ 。她答对了多少题？

In a mathematics contest with ten problems, a student gains $5$ points for a correct answer and loses $2$ points for an incorrect answer. If Olivia answered every problem and her score was $29,$ how many correct answers did she have?

$5$

$6$

$7$

$8$

$9$

知识点：一次方程

难度评级：1290

小提示：

设 $x$ 为答对题数。

Let $x$ be the number of correct answers.

大提示：

那么答错题数是 $10-x$ 。

Then the number of incorrect answers is $10-x.$

解答：

设 $x$ 为答对题数。那么她答错了 $10 - x$ 题。

她的总分为 $5x - 2(10 - x) = 7x - 20\text{。}$

该值等于 $29$ ，解得 $7x - 20 = 29$ $x = 7\text{。}$

所以正确答案是 C。

Let $x$ be the number of correct answers. Then she answered $10 - x$ questions incorrectly.

This gives her a total score of $5x - 2(10 - x) = 7x - 20.$

We know that this equals $29,$ and solving yields $7x - 20 = 29$ $x = 7.$

Thus, C is the correct answer.

18.

Gage 连续 $5$ 天每天滑冰 $1$ 小时 $15$ 分钟，又连续 $3$ 天每天滑冰 $1$ 小时 $30$ 分钟。为了让整个九天期间平均每天滑冰 $85$ 分钟，他第九天需要滑冰多久？

Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?

$1$ 小时

$1$ hr

$1$ 小时 $10$ 分钟

$1$ hr $10$ min

$1$ 小时 $20$ 分钟

$1$ hr $20$ min

$1$ 小时 $40$ 分钟

$1$ hr $40$ min

$2$ 小时

$2$ hr

知识点：平均数单位换算

难度评级：1360

小提示：

把所有滑冰时间都换算成分钟。

Convert all skating times to minutes.

大提示：

求 $9$ 天平均 $85$ 分钟所需的总分钟数。

Find the total minutes needed for a $9$ -day average of $85.$

解答：

Gage 已经滑冰的总时间为 $5 \cdot 75 + 3 \cdot 90 = 645$ 分钟。

若平均每天滑冰 $85$ 分钟，那么 $9$ 天总共需要滑冰 $85 \cdot 9 = 765$ 分钟。

因此 Gage 最后一天需要滑冰 $765 - 645 = 120$ 分钟。 $120$ 分钟等于 $2$ 小时。

所以正确答案是 E。

Gage has skated a total of $5 \cdot 75 + 3 \cdot 90 = 645$ minutes.

For an average of $85$ minutes over $9$ days, Gage must have skated a total of $85 \cdot 9 = 765$ minutes.

This means that Gage must skate $765 - 645 = 120$ minutes on the last day. Note that $120$ minutes is the same as $2$ hours.

Thus, E is the correct answer.

19.

$99$ 和 $999$ 之间有多少个整数恰好含有一个数字 $0$ ？

How many whole numbers between $99$ and $999$ contain exactly one $0?$

$72$

$90$

$144$

$162$

$180$

知识点：数字分类讨论

难度评级：1390

小提示：

这个数是三位数。

The number is a three-digit number.

大提示：

唯一的 $0$ 可以在十位或个位，但不能在百位。

The single $0$ can be in the tens place or the ones place, but not the hundreds place.

解答：

数字 $0$ 可以在十位或个位，这有 $2$ 种选择。

其他每一位各有 $9$ 种选择，所以共有 $2 \cdot 9 \cdot 9 = 162$ 个数。

所以正确答案是 D。

Note that the $0$ digit can either be the tens or the units digit. This gives us $2$ options for this.

There are $9$ options for each of the other digits for a total of $2 \cdot 9 \cdot 9 = 162$ numbers.

Thus, D is the correct answer.

20.

三角形 $XYZ$ 的面积是 $8$ 平方英寸。点 $A$ 和 $B$ 分别是全等线段 $\overline{XY}$ 和 $\overline{XZ}$ 的中点。高 $\overline{XC}$ 平分 $\overline{YZ}$ 。阴影区域的面积（平方英寸）是

The area of triangle $XYZ$ is $8$ square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}.$ Altitude $\overline{XC}$ bisects $\overline{YZ}.$ The area (in square inches) of the shaded region is

$1 \frac{1}{2}$

$2$

$2 \frac{1}{2}$

$3$

$3 \frac{1}{2}$

知识点：相似面积比中点

难度评级：1450

小提示：

高把这个等腰三角形分成两个相等的部分。

The altitude splits the isosceles triangle into two equal halves.

大提示：

过中点的线段形成边长为一半的相似三角形。

The segment through the midpoints creates a similar triangle with half the side lengths.

解答：

因为 $XY=XZ$ ，且 $XC$ 平分 $YZ$ ，所以高 $XC$ 将 $\triangle XYZ$ 分成两个全等三角形。左半部分 $\triangle XYC$ 的面积为 $4$ 。

在 $\triangle XYC$ 中，点 $A$ 是 $XY$ 的中点。过 $A$ 的水平线段与 $XC$ 在半高处相交，因此上方小三角形与 $\triangle XYC$ 相似，比例因子为 $\frac{1}{2}$ 。它的面积是原面积的 $\frac{1}{4}$ ，即 $4$ 的四分之一，也就是 $1$ 。

阴影区域是左半部分剩余的面积，所以面积为 $4-1=3$ 。

所以正确答案是 D。

Since $XY=XZ$ and $XC$ bisects $YZ,$ altitude $XC$ splits $\triangle XYZ$ into two congruent triangles. The left half $\triangle XYC$ has area $4.$

In $\triangle XYC,$ point $A$ is the midpoint of $XY.$ The horizontal segment through $A$ meets $XC$ halfway up, so the small top triangle is similar to $\triangle XYC$ with scale factor $\frac{1}{2}.$ Its area is therefore $\frac{1}{4}$ of $4,$ or $1.$

The shaded region is the rest of the left half, so its area is $4-1=3.$

Thus, D is the correct answer.

21.

Harold 将一枚五美分硬币抛四次。他得到的正面数至少和反面数一样多的概率是

Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is

$\dfrac{5}{16}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{11}{16}$

知识点：二项概率组合

难度评级：1550

小提示：

正面数至少和反面数一样多，表示有 $2$ 、 $3$ 或 $4$ 个正面。

At least as many heads as tails means $2,$ $3,$ or $4$ heads.

大提示：

按选择哪些次抛掷为正面来计数。

Count outcomes by choosing which tosses are heads.

解答：

四次抛硬币共有 $2^4=16$ 个等可能结果。

正面数至少和反面数一样多，意味着得到 $2$ 、 $3$ 或 $4$ 个正面。这样的结果数为 $\begin{aligned} &\binom{4}{2}+\binom{4}{3}+\binom{4}{4} \\ &\quad {}=6+4+1=11 \end{aligned}\text{。}$

因此概率为 $\dfrac{11}{16}$ 。

所以正确答案是 E。

There are $2^4=16$ equally likely outcomes for four coin tosses.

At least as many heads as tails means getting $2,$ $3,$ or $4$ heads. The number of such outcomes is $\begin{aligned} &\binom{4}{2}+\binom{4}{3}+\binom{4}{4} \\ &\quad {}=6+4+1=11. \end{aligned}$

Thus the probability is $\dfrac{11}{16}.$

Thus, E is the correct answer.

22.

六个棱长为一英寸的立方体如图固定在一起。求总表面积，单位为平方英寸。包括顶面、底面和侧面。

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.

$18$

$24$

$26$

$30$

$36$

知识点：表面积立体几何

难度评级：1550

小提示：

从 $6$ 个分开的立方体开始，每个有 $6$ 个暴露面。

Start with $6$ separate cubes, each with $6$ exposed faces.

大提示：

每对粘在一起的面会减少两个暴露面。

Each glued pair of faces removes two exposed faces.

解答：

可以数不暴露的面，从而求出有多少个面贡献表面积。

有三个立方体各有 $1$ 个面不暴露，两个立方体各有 $2$ 个面不暴露，一个立方体有 $3$ 个面不暴露。

这样不暴露的面共有 $3 \cdot 1 + 2 \cdot 2 + 1 \cdot 3 = 10$ 个，于是暴露的面共有 $6 \cdot 6 - 10 = 36 - 10$ $= 26$ 个。

每个暴露面贡献的面积为 $1^2 = 1$ ，所以总表面积为 $1 \cdot 26 = 26$ 。

所以正确答案是 C。

We can count the number of unexposed faces to find how many faces contribute to the surface area.

Three cubes have $1$ face unexposed, two cubes have $2$ faces unexposed, and one cube has $3$ faces unexposed.

This gives us a total of $3 \cdot 1 + 2 \cdot 2 + 1 \cdot 3 = 10$ unexposed faces, which gives us $6 \cdot 6 - 10 = 36 - 10$ $= 26$ exposed faces.

Each exposed face contributes $1^2 = 1$ to the surface area, for a total surface area of $1 \cdot 26 = 26.$

Thus, C is the correct answer.

23.

图中显示了瓷砖地板的一个角。如果整个地板都按这种方式铺设，并且四个角都像这个角一样，那么较深色瓷砖占整个瓷砖地板的几分之几？

A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?

$\dfrac{1}3$

$\dfrac{4}9$

$\dfrac{1}2$

$\dfrac{5}9$

$\dfrac{5}8$

知识点：铺砖面积比

难度评级：1580

小提示：

寻找重复的 $3\times3$ 方块。

Look for a repeating $3\times3$ block.

大提示：

在一个方块中，把两块半深色瓷砖拼成一块完整的深色瓷砖。

In one block, combine pairs of half dark squares into whole dark squares.

解答：

注意有重复的 $3 \times 3$ 区域，并且具有相同图案（可能旋转方向不同）。

在这个区域中，有三个深色单位正方形，另有两个深色三角形合起来形成一个单位正方形。

因此深色区域面积为 $4$ ，整个区域面积为 $9$ 。所求分数为 $\dfrac{4}{9}$ 。

所以正确答案是 B。

Notice that there are repeating $3 \times 3$ regions with the same pattern (they might be rotated differently).

In this region, there are three dark unit squares and two dark triangles that combine to form another unit square.

This makes the area of the darker region $4$ and the whole region $9.$ The desired fraction is then $\dfrac{4}{9}.$

Thus, B is the correct answer.

24.

Miki 有一打大小相同的橙子和一打大小相同的梨。Miki 用榨汁机从 $3$ 个梨中榨出 $8$ 盎司梨汁，从 $2$ 个橙子中榨出 $8$ 盎司橙汁。她用相同数量的梨和橙子制作梨橙混合果汁。混合果汁中梨汁占百分之多少？

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract $8$ ounces of pear juice from $3$ pears and $8$ ounces of orange juice from $2$ oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

$30$

$40$

$50$

$60$

$70$

知识点：混合问题百分数

难度评级：1610

小提示：

选择数量相同且便于计算的梨和橙子。

Use an equal convenient number of pears and oranges.

大提示：

六个梨和六个橙子会让两种果汁量都容易计算。

Six pears and six oranges make both juice amounts easy to compute.

解答：

使用 $6$ 个梨和 $6$ 个橙子，这样两种水果数量相同。

因为 $3$ 个梨榨出 $8$ 盎司， $6$ 个梨榨出 $16$ 盎司。因为 $2$ 个橙子榨出 $8$ 盎司， $6$ 个橙子榨出 $24$ 盎司。

混合果汁总量为 $16+24=40$ 盎司，其中 $16$ 盎司是梨汁。梨汁百分比为 $\frac{16}{40}=40\%$ 。

所以正确答案是 B。

Use $6$ pears and $6$ oranges, which is an equal number of each fruit.

Since $3$ pears make $8$ ounces, $6$ pears make $16$ ounces. Since $2$ oranges make $8$ ounces, $6$ oranges make $24$ ounces.

The blend has $16+24=40$ ounces total, of which $16$ ounces is pear juice. The pear-juice percent is $\frac{16}{40}=40\%.$

Thus, B is the correct answer.

25.

Loki、Moe、Nick 和 Ott 是好朋友。Ott 没有钱，但其他人有钱。Moe 给了 Ott 自己钱数的五分之一，Loki 给了 Ott 自己钱数的四分之一，Nick 给了 Ott 自己钱数的三分之一。每个人给 Ott 的钱数相同。现在 Ott 拥有这群人总钱数的几分之几？

Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group’s money does Ott now have?

$\dfrac{1}{10}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

知识点：分数逆推法

难度评级：1560

小提示：

假设每个朋友都给 Ott 相同且便于计算的金额。

Assume each friend gives Ott the same convenient amount.

大提示：

如果每人给 Ott $\$1$ ，反推出每个朋友原来的钱数。

If each gives Ott $\$1,$ work backward to each friend’s original amount.

解答：

因为只有各人所给钱数占原有钱数的比例重要，所以假设每个人都给 Ott $\$1$ 。这意味着 Moe 原来有 $\$5$ ，Loki 原来有 $\$4$ ，Nick 原来有 $\$3$ 。

转账不会改变这群人的总钱数，即 $\$5 + \$4 + \$3 = \$12\text{。}$ Ott 现在有 $\$3$ 。

因此 Ott 拥有这群人总钱数的 $\dfrac{3}{12} = \dfrac{1}{4}\text{，}$ 也就是四分之一。

所以正确答案是 B。

Because only the fractions matter, suppose each person gave Ott $\$1.$ This means that Moe had $\$5,$ Loki had $\$4,$ and Nick had $\$3$ originally.

The transfers do not change the group’s total amount of money, which is $\$5 + \$4 + \$3 = \$12.$ Ott now has $\$3.$

This means that Ott has $\dfrac{3}{12} = \dfrac{1}{4}$ of the group’s money.

Thus, B is the correct answer.