2002 年 AMC 8 真题（P版）

1 答案 D

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。

English

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 答案 A

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。

English

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 答案 C

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。

English

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 答案 B

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。

English

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 答案 C

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。

English

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 答案 A

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。

English

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 答案 E

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。

English

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 答案 D

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。

English

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 答案 B

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。

English

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 答案 E

\(70\) 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。

English

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 答案 C

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。

English

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 答案 B

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。

English

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 答案 E

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。

English

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

Thus, E is the correct answer.

14 答案 B

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。

English

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 答案 E

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。

English

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 答案 E

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。

English

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 答案 C

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。

English

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 答案 E

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。

English

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 答案 D

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。

English

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 答案 D

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。

English

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 答案 E

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。

English

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 答案 C

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。

English

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 答案 B

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。

English

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 答案 B

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。

English

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 答案 B

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。

English

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.