2000 年 AMC 8 真题（P版）

1 答案 B

Brianna 的年龄是 \(42 \div 2 = 21\) 岁。因此 Caitlin 是 \(21 - 5 = 16\) 岁。

所以正确答案是 B。

English

Brianna is \(42 \div 2 = 21\) years old. Caitlin is therefore \(21 - 5 = 16\) years old.

Thus, B is the correct answer.

2 答案 A

\(0\) 没有倒数，\(1\) 和 \(-1\) 的倒数都是它们本身。

\(2\) 的倒数是 \(\frac{1}{2}\)，但 \(2\) 不小于 \(\frac{1}{2}\)。

因为 \(-2 < -\frac 12\)，所以 \(-2\) 是唯一小于其倒数的选项。

所以正确答案是 A。

English

\(0\) has no reciprocal, and \(1\) and \(-1\) are their own reciprocals.

The reciprocal of \(2\) is \(\frac{1}{2},\) but \(2\) is not less than \(\frac{1}{2}.\)

Therefore, as \(-2 < -\frac 12,\) we know that \(-2\) is the only one of the answer choices that is less than its reciprocal.

Thus, A is the correct answer.

3 答案 D

大于 \(\frac{5}{3}\) 的最小整数是 \(2\)。小于 \(2\pi\) 的最大整数是 \(6\)。

在这个范围内的整数是 \[ 2, 3, 4, 5, 6\text{。}\]

所以正确答案是 D。

English

The smallest whole number greater than \(\frac{5}{3}\) is \(2.\) The greatest whole number less than \(2\pi\) is \(6.\)

The whole numbers within this range are \[ 2, 3, 4, 5, 6. \]

Thus, D is the correct answer.

4 答案 E

唯一显示所有数据点的图是图 E。

所以正确答案是 E。

English

The only graph that shows all the data points is graph E .

Thus, E is the correct answer.

5 答案 C

为了让校长人数最多，假设这段时间的第一年是某位校长任期的最后一年。

接着可以有 \(2\) 位校长完成接下来的 \(6\) 年，然后再有一位校长任最后一年。

总共是 \(4\) 位校长。

所以正确答案是 C。

English

To maximize the number of principals, assume that the first year of this period is the final year of some principal’s term.

Then, there can be \(2\) more principals for \(6\) years, followed by another principal who works the final year.

This is \(4\) principals.

Thus, C is the correct answer.

6 答案 A

可以从大正方形中减去上方的单位正方形、右下角的单位正方形，以及右上方的 \(4 \times 4\) 正方形。

\(ABCD\) 的阴影面积为 \[ 5^2 - 2 \cdot 1^2 - 4^2 = 7\text{。}\]

所以正确答案是 A。

English

We can subtract out the areas of the top unit square, the bottom right unit square, and the top right \(4 \times 4\) square.

The shaded area in \(ABCD\) is therefore \[ 5^2 - 2 \cdot 1^2 - 4^2 = 7. \]

Thus, A is the correct answer.

7 答案 B

负乘积可能来自三个负因数，也可能来自一个负因数和两个正因数。

选三个负因数时，最小乘积为 \[ -8 \cdot -6 \cdot -4 = -192\text{。}\]选一个负因数时，应选最小的负数和最大的两个正数，得到 \[ -8 \cdot 5 \cdot 7 = -280\text{。}\]因为 \(-280<-192\)，所以最小可能乘积是 \(-280\)。

所以正确答案是 B。

English

A negative product comes either from three negative factors or from one negative factor and two positive factors.

With three negative factors, the minimum is \[ -8 \cdot -6 \cdot -4 = -192. \] With one negative factor, use the most negative number and the two largest positive numbers to get \[ -8 \cdot 5 \cdot 7 = -280. \] Since \(-280<-192,\) the minimum possible product is \(-280.\)

Thus, B is the correct answer.

8 答案 D

一个骰子各面的点数和为 \[ 1 + 2 + 3 + 4 + 5 + 6 = 21\text{。}\]因此 \(3\) 个骰子的总点数是 \(3 \cdot 21 = 63\)。

可见点数之和为 \[ 1 + 1 + 2 + 3 + 4 + 5 + 6 = 22\text{。}\]所以不可见点数之和是 \(63 - 22 = 41\)。

所以正确答案是 D。

English

The sum of the numbers on one die is \[ 1 + 2 + 3 + 4 + 5 + 6 = 21. \] Therefore, the sum of the numbers on all \(3\) dice is \(3 \cdot 21 = 63.\)

The visible numbers add up to \[ 1 + 1 + 2 + 3 + 4 + 5 + 6 = 22. \] This makes the sum of the unseen numbers \(63 - 22 = 41.\)

Thus, D is the correct answer.

9 答案 D

\(3\) 位数的 \(5\) 的幂只有 \(125\) 和 \(625\)。这说明标为 \(2\) 的位置填的是 \(2\)。

唯一的 \(3\) 位数且以 \(2\) 开头的 \(2\) 的幂是 \(256\)，所以带框方格填 \(6\)。

所以正确答案是 D。

English

The only \(3\)-digit powers of \(5\) are \(125\) and \(625.\) This means that the \(2\) spot is filled with a \(2.\)

The only \(3\)-digit power of \(2\) beginning with a \(2\) is \(256,\) so the outlined square is filled with a \(6.\)

Thus, D is the correct answer.

10 答案 E

设 Ara 和 Shea 原来的身高为 \(x\)。则 \[ \begin{align*} 1.2x &= 60 \\ x &= 50 \end{align*}\text{。}\]

Shea 长高了 \(10\) 英寸，所以 Ara 长高了 \(10 \div 2 = 5\) 英寸，她现在高 \(50 + 5 = 55\) 英寸。

所以正确答案是 E。

English

Let \(x\) be Ara and Shea’s initial height. Then we get that \[ \begin{align*} 1.2x &= 60 \\ x &= 50. \end{align*} \]

This means that Shea grew \(10\) inches, which means that Ara grew \(10 \div 2 = 5\) inches, making her \(50 + 5 = 55\) inches tall.

Thus, E is the correct answer.

11 答案 C

以 \(1\)、\(2\) 或 \(5\) 结尾的数都满足条件，分别为 \(11\)、\(21\)、\(31\)、\(41\)；\(12\)、\(22\)、\(32\)、\(42\)；以及 \(15\)、\(25\)、\(35\)、\(45\)。这共有 \(12\) 个数。

其余满足条件的数是 \(24\)、\(33\)、\(36\)、\(44\)、\(48\)。以 \(0\) 结尾的数不行，因为除以 \(0\) 没有定义。

因此共有 \(12+5=17\) 个这样的数。

所以正确答案是 C。

English

Numbers ending in \(1,\) \(2,\) or \(5\) all work in the lists \(11,\) \(21,\) \(31,\) and \(41;\) \(12,\) \(22,\) \(32,\) and \(42;\) and \(15,\) \(25,\) \(35,\) and \(45.\) This gives \(12\) numbers.

The remaining working numbers are \(24,\) \(33,\) \(36,\) \(44,\) and \(48.\) Numbers ending in \(0\) do not work because division by \(0\) is undefined.

Thus there are \(12+5=17\) such numbers.

Thus, C is the correct answer.

12 答案 D

墙共有 \(7\) 行，每行高 \(1\) 英尺。

为了使用最少砌块，第 \(1\)、\(3\)、\(5\) 和 \(7\) 行可采用图中底行的模式，每行需要 \(50\) 块砌块。

第 \(2\)、\(4\) 和 \(6\) 行采用图中上行的模式，中间有 \(49\) 块 \(2\) 英尺砌块，两端各有一块 \(1\) 英尺砌块，共 \(51\) 块砌块。

将 \(4\) 行 \(50\) 块和 \(3\) 行 \(51\) 块相加，得到 \[4 \cdot 50 + 3 \cdot 51 = 200 + 153 = 353\] 块砌块。

所以正确答案是 D。

English

The total number of rows in the wall is \(7,\) with each row being \(1\) foot high.

To use the minimum number of bricks, rows \(1,\) \(3,\) \(5,\) and \(7\) will have the same pattern as the bottom row in the picture, which requires \(50\) bricks to construct.

Rows \(2,\) \(4,\) and \(6\) will have the same pattern as the upper row in the picture, which has \(49\) \(2\)-foot bricks in the middle and one \(1\)-foot brick on each end, for a total of \(51\) bricks.

When you add up \(4\) rows of \(50\) bricks and \(3\) rows of \(51\) bricks, you get a total of \[4 \cdot 50 + 3 \cdot 51 = 200 + 153 = 353\] bricks.

Thus, D is the correct answer.

13 答案 C

我们有 \[ \begin{aligned} \angle CAT + \angle ATC \\ \quad {}+ \angle ACT &= 180 \\ 36 + 2 \cdot \angle ATC &= 180 \\ \angle ATC &= 72^{\circ} \end{aligned}\text{。}\]

因为该角被平分，所以 \[\angle RTC = 72 \div 2 = 36^{\circ}\text{。}\]

最后，\[ \begin{aligned} \angle RTC + \angle TCR \\ \quad {}+ \angle CRT &= 180 \\ 36 + 72 + \angle CRT &= 180 \\ \angle CRT &= 72^{\circ} \end{aligned}\text{。}\]

所以正确答案是 C。

English

We get \[ \begin{aligned} \angle CAT + \angle ATC \\ \quad {}+ \angle ACT &= 180 \\ 36 + 2 \cdot \angle ATC &= 180 \\ \angle ATC &= 72^{\circ}. \end{aligned} \]

Due to bisection, we also know that \[\angle RTC = 72 \div 2 = 36^{\circ}.\]

Finally, we see that \[ \begin{aligned} \angle RTC + \angle TCR \\ \quad {}+ \angle CRT &= 180 \\ 36 + 72 + \angle CRT &= 180 \\ \angle CRT &= 72^{\circ}. \end{aligned} \]

Thus, C is the correct answer.

14 答案 D

幂的个位数字只取决于底数的个位数字。

观察可知，\(9\) 的偶次幂个位是 \(1\)，奇次幂个位是 \(9\)。

因此 \(19^{19}\) 个位是 \(9\)，\(99^{99}\) 个位也是 \(9\)。相加后的个位为 \(8\)。

所以正确答案是 D。

English

Note that the units digit of a power depends only upon the units digit of the base.

Experimenting, we get that \(9\) to an even power ends with a \(1\) and to an odd power ends with a \(9.\)

Therefore, \(19^{19}\) ends with a \(9\) and \(99^{99}\) also ends with a \(9.\) Adding them together yields a number that ends in \(8.\)

Thus, D is the correct answer.

15 答案 C

大等边三角形边长为 \(4\)，中等边三角形边长为 \(2\)，小等边三角形边长为 \(1\)。

周长为 \[ \begin{gather*} CB + CD + DE + EF \\ {}+ FG + GA + AB \\ = 4 + 4 + 2 + 2 + 1 \\ + 1 + 1 = 15 \end{gather*}\text{。}\]

所以正确答案是 C。

English

The large equilateral triangle has side length \(4,\) the middle one has side length \(2,\) and the smaller one has side length \(1.\)

The perimeter is therefore \[ \begin{gather*} CB + CD + DE + EF \\ {}+ FG + GA + AB \\ = 4 + 4 + 2 + 2 + 1 \\ + 1 + 1 = 15. \end{gather*} \]

Thus, C is the correct answer.

16 答案 C

长为 \(1000 \div 25 = 40\) 米，周长为 \(1000 \div 10 = 100\) 米。

周长是长与宽之和的 \(2\) 倍。

因此宽为 \[100 \div 2 - 40 = 10\] 米，面积为 \[40 \cdot 10 = 400\] 平方米。

所以正确答案是 C。

English

We can see that the length is \(1000 \div 25 = 40\) meters, and the perimeter is \(1000 \div 10 = 100\) meters.

Note that the perimeter is \(2\) times the sum of the length and width.

This means that the width is \[100 \div 2 - 40 = 10\] meters, and the area is \[40 \cdot 10 = 400\] square meters.

Thus, C is the correct answer.

17 答案 A

计算如下：\[ \begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(\frac{1}{2})^2}{3} - \dfrac{1^2}{(\frac{4}{3})} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3} \end{gather*}\text{。}\]

所以正确答案是 A。

English

We can calculate it as follows. \[ \begin{gather*} [(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)] \\ = [\dfrac{1^2}{2} \otimes 3] - [1 \otimes \dfrac{2^2}{3}] \\ = [\dfrac{1}{2} \otimes 3] - [1 \otimes \dfrac{4}{3}] \\ = \dfrac{(\frac{1}{2})^2}{3} - \dfrac{1^2}{(\frac{4}{3})} \\ = \dfrac{1}{4} \cdot \dfrac{1}{3} - \dfrac{3}{4} \\ = \dfrac{1}{12} - \dfrac{3}{4} \\= -\dfrac{2}{3}. \end{gather*} \]

Thus, A is the correct answer.

18 答案 E

假设这个网格上相邻钉点相距 \(1\) 个单位。

区域 \(I\) 是底为 \(1\)、高为 \(1\) 的平行四边形，所以面积为 \(1 \cdot 1 = 1\)。

区域 \(II\) 可以分成 \(2\) 个三角形，每个底为 \(1\)、高为 \(1\)。面积总和为 \[ 2 \cdot \dfrac{1}{2} \cdot 1 \cdot 1 = 1\text{。}\]因此两个区域面积相同。

每个区域都有 \(2\) 条长度为 \(\sqrt{2}\) 的边。区域 \(I\) 有 \(2\) 条单位边，而区域 \(II\) 只有 \(1\) 条单位边。

区域 \(II\) 的另一条边显然大于 \(1\)，所以区域 \(II\) 的周长更大。

所以正确答案是 E。

English

Assume that the pegs on this grid are separated by \(1\) unit.

Note that region \(I\) is a parallelogram with base \(1\) and height \(1,\) making its area \(1 \cdot 1 = 1.\)

We can split region \(II\) into \(2\) triangles, both with base \(1\) and height \(1.\) This makes the sum of the areas \[ 2 \cdot \dfrac{1}{2} \cdot 1 \cdot 1 = 1. \] This shows that both regions have the same area.

Note that each region has \(2\) sides that are of length \(\sqrt{2}.\) Region \(I\) has \(2\) unit sides, whereas region \(II\) only has \(1.\)

The other side of region \(II\) is clearly greater than \(1,\) which shows that region \(II\) has the greater perimeter.

Thus, E is the correct answer.

19 答案 C

如下图，作一个覆盖图形下半部分的矩形。

于是 \[ [ABCD] = [ABD] + [BCD]\text{。}\]

还知道 \[ [ABD] = [BDEF] - [ABE]\]\[ - [ADF]\text{。}\]

\(ABE\) 和 \(ADF\) 都是四分之一圆，它们合起来形成一个半圆，面积与 \(BCD\) 相同。

因此 \[ [ABD] = [BDEF] - [BCD] \] 并且 \[ [ABCD] = [BCD] + [BDEF]\]\[ - [BCD] \] \[ = [BDEF] = 10 \cdot 5 = 50\text{。}\]

所以正确答案是 C。

English

Create a rectangle that covers the bottom half of the figure as shown below.

Then, we get that \[ [ABCD] = [ABD] + [BCD]. \]

We also know that \[ [ABD] = [BDEF] - [ABE]\]\[ - [ADF]. \]

\(ABE\) and \(ADF\) are both quarter-circles that form a semicircle with the same area as \(BCD.\)

This means that \[ [ABD] = [BDEF] - [BCD] \] and \[ [ABCD] = [BCD] + [BDEF]\]\[ - [BCD] \] \[ = [BDEF] = 10 \cdot 5 = 50. \]

Thus, C is the correct answer.

20 答案 A

一美分硬币的数量必须与 \(102\) 模 \(5\) 的余数相同，所以可能有 \(2\) 枚或 \(7\) 枚一美分硬币。如果有七枚，只剩两枚硬币给五美分硬币、十美分硬币和二十五美分硬币，不可能每种至少一枚。

所以有 \(2\) 枚一美分硬币。剩下 \(7\) 枚硬币价值 \(100\) 分。设五美分硬币、十美分硬币和二十五美分硬币的数量分别为 \(n\)、\(d\)、\(q\)，则 \(n+d+q=7\)，且 \(5n+10d+25q=100\)。

将价值方程除以 \(5\)，再减去硬币数量方程，得到 \(d+4q=13\)。唯一正整数解是 \(q=3\)、\(d=1\)、\(n=3\)。

因此必须有 \(1\) 枚十美分硬币。

所以正确答案是 A。

English

The number of pennies must have the same remainder as \(102\) modulo \(5,\) so there are either \(2\) or \(7\) pennies. Seven pennies would leave only two coins for nickels, dimes, and quarters, impossible because at least one of each type is needed.

So there are \(2\) pennies. The remaining \(7\) coins are worth \(100\) cents. If \(n,\) \(d,\) and \(q\) are the numbers of nickels, dimes, and quarters, then \(n+d+q=7\) and \(5n+10d+25q=100.\)

Dividing the value equation by \(5\) and subtracting the coin-count equation gives \(d+4q=13.\) The only positive solution is \(q=3,\) \(d=1,\) and \(n=3.\)

Thus there must be \(1\) dime.

Thus, A is the correct answer.

21 答案 B

若记录 Keiko 的硬币和 Ephraim 的两枚硬币，共有 \(8\) 个等可能结果。

如果 Keiko 得到正面，Ephraim 必须恰好得到一个正面，这有 \(2\) 种结果。如果 Keiko 得到反面，Ephraim 必须没有正面，这有 \(1\) 种结果。

所以有 \(3\) 个满足条件的结果，共有 \(8\) 个等可能结果，所求概率为 \(\dfrac{3}{8}\)。

所以正确答案是 B。

English

There are \(8\) equally likely outcomes if we record Keiko’s coin and Ephraim’s two coins.

If Keiko gets heads, Ephraim must get exactly one head; this happens in \(2\) outcomes. If Keiko gets tails, Ephraim must get no heads; this happens in \(1\) outcome.

So \(3\) of the \(8\) equally likely outcomes work, and the probability is \(\dfrac{3}{8}.\)

Thus, B is the correct answer.

22 答案 C

原来的表面积为 \[ 6 \cdot 2^2 = 6 \cdot 4 = 24\text{。}\]

注意单位立方体的顶面加上大立方体顶面仍可见的部分，面积与大立方体一个面相同。

因此放在上方的单位立方体只让总表面积增加 \(4\) 个单位正方形，增加量为 \(4\)。

百分比增加为 \[ 100 \cdot \dfrac{4}{24} = 100 \cdot \dfrac{1}{6} \approx 16.7\%\text{。}\]

所以正确答案是 C。

English

The original surface area is just \[ 6 \cdot 2^2 = 6 \cdot 4 = 24. \]

Note that the top face of the unit cube plus the visible area of the top face of the larger cube is the same as the area of one face of the larger cube.

This means that the unit square on top only adds \(4\) unit squares to the total surface area, making the increase \(4.\)

The percent increase is therefore \[ 100 \cdot \dfrac{4}{24} = 100 \cdot \dfrac{1}{6} \approx 16.7\%. \]

Thus, C is the correct answer.

23 答案 B

前四个数的和是 \(4 \cdot 5 = 20\)。后四个数的和是 \(4 \cdot 8 = 32\)。

全部七个数的和是 \(7 \cdot 6\frac{4}{7} = 46\)。共同的数包含在前两个和中。

也就是说，前两个和相加会把每个数计算一次，但共同的数计算两次。

而全部七个数的和只把每个数计算一次。因此用前两个和的总和减去七个数的总和，就得到共同的数。

所以共同的数为 \[ 20 + 32 - 46 = 52 - 46 = 6\text{。}\]

所以正确答案是 B。

English

The sum of the first four numbers is \(4 \cdot 5 = 20.\) The sum of the last four numbers is \(4 \cdot 8 = 32.\)

The sum of all seven numbers is \(7 \cdot 6\frac{4}{7} = 46.\) We know that the number common to both sets is included in both of the first two sums.

This means that the sum of the first two sums includes every number once, except for the common number which is included twice.

The third sum, however, only includes every number once. This means that the sum of the first two sums minus the third sum yields our desired number.

Therefore, the common number is \[ 20 + 32 - 46 = 52 - 46 = 6. \]

Thus, B is the correct answer.

24 答案 D

在 \(\triangle AFG\) 中，两个底角相等，且 \(\angle A=20^\circ\)，所以

\[ \begin{aligned} \angle AFG &= \angle AGF \\ &= \frac{180^\circ-20^\circ}{2} \\ &= 80^\circ \end{aligned}\text{。}\]

因为 \(\angle AFG\) 和 \(\angle BFD\) 构成平角，所以 \(\angle BFD=100^\circ\)。

在 \(\triangle BFD\) 中，\(\angle B+\angle D=180^\circ-100^\circ=80^\circ\)。

所以正确答案是 D。

English

In \(\triangle AFG,\) the two base angles are equal and \(\angle A=20^\circ,\) so

\[ \begin{aligned} \angle AFG &= \angle AGF \\ &= \frac{180^\circ-20^\circ}{2} \\ &= 80^\circ. \end{aligned} \]

Since \(\angle AFG\) and \(\angle BFD\) form a straight angle, \(\angle BFD=100^\circ.\)

In \(\triangle BFD,\) \(\angle B+\angle D=180^\circ-100^\circ=80^\circ.\)

Thus, D is the correct answer.

25 答案 B

设长方形边长为 \(2a\) 和 \(2b\)，所以面积为 \(4ab=72\)，即 \(ab=18\)。

目标三角形外侧的三个直角三角形面积分别为 \(\frac12(2a)(b)=ab\)、\(\frac12(2b)(a)=ab\) 和 \(\frac12(a)(b)=\frac12ab\)。

它们的总面积为 \(\frac52ab=45\)。因此目标三角形面积为 \(72-45=27\)。

所以正确答案是 B。

English

Let the rectangle have side lengths \(2a\) and \(2b,\) so its area is \(4ab=72\) and \(ab=18.\)

The three right triangles outside the desired triangle have areas \(\frac12(2a)(b)=ab,\) \(\frac12(2b)(a)=ab,\) and \(\frac12(a)(b)=\frac12ab.\)

Their total area is \(\frac52ab=45.\) Therefore the desired triangle has area \(72-45=27.\)

Thus, B is the correct answer.