2005 年 AMC 8 真题（P版）

1 答案 B

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。

English

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 答案 C

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。

English

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 答案 D

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。

English

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 答案 C

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。

English

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 答案 B

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。

English

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 答案 C

\(2.00d5\) 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。

English

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 答案 B

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。

English

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 答案 E

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。

English

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 答案 D

\(\triangle ADC\) 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。

English

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 答案 D

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。

English

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 答案 C

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。

English

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 答案 D

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。

English

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 答案 C

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。

English

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 答案 B

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。

English

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 答案 C

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。

English

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 答案 D

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。

English

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 答案 E

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。

English

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 答案 C

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。

English

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 答案 A

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。

English

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 答案 A

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。

English

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 答案 C

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。

English

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 答案 E

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。

English

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 答案 B

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。

English

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 答案 B

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。

English

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 答案 A

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。

English

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.