1995 年 AMC 8 真题（P版）

1 答案 D

这些硬币总共值 \(1 + 5 + 10 + 25 = 41\) 美分。

因为一美元是 \(100\) 美分，所以这相当于一美元的 \(41\%\)。

所以正确答案是 D。

English

The coins total \(1 + 5 + 10 + 25 = 41\) cents.

Since one dollar is \(100\) cents, this is \(41\%\) of a dollar.

Thus, the correct answer is D .

2 答案 C

扎克是 \(15 + 3 = 18\) 岁。

何塞比扎克小 \(4\) 岁，所以何塞是 \(18 - 4 = 14\) 岁。

所以正确答案是 C。

English

Zack is \(15 + 3 = 18\) years old.

Jose is \(4\) years younger, so Jose is \(18 - 4 = 14\).

Thus, the correct answer is C .

3 答案 E

除以 \(\dfrac35\) 等于乘以 \(\dfrac53\)，所以两个运算合起来相当于乘以

\[\dfrac34 \times \dfrac53 = \dfrac54\text{。}\]

因此总效果是乘以 \(\dfrac54\)。

所以正确答案是 E。

English

Dividing by \(\dfrac35\) is the same as multiplying by \(\dfrac53,\) so the two operations together multiply by

\[\dfrac34 \times \dfrac53 = \dfrac54.\]

So the combined effect is multiplying by \(\dfrac54\).

Thus, the correct answer is E .

4 答案 C

本计算 \((6 + 1) \times 2 = 14\)，并把 \(14\) 给苏。

苏计算 \((14 - 1) \times 2 = 26\)。

所以正确答案是 C。

English

Ben computes \((6 + 1) \times 2 = 14\) and gives \(14\) to Sue.

Sue computes \((14 - 1) \times 2 = 26\).

Thus, the correct answer is C .

5 答案 C

整数部分的和为 \(2 + 3 + 4 + 5 = 14\)。

分数部分 \(\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15\) 约为 \(1.28\)，在 \(1\) 和 \(2\) 之间。因此总和在 \(15\) 和 \(16\) 之间，大于它的最小整数是 \(16\)。

所以正确答案是 C。

English

The whole-number parts sum to \(2 + 3 + 4 + 5 = 14\).

The fractions \(\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15\) add to about \(1.28,\) which is between \(1\) and \(2.\) So the total is between \(15\) and \(16,\) and the smallest whole number larger than it is \(16\).

Thus, the correct answer is C .

6 答案 C

正方形 \(I\) 的边长是 \(12 \div 4 = 3\)，正方形 \(II\) 的边长是 \(24 \div 4 = 6\)。

从图中看，\(III\) 的边长是 \(3 + 6 = 9\)，所以它的周长是 \(4 \times 9 = 36\)。

所以正确答案是 C。

English

Square \(I\) has side \(12 \div 4 = 3,\) and square \(II\) has side \(24 \div 4 = 6.\)

From the figure, the side of \(III\) is \(3 + 6 = 9,\) so its perimeter is \(4 \times 9 = 36.\)

Thus, the correct answer is C .

7 答案 B

乘车或骑车的学生占

\[ \begin{aligned} \dfrac12 + \dfrac14 + \dfrac{1}{10} &= \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} \\ &= \dfrac{17}{20} \end{aligned}\text{。}\]

所以步行的比例是 \(1 - \dfrac{17}{20} = \dfrac{3}{20}\)。

所以正确答案是 B。

English

The students who ride make up

\[ \begin{aligned} \dfrac12 + \dfrac14 + \dfrac{1}{10} &= \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} \\ &= \dfrac{17}{20}. \end{aligned} \]

So the fraction who walk is \(1 - \dfrac{17}{20} = \dfrac{3}{20}.\)

Thus, the correct answer is B .

8 答案 D

因为 \(\$1.00\) 是 \(\$1.60\) 的 \(\dfrac{1.00}{1.60} = \dfrac58\)，所以旅行者会得到 \(3000\) 里拉的 \(\dfrac58\)。

也就是 \(\dfrac58 \times 3000 = 1875\) 里拉。

所以正确答案是 D。

English

Since \(\$1.00\) is \(\dfrac{1.00}{1.60} = \dfrac58\) of \(\$1.60,\) the traveler gets \(\dfrac58\) of \(3000\) lire.

That is \(\dfrac58 \times 3000 = 1875\) lire.

Thus, the correct answer is D .

9 答案 C

每个圆的直径为 \(4\)。长方形的短边等于一个直径，所以短边为 \(4\)。

因为以 \(Q\) 为圆心的圆经过 \(P\) 和 \(R\)，三个圆半径都是 \(2\)，长方形长边跨过两个完整直径：\(4 + 4 = 8\)。面积为 \(8 \times 4 = 32\)。

所以正确答案是 C。

English

Each circle has diameter \(4.\) The short side of the rectangle equals one diameter, so it is \(4.\)

Since the circle at \(Q\) passes through \(P\) and \(R,\) all three circles have radius \(2,\) and the long side spans two full diameters: \(4 + 4 = 8.\) The area is \(8 \times 4 = 32.\)

Thus, the correct answer is C .

10 答案 A

夹克折扣节省 \(40\%\) 的 \(\$80 = \$32\)，衬衫折扣节省 \(55\%\) 的 \(\$40 = \$22\)。总共节省 \(\$32 + \$22 = \$54\)。

原总价是 \(\$80 + \$40 = \$120\)，所以节省比例是 \(\dfrac{54}{120} = 0.45 = 45\%\)。

所以正确答案是 A。

English

The jacket discount saves \(40\%\) of \(\$80 = \$32,\) and the shirt discount saves \(55\%\) of \(\$40 = \$22.\) The total saved is \(\$32 + \$22 = \$54.\)

The original total is \(\$80 + \$40 = \$120,\) so the percent saved is \(\dfrac{54}{120} = 0.45 = 45\%.\)

Thus, the correct answer is A .

11 答案 D

该区域的周长为 \(18\) 个街区，所以简和赫克托相遇时合计走了 \(18\) 个街区。因为简的速度是赫克托的两倍，所以简走了 \(12\) 个街区，赫克托走了 \(6\) 个街区。

从底边中点出发，赫克托走 \(6\) 个街区，先到 \(E\)，再向上到 \(D\)；简走 \(12\) 个街区，先到 \(A\)，再向上到 \(B\)，最后沿上边到 \(D\)。所以他们在 \(D\) 相遇。

所以正确答案是 D。

English

The perimeter of the region is \(18\) blocks, so when Jane and Hector meet they have together walked \(18\) blocks. Since Jane walks twice as fast, she covers \(12\) blocks and Hector covers \(6.\)

Starting from the middle of the bottom edge, Hector walks \(6\) blocks (to \(E,\) then up to \(D\)), and Jane walks \(12\) blocks (to \(A,\) up to \(B,\) then across the top to \(D\)). They meet at \(D.\)

Thus, the correct answer is D .

12 答案 E

其他年份都可以：\(90 = 9 \times 10\)、\(91 = 7 \times 13\)、\(92 = 4 \times 23\) 和 \(93 = 3 \times 31\)，都对应合法的月和日。

对 \(1994\)，最后两位只能分解为 \(94 = 2 \times 47\)，而 \(47\) 作为日期太大，\(94\) 或 \(47\) 作为月份也太大。因此 \(1994\) 没有幸运日期。

所以正确答案是 E。

English

Each of the other years works: \(90 = 9 \times 10,\) \(91 = 7 \times 13,\) \(92 = 4 \times 23,\) and \(93 = 3 \times 31,\) each a valid month/day.

For \(1994,\) the last two digits factor only as \(94 = 2 \times 47,\) and \(47\) is too large for a day (and \(94\) or \(47\) is too large for a month). So \(1994\) has no lucky date.

Thus, the correct answer is E .

13 答案 E

在三角形 \(BDE\) 中，\(E\) 和 \(D\) 处的角相等，且 \(\angle B = 90^\circ\)，所以 \(\angle BED = \angle BDE = 45^\circ\)。

因此 \(\angle AED = \angle AEB + \angle BED\) \(= 40^\circ + 45^\circ\) \(= 85^\circ\)。在四边形 \(AEDC\) 中，\(A\) 和 \(C\) 处的角都是 \(90^\circ\)，所以

\[ \begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ \end{aligned}\text{。}\]

所以正确答案是 E。

English

In triangle \(BDE,\) the angles at \(E\) and \(D\) are equal and \(\angle B = 90^\circ,\) so \(\angle BED = \angle BDE = 45^\circ.\)

Then \(\angle AED = \angle AEB + \angle BED\) \(= 40^\circ + 45^\circ\) \(= 85^\circ.\) In quadrilateral \(AEDC,\) the angles at \(A\) and \(C\) are \(90^\circ,\) so

\[ \begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ. \end{aligned} \]

Thus, the correct answer is E .

14 答案 B

整个赛季共有 \(50 + 40 = 90\) 场比赛，\(90\) 的 \(70\%\) 是 \(63\) 场胜利。

球队已经赢了 \(40\) 场，所以还需要赢 \(63 - 40 = 23\) 场。

所以正确答案是 B。

English

The season has \(50 + 40 = 90\) games, and \(70\%\) of \(90\) is \(63\) wins.

The team already has \(40\) wins, so it needs \(63 - 40 = 23\) more.

Thus, the correct answer is B .

15 答案 B

\(\dfrac{4}{37} = 0.\overline{108}\)，循环节长度为 \(3\)。第 \(3\)、\(6\)、\(9\)、\(\ldots\) 位，也就是位置为 \(3\) 的倍数时，数字是 \(8\)。

因为 \(99\) 是 \(3\) 的倍数，所以第 \(99\) 位是 \(8\)，第 \(100\) 位开始下一个循环节，是 \(1\)。

所以正确答案是 B。

English

\(\dfrac{4}{37} = 0.\overline{108},\) repeating with block length \(3.\) The digits in positions \(3,\) \(6,\) \(9,\) \(\ldots\) (multiples of \(3\)) are \(8.\)

Since \(99\) is a multiple of \(3,\) the \(99\)th digit is \(8,\) so the \(100\)th digit starts the next block: it is \(1.\)

Thus, the correct answer is B .

16 答案 C

总工作量为 \(7 \times 3 + 4 \times 5 + 5 \times 9\) \(= 21 + 20 + 45\) \(= 86\) 人天。

所以每人每天的报酬为 \(\$774 \div 86 = \$9\)。巴尔博亚学生的总工作量为 \(20\) 人天，共赚得 \(20 \times \$9 = \$180\)。

所以正确答案是 C。

English

The total student-days are \(7 \times 3 + 4 \times 5 + 5 \times 9\) \(= 21 + 20 + 45\) \(= 86.\)

So each student-day pays \(\$774 \div 86 = \$9.\) Balboa worked \(20\) student-days, earning \(20 \times \$9 = \$180.\)

Thus, the correct answer is C .

17 答案 D

安维尔的六年级人数为 \(11\%\) 乘以 \(100 = 11\) 人；克利奥纳的六年级人数为 \(17\%\) 乘以 \(200 = 34\) 人。

合计为 \(11 + 34 = 45\) 人，占总人数 \(300\) 的 \(\dfrac{45}{300} = 15\%\)。

所以正确答案是 D。

English

Annville has \(11\%\) of \(100 = 11\) sixth graders, and Cleona has \(17\%\) of \(200 = 34\) sixth graders.

Combined, that is \(11 + 34 = 45\) out of \(300\) students, which is \(\dfrac{45}{300} = 15\%.\)

Thus, the correct answer is D .

18 答案 C

四个 L 形区域合计占 \(4 \times \dfrac{3}{16} = \dfrac34\) 的大正方形，所以中心正方形占剩下的 \(\dfrac14\)。

大正方形面积为 \(100 \times 100 = 10000\) 平方英寸，所以中心正方形面积为 \(\dfrac14 \times 10000 = 2500\)，边长为 \(\sqrt{2500} = 50\) 英寸。

所以正确答案是 C。

English

The four L-shaped regions cover \(4 \times \dfrac{3}{16} = \dfrac34\) of the square, so the center square is the remaining \(\dfrac14\) of the total area.

The total area is \(100 \times 100 = 10000\) square inches, so the center square has area \(\dfrac14 \times 10000 = 2500,\) and its side is \(\sqrt{2500} = 50\) inches.

Thus, the correct answer is C .

19 答案 D

图中有 \(2\) 个家庭有 \(1\) 个孩子，\(1\) 个家庭有 \(2\) 个，\(2\) 个家庭有 \(3\) 个，\(2\) 个家庭有 \(4\) 个，\(6\) 个家庭有 \(5\) 个，总共 \(2 + 1 + 2 + 2 + 6 = 13\) 个家庭。

中位数是按顺序排列的第 \(7\) 个值。列出家庭孩子人数后，第 \(7\) 个值是 \(4\)。

所以正确答案是 D。

English

The graph gives \(2\) families with \(1\) child, \(1\) with \(2,\) \(2\) with \(3,\) \(2\) with \(4,\) and \(6\) with \(5,\) for \(2 + 1 + 2 + 2 + 6 = 13\) families.

The median is the \(7\)th value in order. Listing the family sizes, the \(7\)th value is \(4.\)

Thus, the correct answer is D .

20 答案 B

共有 \(6 \times 6 = 36\) 个等可能结果，其中 \(6\) 个是平局，剩下 \(30\) 个结果两数不同。

由对称性，戴安娜的数较大的情况正好是其中一半，即 \(15\) 个。所以概率为 \(\dfrac{15}{36} = \dfrac{5}{12}\)。

所以正确答案是 B。

English

There are \(6 \times 6 = 36\) equally likely outcomes, of which \(6\) are ties, leaving \(30\) outcomes with different numbers.

By symmetry, Diana is larger in exactly half of those, or \(15,\) so the probability is \(\dfrac{15}{36} = \dfrac{5}{12}.\)

Thus, the correct answer is B .

21 答案 B

每个立方体唯一的凸扣都必须插入另一个立方体的孔中才能被隐藏。两个立方体只能共用一个面，因此无法同时隐藏两个凸扣。若用三个立方体隐藏所有凸扣，就需要形成一个三立方体闭环，使每一对立方体都共用一个面；但三个单位立方体不可能两两以面相邻。因此，一个、两个或三个立方体都不行。

四个立方体可以排成一个方形环，每个凸扣插入相邻立方体的孔中，从而外面只露出孔。因此最少需要 \(4\) 个。

所以正确答案是 B。

English

Every cube’s single snap must be plugged into another cube’s hole to be hidden. Two cubes can share only one face, so two cubes cannot hide both snaps. With three cubes, hiding all three snaps would require a three-cube loop in which every pair shares a face, but three unit cubes cannot be pairwise face-adjacent. Thus one, two, or three cubes cannot work.

Four cubes can be arranged in a square ring, each snap fitting into the neighbor’s hole, so only receptacle holes show. The smallest number is \(4.\)

Thus, the correct answer is B .

22 答案 A

质因数分解为 \(6545 = 5 \times 7 \times 11 \times 17\)。要分成两个两位因数，可以配为 \(5 \times 17 = 85\) 和 \(7 \times 11 = 77\)。

这是唯一的两位数配对，它们的和是 \(85 + 77 = 162\)。

所以正确答案是 A。

English

The prime factorization is \(6545 = 5 \times 7 \times 11 \times 17.\) To split into two two-digit factors, pair the primes: \(5 \times 17 = 85\) and \(7 \times 11 = 77.\)

These are the only two-digit pair, and their sum is \(85 + 77 = 162.\)

Thus, the correct answer is A .

23 答案 B

第一位是奇数，有 \(5\) 个选择。第二位是偶数，有 \(5\) 个选择，而且不会与第一位的奇数重复。

第三位可以是剩下 \(8\) 个未用数字中的任意一个，第四位有 \(7\) 个选择。总数为 \(5 \times 5 \times 8 \times 7 = 1400\)。

所以正确答案是 B。

English

The first digit is odd: \(5\) choices. The second is even: \(5\) choices (none of which repeats the odd first digit).

The third digit is any of the \(8\) unused digits, and the fourth is any of the \(7\) remaining. In total, \(5 \times 5 \times 8 \times 7 = 1400.\)

Thus, the correct answer is B .

24 答案 C

因为 \(AB = DC = 12\)，所以 \(AE = 12 - 4 = 8\)。在直角三角形 \(ADE\) 中，\(AD = \sqrt{8^2 + 6^2} = 10\)，因此 \(BC = AD = 10\)。

面积为 \(AB \cdot DE = 12 \cdot 6 = 72\)，同时也等于 \(BC \cdot DF = 10 \cdot DF\)。所以 \(DF = \dfrac{72}{10} = 7.2\)。

所以正确答案是 C。

English

Since \(AB = DC = 12,\) we get \(AE = 12 - 4 = 8.\) In right triangle \(ADE,\) \(AD = \sqrt{8^2 + 6^2} = 10,\) so \(BC = AD = 10.\)

The area is \(AB \cdot DE = 12 \cdot 6 = 72,\) and also \(BC \cdot DF = 10 \cdot DF.\) So \(DF = \dfrac{72}{10} = 7.2.\)

Thus, the correct answer is C .

25 答案 D

考虑一辆 \(6\!:\!00\) 从达拉斯出发、\(11\!:\!00\) 到休斯敦的巴士。它会遇到所有在这段时间内与它同在公路上的开往达拉斯的巴士。

开往达拉斯的巴士每半点从休斯敦出发，行程 \(5\) 小时。与这辆车在路上相遇且不在车站相遇的，是 \(1\!:\!30\)、\(2\!:\!30\)、\(\ldots\)、\(10\!:\!30\) 从休斯敦出发的巴士，共 \(10\) 辆。

所以正确答案是 D。

English

Consider a bus leaving Dallas at \(6\!:\!00,\) arriving in Houston at \(11\!:\!00.\) It meets every Dallas-bound bus that is on the highway during that window.

Dallas-bound buses leave Houston on the half hour and take \(5\) hours. The ones sharing the road (meeting away from a station) are those that left Houston at \(1\!:\!30,\) \(2\!:\!30,\) \(\ldots,\) \(10\!:\!30,\) which is \(10\) buses.

Thus, the correct answer is D .