1989 年 AMC 8 真题（P版）

1 答案 E

把各项配对，使每一对的和都是 \(50\)：\(1 + 49\)、\(11 + 39\)、\(21 + 29\)、\(31 + 19\)、\(41 + 9\)。

一共有 \(5\) 对，所以总和是 \(5 \times 50 = 250\)。

所以正确答案是 E。

English

Pair the terms so each pair sums to \(50\): \(1 + 49,\) \(11 + 39,\) \(21 + 29,\) \(31 + 19,\) and \(41 + 9.\)

There are \(5\) such pairs, so the total is \(5 \times 50 = 250.\)

Thus, the correct answer is E .

2 答案 D

这三个分数分别表示不同数位上的小数：\(\frac{2}{10} = 0.2\)，\(\frac{4}{100} = 0.04\)，\(\frac{6}{1000} = 0.006\)。

相加得 \(0.2 + 0.04 + 0.006 = 0.246\)。

所以正确答案是 D。

English

Each fraction is a decimal in a different place: \(\frac{2}{10} = 0.2,\) \(\frac{4}{100} = 0.04,\) and \(\frac{6}{1000} = 0.006.\)

Adding these gives \(0.2 + 0.04 + 0.006 = 0.246.\)

Thus, the correct answer is D .

3 答案 A

把这些数都写成四位小数：\(0.9900\)、\(0.9099\)、\(0.9000\)、\(0.9090\)、\(0.9009\)。

从左向右比较可知，\(0.9900\) 的百分位数字最大，所以 \(0.99\) 最大。

所以正确答案是 A。

English

Write each number with four decimal places: \(0.9900,\) \(0.9099,\) \(0.9000,\) \(0.9090,\) \(0.9009.\)

Comparing from the left, \(0.9900\) has the largest hundredths digit, so \(0.99\) is the largest.

Thus, the correct answer is A .

4 答案 E

把分子和分母分别取成便于计算的近似数，得到 \(\frac{401}{0.205} \approx \frac{400}{0.2}\)。

由于 \(\frac{400}{0.2} = 2000\)，原数最接近 \(2000\)。

所以正确答案是 E。

English

Round the numerator and denominator: \(\frac{401}{0.205} \approx \frac{400}{0.2}.\)

Since \(\frac{400}{0.2} = 2000,\) the value is closest to \(2000.\)

Thus, the correct answer is E .

5 答案 D

先算括号：\(6 \div 3 = 2\)。再做乘法，\(9 \times 2 = 18\)。

最后 \(-15 + 18 = 3\)。

所以正确答案是 D。

English

Inside the parentheses, \(6 \div 3 = 2.\) Then multiplication comes before addition, so \(9 \times 2 = 18.\)

Finally, \(-15 + 18 = 3.\)

Thus, the correct answer is D .

6 答案 C

从 \(0\) 到 \(20\) 被分成 \(5\) 个相等间隔，所以每个间隔的长度是 \(\frac{20}{5} = 4\)。

\(y\) 距 \(0\) 有 \(3\) 个间隔，所以 \(y = 3 \times 4 = 12\)。

所以正确答案是 C。

English

From \(0\) to \(20\) there are \(5\) equal spaces, so each space is \(\frac{20}{5} = 4.\)

The mark labeled \(y\) is \(3\) spaces from \(0,\) so \(y = 3 \times 4 = 12.\)

Thus, the correct answer is C .

7 答案 D

\(20\) 枚二十五美分硬币和 \(10\) 枚十美分硬币的总价值是 \(\$5.00 + \$1.00 = \$6.00\)。

十枚二十五美分硬币值 \(\$2.50\)，所以 \(n\) 枚十美分硬币必须值 \(\$6.00 - \$2.50 = \$3.50\)。这相当于 \(35\) 枚十美分硬币，因此 \(n = 35\)。

所以正确答案是 D。

English

The value of \(20\) quarters and \(10\) dimes is \(\$5.00 + \$1.00 = \$6.00.\)

Ten quarters are worth \(\$2.50,\) so the \(n\) dimes must supply \(\$6.00 - \$2.50 = \$3.50.\) That is \(35\) dimes, so \(n = 35.\)

Thus, the correct answer is D .

8 答案 E

\(2 \times 3 \times 4 = 24\)，所以原式等于 \(24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}\)。

这等于 \(12 + 8 + 6 = 26\)。

所以正确答案是 E。

English

Since \(2 \times 3 \times 4 = 24,\) distribute it over the sum: \(24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}.\)

This equals \(12 + 8 + 6 = 26.\)

Thus, the correct answer is E .

9 答案 C

男生占每 \(2 + 3 = 5\) 名学生中的 \(2\) 名，也就是全班的 \(\frac{2}{5}\)。全班共有 \(30\) 人这一信息其实不必使用。

化成百分数，\(\frac{2}{5} = 40\%\)。

所以正确答案是 C。

English

Boys make up \(2\) out of every \(2 + 3 = 5\) students, which is \(\frac{2}{5}\) of the class. The total of \(30\) students is not even needed.

As a percent, \(\frac{2}{5} = 40\%.\)

Thus, the correct answer is C .

10 答案 D

钟面上的 \(12\) 个数字把一周分成 \(12\) 等份，每份是 \(\frac{360^\circ}{12} = 30^\circ\)。

七点整时，分针指向 \(12\)，时针指向 \(7\)。沿较短方向相隔 \(5\) 个刻度，所以角度是 \(5 \times 30^\circ = 150^\circ\)。

所以正确答案是 D。

English

The \(12\) numbers divide the clock into \(12\) equal sections of \(\frac{360^\circ}{12} = 30^\circ\) each.

At seven o’clock the hands point to \(12\) and \(7,\) which are \(5\) sections apart the short way, giving \(5 \times 30^\circ = 150^\circ.\)

Thus, the correct answer is D .

11 答案 B

两个图形关于虚线对称，指的是沿这条线把纸折起来后，一个图形恰好落在另一个图形上。关于竖直虚线反射时，图形的左、右位置互换，上、下位置保持不变。

在这个反射下，原图中左上角的小方块会移到右上角，中间那条横杆的倾斜方向会相反，而原本朝右下方伸出的竖杆必须改为朝左下方伸出。只有一个选项同时具备这三个特征。

所以正确答案是 B。

English

Two figures are symmetric with respect to the dashed line when folding the paper along that line makes one coincide with the other. Reflecting across the vertical dashed line swaps left and right while leaving top and bottom unchanged.

Under this reflection the small square in the top-left corner moves to the top-right corner, the straight crossbar reverses its slant, and the stem that points down and to the right must instead point down and to the left. Only one choice has all three of these features.

Thus, the correct answer is B .

12 答案 E

分子是 \(1 - \frac{1}{3} = \frac{2}{3}\)，分母是 \(1 - \frac{1}{2} = \frac{1}{2}\)。

因此 \(\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3}\)。

所以正确答案是 E。

English

The numerator is \(1 - \frac{1}{3} = \frac{2}{3},\) and the denominator is \(1 - \frac{1}{2} = \frac{1}{2}.\)

Dividing gives \(\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3}.\)

Thus, the correct answer is E .

13 答案 A

将分子 \(9\) 除以 \(10\)，得到 \(0.9\)。为了保持分数值不变，分母整体也要除以 \(10\)，即把其中一个因数除以 \(10\)。

把分母中的因数 \(7\) 除以 \(10\)，而另一个因数保持不变，便得到 \(\frac{0.9}{0.7 \times 53}\)。其他选项会使原式改变 \(10\) 倍或 \(100\) 倍。

所以正确答案是 A。

English

To rewrite the numerator \(9\) as \(0.9,\) divide it by \(10.\) To keep the fraction equal, divide the denominator by \(10\) too, which means dividing one of its factors by \(10.\)

Dividing the factor \(7\) by \(10\) gives \(\frac{0.9}{0.7 \times 53},\) which equals the original. Each other choice changes the value by a factor of \(10\) or \(100.\)

Thus, the correct answer is A .

14 答案 C

差最小时，上面的三位数应尽量小，下面要减去的两位数应尽量大。

最大的两位数是 \(96\)，使用数字 \(9\) 和 \(6\)。剩下的数字 \(2\)、\(4\)、\(5\) 能组成的最小三位数是 \(245\)，所以最小差是 \(245 - 96 = 149\)。

所以正确答案是 C。

English

The difference is smallest when the three-digit number (the top) is as small as possible and the two-digit number (subtracted) is as large as possible.

The largest two-digit number is \(96,\) using the digits \(9\) and \(6.\) The smallest three-digit number from the remaining digits \(2,\) \(4,\) \(5\) is \(245.\) So the smallest difference is \(245 - 96 = 149.\)

Thus, the correct answer is C .

15 答案 D

平行四边形的底为 \(10\)，高为 \(8\)，所以面积是 \(10 \times 8 = 80\)。

未涂色三角形 \(ABE\) 的底为 \(AE = AD - ED = 10 - 6 = 4\)，高为 \(8\)，所以面积是 \(\frac{1}{2} \times 4 \times 8 = 16\)。阴影面积是 \(80 - 16 = 64\)。

所以正确答案是 D。

English

The parallelogram has base \(10\) and height \(8,\) so its area is \(10 \times 8 = 80.\)

The unshaded triangle \(ABE\) has base \(AE = AD - ED = 10 - 6 = 4\) and height \(8,\) so its area is \(\frac{1}{2} \times 4 \times 8 = 16.\) The shaded area is \(80 - 16 = 64.\)

Thus, the correct answer is D .

16 答案 A

因为 \(47\) 是奇数，若两个质数的和等于 \(47\)，其中必有一个偶质数。唯一的偶质数是 \(2\)。

于是另一个数必须是 \(47 - 2 = 45\)，但 \(45 = 9 \times 5\)，不是质数。因此没有这样的表示方式。

所以正确答案是 A。

English

Since \(47\) is odd, a sum of two primes equal to \(47\) needs one even prime and one odd prime. The only even prime is \(2.\)

That would require the other number to be \(47 - 2 = 45,\) but \(45 = 9 \times 5\) is not prime. So there is no way.

Thus, the correct answer is A .

17 答案 B

平均数为 \(\frac{6 + 10 + N}{3} = \frac{16 + N}{3}\)。当 \(N = 9\) 时，这个式子为 \(\frac{25}{3} \approx 8.3\)；当 \(N = 17\) 时，它等于 \(11\)。

所以平均数严格介于约 \(8.3\) 和 \(11\) 之间。选项中只有 \(10\) 落在这个范围内。

所以正确答案是 B。

English

The average is \(\frac{6 + 10 + N}{3} = \frac{16 + N}{3}.\) When \(N = 9\) this is \(\frac{25}{3} \approx 8.3,\) and when \(N = 17\) it is \(11.\)

So the average lies strictly between about \(8.3\) and \(11.\) Among the choices, only \(10\) falls in this range.

Thus, the correct answer is B .

18 答案 B

从 \(32\) 开始，按一次倒数键会得到 \(\frac{1}{32}\)。

再按一次，就是对倒数再取倒数，得到 \(\frac{1}{\frac{1}{32}} = 32\)。所以最少需要按 \(2\) 次。

所以正确答案是 B。

English

Pressing the key once changes \(32\) to its reciprocal \(\frac{1}{32}.\)

Pressing it a second time takes the reciprocal again, returning to \(\frac{1}{\frac{1}{32}} = 32.\) So \(2\) presses are enough.

Thus, the correct answer is B .

19 答案 B

图显示的是累计花费，所以六月、七月、八月的花费等于这段时间累计值的增加量。

从图上看，六月开始时累计值略多于 \(2\) 百万美元，八月结束时略多于 \(4.5\) 百万美元，差约为 \(4.5 - 2 = 2.5\) 百万美元。

所以正确答案是 B。

English

The graph gives the total accumulated spending, so the amount spent during June, July, and August equals the accumulated total at the end of August minus the accumulated total at the beginning of June.

The curve is at a bit more than \(2\) million at the beginning of June and a bit more than \(4.5\) million by the end of August. The difference is about \(4.5 - 2 = 2.5\) million.

Thus, the correct answer is B .

20 答案 D

折成立方体后，相对面不能在同一个顶点相遇。由展开图可判断相对面为 \(1\) 与 \(3\)、\(2\) 与 \(5\)、\(4\) 与 \(6\)。

要让三个相邻面的和最大，就从每对相对面中取较大的数 \(3\)、\(5\) 和 \(6\)。它们可以在同一个顶点相遇，和为 \(3 + 5 + 6 = 14\)。

所以正确答案是 D。

English

When the net is folded, the pairs of opposite faces are \(1\) and \(3,\) \(2\) and \(5,\) and \(4\) and \(6.\) Three faces meeting at a corner must come one from each opposite pair.

To maximize the sum, take the larger number from each pair: \(3,\) \(5,\) and \(6.\) These three faces do meet at a corner, and their sum is \(3 + 5 + 6 = 14.\)

Thus, the correct answer is D .

21 答案 D

杰克第一次卖掉 \(25\%\) 后，剩下 \(\frac{3}{4} \times 128 = 96\) 个苹果。第二次又卖掉余数的 \(25\%\)，剩下 \(\frac{3}{4} \times 96 = 72\) 个苹果。

再送给老师 \(1\) 个后，剩下 \(72 - 1 = 71\) 个。

所以正确答案是 D。

English

After selling \(25\%\) to Jill, Jack keeps \(\frac{3}{4} \times 128 = 96\) apples. After selling \(25\%\) of those to June, he keeps \(\frac{3}{4} \times 96 = 72\) apples.

He then gives \(1\) to his teacher, leaving \(72 - 1 = 71.\)

Thus, the correct answer is D .

22 答案 C

字母串 AJHSME 每移动 \(6\) 次回到原顺序，数字串 \(1989\) 每移动 \(4\) 次回到原顺序。

两者第一次同时回到原样是在 \(6\) 和 \(4\) 的最小公倍数 \(12\) 次移动后。因此 AJHSME \(1989\) 第一次出现在第 \(12\) 行。

所以正确答案是 C。

English

The six letters cycle back to their original order AJHSME every \(6\) lines, and the four digits cycle back to \(1989\) every \(4\) lines.

Both happen on the same line at the least common multiple of \(6\) and \(4,\) which is \(12.\) So AJHSME \(1989\) first reappears on line \(12.\)

Thus, the correct answer is C .

23 答案 C

因为每一层都靠齐在同一个角上，所以从每个竖直方向看，露出的阶梯侧面都是 \(3 + 2 + 1 = 6\) 个小正方形面，四个方向共 \(4 \times 6 = 24\) 个侧面。

从上方看，露出的顶面覆盖整个 \(3 \times 3\) 的底面投影，所以顶面积是 \(9\) 平方米。底面不漆，因此总上漆面积是 \(24 + 9 = 33\) 平方米。

所以正确答案是 C。

English

Because each higher layer is flush into one corner, each of the four vertical sides shows a stepped profile of \(3 + 2 + 1 = 6\) exposed square faces, for \(4 \times 6 = 24\) side faces.

Viewed from directly above, the top faces cover the full \(3 \times 3\) footprint, adding \(9\) more faces. The bottom rests on the ground and is not painted, so the total is \(24 + 9 = 33\) square meters.

Thus, the correct answer is C .

24 答案 E

设原正方形边长为 \(4\)。对折后得到一个 \(2 \times 4\) 的双层长方形，再用一条与折痕平行的线把折好的纸切开，分成含折痕的一条和不含折痕的一条。

含折痕的那块展开后是 \(2 \times 4\) 的大长方形，周长为 \(2(2 + 4) = 12\)。另一块展开后成为两个 \(1 \times 4\) 的小长方形，每个周长为 \(2(1 + 4) = 10\)。所求比为 \(\frac{10}{12} = \frac{5}{6}\)。

所以正确答案是 E。

English

Let the square have side \(4.\) Folding in half makes a \(2 \times 4\) stack of two layers. Cutting parallel to the fold splits it into a strip containing the fold and a strip that does not.

The strip with the fold unfolds into the large rectangle, \(2 \times 4,\) with perimeter \(2(2 + 4) = 12.\) The other strip is two separate small rectangles, each \(1 \times 4,\) with perimeter \(2(1 + 4) = 10.\) The ratio is \(\frac{10}{12} = \frac{5}{6}.\)

Thus, the correct answer is E .

25 答案 C

和为偶数需要两个数同奇偶。第一个转盘的偶数为 \(\{4, 8\}\)，奇数为 \(\{3, 5\}\)，所以两种奇偶性的概率都是 \(\frac{2}{4} = \frac{1}{2}\)。第二个转盘的偶数为 \(\{6\}\)，概率为 \(\frac{1}{3}\)；奇数为 \(\{7, 9\}\)，概率为 \(\frac{2}{3}\)。

因此和为偶数的概率是 \(\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}\)。

所以正确答案是 C。

English

The sum is even when both numbers are even or both are odd. The first wheel has evens \(\{4, 8\}\) and odds \(\{3, 5\},\) each with probability \(\frac{2}{4} = \frac{1}{2}.\) The second wheel has even \(\{6\}\) with probability \(\frac{1}{3}\) and odds \(\{7, 9\}\) with probability \(\frac{2}{3}.\)

So the probability of an even sum is \(\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.\)

Thus, the correct answer is C .