2001 年 AMC 8 真题（P版）

1 答案 D

凯西完成工作需要 \(2\cdot300=600\) 秒。因为 \(60\) 秒等于一分钟，所以这是 \(600\div60=10\) 分钟。

所以正确答案是 D。

English

It will take Casey \(2\cdot300=600\) seconds to do the job. Since \(60\) seconds make one minute, this is \(600\div60=10\) minutes.

Thus, D is the correct answer.

2 答案 D

\(24\) 的因数对为 \((1,24)\)、\((2,12)\)、\((3,8)\) 和 \((4,6)\)。

唯一和为 \(11\) 的因数对是 \((3,8)\)，所以较大的数是 \(8\)。

所以正确答案是 D。

English

The factor pairs of \(24\) are \((1,24),\) \((2,12),\) \((3,8),\) and \((4,6)\).

The only pair with sum \(11\) is \((3,8)\), so the larger number is \(8\).

Thus, D is the correct answer.

3 答案 E

安茹有 \(\$63\div3=\$21\)。埃尔伯塔比安茹多 \(\$2\)，所以埃尔伯塔有 \(\$21+\$2=\$23\)。

所以正确答案是 E。

English

Anjou has \(\$63\div3=\$21\). Elberta has \(\$2\) more than Anjou, so Elberta has \(\$21+\$2=\$23\).

Thus, E is the correct answer.

4 答案 E

个位数字必须是 \(2\) 或 \(4\)。若个位是 \(2\)，把其余数字按尽可能小的顺序排列，得到 \(13492\)。若个位是 \(4\)，最小的排列是 \(12394\)，它更小。因此最小的偶数是 \(12394\)，其十位数字是 \(9\)。

所以正确答案是 E。

English

The units digit must be \(2\) or \(4\). If it is \(2\), arranging the remaining digits as small as possible gives \(13492\). If it is \(4\), the smallest arrangement is \(12394\), which is smaller. Therefore the smallest even number is \(12394\), whose tens digit is \(9\).

Thus, E is the correct answer.

5 答案 C

在 \(10\) 秒内，雷声传播了 \(10 \cdot 1088 = 10880\) 英尺。

\(2\) 英里约为 \(2 \cdot 5280 = 10560\) 英尺，接近上面的距离。

所以正确答案是 C。

English

In \(10\) seconds, the thunder was able to travel \(10 \cdot 1088 = 10880\) feet.

Note that \(2\) miles is about \(2 \cdot 5280 = 10560\) feet, which is close to what we found above.

Thus, C is the correct answer.

6 答案 B

第一棵和第四棵之间有 \(3\) 个间隔，因此每个间隔长 \(60 \div 3 = 20\) 英尺。

第一棵和最后一棵之间有 \(5\) 个间隔，所以它们相距 \(5 \cdot 20 = 100\) 英尺。

所以正确答案是 B。

English

There are \(3\) gaps between the first and fourth trees, which means that one gap is \(60 \div 3 = 20\) feet long.

There are \(5\) gaps between the first and last trees. This means that they are \(5 \cdot 20 = 100\) feet apart.

Thus, B is the correct answer.

7 答案 A

风筝面积等于两条对角线的乘积除以 \(2\)。

因此小风筝面积为 \[ \dfrac{6 \cdot 7}{2} = \dfrac{42}{2} = 21\text{。}\]

所以正确答案是 A。

English

Recall that the area of a kite is the product of its diagonals divided by \(2.\)

Therefore, the area of the small kite is \[ \dfrac{6 \cdot 7}{2} = \dfrac{42}{2} = 21. \]

Thus, A is the correct answer.

8 答案 E

长对角线是 \(7\) 个单位，短对角线是 \(6\) 个单位。

在大风筝中，一个单位为 \(3\) 英寸，所以需要的支撑材料总长为 \[ 3(7 + 6) = 3 \cdot 13 = 39\text{。}\]

所以正确答案是 E。

English

The long diagonal is \(7\) units, and the short one is \(6\) units.

In the large kite, one unit is \(3\) inches, so the total amount of bracing material needed is \[ 3(7 + 6) = 3 \cdot 13 = 39. \]

Thus, E is the correct answer.

9 答案 D

整个方格的面积是 \[ 3 \cdot 7 \cdot 3 \cdot 6 = 378\text{。}\]由风筝面积公式，风筝面积是这个面积的一半，所以废料面积为 \(378 \div 2 = 189\)。

所以正确答案是 D。

English

The area of the entire grid would be \[ 3 \cdot 7 \cdot 3 \cdot 6 = 378. \] The area of the kite is one-half this area from the formula for the area of the kite, so the wasted material is \(378 \div 2 = 189.\)

Thus, D is the correct answer.

10 答案 A

四枚州纪念二十五分硬币的面值为 \(\$1\)。

因为 \(2000\%=20\)，收藏者支付面值的 \(20\) 倍，也就是 \(\$20\)。

所以正确答案是 A。

English

Four state quarters have face value \(\$1\).

Since \(2000\%=20\), the collector pays \(20\) times face value, or \(\$20\).

Thus, A is the correct answer.

11 答案 C

因为 \(\overline{DC} \parallel \overline{AB}\)，所以 \(ABCD\) 是梯形。

梯形面积公式为 \[ A = \dfrac{1}{2} (b_1 + b_2) h\text{。}\]

代入 \[ b_1 = DC = 2\text{、}\]\[ b_2 = AB = 4 \] 和 \[ h = CB = 6\text{，}\]可得 \[ A = \dfrac{1}{2} (2 + 4) \cdot 6 \]\[= \dfrac{1}{2} \cdot 6 \cdot 6 = 18\text{。}\]

所以正确答案是 C。

English

We can see that \(ABCD\) is a trapezoid since \(\overline{DC} \parallel \overline{AB}.\)

Recall that the formula for the area of a trapezoid is \[ A = \dfrac{1}{2} (b_1 + b_2) h. \]

Plugging in \[ b_1 = DC = 2, \]\[ b_2 = AB = 4, \] and \[ h = CB = 6, \] we get that \[ A = \dfrac{1}{2} (2 + 4) \cdot 6 \]\[= \dfrac{1}{2} \cdot 6 \cdot 6 = 18. \]

Thus, C is the correct answer.

12 答案 A

计算如下：\[ \begin{align*} (6 \otimes 4) \otimes 3 &= \dfrac{6 + 4}{6 - 4} \otimes 3 \\ &= 5 \otimes 3 \\&= \dfrac{5 + 3}{5 - 3} \\&= 4 \end{align*}\text{。}\]

所以正确答案是 A。

English

We can evaluate it as follows \[ \begin{align*} (6 \otimes 4) \otimes 3 &= \dfrac{6 + 4}{6 - 4} \otimes 3 \\ &= 5 \otimes 3 \\&= \dfrac{5 + 3}{5 - 3} \\&= 4. \end{align*} \]

Thus, A is the correct answer.

13 答案 D

喜欢樱桃派或柠檬派的学生数为 \[ 36 - 12 - 8 - 6 = 10\text{。}\]其中一半喜欢樱桃派，即 \(10 \div 2 = 5\) 人。

樱桃派对应的角度为 \[ 360^{\circ} \cdot \dfrac{5}{36} = 5 \cdot 10^{\circ} = 50^{\circ}\text{。}\]

所以正确答案是 D。

English

The number of students that prefer cherry or lemon pie is \[ 36 - 12 - 8 - 6 = 10. \] Half of these like cherry pie, which is \(10 \div 2 = 5.\)

The number of degrees for cherry pie would then be \[ 360^{\circ} \cdot \dfrac{5}{36} = 5 \cdot 10^{\circ} = 50^{\circ}. \]

Thus, D is the correct answer.

14 答案 C

他有 \(3\) 种肉类选择和 \(4\) 种甜点选择。

他必须从 \(4\) 种蔬菜中选 \(2\) 种。因为顺序不重要，蔬菜选择有 \(\dfrac{4\cdot3}{2}=6\) 种。

因此可能的餐食数为 \(3\cdot6\cdot4=72\)。

所以正确答案是 C。

English

He has \(3\) choices for the meat and \(4\) choices for dessert.

He must choose \(2\) of the \(4\) vegetables. Since order does not matter, there are \(\dfrac{4\cdot3}{2}=6\) vegetable choices.

This gives \(3\cdot6\cdot4=72\) possible meals.

Thus, C is the correct answer.

15 答案 A

克里斯滕加入时，荷马已经削了 \(3 \cdot 4 = 12\) 个土豆，还剩 \(44 - 12 = 32\) 个。

荷马和克里斯滕合起来每分钟削 \(5 + 3 = 8\) 个土豆，因此还需要 \(32 \div 8 = 4\) 分钟削完剩下的。

\(4\) 分钟内，克里斯滕削了 \(4 \cdot 5 = 20\) 个土豆。

所以正确答案是 A。

English

Homer had peeled \(3 \cdot 4 = 12\) potatoes by the time Christen joined him, leaving \(44 - 12 = 32\) potatoes.

Together, Homer and Christen peel \(5 + 3 = 8\) potatoes per minute, taking them \(32 \div 8 = 4\) minutes to peel the rest.

In \(4\) minutes, Christen peeled \(4 \cdot 5 = 20\) potatoes.

Thus, A is the correct answer.

16 答案 E

正方形对折后形成一个 \(4\times2\) 的长方形。

将折叠后的纸沿平行于折痕的方向切成两半，得到两个 \(4\times1\) 的小长方形和一个 \(4\times2\) 的大长方形。

一个小长方形周长与大长方形周长之比为 \[\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56\text{。}\]

所以正确答案是 E。

English

The square is folded in half to create a \(4\times2\) rectangle.

Cutting the folded paper in half parallel to the fold gives two small \(4\times1\) rectangles and one large \(4\times2\) rectangle.

The ratio of the perimeter of a small rectangle to the perimeter of the large rectangle is \[\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56.\]

Thus, E is the correct answer.

17 答案 B

列出的多数增加都是翻倍，也就是增加 \(100\%\)。

在选项中，例外是从 \(2\) 到 \(3\)、从 \(3\) 到 \(4\)、以及从 \(11\) 到 \(12\)。

这些百分比增加分别是 \(\dfrac{300-200}{200}=50\%\)，\(\dfrac{500-300}{300}=66\dfrac23\%\)，以及 \(\dfrac{125000-64000}{64000}\)，约为 \(95\%\)。

最小的是从第 \(2\) 题到第 \(3\) 题。

所以正确答案是 B。

English

Most of the listed increases are doublings, which are \(100\%\) increases.

The exceptions among the answer choices are from \(2\) to \(3\), from \(3\) to \(4\), and from \(11\) to \(12\).

These percent increases are \(\dfrac{300-200}{200}=50\%\), \(\dfrac{500-300}{300}=66\dfrac23\%\), and \(\dfrac{125000-64000}{64000}\), which is about \(95\%\).

The smallest is from question \(2\) to question \(3\).

Thus, B is the correct answer.

18 答案 D

乘积是 \(5\) 的倍数的唯一方法是至少一个骰子掷出 \(5\)。

用补集计数。两个骰子都不掷出 \(5\) 的概率是 \[ \dfrac{5}{6} \cdot \dfrac{5}{6} = \dfrac{25}{36}\text{。}\]

因此至少一个骰子掷出 \(5\) 的概率为 \[ 1 - \dfrac{25}{36} = \dfrac{11}{36}\text{。}\]

所以正确答案是 D。

English

The only way for the product to be a multiple of \(5\) is if at least one of the rolls is \(5.\)

We can do complementary counting. The probability that neither rolls is a \(5\) is \[ \dfrac{5}{6} \cdot \dfrac{5}{6} = \dfrac{25}{36}. \]

Therefore, the probability that at least one roll is a \(5\) is \[ 1 - \dfrac{25}{36} = \dfrac{11}{36}. \]

Thus, D is the correct answer.

19 答案 D

汽车 \(N\) 的速度是汽车 \(M\) 的两倍，所以它在速度轴上的高度应是汽车 \(M\) 的两倍。

因为汽车 \(N\) 以更快速度行驶与汽车 \(M\) 相同的距离，所以它用一半时间完成行程。

因此表示汽车 \(N\) 速度的线段长度是表示汽车 \(M\) 速度的线段长度的一半。

两条线都水平，因为速度恒定。查看给出的图，唯一满足条件的是图 D。

所以正确答案是 D。

English

Car \(N\) travels at twice the speed of car \(M,\) so it is represented by a point that is twice as high on the speed axis as car \(M.\)

Since car \(N\) travels at the same distance as car \(M\) but at a faster speed, it takes half the time to complete the trip.

Therefore, the line representing car \(N\)’s speed is half the length of the line representing car \(M\)’s speed.

Both lines are horizontal because the speeds are constant. Examining the given graphs, the only one that meets these conditions is graph D .

Thus, D is the correct answer.

20 答案 A

因为奎伊只知道卡莱安娜的分数，所以奎伊和卡莱安娜的分数相同，即 \(\text{K} = \text{Q}\)。

马蒂知道他的分数高于卡莱安娜，所以 \(\text{M} \gt \text{K}\)。莎娜知道她的分数低于卡莱安娜，所以 \(\text{S} \lt \text{K}\)。

用 \(\text{Q}\) 替换 \(\text{K}\)，得到 \(\text{S} \lt \text{Q} \lt \text{M}\)。

所以正确答案是 A。

English

Since Quay only knows Kaleana’s score, we get that Quay and Kaleana have the same score, so \(\text{K} = \text{Q}.\)

Marty knows that his score is higher than Kaleana’s, so \(\text{M} \gt \text{K}.\) Shana knows that her score is lower than Kaleana’s, so \(\text{S} \lt \text{K}.\)

Substituting \(\text{Q}\) for \(\text{K},\) we get the following inequality: \(\text{S} \lt \text{Q} \lt \text{M}.\)

Thus, A is the correct answer.

21 答案 D

中位数是第三大的数，即 \(18\)。有两个数小于 \(18\)，两个数大于它。

平均数为 \(15\)，所以五个数的和为 \(5 \cdot 15 = 75\)。要在这个总和下使最大数尽量大，其余数必须尽量小。

小于 \(18\) 的两个数必须是不同的正整数，所以最小为 \(1\) 和 \(2\)。

紧接在 \(18\) 后的数也应尽量小，所以是 \(19\)。

因此剩下的数，也就是最大可能值，为 \[ 75 - 1 - 2 - 18 - 19 = 35\text{。}\]

所以正确答案是 D。

English

The median of the set of numbers is the third largest number, which is \(18.\) There are two numbers less than \(18\) and two numbers greater than it.

The mean of the set is \(15,\) so the sum of all the numbers is \(5 \cdot 15 = 75.\) In order to maximize the largest number with this sum, the other numbers must be as small as possible.

The two numbers less than \(18\) must be positive and distinct, so they must be \(1\) and \(2.\)

The number immediately after \(18\) must also be as small as possible, so it must be \(19.\)

Therefore, the remaining number, the maximum possible value in the set, is \[ 75 - 1 - 2 - 18 - 19 = 35. \]

Thus, D is the correct answer.

22 答案 E

答对 \(20\) 题时，分数是 \(100\)。

答对 \(19\) 题时，根据剩下一题是答错还是未作答，分数是 \(95\) 或 \(96\)。

答对 \(18\) 题时，剩下两题可以贡献 \(0\)、\(1\) 或 \(2\) 分，所以可得 \(90\)、\(91\) 或 \(92\)。

因此列出的分数 \(90,91,92,95\) 都可能，而 \(97\) 不可能。

所以正确答案是 E。

English

With \(20\) correct answers, the score is \(100\).

With \(19\) correct answers, the score is either \(95\) or \(96\), depending on whether the remaining question is incorrect or unanswered.

With \(18\) correct answers, the remaining two questions can contribute \(0\), \(1\), or \(2\) points, giving \(90\), \(91\), or \(92\).

Thus the listed scores \(90,91,92,95\) are possible, while \(97\) is not.

Thus, E is the correct answer.

23 答案 D

以小等边三角形的边长为一个单位，可能的三角形边长模式有：

边长为 \(2,2,2\) 的等边三角形；边长为 \(1,1,1\) 的等边三角形；边长为 \(1,1,\sqrt3\) 的小等腰三角形；以及边长为 \(1,\sqrt3,2\) 的较大等腰三角形。

这四种模式都在图中出现，而且由对称性可知，用六个点中任意三个形成的三角形都属于其中一种。

因此有 \(4\) 个互不全等的三角形。

所以正确答案是 D。

English

Using the side length of the small equilateral triangles as one unit, the possible triangle side-length patterns are:

equilateral with sides \(2,2,2\); equilateral with sides \(1,1,1\); small isosceles with sides \(1,1,\sqrt3\); and larger isosceles with sides \(1,\sqrt3,2\).

These four patterns all occur in the figure, and symmetry shows every triangle formed by three of the six points matches one of them.

Thus there are \(4\) noncongruent triangles.

Thus, D is the correct answer.

24 答案 B

每一半都有 \(3\) 个红、\(5\) 个蓝和 \(8\) 个白三角形。

\(2\) 对红红配对各使用每一半中的 \(2\) 个红三角形，所以每一半剩下的一个红三角形用于 \(2\) 对红白配对。于是所有红三角形都已计算。

\(3\) 对蓝蓝配对各使用每一半中的 \(3\) 个蓝三角形，每一半还剩 \(2\) 个蓝三角形。由于没有更多蓝蓝配对，这 \(4\) 个蓝三角形必须与白三角形配对。

所以每一半中，\(1\) 个白三角形与红色配对，\(2\) 个白三角形与蓝色配对，剩下 \(8-3=5\) 个白三角形与白三角形配对。

所以正确答案是 B。

English

Each half has \(3\) red, \(5\) blue, and \(8\) white triangles.

The \(2\) red-red pairs use \(2\) red triangles from each half, so the remaining red triangle from each half is used in the \(2\) red-white pairs. Thus all red triangles are accounted for.

The \(3\) blue-blue pairs use \(3\) blue triangles from each half, leaving \(2\) blue triangles from each half. Since no more blue-blue pairs occur, those \(4\) blue triangles must pair with white triangles.

So on each half, \(1\) white is used with red and \(2\) whites are used with blue, leaving \(8-3=5\) white triangles to pair with white triangles.

Thus, B is the correct answer.

25 答案 D

一个在 \(4000\)、\(5000\) 或 \(7000\) 范围内的数乘以 \(2\) 或更大的数后，不可能仍是这些四位排列之一。因此较小的数必须以 \(2\) 开头，较大的数必须是它的两倍或三倍。

对于两倍，只有以 \(4\) 结尾的选项可能。但 \(\frac{5724}{2}=2862\)，\(\frac{7254}{2}=3627\)，都没有恰好使用数字 \(2,4,5,7\)。

对于三倍，可能的选项是能被 \(3\) 整除的 \(7245,7425,7542\)。分别除以三得到 \(2415,2475,2514\)，只有 \(2475\) 恰好使用所需数字。

因此 \(7425=3\cdot2475\) 是唯一列出的倍数。

所以正确答案是 D。

English

A number in the \(4000\), \(5000\), or \(7000\) range cannot be multiplied by \(2\) or more and remain one of the given four-digit permutations. So the smaller number must start with \(2\), and the larger number must be either double or triple it.

For doubles, only answer choices ending in \(4\) can work. But \(\frac{5724}{2}=2862\) and \(\frac{7254}{2}=3627\), neither of which uses exactly the digits \(2,4,5,7\).

For triples, the possible answer choices are those divisible by \(3\): \(7245,7425,7542\). Dividing gives \(2415,2475,2514\), and only \(2475\) uses exactly the required digits.

Therefore \(7425=3\cdot2475\) is the unique listed multiple.

Thus, D is the correct answer.