2024 年 AMC 8 真题（P版）

1 答案 B

只需考虑各数的个位。模 \(10\) 计算，原式为 \[ \begin{gathered} 2-2-2-2-2-2=-8,\\ -8\equiv2\pmod{10} \end{gathered}\text{。}\]因此原式的个位数字是 \(2\)。

所以正确答案是 B。

English

Only the ones digits matter. Modulo \(10,\) the expression is \[ \begin{gathered} 2-2-2-2-2-2=-8,\\ -8\equiv2\pmod{10}. \end{gathered} \] Therefore its ones digit is \(2.\)

Thus, B is the correct answer.

2 答案 C

提出公因数 \(11\) 后，\(\dfrac{44}{11}\) 化为 \(4\)，\(\dfrac{110}{44}\) 化为 \(\dfrac{10}{4} = \dfrac{5}{2} = 2.5\)，而 \(\dfrac{44}{1100}\) 化为 \(\dfrac{4}{100} = 0.04\)。因此 \(4 + 2.5 + 0.04 = 6.54\)。

所以正确答案是 C。

English

We can simplify the fractions by taking out the common factor \(11\): \( \dfrac{44}{11} \) simplifies to \( 4 \), \( \dfrac{110}{44} \) simplifies to \( \dfrac{10}{4} = \dfrac{5}{2} = 2.5 \), and \( \dfrac{44}{1100} \) simplifies to \( \dfrac{4}{100} = 0.04 \). Therefore, we have \( 4 + 2.5 + 0.04 = 6.54.\)

Thus, C is the correct answer.

3 答案 E

可见阴影部分由 \(10\) 乘 \(10\) 的正方形中位于 \(9\) 乘 \(9\) 的正方形之外的部分，以及 \(7\) 乘 \(7\) 的正方形中位于 \(4\) 乘 \(4\) 的正方形之外的部分组成：\[ \begin{gathered} 10^2-9^2+7^2-4^2 \\ =100-81+49-16 \\ =52 \end{gathered}\text{。}\]

所以正确答案是 E。

English

The visible shaded region is the part inside the \(10\) by \(10\) square but outside the \(9\) by \(9\) square, together with the part inside the \(7\) by \(7\) square but outside the \(4\) by \(4\) square. Its area is \[ \begin{gathered} 10^2-9^2+7^2-4^2 \\ =100-81+49-16 \\ =52. \end{gathered} \]

Thus, E is the correct answer.

4 答案 E

要求 Yunji 漏掉的数，先求从 \(1\) 到 \(9\) 的总和，再求这个总和与比它小的最大完全平方数之差。从 \(1\) 到 \(9\) 的整数和为 \[ 1 + \ldots + 9 = \dfrac{9(9+1)}{2} = 45 \] 小于 \(45\) 的最大完全平方数是 \(36\)，所以漏掉的数是 \(45 - 36 = 9\)。

所以正确答案是 E。

English

To find the number that Yunji left out, we need to find the sum of the integers from \(1\) to \(9\) and find its difference with the largest perfect square below the sum. We can calculate the sum of the integers from \(1\) to \(9\) as follows: \[ 1 + \ldots + 9 = \dfrac{9(9+1)}{2} = 45 \] The largest perfect square less than \( 45 \) would be \( 36 \) and \( 45 - 36 = 9 \).

Thus, E is the correct answer.

5 答案 B

乘积要成为 \(6\) 的倍数，两个骰子必须共同提供因子 \(2\) 和 \(3\)。选项中的以下各个和都能出现：\[ \begin{gathered} 5=2+3,\quad 7=1+6, \\ 8=2+6,\quad 9=3+6 \end{gathered}\text{。}\]但不可能得到和为 \(6\) 且乘积能被 \(6\) 整除的结果：和为 \(6\) 的骰子点数组合是 \((1,5)\)、\((2,4)\)、\((3,3)\)、\((4,2)\) 和 \((5,1)\)，它们的乘积都不能被 \(6\) 整除。

正确答案是 B。

English

For the product to be a multiple of \(6\), the two dice together must supply a factor of \(2\) and a factor of \(3\). The possible sums among the answer choices can occur as follows: \[ \begin{gathered} 5=2+3,\quad 7=1+6, \\ 8=2+6,\quad 9=3+6. \end{gathered} \] There is no way to get sum \(6\) while also having a product divisible by \(6\): the pairs with sum \(6\) are \((1,5),\) \((2,4),\) \((3,3),\) \((4,2),\) and \((5,1)\), and none have product divisible by \(6\).

Thus, B is the correct answer.

6 答案 D

路线 R 最短，因为它用直线弦替代了弧线部分。路线 Q 最长，因为它在仍然使用弯曲端部的同时，还包含最多的内部穿越距离。

接下来只需比较路线 P 和 S。路线 S 中相关的直线段是横穿冰场的一条对角线，而路线 P 中对应的直线段是同一个直角三角形的一条直角边。对角线比直角边长，所以路线 S 比路线 P 长。

从短到长的顺序为 R、P、S、Q。

所以正确答案是 D。

English

Path R is shortest because it replaces curved arc portions with straight-line chords. Path Q is longest because it includes the most interior crossing distance while still using the curved ends.

It remains to compare paths P and S. The relevant straight piece in path S is a diagonal across the rink, while the corresponding straight piece in path P is a side of the same right triangle. A diagonal is longer than a side, so path S is longer than path P.

The order from shortest to longest is R, P, S, Q.

Thus, D is the correct answer.

7 答案 E

\(2\times2\) 和 \(1\times4\) 瓷砖面积都是 \(4\)。矩形面积为 \(21\)，所以 \(1\times1\) 瓷砖数必须同余于 \(1\pmod4\)。选项中只可能是 \(1\) 或 \(5\)。

不能只用一块 \(1\times1\) 瓷砖。否则大瓷砖覆盖其余 \(20\) 个格子，会有两行的 \(7\) 个格子全由大瓷砖覆盖。但每块 \(2\times2\) 瓷砖在某行覆盖 \(2\) 格，每块 \(1\times4\) 瓷砖在某行覆盖 \(4\) 格，所以一行由大瓷砖覆盖的格数应为偶数，不可能是 \(7\)。

下图说明 \(5\) 块单位瓷砖可以做到。

正确答案是 E。

English

The \(2\times2\) and \(1\times4\) tiles each have area \(4\). Since the rectangle has area \(21\), the number of \(1\times1\) tiles must be congruent to \(1\pmod4\). Among the choices, only \(1\) and \(5\) are possible by area.

It is impossible to use just one \(1\times1\) tile. If the larger tiles covered the other \(20\) cells, then two rows would have all \(7\) cells covered by larger tiles. But each \(2\times2\) tile covers \(2\) cells in any row it meets, and each \(1\times4\) tile covers \(4\) cells in one row, so each row would have an even number of cells covered by larger tiles. A row cannot have \(7\) such cells.

The following tiling shows that \(5\) unit tiles are possible.

Thus, E is the correct answer.

8 答案 D

星期二结束时，Taye 可能有 \(5\) 或 \(4\) 美元。星期三结束时，可能的金额为 \[ 8,10,7,8\text{，}\]所以不同的金额是 \(7,8,10\)。

星期四结束时，这些金额可能变成 \[ 10,14,\quad 11,16,\quad 13,20\text{。}\]共有 \(6\) 种不同的金额。

所以正确答案是 D。

English

After Tuesday, Taye could have \(5\) or \(4\) dollars. After Wednesday, the possible amounts are \[ 8,10,7,8, \] so the distinct amounts are \(7,8,10\).

After Thursday, these can become \[ 10,14,\quad 11,16,\quad 13,20. \] These are \(6\) distinct dollar amounts.

Thus, D is the correct answer.

9 答案 E

设 Maria 有 \(r\) 个红色弹珠。红色是绿色的一半，所以绿色有 \(2r\) 个；蓝色是绿色的两倍，所以蓝色有 \(2(2r) = 4r\) 个。总数为 \(r + 2r + 4r = 7r\)，必须是 \(7\) 的倍数。选项中只有 \(28\) 是 \(7\) 的倍数。

正确答案是 E。

English

We can let \(r\) be the number of red marbles that Maria has. Since Maria has half as many red marbles as green, then we know that she has \(2r\) green marbles. Moreover, since she has twice as many blue marbles as green, then she will have \(2(2r) = 4r\) blue marbles. Adding these together gives us \( r + 2r + 4r = 7r, \) and so the answer must be a multiple of \(7.\) Among the answer choices, only \(28\) is a multiple of \(7.\)

Thus, E is the correct answer.

10 答案 B

从 \(1980\) 年到 \(2030\) 年相隔 \(50\) 年，所以预计到 \(2030\) 年，CO_\(2\) 读数将增加约 \(50 \times 1.515 = 75.75 \approx 76\) ppm。由于 \(1980\) 年的 CO_\(2\) 读数是 \(338\) ppm，所以到 \(2030\) 年的预计读数为 \(338 + 76 = 414\) ppm。

正确答案是 B。

English

There are \(50\) years between \(1980\) and \(2030\), so we can expect the CO_\(2\) reading to increase by \(50 \times 1.515 = 75.75 \approx 76 \) ppm by \(2030\). Since the CO_\(2\) reading in \(1980\) was \(338\) ppm, then we will have \(338 + 76 = 414\) ppm by \(2030\).

Thus, B is the correct answer.

11 答案 D

\(\overline{\rm AB}\) 是水平线段，长度为 \(11-5=6\)。三角形面积为 \(12\)，所以高为 \(\dfrac{2(12)}{6}=4\)。因为 \(y>7\)，所以 \(y\) 必须是 \(7+4=11\)。

正确答案是 D。

English

Consider the base of the triangle to be \(\overline{\rm AB}\) which has length \(11-5=6\). Given that the area of the triangle is \(12\), its height must be of length \(\dfrac{2(12)}{6}=4\). Since \(y>7\), then \(y\) must be \(7+4=11\).

Thus, D is the correct answer.

12 答案 E

设第 \(1\) 个鱼缸有 \(x\) 条孔雀鱼，则第 \(2\) 个鱼缸有 \(x+1\) 条，第 \(3\) 个鱼缸有 \(x+3\) 条，第 \(4\) 个鱼缸有 \(x+6\) 条。由 \(4\) 个鱼缸共有 \(90\) 条孔雀鱼求 \(x\)：\[x + x + 1 + x + 3 + x + 6 = 90\] \[4x+10=90\] \[x=20\text{。}\]

题目问的是第 \(4\) 个鱼缸的数量，不是第 \(1\) 个鱼缸。因此第 \(4\) 个鱼缸有 \(x+6=20+6=26\) 条。

正确答案是 E。

English

Let \(x\) be the number of guppies in the \(1\)st tank. Hence, there are \(x+1\) guppies in the \(2\)nd tank, \(x+3\) guppies in the \(3\)rd tank, and \(x+6\) guppies in the \(4\)th tank. We then use the fact that there are a total of \(90\) guppies in the \(4\) tanks to find \(x\): \[x + x + 1 + x + 3 + x + 6 = 90\] \[4x+10=90\] \[x=20.\]

Note that we are not yet done since we are asked for the number of guppies in the \(4\)th tank and not the \(1\)st. There are \(x+6=20+6=26\) guppies in the \(4\)th tank.

Thus, E is the correct answer.

13 答案 B

所有序列都必须以向上 \((U)\) 开始、以向下 \((D)\) 结束，并且在任何时刻，向下的次数都不能超过此前向上的次数。可行序列为 \[UUUDDD\] \[UUDUDD\] \[UUDDUD\] \[UDUDUD\] \[UDUUDD\] 共有五种可行序列。

所以正确答案是 B。

English

We can deduce from the choices that it is possible to exhaust all possible cases for this problem. Note that all sequences must start with up \((U)\) and end with down \((D)\), and that it should not be possible to go down more times than Buzz has gone up so far. Keeping this in mind, we can arrive at the following possible cases: \[UUUDDD\] \[UUDUDD\] \[UUDDUD\] \[UDUDUD\] \[UDUUDD\] which is a total of five possible sequences.

Thus, B is the correct answer.

14 答案 A

系统地追踪到 \(Z\) 的最短总距离，可以逐一考虑从 \(A\) 到每个城镇的最短距离。例如，从 \(A\) 到城镇 \(X\) 的最短距离显然是 \(5\) 千米。

接着看城镇 \(M\)：先经过城镇 \(X\) 比直接从 \(A\) 过去更短，所以到城镇 \(M\) 的最短路长为 \(7\) 千米。

到 \(Y\) 时，从 \(X\) 来需要 \(15\) 千米，从 \(M\) 来只需 \(13\) 千米，所以从 \(A\) 到 \(Y\) 的最短距离为 \(13\) 千米。

对城镇 \(C\) 做同样的分析，可得从城镇 \(Y\) 来最短，为 \(18\) 千米。

最后到 \(Z\) 可以分别从 \(Y\)、\(C\) 或 \(M\) 来，对应总距离为 \(30\)、\(28\) 和 \(32\)。最短距离是 \(28\) 千米，也就是从 \(A\) 到 \(Z\) 的答案。

所以正确答案是 A。

English

A systematic way of tracking the shortest overall distance to \(Z\) is to consider the shortest distance to get to each town from \(A\). For instance, the shortest distance to get to town \(X\) from \(A\) is \(5\) km, trivially.

Then, for town \(M\), going to town \(X\) first will be shorter compared to going directly from \(A\), so the shortest path to town \(M\) has a length of \(7\) km.

For town \(Y\), it will take us \(15\) km if we come from town \(X\) and only \(13\) km coming from \(M,\) so \(13\) km is the length of shortest path to \(Y\) from \(A\).

Doing the same for town \(C\) will give us \(18\) km as the shortest distance by coming from town \(Y\).

Finally, for town \(Z\), we can either come from town \(Y,\) \(C,\) or \(M\). The total distance if we come from each three towns respectively would be \(30,\) \(28,\) and \(32\). Hence, \(28\) km is the shortest distance from \(A\) to \(Z\).

Thus, A is the correct answer.

15 答案 C

首先注意到 \[\text{FLYFLY} = 1001(\text{FLY})\] 同理，\[\text{BUGBUG}=1001(\text{BUG})\] 所以原方程可化为 \(8 \cdot \text{FLY} = \text{BUG}\)。

因为 \(\text{BUG}\) 是三位数，所以 \(\text{FLY}\le124\)。不超过 \(124\) 且各位数字互异的最大三位数是 \(124\)，但 \(8\cdot124=992\) 中有重复数字，而且还重复使用了数字 \(2\)。

下一个候选数是 \(123\)，且 \(8\cdot123=984\)。六个数字 \(1\)、\(2\)、\(3\)、\(9\)、\(8\) 和 \(4\) 互不相同，所以 \(123\) 是 \(\text{FLY}\) 的最大可能值。

因此 \[ \begin{gathered} \text{FLY} + \text{BUG} = 123 + 984 \\ = 1107 \end{gathered}\text{。}\]

正确答案是 C。

English

Firstly, note that \[\text{FLYFLY} = 1001(\text{FLY})\] and, similarly, \[\text{BUGBUG}=1001(\text{BUG})\] so the equation can be simplified to \(8 \cdot \text{FLY} = \text{BUG}.\)

Because \(\text{BUG}\) has three digits, \(\text{FLY}\le124.\) The largest possible three-digit number at most \(124\) with distinct digits is \(124,\) but \(8\cdot124=992\) repeats a digit and also reuses the digit \(2.\)

The next candidate is \(123,\) and \(8\cdot123=984.\) The six digits \(1,\) \(2,\) \(3,\) \(9,\) \(8,\) and \(4\) are all distinct, so \(123\) is the greatest possible value of \(\text{FLY}.\)

Hence, \[ \begin{gathered} \text{FLY} + \text{BUG} = 123 + 984 \\ = 1107. \end{gathered} \]

Thus, C is the correct answer.

16 答案 D

从 \(1\) 到 \(81\) 共有 \(27\) 个 \(3\) 的倍数。一行或一列的乘积能被 \(3\) 整除，当且仅当它至少含有一个这样的倍数。

设有 \(r\) 行和 \(c\) 列的乘积能被 \(3\) 整除。每个 \(3\) 的倍数都必须位于这 \(r\) 行之一，同时也位于这 \(c\) 列之一，否则它会再制造出一行或一列被标记。因此 \(27\) 个倍数必须装入 \(rc\) 个交叉格中。若 \(r+c\le10\)，则 \(rc\le25\)，格子不够。因此至少要标记 \(11\) 行和列。

可以把 \(25\) 个 \(3\) 的倍数放入一个 \(5\times5\) 区块，再把剩下的 \(2\) 个放在第六列中，且位于其中两行内。这样恰好标记 \(5\) 行和 \(6\) 列，共 \(11\)。

正确答案是 D。

English

There are \(27\) multiples of \(3\) from \(1\) through \(81\). A row or column has product divisible by \(3\) exactly when it contains at least one of these multiples.

Suppose \(r\) rows and \(c\) columns have products divisible by \(3\). Every multiple of \(3\) must lie in one of those \(r\) rows and also in one of those \(c\) columns, or else it would create another marked row or column. Thus the \(27\) multiples must fit in the \(rc\) intersection cells. If \(r+c\le10\), then \(rc\le25\), which is too small. So at least \(11\) rows and columns are needed.

This can be done by placing \(25\) multiples of \(3\) in a \(5\times5\) block, then placing the remaining \(2\) multiples in a sixth column within two of those same rows. Then exactly \(5\) rows and \(6\) columns are marked, for a total of \(11\).

Thus, D is the correct answer.

17 答案 E

先选白王所在方格。若白王在中心，它攻击所有其他方格，黑王有 \(0\) 个可放位置。

若白王在角落，它攻击 \(3\) 个方格，黑王有 \(9-1-3=5\) 个安全格。角落有 \(4\) 个，所以共有 \(4\cdot5=20\) 种。

若白王在边上但不是角落，它攻击 \(5\) 个方格，黑王有 \(9-1-5=3\) 个安全格。这样的边中格有 \(4\) 个，所以共有 \(4\cdot3=12\) 种。

总数为 \(20+12=32\)。

所以正确答案是 E。

English

Count ordered placements by first choosing the square for the white king. If the white king is in the center, it attacks every other square, so there are \(0\) choices for the black king.

If the white king is in a corner, it attacks \(3\) squares, so the black king has \(9-1-3=5\) safe squares. There are \(4\) corner choices, giving \(4\cdot5=20\) placements.

If the white king is on an edge but not a corner, it attacks \(5\) squares, so the black king has \(9-1-5=3\) safe squares. There are \(4\) such edge choices, giving \(4\cdot3=12\) placements.

The total number of placements is \(20+12=32\).

Thus, E is the correct answer.

18 答案 A

设 \(\theta\) 为 \(\angle BOC\) 的度数。

阴影区域的一部分是半径 \(2\) 圆与半径 \(1\) 圆之间的圆环，面积为 \(4\pi-\pi=3\pi\)。另一部分是最大圆的一个扇形减去半径 \(2\) 的圆，面积为 \(\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi)\)。所以阴影面积为 \(3\pi + \dfrac{\theta}{360}(5\pi)\)。

非阴影区域由最小圆和外圆环未阴影部分组成，面积为 \(\pi + \dfrac{360-\theta}{360}(5\pi)\)。

令两部分面积相等并解 \(\theta\)：\[ \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} \] \[ 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) \] \[ \dfrac{2}{5} = 1 - \dfrac{2\theta}{360} \] \[ 2\theta = \dfrac{3}{5}(360) \] \[ \theta = 108\text{。}\]

所以正确答案是 A。

English

Let \(\theta\) be the measure of \(\angle BOC.\)

One component of the shaded region is the area of the circle with radius \(2\) minus the area of the circle with radius \(1.\) This part has area \(4\pi-\pi=3\pi.\) The remaining area is a sector of the biggest circle minus the area of the circle with radius \(2\). This has area \(\dfrac{\theta}{360}(9\pi-4\pi) = \dfrac{\theta}{360}(5\pi).\) Hence, the total area of the shaded region is \(3\pi + \dfrac{\theta}{360}(5\pi).\)

Next, we note that the unshaded region is composed of the smallest circle and the unshaded portion of the outer ring. This will have a total area of \(\pi + \dfrac{360-\theta}{360}(5\pi).\)

Lastly, we equate the area of both regions and solve for \(\theta:\) \[ \begin{gathered} 3\pi + \dfrac{\theta}{360}(5\pi) \\ = \pi + \dfrac{360-\theta}{360}(5\pi) \end{gathered} \] \[ 2\pi = \dfrac{360-\theta-\theta}{360}(5\pi) \] \[ \dfrac{2}{5} = 1 - \dfrac{2\theta}{360} \] \[ 2\theta = \dfrac{3}{5}(360) \] \[ \theta = 108. \]

Thus, A is the correct answer.

19 答案 C

红色鞋有 \(\dfrac{3}{5} \times 15 = 9\) 双，白色鞋有 \(6\) 双。另外，高帮鞋有 \(\dfrac{2}{3} \times 15 = 10\) 双，低帮鞋有 \(5\) 双。为了让红色高帮尽量少，先让全部 \(6\) 双白色鞋都是高帮，还需要 \(10-6=4\) 双红色高帮，所以最小分数为 \(\dfrac{4}{15}\)。

所以正确答案是 C。

English

Jordan has \(\dfrac{3}{5} \times 15 = 9\) pairs of red sneakers and \(6\) pairs of white sneakers. Moreover, \(\dfrac{2}{3} \times 15 = 10\) are high-top and \(5\) are low-top. If we want to minimize the number of red high-top sneakers, then we can set all \(6\) white sneakers to be high-top, leaving \(10-6=4\) red sneakers as high-top. Hence, the fraction of red high-top sneakers would be \(\dfrac{4}{15}\).

Thus, C is the correct answer.

20 答案 D

首先注意到，只有沿着正方形面的对角线才能得到等边三角形，否则三角形至少有一个角与其余的不同。之后就容易逐一列出所有可能的等边三角形：\(\triangle PVT\)、\(\triangle PRT\) 和 \(\triangle PRV\)。

所以正确答案是 D。

English

We first note that we can only form equilateral triangles if we go through the diagonals of the square faces, otherwise at least one angle of the triangle will be different. Afterwards, it is easy to exhaust all possible equilateral triangles that can be formed: \(\triangle PVT,\) \(\triangle PRT,\) and \(\triangle PRV.\)

Thus, D is the correct answer.

21 答案 E

设最初绿色青蛙有 \(g\) 只，黄色青蛙有 \(y\) 只，则 \(g=3y\)。移动后有 \[\dfrac{g+5-3}{y+3-5} = \dfrac{4}{1}\text{。}\]

把 \(3y\) 代入 \(g\)，即可求出原来的黄色青蛙数 \(y\)：\[\dfrac{3y+2}{y-2} = 4\] \[3y+2 = 4(y-2) = 4y-8\] \[y = 10\]

因此最初黄色有 \(10\) 只，绿色有 \(3 \times 10 = 30\) 只。移动后黄色有 \(10+3-5 = 8\) 只，绿色有 \(30+5-3=32\) 只，差为 \(32-8=24\)。

所以正确答案是 E。

English

We can let \(g\) be the number of green frogs and \(y\) be the number of yellow frogs. Initially, we have \(g=3y\). Then, after some frogs moved, we have the following proportion: \[\dfrac{g+5-3}{y+3-5} = \dfrac{4}{1}.\]

Substituting \(3y\) for \(g\) will allow us to determine the number of yellow frogs originally (\(y\)): \[\dfrac{3y+2}{y-2} = 4\] \[3y+2 = 4(y-2) = 4y-8\] \[y = 10\]

Hence, there were \(10\) yellow frogs and \(3 \times 10 = 30\) green frogs initially. After some frogs moved, we now have \(10+3-5 = 8\) yellow frogs and \(30+5-3=32\) green frogs, giving us a difference of \(32-8=24\) between the number of green and yellow frogs.

Thus, E is the correct answer.

22 答案 B

胶带占据一个外半径为 \(2\) 英寸、内半径为 \(1\) 英寸的圆环，所以它的截面积为 \[ \pi(2^2-1^2)=3\pi\text{。}\]

胶带展开后，同一个截面成为厚 \(0.015\) 英寸、长 \(L\) 英寸的长方形。因此 \[ 0.015L=3\pi\text{，}\]所以 \(L=200\pi\approx628\) 英寸。四舍五入到最接近的 \(100\) 英寸整数倍，胶带长约 \(600\) 英寸。

所以正确答案是 B。

English

The tape occupies an annulus with outer radius \(2\) inches and inner radius \(1\) inch, so its cross-sectional area is \[ \pi(2^2-1^2)=3\pi. \]

When unrolled, this same cross section is a rectangle of thickness \(0.015\) inches and length \(L.\) Thus \[ 0.015L=3\pi, \] so \(L=200\pi\approx628\) inches. Rounded to the nearest \(100\) inches, this is \(600\) inches.

Thus, B is the correct answer.

23 答案 C

若线段水平差为 \(a\)、竖直差为 \(b\)，它穿过 \(a\) 条竖直网格线和 \(b\) 条水平网格线，但在格点处的穿越会被重复计算。因此内部被线段穿过的方格数为 \[ a+b-\gcd(a,b)\text{。}\]

这里坐标差为 \(3000\) 和 \(5000\)，且 \(\gcd(3000,5000)=1000\)。所以方格数为 \[ 3000+5000-1000=7000\text{。}\]

正确答案是 C。

English

For a segment whose endpoint differences are \(a\) horizontally and \(b\) vertically, the segment crosses \(a\) vertical grid lines and \(b\) horizontal grid lines, but crossings at lattice points are counted twice. Therefore the number of cells whose interiors are intersected is \[ a+b-\gcd(a,b). \]

Here the endpoint differences are \(3000\) and \(5000\), and \(\gcd(3000,5000)=1000\). Thus the number of cells colored is \[ 3000+5000-1000=7000. \]

Thus, C is the correct answer.

24 答案 B

每座山都是 \(45^\circ\text{-}45^\circ\text{-}90^\circ\) 三角形。高为 \(x\) 的这种直角等腰三角形面积为 \(x^2\)，因为两条直角边都长 \(x\sqrt2\)。

两座大山的面积分别为 \(8^2=64\) 和 \(12^2=144\)。重叠部分也是 \(45^\circ\text{-}45^\circ\text{-}90^\circ\) 三角形，高为 \(h\)，面积为 \(h^2\)。因此 \[ 64+144-h^2=183\text{，}\]所以 \(h^2=25\)，进而 \(h=5\)。

正确答案是 B。

English

Each mountain is a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle. A right isosceles triangle with height \(x\) has area \(x^2\), since its two perpendicular sides each have length \(x\sqrt2\).

The two large mountains have areas \(8^2=64\) and \(12^2=144\). Their overlap is also a \(45^\circ\text{-}45^\circ\text{-}90^\circ\) triangle with height \(h\), so its area is \(h^2\). Thus \[ 64+144-h^2=183, \] so \(h^2=25\), and \(h=5\).

Thus, B is the correct answer.

25 答案 C

前 \(8\) 名乘客坐下后，还有 \(4\) 个空座分布在 \(12\) 个座位中，所以空座集合共有 \(\binom{12}{4}=495\) 种等可能情况。

用补集计数，即没有任何一排有两个相邻空座。在一排三个座位中，不含相邻空座的方案所含空座数可以是 \(0\)、\(1\) 或 \(2\)，对应的方案数分别是 \(1\)、\(3\) 和 \(1\)。因此，我们需要求下式中 \(x^4\) 的系数：\[ (1+3x+x^2)^4\text{。}\]这个系数是 \[ \begin{gathered} 3^4+4\cdot3\cdot3^2+\binom42 \\ =81+108+6 \\ =195 \end{gathered}\text{。}\]

因此至少有一排存在相邻空座的概率为 \[ \frac{495-195}{495}=\frac{20}{33}\text{。}\]

所以正确答案是 C。

English

After the first \(8\) passengers sit, there are \(4\) empty seats among the \(12\) seats, so there are \(\binom{12}{4}=495\) equally likely sets of empty seats.

Count the complement, where no row has two adjacent empty seats. In a row of three seats, the possible empty-seat patterns with no adjacent empty seats have sizes \(0,\) \(1,\) and \(2,\) with \(1,\) \(3,\) and \(1\) choices respectively. So we need the coefficient of \(x^4\) in \[ (1+3x+x^2)^4. \] This coefficient is \[ \begin{gathered} 3^4+4\cdot3\cdot3^2+\binom42 \\ =81+108+6 \\ =195. \end{gathered} \]

Therefore the probability that at least one row has adjacent empty seats is \[ \frac{495-195}{495}=\frac{20}{33}. \]

Thus, C is the correct answer.