2002B AMC 10 真题（P版）

1 答案 E

因为 \(6^{2002} = 2^{2002}\cdot 3^{2002}\)，所以 \[\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}\text{。}\]

所以正确答案是 E。

English

Since \(6^{2002} = 2^{2002}\cdot 3^{2002},\) the ratio becomes \[\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.\]

Thus, the correct answer is E.

2 答案 C

直接代入：\[(2, 4, 6) = \dfrac{2\cdot4\cdot6}{2+4+6} = \dfrac{48}{12} = 4\text{。}\]

所以正确答案是 C。

English

Substituting directly, \[(2, 4, 6) = \dfrac{2\cdot4\cdot6}{2+4+6} = \dfrac{48}{12} = 4.\]

Thus, the correct answer is C.

3 答案 A

平均数可以写成 \[ \begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111 \end{aligned}\text{。}\]

把这九个各位均为一的数按列相加，得到 \(M = 123456789\)。

\(M\) 唯一缺少的数字是 \(0\)。

所以正确答案是 A。

English

The mean is \[ \begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111. \end{aligned} \]

Adding these nine repunits column by column gives \(M = 123456789.\)

The only digit missing from \(M\) is \(0.\)

Thus, the correct answer is A.

4 答案 D

从前两项提出 \(3x-2\)，得 \[ \begin{aligned} &(3x - 2)\big[(4x + 1) - 4x\big] + 1 \\ &= (3x - 2)(1) + 1 \\ &= 3x - 1 \end{aligned}\text{。}\]

当 \(x = 4\) 时，表达式等于 \(3\cdot4 - 1 = 11\)。

所以正确答案是 D。

English

Factoring \(3x-2\) from the first two terms, \[ \begin{aligned} &(3x - 2)\big[(4x + 1) - 4x\big] + 1 \\ &= (3x - 2)(1) + 1 \\ &= 3x - 1. \end{aligned} \]

At \(x = 4,\) this equals \(3\cdot4 - 1 = 11.\)

Thus, the correct answer is D.

5 答案 E

两个小圆沿大圆的一条直径排列，所以大圆直径为 \(2\cdot3 + 2\cdot2 = 10\)，半径为 \(5\)。

阴影面积为 \[ \begin{aligned} &\pi(5^2) - \pi(3^2) - \pi(2^2) \\ &= \pi(25 - 9 - 4) \\ &= 12\pi \end{aligned}\text{。}\]

所以正确答案是 E。

English

The two small circles line up along a diameter of the big circle, so that diameter is \(2\cdot3 + 2\cdot2 = 10\) and the large radius is \(5.\)

The shaded region is the large disk with the two small disks removed: \[ \begin{aligned} &\pi(5^2) - \pi(3^2) - \pi(2^2) \\ &= \pi(25 - 9 - 4) \\ &= 12\pi. \end{aligned} \]

Thus, the correct answer is E.

6 答案 B

分解得 \(n^2 - 3n + 2 = (n-1)(n-2)\)。

当 \(n \ge 4\) 时，两个因数都大于 \(1\)，乘积为合数。当 \(n = 1\) 或 \(n = 2\) 时，原式为 \(0\)；当 \(n = 3\) 时，原式为 \((2)(1) = 2\)，是质数。

所以恰好有一个 \(n\) 可行。

所以正确答案是 B。

English

Factor as \(n^2 - 3n + 2 = (n-1)(n-2).\)

For \(n \ge 4,\) both factors exceed \(1,\) so the product is composite. For \(n = 1\) and \(n = 2\) the value is \(0,\) and for \(n = 3\) the value is \((2)(1) = 2,\) which is prime.

So exactly one value of \(n\) works.

Thus, the correct answer is B.

7 答案 E

和 \(\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n\) 大于 \(0\)，并且小于 \(\dfrac12 + \dfrac13 + \dfrac17 + 1 \lt 2\)，所以作为整数只能等于 \(1\)。

因为 \(\dfrac12 + \dfrac13 + \dfrac17 = \dfrac{41}{42}\)，所以 \(\dfrac1n = \dfrac{1}{42}\)，即 \(n = 42\)。

具体说，\(2\)、\(3\)、\(6\)、\(7\) 都整除 \(42\)，而 \(n = 42\) 不大于 \(84\)，所以 \(n \gt 84\) 为假。

所以正确答案是 E。

English

The sum \(\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n\) is greater than \(0\) and less than \(\dfrac12 + \dfrac13 + \dfrac17 + 1 \lt 2,\) so as an integer it must equal \(1.\)

Since \(\dfrac12 + \dfrac13 + \dfrac17 = \dfrac{41}{42},\) we need \(\dfrac1n = \dfrac{1}{42},\) so \(n = 42.\)

Then \(2,\) \(3,\) \(6,\) and \(7\) all divide \(42,\) but \(n = 42\) is not greater than \(84.\) So the false statement is \(n \gt 84.\)

Thus, the correct answer is E.

8 答案 D

\(31\) 天等于 \(4\) 周又 \(3\) 天，所以恰好是每月第 \(1\)、第 \(2\)、第 \(3\) 日所对应的三个星期几会出现五次。

七月要有五个星期一，星期一就必须是七月第 \(1\)、第 \(2\) 或第 \(3\) 日。三种情况下，八月 \(1\) 日分别是星期二、星期三或星期四，而八月出现五次的三个星期几都包括星期四。

也可以直接看出：七月有 \(31\) 天，所以八月 \(1\) 日与七月 \(4\) 日是同一星期几。无论星期一是七月第 \(1\)、第 \(2\) 还是第 \(3\) 日，八月出现五次的三个星期几中总有星期四。

所以正确答案是 D。

English

A \(31\)-day month is \(4\) weeks plus \(3\) extra days, so exactly the weekdays of the \(1\)st, \(2\)nd, and \(3\)rd of the month occur five times.

For July to have five Mondays, Monday must be one of July \(1,\) \(2,\) or \(3.\) In all three cases August \(1\) lands on a Tuesday, Wednesday, or Thursday, and the common weekday among the resulting five-time days is Thursday.

More directly, since July has \(31\) days, August \(1\) is the same weekday as July \(4.\) With Monday on July \(1,\) \(2,\) or \(3,\) the three five-time weekdays of August always include Thursday.

Thus, the correct answer is D.

9 答案 D

字母顺序为 \(A, M, O, S, U\)。以 \(A\)、\(M\)、\(O\)、\(S\) 开头的单词先出现，占据从第 \(1\) 位到第 \(96\) 位；每个开头有 \(24\) 个。

以 \(U\) 开头的单词占据第 \(97\) 到第 \(120\) 位。在这一组中，\(USAMO\) 是第 \(19\) 个，所以总位置为 \(96 + 19 = 115\)。

所以正确答案是 D。

English

The alphabetical order of the letters is \(A, M, O, S, U.\) Words beginning with \(A,\) \(M,\) \(O,\) or \(S\) fill positions \(1\) through \(96\) (four choices of first letter, \(24\) each).

Words beginning with \(U\) occupy positions \(97\)–\(120.\) Listing them alphabetically, \(USAMO\) is the \(19\)th such word, so it occupies position \(96 + 19 = 115.\)

Thus, the correct answer is D.

10 答案 C

根为 \(a\)、\(b\)。由韦达定理，\(a + b = -a\)，且 \(ab = b\)。

先看 \(ab = b\)。因为 \(b \ne 0\)，可得 \(a = 1\)。再用 \(a + b = -a\)，得到 \(1 + b = -1\)，所以 \(b = -2\)。

因此 \((a, b) = (1, -2)\)，所以正确答案是 C。

English

Since the roots are \(a\) and \(b,\) Vieta’s formulas give \(a + b = -a\) and \(ab = b.\)

From \(ab = b\) with \(b \ne 0,\) we get \(a = 1.\) Then \(a + b = -a\) gives \(1 + b = -1,\) so \(b = -2.\)

Thus \((a, b) = (1, -2),\) and the correct answer is C.

11 答案 B

设三个整数为 \(n - 1\)、\(n\)、\(n + 1\)。它们的乘积为 \(n(n^2 - 1)\)，和为 \(3n\)，因此 \[n(n^2 - 1) = 8(3n) = 24n\text{。}\]

因为 \(n \ne 0\)，可得 \(n^2 - 1 = 24\)，于是 \(n^2 = 25\)，从而 \(n = 5\)。

三个整数为 \(4\)、\(5\)、\(6\)，平方和为 \(4^2 + 5^2 + 6^2 = 16 + 25 + 36\) \(= 77\)。

所以正确答案是 B。

English

Let the integers be \(n - 1,\) \(n,\) \(n + 1.\) Their product is \(n(n^2 - 1)\) and their sum is \(3n,\) so \[n(n^2 - 1) = 8(3n) = 24n.\]

Since \(n \ne 0,\) we get \(n^2 - 1 = 24,\) so \(n^2 = 25\) and \(n = 5.\)

The three integers are \(4,\) \(5,\) and \(6,\) and \(4^2 + 5^2 + 6^2 = 16 + 25 + 36\) \(= 77.\)

Thus, the correct answer is B.

12 答案 E

交叉相乘得 \((x - 1)(x - 6) = (x - 2)(x - k)\)，展开为 \[ \begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k \end{aligned}\text{。}\]

约去 \(x^2\) 后得到 \((k - 5)x = 2k - 6\)。当 \(k\) 等于 \(1\)、\(2\)、\(3\) 或 \(4\) 时，都能得到一个有效的 \(x\)，且它既不是使分母为零的 \(2\)，也不是 \(6\)。当 \(k=5\) 时，方程变为 \(0\cdot x=4\)，无解。

所以正确答案是 E。

English

Cross multiplying gives \((x - 1)(x - 6) = (x - 2)(x - k),\) which expands to \[ \begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k. \end{aligned} \]

Cancelling \(x^2\) leaves \((k - 5)x = 2k - 6.\) For \(k\) equal to \(1,\) \(2,\) \(3,\) or \(4,\) this gives a valid value of \(x\) that is neither excluded denominator value \(2\) nor \(6.\) For \(k=5,\) the equation instead becomes \(0\cdot x=4,\) which has no solution.

Thus, the correct answer is E.

13 答案 D

分组因式分解：\[ \begin{aligned} &8xy - 12y + 2x - 3 \\ &= 4y(2x - 3) + (2x - 3) \\ &= (4y + 1)(2x - 3) \end{aligned}\text{。}\]

要使它对所有 \(y\) 都等于 \(0\)，不能依靠含 \(y\) 的因子为零，所以必须有 \(2x - 3 = 0\)，得到 \(x = \dfrac32\)。

所以正确答案是 D。

English

Grouping and factoring, \[ \begin{aligned} &8xy - 12y + 2x - 3 \\ &= 4y(2x - 3) + (2x - 3) \\ &= (4y + 1)(2x - 3). \end{aligned} \]

For this to equal \(0\) for all \(y,\) the factor that depends on \(y\) cannot be forced to zero, so we need \(2x - 3 = 0,\) giving \(x = \dfrac32.\)

Thus, the correct answer is D.

14 答案 B

因为 \(25 = 5^2\)、\(64 = 2^6\)，所以 \(25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150}\)，从而 \[N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}\text{。}\]

写成 \(2^{75} = 2^{64}\cdot 2^{11}\)，可得 \(N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048\)。

因此 \(N\) 是 \(2048\) 后接 \(64\) 个零，数字和为 \(2 + 0 + 4 + 8 = 14\)。

所以正确答案是 B。

English

Since \(25 = 5^2\) and \(64 = 2^6,\) we have \(25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150},\) so \[N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.\]

Writing \(2^{75} = 2^{64}\cdot 2^{11},\) we get \(N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.\)

So \(N\) is \(2048\) followed by \(64\) zeros, and its digit sum is \(2 + 0 + 4 + 8 = 14.\)

Thus, the correct answer is B.

15 答案 E

数 \(A - B\) 与 \(A + B\) 相差 \(2B\)，所以奇偶性相同。因为它们是质数，两者都必须是奇数，这迫使 \(A\) 与 \(B\) 奇偶性相反。

因为 \(2\) 是唯一的偶质数，所以要么 \(A=2\)，要么 \(B=2\)。第一种情况不可能，因为正质数 \(B\) 会使 \(A-B<0\)。因此 \(B=2\)。

现在 \(A - 2\)、\(A\)、\(A + 2\) 是三个质数。任意三个相差 \(2\) 的整数中必有一个能被 \(3\) 整除，所以这个数本身必须是 \(3\)。唯一的正数情形是 \(3\)、\(5\)、\(7\)。

这四个质数是 \(2\)、\(3\)、\(5\)、\(7\)，它们的和为 \(17\)，也是质数。

所以正确答案是 E。

English

The numbers \(A - B\) and \(A + B\) differ by \(2B,\) so they have the same parity. Being prime, they must both be odd, which forces \(A\) and \(B\) to have opposite parity.

Since \(2\) is the only even prime, either \(A=2\) or \(B=2.\) The first case is impossible because the positive prime \(B\) would make \(A-B<0.\) Hence \(B=2.\)

Now \(A - 2,\) \(A,\) and \(A + 2\) are three primes. One of any three integers spaced \(2\) apart is divisible by \(3,\) so that member must itself be \(3.\) The only positive possibility is \(3,\) \(5,\) \(7.\)

The four primes are \(2,\) \(3,\) \(5,\) \(7,\) and their sum is \(17,\) which is prime.

Thus, the correct answer is E.

16 答案 D

设 \(\dfrac{n}{20 - n} = k^2\)，其中 \(k \ge 0\) 是整数。于是 \[n = \dfrac{20k^2}{k^2 + 1}\text{。}\]

因为 \(k^2\) 与 \(k^2 + 1\) 互质，所以 \(k^2 + 1\) 必须整除 \(20\)。这只在 \(k = 0\)、\(1\)、\(2\)、\(3\) 时发生，对应 \(k^2 + 1 = 1\)、\(2\)、\(5\)、\(10\)。

对应的整数值为 \(n = 0\)、\(10\)、\(16\)、\(18\)，因此共有 \(4\) 个这样的 \(n\)。

所以正确答案是 D。

English

Suppose \(\dfrac{n}{20 - n} = k^2\) for some integer \(k \ge 0.\) Solving, \[n = \dfrac{20k^2}{k^2 + 1}.\]

Since \(k^2\) and \(k^2 + 1\) share no common factor, \(k^2 + 1\) must divide \(20.\) This happens only for \(k = 0,\) \(1,\) \(2,\) \(3,\) giving \(k^2 + 1 = 1,\) \(2,\) \(5,\) \(10.\)

The corresponding values \(n = 0,\) \(10,\) \(16,\) \(18\) are all integers, so there are \(4\) such \(n.\)

Thus, the correct answer is D.

17 答案 C

将八边形放在坐标轴上，使水平边和竖直边的长度为 \(2\)，每条斜边的水平位移为 \(\sqrt2\)，竖直位移也为 \(\sqrt2\)。\[ \begin{aligned} A &= (\sqrt2, 0), \\ D &= (2 + 2\sqrt2, 2 + \sqrt2), \\ G &= (0, 2 + \sqrt2) \end{aligned}\text{。}\]

因为 \(D\) 和 \(G\) 高度同为 \(2 + \sqrt2\)，所以 \(DG\) 水平，长度为 \(2 + 2\sqrt2\)，从 \(A\) 到这条水平线的高为 \(2 + \sqrt2\)。

因此 \[ \begin{aligned} [\triangle ADG] &= \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) \\ &= (1 + \sqrt2)(2 + \sqrt2) \\ &= 4 + 3\sqrt2 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Set the octagon on coordinate axes with the axis-aligned sides of length \(2\) and each slanted side spanning \(\sqrt2\) horizontally and \(\sqrt2\) vertically. Then \[ \begin{aligned} A &= (\sqrt2, 0), \\ D &= (2 + 2\sqrt2, 2 + \sqrt2), \\ G &= (0, 2 + \sqrt2). \end{aligned} \]

Since \(D\) and \(G\) share the height \(2 + \sqrt2,\) segment \(DG\) is horizontal with length \(2 + 2\sqrt2,\) and the height from \(A\) up to that level is \(2 + \sqrt2.\)

Therefore \[ \begin{aligned} [\triangle ADG] &= \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) \\ &= (1 + \sqrt2)(2 + \sqrt2) \\ &= 4 + 3\sqrt2. \end{aligned} \]

Thus, the correct answer is C.

18 答案 D

任意两个不同圆最多相交于 \(2\) 个点。四个圆的配对数为 \(\binom{4}{2} = 6\)，所以最多有 \(6\cdot 2 = 12\) 个交点。

这个最大值可以达到：安排四个圆，使每一对圆都相交于两个不同的点，就会得到 \(12\) 个交点。

所以正确答案是 D。

English

Any two distinct circles intersect in at most \(2\) points. There are \(\binom{4}{2} = 6\) pairs of circles, giving at most \(6\cdot 2 = 12\) intersection points.

This maximum is achievable by a configuration where every pair of circles crosses twice, so the answer is \(12.\)

Thus, the correct answer is D.

19 答案 C

设 \(d = a_2 - a_1\)。因为 \(a_{k+100} = a_k + 100d\)，第二组的一百项相对于第一组对应项的增量都相同，所以第二组的总和比第一组多 \(100\cdot 100 d\)：\[ \begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d \end{aligned}\text{。}\]

因此 \(200 = 100 + 10000d\)，解得 \(d = \dfrac{100}{10000} = 0.01\)。

所以正确答案是 C。

English

Let \(d = a_2 - a_1.\) Then \(a_{k+100} = a_k + 100d,\) so the second block sum is the first plus \(100\cdot 100 d:\) \[ \begin{aligned} &a_{101} + \cdots + a_{200} \\ &= (a_1 + \cdots + a_{100}) \\ &\quad {}+ 10000d. \end{aligned} \]

Therefore \(200 = 100 + 10000d,\) giving \(d = \dfrac{100}{10000} = 0.01.\)

Thus, the correct answer is C.

20 答案 B

将方程改写为 \(a + 8c = 4 + 7b\) 和 \(8a - c = 7 - 4b\)。两式平方后相加：\[ \begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2 \end{aligned}\text{。}\]

左边展开为 \(65a^2 + 65c^2\)，\(ac\) 项抵消；右边展开为 \(65 + 65b^2\)，\(b\) 项抵消。因此 \[65(a^2 + c^2) = 65(1 + b^2)\text{，}\]所以 \(a^2 - b^2 + c^2 = 1\)。

所以正确答案是 B。

English

Rewrite the equations as \(a + 8c = 4 + 7b\) and \(8a - c = 7 - 4b.\) Squaring both and adding, \[ \begin{aligned} &(a + 8c)^2 + (8a - c)^2 \\ &= (4 + 7b)^2 \\ &\quad {}+ (7 - 4b)^2. \end{aligned} \]

The left side expands to \(65a^2 + 65c^2\) (the \(ac\) terms cancel), and the right side expands to \(65 + 65b^2\) (the \(b\) terms cancel). So \[65(a^2 + c^2) = 65(1 + b^2),\] giving \(a^2 - b^2 + c^2 = 1.\)

Thus, the correct answer is B.

21 答案 B

设 Andy 的草坪面积为 \(A\)，则 Beth 的面积为 \(\dfrac{A}{2}\)，Carlos 的面积为 \(\dfrac{A}{3}\)。再设 Carlos 的速度为 \(R\)，Beth 和 Andy 的速度分别为 \(2R\)、\(3R\)。

三人所需时间分别为 \[ \begin{aligned} &\text{Andy: } \dfrac{A}{3R}, \\ &\text{Beth: } \dfrac{\frac{A}{2}}{2R} = \dfrac{A}{4R}, \\ &\text{Carlos: } \dfrac{\frac{A}{3}}{R} = \dfrac{A}{3R} \end{aligned}\text{。}\]

其中最短的是 \(\dfrac{A}{4R}\)，所以 Beth 最先完成。

所以正确答案是 B。

English

Let Andy’s lawn have area \(A,\) so Beth’s is \(\dfrac{A}{2}\) and Carlos’ is \(\dfrac{A}{3}.\) Let Carlos mow at rate \(R,\) so Beth mows at \(2R\) and Andy at \(3R.\)

The times are \[ \begin{aligned} &\text{Andy: } \dfrac{A}{3R}, \\ &\text{Beth: } \dfrac{\frac{A}{2}}{2R} = \dfrac{A}{4R}, \\ &\text{Carlos: } \dfrac{\frac{A}{3}}{R} = \dfrac{A}{3R}. \end{aligned} \]

Since \(\dfrac{A}{4R}\) is the smallest, Beth finishes first.

Thus, the correct answer is B.

22 答案 B

设 \(OM = a\)、\(ON = b\)，则 \(OX = 2a\)、\(OY = 2b\)。由 \(O\) 处直角得 \[19^2 = (2a)^2 + b^2\] 和 \[22^2 = a^2 + (2b)^2\text{。}\]

两式相加得 \(5(a^2 + b^2) = 19^2 + 22^2 = 845\)，所以 \(a^2 + b^2 = 169\)，并且 \(MN = \sqrt{a^2 + b^2} = 13\)。

由于 \(\triangle XOY \sim \triangle MON\)，相似比为 \(2\)，所以 \(XY = 2\cdot MN = 26\)。

所以正确答案是 B。

English

Let \(OM = a\) and \(ON = b,\) so \(OX = 2a\) and \(OY = 2b.\) The right angle at \(O\) gives \[19^2 = (2a)^2 + b^2\] and \[22^2 = a^2 + (2b)^2.\]

Adding these, \(5(a^2 + b^2) = 19^2 + 22^2 = 845,\) so \(a^2 + b^2 = 169\) and \(MN = \sqrt{a^2 + b^2} = 13.\)

Since \(\triangle XOY \sim \triangle MON\) with ratio \(2,\) we have \(XY = 2\cdot MN = 26.\)

Thus, the correct answer is B.

23 答案 D

令 \(n = 1\)，得到 \(a_{m+1} = a_m + a_1 + m\) \(= a_m + (m + 1)\)，所以 \(a_{m+1} - a_m = m + 1\)。

从 \(m = 1\) 加到 \(11\)，有 \[ \begin{aligned} a_{12} - a_1 &= 2 + 3 + \cdots + 12 \\ &= \dfrac{12\cdot 13}{2} - 1 \\ &= 77 \end{aligned}\text{。}\]

因此 \(a_{12} = 1 + 77 = 78\)。

所以正确答案是 D。

English

Setting \(n = 1,\) we get \(a_{m+1} = a_m + a_1 + m\) \(= a_m + (m + 1),\) so \(a_{m+1} - a_m = m + 1.\)

Summing from \(m = 1\) to \(11,\) \[ \begin{aligned} a_{12} - a_1 &= 2 + 3 + \cdots + 12 \\ &= \dfrac{12\cdot 13}{2} - 1 \\ &= 77. \end{aligned} \]

Therefore \(a_{12} = 1 + 77 = 78.\)

Thus, the correct answer is D.

24 答案 D

把圆心 \(O\) 放在高度 \(20\) 处。最低点 \(A\) 的高度为 \(0\)，乘客到达的高度为 \(10\)，也就是在圆心下方 \(10\) 英尺处。

从乘客所在点向摩天轮的竖直直径作水平线段。所得直角三角形的竖直直角边长为 \(10\)，斜边即半径，长为 \(20\)。直角边是斜边的一半，所以乘客所在半径与竖直向下方向成 \(60^\circ\)。

摩天轮在 \(60\) 秒内转过 \(360^\circ\)，所以转过 \(60^\circ\) 需要 \(\dfrac{60}{360}\cdot 60 = 10\) 秒。

所以正确答案是 D。

English

Put the center \(O\) at height \(20.\) The bottom \(A\) is at height \(0,\) and the rider reaches height \(10,\) which is \(10\) feet below the center.

Draw a horizontal segment from the rider to the wheel’s vertical diameter. The resulting right triangle has a vertical leg of length \(10\) and a hypotenuse (the radius) of length \(20.\) That leg is half the hypotenuse, so the radius to the rider makes \(60^\circ\) with the downward vertical.

The wheel turns \(360^\circ\) in \(60\) seconds, so turning \(60^\circ\) takes \(\dfrac{60}{360}\cdot 60 = 10\) seconds.

Thus, the correct answer is D.

25 答案 A

设原列表有 \(n\) 个整数，平均数为 \(m\)，则总和为 \(mn\)。加入 \(15\) 后，\[ \begin{aligned} &(m + 2)(n + 1) \\ &= mn + 15 \\ &\implies m + 2n = 13 \end{aligned}\text{。}\]

再加入 \(1\) 后，\[ \begin{aligned} &(m + 1)(n + 2) \\ &= mn + 16 \\ &\implies 2m + n = 14 \end{aligned}\text{。}\]

解方程组 \(m + 2n = 13\)、\(2m + n = 14\)，得 \(m = 5\)、\(n = 4\)，所以原列表有四个整数。

所以正确答案是 A。

English

Let the original list have \(n\) integers with mean \(m,\) so its sum is \(mn.\) Appending \(15\) gives \[ \begin{aligned} &(m + 2)(n + 1) \\ &= mn + 15 \\ &\implies m + 2n = 13. \end{aligned} \]

Appending \(1\) to that enlarged list gives \[ \begin{aligned} &(m + 1)(n + 2) \\ &= mn + 16 \\ &\implies 2m + n = 14. \end{aligned} \]

Solving \(m + 2n = 13\) and \(2m + n = 14\) yields \(m = 5\) and \(n = 4.\)

Thus, the correct answer is A.