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AMC10 2003 B

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AMC10 · 2003 (B)

Q1
Which of the following is the same as \[ \frac{2 -4 + 6 -8 + 10 -12 + 14}{3 -6 + 9 -12 + 15 -18 + 21} \]?
以下哪个与 \[ \frac{2 -4 + 6 -8 + 10 -12 + 14}{3 -6 + 9 -12 + 15 -18 + 21} \] 相同？
Correct Answer: C
We have \[ \frac{2 -4 + 6 -8 + 10 -12 + 14}{3 -6 + 9 -12 + 15 -18 + 21} = \frac{2(1 -2 + 3 -4 + 5 -6 + 7)}{3(1 -2 + 3 -4 + 5 -6 + 7)} = \frac{2}{3}. \]
我们有 \[ \frac{2 -4 + 6 -8 + 10 -12 + 14}{3 -6 + 9 -12 + 15 -18 + 21} = \frac{2(1 -2 + 3 -4 + 5 -6 + 7)}{3(1 -2 + 3 -4 + 5 -6 + 7)} = \frac{2}{3}. \]
Q2
Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $1 more than a pink pill, and Al’s pills cost a total of $546 for the two weeks. How much does one green pill cost?
Al 得了代数炎，必须每天服用一颗绿色药丸和一颗粉色药丸，持续两周。绿色药丸的价格比粉色药丸贵 1 美元，Al 的药丸两周总共花费 546 美元。一颗绿色药丸多少钱？
Correct Answer: D
The cost of each day’s pills is $546/14 = 39$ dollars. If $x$ denotes the cost of one green pill, then $x + (x -1) = 39$, so $x = 20$.
每天药丸的花费是 $546/14 = 39$ 美元。如果 $x$ 表示一颗绿色药丸的花费，则 $x + (x -1) = 39$，所以 $x = 20$。
Q3
The sum of 5 consecutive even integers is 4 less than the sum of the first 8 consecutive odd counting numbers. What is the smallest of the even integers?
5 个连续偶整数的和比前 8 个连续奇数的和少 4。求这些偶整数中最小的那个。
Correct Answer: B
Let $n$ be the smallest of the even integers. Since $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64$, we have $60 = n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 5n + 20$, so $n = 8$.
设 $n$ 是最小的偶整数。因为 $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64$，所以 $60 = n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 5n + 20$，因此 $n = 8$。
Q4
Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $1 each, begonias $1.50 each, cannas $2 each, dahlias $2.50 each, and Easter lilies $3 each. What is the least possible cost, in dollars, for her garden? [Diagram shows a rectangle divided into regions with side lengths labeled 1, 5, 4, 7, 3, 3, 5, 6]
Rose 在她的矩形花坛的每个矩形区域中种不同种类的花。图中标示了花坛矩形区域的边长，单位为英尺。她每个区域每平方英尺种一朵花。紫菀每朵 1 美元，秋海棠每朵 1.50 美元，美人蕉每朵 2 美元，大丽花每朵 2.50 美元，复活节百合每朵 3 美元。她的花园的最低可能花费是多少美元？[图示一个矩形分为区域，边长标示为 1, 5, 4, 7, 3, 3, 5, 6]
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Correct Answer: A
To minimize the cost, Rose should place the most expensive flowers in the smallest region, the next most expensive in the second smallest, etc. The areas of the regions are shown in the figure, so the minimal total cost, in dollars, is $(3)(4) + (2.5)(6) + (2)(15) + (1.5)(20) + (1)(21) = 108$.
为了最小化花费，Rose 应该将最贵的花种在最小区域，其次最贵的种在第二小区域，依此类推。图中区域面积已示，最小总花费为 $(3)(4) + (2.5)(6) + (2)(15) + (1.5)(20) + (1)(21) = 108$ 美元。
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Q5
Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath he cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that no grass is missed. He walks at the rate of 5000 feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?
Moe 用割草机修剪他的 90 英尺乘 150 英尺的矩形草坪。割草宽度为 28 英寸，但他每条割草重叠 4 英寸以确保不漏草。他推着割草机行走速度为每小时 5000 英尺。Moe 修剪草坪需要多长时间（小时）？以下哪个最接近？
Correct Answer: C
The area of the lawn is $90 \cdot 150 = 13,500$ ft$^2$. Moe cuts about two square feet for each foot he pushes the mower forward, so he cuts $2(5000) = 10,000$ ft$^2$ per hour. Therefore, it takes about $13,500 / 10,000 = 1.35$ hours.
草坪面积为 $90 \cdot 150 = 13,500$ ft$^2$。Moe 每推进一步割大约两平方英尺，所以每小时割 $2(5000) = 10,000$ ft$^2$。因此，需要约 $13,500 / 10,000 = 1.35$ 小时。
Q6
Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is 4 : 3. The horizontal length of a “27-inch” television screen is closest, in inches, to which of the following? [Diagram shows a rectangle labeled “Length” (horizontal), “Height” (vertical), and “Diagonal”]
许多电视屏幕是长宽比为 4 : 3 的矩形，以对角线长度来测量尺寸。“27英寸”电视屏幕的水平长度（英寸）最接近于下列哪个选项？[图示一个矩形，标有“Length”（水平）、“Height”（垂直）和“Diagonal”]
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Correct Answer: D
(D) The height, length, and diagonal are in the ratio $3:4:5$. The length of the diagonal is $27$, so the horizontal length is $\frac{4}{5}(27)=21.6$ inches.
（D）高度、长度和对角线的比为 $3:4:5$。对角线长度为 $27$，因此水平长度为 $\frac{4}{5}(27)=21.6$ 英寸。
Q7
The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. For example, $\lfloor 3 \rfloor = 3$, and $\lfloor 9/2 \rfloor = 4$. Compute $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor$.
符号 $\lfloor x \rfloor$ 表示不大于 $x$ 的最大整数。例如，$\lfloor 3 \rfloor = 3$，$\lfloor 9/2 \rfloor = 4$。计算 $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor$。
Correct Answer: B
(B) The first three values in the sum are 1, the next five are 2, the next seven are 3, and the final one is 4 for a total of $3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38.$
（B）和式中的前三个值是 1，接下来的五个是 2，再接下来的七个是 3，最后一个是 4，因此总计为 $3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38.$
Q8
The second and fourth terms of a geometric sequence are 2 and 6. Which of the following is a possible first term?
一个几何数列的第二项和第四项分别为 2 和 6。下列哪项可能是首项？
Correct Answer: B
(B) Let the sequence be denoted $a, ar, ar^2, ar^3, \ldots$, with $ar = 2$ and $ar^3 = 6$. Then $r^2 = 3$ and $r = \sqrt{3}$ or $r = -\sqrt{3}$. Therefore $a = \frac{2\sqrt{3}}{3}$ or $a = -\frac{2\sqrt{3}}{3}$.
（B）设数列表示为 $a, ar, ar^2, ar^3, \ldots$，且 $ar = 2$、$ar^3 = 6$。则 $r^2 = 3$，所以 $r = \sqrt{3}$ 或 $r = -\sqrt{3}$。因此 $a = \frac{2\sqrt{3}}{3}$ 或 $a = -\frac{2\sqrt{3}}{3}$。
Q9
Find the value of $x$ that satisfies the equation $25^{-2} = 548/x \div 526/x \cdot 2517/x$.
求满足方程 $25^{-2} = 548/x \div 526/x \cdot 2517/x$ 的 $x$ 的值。
Correct Answer: B
(B) Write all the terms with the common base 5. Then $5^{-4}=25^{-2}=\dfrac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}=\dfrac{5^{48/x}}{5^{26/x}\cdot5^{34/x}}=5^{(48-26-34)/x}=5^{-12/x}.$ It follows that $-\dfrac{12}{x}=-4$, so $x=3$. OR First write $25$ as $5^2$. Raising both sides to the $x$ power gives $5^{-4x}=\dfrac{5^{48}}{5^{26}5^{34}}=5^{48-26-34}=5^{-12}.$
（B）把所有项都写成以 5 为底的形式。则 $5^{-4}=25^{-2}=\dfrac{5^{48/x}}{5^{26/x}\cdot25^{17/x}}=\dfrac{5^{48/x}}{5^{26/x}\cdot5^{34/x}}=5^{(48-26-34)/x}=5^{-12/x}.$ 因此 $-\dfrac{12}{x}=-4$，所以 $x=3$。 或者 先把 $25$ 写成 $5^2$。两边同取 $x$ 次幂，得 $5^{-4x}=\dfrac{5^{48}}{5^{26}5^{34}}=5^{48-26-34}=5^{-12}.$
Q10
Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?
内布拉斯加州（AMC 的家乡）改变了其车牌方案。旧车牌由一个字母后跟四个数字组成。新车牌由三个字母后跟三个数字组成。新车牌可能数量是旧车牌的多少倍？
Correct Answer: C
(C) In the old scheme $26 \times 10^4$ different plates could be constructed. In the new scheme $26^3 \times 10^3$ different plates can be constructed. There are $\dfrac{26^3 \times 10^3}{26 \times 10^4}=\dfrac{26^2}{10}$ times as many possible plates with the new scheme.
（C）在旧方案中，可以构造 $26 \times 10^4$ 种不同的车牌。在新方案中，可以构造 $26^3 \times 10^3$ 种不同的车牌。因此，新方案下可用车牌数量是 $\dfrac{26^3 \times 10^3}{26 \times 10^4}=\dfrac{26^2}{10}$ 倍。
Q11
A line with slope 3 intersects a line with slope 5 at the point (10, 15). What is the distance between the x-intercepts of these two lines?
一条斜率为 3 的直线与一条斜率为 5 的直线相交于点 (10, 15)。这两条直线的 x 轴截距之间的距离是多少？
Correct Answer: A
(A) The two lines have equations $y-15=3(x-10)$ and $y-15=5(x-10)$. The $x$-intercepts, obtained by setting $y=0$ in the respective equations, are 5 and 7. The distance between the points $(5,0)$ and $(7,0)$ is 2.
（A）两条直线的方程为 $y-15=3(x-10)$ 和 $y-15=5(x-10)$。 将各自方程中的 $y=0$，得到 $x$ 轴截距分别为 5 和 7。点 $(5,0)$ 与 $(7,0)$ 之间的距离为 2。
Q12
Al, Betty, and Clare split $1000 among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of $1500. Betty and Clare have both doubled their money, whereas Al has managed to lose $100. What was Al’s original portion?
Al、Betty 和 Clare 将 1000 美元分给三人，以不同方式投资。每人初始金额不同。一年后他们总共有 1500 美元。Betty 和 Clare 的钱都翻倍了，而 Al 亏了 100 美元。Al 的初始金额是多少？
Correct Answer: C
(C) Denote the original portions for Al, Betty, and Clare as $a$, $b$, and $c$, respectively. Then $a+b+c=1000$ and $a-100+2(b+c)=1500$. Substituting $b+c=1000-a$ in the second equation, we have $a-100+2(1000-a)=1500$. This yields $a=400$, which is Al’s original portion. Note that although we know that $b+c=600$, we have no way of determining either $b$ or $c$.
（C）分别用 $a$、$b$、$c$ 表示 Al、Betty 和 Clare 的原始份额，则 $a+b+c=1000$ 且 $a-100+2(b+c)=1500$。 将 $b+c=1000-a$ 代入第二个方程，得到 $a-100+2(1000-a)=1500$。 由此可得 $a=400$，这就是 Al 的原始份额。 注意，尽管我们知道 $b+c=600$，但无法确定 $b$ 或 $c$ 的具体值。
Q13
Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$. For example, $\clubsuit(8) = 8$ and $\clubsuit(123) = 1 + 2 + 3 = 6$. For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x)) = 3$?
设 $\clubsuit(x)$ 表示正整数 $x$ 的各位数字之和。例如，$\clubsuit(8) = 8$，$\clubsuit(123) = 1 + 2 + 3 = 6$。有且仅有几个两位数 $x$ 满足 $\clubsuit(\clubsuit(x)) = 3$？
Correct Answer: E
(E) Let $y=\clubsuit(x)$. Since $x\le 99$, we have $y\le 18$. Thus if $\clubsuit(y)=3$, then $y=3$ or $y=12$. The 3 values of $x$ for which $\clubsuit(x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit(x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are 10 values in all.
（E）令 $y=\clubsuit(x)$。由于 $x\le 99$，所以 $y\le 18$。因此若 $\clubsuit(y)=3$，则 $y=3$ 或 $y=12$。使得 $\clubsuit(x)=3$ 的 $x$ 有 3 个：12、21、30；使得 $\clubsuit(x)=12$ 的 $x$ 有 7 个：39、48、57、66、75、84、93。总共有 10 个取值。
Q14
Given that $3^8 \cdot 5^2 = ab$, where both $a$ and $b$ are positive integers, find the smallest possible value for $a + b$.
已知 $3^8 \cdot 5^2 = ab$，其中 $a$ 和 $b$ 均为正整数，求 $a + b$ 的最小可能值。
Correct Answer: D
(D) Since \(a\) must be divisible by 5, and \(3^8\cdot 5^2\) is divisible by \(5^2\) but not by \(5^3\), we have \(b\le 2\). If \(b=1\), then \[ a^b=(3^8 5^2)^1=(164,025)^1 \] and \(a+b=164,026\). If \(b=2\), then \[ a^b=(3^4 5)^2=405^2 \] so \(a+b=407\), which is the smallest value.
（D）由于 \(a\) 必须能被 5 整除，并且 \(3^8\cdot 5^2\) 能被 \(5^2\) 整除但不能被 \(5^3\) 整除，所以 \(b\le 2\)。若 \(b=1\)，则 \[ a^b=(3^8 5^2)^1=(164,025)^1 \] 且 \(a+b=164,026\)。 若 \(b=2\)，则 \[ a^b=(3^4 5)^2=405^2 \] 因此 \(a+b=407\)，这是最小值。
Q15
There are 100 players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest 28 players are given a bye, and the remaining 72 players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is
单打网球锦标赛有 100 名选手。采用单淘汰制，输掉比赛的选手被淘汰。第一轮，最强的 28 名选手轮空，其余 72 名选手两两配对比赛。每轮后，剩余选手进入下一轮。比赛继续直到仅剩一名不败选手。总共进行了多少场比赛？
Correct Answer: E
(E) In the first round $100-64=36$ players are eliminated, one per match. In the second round there are 32 matches, in the third 16, then 8, 4, 2, and 1. The total number of matches is: $36+32+16+8+4+2+1=99$. Note that 99 is divisible by 11, but 99 does not satisfy any of the other conditions given as answer choices.
（E）第一轮中有 $100-64=36$ 名选手被淘汰，每场比赛淘汰一人。第二轮有 32 场比赛，第三轮 16 场，然后分别是 8、4、2 和 1 场。比赛总场数为： $36+32+16+8+4+2+1=99$。 注意，99 能被 11 整除，但 99 不满足答案选项中给出的其他任何条件。
Q16
A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year 2003?
一家餐厅提供三种甜点，小菜的数量恰好是主菜数量的两倍。一顿晚餐包括一道小菜、一道主菜和一道甜点。餐厅应该提供多少最少的主菜数量，使得顾客在2003年每天都能吃到不同的晚餐？
Correct Answer: E
(E) Let \(m\) denote the number of main courses needed to meet the requirement. Then the number of dinners available is \(3 \cdot m \cdot 2m = 6m^{2}\). Thus \(m^{2}\) must be at least \(365/6 \approx 61\). Since \(7^{2} = 49 < 61 < 64 = 8^{2}\), 8 main courses is enough, but 7 is not.
（E）令 \(m\) 表示满足要求所需的主菜数量。则可供选择的晚餐数量为 \(3 \cdot m \cdot 2m = 6m^{2}\)。因此 \(m^{2}\) 至少应为 \(365/6 \approx 61\)。由于 \(7^{2}=49<61<64=8^{2}\)，所以需要 8 道主菜才足够，而 7 道不够。
Q17
An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies 75% of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius? (Note: A cone with radius $r$ and height $h$ has volume $\pi r^2 h /3$, and a sphere with radius $r$ has volume $4\pi r^3 /3$.)
一个冰淇淋甜筒由一个香草冰淇淋球和一个直径与球相同的圆锥组成。如果冰淇淋融化，它正好充满圆锥。假设融化的冰淇淋体积占冷冻冰淇淋体积的75%。圆锥的高度与其半径的比是多少？（注：半径为$r$、高度为$h$的圆锥体积为$\pi r^2 h /3$，半径为$r$的球体积为$4\pi r^3 /3$。）
Correct Answer: B
(B) Let \(r\) be the radius of the sphere and cone, and let \(h\) be the height of the cone. Then the conditions of the problem imply that \[ \frac{3}{4}\left(\frac{4}{3}\pi r^{3}\right)=\frac{1}{3}\pi r^{2}h,\ \text{so }h=3r. \] Therefore, the ratio of \(h\) to \(r\) is \(3:1\).
（B）设 \(r\) 为球和圆锥的半径，设 \(h\) 为圆锥的高。则题目的条件意味着 \[ \frac{3}{4}\left(\frac{4}{3}\pi r^{3}\right)=\frac{1}{3}\pi r^{2}h,\ \text{因此 }h=3r. \] 因此，\(h\) 与 \(r\) 的比为 \(3:1\)。
Q18
What is the largest integer that is a divisor of $(n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$ for all positive even integers $n$?
对于所有正偶数$n$，什么最大的整数能整除$(n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$？
Correct Answer: D
(D) Among five consecutive odd numbers, at least one is divisible by 3 and exactly one is divisible by 5, so the product is always divisible by 15. The cases $n = 2$, $n = 10$, and $n = 12$ demonstrate that no larger common divisor is possible, since 15 is the greatest common divisor of $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$, $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$, and $13 \cdot 15 \cdot 17 \cdot 19 \cdot 21$.
(D) 在五个连续的奇数中，至少有一个能被 3 整除，并且恰好有一个能被 5 整除，因此它们的乘积总能被 15 整除。取 $n = 2$、$n = 10$、$n = 12$ 的情形可说明不可能有更大的公因数，因为 15 是 $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$、$11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ 和 $13 \cdot 15 \cdot 17 \cdot 19 \cdot 21$ 的最大公因数。
Q19
Three semicircles of radius 1 are constructed on diameter AB of a semicircle of radius 2. The centers of the small semicircles divide AB into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? [Diagram shows large semicircle radius 2 on diameter AB, with three small semicircles of radius 1 inside it along AB]
在大半径为2的半圆的直径AB上，构造了三个半径为1的半圆。小半圆的圆心将AB分成四个等长的线段，如图所示。阴影区域是大半圆内但在小半圆外的面积是多少？[图示：在直径AB上的大半圆半径2，沿AB在其内部有三个半径1的小半圆]
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Correct Answer: E
(E) The area of the larger semicircle is $\frac{1}{2}\pi(2)^2=2\pi.$ The region deleted from the larger semicircle consists of five congruent sectors and two equilateral triangles. The area of each of the sectors is $\frac{1}{6}\pi(1)^2=\frac{\pi}{6}$ and the area of each triangle is $\frac{1}{2}\cdot 1\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4},$ so the area of the shaded region is $2\pi-5\cdot\frac{\pi}{6}-2\cdot\frac{\sqrt{3}}{4}=\frac{7}{6}\pi-\frac{\sqrt{3}}{2}.$
（E）较大半圆的面积为 $\frac{1}{2}\pi(2)^2=2\pi.$ 从较大半圆中删去的区域由五个全等扇形和两个等边三角形组成。每个扇形的面积为 $\frac{1}{6}\pi(1)^2=\frac{\pi}{6}$ 每个三角形的面积为 $\frac{1}{2}\cdot 1\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4},$ 因此阴影部分的面积为 $2\pi-5\cdot\frac{\pi}{6}-2\cdot\frac{\sqrt{3}}{4}=\frac{7}{6}\pi-\frac{\sqrt{3}}{2}.$
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Q20
In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of △AEB. [Diagram shows rectangle ABCD with AB = 5 (top), BC = 3 (right side), points F and G on CD with DF = 1, GC = 2]
在矩形ABCD中，AB = 5，BC = 3。点F和G在CD上，使得DF = 1，GC = 2。直线AF和BG相交于E。求△AEB的面积。[图示：矩形ABCD，AB = 5（顶部），BC = 3（右侧），点F和G在CD上，DF = 1，GC = 2]
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Correct Answer: D
(D) Let $H$ be the foot of the perpendicular from $E$ to $\overline{DC}$. Since $CD=AB=5$, $FG=2$, and $\triangle FEG$ is similar to $\triangle AEB$, we have \[ \frac{EH}{EH+3}=\frac{2}{5}, \] so $5EH=2EH+6$, and $EH=2$. Hence \[ \text{Area}(\triangle AEB)=\frac{1}{2}(2+3)\cdot 5=\frac{25}{2}. \]
（D）设 $H$ 为从 $E$ 到 $\overline{DC}$ 的垂足。由于 $CD=AB=5$，$FG=2$，且 $\triangle FEG$ 与 $\triangle AEB$ 相似，我们有 \[ \frac{EH}{EH+3}=\frac{2}{5}, \] 所以 $5EH=2EH+6$，并且 $EH=2$。因此 \[ \text{Area}(\triangle AEB)=\frac{1}{2}(2+3)\cdot 5=\frac{25}{2}. \]
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Q21
A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?
一个袋子里有2颗红色珠子和2颗绿色珠子。你伸手进袋子取出颗珠子，无论取出的是什么颜色的珠子，都用一颗红色珠子替换回去。经过三次这样的替换后，袋子里所有珠子都是红色的概率是多少？
Correct Answer: C
(C) The beads will all be red at the end of the third draw precisely when two green beads are chosen in the three draws. If the first bead drawn is green, then there will be one green and three red beads in the bag before the second draw. So the probability that green beads are drawn in the first two draws is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that a green bead is chosen, then a red bead, and then a green bead, is $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{32}$. Finally, the probability that a red bead is chosen then two green beads is $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{16}$. The sum of these probabilities is $\frac{1}{8}+\frac{3}{32}+\frac{1}{16}=\frac{9}{32}$.
（C）在第三次抽取结束时，珠子全为红色当且仅当三次抽取中恰好抽到了两颗绿色珠子。若第一次抽到的是绿色珠子，则在第二次抽取前袋中将有一颗绿色珠子和三颗红色珠子。因此，前两次抽取都抽到绿色珠子的概率为 $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$。 先抽到一颗绿色珠子，再抽到一颗红色珠子，然后抽到一颗绿色珠子的概率为 $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4}=\frac{3}{32}$。 最后，先抽到一颗红色珠子，然后再抽到两颗绿色珠子的概率为 $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{16}$。 这些概率之和为 $\frac{1}{8}+\frac{3}{32}+\frac{1}{16}=\frac{9}{32}$。
Q22
A clock chimes once at 30 minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February 26, 2003, on what date will the 2003rd chime occur?
一个钟在每小时30分时敲一次，在整点时按照小时数敲响。例如，下午1点敲1下，正午和午夜敲12下。从2003年2月26日上午11:15开始，第2003次敲响将在哪一天发生？
Correct Answer: B
(B) In any twelve-hour period, there are 12 half-hour chimes and $1+2+3+\cdots+12=78$ on-the-hour chimes. Hence, a twelve-hour period results in 90 chimes. Dividing 2003 by 90 yields a quotient of 22.25. Therefore the 2003rd chime will occur a little more than 11 days later, on March 9.
（B）在任意一个 12 小时的时间段内，有 12 次半点报时，以及整点报时次数为 $1+2+3+\cdots+12=78$。因此，一个 12 小时的时间段共计会有 90 次报时。用 2003 除以 90 得到商为 22.25。因此，第 2003 次报时将发生在 11 天多一点之后，也就是 3 月 9 日。
Q23
A regular octagon ABCDEFGH has an area of one square unit. What is the area of the rectangle ABEF? [Diagram shows regular octagon labeled A B C D E F G H clockwise]
一个正八边形ABCDEFGH的面积为一平方单位。矩形ABEF的面积是多少？[图示为顺时针标记A B C D E F G H的正八边形]
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Correct Answer: D
(D) Let $O$ be the intersection of the diagonals of $ABEF$. Since the octagon is regular, $\triangle AOB$ has area $1/8$. Since $O$ is the midpoint of $AE$, $\triangle OAB$ and $\triangle BOE$ have the same area. Thus $\triangle ABE$ has area $1/4$, so $ABEF$ has area $1/2$.
(D) 设 $O$ 为 $ABEF$ 的对角线交点。由于该八边形是正八边形，$\triangle AOB$ 的面积为 $1/8$。因为 $O$ 是 $AE$ 的中点，$\triangle OAB$ 与 $\triangle BOE$ 的面积相同。因此 $\triangle ABE$ 的面积为 $1/4$，所以 $ABEF$ 的面积为 $1/2$。
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Q24
The first four terms in an arithmetic sequence are x + y, x − y, xy, and x/y, in that order. What is the fifth term?
一个等差数列的前四项依次为 x + y, x − y, xy, 和 x/y。第几项是多少？
Correct Answer: E
(E) Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus $$x-3y=xy \text{ and } x-5y=\frac{x}{y},$$ so $xy-5y^2=x$. Combining these equations we obtain $$(x-3y)-5y^2=x \text{ and, therefore, } -3y-5y^2=0.$$ Since $y$ cannot be $0$, we have $y=-3/5$, and it follows that $x=-9/8$. The fifth term in the sequence is $x-7y=123/40$.
（E）由于前两项之差为 $-2y$，该数列的第三项和第四项必为 $x-3y$ 与 $x-5y$。因此 $$x-3y=xy \text{ 且 } x-5y=\frac{x}{y},$$ 所以 $xy-5y^2=x$。联立这些方程可得 $$(x-3y)-5y^2=x \text{，因此 } -3y-5y^2=0。$$ 由于 $y$ 不能为 $0$，故 $y=-3/5$，从而 $x=-9/8$。数列的第五项为 $x-7y=123/40$。
Q25
How many distinct four-digit numbers are divisible by 3 and have 23 as their last two digits?
有多少个不同的四位数能被3整除且末两位数是23？
Correct Answer: B
(B) A number is divisible by 3 if and only if the sum of its digits is divisible by 3. So a four-digit number $ab23$ is divisible by 3 if and only if the two-digit number $ab$ leaves a remainder of 1 when divided by 3. There are 90 two-digit numbers, of which $90/3 = 30$ leave a remainder of 1 when divided by 3.
（B）一个数能被 3 整除，当且仅当它的各位数字之和能被 3 整除。因此，四位数 $ab23$ 能被 3 整除，当且仅当两位数 $ab$ 除以 3 的余数为 1。两位数共有 90 个，其中有 $90/3 = 30$ 个除以 3 的余数为 1。