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AMC8 1996

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AMC8 · 1996

Q1
How many positive factors of 36 are also multiples of 4?
36 有多少个正因数同时也是 4 的倍数?
Correct Answer: B
The positive factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Of these, 4, 12, and 36 are also multiples of 4.
36 的正因数是 1、2、3、4、6、9、12、18 和 36。其中,4、12 和 36 也是 4 的倍数。
Q2
José, Thuy, and Kareem each start with the number 10. José subtracts 1 from the number 10, doubles his answer, and then adds 2. Thuy doubles the number 10, subtracts 1 from her answer, then adds 2. Kareem subtracts 1 from the number 10, adds 2 to his answer, and then doubles the result. Who gets the largest final answer?
José、Thuy 和 Kareem 都从数字 10 开始。José 从 10 中减去 1,将结果加倍,然后加 2。Thuy 将 10 加倍,从结果中减去 1,然后加 2。Kareem 从 10 中减去 1,将结果加 2,然后加倍。谁得到最大的最终结果?
Correct Answer: C
Starting with 10: José computes, consecutively, 9, then 18, and finally 20. Thuy computes, consecutively, 20, then 19, and finally 21. Kareem computes, consecutively, 9, then 11, and finally 22. Thus Kareem gets the largest final answer.
从 10 开始:José 依次计算 9,然后 18,最后 20。Thuy 依次计算 20,然后 19,最后 21。Kareem 依次计算 9,然后 11,最后 22。因此 Kareem 得到最大的最终结果。
Q3
The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares). The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on. After all 64 numbers are written, the sum of the numbers in the four corners will be
从 1 到 64 的 64 个整数,每个方格一个,写在一个棋盘上(8×8 的 64 个方格)。前 8 个数字按顺序写在第一行,接下来的 8 个写在第二行,依此类推。写完所有 64 个数字后,四角数字的和是
Correct Answer: A
The first row is 1, 2, 3, …, 7, 8 and the last row is 57, 58, 59, …, 63, 64. Thus the four corner numbers are 1, 8, 57, and 64, and their sum is 130.
第一行是 1、2、3、…、7、8,最后一行是 57、58、59、…、63、64。因此四个角上的数字是 1、8、57 和 64,它们的和是 130。
solution
Q4
\(\frac{2+4+6+\cdots+34}{3+6+9+\cdots+51} =\)
\(\frac{2+4+6+\cdots+34}{3+6+9+\cdots+51} =\)
stem
Correct Answer: B
\(2 + 4 + 6 + \cdots + 34 = 2(1 + 2 + 3 + \cdots + 17)\) \(3 + 6 + 9 + \cdots + 51 = 3(1 + 2 + 3 + \cdots + 17)\) so the ratio is \(\frac{2}{3}\).
\(2 + 4 + 6 + \cdots + 34 = 2(1 + 2 + 3 + \cdots + 17)\) \(3 + 6 + 9 + \cdots + 51 = 3(1 + 2 + 3 + \cdots + 17)\) 所以比值为 \(\frac{2}{3}\)。
Q5
The letters P, Q, R, S, and T represent numbers located on the number line as shown. Which of the following expressions represents a negative number?
字母 P、Q、R、S 和 T 表示数轴上所示的位置。以下哪个表达式表示一个负数?
Correct Answer: A
Since P is to the left of Q, $P - Q$ is negative. All other expressions are positive.
由于 P 在 Q 的左侧,$P - Q$ 是负数。所有其他表达式都是正数。
Q6
What is the smallest result that can be obtained by the following process? • Choose three different numbers from the set $\{3, 5, 7, 11, 13, 17\}$. • Add two of these numbers. • Multiply their sum by the third number.
通过以下过程能得到的最小结果是多少? • 从集合 $\{3, 5, 7, 11, 13, 17\}$ 中选择三个不同的数。 • 将其中两个数相加。 • 将它们的和乘以第三个数。
Correct Answer: C
Answer (C): To obtain the smallest result, use the three smallest numbers. This yields three choices: $3(5+7)=36,\quad 5(3+7)=50,\quad 7(3+5)=56.$ Thus $36$ is the smallest result.
答案(C):要得到最小的结果,使用三个最小的数。这样有三种选择: $3(5+7)=36,\quad 5(3+7)=50,\quad 7(3+5)=56.$ 因此,$36$ 是最小的结果。
Q7
Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, then in how many months from that time will they have the same number of goldfish?
Brent的金鱼每月增加到四倍(数量变为原来的四倍),Gretel的金鱼每月增加到两倍。如果Brent有4条金鱼时Gretel有128条金鱼,那么从那时起多少个月后它们金鱼数量相同?
Correct Answer: B
Month 0: Brent 4, Gretel 128; Month 1: 16,256; Month 2:64,512; Month 3:256,1024; Month 4:1024,2048; Month 5:4096,4096. Equal in 5 months.
第0个月:Brent 4, Gretel 128;第1个月:16,256;第2个月:64,512;第3个月:256,1024;第4个月:1024,2048;第5个月:4096,4096。5个月后相等。
Q8
Points A and B are 10 units apart. Points B and C are 4 units apart. Points C and D are 3 units apart. If A and D are as close as possible, then the number of units between them is
点A和B相距10个单位。点B和C相距4个单位。点C和D相距3个单位。如果A和D尽可能接近,那么它们之间的单位数是
Correct Answer: B
The shortest distance occurs when the four points are collinear with B-C-D towards A, so AD = AB - BC - CD = 10 - 4 - 3 = 3.
最短距离发生在四点共线,B-C-D朝向A时,所以AD = AB - BC - CD = 10 - 4 - 3 = 3。
Q9
If 5 times a number is 2, then 100 times the reciprocal of the number is
如果一个数的5倍是2,那么该数的倒数的100倍是
Correct Answer: D
Let n be the number. 5n=2 so n=2/5. Reciprocal 5/2, 100*(5/2)=250.
设该数为n。5n=2,所以n=2/5。倒数5/2,100*(5/2)=250。
Q10
When Walter drove up to the gasoline pump, he noticed that his gasoline tank was $\frac{1}{8}$ full. He purchased 7.5 gallons of gasoline for \$10. With this additional gasoline, his gasoline tank was then $\frac{5}{8}$ full. The number of gallons of gasoline his tank holds when it is full is
当Walter开车到加油泵时,他注意到他的油箱有 $\frac{1}{8}$ 满。他买了7.5加仑汽油,花了\$10。加了这部分汽油后,他的油箱有 $\frac{5}{8}$ 满。他的油箱满时容纳的加仑数是
Correct Answer: D
Answer (D): The gasoline tank going from $\frac{1}{8}$ to $\frac{5}{8}$ full represents an increase of $\frac{5}{8}-\frac{1}{8}=\frac{1}{2}$ tank. Since half a tank is 7.5 gallons, it follows that a full tank is $2\times 7.5=15$ gallons.
答案(D):汽油箱从 $\frac{1}{8}$ 满到 $\frac{5}{8}$ 满表示增加了 $\frac{5}{8}-\frac{1}{8}=\frac{1}{2}$ 箱。因为半箱是 7.5 加仑,所以一整箱是 $2\times 7.5=15$ 加仑。
solution
Q11
Let $x$ be the number $0.\underbrace{0000\dots0000}_{1996\ zeros}1$, where there are 1996 zeros after the decimal point. Which of the following expressions represents the largest number?
设 $x$ 为小数 $0.\underbrace{0000\dots0000}_{1996\ zeros}1$,其中小数点后有 1996 个零。以下哪个表达式表示最大的数?
Correct Answer: D
Answer (D): Since $x$ is near zero, $3+x$ and $3-x$ are near $3$. Also $3\cdot x$ and $\frac{x}{3}$ are near zero. However, $\frac{3}{x}$ is the number $$ 3\,0000\ldots 0000, $$ which is a $3$ followed by $1997$ zeros. This number is much larger than any of the other alternatives.
答案(D):由于 $x$ 接近于 $0$,所以 $3+x$ 和 $3-x$ 都接近于 $3$。同时,$3\cdot x$ 和 $\frac{x}{3}$ 都接近于 $0$。然而,$\frac{3}{x}$ 是下面这个数 $$ 3\,0000\ldots 0000, $$ 它是一个 $3$ 后面跟着 $1997$ 个 $0$。这个数远大于其他任何选项。
Q12
The number should be removed from the list 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 so that the average of the remaining numbers is 6.1?
从列表 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 中移除哪个数,使得剩余数字的平均数为 6.1?
Correct Answer: B
Answer (B): The sum of the eleven numbers is 66. For the average of ten numbers to be 6.1, the sum of the ten numbers must be $10 \times 6.1 = 61$. Thus, remove the 5.
答案(B):这 11 个数的和是 66。要使 10 个数的平均数为 6.1,这 10 个数的和必须是 $10 \times 6.1 = 61$。因此,去掉 5。
Q13
In the fall of 1996, a total of 800 students participated in an annual school clean-up day. The organizers of the event expect that in each of the years 1997, 1998, and 1999, participation will increase by 50% over the previous year. The number of participants the organizers expect in the fall of 1999 is
1996 年秋季共有 800 名学生参加年度学校清洁日。组织者预计 1997、1998 和 1999 年每年的参与人数将比前一年增加 50%。组织者预计 1999 年秋季的参与人数为
Correct Answer: E
Answer (E): Since 50% of 800 is 400, the organizers expect $800 + 400 = 1200$ participants in 1997. Since 50% of 1200 is 600, $1200 + 600 = 1800$ are expected in 1998. Since 50% of 1800 is 900, $1800 + 900 = 2700$ are expected in 1999.
答案(E):由于800的50%是400,组织者预计1997年有$800 + 400 = 1200$名参与者。由于1200的50%是600,预计1998年有$1200 + 600 = 1800$名。由于1800的50%是900,预计1999年有$1800 + 900 = 2700$名。
Q14
Six different digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are placed in the squares in the figure shown so that the sum of the entries in the vertical column is 23 and the sum of the entries in the horizontal row is 12. The sum of the six digits used is
从集合 $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ 中选取六个不同的数字放置在图示的方格中,使得垂直列的数字和为 23,水平行的数字和为 12。所用六个数字的和为
stem
Correct Answer: B
The column must include 7,8,9 but adjusted; digits 1,2,3,6,8,9 sum to 29.
列必须包括 7,8,9 但需调整;数字 1,2,3,6,8,9 和为 29。
solution
Q15
The remainder when the product $1492 \cdot 1776 \cdot 1812 \cdot 1996$ is divided by 5 is
$1492 \cdot 1776 \cdot 1812 \cdot 1996$ 的乘积除以 5 的余数是
Correct Answer: E
Answer (E): The last digit of a product is determined by the product of the last digits of the factors. Since $2\cdot 6\cdot 2\cdot 6=144$, the last digit of the product is $4$. Since multiples of $5$ end in $0$ or $5$, any number with last digit $4$ leaves a remainder of $4$ when divided by $5$.
答案(E):一个乘积的末位数字由各个因数末位数字的乘积决定。因为 $2\cdot 6\cdot 2\cdot 6=144$,所以该乘积的末位数字是 $4$。由于 $5$ 的倍数的末位是 $0$ 或 $5$,任何末位为 $4$ 的数除以 $5$ 都会余 $4$。
Q16
$1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + 13 - \dots + 1992 + 1993 - 1994 - 1995 + 1996 = \dots$
$1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + 13 - \dots + 1992 + 1993 - 1994 - 1995 + 1996 = \dots$
Correct Answer: C
Answer (C): Combining in groups of four yields $1-2-3+4=0,\qquad 5-6-7+8=0,\qquad 9-10-11+12=0$ and so on. Since there are 499 groups of four in 1996, it follows that the sum is zero.
答案(C):按每四项一组相加可得 $1-2-3+4=0,\qquad 5-6-7+8=0,\qquad 9-10-11+12=0$ 依此类推。由于 1996 中有 499 组四项,因此总和为 0。
Q17
Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates $(2, 2)$. What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR$?
图 $OPQR$ 是一个正方形。点 $O$ 是原点,点 $Q$ 的坐标为 $(2, 2)$。求点 $T$ 的坐标,使得三角形 $PQT$ 的面积等于正方形 $OPQR$ 的面积。
stem
Correct Answer: C
Answer (C): Since $OPQR$ is a square and point $Q$ has coordinates $(2,2)$, it follows that point $P$ has coordinates $(2,0)$ and point $R$ has coordinates $(0,2)$. Thus the side of square $OPQR$ is $2$ and the area of the $2$ by $2$ square $OPQR$ is $2^2 = 4$. The area of triangle $PQT$ is $\frac{1}{2}\cdot PT \cdot PQ = \frac{1}{2}\cdot PT \cdot 2 = PT$. Since the area of the triangle equals the area of the square, $PT = 4$. Thus the point $T$ has coordinates $(-2,0)$.
答案(C):由于 $OPQR$ 是一个正方形,且点 $Q$ 的坐标为 $(2,2)$,可知点 $P$ 的坐标为 $(2,0)$,点 $R$ 的坐标为 $(0,2)$。因此,正方形 $OPQR$ 的边长为 $2$,这个 $2\times 2$ 的正方形 $OPQR$ 的面积为 $2^2 = 4$。三角形 $PQT$ 的面积为 $\frac{1}{2}\cdot PT \cdot PQ = \frac{1}{2}\cdot PT \cdot 2 = PT$。由于三角形的面积等于正方形的面积,所以 $PT = 4$。因此点 $T$ 的坐标为 $(-2,0)$。
Q18
Ana's monthly salary was $2000 in May. In June she received a 20% raise. In July she received a 20% pay cut. After the two changes in June and July, Ana's monthly salary was
Ana 五月份的月薪是 2000 美元。六月份她涨薪 20%。七月份她减薪 20%。经过六月和七月的两次变化后,Ana 的月薪是
Correct Answer: A
Answer (A): After the first change, Ana's salary was $\$2000 + 0.20(\$2000) = \$2400$. After the second change, Ana's salary was $\$2400 - 0.20(\$2400) = \$1920$.
答案(A):第一次变化后,Ana 的工资为 $\$2000 + 0.20(\$2000) = \$2400$。第二次变化后,Ana 的工资为 $\$2400 - 0.20(\$2400) = \$1920$。
Q19
The pie charts at the right indicate the percent of students who prefer golf, bowling, or tennis at East Junior High School and West Middle School. The total number of students at East is 2000 and at West, 2500. In the two schools combined, the percent of students who prefer tennis is
右边的饼图显示了东初中和西中学的学生偏好高尔夫、保龄球或网球的百分比。东校总学生数 2000 人,西校 2500 人。两校合计,偏好网球的学生百分比是
stem
Correct Answer: C
Answer (C): The first pie chart shows that 22% of 2000, or 440 students at East prefer tennis. The second chart shows that 40% of 2500, or 1000 students at West prefer tennis. Thus 1440 of the total of 4500 students prefer tennis. This gives $\frac{1440}{4500}=0.32$, or 32% that prefer tennis.
答案(C):第一个饼图显示,在东校区的2000名学生中有22%(即440名)更喜欢网球。第二个图表显示,在西校区的2500名学生中有40%(即1000名)更喜欢网球。因此,总共4500名学生中有1440名更喜欢网球。由此可得 $\frac{1440}{4500}=0.32$,即喜欢网球的占32%。
Q20
Suppose there is a special key on a calculator that replaces the number $x$ currently displayed with the number given by the formula $\frac{1}{1-x}$. For example, if the calculator is displaying 2 and the special key is pressed, then the calculator will display $-1$ since $\frac{1}{1-2} = -1$. Now suppose that the calculator is displaying 5. After the special key is pressed 100 times in a row, the calculator will display
假设计算器上有一个特殊按键,将当前显示的数字 $x$ 替换为公式 $\frac{1}{1-x}$ 给出的数字。例如,如果计算器显示 2,按下特殊键后,将显示 $-1$,因为 $\frac{1}{1-2} = -1$。现在假设计算器显示 5。连续按下特殊键 100 次后,计算器将显示
Correct Answer: A
Answer (A): After the special key is pressed once, the calculator display reads $-0.25$ since $1/(1-5)=1/(-4)=-0.25$. If the key is pressed again, the calculator display reads $0.8$ since $1/(1-(-0.25))=1/(1.25)=0.8$. If the key is pressed a third time, the calculator display reads $5$, since $1/(1-0.8)=1/(0.2)=5$. Thus pressing the special key three times returns to the original calculator display. The calculator display will continue to cycle through the three answer $-0.25$, $0.8$, and $5$. Since $100$ is $1$ more than a multiple of $3$, the calculator display will be $-0.25$.
答案(A):按一次特殊按键后,计算器显示为 $-0.25$,因为 $1/(1-5)=1/(-4)=-0.25$。再按一次按键后,计算器显示为 $0.8$,因为 $1/(1-(-0.25))=1/(1.25)=0.8$。第三次按下按键时,计算器显示为 $5$,因为 $1/(1-0.8)=1/(0.2)=5$。因此,连续按三次特殊按键会回到最初的计算器显示。计算器显示将继续在 $-0.25$、$0.8$ 和 $5$ 这三个数之间循环。由于 $100$ 比 $3$ 的一个倍数多 $1$,所以计算器显示将是 $-0.25$。
Q21
How many subsets containing three different numbers can be selected from the set $\{89, 95, 99, 132, 166, 173\}$ so that the sum of the three numbers is even?
从集合 \{89, 95, 99, 132, 166, 173\} 中能选出多少个包含三个不同数字的子集,使得三个数的和为偶数?
Correct Answer: D
Answer (D): The sum of three numbers is even if all three numbers are even, or if two numbers are odd and one is even. Since there are only two even numbers in the set, it follows that the three numbers must include two odd numbers and one even. The possibilities are: \[ \begin{array}{ccc} \{89,95,132\} & \{89,99,132\} & \{89,173,132\} \\ \{89,95,166\} & \{89,96,166\} & \{89,173,166\} \\ \\ \{95,99,132\} & \{95,173,132\} & \{99,173,132\} \\ \{95,99,166\} & \{95,173,166\} & \{99,173,166\} \end{array} \] Thus there are 12 possibilities.
答案(D):三个数的和为偶数,当且仅当这三个数全为偶数,或者两个为奇数、一个为偶数。由于该集合中只有两个偶数,因此这三个数必须包含两个奇数和一个偶数。可能的组合为: \[ \begin{array}{ccc} \{89,95,132\} & \{89,99,132\} & \{89,173,132\} \\ \{89,95,166\} & \{89,96,166\} & \{89,173,166\} \\ \\ \{95,99,132\} & \{95,173,132\} & \{99,173,132\} \\ \{95,99,166\} & \{95,173,166\} & \{99,173,166\} \end{array} \] 因此共有 12 种可能。
Q22
The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle $ABC$ is
相邻点之间的水平和垂直距离均为1单位。三角形 $ABC$ 的面积是
stem
Correct Answer: B
Answer (B): From the total area of 12, subtract the area of the four surrounding polygons whose areas are indicated in the diagram. Thus the area of the remaining triangle $ABC$ is $$12-6-3-0.5-2=0.5=\frac{1}{2}$$ Note. Points whose coordinates are integers are called lattice points. According to Pick’s Theorem, if there are $I$ lattice points in the interior of a triangle and $B$ lattice points on the boundary, then the area of the triangle is $I+\frac{B}{2}-1$. In this problem, $I=0$ and $B=3$. Therefore the area of the triangle is $0+\frac{3}{2}-1=\frac{1}{2}$.
答案(B):从总面积 12 中减去图中标明的四个周围多边形的面积。因此剩余三角形 $ABC$ 的面积为 $$12-6-3-0.5-2=0.5=\frac{1}{2}$$ 注:坐标为整数的点称为格点。根据皮克定理,若三角形内部有 $I$ 个格点、边界上有 $B$ 个格点,则三角形面积为 $I+\frac{B}{2}-1$。在本题中,$I=0$ 且 $B=3$。因此三角形面积为 $0+\frac{3}{2}-1=\frac{1}{2}$。
solution
Q23
The manager of a company planned to distribute a \$50 bonus to each employee from the company fund, but the fund contained \$5 less than what was needed. Instead the manager gave each employee a \$45 bonus and kept the remaining \$95 in the company fund. The amount of money in the company fund before any bonuses were paid was
公司经理计划从公司基金中向每位员工发放50美元奖金,但基金比所需少5美元。于是经理改发给每位员工45美元奖金,并将剩余的95美元留在公司基金中。在发放任何奖金前,公司基金中的金额是
Correct Answer: E
Answer (E): Had there been \$5 more in the company fund, there would have been \$95 + \$5 = \$100 which would have been enough to give each employee another \$5. Thus there are $\frac{\$100}{\$5}=20$ employees. So the company fund contained $20\cdot\$45+\$95=\$995$.
答案(E):如果公司基金多 \$5,那么就会有 \$95 + \$5 = \$100,这就足够给每位员工再多发 \$5。因此员工人数为 $\frac{\$100}{\$5}=20$ 人。所以公司基金共有 $20\cdot\$45+\$95=\$995$。
Q24
The measure of angle $ABC$ is $50^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$. The measure of angle $ADC$ is
角 $ABC$ 的度数为 $50^\circ$,$\overline{AD}$ 平分角 $BAC$,$\overline{DC}$ 平分角 $BCA$。角 $ADC$ 的度数是
stem
Correct Answer: C
Answer (C): Since the sum of the measures of the angles of a triangle is 180°, in triangle $ABC$ it follows that $\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ.$ The measures of angles $DAC$ and $DCA$ are half that of angles $BAC$ and $BCA$, respectively, so $\angle DAC+\angle DCA=\frac{130^\circ}{2}=65^\circ.$ In triangle $ACD$, we have $\angle ADC=180^\circ-65^\circ=115^\circ.$
答案(C):由于三角形内角和为 $180^\circ$,在三角形 $ABC$ 中有 $\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ.$ 角 $DAC$ 和 $DCA$ 的度数分别是角 $BAC$ 和 $BCA$ 的一半,因此 $\angle DAC+\angle DCA=\frac{130^\circ}{2}=65^\circ.$ 在三角形 $ACD$ 中,有 $\angle ADC=180^\circ-65^\circ=115^\circ.$
Q25
A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region?
从圆形区域内随机选择一点。该点比到区域边界的距离更靠近区域中心的概率是多少?
Correct Answer: A
Answer (A): Suppose that the circle has radius 1. Then, being closer to the center of the region than the boundary of the region (the circle) would mean the chosen point must be inside the circle of radius $\frac{1}{2}$ with the same center as the larger circle of radius 1. The area of the smaller region is $\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$, and the area of the total region is $\pi(1)^2=\pi$. Since the area of the smallest region is $\frac{1}{4}$ of the area of the total region, the required probability is $\frac{1}{4}$. Note. The odds that the point is closer to the center are 1:3.
答案(A):假设圆的半径为 1。那么,“离该区域的中心比离该区域的边界(圆周)更近”意味着所选点必须位于半径为 $\frac{1}{2}$ 的圆内,并且该小圆与半径为 1 的大圆同心。较小区域的面积为 $\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}$,而整个区域的面积为 $\pi(1)^2=\pi$。由于最小区域的面积是总区域面积的 $\frac{1}{4}$,所求概率为 $\frac{1}{4}$。 注:该点更靠近中心的赔率为 1:3。
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