/

AMC8 2001

You are not logged in. After submit, your report may not be available on other devices. Login

AMC8 · 2001

Q1
Casey's shop class is making a golf trophy. He has to paint 300 dimples on a golf ball. If it takes him 2 seconds to paint one dimple, how many minutes will he need to do his job?
凯西的商店课正在制作一个高尔夫奖杯。他必须在高尔夫球上画300个坑坑。如果他画一个坑坑需要2秒，他需要多少分钟来完成他的工作？
stem
Correct Answer: D
At 2 seconds per dimple, it takes $300 \times 2 = 600$ seconds to paint them. Since there are 60 seconds in a minute, he will need $600 \div 60 = 10$ minutes.
每个坑坑2秒，所以画它们需要 $300 \times 2 = 600$ 秒。因为1分钟有60秒，他需要 $600 \div 60 = 10$ 分钟。
Q2
I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
我在想两个整数。它们的乘积是24，它们的和是11。较大的数是多少？
Correct Answer: D
Since their sum is to be 11, only positive factors need to be considered. Number pairs whose product is 24 are $(1, 24)$, $(2, 12)$, $(3, 8)$ and $(4, 6)$. The sum of the third pair is 11, so the numbers are 3 and 8. The larger one is 8.
因为它们的和是11，只需考虑正因数即可。乘积为24的数对是 $(1, 24)$、$(2, 12)$、$(3, 8)$ 和 $(4, 6)$。第三对的和是11，所以这些数是3和8。较大的是8。
Q3
Granny Smith has $63. Elberta has $2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have?
Granny Smith 有 $63。Elberta 比 Anjou 多 $2，Anjou 有 Granny Smith 三分之一那么多。Elberta 有多少美元？
Correct Answer: E
(E) Anjou has one-third as much money as Granny Smith, so Anjou has \$21. Elberta has \$2 more than Anjou, and $\$21+\$2=\$23$.
（E）Anjou 的钱是 Granny Smith 的三分之一，所以 Anjou 有 \$21。Elberta 比 Anjou 多 \$2，因此 $\$21+\$2=\$23$。
Q4
The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
数字1、2、3、4和9各用一次，形成可能的最小的偶五位数。十分位上的数字是
Correct Answer: E
To make the number as small as possible, the smaller digits are placed in the higher-value positions. To make the number even, the larger even digit 4 must be the units digit. The smallest possible even number is 12394 and 9 is in the tens place.
为了使数字尽可能小，将较小的数字放在高位。为了使数字为偶，较大的偶数字4必须是个位数字。可能的最小偶数是12394，十分位是9。
Q5
On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.
在一个漆黑暴风雨的夜晚，史努比突然看到一道闪电。10秒后他听到了雷声。声速是1088英尺每秒，1英里是5280英尺。估计到最近的半英里，史努比距离闪电有多远。
stem
Correct Answer: C
Use the formula $d = rt$ (distance equals rate times time): 1088 feet per second $\times 10$ seconds = 10880 feet, which is just 320 feet more than two miles. Therefore, Snoopy is just about two miles from the flash of lightning.
使用公式 $d = rt$（距离等于速率乘时间）：1088英尺每秒 $\times 10$ 秒 = 10880英尺，这刚好比两英里多320英尺。因此，史努比距离闪电大约两英里。
Q6
Six trees are equally spaced along one side of a straight road. The distance from the first tree to the fourth is 60 feet. What is the distance in feet between the first and last trees?
六棵树等间距地沿着一条直路的单侧排列。从第一棵树到第四棵树的距离是60英尺。第一棵树和最后一棵树之间的距离是多少英尺？
Correct Answer: B
There are three spaces between the first tree and the fourth tree, so the distance between adjacent trees is 20 feet. There are 6 trees with five of these 20-foot spaces, so the distance between the first and last trees is 100 feet.
第一棵树和第四棵树之间有三个间距，所以相邻树之间的距离是20英尺。有6棵树，有五个这样的20英尺间距，所以第一棵树和最后一棵树之间的距离是100英尺。
solution
Q7
What is the number of square inches in the area of the small kite?
小风筝的面积有多少平方英寸？
stem
Correct Answer: A
(A) The area is made up of two pairs of congruent triangles. The top two triangles can be arranged to form a $2 \times 3$ rectangle. The bottom two triangles can be arranged to form a $5 \times 3$ rectangle. The kite’s area is $6 + 15 = 21$ square inches.
（A）该面积由两对全等三角形组成。上面的两个三角形可以重新排列组成一个 $2 \times 3$ 的长方形。下面的两个三角形可以重新排列组成一个 $5 \times 3$ 的长方形。风筝的面积为 $6 + 15 = 21$ 平方英寸。
solution
Q8
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?
Genevieve在大风筝上加了支撑，形成一个连接风筝对角线的十字。她需要多少英寸的支撑材料？
stem
Correct Answer: E
(E) The small kite is 6 inches wide and 7 inches high, so the larger kite is 18 inches wide and 21 inches high. The amount of bracing needed is $18 + 21 = 39$ inches.
（E）小风筝宽 6 英寸、高 7 英寸，所以大风筝宽 18 英寸、高 21 英寸。所需的支撑条长度为 $18 + 21 = 39$ 英寸。
solution
Q9
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?
大风筝覆盖着金箔。箔是从刚好覆盖整个网格的矩形片裁剪的。从四个角切掉的废料有多少平方英寸？
stem
Correct Answer: D
(D) The upper corners can be arranged to form a $6 \times 9$ rectangle and the lower corners can be arranged to form a $15 \times 9$ rectangle. The total area is $54 + 135 = 189$ square inches. (Note that the kite’s area is also 189 square inches.)
（D）上方的角可以重新排列组成一个 $6 \times 9$ 的长方形，下方的角可以重新排列组成一个 $15 \times 9$ 的长方形。总面积为 $54 + 135 = 189$ 平方英寸。（注意：风筝的面积也是 189 平方英寸。）
solution
Q10
A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
一位收藏家提出以面值的2000%购买州立25美分硬币。以这个价格，Bryden的四个州立25美分硬币能得到多少钱？
Correct Answer: A
2000% = 20.00, so the quarters are worth 20 times their face value. That makes the total value 20(4)($0.25) = $20.
2000%=20.00，所以这些硬币价值是面值的20倍。总价值是20(4)($0.25)=$20。
Q11
Points A, B, C, and D have these coordinates: A(3,2), B(3,-2), C(-3,-2) and D(-3,0). The area of quadrilateral ABCD is
点 A、B、C 和 D 的坐标分别是：A(3,2)、B(3,-2)、C(-3,-2) 和 D(-3,0)。四边形 ABCD 的面积是
stem
Correct Answer: C
The lower part is a 6 × 2 rectangle with area 12. The upper part is a triangle with base 6 and altitude 2 with area 6. The total area is 12 + 6 = 18.
下部是一个 $6 \times 2$ 的矩形，面积为 12。上部是一个底为 6、高为 2 的三角形，面积为 6。总面积为 $12 + 6 = 18$。
solution
Q12
If $a \otimes b = \frac{a+b}{a-b}$, then $(6 \otimes 4) \otimes 3 =$
如果 $a \otimes b = \frac{a+b}{a-b}$，则 $(6 \otimes 4) \otimes 3 =$
Correct Answer: A
(A) $6\otimes 4=\dfrac{6+4}{6-4}=\dfrac{10}{2}=5$, and $5\otimes 3=\dfrac{5+3}{5-3}=\dfrac{8}{2}=4$. Note: $(6\otimes 4)\otimes 3\ne 6\otimes (4\otimes 3)$. Does $(6\otimes 4)\otimes 3=3\otimes (6\otimes 4)$?
（A）$6\otimes 4=\dfrac{6+4}{6-4}=\dfrac{10}{2}=5$，并且 $5\otimes 3=\dfrac{5+3}{5-3}=\dfrac{8}{2}=4$。 注意：$(6\otimes 4)\otimes 3\ne 6\otimes (4\otimes 3)$。那么 $(6\otimes 4)\otimes 3=3\otimes (6\otimes 4)$ 吗？
Q13
Of the 36 students in Richelle's class, 12 prefer chocolate pie, 8 prefer apple, and 6 prefer blueberry. Half of the remaining students prefer cherry pie and half prefer lemon. For Richelle's pie graph showing this data, how many degrees should she use for cherry pie?
Richelle 班上有 36 名学生，其中 12 名喜欢巧克力派，8 名喜欢苹果派，6 名喜欢蓝莓派。剩下的学生有一半喜欢樱桃派，一半喜欢柠檬派。在 Richelle 的派图中，樱桃派应该使用多少度？
stem
Correct Answer: D
Since $12 + 8 + 6 = 26$， there are $36 - 26 = 10$ children who prefer cherry or lemon pie. These ten are divided into equal parts of 5 each. $$\frac{5}{36} \times 360^\circ = 5 \times 10^\circ = 50^\circ.$$
因为 $12 + 8 + 6 = 26$，所以有 $36 - 26 = 10$ 名学生喜欢樱桃派或柠檬派。这 10 人平分，每种各 5 人。$$\frac{5}{36} \times 360^\circ = 5 \times 10^\circ = 50^\circ$$。
Q14
Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might he choose? Meat: beef, chicken, pork Vegetables: baked beans, corn, potatoes, tomatoes Dessert: brownies, chocolate cake, chocolate pudding, ice cream
Tyler 在自助餐队伍中选择一种肉类、两种不同的蔬菜和一种甜点。如果食物项目的顺序不重要，他可以选择多少种不同的餐点？肉类：牛肉、鸡肉、猪肉 蔬菜：烤豆子、玉米、土豆、西红柿 甜点：布朗尼、巧克力蛋糕、巧克力布丁、冰激凌
Correct Answer: C
(C) There are 3 choices for the meat and 4 for dessert. There are 6 ways to choose the two vegetables. The first vegetable may be chosen in 4 ways and the second in 3 ways. This would seem to make 12 ways, but since the order is not important the 12 must be divided by 2. Otherwise, for example, both tomatoes/corn and corn/tomatoes would be included. The 6 choices are beans/corn, beans/potatoes, beans/tomatoes, corn/potatoes, corn/tomatoes and potatoes/tomatoes. The answer is $3(4)(6)=72$.
（C）肉类有 3 种选择，甜点有 4 种选择。 选择两种蔬菜有 6 种方法。第一种蔬菜有 4 种选法，第二种有 3 种选法。看起来共有 12 种，但由于顺序不重要，12 必须除以 2。否则，例如，tomatoes/corn 和 corn/tomatoes 都会被算作不同情况。6 种选择是 beans/corn、beans/potatoes、beans/tomatoes、corn/potatoes、corn/tomatoes 和 potatoes/tomatoes。 答案是 $3(4)(6)=72$。
Q15
Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?
Homer 以每分钟 3 个土豆的速度开始剥 44 个土豆的堆。四分钟后 Christen 加入他，以每分钟 5 个土豆的速度剥。当他们完成时，Christen 剥了多少个土豆？
Correct Answer: A
(A) After 4 minutes Homer had peeled 12 potatoes. When Christen joined him, the combined rate of peeling was 8 potatoes per minute, so the remaining 32 potatoes required 4 minutes to peel. In these 4 minutes Christen peeled 20 potatoes. OR \[ \begin{array}{c|c|c|c} \text{minute} & \text{Homer} & \text{Christen} & \text{running total}\\ \hline 1 & 3 & & 3\\ 2 & 3 & & 6\\ 3 & 3 & & 9\\ 4 & 3 & & 12\\ \hline 5 & 3 & 5 & 20\\ 6 & 3 & 5 & 28\\ 7 & 3 & 5 & 36\\ 8 & 3 & 5 & 44\\ \hline \text{Totals} & 24 & 20 & \end{array} \]
（A）4分钟后，Homer 已经削了12个土豆。当 Christen 加入后，两人合计的削皮速度是每分钟8个土豆，因此剩下的32个土豆需要4分钟削完。在这4分钟里，Christen 削了20个土豆。 或者 \[ \begin{array}{c|c|c|c} \text{分钟} & \text{Homer} & \text{Christen} & \text{累计总数}\\ \hline 1 & 3 & & 3\\ 2 & 3 & & 6\\ 3 & 3 & & 9\\ 4 & 3 & & 12\\ \hline 5 & 3 & 5 & 20\\ 6 & 3 & 5 & 28\\ 7 & 3 & 5 & 36\\ 8 & 3 & 5 & 44\\ \hline \text{合计} & 24 & 20 & \end{array} \]
solution
Q16
A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
一张边长4英寸的正方形纸张，沿垂直方向对折。然后两层纸都沿平行于折痕的方向切成两半。形成了三个新矩形，一个大的和两个小的。其中一个小矩形的周长与大矩形的周长之比是多少？
stem
Correct Answer: E
The dimensions of the new rectangles are shown. The perimeter of a small rectangle is $4 + 1 + 4 + 1 = 10$ inches and for the large one it is $4 + 2 + 4 + 2 = 12$ inches. The ratio is $10/12 = 5/6$.
新矩形的尺寸如图所示。小矩形的周长是$4 + 1 + 4 + 1 = 10$英寸，大矩形的周长是$4 + 2 + 4 + 2 = 12$英寸。比值为$10/12 = 5/6$。
solution
Q17
For the game show *Who Wants To Be A Millionaire?*, the dollar values of each question are shown in the following table (where K = 1000). Between which two questions is the percent increase of the value the smallest? [Table: Question 1 to 15 values: 100,200,300,500,1K,2K,4K,8K,16K,32K,64K,125K,250K,500K,1000K]
在游戏节目《谁想成为百万富翁？》中，各题的美元价值如以下表格所示（其中K=1000）。哪两题之间的价值百分比增长最小？[表格：第1至15题价值：100,200,300,500,1K,2K,4K,8K,16K,32K,64K,125K,250K,500K,1000K]
stem
Correct Answer: B
(B) The percent increase from $a$ to $b$ is given by $\dfrac{b-a}{a}(100\%)$ For example, the percent increase for the first two questions is $\dfrac{200-100}{100}(100\%)=100\%$ Each time the amount doubles there is a 100% increase. The only exceptions in this game are 2 to 3 (50%), 3 to 4 ($66\dfrac{2}{3}\%$) and 11 to 12 (about 95%). The answer is (B). OR $\begin{array}{r|rrrrrrrrrrrrrrr} \text{Question} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \text{Value} & 100 & 200 & 300 & 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} & 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K}\\ \%\ \text{Increase} & \text{—} & 100 & 50 & 66.7 & 100 & 100 & 100 & 100 & 100 & 100 & 100 & 95 & 100 & 100 & 100 \end{array}$
（B）从$a$增加到$b$的百分比增幅为 $\dfrac{b-a}{a}(100\%)$ 例如，前两题的百分比增幅是 $\dfrac{200-100}{100}(100\%)=100\%$ 每当数值翻倍，就会增加100%。这个游戏中唯一的例外是：2到3（50%）、3到4（$66\dfrac{2}{3}\%$）以及11到12（约95%）。答案是（B）。 或者 $\begin{array}{r|rrrrrrrrrrrrrrr} \text{题号} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \text{数值} & 100 & 200 & 300 & 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} & 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K}\\ \%\ \text{增幅} & \text{—} & 100 & 50 & 66.7 & 100 & 100 & 100 & 100 & 100 & 100 & 100 & 95 & 100 & 100 & 100 \end{array}$
Q18
Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5?
掷两个骰子。两个数字之积是5的倍数的概率是多少？
Correct Answer: D
(D) There would be $6 \times 6 = 36$ entries in the table if it were complete, but only the 11 entries that are multiples of 5 are shown. The probability of getting a multiple of 5 is $11/36$. \[ \begin{array}{c|cccccc} \times & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & & & & & 5 & \\ 2 & & & & & 10 & \\ 3 & & & & & 15 & \\ 4 & & & & & 20 & \\ 5 & 5 & 10 & 15 & 20 & 25 & 30 \\ 6 & & & & & 30 & \\ \end{array} \]
(D) 如果这个表格是完整的，那么应有 $6 \times 6 = 36$ 个条目，但这里只显示了 11 个是 5 的倍数的条目。得到 5 的倍数的概率是 $11/36$。 \[ \begin{array}{c|cccccc} \times & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & & & & & 5 & \\ 2 & & & & & 10 & \\ 3 & & & & & 15 & \\ 4 & & & & & 20 & \\ 5 & 5 & 10 & 15 & 20 & 25 & 30 \\ 6 & & & & & 30 & \\ \end{array} \]
solution
Q19
Car M traveled at a constant speed for a given time. This is shown by the dashed line. Car N traveled at twice the speed for the same distance. If Car N's speed and time are shown as solid line, which graph illustrates this?
汽车M以恒定速度行驶了一段时间。这由虚线所示。汽车N以两倍速度行驶相同的距离。如果汽车N的速度和时间由实线所示，哪张图示意了这一点？
Correct Answer: D
The second car travels the same distance at twice the speed; therefore, it needs half the time required for the first car. Graph D shows this relationship.
第二辆车以两倍速度行驶相同距离；因此，它需要第一辆车所需时间的一半。图D显示了这种关系。
Q20
Kaleana shows her test score to Quay, Marty and Shana, but the others keep theirs hidden. Quay thinks, "At least two of us have the same score." Marty thinks, "I didn't get the lowest score." Shana thinks, "I didn't get the highest score." List the scores from lowest to highest for Marty (M), Quay (Q) and Shana (S).
Kaleana向Quay、Marty和Shana展示她的考试分数，但其他人隐藏了他们的分数。Quay想：“我们至少有两个人的分数相同。” Marty想：“我没有得到最低分。” Shana想：“我没有得到最高分。” 将Marty (M)、Quay (Q)和Shana (S)的分数从最低到最高列出。
Correct Answer: A
Quay indicates that she has the same score as Kaleana. Marty’s statement indicates that her score is higher than Kaleana’s, and Shana’s statement indicates that her score is lower than Kaleana’s. The sequence S,Q,M is the correct one.
Quay表示她和Kaleana分数相同。Marty的陈述表示她的分数高于Kaleana，Shana的陈述表示她的分数低于Kaleana。序列S,Q,M是正确的。
Q21
The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
一个由五个不同正整数组成的集合的平均数是15。中位数是18。这五个整数中最大的数可能的最大值是
Correct Answer: D
The sum of all five numbers is 5(15)=75. Let the numbers be W, X, 18, Y and Z in increasing order. For Z to be as large as possible, make W, X and Y as small as possible. The smallest possible values are W = 1, X = 2 and Y = 19. Then the sum of W, X, 18 and Y is 40, and the difference, 75 −40 = 35, is the largest possible value of Z.
这五个数的和是$5\times15=75$。设这些数按升序为$W,X,18,Y,Z$。要使$Z$尽可能大，要使$W,X,Y$尽可能小。可能的最小值为$W=1,X=2,Y=19$。则$W+X+18+Y=40$，差值为$75-40=35$，即$Z$的最大可能值。
Q22
On a twenty-question test, each correct answer is worth 5 points, each unanswered question is worth 1 point and each incorrect answer is worth 0 points. Which of the following scores is NOT possible?
在一份20题的测试中，每题正确得5分，每题未答得1分，每题错误得0分。以下哪个分数是不可能的？
Correct Answer: E
To get a score in the 90s, a student must get 18 or 19 correct answers. If the number is 18, then the other two questions are worth 0+0, 0+1, 1+0 or 1+1, producing total scores of 90, 91 or 92. If the number correct is 19, then the total is 95+0 or 95+1. Therefore, the only possible scores in the 90s are 90, 91, 92, 95 and 96. This leaves 97 as an impossible score.
要得到90分的成绩，学生必须答对18或19题。如果答对18题，则剩下两题得分组合为0+0、0+1、1+0或1+1，总分为90、91或92。如果答对19题，则总分为95+0或95+1，即95或96。因此90分的可能分数只有90、91、92、95、96，97分不可能。
Q23
Points R, S and T are vertices of an equilateral triangle, and points X, Y and Z are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?
点$R,S,T$是一个等边三角形的顶点，点$X,Y,Z$是其边的中点。用这六个点中的任意三个作为顶点，可以画出多少个非全等的三角形？
stem
Correct Answer: D
There are four noncongruent triangles.
有四个非全等的三角形。
solution
Q24
Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
这个图形的每一半由3个红三角形、5个蓝三角形和8个白三角形组成。当上半部分沿中心线向下折叠时，有2对红三角形重合，3对蓝三角形重合。有2对红白对。有多少对白三角形重合？
stem
Correct Answer: B
All six red triangles are accounted for, so the two unmatched upper blue triangles must coincide with lower white triangles. Since one lower white triangle is matched with a red triangle and two are matched with blue triangles, there are five left and these must match with upper white triangles.
所有六个红三角形都已匹配，因此上半部分的两个未匹配蓝三角形必须与下半部分的白色三角形重合。下半部分有一个白三角形与红三角形匹配，两个与蓝三角形匹配，还剩五个，这些必须与上半部分的白三角形匹配。
solution
Q25
There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5 and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?
有24个四位数，它们恰好各使用一次数字2、4、5、7。其中只有一个四位数是另一个的倍数。以下哪个是它？
Correct Answer: D
(D) Six of the 24 numbers are in the 2000s, six in the 4000s, six in the 5000s and six in the 7000s. Doubling and tripling numbers in the 2000s produce possible solutions, but any multiple of those in the other sets is larger than 8000. Units digits of the numbers are 2, 4, 5 and 7, so their doubles will end in 4, 8, 0 and 4, respectively. Choice (A) 5724 ends in 4 but $5724/2=2862$, not one of the 24 numbers. Likewise, choice (C) 7254 produces $7254/2=3627$, also not one of the numbers. When the units digits are tripled the resulting units digits are 6, 2, 5 and 1 and choices (B) 7245, (D) 7425 and (E) 7542 are possibilities. Division by 3 yields 2415, 2475 and 2514 respectively. Only the second of these numbers is one of the 24 given numbers. Choice (D) is correct.
（D）这 24 个数中，有 6 个在 2000 段，6 个在 4000 段，6 个在 5000 段，6 个在 7000 段。将 2000 段中的数加倍或三倍会产生可能的解，但对其他几组中的数取倍数都会大于 8000。 这些数的个位数字是 2、4、5 和 7，因此它们的两倍的个位分别是 4、8、0 和 4。选项（A）5724 的个位是 4，但 $5724/2=2862$，不是这 24 个数之一。同样，选项（C）7254 有 $7254/2=3627$，也不是这些数之一。将个位数字乘以 3 后得到的个位数字是 6、2、5 和 1，因此选项（B）7245、（D）7425 和（E）7542 都有可能。分别除以 3 得到 2415、2475 和 2514。只有第二个数是所给 24 个数之一。因此选项（D）正确。