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AMC8 2002

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AMC8 · 2002

Q1
A circle and two distinct lines are drawn on a sheet of paper. What is the largest possible number of points of intersection of these figures?
一张纸上画了一个圆和两条不同的直线。这些图形相交点的最大可能数量是多少？
Correct Answer: D
Two distinct lines can intersect in one point whereas a line can intersect a circle in two points. The maximum number 5 can be achieved if the lines and circle are arranged as shown. Note that the lines could also meet outside the circle for the same result. (Other arrangements of the lines and circle can produce 0, 1, 2, 3, or 4 points of intersection.)
两条不同的直线可以相交于一点，而一条直线可以与圆相交于两点。如果直线和圆按如图所示排列，可以实现最大数量5。注意，直线也可以在圆外相交，得到相同结果。（直线和圆的其他排列可以产生0、1、2、3或4个相交点。）
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Q2
How many different combinations of \$5 bills and \$2 bills can be used to make a total of \$17? Order does not matter in this problem.
用5美元钞票和2美元钞票凑成总计17美元，有多少种不同的组合？本题中顺序无关。
stem
Correct Answer: A
(A) Since the total \$17 is odd, there must be an odd number of \$5 bills. One \$5 bill plus six \$2 bills is a solution, as is three \$5 bills plus one \$2 bill. Five \$5 bills exceeds \$17, so these are the only two combinations that work.
（A）因为总额 \$17 是奇数，所以 \$5 钞票的张数必须是奇数。用 1 张 \$5 加 6 张 \$2 是一种解，3 张 \$5 加 1 张 \$2 也是一种解。5 张 \$5 已经超过 \$17，所以只有这两种组合可行。
Q3
What is the smallest possible average of four distinct positive even integers?
四个不同的正偶整数的最小可能平均数是多少？
Correct Answer: C
The smallest average will occur when the numbers are as small as possible. The four smallest distinct positive even integers are 2, 4, 6, and 8 and their average is 5. Note: These numbers form an arithmetic sequence. The average of the numbers in any arithmetic sequence is the average of the first and last terms.
为了使平均数最小，数字应尽可能小。四个最小的不同正偶整数是2、4、6和8，它们的平均数是5。注意：这些数字构成一个等差数列。任何等差数列中数字的平均数等于首尾项的平均数。
Q4
The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
年份2002是一个回文数（从左到右和从右到左读相同）。2002年之后下一个回文年的各位数字乘积是多少？
Correct Answer: B
The next palindrome is 2112. The product of its digits is $2 \cdot 1 \cdot 1 \cdot 2 = 4$.
下一个回文年是2112。其各位数字乘积是$2 \cdot 1 \cdot 1 \cdot 2 = 4$。
Q5
Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?
卡洛斯·蒙塔多于2002年11月9日星期六出生。706天后，卡洛斯是星期几？
Correct Answer: C
Since 706 days is 700 plus 6 days, it is 100 weeks plus 6 days. Friday is 6 days after Saturday.
706天是700天加6天，即100周加6天。星期六后6天是星期五。
Q6
A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. One of these graphs shows the volume of water in the birdbath during the filling time and continuing into the overflow time. Which one is it?
一个鸟浴设计成会溢水以实现自清洁。水以每分钟20毫升的速率流入，以每分钟18毫升的速率排出。其中一个图表显示了鸟浴在填充时间及溢水时间的水量变化。哪一个是正确的？
stem
Correct Answer: A
Initially, volume increases with time as shown by graphs A, C, and E. But once the birdbath is full, the volume remains constant as the birdbath overflows. Only graph A shows both features.
最初，水量随时间增加，如图A、C和E所示。但一旦鸟浴满了，水量保持恒定，因为鸟浴溢水。只有图A显示了这两个特征。
Q7
The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
Sawyer夫人的班级学生进行五种糖果的味觉测试。每位学生选择一种糖果。他们的偏好柱状图如图所示。她班上有多少百分比的学生选择了糖果E？
stem
Correct Answer: E
There are $6 + 8 + 4 + 2 + 5 = 25$ students. Of the 25 students 5 prefer candy E and $\frac{5}{25} = \frac{20}{100} = 20\%$.
有 $6 + 8 + 4 + 2 + 5 = 25$ 名学生。在25名学生中，有5名喜欢糖果E，$\frac{5}{25} = \frac{20}{100} = 20\%$。
Q8
How many of his European stamps were issued in the '80s?
他的欧洲邮票中有多少是80年代发行的？
stem
Correct Answer: D
There are 15 French stamps and 9 Spanish stamps issued in the '80s. So there are $15 + 9 = 24$ European stamps listed in the table in the '80s.
80年代发行的法国邮票有15张，西班牙邮票有9张。所以表中80年代的欧洲邮票有 $15 + 9 = 24$ 张。
Q9
His South American stamps issued before the '70s cost him
他在70年代之前发行的南美邮票花费了他
Correct Answer: B
His South American stamps issued before the '70s include $4 + 7 = 11$ from Brazil that cost $11 \times \$0.06 = \$0.66$ and $6 + 4 = 10$ from Peru that cost $10 \times \$0.04 = \$0.40$. Their total cost is $0.66 + 0.40 = \$1.06$.
他在70年代之前发行的南美邮票包括巴西的 $4 + 7 = 11$ 张，花费 $11 \times \$0.06 = \$0.66$，秘鲁的 $6 + 4 = 10$ 张，花费 $10 \times \$0.04 = \$0.40$。总花费是 $0.66 + 0.40 = \$1.06$。
Q10
The average price of his '70s stamps is closest to
他的70年代邮票的平均价格最接近
Correct Answer: E
The '70s stamps cost: Brazil, $12(\$$0.06) = \$0.72$; Peru, $6(\$$0.04) = \$0.24$; France, $12(\$$0.06) = \$0.72$; Spain, $13(\$$0.05) = \$0.65$. The total is $2.33$ for the 43 stamps and the average price is $\frac{\$$2.33}{43} \approx \$0.054 \approx 5.5¢$.
70年代邮票花费：巴西，$12(\$$0.06) = \$0.72$；秘鲁，$6(\$$0.04) = \$0.24$；法国，$12(\$$0.06) = \$0.72$；西班牙，$13(\$$0.05) = \$0.65$。43张邮票总计 $2.33$，平均价格 $\frac{\$$2.33}{43} \approx \$0.054 \approx 5.5¢$。
Q11
A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
一个由相同正方形瓦片构成的正方形序列。每个正方形的边长比前一个正方形的边长多一个瓦片长度。展示了前三个正方形。第七个正方形比第六个正方形多需要多少瓦片？
stem
Correct Answer: C
(C) To build the second square from the first, add 3 tiles. To build the third from the second, add 5 tiles. The pattern of adding an odd number of tiles continues. For the fourth square, add 7; for the fifth, add 9; for the sixth, add 11 and for the seventh, add 13.
（C）要从第一个正方形建成第二个，需要加 3 块瓷砖。要从第二个建成第三个，需要加 5 块瓷砖。继续遵循每次增加奇数块瓷砖的规律。第四个正方形加 7 块；第五个加 9 块；第六个加 11 块；第七个加 13 块。
Q12
A board game spinner is divided into three regions labeled A, B and C. The probability of the arrow stopping on region A is $\frac{1}{3}$ and on region B is $\frac{1}{2}$. The probability of the arrow stopping on region C is
一个棋盘游戏转盘分为三个区域，标有A、B和C。箭头停在区域A的概率为$\frac{1}{3}$，停在区域B的概率为$\frac{1}{2}$。箭头停在区域C的概率是
stem
Correct Answer: B
Since the sum of the three probabilities is 1, the probability of stopping on region $C$ is $1 - \frac{1}{3} - \frac{1}{2} = \frac{6}{6} - \frac{2}{6} - \frac{3}{6} = \frac{1}{6}$.
由于三个概率之和为1，停在区域$C$的概率为$1 - \frac{1}{3} - \frac{1}{2} = \frac{6}{6} - \frac{2}{6} - \frac{3}{6} = \frac{1}{6}$。
Q13
For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
伯特生日时得到一个满载时可容纳125颗果冻豆的盒子。几周后，嘉莉得到一个更大的满载果冻豆的盒子。她的盒子高度、宽度和长度都是伯特的盒子的两倍。大约，嘉莉得到了多少颗果冻豆？
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Correct Answer: E
Since the exact dimensions of Bert's box do not matter, assume the box is $1 \times 2 \times 3$. Its volume is 6. Carrie's box is $2 \times 4 \times 6$, so its volume is 48 or 8 times the volume of Bert's box. Carrie has approximately $8(125) = 1000$ jellybeans.
由于伯特盒子的确切尺寸无关紧要，假设盒子为$1 \times 2 \times 3$。其体积为6。嘉莉的盒子为$2 \times 4 \times 6$，体积为48，即伯特盒子体积的8倍。嘉莉大约有$8(125) = 1000$颗果冻豆。
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Q14
A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off these sale prices and claims that the final price of these items is 50% off the original price. The total discount is
一个商人对一大群商品打7折。后来，商人再对这些折后价格打8折，并声称这些商品的最终价格是原价的50% off。总折扣是
Correct Answer: B
The first discount means that the customer will pay 70% of the original price. The second discount means a selling price of 80% of the discounted price. Because $0.80(0.70) = 0.56 = 56\%$, the customer pays 56% of the original price and thus receives a 44% discount.
第一次折扣意味着顾客支付原价的70%。第二次折扣意味着销售价为折后价格的80%。因为$0.80(0.70) = 0.56 = 56\%$，顾客支付原价的56%，因此获得44%的折扣。
Q15
Which of the following polygons has the largest area?
下列哪个多边形面积最大？
stem
Correct Answer: E
(E) Areas may be found by dividing each polygon into triangles and squares as shown. Note: Pick’s Theorem may be used to find areas of geoboard polygons. If $I$ is the number of dots inside the figure, $B$ is the number of dots on the boundary and $A$ is the area, then $A = I + \frac{B}{2} - 1$. Geoboard figures in this problem have no interior points, so the formula simplifies to $A = \frac{B}{2} - 1$. For example, in polygon $D$ the number of boundary points is 11 and $\frac{11}{2} - 1 = 4\frac{1}{2}$.
（E）如图所示，可以通过把每个多边形分割成三角形和正方形来求面积。 注：可以用皮克定理来求钉板（geoboard）多边形的面积。若 $I$ 为图形内部点的个数，$B$ 为边界上的点的个数，$A$ 为面积，则 $A = I + \frac{B}{2} - 1$。本题中的钉板图形没有内部点，因此公式可简化为 $A = \frac{B}{2} - 1$。例如，在多边形 $D$ 中，边界点的个数为 11，且 $\frac{11}{2} - 1 = 4\frac{1}{2}$。
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Q16
Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each triangle. Which one of the following is true?
在3-4-5直角三角形的各边上构造直角等腰三角形，如图所示。大写字母表示每个三角形的面积。以下哪一个是正确的？
stem
Correct Answer: E
(E) The areas are $W=\frac{1}{2}(3)(4)=6$, $X=\frac{1}{2}(3)(3)=4\frac{1}{2}$, $Y=\frac{1}{2}(4)(4)=8$ and $Z=\frac{1}{2}(5)(5)=12\frac{1}{2}$. Therefore, (E) is correct. $X+Y=4\frac{1}{2}+8=12\frac{1}{2}=Z$. The other choices are incorrect.
（E）这些面积分别为：$W=\frac{1}{2}(3)(4)=6$，$X=\frac{1}{2}(3)(3)=4\frac{1}{2}$，$Y=\frac{1}{2}(4)(4)=8$，以及$Z=\frac{1}{2}(5)(5)=12\frac{1}{2}$。因此，（E）是正确的。$X+Y=4\frac{1}{2}+8=12\frac{1}{2}=Z$。其他选项是不正确的。
Q17
In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
在一场有十道题的数学竞赛中，学生答对一题得5分，答错一题扣2分。如果Olivia回答了所有题目，她的得分为29，她答对了几题？
Correct Answer: C
(C) Olivia solved at least 6 correctly to have scored over 25. Her score for six correct would be $6(+5)+4(-2)=22$, which is too low. If she answered 7 correctly, her score would be $7(+5)+3(-2)=29$, and this was her score. The correct choice is (C).
（C）Olivia 至少答对了 6 题才能得分超过 25。她若答对 6 题，得分为 $6(+5)+4(-2)=22$，这太低了。如果她答对 7 题，得分为 $7(+5)+3(-2)=29$，而这正是她的得分。正确选项是（C）。
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Q18
Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?
Gage连续5天每天滑冰1小时15分钟，又连续3天每天滑冰1小时30分钟。为了整个9天平均每天滑冰85分钟，第九天他需要滑冰多长时间？
Correct Answer: E
In 5 days, Gage skated for 5 × 75 = 375 minutes, and in 3 days he skated for 3×90 = 270 minutes. So, in 8 days he skated for 375+270 = 645 minutes. To average 85 minutes per day for 9 days he must skate 9×85 = 765 minutes, so he must skate 765 −645 = 120 minutes = 2 hours the ninth day.
5天滑冰总时间5 × 75 = 375分钟，3天滑冰3×90 = 270分钟。8天总共滑冰375+270 = 645分钟。要9天平均85分钟/天，总需滑冰9×85 = 765分钟，所以第九天需滑冰765 −645 = 120分钟=2小时。
Q19
How many whole numbers between 99 and 999 contain exactly one 0?
99和999之间有多少个整数恰好包含一个0？
Correct Answer: D
Numbers with exactly one zero have the form a0b or ab0, where the blanks are not zeros. There are (9 · 1 · 9) + (9 · 9 · 1) = 81 + 81 = 162 such numbers.
恰好一个0的数形式为a0b或ab0，其中空白处不为零。有(9 · 1 · 9) + (9 · 9 · 1) = 81 + 81 = 162个数。
Q20
The area of triangle $XYZ$ is 8 square inches. Points $A$ and $B$ are midpoints of congruent segments $\overline{XY}$ and $\overline{XZ}$. Altitude $\overline{XC}$ bisects $\overline{YZ}$. The area (in square inches) of the shaded region is
三角形$XYZ$的面积是8平方英寸。点$A$和$B$分别是全等线段$\overline{XY}$和$\overline{XZ}$的中点。高$\overline{XC}$平分$\overline{YZ}$。阴影区域的面积（平方英寸）是
stem
Correct Answer: D
(D) Segments $\overline{AD}$ and $\overline{BE}$ are drawn perpendicular to $\overline{YZ}$. Segments $\overline{AB}$, $\overline{AC}$ and $\overline{BC}$ divide $\triangle XYZ$ into four congruent triangles. Vertical line segments $\overline{AD}$, $\overline{XC}$ and $\overline{BE}$ divide each of these in half. Three of the eight small triangles are shaded, or $\frac{3}{8}$ of $\triangle XYZ$. The shaded area is $\frac{3}{8}(8)=3$.
(D) 线段 $\overline{AD}$ 和 $\overline{BE}$ 作垂直于 $\overline{YZ}$。线段 $\overline{AB}$、$\overline{AC}$ 和 $\overline{BC}$ 将 $\triangle XYZ$ 分成四个全等三角形。竖直线段 $\overline{AD}$、$\overline{XC}$ 和 $\overline{BE}$ 将其中每个再平分。8 个小三角形中有 3 个被阴影涂色，即占 $\triangle XYZ$ 的 $\frac{3}{8}$。阴影面积为 $\frac{3}{8}(8)=3$。
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Q21
Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is
Harold 抛掷一枚五分镍币四次。他得到至少与反面一样多的正面的概率是
stem
Correct Answer: E
There are 16 possible outcomes: HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, THHT, TTHH and HTTT, THTT, TTHT, TTTH, TTTT. The first eleven have at least as many heads as tails. The probability is $\frac{11}{16}$.
总共有16种可能结果：HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, THHT, TTHH 和 HTTT, THTT, TTHT, TTTH, TTTT。前11种结果正面的数量至少与反面一样多。概率是 $\frac{11}{16}$。
Q22
Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom and sides.
六个每个边长一英寸的立方体被粘在一起，如图所示。求总表面积（平方英寸）。包括顶部、底部和侧面。
stem
Correct Answer: C
(C) When viewed from the top and bottom, there are 4 faces exposed; from the left and right sides, there are 4 faces exposed and from the front and back, there are 5 faces exposed. The total is $4 + 4 + 4 + 4 + 5 + 5 = 26$ exposed faces.
（C）从上面和下面看，有 4 个面暴露；从左侧和右侧看，有 4 个面暴露；从前面和后面看，有 5 个面暴露。总数为 $4 + 4 + 4 + 4 + 5 + 5 = 26$ 个暴露的面。
Q23
A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?
展示了一个铺砖地板的一个角落。如果整个地板都以这种方式铺砖，并且四个角落看起来都像这个，那么铺砖地板中有多少分数是深色瓷砖？
stem
Correct Answer: B
The 6 × 6 square in the upper left-hand region is tessellated, so finding the proportion of darker tiles in this region will answer the question. The top left-hand corner of this region is a 3 × 3 square that has $3 + 2(\frac{1}{2}) = 4$ darker tiles. So $\frac{4}{9}$ of the total area will be made of darker tiles.
左上角的6 × 6正方形区域是镶嵌的，所以找出这个区域中深色瓷砖的比例就能回答问题。这个区域的左上角是一个3 × 3正方形，其中有 $3 + 2(\frac{1}{2}) = 4$ 个深色瓷砖。所以总面积的 $\frac{4}{9}$ 是深色瓷砖。
solution
Q24
Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
Miki 有一打大小相同的橙子和一打大小相同的梨。Miki 用榨汁机从3个梨中榨出8盎司梨汁，从2个橙子中榨出8盎司橙汁。她用等数量的梨和橙子制作梨橙汁混合饮料。这个混合饮料中有百分之多少是梨汁？
stem
Correct Answer: B
Use 6 pears to make 16 oz of pear juice and 6 oranges to make 24 oz of orange juice for a total of 40 oz of juice. The percent of pear juice is $\frac{16}{40} = \frac{4}{10} = 40\%$.
用6个梨制作16盎司梨汁，用6个橙子制作24盎司橙汁，总共40盎司汁。梨汁的百分比是 $\frac{16}{40} = \frac{4}{10} = 40\%$ 。
Q25
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
Loki、Moe、Nick 和 Ott 是好朋友。Ott 没有钱，但其他人有。Moe 给了 Ott 他钱的一第五，Loki 给了 Ott 他钱的一第四，Nick 给了 Ott 他钱的三分之一。每人都给了 Ott 等量的钱。Ott 现在拥有团体总钱款的几分之几？
Correct Answer: B
(B) Only the fraction of each friend’s money is important, so we can assume any convenient amount is given to Ott. Suppose that each friend gave Ott \$1. If this is so, then Moe had \$5 originally and now has \$4, Loki had \$4 and now has \$3, and Nick had \$3, and now has \$2. The four friends now have \$4 + \$3 + \$2 + \$3 = \$12, so Ott has $\frac{3}{12}=\frac{1}{4}$ of the group’s money. This same reasoning applies to any amount of money.
（B）每个朋友的钱所占的比例才是关键，因此我们可以假设给 Ott 的金额是任意方便的数。假设每个朋友都给了 Ott \$1。这样的话，Moe 原来有 \$5，现在有 \$4；Loki 原来有 \$4，现在有 \$3；Nick 原来有 \$3，现在有 \$2。四个朋友现在共有 \$4 + \$3 + \$2 + \$3 = \$12，所以 Ott 拥有该小组资金的 $\frac{3}{12}=\frac{1}{4}$。同样的推理适用于任何金额。