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AMC8 1997

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AMC8 · 1997

Q1
\(\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =\)
\(\frac{1}{10} + \frac{9}{100} + \frac{9}{1000} + \frac{7}{10000} =\)
Correct Answer: C
Convert to decimals: \(0.1 + 0.09 + 0.009 + 0.0007 = 0.1997\).
转换为小数：\(0.1 + 0.09 + 0.009 + 0.0007 = 0.1997\)。
Q2
Ahn chooses a two-digit number, subtracts it from 200, and doubles the result. What is the largest number Ahn can get?
Ahn 选择一个两位数，从 200 中减去它，然后将结果加倍。Ahn 能得到的最大数是多少？
Correct Answer: D
To maximize \(2 \times (200 - n)\), minimize the two-digit \(n = 10\), so \(2 \times 190 = 380\).
要最大化 \(2 \times (200 - n)\)，最小化两位数 \(n = 10\)，因此 \(2 \times 190 = 380\)。
Q3
Which of the following numbers is the largest?
下列哪个数最大？
Correct Answer: B
Write to four decimal places: 0.9700, 0.9790, 0.9709, 0.9070, 0.9089. The largest is 0.9790.
写成四位小数：0.9700, 0.9790, 0.9709, 0.9070, 0.9089。最大的是 0.9790。
Q4
Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If Julie speaks at the ideal rate, which of the following number of words would be appropriate length for her speech?
Julie 正在为她的班级准备演讲。她的演讲必须持续在半小时到四分之三小时之间。理想的演讲速度是每分钟 150 词。如果 Julie 以理想速度演讲，下列哪个词数适合她的演讲长度？
Correct Answer: E
Half hour: \(30 \times 150 = 4500\) words; three-quarters hour: \(45 \times 150 = 6750\) words. Only 5650 is in between.
半小时：\(30 \times 150 = 4500\) 词；四分之三小时：\(45 \times 150 = 6750\) 词。只有 5650 在之间。
Q5
There are many two-digit multiples of 7, but only two of the multiples have a digit sum of 10. The sum of these two multiples of 7 is
有许多两位数的 7 的倍数，但只有两个倍数的数字和为 10。这两个 7 的倍数的和是
Correct Answer: A
The two-digit multiples of 7 with digit sum 10 are 28 and 91. Their sum is 119.
数字和为 10 的两位数 7 的倍数是 28 和 91。它们的和是 119。
Q6
In the number 74982.1035 the value of the place occupied by the digit 9 is how many times as great as the value of the place occupied by the digit 3?
在数字 74982.1035 中，数字 9 所在位置的值是数字 3 所在位置的值的多少倍？
Correct Answer: C
The digit 9 is in the hundreds place (value 100), the digit 3 is in the thousandths place (value 0.001). \(100 / 0.001 = 100{,}000\). Each place shift is a factor of 10, and there are 5 shifts.
数字 9 在百位（值 100），数字 3 在千分位（值 0.001）。\(100 / 0.001 = 100{,}000\)。每位移动一次是 10 的倍数，总共 5 次移动。
solution
Q7
The area of the smallest square that will contain a circle of radius 4 is
能容纳半径为 4 的圆的最小正方形的面积是
Correct Answer: D
The smallest square has side length equal to the diameter, 8, so area \(8^2 = 64\).
最小正方形的边长等于直径 8，因此面积 \(8^2 = 64\)。
solution
Q8
Walter gets up at 6:30 am, catches the school bus at 7:30 am, has 6 classes that last 50 minutes each, has 30 minutes for lunch, and has 2 hours additional time at school. He takes the bus that arrives at 4:00 pm. How many minutes has he spent on the bus?
沃尔特早上 6:30 起床，7:30 坐校车，上午有 6 节课每节 50 分钟，午餐 30 分钟，学校还有 2 小时额外时间。他坐的校车下午 4:00 到家。他总共在校车上花了多少分钟？
Correct Answer: B
Total time from 7:30 AM to 4:00 PM is 8.5 hours = 510 minutes. School time: classes 300 min + lunch 30 + 2 hours 120 = 450 min. Bus time: 510 - 450 = 60 min.
从早上 7:30 到下午 4:00 总共 8.5 小时 = 510 分钟。学校时间：上课 300 分钟 + 午餐 30 + 2 小时 120 = 450 分钟。校车时间：510 - 450 = 60 分钟。
Q9
Three students, with different names, line up single file. What is the probability that they are in alphabetical order from front-to-back?
三个学生，名字不同，排成一列单队。从前到后按字母顺序排列的概率是多少？
Correct Answer: C
Answer (B): Walter is gone from 7:30 AM until 4:00 PM, a total of 8 hours and 30 minutes. He is in class for $6(50\ \text{min.})=300\ \text{min.}$ or 5 hrs., at lunch for $\frac{1}{2}$ hr., and has 2 hours additional time. His total time at school is 7 hrs. and 30 min., so he was on the bus for 1 hour or 60 minutes.
答案（B）：Walter 从上午 7:30 到下午 4:00 不在家，一共 8 小时 30 分钟。他上课时间为 $6(50\ \text{分钟})=300\ \text{分钟}$，即 5 小时；午餐时间为 $\frac{1}{2}$ 小时；另外还有 2 小时的额外时间。他在学校的总时间是 7 小时 30 分钟，所以他在公交车上的时间是 1 小时，即 60 分钟。
Q10
What fraction of this square region is shaded? Stripes are equal in width, and the figure is drawn to scale.
这个正方形区域中有多少分数被涂黑？条纹宽度相等，且图形按比例绘制。
stem
Correct Answer: C
The figure consists of 36 small squares, 21 shaded, \(\frac{21}{36} = \frac{7}{12}\).
图形由 36 个小正方形组成，21 个被涂黑，\(\frac{21}{36} = \frac{7}{12}\)。
solution
Q11
Let \(d(N)\) mean the number of positive divisors of \(N\). For example, \(d(3) = 2\). Find the value of picture.
设 $d(N)$ 表示 $N$ 的正因数个数。例如，$d(3)=2$。求图中表达式的值。
Correct Answer: A
Answer (A): Both 1 and 11 divide 11, so $[11]=2$, and since 1, 2, 4, 5, 10, and 20 divide 20, then $[20]=6$. The inner expression, $[11]\times[20]=2\times6=12$. Finally, $[12]=6$ because 1, 2, 3, 4, 6, and 12 divide 12.
答案（A）：1 和 11 都整除 11，所以 $[11]=2$；并且 1、2、4、5、10、20 都整除 20，因此 $[20]=6$。内层表达式为 $[11]\times[20]=2\times6=12$。最后，$[12]=6$，因为 1、2、3、4、6、12 都整除 12。
solution
Q12
\(\angle 1 + \angle 2 = 180^\circ\) \(\angle 3 = \angle 4\) Find \(\angle 4\) (figure with angles 40° and 70° shown).
$\angle1+\angle2=180^\circ$ $\angle3=\angle4$ 求 $\angle4$（图中标有 $40^\circ$ 和 $70^\circ$）。
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Correct Answer: D
Triangle sum: \(40^\circ + 70^\circ + \angle 1 = 180^\circ\), so \(\angle 1 = 70^\circ\), \(\angle 2 = 110^\circ\). Then \(110^\circ + 2\angle 4 = 180^\circ\), \(\angle 4 = 35^\circ\).
三角形内角和：$40^\circ+70^\circ+\angle1=180^\circ$，故 $\angle1=70^\circ$，$\angle2=110^\circ$。然后 $110^\circ+2\angle4=180^\circ$，$\angle4=35^\circ$。
Q13
Three bags of jelly beans contain 26, 28, and 30 beans. The ratios of yellow beans to all beans in each of these bags are 50%, 25%, and 20%, respectively. All three bags of candy are dumped into one bowl. Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?
三个果冻豆袋子分别含 26、28 和 30 颗豆子。各袋中黄色豆子与总豆子的比例分别为 50%、25% 和 20%。将三个袋子全部倒入一个碗中。碗中黄色果冻豆与总豆子的比例最接近以下哪一项？
Correct Answer: A
Answer (A): There is a total of $26+28+30=84$ jelly beans: $50\%$ of $26=13$ $25\%$ of $28=7$ $20\%$ of $30=\underline{6}$ $\underline{26}$ $\dfrac{26}{84}=0.3095\approx 31\%$
答案（A）：一共有 $26+28+30=84$ 颗果冻豆： $26$ 的 $50\%=13$ $28$ 的 $25\%=7$ $30$ 的 $20\%=\underline{6}$ $\underline{26}$ $\dfrac{26}{84}=0.3095\approx 31\%$
Q14
There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?
有一个五个正整数的集合，其平均数（均值）为 5，中位数为 5，且唯一众数为 8。求该集合中最大整数与最小整数之差。
Correct Answer: D
Answer (D): Since 5 is the median, there must be two integers greater than 5 and two less than 5. The five integers may be arranged this way: $\_\ \_\ 5\ 8\ 8$. Since the mean is 5, the sum of all five integers is 25. The numbers 5, 8, and 8 total 21, leaving 4 for the sum of the first two. They can't both be 2 since 8 is the only mode, so they are 1 and 3. The difference between 8 and 1 is 7.
答案（D）：由于 5 是中位数，必须有两个整数大于 5，两个小于 5。这五个整数可以按如下方式排列：$\_\ \_\ 5\ 8\ 8$。由于平均数是 5，这五个整数的总和是 25。数 5、8 和 8 的和为 21，因此前两个数的和为 4。它们不可能都是 2，因为 8 是唯一的众数，所以它们是 1 和 3。8 和 1 的差是 7。
Q15
Each side of the large square in the figure is trisected. The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is
图中大正方形的每条边被三分等分。内接正方形的顶点位于这些三分点上，如图所示。内接正方形面积与大正方形面积之比为
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Correct Answer: B
Answer (B): Taking the side of the large square to be 3 inches gives an area of 9 square inches. Each of the four right triangles has an area of $\frac{1}{2}(2)(1)=1$ sq. inch. The area of the inscribed square is the area of the large square minus the area of the four right triangles, that is, $9-4(1)=5$ sq. inches. The desired ratio is $\frac{5}{9}$.
答案（B）：设大正方形的边长为 3 英寸，则面积为 9 平方英寸。四个直角三角形中每一个的面积为 $\frac{1}{2}(2)(1)=1$ 平方英寸。内接正方形的面积等于大正方形的面积减去四个直角三角形的面积，即 $9-4(1)=5$ 平方英寸。所求比值为 $\frac{5}{9}$。
Q16
Penni Precisely buys $100 worth of stock in each of three companies: Alabama Almonds, Boston Beans, and California Cauliflower. After one year, AA was up 20%, BB was down 25%, and CC was unchanged. For the second year, AA was down 20% from the previous year, BB was up 25% from the previous year, and CC was unchanged. If A, B, and C are the final value of the stock, then
Penni Precisely 在三家公司：Alabama Almonds、Boston Beans 和 California Cauliflower 各购买了价值 $100 的股票。一年后，AA 上涨 20%，BB 下跌 25%，CC 不变。第二年，AA 较前一年下跌 20%，BB 较前一年上涨 25%，CC 不变。如果 A、B 和 C 是股票的最终价值，则
Correct Answer: E
Answer (E): At the end of two years, stock values are: Stock AA: \$100(1.2)(0.8) = \$96 Stock BB: \$100(0.75)(1.25) = \$93.75 stock CC: \$100(1)(1) = \$100 So, $B < A < C$.
答案（E）：两年末，各股票价值为： 股票 AA：\$100(1.2)(0.8) = \$96 股票 BB：\$100(0.75)(1.25) = \$93.75 股票 CC：\$100(1)(1) = \$100 因此，$B < A < C$。
Q17
A cube has eight vertices (corners) and twelve edges. A segment, such as x, which joins two vertices not joined by an edge is called a diagonal. Segment y is also a diagonal. How many diagonals does a cube have?
一个立方体有八个顶点（角）和十二条边。连接两个不相邻顶点的线段（如 x）称为对角线。线段 y 也是对角线。立方体有几个对角线？
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Correct Answer: E
Answer (E): There are two diagonals, such as $x$, in each of the six faces for a total of twelve face diagonals. There are also four space diagonals, such as $y$, which are within the cube. This makes a total of 16.
答案（E）：立方体的每个面上有两条对角线，例如 $x$。六个面一共有 12 条面内对角线。立方体内部还有 4 条空间对角线，例如 $y$。因此总共有 16 条对角线。
solution
Q18
At the grocery store last week, small boxes of facial tissue were priced at 4 boxes for $5. This week they are on sale at 5 boxes for $4. The percent decrease in the price per box during the sale was closest to
上周杂货店，小盒面巾纸的价格是 4 盒 $5。本周特价 5 盒 $4。每盒价格在促销期间的百分比降幅最接近
Correct Answer: B
Answer (B): Last week one box cost \$1.25; this week one box costs \$0.80. The decrease, \$0.45, compared to original price of \$1.25, is a decrease of $\frac{0.45}{1.25}=0.36$ or 36%, so choice (B) is closest.
答案（B）：上周一盒的价格是 \$1.25；这周一盒的价格是 \$0.80。降价幅度为 \$0.45，相对于原价 \$1.25，降幅为 $\frac{0.45}{1.25}=0.36$，即 36%，所以选项（B）最接近。
Q19
If the product \(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} = 9\), what is the sum of a and b?
如果乘积 \(\frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6} = 9\)，a 和 b 的和是多少？
Correct Answer: D
Answer (D): Since $\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b}=9$, we see that $\frac{a}{2}=9$. Thus, $a=18$, $b=17$, and $a+b=35$.
答案（D）：因为 $\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdots\frac{a}{b}=9$，所以可知 $\frac{a}{2}=9$。因此，$a=18$，$b=17$，且 $a+b=35$。
Q20
A pair of 8-sided dice have sides numbered 1 through 8. Each side has the same probability of landing face up. The probability that the product of the two numbers on the sides that land face-up exceeds 36 is
一对 8 面骰子，面编号 1 到 8。每面朝上的概率相同。两个朝上面数字的乘积超过 36 的概率是
Correct Answer: A
Answer (A): There are 64 equally likely possibilities for the numbers on the two dice. Of these, only (5, 8), (6, 7), (6, 8), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), and (8, 8) give products exceeding 36, so the probability of this occurring is \(\frac{10}{64}=\frac{5}{32}\).
答案（A）：两个骰子上的数字共有 64 种等可能的情况。其中，只有 (5, 8)、(6, 7)、(6, 8)、(7, 6)、(7, 7)、(7, 8)、(8, 5)、(8, 6)、(8, 7) 和 (8, 8) 的乘积大于 36，因此该事件发生的概率为 $\frac{10}{64}=\frac{5}{32}$。
Q21
Each corner cube is removed from this 3 cm × 3 cm × 3 cm cube. The surface area of the remaining figure is
从这个 $3\ cm \times 3\ cm \times 3\ cm$ 的正方体中移除了每个角上的小正方体。剩余图形的表面积是
stem
Correct Answer: D
Answer (D): In terms of the original cube, three square centimeters are lost from each corner, but three new squares are added as the sides of a cavity in that corner. The total area remains at 54 square centimeters.
答案（D）：就原立方体而言，每个角会减少 $3$ 平方厘米的面积，但在该角形成的凹槽会新增 $3$ 个正方形面作为其侧面。因此，总表面积仍为 $54$ 平方厘米。
solution
Q22
A two-inch cube (2 × 2 × 2) of silver weighs 3 pounds and is worth $200. How much is a three-inch cube of silver worth?
一个 2 英寸的银立方体（$2 \times 2 \times 2$）重 3 磅，价值 200 美元。一个 3 英寸的银立方体价值多少？
Correct Answer: E
Answer (E): The volume of a two-inch cube is $2^3=8$ cu. inches, while that of a three-inch cube is $27$ cu. inches. Therefore, the weight and value of the larger cube is $\frac{27}{8}$ times that of the smaller. \$200$\left(\frac{27}{8}\right)$ = \$675. Note: The actual weight of the cubes is not needed to solve the problem.
答案（E）：边长为2英寸的立方体体积为 $2^3=8$ 立方英寸，而边长为3英寸的立方体体积为 $27$ 立方英寸。因此，大立方体的重量和价值是小立方体的 $\frac{27}{8}$ 倍。 \$200$\left(\frac{27}{8}\right)$ = \$675。 注：解决该问题不需要知道立方体的实际重量。
Q23
There are positive integers that have these properties: I. The sum of the squares of their digits is 50, and II. Each digit is larger than the one on its left. The product of the digits of the largest integer with both properties is
存在具有以下性质的正整数：I. 其各位数字平方和为 50，且 II. 每位数字都大于其左边的数字。具有两者性质的最大整数的各位数字乘积是
Correct Answer: C
Answer (C): To meet the first condition, numbers which sum to 50 must be chosen from the set of squares $\{1,4,9,16,25,36,49\}$. To meet the second condition, the squares selected must be different. Consequently, there are three possibilities: $1+49$, $1+4+9+36$, and $9+16+25$. These correspond to the integers 17, 1236, and 345, respectively. The largest is 1236, and the product of its digits is $1\cdot 2\cdot 3\cdot 6=36$.
答案（C）：为满足第一个条件，必须从平方数集合 $\{1,4,9,16,25,36,49\}$ 中选出和为 50 的数。为满足第二个条件，所选的平方数必须互不相同。因此，有三种可能：$1+49$、$1+4+9+36$、以及 $9+16+25$。它们分别对应整数 17、1236 和 345。最大的是 1236，其各位数字的乘积为 $1\cdot 2\cdot 3\cdot 6=36$。
Q24
Diameter ACE is divided at C in the ratio 2 : 3. The two semicircles, ABC and CDE, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is
直径 ACE 在 C 点按 2 : 3 的比例分割。两个半圆 ABC 和 CDE 将圆形区域分成上部（阴影）区域和下部区域。上部区域面积与下部区域面积的比值为
stem
Correct Answer: C
Answer (C): Let the diameter of the large circle equal 10. Then the ratio is: $\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ R+S\end{array}\right)-\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ S\end{array}\right)+\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T\end{array}\right)$ $\dfrac{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 2^{2}+\dfrac{1}{2}\pi 3^{2}}{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 3^{2}+\dfrac{1}{2}\pi 2^{2}}=\dfrac{15\pi}{10\pi}=\dfrac{3}{2}\ \text{or}\ 3:2$ $\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T+U\end{array}\right)-\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ T\end{array}\right)+\left(\begin{array}{c}\text{Area of}\\ \text{semicircle}\\ S\end{array}\right)$
答案（C）：设大圆的直径为 10，则比值为： $\left(\begin{array}{c}\text{半圆}\\ R+S\text{ 的面积}\end{array}\right)-\left(\begin{array}{c}\text{半圆}\\ S\text{ 的面积}\end{array}\right)+\left(\begin{array}{c}\text{半圆}\\ T\text{ 的面积}\end{array}\right)$ $\dfrac{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 2^{2}+\dfrac{1}{2}\pi 3^{2}}{\dfrac{1}{2}\pi 5^{2}-\dfrac{1}{2}\pi 3^{2}+\dfrac{1}{2}\pi 2^{2}}=\dfrac{15\pi}{10\pi}=\dfrac{3}{2}\ \text{或}\ 3:2$ $\left(\begin{array}{c}\text{半圆}\\ T+U\text{ 的面积}\end{array}\right)-\left(\begin{array}{c}\text{半圆}\\ T\text{ 的面积}\end{array}\right)+\left(\begin{array}{c}\text{半圆}\\ S\text{ 的面积}\end{array}\right)$
solution
Q25
All of the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the rightmost digit (the units digit) of the product?
将从 2 到 98（包含）的所有偶数相乘，但排除那些末位为 0 的数。乘积的最右边数字（个位数）是多少？
Correct Answer: D
Answer (D): If the numbers 2, 4, 6, and 8 are multiplied, the product is 384, so 4 is the final digit of the product of a set of numbers ending in 2, 4, 6, and 8. Since there are ten such sets of numbers, the final digit of the overall product is the same as the final digit of $4^{10}$. Now, $4^{10}=(4^2)^5=16^5$. Next, consider $6^5$. Since any number of 6’s multiply to give 6 as the final digit, the final digit of the required product is 6.
答案（D）：如果把 2、4、6 和 8 相乘，积为 384，因此以 2、4、6、8 结尾的一组数的乘积末位是 4。由于有十组这样的数，整体乘积的末位与 $4^{10}$ 的末位相同。现在，$4^{10}=(4^2)^5=16^5$。接着考虑 $6^5$。因为任意多个以 6 结尾的数相乘，其末位仍为 6，所以所求乘积的末位是 6。