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AMC8 1992

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AMC8 · 1992

Q1
\frac{10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1}{1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9} =
\frac{10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1}{1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9} =
Correct Answer: B
Answer (B): Group the numerator in pairs from the left, and group the denominator in pairs from the left: $\frac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}$ Numerator: $(10-9)+(8-7)+(6-5)+(4-3)+(2-1)=5\times 1$ Denominator: $(1-2)+(3-4)+(5-6)+(7-8)+9=4\times(-1)+9$ Hence the answer is $\frac{5(1)}{4(-1)+9}=\frac{5}{5}=1$.
将分子和分母重新分组为正项和负项，\frac{(10+8+6+4+2)-(9+7+5+3+1)}{(1+3+5+7+9)-(2+4+6+8)} = \frac{30-25}{25-20} = \frac{5}{5} = 1。答案（B）：将分子从左到右两两分组，将分母也从左到右两两分组： $\frac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}$ 分子：$(10-9)+(8-7)+(6-5)+(4-3)+(2-1)=5\times 1$ 分母：$(1-2)+(3-4)+(5-6)+(7-8)+9=4\times(-1)+9$ 因此结果为 $\frac{5(1)}{4(-1)+9}=\frac{5}{5}=1$。
Q2
Which of the following is not equal to \frac{5}{4}?
下列哪个不等于 \frac{5}{4}？
Correct Answer: D
Answer (D): $1\frac{1}{5}=\frac{6}{5}\neq\frac{5}{4}.$
答案（D）：$1\frac{1}{5}=\frac{6}{5}\neq\frac{5}{4}.$
Q3
What is the largest difference that can be formed by subtracting two numbers chosen from the set \{-16, -4, 0, 2, 4, 12\}
从集合 \{-16, -4, 0, 2, 4, 12\} 中选择两个数相减，能形成的最大的差是多少？
Correct Answer: D
Answer (D): to obtain the largest difference, subtract the smallest number, −16, from the largest number, 12. Thus 12 − (−16) = 28.
答案 (D)：要得到最大差值，从最大数12减去最小数−16。这样12 − (−16) = 28。
solution
Q4
During the softball season, Judy had 35 hits. Among her hits were 1 home run, 1 triple and 5 doubles. The rest of her hits were singles. What percent of her hits were singles?
在垒球赛季中，朱迪有35次安打。其中1次全垒打、1次三垒打和5次二垒打。其余的安打是单打。她单打占安打总数的百分之多少？
Correct Answer: E
Answer (E): Judy had a total of 35 hits, of which \(35 - (1 + 1 + 5) = 28\) were singles. Thus \(\frac{28}{35} = \frac{4}{5}\) or 80% were singles.
答案（E）：Judy 一共有 35 次安打，其中 \(35-(1+1+5)=28\) 次是一垒安打。因此 \(\frac{28}{35}=\frac{4}{5}\)，即 80% 是一垒安打。
Q5
A circle of diameter 1 is removed from a 2£3 rectangle, as shown. Which whole number is closest to the area of the shaded region?
从2×3矩形中移除一个直径为1的圆，如图所示。阴影区域的面积最接近哪个整数？
stem
Correct Answer: E
Answer (E): The area of the circle is between 1/2 and 1. To see this, draw squares around and inside the circle. The area of the large square is 1, the area of the small square is 1/2, and the circle fits between the two squares. The area of the rectangle with the circle removed is therefore between 5 and 5.5, so the whole number closest to this area is 5.
答案（E）：圆的面积介于 $1/2$ 和 $1$ 之间。为说明这一点，在圆的外部和内部分别画一个正方形。大正方形的面积是 $1$，小正方形的面积是 $1/2$，而圆位于这两个正方形之间。因此，把圆从矩形中去掉后剩余部分的面积介于 $5$ 和 $5.5$ 之间，所以最接近该面积的整数是 $5$。
solution
Q6
See the picture:
见图：
Correct Answer: D
Answer (D): $(1 + 3 - 4) + (2 + 5 - 6) = 0 + 1 = 1.$
答案（D）：$(1 + 3 - 4) + (2 + 5 - 6) = 0 + 1 = 1。$
Q7
The digit-sum of 998 is 9 + 9 + 8 = 26. How many 3-digit whole numbers, whose digit-sum is 26, are even?
998 的数位和是 9 + 9 + 8 = 26。有多少个三位整数的数位和是 26 且是偶数？
Correct Answer: A
The only 3-digit whole numbers with a digit-sum of 26 are 899, 989 and 998. Of these, only 998 is even. Thus there is only one such number.
数位和为 26 的唯一三位整数是 899、989 和 998。其中只有 998 是偶数。因此只有一个这样的数。
Q8
A store owner bought 1500 pencils at $0.10 each. If he sells them for $0.25 each, how many of them must he sell to make a profit of exactly $100.00?
一家店主以每支 0.10 美元的价格买了 1500 支铅笔。如果他以每支 0.25 美元的价格出售，要获得恰好 100.00 美元的利润，他必须卖出多少支？
Correct Answer: C
Answer (C): Since he bought 1500 pencils at \$0.10 each, he paid $1500\times\$0.10=\$150$. To make \$100 profit he must take in $\$150+\$100=\$250$. Therefore, selling the pencils for \$0.25 each, he must sell $\$250\div\$0.25=1000$ pencils.
答案（C）：由于他以每支 \$0.10 的价格买了 1500 支铅笔，他一共支付了 $1500\times\$0.10=\$150$。要获得 \$100 的利润，他的总收入必须达到 $\$150+\$100=\$250$。因此，如果以每支 \$0.25 的价格卖出，他必须卖出 $\$250\div\$0.25=1000$ 支铅笔。
Q9
The population of a small town is 480. The graph indicates the number of females and males in the town, but the vertical scale-values are omitted. How many males live in the town?
一个小城镇的人口是 480。图表显示了镇上女性和男性的数量，但纵轴刻度值被省略了。镇上有多少男性？
stem
Correct Answer: B
The ratio of males to the total population is 1 to 3. Thus, there are 1/3 of 480, or 160 males in the town.
男性的比例与总人口之比是 1:3。因此，有 480 的 1/3，即 160 名男性。
Q10
An isosceles right triangle with legs of length 8 is partitioned into 16 congruent triangles as shown. The shaded area is
一个腿长为 8 的等腰直角三角形被分割成 16 个全等的三角形，如图所示。阴影部分的面积是
stem
Correct Answer: B
Answer (B): The area of each of the small shaded triangles is $\frac{1}{2}\times 2\times 2=2$. There are ten of these, so the shaded area is $2\times 10=20$.
答案（B）：每个小的阴影三角形的面积是 $\frac{1}{2}\times 2\times 2=2$。这样的三角形有 10 个，所以阴影部分的面积是 $2\times 10=20$。
Q11
The bar graph shows the results of a survey on color preferences. What percent preferred blue?
柱状图显示了一项关于颜色偏好的调查结果。有百分之多少人喜欢蓝色？
stem
Correct Answer: B
Answer (B): The total frequency for all colors is $50 + 60 + 40 + 60 + 40 = 250$. The frequency for blue is 60. Thus the percent that preferred blue is $60/250$, or 24%.
答案（B）：所有颜色的总频数为 $50 + 60 + 40 + 60 + 40 = 250$。蓝色的频数为 60。因此，偏好蓝色的百分比为 $60/250$，即 24%。
Q12
The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first 30,000 miles the car traveled. For how many miles were each tire used?
一辆车的五个轮胎（四个公路轮胎和一个全尺寸备用胎）在汽车行驶的前30,000英里中进行了轮换，使得每个轮胎的使用里程相同。每个轮胎使用了多少英里？
Correct Answer: C
Answer (C): The total number of miles of wear is 30,000 × 4 = 120,000. Since the wear is shared equally by each of the 5 tires, each tire traveled 120,000÷5 = 24,000 miles. OR Since each of the tires was on for \[ \frac{4}{5} \] of the driving, it follows that each was used \[ \frac{4}{5} \times 30,000 = 24,000 \] miles.
答案 (C)：总磨损里程为30,000 × 4 = 120,000。由5个轮胎均等分担，每轮胎行驶120,000÷5 = 24,000英里。 或者 每个轮胎使用时间占驾驶时间的 \[ \frac{4}{5} \]，因此每个轮胎使用 \[ \frac{4}{5} \times 30,000 = 24,000 \] 英里。
Q13
Five test scores have a mean (average score) of 90, a median (middle score) of 91 and a mode (most frequent score) of 94. The sum of the two lowest test scores is
五个考试成绩的平均分（均值）为90，中位数（中间成绩）为91，众数（最频繁成绩）为94。两个最低考试成绩的和是
Correct Answer: B
Answer (B): If the mean is 90, then the sum of all five scores is 5 × 90 = 450. Since the median of the five scores is 91, at least one score must be 91 and two other scores must be greater than or equal to 91. Since 94 is the mode, there are two scores of 94. The sum of the remaining scores must equal 450 − (94 + 94 + 91) = 171.
答案 (B)：如果平均数为90，则五分总和为5 × 90 = 450。中位数为91，因此至少有一个分数为91，且另外两个分数大于或等于91。94为众数，因此有两个94。剩余分数之和须为450 − (94 + 94 + 91) = 171。
Q14
When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is
向一个三分之一满的容器中加入4加仑后，容器正好是一半满。容器的容量（加仑）是
Correct Answer: D
Answer (D): Since 4 gallons is the difference between being 1/3 full and 1/2 full, it follows that 4 gallons is \[ \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \] of the capacity of the tank. Thus the capacity of the tank must be 24 gallons.
答案 (D)：4加仑是从1/3满到1/2满的差，因此4加仑是罐体容量的 \[ \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]。因此罐体容量须为24加仑。
Q15
What is the 1992nd letter in this sequence?
这个序列中的第1992个字母是什么？
Correct Answer: C
Answer (C): The pattern repeats every 9 letters. Dividing 1992 by 9 yields a remainder of 3. Therefore, the 1992nd letter corresponds to the third letter in the sequence, which is C.
答案 (C)：模式每9个字母重复一次。1992除以9余3。因此，第1992个字母对应序列中的第三个字母，即C。
Q16
Which cylinder has twice the volume of the cylinder shown to the right?
哪个圆柱体的体积是右侧所示圆柱体的两倍？
stem
Correct Answer: B
Answer (B): Cylinder (B) can be obtained by stacking one copy of the given cylinder on top of another. The formula for the volume of a cylinder with radius r and height h is V = πr²h. Use this to show that none of the other cylinders has twice the volume of the given cylinder: Cylinder | Volume Given: | π × 10² × 5 = 500π (A): | π × 20² × 5 = 2000π (C): | π × 5² × 20 = 500π (D): | π × 20² × 10 = 4000π Note. If the radius remains the same and the height is doubled, then the volume will double as in (B). Doubling the radius while the height remains the same will multiply the volume by 4, as in (A).
答案 (B)：圆柱 (B) 可由两个给定圆柱叠加得到。圆柱体积公式为 V = πr²h。用此显示其他圆柱体积不是给定圆柱的两倍： 圆柱 | 体积 给定 | π × 10² × 5 = 500π (A) | π × 20² × 5 = 2000π (C) | π × 5² × 20 = 500π (D) | π × 20² × 10 = 4000π 注：半径不变高度加倍则体积加倍，如(B)。半径加倍高度不变则体积乘4，如(A)。
solution solution
Q17
The sides of a triangle have lengths 6, 5, 10, and s, where s is a whole number. What is the smallest possible value of s?
一个三角形的边长为6、5、10和s，其中s是整数。s的最小可能值为多少？
stem
Correct Answer: B
Answer (B): For any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Thus 6.5 + s must be greater than 10. The smallest such whole number for s is 4.
答案 (B)：任意三角形，两边之和大于第三边。因此6.5 + s > 10。s的最小整数为4。
Q18
On a trip, a car traveled 80 miles in an hour and a half, then was stopped in traffic for 30 minutes, then traveled 100 miles during the next 2 hours. What was the car's average speed in miles per hour for the 4-hour trip?
在一趟旅行中，一辆汽车在1.5小时内行驶了80英里，然后在交通堵塞中停了30分钟，然后在接下来的2小时内行驶了100英里。这4小时旅行的汽车平均速度是多少英里/小时？
Correct Answer: A
Answer (A): During the 4 hours, the car traveled a total of 80 + 0 + 100 = 180 miles for an average speed of 180/4 = 45 miles per hour.
答案 (A)：4小时内，汽车总行驶80 + 0 + 100 = 180英里，平均速度180/4 = 45英里/小时。
Q19
The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?
州际高速公路上第5个出口和第26个出口之间的距离是118英里。如果任意两个出口相距至少5英里，那么在第5个和第26个出口之间，两个连续出口之间可能的最大英里数是多少？
Correct Answer: C
Answer (C): There are 21 segments between the 5th and 26th exits. Using the minimum length of 5 miles, 20 segments would yield 20 × 5 = 100 miles. This leaves 118 − 100 = 18 miles for the other segment. OR There are 21 segments between the 5th and 26th exits. If each segment were its minimal 5 miles length, then the total distance between the 5th and 26th exits would be 105 miles. Since 118 − 105 = 13, all additional miles could occur between one pair of consecutive exits. Such a pair would be 5 + 13 = 18 miles apart.
答案 (C)：第5和第26出口间有21段路段。用最小长度5英里，20段为20 × 5 = 100英里。剩余118 − 100 = 18英里给另一段。 或者 第5和第26出口间有21段。若每段均为最小5英里，则总距离105英里。118 − 105 = 13，因此额外13英里可集中在相邻一对出口间，该对相距5 + 13 = 18英里。
Q20
Which pattern of identical squares could NOT be folded along the lines shown to form a cube?
哪种相同正方形的图案无法沿所示线折叠成一个立方体？
Correct Answer: D
Answer (D): Any attempt to fold the squares would result in square 1 being superimposed on square 2. Have students cut and fold the other four patterns into cubes.
答案 (D)：折叠方块时，方块1总会叠在方块2上。让学生剪切并折叠其他四个图案成立方体。
solution
Q21
Northside's Drum and Bugle Corps raised money for a trip. The drummers and bugle players kept separate sales records. According to the double bar graph, in what month did one group's sale exceed the other's by the greatest percent?
Northside的鼓乐队为旅行筹集资金。鼓手和小号手保持单独的销售记录。根据双柱状图，在哪个月一个小组的销售额超过另一个小组的百分比最大？
stem
Correct Answer: B
Answer (B): Compute the ratios for each month: \[ \begin{array}{lccccc} \text{Month} & \text{Drums} & \text{Bugles} & \text{Diff.} & \text{Diff.:Lower} & \% \text{ exc.} \\ \text{Jan:} & 7 & 9 & 2 & 2\!:\!9 & 29\% \\ \text{Feb:} & 5 & 3 & 2 & 2\!:\!3 & 67\% \\ \text{Mar:} & 9 & 6 & 3 & 3\!:\!6 & 50\% \\ \text{Apr:} & 9 & 12 & 3 & 3\!:\!9 & 33\% \\ \text{May:} & 8 & 10 & 2 & 2\!:\!8 & 25\% \\ \end{array} \] Thus the percent is greatest in February. Note. Students can estimate the required ratio by visually comparing the difference between the columns to the shorter column.
答案（B）：计算每个月的比率： \[ \begin{array}{lccccc} \text{月份} & \text{鼓} & \text{号} & \text{差} & \text{差:较小者} & \text{超出百分比} \\ \text{一月:} & 7 & 9 & 2 & 2\!:\!9 & 29\% \\ \text{二月:} & 5 & 3 & 2 & 2\!:\!3 & 67\% \\ \text{三月:} & 9 & 6 & 3 & 3\!:\!6 & 50\% \\ \text{四月:} & 9 & 12 & 3 & 3\!:\!9 & 33\% \\ \text{五月:} & 8 & 10 & 2 & 2\!:\!8 & 25\% \\ \end{array} \] 因此，百分比在二月最大。 注：学生可以通过目测比较两列之间的差与较短一列的大小，来估计所需的比率。
Q22
Eight 1 × 1 square tiles are arranged as shown so their outside edges form a polygon with a perimeter of 14 units. Two additional tiles of the same size are added to the figure so that at least one side of each tile is shared with a side of one of the squares in the original figure. Which of the following could be the perimeter of the new figure?
八个1×1正方形瓷砖按所示排列，它们的外部边缘形成一个周长为14单位的的多边形。添加两个相同大小的瓷砖，使得每个瓷砖至少有一边与原图中一个正方形的边共享。新图形的周长可能是以下哪一个？
stem
Correct Answer: C
Answer (C): When a new tile is added to the original figure, it may have one or two sides in common with the given tiles, as shown. When a tile shares one side, the original perimeter is increased by 2. When a tile shares two sides, there is no change in the perimeter. By adding two tiles, the only possible changes to the perimeter are increases of 0, 2 or 4. Hence, the possible value of the perimeter are 14, 16, or 18.
答案 (C)：向原图添加新瓦片时，新瓦片可能与现有瓦片共一边或两边。如图所示。共一边时，周长增加2；共两边时，周长不变。添加两块瓦片，周长仅可能增加0、2或4。因此可能周长为14、16或18。
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Q23
If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is
抛掷两个骰子，两个骰子顶面数字之积大于10的概率是
Correct Answer: B
Answer (B): Make a table and fill the products greater than 10. \[ \begin{array}{c|cccccc} \times & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1: & 1 & 2 & 3 & 4 & 5 & 6 \\ 2: & 2 & 4 & 6 & 8 & 10 & 12 \\ 3: & 3 & 6 & 9 & 12 & 15 & 18 \\ 4: & 4 & 8 & 12 & 16 & 20 & 24 \\ 5: & 5 & 10 & 15 & 20 & 25 & 30 \\ 6: & 6 & 12 & 18 & 24 & 30 & 36 \\ \end{array} \] Since there are 17 such products out of a possible 36 products, the probability is \[ \frac{17}{36} \].
答案 (B)：制表并填入大于10的乘积。 \[ \begin{array}{c|cccccc} \times & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1: & 1 & 2 & 3 & 4 & 5 & 6 \\ 2: & 2 & 4 & 6 & 8 & 10 & 12 \\ 3: & 3 & 6 & 9 & 12 & 15 & 18 \\ 4: & 4 & 8 & 12 & 16 & 20 & 24 \\ 5: & 5 & 10 & 15 & 20 & 25 & 30 \\ 6: & 6 & 12 & 18 & 24 & 30 & 36 \\ \end{array} \] 36个乘积中有17个大于10，因此概率为 \[ \frac{17}{36} \]。
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Q24
Four circles of radius 3 are arranged as shown. Their centers are the vertices of a square. The area of the shaded region is closest to
四个半径为3的圆按所示排列。它们的圆心是正方形的顶点。阴影区域的面积最接近
stem
Correct Answer: A
Answer (A): The four quarter-circles that lie inside the square have a total area equal to the area of one of the circles, $9\pi$. The length of a side of the square is equal to two radii, $6$, and thus the square has area $36$. The difference is $36-9\pi<36-9(3)=9$, so it is closest to $7.7$. (the area, to one decimal place, is $7.7$.)
答案（A）：正方形内的四个四分之一圆的总面积等于其中一个圆的面积，即 $9\pi$。正方形的边长等于两个半径之和，为 $6$，因此正方形的面积是 $36$。两者的差为 $36-9\pi<36-9(3)=9$，所以最接近 $7.7$。（该面积保留一位小数为 $7.7$。）
Q25
One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, etc. After how many pourings does exactly one tenth of the original water remain?
从一个满的容器中倒出一半的水。然后倒出剩余水的三分之一。继续这个过程：第三次倒出剩余水的四分之一，第四次倒出剩余水的五分之一，等等。经过多少次倒水后，剩余的正好是原来水的十分之一？
Correct Answer: D
Answer (D): After the first pouring, $\frac{1}{2}$ remains. After the second pouring $\frac{1}{2}\times\frac{2}{3}$ remains. After the third pouring $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}$ remains. How many pouring until $\frac{1}{10}$ remains? $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\times\frac{9}{10}=\frac{1}{10}$ indicates 9 pourings.
答案（D）：第一次倒出后，剩下 $\frac{1}{2}$。第二次倒出后，剩下 $\frac{1}{2}\times\frac{2}{3}$。第三次倒出后，剩下 $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}$。要倒出多少次才会剩下 $\frac{1}{10}$？ $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}\times\frac{8}{9}\times\frac{9}{10}=\frac{1}{10}$ 这表示需要倒出 9 次。
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