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AMC8 1993

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AMC8 · 1993

Q1
Which pair of numbers does NOT have a product equal to 36?
以下哪一对数的乘积不等于36？
Correct Answer: C
\frac{1}{2} \times (-72) = -36.
\frac{1}{2} \times (-72) = -36。
Q2
When the fraction 49/84 is expressed in simplest form, then the sum of the numerator and the denominator will be
将分数49/84化成最简形式后，分子和分母的和是
Correct Answer: C
$\frac{49}{84}=\frac{7\times7}{7\times12}=\frac{7}{12}.$ The sum of the numerator and the denominator is $7+12=19$.
$\frac{49}{84}=\frac{7\times7}{7\times12}=\frac{7}{12}$。 分子与分母之和为 $7+12=19$。
Q3
Which of the following numbers has the largest prime factor?
以下哪个数的最大质因数最大？
Correct Answer: B
39 = 3×13, 51 = 3×17, 77 = 7×11, 91 = 7×13, 121 = 11². Largest prime factor is 17.
39 = 3×13, 51 = 3×17, 77 = 7×11, 91 = 7×13, 121 = 11²。最大质因数是17。
Q4
1000 × 1993 × 0.1993 × 10 =
1000 × 1993 × 0.1993 × 10 =
Correct Answer: E
Answer (E): $1000 \times 1993 \times 0.1993 \times 10 = ((1000 \times 10) \times 0.1993) \times 1993$ $= (10{,}000 \times 0.1993) \times 1993$ $= 1993 \times 1993 = (1993)^2.$
答案（E）： $1000 \times 1993 \times 0.1993 \times 10 = ((1000 \times 10) \times 0.1993) \times 1993$ $= (10{,}000 \times 0.1993) \times 1993$ $= 1993 \times 1993 = (1993)^2.$
Q5
Which one of the following bar graphs could represent the data from the circle graph?
以下哪个条形图可能代表圆图中的数据？
stem
Correct Answer: C
The unshaded area is half the total, and each of the shaded areas is one fourth of the total. This is represented in bar graph (C).
未着色区域占总面积的一半，着色区域各占总面积的四分之一。这在条形图(C)中表现出来。
Q6
A can of soup can feed 3 adults or 5 children. If there are 5 cans of soup and 15 children are fed, then how many adults would the remaining soup feed?
一罐汤可以喂饱 3 个成人或 5 个孩子。如果有 5 罐汤并喂饱了 15 个孩子，那么剩下的汤可以喂饱多少成人？
Correct Answer: B
Three cans of soup are needed for 15 children, so the remaining 2 cans of soup will feed $2\times3=6$ adults.
15 个孩子需要 3 罐汤，因此 1 罐汤相当于供 5 个孩子；剩余 2 罐汤可供 $2\times3=6$ 个成年人。
Q7
3³ + 3³ + 3³ =
$3^3 + 3^3 + 3^3 =$
Correct Answer: A
\[3^3+3^3+3^3=3(3^3)=3(3\times3\times3)=3\times3\times3\times3=3^4.\] OR \[3^3+3^3+3^3=27+27+27=81=9\times9=3\times3\times3\times3=3^4.\]
\[3^3+3^3+3^3=3(3^3)=3(3\times3\times3)=3\times3\times3\times3=3^4.\] 或 \[3^3+3^3+3^3=27+27+27=81=9\times9=3\times3\times3\times3=3^4.\]
Q8
To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains 60 pills, then the supply of medicine will last approximately
为了控制血压，吉尔的祖母每隔一天吃半片药。如果一盒药含有 60 片，那么这盒药大约能用
Correct Answer: D
Since she takes one half of a pill every other day, one pill will last 4 days. Hence 60 pills will last $60\times4=240$ days, or about 8 months.
她每隔一天吃半片药，因此 1 片药可吃 4 天。于是 60 片可吃 $60\times4=240$ 天，约 8 个月。
Q9
Consider the operation * defined by the following table: [table: 1*1=2,1*2=4,1*3=1,1*4=3; 2*?= table incomplete but from text; example 3*2=1]. Then (2 * 4) * (1 * 3) =
考虑由下表定义的操作 *：[表格：1*1=2,1*2=4,1*3=1,1*4=3; 2*?=表格不完整但从文本；例如3*2=1]。那么 (2 * 4) * (1 * 3) =
stem
Correct Answer: D
Substituting the values from the table yields \[(2\ast4)\ast(1\ast3)=3\ast3=4.\] Query. Would evaluating the products \[((2\ast4)\ast1)\ast3,\ (2\ast(4\ast1))\ast3,\ 2\ast((4\ast1)\ast3)\text{ and }2\ast(4\ast(1\ast3)),\] yield the same result?
把表中的取值代入： \[(2\ast4)\ast(1\ast3)=3\ast3=4.\] 问题：计算 \[((2\ast4)\ast1)\ast3,\ (2\ast(4\ast1))\ast3,\ 2\ast((4\ast1)\ast3)\text{ 与 }2\ast(4\ast(1\ast3))\] 会得到相同结果吗？
Q10
This line graph represents the price of a trading card during the first 6 months of 1993.
这条线图表示了 1993 年前 6 个月一张交易卡的价格。
stem
Correct Answer: B
Answer (B): The graph shows the following changes in the price of the card: \[ \begin{array}{lll} \textit{Jan}: & \$2.50 \text{ to } \$2.00 & \textbf{drop\ of\ } \$0.50 \\ \textit{Feb}: & \$2.00 \text{ to } \$4.00 & \textbf{rise\ of\ } \$2.00 \\ \textit{Mar}: & \$4.00 \text{ to } \$1.50 & \textbf{drop\ of\ } \$2.50 \\ \textit{Apr}: & \$1.50 \text{ to } \$4.50 & \textbf{rise\ of\ } \$3.00 \\ \textit{May}: & \$4.50 \text{ to } \$3.00 & \textbf{drop\ of\ } \$1.50 \\ \textit{Jun}: & \$3.00 \text{ to } \$1.00 & \textbf{drop\ of\ } \$2.00 \\ \end{array} \] The greatest drop occurred during March.
答案（B）：该图表显示卡片价格的以下变化： \[ \begin{array}{lll} \textit{1月}: & \$2.50 \text{ 到 } \$2.00 & \textbf{下降\ } \$0.50 \\ \textit{2月}: & \$2.00 \text{ 到 } \$4.00 & \textbf{上升\ } \$2.00 \\ \textit{3月}: & \$4.00 \text{ 到 } \$1.50 & \textbf{下降\ } \$2.50 \\ \textit{4月}: & \$1.50 \text{ 到 } \$4.50 & \textbf{上升\ } \$3.00 \\ \textit{5月}: & \$4.50 \text{ 到 } \$3.00 & \textbf{下降\ } \$1.50 \\ \textit{6月}: & \$3.00 \text{ 到 } \$1.00 & \textbf{下降\ } \$2.00 \\ \end{array} \] 最大跌幅发生在3月。
Q11
Consider this histogram of the scores for 81 students taking a test: The median is in the interval labeled
考虑这81名学生考试成绩的直方图：中位数位于标记的区间
stem
Correct Answer: C
Since 81 took the test, the median (middle) score is 41st. The test interval containing the 41st score is labeled 70.
共有 81 人参加测试，中位数（居中的成绩）是第 41 个。包含第 41 个成绩的区间标为 70。
Q12
See the picture:
看图片：
stem
Correct Answer: E
Answer (E): The six permutations of $+$, $-$ and $\times$ yield these results: $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$ $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$ $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$ $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$ $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$ $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19.$ The only result listed is 19.
答案（E）：$+$、$-$ 和 $\times$ 的六种排列会得到以下结果： $5 \times 4 + 6 - 3 = 20 + 6 - 3 = 23$ $5 \times 4 - 6 + 3 = 20 - 6 + 3 = 17$ $5 + 4 \times 6 - 3 = 5 + 24 - 3 = 26$ $5 - 4 \times 6 + 3 = 5 - 24 + 3 = -16$ $5 + 4 - 6 \times 3 = 5 + 4 - 18 = -9$ $5 - 4 + 6 \times 3 = 5 - 4 + 18 = 19.$ 所列出的结果中只有 19。
Q13
The word "HELP" in block letters is painted in black with strokes 1 unit wide on a 5 by 15 rectangular white sign with dimensions as shown. The area of the white portion of the sign, in square units, is
单词“HELP”用1单位宽的黑线条在5×15的白色矩形标志上涂成块字母，尺寸如图所示。标志白色部分的面积（平方单位）是
stem
Correct Answer: D
Compute the area of the black letters and subtract it from $5\times15=75$, the total area of the sign: $H$: $2(1\times5)+1\times1=11$. $E$: $1\times5+3(2\times1)=11$. $L$: $1\times5+1\times2=7$. $P$: $1\times5+2(1\times1)+1\times3=10$. The area of the white portion is $75-(11+11+7+10)=36$.
先求黑色字母面积，再用总面积 $5\times15=75$ 相减： $H$：$2(1\times5)+1\times1=11$。 $E$：$1\times5+3(2\times1)=11$。 $L$：$1\times5+1\times2=7$。 $P$：$1\times5+2(1\times1)+1\times3=10$。 白色部分面积为 $75-(11+11+7+10)=36$。
solution
Q14
The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers 1, 2, 3. Then A + B =
表中的九个方格要填写，使得每行和每列都包含数字1、2、3。然后 A + B =
stem
Correct Answer: C
Only 3’s can complete a $2\times2$ square whose diagonal is given. If two entries in a row or column are known, the third is determined. Use this to complete the table; thus $A=1$ and $B=3$. Therefore, $A+B=4$.
给定对角线的 $2\times2$ 方格只能用 3 来补全。若某一行或某一列已知两个数，则第三个数被确定。按此补全表格得到 $A=1$，$B=3$，因此 $A+B=4$。
solution
Q15
The arithmetic mean (average) of four numbers is 85. If the largest of these numbers is 97, then the mean of the remaining three numbers is
四个数的算术平均数（平均值）是85。其中最大的数是97，则其余三个数的平均数是
Correct Answer: A
Sum 340, remaining 243, mean 81.
总和340，其余243，平均81。
Q16
\frac{1}{1 + \frac{1}{2 + \frac{1}{3}}} =
\frac{1}{1 + \frac{1}{2 + \frac{1}{3}}} =
Correct Answer: C
Answer (C): Using the common denominators and simplifying yield $\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3}}}=\dfrac{1}{1+\dfrac{1}{\dfrac{7}{3}}}=\dfrac{1}{1+\dfrac{3}{7}}=\dfrac{1}{\dfrac{10}{7}}=\dfrac{7}{10}.$
答案（C）：使用公分母并化简得到 $\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{3}}}=\dfrac{1}{1+\dfrac{1}{\dfrac{7}{3}}}=\dfrac{1}{1+\dfrac{3}{7}}=\dfrac{1}{\dfrac{10}{7}}=\dfrac{7}{10}.$
Q17
Square corners, 5 units on a side, are removed from a 20 unit by 30 unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is
从一张20单位×30单位的矩形硬纸板上切掉四个边长为5单位的正方形角，然后将边折起形成一个开口盒子。盒子内部表面积是（平方单位）
stem
Correct Answer: B
The interior (or exterior) has the same surface area as one side of the sheet of cardboard after the corners have been removed. The area of the sheet is $30\times20=600$ and the area of each square corner removed is $5\times5=25$, so the answer is $600-(4\times25)=500$.
盒子的内表面（或外表面）面积等于剪去四个角后纸板一面的面积。纸板面积为 $30\times20=600$，每个剪去的正方形角面积为 $5\times5=25$，所以面积为 $600-4\times25=500$。
solution
Q18
The rectangle shown has length AC = 32, width AE = 20, and B and F are midpoints of AC and AE, respectively. The area of the quadrilateral ABDF is
如图所示的矩形，长AC=32，宽AE=20，B和F分别是AC和AE的中点。四边形ABDF的面积是
stem
Correct Answer: A
Answer (A): Rectangle $ACDE$ has area $32 \times 200 = 640$. Triangle $BCD$ has area $(16 \times 20)/2 = 160$, and triangle $DEF$ has area $(10 \times 32)/2 = 160$. The remaining area, $ABDF$, is $640 - (160 + 160) = 320$.
答案（A）：矩形 $ACDE$ 的面积为 $32 \times 200 = 640$。三角形 $BCD$ 的面积为 $(16 \times 20)/2 = 160$，三角形 $DEF$ 的面积为 $(10 \times 32)/2 = 160$。剩余部分 $ABDF$ 的面积为 $640 - (160 + 160) = 320$。
solution
Q19
(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193) =
$(1901 + 1902 + 1903 + \dots + 1993) - (101 + 102 + 103 + \dots + 193) =
Correct Answer: A
English (LaTeX-ready): Answer (A): Each number in the first set of numbers is 1800 more than the corresponding number in the second set: $\begin{array}{cccc} 1901, & 1902, & 1903, & \ldots,\ 1993 \\ -101, & -102, & -103, & \ldots,\ -193 \\ 1800, & 1800, & 1800, & \ldots,\ 1800 \end{array}$ Thus the sum of the first set of numbers $93 \times 1800 = 167{,}400$ more than the sum of the second set.
中文（翻译）： 答案（A）：第一组中的每个数都比第二组中对应的数大 $1800$： $\begin{array}{cccc} 1901, & 1902, & 1903, & \ldots,\ 1993 \\ -101, & -102, & -103, & \ldots,\ -193 \\ 1800, & 1800, & 1800, & \ldots,\ 1800 \end{array}$ 因此，第一组数的和比第二组数的和多 $93 \times 1800 = 167{,}400$。
Q20
When $10^{93} - 93$ is expressed as a single whole number, the sum of the digits is
当$10^{93} - 93$表示为单个整数时，其各位数字之和是
Correct Answer: D
Since $10^{93}=1\underbrace{00\cdots00}_{93\text{ zeros}}$, we have \[10^{93}-93=\underbrace{99\cdots9}_{91\text{ nines}}07.\] Thus the sum of the digits is $(91\times9)+7=826$.
因为 $10^{93}=1\underbrace{00\cdots00}_{93\text{ 个零}}$，所以 \[10^{93}-93=\underbrace{99\cdots9}_{91\text{ 个 9}}07.\] 因此各位数字之和为 $(91\times9)+7=826$。
Q21
If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by
如果一个长方形的长增加20%，宽增加50%，那么面积增加了
Correct Answer: D
Answer (D): When a problem indicates a general result, then it must hold for any specific case. Therefore, suppose the original rectangle is 10 by 10 with area 100. The new length is 10 + 2 = 12 and the new width is 10 + 5 = 15. Hence the new area is 12 × 15 = 180 for an increase of 80%.
答案（D）：当题目给出一个一般性结论时，它必须对任何特定情况都成立。因此，设原来的矩形为 10×10，面积为 100。新长度为 10 + 2 = 12，新宽度为 10 + 5 = 15。因此新面积为 12×15 = 180，增加了 80%。
Q22
Pat Peano has plenty of 0’s, 1’s, 3’s, 4’s, 5’s, 6’s, 7’s, 8’s and 9’s, but he has only twenty-two 2’s. How far can he number the pages of his scrapbook with these digits?
Pat Peano 有大量的 0、1、3、4、5、6、7、8 和 9，但他只有二十二个 2。他能用这些数字为他的剪贴簿编号到多少页？
Correct Answer: D
Answer (D): Ten 2’s are needed in the unit’s place in counting to 100 and ten more 2’s are used in the ten’s place. With the remaining two 2’s he can number 102 and 112 and continue all the way to 119 before needing another 2.
答案（D）：在从 1 数到 100 的过程中，个位需要用到 10 个“2”，十位还需要再用 10 个“2”。剩下的 2 个“2”可以用来编号 102 和 112，并且还能一直继续编号到 119，之后才需要再用一个“2”。
Q23
Five runners, P, Q, R, S, T, have a race, and P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have finished third in the race?
五个跑步者 P、Q、R、S、T 参加比赛，P 打败 Q，P 打败 R，Q 打败 S，T 在 P 之后 Q 之前结束。谁不可能获得第三名？
Correct Answer: C
Answer (C): Since $P$, $T$ and $Q$ must finish in front of $S$, $S$ cannot be third. Since $P$ is the winner, $P$ cannot be third. Thus the only possible orders are $PRTQS$, $PTRQS$, $PTQRS$ and $PTQSR$, which show that anyone except $P$ and $S$ could finish third.
答案（C）：由于 $P$、$T$ 和 $Q$ 必须排在 $S$ 前面，$S$ 不可能获得第三名。由于 $P$ 是获胜者，$P$ 不可能获得第三名。因此，唯一可能的顺序是 $PRTQS$、$PTRQS$、$PTQRS$ 和 $PTQSR$，这表明除了 $P$ 和 $S$ 之外，其他人都可能获得第三名。
Q24
What number is directly above 142 in this array of numbers? [triangular array 1 to ...]
在这个数字阵列中，142 正上方的数字是什么？[三角阵列 1 到 ...]
stem
Correct Answer: C
Rows end squares: line with 142 ends 144, above ends 121, so above 142 is 120.
行以平方数结束：包含 142 的行以 144 结束，其上一行以 121 结束，因此 142 上方是 120。
solution
Q25
A checkerboard consists of one-inch squares. A square card, 1.5 inches on a side, is placed on the board so that it covers part or all of the area of each of n squares. The maximum possible value of n is
一个棋盘由一英寸见方的方格组成。一个边长 1.5 英寸的方形卡片放在棋盘上，使得它覆盖了 n 个方格的部分或全部区域。n 的最大可能值是
Correct Answer: E
Answer (E): Using the Pythagorean Theorem, the length of the diagonal of the card is $\sqrt{(1.5)^2 + (1.5)^2} = \sqrt{4.5} \approx 2.1$. This is longer than 2, the length of two adjacent squares. The figure shows 12 squares being touched.
答案（E）：使用勾股定理，卡片对角线的长度为 $\sqrt{(1.5)^2 + (1.5)^2} = \sqrt{4.5} \approx 2.1$。这比 2（两个相邻方格的边长之和）更长。图中显示有 12 个方格被触及。
solution