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AMC8 1998

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AMC8 · 1998

Q1
For $x = 7$, which of the following is smallest?
当 $x = 7$ 时，下列哪个最小？
Correct Answer: B
Only $\frac{6}{7}$(A) and $\frac{6}{8}$(B) are less than 1. For fractions: If the numerators are equal, the smaller fraction will have a larger denominator. Therefore, $\frac{6}{8}$ is smaller than $\frac{6}{7}$(A).
只有 $\frac{6}{7}$(A) 和 $\frac{6}{8}$(B) 小于 1。对于分数：如果分子相同，分母越大的分数越小。因此，$\frac{6}{8}$ 比 $\frac{6}{7}$(A) 小。
Q2
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stem
Correct Answer: E
$3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2$.
$3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2$。
Q3
What is $\frac{3}{8} + \frac{7}{8} \div \frac{4}{5}$?
计算 $\frac{3}{8} + \frac{7}{8} \div \frac{4}{5}$ 等于多少？
stem
Correct Answer: B
$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$. Therefore, $\frac{5}{4} \div \frac{4}{5} = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16}$.
$\frac{3}{8} + \frac{7}{8} = \frac{10}{8} = \frac{5}{4}$。因此，$\frac{5}{4} \div \frac{4}{5} = \frac{5}{4} \cdot \frac{5}{4} = \frac{25}{16}$。
Q4
How many triangles are in this figure? (Some triangles may overlap other triangles.)
这幅图中有多少个三角形？（有些三角形可能与其他三角形重叠。）
stem
Correct Answer: E
The triangles are $ABE, ACE, BCD, BCE, BDE$.
三角形有 $ABE, ACE, BCD, BCE, BDE$。
solution
Q5
Which of the following numbers is largest?
下列哪个数最大？
Correct Answer: B
Answer (B): (A) $9.12344 = 9.12344000\ldots,$ (B) $9.12\overline{34} = 9.12344444\ldots,$ (C) $9.\overline{1234} = 9.12343434\ldots,$ (D) $9.1\overline{234} = 9.12342342\ldots,$ (E) $9.\overline{1234} = 9.12341234\ldots.$
答案（B）： (A) $9.12344 = 9.12344000\ldots,$ (B) $9.12\overline{34} = 9.12344444\ldots,$ (C) $9.\overline{1234} = 9.12343434\ldots,$ (D) $9.1\overline{234} = 9.12342342\ldots,$ (E) $9.\overline{1234} = 9.12341234\ldots.$
Q6
Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
点在水平和垂直方向上相距一单位。多项式围成的平方单位数是
stem
Correct Answer: B
Slide triangle $A$ down to fill in triangle $B$. The resulting $2 \times 3$ rectangle has area 6.
将三角形 $A$ 向下滑动填充三角形 $B$。得到的 $2 \times 3$ 矩形面积为 6。
solution
Q7
$100 \times 19.98 \times 1.998 \times 1000 =$
$100 \times 19.98 \times 1.998 \times 1000 = $
Correct Answer: D
Use the associative property to group as follows: $(100 \times 19.98) \times (1.998 \times 1000) = 1998 \times 1998 = (1998)^2$.
使用结合律分组如下：$(100 \times 19.98) \times (1.998 \times 1000) = 1998 \times 1998 = (1998)^2$。
Q8
A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days?
一个儿童戏水池中有 200 加仑水。如果水每天蒸发 0.5 加仑，且没有其他水加入或移除，30 天后池中水有多少加仑？
Correct Answer: C
$200 - 0.5(30) = 200 - 15 = 185$ gallons.
$200 - 0.5(30) = 200 - 15 = 185$ 加仑。
Q9
For a sale, a store owner reduces the price of a $10 scarf by 20%. Later the price is lowered again, this time by one-half the reduced price. The price is now
店主为促销将一条 10 美元的围巾降价 20%。后来价格再次降低，这次是降低已降价的一半。现在的价格是
Correct Answer: C
First: $10 - 0.2(10) = 8$. Second: $8 - 0.5(8) = 4$. Or 40% of original: $0.4(10)=4$.
第一次：$10 - 0.2(10) = 8$。第二次：$8 - 0.5(8) = 4$。或原价的 40%：$0.4(10)=4$。
Q10
Each of the letters W, X, Y, and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If $\frac{W}{X} - \frac{Y}{Z} = 1$, then the sum of W and Y is
字母 W、X、Y 和 Z 各代表集合 {1, 2, 3, 4} 中的不同整数，不一定按此顺序。如果 $\frac{W}{X} - \frac{Y}{Z} = 1$，则 W 和 Y 的和是
Correct Answer: E
The only arrangement that produces a whole number is $\frac{3}{1} - \frac{4}{2} = 1$. Therefore, $W + Y = 3 + 4 = 7$.
唯一产生整数的排列是 $\frac{3}{1} - \frac{4}{2} = 1$。因此，$W + Y = 3 + 4 = 7$。
Q11
Harry has 3 sisters and 5 brothers. His sister Harriet has S sisters and B brothers. What is the product of S and B?
Harry 有 3 个姐妹和 5 个兄弟。他的姐姐 Harriet 有 S 个姐妹和 B 个兄弟。S 和 B 的乘积是多少？
Correct Answer: C
Harry has 3 sisters and 5 brothers, so there are 3 girls and 6 boys total. Harriet has 2 sisters and 6 brothers. $2 \times 6 = 12$.
Harry 有 3 个姐妹和 5 个兄弟，所以总共有 3 个女孩和 6 个男孩。Harriet 有 2 个姐妹和 6 个兄弟。$2 \times 6 = 12$。
Q12
$2(1 - \frac{1}{2}) + 3(1 - \frac{1}{3}) + 4(1 - \frac{1}{4}) + \dots + 10(1 - \frac{1}{10}) =$
$2(1 - \frac{1}{2}) + 3(1 - \frac{1}{3}) + 4(1 - \frac{1}{4}) + \dots + 10(1 - \frac{1}{10}) =$
Correct Answer: A
Answer (A): $2\left(1-\frac{1}{2}\right)+3\left(1-\frac{1}{3}\right)+4\left(1-\frac{1}{4}\right)+\cdots+10\left(1-\frac{1}{10}\right)=$ $2\left(\frac{1}{2}\right)+3\left(\frac{2}{3}\right)+4\left(\frac{3}{4}\right)+\cdots+10\left(\frac{9}{10}\right)=$ $1+2+3+\cdots+9=45.$
答案（A）： $2\left(1-\frac{1}{2}\right)+3\left(1-\frac{1}{3}\right)+4\left(1-\frac{1}{4}\right)+\cdots+10\left(1-\frac{1}{10}\right)=$ $2\left(\frac{1}{2}\right)+3\left(\frac{2}{3}\right)+4\left(\frac{3}{4}\right)+\cdots+10\left(\frac{9}{10}\right)=$ $1+2+3+\cdots+9=45.$
Q13
What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale.)
阴影正方形的面积与大正方形的面积之比是多少？（图形按比例绘制。）
stem
Correct Answer: C
Divide the square into 16 smaller squares. The shaded square is 2 small squares. Ratio $2/16 = 1/8$.
将正方形分成 16 个小正方形。阴影正方形是 2 个小正方形。比例 $2/16 = 1/8$。
Q14
At Annville Junior High School, 30% of the students in the Math Club are in the Science Club, and 80% of the students in the Science Club are in the Math Club. There are 15 students in the Science Club. How many students are in the Math Club?
在 Annville 初中，数学俱乐部 30% 的学生在科学俱乐部，科学俱乐部 80% 的学生在数学俱乐部。科学俱乐部有 15 名学生。数学俱乐部有多少学生？
Correct Answer: E
Answer (E): Since 80% of the Science Club members are also in the Math Club, there are $0.8(15)=12$ students common to both clubs. Because 30% of the students in the Math Club are also in the Science Club, there are $12\div 0.3=40$ students in the Math Club.
答案（E）：由于科学俱乐部成员中有 80% 同时参加数学俱乐部，所以两者共有的学生人数为 $0.8(15)=12$。又因为数学俱乐部学生中有 30% 也参加科学俱乐部，所以数学俱乐部的学生人数为 $12\div 0.3=40$。
solution
Q15
Estimate the population of Nisos in the year 2050.
估计 2050 年 Nisos 的总人口。
Correct Answer: D
Answer (D): In the year 1998 the population is 200. In 2023 the population will be $200(3)=600$. In 2048 the population will be $600(3)=1800$. In 2050 the population will be about 2000.
答案（D）： 在 1998 年，人口为 200。 在 2023 年，人口将为 $200(3)=600$。 在 2048 年，人口将为 $600(3)=1800$。 在 2050 年，人口将约为 2000。
Q16
Estimate the year in which the population of Nisos will be approximately 6,000.
估计 Nisos 人口大约达到 6000 年的年份。
Correct Answer: B
Answer (B): \[ \begin{array}{c c} \text{Year} & \text{Population}\\ 1998 & 200\\ 2023 & 600\\ 2048 & 1{,}800\\ 2073 & 5{,}400\\ 2098 & 16{,}200 \end{array} \] Of the choices available the year 2075 is the best estimate.
答案（B）： \[ \begin{array}{c c} \text{年份} & \text{人口}\\ 1998 & 200\\ 2023 & 600\\ 2048 & 1{,}800\\ 2073 & 5{,}400\\ 2098 & 16{,}200 \end{array} \] 在可选项中，2075 年是最佳估计。
Q17
In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
大约从 1998 年起多少年后，Nisos 的人口将达到 Queen Irene 宣称岛屿所能支持的数量？
Correct Answer: C
Answer (C): \[ \begin{array}{r|r|r} \text{Year} & \text{Population} & \text{Area Needed} \\ \hline 1998 & 200 & 300 \\ 2023 & 600 & 900 \\ 2048 & 1{,}800 & 2{,}700 \\ 2073 & 5{,}400 & 8{,}100 \\ 2098 & 16{,}200 & 24{,}300 \\ \end{array} \] The Isles can support $24{,}900 \div 1.5 = 16{,}600$ people. The chart shows that this will happen about the year 2098, or in about 100 years.
答案（C）： \[ \begin{array}{r|r|r} \text{年份} & \text{人口} & \text{所需面积} \\ \hline 1998 & 200 & 300 \\ 2023 & 600 & 900 \\ 2048 & 1{,}800 & 2{,}700 \\ 2073 & 5{,}400 & 8{,}100 \\ 2098 & 16{,}200 & 24{,}300 \\ \end{array} \] 这些岛屿可容纳的人口为 $24{,}900 \div 1.5 = 16{,}600$ 人。图表显示，这将发生在大约 2098 年，也就是大约 100 年后。
Q18
As indicated by the diagram at the right, a rectangular piece of paper is folded bottom to top, then left to right, and finally, a hole is punched at X. What does the paper look like when unfolded?
如右图所示，一张矩形纸从下折到上，然后从左折到右，最后在 X 处打一个洞。展开纸后是什么样子？
stem
Correct Answer: B
Answer (B): The folded rectangle appears in the upper right corner of the sheet of paper, and the hole is punched in its upper left corner. Only Figure (B) has a hole in the upper left corner of the upper right rectangle of the unfolded sheet.
答案（B）：折叠后的矩形出现在纸张的右上角，孔打在该矩形的左上角。只有图（B）在展开后的纸张右上角矩形的左上角有一个孔。
solution
Q19
Tamika selects two different numbers at random from the set {8, 9, 10} and adds them. Carlos takes two different numbers at random from the set {3, 5, 6} and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?
Tamika 从集合 {8, 9, 10} 中随机选两个不同的数相加。Carlos 从集合 {3, 5, 6} 中随机选两个不同的数相乘。Tamika 的结果大于 Carlos 的结果的概率是多少？
Correct Answer: A
Answer (A): Tamika can get the numbers $8+9=17$, $8+10=18$, or $9+10=19$. Carlos can get $3\times 5=15$, $3\times 6=18$, or $5\times 6=30$. The possible ways to pair these are: (17, 15), (17, 18), (17, 30), (18, 15), (18, 18), (18, 30), (19, 15), (19, 18), (19, 30). Four of these nine pairs show Tamika with a higher result, so the probability is $4/9$.
答案（A）：Tamika 可能得到的数是 $8+9=17$、$8+10=18$ 或 $9+10=19$。Carlos 可能得到 $3\times 5=15$、$3\times 6=18$ 或 $5\times 6=30$。这些结果的配对方式为：(17, 15)、(17, 18)、(17, 30)、(18, 15)、(18, 18)、(18, 30)、(19, 15)、(19, 18)、(19, 30)。这九种配对中有四种显示 Tamika 的结果更大，因此概率为 $4/9$。
Q20
Let $PQRS$ be a square piece of paper. $P$ is folded onto $R$ and then $Q$ is folded onto $S$. The area of the resulting figure is 9 square inches. Find the perimeter of square $PQRS$.
设 $PQRS$ 是一张正方形纸。将 $P$ 折到 $R$，然后将 $Q$ 折到 $S$。所得图形的面积是 9 平方英寸。求正方形 $PQRS$ 的周长。
stem
Correct Answer: D
Answer (D): After folding the square twice the resulting figure is an isosceles triangle with area 9 square inches. Since there are 4 such congruent triangles in the square, the area of the square is 36 square inches. Therefore, the sides of $PQRS$ are 6 inches, and the perimeter is 24 inches.
答案（D）：将正方形对折两次后，得到的图形是一个面积为 9 平方英寸的等腰三角形。由于在正方形中有 4 个这样的全等三角形，所以正方形的面积是 36 平方英寸。因此，$PQRS$ 的边长为 6 英寸，周长为 24 英寸。
Q21
A $4 \times 4 \times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?
一个 $4 \times 4 \times 4$ 的立方体盒子装有 64 个完全填满盒子的小立方体。其中有多少个小立方体触及盒子的侧面或底部？
Correct Answer: B
Answer (B): The $2 \times 2 \times 3$ core contains all of the small cubes that do not touch a side or the bottom. These 12 cubes are subtracted from 64 to leave 52.
答案（B）：$2 \times 2 \times 3$ 的核心包含所有不接触侧面或底面的 小立方体。这 12 个立方体从 64 中减去，剩下 52。
solution
Q22
Terri produces a sequence ... starts with positive integer, rules: <10 *9; even>9 /2; odd>9 -5. Sequence begins 98,49,... find 98th term.
Terri 生成一个序列……从正整数开始，规则：<10 *9；偶数>9 /2；奇数>9 -5。序列从 98,49,... 开始，求第 98 项。
Correct Answer: D
Answer (D): The sequence is 98, 49, 44, 22, 11, 6, 54, 27, 22, . . . . After 3 terms the cycle (22, 11, 6, 54, 27) is repeated. The 98th term is the fifth term of the cycle, and this is 27.
答案（D）：数列为 98，49，44，22，11，6，54，27，22，…… 在前 3 项之后，循环（22，11，6，54，27）重复出现。第 98 项是该循环的第 5 项，因此为 27。
Q23
If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle?
如果图中的模式继续，第八个三角形的内部有几分之几被涂黑？
stem
Correct Answer: C
Answer (C): \[ \begin{array}{c c c} \text{Step} & \text{Number of triangles} & \text{Number of shaded triangles} \\ 1 & 1 & 0 \\ 2 & 4 & 0+1=1 \\ 3 & 9 & 1+2=3 \\ 4 & 16 & 1+2+3=6 \\ 5 & 25 & 1+2+3+4=10 \\ 6 & 36 & 1+2+3+4+5=15 \\ 7 & 49 & 1+2+3+4+5+6=21 \\ 8 & 64 & 1+2+3+4+5+6+7=28 \\ \end{array} \] The ratio at step eight: $\frac{28}{64}=\frac{7}{16}$.
答案（C）： \[ \begin{array}{c c c} \text{步骤} & \text{三角形的数量} & \text{涂色三角形的数量} \\ 1 & 1 & 0 \\ 2 & 4 & 0+1=1 \\ 3 & 9 & 1+2=3 \\ 4 & 16 & 1+2+3=6 \\ 5 & 25 & 1+2+3+4=10 \\ 6 & 36 & 1+2+3+4+5=15 \\ 7 & 49 & 1+2+3+4+5+6=21 \\ 8 & 64 & 1+2+3+4+5+6+7=28 \\ \end{array} \] 第八步的比值：$\frac{28}{64}=\frac{7}{16}$。
Q24
A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and hades square 3, skip two squares and shades square 6, ships 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column. What is the number of the shaded square that ¯rst achieves this result?
一个有 8 列的矩形棋盘，方格从左上角开始编号，从左到右，行一是 1 到 8，行二是 9 到 16，依此类推。学生涂黑方格 1，然后跳过一个涂黑 3，跳过两个涂黑 6，跳过三个涂黑 10，并以此类推，直到每列至少有一个涂黑方格。第一个实现此结果的涂黑方格编号是多少？
stem
Correct Answer: E
Answer (E): The numbers in the first column all have remainders of 1 when divided by 8, those of the second column have remainders of 2 when divided by 8, and so on. We need to find numbered squares so that each remainder 0 through 7 appears at least once. The squares that are shaded are numbered 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, and the remainders upon dividing by 8 are 1, 3, 6, 2, 7, 5, 4, 4, 5, 7, 2, 6, 3, 1, 0. Thus, we must shaded square 120 to obtain the first shaded square in the last column.
答案（E）：第一列中的数除以$8$都余$1$，第二列中的数除以$8$都余$2$，依此类推。我们需要找到一些编号的方格，使得余数$0$到$7$每个至少出现一次。被涂色的方格编号为$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120$，它们分别除以$8$的余数为$1, 3, 6, 2, 7, 5, 4, 4, 5, 7, 2, 6, 3, 1, 0$。因此，我们必须将方格$120$涂色，才能得到最后一列中的第一个被涂色方格。
Q25
Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has \$36 when they begin and $36 when they end, what is the total amount that all three friends have?
三位慷慨的朋友，各有一些现金，按以下方式重新分配金钱：Ami 给 Jan 和 Toy 足够的钱，使她们各自的金额翻倍。Jan 然后给 Ami 和 Toy 足够的钱，使她们各自的金额翻倍。最后，Toy 给 Ami 和 Jan 足够的钱，使她们各自的金额翻倍。如果 Toy 开始和结束时都有 \$36，他们三人总共有多少钱？
Correct Answer: D
Answer (D): Since Toy begins with \$36 and her amount is doubled in the first two exchanges, her amounts are \$36, \$72, \$144, and \$36. This means that she gave away \$108, and this is exactly enough to double the amounts of Ami and Jan. So, the total must be $2(\$108)+\$36=\$252$.
答案（D）：由于 Toy 一开始有 \$36，并且在前两次交换中她的金额都翻倍，所以她的金额分别是 \$36、\$72、\$144 和 \$36。这意味着她送出了 \$108，而这恰好足以使 Ami 和 Jan 的金额都翻倍。因此，总额必为 $2(\$108)+\$36=\$252$。