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AMC8 2008

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AMC8 · 2008

Q1
Susan had $50$ to spend at the carnival. She spent $12$ on food and twice as much on rides. How many dollars did she have left to spend?
Susan 在嘉年华上有 $50$ 美元可以花。她花了 $12$ 买食物，在游乐设施上花了这两倍的钱。她还剩下多少美元可以花？
Correct Answer: B
If Susan spent 12 dollars, then twice that much on rides, then she spent $12+12 \times 2=36$ dollars in total. We subtract $36$ from $50$ to get $\boxed{\textbf{(B)}\ 14}$
Susan 花了 12 美元，那么在游乐设施上花了两倍，即 $12+12 \times 2=36$ 美元总共。从 $50$ 中减去 $36$ 得到 $\boxed{\textbf{(B)}\ 14}$
Q2
The ten-letter code BEST OF LUCK represents the ten digits $0-9$, in order. What $4$-digit number is represented by the code word CLUE?
十个字母的密码 BEST OF LUCK 代表数字 $0-9$，按顺序对应。密码单词 CLUE 表示的四位数是？
Correct Answer: A
Answer (A): Because the key to the code starts with zero, all the letters represent numbers that are one less than their position. Using the key, C is $9-1=8$, and similarly L is 6, U is 7, and E is 1. \[ \begin{array}{cccccccccc} \text{BEST} & & \text{OF} & & \text{LUCK} \\ 0 1 2 3 & & 4 5 & & 6 7 8 9 \end{array} \] CLUE = 8671
答案（A）：因为密码的关键字从零开始，所以所有字母代表的数字比它们的位置少一。根据密钥，C 是 $9-1=8$，同理，L 是 6，U 是 7，E 是 1。 \[ \begin{array}{cccccccccc} \text{BEST} & & \text{OF} & & \text{LUCK} \\ 0 1 2 3 & & 4 5 & & 6 7 8 9 \end{array} \] CLUE = 8671
Q3
If February is a month that contains Friday the $13$th, what day of the week is February $1$?
如果二月包含星期五的 $13$ 号，那么二月 $1$ 号是星期几？
Correct Answer: A
Answer (A): A week before the $13^{th}$ is the $6^{th}$, which is the first Friday of the month. Counting back from that, the $5^{th}$ is a Thursday, the $4^{th}$ is a Wednesday, the $3^{rd}$ is a Tuesday, the $2^{nd}$ is a Monday, and the $1^{st}$ is a Sunday.
答案（A）：$13^{th}$ 的前一周是 $6^{th}$，也就是这个月的第一个星期五。从那天往前推，$5^{th}$ 是星期四，$4^{th}$ 是星期三，$3^{rd}$ 是星期二，$2^{nd}$ 是星期一，而 $1^{st}$ 是星期日。
Q4
In the figure, the outer equilateral triangle has area $16$, the inner equilateral triangle has area $1$, and the three trapezoids are congruent. What is the area of one of the trapezoids?
在图中，外层等边三角形的面积是 $16$，内层等边三角形的面积是 $1$，三个梯形全等。其中一个梯形的面积是多少？
stem
Correct Answer: C
Answer (C): The area of the outer triangle with the inner triangle removed is $16-1=15$, the total area of the three congruent trapezoids. Each trapezoid has area $\frac{15}{3}=5$.
答案（C）：外三角形去掉内三角形后的面积为 $16-1=15$，这就是三个全等梯形的总面积。每个梯形的面积为 $\frac{15}{3}=5$。
Q5
Barney Schwinn notices that the odometer on his bicycle reads $1441$, a palindrome, because it reads the same forward and backward. After riding $4$ more hours that day and $6$ the next, he notices that the odometer shows another palindrome, $1661$. What was his average speed in miles per hour?
Barney Schwinn 注意到他的自行车里程表显示 $1441$，这是一个回文数，因为正读反读都一样。那天他又骑了 $4$ 小时，第二天骑了 $6$ 小时后，他注意到里程表又显示另一个回文数 $1661$。他的平均速度是多少英里每小时？
Correct Answer: E
Answer (E): Barney rides $1661-1441=220$ miles in $10$ hours, so his average speed is $\frac{220}{10}=22$ miles per hour.
答案（E）：Barney 行驶了 $1661-1441=220$ 英里，用了 $10$ 小时，所以他的平均速度是 $\frac{220}{10}=22$ 英里/小时。
Q6
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?
在图中，灰色正方形的面积与白色正方形的面积之比是多少？
stem
Correct Answer: D
Answer (D): After subdividing the central gray square as shown, 6 of the 16 congruent squares are gray and 10 are white. Therefore, the ratio of the area of the gray squares to the area of the white squares is $6:10$ or $3:5$.
答案（D）：如图将中央的灰色正方形细分后，16个全等小正方形中有6个是灰色，10个是白色。因此，灰色小正方形的面积与白色小正方形的面积之比为$6:10$，即$3:5$。
Q7
If $\frac{3}{5} = \frac{M}{45} = \frac{60}{N}$, what is $M + N$?
如果 $\frac{3}{5} = \frac{M}{45} = \frac{60}{N}$，那么 $M + N$ 是多少？
Correct Answer: E
Answer (E): Note that $\frac{M}{45}=\frac{3}{5}=\frac{3\cdot 9}{5\cdot 9}=\frac{27}{45}$, so $M=27$. Similarly, $\frac{60}{N}=\frac{3}{5}=\frac{3\cdot 20}{5\cdot 20}=\frac{60}{100}$, so $N=100$. The sum $M+N=27+100=127$.
答案（E）：注意到 $\frac{M}{45}=\frac{3}{5}=\frac{3\cdot 9}{5\cdot 9}=\frac{27}{45}$，所以 $M=27$。同理，$\frac{60}{N}=\frac{3}{5}=\frac{3\cdot 20}{5\cdot 20}=\frac{60}{100}$，所以 $N=100$。因此 $M+N=27+100=127$。
Q8
Candy sales of the Boosters Club for January through April are shown. What were the average sales per month in dollars?
助推俱乐部一月至四月的糖果销售额如图所示。每月平均销售额（美元）是多少？
stem
Correct Answer: D
Answer (D): The sales in the 4 months were \$100, \$60, \$40 and \$120. The average sales were $\frac{100+60+40+120}{4}=\frac{320}{4}=\$80$.
答案（D）：这4个月的销售额分别是 \$100、\$60、\$40 和 \$120。平均销售额为 $\frac{100+60+40+120}{4}=\frac{320}{4}=\$80$。
Q9
In $2005$ Tycoon Tammy invested $\$100$ for two years. During the first year her investment suffered a $15\%$ loss, but during the second year the remaining investment showed a $20\%$ gain. Over the two-year period, what was the change in Tammy's investment?
2005 年，Tycoon Tammy 投资了 $\$100$ 两年。第一年她的投资损失了 $15\%$，但第二年剩余投资获得了 $20\%$ 的收益。在两年期间，Tammy 的投资变化是多少？
Correct Answer: D
Answer (D): At the end of the first year, Tammy’s investment was 85% of the original amount, or \$85. At the end of the second year, she had 120% of her first year’s final amount, or $120\%\text{ of }\$85 = 1.2(\$85) = \$102$. Over the two-year period, Tammy’s investment changed from \$100 to \$102, so she gained 2%.
答案（D）：第一年末，Tammy 的投资变为原始金额的 85%，即 \$85。第二年末，她拥有第一年末金额的 120%，即 $120\%\text{ 的 }\$85 = 1.2(\$85) = \$102$。两年期间，Tammy 的投资从 \$100 变为 \$102，因此她收益了 2%。
Q10
The average age of the $6$ people in Room A is $40$. The average age of the $4$ people in Room B is $25$. If the two groups are combined, what is the average age of all the people?
A 室 $6$ 人的平均年龄是 $40$。B 室 $4$ 人的平均年龄是 $25$。如果将两组人合并，所有人的平均年龄是多少？
Correct Answer: D
Answer (D): The sum of the ages of the 6 people in Room A is $6\times 40=240$. The sum of the ages of the 4 people in Room B is $4\times 25=100$. The sum of the ages of the 10 people in the combined group is $100+240=340$, so the average age of all the people is $\frac{340}{10}=34$.
答案（D）：A房间中6个人的年龄和为 $6\times 40=240$。B房间中4个人的年龄和为 $4\times 25=100$。合并后10个人的年龄总和为 $100+240=340$，因此所有人的平均年龄为 $\frac{340}{10}=34$。
Q11
Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
林肯中学八年级的39名学生每人都有狗或猫或既有狗又有猫。有20名学生有狗，26名学生有猫。有多少名学生既有狗又有猫？
Correct Answer: A
The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{\textbf{(A)}\ 7}$.
两个集合的并集等于每个集合的大小之和减去它们的交集。既有狗又有猫的学生数量是$20+26-39=\boxed{\textbf{(A)}\ 7}$。
Q12
A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
一个球从3米高处落下。第一次弹起时升到2米高。它不断落下和弹起，每次弹起高度是前一次的$\frac{2}{3}$。在第几次弹起时它不会升到0.5米高？
stem
Correct Answer: C
Answer (C): The table gives the height of each bounce. \[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Bounce} & 1 & 2 & 3 & 4 & 5\\ \hline \text{Height in Meters} & 2 & \frac{2}{3}\cdot 2=\frac{4}{3} & \frac{2}{3}\cdot \frac{4}{3}=\frac{8}{9} & \frac{2}{3}\cdot \frac{8}{9}=\frac{16}{27} & \frac{2}{3}\cdot \frac{16}{27}=\frac{32}{81}\\ \hline \end{array} \] Because $\frac{16}{27}>\frac{16}{32}=\frac{1}{2}$ and $\frac{32}{81}<\frac{32}{64}=\frac{1}{2}$, the ball first rises to less than $0.5$ meters on the fifth bounce. Note: Because all the fractions have odd denominators, it is easier to double the numerators than to halve the denominators. So compare $\frac{16}{27}$ and $\frac{32}{81}$ to their numerators' fractional equivalents of $\frac{1}{2}$, $\frac{16}{32}$ and $\frac{32}{64}$.
答案（C）：下表给出了每次弹跳的高度。 \[ \begin{array}{|c|c|c|c|c|c|} \hline \text{弹跳次数} & 1 & 2 & 3 & 4 & 5\\ \hline \text{高度（米）} & 2 & \frac{2}{3}\cdot 2=\frac{4}{3} & \frac{2}{3}\cdot \frac{4}{3}=\frac{8}{9} & \frac{2}{3}\cdot \frac{8}{9}=\frac{16}{27} & \frac{2}{3}\cdot \frac{16}{27}=\frac{32}{81}\\ \hline \end{array} \] 因为 $\frac{16}{27}>\frac{16}{32}=\frac{1}{2}$ 且 $\frac{32}{81}<\frac{32}{64}=\frac{1}{2}$，所以小球第一次反弹到低于 $0.5$ 米是在第 5 次弹跳。 注：由于这些分数的分母都是奇数，把分子加倍比把分母减半更容易。因此，将 $\frac{16}{27}$ 和 $\frac{32}{81}$ 与它们在“分子不变、分母取相应比较值”下的等价比较：$\frac{1}{2}$、$\frac{16}{32}$ 和 $\frac{32}{64}$ 进行比较。
Q13
Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than $100$ pounds or more than $150$ pounds. So the boxes are weighed in pairs in every possible way. The results are $122$, $125$ and $127$ pounds. What is the combined weight in pounds of the three boxes?
哈曼先生需要知道他要邮寄的三个盒子的总重量（磅）。但是，可用的秤不准确，无法称量小于100磅或大于150磅的重量。因此，盒子以每种可能的方式成对称重。结果分别是122、125和127磅。三个盒子的总重量是多少磅？
Correct Answer: C
Answer (C): Because each box is weighed two times, once with each of the other two boxes, the total $122+125+127=374$ pounds is twice the combined weight of the three boxes. The combined weight is $\frac{374}{2}=187$ pounds.
答案（C）：因为每个箱子被称重两次（分别与另外两个箱子各称一次），总和 $122+125+127=374$ 磅是三个箱子总重量的两倍。三个箱子的总重量为 $\frac{374}{2}=187$ 磅。
Q14
Three As, three Bs and three Cs are placed in the nine spaces so that each row and column contain one of each letter. If A is placed in the upper left corner, how many arrangements are possible?
在九个空格中放置三个A、三个B和三个C，使得每行和每列包含每个字母各一个。如果A放在左上角，可能的排列有多少种？
stem
Correct Answer: C
Answer (C): There are only two possible spaces for the B in row 1 and only two possible spaces for the A in row 2. Once these are placed, the entries in the remaining spaces are determined. The four arrangements are:
答案（C）：第 1 行中 B 只有两个可能的位置，第 2 行中 A 也只有两个可能的位置。一旦把它们放好，其余空格中的条目就被确定了。 四种排列为：
Q15
In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
在特蕾莎的前8场篮球比赛中，她得分分别是7、4、3、6、8、3、1和5分。在第九场比赛中，她得分少于10分，且九场比赛的场均得分是整数。同样，在第十场比赛中，她得分少于10分，且十场比赛的场均得分也是整数。第九场和第十场比赛得分的乘积是多少？
stem
Correct Answer: B
Answer (B): The sum of the points Theresa scored in the first $8$ games is $37$. After the ninth game, her point total must be a multiple of $9$ between $37$ and $37+9=46$, inclusive. The only such point total is $45=37+8$, so in the ninth game she scored $8$ points. Similarly, the next point total must be a multiple of $10$ between $45$ and $45+9=54$. The only such point total is $50=45+5$, so in the tenth game she scored $5$ points. The product of the number of points scored in Theresa’s ninth and tenth games is $8\cdot 5=40$.
答案（B）：Theresa 在前 $8$ 场比赛中得到的总分是 $37$。打完第九场后，她的总分必须是介于 $37$ 和 $37+9=46$（含）之间的 $9$ 的倍数。唯一满足的总分是 $45=37+8$，因此第九场她得了 $8$ 分。同理，下一次总分必须是介于 $45$ 和 $45+9=54$ 之间的 $10$ 的倍数。唯一满足的总分是 $50=45+5$，因此第十场她得了 $5$ 分。Theresa 第九场和第十场得分的乘积为 $8\cdot 5=40$。
Q16
A shape is created by joining seven unit cubes, as shown. What is the ratio of the volume in cubic units to the surface area in square units?
通过连接七个单位立方体形成的形状，如图所示。体积（立方单位）与表面积（平方单位）的比率为多少？
stem
Correct Answer: D
Answer (D): The volume is $7 \times 1 = 7$ cubic units. Six of the cubes have $5$ square faces exposed. The middle cube has no face exposed. So the total surface area of the figure is $5 \times 6 = 30$ square units. The ratio of the volume to the surface area is $7:30$.
答案（D）：体积为 $7 \times 1 = 7$ 立方单位。其中有 $6$ 个小立方体各有 $5$ 个正方形面暴露在外。中间的小立方体没有任何面暴露。因此该立体的总表面积为 $5 \times 6 = 30$ 平方单位。体积与表面积之比为 $7:30$。
Q17
Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of $50$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
奥斯本女士让班上的每个学生画一个周长为 $50$ 单位的整数边长矩形。所有学生都计算了自己画的矩形的面积。最大可能面积与最小可能面积的差是多少？
Correct Answer: D
Answer (D): The formula for the perimeter of a rectangle is $2l+2w$, so $2l+2w=50$, and $l+w=25$. Make a chart of the possible widths, lengths, and areas, assuming all the widths are shorter than all the lengths. $\begin{array}{|c|cccccccccccc|} \hline \text{Width} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{Length} & 24 & 23 & 22 & 21 & 20 & 19 & 18 & 17 & 16 & 15 & 14 & 13 \\ \hline \text{Area} & 24 & 46 & 66 & 84 & 100 & 114 & 126 & 136 & 144 & 150 & 154 & 156 \\ \hline \end{array}$ The largest possible area is $13\times 12=156$ and the smallest is $1\times 24=24$, for a difference of $156-24=132$ square units. Note: The product of two numbers with a fixed sum increases as the numbers get closer together. That means, given the same perimeter, the square has a larger area than any rectangle, and a rectangle with a shape closest to a square will have a larger area than other rectangles with equal perimeters.
答案（D）：长方形周长的公式是 $2l+2w$，所以 $2l+2w=50$，并且 $l+w=25$。在假设所有宽都小于所有长的情况下，制作一个可能的宽、长与面积的表格。 $\begin{array}{|c|cccccccccccc|} \hline \text{宽} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{长} & 24 & 23 & 22 & 21 & 20 & 19 & 18 & 17 & 16 & 15 & 14 & 13 \\ \hline \text{面积} & 24 & 46 & 66 & 84 & 100 & 114 & 126 & 136 & 144 & 150 & 154 & 156 \\ \hline \end{array}$ 最大的可能面积是 $13\times 12=156$，最小的是 $1\times 24=24$，因此差值为 $156-24=132$ 平方单位。 注意：当两个数的和固定时，它们的乘积会随着这两个数彼此越接近而增大。这意味着在周长相同的情况下，正方形的面积比任何长方形都大；而形状越接近正方形的长方形，其面积也会比其他周长相同的长方形更大。
Q18
Two circles that share the same center have radii $10$ meters and $20$ meters. An aardvark runs along the path shown, starting at A and ending at K. How many meters does the aardvark run?
两个同心圆，半径分别为 $10$ 米和 $20$ 米。土豚沿着图示路径从 A 点跑到 K 点。土豚跑了多少米？
stem
Correct Answer: E
Answer (E): The length of first leg of the aardvark’s trip is $\frac{1}{4}(2\pi \times 20)=10\pi$ meters. The third and fifth legs are each $\frac{1}{4}(2\pi \times 10)=5\pi$ meters long. The second and sixth legs are each $10$ meters long, and the length of the fourth leg is $20$ meters. The length of the total trip is $10\pi+5\pi+5\pi+10+10+20=20\pi+40$ meters.
答案（E）：食蚁兽行程第一段的长度为 $\frac{1}{4}(2\pi \times 20)=10\pi$ 米。第三段和第五段的长度各为 $\frac{1}{4}(2\pi \times 10)=5\pi$ 米。第二段和第六段各为 $10$ 米，第四段的长度为 $20$ 米。全程长度为 $10\pi+5\pi+5\pi+10+10+20=20\pi+40$ 米。
Q19
Eight points are spaced at intervals of one unit around a $2\times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the points are one unit apart?
八个点围绕 $2\times 2$ 正方形以一单位间隔等距放置，如图所示。从 8 个点中随机选择两个点。这两个点相距一单位的概率是多少？
stem
Correct Answer: B
Answer (B): Choose two points. Any of the 8 points can be the first choice, and any of the 7 other points can be the second choice. So there are $8 \times 7 = 56$ ways of choosing the points in order. But each pair of points is counted twice, so there are $\frac{56}{2} = 28$ possible pairs. Label the eight points as shown. Only segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EF}$, $\overline{FG}$, $\overline{GH}$ and $\overline{HA}$ are 1 unit long. So 8 of the 28 possible segments are 1 unit long, and the probability that the points are one unit apart is $\frac{8}{28}=\frac{2}{7}$.
答案（B）：选取两点。8 个点中的任意一个都可以作为第一个选择，其余 7 个点中的任意一个都可以作为第二个选择。所以按顺序选点共有 $8 \times 7 = 56$ 种方法。但每一对点被计算了两次，因此共有 $\frac{56}{2} = 28$ 对可能的点对。 按图所示给八个点标记。只有线段 $\overline{AB}$、$\overline{BC}$、$\overline{CD}$、$\overline{DE}$、$\overline{EF}$、$\overline{FG}$、$\overline{GH}$ 和 $\overline{HA}$ 的长度为 1 个单位。因此，在 28 条可能的线段中有 8 条长度为 1 个单位，两点相距 1 个单位的概率为 $\frac{8}{28}=\frac{2}{7}$。
solution
Q20
The students in Mr. Neatkin's class took a penmanship test. Two-thirds of the boys and $\frac{3}{4}$ of the girls passed the test, and an equal number of boys and girls passed the test. What is the minimum possible number of students in the class?
尼特金先生班上的学生参加了一次笔迹测试。男生中有 $\frac{2}{3}$ 和女生中有 $\frac{3}{4}$ 通过了测试，并且通过测试的男生和女生人数相等。班上学生的最小可能人数是多少？
Correct Answer: B
Answer (B): Because $\frac{2}{3}$ of the boys passed, the number of boys in the class is a multiple of 3. Because $\frac{3}{4}$ of the girls passed, the number of girls in the class is a multiple of 4. Set up a chart and compare the number of boys who passed with the number of girls who passed to find when they are equal. \[ \begin{array}{|c|c|} \hline \text{Total boys} & \text{Boys passed} \\ \hline 3 & 2 \\ \hline 6 & 4 \\ \hline 9 & 6 \\ \hline \end{array} \qquad \begin{array}{|c|c|} \hline \text{Total girls} & \text{Girls passed} \\ \hline 4 & 3 \\ \hline 8 & 6 \\ \hline \end{array} \] The first time the number of boys who passed equals the number of girls who passed is when they are both 6. The minimum possible number of students is $9+8=17$.
答案（B）：因为男生中有$\frac{2}{3}$通过，所以班里男生人数是3的倍数。因为女生中有$\frac{3}{4}$通过，所以班里女生人数是4的倍数。列一个表，比较通过的男生人数与通过的女生人数，找出它们相等时的情况。 \[ \begin{array}{|c|c|} \hline \text{男生总数} & \text{通过的男生} \\ \hline 3 & 2 \\ \hline 6 & 4 \\ \hline 9 & 6 \\ \hline \end{array} \qquad \begin{array}{|c|c|} \hline \text{女生总数} & \text{通过的女生} \\ \hline 4 & 3 \\ \hline 8 & 6 \\ \hline \end{array} \] 男生通过人数第一次等于女生通过人数是在两者都为6时。学生人数的最小可能值是$9+8=17$。
Q21
Jerry cuts a wedge from a $6$-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
杰瑞从一个6厘米高的博洛尼亚香肠圆柱体中切下一个楔形块，如虚线所示。哪个选项最接近他的楔形块的体积（立方厘米）？
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Correct Answer: C
Answer (C): Using the formula for the volume of a cylinder, the bologna has volume $ \pi r^2 h = \pi \times 4^2 \times 6 = 96\pi $. The cut divides the bologna in half. The half-cylinder will have volume $ \frac{96\pi}{2} = 48\pi \approx 151\ \text{cm}^3 $. Note: The value of $\pi$ is slightly greater than 3, so to estimate the volume multiply $48(3)=144\ \text{cm}^3$. The product is slightly less than and closer to answer C than any other answer.
答案（C）：使用圆柱体体积公式，这块博洛尼亚香肠的体积为 $ \pi r^2 h = \pi \times 4^2 \times 6 = 96\pi $。切一刀把香肠分成两半。半圆柱的体积为 $ \frac{96\pi}{2} = 48\pi \approx 151\ \text{cm}^3 $。 注：$\pi$ 的值略大于 3，因此估算体积时可用 $48(3)=144\ \text{cm}^3$。该结果略小于真实值，并且比其他选项更接近答案 C。
Q22
For how many positive integer values of $n$ are both $\frac{n}{3}$ and $3n$ three-digit whole numbers?
有且仅有几个正整数 $n$ 使得 $\frac{n}{3}$ 和 $3n$ 都是三位数的整数？
Correct Answer: A
Answer (A): Because $\frac{n}{3}$ is at least 100 and is an integer, $n$ is at least 300 and is a multiple of 3. Because $3n$ is at most 999, $n$ is at most 333. The possible values of $n$ are 300, 303, 306, ..., 333 = $3 \cdot 100, 3 \cdot 101, 3 \cdot 102, ..., 3 \cdot 111$, so the number of possible values is $111 - 100 + 1 = 12$.
答案（A）：因为 $\frac{n}{3}$ 至少为 100 且为整数，所以 $n$ 至少为 300 且是 3 的倍数。因为 $3n$ 至多为 999，所以 $n$ 至多为 333。$n$ 的可能取值为 300、303、306、…、333，即 $3 \cdot 100, 3 \cdot 101, 3 \cdot 102, …, 3 \cdot 111$，因此可能取值的个数为 $111 - 100 + 1 = 12$。
Q23
In square $ABCE$, $AF = 2FE$ and $CD = 2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$?
在正方形 $ABCE$ 中，$AF = 2FE$ 且 $CD = 2DE$。$ riangle BFD$ 的面积与正方形 $ABCE$ 的面积之比是多少？
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Correct Answer: C
Answer (C): Because the answer is a ratio, it does not depend on the side length of the square. Let $AF=2$ and $FE=1$. That means square $ABCE$ has side length $3$ and area $3^2=9$ square units. The area of $\triangle BAF$ is equal to the area of $\triangle BCD=\frac{1}{2}\cdot 3\cdot 2=3$ square units. Triangle $DEF$ is an isosceles right triangle with leg lengths $DE=FE=1$. The area of $\triangle DEF$ is $\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$ square units. The area of $\triangle BFD$ is equal to the area of the square minus the areas of the three right triangles: $9-(3+3+\frac{1}{2})=\frac{5}{2}$. So the ratio of the area of $\triangle BFD$ to the area of square $ABCE$ is $\frac{\frac{5}{2}}{9}=\frac{5}{18}$.
答案（C）：因为答案是一个比值，所以它不依赖于正方形的边长。令 $AF=2$ 且 $FE=1$。这意味着正方形 $ABCE$ 的边长为 $3$，面积为 $3^2=9$ 平方单位。$\triangle BAF$ 的面积等于 $\triangle BCD$ 的面积，即 $\frac{1}{2}\cdot 3\cdot 2=3$ 平方单位。三角形 $DEF$ 是等腰直角三角形，两条直角边长为 $DE=FE=1$。$\triangle DEF$ 的面积为 $\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$ 平方单位。$\triangle BFD$ 的面积等于正方形的面积减去三个直角三角形的面积：$9-(3+3+\frac{1}{2})=\frac{5}{2}$。因此，$\triangle BFD$ 的面积与正方形 $ABCE$ 的面积之比为 $\frac{\frac{5}{2}}{9}=\frac{5}{18}$。
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Q24
Ten tiles numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
十块编号为1到10的瓷砖面朝下放置。随机翻开一块瓷砖，并掷一颗骰子。瓷砖和骰子上的数字乘积为完全平方的概率是多少？
Correct Answer: C
Answer (C): There are $10\times 6=60$ possible pairs. The squares less than $60$ are $1,4,9,16,25,36$ and $49$. The possible pairs with products equal to the given squares are $(1,1),(2,2),(1,4),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6)$ and $(9,4)$. So the probability is $\frac{11}{60}$.
答案（C）：共有 $10\times 6=60$ 种可能的数对。小于 $60$ 的完全平方数为 $1,4,9,16,25,36$ 和 $49$。乘积等于这些平方数的可能数对为 $(1,1),(2,2),(1,4),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6)$ 和 $(9,4)$。因此概率为 $\frac{11}{60}$。
Q25
Margie's winning art design is shown. The smallest circle has radius $2$ inches, with each successive circle's radius increasing by $2$ inches. Approximately what percent of the design is black?
玛吉获奖的艺术设计如图所示。最小的圆半径为2英寸，每一个后续圆的半径增加2英寸。设计中大约百分之多少是黑色的？
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Correct Answer: A
Answer (A): $\begin{array}{|c|c|c|} \hline \text{Circle #} & \text{Radius} & \text{Area} \\ \hline 1 & 2 & 4\pi \\ \hline 2 & 4 & 16\pi \\ \hline 3 & 6 & 36\pi \\ \hline 4 & 8 & 64\pi \\ \hline 5 & 10 & 100\pi \\ \hline 6 & 12 & 144\pi \\ \hline \end{array}$ The total black area is $4\pi + (36-16)\pi + (100-64)\pi = 60\pi\ \text{in}^2.$ So the percent of the design that is black is $100 \times \frac{60\pi}{144\pi} = 100 \times \frac{5}{12}$ or about $42\%.$
答案（A）： $\begin{array}{|c|c|c|} \hline \text{圆编号} & \text{半径} & \text{面积} \\ \hline 1 & 2 & 4\pi \\ \hline 2 & 4 & 16\pi \\ \hline 3 & 6 & 36\pi \\ \hline 4 & 8 & 64\pi \\ \hline 5 & 10 & 100\pi \\ \hline 6 & 12 & 144\pi \\ \hline \end{array}$ 黑色部分的总面积为 $4\pi + (36-16)\pi + (100-64)\pi = 60\pi\ \text{in}^2.$ 因此，该图案中黑色所占的百分比为 $100 \times \frac{60\pi}{144\pi} = 100 \times \frac{5}{12}$，约为 $42\%.$