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AMC8 2007

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AMC8 · 2007

Q1
Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of 10 hours per week helping around the house for 6 weeks. For the first 5 weeks she helps around the house for 8, 11, 7, 12 and 10 hours. How many hours must she work during the final week to earn the tickets?
Theresa的父母同意如果她在6周内每周平均帮助家里做10小时家务，就给她买去看她最喜欢的乐队的门票。前5周她帮助家里的时间分别是8、11、7、12和10小时。最后一周她必须工作多少小时才能拿到门票？
Correct Answer: D
(D) The first 5 weeks Theresa works a total of $8+11+7+12+10=48$ hours. She has promised to work $6\times 10=60$ hours. She must work $60-48=12$ hours during the final week.
（D）前5周 Theresa 一共工作了 $8+11+7+12+10=48$ 小时。她承诺要工作 $6\times 10=60$ 小时。因此她在最后一周必须工作 $60-48=12$ 小时。
Q2
Six-hundred fifty students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
有650名学生接受了关于意面偏好的调查。选择有lasagna、manicotti、ravioli和spaghetti。调查结果显示在条形图中。喜欢spaghetti的学生人数与喜欢manicotti的学生人数之比是多少？
stem
Correct Answer: E
(E) The ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti is $\frac{250}{100}=\frac{5}{2}$.
（E）喜欢意大利面（spaghetti）的学生人数与喜欢意式烤通心粉（manicotti）的学生人数之比为 $\frac{250}{100}=\frac{5}{2}$。
Q3
What is the sum of the two smallest prime factors of 250?
250的两个最小质因数的和是多少？
Correct Answer: C
(C) The prime factorization of 250 is $2\cdot5\cdot5\cdot5$. The sum of 2 and 5 is 7.
（C）250 的质因数分解是 $2\cdot5\cdot5\cdot5$。2 和 5 的和是 7。
Q4
A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?
一个鬼屋有六个窗户。Georgie the Ghost可以通过一个窗户进入房子并从另一个不同窗户离开，有多少种方式？
Correct Answer: D
(D) Georgie has 6 choices for the window in which to enter. After entering, Georgie has 5 choices for the window from which to exit. So altogether there are $6 \times 5 = 30$ different ways for Georgie to enter one window and exit another.
(D) Georgie 有 6 种选择可以从某个窗户进入。进入后，Georgie 有 5 种选择可以从某个窗户离开。因此总共有 $6 \times 5 = 30$ 种不同的方法让 Georgie 从一个窗户进入并从另一个窗户离开。
Q5
Chandler wants to buy a \$500 mountain bike. For his birthday, his grandparents send him \$50, his aunt sends him \$35 and his cousin gives him \$15. He earns \$16 per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?
Chandler想买一辆500美元的山地车。他的生日时，祖父母给了他50美元，姑姑给了他35美元，表亲给了他15美元。他每周做报纸递送赚16美元。他将使用所有的生日钱和报纸递送赚的钱。需要多少周他才能买得起山地车？
Correct Answer: B
(B) For his birthday, Chandler gets $50+35+15=100$ dollars. Therefore, he needs $500-100=400$ dollars more. It will take Chandler $400\div16=25$ weeks to earn 400 dollars, so he can buy his bike after 25 weeks.
（B）在生日时，Chandler 得到 $50+35+15=100$ 美元。因此，他还需要 $500-100=400$ 美元。Chandler 需要 $400\div16=25$ 周才能赚到 400 美元，所以他可以在 25 周后买到他的自行车。
Q6
The average cost of a long-distance call in the USA in 1985 was 41 cents per minute, and the average cost of a long-distance call in the USA in 2005 was 7 cents per minute. Find the approximate percent decrease in the cost per minute of a long-distance call.
1985年美国长途电话的平均费用是每分钟41美分，2005年是每分钟7美分。求长途电话每分钟费用的大致百分比减少量。
Correct Answer: E
(E) The difference in the cost of a long-distance call per minute from 1985 to 2005 was $41-7=34$ cents. The percent decrease is $100\times\frac{34}{41}\approx 100\times\frac{32}{40}=100\times\frac{8}{10}=80\%$.
（E）1985 年到 2005 年长途电话每分钟费用的差额为 $41-7=34$ 美分。降幅百分比为 $100\times\frac{34}{41}\approx 100\times\frac{32}{40}=100\times\frac{8}{10}=80\%$。
Q7
The average age of 5 people in a room is 30 years. An 18-year-old person leaves the room. What is the average age of the four remaining people?
房间里有5个人，平均年龄30岁。一个18岁的人离开了房间。剩下4人的平均年龄是多少？
Correct Answer: D
(D) Originally the sum of the ages of the people in the room is $5 \times 30 = 150$. After the 18-year-old leaves, the sum of the ages of the remaining people is $150 - 18 = 132$. So the average age of the four remaining people is $\frac{132}{4} = 33$ years.
（D）最初房间里所有人的年龄总和为 $5 \times 30 = 150$。18 岁的人离开后，剩下人的年龄总和为 $150 - 18 = 132$。因此，剩下四个人的平均年龄为 $\frac{132}{4} = 33$ 岁。
Q8
In trapezoid ABCD, $\overline{AD}$ is perpendicular to $\overline{DC}$, $AD = AB = 3$, and $DC = 6$. In addition, $E$ is on $\overline{DC}$, and $BE$ is parallel to $\overline{AD}$. Find the area of $\triangle BEC$.
梯形ABCD中，$\overline{AD}$ 与 $\overline{DC}$ 垂直，$AD = AB = 3$，$DC = 6$。此外，$E$ 在 $\overline{DC}$ 上，$BE$ 平行于 $\overline{AD}$。求 $\triangle BEC$ 的面积。
stem
Correct Answer: B
(B) Note that $ABED$ is a square with side $3$. Subtract $DE$ from $DC$, to find that $EC$, the base of $\triangle BEC$, has length $3$. The area of $\triangle BEC$ is $\frac{1}{2}\cdot 3 \cdot 3=\frac{9}{2}=4.5$.
（B）注意到$ABED$是一个边长为$3$的正方形。用$DC$减去$DE$，可知$\overline{EC}$（$\triangle BEC$的底边）长度为$3$。$\triangle BEC$的面积为$\frac{1}{2}\cdot 3 \cdot 3=\frac{9}{2}=4.5$。
Q9
To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?
要完成下面的网格，每个数字1到4必须在每行和每列中各出现一次。右下角的方格应该是多少？
stem
Correct Answer: B
(B) The number in the last column of the second row must be 1 because there are already a 2 and a 3 in the second row and a 4 in the last column. By similar reasoning, the number above the 1 must be 3. So the number in the lower right-hand square must be 2. This is not the only way to find the solution. \[ \begin{array}{|c|c|c|c|} \hline 1 & & 2 & 3 \\ \hline 2 & 3 & & 1 \\ \hline & & & 4 \\ \hline & & & 2 \\ \hline \end{array} \] The completed square is \[ \begin{array}{|c|c|c|c|} \hline 1 & 4 & 2 & 3 \\ \hline 2 & 3 & 4 & 1 \\ \hline 3 & 2 & 1 & 4 \\ \hline 4 & 1 & 3 & 2 \\ \hline \end{array} \]
（B）第二行最后一列的数字必须是 1，因为第二行已经有了 2 和 3，而最后一列已经有了 4。用类似的推理，1 上面的数字必须是 3。因此右下角的方格必须是 2。这并不是找到解的唯一方法。 \[ \begin{array}{|c|c|c|c|} \hline 1 & & 2 & 3 \\ \hline 2 & 3 & & 1 \\ \hline & & & 4 \\ \hline & & & 2 \\ \hline \end{array} \] 完成后的方阵为 \[ \begin{array}{|c|c|c|c|} \hline 1 & 4 & 2 & 3 \\ \hline 2 & 3 & 4 & 1 \\ \hline 3 & 2 & 1 & 4 \\ \hline 4 & 1 & 3 & 2 \\ \hline \end{array} \]
Q10
For any positive integer $n$, define $n$ to be the sum of the positive factors of $n$. For example, $6 = 1 + 2 + 3 + 6 = 12$. Find $11$.
对于任意正整数 $n$，定义 $\sigma(n)$ 为 $n$ 的所有正因数的和。例如，$\sigma(6) = 1 + 2 + 3 + 6 = 12$。求 $\sigma(11)$。
Correct Answer: D
(D) First calculate \(\boxed{11}=1+11=12\). So \[ [\boxed{11}]=[\boxed{12}]=1+2+3+4+6+12=28 \]
（D）先计算 \(\boxed{11}=1+11=12\)。所以 \[ [\boxed{11}]=[\boxed{12}]=1+2+3+4+6+12=28 \]
Q11
Tiles I, II, III and IV are translated so one tile coincides with each of the rectangles A, B, C and D. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle C?
瓦片 I、II、III 和 IV 被平移，使得每个瓦片与矩形 A、B、C 和 D 中的一个重合。在最终排列中，任何两个相邻瓦片共用的边的两个数字必须相同。哪块瓦片被平移到矩形 C？
stem
Correct Answer: D
(D) Because Tile III has a 0 on the bottom edge and there is no 0 on any other tile, Tile III must be placed on C or D. Because Tile III has a 5 on the right edge and there is no 5 on any other tile, Tile III must be placed on the right, on D. Because Tile III has a 1 on the left edge and only Tile IV has a 1 on the right edge, Tile IV must be placed to the left of Tile III, that is, on C.
（D）由于第III块瓷片的下边缘是0，并且其他任何瓷片上都没有0，因此第III块必须放在C或D上。由于第III块瓷片的右边缘是5，并且其他任何瓷片上都没有5，因此第III块必须放在右侧，即D上。由于第III块瓷片的左边缘是1，并且只有第IV块瓷片的右边缘是1，因此第IV块必须放在第III块的左边，也就是放在C上。
solution
Q12
A unit hexagram is composed of a regular hexagon of side length 1 and its 6 equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?
一个单位六芒星由边长为 1 的正六边形及其 6 个等边三角形扩展组成，如图所示。扩展部分的面积与原始六边形面积的比率为多少？
stem
Correct Answer: A
(A) Use diagonals to cut the hexagon into 6 congruent triangles. Because each exterior triangle is also equilateral and shares an edge with an internal triangle, each exterior triangle is congruent to each interior triangle. Therefore, the ratio of the area of the extensions to the area of the hexagon is 1:1.
（A）用对角线把六边形分成 6 个全等的三角形。由于每个外侧三角形也是等边三角形，并且与一个内部三角形共享一条边，因此每个外侧三角形都与每个内部三角形全等。因此，延伸部分的面积与六边形面积之比为 1:1。
solution
Q13
Sets A and B, shown in the Venn diagram, have the same number of elements. Their union has 2007 elements and their intersection has 1001 elements. Find the number of elements in A.
集合 A 和 B 如维恩图所示，具有相同数量的元素。它们的并集有 2007 个元素，交集有 1001 个元素。求集合 A 中的元素个数。
stem
Correct Answer: C
(C) Let $C$ denote the set of elements that are in $A$ but not in $B$. Let $D$ denote the set of elements that are in $B$ but not in $A$. Because sets $A$ and $B$ have the same number of elements, the number of elements in $C$ is the same as the number of elements in $D$. This number is half the number of elements in the union of $A$ and $B$ minus the intersection of $A$ and $B$. That is, the number of elements in each of $C$ and $D$ is \[ \frac{1}{2}(2007-1001)=\frac{1}{2}\cdot 1006=503. \] Adding the number of elements in $A$ and $B$ to the number in $A$ but not in $B$ gives $1001+503=1504$ elements in $A$.
（C）设 $C$ 表示属于 $A$ 但不属于 $B$ 的元素集合。设 $D$ 表示属于 $B$ 但不属于 $A$ 的元素集合。由于集合 $A$ 和 $B$ 的元素个数相同，$C$ 中元素的个数与 $D$ 中元素的个数相同。这个数等于 $A$ 与 $B$ 的并集的元素个数减去交集的元素个数后的一半。也就是说，$C$ 和 $D$ 中每个集合的元素个数为 \[ \frac{1}{2}(2007-1001)=\frac{1}{2}\cdot 1006=503. \] 将 $A$ 与 $B$ 的元素个数加上“在 $A$ 中但不在 $B$ 中”的元素个数，得到 $A$ 中共有 $1001+503=1504$ 个元素。
solution
Q14
The base of isosceles △ABC is 24 and its area is 60. What is the length of one of the congruent sides?
等腰三角形 △ABC 的底边长为 24，其面积为 60。求一对全等边的长度。
Correct Answer: C
(C) Let $BD$ be the altitude from $B$ to $AC$ in $\triangle ABC$. Then $60=$ the area of $\triangle ABC=\frac{1}{2}\cdot 24\cdot BD$, so $BD=5$. Because $\triangle ABC$ is isosceles, $\triangle ABD$ and $\triangle CBD$ are congruent right triangles. This means that $AD=DC=\frac{24}{2}=12$. Applying the Pythagorean Theorem to $\triangle ABD$ gives $$ AB^2=5^2+12^2=169=13^2, $$ so $AB=13$.
（C）设$BD$为$\triangle ABC$中从$B$到$AC$的高。 则$60=$ $\triangle ABC$的面积$=\frac{1}{2}\cdot 24\cdot BD$，所以$BD=5$。因为$\triangle ABC$是等腰三角形，$\triangle ABD$与$\triangle CBD$是全等的直角三角形。这意味着$AD=DC=\frac{24}{2}=12$。对$\triangle ABD$应用勾股定理得 $$ AB^2=5^2+12^2=169=13^2, $$ 所以$AB=13$。
Q15
Let a, b and c be numbers with 0 < a < b < c. Which of the following is impossible?
设 a、b 和 c 是满足 0 < a < b < c 的数。以下哪个是不可能的？
Correct Answer: A
(A) Because $b<c$ and $0<a$, adding corresponding sides of the inequalities gives $b<a+c$, so (A) is impossible. To see that the other choices are possible, consider the following choices for $a$, $b$, and $c$: (B) and (C): $a=1$, $b=2$, and $c=4$; (D): $a=\dfrac{1}{3}$, $b=1$, and $c=2$; (E): $a=\dfrac{1}{2}$, $b=1$, and $c=2$.
（A）因为 $b<c$ 且 $0<a$，将不等式对应边相加可得 $b<a+c$，所以（A）不可能。为了说明其他选项是可能的，考虑下面这些 $a$、$b$、$c$ 的取值： （B）和（C）：$a=1$，$b=2$，$c=4$； （D）：$a=\dfrac{1}{3}$，$b=1$，$c=2$； （E）：$a=\dfrac{1}{2}$，$b=1$，$c=2$。
Q16
Amanda Reckonwith draws five circles with radii 1, 2, 3, 4 and 5. Then for each circle she plots the point (C, A), where C is its circumference and A is its area. Which of the following could be her graph?
Amanda Reckonwith 画了五个半径分别为 1、2、3、4 和 5 的圆。然后对于每个圆，她绘制点 (C, A)，其中 C 是其周长，A 是其面积。以下哪一个可能是她的图像？
Correct Answer: A
(A) The circumferences of circles with radii 1 through 5 are $2\pi$, $4\pi$, $6\pi$, $8\pi$ and $10\pi$, respectively. Their areas are, respectively, $\pi$, $4\pi$, $9\pi$, $16\pi$ and $25\pi$. The points $(2\pi,\pi)$, $(4\pi,4\pi)$, $(6\pi,9\pi)$, $(8\pi,16\pi)$ and $(10\pi,25\pi)$ are graphed in (A). It is the only graph of an increasing quadratic function, called a parabola.
(A) 半径为 1 到 5 的圆的周长分别是 $2\pi$、$4\pi$、$6\pi$、$8\pi$ 和 $10\pi$。它们的面积分别是 $\pi$、$4\pi$、$9\pi$、$16\pi$ 和 $25\pi$。点 $(2\pi,\pi)$、$(4\pi,4\pi)$、$(6\pi,9\pi)$、$(8\pi,16\pi)$ 和 $(10\pi,25\pi)$ 被绘制在 (A) 中。这是一个递增的二次函数的唯一图像，称为抛物线。
solution
Q17
A mixture of 30 liters of paint is 25% red tint, 30% yellow tint and 45% water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?
一种 30 升的油漆混合物含有 25% 红色颜料、30% 黄色颜料和 45% 水。在原混合物中加入 5 升黄色颜料。新混合物中黄色颜料的百分比是多少？
Correct Answer: C
(C) There are 0.30(30) = 9 liters of yellow tint in the original 30-liter mixture. After adding 5 liters of yellow tint, 14 of the 35 liters of the new mixture are yellow tint. The percent of yellow tint in the new mixture is $100\times\frac{14}{35}=100\times\frac{2}{5}$ or 40%.
(C) 原来 30 升的混合液中，黄色染料占 $0.30(30)=9$ 升。加入 5 升黄色染料后，新混合液共有 35 升，其中 14 升是黄色染料。新混合液中黄色染料所占百分比为 $100\times\frac{14}{35}=100\times\frac{2}{5}$，即 40%。
Q18
The product of the two 99-digit numbers 303,030,303, ..., 030,303 and 505,050,505, ..., 050,505 has thousands digit A and units digit B. What is the sum of A and B?
两个 99 位数 303,030,303, ..., 030,303 和 505,050,505, ..., 050,505 的乘积，千位数字为 A，个位数字为 B。A 和 B 的和是多少？
Correct Answer: D
(D) To find $A$ and $B$, it is sufficient to consider only $303 \cdot 505$, because 0 is in the thousands place in both factors. So $A = 3$ and $B = 5$, and the sum is $A + B = 3 + 5 = 8$.
（D）要找出 $A$ 和 $B$，只需考虑 $303 \cdot 505$，因为两个因数的千位上都是 0。 因此 $A = 3$、$B = 5$，所以它们的和为 $A + B = 3 + 5 = 8$。
solution
Q19
Pick two consecutive positive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
挑选两个连续的正整数，它们的和小于 100。将这两个整数平方，然后求平方差。以下哪一个可能是该差？
Correct Answer: C
(C) One of the squares of two consecutive integers is odd and the other is even, so their difference must be odd. This eliminates A, B and D. The largest consecutive integers that have a sum less than 100 are 49 and 50, whose squares are 2401 and 2500, with a difference of 99. Because the difference of the squares of consecutive positive integers increases as the integers increase, the difference cannot be 131. The difference between the squares of 40 and 39 is 79.
（C）两个相邻整数的平方，一个为奇数另一个为偶数，因此它们的差必为奇数。这排除了 A、B 和 D。和小于 100 的最大一对相邻整数是 49 和 50，它们的平方分别是 2401 和 2500，差为 99。由于相邻正整数平方的差会随着整数增大而增大，因此差不可能是 131。$40$ 与 $39$ 的平方之差是 79。
Q20
Before district play, the Unicorns had won 45% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?
在地区赛前，独角兽队篮球比赛胜率为 45%。地区赛中，他们赢了 6 场，输了 2 场，整个赛季胜率为一半。他们总共打了多少场比赛？
Correct Answer: A
(A) Because $45\%$ is the same as the simplified fraction $\frac{9}{20}$, the Unicorns won 9 games for each 20 games they played. This means that the Unicorns must have played some multiple of 20 games before district play. The table shows the possibilities that satisfy the conditions in the problem. Only when the Unicorns played 40 games before district play do they finish winning half of their games. So the Unicorns played $24+24=48$ games.
（A）因为$45\%$与最简分数$\frac{9}{20}$相同，独角兽队每打20场就赢9场。这意味着独角兽队在分区赛开始前所打的场次必须是20的某个倍数。下表给出了满足题目条件的可能情况。 只有当独角兽队在分区赛前打了40场时，他们最终才会赢下全部比赛的一半。因此独角兽队一共打了$24+24=48$场比赛。
solution
Q21
Two cards are dealt from a deck of four red cards labeled A, B, C, D and four green cards labeled A, B, C, D. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?
从一副牌中抽出两张牌，这副牌有四张标有A、B、C、D的红牌和四张标有A、B、C、D的绿牌。获胜对子是两张相同颜色或两张相同字母的牌。抽出获胜对子的概率是多少？
Correct Answer: D
(D) After the first card is dealt, there are seven left. The three cards with the same color as the initial card are winners and so is the card with the same letter but a different color. That means four of the remaining seven cards form winning pairs with the first card, so the probability of winning is $\frac{4}{7}$.
（D）发出第一张牌后，还剩下七张牌。与初始牌颜色相同的三张牌都算赢，另外那张字母相同但颜色不同的牌也算赢。这意味着剩下的七张牌中有四张与第一张牌组成获胜配对，因此获胜的概率为 $\frac{4}{7}$。
Q22
A lemming sits at a corner of a square with side length 10 meters. The lemming runs 6.2 meters along a diagonal toward the opposite corner. It stops, makes a 90° right turn and runs 2 more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
一只旅鼠坐在边长10米的正方形的角上。它沿着对角线向对角跑了6.2米。然后停下，向右转90°，再跑2米。科学家测量旅鼠到正方形每条边的最近距离。这些四个距离的平均值是多少米？
Correct Answer: C
(C) Wherever the lemming is inside the square, the sum of the distances to the two horizontal sides is 10 meters and the sum of the distances to the two vertical sides is 10 meters. Therefore the sum of all four distances is 20 meters, and the average of the four distances is $\frac{20}{4}=5$ meters.
（C）无论旅鼠在正方形内的什么位置，到两条水平边的距离之和是10米，到两条竖直边的距离之和也是10米。因此四个距离的总和是20米，这四个距离的平均值为 $\frac{20}{4}=5$ 米。
Q23
What is the area of the shaded pinwheel shown in the 5 × 5 grid?
5×5网格中所示的阴影风车图案的面积是多少？
stem
Correct Answer: B
(B) Find the area of the unshaded portion of the $5\times 5$ grid, then subtract the unshaded area from the total area of the grid. The unshaded triangle in the middle of the top of the $5\times 5$ grid has a base of $3$ and an altitude of $\frac{5}{2}$. The four unshaded triangles have a total area of $4\times \frac{1}{2}\times 3\times \frac{5}{2}=15$ square units. The four corner squares are also unshaded, so the shaded pinwheel has an area of $25-15-4=6$ square units.
（B）先求出 $5\times 5$ 方格中未阴影部分的面积，然后用方格的总面积减去未阴影面积。位于 $5\times 5$ 方格顶部中间的未阴影三角形，底为 $3$，高为 $\frac{5}{2}$。四个未阴影三角形的总面积为 $4\times \frac{1}{2}\times 3\times \frac{5}{2}=15$ 平方单位。四个角上的小正方形也未着阴影，因此阴影风车形的面积为 $25-15-4=6$ 平方单位。
Q24
A bag contains four pieces of paper, each labeled with one of the digits 1, 2, 3 or 4, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?
一个袋子里有四张纸，每张标有一个数字1、2、3或4，没有重复。从中不放回地抽三张纸，依次构造一个三位数。这个三位数是3的倍数的概率是多少？
Correct Answer: C
(C) A number is a multiple of three when the sum of its digits is a multiple of 3. If the number has three distinct digits drawn from the set $\{1,2,3,4\}$, then the sum of the digits will be a multiple of three when the digits are $\{1,2,3\}$ or $\{2,3,4\}$. That means the number formed is a multiple of three when, after the three draws, the number remaining in the bag is $1$ or $4$. The probability of this occurring is $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$.
（C）当一个数的各位数字之和是 $3$ 的倍数时，这个数就是 $3$ 的倍数。若该数由集合 $\{1,2,3,4\}$ 中抽取的三个互不相同的数字组成，则当数字为 $\{1,2,3\}$ 或 $\{2,3,4\}$ 时，数字和为 $3$ 的倍数。因此，当抽取三个数字后袋中剩下的数字是 $1$ 或 $4$ 时，所组成的数是 $3$ 的倍数。该事件发生的概率为 $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$。
Q25
On the dart board shown in the figure, the outer circle has radius 6 and the inner circle has radius 3. Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values in the regions. What is the probability that the score is odd?
图示飞镖盘外圆半径6，内圆半径3。三条半径将每个圆分成三个相等区域，显示分数。飞镖击中某区域的概率与该区域面积成正比。两支飞镖击中该盘，分数为两个区域分数的和。分数为奇数的概率是多少？
stem
Correct Answer: B
(B) The outer circle has area $36\pi$ and the inner circle has area $9\pi$, making the area of the outer ring $36\pi-9\pi=27\pi$. So each region in the outer ring has area $\frac{27\pi}{3}=9\pi$, and each region in the inner circle has area $\frac{9\pi}{3}=3\pi$. The probability of hitting a given region in the inner circle is $\frac{3\pi}{36\pi}=\frac{1}{12}$, and the probability of hitting a given region in the outer ring is $\frac{9\pi}{36\pi}=\frac{1}{4}$. For the score to be odd, one of the numbers must be 1 and the other number must be 2. The probability of hitting a 1 is \[ \frac{1}{4}+\frac{1}{4}+\frac{1}{12}=\frac{7}{12}, \] and the probability of hitting a 2 is \[ 1-\frac{7}{12}=\frac{5}{12}. \] Therefore, the probability of hitting a 1 and a 2 in either order is \[ \frac{7}{12}\cdot\frac{5}{12}+\frac{5}{12}\cdot\frac{7}{12}=\frac{70}{144}=\frac{35}{72}. \]
（B）外圆面积为 $36\pi$，内圆面积为 $9\pi$，因此外环面积为 $36\pi-9\pi=27\pi$。所以外环中每个区域的面积为 $\frac{27\pi}{3}=9\pi$，内圆中每个区域的面积为 $\frac{9\pi}{3}=3\pi$。击中内圆某一给定区域的概率为 $\frac{3\pi}{36\pi}=\frac{1}{12}$，击中外环某一给定区域的概率为 $\frac{9\pi}{36\pi}=\frac{1}{4}$。要使得得分为奇数，其中一个数必须是 1，另一个数必须是 2。击中 1 的概率是 \[ \frac{1}{4}+\frac{1}{4}+\frac{1}{12}=\frac{7}{12}, \] 击中 2 的概率是 \[ 1-\frac{7}{12}=\frac{5}{12}. \] 因此，以任意顺序击中一个 1 和一个 2 的概率为 \[ \frac{7}{12}\cdot\frac{5}{12}+\frac{5}{12}\cdot\frac{7}{12}=\frac{70}{144}=\frac{35}{72}. \]