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AMC8 2006

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AMC8 · 2006

Q1
Mindy made three purchases for \$1.98, \$5.04 and \$9.89. What was her total, to the nearest dollar?
Mindy 进行了三次购买，价格分别为 \$1.98、\$5.04 和 \$9.89。她的总花费，四舍五入到最接近的美元是多少？
Correct Answer: D
The three prices round to \$2, \$5, and \$10, which have a sum of $\boxed{\textbf{(D)}\ 17}$.
这三个价格四舍五入分别为 \$2、 \$5 和 \$10，总和为 $\boxed{\textbf{(D)}\ 17}$。
Q2
On the AMC 8 contest Billy answers 13 questions correctly, answers 7 questions incorrectly and doesn't answer the last 5. What is his score?
在 AMC 8 竞赛中，Billy 正确回答了 13 道题，错误回答了 7 道题，没有回答最后 5 道题。他的得分是多少？
Correct Answer: C
(C) On the AMC 8 a student's score is the number of problems answered correctly. So Billy's score is 13. Because there is no penalty for guessing, if he wants to increase his score, he probably should fill in the last five answers.
（C）在 AMC 8 中，学生的得分等于答对题目的数量。因此，比利的得分是 13。由于猜测不扣分，如果他想提高分数，他很可能应该把最后五题的答案都填上。
Q3
Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?
Elisa 在泳池中游泳。她刚开始时，25 分钟完成 10 圈。现在她 24 分钟完成 12 圈。她提高了多少分钟的单圈时间？
Correct Answer: A
(A) When Elisa started, she completed a lap in $\frac{25}{10}=2.5$ minutes. Now she can complete a lap in $\frac{24}{12}=2$ minutes. She has improved her lap time by $2.5-2=0.5$ or $\frac{1}{2}$ minute.
（A）当 Elisa 刚开始时，她跑完一圈用时 $\frac{25}{10}=2.5$ 分钟。现在她跑完一圈用时 $\frac{24}{12}=2$ 分钟。她的单圈用时提高了 $2.5-2=0.5$ 分钟，即 $\frac{1}{2}$ 分钟。
Q4
Initially, a spinner points west. Chenille moves it clockwise $2 \frac{1}{4}$ revolutions and then counterclockwise $3 \frac{1}{4}$ revolutions. In what direction does the spinner point after the two moves?
最初，转盘指向西。Chenille 将其顺时针转动 $2 \frac{1}{4}$ 圈，然后逆时针转动 $3 \frac{1}{4}$ 圈。转盘最终指向哪个方向？
stem
Correct Answer: B
(B) Ignore the number of complete revolutions because they do not affect direction. One-fourth of the distance around the circle clockwise from west is north. Three-fourths of the distance counterclockwise around the circle from north is east. Chenille’s spinner points east.
（B）忽略完整转了多少圈，因为这不会影响方向。从西向顺时针沿圆周走四分之一圈会到达北。从北向逆时针沿圆周走四分之三圈会到达东。Chenille 的转盘指向东。
Q5
Points A, B, C and D are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
点 A、B、C 和 D 是大正方形边的中点。如果大正方形面积为 60，小正方形的面积是多少？
stem
Correct Answer: D
(D) Divide the larger square into 8 congruent triangles, as shown, 4 of which make up the smaller square. The area of the smaller square is $\frac{4}{8}$ or $\frac{1}{2}$ of the area of the larger square, so the area of the smaller square is equal to 30.
（D）如图所示，将较大的正方形分成 8 个全等三角形，其中 4 个组成较小的正方形。 较小正方形的面积是较大正方形面积的 $\frac{4}{8}$，即 $\frac{1}{2}$，因此较小正方形的面积等于 30。
Q6
The letter T is formed by placing two 2 inch × 4 inch rectangles next to each other, as shown. What is the perimeter of the T, in inches?
字母 T 是由两个 2 英寸 × 4 英寸的矩形并排放置形成的，如图所示。T 的周长是多少英寸？
stem
Correct Answer: C
If the two rectangles were seperate, the perimeter would be $2(2(2+4)=24$. It easy to see that their connection erases 2 from each of the rectangles, so the final perimeter is $24-2 \times 2 = \boxed{\textbf{(C)}\ 20}$.
如果两个矩形是分开的，周长将是 $2(2(2+4)=24$。很容易看出它们的连接消除了每个矩形的 2，所以最终周长是 $24-2 \times 2 = \boxed{\textbf{(C)}\ 20}$。
solution
Q7
Circle X has a radius of $\pi$. Circle Y has a circumference of $8\pi$. Circle Z has an area of $9\pi$. List the circles in order from smallest to largest radius.
圆 X 的半径是 $\pi$。圆 Y 的周长是 $8\pi$。圆 Z 的面积是 $9\pi$。将这些圆按半径从小到大排序。
Correct Answer: B
(B) Because circumference $C = 2\pi r$ and circle $Y$ has circumference $8\pi$, its radius is $\frac{8\pi}{2\pi} = 4$. Because area $A = \pi r^2$ and circle $Z$ has area $9\pi$, its radius is $\sqrt{9} = 3$. Ordering the radii gives $3 < \pi < 4$, so the circles in ascending order of radii length are $Z$, $X$ and $Y$.
（B）因为周长 $C = 2\pi r$，且圆 $Y$ 的周长为 $8\pi$，所以它的半径是 $\frac{8\pi}{2\pi} = 4$。因为面积 $A = \pi r^2$，且圆 $Z$ 的面积为 $9\pi$，所以它的半径是 $\sqrt{9} = 3$。将半径大小排序得到 $3 < \pi < 4$，因此按半径从小到大排列的圆依次是 $Z$、$X$、$Y$。
Q8
The table shows some of the results of a survey by radio station KAMC. What percentage of the males surveyed listen to the station?
该表格显示了电台 KAMC 的一项调查的部分结果。被调查的男性中有多少百分比听这个电台？
stem
Correct Answer: E
(E) Because $200-96=104$ of those surveyed were male, $104-26=78$ of those surveyed are male listeners. \[ \begin{array}{c|c|c|c} & \text{Listen} & \text{Don't Listen} & \text{Total}\\ \hline \text{Male} & 78 & 26 & 104\\ \text{Female} & 58 & 38 & 96\\ \hline \text{Total} & 136 & 64 & 200 \end{array} \] The percentage of males surveyed who listen to KAMC is $\frac{78}{104}\times100\%=75\%$.
（E）因为被调查的$200-96=104$人是男性，且$104-26=78$人是收听者，所以被调查的男性收听者有$78$人。 \[ \begin{array}{c|c|c|c} & \text{收听} & \text{不收听} & \text{总计}\\ \hline \text{男性} & 78 & 26 & 104\\ \text{女性} & 58 & 38 & 96\\ \hline \text{总计} & 136 & 64 & 200 \end{array} \] 被调查男性中收听 KAMC 的百分比为$\frac{78}{104}\times100\%=75\%$。
Q9
What is the product of $\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{2006}{2005}$?
计算 $\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \dots \times \frac{2006}{2005}$ 的积？
Correct Answer: C
(C) Note that in each fraction, the numerator is the same as the denominator in the next fraction, so they divide. The product of $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}=\frac{2006}{2}=1003.$
（C）注意在每个分数中，分子与下一个分数的分母相同，因此可以相互约去。乘积 $\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdots\times\frac{2006}{2005}=\frac{2006}{2}=1003.$
Q10
Jorge's teacher asks him to plot all the ordered pairs $(w, l)$ of positive integers for which $w$ is the width and $l$ is the length of a rectangle with area 12. What should his graph look like?
Jorge 的老师让他绘制所有正整数有序对 $(w, l)$，其中 $w$ 是矩形的宽度，$l$ 是长度，面积为 12。他的图形应该是什么样的？
Correct Answer: A
(A) When the area of a rectangle is 12 square units and the sides are integers, the factors of 12 are the possible lengths of the sides. In point form, the side lengths could be (1, 12), (2, 6), (3, 4), (4, 3), (6, 2) and (12, 1). Only graph A fits these points.
（A）当一个长方形的面积为 12 平方单位且边长为整数时，12 的因数就是边长可能的取值。用点的形式表示，边长可以是 (1, 12)、(2, 6)、(3, 4)、(4, 3)、(6, 2) 和 (12, 1)。只有图 A 符合这些点。
Q11
How many two-digit numbers have digits whose sum is a perfect square?
有多少个两位数的各位数字之和是完全平方数？
Correct Answer: C
(C) The sum of the digits of a two-digit number is at most 9 + 9 = 18. This means the only possible perfect square sums are 1, 4, 9 and 16. Each square has the following two-digit possibilities: 1 : 10 4 : 40, 31, 22, 13 9 : 90, 81, 72, 63, 54, 45, 36, 27, 18 16 : 97, 88, 79 There are 17 two-digit numbers in all.
(C) 两位数的各位数字之和最大为 $9+9=18$。因此，可能的完全平方和只有 $1,4,9,16$。每个平方对应的两位数情况如下： 1：10 4：40，31，22，13 9：90，81，72，63，54，45，36，27，18 16：97，88，79 总共有 17 个两位数。
Q12
Antonette gets 70% on a 10-problem test, 80% on a 20-problem test and 90% on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?
Antonette 在一份10题的测试中得70%，在20题测试中得80%，在30题测试中得90%。如果将三份测试合并成一份60题的测试，她的总分百分比最接近于多少？
Correct Answer: D
(D) Note that 70% of 10 is 7, 80% of 20 is 16 and 90% of 30 is 27. Antonette answers 7 + 16 + 27 = 50 problems correctly out of 60 problems in all. Her overall score is $\frac{50}{60}$ or 83.$\bar{3}$%.
（D）注意：10 的 70% 是 7，20 的 80% 是 16，30 的 90% 是 27。Antonette 在总共 60 题中答对了 $7+16+27=50$ 题。她的总得分是 $\frac{50}{60}$，即 83.$\bar{3}$%。
Q13
Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike. He bikes at a uniform rate of 16 miles per hour. They both bike on the same 62-mile route between Escanaba and Marquette. At what time in the morning do they meet?
Cassie 在上午8:30 从 Escanaba 出发骑车前往 Marquette，匀速12英里/小时。Brian 在上午9:00 从 Marquette 出发骑车前往 Escanaba，匀速16英里/小时。他们在 Escanaba 和 Marquette 之间相同的62英里路线上骑行。他们上午几点相遇？
Correct Answer: D
(D) Between 8:30 and 9:00 AM Cassie travels 6 miles. At 9:00 Cassie and Brian are only 56 miles apart. After 9:00, because they are both biking towards each other, the distance between them decreases at the rate of $12+16=28$ miles per hour. At that rate, it will take them $\frac{56}{28}=2$ hours to meet. So they will meet at 11:00 AM.
（D）在上午 8:30 到 9:00 之间，Cassie 行驶了 6 英里。到 9:00 时，Cassie 和 Brian 相距只有 56 英里。9:00 之后，因为他们都朝着彼此骑行，两人之间的距离以每小时 $12+16=28$ 英里的速度减少。按这个速度，他们需要 $\frac{56}{28}=2$ 小时才能相遇。所以他们会在上午 11:00 相遇。
Q14
If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?
如果 Bob 和 Chandra 都读完整本书，Bob 比 Chandra 多花多少秒阅读？
Correct Answer: B
(B) Bob takes $45-30=15$ more seconds per page than Chandra. So the difference in their total reading times is $760\cdot 15=11{,}400$ seconds. Bob will spend $11{,}400$ more seconds reading than Chandra.
（B）Bob 每页比 Chandra 多用 $45-30=15$ 秒。因此他们总阅读时间的差为 $760\cdot 15=11{,}400$ 秒。Bob 的阅读时间将比 Chandra 多 $11{,}400$ 秒。
Q15
Chandra and Bob, who each have a copy of the book, decide that they can save time by "team reading" the novel. In this scheme, Chandra will read from page 1 to a certain page and Bob will read from the next page through page 760, finishing the book. When they are through they will tell each other about the part they read. What is the last page that Chandra should read so that she and Bob spend the same amount of time reading the novel?
Chandra 和 Bob 各有一本相同的书，他们决定通过“团队阅读”来节省时间。在这种方案中，Chandra 从第1页读到某一页，Bob 从下一页读到第760页，完成整书。完成后他们互相讲述自己读的部分。为了让他们阅读小说所花时间相同，Chandra 应该读到最后一页是第几页？
Correct Answer: C
(C) The ratio of time it takes Bob to read a page to the time it takes Chandra to read a page is $45\!:\!30$ or $3\!:\!2$, so Bob should read $\frac{2}{3}$ of the number of pages that Chandra reads. Divide the book into $5$ parts, each with $\frac{760}{5}=152$ pages. Chandra will read the first $3\cdot152=456$ pages, while Bob reads the last $2\cdot152=304$ pages.
（C）鲍勃读一页所需时间与钱德拉读一页所需时间之比为 $45\!:\!30$，即 $3\!:\!2$，因此鲍勃应读钱德拉阅读页数的 $\frac{2}{3}$。把全书分成 $5$ 份，每份有 $\frac{760}{5}=152$ 页。钱德拉读前 $3\cdot152=456$ 页，而鲍勃读后 $2\cdot152=304$ 页。
Q16
Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?
在钱德拉和鲍勃开始阅读之前，爱丽丝说她想和他们一起组队阅读。如果他们将书分成三部分，使得每个人阅读的时间相同，那么每个人需要阅读多少秒？
Correct Answer: E
(E) The least common multiple of 20, 45 and 30 is $2^2 \cdot 3^2 \cdot 5 = 180$. Using the LCM, in 180 seconds Alice reads $\frac{180}{20} = 9$ pages, Chandra reads $\frac{180}{30} = 6$ pages and Bob reads $\frac{180}{45} = 4$ pages. Together they read a total of 19 pages in 180 seconds. The total number of seconds each reads is $\frac{760}{19} \cdot 180 = 7200$.
（E）20、45 和 30 的最小公倍数是 $2^2 \cdot 3^2 \cdot 5 = 180$。利用最小公倍数，在 180 秒内，Alice 读了 $\frac{180}{20} = 9$ 页，Chandra 读了 $\frac{180}{30} = 6$ 页，Bob 读了 $\frac{180}{45} = 4$ 页。他们一共在 180 秒内读了 19 页。则每个人读书的总秒数为 $\frac{760}{19} \cdot 180 = 7200$。
Q17
Jeff rotates spinners P, Q and R and adds the resulting numbers. What is the probability that his sum is an odd number?
杰夫旋转转盘 P、Q 和 R，并将结果数字相加。他的总和是奇数的概率是多少？
stem
Correct Answer: B
(B) Because the sum of a number from spinner Q and a number from spinner R is always odd, the sum of the numbers on the three spinners will be odd exactly when the number from spinner P is even. Because 2 is the only even number on spinner P, the probability of getting an odd sum is $\frac{1}{3}$.
（B）因为从转盘 Q 取到的数与从转盘 R 取到的数之和总是奇数，所以三个转盘上的数之和为奇数，当且仅当从转盘 P 取到的数是偶数。由于转盘 P 上唯一的偶数是 2，因此得到奇数和的概率是 $\frac{1}{3}$。
Q18
A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?
一个边长为 3 英寸的立方体由 27 个边长为 1 英寸的小立方体构成。其中 19 个小立方体是白色的，8 个是黑色的。如果 8 个黑色小立方体放置在大立方体的角上，那么大立方体表面积中白色部分占的几分之几？
Correct Answer: D
(D) Four black and five white squares are visible on each of the six faces of the cube. So $\frac{5}{9}$ of the surface will be white.
(D) 立方体的六个面上每个面都能看到四个黑色方格和五个白色方格。因此，表面积中有 $\frac{5}{9}$ 将是白色。
solution
Q19
Triangle ABC is an isosceles triangle with AB = BC. Point D is the midpoint of both BC and AE, and CE is 11 units long. Triangle ABD is congruent to triangle ECD. What is the length of BD?
三角形 ABC 是等腰三角形，AB = BC。点 D 是 BC 和 AE 的中点，CE 长 11 个单位。三角形 ABD 与三角形 ECD 全等。BD 的长度是多少？
stem
Correct Answer: D
(D) Because triangles $ABD$ and $ECD$ are congruent, and triangle $ABC$ is isosceles, $EC = AB = BC = 11$. That means $BD = \frac{11}{2}$ or $5.5$.
（D）因为三角形 $ABD$ 与 $ECD$ 全等，且三角形 $ABC$ 是等腰三角形，所以 $EC = AB = BC = 11$。这意味着 $BD = \frac{11}{2}$，即 $5.5$。
Q20
A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?
一个单打锦标赛有六名选手。每位选手只与其他选手各对战一次，没有平局。如果海伦赢了 4 场比赛，伊内斯赢了 3 场，珍妮特赢了 2 场，肯德拉赢了 2 场，拉拉赢了 2 场，那么莫妮卡赢了多少场比赛？
Correct Answer: C
(C) Each of the six players played 5 games, and each game involved two players. So there were $\frac{6\cdot 5}{2}=15$ games. Helen, Ines, Janet, Kendra and Lara won a total of $4+3+2+2+2=13$ games, so Monica won $15-13=2$ games.
（C）六名选手每人打了5场比赛，每场比赛有两名选手参与。因此总共有 $\frac{6\cdot 5}{2}=15$ 场比赛。Helen、Ines、Janet、Kendra 和 Lara 一共赢了 $4+3+2+2+2=13$ 场，所以 Monica 赢了 $15-13=2$ 场。
Q21
An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. The aquarium is filled with water to a depth of 37 cm. A rock with volume 1000 cm³ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
一个水族箱有一个长方形底面，长100厘米，宽40厘米，高50厘米。水族箱中注水至37厘米深。然后放入一块体积为1000立方厘米的石头，完全浸没。问水位上升多少厘米？
Correct Answer: A
(A) Using the volume formula $lwh = V$, the volume of water in the aquarium is $100 \times 40 \times 37 = 148,000\ \text{cm}^3$. When the rock is put in, the water and the rock will occupy a box-shaped region with volume $148,000 + 1000 = 149,000\ \text{cm}^3$. The volume of the water and the rock is $100 \times 40 \times h$, where $h$ is the new height of the water. The new volume $= 4000h = 149,000\ \text{cm}^3$, so the new height is $$ h=\frac{149000}{4000}=37.25\ \text{cm}. $$ After adding the rock, the water rises $37.25 - 37 = 0.25\ \text{cm}$.
（A）使用体积公式 $lwh = V$，鱼缸中水的体积为 $100 \times 40 \times 37 = 148,000\ \text{cm}^3$。当把石头放入后，水和石头将占据一个长方体区域，其体积为 $148,000 + 1000 = 149,000\ \text{cm}^3$。水和石头的总体积为 $100 \times 40 \times h$，其中 $h$ 是水面新的高度。新的体积满足 $4000h = 149,000\ \text{cm}^3$，所以新的高度为 $$ h=\frac{149000}{4000}=37.25\ \text{cm}. $$ 加入石头后，水面上升 $37.25 - 37 = 0.25\ \text{cm}$。
Q22
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
在底部一排单元格中放置三个不同的个位正整数。相邻单元格中的数字相加，将和放在它们上面的单元格中。在第二行，继续同样的过程，得到顶部的单元格中的数字。顶部单元格中可能的最大数和最小数的差是多少？
stem
Correct Answer: D
(D) If the lower cells contain $A$, $B$ and $C$, then the second row will contain $A+B$ and $B+C$, and the top cell will contain $A+2B+C$. To obtain the smallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top number will be $7$. For the largest sum, place $9$ in the center cell and $7$ and $8$ in the outer ones. This top number will be $33$. The difference is $33-7=26$.
（D）如果底部的三个格子分别为 $A$、$B$ 和 $C$，那么第二行将为 $A+B$ 和 $B+C$，顶部格子将为 $A+2B+C$。要得到最小和，把 $1$ 放在中间格，把 $2$ 和 $3$ 放在两侧格。顶部数字将是 $7$。要得到最大和，把 $9$ 放在中间格，把 $7$ 和 $8$ 放在两侧格。顶部数字将是 $33$。差为 $33-7=26$。
Q23
A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?
一个盒子中有金币。如果金币平均分给六个人，分余4枚。如果平均分给五个人，分余3枚。盒子中满足这两个条件的最小金币数是多少？平均分给七个人时，余多少枚？
Correct Answer: A
(A) The counting numbers that leave a remainder of 4 when divided by 6 are 4, 10, 16, 22, 28, 34, .... The counting numbers that leave a remainder of 3 when divided by 5 are 3, 8, 13, 18, 23, 28, 33, .... So 28 is the smallest possible number of coins that meets both conditions. Because $4\times 7=28$, there are no coins left when they are divided among seven people.
（A）把一个数除以 6 余 4 的正整数有 4，10，16，22，28，34，……；把一个数除以 5 余 3 的正整数有 3，8，13，18，23，28，33，……。因此，满足两个条件的最小硬币数是 28。因为 $4\times 7=28$，把这些硬币分给 7 个人时不会剩下硬币。
Q24
In the multiplication problem below, A, B, C and D are different digits. What is A + B?
在下面的乘法问题中，A、B、C和D是不同的数字。A + B是多少？
stem
Correct Answer: A
(A) We can decompose $CDCD$ into $CD \times 100 + CD = CD(101)$. That means that $A = 1$ and $B = 0$. The sum is $1 + 0 = 1$.
（A）我们可以将$CDCD$分解为$CD \times 100 + CD = CD(101)$。这意味着$A = 1$且$B = 0$。其和为$1 + 0 = 1$。
Q25
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
Barry在3张卡片的每面写了一个不同的数字，并如图所示将卡片放在桌上。三张卡片上两面的数字之和相等。隐藏面的三个数字是素数。隐藏素数的平均数是多少？
stem
Correct Answer: B
(B) There are one odd and two even numbers showing. Because all primes other than 2 are odd and the sum of an even number and an odd number is odd, the common sum must be odd. That means 2 must be opposite 59 and the common sum is $2+59=61$. The other two hidden numbers are $61-44=17$ and $61-38=23$. The average of 2, 17 and 23 is $\frac{2+17+23}{3}=\frac{42}{3}=14$.
(B) 现在显示的是一个奇数和两个偶数。因为除了 2 以外的所有质数都是奇数，而偶数与奇数之和为奇数，所以公共和必定是奇数。这意味着 2 必须与 59 相对，公共和为 $2+59=61$。另外两个隐藏的数字是 $61-44=17$ 和 $61-38=23$。2、17 和 23 的平均数是 $\frac{2+17+23}{3}=\frac{42}{3}=14$。