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AMC8 2016

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AMC8 · 2016

Q1
The longest professional tennis match ever played lasted a total of 11 hours and 5 minutes. How many minutes was this?
历史上最长的职业网球比赛总共持续了11小时5分钟。这是多少分钟？
stem
Correct Answer: C
There are 60 minutes in 1 hour, so 11 hours plus 5 minutes is equal to $11\cdot 60 + 5 = 665$ minutes.
1 小时有 60 分钟，因此 11 小时加 5 分钟等于 $11\cdot 60 + 5 = 665$ 分钟。
Q2
In rectangle ABCD, AB = 6 and AD = 8. Point M is the midpoint of AD. What is the area of $\triangle AMC$?
在矩形ABCD中，AB = 6，AD = 8。点M是AD的中点。$ riangle AMC$的面积是多少？
Correct Answer: A
The area of triangle ACD is $\tfrac{1}{2}\cdot 8\cdot 6=24$. The area of triangle MCD is $\tfrac{1}{2}\cdot 4\cdot 6=12$. So the area of triangle AMC is $24-12=12$.
三角形 ACD 的面积为 $\tfrac{1}{2}\cdot 8\cdot 6=24$。三角形 MCD 的面积为 $\tfrac{1}{2}\cdot 4\cdot 6=12$。因此三角形 AMC 的面积是 $24-12=12$。
Q3
Four students take an exam. Three of their scores are 70, 80, and 90. If the average of their four scores is 70, then what is the remaining score?
四个学生参加考试。其中三个分数是70、80和90。如果四个分数的平均分是70，那么剩下的分数是多少？
Correct Answer: A
The given scores 70, 80, and 90 are a total of 30 above the stated average. Thus the remaining score is 30 points below the average, and $70-30=40$.
给出的分数 70、80 和 90 相对于所给平均分总计高出 30 分。因此剩下那次分数应比平均分低 30 分，于是 $70-30=40$。
Q4
When Cheenu was a boy he could run 15 miles in 3 hours and 30 minutes. As an old man he can now walk 10 miles in 4 hours. How many minutes longer does it take for him to walk a mile now compared to when he was a boy?
Cheenu小时候能在3小时30分钟内跑15英里。现在作为老人，他能在4小时内走10英里。现在走一英里比小时候多花多少分钟？
Correct Answer: B
As a boy it took Cheenu 3 hours and 30 minutes, which is 210 minutes, to go 15 miles. That is a rate of $210 \div 15 = 14$ minutes per mile. As an old man it takes him 4 hours, or 240 minutes, to travel 10 miles. That is a rate of $240 \div 10 = 24$ minutes per mile. It takes him $24 - 14 = 10$ minutes more to walk a mile as an old man.
作为男孩，Cheenu 走15英里用了3小时30分钟，即210分钟。速度是每英里 $210 \div 15 = 14$ 分钟。作为老年人，他走10英里用了4小时，即240分钟。速度是每英里 $240 \div 10 = 24$ 分钟。作为老年人，走一英里比从前多用 $24 - 14 = 10$ 分钟。
Q5
The number $N$ is a two-digit number. -- When $N$ is divided by 9, the remainder is 1. -- When $N$ is divided by 10, the remainder is 3. What is the remainder when $N$ is divided by 11?
数$N$是一个两位数。 -- 当$N$除以9时，余数是1。 -- 当$N$除以10时，余数是3。 $N$除以11的余数是多少？
Correct Answer: E
The two-digit numbers that leave a remainder of 3 when divided by 10 are: 13, 23, 33, 43, 53, 63, 73, 83, 93. The two-digit numbers that leave a remainder of 1 when divided 9 are: 10, 19, 28, 37, 46, 55, 64, 73, 82, 91. Among these two sets, 73 is the only common number. When 73 is divided by 11 the remainder is 7.
被 10 除余 3 的两位数有：13、23、33、43、53、63、73、83、93。被 9 除余 1 的两位数有：10、19、28、37、46、55、64、73、82、91。在这两组数中，唯一的公共数是 73。将 73 除以 11，余数为 7。
Q6
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names? *(Bar graph with horizontal axis labeled name length: 3,4,5,6,7 and vertical axis frequency; frequencies: 7 at 3, 3 at 4, 1 at 5, 4 at 6, 4 at 7)*
下图柱状图表示19个人的名字长度（以字母数计）。这些名字的长度中位数是多少？*(柱状图横轴标签为名字长度：3,4,5,6,7，纵轴为频数；频数：3处为7，4处为3，5处为1，6处为4，7处为4)*
stem
Correct Answer: B
Answer (B): The 19 name lengths are 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 6, 6, 7, 7, 7, 7. The tenth value, 4, is the median.
答案 (B)：19 个名字长度分别为 3、3、3、3、3、3、3、4、4、4、5、6、6、6、6、7、7、7、7。第十个数 4 是中位数。
Q7
Which of the following numbers is not a perfect square?
下列哪个数不是完全平方数？
Correct Answer: B
The numbers $1^{2016}$, $3^{2018}$, $5^{2020}$ have even exponents and hence are squares. The number $2^{2017}$ is not a perfect square because it is twice a square $2(2^{1008})^2$. Since $4^{2019} = (2^{2})^{2019} = 2^{4038}$, it is also a perfect square.
数 $1^{2016}$、$3^{2018}$ 和 $5^{2020}$ 的指数为偶数，因此它们都是完全平方数。数 $2^{2017}$ 不是完全平方数，因为它等于一个平方数的两倍：$2(2^{1008})^2$。由于 $4^{2019}=(2^{2})^{2019}=2^{4038}$，它也是一个完全平方数。
Q8
Find the value of the expression $100 - 98 + 96 - 94 + 92 - 90 + \cdots + 8 - 6 + 4 - 2$.
求表达式的值：$100 - 98 + 96 - 94 + 92 - 90 + \cdots + 8 - 6 + 4 - 2$。
Correct Answer: C
Evaluate the expression by grouping as follows: $(100-98)+(96-94)+\cdots+(8-6)+(4-2)=2+2+\cdots+2+2=2\cdot25=50$.
按如下方式分组计算该表达式：$(100-98)+(96-94)+\cdots+(8-6)+(4-2)=2+2+\cdots+2+2=2\cdot25=50$。
Q9
What is the sum of the distinct prime integer divisors of 2016?
2016的互异素整数因数的和是多少？
Correct Answer: B
The prime factorization of 2016 is $2016=(2^5)(3^2)(7)$, so the distinct prime divisors of 2016 are $2,3,$ and $7$, and their sum is $2+3+7=12$.
2016 的素因数分解为：$2016=(2^5)(3^2)(7)$，因此 2016 的不同素因数是 $2,3,7$，它们的和为 $2+3+7=12$。
Q10
Suppose that $a * b$ means $3a - b$. What is the value of $x$ if $2*(5*x) = 1$?
假设$a * b$表示$3a - b$。若$2*(5*x) = 1$，则$x$的值是多少？
Correct Answer: D
Since $2 \ast (5 \ast x) = 1$, it follows that $6-(5 \ast x) = 1$, and so $5 \ast x = 5$. Applying the formula again, $15-x = 5$, and therefore $x = 10$.
由于 $2 \ast (5 \ast x) = 1$，于是 $6-(5 \ast x) = 1$，从而 $5 \ast x = 5$。再次应用该公式，$15-x = 5$，因此 $x = 10$。
Q11
Determine how many two-digit numbers satisfy the following property: When the number is added to the number obtained by reversing its digits, the sum is 132.
确定有多少个两位数满足以下性质：将该数与其数字反转后得到的数相加，和为132。
Correct Answer: B
Let $ab$ be the two-digit number. Then $132=(10a+b)+(10b+a)=11(a+b)$. Thus $a+b=12$. The possible numbers are: $39, 93, 48, 84, 57, 75,$ and $66$. There are seven two-digit numbers that meet this criterion.
设两位数为 $ab$。则 $132=(10a+b)+(10b+a)=11(a+b)$，因此 $a+b=12$。可能的数有：$39, 93, 48, 84, 57, 75, 66$。共有七个符合条件的两位数。
Q12
Jefferson Middle School has the same number of boys and girls. Three-fourths of the girls and two-thirds of the boys went on a field trip. What fraction of the students on the field trip were girls?
杰斐逊中学男孩和女孩数量相同。\frac{3}{4}的女孩和\frac{2}{3}的男孩参加了实地考察旅行。实地考察旅行中的学生有女孩占几分之几？
Correct Answer: B
Converting the given fractions to the same denominator, we see that $9/12$ of the girls and $8/12$ of the boys went on the trip. So the ratio of the number of girls to the number of boys was $9:8$, and it follows that $9/17$ of the students on the trip were girls.
将给定分数通分后，有 $9/12$ 的女生和 $8/12$ 的男生参加了旅行。因此女生人数与男生人数的比是 $9:8$，从而参加旅行的学生中有 $9/17$ 是女生。
Q13
Two different numbers are randomly selected from the set $\{-2, -1, 0, 3, 4, 5\}$ and multiplied together. What is the probability that the product is 0?
从集合$\{-2, -1, 0, 3, 4, 5\}$中随机选取两个不同的数相乘。乘积为0的概率是多少？
Correct Answer: D
There are $6\cdot 5=30$ possible pairs of numbers. For a product to be $0$, either the first factor or the second factor must be $0$, so there are $1\cdot 5+5\cdot 1=10$ such products. The desired probability is $\frac{10}{30}=\frac{1}{3}$.
共有 $6\cdot 5=30$ 种可能的数对。要使乘积为 $0$，要么第一个因数为 $0$，要么第二个因数为 $0$，因此共有 $1\cdot 5+5\cdot 1=10$ 种这样的乘积。所求概率为 $\frac{10}{30}=\frac{1}{3}$。
Q14
Karl's car uses a gallon of gas every 35 miles, and his gas tank holds 14 gallons when it is full. One day Karl started with a full tank of gas, drove 350 miles, bought 8 gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day?
Karl的小汽车每35英里耗油1加仑，其油箱满时可装14加仑。有一天Karl从满油箱开始，开车350英里，买了8加仑油，继续开车到目的地。到达时油箱半满。那天Karl总共开了多少英里？
Correct Answer: A
In driving 350 miles, Karl used $350/35=10$ gallons of gas, so he had $14-10=4$ gallons left in his tank. After buying 8 more gallons, he had $4+8=12$ gallons. When he arrived at his destination, he had $14/2=7$ gallons left, so he used an additional $12-7=5$ gallons. This let him drive an additional $5\cdot 35=175$ miles, so he drove a total of $350+175=525$ miles.
行驶了 350 英里，Karl 用了 $350/35=10$ 加仑汽油，因此油箱还剩 $14-10=4$ 加仑。再加了 8 加仑后，他有 $4+8=12$ 加仑。到达目的地时，他还剩 $14/2=7$ 加仑，所以又用了 $12-7=5$ 加仑。这使他又行驶了 $5\cdot 35=175$ 英里，因此总共行驶了 $350+175=525$ 英里。
Q15
What is the largest power of 2 that is a divisor of $13^4 - 11^4$?
$13^4 - 11^4$的最大2的幂次除数是多少？
Correct Answer: C
First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$. Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$. Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{\textbf{(C)}\ 32}$.
首先，使用平方差公式$13^4 - 11^4 = (13^2)^2 - (11^2)^2 = (13^2 + 11^2)(13^2 - 11^2)$。再次使用平方差并化简，得$(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$。右侧不含2的因子，因此$13^4 - 11^4$的最大2的幂次除数为$\boxed{\textbf{(C)}\ 32}$。
Q16
Annie and Bonnie are running laps around a 400-meter oval track. They started together, but Annie has pulled ahead, because she runs 25% faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?
Annie 和 Bonnie 在一个 400 米椭圆跑道上跑圈。她们一起起跑，但 Annie 已经领先，因为她的速度比 Bonnie 快 25%。当她第一次超过 Bonnie 时，Annie 会跑完多少圈？
stem
Correct Answer: D
Let N be the number of laps run by Annie when she passes Bonnie for the first time. The number of laps run by Bonnie is N-1. Then $\frac{N}{N-1}=1.25=\frac{5}{4}$. So $N=5$.
设 $N$ 为安妮第一次追上邦妮时安妮跑过的圈数。此时邦妮跑过的圈数为 $N-1$。于是有 $\frac{N}{N-1}=1.25=\frac{5}{4}$，所以 $N=5$。
Q17
An ATM password at Fred's Bank is composed of four digits from 0 to 9, with repeated digits allowable. If no password may begin with the sequence 9, 1, 1, then how many passwords are possible?
Fred 银行的 ATM 密码由 0 到 9 的四位数字组成，允许重复数字。如果密码不得以序列 9,1,1 开头，则可能有多少个密码？
Correct Answer: D
If there were no restrictions, then $10^4$ passwords would be possible. Among these, 10 passwords begin with 9 1 1, and have 10 options for the fourth digit. Thus $10^4 - 10 = 9990$ passwords satisfy the condition.
若没有任何限制，则可能的密码有 $10^4$ 个。在这些密码中，有 10 个以 9 1 1 开头，并且第四位有 10 种选择。因此满足条件的密码共有 $10^4 - 10 = 9990$ 个。
Q18
In an All-Area track meet, 216 sprinters enter a 100-meter dash competition. The track has 6 lanes, so only 6 sprinters can compete at a time. At the end of each race the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
在全区田径比赛中，有 216 名短跑运动员参加 100 米短跑比赛。跑道有 6 条道，因此每次只能有 6 名运动员比赛。每场比赛结束后，五名非获胜者被淘汰，获胜者将在稍后的比赛中再次参赛。需要多少场比赛才能确定冠军短跑运动员？
Correct Answer: C
Divide the 216 sprinters into 36 groups of 6. Run 36 races to eliminate 180 sprinters, leaving 36 winners. Divide the 36 winners into 6 groups of 6, run 6 races to eliminate 30 sprinters, leaving 6 winners. Finally run the last race to determine the champion. The number of races run is $36+6+1 = 43$.
将216名短跑选手分成36组，每组6人。进行36场比赛淘汰180名选手，剩下36名获胜者。把这36名获胜者再分成6组，每组6人，再进行6场比赛淘汰30名选手，剩下6名获胜者。最后再进行一场决赛以确定冠军。比赛总场数为 $36+6+1 = 43$。
Q19
The sum of 25 consecutive even integers is 10,000. What is the largest of these 25 consecutive even integers?
25 个连续偶数的和是 10,000。这些 25 个连续偶数中最大的是多少？
Correct Answer: E
The average of the 25 even integers is $10000/25=400$. So 12 consecutive even integers will be larger than 400 and 12 consecutive even integers will be smaller than 400. The sum $376+378+\cdots+398+400+402+\cdots+422+424=10000$. The largest of these numbers is $424$.
这25个偶整数的平均数是 $10000/25=400$。因此有12个连续偶数大于400，12个连续偶数小于400。于是 $376+378+\cdots+398+400+402+\cdots+422+424=10000$。其中最大的数是 $424$。
Q20
The least common multiple of $a$ and $b$ is 12, and the least common multiple of $b$ and $c$ is 15. What is the least possible value of the least common multiple of $a$ and $c$?
$a$ 和 $b$ 的最小公倍数是 12，$b$ 和 $c$ 的最小公倍数是 15。$a$ 和 $c$ 的最小公倍数的最小可能值是多少？
Correct Answer: A
If b = 1, then a = 12 and c = 15, and the least common multiple of a and c is 60. If b > 1, then any prime factor of b must also be a factor of both 12 and 15, and thus the only possible value is b = 3. In this case, a must be a multiple of 4 and a divisor of 12, so a = 4 or a = 12. Similarly, c must be a multiple of 5 and a divisor of 15, so c = 5 or c = 15. It follows that the least common multiple of a and c must be a multiple of 20. When a = 4, b = 3, and c = 5, the least common multiple of a and c is exactly 20.
若 b = 1，则 a = 12 且 c = 15，此时 a 与 c 的最小公倍数为 60。若 b > 1，则 b 的任一质因数都必须同时整除 12 和 15，因此 b 只能取 3。在这种情况下，a 必须是 4 的倍数且为 12 的约数，所以 a = 4 或 a = 12。类似地，c 必须是 5 的倍数且为 15 的约数，所以 c = 5 或 c = 15。于是 a 与 c 的最小公倍数必须是 20 的倍数。当 a = 4、b = 3、c = 5 时，a 与 c 的最小公倍数恰为 20。
Q21
A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
一个盒子里有3个红筹码和2个绿筹码。从盒子里随机抽取筹码，一次抽一个，不放回，直到抽到所有3个红筹码或抽到两个绿筹码为止。抽到3个红筹码的概率是多少？
Correct Answer: B
Consider drawing all five chips and listing the 10 possible outcomes: RRRGG, RRGRG, RGRRG, GRRRG, GGRRR, GRGRR, RGGRR, GRRGR, RGRGR, RRGGR. All 10 of these outcomes are equally likely. The outcomes that end in G correspond to the outcomes where the 3 reds are drawn and the outcomes that end in R correspond to the outcomes where the 2 greens are drawn. The probability that the 3 reds are drawn is $\frac{4}{10}=\frac{2}{5}$.
考虑把五个筹码全部抽完，并列出10种可能的结果：RRRGG、RRGRG、RGRRG、GRRRG、GGRRR、GRGRR、RGGRR、GRRGR、RGRGR、RRGGR。这10种结果等可能。以 G 结尾的结果对应于抽到3个红筹的情形，而以 R 结尾的结果对应于抽到2个绿筹的情形。抽到3个红筹的概率是 $\frac{4}{10}=\frac{2}{5}$。
Q22
Rectangle DEFA below is a 3×4 rectangle with DC = CB = BA = 1. The area of the “bat wings” (the shaded area) is
下面的矩形DEFA是一个3×4的矩形，且DC = CB = BA = 1。“蝙蝠翅膀”（阴影区域）的面积是
stem
Correct Answer: C
The area of triangle BCE is $\tfrac12(1)(4)=2$. Triangles CBH and EFH are similar. Since $CB=\tfrac13 EF$, it follows that $IH=\tfrac13 GH=\tfrac14 IG=1$. The area of triangle CBH is $\tfrac12$, so the area of triangle ECH is $2-\tfrac12=\tfrac32$. Thus the batwing's area is $3$.
三角形 BCE 的面积为 $\tfrac12(1)(4)=2$。三角形 CBH 与 EFH 相似。由于 $CB=\tfrac13 EF$， 因此 $IH=\tfrac13 GH=\tfrac14 IG=1$。 三角形 CBH 的面积为 $\tfrac12$，所以三角形 ECH 的面积是 $2-\tfrac12=\tfrac32$。因此“蝙蝠翼”图形的面积为 $3$。
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Q23
Two congruent circles centered at points A and B each pass through the other’s center. The line containing both A and B is extended to intersect the circles at points C and D. The two circles intersect at two points, one of which is E. What is the degree measure of $\angle CED$?
两个全等的圆分别以点A和B为中心，每个圆都经过对方的中心。包含A和B的直线延长，与圆相交于点C和D。两个圆相交于两点，其中之一是E。$\angle CED$ 的度量是多少度？
Correct Answer: C
We know $\triangle AEB$ is equilateral since each of its sides is a radius of one of the congruent circles. Thus the measure of $\angle AEB$ is $60^\circ$. Since $DB$ is a diameter of circle $A$ and $AC$ is a diameter of circle $B$, it follows that $\angle DEB$ and $\angle AEC$ are both right angles. Therefore the degree measure of $\angle DEC$ is $90^\circ+90^\circ-60^\circ=120^\circ$.
因为两个全等圆的半径构成了 $\triangle AEB$ 的三条边，故 $\triangle AEB$ 为等边三角形，因此 $\angle AEB=60^\circ$。又因为 $DB$ 是以 $A$ 为圆心的圆的直径，$AC$ 是以 $B$ 为圆心的圆的直径，所以 $\angle DEB$ 和 $\angle AEC$ 均为直角。于是 $\angle DEC$ 的度数为 $90^\circ+90^\circ-60^\circ=120^\circ$.
Q24
The digits 1, 2, 3, 4, and 5 are each used once to write a five-digit number PQRST. The three-digit number PQR is divisible by 4, the three-digit number QRS is divisible by 5, and the three-digit number RST is divisible by 3. What is P?
数字1、2、3、4和5各使用一次，写成五位数PQRST。三位数PQR能被4整除，三位数QRS能被5整除，三位数RST能被3整除。P是多少？
Correct Answer: A
Since QRS is divisible by 5, we know that S = 5. Since PQR is divisible by 4, we know that QR is 12, 32, or 24. So RST will be either 25T or 45T and divisible by 3. Using the available digits, 453 is the only number that is divisible by 3. So T = 3, R = 4, and P = 1.
由于 QRS 能被 5 整除，所以 S = 5。由于 PQR 能被 4 整除，所以 QR 为 12、32 或 24。因此 RST 将是 25T 或 45T，且需能被 3 整除。使用可用的数字，453 是唯一能被 3 整除的数。因此 T = 3，R = 4，P = 1。
Q25
A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?
一个半圆内接于一个底边长16、高15的等腰三角形中，使得半圆的直径包含在三角形的底边上，如图所示。求半圆的半径。
stem
Correct Answer: B
Let O be the midpoint of base AB of triangle ABC and the center of the semicircle. Triangle OBC is a right triangle with $OB=8$ and $OC=15$, and so, by the Pythagorean Theorem, $BC=17$. Let E be the point where the semicircle intersects $BC$, so radius $OE$ is perpendicular to $BC$. Then triangles $OEB$ and $COB$ are similar, and therefore $\frac{OE}{CO}=\frac{OB}{CB}$. Hence, $\frac{OE}{15}=\frac{8}{17}$ and so $OE=\frac{120}{17}$.
设 O 为底边 AB 的中点，也是半圆的圆心。三角形 OBC 为直角三角形，且 $OB=8$, $OC=15$，由勾股定理得 $BC=17$。设 E 为半圆与 $BC$ 的交点，则半径 $OE$ 垂直于 $BC$。于是三角形 $OEB$ 与 $COB$ 相似，因此 $\frac{OE}{CO}=\frac{OB}{CB}$。故 $\frac{OE}{15}=\frac{8}{17}$，从而 $OE=\frac{120}{17}$。
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