AMC8--2011年真题解析

时间：2020-05-05 18:05:49 来源：作者：网络

原标题：AMC8 -- 2011年真题解析

Problem 1

Answer: E

Solution:

The three apples cost 3*50=150 cents. The change Margie receives is 500-150=350 cents.

Problem 2

Answer: E

Solution:

The area of a rectangle is given by the formula length times width. Karl's garden is 20*45=900 square feet and Makenna's garden is 25*40=1000 square feet. Makenna's garden is larger by 1000-900=100 square feet.

Problem 3

Answer: D

Solution:

One way of approaching this is drawing the next circle of boxes around the current square.We can now count the number of black and white tiles; 32 black tiles and 17 white tiles. This means the answer is 32:17.

Problem 4

Answer: C

Solution:

First, put the numbers in increasing order. 0,0, 1,1,2,2,3,3,3. The mean is 15/9; The median is 2, and the mode is 3 .

Problem 5

Answer: D

Solution:

There are 60 minutes in an hour. 2011/60=33 hours and 31 minutes. There are 24 hours in a day, so the time is 9 hours and 31 minutes after midnight on January 2, 2011. (midnight of Jan. 1st is 0:00 of Jan.1st)

Problem 6

Answer: D

Solution:

The number of adults who own both cars and motorcycles is 331+45-351=25. Out of the 331 car owners, 25 of them own motorcycles, 331-25=306 of them don't.

Problem 7

Answer: C

Solution:

Assume that the area of each square is 1. Then, the area of the bolded region in the top left square is 1/4. The area of the top right bolded region is 1/8. The area of the bottom left bolded region is 3/8. And the area of the bottom right bolded region is 1/4. Add the four fractions: 1/4+1/8+3/8+1/4=1. The four squares together have an area of 4, so the percentage bolded is 1/4*100=25%.

Problem 8

Answer: B

Solution:

By adding a number from Bag A and a number from Bag B together, the values we can get are 3,5,7,5,7,9,7,9,11. Therefore the number of different values is 5.

Problem 9

Answer: E

Solution:

We observe the graph and see that the shape of the graph does not matter. We only want the total time it took Carmen and the total distance she traveled. Based on the graph, Carmen traveled 35 miles for 7 hours. Therefore, her average speed is 5.

Problem 10

Answer: C

Solution:

Let x be the number of miles you ride. The number of miles you ride after the first half mile is x-0.5 We can write this equation: 10=2.4+0.2*(x-0.5)/0.1+2 ; thus, x=3.3

Problem 11

Answer: A

Solution:

Average the differences between each day. We get 10,-10, 20, 30, -20. . We find the average of this list to get 6.

Problem 12

Answer: B

Solution:

If we seat Angie first, there would be only one out of three ways Carlos can sit across from Angie. So the final answer is 1/3.

Problem 13

Answer: C

Solution:

The overlap length is 5, so the shaded area is 5*15. The area of the whole shape is 25*15. . The fraction (5*15)/(25*15) reduces to 1/5 or 20%.

Problem 14

Answer: C

Solution:

At Colfax Middle School, there are 4/9 * 270=120 girls. At Winthrop Middle School, there are 5/9*180=100 girls. The ratio of girls to the total number of students is (120+100)/(270+180)=22/45.

Problem 15

Answer: D

Solution:

4^5 * 5^10=2^10 * 5^10=10^10. That is one 1 followed by ten 0's, which is 11 digits.

Problem 16

Answer: C

Solution:

25-25-30: Draw the altitude for the side with length 30 on the 25-25-30 triangle. Thus, we have two 15-height-25 right triangles. By the Pythagorean theorem, we have the height is 20.

25-25-40: In the same way, we get two 15-20-25 right triangle.

Thus, the ratio of A/B = (1/2*30*20)/(1/2*40*15)=1/1.

Problem 17

Answer: A

Solution:

The prime factorization of 588 is 2*2*3*7*7; we can see w=2, x=1,and z=2. Because 5^0=1,y=0. 2w+3x+5y+7z=4+3+0+14=21.

roblem 18

Answer: D

Solution:

There are 6*6=36 ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining 36-6=30 ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are 30/2=15 ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is (15+6)/36=7/12.

Problem 19

Answer: D

Solution:

The number of rectangles with just one section is 3. The number of rectangles with two sections is 5. There are none with only three sections. The number of rectangles with four sections is 4. 3+5+3=11.

Problem 20

Answer: D

Solution:

If you draw altitudes from A and B to CD the trapezoid will be divided into two right triangles and a rectangle. Using Pythagorean theorem, the sides of left right triangle are 15,12,9; the sides of other right triangle are 12, 16, 20. Thus, DC=9+50+16=75.

The area of the trapezoid is (50+75)*12/2=750.

Problem 21

Answer: C

Solution:

If at least half the guesses are too low, then Norb's age must be greater than 36. If two of the guesses are off by one, then his age is in between two guesses whose difference is 2. It could be 37or 48. Lastly, Norb's age is a prime number so the answer must be 37.

中文解析： “At least half of you guessed too low". 说明前一半的数24， 28， 30，32，36都比答案小。 ”two of you are off by one": 告诉我们有两个人的猜测只比答案差1，那有可能是36和38中间的37，也有可能是47和49之间的48. 但是“my age is a prime number"，答案必须是一个质数，则是37.

Problem 22

Answer: D

Solution:

We want the tens digit So, we take 7^2011(mod 100) . That is congruent to 7^11(mod 100). . From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 4.

中文解析：类似Units digit，我们关注7的n次方的后两位的规律， 7的1，2，3，4，5，6,......次方的后两位分别是: 7, 49, 43, 01； 07, 49,...... 可以看出每4个一组重复的规律。 2011=502*4+3，因此7的2011次方，应该是43. 十位数是4.

Problem 23

Answer: D

Solution:

Case 1: The last digit is 5. The number of integers is this case is 4*4*3*1=48.

Case 2: The last digit is 0. The number of integers in this case is 1*3*4*3=36.

Therefore, the answer is 48+36=84.

中文解析：这个4位数只能从{0，1，2，3，4，5}中选择不同的数字，必须包含数字5. 因为是5的倍数，个位数必须是5或者0.

分两种情况讨论:

Case 1: 个位数是5：千位数不能是0，不能是5，可以是1，2，3，4中的任意一个，有4种选择；百位数有4种选择，十位数有3种选择，个位数必须是5. 这种情况下，4*4*3*1=48.

Case 2: 各位数十0：除个位数外，必须有一个数字是5，如果千位数是5，则百位数有4种选择，十位数有3种选择，则有：1*4*3*1=12. 同理，也可以是百位数是5，则： 4*1*3*1=12，也可以是十位数是5，则4*3*1*1=12. 因此，再个位数是0的情况下，共有12+12+12=36种选择。

Case 1 和 Case 2 合起来，共 48+36=84.

Problem 24

Answer: A

Solution:

For the sum of two numbers to be odd, one must be odd and the other must be even, There are only 2 is a prime number in even number. So our only combination could be 2 and 9999. However, 9999 is clearly divisible by 3. So the number of ways 10001 can be written as the sum of two primes is 0.