经验教程|AMC8 -- 1990年真题解析( 二 )

In the second case, the common sum would be (10+11+12+13+14+15)/3=25, so 11 must be paired with 14, which it isn’t. Thus, the only possibility is the first case and the sum of the six numbers is 81.
Problem 12经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: B
Solution:
For each choice of the thousands digit, there are 6 numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need 5 more numbers.
We can just list from here:5247, 5274, 5427, 5472, 5724.
Problem 13经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: C
Solution:
After the first ounce, there are 3.5 ounces left. Since each additional ounce or fraction of an ounce adds 22 cents to the total cost, we need to add 4*22 to the cost for the first ounce.
So, the total price is 30+4*22=118 cents.
Problem 14经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: B
Solution:
The total number of balls in the bag must be 4*6=24. so there are 24-6=18 green balls .
Problem 15经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: E
Solution:
Since the area of the whole figure is 100, each square has an area of 25 and the side length is 5.There are 10 sides of this length, so the perimeter is 10*5=50.
Problem 16经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: D
Solution:
In the middle, we have …+1010-1000+990-…, If we match up the back with the front, and then do the same for the rest, we get pairs with 2000 and -2000, so these will cancel out. In the middle, we have 2000-1000 which doesn't cancel, but gives us 1000.
Problem 17经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: A
Solution:
This is a 1 yard by 20 yard by 1/12 yard sidewalk, so its volume in yards is 1*20*1/12=5/3. Since concrete must be ordered in a whole number of cubic yards, we need 2.
Problem 18经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: C
Solution:
In addition to the original 12 edges, each original vertex contributes 3 new edges.
There are 8 original vertices, so there are 12+3*8=36 edges in the new figure.
Problem 19经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: B
Solution:
p is a person seated, o is an empty seat. The pattern of seating that results in the fewest occupied seats is opoopoopoo...po we can group the seats in 3s opo opo opo ... opo
there are a total of 40 groups.
Problem 20经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: A
Solution:
Let s be the sum of all the incomes but the largest one. For the actual data, the mean is (s+98000)/1000, and for the incorrect data the mean is (s+980000)/1000. The difference is 882.
Problem 21经验教程|AMC8 -- 1990年真题解析
文章图片

Answer: B
Solution:
We just use the definition to find the first number is 1/4.


推荐阅读