考题23-25 2014 AMC 8

AMC 8数学竞赛专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8数学竞赛试题及答案吧：

Problem 23

Three members of the Euclid Middle School girls' softball team had the following conversation.

Ashley: I just realized that our uniform numbers are all -digit primes.

Brittany : And the sum of your two uniform numbers is the date of my birthday earlier this month.

Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.

Ashley: And the sum of your two uniform numbers is today's date.

What number does Caitlin wear?

Solution

The maximum amount of days any given month can have is 31, and the smallest two digit primes are 11, 13, and 17. There are a few different sums that can be deduced from the following numbers, which are 24, 30, and 28, all of which represent the three days. Therefore, since Brittany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to 24. Similarly, Caitlin says that the other two people's uniform numbers is later, so the sum must add up to 30. This leaves 28 as today's date. From this, Caitlin was referring to the uniform wearers 13 and 17, telling us that her number is 11, giving our solution as 

Problem 24

One day the Beverage Barn sold  cans of soda to  customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

Solution

In order to maximize the median, we need to make the first half of the numbers as small as possible. Since there are  people, the median will be the average of the  and  largest amount of cans per person. To minimize the first 49, they would each have one can. Subtracting these  cans from the  cans gives us  cans left to divide among  people. Taking  gives us  and a remainder of . Seeing this, the largest number of cans the  person could have is , which leaves  to the rest of the people. The average of  and  is . Thus our answer is .

Problem 25

A straight one-mile stretch of highway,  feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at  miles per hour, how many hours will it take to cover the one-mile stretch?

Note:  mile =  feet

Solution

There are two possible interpretations of the problem: that the road as a whole is  feet wide, or that each lane is  feet wide. Both interpretations will arrive at the same result. However, let us stick with the first interpretation for simplicity. Each lane must then be feet wide, so Robert must be riding his bike in semicircles with radius  feet and diameter  feet. Since the road is  feet long, over the whole mile, Robert rides  semicircles in total. Were the semicircles full circles, their circumference would be  feet; as it is, the circumference of each is half that, or  feet. Therefore, over the stretch of highway, Robert rides a total of  feet, equivalent to  miles. Robert rides at 5 miles per hour, so divide the  miles by  mph (because t = d/r time = distance/rate) to arrive at  hours.

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