首页> 重点归纳 > 考题7-8 2014 AMC 8

考题7-8 2014 AMC 8

2018-08-04 重点归纳

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 7

There are four more girls than boys in Ms. Raub's class of $28$ students. What is the ratio of number of girls to the number of boys in her class?

$AMC 8数学竞赛试题及答案$

Solution

We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$ . Since the total number of students is equal to $28$ , we get $x+x-4=28$ . Solving this equation, we get $x=16$ . There are $16-4=12$ boys in our class, and our answer is $AMC 8数学竞赛试题及答案$ .

Problem 8

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?

$AMC 8数学竞赛试题及答案$

Solution

A number is divisible by $11$ if the difference between the sum of the digits in the odd-numbered slots (e.g. the ones slot, the hundreds slot, etc.) and the sum in the even-numbered slots (e.g. the tens slot, the thousands slot) is a multiple of $11$ . So $1 + 2 - A$ is equivalent to $0\pmod{11}$ . Clearly $1+2-A$ cannot be equal to $11$ or any multiple of $11$ greater than that. Also, if the expression $1+2-A$ is to be equal to a negative multiple of $A$ , $A$ must be 14 or greater, which violates the condition that A is a digit. So $AMC 8数学竞赛试题及答案$ .