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考题1-2 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的官方真题以及官方解答吧：

Problem 1

Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry’s answer is $H$ and Terry’s answer is $T$ , what is $H-T$ ?

$考题1-2 2014 AMC 8$

Solution

We have $H=8-7=1$ and $考题1-2 2014 AMC 8$ . Clearly $1-11=-10$ , so our answer is $\boxed{\textbf{(A)}-10}$ .

Problem 2

Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?

$考题1-2 2014 AMC 8$

Solution

The fewest amount of coins that can be used is 2 (a quarter and a dime). The greatest amount is 7, if he only uses nickels. Therefore we have $7-2=\boxed{\textbf{(E)}~5}$ .