首页> 重点归纳 > AMC数学竞赛真题2016年10A 3-4

AMC数学竞赛真题2016年10A 3-4

2018-10-23 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校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AMC10数学竞赛真题以及官方解答吧：

Problem 3

For every dollar Ben spent on bagels, David spent $25$ cents less. Ben paid $$12.50$ more than David. How much did they spend in the bagel store together?

$\textbf{(A)}\ $37.50 \qquad\textbf{(B)}\ $50.00\qquad\textbf{(C)}\ $87.50\qquad\textbf{(D)}\ $90.00\qquad\textbf{(E)}\ $92.50$

Solution

If Ben paid $$ 12.50$ more than David, then he paid $12.5/.25=$ 50.00$ . Thus, David paid $$ 37.50$ , and they spent $50.00+37.50 =$ 87.50 \implies \boxed{\textbf{(C) }$ 87.50}$ .

Problem 4

The remainder can be defined for all real numbers $x$ and $y$ with $y \neq 0$ by

where $\left \lfloor \tfrac{x}{y} \right \rfloor$ denotes the greatest integer less than or equal to $\tfrac{x}{y}$ . What is the value of $\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )$ ?

$\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}$

Solution

The value, by definition, is

amc数学竞赛

Solution 2

Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is $\boxed{B}$