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AMC数学竞赛真题2017年10B 1-2

2018-08-29 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 1

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary's number?

$amc10$

Solution

Solution 1

Let her $2$ -digit number be $x$ . Multiplying by $3$ makes it a multiple of $3$ , meaning that the sum of its digits is divisible by $3$ . Adding on $11$ increases the sum of the digits by $1+1 = 2,$ and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be $2$ more than a multiple of $3$ . There are two such numbers between $71$ and $75$ : $71$ and $74.$ Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed: $\[\]$ For $71,$ we reverse the digits, resulting in $17.$ Subtracting $11$ , we get $6.$ We can already see that dividing this by $3$ will not be a two-digit number, so $71$ does not meet our requirements. $\[\]$ Therefore, the answer must be the reversed steps applied to $74.$ We have the following: $\[\]$ $美国数学竞赛$ $\[\]$ Therefore, our answer is $\boxed{\bold{(B)} 12}$ .

Solution 2

Working backwards, we reverse the digits of each number from $71$ ~ $75$ and subtract $11$ from each, so we have $amc数学竞赛$ The only numbers from this list that are divisible by $3$ are $6$ and $36$ . We divide both by $3$ , yielding $2$ and $12$ . Since $2$ is not a two-digit number, the answer is $\boxed{\textbf{(B)}\ 12}$ .

Solution 3

You can just plug in the numbers to see which one works. When you get to $12$ , you multiply by $3$ and add $11$ to get $47$ . When you reverse the digits of $47$ , you get $74$ , which is within the given range. Thus, the answer is $amc真题$ .

Problem 2

Sofia ran $5$ laps around the $400$ -meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

$\textbf{(A)}\ \text{5 minutes and 35 seconds}\qquad\textbf{(B)}\ \text{6 minutes and 40 seconds}\qquad\textbf{(C)}\ \text{7 minutes and 5 seconds}\qquad\textbf{(D)}\ \text{7 minutes and 25 seconds}$ $\qquad\textbf{(E)}\ \text{8 minutes and 10 seconds}$

Solution

If Sofia ran the first $100$ meters of each lap at $4$ meters per second and the remaining $300$ meters of each lap at $5$ meters per second, then she took $\frac{100}{4}+\frac{300}{5}=25+60=85$ seconds for each lap. Because she ran $5$ laps, she took a total of $5 \cdot 85=425$ seconds, or $7$ minutes and $5$ seconds. The answer is $\boxed{\textbf{(C)}\ \text{7 minutes and 5 seconds}}$ .