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AMC数学竞赛真题2017年10A 9-10

2018-08-27 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 9

Minnie rides on a flat road at $20$ kilometers per hour (kph), downhill at $30$ kph, and uphill at $5$ kph. Penny rides on a flat road at $30$ kph, downhill at $40$ kph, and uphill at $10$ kph. Minnie goes from town $A$ to town $B$ , a distance of $10$ km all uphill, then from town $B$ to town $C$ , a distance of $15$ km all downhill, and then back to town $A$ , a distance of $20$ km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the $45$ -km ride than it takes Penny?

$AMC10数学竞赛真题$

Solution

The distance from town $A$ to town $B$ is $10$ km uphill, and since Minnie rides uphill at a speed of $5$ kph, it will take her $2$ hours. Next, she will ride from town $B$ to town $C$ , a distance of $15$ km all downhill. Since Minnie rides downhill at a speed of $30$ kph, it will take her half an hour. Finally, she rides from town $C$ back to town $A$ , a flat distance of $20$ km. Minnie rides on a flat road at $20$ kph, so this will take her $1$ hour. Her entire trip takes her $3.5$ hours. Secondly, Penny will go from town $A$ to town $C$ , a flat distance of $20$ km. Since Penny rides on a flat road at $30$ kph, it will take her $\frac{2}{3}$ of an hour. Next Penny will go from town $C$ to town $B$ , which is uphill for Penny. Since Penny rides at a speed of $10$ kph uphill, and town $C$ and $B$ are $15$ km apart, it will take her $1.5$ hours. Finally, Penny goes from Town $B$ back to town $A$ , a distance of $10$ km downhill. Since Penny rides downhill at $40$ kph, it will only take her $\frac{1}{4}$ of an hour. In total, it takes her $29/12$ hours, which simplifies to $2$ hours and $25$ minutes. Finally, Penny's $2$ Hour $25$ Minute trip was $\boxed{\textbf{(C)}\ 65}$ minutes less than Minnie's $3$ Hour $30$ Minute Trip

Problem 10

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

$AMC10$

Solution

The triangle inequality generalizes to all polygons, so $x < 3+7+15$ and $x+3+7>15$ to get $5<x<25$ . Now, we know that there are $19$ numbers between $5$ and $25$ exclusive, but we must subtract $2$ to account for the 2 lengths already used that are between those numbers, which gives $AMC10数学竞赛真题及解答$