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考题9-10 2015 AMC 10A

2018-08-06 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

Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

$考题9-10 2015 AMC 10A$

Solution

Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told $r_2=\frac{11r_1}{10}$ $\pi r_1^2h_1=\pi r_2^2h_2$ Substituting the first equation into the second and dividing both sides by $\pi$ , we getTherefore, $\boxed{\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}}$

Problem 10

How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$ .

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

The first thing one would want to do is see a possible value that works and then stem off of it. For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since:

$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed because of the $cd$ .

We get the same problem if we start with a $d$ , since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$ .

If we start with a $b$ , the next letter would have to be a $d$ , and since we can put an $a$ next to it and then a $c$ after that, this configuration works. The same approach applies if we start with a $c$ .

So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$ , giving us:

$1 + 1 = \boxed{\textbf{(C)}\ 2}$