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考题23-25 2017 AMC 8

2018-08-06 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 23

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution

It is well known that Distance=Speed $\cdot$ Time. In the question, we want distance. From the question, we have that the time is $60\text{mins}$ (One hour). By the equation derived from Distance=Speed $\cdot$ Time, we have Speed=Distance/Time, so the speed is $1$ mile/ $x$ mins. Because we want the distance, we multiply the time and speed together yielding $60\text{mins}\cdot \frac{1\text{mile}}{x\text{mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are:

$考题23-25 2017 AMC 8$

We then start our trial and error: The factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$ . We plug each of those numbers in for $x$ , and we get that $x$ is 5, so

$考题23-25 2017 AMC 8$

Problem 24

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

Solution 1

In $360$ days, there are

$考题23-25 2017 AMC 8$

days without calls. Note that in the last five days of the year, day $361$ and $362$ also do not have any calls, as they are not multiples of $3$ , $4$ , or $5$ . Thus our answer is $144+2 = \boxed{\textbf{(D)}\ 146}$ .

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Alternatively, there are $365\cdot \frac45 \cdot \frac 34 \cdot \frac23=\boxed{146}$ days without calls. Multiplying the fractions in this order prevents partial days, as $365$ is a multiple of $5$ , $365\cdot \frac45$ is a multiple of $4$ and $365\cdot \frac45 \cdot \frac 34$ is a multiple of $3$ .

Solution 2

We use Principle of Inclusion and Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is

$\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor$

To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days. This number is

$\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor$

We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor$ . Therefore, our answer is

$365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor$

$= 365 - 285+72 - 6 = \boxed{\textbf{(D) } 146}$ .

Problem 25

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$ . Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?

$考题23-25 2017 AMC 8$

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

Solution

Let the centers of the circles containing arcs $\overarc{SR}$ and $\overarc{TR}$ be $S'$ and $T'$ , respectively. Extend $\overline{US}$ and $\overline{UT}$ to $S'$ and $T'$ , and connect point $S'$ with point $T'$ .

$考题23-25 2017 AMC 8$

We can clearly see that $\triangle US'T'$ is an equilateral triangle, because two of its angles are $60^\circ$ , which is the degree measure of $\frac{1}{6}$ a circle. The area of the figure is equal to $[\triangle US'T']$ minus the combined area of the $2$ sectors of the circles. Using the area formula for an equilateral triangle, $\frac{a^2\sqrt{3}}{4},$ where $a$ is the side length of the equilateral triangle, $[\triangle US'T']$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2\cdot\frac16\cdot\pi r^2$ , which is $\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.$ Thus, our final answer is $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$