2010 AMC 8 详解 | LIVE by Po-Shen Loh

1.

欧几里得中学的数学老师有热尔曼小姐、牛顿先生和杨夫人。今年参加 AMC $8$ 竞赛的学生中，热尔曼小姐班有 $11$ 人，牛顿先生班有 $8$ 人，杨夫人班有 $9$ 人。欧几里得中学共有多少名学生参加竞赛？

At Euclid Middle School the mathematics teachers are Miss Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Miss Germain’s class, $8$ students in Mr. Newton’s class, and $9$ students in Mrs. Young’s class taking the AMC $8$ Contest this year. How many mathematics students at Euclid Middle School are taking the contest?

$\ 26$

$\ 27$

$\ 28$

$\ 29$

$\ 30$

知识点：基本计数

难度评级：370

小提示：

题目没有描述班级重叠，所以总数直接相加

No class overlap is described, so the total is a direct sum.

大提示：

把三个班的人数相加

Add the three class counts.

解答：

学生总数为 $11+8+9 = 28$ 。

所以正确答案是 C。

There are $11+8+9 = 28$ students.

Therefore, the answer is C .

2.

对于正整数 $a$ 、 $b$ ，定义 $a \ast b = \dfrac{a\times b}{a+b}$ 。那么 $5 \ast 10$ 是多少？

If $a \ast b = \dfrac{a\times b}{a+b}$ for $a,$ $b$ positive integers, then what is $5 \ast 10?$

$\ \dfrac{3}{10}$

$\ 1$

$\ 2$

$\ \dfrac{10}{3}$

$\ 50$

知识点：自定义运算分数

难度评级：610

小提示：

把分母相加后，化简 $\frac{50}{15}$

Reduce $\frac{50}{15}$ after adding the denominator.

大提示：

将给定的两个数代入 $a\ast b$ 的定义

Substitute the two given numbers into the definition of $a\ast b$ .

解答：

由 $a\ast b$ 的定义， $5 \ast 10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}\text{。}$

所以正确答案是 D。

Given our definition of $a\ast b,$ we have $5 \ast 10 = \dfrac{5 \times 10}{5 + 10} = \dfrac{50}{15} = \dfrac{10}{3}.$

Thus, the answer is D .

3.

图中显示一年中前十个月五加仑汽油的价格。最高价格比最低价格高百分之多少？

The graph shows the price of five gallons of gasoline during the first ten months of the year. By what percent is the highest price more than the lowest price?

$\ 50$

$\ 62$

$\ 70$

$\ 89$

$\ 100$

知识点：数据与图表解读百分数

难度评级：790

小提示：

高出百分之多少表示增加量除以最低价格

Percent more means the increase divided by the lowest price.

大提示：

从图中读出最高和最低的柱形高度

Read the highest and lowest bar heights from the graph.

解答：

最高价格为 $17$ ，最低价格为 $10$ 。因此百分数为 $(\dfrac{17}{10}-1)\cdot 100 = 70\text{。}$

所以正确答案是 C。

The highest price is $17$ and the lowest price is $10.$ This means the percent is $(\dfrac{17}{10}-1)\cdot 100 = 70.$

Thus, the answer is C .

4.

数 $2$ ， $3$ ， $0$ ， $3$ ， $1$ ， $4$ ， $0$ ， $3$ 的平均数、中位数和众数之和是多少？

What is the sum of the mean, median, and mode of the numbers $2,$ $3,$ $0,$ $3,$ $1,$ $4,$ $0,$ $3?$

$\ 6.5$

$\ 7$

$\ 7.5$

$\ 8.5$

$\ 9$

知识点：平均数中位数（数据）众数

难度评级：720

小提示：

平均数使用全部八个数，中位数使用中间两个数

The mean uses all eight numbers, while the median uses the two middle numbers.

大提示：

先把列表按从小到大排列，再找中位数和众数

Put the list in increasing order before finding the median and mode.

解答：

重新排序后的列表是 $0,0,1,2,3,3,3,4$ 。

中位数是中间两个数的平均数，即 $\dfrac{2+3}2 = 2.5$ 。众数是 $3$ ，因为 $3$ 出现次数最多。平均数为 $\dfrac{0+0+1+2+3+3+3+4}{8}$ $= \dfrac{16}{8} = 2\text{。}$

它们的和为 $2.5+3+2 = 7.5$ 。

所以正确答案是 C。

The list reordered is $0,0,1,2,3,3,3,4.$

The median is the mean of the middle two numbers, which would be $\dfrac{2+3}2 = 2.5.$ The mode is $3$ since $3$ appears the most. The mean is $\dfrac{0+0+1+2+3+3+3+4}{8}$ $= \dfrac{16}{8} = 2.$

Their sum is $2.5+3+2 = 7.5.$

Thus, the answer is C .

5.

爱丽丝需要更换厨房里一个位于天花板下方 $10$ 厘米处的灯泡。天花板离地面 $2.4$ 米。爱丽丝身高 $1.5$ 米，并且手能伸到头顶上方 $46$ 厘米。站在凳子上时，她刚好能够到灯泡。凳子的高度是多少厘米？

Alice needs to replace a light bulb located $10$ centimeters below the ceiling in her kitchen. The ceiling is $2.4$ meters above the floor. Alice is $1.5$ meters tall and can reach $46$ centimeters above the top of her head. Standing on a stool, she can just reach the light bulb. What is the height of the stool, in centimeters?

$\ 32$

$\ 34$

$\ 36$

$\ 38$

$\ 40$

知识点：单位换算

难度评级：770

小提示：

凳子补上爱丽丝的身高、伸手高度和灯泡距天花板的距离后剩下的高度差

The stool covers the remaining gap after Alice’s height, reach, and the bulb offset.

大提示：

先把天花板高度和爱丽丝的身高都换算成厘米

Convert the ceiling and Alice’s height to centimeters first.

解答：

天花板高度为 $2.4 \cdot 100 = 240$ 厘米。减去已知的各段高度，得到 $240 - 150 - 46 - 10 = 34\text{。}$ 因此凳子的高度是 $34$ 厘米。所以正确答案是 B。

We know the height of the ceiling is $2.4 \cdot 100 = 240$ cm. Subtracting out all the given values, we get $240 - 150 - 46 - 10 = 34.$ Therefore, the height of the stool is $34$ cm. Thus, B is the correct answer.

6.

下列哪个图形有最多条对称轴？

Which of the following figures has the greatest number of lines of symmetry?

等边三角形

equilateral triangle

非正方形的菱形

non-square rhombus

非正方形的长方形

non-square rectangle

等腰梯形

isosceles trapezoid

正方形

square

知识点：变换对称性

难度评级：560

小提示：

正方形既有对角线对称轴，也有连接对边中点的对称轴

A square has both diagonal and midpoint-to-midpoint symmetry lines.

大提示：

分别数出每种图形的反射对称轴

Count reflection lines for each named shape.

解答：

每条对称轴都必须经过图形的中心，否则中心会只在一侧，反射后无法重合。

如果对称轴经过某条边，它必须经过这条边的中点，并且与这条边垂直，这样反射后该边两侧长度相同且仍落在自身上。

如果对称轴经过某个顶点，它必须经过该角的角平分线，并且两侧边长相同，反射后角才会重合。

现在逐个考察这些图形。等边三角形只有 $3$ 条经过顶点或边中点的直线，所以至多有 $3$ 条对称轴。非正方形的菱形只有两条对称轴，因为只有经过相对顶点的直线可行，而经过边中点的直线与该边不垂直。非正方形的长方形也只有两条对称轴，它们经过对边的中点；经过顶点的直线不平分该角，所以不可行。等腰梯形只有 $1$ 条对称轴，它经过一组对边的中点。正方形有 $4$ 条对称轴，分别经过相对顶点和相对边的中点。

因此正方形有最多条对称轴。

所以正确答案是 E。

First, each line of symmetry must go through the center of the shape. This would ensure that the center of the shape isn’t on only one side, as that would make it asymmetric.

Next, if a line goes through any side, it must go through its midpoint, to ensure that after a reflection over this line, the same amount of the line is on both sides. Moreover, it must be perpendicular, to ensure that the line when reflected stays on itself.

Similarly, if a line goes through any corner, it must go through its angle bisector, to ensure that after a reflection over this line, the angle of the line is the same after reflection. Moreover, it must have the same side length on both sides.

Now, let’s look at each of the shapes. An equilateral triangle has $3$ lines that intersect a corner or a midpoint, so it has at most $3$ symmetry lines. A non-square rhombus only has two symmetry lines, as only the lines that go through the corners work, but not the ones through the midpoints as they would not intersect perpendicularly. A non-square rectangle only has two symmetry lines as it has symmetry lines through the midpoints of opposite sides, but not through the corners since it doesn’t bisect the angle. An isosceles trapezoid only has $1$ symmetry line, that goes through the midpoints of opposite sides. A square has $4$ symmetry lines, which go through opposite corners and opposite midpoints.

A square therefore has the most symmetry lines.

Thus, the answer is E .

7.

只使用一美分、五美分、十美分和二十五美分硬币，弗雷迪最少需要多少枚硬币，才能支付小于一美元的任意金额？

Using only pennies, nickels, dimes, and quarters, what is the smallest number of coins Freddie would need so he could pay any amount of money less than a dollar?

$\ 6$

$\ 10$

$\ 15$

$\ 25$

$\ 99$

知识点：钱币最优化

难度评级：1350

小提示：

能支付从零到 $24$ 美分的所有金额后，二十五美分硬币可以把范围扩展到 $99$ 美分

After making every amount through $24$ cents, quarters can extend the range to $99$ cents.

大提示：

先考虑支付 $4$ 、 $9$ 、 $24$ 美分等小金额所必须的硬币

First force the coins needed to make small amounts like $4$ , $9$ , and $24$ cents.

解答：

要支付 $4$ 美分，至少需要四枚一美分硬币。有了这四枚后，为了能支付 $5$ 美分，下一枚硬币的面值至多为 $5$ 美分，最好选一枚五美分硬币。这五枚硬币总值只有 $9$ 美分。下一枚硬币的面值至多为 $10$ 美分；即使选一枚十美分硬币，总值也只有 $19$ 美分。因此还需要一枚面值至多为 $20$ 美分的硬币，所以在不产生金额缺口的情况下使用二十五美分硬币之前，至少需要七枚硬币。

若总共只有九枚硬币，在必需的七枚之后最多只剩两枚。前七枚的总值至多为 $4+5+10+10=29$ 美分，再加两枚二十五美分硬币也只有 $79$ 美分，所以九枚硬币不能支付从一美分到 $99$ 美分的每个金额。

十枚硬币确实可以做到：四枚一美分硬币、一枚五美分硬币、两枚十美分硬币和三枚二十五美分硬币，可以支付从 $1$ 到 $99$ 美分的每个金额。因此最少需要 $10$ 枚。

所以正确答案是 B。

At least four pennies are necessary to pay $4$ cents. After those four pennies, the next coin must be worth at most $5$ cents so that $5$ cents can be paid; a nickel is the best choice. Those five coins total only $9$ cents. The next coin must be worth at most $10$ cents, and even a dime brings the total to only $19$ cents. One more coin worth at most $20$ cents is therefore necessary, so at least seven coins are needed before quarters can be used without leaving a gap.

With only nine coins total, at most two coins could remain after those required seven. The first seven can total at most $4+5+10+10=29$ cents, and two more quarters would bring the total to only $79$ cents, so nine coins cannot pay every amount through $99$ cents.

Ten coins do work: four pennies, one nickel, two dimes, and three quarters can make every amount from $1$ through $99$ cents. Therefore the minimum is $10$ .

Therefore, the answer is B .

8.

艾米丽沿一条笔直长路骑自行车时，看到埃默森在她前方 $\frac{1}{2}$ 英里处同方向滑行。她超过他后，可以从后视镜里看到他，直到他落在她后方 $\frac{1}{2}$ 英里处。艾米丽以每小时 $12$ 英里的恒定速度骑行，埃默森以每小时 $8$ 英里的恒定速度滑行。艾米丽能看到埃默森多少分钟？

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $\frac{1}{2}$ mile in front of her. After she passes him, she can see him in her rear view mirror until he is $\frac{1}{2}$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson?

$\ 6$

$\ 8$

$\ 12$

$\ 15$

$\ 16$

知识点：相对速度路程、速度与时间单位换算

难度评级：1100

小提示：

在可见的时间段内，艾米丽相对埃默森总共多前进 $1$ 英里

During the visible interval Emily gains a total of $1$ mile on Emerson.

大提示：

使用艾米丽相对于埃默森的速度

Use Emily’s speed relative to Emerson’s speed.

解答：

设 $d$ 为艾米丽领先埃默森的距离。艾米丽能看到埃默森当且仅当 $-\dfrac 12 \leq d \leq \frac 12$ 。其中 $t$ 以小时为单位；令 $t= 0$ 时 $d = -\frac 12$ 。根据两人的相对速度， $d = (12-8)t - \frac 12$ 。当 $d = \frac 12$ 时， $\frac 12 = 4t - \frac 12 \implies 1 = 4t$ $\implies t = 0.25\text{。}$ 因此经过 $0.25$ 小时，也就是 $60\cdot0.25 = 15$ 分钟。

所以正确答案是 D。

Let $d$ be how far Emily is ahead of Emerson. Emily sees Emerson if $-\dfrac 12 \leq d \leq \frac 12.$ Suppose at $t= 0,$ where $t$ is in hours, that $d = -\frac 12.$ Then, $d = (12-8)t - \frac 12 .$ Since we must find where $d = \frac 12,$ we find the time where $\frac 12 = 4t - \frac 12 \implies 1 = 4t$ $\implies t = 0.25.$ Since $0.25$ hours passed, we know that $60\cdot0.25 = 15$ minutes passed.

Therefore, the answer is D .

9.

瑞安在一份有 $25$ 道题的测试中答对了 $80\%$ ，在一份有 $40$ 道题的测试中答对了 $90\%$ ，在一份有 $10$ 道题的测试中答对了 $70\%$ 。所有题目中瑞安答对了百分之多少？

Ryan got $80\%$ of the problems correct on a $25$ -problem test, $90\%$ on a $40$ -problem test, and $70\%$ on a $10$ -problem test. What percent of all the problems did Ryan answer correctly?

$\ 63$

$\ 75$

$\ 80$

$\ 84$

$\ 86$

知识点：百分数

难度评级：960

小提示：

用答对总题数除以总题数

Divide the total correct answers by the total number of problems.

大提示：

将每份测试的百分比转换成答对题数

Convert each test percentage into a number of correct answers.

解答：

三份测试共有 $25+40+10 = 75$ 题。

第一份测试答对 $0.8\cdot25 = 20$ 题。

第二份测试答对 $0.9\cdot40 = 36$ 题。

第三份测试答对 $0.7\cdot10 = 7$ 题。

因此他总共答对 $20+36+7=63$ 题，答对比例为 $\dfrac{63}{75} = 84\%$ 。

所以正确答案是 D。

There were a total of $25+40+10 = 75$ problems.

On the first test, he solved $0.8\cdot25 = 20$ problems.

On the second test, he solved $0.9\cdot40 = 36$ problems.

On the third test, he solved $0.7\cdot10 = 7$ problems.

Therefore, he solved a total of $20+36+7=63.$ This means the fraction he solved is $\dfrac{63}{75} = 84\%.$

Therefore, the answer is D .

10.

如图摆放时，六片圆形意大利辣香肠正好横跨一个 $12$ 英寸披萨的直径。如果总共 $24$ 片辣香肠无重叠地放在这个披萨上，披萨面积的几分之几被辣香肠覆盖？

Six pepperoni circles will exactly fit across the diameter of a $12$ -inch pizza when placed as shown. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?

$\ \dfrac 12$

$\ \dfrac 23$

$\ \dfrac 34$

$\ \dfrac 56$

$\ \dfrac 78$

知识点：圆面积面积比长度、面积与体积的缩放关系

难度评级：1030

小提示：

面积按直径比例的平方缩放，然后乘以 $24$ 片

Area scales with the square of the diameter, then multiply by $24$ pepperoni circles.

大提示：

六个小圆直径等于披萨直径

Six small diameters equal the pizza diameter.

解答：

每片圆形辣香肠的直径是大圆直径的 $\dfrac{1}{6}$ ，所以它的面积是总面积的 $(\frac 16)^2 = \frac 1{36}$ 。

因为有 $24$ 片辣香肠，所以覆盖面积为 $24 \cdot \dfrac 1{36} = \frac 23$ 。

所以正确答案是 B。

Each circle has $\dfrac{1}{6}$ the diameter of the large circle, so it has $(\frac 16)^2 = \frac 1{36}$ of the total area.

Since there are $24$ pepperoni, they take up $24 \cdot \dfrac 1{36} = \frac 23$ of the area.

Therefore, the answer is B .

11.

一棵树的树顶比另一棵树的树顶高 $16$ 英尺。两棵树的高度之比为 $3:4$ 。较高的树高多少英尺？

The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4.$ In feet, how tall is the taller tree?

$\ 48$

$\ 64$

$\ 80$

$\ 96$

$\ 112$

知识点：比与比例一次方程

难度评级：900

小提示：

较高的树包含四份这样的相等部分

The taller tree has four of those equal parts.

大提示：

比值 $3:4$ 表示高度差是一份

The ratio $3:4$ means the height difference is one ratio part.

解答：

设较高和较矮的树高分别为 $b$ 、 $s$ 。则 $b= s+16$ ，且 $s = 0.75b$ 。代入得 $b = 0.75b + 16 \implies 0.25b = 16$ $\implies b = 64\text{。}$

所以正确答案是 B。

Let $b,$ $s$ be the heights of bigger and smaller trees respectively. Then, $b= s+16$ and $s = 0.75b.$ If we substitute, we get $b = 0.75b + 16 \implies 0.25b = 16$ $\implies b = 64.$

Thus, the answer is B .

12.

一个大袋子里有 $500$ 个球，其中 $80\%$ 是红球，其余是蓝球。必须拿走多少个红球，才能使剩余球中 $75\%$ 是红球？

Of the $500$ balls in a large bag, $80\%$ are red and the rest are blue. How many of the red balls must be removed so that $75\%$ of the remaining balls are red?

$\ 25$

$\ 50$

$\ 75$

$\ 100$

$\ 150$

知识点：百分数

难度评级：1120

小提示：

如果红球占 $75\%$ ，那么蓝球占剩余袋子的 $25\%$

If red balls are $75\%$ , then blue balls are $25\%$ of the remaining bag.

大提示：

蓝球的数量保持不变

The number of blue balls stays fixed.

解答：

若 $r$ 是红球数，则 $r = 0.8\cdot500 = 400\text{。}$ 因此若 $b$ 是蓝球数，则 $b = 500-400 = 100\text{。}$

拿走一些球后，如果红球占 $75\%$ ，那么蓝球占 $25\%$ 。蓝球数量不变，所以剩余球总数为 $\dfrac{100}{0.25} = 400$ 。这表示总球数减少了 $500-400 = 100$ ，也就是拿走了一百个红球。

所以正确答案是 D。

If $r$ is the number of red balls, then $r = 0.8\cdot500 = 400.$ Therefore, if $b$ is the number of blue balls, then $b = 500-400 = 100.$

If there are $75\%$ red balls after removing balls, then there are $25\%$ blue balls. This means the total number of balls is $\dfrac{100}{0.25} = 400.$ This means the total number of balls decreased by $500-400 = 100.$

Thus, the answer is D .

13.

一个三角形的三边长度（单位为英寸）是三个连续整数。最短边长度是周长的 $30\%$ 。最长边长度是多少？

The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is $30\%$ of the perimeter. What is the length of the longest side?

$\ 7$

$\ 8$

$\ 9$

$\ 10$

$\ 11$

知识点：周长一次方程百分数

难度评级：1060

小提示：

使用最短边为周长 $30\%$ 这个条件

Use the condition that the shortest side is $30\%$ of the perimeter.

大提示：

设三条边长是从最短边开始的三个连续整数

Let the side lengths be three consecutive integers starting with the shortest.

解答：

设最短边长为 $s$ 。三边长度为 $s$ 、 $s+1$ 、 $s+2$ ，周长为 $3s+3$ 。因为 $s = 0.3(3s+3)$ ，所以 $s = 0.9s + 0.9$ 。因此 $0.1s = 0.9$ ，得到 $s = 9$ ，最长边为 $s + 2 = 11$ 。

所以正确答案是 E。

Let $s$ be the smallest length. Then, all the side lengths are $s,$ $s+1,$ $s+2.$ This would make the perimeter equal to $3s+3.$ Since $s = 0.3(3s+3),$ then $s = 0.9s + 0.9 .$ This makes $0.1s = 0.9,$ so $s = 9$ which makes the longest side length $s + 2 = 11.$

Thus, the answer is E .

14.

$2010$ 的质因数之和是多少？

What is the sum of the prime factors of $2010?$

$\ 67$

$\ 75$

$\ 77$

$\ 201$

$\ 210$

知识点：质因数分解

难度评级：960

小提示：

去掉 $2$ 、 $5$ 和 $3$ 后，剩下的因数是质数

After removing $2$ , $5$ , and $3$ , the remaining factor is prime.

大提示：

利用 $2010$ 能被 $10$ 和 $3$ 整除来分解质因数

Factor $2010$ using divisibility by $10$ and by $3$ .

解答：

首先，质数 $2$ 、 $5$ 是 $2010$ 的因数，因为它是 $10$ 的倍数。 $2010$ 除以 $10$ 得 $201$ 。接着， $3$ 是 $201$ 的因数，因为 $201$ 的数位和是 $3$ 的倍数。再除以 $3$ 得到质数 $67$ 。因此质因数为 $2$ 、 $3$ 、 $5$ 、 $67$ ，它们的和为 $77$ 。

所以正确答案是 C。

First, the primes $2,$ $5$ are factors of $2010$ since it is a multiple of $10.$ Dividing $2010$ by $10$ is $201.$ Then, $3$ is a factor of $201$ since the digit sum of $201$ is a multiple of $3.$ Dividing this by $3$ yields the prime number $67.$ This means the prime factors are $2,$ $3,$ $5,$ $67,$ which makes their sum $77.$

Thus, the correct answer is C .

15.

一个罐子里有 $5$ 种不同颜色的软糖。 $30\%$ 是蓝色， $20\%$ 是棕色， $15\%$ 是红色， $10\%$ 是黄色，另外 $30$ 颗是绿色。如果把一半蓝色软糖换成棕色软糖，那么会有多少颗棕色软糖？

A jar contains $5$ different colors of gum drops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and the other $30$ gum drops are green. If half of the blue gum drops are replaced by brown gum drops, how many gum drops will be brown?

$\ 35$

$\ 36$

$\ 42$

$\ 48$

$\ 64$

知识点：百分数

难度评级：1200

小提示：

用绿色软糖数量求总数，再把一半蓝色数量转到棕色

Use the green count to find the total, then move half the blue count to brown.

大提示：

先求罐子中绿色软糖占百分之多少

Find what percent of the jar is green.

解答：

绿色所占百分比是 $100\%$ 减去其他颜色百分比之和： $100\%-30\%-20\%-$ $15\%-10\% = 25\%$

因为 $30$ 颗绿色软糖是总数的 $25\%$ ，所以软糖总数为 $\dfrac { 30}{0.25} = 120$ 。

一开始棕色软糖有 $0.2\cdot120 = 24$ 颗，蓝色软糖有 $0.3\cdot120 = 36$ 颗。把一半蓝色换成棕色会增加 $\dfrac{36}{2} = 18$ 颗棕色软糖。因此棕色软糖共有 $24+18= 42$ 颗。

所以正确答案是 C。

Since we have percentages for every color except green, the percent of green is $100\%$ minus the sum of the other colors. This would make the percent of green equal to $100\%-30\%-20\%-$ $15\%-10\% = 25\%$

Since we know $30$ gum drops is $25\%,$ we know that the total number of gum drops is $\dfrac { 30}{0.25} = 120.$

This means there are $0.2\cdot120 = 24$ brown gum drops to start and $0.3\cdot120 = 36$ blue gum drops. If half of the blue gum drops are turned to brown, then $\dfrac{36}{2} = 18$ more brown gum drops are added. Therefore, we have $24+18= 42$ brown gum drops.

Thus, the answer is C .

16.

一个正方形和一个圆的面积相同。正方形边长与圆半径之比是多少？

A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?

$\ \dfrac{\sqrt{\pi}}{2}$

$\ \sqrt{\pi}$

$\ \pi$

$\ 2\pi$

$\ \pi^{2}$

知识点：圆面积正方形（几何）

难度评级：1000

小提示：

设 $s^2=\pi r^2$ ，求 $\frac{s}{r}$

Set $s^2=\pi r^2$ and solve for $\frac{s}{r}$ .

大提示：

分别用边长和半径表示正方形面积与圆面积

Write the square area and circle area in terms of side length and radius.

解答：

设正方形边长为 $s$ ，圆半径为 $r$ 。由于面积相同， $s^2 = \pi r^2$ 。因此 $(\dfrac sr)^2 = \pi$ ，所以 $\frac sr = \sqrt{ \pi}$ 。

所以正确答案是 B。

Let $s$ be the side length of the square and let $r$ be the radius of the circle. Then, since they have the same areas, $s^2 = \pi r^2 .$ This means $(\dfrac sr)^2 = \pi,$ so $\frac sr = \sqrt{ \pi} .$

Thus, the answer is B .

17.

图中八边形由 $10$ 个单位正方形组成。 $\overline{PQ}$ 下方部分是一个单位正方形和一个底为 $5$ 的三角形。如果 $\overline{PQ}$ 平分这个八边形的面积，那么 $\dfrac{XQ}{QY}$ 是多少？

The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5.$ If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}?$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

知识点：面积分割三角形面积

难度评级：1670

小提示：

在 $PQ$ 下方，减去单位正方形后得到一个三角形面积方程

Below $PQ$ , subtract the unit square to get a triangle area equation.

大提示：

直线 $PQ$ 将八边形面积分成相等的两部分

The line $PQ$ cuts the octagon into two equal areas.

解答：

因为 $PQ$ 平分面积，直线下方的面积为 $5$ 。去掉右侧的单位正方形后，底部是一个底为 $5$ 、面积为 $4$ 的三角形。设这个三角形的底为 $PZ$ 。

面积为 $4$ ，所以 $\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4$ $\implies ZQ = 1.6\text{。}$

因此 $QY = QZ-1 = 0.6$ ，且 $XQ = 2-QZ = 0.4$ 。于是 $\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23$ 。

所以正确答案是 D。

Since $PQ$ bisects the area, the area under the line is $5.$ Removing the square on the right makes the bottom a triangle of base $5$ with area $4.$ Let the base of this triangle be $PZ.$

The area being $4$ means $\dfrac{(PZ)(ZQ)}{2}=\frac{5(ZQ)}{2} = 4$ $\implies ZQ = 1.6.$

Therefore, $QY = QZ-1 = 0.6$ and $XQ = 2-QZ = 0.4.$ This would make $\dfrac{XQ}{QY} = \dfrac {0.4}{0.6} = \dfrac 23.$

Thus, the answer is D .

18.

一扇装饰窗由一个长方形和两端的半圆组成。 $AD$ 与 $AB$ 的比为 $3:2$ ，且 $AB=30$ 英寸。长方形面积与两个半圆面积之和的比是多少？

A decorative window is made up of a rectangle with semicircles on either end. The ratio of $AD$ to $AB$ is $3:2,$ and $AB=30$ inches. What is the ratio of the area of the rectangle to the combined areas of the semicircles?

$\ 2:3$

$\ 3:2$

$\ 6:\pi$

$\ 9:\pi$

$\ 30 :\pi$

知识点：圆面积矩形面积比

难度评级：1240

小提示：

用比值 $AD:AB=3:2$ ，将长方形高度用 $AB$ 表示

Use the ratio $AD:AB=3:2$ to express the rectangle height in terms of $AB$ .

大提示：

两个半圆合起来是一个整圆

The two semicircles combine to one full circle.

解答：

两个半圆合起来是一个直径为 $d=30$ 的圆，其半径为 $\dfrac{d}{2}$ 。因此两个半圆的总面积为 $(\dfrac{d}{2})^2 \cdot \pi = \dfrac{\pi d^2}{4}\text{。}$

因为 $AD:AB=3:2$ 且 $AB=d$ ，所以 $AD = \dfrac 32 d$ 。长方形面积为 $\dfrac 32 d^2$ 。所求比值为 $\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}$ 。

所以正确答案是 C。

Combining the semicircles would make a circle of diameter $d=30.$ This would make the radius equal to $\dfrac{d}{2}.$ Therefore, the combined area of the semicircles is $(\dfrac{d}{2})^2 \cdot \pi = \dfrac{\pi d^2}{4}.$

Side $AD = \dfrac 32 d$ because $AD:AB=3:2$ and $AB=d.$ The area of the rectangle is therefore $\dfrac 32 d^2.$ The ratio of the area of the rectangle to the area of the semicircles is $\dfrac{\dfrac 32 d^2}{\dfrac{\pi d^2}{4}} = \dfrac{6}{\pi}.$

Thus, the answer is C .

19.

图中的两个圆有相同的圆心 $C$ 。弦 $\overline{AD}$ 在 $B$ 点与内圆相切， $AC$ 长 $10$ ，且弦 $\overline{AD}$ 长 $16$ 。两个圆之间区域的面积是多少？

The two circles pictured have the same center $C.$ Chord $\overline{AD}$ is tangent to the inner circle at $B,$ $AC$ is $10,$ and chord $\overline{AD}$ has length $16.$ What is the area between the two circles?

$\ 36 \pi$

$\ 49 \pi$

$\ 64 \pi$

$\ 81 \pi$

$\ 100 \pi$

知识点：圆面积弦勾股定理

难度评级：1390

小提示：

半条弦、内圆半径和外圆半径组成一个直角三角形

Half the chord, the inner radius, and the outer radius form a right triangle.

大提示：

指向切点的半径垂直于切线

The radius to a tangent point is perpendicular to the chord.

解答：

两圆之间的面积等于大圆面积减去小圆面积，即 $(AC)^2\pi - (CB)^2 \pi$ $= \pi(AC^2 - CB^2)\text{。}$ 由勾股定理， $AC^2 - CB^2 = AB^2\text{。}$ 因此只需求 $AB^2 \pi$ 。

因为 $AB$ 是 $AD$ 的一半，所以 $AB = 8$ 。于是 $AB^2 \pi = 64 \pi$ 。

所以正确答案是 C。

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be $(AC)^2\pi - (CB)^2 \pi$ $= \pi(AC^2 - CB^2).$ By the Pythagorean Theorem, we can get $AC^2 - CB^2 = AB^2.$ Therefore, we need to find $AB^2 \pi.$

Since $AB$ is half of $AD,$ we get $AB = 8.$ This makes $AB^2 \pi = 64 \pi.$

Thus, the answer is C .

20.

一个房间里， $\frac{2}{5}$ 的人戴手套， $\frac{3}{4}$ 的人戴帽子。房间里至少有多少人同时戴帽子和手套？

In a room, $\frac{2}{5}$ of the people are wearing gloves, and $\frac{3}{4}$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and gloves?

$\ 3$

$\ 5$

$\ 8$

$\ 15$

$\ 20$

知识点：容斥原理最小公倍数

难度评级：1610

小提示：

要最小化重叠，如果可能，让每个人都至少属于两个群体之一

Minimize overlap by making everyone counted by at least one of the two groups if possible.

大提示：

总人数必须使两个分数对应的人数都是整数

The total number of people must make both fractions whole numbers.

解答：

因为房间中 $\dfrac 25$ 的人戴手套，所以总人数必须是 $5$ 的倍数。因为 $\dfrac 34$ 的人戴帽子，所以总人数必须是 $4$ 的倍数。因此总人数必须是 $20$ 的倍数。

另外，由容斥原理可以写出下面的关系：同时戴两者的比例 = 戴手套的比例 + 戴帽子的比例 − 至少戴其中一种的比例。

所求比例等于 $\dfrac{2}{5} + \dfrac 34 -$ 至少戴一种物品的比例。要使同时戴两者的人数最少，就让至少戴一种物品的比例尽可能大，最多为 $1$ 。所以同时戴两者的比例至少为 $\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}$ 。

取最小的正总人数 $20$ ，在这 $20$ 人中，同时戴两者的人数为 $3$ 。

所以正确答案是 A。

Since our room has $\dfrac 25$ of the people wearing gloves, the number of people must be a multiple of $5.$ Since our room has $\dfrac 34$ of the people wearing hats, the number of people must be a multiple of $4.$ Therefore, the people in the room must be a multiple of $20.$

Now, we can also use the following formula by the principle of inclusion exclusion: Fraction of people wearing both = Fraction of people wearing gloves + Fraction of people wearing hats - Fraction of people wearing either.

This makes our desired fraction equal to $\dfrac{2}{5} + \dfrac 34 -$ Fraction of people who wear either. If we wish to minimize the number who wear both, we maximize the fraction of people who wear either, up to $1.$ Therefore, the fraction of people that wear both is $\dfrac{2}{5} + \dfrac 34- 1 = \dfrac 3{20}.$

Since our number is a (positive) multiple of $20,$ we have the number of people wearing both as $3$ if we choose to have just $20$ people.

Therefore, A is the correct answer.

21.

惠很爱读书。她买了一本畅销书《数学真美》。第一天，惠读了全书页数的 $\frac{1}{5}$ 又多读 $12$ 页；第二天，她读了剩余页数的 $\frac{1}{4}$ 又多读 $15$ 页；第三天，她读了剩余页数的 $\frac{1}{3}$ 又多读 $18$ 页。随后她发现只剩 $62$ 页要读，并在次日读完。这本书共有多少页？

Hui is an avid reader. She bought a copy of the bestseller Math is Beautiful. On the first day, Hui read $\frac{1}{5}$ of the pages plus $12$ more, and on the second day she read $\frac{1}{4}$ of the remaining pages plus $15$ pages. On the third day she read $\frac{1}{3}$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?

$\ 120$

$\ 180$

$\ 240$

$\ 300$

$\ 360$

知识点：分数逆推法

难度评级：1520

小提示：

先倒推额外读的页数，再倒推每天读掉的分数

Undo the extra pages first, then undo the fraction read each day.

大提示：

从第三天后剩下的 $62$ 页倒推

Work backward from the $62$ pages left after the third day.

解答：

第三天后剩下 $62$ 页。在读最后额外的 $18$ 页之前，她还剩 $80$ 页。这是当时剩余页数的 $\dfrac 23$ ，所以第三天开始前她剩下 $120$ 页。

在读第二天额外的 $15$ 页之前，她还剩 $120+15=135$ 页。这是当时剩余页数的 $\dfrac 34$ ，所以第二天开始前她剩下 $180$ 页。

在读第一天额外的 $12$ 页之前，她还剩 $180+12=192$ 页。这是当时剩余页数的 $\dfrac 45$ ，所以第一天开始前她有 $240$ 页，也就是全书 $240$ 页。

所以正确答案是 C。

The pages left after the third day is $62.$ Before reading the last $18$ pages, she had $80$ pages left. This is $\dfrac 23$ of the pages remaining, so she had $120$ pages left before the third day.

Before reading the $15$ pages, she had $120+15=135$ pages left. This is $\dfrac 34$ of the pages remaining, so she had $180$ pages left before the second day.

Before reading the $12$ pages, she had $180+12=192$ pages left. This is $\dfrac 45$ of the pages remaining, so she had $240$ pages left before the first day, making the book $240$ pages.

Thus, the answer is C .

22.

一个三位数的百位数字比个位数字大 $2$ 。将这个三位数的数字倒序排列，再用原三位数减去倒序后的数。结果的个位数字是多少？

The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?

$0$

$2$

$4$

$6$

$8$

知识点：位值数字

难度评级：1260

小提示：

倒序数相减时，十位数字会抵消

The tens digit cancels when the reversed number is subtracted.

大提示：

用个位数字表示原数，并让百位数字比个位数字大二

Represent the original number with hundreds digit two more than the units digit.

解答：

设个位数字为 $u$ ，十位数字为 $t$ 。则百位数字为 $u+2$ 。原数为 $100(u+2) +10t+u$ $=101u+200+10t\text{，}$ 倒序数为 $100u +10t+(u+2)$ $=101u+2+10t\text{。}$ 两数之差为 $(101u+10t+200)-$ $(101u+10t+2)=198\text{。}$ 因此结果的个位数字是 $8$ 。

所以正确答案是 E。

Let the units digit be $u,$ and let the tens digit be $t.$ This makes the hundreds digit be $u+2.$ This makes the number equal to $100(u+2) +10t+u$ $=101u+200+10t$ and the reversed number is $100u +10t+(u+2)$ $=101u+2+10t.$ This makes the difference equal to $(101u+10t+200)-$ $(101u+10t+2)=198.$ This makes the units digit $8.$

Therefore, the answer is E .

23.

半圆 $POQ$ 和 $ROS$ 都经过圆心 $O$ 。两个半圆面积之和与圆 $O$ 面积之比是多少？

Semicircles $POQ$ and $ROS$ pass through the center of circle $O.$ What is the ratio of the combined areas of the two semicircles to the area of circle $O?$

$\dfrac{\sqrt 2}{4}$

$\dfrac{1}{2}$

$\dfrac{2}{\pi}$

$\dfrac{2}{3}$

$\dfrac{\sqrt 2}{2}$

知识点：圆面积勾股定理面积比

难度评级：1540

小提示：

两个半圆合起来的面积等于一个半径为 $1$ 的圆的面积

The two semicircles together have the area of one circle of radius $1$ .

大提示：

从坐标求圆 $O$ 的半径

Find the radius of circle $O$ from the coordinates.

解答：

每个半圆的面积是 $\pi \dfrac {r^2}{2}$ 。它们的半径都是 $1$ ，所以两个半圆的总面积为 $\pi \frac{1}{2} + \pi \frac 12 = \pi$ 。

大圆半径等于 $OQ$ 的长度，即 $\sqrt{1^2+1^2} = \sqrt 2$ 。它的面积为 $\pi r^2 = \pi(\sqrt{2})^2 = 2\pi$ 。

因此比值为 $\dfrac{\pi}{2\pi} = \frac 12$ 。

所以正确答案是 B。

The area of each of the semicircles is $\pi \dfrac {r^2}{2} .$ Each of them has a radius of $1,$ so their combined area is $\pi \frac{1}{2} + \pi \frac 12 = \pi.$

Next, the radius of the larger circle is equal to the length of $OQ,$ which is equal to $\sqrt{1^2+1^2} = \sqrt 2.$ Its area is $\pi r^2 = \pi(\sqrt{2})^2 = 2\pi.$

This means the ratio is $\dfrac{\pi}{2\pi} = \frac 12.$

Thus, the answer is B .

24.

三个数 $10^8$ 、 $5^{12}$ 和 $2^{24}$ 的正确大小顺序是什么？

What is the correct ordering of the three numbers, $10^8,$ $5^{12},$ and $2^{24}?$

2^{24} < 10^8 < 5^{12}

2^{24} < 5^{12} < 10^8

5^{12} < 2^{24} < 10^8

10^8 < 5^{12} < 2^{24}

10^8 < 2^{24} < 5^{12}

知识点：指数

难度评级：1480

小提示：

比较 $2^{24}$ 、 $10^8=(2\cdot5)^8$ 和 $5^{12}$

Compare $2^{24}$ , $10^8=(2\cdot5)^8$ , and $5^{12}$ .

大提示：

把幂改写成带有可比较的八次方因子的形式

Rewrite the powers so they have comparable eighth-power factors.

解答：

首先， $2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8\text{。}$

接着， $10^8 = (2^8)(5^8) =$ $(4^4)(5^8) < (5^4)(5^8) = 5^{12}\text{。}$ 因此 $2^{24} < 10^8 < 5^{12}$ 。

所以正确答案是 A。

First, we get $2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.$

Next, we get $10^8 = (2^8)(5^8) =$ $(4^4)(5^8) < (5^4)(5^8) = 5^{12}.$ This means $2^{24} < 10^8 < 5^{12} .$

Thus, the answer is A .

25.

乔每天在学校爬一段 $6$ 级的楼梯。乔每次可以爬 $1$ 、 $2$ 或 $3$ 级。例如，乔可以先爬 $3$ 级，再爬 $1$ 级，再爬 $2$ 级。乔有多少种爬完楼梯的方法？

Every day at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1,$ $2,$ or $3$ at a time. For example, Jo could climb $3,$ then $1,$ then $2.$ In how many ways can Jo climb the stairs?

$\ 13$

$\ 18$

$\ 20$

$\ 22$

$\ 24$

知识点：分拆与有序分拆递推计数

难度评级：1540

小提示：

第一步可以爬 $1$ 、 $2$ 或 $3$ 级，这会给出递推关系

The first step can be $1$ , $2$ , or $3$ stairs, giving a recurrence.

大提示：

设 $w_n$ 为爬 $n$ 级楼梯的方法数

Let $w_n$ be the number of ways to climb $n$ stairs.

解答：

设 $w_n$ 为爬 $n$ 级楼梯的方法数。对于 $n\ge4$ ，第一步可以爬 $1$ 、 $2$ 或 $3$ 级，所以 $w_n=w_{n-1}+w_{n-2}+w_{n-3}$ 。

有 $w_1=1$ 、 $w_2=2$ 、 $w_3=4$ 。因此 $w_4=4+2+1=7$ 、 $w_5=7+4+2=13$ 、 $w_6=13+7+4=24$ 。

所以正确答案是 E。

Let $w_n$ be the number of ways to climb $n$ stairs. For $n\ge4$ , the first step can be $1$ , $2$ , or $3$ stairs, so $w_n=w_{n-1}+w_{n-2}+w_{n-3}$ .

We have $w_1=1$ , $w_2=2$ , and $w_3=4$ . Therefore $w_4=4+2+1=7$ , $w_5=7+4+2=13$ , and $w_6=13+7+4=24$ .

Thus, the answer is E .