2009 AMC 8 详解 | LIVE by Po-Shen Loh

1.

布里奇特在杂货店买了一袋苹果。她把一半苹果给了安。然后她给了凯茜 $3$ 个苹果，自己留下 $4$ 个苹果。布里奇特买了多少个苹果？

Bridget bought a bag of apples at the grocery store. She gave half of the apples to Ann. Then she gave Cassie $3$ apples, keeping $4$ apples for herself. How many apples did Bridget buy?

$3$

$4$

$7$

$11$

$14$

知识点：逆推法

难度评级：370

小提示：

从布里奇特自己留下和给凯茜的苹果数倒推

Work backward from the apples Bridget kept and gave to Cassie.

大提示：

这些苹果就是没有给安的那一半

Those apples were the half not given to Ann.

解答：

可以倒推，从布里奇特自己留下的 $4$ 个苹果开始。加上她给凯茜的 $3$ 个苹果，这一半共有 $7$ 个苹果。

因为她把原来苹果的一半给了安，所以再乘以 $2$ 。 $7 \cdot 2 = 14$ ，因此布里奇特一开始有 $14$ 个苹果。

所以正确答案是 E。

We can work backwards, starting with the $4$ apples that Bridget kept for herself. Adding the $3$ apples that she gave Cassie, she now has $7$ apples.

Finally, we multiply this value by $2$ since she gave half of her initial apples to Ann. $7 \cdot 2 = 14,$ so Bridget started off with $14$ apples.

Thus, E is the correct answer.

2.

当地汽车经销店平均每卖出 $4$ 辆跑车，就会卖出 $7$ 辆轿车。经销店预测下个月会卖出 $28$ 辆跑车。那么预计会卖出多少辆轿车？

On average, for every $4$ sports cars sold at the local car dealership, $7$ sedans are sold. The dealership predicts that it will sell $28$ sports cars next month. How many sedans does it expect to sell?

$7$

$32$

$35$

$49$

$112$

知识点：比与比例

难度评级：450

小提示：

使用 $4$ 辆跑车对应 $7$ 辆轿车的比例

Use the ratio $4$ sports cars to $7$ sedans.

大提示：

从 $4$ 辆跑车到 $28$ 辆跑车是乘以 $7$

Going from $4$ to $28$ sports cars multiplies by $7.$

解答：

列出比例 $\dfrac{4}{7} = \dfrac{28}{x}\text{。}$ 交叉相乘得 $4x = 7 \cdot 28$ 因而 $x = 49\text{。}$

所以正确答案是 D。

Set up the proportion $\dfrac{4}{7} = \dfrac{28}{x}.$ Cross-multiplying gives $4x = 7 \cdot 28$ and hence $x = 49.$

Thus, D is the correct answer.

3.

图像显示苏珊娜骑自行车的恒定速度。如果她以同样速度一共骑半小时，她会骑多少英里？

The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of half an hour at the same speed, how many miles will she have ridden?

$5$

$5.5$

$6$

$6.5$

$7$

知识点：速率数据与图表解读

难度评级：560

小提示：

从图像上读出一个清楚的点

Read one clear point from the graph.

大提示：

如果 $15$ 分钟骑 $3$ 英里，那么 $30$ 分钟是两倍时间

If $15$ minutes gives $3$ miles, then $30$ minutes is twice as long.

解答：

从图中可知，苏珊娜骑了 $3$ 英里，用时 $15$ 分钟。因此在 $30$ 分钟内，她会骑 $6$ 英里。

所以正确答案是 C。

From the graph, we can see that Suzanna rides $3$ miles in $15$ minutes. This means that in $30$ minutes, she will have ridden $6$ miles.

Thus, C is the correct answer.

4.

下面显示的五块拼片可以拼成下方五个图形中的四个。哪一个图形不能拼成？

The five pieces shown below can be arranged to form four of the five figures below. Which figure cannot be formed?

知识点：铺砖

难度评级：720

小提示：

找出这五块拼片中必须出现在最终图形里的一个特征

Look for a feature in the five pieces that must appear in the final figure.

大提示：

那块长 $5$ 个小方格的拼片必须能作为一段连续部分放进去

The long $5$ -square piece must fit as an unbroken segment somewhere.

解答：

注意选项 B 中没有任何长 $5$ 个方格的连续线段。这说明那块长 $5$ 个方格的拼片无法放入该图形。

所以正确答案是 B。

Note that option B does not have any segments in it that are $5$ blocks long. This means that it is impossible to arrange the $5$ block long piece to fit within the figure.

Thus, B is the correct answer.

5.

一个数列以 $1$ 、 $2$ 、 $3$ 开始。数列的第四个数是前三个数之和，即 $1 + 2 + 3 = 6$ 。同样地，第四个数之后的每个数都是它前面三个数之和。这个数列的第八个数是多少？

A sequence of numbers starts with $1,$ $2,$ and $3.$ The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1 + 2 + 3 = 6.$ In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

$11$

$20$

$37$

$68$

$99$

知识点：递推

难度评级：770

小提示：

一项一项地生成这个数列

Generate the sequence one term at a time.

大提示：

每个新项都是前面三项之和

Each new term is the sum of the previous three terms.

解答：

数列开始为 $1$ 、 $2$ 、 $3$ 。

接下来的项为 $1+2+3=6$ 、 $2+3+6=11$ 、 $3+6+11=20$ 、 $6+11+20=37$ 、 $11+20+37=68$ 。

所以正确答案是 D。

The sequence begins $1,$ $2,$ $3.$

The next terms are $1+2+3=6,$ $2+3+6=11,$ $3+6+11=20,$ $6+11+20=37,$ and $11+20+37=68.$

Thus, D is the correct answer.

6.

史蒂夫的空游泳池装满时可容纳 $24{,}000$ 加仑水。它将由 $4$ 根水管注水，每根水管每分钟供水 $2.5$ 加仑。装满史蒂夫的游泳池需要多少小时？

Steve’s empty swimming pool will hold $24{,}000$ gallons of water when full. It will be filled by $4$ hoses, each of which supplies $2.5$ gallons of water per minute. How many hours will it take to fill Steve’s pool?

$40$

$42$

$44$

$46$

$48$

知识点：速率单位换算

难度评级：870

小提示：

先求四根水管合起来的注水速度

First find the combined filling rate of the four hoses.

大提示：

求出总分钟数后，再换算成小时

Convert minutes to hours after finding the total number of minutes.

解答：

$4$ 根水管合起来每分钟注入 $2.5 \cdot 4 = 10$ 加仑水。

要注满 $24{,}000$ 加仑，水管需要 $24{,}000 \div 10 = 2{,}400$ 分钟。

$2{,}400$ 分钟等于 $2{,}400 \div 60 = 40$ 小时。

所以正确答案是 A。

The $4$ hoses together fill the pool with $2.5 \cdot 4 = 10$ gallons of water per minute.

To fill $24{,}000$ gallons, it will take the hoses $24{,}000 \div 10 = 2{,}400$ minutes to fill the pool.

$2{,}400$ minutes is the same as $2{,}400 \div 60 = 40$ hours.

Thus, A is the correct answer.

7.

三角形地块 $ACD$ 位于阿斯彭路、布朗路和一条铁路之间。主街为东西方向，铁路为南北方向。图中的数字表示距离，单位为英里。铁路轨道的宽度可以忽略。地块 $ACD$ 的面积是多少平方英里？

The triangular plot of land $ACD$ lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land $ACD?$

$2$

$3$

$4.5$

$6$

$9$

知识点：三角形面积

难度评级：900

小提示：

用 $CD$ 作为三角形 $ACD$ 的底

Use $CD$ as a base for triangle $ACD.$

大提示：

从 $A$ 到铁路的垂直距离由 $AB$ 表示

The perpendicular distance from $A$ to the railroad is shown by $AB.$

解答：

三角形 $ADC$ 的底边是 $CD$ ，长度为 $3$ 。高也是 $3$ 。

因此 $ADC$ 的面积为 $\dfrac{1}{2} \cdot 3 \cdot 3 = 4.5$ 。

所以正确答案是 C。

The base of $ADC$ is $CD,$ which is $3.$ The altitude is $3$ as well.

Therefore, the area of $ADC$ is $\dfrac{1}{2} \cdot 3 \cdot 3 = 4.5.$

Thus, C is the correct answer.

8.

一个长方形的长增加 $10\%$ ，宽减少 $10\%$ 。新面积是原面积的百分之多少？

The length of a rectangle is increased by $10\%$ and the width is decreased by $10\%.$ What percent of the old area is the new area?

$90$

$99$

$100$

$101$

$110$

知识点：面积百分数

难度评级：920

小提示：

将长和宽的缩放因子相乘

Multiply the scale factors for length and width.

大提示：

新面积是原面积的 $1.10\cdot0.90$ 倍

The new area is $1.10\cdot0.90$ times the old area.

解答：

设原来的长和宽为 $l$ 和 $w$ ，则原面积为 $lw$ 。

新的长为 $1.1l$ ，新的宽为 $0.9w$ 。因此新面积为 $1.1\cdot0.9lw=0.99lw$ 。

这说明新面积是原面积的 $99\%$ 。

所以正确答案是 B。

Let the old length and width be $l$ and $w,$ so the old area is $lw.$

The new length is $1.1l,$ and the new width is $0.9w.$ Thus the new area is $1.1\cdot0.9lw=0.99lw.$

This shows that the new area is $99\%$ of the old area.

Thus, B is the correct answer.

9.

在一个等边三角形的一边上作一个正方形。在正方形的一条不相邻的边上作一个正五边形，如图所示。在五边形的一条不相邻的边上作一个正六边形。用同样方式继续作正多边形，直到作出正八边形。所得图形的外边界有多少条边？

Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a regular hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?

$21$

$23$

$25$

$27$

$29$

知识点：正多边形基本计数

难度评级：1120

小提示：

只数每个多边形留在外边界上的边

Count only the outside sides of each polygon.

大提示：

中间的多边形各共享两条边，两端的多边形各共享一条边

Middle polygons share two sides; the two end polygons share one side each.

解答：

三角形和八边形在链的两端，所以各有一条边成为共享边而不在外边界上。正方形、五边形、六边形和七边形在中间，所以各有两条边成为共享边。

所得图形有 $\begin{aligned} &(3-1)+(4-2) \\ &\quad {}+(5-2)+(6-2) \\ &\quad {}+(7-2)+(8-1)=23 \end{aligned}$ 条边。

所以正确答案是 B。

The triangle and octagon are at the ends of the chain, so each loses one side to a shared edge. The square, pentagon, hexagon, and heptagon are in the middle, so each loses two sides to shared edges.

The resulting polygon has $\begin{aligned} &(3-1)+(4-2) \\ &\quad {}+(5-2)+(6-2) \\ &\quad {}+(7-2)+(8-1)=23 \end{aligned}$ sides.

Thus, B is the correct answer.

10.

在一个由 $64$ 个单位正方形组成的棋盘上，随机选择一个单位正方形。所选正方形不接触棋盘外边缘的概率是多少？

On a checkerboard composed of $64$ unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?

$\dfrac{1}{16}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{49}{64}$

知识点：基本概率

难度评级：980

小提示：

只有严格位于边框内部的正方形不接触外边缘

Only the squares strictly inside the border do not touch the edge.

大提示：

一个 $8\times8$ 棋盘有 $6\times6$ 的内部区域

An $8\times8$ board has a $6\times6$ interior.

解答：

内部正方形共有 $(8 - 2)^2 = 6^2 = 36$ 个。

因此选中其中一个的概率是 $\dfrac{36}{64} = \dfrac{9}{16}\text{。}$

所以正确答案是 D。

There are $(8 - 2)^2 = 6^2 = 36$ squares on the interior.

This means that the probability of choosing one of these squares is $\dfrac{36}{64} = \dfrac{9}{16}.$

Thus, D is the correct answer.

11.

阿马科中学的书店出售铅笔，每支铅笔价格为整数美分。一些七年级学生每人买了一支铅笔，总共支付 $1.43$ 美元。 $30$ 名六年级学生中的一些人每人买了一支铅笔，总共支付 $1.95$ 美元。买铅笔的六年级学生比七年级学生多多少人？

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?

$1$

$2$

$3$

$4$

$5$

知识点：最大公约数质因数分解

难度评级：1290

小提示：

铅笔价格必须同时整除两个以美分表示的总价

The pencil price must divide both total costs in cents.

大提示：

利用最多只有 $30$ 名六年级学生买铅笔，排除 $1$ 美分的价格

Use the fact that at most $30$ sixth graders bought pencils to rule out a $1$ -cent price.

解答：

买铅笔的七年级学生人数等于 $143$ 除以一支铅笔的价格。同样，买铅笔的六年级学生人数等于 $195$ 除以一支铅笔的价格。

因此铅笔价格同时整除 $143$ 和 $195$ 。质因数分解得 $143 = 11 \cdot 13$ 和 $195 = 3 \cdot 5 \cdot 13\text{。}$ 同时整除 $143$ 和 $195$ 的正整数只有 $1$ 和 $13$ 。

如果一支铅笔价格是 $1$ 美分，那么会有 $195$ 名六年级学生买铅笔，这是不可能的。因此一支铅笔价格是 $13$ 美分。

于是 $143 \div 13 = 11$ 名七年级学生买铅笔， $195 \div 13 = 15$ 名六年级学生买铅笔。所以六年级学生比七年级学生多 $4$ 人。

所以正确答案是 D。

The number of seventh graders that bought a pencil is $143$ divided by the price of a pencil. Similarly, the number of sixth graders that bought a pencil is $195$ divided by the price of a pencil.

This means that the price of a pencil divides both $143$ and $195.$ Prime factorizing, we get $143 = 11 \cdot 13$ and $195 = 3 \cdot 5 \cdot 13.$ The only numbers that divide both $143$ and $195$ are $1$ and $13.$

If $1$ cent was the price of the pencil, that means $195$ sixth graders bought pencils, which is not possible. Therefore, the price of a pencil is $13$ cents.

This means that $143 \div 13 = 11$ seventh graders bought a pencil, and $195 \div 13 = 15$ sixth graders bought a pencil. Therefore, $4$ more sixth graders than seventh graders bought pencils.

Thus, D is the correct answer.

12.

如图所示的两个转盘各转一次，并各自落在一个编号扇区上。两个扇区中的数字之和为质数的概率是多少？

The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{7}{9}$

$\dfrac{5}{6}$

知识点：基本概率质数

难度评级：1120

小提示：

列一个 $3\times3$ 表格记录所有可能的和

Make a $3\times3$ table of possible sums.

大提示：

在所有可能结果中，只有和为 $9$ 的情况不是质数

Only sums of $9$ are not prime among the possible outcomes.

解答：

可以求出每种可能结果中两个数字的和。

$\begin{gather*} 1 + 2 = 3 \\ 1 + 4 = 5 \\ 1 + 6 = 7 \\ 3 + 2 = 5 \\ 3 + 4 = 7 \\ 3 + 6 = 9 \\ 5 + 2 = 7 \\ 5 + 4 = 9 \\ 5 + 6 = 11 \end{gather*}\text{。}$

只有 $2$ 个结果的和不是质数，也就是和为 $9$ 的两种情况。因此和为质数的概率是 $\dfrac{7}{9}$ 。

所以正确答案是 D。

We can find the sum of the two numbers in every possible outcome.

$\begin{gather*} 1 + 2 = 3 \\ 1 + 4 = 5 \\ 1 + 6 = 7 \\ 3 + 2 = 5 \\ 3 + 4 = 7 \\ 3 + 6 = 9 \\ 5 + 2 = 7 \\ 5 + 4 = 9 \\ 5 + 6 = 11. \end{gather*}$

There are only $2$ outcomes where the sum is not prime (the two instances when the sum is $9$ ). Therefore, the probability that the sum is prime is $\dfrac{7}{9}.$

Thus, D is the correct answer.

13.

一个三位整数包含数字 $1$ 、 $3$ 、 $5$ 各一次。这个整数能被 $5$ 整除的概率是多少？

A three-digit integer contains one of each of the digits $1,$ $3,$ and $5.$ What is the probability that the integer is divisible by $5?$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

知识点：基本概率整除性

难度评级：930

小提示：

一个数能被 $5$ 整除，当且仅当末位数字是 $0$ 或 $5$

A number is divisible by $5$ exactly when its last digit is $0$ or $5.$

大提示：

对数字 $1$ 、 $3$ 、 $5$ ，每个数字作为末位的可能性相同

With digits $1,$ $3,$ and $5,$ each digit is equally likely to be last.

解答：

这个数以 $1$ 、 $3$ 或 $5$ 结尾的可能性相同。

只有当末位是 $5$ 时，它才能被 $5$ 整除，概率为 $\dfrac{1}{3}$ 。

所以正确答案是 B。

The number is equally likely to end in $1,$ $3,$ or $5.$

It is divisible by $5$ only if the last digit is $5,$ which happens with probability $\dfrac{1}{3}.$

Thus, B is the correct answer.

14.

奥斯汀和坦普尔相距 $50$ 英里，由 $35$ 号州际公路相连。邦妮从奥斯汀开车到她女儿在坦普尔的家，平均速度为每小时 $60$ 英里。她把车留给女儿后，沿同一路线坐公交车返回奥斯汀，返程平均速度为每小时 $40$ 英里。整个往返行程的平均速度是多少英里每小时？

Austin and Temple are $50$ miles apart along Interstate $35.$ Bonnie drove from Austin to her daughter’s house in Temple, averaging $60$ miles per hour. Leaving the car with her daughter, Bonnie rode a bus back to Austin along the same route and averaged $40$ miles per hour on the return trip. What was the average speed for the round trip, in miles per hour?

$46$

$48$

$50$

$52$

$54$

知识点：调和平均数路程、速度与时间

难度评级：1290

小提示：

平均速度等于总路程除以总时间

Average speed is total distance divided by total time.

大提示：

两段 $50$ 英里的路程所用时间不同

The two $50$ -mile trips take different amounts of time.

解答：

从奥斯汀到坦普尔用时 $50 \div 60 = \dfrac{5}{6}$ 小时。从坦普尔到奥斯汀用时 $50 \div 40 = \dfrac{5}{4}$ 小时。因此往返总时间为 $\dfrac{5}{6} + \dfrac{5}{4} = \dfrac{25}{12}$ 小时。

往返总路程为 $2 \cdot 50 = 100$ 英里。因此往返平均速度为 $100 \div \dfrac{25}{12} = 48$ 英里每小时。

所以正确答案是 B。

The trip from Austin to Temple took $50 \div 60 = \dfrac{5}{6}$ hours. The trip from Temple to Austin took $50 \div 40 = \dfrac{5}{4}$ hours. This means that the total time for the round trip was $\dfrac{5}{6} + \dfrac{5}{4} = \dfrac{25}{12}$ hours.

The total distance of the round trip was $2 \cdot 50 = 100$ miles. Therefore, the average speed for the round trip was $100 \div \dfrac{25}{12} = 48$ miles per hour.

Thus, B is the correct answer.

15.

一份可做 $5$ 份热巧克力的食谱需要 $2$ 块巧克力、 $\dfrac{1}{4}$ 杯糖、 $1$ 杯水和 $4$ 杯牛奶。乔丹有 $5$ 块巧克力、 $2$ 杯糖、很多水和 $7$ 杯牛奶。如果她保持相同的配料比例，最多能做多少份热巧克力？

A recipe that makes $5$ servings of hot chocolate requires $2$ squares of chocolate, $\dfrac{1}{4}$ cup sugar, $1$ cup water and $4$ cups milk. Jordan has $5$ squares of chocolate, $2$ cups of sugar, lots of water, and $7$ cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make?

$5 \dfrac{1}{8}$

$6 \dfrac{1}{4}$

$7 \dfrac{1}{2}$

$8 \dfrac{3}{4}$

$9 \dfrac{7}{8}$

知识点：比与比例

难度评级：1220

小提示：

比较每种配料分别能支持多少批食谱

Compare how many recipe batches each ingredient can support.

大提示：

能支持批数最少的配料限制了份数

The ingredient that supports the fewest batches limits the servings.

解答：

需要找出哪种配料是限制因素。

乔丹的巧克力足够做 $5\div2=\dfrac{5}{2}$ 批，糖足够做 $2\div\dfrac{1}{4}=8$ 批，牛奶足够做 $7\div4=\dfrac{7}{4}$ 批。

牛奶是限制因素，所以乔丹可以做 $5\cdot\dfrac{7}{4}=\dfrac{35}{4}=8\dfrac{3}{4}$ 份。

所以正确答案是 D。

We need to find which ingredient is the limiting factor.

Jordan has enough chocolate for $5\div2=\dfrac{5}{2}$ batches, enough sugar for $2\div\dfrac{1}{4}=8$ batches, and enough milk for $7\div4=\dfrac{7}{4}$ batches.

The milk is limiting, so Jordan can make $5\cdot\dfrac{7}{4}=\dfrac{35}{4}=8\dfrac{3}{4}$ servings.

Thus, D is the correct answer.

16.

有多少个 $3$ 位正整数，其各位数字的乘积等于 $24$ ？

How many $3$ -digit positive integers have digits whose product equals $24?$

$12$

$15$

$18$

$21$

$24$

知识点：数字多重集排列分类讨论

难度评级：1350

小提示：

列出乘积为 $24$ 的无序数字三元组

List the unordered triples of digits whose product is $24.$

大提示：

三个数字互不相同的三元组有 $6$ 种排列，而 $(2,2,6)$ 有 $3$ 种排列

Triplets with three distinct digits have $6$ permutations, while $(2,2,6)$ has $3.$

解答：

小于 $10$ 且乘积为 $24$ 的整数三元组只有 $(1, 3, 8), (1, 4, 6), (2, 2, 6)\text{，}$ $(2, 3, 4)\text{。}$

有 $3$ 个数字互不相同的三元组，每个可以重新排列成 $6$ 个不同的 $3$ 位正整数。另一个三元组可以排列成 $3$ 个不同的 $3$ 位正整数。

总数为 $3 \cdot 6 + 3 = 21$ 。

所以正确答案是 D。

The only triples of integers less than $10$ that multiply to $24$ are $(1, 3, 8), (1, 4, 6), (2, 2, 6),$ $(2, 3, 4).$

The triples with $3$ distinct numbers can be rearranged to form $6$ distinct $3$ -digit positive integers. The other triple can be arranged to form $3$ distinct $3$ -digit positive integers.

This leaves a total of $3 \cdot 6 + 3 = 21$ integers.

Thus, D is the correct answer.

17.

正整数 $x$ 和 $y$ 是满足下列条件的两个最小正整数： $360$ 与 $x$ 的乘积是平方数，且 $360$ 与 $y$ 的乘积是立方数。 $x$ 与 $y$ 的和是多少？

The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y?$

$80$

$85$

$115$

$165$

$610$

知识点：质因数分解完全平方数完全幂

难度评级：1400

小提示：

将 $360$ 分解为质因数

Factor $360$ into primes.

大提示：

平方数要求指数为偶数，立方数要求指数为 $3$ 的倍数

Make exponents even for a square and multiples of $3$ for a cube.

解答：

一个数是完全平方数时，质因数分解中每个指数都必须是偶数。一个数是立方数时，指数都必须能被 $3$ 整除。

分解 $360$ ，得 $360 = 2^3 \cdot 3^2 \cdot 5\text{。}$ 要让 $360x$ 成为完全平方数且 $x$ 最小， $x$ 必须提供一个因子 $2$ 和一个因子 $5$ 。因此可取 $x = 10$ 。

要让 $360y$ 成为立方数， $y$ 必须提供一个因子 $3$ 和两个因子 $5$ 。因此可取 $y = 75$ ，所以 $x + y = 85$ 。

所以正确答案是 B。

For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by $3.$

We can factor $360$ to get $360 = 2^3 \cdot 3^2 \cdot 5.$ For $360x$ to be a perfect square and $x$ to be minimized, $x$ must have one factor of $2$ and one factor of $5.$ Therefore, we can let $x = 10.$

For $360y$ to be a cube, $y$ must have one factor of $3$ and two factors of $5.$ Therefore, we can let $y = 75,$ suggesting $x + y = 85.$

Thus, B is the correct answer.

18.

图中表示一个 $7$ 英尺乘 $7$ 英尺的地板，由面积为 $1$ 平方英尺的阴影瓷砖和非阴影瓷砖铺成。注意四个角是非阴影瓷砖。如果一个 $15$ 英尺乘 $15$ 英尺的地板也按同样方式铺，需要多少块非阴影瓷砖？

The diagram represents a $7$ -foot-by- $7$ -foot floor that is tiled with $1$ -square-foot shaded tiles and unshaded tiles. Notice that the corners have unshaded tiles. If a $15$ -foot-by- $15$ -foot floor is to be tiled in the same manner, how many unshaded tiles will be needed?

$49$

$57$

$64$

$96$

$126$

知识点：找规律完全平方数

难度评级：1310

小提示：

观察 $7\times7$ 地板中非阴影瓷砖的模式

Look for the pattern in the unshaded tiles of the $7\times7$ floor.

大提示：

对于奇数边长 $2k-1$ ，非阴影瓷砖数遵循 $k^2$

For an odd side length $2k-1,$ the unshaded tile count follows $k^2.$

解答：

在 $7\times7$ 的例子中，有 $4$ 行各含 $4$ 块非阴影瓷砖，所以共有 $4^2=16$ 块非阴影瓷砖。

对 $15\times15$ 且模式相同的地板，会有 $8$ 行各含 $8$ 块非阴影瓷砖。

因此非阴影瓷砖数为 $8^2=64$ 。

所以正确答案是 C。

In the $7\times7$ example, there are $4$ rows that contain $4$ unshaded tiles, for $4^2=16$ unshaded tiles.

For a $15\times15$ floor with the same pattern, there will be $8$ such rows with $8$ unshaded tiles each.

Thus the number of unshaded tiles is $8^2=64.$

Thus, C is the correct answer.

19.

一个等腰三角形的两个角分别为 $70^\circ$ 和 $x^\circ$ 。 $x$ 的三个可能值之和是多少？

Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ.$ What is the sum of the three possible values of $x?$

$95$

$125$

$140$

$165$

$180$

知识点：等腰三角形角度和分类讨论

难度评级：1310

小提示：

已知的 $70^\circ$ 角和 $x^\circ$ 角在等腰三角形中有三种可能位置

There are three ways the known $70^\circ$ angle and the $x^\circ$ angle can sit in an isosceles triangle.

大提示：

分别考虑它们是否为相等的角，或者其中一个是否为顶角

Consider whether they are the equal angles, or whether one is the vertex angle.

解答：

所有可能情形如下图所示。

在第一种情形中，由等腰三角形性质可得 $x = 70$ 。

在第二种情形中，有 $70 \cdot 2 + x = 180\text{，}$ 因此 $x = 40$ 。

在第三种情形中，有 $2x + 70 = 180\text{，}$ 因此 $x = 55$ 。

这些值之和为 $70 + 40 + 55 = 165\text{。}$

所以正确答案是 D。

All the following possibilities are shown below.

In the first scenario, we get $x = 70$ by the properties of the isosceles triangle.

In the second scenario, we get that $70 \cdot 2 + x = 180,$ from which we get that $x = 40.$

From the third scenario, we get that $2x + 70 = 180,$ from which we get that $x = 55.$

The sum of these values yields $70 + 40 + 55 = 165.$

Thus, D is the correct answer.

20.

从下方阵列中的八个点里选三个点作为顶点，可以形成多少个不全等三角形？

How many non-congruent triangles have vertices at three of the eight points in the array shown below?

$5$

$6$

$7$

$8$

$9$

知识点：全等（几何）系统列举对称性

难度评级：1470

小提示：

利用对称性，避免逐个统计所有三角形

Use symmetry to avoid counting every triangle separately.

大提示：

从两行点阵中选顶点后，按边长给三角形分类

Classify triangles by their side lengths after choosing vertices from the two-row array.

解答：

由对称性，只需列出每种可能三角形形状的一个代表。

一份完整的不全等三角形代表列表是 $ABE$ 、 $ABG$ 、 $ACE$ 、 $ABH$ 、 $AEH$ 、 $ACF$ 、 $ACH$ 和 $ADF$ 。

八个点形成的其他三角形都与这 $8$ 个三角形中的某一个全等。

所以正确答案是 D。

By symmetry, it is enough to list one representative of each possible triangle shape.

One complete list of non-congruent possibilities is $ABE,$ $ABG,$ $ACE,$ $ABH,$ $AEH,$ $ACF,$ $ACH,$ and $ADF.$

Every other triangle formed from the eight points is congruent to one of these $8$ triangles.

Thus, D is the correct answer.

21.

安迪和贝瑟妮有一个 $40$ 行 $75$ 列的数字矩形阵列。安迪把每一行的数相加，他得到的 $40$ 个行和的平均数是 $A$ 。贝瑟妮把每一列的数相加，她得到的 $75$ 个列和的平均数是 $B$ 。 $\dfrac{A}{B}$ 的值是多少？

Andy and Bethany have a rectangular array of numbers with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A.$ Bethany adds the numbers in each column. The average of her $75$ sums is $B.$ What is the value of $\dfrac{A}{B}?$

$\dfrac{64}{225}$

$\dfrac{8}{15}$

$1$

$\dfrac{15}{8}$

$\dfrac{225}{64}$

知识点：双重计数平均数

难度评级：1340

小提示：

设 $S$ 为阵列中所有数的和

Let $S$ be the sum of all entries in the array.

大提示：

安迪的平均行和为 $\frac{S}{40}$ ，贝瑟妮的平均列和为 $\frac{S}{75}$

Andy has average row sum $\frac{S}{40},$ while Bethany has average column sum $\frac{S}{75}.$

解答：

设 $S$ 是阵列中所有数的和。

安迪的 $40$ 个行和加起来也是 $S$ ，所以平均数为 $A=\dfrac{S}{40}$ 。贝瑟妮的 $75$ 个列和加起来也是 $S$ ，所以平均数为 $B=\dfrac{S}{75}$ 。

因此 $\dfrac{A}{B}=\dfrac{\frac{S}{40}}{\frac{S}{75}}=\dfrac{75}{40}=\dfrac{15}{8}$ 。

所以正确答案是 D。

Let $S$ be the sum of all numbers in the array.

Andy’s $40$ row sums also add to $S,$ so their average is $A=\dfrac{S}{40}.$ Bethany’s $75$ column sums also add to $S,$ so their average is $B=\dfrac{S}{75}.$

Therefore $\dfrac{A}{B}=\dfrac{\frac{S}{40}}{\frac{S}{75}}=\dfrac{75}{40}=\dfrac{15}{8}.$

Thus, D is the correct answer.

22.

从 $1$ 到 $1000$ 之间，有多少个整数不含数字 $1$ ？

How many whole numbers between $1$ and $1000$ do not contain the digit $1?$

$512$

$648$

$720$

$728$

$800$

知识点：数字乘法原理分类讨论

难度评级：1400

小提示：

按位数分类计数，或者把数补前导零

Count by number of digits, or pad numbers with leading zeroes.

大提示：

三个补零后的数位各有 $9$ 种选择（禁止使用数字 $1$ ），但要排除 $000$

For three padded digits, each position has $9$ choices if digit $1$ is forbidden, but exclude $000.$

解答：

可以按位数分类。

一位数共有 $8$ 个不含数字 $1$ 。

两位数共有 $8 \cdot 9 = 72$ 个不含数字 $1$ 。

三位数共有 $8 \cdot 9 \cdot 9 = 648$ 个不含数字 $1$ 。

因此不含数字 $1$ 的数共有 $8 + 72 + 648 = 728$ 个。

所以正确答案是 D。

We can case on the number of digits.

There are $8$ one digit numbers excluding $1.$

There are $8 \cdot 9 = 72$ two digit numbers that lack the digit $1.$

There are $8 \cdot 9 \cdot 9 = 648$ three digit numbers that do not include $1.$

This yields a total of $8 + 72 + 648 = 728$ numbers that do not contain the digit $1.$

Thus, D is the correct answer.

23.

在学校最后一天，旺德福老师给班上的学生发果冻豆。她给每个男生的果冻豆数等于班上男生人数；给每个女生的果冻豆数等于班上女生人数。她带了 $400$ 颗果冻豆，发完后剩下六颗。班上男生比女生多两人。她班上有多少名学生？

On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

$26$

$28$

$30$

$32$

$34$

知识点：二次方程因式分解

难度评级：1600

小提示：

设女生人数为 $g$ ，则男生人数为 $g+2$

Let the number of girls be $g,$ so the number of boys is $g+2.$

大提示：

发出的果冻豆总数为 $400-6$

The total jelly beans given out is $400-6.$

解答：

设班上男生人数为 $b$ ，女生人数为 $g$ 。由题意， $b = g + 2$ 。

如果每个男生得到 $b$ 颗果冻豆，那么给所有男生共发出 $b^2$ 颗。同样，给所有女生共发出 $g^2$ 颗。

因此 $\begin{gather*} b^2 + g^2 = 400 - 6 \\ (g + 2)^2 + g^2 = 394 \\ 2g^2 + 4g + 4 = 394 \\ g^2 + 2g - 195 = 0 \\ (g + 15)(g - 13) = 0 \end{gather*}\text{。}$ 因为 $g$ 不能为负，所以 $g = 13$ 。这意味着 $b = 15$ ，所以 $b + g = 28$ 。

所以正确答案是 B。

Let $b$ be the number of boys in the class and $g$ be the number of girls. From the problem, we get that $b = g + 2.$

If each boy gets $b$ jelly beans, then Mrs. Wonderful will give out a total of $b^2$ jelly beans to all the boys. Similarly, she will give out $g^2$ jelly beans to all the girls.

Therefore, $\begin{gather*} b^2 + g^2 = 400 - 6 \\ (g + 2)^2 + g^2 = 394 \\ 2g^2 + 4g + 4 = 394 \\ g^2 + 2g - 195 = 0 \\ (g + 15)(g - 13) = 0. \end{gather*}$ Since $g$ cannot be negative, we get that $g = 13.$ This means that $b = 15,$ so $b + g = 28.$

Thus, B is the correct answer.

24.

字母 $A$ 、 $B$ 、 $C$ 、 $D$ 表示互不相同的数字。如果 $\begin{array}{ccc} &A &B \\ + &C &A \\ \hline &D &A \end{array}$ 且 $\begin{array}{ccc} &A &B \\ - &C &A \\ \hline &&A \end{array}$ ，那么 $D$ 表示哪个数字？

The letters $A,$ $B,$ $C$ and $D$ all represent different digits. If $\begin{array}{ccc} &A &B \\ + &C &A \\ \hline &D &A \end{array}$ and $\begin{array}{ccc} &A &B \\ - &C &A \\ \hline &&A \end{array},$ what digit does $D$ represent?

$5$

$6$

$7$

$8$

$9$

知识点：数字谜位值

难度评级：1630

小提示：

先使用加法的个位

Use the ones column of the addition first.

大提示：

减法会迫使一个涉及 $A$ 的借位关系

The subtraction forces a borrowing relation involving $A.$

解答：

从加法的个位可知， $A+B$ 的个位仍是 $A$ ，所以 $B=0$ 。

在减法 $AB-CA=A$ 中，现在有 $A0-CA=A$ 。个位必须借位，所以 $10-A=A$ ，得 $A=5$ 。

于是 $50-C5=5$ ，所以 $C=4$ 。在加法中， $50+45=95$ ，所以 $D=9$ 。

所以正确答案是 E。

From the ones column of the addition, $A+B$ ends in $A,$ so $B=0.$

In the subtraction $AB-CA=A,$ we now have $A0-CA=A.$ The ones column requires a borrow, so $10-A=A,$ giving $A=5.$

Then $50-C5=5,$ so $C=4.$ In the addition, $50+45=95,$ so $D=9.$

Thus, E is the correct answer.

25.

一个体积为一立方英尺的立方体被三次平行于顶面的切割分成四块。第一刀距离顶面 $\dfrac{1}{2}$ 英尺。第二刀在第一刀下方 $\dfrac{1}{3}$ 英尺处，第三刀在第二刀下方 $\dfrac{1}{17}$ 英尺处。从上到下四块标为 $A$ 、 $B$ 、 $C$ 、 $D$ 。然后按 $C$ 、 $B$ 、 $A$ 、 $D$ 的顺序首尾相接粘在一起，形成如图所示的长立体。这个立体的总表面积是多少平方英尺？

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\dfrac{1}{2}$ foot from the top face. The second cut is $\dfrac{1}{3}$ foot below the first cut, and the third cut is $\dfrac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled $A,$ $B,$ $C,$ and $D.$ The pieces are then glued together end to end in the order $C,$ $B,$ $A,$ $D$ to make a long solid as shown. What is the total surface area of this solid in square feet?

$6$

$7$

$\dfrac{419}{51}$

$\dfrac{158}{17}$

$11$

知识点：表面积立体几何

难度评级：1620

小提示：

从六个方向观察来思考表面积

Think of surface area by looking from the six directions.

大提示：

从任一侧面看，这四块会叠回一个单位立方体的侧视图

The four pieces stack back to a unit cube when viewed from either side.

解答：

从六个坐标方向观察这个立体。

从任一端观察时，外露竖直面的总高度等于 $A$ 块的厚度，即 $\dfrac{1}{2}$ 英尺。因此每个端向视图的面积为 $\dfrac{1}{2}$ 平方英尺。

从每个侧面看，四块拼成原单位立方体的侧视图，所以每个侧视图面积为 $1$ 平方英尺。

从上方和下方看，每个视图都显示四个 $1$ 乘 $1$ 的面，所以每个面积为 $4$ 平方英尺。

总表面积为 $\dfrac{1}{2}+\dfrac{1}{2}+1+1+4+4=11$ 平方英尺。

所以正确答案是 E。

Look at the solid from the six coordinate directions.

Viewed from either end, the exposed vertical faces have total height equal to the thickness of piece $A,$ namely $\dfrac{1}{2}$ foot. Thus each end view has area $\dfrac{1}{2}$ square foot.

From each side, the four pieces stack to the side view of the original unit cube, so each side view has area $1$ square foot.

From the top and bottom, each view shows four $1$ -by- $1$ faces, so each has area $4$ square feet.

The total surface area is $\dfrac{1}{2}+\dfrac{1}{2}+1+1+4+4=11$ square feet.

Thus, E is the correct answer.