1999 AMC 8 详解 | LIVE by Po-Shen Loh

1.

$(6 \ ? \ 3) + 4 - (2 - 1) = 5$ 。要使这个等式成立， $6$ 和 $3$ 之间的问号应替换为

$(6 \ ? \ 3) + 4 - (2 - 1) = 5.$ To make this statement true, the question mark between the $6$ and the $3$ should be replaced by

$\div$

$\times$

$+$

$-$

以上都不是

None of these

知识点：运算顺序

难度评级：370

小提示：

先化简不含问号的部分

Simplify the part without the question mark first

大提示：

找出能使 $6\ ?\ 3=2$ 的运算

Find the operation that makes $6\ ?\ 3=2$

解答：

我们有 $4 - (2 - 1) = 4 - 1 = 3\text{。}$

因此 $(6 \ ? \ 3) + 3 = 5$ $6 \ ? \ 3 = 2\text{。}$

唯一可行的选择是 $\div$ 。

所以正确答案是 A。

We have that $4 - (2 - 1) = 4 - 1 = 3.$

Then $(6 \ ? \ 3) + 3 = 5$ $6 \ ? \ 3 = 2.$

The only choice that works is $\div.$

Thus, A is the correct answer.

2.

$10$ 点整时，钟表两根指针形成的较小角的度数是多少？

What is the degree measure of the smaller angle formed by the hands of a clock at $10$ o’clock?

$30$

$45$

$60$

$75$

$90$

知识点：时钟

难度评级：450

小提示：

每相邻两个小时刻度相差 $30^\circ$

Each hour mark is $30^\circ$ apart

大提示：

$10$ 点整时，两根指针相隔两个小时刻度

At $10$ o’clock, the hands are two hour marks apart

解答：

$10$ 点整时，时针指向 $10$ ，分针指向 $12$ 。

两根指针相隔整个圆的 $\frac{1}{6}$ 。所以角度为 $360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}\text{。}$

所以正确答案是 C。

At $10$ o’clock, we have that the hour hand is at $10$ and the minute hand is at $12.$

This means that the hands are a $\frac{1}{6}$ of the entire circle apart. This is equal to $360^{\circ} \cdot \dfrac{1}{6} = 60^{\circ}.$

Thus, C is the correct answer.

3.

下列哪一组三个数的和不等于 $1$ ？

Which triplet of numbers has a sum NOT equal to $1?$

$(\frac12,\frac13,\frac16)$

$(2,-2,1)$

$(0.1,0.3,0.6)$

$(1.1,-2.1,1.0)$

$(-\frac32,-\frac52,5)$

知识点：分数小数

难度评级：560

小提示：

先计算看起来最容易的和

Compute the easiest-looking sums first

大提示：

只有一组选项的总和不是 $1$

Only one listed triplet fails to total $1$

解答：

A 等于 $\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1\text{。}$

B 化简为 $2 - 2 + 1 = 1\text{。}$

C 的和是 $0.1 + 0.3 + 0.6 = 1\text{。}$

但 D 的和为 $1.1 - 2.1 + 1.0 = 0\text{。}$

为确认起见，E 的和为 $-\dfrac{8}{2} + 5 = 5 - 4 = 1\text{。}$

所以正确答案是 D。

We have that A is the same as $\dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = 1.$

B reduces to $2 - 2 + 1 = 1.$

C adds up to $0.1 + 0.3 + 0.6 = 1.$

D however adds to $1.1 - 2.1 + 1.0 = 0.$

To make sure, we check that E has a sum of $-\dfrac{8}{2} + 5 = 5 - 4 = 1.$

Thus, D is the correct answer.

4.

图中显示了骑车人阿尔韦托（Alberto）和比约恩（Bjorn）骑行的英里数。四小时后，阿尔韦托大约比比约恩多骑了多少英里？

The diagram shows the miles traveled by bikers Alberto and Bjorn. After four hours, about how many more miles has Alberto biked than Bjorn?

$15$

$20$

$25$

$30$

$35$

知识点：数据与图表解读

难度评级：660

小提示：

读出 $4$ 小时时两人的距离

Read both distances at $4$ hours

大提示：

用阿尔韦托的距离减去比约恩的距离

Subtract Bjorn’s distance from Alberto’s

解答：

从图中可见， $4$ 小时后阿尔韦托骑了 $60$ 英里，比约恩骑了 $45$ 英里。

两者的距离差是 $60 - 45 = 15$ 英里。

所以正确答案是 A。

Looking at the graph, we have that Alberto has biked $60$ miles and Bjorn $45$ miles after $4$ hours.

The difference between the two distances is $60 - 45 = 15$ miles.

Thus, A is the correct answer.

5.

一个长 $50$ 英尺、宽 $10$ 英尺的长方形花园被栅栏围住。为了在使用同样栅栏的情况下扩大花园，花园改成正方形。面积增加了多少平方英尺？

A rectangular garden $50$ feet long and $10$ feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?

$100$

$200$

$300$

$400$

$500$

知识点：周长面积

难度评级：770

小提示：

周长保持不变

Keep the perimeter the same

大提示：

正方形边长是原周长的四分之一

The square side length is one-fourth of the old perimeter

解答：

当前栅栏的周长为 $2(50 + 10) = 2 \cdot 60 = 120$ 英尺。如果四条边都相等，那么每条边长为 $120 \div 4 = 30$ 英尺。因此正方形面积为 $30^2 = 900$ 平方英尺。原花园面积为 $50 \cdot 10 = 500$ 平方英尺。

因此面积差是 $900 - 500 = 400$ 平方英尺。

所以正确答案是 D。

The current length of the fence is $2(50 + 10) = 2 \cdot 60 = 120$ feet. If all the sides become the same, then each side length is $120 \div 4 = 30$ feet. As such, the area of the square is then $30^2 = 900$ square feet. Note that the original area of the garden is $50 \cdot 10 = 500$ square feet.

Therefore, the difference is $900 - 500 = 400$ square feet.

Thus, D is the correct answer.

6.

博、科、弗洛、乔和莫的钱数各不相同。乔和博都没有弗洛的钱多。博和科都比莫钱多。乔比莫钱多，但比博钱少。谁的钱最少？

Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?

博

Bo

科

Coe

弗洛

Flo

乔

Jo

莫

Moe

知识点：逻辑推理不等式

难度评级：820

小提示：

排除所有已知比别人钱多的人

Eliminate anyone known to have more than someone else

大提示：

用每个比较关系找出谁的钱比莫多

Use each comparison to show who is above Moe

解答：

设 $B, C, F, J$ 和 $M$ 分别表示博、科、弗洛、乔和莫的钱数。

那么 $J \lt F,\quad B \lt F\text{。}$

还知道 $B \gt M,\quad C \gt M\text{。}$

最后，题目给出 $M \lt J \lt B\text{。}$

由第一组和第二组不等式可知 $F \gt M$ 。

因此每个人的钱都比莫多。

所以正确答案是 E。

Let $B, C, F, J,$ and $M$ represent the amount of money that Bo, Coe, Flo, Jo, and Moe have respectively.

Then $J \lt F,\quad B \lt F.$

We also have $B \gt M,\quad C \gt M.$

Finally, we are given $M \lt J \lt B.$

From the first and second inequalities, we can get that $F \gt M.$

This shows that everyone has more money than Moe.

Thus, E is the correct answer.

7.

一条高速公路的第三个出口在 $40$ 英里标处，第十个出口在 $160$ 英里标处。一个服务中心位于从第三个出口到第十个出口路程的四分之三处。这个服务中心应在多少英里标处？

The third exit on a highway is located at milepost $40$ and the tenth exit is at milepost $160.$ There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this service center?

$90$

$100$

$110$

$120$

$130$

知识点：分数

难度评级：900

小提示：

先求 $40$ 和 $160$ 两个英里标之间的距离

First find the distance between mileposts $40$ and $160$

大提示：

从 $40$ 英里标向前走这段距离的 $\frac{3}{4}$

Move $\frac{3}{4}$ of that distance past milepost $40$

解答：

第三个和第十个出口之间相差 $160 - 40 = 120$ 英里。

这意味着全程的 $\frac{3}{4}$ 处位于 $40 + \dfrac{3}{4} \cdot 120 = 130$ 英里标处。

所以正确答案是 E。

There are $160 - 40 = 120$ miles between the third and tenth exits.

This means $\frac{3}{4}$ of the way is at the $40 + \dfrac{3}{4} \cdot 120 = 130$ milepost.

Thus, E is the correct answer.

8.

六个正方形的正反两面都涂了颜色（ $R$ 表示红色， $B$ 表示蓝色， $O$ 表示橙色， $Y$ 表示黄色， $G$ 表示绿色， $W$ 表示白色）。它们按图所示铰接在一起，然后折成一个立方体。与白色面相对的面是

Six squares are colored, front and back ( $R$ = red, $B$ = blue, $O$ = orange, $Y$ = yellow, $G$ = green, and $W$ = white). They are hinged together as shown, then folded to form a cube. The face opposite the white face is

$B$

$G$

$O$

$R$

$Y$

知识点：展开图（立体几何）正方体

难度评级：960

小提示：

围绕标有 $Y$ 的正方形来折叠

Fold around the square marked $Y$

大提示：

跟踪哪个面会落到 $W$ 的对面

Track which face lands opposite $W$

解答：

设立方体的黄色面朝上。

这时白色、绿色和橙色面可以向下折。

因为蓝色面连在绿色面上，所以它会向后折。

因此蓝色面最终会与白色面相对。

所以正确答案是 A。

Consider the cube with the yellow face facing upwards.

Then we can fold the white, green, and orange faces down.

Since the blue face is attached to the green face, it will end up folding backwards.

This means that the blue face will end up facing opposite the white face.

Thus, A is the correct answer.

9.

如图，三个花坛互相重叠。花坛 $A$ 有 $500$ 株植物，花坛 $B$ 有 $450$ 株植物，花坛 $C$ 有 $350$ 株植物。花坛 $A$ 和 $B$ 共有 $50$ 株植物，花坛 $A$ 和 $C$ 共有 $100$ 株植物。植物总数是

Three flower beds overlap as shown. Bed $A$ has $500$ plants, bed $B$ has $450$ plants, and bed $C$ has $350$ plants. Beds $A$ and $B$ share $50$ plants, while beds $A$ and $C$ share $100.$ The total number of plants is

$850$

$1000$

$1150$

$1300$

$1450$

知识点：容斥原理韦恩图

难度评级：1020

小提示：

先把三个花坛的数量相加

Add the three bed counts first

大提示：

减去被重复计算的重叠部分

Subtract the overlaps that were counted twice

解答：

有 $50 + 100 = 150$ 株植物在两个花坛中，且没有植物同时在三个花坛中。

所以植物总数为 $500 + 450 + 350 - 150 = 1150\text{。}$ 减去重叠部分是为了去掉被计算两次的植物。

所以正确答案是 C。

Note there are $50 + 100 = 150$ plants that are in two beds and there are no plants in all three beds.

The total number of plants is then $500 + 450 + 350 - 150 = 1150.$ We subtract to get rid of the plants that we counted twice.

Thus, C is the correct answer.

10.

一个交通灯的完整周期为 $60$ 秒。每个周期中绿灯 $25$ 秒，黄灯 $5$ 秒，红灯 $30$ 秒。在随机选取的一个时刻，交通灯不是绿灯的概率是多少？

A complete cycle of a traffic light takes $60$ seconds. During each cycle the light is green for $25$ seconds, yellow for $5$ seconds, and red for $30$ seconds. At a randomly chosen time, what is the probability that the light will NOT be green?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

$\dfrac{7}{12}$

知识点：几何概率对立事件概率

难度评级：980

小提示：

不是绿灯就是红灯或黄灯

Not green means red or yellow

大提示：

用非绿灯时间除以总周期时间

Use time not green over total cycle time

解答：

在一个周期中，交通灯不是绿灯的时间为 $30 + 5 = 35$ 秒。

所以不是绿灯的概率为 $\dfrac{35}{60} = \dfrac{7}{12}\text{。}$

所以正确答案是 E。

During a given cycle, the light is not green for $30 + 5 = 35$ seconds.

Then the probability that it is not green is $\dfrac{35}{60} = \dfrac{7}{12}.$

Thus, E is the correct answer.

11.

将 $1$ 、 $4$ 、 $7$ 、 $10$ 和 $13$ 这五个数分别填入五个方格中，使横排三个数之和等于竖列三个数之和。横排或竖列的和最大可能是多少？

Each of the five numbers $1,$ $4,$ $7,$ $10,$ and $13$ is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is

$20$

$21$

$22$

$24$

$30$

知识点：最优化双重计数

难度评级：1140

小提示：

中间方格会在两个和中都被计算

The center square is counted in both sums

大提示：

让中间数尽可能大，再把重复计算后的总和平均分成两份

Maximize the center number, then split the doubled total equally

解答：

设中心方格中的数为 $x$ 。横排之和加上竖列之和时，其他数各计算一次，而中心数计算两次。

所以公共和的两倍是 $1+4+7+10+13+x$ $=35+x$ 。当 $x=13$ 时，这个值最大。

因此公共和至多为 $\dfrac{35+13}{2}=24$ ，而且可以达到：把 $13$ 放在中心，把 $1$ 和 $10$ 放在一排的两端，把 $4$ 和 $7$ 放在另一排的两端。两个和都是 $24$ 。

所以正确答案是 D。

Let $x$ be the number in the center. The horizontal sum plus the vertical sum counts every number once, except the center number is counted twice.

So twice the common sum is $1+4+7+10+13+x$ $=35+x$ . This is largest when $x=13$ .

The largest common sum is therefore at most $\dfrac{35+13}{2}=24$ , and it is attainable: put $13$ in the center, $1$ and $10$ at the ends of one row, and $4$ and $7$ at the ends of the other. Both sums are $24$ .

Thus, D is the correct answer.

12.

中学小将队赢的场数与输的场数之比为 $\frac{11}{4}$ ，没有平局。四舍五入到最接近的整数百分比，这支队输了百分之多少的比赛？

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $\frac{11}{4}.$ To the nearest whole percent, what percent of its games did the team lose?

$24\%$

$27\%$

$36\%$

$45\%$

$73\%$

知识点：比与比例百分数

难度评级：1120

小提示：

把 $11$ 胜 $4$ 负看作一组

Use $11$ wins and $4$ losses as one batch

大提示：

输掉的比例是输的场数除以总场数

The lost fraction is losses over total games

解答：

可以把这支队看成每 $15$ 场比赛中赢 $11$ 场、输 $4$ 场。

输掉的百分比为 $\dfrac4{15}\cdot100\%\approx 27\%$ 。

所以正确答案是 B。

The team can be viewed as winning $11$ games and losing $4$ games in each batch of $15$ games.

The percent lost is $\dfrac4{15}\cdot100\%\approx 27\%$ .

Thus, B is the correct answer.

13.

一个计算机科学营有 $40$ 名成员，平均年龄为 $17$ 岁。其中有 $20$ 个女孩、 $15$ 个男孩和 $5$ 个成年人。如果女孩的平均年龄为 $15$ 岁，男孩的平均年龄为 $16$ 岁，那么成年人的平均年龄是多少？

The average age of the $40$ members of a computer science camp is $17$ years. There are $20$ girls, $15$ boys, and $5$ adults. If the average age of the girls is $15$ and the average age of the boys is $16,$ what is the average age of the adults?

$26$

$27$

$28$

$29$

$30$

知识点：平均数

难度评级：1150

小提示：

把每个平均年龄转化为年龄总和

Convert each average into a total age

大提示：

从全营总年龄中减去女孩和男孩的总年龄

Subtract the girls’ and boys’ totals from the camp total

解答：

全营所有人的年龄总和为 $40 \cdot 17 = 680$ 。

女孩年龄总和为 $20 \cdot 15 = 300$ ，男孩年龄总和为 $15 \cdot 16 = 240$ 。

因此成年人的年龄总和为 $680 - 300 - 240 = 140$ 。

所以成年人的平均年龄为 $140 \div 5 = 28\text{。}$

所以正确答案是 C。

The sum of the ages of everybody at the camp is $40 \cdot 17 = 680.$

The sum of the ages of the girls is $20 \cdot 15 = 300$ and of the boys is $15 \cdot 16 = 240.$

As such, we know that the ages of the adults must be $680 - 300 - 240 = 140.$

And therefore, the average age of the adults is then $140 \div 5 = 28.$

Thus, C is the correct answer.

14.

在梯形 $ABCD$ 中，边 $AB$ 和 $CD$ 相等。 $ABCD$ 的周长是

In trapezoid $ABCD,$ the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is

$27$

$30$

$32$

$34$

$48$

知识点：梯形勾股定理

难度评级：1220

小提示：

从 $B$ 作一条高

Drop an altitude from $B$

大提示：

两侧直角三角形的水平直角边都是 $4$

The two side right triangles have horizontal leg $4$

解答：

设从 $B$ 到 $\overline{AD}$ 的高交 $\overline{AD}$ 于 $H$ 。

于是我们有 $AH = \dfrac{16 - 8}{2} = 4\text{，}$ 因为 $AB = CD$ 。

于是 $AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\text{。}$

周长为 $8 + 16 + 2 \cdot 5 = 34\text{。}$

所以正确答案是 D。

Let $H$ be where the altitude from $B$ to $\overline{AD}$ intersects $\overline{AD}.$

Then we have that $AH = \dfrac{16 - 8}{2} = 4,$ since $AB = CD.$

We then have that $AB = \sqrt{4^2 + 3^2} = \sqrt{25} = 5.$

Then the perimeter is $8 + 16 + 2 \cdot 5 = 34.$

Thus, D is the correct answer.

15.

平原镇的自行车牌照每个含有三个字母。第一个字母从集合 $\{C,H,L,P,R\}$ 中选，第二个从 $\{A,I,O\}$ 中选，第三个从 $\{D,M,N,T\}$ 中选。

当平原镇需要更多牌照时，他们增加了两个新字母。这两个新字母可以都加入同一个集合，也可以分别加入两个不同集合。加入两个字母后，最多能多制作多少个额外牌照？

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set $\{C,H,L,P,R\},$ the second from $\{A,I,O\},$ and the third from $\{D,M,N,T\}.$

When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of additional license plates that can be made by adding two letters?

$24$

$30$

$36$

$40$

$60$

知识点：乘法原理最优化分类讨论

难度评级：1330

小提示：

原来的牌照数是 $5\cdot3\cdot4$

The old number of plates is $5\cdot3\cdot4$

大提示：

要让乘积最大，应让三个因子尽量接近平衡

To maximize a product, make the three factors as balanced as possible

解答：

目前一共可以制作 $5 \cdot 3 \cdot 4 = 60$ 个牌照。

如果两个字母都加入第一个集合，那么可以制作 $7 \cdot 3 \cdot 4 = 84$ 个牌照。

如果两个字母都加入第二个集合，那么有 $5 \cdot 5 \cdot 4 = 100$ 个牌照。

如果两个字母都加入第三个集合，那么有 $5 \cdot 3 \cdot 6 = 90$ 种选择。

如果一个字母加入第一个集合，另一个加入第二个集合，那么有 $6 \cdot 4 \cdot 4 = 96$ 个牌照。

如果另一个字母改为加入第三个集合，那么可以得到 $6 \cdot 3 \cdot 5 = 90$ 个牌照。

最后，如果两个字母分别加入第二个和第三个集合，那么有 $5 \cdot 4 \cdot 5 = 100$ 个牌照。

最大可以达到 $100$ 个牌照，因此额外增加 $100 - 60 = 40$ 个。

所以正确答案是 D。

There are currently $5 \cdot 3 \cdot 4 = 60$ license plates that can be made.

If both letters are added to the first set, then there are $7 \cdot 3 \cdot 4 = 84$ possible plates.

If they are both added to the second, there are $5 \cdot 5 \cdot 4 = 100$ plates.

If they are added to the third, there are $5 \cdot 3 \cdot 6 = 90$ choices.

If one is added to the first set and the other to the second set, there are $6 \cdot 4 \cdot 4 = 96$ plates.

If the other is added to the third set, we get $6 \cdot 3 \cdot 5 = 90$ possible plates.

Finally, if the letters are added to the second and third sets, there are $5 \cdot 4 \cdot 5 = 100$ plates.

We see that $100$ is the greatest number of plates that we can achieve. This is an additional $100 - 60 = 40$ plates.

Thus, D is the correct answer.

16.

托里的数学测试有 $75$ 道题： $10$ 道算术题、 $30$ 道代数题和 $35$ 道几何题。她答对了 $70\%$ 的算术题、 $40\%$ 的代数题和 $60\%$ 的几何题，但因为答对题数少于 $60\%$ ，所以没有通过测试。

她还需要多答对多少道题，才能达到 $60\%$ 的及格分数？

Tori’s mathematics test had $75$ problems: $10$ arithmetic, $30$ algebra, and $35$ geometry problems. Although she answered $70\%$ of the arithmetic, $40\%$ of the algebra, and $60\%$ of the geometry problems correctly, she did not pass the test because she got less than $60\%$ of the problems right.

How many more problems would she have needed to answer correctly to earn a $60\%$ passing grade?

$1$

$5$

$7$

$9$

$11$

知识点：百分数

难度评级：1230

小提示：

按类别计算托里目前答对了多少题

Compute Tori’s current correct answers by category

大提示：

将这个总数与 $75$ 的 $60\%$ 比较

Compare that total with $60\%$ of $75$

解答：

她答对了 $10 \cdot 0.7 = 7$ 道算术题，答对了 $30 \cdot 0.4 = 12$ 道代数题，还答对了 $35 \cdot 0.6 = 21$ 道几何题。

因此她总共答对 $7 + 12 + 21 = 40$ 道题。

要达到 $60\%$ ，托里必须答对 $75 \cdot 0.6 = 45$ 道题。所以她还需要多答对 $45 - 40 = 5$ 道题。

所以正确答案是 B。

She answered $10 \cdot 0.7 = 7$ arithmetic questions correctly. She answered $30 \cdot 0.4 = 12$ algebra ones correctly. She also got $35 \cdot 0.6 = 21$ geometry questions correctly.

This means she got a total of $7 + 12 + 21 = 40$ questions correct.

As such, in order to get a $60\%,$ Tori must have answered $75 \cdot 0.6 = 45$ questions correctly. This means that she would have needed to answer an additional $45 - 40 = 5$ questions correctly.

Thus, B is the correct answer.

17.

第 $17$ 、 $18$ 和 $19$ 题参考以下内容：

给一群人做饼干

中央中学有 $108$ 名参加 AMC $8$ 的学生，他们晚上聚在一起讨论题目，并且平均每人吃两个饼干。今年沃尔特和格蕾特要烤邦妮牌美味条形饼干。这个食谱每盘做 $15$ 个饼干，配料包括： $1\dfrac{1}{2}$ 杯面粉、 $2$ 个鸡蛋、 $3$ 汤匙黄油、 $\dfrac{3}{4}$ 杯糖和 $1$ 包巧克力粒。他们只做完整份食谱，不做部分份。

沃尔特可以按半打购买鸡蛋。为了做足够的饼干，他应该买多少半打鸡蛋？（可能会剩下一些鸡蛋和一些饼干。）

Problems $17,$ $18,$ and $19$ refer to the following:

Cookies For a Crowd

At Central Middle School the $108$ students who take the AMC $8$ meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonnie’s Best Bar Cookies this year. Their recipe, which makes a pan of $15$ cookies, lists these items: $1\dfrac{1}{2}$ cups of flour, $2$ eggs, $3$ tablespoons butter, $\dfrac{3}{4}$ cups sugar, and $1$ package of chocolate drops. They will make only full recipes, not partial recipes.

Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may be left over.)

$1$

$2$

$5$

$7$

$15$

知识点：取整函数单位换算

难度评级：1250

小提示：

先求做 $216$ 个饼干需要多少完整份食谱

Find how many full recipes are needed for $216$ cookies

大提示：

每份完整食谱用 $2$ 个鸡蛋

Each full recipe uses $2$ eggs

解答：

因为学生平均每人吃 $2$ 个饼干，他们总共会吃 $108 \cdot 2 = 216$ 个饼干。

每份食谱做 $15$ 个饼干，所以需要 $\left\lceil\dfrac{216}{15}\right\rceil = 15$ 份完整食谱，才能做够。

每盘需要 $2$ 个鸡蛋，所以需要 $15 \cdot 2 = 30$ 个鸡蛋。半打是 $6$ 个鸡蛋，因此需要 $30 \div 6 = 5$ 个半打。

所以正确答案是 C。

Since the students eat an average of $2$ cookies each, they will eat a total of $108 \cdot 2 = 216$ cookies.

Each recipe makes $15$ cookies, which means we need $\left\lceil\dfrac{216}{15}\right\rceil = 15$ full recipes to make enough cookies.

Each pan requires $2$ eggs, which means we need $15 \cdot 2 = 30$ eggs. There are $6$ eggs in a half-dozen, so we need $30 \div 6 = 5$ half-dozens.

Thus, C is the correct answer.

18.

他们得知同一晚有一场大型音乐会，出席人数会减少 $25\%$ 。他们应该为这个较小的聚会做多少份饼干食谱？

They learn that a big concert is scheduled for the same night and attendance will be down $25\%.$ How many recipes of cookies should they make for their smaller party?

$6$

$8$

$9$

$10$

$11$

知识点：百分数取整函数

难度评级：1310

小提示：

减少 $25\%$ 后还剩 $75\%$ 的出席人数

A $25\%$ drop leaves $75\%$ attendance

大提示：

所需食谱份数要向上取整

Round the needed number of recipes up

解答：

现在聚会上只有 $108 \cdot \left(1 - \dfrac{1}{4}\right) = 108 \cdot \dfrac{3}{4} = 81$ 人。这意味着需要 $\left\lceil \dfrac{81 \cdot 2}{15} \right\rceil = \left\lceil \dfrac{162}{15} \right\rceil = 11$ 份食谱。

所以正确答案是 E。

There will now only be $108 \cdot \left(1 - \dfrac{1}{4}\right) = 108 \cdot \dfrac{3}{4} = 81$ people at the party. This means we need $\left\lceil \dfrac{81 \cdot 2}{15} \right\rceil = \left\lceil \dfrac{162}{15} \right\rceil = 11$ recipes.

Thus, E is the correct answer.

19.

鼓手生病了，音乐会取消了。沃尔特和格蕾特必须烤出足够的饼干，共计 $216$ 块。每条黄油相当于 $8$ 汤匙。需要多少条黄油？（当然，可能会剩下一些黄油。）

The drummer gets sick. The concert is cancelled. Walter and Gretel must make enough pans of cookies to supply $216$ cookies. There are $8$ tablespoons in a stick of butter. How many sticks of butter will be needed? (Some butter may be left over, of course.)

$5$

$6$

$7$

$8$

$9$

知识点：单位换算取整函数

难度评级：1340

小提示：

先把需要的盘数向上取整

First round the number of pans up

大提示：

将黄油汤匙数换算成条数并向上取整

Convert tablespoons of butter to sticks and round up

解答：

要做 $216$ 个饼干，他们必须做 $\left\lceil \dfrac{216}{15} \right\rceil = 15$ 盘。每盘需要 $3$ 汤匙黄油，所以所有饼干共需要 $15 \cdot 3 = 45$ 汤匙。

因此他们需要 $\left\lceil \dfrac{45}{8} \right\rceil = 6$ 条黄油。

所以正确答案是 B。

To make $216$ cookies, they have to make $\left\lceil \dfrac{216}{15} \right\rceil = 15$ pans. Since each pan requires $3$ tablespoons of butter, all the pans will need $15 \cdot 3 = 45$ tablespoons.

They will then need $\left\lceil \dfrac{45}{8} \right\rceil = 6$ sticks of butter.

Thus, B is the correct answer.

20.

图 $1$ 称为“堆叠图”。数字表示每个位置上堆了多少个立方体。图 $2$ 显示这些立方体，图 $3$ 显示从正面看到的堆叠立方体视图。

下列哪一个是图 $4$ 中堆叠图的正面视图？

Figure $1$ is called a “stack map.” The numbers tell how many cubes are stacked in each position. Fig. $2$ shows these cubes, and Fig. $3$ shows the view of the stacked cubes as seen from the front.

Which of the following is the front view for the stack map in Fig. $4?$

知识点：立体几何

难度评级：1410

小提示：

正面视图在每一列只保留较高的堆

A front view keeps only the taller stack in each column

大提示：

对每一列，取前后两个高度中的较大值

For each column, take the larger front/back height

解答：

注意某一列在正面视图中的高度，是前后两个位置高度的最大值。

左列中，后面的立方体堆较高，高度为 $2$ 。

中列中，前面的立方体堆较高，高度为 $3$ 。

右列的高度为 $4$ 。

所以正确答案是 B。

Note that the height of the column is the maximum height of the front and back columns.

In the left column, we have that the back column is taller with height $2.$

In the middle column, we have that the front column is taller with height $3.$

Finally, the right column has height $4.$

Thus, B is the correct answer.

21.

角 $A$ 的度数是

The degree measure of angle $A$ is

$20$

$30$

$35$

$40$

$45$

知识点：导角

难度评级：1520

小提示：

在两个交点处使用补角关系

Use supplementary angles at the two intersections

大提示：

再在靠近 $40^\circ$ 角的三角形中使用三角形内角和

Then use a triangle angle sum near the $40^\circ$ angle

解答：

如下标记顶点。

由补角关系， $\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}\text{。}$ 同样由补角关系， $\angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ}\text{。}$ 利用三角形内角和为 $180^{\circ}$ ，得到 $\begin{aligned} \angle ECD &= 180^{\circ} - 70^{\circ} - 40^{\circ} \\ &= 70^{\circ} \end{aligned}\text{。}$ 再由对顶角相等， $\angle ACB = \angle ECD = 70^{\circ}\text{。}$ 最后在三角形中使用内角和，得到 $\angle A = 180^{\circ} - 80^{\circ} - 70^{\circ} = 30^{\circ}\text{。}$

所以正确答案是 B。

Label the vertices as below.

We then have that $\angle ABC = 180^{\circ} - 100^{\circ} = 80^{\circ}$ by supplementary angles. Then we have $\angle CED = 180^{\circ} - 110^{\circ} = 70^{\circ}$ again by supplementary angles. Using the sum of the interior angles of a triangle is $180^{\circ},$ we get $\begin{aligned} \angle ECD &= 180^{\circ} - 70^{\circ} - 40^{\circ} \\ &= 70^{\circ}. \end{aligned}$ Then, using vertical angles, we have $\angle ACB = \angle ECD = 70^{\circ}.$ Finally, using the sum of the interior angles of a triangle, we get $\angle A = 180^{\circ} - 80^{\circ} - 70^{\circ} = 30^{\circ}.$

Thus, B is the correct answer.

22.

在一个遥远的地方，三条鱼可以换两个面包，一个面包可以换四袋米。一条鱼值多少袋米？

In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?

$\dfrac{3}{8}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$2\dfrac{2}{3}$

$3\dfrac{1}{3}$

知识点：比与比例单位换算

难度评级：1180

小提示：

把每个交换关系写成价值等式

Turn each trade into a value equation

大提示：

先把鱼换成面包，再把面包换成米

Convert fish to bread, then bread to rice

解答：

三条鱼值两个面包，所以一条鱼值 $\dfrac23$ 个面包。

一条面包值 $4$ 袋米，所以一条鱼值 $\dfrac23\cdot4=\dfrac83=2\dfrac23$ 袋米。

所以正确答案是 D。

Three fish are worth two loaves of bread, so one fish is worth $\dfrac23$ of a loaf.

One loaf is worth $4$ bags of rice, so one fish is worth $\dfrac23\cdot4=\dfrac83=2\dfrac23$ bags of rice.

Thus, D is the correct answer.

23.

正方形 $ABCD$ 的边长为 $3$ 。线段 $CM$ 和 $CN$ 将正方形面积分成三个相等部分。线段 $CM$ 有多长？

Square $ABCD$ has sides of length $3.$ Segments $CM$ and $CN$ divide the square’s area into three equal parts. How long is segment $CM?$

$\sqrt{10}$

$\sqrt{12}$

$\sqrt{13}$

$\sqrt{14}$

$\sqrt{15}$

知识点：三角形面积勾股定理

难度评级：1520

小提示：

三个区域各占正方形面积的三分之一

Each of the three regions has area one-third of the square

大提示：

用 $\triangle BMC$ 的面积求 $BM$

Use the area of $\triangle BMC$ to find $BM$

解答：

正方形面积为 $3^2 = 9$ ，所以每个区域的面积是 $9 \div 3 = 3$ 。

因此 $\triangle BMC$ 的面积为 $3$ ，所以 $\dfrac{1}{2} \cdot 3 \cdot BM = 3$ $BM = 2\text{。}$

因为 $\triangle BMC$ 是直角三角形， $CM = \sqrt{2^2 + 3^2} = \sqrt{13}\text{。}$

所以正确答案是 C。

The area of the square is $3^2 = 9,$ which means that the area of one region is $9 \div 3 = 3.$

This means the area of $\triangle BMC$ is $3,$ which means that $\dfrac{1}{2} \cdot 3 \cdot BM = 3$ $BM = 2.$

Since $\triangle BMC$ is right, we have $CM = \sqrt{2^2 + 3^2} = \sqrt{13}.$

Thus, C is the correct answer.

24.

$1999^{2000}$ 除以 $5$ 的余数是

When $1999^{2000}$ is divided by $5,$ the remainder is

$4$

$3$

$2$

$1$

$0$

知识点：模运算个位数字

难度评级：1470

小提示：

先将 $1999$ 对 $5$ 取模

Reduce $1999$ modulo $5$

大提示：

对 $5$ 取模时， $-1$ 的各个偶次幂余数都相同

Even powers of $-1$ modulo $5$ have the same remainder

解答：

要求除以 $5$ 的余数，只需要关注个位数。

因此只需观察个位数 $9$ 的幂的规律。

$9$ 的幂的个位数在 $9$ 和 $1$ 之间交替： $9, 81, 729, \cdots\text{。}$ 因为指数是偶数，所以 $1999^{2000}$ 的个位数为 $1$ 。

除以 $5$ 的余数就是 $1$ ，因为它比 $10$ 的倍数多 $1$ 。

所以正确答案是 D。

Note that to find the remainder when divided by $5,$ we only care about the units digit.

This means we only have to observe how the powers of the units digit work, namely the powers of $9.$

Then, looking at powers of $9,$ we see that the units digit alternates between $9$ and $1:$ $9, 81, 729, \cdots.$ This means that $1999^{2000}$ ends in a $1$ since the power is even.

The remainder when divided by $5$ is then $1,$ since it is $1$ more than a multiple of $10.$

Thus, D is the correct answer.

25.

点 $B$ 、 $D$ 和 $J$ 是直角三角形 $ACG$ 各边的中点。点 $K$ 、 $E$ 和 $I$ 是三角形 $JDG$ 各边的中点，依此类推。如果这种分割并涂阴影的过程进行 $100$ 次（图中显示了前三次），且 $AC = CG = 6$ ，那么阴影三角形的总面积最接近

Points $B,$ $D,$ and $J$ are midpoints of the sides of right triangle $ACG.$ Points $K,$ $E,$ $I$ are midpoints of the sides of triangle $JDG,$ etc. If the dividing and shading process is done $100$ times (the first three are shown) and $AC = CG = 6,$ then the total area of the shaded triangles is nearest

$6$

$7$

$8$

$9$

$10$

知识点：等比数列相似

难度评级：1620

小提示：

阴影三角形的面积形成一个等比数列

The shaded triangles form a geometric area pattern

大提示：

每个新的阴影三角形面积是前一个阴影面积的四分之一

Each new shaded triangle has one-fourth the previous shaded area

解答：

第一个阴影三角形的两条直角边为 $3$ 和 $3$ ，所以面积为 $\dfrac12\cdot3\cdot3=\dfrac92$ 。

之后每一步，相关三角形的边长变为原来的一半，因此阴影面积乘以 $\left(\dfrac12\right)^2=\dfrac14$ 。

许多步后的阴影总面积非常接近等比和 $\begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-\frac{1}{4}} \\ {}= 6 \end{gathered}\text{。}$

$100$ 步后被省略的尾项极小，所以总面积最接近 $6$ 。

所以正确答案是 A。

The first shaded triangle has legs $3$ and $3$ , so its area is $\dfrac12\cdot3\cdot3=\dfrac92$ .

At each later step, the relevant triangle has half the leg length, so the shaded area is multiplied by $\left(\dfrac12\right)^2=\dfrac14$ .

The total shaded area after many steps is therefore very close to the geometric sum $\begin{gathered} \frac92\left(1+\frac14+\frac1{16}+\cdots\right) \\ {}= \frac92\cdot\frac1{1-\frac{1}{4}} \\ {}= 6. \end{gathered}$

After $100$ steps the omitted tail is tiny, so the total area is nearest $6$ .

Thus, A is the correct answer.