1986 AMC 8 详解 | LIVE by Po-Shen Loh

1.

$1861$ 年七月，印度乞拉朋齐降雨 $366$ 英寸。那个月平均每小时降雨多少英寸？

In July $1861,$ $366$ inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

$\dfrac{366}{31 \times 24}$

$\dfrac{366 \times 31}{24}$

$\dfrac{366 \times 24}{31}$

$\dfrac{31 \times 24}{366}$

$366 \times 31 \times 24$

知识点：速率单位换算

难度评级：660

小提示：

平均速率等于总降雨量除以这个月的总小时数

The average rate is the total rainfall divided by the total number of hours in the month

大提示：

七月有 $31$ 天，每天有 $24$ 小时

July has $31$ days, and each day has $24$ hours

解答：

平均每小时降雨量等于总降雨量除以总小时数。七月有 $31$ 天，即 $31 \times 24$ 小时，所以平均值是 $\dfrac{366}{31 \times 24} = \dfrac{61}{124}$ 英寸每小时。

所以正确答案是 A。

The average rainfall per hour equals the total rainfall divided by the total number of hours. July has $31$ days, or $31 \times 24$ hours, so the average is $\dfrac{366}{31 \times 24} = \dfrac{61}{124}$ inches per hour.

Thus, the correct answer is A .

2.

下列哪个数的倒数最大？

Which of the following numbers has the largest reciprocal?

$\dfrac13$

$\dfrac25$

$1$

$5$

$1986$

知识点：分数

难度评级：560

小提示：

正数越小，它的倒数越大

The reciprocal of a positive number is largest when the number itself is smallest

大提示：

比较这五个数，找出最小的那个

Compare the five numbers and find the smallest one

解答：

较大的正数倒数较小，较小的正数倒数较大。列出的数中最小的是 $\dfrac13$ ，它的倒数 $3$ 最大。

所以正确答案是 A。

A large positive number has a small reciprocal, and a small positive number has a large reciprocal. The smallest number listed is $\dfrac13,$ whose reciprocal $3$ is the largest.

Thus, the correct answer is A .

3.

从集合 $\{7, 25, -1, 12, -3\}$ 中选三个不同的数相加，可能得到的最小和是

The smallest sum one could get by adding three different numbers from the set $\{7, 25, -1, 12, -3\}$ is

$-3$

$-1$

$3$

$5$

$21$

知识点：最优化

难度评级：660

小提示：

要让和尽可能小，选三个最小的数

To make the sum as small as possible, choose the three smallest numbers

大提示：

三个最小的数是 $-3$ 、 $-1$ 和 $7$

The three smallest numbers are $-3,$ $-1,$ and $7$

解答：

集合中三个最小的数是 $-3$ 、 $-1$ 和 $7$ 。它们的和是 $-3 + (-1) + 7 = 3$ 。

所以正确答案是 C。

The three smallest numbers in the set are $-3,$ $-1,$ and $7.$ Their sum is $-3 + (-1) + 7 = 3.$

Thus, the correct answer is C .

4.

乘积 $(1.8)(40.3 + 0.07)$ 最接近

The product $(1.8)(40.3 + 0.07)$ is closest to

$7$

$42$

$74$

$84$

$737$

知识点：估算

难度评级：730

小提示：

估算各因数： $1.8$ 接近 $2$ ， $40.3 + 0.07$ 接近 $40$

Round each factor: $1.8$ is near $2$ and $40.3 + 0.07$ is near $40$

大提示：

先估算 $2 \times 40$ ，再因为 $1.8 \lt 2$ 稍微向下调整

Estimate $2 \times 40$ and adjust slightly downward since $1.8 \lt 2$

解答：

第二个因数 $40.3 + 0.07 = 40.37$ 约为 $40$ ，而 $1.8$ 约为 $2$ 。快速估算为 $2(40) - 0.2(40) = 80 - 8 = 72$ ，所以乘积最接近 $74$ 。

所以正确答案是 C。

The second factor $40.3 + 0.07 = 40.37$ is about $40,$ and $1.8$ is about $2.$ A quick estimate is $2(40) - 0.2(40) = 80 - 8 = 72,$ so the product is closest to $74.$

Thus, the correct answer is C .

5.

一场比赛从某天中午开始， $1000$ 分钟后结束。比赛在什么时间结束？

A contest began at noon one day and ended $1000$ minutes later. At what time did the contest end?

晚上 $10\!:\!00$

$10\!:\!00$ p.m.

午夜

midnight

凌晨 $2\!:\!30$

$2\!:\!30$ a.m.

凌晨 $4\!:\!40$

$4\!:\!40$ a.m.

早上 $6\!:\!40$

$6\!:\!40$ a.m.

知识点：日期与时间单位换算

难度评级：730

小提示：

把 $1000$ 分钟除以 $60$ ，化成小时和分钟

Convert $1000$ minutes into hours and minutes by dividing by $60$

大提示：

$1000$ 分钟是 $16$ 小时 $40$ 分钟；把它加到中午

$1000$ minutes is $16$ hours and $40$ minutes; add that to noon

解答：

因为 $1000$ 分钟 $= \dfrac{1000}{60}$ 小时 $= 16\dfrac23$ 小时 $= 16$ 小时 $40$ 分钟，所以比赛在中午后 $16$ 小时 $40$ 分钟结束，即凌晨 $4\!:\!40$ 。

所以正确答案是 D。

Since $1000$ minutes $= \dfrac{1000}{60}$ hours $= 16\dfrac23$ hours $= 16$ hours $40$ minutes, the contest ended $16$ hours $40$ minutes past noon, which is $4\!:\!40$ a.m.

Thus, the correct answer is D .

6.

下列表达式的值是多少？

$\frac{2}{1 - \frac{2}{3}}$

What is the value of the following expression?

$\frac{2}{1 - \frac{2}{3}}$

$-3$

$-\dfrac43$

$\dfrac23$

$2$

$6$

知识点：分数运算顺序

难度评级：660

小提示：

先化简分母 $1 - \dfrac23$

First simplify the denominator $1 - \dfrac23$

大提示：

除以 $\dfrac13$ 等于乘以 $3$

Dividing by $\dfrac13$ is the same as multiplying by $3$

解答：

分母是 $1 - \dfrac23 = \dfrac13$ ，所以表达式是 $\dfrac{2}{\frac{1}{3}} = 2 \times 3 = 6$ 。

所以正确答案是 E。

The denominator is $1 - \dfrac23 = \dfrac13,$ so the expression is $\dfrac{2}{\frac{1}{3}} = 2 \times 3 = 6.$

Thus, the correct answer is E .

7.

$\sqrt{8}$ 和 $\sqrt{80}$ 之间有多少个整数？

How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}?$

$5$

$6$

$7$

$8$

$9$

知识点：区间内整数计数完全平方数

难度评级：820

小提示：

找出刚大于 $8$ 和刚小于 $80$ 的完全平方数

Find perfect squares just above $8$ and just below $80$

大提示：

$\sqrt{8}$ 在 $2$ 和 $3$ 之间， $\sqrt{80}$ 在 $8$ 和 $9$ 之间

$\sqrt{8}$ is between $2$ and $3,$ and $\sqrt{80}$ is between $8$ and $9$

解答：

因为 $2 \lt \sqrt{8} \lt 3$ ，且 $8 \lt \sqrt{80} \lt 9$ ，所以严格在 $\sqrt{8}$ 和 $\sqrt{80}$ 之间的整数是 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $7$ 、 $8$ 。共有六个。

所以正确答案是 B。

Since $2 \lt \sqrt{8} \lt 3$ and $8 \lt \sqrt{80} \lt 9,$ the whole numbers strictly between $\sqrt{8}$ and $\sqrt{80}$ are $3,$ $4,$ $5,$ $6,$ $7,$ $8.$ There are six of them.

Thus, the correct answer is B .

8.

在所示乘法中， $B$ 表示一个数字。 $B$ 的值是多少？

$\begin{array}{r} B2 \\ \times\, 7B \\ \hline 6396 \end{array}$

In the multiplication shown, $B$ represents a digit. What is the value of $B?$

$\begin{array}{r} B2 \\ \times\, 7B \\ \hline 6396 \end{array}$

$3$

$5$

$6$

$7$

$8$

知识点：数字谜个位数字

难度评级：930

小提示：

乘积的个位是 $6$ ，它来自 $B \times 2$

The units digit of the product is $6,$ and it comes from $B \times 2$

大提示：

$B \times 2$ 以 $6$ 结尾时 $B = 3$ 或 $B = 8$ ；乘积很大，所以检验哪个值可行

$B \times 2$ ends in $6$ when $B = 3$ or $B = 8;$ the product is large, so test which value works

解答：

乘积的个位来自 $B \times 2$ ，只有当 $B = 3$ 或 $B = 8$ 时才以 $6$ 结尾。由于乘积 $6396$ 大于 $6000$ ， $B$ 必须是 $8$ ：确实 $82 \times 78 = 6396$ 。

所以正确答案是 E。

The units digit of the product comes from $B \times 2,$ which ends in $6$ only when $B = 3$ or $B = 8.$ Since the product $6396$ exceeds $6000,$ $B$ must be $8:$ indeed $82 \times 78 = 6396.$

Thus, the correct answer is E .

9.

只沿图示路径和方向，从 $M$ 到 $N$ 有多少条不同路线？

Using only the paths and the directions shown, how many different routes are there from $M$ to $N?$

$2$

$3$

$4$

$5$

$6$

知识点：基本计数系统列举

难度评级：960

小提示：

按第一步分类列出路线：从 $M$ 可以到 $A$ 或 $B$

List the routes by the first step: from $M$ you can go to $A$ or to $B$

大提示：

仔细沿箭头走；从 $B$ 可以到 $A$ 、 $C$ 或直接到 $N$

Follow the arrows carefully; from $B$ you can reach $A,$ $C,$ or $N$ directly

解答：

沿箭头可行的路线是 $MADCN$ 、 $MACN$ 、 $MBADCN$ 、 $MBACN$ 、 $MBCN$ 和 $MBN$ 。共有六条。

所以正确答案是 E。

Following the arrows, the possible routes are $MADCN,$ $MACN,$ $MBADCN,$ $MBACN,$ $MBCN,$ and $MBN.$ There are six of them.

Thus, the correct answer is E .

10.

一幅宽 $3$ 英尺的画挂在一面宽 $19$ 英尺的墙正中央。画最近的一边距离墙端多少英尺？

A picture $3$ feet across is hung in the center of a wall that is $19$ feet wide. How many feet from the end of the wall is the nearest edge of the picture?

$1\dfrac12$

$8$

$9\dfrac12$

$16$

$22$

知识点：对称性

难度评级：730

小提示：

画在正中央，所以两侧空隙相等

The picture is centered, so the two gaps on the sides are equal

大提示：

用墙宽减去画宽，再把剩余长度平均分到两侧

Subtract the picture width from the wall width, then split the remainder between the two sides

解答：

画留下 $19 - 3 = 16$ 英尺的墙面，平均分成两侧各 $8$ 英尺的空隙。所以画最近的一边离墙端 $8$ 英尺。

所以正确答案是 B。

The picture leaves $19 - 3 = 16$ feet of wall, split equally into two gaps of $8$ feet each. So the nearest edge of the picture is $8$ feet from the end of the wall.

Thus, the correct answer is B .

11.

如果 $A * B$ 表示 $\dfrac{A + B}{2}$ ，那么 $(3 * 5) * 8$ 等于

If $A * B$ means $\dfrac{A + B}{2},$ then $(3 * 5) * 8$ is

$6$

$8$

$12$

$16$

$30$

知识点：自定义运算

难度评级：820

小提示：

运算 $A * B$ 给出 $A$ 和 $B$ 的平均数

The operation $A * B$ gives the average of $A$ and $B$

大提示：

先计算 $3 * 5$ ，再把结果和 $8$ 结合

First compute $3 * 5,$ then combine that result with $8$

解答：

首先， $3 * 5 = \dfrac{3 + 5}{2} = 4$ 。然后 $4 * 8 = \dfrac{4 + 8}{2} = 6$ 。

所以正确答案是 A。

First, $3 * 5 = \dfrac{3 + 5}{2} = 4.$ Then $4 * 8 = \dfrac{4 + 8}{2} = 6.$

Thus, the correct answer is A .

12.

图表显示了一个数学班 $30$ 名学生在最近两次测验中的成绩分布。例如，正好有一名学生在测验 $1$ 中得“D”，在测验 $2$ 中得“C”（圈出的格子）。有百分之多少的学生两次测验得到了相同成绩？

The table shown displays the grade distribution of the $30$ students in a mathematics class on the last two tests. For example, exactly one student received a ‘D’ on Test $1$ and a ‘C’ on Test $2$ (the circled entry). What percent of the students received the same grade on both tests?

$12\%$

$25\%$

$33\dfrac13\%$

$40\%$

$50\%$

知识点：数据与图表解读百分数

难度评级：860

小提示：

两次成绩相同的学生位于表格从左上到右下的对角线上

A student with the same grade on both tests lies on the diagonal running from the top-left to the bottom-right of the table

大提示：

把对角线上的数相加，再除以 $30$ ，并转成百分数

Add the diagonal entries, then divide by $30$ and convert to a percent

解答：

学生两次测验成绩相同，正对应主对角线上的格子。这些数是 $2 + 4 + 5 + 1 + 0 = 12$ 。

所以比例是 $\dfrac{12}{30} = \dfrac{4}{10} = 40\%$ 。

所以正确答案是 D。

A student received the same grade on both tests exactly when counted on the main diagonal. Those entries are $2 + 4 + 5 + 1 + 0 = 12.$

So the fraction is $\dfrac{12}{30} = \dfrac{4}{10} = 40\%.$

Thus, the correct answer is D .

13.

图中多边形的周长是

The perimeter of the polygon shown is

$14$

$20$

$28$

$48$

无法由给定信息确定

cannot be determined from the information given

知识点：周长矩形

难度评级：960

小提示：

所有角都是直角，所以水平边和竖直边都可以分别重新组合

All the angles are right angles, so the horizontal edges and the vertical edges can each be regrouped

大提示：

向右和向左的水平边都各自跨过总宽 $8$ ，向上和向下的竖直边也都各自跨过总高 $6$

The rightward and leftward horizontal edges each span the full width $8,$ and the upward and downward vertical edges each span the full height $6$

解答：

因为每个角都是直角，把边平移重组可见，水平边总共走过宽度两次，竖直边走过高度两次。因此周长等于完整 $8$ 乘 $6$ 矩形的周长，即 $2(8 + 6) = 28$ 。

答案不取决于缺口具体切在哪里。

所以正确答案是 C。

Because every angle is a right angle, sliding the edges shows that the horizontal edges together traverse the width twice and the vertical edges traverse the height twice. So the perimeter equals that of the full $8$ by $6$ rectangle, $2(8 + 6) = 28.$

The answer does not depend on exactly where the notch is cut.

Thus, the correct answer is C .

14.

如果 $200 \le a \le 400$ 且 $600 \le b \le 1200$ ，那么商 $\dfrac{b}{a}$ 的最大值是

If $200 \le a \le 400$ and $600 \le b \le 1200,$ then the largest value of the quotient $\dfrac{b}{a}$ is

$\dfrac32$

$3$

$6$

$300$

$600$

知识点：不等式最优化

难度评级：820

小提示：

要让商最大，分子应尽可能大，分母应尽可能小

A quotient is largest when the numerator is as large as possible and the denominator as small as possible

大提示：

取最大的 $b$ 和最小的 $a$

Use the largest $b$ and the smallest $a$

解答：

当 $b$ 取最大值、 $a$ 取最小值时，商 $\dfrac{b}{a}$ 最大，得到 $\dfrac{1200}{200} = 6$ 。

所以正确答案是 C。

The quotient $\dfrac{b}{a}$ is largest with the biggest $b$ and smallest $a,$ giving $\dfrac{1200}{200} = 6.$

Thus, the correct answer is C .

15.

阿贾克斯折扣店的促销价比原价低 $50\%$ 。周六在促销价基础上再打 $20\%$ 折扣。一件原价 $\$180$ 的外套周六价格是多少？

Sale prices at the Ajax Outlet Store are $50\%$ below original prices. On Saturdays an additional discount of $20\%$ off the sale price is given. What is the Saturday price of a coat whose original price is $\$180?$

$\$54$

$\$72$

$\$90$

$\$108$

$\$110$

知识点：百分数

难度评级：860

小提示：

打 $50\%$ 折扣后剩下原价的一半

A $50\%$ discount leaves half the original price

大提示：

再取促销价的 $80\%$ ，表示额外打掉 $20\%$

Then take $80\%$ of that sale price to apply the extra $20\%$ off

解答：

促销价是 $\$180$ 的 $50\%$ ，即 $\$90$ 。周六价格再减 $20\%$ ，留下 $\$90$ 的 $80\%$ ，即 $\$72$ 。

所以正确答案是 B。

The sale price is $50\%$ of $\$180,$ or $\$90.$ The Saturday price takes another $20\%$ off, leaving $80\%$ of $\$90,$ which is $\$72.$

Thus, the correct answer is B .

16.

一幅条形图显示了一家快餐连锁店在每个季节卖出的汉堡数量。但是，表示冬季销量的条形被一块污迹遮住了。如果这家连锁店全年汉堡销量中正好有 $25\%$ 是秋季卖出的，那么冬季卖出多少百万个汉堡？

A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar indicating the number sold during the winter is covered by a smudge. If exactly $25\%$ of the chain’s hamburgers are sold in the fall, how many million hamburgers are sold in the winter?

$2.5$

$3$

$3.5$

$4$

$4.5$

知识点：百分数

难度评级：960

小提示：

秋季销量 $4$ 百万个是全年总量的 $25\%$ ，所以先求全年总量

The fall sales of $4$ million are $25\%$ of the yearly total, so find the yearly total first

大提示：

从全年总量中减去春、夏、秋三季销量

Subtract the spring, summer, and fall sales from the yearly total

解答：

如果秋季 $4$ 百万个是全年总量的 $25\%$ ，那么全年总量是 $16$ 百万个。

冬季销量是 $16 - (4.5 + 5 + 4) = 2.5$ 百万个。

所以正确答案是 A。

If the fall sales of $4$ million are $25\%$ of the yearly total, then the yearly total is $16$ million.

The winter sales are $16 - (4.5 + 5 + 4) = 2.5$ million.

Thus, the correct answer is A .

17.

设 $o$ 是一个奇整数， $n$ 是任意整数。关于整数 $o^2 + no$ ，下列哪项总为真？

Let $o$ be an odd whole number and let $n$ be any whole number. Which of the following statements about the whole number $o^2 + no$ is always true?

它总是奇数

it is always odd

它总是偶数

it is always even

只有当 $n$ 为偶数时它才是偶数

it is even only if $n$ is even

只有当 $n$ 为奇数时它才是奇数

it is odd only if $n$ is odd

只有当 $n$ 为偶数时它才是奇数

it is odd only if $n$ is even

知识点：奇偶性因式分解

难度评级：1000

小提示：

因式分解： $o^2 + no = o(o + n)$

Factor: $o^2 + no = o(o + n)$

大提示：

由于 $o$ 是奇数， $o(o + n)$ 的奇偶性与 $o + n$ 相同

Since $o$ is odd, the parity of $o(o + n)$ matches the parity of $o + n$

解答：

因式分解得 $o^2 + no = o(o + n)$ 。因为 $o$ 是奇数，这个乘积为奇数当且仅当 $o + n$ 为奇数，而这只在 $n$ 为偶数时发生。当 $n$ 为奇数时， $o + n$ 为偶数，乘积为偶数。

所以这个数只有当 $n$ 为偶数时才是奇数。

所以正确答案是 E。

Factor $o^2 + no = o(o + n).$ Because $o$ is odd, the product is odd exactly when $o + n$ is odd, which happens only when $n$ is even. When $n$ is odd, $o + n$ is even and the product is even.

So the number is odd only if $n$ is even.

Thus, the correct answer is E .

18.

一个矩形牧场要用一段 $100$ 米石墙作为第四边，只在另外三边围栅栏。栅栏上每隔 $12$ 米立一个柱子，包括栅栏与石墙相接处的两个柱子。围出一个 $36$ 米乘 $60$ 米的区域，最少需要多少根柱子？

A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence, including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60$ m?

$11$

$12$

$13$

$14$

$16$

知识点：周长最优化植树问题

难度评级：1030

小提示：

要用最少柱子，让 $60$ 米边靠墙，这样需要围的长度最短

To use the fewest posts, let the $60$ meter side lie along the wall, so the fenced length is as short as possible

大提示：

栅栏路径长 $36 + 60 + 36 = 132$ 米；每隔 $12$ 米放一根柱子，并计算两端

The fenced path is $36 + 60 + 36 = 132$ meters; place a post every $12$ meters, counting both ends

解答：

当石墙作为较长的 $60$ 米边时，用的柱子最少；此时栅栏覆盖两条 $36$ 米边和一条 $60$ 米边，总路径长 $36 + 60 + 36 = 132$ 米。

每隔 $12$ 米放一根柱子并包括两端，共用 $\dfrac{132}{12} + 1 = 12$ 根柱子。

所以正确答案是 B。

The fewest posts are used when the wall serves as the longer $60$ meter side, so the fence covers two $36$ meter sides and one $60$ meter side, a path of $36 + 60 + 36 = 132$ meters.

Placing a post every $12$ meters, including both ends, uses $\dfrac{132}{12} + 1 = 12$ posts.

Thus, the correct answer is B .

19.

旅行开始时，里程表读数为 $56{,}200$ 英里。司机给油箱加满了 $6$ 加仑汽油。旅途中，当里程表读数为 $56{,}560$ 时，司机又给油箱加了 $12$ 加仑汽油。旅行结束时，司机又加了 $20$ 加仑汽油；此时里程表读数为 $57{,}060$ 。按最接近的十分之一计算，整段旅行中汽车平均每加仑行驶多少英里？

At the beginning of a trip, the mileage odometer read $56{,}200$ miles. The driver filled the gas tank with $6$ gallons of gasoline. During the trip, the driver filled the tank again with $12$ gallons of gasoline when the odometer read $56{,}560.$ At the end of the trip, the driver filled the tank again with $20$ gallons of gasoline; the odometer read $57{,}060.$ To the nearest tenth, what was the car’s average miles-per-gallon for the entire trip?

$22.5$

$22.6$

$24.0$

$26.9$

$27.5$

知识点：速率

难度评级：1060

小提示：

旅途中用掉的油只包括出发后补加的油： $12 + 20$ 加仑

The gas used during the trip is only what was added after the start: $12 + 20$ gallons

大提示：

用总行驶英里数除以旅途中用掉的加仑数

Divide the total miles driven by the gallons used during the trip

解答：

旅行距离是 $57{,}060 - 56{,}200 = 860$ 英里。最初的 $6$ 加仑只是出发前加满油箱；旅途中真正用掉的是后来为补回油量所加的 $12 + 20 = 32$ 加仑。

所以平均值是 $\dfrac{860}{32} \approx 26.9$ 英里每加仑。

所以正确答案是 D。

The trip was $57{,}060 - 56{,}200 = 860$ miles. The initial $6$ gallons only topped off the tank before the trip; the gas actually used during the trip is the $12 + 20 = 32$ gallons later added to replace it.

So the average is $\dfrac{860}{32} \approx 26.9$ miles per gallon.

Thus, the correct answer is D .

20.

表达式

$\frac{(304)^5}{(29.7)(399)^4}$

的值最接近

The value of the expression

$\frac{(304)^5}{(29.7)(399)^4}$

is closest to

$0.003$

$0.03$

$0.3$

$3$

$30$

知识点：估算指数

难度评级：980

小提示：

把数字取近似： $304 \approx 300$ ， $29.7 \approx 30$ ， $399 \approx 400$

Round the numbers: $304 \approx 300,$ $29.7 \approx 30,$ and $399 \approx 400$

大提示：

化简 $\dfrac{300^5}{30 \cdot 400^4} = 10 \left(\dfrac34\right)^4$

Simplify $\dfrac{300^5}{30 \cdot 400^4} = 10 \left(\dfrac34\right)^4$

解答：

估算：

$\begin{aligned} \dfrac{300^5}{30 \cdot 400^4} &= 10 \left(\dfrac{300}{400}\right)^4 \\ &= 10 \left(\dfrac34\right)^4 \\ &= 10 \cdot \dfrac{81}{256} \approx 3 \end{aligned}\text{。}$

所以正确答案是 D。

Estimating,

$\begin{aligned} \dfrac{300^5}{30 \cdot 400^4} &= 10 \left(\dfrac{300}{400}\right)^4 \\ &= 10 \left(\dfrac34\right)^4 \\ &= 10 \cdot \dfrac{81}{256} \approx 3. \end{aligned}$

Thus, the correct answer is D .

21.

在所示 T 形图中，分别把八个带字母的相同正方形中的一个与四个阴影正方形放在一起。所得图形中有多少个可以折成一个没有顶盖的立方体盒子？

Suppose one of the eight lettered identical squares is included with the four shaded squares in the T-shaped figure shown. How many of the resulting figures can be folded into a topless cubical box?

$2$

$3$

$4$

$5$

$6$

知识点：展开图（立体几何）分类讨论

难度评级：1220

小提示：

无顶盒有 $5$ 个面，所以四个阴影正方形加上一个新增正方形必须能折成缺少顶面的立方体

A topless box has $5$ faces, so the four shaded squares plus the one added square must fold into a cube missing its top

大提示：

逐个尝试带字母的正方形作为第五个面；若折叠后两个面重合，该位置失败

Try each lettered square as the fifth face; a placement fails when folding would force two faces into the same spot

解答：

四个阴影正方形折成开口盒子的四个面，新增正方形提供第五个面。想象折叠过程，正方形 $A$ 、 $E$ 、 $H$ 、 $B$ 、 $D$ 和 $F$ 都能补成一个有效的无顶盒。

正方形 $C$ 和 $G$ 不行，因为折叠后会迫使四个面在同一个角相遇。因此有 $6$ 个有效图形。

所以正确答案是 E。

The four shaded squares fold into four faces of an open box, and the added square supplies the fifth face. Picturing the folds, the squares $A,$ $E,$ $H,$ $B,$ $D,$ and $F$ each complete a valid topless box.

The squares $C$ and $G$ do not work, because folding would force four faces to meet at a single corner. That leaves $6$ valid figures.

Thus, the correct answer is E .

22.

艾伦、贝丝、卡洛斯和戴安娜正在讨论本评分期数学课可能得到的成绩。艾伦说：“如果我得 A，那么贝丝会得 A。”贝丝说：“如果我得 A，那么卡洛斯会得 A。”卡洛斯说：“如果我得 A，那么戴安娜会得 A。”这些话都是真的，但只有两名学生得了 A。哪两名学生得了 A？

Alan, Beth, Carlos, and Diana were discussing their possible grades in mathematics class this grading period. Alan said, “If I get an A, then Beth will get an A.” Beth said, “If I get an A, then Carlos will get an A.” Carlos said, “If I get an A, then Diana will get an A.” All of these statements were true, but only two of the students received an A. Which two received A’s?

艾伦、贝丝

Alan, Beth

贝丝、卡洛斯

Beth, Carlos

卡洛斯、戴安娜

Carlos, Diana

艾伦、戴安娜

Alan, Diana

贝丝、戴安娜

Beth, Diana

知识点：逻辑推理

难度评级：1060

小提示：

如果链条中较早的学生得 A，那么他后面的所有学生也必须得 A

If a student earlier in the chain gets an A, everyone after them in the chain must also get an A

大提示：

要正好有两个 A，这两个必须是链条最后的学生

For exactly two A’s, the two must be the last students in the chain

解答：

这些陈述形成一条链：艾伦得 A 会推出贝丝得 A，贝丝得 A 会推出卡洛斯得 A，卡洛斯得 A 会推出戴安娜得 A。如果艾伦得 A，则四人都得 A；如果贝丝得 A，则三人得 A。

正好有两个 A 的唯一方式是链条最后两人得 A，即卡洛斯和戴安娜。

所以正确答案是 C。

The statements form a chain: Alan’s A forces Beth’s, Beth’s forces Carlos’s, and Carlos’s forces Diana’s. If Alan got an A, all four would; if Beth got an A, three would.

The only way to have exactly two A’s is for them to be the last two in the chain, Carlos and Diana.

Thus, the correct answer is C .

23.

大圆的直径是 $AC$ 。两个小圆的圆心在 $AC$ 上，并且在大圆圆心 $O$ 处相切。如果每个小圆半径为 $1$ ，阴影区域面积与一个小圆面积之比是多少？

The large circle has diameter $AC.$ The two small circles have their centers on $AC$ and just touch at $O,$ the center of the large circle. If each small circle has radius $1,$ what is the value of the ratio of the area of the shaded region to the area of one of the small circles?

在 $\dfrac12$ 和 $1$ 之间

between $\dfrac12$ and $1$

$1$

在 $1$ 和 $\dfrac32$ 之间

between $1$ and $\dfrac32$

在 $\dfrac32$ 和 $2$ 之间

between $\dfrac32$ and $2$

无法由给定信息确定

cannot be determined from the information given

知识点：圆面积面积分割

难度评级：1090

小提示：

大圆半径是 $2$ ，因为 $AC$ 等于两个小圆的直径之和

The large circle has radius $2,$ since $AC$ equals two small diameters

大提示：

由对称性，阴影区域是大圆面积减去两个小圆面积后的一半

By symmetry the shaded region is half the difference between the large circle’s area and the two small circles’ area

解答：

大圆半径为 $2$ ，所以面积是 $\pi(2)^2 = 4\pi$ ，每个小圆面积是 $\pi$ 。

由对称性，阴影区域是面积差的一半： $\dfrac12(4\pi - 2\pi) = \pi$ 。它与一个小圆面积 $\pi$ 的比值是 $1$ 。

所以正确答案是 B。

The large circle has radius $2,$ so its area is $\pi(2)^2 = 4\pi,$ and each small circle has area $\pi.$

By symmetry the shaded region is half the difference of the areas: $\dfrac12(4\pi - 2\pi) = \pi.$ The ratio of this to one small circle’s area $\pi$ is $1.$

Thus, the correct answer is B .

24.

金氏初中的 $600$ 名学生午餐时被平均分成三组。每组在不同时间吃午餐。电脑随机把每名学生分到三组之一。三个朋友艾尔、鲍勃和卡罗尔被分到同一午餐组的概率大约是

The $600$ students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of the three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately

$\dfrac{1}{27}$

$\dfrac19$

$\dfrac18$

$\dfrac16$

$\dfrac13$

知识点：条件概率无放回抽样

难度评级：1120

小提示：

固定艾尔的组；再求鲍勃也进同一组的概率

Fix Al’s group; then find the chance that Bob lands in that same group

大提示：

固定艾尔的组后，鲍勃进入该组的概率是 $\frac{199}{599}$ ，接着卡罗尔进入该组的概率是 $\frac{198}{598}$

After Al’s group is fixed, Bob’s chance is $\frac{199}{599}$ and then Carol’s is $\frac{198}{598}$

解答：

不论艾尔在哪一组，鲍勃进入同一组的概率为 $\dfrac{199}{599}$ 。如果鲍勃进入该组，卡罗尔也进入该组的概率为 $\dfrac{198}{598}$ 。

所以三人在同一组的概率约为 $\dfrac{199}{599} \cdot \dfrac{198}{598} \approx 0.110$ ，最接近 $\dfrac19$ 。

所以正确答案是 B。

Whatever group Al is in, Bob joins that same group with probability $\dfrac{199}{599}.$ If Bob does, Carol joins them with probability $\dfrac{198}{598}.$

Thus all three share a group with probability $\dfrac{199}{599} \cdot \dfrac{198}{598} \approx 0.110,$ closest to $\dfrac19.$

Thus, the correct answer is B .

25.

下列哪一组整数的平均数最大？

Which of the following sets of whole numbers has the largest average?

$1$ 到 $101$ 之间的 $2$ 的倍数

multiples of $2$ between $1$ and $101$

$1$ 到 $101$ 之间的 $3$ 的倍数

multiples of $3$ between $1$ and $101$

$1$ 到 $101$ 之间的 $4$ 的倍数

multiples of $4$ between $1$ and $101$

$1$ 到 $101$ 之间的 $5$ 的倍数

multiples of $5$ between $1$ and $101$

$1$ 到 $101$ 之间的 $6$ 的倍数

multiples of $6$ between $1$ and $101$

知识点：等差数列平均数

难度评级：1060

小提示：

对于等间隔的数，平均数等于最小数和最大数的平均数

For evenly spaced numbers, the average equals the average of the smallest and largest

大提示：

设 $a_1$ 和 $a_n$ 分别为首项和末项；对每组计算 $\dfrac{a_1 + a_n}{2}$ ，再进行比较

Let $a_1$ and $a_n$ be the first and last terms; compute $\dfrac{a_1 + a_n}{2}$ for each set and compare

解答：

对一组等间隔整数，平均数等于最小数与最大数的平均数。各组平均数为：A： $\dfrac{2 + 100}{2} = 51$ ，B： $\dfrac{3 + 99}{2} = 51$ ，C： $\dfrac{4 + 100}{2} = 52$ ，D： $\dfrac{5 + 100}{2} = 52.5$ ，E： $\dfrac{6 + 96}{2} = 51$ 。

最大平均数是 $52.5$ ，来自 $5$ 的倍数。

所以正确答案是 D。

For a set of evenly spaced whole numbers, the average is the average of the smallest and largest. The averages are: A: $\dfrac{2 + 100}{2} = 51,$ B: $\dfrac{3 + 99}{2} = 51,$ C: $\dfrac{4 + 100}{2} = 52,$ D: $\dfrac{5 + 100}{2} = 52.5,$ E: $\dfrac{6 + 96}{2} = 51.$

The largest average is $52.5,$ from the multiples of $5.$

Thus, the correct answer is D .