1994 AMC 8 详解 | LIVE by Po-Shen Loh

1.

下列哪一个最大？

Which of the following is the largest?

$\dfrac{1}{3}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{7}{24}$

知识点：分数

难度评级：560

小提示：

把五个分数都化成同一个分母

Rewrite all five fractions over one common denominator

大提示：

这些分数的最小公分母是 $24$

The least common denominator of these fractions is $24$

解答：

通分到分母 $24$ ，这些分数分别是 $\dfrac{8}{24}$ 、 $\dfrac{6}{24}$ 、 $\dfrac{9}{24}$ 、 $\dfrac{10}{24}$ 和 $\dfrac{7}{24}$ 。

最大的分子是 $10$ ，所以 $\dfrac{5}{12} = \dfrac{10}{24}$ 最大。

所以正确答案是 D。

Over the common denominator $24$ , the fractions are $\dfrac{8}{24},$ $\dfrac{6}{24},$ $\dfrac{9}{24},$ $\dfrac{10}{24},$ and $\dfrac{7}{24}.$

The largest numerator is $10$ , so $\dfrac{5}{12} = \dfrac{10}{24}$ is the largest.

Thus, the correct answer is D .

2.

下列表达式的值是多少？

$\begin{aligned} &\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} \\ &\quad {}+ \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} \\ &\quad {}+ \frac{9}{10} + \frac{55}{10} \end{aligned}$

What is the value of the following expression?

$\begin{aligned} &\frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} \\ &\quad {}+ \frac{5}{10} + \frac{6}{10} + \frac{7}{10} + \frac{8}{10} \\ &\quad {}+ \frac{9}{10} + \frac{55}{10} \end{aligned}$

$4\dfrac{1}{2}$

$6.4$

$9$

$10$

$11$

知识点：分数

难度评级：560

小提示：

所有项的分母都是 $10$ ，所以先把分子相加

All the terms share the denominator $10$ , so add the numerators first

大提示：

$1 + 2 + \cdots + 9 = 45$ ，再加上 $55$

$1 + 2 + \cdots + 9 = 45,$ and then add the $55$

解答：

分子之和为 $1 + 2 + \cdots + 9 + 55$ $= 45 + 55$ $= 100$ 。

所以总和是 $\dfrac{100}{10} = 10$ 。

所以正确答案是 D。

The numerators sum to $1 + 2 + \cdots + 9 + 55$ $= 45 + 55$ $= 100.$

So the total is $\dfrac{100}{10} = 10.$

Thus, the correct answer is D .

3.

Maria 每天必须工作 $8$ 小时，这不包括她 $45$ 分钟的午餐时间。如果她上午 $7{:}25$ 开始工作，并在中午吃午餐，那么她的工作日将在什么时候结束？

Each day Maria must work $8$ hours. This does not include the $45$ minutes she takes for lunch. If she begins working at $7{:}25$ A.M. and takes her lunch break at noon, then her working day will end at

下午 $3{:}40$

$3{:}40$ P.M.

下午 $3{:}55$

$3{:}55$ P.M.

下午 $4{:}10$

$4{:}10$ P.M.

下午 $4{:}25$

$4{:}25$ P.M.

下午 $4{:}40$

$4{:}40$ P.M.

知识点：时钟

难度评级：660

小提示：

整个工作日经过 $8$ 小时工作时间加上 $45$ 分钟午餐时间

The workday spans $8$ hours of work plus the $45$ -minute lunch break

大提示：

先把 $8$ 小时加到上午 $7{:}25$ ，再加 $45$ 分钟

Add $8$ hours to $7{:}25$ A.M., then add $45$ more minutes

解答：

上午 $7{:}25$ 之后八小时是下午 $3{:}25$ 。

再加上 $45$ 分钟午餐时间，结束时间为下午 $4{:}10$ 。

所以正确答案是 C。

Eight hours after $7{:}25$ A.M. is $3{:}25$ P.M.

Adding the $45$ -minute lunch break gives an ending time of $4{:}10$ P.M.

Thus, the correct answer is C .

4.

右图绕中心顺时针旋转 $120°$ 后，结果是哪一个？

Which of the following represents the result when the figure shown at the right is rotated clockwise $120°$ about its center?

知识点：变换

难度评级：660

小提示：

顺时针旋转 $120°$ 会把三个图形各自移动到顺时针方向的下一个位置

A clockwise turn of $120°$ moves each of the three shapes to the next position in the clockwise direction.

大提示：

正方形仍是正方形，圆仍是圆；只有位置发生变化

The square must stay a square and the circle a circle; only their positions change.

解答：

顺时针旋转 $120°$ 后，每个图形移动到顺时针方向的下一个位置：上方的三角形移到右下，右下的正方形移到左下，左下的圆移到上方。

因此结果应为圆在上方，正方形在左下，三角形在右下，且每个图形形状保持不变。

所以正确答案是 B。

A clockwise turn of $120°$ sends each shape to the next position clockwise: the triangle at the top moves to the lower right, the square at the lower right moves to the lower left, and the circle at the lower left moves to the top.

The result therefore has a circle on top, a square at the lower left, and a triangle at the lower right, with each shape keeping its original form.

Thus, the correct answer is B .

5.

已知 $1$ 英里 $= 8$ 弗隆，且 $1$ 弗隆 $= 40$ 杆，那么一英里有多少杆？

Given that $1$ mile $= 8$ furlongs and $1$ furlong $= 40$ rods, the number of rods in one mile is

$5$

$320$

$660$

$1760$

$5280$

知识点：单位换算

难度评级：560

小提示：

先把英里换成弗隆，再把弗隆换成杆

Convert miles to furlongs, then furlongs to rods

大提示：

一英里是 $8$ 弗隆，每弗隆是 $40$ 杆

One mile is $8$ furlongs, and each furlong is $40$ rods

解答：

一英里是 $8$ 弗隆，每弗隆是 $40$ 杆，所以一英里是 $8 \times 40 = 320$ 杆。

所以正确答案是 B。

One mile is $8$ furlongs, and each furlong is $40$ rods, so one mile is $8 \times 40 = 320$ rods.

Thus, the correct answer is B .

6.

任意六个连续正整数的乘积的个位数字是

The unit’s digit (one’s digit) of the product of any six consecutive positive whole numbers is

$0$

$2$

$4$

$6$

$8$

知识点：个位数字整除性

难度评级：730

小提示：

任意六个连续整数中至少有一个 $5$ 的倍数，也至少有一个偶数

Among any six consecutive numbers there is at least one multiple of $5$ and at least one even number

大提示：

$5$ 的倍数乘以偶数一定是 $10$ 的倍数

A multiple of $5$ times an even number is a multiple of $10$

解答：

任意六个连续正整数中至少包含一个 $5$ 的倍数和一个偶数，所以它们的乘积是 $5 \times 2 = 10$ 的倍数。

$10$ 的倍数个位数字总是 $0$ 。

所以正确答案是 A。

Any six consecutive whole numbers include at least one multiple of $5$ and at least one even number, so their product is a multiple of $5 \times 2 = 10.$

A multiple of $10$ always ends in $0$ .

Thus, the correct answer is A .

7.

如果 $\angle A = 60°$ ， $\angle E = 40°$ ，且 $\angle C = 30°$ ，那么 $\angle BDC =$

If $\angle A = 60°,$ $\angle E = 40°,$ and $\angle C = 30°,$ then $\angle BDC =$

$40°$

$50°$

$60°$

$70°$

$80°$

知识点：导角角度和

难度评级：920

小提示：

在三角形 $ABE$ 中， $A$ 和 $E$ 处的角决定 $\angle ABE$

In triangle $ABE$ , the angles at $A$ and $E$ determine the angle $\angle ABE$

大提示：

因为 $D$ 在 $EB$ 上，且 $A, B, C$ 共线，所以 $\angle DBC$ 是 $\angle ABE$ 的补角；再使用三角形 $BDC$

Since $D$ lies on $EB$ and $A, B, C$ are collinear, $\angle DBC$ is the supplement of $\angle ABE$ ; then use triangle $BDC$

解答：

在三角形 $ABE$ 中， $\angle ABE = 180°$ $- (60° + 40°)$ $= 80°$ 。

因为 $A, B, C$ 共线且 $D$ 在 $EB$ 上，所以 $\angle DBC = 180° - 80° = 100°$ 。

在三角形 $BDC$ 中， $\angle BDC = 180°$ $- (100° + 30°)$ $= 50°$ 。

所以正确答案是 B。

In triangle $ABE,$ $\angle ABE = 180°$ $- (60° + 40°)$ $= 80°.$

Since $A, B, C$ are collinear and $D$ lies on segment $EB,$ the angle $\angle DBC = 180° - 80° = 100°.$

In triangle $BDC,$ $\angle BDC = 180°$ $- (100° + 30°)$ $= 50°.$

Thus, the correct answer is B .

8.

有多少个三位正整数的数位和等于 $25$ ？

For how many three-digit whole numbers does the sum of the digits equal $25?$

$2$

$4$

$6$

$8$

$10$

知识点：数字多重集排列

难度评级：920

小提示：

最大的三位数数位和是 $9 + 9 + 9 = 27$ ，所以这些数字必须接近全是 $9$

The largest possible digit sum is $9 + 9 + 9 = 27$ , so the digits must be close to all $9$ s

大提示：

找出和为 $25$ 的数字组合，再数每种组合的排列数

Find the digit combinations that sum to $25$ , then count how many arrangements each has

解答：

数位和为 $25$ ，而最大可能和是 $27$ ，所以至少有一个数字是 $9$ 。可能的数字组是 $\{9, 9, 7\}$ 和 $\{9, 8, 8\}$ 。

每组都有 $3$ 种不同排列： $997$ 、 $979$ 、 $799$ 以及 $988$ 、 $898$ 、 $889$ ，共 $6$ 个数。

所以正确答案是 C。

Since the digits sum to $25$ and the maximum is $27$ , at least one digit is $9$ . The possible digit sets are $\{9, 9, 7\}$ and $\{9, 8, 8\}.$

Each set has $3$ distinct arrangements ( $997,$ $979,$ $799$ and $988,$ $898,$ $889$ ), giving $6$ numbers in all.

Thus, the correct answer is C .

9.

一位购物者购买一件 $\$100$ 的外套，该外套正在减价 $20\%$ 出售。使用优惠券后，又从折后价中减去 $\$5$ 。最终售价再加 $8\%$ 的销售税。购物者为这件外套总共支付

A shopper buys a $\$100$ coat on sale for $20\%$ off. An additional $\$5$ is taken off the sale price by using a discount coupon. A sales tax of $8\%$ is paid on the final selling price. The total amount the shopper pays for the coat is

$\$81.00$

$\$81.40$

$\$82.00$

$\$82.08$

$\$82.40$

知识点：百分数

难度评级：860

小提示：

先应用 $20\%$ 折扣，再减去 $\$5$ 优惠券

Apply the $20\%$ discount first, then subtract the $\$5$ coupon

大提示：

把得到的价格乘以 $1.08$ ，以加入 $8\%$ 的税

Multiply the resulting price by $1.08$ to add the $8\%$ tax

解答：

$20\%$ 折扣使价格降到 $\$80$ ， $\$5$ 优惠券又使它降到 $\$75$ 。

加上 $8\%$ 的税后，总价为 $1.08 \times \$75 = \$81.00$ 。

所以正确答案是 A。

The $20\%$ discount lowers the price to $\$80,$ and the $\$5$ coupon reduces it to $\$75.$

Adding $8\%$ tax gives $1.08 \times \$75 = \$81.00.$

Thus, the correct answer is A .

10.

有多少个正整数 $N$ 满足 $(N \gt 0)$ ，使表达式

$\frac{36}{N+2}$

是整数？

For how many positive integer values of $N$ $(N \gt 0)$ is the expression

$\frac{36}{N+2}$

an integer?

$7$

$8$

$9$

$10$

$12$

知识点：整除性因数

难度评级：860

小提示：

$\dfrac{36}{N+2}$ 是整数，当且仅当 $N+2$ 是 $36$ 的因数

$\dfrac{36}{N+2}$ is an integer exactly when $N+2$ is a divisor of $36$

大提示：

因为 $N \gt 0$ ，所以只计算 $36$ 的大于 $2$ 的因数

Since $N \gt 0,$ only divisors of $36$ that are greater than $2$ count

解答：

表达式为整数时， $N+2$ 必须整除 $36$ 。 $36$ 的正因数是 $1$ 、 $2$ 、 $3$ 、 $4$ 、 $6$ 、 $9$ 、 $12$ 、 $18$ 、 $36$ 。

因为 $N \gt 0$ ，所以 $N + 2 \gt 2$ ，可用的因数是 $3$ 、 $4$ 、 $6$ 、 $9$ 、 $12$ 、 $18$ 、 $36$ ，共有 $7$ 个值。

所以正确答案是 A。

The expression is an integer when $N+2$ divides $36.$ The divisors of $36$ are $1,$ $2,$ $3,$ $4,$ $6,$ $9,$ $12,$ $18,$ $36.$

Because $N \gt 0,$ we need $N + 2 \gt 2,$ leaving the divisors $3,$ $4,$ $6,$ $9,$ $12,$ $18,$ $36$ : that is $7$ values.

Thus, the correct answer is A .

11.

去年夏天有 $100$ 名学生参加篮球营。其中有 $52$ 名男生和 $48$ 名女生。另外， $40$ 名学生来自 Jones Middle School， $60$ 名来自 Clay Middle School。有二十名女生来自 Jones Middle School。来自 Clay Middle School 的男生有多少名？

Last summer $100$ students attended basketball camp. Of those attending, $52$ were boys and $48$ were girls. Also, $40$ students were from Jones Middle School and $60$ were from Clay Middle School. Twenty of the girls were from Jones Middle School. How many of the boys were from Clay Middle School?

$20$

$32$

$40$

$48$

$52$

知识点：逻辑推理

难度评级：820

小提示：

先求有多少名女生来自 Clay Middle School

First find how many girls came from Clay Middle School

大提示：

用 Clay 的总人数 $60$ 减去 Clay 女生人数，得到 Clay 男生人数

Subtract the Clay girls from the $60$ Clay students to get the Clay boys

解答：

共有 $48$ 名女生，其中 $20$ 名来自 Jones，所以来自 Clay 的女生有 $48 - 20 = 28$ 名。

Clay 总共有 $60$ 名学生，所以来自 Clay 的男生人数是 $60 - 28 = 32$ 。

所以正确答案是 B。

Since $48$ girls attended and $20$ were from Jones, $48 - 20 = 28$ girls were from Clay.

Clay had $60$ students total, so the number of Clay boys is $60 - 28 = 32.$

Thus, the correct answer is B .

12.

图中的三个大正方形大小相同。各线段与正方形边的交点都是边的中点。三个正方形中的阴影面积如何比较？

Each of the three large squares shown is the same size. Segments that intersect the sides of the squares intersect at the midpoints of the sides. How do the shaded areas of these squares compare?

三个阴影面积都相等。

The shaded areas in all three are equal.

只有 $I$ 和 $II$ 的阴影面积相等。

Only the shaded areas of $I$ and $II$ are equal.

只有 $I$ 和 $III$ 的阴影面积相等。

Only the shaded areas of $I$ and $III$ are equal.

只有 $II$ 和 $III$ 的阴影面积相等。

Only the shaded areas of $II$ and $III$ are equal.

$I$ 、 $II$ 和 $III$ 的阴影面积都不同。

The shaded areas of $I,$ $II,$ and $III$ are all different.

知识点：面积分割面积比

难度评级：820

小提示：

把每个图形切成若干相等小块，求每个大正方形中阴影部分所占的比例

Find what fraction of each large square is shaded by cutting each figure into equal smaller pieces.

大提示：

把正方形 $I$ 分成 $8$ 个相等三角形，正方形 $II$ 分成 $4$ 个相等小格，正方形 $III$ 分成 $16$ 个相等三角形，再数阴影块数

Divide square $I$ into $8$ equal triangles, square $II$ into $4$ equal cells, and square $III$ into $16$ equal triangles, then count the shaded pieces in each.

解答：

在正方形 $II$ 中，阴影占 $1$ 个小格，总共有 $4$ 个相等小格，所以阴影面积是 $\tfrac14$ 。

正方形 $I$ 可分成 $8$ 个相等三角形，其中 $2$ 个涂色；正方形 $III$ 可分成 $16$ 个相等三角形，其中 $4$ 个涂色。这些比例都是 $\tfrac{2}{8} = \tfrac{4}{16} = \tfrac14$ 。

因为三个图形的阴影面积都占 $\tfrac14$ ，所以阴影面积全都相等。

所以正确答案是 A。

In square $II,$ $1$ of the $4$ equal cells is shaded, so $\tfrac14$ of it is shaded.

Square $I$ breaks into $8$ equal triangles with $2$ shaded, and square $III$ breaks into $16$ equal triangles with $4$ shaded; each of these equals $\tfrac{2}{8} = \tfrac{4}{16} = \tfrac14.$

Since every figure has exactly $\tfrac14$ shaded, the shaded areas are all equal.

Thus, the correct answer is A .

13.

位于 $\dfrac{1}{6}$ 和 $\dfrac{1}{4}$ 正中间的数是

The number halfway between $\dfrac{1}{6}$ and $\dfrac{1}{4}$ is

$\dfrac{1}{10}$

$\dfrac{1}{5}$

$\dfrac{5}{24}$

$\dfrac{7}{24}$

$\dfrac{5}{12}$

知识点：平均数分数

难度评级：660

小提示：

两个数正中间的数就是它们的平均数

The number halfway between two numbers is their average

大提示：

先把 $\dfrac16$ 和 $\dfrac14$ 相加，再把和除以 $2$

Add $\dfrac16$ and $\dfrac14,$ then divide the sum by $2$

解答：

两个数正中间的数是它们的平均数： $\dfrac{\frac16 + \frac14}{2}$ 。

因为 $\dfrac16 + \dfrac14 = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12}$ ，所以平均数是 $\dfrac{5}{12} \div 2 = \dfrac{5}{24}$ 。

所以正确答案是 C。

The number halfway between two values is their average: $\dfrac{\frac16 + \frac14}{2}.$

Since $\dfrac16 + \dfrac14 = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12},$ the average is $\dfrac{5}{12} \div 2 = \dfrac{5}{24}.$

Thus, the correct answer is C .

14.

pairball 每次可以由两个孩子玩。 $90$ 分钟内始终只有两个孩子同时在玩，五个孩子轮流上场，使每个人玩的时间相同。每个孩子玩了多少分钟？

Two children at a time can play pairball. For $90$ minutes, with only two children playing at one time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

$9$

$10$

$18$

$20$

$36$

知识点：速率

难度评级：820

小提示：

两个孩子玩满 $90$ 分钟，所以总游戏时间是 $2 \times 90$ 个“孩子分钟”

With two children playing for the whole $90$ minutes, the total playing time is $2 \times 90$ child-minutes

大提示：

把总游戏时间平均分给 $5$ 个孩子

Divide the total playing time equally among the $5$ children

解答：

任意时刻都有两个孩子在玩，共 $90$ 分钟，所以总游戏时间是 $2 \times 90 = 180$ 个孩子分钟。

平均分给 $5$ 个孩子，每个孩子玩 $\dfrac{180}{5} = 36$ 分钟。

所以正确答案是 E。

Two children play at every moment for $90$ minutes, so the total playing time is $2 \times 90 = 180$ child-minutes.

Split equally among $5$ children, each plays $\dfrac{180}{5} = 36$ minutes.

Thus, the correct answer is E .

15.

如果这条路径按相同规律继续，那么从点 $425$ 到点 $427$ 的箭头序列是哪一个？

If this path is to continue in the same pattern, then which sequence of arrows goes from point $425$ to point $427?$

知识点：模运算找规律

难度评级：910

小提示：

箭头按长度为 $4$ 的周期重复，所以某点出发的箭头只取决于它除以 $4$ 的余数

The arrows repeat in a cycle of length $4$ , so a point’s outgoing arrows depend only on its remainder when divided by $4$ .

大提示：

因为 $425$ 和 $427$ 除以 $4$ 的余数分别是 $1$ 和 $3$ ，所以箭头与从点 $1$ 到点 $3$ 的箭头相同

Since $425$ and $427$ leave remainders $1$ and $3$ when divided by $4$ , the arrows match those going from point $1$ to point $3$ .

解答：

这个模式每 $4$ 个点重复一次，所以从某点出发的箭头只取决于它除以 $4$ 的余数。

因为 $425 = 4(106) + 1$ ，且 $427 = 4(106) + 3$ ，所以从 $425$ 到 $427$ 的路径与从点 $1$ 到点 $3$ 相同：从 $1$ 向上到 $2$ ，再向右到 $3$ 。

所以正确答案是 A。

The pattern repeats every $4$ points, so the arrows leaving a point depend only on its remainder upon division by $4.$

Because $425 = 4(106) + 1$ and $427 = 4(106) + 3,$ the path from $425$ to $427$ looks just like the path from point $1$ to point $3$ : from $1$ it goes up to $2,$ then right to $3.$

Thus, the correct answer is A .

16.

一个正方形的周长是另一个正方形周长的 $3$ 倍。较大正方形的面积是较小正方形面积的多少倍？

The perimeter of one square is $3$ times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

$2$

$3$

$4$

$6$

$9$

知识点：相似面积比

难度评级：820

小提示：

如果周长是 $3$ 倍，那么边长也是 $3$ 倍

If the perimeter is $3$ times as large, the side length is also $3$ times as large

大提示：

面积按边长比例的平方缩放

Area scales as the square of the ratio of side lengths

解答：

正方形的周长与边长成正比，所以较大正方形的边长是较小正方形的 $3$ 倍。

面积等于边长的平方，所以较大面积是较小面积的 $3^2 = 9$ 倍。

所以正确答案是 E。

A square’s perimeter is proportional to its side, so the larger square has side length $3$ times the smaller one.

Area is the side squared, so the larger area is $3^2 = 9$ times the smaller.

Thus, the correct answer is E .

17.

Pauline Bunyan 第一小时可以铲 $20$ 立方码雪，第二小时铲 $19$ 立方码，第三小时铲 $18$ 立方码，依此类推，每小时都比上一小时少铲一立方码。如果她的车道宽 $4$ 码、长 $10$ 码，并被 $3$ 码深的雪覆盖，那么她铲干净大约需要多少小时？

Pauline Bunyan can shovel snow at the rate of $20$ cubic yards for the first hour, $19$ cubic yards for the second, $18$ for the third, etc., always shoveling one cubic yard less per hour than the previous hour. If her driveway is $4$ yards wide, $10$ yards long, and covered with snow $3$ yards deep, then the number of hours it will take her to shovel it clean is closest to

$4$

$5$

$6$

$7$

$12$

知识点：体积等差数列

难度评级：1000

小提示：

先求雪的总体积： $4 \times 10 \times 3$ 立方码

First find the total volume of snow: $4 \times 10 \times 3$

大提示：

逐小时累加 $20 + 19 + 18 + \cdots$ ，直到累计量达到雪的体积

Add $20 + 19 + 18 + \cdots$ hour by hour until the running total reaches the volume

解答：

雪的体积是 $4 \times 10 \times 3 = 120$ 立方码。

累计铲雪量是 $20 + 19 + 18 + 17$ $+ 16 + 15 + 14 = 119$ 立方码，也就是 $7$ 小时的工作量。第八小时的铲雪速度是每小时 $13$ 立方码，因此剩下的一立方码需要 $\dfrac{1}{13}$ 小时。准确时间是 $7\dfrac{1}{13}$ 小时，最接近 $7$ 小时。

所以正确答案是 D。

The volume of snow is $4 \times 10 \times 3 = 120$ cubic yards.

The amounts shoveled add up to $20 + 19 + 18 + 17$ $+ 16 + 15 + 14 = 119$ after $7$ hours. In the eighth hour her rate is $13$ cubic yards per hour, so the last cubic yard takes $\dfrac{1}{13}$ hour. The exact time is therefore $7\dfrac{1}{13}$ hours, which is closest to $7$ .

Thus, the correct answer is D .

18.

Mike 离家后，先在城市交通中缓慢向东行驶。到达高速公路后，他更快地继续向东行驶，直到到达购物中心并停车。他在购物中心购物一小时。Mike 沿原路回家，先在高速公路上快速向西行驶，然后在城市交通中缓慢行驶。每个图的纵轴表示离家的距离，横轴表示离家后经过的时间。哪个图最能表示 Mike 的行程？

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike’s trip?

知识点：数据与图表解读路程、速度与时间

难度评级：920

小提示：

外出时离家的距离增加，购物一小时期间保持不变，回家时距离减少

Distance from home increases on the way out, stays constant during the hour of shopping, and decreases on the way back.

大提示：

城市慢速行驶对应较平缓的斜率，高速快速行驶对应较陡的斜率；在购物中心停车是一段水平线

Slow city driving is a gentle (shallow) slope and fast highway driving is a steep slope; the stop at the mall is a flat, horizontal segment.

解答：

去程中 Mike 先慢速行驶，对应平缓上升；再快速行驶，对应较陡上升。购物一小时期间离家距离不变，对应水平线段。

回程中他先快速行驶，对应较陡下降；再慢速行驶，对应较平缓下降。只有先慢后快上升、顶部水平、再先快后慢下降的图符合。

所以正确答案是 B。

On the way out Mike drives slowly (a shallow slope) then rapidly (a steep slope), so the distance rises slowly and then more steeply. During the hour at the mall the distance stays the same, a flat segment.

On the return he drives rapidly (steep) then slowly (shallow), so the distance falls steeply and then more gently. Only the graph with a slow-then-fast rise, a flat top, and a fast-then-slow fall fits.

Thus, the correct answer is B .

19.

在一个 $4$ 乘 $4$ 的正方形外侧，以正方形的四条边为直径作四个半圆。另一个正方形 $ABCD$ 的边分别与原正方形对应边平行，并且 $ABCD$ 的每一边都与一个半圆相切。正方形 $ABCD$ 的面积是

Around the outside of a $4$ by $4$ square, construct four semicircles with the four sides of the square as their diameters. Another square, $ABCD,$ has its sides parallel to the corresponding sides of the original square, and each side of $ABCD$ is tangent to one of the semicircles. The area of the square $ABCD$ is

$16$

$32$

$36$

$48$

$64$

知识点：正方形（几何）圆

难度评级：980

小提示：

每个半圆以这个 $4$ 乘 $4$ 正方形的一条边为直径，所以半径是 $2$

Each semicircle has a side of the $4$ by $4$ square as its diameter, so its radius is $2$

大提示：

$ABCD$ 的一条边等于原正方形边长加上两个半径

A side of $ABCD$ equals the original side length plus two radii

解答：

每个半圆建在长度为 $4$ 的边上，所以半径为 $2$ 。每个半圆从原正方形的一边向外突出一个半径。

$ABCD$ 的每条边等于原边长加两个半径： $4 + 2(2) = 8$ 。所以 $ABCD$ 的面积是 $8^2 = 64$ 。

所以正确答案是 E。

Each semicircle is built on a side of length $4,$ so its radius is $2.$ A semicircle bulges out from the middle of each side by that radius.

Each side of $ABCD$ is the original side plus two radii: $4 + 2(2) = 8.$ So the area of $ABCD$ is $8^2 = 64.$

Thus, the correct answer is E .

20.

从下列集合中选出四个不同数字 $W$ ， $X$ ， $Y$ 和 $Z$ ：

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\text{。}$

如果要使 $\dfrac{W}{X} + \dfrac{Y}{Z}$ 尽可能小，那么 $\dfrac{W}{X} + \dfrac{Y}{Z}$ 必须等于

Let $W,$ $X,$ $Y,$ and $Z$ be four different digits selected from the set

$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

If the sum $\dfrac{W}{X} + \dfrac{Y}{Z}$ is to be as small as possible, then $\dfrac{W}{X} + \dfrac{Y}{Z}$ must equal

$\dfrac{2}{17}$

$\dfrac{3}{17}$

$\dfrac{17}{72}$

$\dfrac{25}{72}$

$\dfrac{13}{36}$

知识点：分数最优化

难度评级：1000

小提示：

要让每个分数小，应使用最小的数字作分子、最大的数字作分母

To make each fraction small, use the smallest digits as numerators and the largest as denominators

大提示：

比较 $\dfrac18 + \dfrac29$ 和 $\dfrac19 + \dfrac28$ ，看哪一个更小

Compare $\dfrac18 + \dfrac29$ with $\dfrac19 + \dfrac28$ to see which is smaller

解答：

小分子和大分母会产生较小分数，所以用 $1$ 和 $2$ 作分子，用 $8$ 和 $9$ 作分母。

把较大的分子配给较大的分母，得到 $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72}$ ，它小于 $\dfrac19 + \dfrac28 = \dfrac{26}{72}$ 。所以最小和是 $\dfrac{25}{72}$ 。

所以正确答案是 D。

Small numerators and large denominators make small fractions, so use $1$ and $2$ as numerators and $8$ and $9$ as denominators.

Pairing the larger numerator with the larger denominator gives $\dfrac18 + \dfrac29 = \dfrac{9 + 16}{72} = \dfrac{25}{72},$ which is smaller than $\dfrac19 + \dfrac28 = \dfrac{26}{72}.$ So the minimum sum is $\dfrac{25}{72}.$

Thus, the correct answer is D .

21.

一台口香糖球机中有 $9$ 个红色、 $7$ 个白色和 $8$ 个蓝色口香糖球。为了保证得到四个同色口香糖球，一个人至少必须购买多少个口香糖球？

A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

$8$

$9$

$10$

$12$

$18$

知识点：抽屉原理

难度评级：980

小提示：

考虑最坏情况：最多能抽到多少个球而每种颜色都不超过三个？

Consider the worst case: how many gumballs can you draw with at most three of each color?

大提示：

每种颜色都至少有三个，所以你可能先抽到每种颜色各三个；下一个就必然形成四个同色

Each color has at least three available, so you could draw three of each before the next one forces a fourth

解答：

最坏情况下，一个人可以先抽到 $3$ 个红球、 $3$ 个白球和 $3$ 个蓝球，共 $9$ 个球，还没有四个同色。

第十个球必须与其中某一种颜色相同，从而得到四个同色。因此需要 $10$ 个球。

所以正确答案是 C。

In the worst case, a person could draw $3$ red, $3$ white, and $3$ blue, which is $9$ gumballs, without yet having four of any color.

The next (tenth) gumball must match one of these colors, giving four of that color. So $10$ gumballs are needed.

Thus, the correct answer is C .

22.

右图中的两个转盘各转一次，并把得到的两个数相加。两个数之和为偶数的概率是

The two wheels shown at the right are spun and the two resulting numbers are added. The probability that the sum of the two numbers is even is

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{5}{12}$

$\dfrac{4}{9}$

知识点：基本概率奇偶性

难度评级：1060

小提示：

第一个转盘中 $3$ 占一半， $1$ 和 $2$ 各占四分之一；第二个转盘被平均分成三份

On the first wheel $3$ covers half and $1$ and $2$ each cover a quarter; the second wheel is divided into equal thirds

大提示：

和为偶数当且仅当两个数同为偶数或同为奇数

The sum is even when both numbers are even or both are odd

解答：

第一个转盘中， $P(1) = \dfrac14$ ， $P(2) = \dfrac14$ ， $P(3) = \dfrac12$ 。第二个转盘中， $4$ 、 $5$ 、 $6$ 各自的概率为 $\dfrac13$ 。

和为偶数需要两个数同奇偶。两数都为奇数时，第一个数是 $1$ 或 $3$ ，概率为 $\left(\dfrac34\right)$ ；第二个数是 $5$ ，概率为 $\left(\dfrac13\right)$ ，所以概率为 $\dfrac34 \cdot \dfrac13 = \dfrac14$ 。两数都为偶数时，第一个数是 $2$ ，概率为 $\left(\dfrac14\right)$ ；第二个数是 $4$ 或 $6$ ，概率为 $\left(\dfrac23\right)$ ，所以概率为 $\dfrac14 \cdot \dfrac23 = \dfrac16$ 。

总概率是 $\dfrac14 + \dfrac16 = \dfrac{5}{12}$ 。

所以正确答案是 D。

On the first wheel, $P(1) = \dfrac14,$ $P(2) = \dfrac14,$ and $P(3) = \dfrac12.$ On the second wheel, each of $4,$ $5,$ $6$ has probability $\dfrac13.$

The sum is even when both numbers are odd or both are even. Both odd: the first is $1$ or $3$ $\left(\dfrac34\right)$ and the second is $5$ $\left(\dfrac13\right),$ giving $\dfrac34 \cdot \dfrac13 = \dfrac14.$ Both even: the first is $2$ $\left(\dfrac14\right)$ and the second is $4$ or $6$ $\left(\dfrac23\right),$ giving $\dfrac14 \cdot \dfrac23 = \dfrac16.$

The total probability is $\dfrac14 + \dfrac16 = \dfrac{5}{12}.$

Thus, the correct answer is D .

23.

如果 $X$ ， $Y$ 和 $Z$ 是不同数字，那么下列加法中可能得到的最大 $3$ 位和

$\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}$

具有哪种形式？

If $X,$ $Y,$ and $Z$ are different digits, then the largest possible $3$ -digit sum for

$\begin{array}{cr} & XXX \\ & YX \\ + & X \\ \hline \end{array}$

has the form

$XXY$

$XYZ$

$YYX$

$YYZ$

$ZZY$

知识点：数字谜位值

难度评级：1090

小提示：

按列写出加法；如果 $X$ 太大，百位会进位，结果就变成四位数

Write the addition in columns; if $X$ is too large the hundreds place carries and the answer gains a fourth digit.

大提示：

先取能使和仍为三位数的最大 $X$ ，再选 $Y$ 使总和尽可能大，并看结果数字对应哪些字母

Take the largest $X$ that keeps the sum to three digits, then choose $Y$ to make the total as large as possible and see which letters the resulting digits match.

解答：

和的百位来自 $X$ 加上可能的进位，所以若 $X = 9$ ，和会变成四位数。允许的最大值是 $X = 8$ 。

为了使和最大，取 $Y = 9$ ：此时 $888 + 98 + 8 = 994$ 。它的数字是 $9$ 、 $9$ 、 $4$ ；因为 $9 = Y$ ，而 $4$ 是新数字 $Z$ ，所以和的形式是 $YYZ$ 。

所以正确答案是 D。

The hundreds digit of the sum comes from $X$ plus any carry, so if $X = 9$ the sum would spill over into four digits. The largest allowed value is $X = 8.$

To make the sum as large as possible, take $Y = 9:$ then $888 + 98 + 8 = 994.$ Its digits are $9,$ $9,$ $4;$ since $9 = Y$ and $4$ is a new digit $Z,$ the sum has the form $YYZ.$

Thus, the correct answer is D .

24.

一个 $2$ 乘 $2$ 的正方形被分成四个 $1$ 乘 $1$ 的小正方形。每个小正方形要涂成绿色或红色。共有多少种不同涂法，使得没有任何绿色小正方形的上边或右边与红色小正方形相邻？绿色小正方形可以少到零个，也可以多到四个。

A $2$ by $2$ square is divided into four $1$ by $1$ squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.

$4$

$6$

$7$

$8$

$16$

知识点：有限制的排列分类讨论

难度评级：1150

小提示：

如果一个小正方形是绿色，那么它正上方和正右方的小正方形也必须是绿色

If a square is green, then the square directly above it and the square directly to its right must also be green

大提示：

这迫使绿色方格集中在右上方向；按绿色方格的个数分类计数

This forces the green squares to cluster toward the top-right; count by how many squares are green

解答：

条件表示绿色方格不能在上方或右方紧邻红色方格，因此任何绿色方格都会迫使它上方和右方的方格也为绿色。绿色方格必须向右上角聚集。

合法涂法为：全红；只有右上角为绿；整行上排为绿；整列右排为绿；除左下角外全绿；全绿。共 $6$ 种。

所以正确答案是 B。

The rule says a green square cannot have a red square on its top or right side, so any green square forces the squares above and to its right to be green as well. The green squares must therefore cluster toward the top-right corner.

The valid colorings are: all four red; only the top-right green; the whole top row green; the whole right column green; all green except the bottom-left; and all four green. That is $6$ colorings.

Thus, the correct answer is B .

25.

求下列乘积结果的各位数字之和：

$\underbrace{9999\cdots99}_{94} \times \underbrace{4444\cdots44}_{94}$

也就是由 $94$ 个九组成的数乘以由 $94$ 个四组成的数。

Find the sum of the digits in the answer to

$\underbrace{9999\cdots99}_{94} \times \underbrace{4444\cdots44}_{94}$

where a string of $94$ nines is multiplied by a string of $94$ fours.

$846$

$855$

$945$

$954$

$1072$

知识点：数字找规律小情形

难度评级：1200

小提示：

先试小例子： $9 \times 4$ 、 $99 \times 44$ 、 $999 \times 444$

Try small cases first: $9 \times 4,$ $99 \times 44,$ $999 \times 444$

大提示：

乘积是一串 $4$ ，接着一个 $3$ ，再接一串 $5$ ，最后一个 $6$ ；数一数每种数字有多少个

The product is a block of $4$ s, then a $3,$ then a block of $5$ s, then a $6$ ; count how many of each

解答：

小例子显示规律： $99 \times 44 = 4356$ 和 $999 \times 444 = 443556$ 。一般地，由 $n$ 个九组成的数乘以由 $n$ 个四组成的数，结果为 $(n-1)$ 个四，然后一个 $3$ ，再 $(n-1)$ 个五，最后一个 $6$ 。

当 $n = 94$ 时，乘积含有 $93$ 个四、一个 $3$ 、 $93$ 个五和一个 $6$ 。数字和为 $93(4) + 3$ $+ 93(5) + 6$ $= 93(9) + 9$ $= 94(9)$ $= 846$ 。

所以正确答案是 A。

Small cases show the pattern: $99 \times 44 = 4356$ and $999 \times 444 = 443556.$ In general, a string of $n$ nines times a string of $n$ fours gives $(n-1)$ fours, then a $3,$ then $(n-1)$ fives, then a $6.$

For $n = 94,$ the product has $93$ fours, one $3,$ $93$ fives, and one $6.$ The digit sum is $93(4) + 3$ $+ 93(5) + 6$ $= 93(9) + 9$ $= 94(9)$ $= 846.$

Thus, the correct answer is A .