2014 AMC 8 详解 | LIVE by Po-Shen Loh

1.

Harry 和 Terry 都被要求计算 $8-(2+5)$ 。Harry 得到正确答案。Terry 忽略括号，计算了 $8-2+5$ 。如果 Harry 的答案是 $H$ ，Terry 的答案是 $T$ ，那么 $H-T$ 是多少？

Harry and Terry are each told to calculate $8-(2+5).$ Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5.$ If Harry’s answer is $H$ and Terry’s answer is $T,$ what is $H-T?$

$-10$

$-6$

$0$

$6$

$10$

知识点：运算顺序

难度评级：370

小提示：

Harry 先计算括号里面

Compute inside the parentheses first for Harry.

大提示：

将 Harry 的结果与 Terry 从左到右计算的结果比较

Compare Harry’s result with Terry’s left-to-right calculation.

解答：

Harry 正确计算： $\begin{align*} 8 - (2 + 5) &= 8 - 7 \\ &= 1 \end{align*}\text{。}$ Terry 错误计算： $\begin{align*} 8 - 2 + 5 &= 6 + 5 \\ &= 11 \end{align*}\text{。}$ 因此 $\begin{align*} H - T &= 1 - 11 \\ &= -10 \end{align*}\text{。}$

所以正确答案是 A。

Harry calculates it correctly as follows: $\begin{align*} 8 - (2 + 5) &= 8 - 7 \\ &= 1. \end{align*}$ Terry calculates it incorrectly as follows: $\begin{align*} 8 - 2 + 5 &= 6 + 5 \\ &= 11. \end{align*}$ Therefore: $\begin{align*} H - T &= 1 - 11 \\ &= -10. \end{align*}$

Thus, A is the correct answer.

2.

Paul 欠 Paula $35$ 美分，并且口袋里有 $5$ 美分、 $10$ 美分和 $25$ 美分硬币可用来付钱。他可以用来支付的硬币数最大值与最小值之差是多少？

Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?

$1$

$2$

$3$

$4$

$5$

知识点：钱币最优化

难度评级：450

小提示：

用最小的硬币可以使硬币数量最大

Use the smallest coins to maximize the number of coins.

大提示：

用一枚二十五美分和一枚十美分硬币可以使硬币数量最小

Use a quarter and a dime to minimize the number of coins.

解答：

要使用最多硬币，Paul 可以全用 $5$ 美分硬币，得到 $\frac{35}{5}=7$ 枚。

要使用最少硬币，他可以用一枚 $25$ 美分和一枚 $10$ 美分硬币，共 $2$ 枚。差为 $7-2=5$ 。

所以正确答案是 E。

To use the largest number of coins, Paul would use only $5$ -cent coins, which gives $\frac{35}{5}=7$ coins.

To use the smallest number of coins, Paul would use a $25$ -cent coin and a $10$ -cent coin, for a total of $2$ coins. The difference is $7-2=5$ .

Thus, E is the correct answer.

3.

Isabella 有一周时间为学校作业读完一本书。前三天她平均每天读 $36$ 页，接下来三天平均每天读 $44$ 页。最后一天她读了 $10$ 页读完这本书。这本书有多少页？

Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?

$240$

$250$

$260$

$270$

$280$

知识点：平均数

难度评级：560

小提示：

将每段平均数转换成总页数

Convert each average into a total number of pages.

大提示：

将前三天、接下来三天和最后一天相加

Add the first three days, next three days, and last day.

解答：

前三天 Isabella 读了 $36 \cdot 3 = 108$ 页。

接下来三天 Isabella 读了 $44 \cdot 3 = 132$ 页。

因此她一共读了 $108 + 132 + 10 = 250$ 页。

所以正确答案是 B。

During the first three days, Isabella read $36 \cdot 3 = 108$ pages.

During the next three days, Isabella read $44 \cdot 3 = 132$ pages.

Therefore, she read a total of $108 + 132 + 10 = 250$ pages.

Thus, B is the correct answer.

4.

两个质数之和为 $85$ 。这两个质数的乘积是多少？

The sum of two prime numbers is $85.$ What is the product of these two prime numbers?

$85$

$91$

$115$

$133$

$166$

知识点：质数奇偶性

难度评级：770

小提示：

两个质数之和为奇数时，必须包含唯一的偶质数

An odd sum of two primes must include the only even prime.

大提示：

从 $85$ 中减去那个偶质数

Subtract that even prime from $85$ .

解答：

两个奇质数之和是偶数。因为 $85$ 是奇数，所以其中一个质数必须是唯一的偶质数 $2$ 。

另一个质数是 $85-2=83$ ，所以乘积为 $2\cdot83=166$ 。

所以正确答案是 E。

The sum of two odd primes is even. Since $85$ is odd, one of the primes must be the only even prime, $2$ .

The other prime is $85-2=83$ , so the product is $2\cdot83=166$ .

Thus, E is the correct answer.

5.

Margie 的汽车每加仑汽油可以行驶 $32$ 英里，现在汽油价格为每加仑 $\$4$ 。Margie 用 $\$20$ 的汽油可以行驶多少英里？

Margie’s car can go $32$ miles on a gallon of gas, and gas currently costs $\$4$ per gallon. How many miles can Margie drive on $\$20$ worth of gas?

$64$

$128$

$160$

$320$

$640$

知识点：速率钱币

难度评级：450

小提示：

先求 $\$20$ 可以买多少加仑汽油

First find how many gallons $\$20$ buys.

大提示：

将加仑数乘以每加仑 $32$ 英里

Multiply the gallons by $32$ miles per gallon.

解答：

用 $\$20$ ，Margie 可以买 $\frac{20}{4}=5$ 加仑汽油。

每加仑可行驶 $32$ 英里，所以她可以行驶 $5\cdot32=160$ 英里。

所以正确答案是 C。

With $\$20$ , Margie can buy $\frac{20}{4}=5$ gallons of gas.

At $32$ miles per gallon, she can drive $5\cdot32=160$ miles.

Thus, C is the correct answer.

6.

六个长方形的共同底宽都是 $2$ ，长度分别为 $1$ ， $4$ ， $9$ ， $16$ ， $25$ 和 $36$ 。这六个长方形面积之和是多少？

Six rectangles each with a common base width of $2$ have lengths of $1,$ $4,$ $9,$ $16,$ $25,$ and $36.$ What is the sum of the areas of the six rectangles?

$91$

$93$

$162$

$182$

$202$

知识点：矩形面积

难度评级：720

小提示：

每个长方形面积等于 $2$ 乘以其列出的长度

Each rectangle has area $2$ times its listed length.

大提示：

先把长度相加，再乘以 $2$

Add the lengths first, then multiply by $2$ .

解答：

每个长方形面积是 $2$ 乘对应长度。可以先把共同因子 $2$ 提出来，只需求长度之和，即 $91$ 。

因此面积之和为 $2 \cdot 91 = 182$ 。

所以正确答案是 D。

To find the area of each rectangle we multiply $2$ by their respective lengths. This means that we can factor out the $2$ and we are left with the sum of the lengths. Adding together the lengths, we get $91.$

Therefore, the sum of the areas is $2 \cdot 91 = 182.$

Thus, D is the correct answer.

7.

Raub 女士班上有 $28$ 名学生，女生比男生多四人。班上女生人数与男生人数之比是多少？

There are four more girls than boys in Ms. Raub’s class of $28$ students. What is the ratio of the number of girls to the number of boys in her class?

$3 : 4$

$4 : 3$

$3 : 2$

$7 : 4$

$2 : 1$

知识点：比与比例一次方程

难度评级：770

小提示：

设男生人数为 $b$

Let the number of boys be $b$ .

大提示：

女生人数为 $b+4$ ，总人数为 $28$

Then the number of girls is $b+4$ , and the total is $28$ .

解答：

设班上男生人数为 $x$ ，则女生有 $x + 4$ 人。因为共有 $28$ 名学生， $\begin{align*} x + x + 4 &= 28\\ 2x&=24 \\ x&=12 \end{align*}\text{。}$ 因此女生与男生之比为 $16 : 12 = 4 : 3$ 。

所以正确答案是 B。

Let $x$ be the number of boys in the class. This means that there are $x + 4$ girls. As there are $28$ students, we know that: $\begin{align*} x + x + 4 &= 28\\ 2x&=24 \\ x&=12. \end{align*}$ Therefore, the ratio of girls to boys is $16 : 12 = 4 : 3.$

Thus, B is the correct answer.

8.

初中数学俱乐部的十一名成员每人支付了相同的金额，请一位嘉宾在数学俱乐部的会议上讲解解题方法。他们共付给这位嘉宾 $\$\underline{1}\underline{A}\underline{2}$ 。这个 $3$ 位数中缺失的数字 $A$ 是多少？

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\$\underline{1}\underline{A}\underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?

$0$

$1$

$2$

$3$

$4$

知识点：整除性数字

难度评级：960

小提示：

$11$ 名成员支付的总额必须能被 $11$ 整除

The total paid by $11$ members must be divisible by $11$ .

大提示：

对三位数使用 $11$ 的交错和整除判定

For a three-digit number, use the alternating-sum test for divisibility by $11$ .

解答：

因为 $11$ 个人支付相同金额，所以总额能被 $11$ 整除。因此 $\underline{1} \underline{A} \underline{2}$ 必须能被 $11$ 整除。

$11$ 的整除规则是：交错位数字和之差若能被 $11$ 整除，则原数能被 $11$ 整除。

对 $\underline{1} \underline{A} \underline{2}$ ，这个差为 $1 + 2 - A = 3 - A\text{。}$ 要让它能被 $11$ 整除，只能有 $A = 3$ 。

所以正确答案是 D。

Note that since $11$ people paid the same amount, then the resulting sum is divisible by $11.$ Therefore, $\underline{1} \underline{A} \underline{2}$ must be divisible by $11.$

Remember the divisibility rule for $11$ : if we take the difference of the sums of alternating digits, and this difference is divisible by $11,$ the whole number is divisible by $11.$

For $\underline{1} \underline{A} \underline{2},$ the aforementioned difference is $1 + 2 - A = 3 - A.$ The only way for this to be divisible by $11$ is if $A = 3.$

Thus, D is the correct answer.

9.

在 $\triangle ABC$ 中， $D$ 是边 $\overline{AC}$ 上一点，使得 $BD=DC$ ，且 $\angle BCD$ 为 $70^\circ$ 。 $\angle ADB$ 的度数是多少？

In $\triangle ABC,$ $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ.$ What is the degree measure of $\angle ADB?$

$100$

$120$

$135$

$140$

$150$

知识点：等腰三角形导角

难度评级：900

小提示：

因为 $BD=DC$ ，三角形 $BCD$ 是等腰三角形

Since $BD=DC$ , triangle $BCD$ is isosceles.

大提示：

先求 $\angle BDC$ ，再使用点 $D$ 处的平角

Find $\angle BDC$ , then use the straight angle at $D$ .

解答：

因为 $BD=DC$ ，三角形 $BDC$ 是等腰三角形，所以 $\angle DBC=\angle BCD=70^\circ$ 。

因此 $\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ\end{aligned}\text{。}$ 又因为 $A,D,C$ 共线， $\angle ADB$ 与 $\angle BDC$ 构成平角，所以 $\angle ADB=180^\circ-40^\circ=140^\circ$ 。

所以正确答案是 D。

Since $BD=DC$ , triangle $BDC$ is isosceles, so $\angle DBC=\angle BCD=70^\circ$ .

Thus $\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ.\end{aligned}$ Because $A,D,C$ are collinear, $\angle ADB$ and $\angle BDC$ form a straight angle, so $\angle ADB=180^\circ-40^\circ=140^\circ$ .

Thus, D is the correct answer.

10.

第一届 AMC $8$ 于 $1985$ 年举行，此后每年举行一次。Samantha 满 $12$ 岁的那一年参加了第七届 AMC $8$ 。Samantha 出生于哪一年？

The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8.$ In what year was Samantha born?

$1979$

$1980$

$1981$

$1982$

$1983$

知识点：年龄问题日期与时间

难度评级：770

小提示：

第七届 AMC $8$ 比第一届 AMC $8$ 晚六年

The seventh AMC $8$ was six years after the first AMC $8.$

大提示：

从那一年减去 Samantha 的年龄

Subtract Samantha’s age from that year.

解答：

第七届 AMC $8$ 是第一届之后 $6$ 年举行的。因此 Samantha 参加的是 $1985 + 6 = 1991$ 年的竞赛。

她那年 $12$ 岁，所以出生年份为 $1991 - 12 = 1979$ 。

所以正确答案是 A。

The seventh AMC $8$ would have been administered $6$ years after the first one. Therefore, Samantha took it in $1985 + 6 = 1991.$

This means that Samantha was born $12$ years prior in $1991 - 12 = 1979.$

Thus, A is the correct answer.

11.

Jack 想从自己家骑车到 Jill 家，Jill 家位于 Jack 家以东三个街区、以北两个街区。每骑过一个街区后，Jack 可以继续向东或向北，但他需要避开离他家以东一个街区、以北一个街区的危险路口。若他总共骑五个街区到达 Jill 家，有多少种路线？

Jack wants to bike from his house to Jill’s house, which is located three blocks east and two blocks north of Jack’s house. After biking each block, Jack can continue either east or north, but he needs to avoid a dangerous intersection one block east and one block north of his house. In how many ways can he reach Jill’s house by biking a total of five blocks?

$4$

$5$

$6$

$8$

$10$

知识点：格路补集计数

难度评级：1100

小提示：

任何最短路径都用三步向东和两步向北

Any shortest path uses three east moves and two north moves.

大提示：

只统计避开先走一东一北后到达的路口的路径

Count or list only the paths that avoid the intersection after one east and one north move.

解答：

用 E 表示向东一个街区，N 表示向北一个街区。最短路线使用三个 E 和两个 N。为了避开离 Jack 家以东一格、以北一格的危险路口，前两步必须是 EE 或 NN。

可能路线为 $EEENN$ 、 $EENEN$ 、 $EENNE$ 和 $NNEEE$ ，共 $4$ 条路线。

所以正确答案是 A。

Let E represent traveling one block east and N represent traveling one block north. A shortest route uses three E moves and two N moves. To avoid the dangerous intersection one block east and one block north of Jack’s house, the first two moves must be either EE or NN.

The possible routes are $EEENN$ , $EENEN$ , $EENNE$ , and $NNEEE$ , for $4$ routes.

Thus, A is the correct answer.

12.

一本杂志刊登了三位名人的照片，以及这三位名人婴儿时期的三张照片。婴儿照片没有标明对应的名人。读者被要求将每位名人与正确的婴儿照片配对。若读者随机猜测，三组都配对正确的概率是多少？

A magazine printed photos of three celebrities along with three photos of the celebrities as babies. The baby pictures did not identify the celebrities. Readers were asked to match each celebrity with the correct baby picture. What is the probability that a reader guessing at random will match all three correctly?

$\dfrac{1}{9}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

知识点：基本概率排列

难度评级：940

小提示：

将三张婴儿照配给三位名人共有 $3!$ 种方式

There are $3!$ ways to match the babies to celebrities.

大提示：

只有一种配对完全正确

Only one matching is completely correct.

解答：

读者给三位名人配对共有 $3! = 6$ 种方式。其中只有一种是正确配对。因此随机猜对全部三组的概率是 $\frac{1}{6}$ 。

所以正确答案是 B。

Notice that there are $3! = 6$ total ways that a reader could match the celebrities. However, only one of these is the correct matching. Therefore, the probability that the reader guesses it correctly is $\frac{1}{6}.$

Thus, B is the correct answer.

13.

如果 $n$ 和 $m$ 是整数，且 $n^2 + m^2$ 是偶数，那么下列哪一项是不可能的？

If $n$ and $m$ are integers and $n^2 + m^2$ is even, which of the following is impossible?

$n$ 和 $m$ 都是偶数

$n$ and $m$ are even

$n$ 和 $m$ 都是奇数

$n$ and $m$ are odd

$n + m$ 是偶数

$n + m$ is even

$n + m$ 是奇数

$n + m$ is odd

以上都不是不可能的

none of these are impossible

知识点：奇偶性

难度评级：1100

小提示：

一个平方数与其底数整数有相同奇偶性

A square has the same parity as its base integer.

大提示：

两个平方数之和为偶数时，两个整数奇偶性相同

For a sum of two squares to be even, the two integers have the same parity.

解答：

因为 $n^2 + m^2$ 是偶数，所以 $n^2$ 与 $m^2$ 要么都是奇数，要么都是偶数。

若它们都是奇数，则 $n$ 和 $m$ 都是奇数。若它们都是偶数，则 $n$ 和 $m$ 都是偶数。只要 $n$ 和 $m$ 同奇偶，它们的和都是偶数。

因此 $n + m$ 不可能是奇数。

所以正确答案是 D。

Since $n^2 + m^2$ is even, either $n^2$ and $m^2$ are both odd, or both even.

If they are both odd, then $n$ and $m$ are both odd. If they are both even, then $n$ and $m$ are both even. If $n$ and $m$ are both odd or even, their sum will always be even.

Therefore, $n + m$ is never odd.

Thus, D is the correct answer.

14.

长方形 $ABCD$ 和直角三角形 $DCE$ 面积相同。它们拼在一起形成如图所示的梯形。 $DE$ 是多少？

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE?$

$12$

$13$

$14$

$15$

$16$

知识点：面积勾股定理

难度评级：1140

小提示：

将三角形面积设为等于长方形面积

Set the triangle area equal to the rectangle area.

大提示：

求出 $CE$ 后，使用 $5$ - $12$ - $13$ 直角三角形

After finding $CE$ , use the $5$ - $12$ - $13$ right triangle.

解答：

长方形 $ABCD$ 的面积为 $5 \cdot 6 = 30\text{。}$

而这个三角形的面积为 $\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*}$ 因此 $\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12 \end{align*}\text{。}$ 再由勾股定理， $\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13 \end{align*}\text{。}$

所以正确答案是 B。

The area of $ABCD$ is $5 \cdot 6 = 30.$

The area of $\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*}$ Therefore: $\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12. \end{align*}$ Then using the Pythagorean theorem we get that $\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13. \end{align*}$

Thus, B is the correct answer.

15.

圆心为 $O$ 的圆周被分成 $12$ 段相等的弧，并如图标上字母 $A$ 到 $L$ 。角 $x$ 与 $y$ 的和是多少度？

The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y?$

$75$

$80$

$90$

$120$

$150$

知识点：弧等腰三角形导角

难度评级：1220

小提示：

相邻两个字母在圆心处截出的角为 $30^\circ$

Each adjacent pair of letters cuts off $30^\circ$ at the center.

大提示：

两个相关三角形都是等腰三角形，因为它们的边是半径

The two relevant triangles are isosceles because their sides are radii.

解答：

$12$ 段弧等分圆周，所以每段弧对应的圆心角为 $\frac{360^{\circ}}{12} = 30^{\circ}$ 。

$\angle AOE$ 跨过 $4$ 段弧，所以 $\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}\text{。}$ 类似地， $\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}\text{。}$ 我们还知道这两个三角形都是等腰三角形，因为它们各有两条边是半径。因此 $x = \dfrac{180 - 120}{2} = 30^{\circ}$ 且 $y = \dfrac{180 - 60}{2} = 60^{\circ}\text{。}$ 所以 $x + y = 90^{\circ}$ 。

所以正确答案是 C。

Note that each of the $12$ arcs splits the circle evenly, so they each cover $\frac{360^{\circ}}{12} = 30^{\circ}.$

$\angle AOE$ spans $4$ of these arcs, so $\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}.$ Similarly, $\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}.$ We also know that both triangles are isosceles since two of their sides are radii. Therefore, $x = \dfrac{180 - 120}{2} = 30^{\circ}$ and $y = \dfrac{180 - 60}{2} = 60^{\circ}.$ Therefore, $x + y = 90^{\circ}.$

Thus, C is the correct answer.

16.

“Middle School Eight” 篮球联盟有 $8$ 支球队。每个赛季，每支球队与联盟中其他每支球队各打两场比赛（一主一客），并且每支球队还与非联盟对手打 $4$ 场比赛。一个赛季中涉及 “Middle School Eight” 球队的比赛总数是多少？

The “Middle School Eight” basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the “Middle School Eight” teams?

$60$

$88$

$96$

$144$

$160$

知识点：基本计数

难度评级：1170

小提示：

将非联盟比赛与联盟内部比赛分开统计

Count non-conference games separately from conference games.

大提示：

对联盟比赛，统计每支球队对其他七队的主场比赛

For conference games, count each team’s home games against the other seven teams.

解答：

每支球队有 $4$ 场非联盟比赛，所以涉及 Middle School Eight 球队与非联盟对手的比赛有 $8\cdot4=32$ 场。

联盟内部，这 $8$ 支球队中的每支都有 $7$ 场对阵其他球队的主场比赛，所以联盟比赛有 $8\cdot7=56$ 场。总数为 $32+56=88$ 。

所以正确答案是 B。

Each team plays $4$ non-conference games, for $8\cdot4=32$ games involving Middle School Eight teams and non-conference opponents.

Within the conference, each of the $8$ teams has $7$ home games against the other teams, so there are $8\cdot7=56$ conference games. The total is $32+56=88$ .

Thus, B is the correct answer.

17.

George 步行 $1$ 英里去学校。他每天同一时间离家，以每小时 $3$ 英里的稳定速度行走，并且刚好在上课开始时到达。

今天他被宜人的天气分心，前 $\frac{1}{2}$ 英里只以每小时 $2$ 英里的速度行走。为了今天仍然刚好在上课开始时到达，George 最后 $\frac{1}{2}$ 英里必须以每小时多少英里的速度跑？

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins.

Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?

$4$

$6$

$8$

$10$

$12$

知识点：路程、速度与时间

难度评级：1240

小提示：

求平常以每小时 $3$ 英里的速度走一英里所需的时间

Find the usual time for one mile at $3$ mph.

大提示：

减去今天走前半英里已经花的时间

Subtract the time already spent walking the first half mile.

解答：

George 平常走 $1$ 英里时速度为每小时 $3$ 英里，需要 $\dfrac{1}{3}$ 小时，也就是 $20$ 分钟。

今天他走前 $\dfrac{1}{2}$ 英里时，速度为每小时 $2$ 英里，用时 $\dfrac{1}{2}\div2=\dfrac{1}{4}$ 小时，也就是 $15$ 分钟。他还剩 $5$ 分钟，即 $\dfrac{1}{12}$ 小时，来走完 $\dfrac{1}{2}$ 英里。

所需速度为 $\dfrac{1}{2}\div\dfrac{1}{12}=6$ 英里每小时。

所以正确答案是 B。

If George normally walks $1$ mile at $3$ miles per hour, it takes him $\dfrac{1}{3}$ hour, or $20$ minutes, to get to school.

Today he walked the first $\dfrac{1}{2}$ mile at $2$ miles per hour, taking $\dfrac{1}{2}\div2=\dfrac{1}{4}$ hour, or $15$ minutes. He has $5$ minutes, which is $\dfrac{1}{12}$ hour, left to cover $\dfrac{1}{2}$ mile.

His required speed is $\dfrac{1}{2}\div\dfrac{1}{12}=6$ miles per hour.

Thus, B is the correct answer.

18.

昨天 City Hospital 有四个孩子出生。假设每个孩子是男孩或女孩的可能性相同。下列哪种结果最可能发生？

Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?

$4$ 个都是男孩

all $4$ are boys

$4$ 个都是女孩

all $4$ are girls

$2$ 个女孩和 $2$ 个男孩

$2$ are girls and $2$ are boys

$3$ 个是一种性别， $1$ 个是另一种性别

$3$ are of one gender and $1$ is of the other gender

所有这些结果可能性相同

all of these outcomes are equally likely

知识点：二项概率组合

难度评级：1170

小提示：

按男孩和女孩的人数分类列出结果

List outcomes by the number of boys and girls.

大提示：

最可能的类别是排列方式最多的类别

The most likely category is the one with the most arrangements.

解答：

四个孩子按出生顺序共有 $2^4=16$ 个等可能结果。各类别数量为：全是男孩 $1$ 种；全是女孩 $1$ 种；两男两女 $\binom{4}{2}=6$ 种；三人一种性别、一人另一种性别 $2\binom{4}{1}=8$ 种。

最大数量是 $8$ ，所以最可能的结果是三个孩子是一种性别，一个孩子是另一种性别。

所以正确答案是 D。

There are $2^4=16$ equally likely birth-order outcomes. The counts by category are: all boys, $1$ ; all girls, $1$ ; two boys and two girls, $\binom{4}{2}=6$ ; and three of one gender and one of the other, $2\binom{4}{1}=8$ .

The largest count is $8$ , so the most likely outcome is three children of one gender and one of the other.

Thus, D is the correct answer.

19.

一个边长为 $3$ 英寸的立方体要由 $27$ 个边长为 $1$ 英寸的小立方体组成。其中二十一个小立方体涂成红色， $6$ 个涂成白色。

如果构造这个 $3$ 英寸立方体时让可见白色表面积尽可能小，那么白色占总表面积的几分之几？

A cube with $3$ -inch edges is to be constructed from $27$ smaller cubes with $1$ -inch edges. Twenty-one of the cubes are colored red and $6$ are colored white.

If the $3$ -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?

$\dfrac{5}{54}$

$\dfrac{1}{9}$

$\dfrac{5}{27}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

知识点：正方体表面积最优化

难度评级：1410

小提示：

将一个白色小立方体藏在 $3\times3\times3$ 立方体的中心

Hide one white cube in the center of the $3\times3\times3$ cube.

大提示：

将其余白色小立方体放在各面中心，使每个只露出一个白色面

Put the remaining white cubes in face centers to expose only one white face each.

解答：

为了最小化可见白色面积，把一个白色小立方体放在大立方体中心，这样没有任何面可见。把另外五个白色小立方体放在大立方体的面中心位置，每个只贡献一个可见白色小正方形。

可见白色表面积为 $5$ 平方英寸。边长 $3$ 英寸的立方体总表面积为 $6\cdot3^2=54$ 平方英寸，所以白色部分占 $\dfrac{5}{54}$ 。

所以正确答案是 A。

To minimize visible white area, place one white cube in the center of the large cube, where no faces are visible. Place each of the other five white cubes at the center of a face of the large cube, where each contributes only one visible white square.

The visible white surface area is therefore $5$ square inches. The total surface area of the $3$ -inch cube is $6\cdot3^2=54$ square inches, so the fraction that is white is $\dfrac{5}{54}$ .

Thus, A is the correct answer.

20.

长方形 $ABCD$ 的边长为 $CD=3$ 、 $DA=5$ 。半径为 $1$ 的圆以 $A$ 为圆心，半径为 $2$ 的圆以 $B$ 为圆心，半径为 $3$ 的圆以 $C$ 为圆心。下列哪一个最接近长方形内但三个圆外的区域面积？

Rectangle $ABCD$ has sides $CD=3$ and $DA=5.$ A circle of radius $1$ is centered at $A,$ a circle of radius $2$ is centered at $B,$ and a circle of radius $3$ is centered at $C.$ Which of the following is closest to the area of the region inside the rectangle but outside all three circles?

$3.5$

$4.0$

$4.5$

$5.0$

$5.5$

知识点：扇形面积估算

难度评级：1340

小提示：

在长方形内，每个圆都贡献一个四分之一圆扇形

Inside the rectangle, each circle contributes a quarter-circle sector.

大提示：

从长方形面积中减去扇形总面积，并近似 $\pi$

Subtract the total sector area from the rectangle area and approximate $\pi$ .

解答：

三个圆在长方形内的部分都是四分之一圆，半径分别为 $1$ 、 $2$ 、 $3$ 。它们的总面积为 $\dfrac{\pi}{4}+\pi+\dfrac{9\pi}{4}=\dfrac{7\pi}{2}\text{。}$

用 $\pi\approx\dfrac{22}{7}$ ，这个总面积约为 $11$ 。长方形面积为 $3\cdot5=15$ ，所以所求面积约为 $15-11=4$ 。

所以正确答案是 B。

The parts of the three circles inside the rectangle are quarter-circles with radii $1$ , $2$ , and $3$ . Their total area is $\dfrac{\pi}{4}+\pi+\dfrac{9\pi}{4}=\dfrac{7\pi}{2}.$

Using $\pi\approx\dfrac{22}{7}$ , this total is about $11$ . The rectangle has area $3\cdot5=15$ , so the desired area is about $15-11=4$ .

Thus, B is the correct answer.

21.

$7$ 位数 $\underline{74A52B1}$ 和 $\underline{326AB4C}$ 都是 $3$ 的倍数。下列哪一个可能是 $C$ 的值？

The $7$ -digit numbers $\underline{74A52B1}$ and $\underline{326AB4C}$ are each multiples of $3$ . Which of the following could be the value of $C$ ?

$1$

$2$

$3$

$5$

$8$

知识点：整除性数字

难度评级：1300

小提示：

对两个数都使用被 $3$ 整除的判定

Use the divisibility-by- $3$ test on both numbers.

大提示：

比较第一个数对 $A+B$ 的要求与第二个数的要求

Compare what the first number says about $A+B$ with what the second number needs.

解答：

为使 $\underline{74A52B1}$ 能被 $3$ 整除，数字和 $19+A+B$ 必须是 $3$ 的倍数。因此 $A+B$ 比 $3$ 的某个倍数小 $1$ 。

为使 $\underline{326AB4C}$ 能被 $3$ 整除，数字和 $15+A+B+C$ 必须是 $3$ 的倍数。因为 $15$ 已经能被 $3$ 整除，所以 $A+B+C$ 必须能被 $3$ 整除。

因此 $C$ 必须比 $3$ 的某个倍数大 $1$ 。选项中只有 $1$ 符合这个条件。

所以正确答案是 A。

For $\underline{74A52B1}$ to be divisible by $3$ , the digit sum $19+A+B$ must be a multiple of $3$ . Hence $A+B$ is $1$ less than a multiple of $3$ .

For $\underline{326AB4C}$ to be divisible by $3$ , the digit sum $15+A+B+C$ must be a multiple of $3$ . Since $15$ is already divisible by $3$ , $A+B+C$ must be divisible by $3$ .

Therefore $C$ must be $1$ more than a multiple of $3$ . Among the answer choices, only $1$ has that form.

Thus, A is the correct answer.

22.

一个 $2$ 位数满足：其两个数字的乘积加上两个数字的和等于这个数本身。这个数的个位数字是什么？

A $2$ -digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the units digit of the number?

$1$

$3$

$5$

$7$

$9$

知识点：数字代数变形

难度评级：1030

小提示：

设十位数字为 $a$ ，个位数字为 $b$

Let the tens digit be $a$ and the units digit be $b$ .

大提示：

将条件翻译为 $ab+a+b=10a+b$

Translate the condition into $ab+a+b=10a+b$ .

解答：

设这个数为 $10a+b$ ，其中 $a$ 是十位数字， $b$ 是个位数字。条件给出 $ab+a+b=10a+b\text{。}$

两边消去 $b$ ，得 $ab+a=10a$ ，所以 $ab=9a$ 。由于 $a\neq0$ ，得到 $b=9$ 。

所以正确答案是 E。

Let the number be $10a+b$ , where $a$ is the tens digit and $b$ is the units digit. The condition gives $ab+a+b=10a+b.$

Canceling $b$ from both sides gives $ab+a=10a$ , so $ab=9a$ . Since $a\neq0$ , we get $b=9$ .

Thus, E is the correct answer.

23.

Euclid Middle School 女子垒球队的三名队员进行了以下对话。

Ashley：我刚意识到我们的球衣号码都是 $2$ 位质数。

Bethany：而且你们两个球衣号码之和是我本月早些时候生日的日期。

Caitlin：真巧。你们两个球衣号码之和是我本月晚些时候生日的日期。

Ashley：而你们两个球衣号码之和是今天的日期。

Caitlin 穿几号球衣？

Three members of the Euclid Middle School girls’ softball team had the following conversation.

Ashley: I just realized that our uniform numbers are all $2$ -digit primes.

Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.

Caitlin: That’s funny. The sum of your two uniform numbers is the date of my birthday later this month.

Ashley: And the sum of your two uniform numbers is today’s date.

What number does Caitlin wear?

$11$

$13$

$17$

$19$

$23$

知识点：质数逻辑推理分类讨论

难度评级：1610

小提示：

两个两位质数之和必须是一个月份中的日期

Pair sums of two two-digit primes must be dates in a month.

大提示：

利用“早些时候、今天、晚些时候”的顺序判断哪个球衣号码最小

Use the earlier-today-later order to determine which uniform number is smallest.

解答：

任意两人的号码和必须是月份中的日期，所以至多为 $31$ 。两个两位质数之和不超过 $31$ 的可能有 $11+13=24$ 、 $11+17=28$ 、 $11+19=30$ 和 $13+17=30$ 。

三个日期分别是早些时候、今天和晚些时候，所以必须互不相同且递增。唯一可能的一组三个不同日期是 $24,28,30$ ，对应号码 $11$ 、 $13$ 、 $17$ 。

Bethany 的日期是 Ashley 加 Caitlin，Ashley 的日期是 Bethany 加 Caitlin，Caitlin 的晚些时候日期是 Ashley 加 Bethany。因此 Caitlin 的号码是三者中最小的，即 $11$ 。

所以正确答案是 A。

Any pair-sum must be a date in the month, so it is at most $31$ . The possible sums of two two-digit primes that are at most $31$ are $11+13=24$ , $11+17=28$ , $11+19=30$ , and $13+17=30$ .

The three dates are earlier, today, and later, so they must be distinct and increasing. The only possible set of three distinct dates is $24,28,30$ , coming from the numbers $11$ , $13$ , and $17$ .

Bethany’s date is Ashley plus Caitlin, Ashley’s date is Bethany plus Caitlin, and Caitlin’s later date is Ashley plus Bethany. This means Caitlin’s number is the smallest of the three, namely $11$ .

Thus, A is the correct answer.

24.

某天 Beverage Barn 卖出了 $252$ 罐汽水，购买者共有 $100$ 位，并且每位顾客至少买了一罐。当天每位顾客购买罐数的中位数最大可能是多少？

One day the Beverage Barn sold $252$ cans of soda to $100$ customers, and every customer bought at least one can of soda. What is the maximum possible median number of cans of soda bought per customer on that day?

$2.5$

$3.0$

$3.5$

$4.0$

$4.5$

知识点：中位数（数据）最优化

难度评级：1560

小提示：

中位数是按顺序排列后第 $50$ 项和第 $51$ 项的平均数

The median is the average of the $50$ th and $51$ st ordered purchases.

大提示：

让前 $49$ 位顾客购买数尽可能小

Make the first $49$ purchases as small as possible.

解答：

将 $100$ 位顾客的购买数从小到大排列。中位数是第 $50$ 项和第 $51$ 项的平均数。为了最大化它，让前 $49$ 项尽可能小，都等于 $1$ 。

如果第 $50$ 项至少为 $4$ ，那么总数至少为 $49+51\cdot4=253$ ，超过二百五十二。所以第 $50$ 项至多为 $3$ 。

这个最大值可以达到：取 $49$ 项等于 $1$ ，第 $50$ 项等于 $3$ ，最后 $50$ 项都等于 $4$ 。总数为 $49+3+50\cdot4=252$ ，中位数为 $\frac{3+4}{2}=3.5$ 。

所以正确答案是 C。

Order the $100$ purchases from least to greatest. The median is the average of the $50$ th and $51$ st entries. To maximize it, make the first $49$ entries as small as possible, so set them all equal to $1$ .

If the $50$ th entry were at least $4$ , then the total would be at least $49+51\cdot4=253$ , too many cans. So the $50$ th entry is at most $3$ .

This maximum is attainable: use $49$ entries equal to $1$ , the $50$ th entry equal to $3$ , and the final $50$ entries equal to $4$ . The total is $49+3+50\cdot4=252$ , and the median is $\frac{3+4}{2}=3.5$ .

Thus, C is the correct answer.

25.

一段笔直的一英里高速公路关闭，路宽 $40$ 英尺。Robert 沿图示由半圆组成的路径骑自行车。如果他以每小时 $5$ 英里的速度骑行，走完这一英里路段需要多少小时？

注意： $1$ 英里等于 $5280$ 英尺

A straight one-mile stretch of highway, $40$ feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at $5$ miles per hour, how many hours will it take to cover the one-mile stretch?

Note: $1$ mile = $5280$ feet

$\dfrac{\pi}{11}$

$\dfrac{\pi}{10}$

$\dfrac{\pi}{5}$

$\dfrac{2\pi}{5}$

$\dfrac{2\pi}{3}$

知识点：弧单位换算路程、速度与时间

难度评级：1460

小提示：

每个半圆让 Robert 沿公路前进 $40$ 英尺

Each semicircle advances Robert $40$ feet along the highway.

大提示：

比较半圆弧长与其直径

Compare the length of a semicircle to its diameter.

解答：

每个半圆让 Robert 沿公路前进 $40$ 英尺，所以一英里路段需要 $\frac{5280}{40}=132$ 个半圆。

每个半圆直径为 $40$ 英尺，因此弧长为 $20\pi$ 英尺。Robert 骑行总距离为 $132\cdot20\pi=2640\pi$ 英尺，也就是 $\dfrac{2640\pi}{5280}=\dfrac{\pi}{2}$ 英里。

速度为每小时 $5$ 英里，所以时间为 $\dfrac{\pi}{2}\div5=\dfrac{\pi}{10}$ 小时。

所以正确答案是 B。

Each semicircle advances Robert $40$ feet along the highway, so a one-mile stretch requires $\frac{5280}{40}=132$ semicircles.

Each semicircle has diameter $40$ feet, so its arc length is $20\pi$ feet. The total distance Robert rides is $132\cdot20\pi=2640\pi$ feet, which is $\dfrac{2640\pi}{5280}=\dfrac{\pi}{2}$ miles.

At $5$ miles per hour, the time is $\dfrac{\pi}{2}\div5=\dfrac{\pi}{10}$ hours.

Thus, B is the correct answer.