2004 AMC 8 真题与在线限时测验

计时

40:00

1.

在地图上， $12$ 厘米的长度表示 $72$ 千米。那么 $17$ 厘米的长度表示多少千米？

On a map, a $12$ -centimeter length represents $72$ kilometers. How many kilometers does a $17$ -centimeter length represent?

$6$

$102$

$204$

$864$

$1224$

答案：B

知识点：比与比例

难度评级：370

小提示：

先求 $1$ 厘米表示多少千米

Find how many kilometers $1$ centimeter represents.

大提示：

将这个比例乘以 $17$

Multiply that scale by $17.$

解答：

$1$ 厘米表示 $72 \div 12 = 6$ 千米。因此 $17$ 厘米表示 $6 \cdot 17 = 102$ 千米。

所以正确答案是 B。

Note that $1$ cm represents $72 \div 12 = 6$ kilometers. This means that $17$ cm represents $6 \cdot 17 = 102$ kilometers.

Thus, B is the correct answer.

2.

重新排列 $2004$ 中的四个数字，可以组成多少个不同的四位数？

How many different four-digit numbers can be formed by rearranging the four digits in $2004?$

$4$

$6$

$16$

$24$

$81$

答案：B

知识点：多重集排列有限制的排列

难度评级：450

小提示：

四位数不能以 $0$ 开头

A four-digit number cannot start with $0.$

大提示：

先选择哪个非零数字放在首位，再放置另一个非零数字

Choose which nonzero digit comes first, then place the other nonzero digit.

解答：

有 $2$ 个非零数字可以作为千位数字。

选定后，需要安排另外 $3$ 个数字。另一个非零数字有 $3$ 个位置可放。

因此共有 $2 \cdot 3 = 6$ 个可能的数。

所以正确答案是 B。

Note that there are $2$ non-zero digits that could be the thousands digit.

After choosing that, we need to arrange the other $3$ digits. There are $3$ spots for the other non-zero digit.

This gives us $2 \cdot 3 = 6$ possible numbers.

Thus, B is the correct answer.

3.

十二个朋友在奥斯卡超大份牡蛎餐厅共进晚餐，每人点了一份餐。餐量非常大，足够 $18$ 个人吃。如果他们共享食物，为了刚好够这 $12$ 个人吃，他们本应点多少份餐？

Twelve friends met for dinner at Oscar’s Overstuffed Oyster House, and each ordered one meal. The portions were so large, there was enough food for $18$ people. If they shared, how many meals should they have ordered to have just enough food for the $12$ of them?

$8$

$9$

$10$

$15$

$18$

答案：A

知识点：比与比例

难度评级：560

小提示：

$12$ 份餐足够 $18$ 个人吃

The $12$ meals made enough food for $18$ people.

大提示：

先求每人需要多少份餐，再按 $12$ 个人缩放

Find how many meals are needed per person, then scale to $12$ people.

解答：

$12$ 份餐能喂饱 $18$ 个人，所以一份餐能喂饱 $\dfrac{18}{12} = \dfrac{3}{2}$ 个人。

因此所需餐数为 $12 \div \dfrac{3}{2} = 8$ ，正好够 $12$ 个人吃。

所以正确答案是 A。

Note that $12$ meals feed $18$ people. This means that one meal feeds $\dfrac{18}{12} = \dfrac{3}{2}$ people.

This means that they need $12 \div \dfrac{3}{2} = 8$ meals for $12$ people.

Thus, A is the correct answer.

4.

汉密尔顿老师的八年级班想参加一年一度的三人篮球队锦标赛。兰斯、萨莉、乔伊和弗雷德被选为队员。三名首发队员有多少种选择方式？

Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. Lance, Sally, Joy, and Fred are chosen for the team. In how many ways can the three starters be chosen?

$2$

$4$

$6$

$8$

$10$

答案：B

知识点：组合补集计数

难度评级：660

小提示：

选择三名首发等同于选择谁不首发

Choosing the three starters is the same as choosing who sits out.

大提示：

有 $4$ 名可能不首发的队员

There are $4$ possible players who could be the alternate.

解答：

若有 $3$ 名首发，则有一人不首发。选择不首发的人就确定了首发。

不首发的人有 $4$ 种选择。

所以正确答案是 B。

If there are $3$ starters, then one person must not be starting. Choosing the person who doesn’t start determines the starters.

There are $4$ choices for the person who doesn’t start.

Thus, B is the correct answer.

5.

汉密尔顿老师的八年级班想参加一年一度的三人篮球队锦标赛。每场比赛的失败队伍会被淘汰。如果有十六支队伍参赛，需要打多少场比赛才能决出冠军？

Ms. Hamilton’s eighth-grade class wants to participate in the annual three-person-team basketball tournament. The losing team of each game is eliminated from the tournament. If sixteen teams compete, how many games will be played to determine the winner?

$4$

$7$

$8$

$15$

$16$

答案：D

知识点：基本计数不变量

难度评级：730

小提示：

每场比赛恰好淘汰一支队伍

Each game eliminates exactly one team.

大提示：

锦标赛在只剩一支队伍时结束

The tournament ends when all but one team have been eliminated.

解答：

每场比赛后都会淘汰一支队伍。要只剩下一支队伍，必须淘汰 $15$ 支队伍。

因此必须打 $15$ 场比赛。

所以正确答案是 D。

Note that after every game, one team gets eliminated. For there to be one team remaining, $15$ teams must have been eliminated.

This means that $15$ games had to have been played.

Thus, D is the correct answer.

6.

萨莉投篮 $20$ 次后，命中率为 $55\%$ 。她又投 $5$ 次后，命中率提高到 $56\%$ 。最后 $5$ 次投篮中她命中了几次？

After Sally takes $20$ shots, she has made $55\%$ of her shots. After she takes $5$ more shots, she raises her percentage to $56\%.$ How many of the last $5$ shots did she make?

$1$

$2$

$3$

$4$

$5$

答案：C

知识点：百分数一次方程

难度评级：870

小提示：

先求萨莉前 $20$ 次命中了多少次

Find how many shots Sally had made after $20$ shots.

大提示：

再求她 $25$ 次投篮后的总命中数

Find how many total made shots she had after $25$ shots.

解答：

萨莉在最初的 $20$ 次投篮中命中了 $20 \times 0.55 = 11$ 次。于是 $\dfrac{11 + x}{25} = 0.56\text{，}$ 所以 $11 + x = 14$ 且 $x = 3$ 。

所以正确答案是 C。

Sally made $20 \times 0.55 = 11$ of her first $20$ shots. Then we get that $\dfrac{11 + x}{25} = 0.56,$ which tells us that $11 + x = 14$ and $x = 3.$

Thus, C is the correct answer.

7.

运动员的目标心率（每分钟心跳次数）是理论最大心率的 $80\%$ 。最大心率由 $220$ 减去运动员年龄（岁）得到。一个 $26$ 岁运动员的目标心率最接近多少？

An athlete’s target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete’s age, in years, from $220.$ To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?

$134$

$155$

$176$

$194$

$243$

答案：B

知识点：百分数估算

难度评级：900

小提示：

先计算 $220-26$

First compute $220-26.$

大提示：

取这个最大心率的 $80\%$ ，再四舍五入

Take $80\%$ of that maximum heart rate and round.

解答：

这名运动员的最大心率为 $220 - 26 = 194$ 。目标心率为 $194 \times 0.8 \approx 155$ 。

所以正确答案是 B。

The maximum heart rate for this athlete would be $220 - 26 = 194.$ Then the target heart rate would be $194 \times 0.8 \approx 155.$

Thus, B is the correct answer.

8.

数字和为 $7$ 的两位正整数有多少个？

Find the number of two-digit positive integers whose digits total $7.$

$6$

$7$

$8$

$9$

$10$

答案：B

知识点：数字系统列举

难度评级：930

小提示：

十位数字不能是 $0$

The tens digit cannot be $0.$

大提示：

一旦选定十位数字，个位数字就确定了

Once the tens digit is chosen, the ones digit is forced.

解答：

十位数字可以从 $1$ 到 $7$ ，而这个数字会确定个位数字。

因此共有 $7$ 个数。

所以正确答案是 B。

Note that the tens digit can range from $1$ to $7,$ and this digit determines the units digit.

Therefore, there are $7$ numbers.

Thus, B is the correct answer.

9.

一个列表中五个数的平均数是 $54$ 。前两个数的平均数是 $48$ 。后三个数的平均数是多少？

The average of the five numbers in a list is $54.$ The average of the first two numbers is $48.$ What is the average of the last three numbers?

$55$

$56$

$57$

$58$

$59$

答案：D

知识点：平均数

难度评级：1020

小提示：

把每个平均数转化为总和

Turn each average into a sum.

大提示：

从五个数总和中减去前两个数的总和

Subtract the sum of the first two numbers from the sum of all five.

解答：

$5$ 个数的总和是 $54 \cdot 5 = 270$ 。前 $2$ 个数的总和是 $48 \cdot 2 = 96$ 。

后 $3$ 个数的总和是 $270 - 96 = 174$ ，所以平均数是 $174 \div 3 = 58$ 。

所以正确答案是 D。

The sum of all $5$ numbers is $54 \cdot 5 = 270.$ The sum of the first $2$ numbers is $48 \cdot 2 = 96.$

The sum of the last $3$ numbers is $270 - 96 = 174.$ The average is therefore $174 \div 3 = 58.$

Thus, D is the correct answer.

10.

“能手”亚伦星期一帮邻居 $1 \frac{1}{4}$ 小时，星期二帮 $50$ 分钟，星期三上午从 $8:20$ 到 $10:45$ ，星期五帮半小时。他每小时得到 $\$3$ 。他这一周赚了多少钱？

Handy Aaron helped a neighbor $1 \frac{1}{4}$ hours on Monday, $50$ minutes on Tuesday, from $8:20$ to $10:45$ on Wednesday morning, and a half-hour on Friday. He is paid $\$3$ per hour. How much did he earn for the week?

$\$8$

$\$9$

$\$10$

$\$12$

$\$15$

答案：E

知识点：单位换算速率

难度评级：1000

小提示：

把所有工作时间都换算成分钟

Convert all the work times to minutes.

大提示：

求出总分钟数后，先换算成小时，再乘以 $\$3$

After finding total minutes, convert to hours before multiplying by $\$3.$

解答：

亚伦星期一工作 $75$ 分钟，星期二工作 $50$ 分钟，星期三工作 $145$ 分钟，星期五工作 $30$ 分钟。

总时间为 $75+50+145+30=300$ 分钟，即 $5$ 小时。

按每小时 $\$3$ 计算，他赚了 $5\cdot3=15$ 美元。

所以正确答案是 E。

Aaron worked $75$ minutes on Monday, $50$ minutes on Tuesday, $145$ minutes on Wednesday, and $30$ minutes on Friday.

The total is $75+50+145+30=300$ minutes, or $5$ hours.

At $\$3$ per hour, he earned $5\cdot3=15$ dollars.

Thus, E is the correct answer.

11.

数字 $-2$ 、 $4$ 、 $6$ 、 $9$ 和 $12$ 按以下规则重新排列：

$1$ 。最大数不在第一位，但在前三位之一。

$2$ 。最小数不在最后一位，但在后三位之一。

$3$ 。中位数既不在第一位，也不在最后一位。

第一位和最后一位数字的平均数是多少？

The numbers $-2,$ $4,$ $6,$ $9$ and $12$ are rearranged according to these rules:

$1.$ The largest isn’t first, but it is in one of the first three places.

$2.$ The smallest isn’t last, but it is in one of the last three places.

$3.$ The median isn’t first or last.

What is the average of the first and last numbers?

$3.5$

$5$

$6.5$

$7.5$

$8$

答案：C

知识点：逻辑推理平均数

难度评级：1060

小提示：

用规则确定最大数、最小数和中位数的位置

Use the rules to locate the largest, smallest, and median numbers.

大提示：

第一位和最后一位由剩下两个数填入

The first and last positions are filled by the two remaining numbers.

解答：

最大数、最小数和中位数都不能在第一位或最后一位。

因此第一位和最后一位是 $4$ 和 $9$ ，顺序不限。它们的平均数为 $(4 + 9) \div 2 = 13 \div 2 = 6.5\text{。}$

所以正确答案是 C。

Note that the largest, smallest, and median numbers cannot be the first or last number.

This means that the first and last numbers are $4$ and $9$ in some order. Their average is $(4 + 9) \div 2 = 13 \div 2 = 6.5.$

Thus, C is the correct answer.

12.

妮基通常让手机保持开机。如果手机开着但她实际上没有使用，电池能持续 $24$ 小时。如果她一直使用，电池只能持续 $3$ 小时。自上次充电以来，她的手机已经开机 $9$ 小时，其中她使用了 $60$ 分钟。如果她不再通话但保持手机开机，电池还能持续多少小时？

Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the battery will last for $24$ hours. If she is using it constantly, the battery will last for only $3$ hours. Since the last recharge, her phone has been on $9$ hours, and during that time she has used it for $60$ minutes. If she doesn’t talk any more but leaves the phone on, how many more hours will the battery last?

$7$

$8$

$11$

$14$

$15$

答案：B

知识点：速率分数

难度评级：1170

小提示：

通话一小时消耗的电量相当于好几个待机小时

One hour of talking uses as much battery as several idle hours.

大提示：

手机待机 $8$ 小时，使用 $1$ 小时

The phone was idle for $8$ hours and used for $1$ hour.

解答：

不使用时，手机每小时消耗 $\frac{1}{24}$ 的电量。使用时，每小时消耗 $\frac{1}{3}$ 的电量。

妮基的手机开机 $9$ 小时，其中 $8$ 小时待机， $1$ 小时通话。

因此手机已消耗 $\frac{2}{3}$ 的电量。剩余 $\frac{1}{3}$ 电量在不使用时还能持续 $8$ 小时。

所以正确答案是 B。

When not in use, her cell phone uses up $\frac{1}{24}$ of its battery per hour. When it is in use, it uses up $\frac{1}{3}$ of its battery per hour.

Niki’s phone has been on for $9$ hours, with $8$ of those hours being idle and $1$ hour being used to talk on the phone.

This means that the phone has used up $\frac{2}{3}$ of its battery. In order to drain the remaining $\frac{1}{3}$ of the battery, the phone can last for $8$ more hours without being used.

Thus, B is the correct answer.

13.

艾米、比尔和赛琳是年龄不同的朋友。以下陈述中恰好只有一个是真的。

$\mathrm{I}$ 。比尔年龄最大。

$\mathrm{II}$ 。艾米不是年龄最大的。

$\mathrm{III}$ 。赛琳不是年龄最小的。

请按从年龄最大到最小排列这些朋友。

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.

$\mathrm{I}.$ Bill is the oldest.

$\mathrm{II}.$ Amy is not the oldest.

$\mathrm{III}.$ Celine is not the youngest.

Rank the friends from the oldest to youngest.

比尔、艾米、赛琳

Bill, Amy, Celine

艾米、比尔、赛琳

Amy, Bill, Celine

赛琳、艾米、比尔

Celine, Amy, Bill

赛琳、比尔、艾米

Celine, Bill, Amy

艾米、赛琳、比尔

Amy, Celine, Bill

答案：E

知识点：逻辑推理分类讨论

难度评级：1190

小提示：

如果比尔最大，检查会有多少个陈述为真

If Bill were oldest, check how many statements would be true.

大提示：

排除谁最大后，使用恰好只有一个陈述为真的条件

After ruling out the oldest person, use the fact that exactly one statement is true.

解答：

如果比尔最大，那么陈述 $\mathrm{I}$ 和 $\mathrm{II}$ 都为真，所以比尔不是最大的。

如果赛琳最大，那么陈述 $\mathrm{II}$ 和 $\mathrm{III}$ 都为真，所以赛琳也不是最大的。

因此艾米最大。陈述 $\mathrm{I}$ 和 $\mathrm{II}$ 为假，所以陈述 $\mathrm{III}$ 必须是唯一为真的陈述。因此赛琳不是最小，比尔最小。

顺序是艾米、赛琳、比尔。

所以正确答案是 E。

If Bill were oldest, then statements $\mathrm{I}$ and $\mathrm{II}$ would both be true, so Bill is not oldest.

If Celine were oldest, then statements $\mathrm{II}$ and $\mathrm{III}$ would both be true, so Celine is not oldest.

Therefore Amy is oldest. Statements $\mathrm{I}$ and $\mathrm{II}$ are false, so statement $\mathrm{III}$ must be the single true statement. Thus Celine is not youngest, leaving Bill youngest.

The order is Amy, Celine, Bill.

Thus, E is the correct answer.

14.

下面钉板四边形围成的面积是多少？

What is the area enclosed by the geoboard quadrilateral below?

$15$

$18\frac{1}{2}$

$22\frac{1}{2}$

$27$

$41$

答案：C

知识点：格点面积分割

难度评级：1190

小提示：

把四边形放入一个容易处理的外接正方形中

Put the quadrilateral inside an easy surrounding square.

大提示：

从外接正方形中减去外侧的长方形和直角三角形

Subtract the outside rectangles and right triangles from the surrounding square.

解答：

将四边形放入外接的 $10\times10$ 正方形中，其面积为 $100$ 。

五个外侧部分的面积分别为 $15$ 、 $\frac12\cdot6\cdot7=21$ 、 $\frac12\cdot1\cdot3=\frac32$ 、 $\frac12\cdot4\cdot5=10$ 、 $\frac12\cdot6\cdot10=30$ 。

外侧面积为 $15+21+\frac32+10+30=77\frac12$ ，所以四边形面积为 $100-77\frac12=22\frac12$ 。

所以正确答案是 C。

Place the quadrilateral inside the surrounding $10\times10$ square, whose area is $100.$

The five outside pieces have areas $15,$ $\frac12\cdot6\cdot7=21,$ $\frac12\cdot1\cdot3=\frac32,$ $\frac12\cdot4\cdot5=10,$ and $\frac12\cdot6\cdot10=30.$

The outside area is $15+21+\frac32+10+30=77\frac12,$ so the quadrilateral area is $100-77\frac12=22\frac12.$

Thus, C is the correct answer.

15.

十三块阴影六边形瓷砖和六块未阴影六边形瓷砖组成下图。如果用与其他瓷砖大小和形状相同的未阴影瓷砖加上一圈边框形成新图形，新图形中未阴影瓷砖总数与阴影瓷砖总数之差是多少？

Thirteen shaded and six unshaded hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of unshaded tiles with the same size and shape as the others, what will be the difference between the total number of unshaded tiles and the total number of shaded tiles in the new figure?

$5$

$7$

$11$

$12$

$18$

答案：C

知识点：铺砖找规律

难度评级：1220

小提示：

原图有一个中心瓷砖、一圈 $6$ 块和一圈 $12$ 块

The original figure has one center tile, a ring of $6,$ and a ring of $12.$

大提示：

下一圈六边形边框有 $18$ 块瓷砖

The next hexagonal border has $18$ tiles.

解答：

原图有 $13$ 块阴影瓷砖和 $6$ 块未阴影瓷砖。

新边框是围绕图形的下一圈六边形环，有 $18$ 块未阴影瓷砖。

新图形有 $6+18=24$ 块未阴影瓷砖和 $13$ 块阴影瓷砖，所以差为 $24-13=11$ 。

所以正确答案是 C。

The original figure has $13$ shaded tiles and $6$ unshaded tiles.

The new border is the next hexagonal ring around the figure, which has $18$ unshaded tiles.

The new figure has $6+18=24$ unshaded tiles and $13$ shaded tiles, so the difference is $24-13=11.$

Thus, C is the correct answer.

16.

两个 $600$ 毫升的水壶中装有橙汁。一个水壶装了 $\frac 13$ 满，另一个装了 $\frac 25$ 满。向每个水壶中加水直到装满，然后把两个水壶都倒入一个大容器。大容器中混合液的橙汁占几分之几？

Two $600$ mL pitchers contain orange juice. One pitcher is $\frac 13$ full and the other pitcher is $\frac 25$ full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?

$\dfrac{1}{8}$

$\dfrac{3}{16}$

$\dfrac{11}{30}$

$\dfrac{11}{19}$

$\dfrac{11}{15}$

答案：C

知识点：混合问题分数

难度评级：1160

小提示：

先计算加水前每个水壶中的橙汁量

Compute the orange juice in each pitcher before water is added.

大提示：

两个水壶装满后，混合液总量为 $1200$ 毫升

After filling both pitchers, the total mixture volume is $1200$ mL.

解答：

第一个水壶含有 $600 \cdot \dfrac{1}{3} = 200$ 毫升橙汁。第二个含有 $600 \cdot \dfrac{2}{5} = 240$ 毫升。

大容器中共有 $200 + 240 = 440$ 毫升橙汁。混合液总量为 $2 \cdot 600 = 1200$ 毫升。

因此橙汁所占比例为 $\dfrac{440}{1200} = \dfrac{11}{30}$ 。

所以正确答案是 C。

The first pitcher contains $600 \cdot \dfrac{1}{3} = 200$ mL of orange juice. The second one has $600 \cdot \dfrac{2}{5} = 240$ mL.

The large container then has $200 + 240 = 440$ mL of orange juice. The total amount of mixture is $2 \cdot 600 = 1200$ mL.

Then the fraction of orange juice is $\dfrac{440}{1200} = \dfrac{11}{30}.$

Thus, C is the correct answer.

17.

三个朋友共有 $6$ 支相同的铅笔，每个人至少有一支。共有多少种可能？

Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?

$1$

$3$

$6$

$10$

$12$

答案：D

知识点：隔板法分类讨论

难度评级：1220

小提示：

先给每个朋友一支铅笔

Give each friend one pencil first.

大提示：

数出剩下 $3$ 支铅笔分给 $3$ 个朋友的方式

Count the ways to distribute the remaining $3$ pencils among the $3$ friends.

解答：

先给每个朋友一支铅笔。还剩 $3$ 支铅笔要分给 $3$ 个朋友。

如果一个朋友得到全部剩下的 $3$ 支，有 $3$ 种方式。如果剩下的铅笔按 $2$ 和 $1$ 分，有 $3\cdot2=6$ 种方式。如果每个朋友再各得一支，有 $1$ 种方式。

总方式数为 $3+6+1=10$ 。

所以正确答案是 D。

First give each friend one pencil. Then $3$ pencils remain to distribute among the $3$ friends.

If one friend gets all $3$ remaining pencils, there are $3$ ways. If the remaining pencils split as $2$ and $1,$ there are $3\cdot2=6$ ways. If each friend gets one more, there is $1$ way.

The total number of ways is $3+6+1=10.$

Thus, D is the correct answer.

18.

五个朋友参加飞镖比赛。每人向同一个圆形靶子投两支飞镖，每个人的得分是击中区域分数之和。靶子区域的分数是整数 $1$ 到 $10$ ，十次投掷击中的区域分值互不相同。得分为：爱丽丝 $16$ 分，本 $4$ 分，辛迪 $7$ 分，戴夫 $11$ 分，埃伦 $17$ 分。谁击中了分值为 $6$ 的区域？

Five friends compete in a dart-throwing contest. Each one has two darts to throw at the same circular target, and each individual’s score is the sum of the scores in the target regions that are hit. The scores for the target regions are the whole numbers $1$ through $10.$ Each throw hits the target in a region with a different value. The scores are: Alice $16$ points, Ben $4$ points, Cindy $7$ points, Dave $11$ points, and Ellen $17$ points. Who hits the region worth $6$ points?

爱丽丝

Alice

本

Ben

辛迪

Cindy

戴夫

Dave

埃伦

Ellen

答案：A

知识点：逻辑推理分类讨论

难度评级：1400

小提示：

本的 $4$ 分只有一种不同正分值配对方式

Ben’s score of $4$ has only one possible pair of distinct positive scores.

大提示：

用每个被迫确定的配对，从剩余分数中排除数字

Use each forced pair to eliminate numbers from the remaining scores.

解答：

本得到 $4$ 分的唯一方式是 $1$ 和 $3$ ，因为他不能两次击中 $2$ 。

辛迪的得分可能来自 $1 + 6,\quad 2 + 5,\quad 3 + 4\text{。}$ 本已经击中了 $1$ 和 $3$ ，所以辛迪必须击中 $2$ 和 $5$ 。

同理，戴夫必须击中 $4$ 和 $7$ 。最后，因为 $7$ 已经被使用，爱丽丝只能击中 $6$ 和 $10$ ，埃伦击中 $8$ 和 $9$ 。

所以正确答案是 A。

The only way to get Ben’s score of $4$ is with a $1$ and $3$ since he can’t hit $2$ twice.

Cindy can achieve her score with $1 + 6,\quad 2 + 5,\quad 3 + 4.$ Ben already hit $1$ and $3,$ so Cindy must have hit $2$ and $5.$

Similarly, Dave must have hit $4$ and $7.$ Finally, since $7$ is already used, Alice is forced to have hit $6$ and $10$ with Ellen hitting $8$ and $9.$

Thus, A is the correct answer.

19.

一个大于 $2$ 的整数，分别除以下各数时余数都是 $2$ ： $3$ 、 $4$ 、 $5$ 和 $6$ 。最小的这样一个数位于下列哪两个数之间？

A whole number larger than $2$ leaves a remainder of $2$ when divided by each of the numbers $3,$ $4,$ $5,$ and $6.$ The smallest such number lies between which two numbers?

$40$ 和 $49$

$40$ and $49$

$60$ 和 $79$

$60$ and $79$

$100$ 和 $129$

$100$ and $129$

$210$ 和 $249$

$210$ and $249$

$320$ 和 $369$

$320$ and $369$

答案：B

知识点：最小公倍数模运算

难度评级：1220

小提示：

如果这个数余 $2$ ，就减去 $2$

If the number leaves remainder $2,$ subtract $2.$

大提示：

结果必须能被 $3$ 、 $4$ 、 $5$ 、 $6$ 整除

The result must be divisible by $3,$ $4,$ $5,$ and $6.$

解答：

设这个数为 $x$ 。那么 $x - 2$ 能被 $3$ 、 $4$ 、 $5$ 和 $6$ 整除。

这些数的最小公倍数是 $60$ ，所以 $x$ 为 $62$ 。

所以正确答案是 B。

Let $x$ be the number. Then $x - 2$ is divisible by $3,$ $4,$ $5,$ and $6.$

The least common multiple of these numbers is $60,$ which makes $x$ $62.$

Thus, B is the correct answer.

20.

一个房间里三分之二的人坐在四分之三的椅子上。剩下的人站着。如果有 $6$ 把空椅子，房间里有多少人？

Two-thirds of the people in a room are seated in three-fourths of the chairs. The rest of the people are standing. If there are $6$ empty chairs, how many people are in the room?

$12$

$18$

$24$

$27$

$36$

答案：D

知识点：分数比与比例

难度评级：1230

小提示：

空椅子是所有椅子的剩余四分之一

The empty chairs are the remaining one-fourth of the chairs.

大提示：

坐着的人是所有人的三分之二

The seated people are two-thirds of all the people.

解答：

因为 $\frac{3}{4}$ 的椅子被坐了，所以 $6$ 把空椅子是所有椅子的 $\frac{1}{4}$ 。因此共有 $24$ 把椅子。

坐着的人数是 $\frac{3}{4}$ 乘 $24$ ，即 $18$ 。

这 $18$ 个坐着的人是所有人的 $\frac{2}{3}$ ，所以总人数为 $18\div\frac23=27$ 。

所以正确答案是 D。

Since $\frac{3}{4}$ of the chairs are occupied, the $6$ empty chairs are $\frac{1}{4}$ of all the chairs. Thus there are $24$ chairs.

The number of seated people is $\frac{3}{4}$ of $24,$ which is $18.$

Those $18$ seated people are $\frac{2}{3}$ of all the people, so the total number of people is $18\div\frac23=27.$

Thus, D is the correct answer.

21.

转盘 $A$ 和 $B$ 被旋转。在每个转盘上，指针落在每个数字上的可能性相同。两个转盘数字的乘积为偶数的概率是多少？

Spinners $A$ and $B$ are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners’ numbers is even?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

答案：D

知识点：基本概率对立事件概率奇偶性

难度评级：1290

小提示：

考虑补事件更容易：乘积为奇数

It is easier to count the complement: an odd product.

大提示：

乘积为奇数要求两个转盘结果都为奇数

An odd product requires both spinner results to be odd.

解答：

要使乘积为偶数，至少一个转盘必须落在偶数上。

使用补集计数，计算两个转盘都落在奇数上的概率。

这个概率为 $\dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{3}\text{。}$

因此至少一个为偶数的概率为 $1 - \dfrac{1}{3} = \dfrac{2}{3}\text{。}$

所以正确答案是 D。

For the product to be even, at least one of the spinners must land on an even number.

We can use complementary counting and calculate the probability of both spinners landing on odds.

This happens with a probability of $\dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1}{3}.$

Then the probability of landing on at least one even is $1 - \dfrac{1}{3} = \dfrac{2}{3}.$

Thus, D is the correct answer.

22.

一个聚会上只有单身女性和已婚男性及其妻子。随机选择一名女性，她是单身的概率为 $\dfrac{2}{5}$ 。房间里的人中，已婚男性占几分之几？

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is $\dfrac{2}{5}.$ What fraction of the people in the room are married men?

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{2}{5}$

$\dfrac{5}{12}$

$\dfrac{3}{5}$

答案：B

知识点：比与比例

难度评级：1450

小提示：

使用一个符合 $\frac{2}{5}$ 比例的方便女性人数

Use a convenient number of women matching the ratio $\frac{2}{5}.$

大提示：

已婚女性和已婚男性人数相等

Married women and married men come in equal numbers.

解答：

为了便于按比例计算，设房间里有 $5$ 名女性。那么单身女性有 $5 \cdot \dfrac{2}{5} = 2$ 名。

因此已婚女性有 $5 - 2 = 3$ 名，也就是已婚男性的人数。

房间里共有 $5 + 3 = 8$ 人。已婚男性所占比例为 $\dfrac{3}{8}$ 。

所以正确答案是 B。

For a convenient ratio model, let there be $5$ women in the room. Then there are $5 \cdot \dfrac{2}{5} = 2$ single women.

This means that there are $5 - 2 = 3$ married women, which is also the number of married men.

There are a total of $5 + 3 = 8$ people in the room. The fraction of married men is $\dfrac{3}{8}.$

Thus, B is the correct answer.

23.

特丝沿长方形街区 $JKLM$ 逆时针跑步。她住在角 $J$ 。哪张图可能表示她到家的直线距离？

Tess runs counterclockwise around rectangular block $JKLM.$ She lives at corner $J.$ Which graph could represent her straight-line distance from home?

答案：D

知识点：数据与图表解读

难度评级：1450

小提示：

跟踪她在长方形每条边上离角 $J$ 的距离

Track the distance from corner $J$ at each side of the rectangle.

大提示：

距离在对角 $L$ 处最大

The distance is greatest at the opposite corner $L.$

解答：

特丝从 $J$ 出发跑向 $K$ 时，她到家的距离不断增加。

从 $K$ 到 $L$ ，距离继续增加，但变化方式不同，并在对角 $L$ 达到最大。

从 $L$ 到 $M$ ，再从 $M$ 回到 $J$ ，距离逐渐减小到 $0$ ，并在 $M$ 处再次改变变化方式。

只有图 $D$ 先以一种方式增加，再改变增加方式；随后改变减少方式，最后减小到 $0$ 。

所以正确答案是 D。

As Tess runs from $J$ to $K,$ her distance from home increases.

From $K$ to $L,$ her distance continues to increase, but with a different shape, reaching its maximum at the opposite corner $L.$

From $L$ to $M$ and then from $M$ back to $J,$ her distance decreases back to $0,$ again changing behavior at $M.$

Graph $D$ is the only graph with this increase, changed-rate increase, changed-rate decrease, and final decrease to $0.$

Thus, D is the correct answer.

24.

图中， $ABCD$ 是长方形， $EFGH$ 是平行四边形。利用图中给出的测量值，求线段 $d$ 的长度；该线段垂直于 $\overline{HE}$ 和 $\overline{FG}$ 。

In the figure, $ABCD$ is a rectangle and $EFGH$ is a parallelogram. Using the measurements given in the figure, what is the length $d$ of the segment that is perpendicular to $\overline{HE}$ and $\overline{FG}?$

$6.8$

$7.1$

$7.6$

$7.8$

$8.1$

答案：C

知识点：平行四边形面积分割勾股定理

难度评级：1540

小提示：

求长方形面积并减去四个角上的三角形

Find the area of the rectangle and subtract the four corner triangles.

大提示：

用勾股定理求平行四边形底边 $HE$

Use the Pythagorean theorem to get the parallelogram base $HE.$

解答：

长方形边长为 $10$ 和 $8$ ，所以面积为 $80$ 。

四个角上三角形的总面积为 $\begin{aligned} &2\cdot\frac12\cdot3\cdot4 \\ &\quad{}+2\cdot\frac12\cdot6\cdot5 \\ &=12+30=42 \end{aligned}\text{。}$ 因此平行四边形 $EFGH$ 的面积为 $80-42=38$ 。

线段 $HE$ 长为 $5$ ，这可由 $3-4-5$ 直角三角形得到。因为平行四边形面积为 $HE\cdot d$ ，所以 $5d=38$ ，从而 $d=7.6$ 。

所以正确答案是 C。

The rectangle has side lengths $10$ and $8,$ so its area is $80.$

The four corner triangles have total area $\begin{aligned} &2\cdot\frac12\cdot3\cdot4 \\ &\quad{}+2\cdot\frac12\cdot6\cdot5 \\ &=12+30=42. \end{aligned}$ Thus parallelogram $EFGH$ has area $80-42=38.$

Segment $HE$ has length $5$ by the $3-4-5$ right triangle. Since the parallelogram area is $HE\cdot d,$ we have $5d=38,$ so $d=7.6.$

Thus, C is the correct answer.

25.

两个 $4 \times 4$ 正方形成直角相交，并且每个正方形都平分另一个正方形被其穿过的两条边，如图所示。圆的直径是两个交点之间的线段。从两个正方形中去掉圆后形成的阴影区域面积是多少？

Two $4 \times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

$16-4\pi$

$16-2\pi$

$28-4\pi$

$28-2\pi$

$32-2\pi$

答案：D

知识点：圆面积容斥原理

难度评级：1580

小提示：

求两个正方形并集覆盖的面积

Find the area covered by the union of the two squares.

大提示：

圆的直径是 $2\times2$ 重叠正方形的对角线

The circle diameter is the diagonal of the $2\times2$ overlap square.

解答：

两个 $4\times4$ 正方形总面积为 $32$ ，但它们的重叠部分是一个 $2\times2$ 、面积为 $4$ 的正方形。因此两个正方形并集的面积为 $32-4=28$ 。

圆的直径是这个 $2\times2$ 重叠正方形的对角线，所以直径为 $2\sqrt2$ ，半径为 $\sqrt2$ 。

圆面积为 $\pi(\sqrt2)^2=2\pi$ ，所以阴影面积为 $28-2\pi$ 。

所以正确答案是 D。

The two $4\times4$ squares have total area $32,$ but their overlap is a $2\times2$ square with area $4.$ Thus the area covered by the union of the two squares is $32-4=28.$

The circle’s diameter is the diagonal of that $2\times2$ overlap square, so the diameter is $2\sqrt2$ and the radius is $\sqrt2.$

The circle area is $\pi(\sqrt2)^2=2\pi,$ so the shaded area is $28-2\pi.$

Thus, D is the correct answer.