1995 AMC 8 详解 | LIVE by Po-Shen Loh

1.

沃尔特口袋里正好有一枚一美分硬币、一枚五美分硬币、一枚十美分硬币和一枚二十五美分硬币。他口袋里的钱是一美元的百分之几？

Walter has exactly one penny, one nickel, one dime and one quarter in his pocket. What percent of one dollar is in his pocket?

$4\%$

$25\%$

$40\%$

$41\%$

$59\%$

知识点：钱币百分数

难度评级：370

小提示：

把这些硬币的面值按美分相加

Add up the values of the coins in cents

大提示：

$1 + 5 + 10 + 25$ 美分，而 $100$ 美分是一美元

$1 + 5 + 10 + 25$ cents, and $100$ cents make one dollar

解答：

这些硬币总共值 $1 + 5 + 10 + 25 = 41$ 美分。

因为一美元是 $100$ 美分，所以这相当于一美元的 $41\%$ 。

所以正确答案是 D。

The coins total $1 + 5 + 10 + 25 = 41$ cents.

Since one dollar is $100$ cents, this is $41\%$ of a dollar.

Thus, the correct answer is D .

2.

何塞比扎克小 $4$ 岁。扎克比伊内兹大 $3$ 岁。伊内兹 $15$ 岁。何塞几岁？

Jose is $4$ years younger than Zack. Zack is $3$ years older than Inez. Inez is $15$ years old. How old is Jose?

$8$

$11$

$14$

$16$

$22$

知识点：年龄问题

难度评级：370

小提示：

从伊内兹算到扎克，再从扎克算到何塞

Work from Inez up to Zack, then down to Jose

大提示：

扎克 $= 15 + 3$ ；然后何塞 $=$ 扎克 $- 4$

Zack $= 15 + 3;$ then Jose $=$ Zack $- 4$

解答：

扎克是 $15 + 3 = 18$ 岁。

何塞比扎克小 $4$ 岁，所以何塞是 $18 - 4 = 14$ 岁。

所以正确答案是 C。

Zack is $15 + 3 = 18$ years old.

Jose is $4$ years younger, so Jose is $18 - 4 = 14$ .

Thus, the correct answer is C .

3.

下列哪种运算对一个数的作用，与先乘以 $\dfrac34$ 再除以 $\dfrac35$ 相同？

Which of the following operations has the same effect on a number as multiplying by $\dfrac34$ and then dividing by $\dfrac35?$

除以 $\dfrac43$

dividing by $\dfrac43$

除以 $\dfrac{9}{20}$

dividing by $\dfrac{9}{20}$

乘以 $\dfrac{9}{20}$

multiplying by $\dfrac{9}{20}$

除以 $\dfrac54$

dividing by $\dfrac54$

乘以 $\dfrac54$

multiplying by $\dfrac54$

知识点：分数

难度评级：560

小提示：

除以 $\dfrac35$ 等于乘以 $\dfrac53$

Dividing by $\dfrac35$ is the same as multiplying by $\dfrac53$

大提示：

把 $\dfrac34 \times \dfrac53$ 合并成一个分数

Combine $\dfrac34 \times \dfrac53$ into a single fraction

解答：

除以 $\dfrac35$ 等于乘以 $\dfrac53$ ，所以两个运算合起来相当于乘以

$\dfrac34 \times \dfrac53 = \dfrac54\text{。}$

因此总效果是乘以 $\dfrac54$ 。

所以正确答案是 E。

Dividing by $\dfrac35$ is the same as multiplying by $\dfrac53,$ so the two operations together multiply by

$\dfrac34 \times \dfrac53 = \dfrac54.$

So the combined effect is multiplying by $\dfrac54$ .

Thus, the correct answer is E .

4.

老师告诉全班：“想一个数，给它加 $1$ ，再把结果加倍。把答案给你的搭档。搭档从得到的数中减去 $1$ ，再把结果加倍，得到自己的答案。”本想的是 $6$ ，并把他的答案给苏。苏的答案应是多少？

A teacher tells the class, “Think of a number, add $1$ to it, and double the result. Give the answer to your partner. Partner, subtract $1$ from the number you are given and double the result to get your answer.” Ben thinks of $6,$ and gives his answer to Sue. What should Sue’s answer be?

$18$

$24$

$26$

$27$

$30$

知识点：运算顺序

难度评级：450

小提示：

先算本的答案：给 $6$ 加 $1$ ，再加倍

First compute Ben’s answer: add $1$ to $6,$ then double

大提示：

苏用本给的数，先减 $1$ ，再把结果加倍

Sue takes Ben’s number, subtracts $1,$ and doubles that

解答：

本计算 $(6 + 1) \times 2 = 14$ ，并把 $14$ 给苏。

苏计算 $(14 - 1) \times 2 = 26$ 。

所以正确答案是 C。

Ben computes $(6 + 1) \times 2 = 14$ and gives $14$ to Sue.

Sue computes $(14 - 1) \times 2 = 26$ .

Thus, the correct answer is C .

5.

求大于下列和的最小整数：

$2\tfrac12 + 3\tfrac13 + 4\tfrac14 + 5\tfrac15\text{。}$

Find the smallest whole number that is larger than the sum

$2\tfrac12 + 3\tfrac13 + 4\tfrac14 + 5\tfrac15.$

$14$

$15$

$16$

$17$

$18$

知识点：分数估算

难度评级：660

小提示：

分别相加整数部分和分数部分

Add the whole-number parts and the fraction parts separately

大提示：

分数 $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ 的和略大于 $1$

The fractions $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ add to a little more than $1$

解答：

整数部分的和为 $2 + 3 + 4 + 5 = 14$ 。

分数部分 $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ 约为 $1.28$ ，在 $1$ 和 $2$ 之间。因此总和在 $15$ 和 $16$ 之间，大于它的最小整数是 $16$ 。

所以正确答案是 C。

The whole-number parts sum to $2 + 3 + 4 + 5 = 14$ .

The fractions $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ add to about $1.28,$ which is between $1$ and $2.$ So the total is between $15$ and $16,$ and the smallest whole number larger than it is $16$ .

Thus, the correct answer is C .

6.

图形 $I$ 、 $II$ 和 $III$ 都是正方形。 $I$ 的周长是 $12$ ， $II$ 的周长是 $24$ 。 $III$ 的周长是

Figures $I,$ $II$ and $III$ are squares. The perimeter of $I$ is $12$ and the perimeter of $II$ is $24.$ The perimeter of $III$ is

$9$

$18$

$36$

$72$

$81$

知识点：周长正方形（几何）

难度评级：770

小提示：

正方形周长是边长的 $4$ 倍，所以先求 $I$ 和 $II$ 的边长

The perimeter of a square is $4$ times its side, so find the side of $I$ and of $II$

大提示：

从图中看， $III$ 的边长等于 $I$ 的边长加 $II$ 的边长

From the figure, the side of $III$ equals the side of $I$ plus the side of $II$

解答：

正方形 $I$ 的边长是 $12 \div 4 = 3$ ，正方形 $II$ 的边长是 $24 \div 4 = 6$ 。

从图中看， $III$ 的边长是 $3 + 6 = 9$ ，所以它的周长是 $4 \times 9 = 36$ 。

所以正确答案是 C。

Square $I$ has side $12 \div 4 = 3,$ and square $II$ has side $24 \div 4 = 6.$

From the figure, the side of $III$ is $3 + 6 = 9,$ so its perimeter is $4 \times 9 = 36.$

Thus, the correct answer is C .

7.

在克洛弗维尤初中，一半学生乘校车回家，四分之一乘汽车回家，十分之一骑自行车回家。其余学生步行回家。步行回家的学生占全体的几分之几？

At Clover View Junior High, one half of the students go home on the school bus. One fourth go home by automobile. One tenth go home on their bicycles. The rest walk home. What fractional part of the students walk home?

$\dfrac{1}{16}$

$\dfrac{3}{20}$

$\dfrac13$

$\dfrac{17}{20}$

$\dfrac{9}{10}$

知识点：分数

难度评级：730

小提示：

给这些分数通分再相加

Give the fractions a common denominator to add them

大提示：

用 $1$ 这个整体减去乘车或骑车回家的总比例

Subtract the total who ride from $1$ (the whole group)

解答：

乘车或骑车的学生占

$\begin{aligned} \dfrac12 + \dfrac14 + \dfrac{1}{10} &= \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} \\ &= \dfrac{17}{20} \end{aligned}\text{。}$

所以步行的比例是 $1 - \dfrac{17}{20} = \dfrac{3}{20}$ 。

所以正确答案是 B。

The students who ride make up

$\begin{aligned} \dfrac12 + \dfrac14 + \dfrac{1}{10} &= \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} \\ &= \dfrac{17}{20}. \end{aligned}$

So the fraction who walk is $1 - \dfrac{17}{20} = \dfrac{3}{20}.$

Thus, the correct answer is B .

8.

一位在意大利旅行的美国人想把美元兑换成意大利里拉。如果 $3000$ 里拉 $= \$1.60$ ，那么用 $\$1.00$ 可以兑换多少里拉？

An American traveling in Italy wishes to exchange American money (dollars) for Italian money (lire). If $3000$ lire $= \$1.60,$ how many lire will the traveler receive in exchange for $\$1.00?$

$180$

$480$

$1800$

$1875$

$4875$

知识点：比与比例

难度评级：820

小提示：

用给定汇率求出一美元对应多少里拉

Find how many lire equal one dollar using the given rate

大提示：

$\$1.00$ 是 $\$1.60$ 的 $\dfrac{1.00}{1.60}$ ，所以取 $3000$ 的这个比例

$\$1.00$ is $\dfrac{1.00}{1.60}$ of $\$1.60,$ so take that fraction of $3000$

解答：

因为 $\$1.00$ 是 $\$1.60$ 的 $\dfrac{1.00}{1.60} = \dfrac58$ ，所以旅行者会得到 $3000$ 里拉的 $\dfrac58$ 。

也就是 $\dfrac58 \times 3000 = 1875$ 里拉。

所以正确答案是 D。

Since $\$1.00$ is $\dfrac{1.00}{1.60} = \dfrac58$ of $\$1.60,$ the traveler gets $\dfrac58$ of $3000$ lire.

That is $\dfrac58 \times 3000 = 1875$ lire.

Thus, the correct answer is D .

9.

如图，三个全等圆的圆心分别为 $P$ 、 $Q$ 和 $R$ ，并与长方形 $ABCD$ 的边相切。以 $Q$ 为圆心的圆直径为 $4$ ，并经过点 $P$ 和 $R$ 。长方形的面积是

Three congruent circles with centers $P,$ $Q$ and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R.$ The area of the rectangle is

$16$

$24$

$32$

$64$

$128$

知识点：相切圆矩形

难度评级：960

小提示：

每个圆的直径都是 $4$ ，所以长方形的短边等于 $4$

The diameter of each circle is $4,$ so the short side of the rectangle equals $4$

大提示：

长边跨过两个完整直径

The long side spans two full diameters

解答：

每个圆的直径为 $4$ 。长方形的短边等于一个直径，所以短边为 $4$ 。

因为以 $Q$ 为圆心的圆经过 $P$ 和 $R$ ，三个圆半径都是 $2$ ，长方形长边跨过两个完整直径： $4 + 4 = 8$ 。面积为 $8 \times 4 = 32$ 。

所以正确答案是 C。

Each circle has diameter $4.$ The short side of the rectangle equals one diameter, so it is $4.$

Since the circle at $Q$ passes through $P$ and $R,$ all three circles have radius $2,$ and the long side spans two full diameters: $4 + 4 = 8.$ The area is $8 \times 4 = 32.$

Thus, the correct answer is C .

10.

一件夹克和一件衬衫原价分别为 $\$80$ 和 $\$40$ 。促销期间，克里斯以 $40\%$ 折扣买了 $\$80$ 的夹克，又以 $55\%$ 折扣买了 $\$40$ 的衬衫。节省的总金额是原总价的百分之几？

A jacket and a shirt originally sold for $\$80$ and $\$40,$ respectively. During a sale Chris bought the $\$80$ jacket at a $40\%$ discount and the $\$40$ shirt at a $55\%$ discount. The total amount saved was what percent of the total of the original prices?

$45\%$

$47\tfrac12\%$

$50\%$

$79\tfrac16\%$

$95\%$

知识点：百分数

难度评级：930

小提示：

分别求出两件商品各节省了多少美元

Find the dollars saved on each item separately

大提示：

用总节省金额除以原总价 $\$120$

Divide the total saved by the total original price of $\$120$

解答：

夹克折扣节省 $40\%$ 的 $\$80 = \$32$ ，衬衫折扣节省 $55\%$ 的 $\$40 = \$22$ 。总共节省 $\$32 + \$22 = \$54$ 。

原总价是 $\$80 + \$40 = \$120$ ，所以节省比例是 $\dfrac{54}{120} = 0.45 = 45\%$ 。

所以正确答案是 A。

The jacket discount saves $40\%$ of $\$80 = \$32,$ and the shirt discount saves $55\%$ of $\$40 = \$22.$ The total saved is $\$32 + \$22 = \$54.$

The original total is $\$80 + \$40 = \$120,$ so the percent saved is $\dfrac{54}{120} = 0.45 = 45\%.$

Thus, the correct answer is A .

11.

简（Jane）走任意距离所用时间是赫克托（Hector）走同样距离所用时间的一半。他们从图中所示 $18$ 个街区区域外侧出发，沿相反方向行走。他们第一次相遇时，最接近哪个点？

Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the $18$ -block area as shown. When they meet for the first time, they will be closest to

$A$

$B$

$C$

$D$

$E$

知识点：路程、速度与时间周长相对速度

难度评级：1030

小提示：

第一次相遇时，他们合起来走完了整个 $18$ 街区的环路

When they first meet, together they have covered the full $18$ -block loop

大提示：

简的速度是赫克托的两倍，所以简走 $12$ 个街区，赫克托走 $6$ 个街区

Jane walks twice as fast, so she covers $12$ blocks while Hector covers $6$

解答：

该区域的周长为 $18$ 个街区，所以简和赫克托相遇时合计走了 $18$ 个街区。因为简的速度是赫克托的两倍，所以简走了 $12$ 个街区，赫克托走了 $6$ 个街区。

从底边中点出发，赫克托走 $6$ 个街区，先到 $E$ ，再向上到 $D$ ；简走 $12$ 个街区，先到 $A$ ，再向上到 $B$ ，最后沿上边到 $D$ 。所以他们在 $D$ 相遇。

所以正确答案是 D。

The perimeter of the region is $18$ blocks, so when Jane and Hector meet they have together walked $18$ blocks. Since Jane walks twice as fast, she covers $12$ blocks and Hector covers $6.$

Starting from the middle of the bottom edge, Hector walks $6$ blocks (to $E,$ then up to $D$ ), and Jane walks $12$ blocks (to $A,$ up to $B,$ then across the top to $D$ ). They meet at $D.$

Thus, the correct answer is D .

12.

一个“幸运”年份是指至少有一个日期按“月/日/年”的形式书写时，满足月份数乘以日期数等于年份的最后两位。例如， $1956$ 是幸运年份，因为日期 $7/8/56$ 满足 $7 \times 8 = 56$ 。下列哪一年不是幸运年份？

A lucky year is one in which at least one date, when written in the form month/day/year, has the following property: the product of the month times the day equals the last two digits of the year. For example, $1956$ is a lucky year because it has the date $7/8/56$ and $7 \times 8 = 56.$ Which of the following is NOT a lucky year?

$1990$

$1991$

$1992$

$1993$

$1994$

知识点：因数分类讨论

难度评级：1120

小提示：

对每一年，尝试把最后两位写成合法的月份 $\times$ 日期

For each year, try to write its last two digits as (month) $\times$ (day) with a valid month and day

大提示：

如果最后两位不能写成一个不超过 $12$ 的数乘以一个不超过 $31$ 的数，这一年就不是幸运年份

A year fails only when the last two digits cannot be a product of a number at most $12$ and a number at most $31$

解答：

其他年份都可以： $90 = 9 \times 10$ 、 $91 = 7 \times 13$ 、 $92 = 4 \times 23$ 和 $93 = 3 \times 31$ ，都对应合法的月和日。

对 $1994$ ，最后两位只能分解为 $94 = 2 \times 47$ ，而 $47$ 作为日期太大， $94$ 或 $47$ 作为月份也太大。因此 $1994$ 没有幸运日期。

所以正确答案是 E。

Each of the other years works: $90 = 9 \times 10,$ $91 = 7 \times 13,$ $92 = 4 \times 23,$ and $93 = 3 \times 31,$ each a valid month/day.

For $1994,$ the last two digits factor only as $94 = 2 \times 47,$ and $47$ is too large for a day (and $94$ or $47$ is too large for a month). So $1994$ has no lucky date.

Thus, the correct answer is E .

13.

图中， $\angle A$ 、 $\angle B$ 和 $\angle C$ 都是直角。如果 $\angle AEB = 40^\circ$ ，且 $\angle BED = \angle BDE$ ，那么 $\angle CDE =$

In the figure, $\angle A,$ $\angle B$ and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE,$ then $\angle CDE =$

$75^\circ$

$80^\circ$

$85^\circ$

$90^\circ$

$95^\circ$

知识点：导角角度和

难度评级：1150

小提示：

在三角形 $BDE$ 中， $E$ 和 $D$ 处的角相等，且 $\angle B = 90^\circ$

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ$

大提示：

$\angle AED = \angle AEB + \angle BED$ ；再用四边形 $AEDC$ 的内角和 $360^\circ$

$\angle AED = \angle AEB + \angle BED;$ then use quadrilateral $AEDC,$ whose angles sum to $360^\circ$

解答：

在三角形 $BDE$ 中， $E$ 和 $D$ 处的角相等，且 $\angle B = 90^\circ$ ，所以 $\angle BED = \angle BDE = 45^\circ$ 。

因此 $\angle AED = \angle AEB + \angle BED$ $= 40^\circ + 45^\circ$ $= 85^\circ$ 。在四边形 $AEDC$ 中， $A$ 和 $C$ 处的角都是 $90^\circ$ ，所以

$\begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ \end{aligned}\text{。}$

所以正确答案是 E。

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ,$ so $\angle BED = \angle BDE = 45^\circ.$

Then $\angle AED = \angle AEB + \angle BED$ $= 40^\circ + 45^\circ$ $= 85^\circ.$ In quadrilateral $AEDC,$ the angles at $A$ and $C$ are $90^\circ,$ so

$\begin{aligned} \angle CDE &= 360^\circ - 90^\circ - 90^\circ \\ &\quad {}- 85^\circ \\ &= 95^\circ. \end{aligned}$

Thus, the correct answer is E .

14.

一支球队在前 $50$ 场比赛中赢了 $40$ 场。为了使整个赛季的胜率正好为 $70\%$ ，这支球队在剩下的 $40$ 场比赛中必须赢多少场？

A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70\%$ of its games for the season?

$20$

$23$

$28$

$30$

$35$

知识点：百分数

难度评级：930

小提示：

整个赛季共有 $50 + 40 = 90$ 场比赛；先求其中的 $70\%$

The season has $50 + 40 = 90$ games; find $70\%$ of that

大提示：

从所需总胜场中减去已经赢的 $40$ 场

Subtract the $40$ games already won from the needed total

解答：

整个赛季共有 $50 + 40 = 90$ 场比赛， $90$ 的 $70\%$ 是 $63$ 场胜利。

球队已经赢了 $40$ 场，所以还需要赢 $63 - 40 = 23$ 场。

所以正确答案是 B。

The season has $50 + 40 = 90$ games, and $70\%$ of $90$ is $63$ wins.

The team already has $40$ wins, so it needs $63 - 40 = 23$ more.

Thus, the correct answer is B .

15.

$\dfrac{4}{37}$ 的小数形式中，小数点右边第 $100$ 位数字是什么？

What is the $100$ th digit to the right of the decimal point in the decimal form of $\dfrac{4}{37}?$

$0$

$1$

$2$

$7$

$8$

知识点：循环小数模运算

难度评级：1060

小提示：

把 $\dfrac{4}{37}$ 写成循环小数，并找出循环节长度

Write $\dfrac{4}{37}$ as a repeating decimal and find the length of the repeating block

大提示：

循环节长度为 $3$ ；用 $100$ 除以 $3$ 的余数确定位置

The block has length $3;$ use the remainder of $100$ divided by $3$

解答：

$\dfrac{4}{37} = 0.\overline{108}$ ，循环节长度为 $3$ 。第 $3$ 、 $6$ 、 $9$ 、 $\ldots$ 位，也就是位置为 $3$ 的倍数时，数字是 $8$ 。

因为 $99$ 是 $3$ 的倍数，所以第 $99$ 位是 $8$ ，第 $100$ 位开始下一个循环节，是 $1$ 。

所以正确答案是 B。

$\dfrac{4}{37} = 0.\overline{108},$ repeating with block length $3.$ The digits in positions $3,$ $6,$ $9,$ $\ldots$ (multiples of $3$ ) are $8.$

Since $99$ is a multiple of $3,$ the $99$ th digit is $8,$ so the $100$ th digit starts the next block: it is $1.$

Thus, the correct answer is B .

16.

三所中学的学生参加了一个暑期项目。艾伦中学的七名学生工作了 $3$ 天。巴尔博亚中学的四名学生工作了 $5$ 天。卡弗中学的五名学生工作了 $9$ 天。学生工作的总报酬为 $\$774$ 。假设每名学生每天获得相同报酬，那么巴尔博亚中学的学生总共赚了多少？

Students from three middle schools worked on a summer project. Seven students from Allen School worked for $3$ days. Four students from Balboa School worked for $5$ days. Five students from Carver School worked for $9$ days. The total amount paid for the students’ work was $\$774.$ Assuming each student received the same amount for a day’s work, how much did the students from Balboa School earn altogether?

$\$9.00$

$\$48.38$

$\$180.00$

$\$193.50$

$\$258.00$

知识点：速率

难度评级：1090

小提示：

计算三所学校学生合计的工作量（以“人天”为单位）

Compute the total number of student-days worked across all three schools

大提示：

用 $\$774$ 除以总人天数，得到每人每天的报酬

Divide $\$774$ by the total student-days to get the pay per student-day

解答：

总工作量为 $7 \times 3 + 4 \times 5 + 5 \times 9$ $= 21 + 20 + 45$ $= 86$ 人天。

所以每人每天的报酬为 $\$774 \div 86 = \$9$ 。巴尔博亚学生的总工作量为 $20$ 人天，共赚得 $20 \times \$9 = \$180$ 。

所以正确答案是 C。

The total student-days are $7 \times 3 + 4 \times 5 + 5 \times 9$ $= 21 + 20 + 45$ $= 86.$

So each student-day pays $\$774 \div 86 = \$9.$ Balboa worked $20$ student-days, earning $20 \times \$9 = \$180.$

Thus, the correct answer is C .

17.

下表给出安维尔和克利奥纳两所小学各年级学生所占百分比：

	$K$	$1$	$2$	$3$	$4$	$5$	$6$
安维尔	$16\%$	$15\%$	$15\%$	$14\%$	$13\%$	$16\%$	$11\%$
克利奥纳	$12\%$	$15\%$	$14\%$	$13\%$	$15\%$	$14\%$	$17\%$

安维尔有 $100$ 名学生，克利奥纳有 $200$ 名学生。两所学校合计，有百分之几的学生在 $6$ 年级？

The table below gives the percent of students in each grade at Annville and Cleona elementary schools:

	$K$	$1$	$2$	$3$	$4$	$5$	$6$
Annville	$16\%$	$15\%$	$15\%$	$14\%$	$13\%$	$16\%$	$11\%$
Cleona	$12\%$	$15\%$	$14\%$	$13\%$	$15\%$	$14\%$	$17\%$

Annville has $100$ students and Cleona has $200$ students. In the two schools combined, what percent of the students are in grade $6?$

$12\%$

$13\%$

$14\%$

$15\%$

$28\%$

知识点：百分数平均数

难度评级：980

小提示：

分别求每所学校的 $6$ 年级学生人数

Find the number of grade- $6$ students at each school separately

大提示：

用两校合计的 $6$ 年级人数除以总人数 $300$

Divide the combined grade- $6$ count by the $300$ total students

解答：

安维尔的六年级人数为 $11\%$ 乘以 $100 = 11$ 人；克利奥纳的六年级人数为 $17\%$ 乘以 $200 = 34$ 人。

合计为 $11 + 34 = 45$ 人，占总人数 $300$ 的 $\dfrac{45}{300} = 15\%$ 。

所以正确答案是 D。

Annville has $11\%$ of $100 = 11$ sixth graders, and Cleona has $17\%$ of $200 = 34$ sixth graders.

Combined, that is $11 + 34 = 45$ out of $300$ students, which is $\dfrac{45}{300} = 15\%.$

Thus, the correct answer is D .

18.

在这个 $100$ 英寸乘 $100$ 英寸的正方形中，四个全等 L 形区域中每一个的面积都是总面积的 $\dfrac{3}{16}$ 。中心正方形的边长是多少英寸？

The area of each of the four congruent L-shaped regions of this $100$ -inch by $100$ -inch square is $\dfrac{3}{16}$ of the total area. How many inches long is the side of the center square?

$25$

$44$

$50$

$62$

$75$

知识点：正方形（几何）面积

难度评级：980

小提示：

四个 L 形区域合起来占正方形的 $4 \times \dfrac{3}{16}$

The four L-shaped regions together make up $4 \times \dfrac{3}{16}$ of the square

大提示：

中心正方形是剩下的面积比例；它的边长是面积的平方根

The center square is the remaining fraction of the area; its side is the square root of its area

解答：

四个 L 形区域合计占 $4 \times \dfrac{3}{16} = \dfrac34$ 的大正方形，所以中心正方形占剩下的 $\dfrac14$ 。

大正方形面积为 $100 \times 100 = 10000$ 平方英寸，所以中心正方形面积为 $\dfrac14 \times 10000 = 2500$ ，边长为 $\sqrt{2500} = 50$ 英寸。

所以正确答案是 C。

The four L-shaped regions cover $4 \times \dfrac{3}{16} = \dfrac34$ of the square, so the center square is the remaining $\dfrac14$ of the total area.

The total area is $100 \times 100 = 10000$ square inches, so the center square has area $\dfrac14 \times 10000 = 2500,$ and its side is $\sqrt{2500} = 50$ inches.

Thus, the correct answer is C .

19.

图中显示乔丹老师英语班学生家庭中孩子人数的分布。这个分布中，每个家庭孩子人数的中位数是

The graph shows the distribution of the number of children in the families of the students in Ms. Jordan’s English class. The median number of children in the family for this distribution is

$1$

$2$

$3$

$4$

$5$

知识点：中位数（数据）数据与图表解读

难度评级：960

小提示：

中位数是把所有家庭中的孩子人数按顺序列出后的中间值

The median is the middle value when all the family sizes are listed in order

大提示：

先数家庭总数，再找中间位置是哪一个

Count the total number of families, then find which position is the middle one

解答：

图中有 $2$ 个家庭有 $1$ 个孩子， $1$ 个家庭有 $2$ 个， $2$ 个家庭有 $3$ 个， $2$ 个家庭有 $4$ 个， $6$ 个家庭有 $5$ 个，总共 $2 + 1 + 2 + 2 + 6 = 13$ 个家庭。

中位数是按顺序排列的第 $7$ 个值。列出家庭孩子人数后，第 $7$ 个值是 $4$ 。

所以正确答案是 D。

The graph gives $2$ families with $1$ child, $1$ with $2,$ $2$ with $3,$ $2$ with $4,$ and $6$ with $5,$ for $2 + 1 + 2 + 2 + 6 = 13$ families.

The median is the $7$ th value in order. Listing the family sizes, the $7$ th value is $4.$

Thus, the correct answer is D .

20.

戴安娜和阿波罗各掷一颗标准骰子，随机得到 $1$ 到 $6$ 中的一个数。戴安娜的数大于阿波罗的数的概率是多少？

Diana and Apollo each roll a standard die obtaining a number at random from $1$ to $6.$ What is the probability that Diana’s number is larger than Apollo’s number?

$\dfrac13$

$\dfrac{5}{12}$

$\dfrac49$

$\dfrac{17}{36}$

$\dfrac12$

知识点：骰子（概率）对称性

难度评级：1120

小提示：

在 $36$ 个等可能结果中，有一些是平局；先去掉这些

Of the $36$ equally likely outcomes, some are ties; remove those

大提示：

由对称性，戴安娜较大的情况正好占非平局结果的一半

By symmetry, Diana is larger in exactly half of the non-tie outcomes

解答：

共有 $6 \times 6 = 36$ 个等可能结果，其中 $6$ 个是平局，剩下 $30$ 个结果两数不同。

由对称性，戴安娜的数较大的情况正好是其中一半，即 $15$ 个。所以概率为 $\dfrac{15}{36} = \dfrac{5}{12}$ 。

所以正确答案是 B。

There are $6 \times 6 = 36$ equally likely outcomes, of which $6$ are ties, leaving $30$ outcomes with different numbers.

By symmetry, Diana is larger in exactly half of those, or $15,$ so the probability is $\dfrac{15}{36} = \dfrac{5}{12}.$

Thus, the correct answer is B .

21.

一个塑料拼接立方体的一面有一个凸出的连接扣，另外五面有接受连接扣的孔。最少需要多少个这样的立方体拼接在一起，才能使外面只露出孔？

A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?

$3$

$4$

$5$

$6$

$8$

知识点：立体几何

难度评级：1170

小提示：

每个立方体都有一个凸扣，必须藏进另一个立方体的孔中

Each cube has one snap that must be hidden inside another cube’s hole

大提示：

试着把立方体排成一个小环，使每个凸扣都插入相邻立方体

Try arranging the cubes in a small ring so every snap plugs into a neighbor

解答：

每个立方体唯一的凸扣都必须插入另一个立方体的孔中才能被隐藏。两个立方体只能共用一个面，因此无法同时隐藏两个凸扣。若用三个立方体隐藏所有凸扣，就需要形成一个三立方体闭环，使每一对立方体都共用一个面；但三个单位立方体不可能两两以面相邻。因此，一个、两个或三个立方体都不行。

四个立方体可以排成一个方形环，每个凸扣插入相邻立方体的孔中，从而外面只露出孔。因此最少需要 $4$ 个。

所以正确答案是 B。

Every cube’s single snap must be plugged into another cube’s hole to be hidden. Two cubes can share only one face, so two cubes cannot hide both snaps. With three cubes, hiding all three snaps would require a three-cube loop in which every pair shares a face, but three unit cubes cannot be pairwise face-adjacent. Thus one, two, or three cubes cannot work.

Four cubes can be arranged in a square ring, each snap fitting into the neighbor’s hole, so only receptacle holes show. The smallest number is $4.$

Thus, the correct answer is B .

22.

数 $6545$ 可以写成一对正的两位数的乘积。这一对数的和是多少？

The number $6545$ can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?

$162$

$172$

$173$

$174$

$222$

知识点：质因数分解因数

难度评级：1170

小提示：

求 $6545$ 的质因数分解

Find the prime factorization of $6545$

大提示：

把质因数分成两个都在 $10$ 到 $99$ 之间的数

Group the prime factors into two numbers that are each between $10$ and $99$

解答：

质因数分解为 $6545 = 5 \times 7 \times 11 \times 17$ 。要分成两个两位因数，可以配为 $5 \times 17 = 85$ 和 $7 \times 11 = 77$ 。

这是唯一的两位数配对，它们的和是 $85 + 77 = 162$ 。

所以正确答案是 A。

The prime factorization is $6545 = 5 \times 7 \times 11 \times 17.$ To split into two two-digit factors, pair the primes: $5 \times 17 = 85$ and $7 \times 11 = 77.$

These are the only two-digit pair, and their sum is $85 + 77 = 162.$

Thus, the correct answer is A .

23.

有多少个四位整数满足：最左边的数字是奇数，第二个数字是偶数，且四个数字都不同？

How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?

$1120$

$1400$

$1800$

$2025$

$2500$

知识点：乘法原理

难度评级：1220

小提示：

按顺序数每个数字位置的选择数：第一位、第二位、第三位、第四位

Count choices for each digit position in turn: first, second, third, fourth

大提示：

第一位有 $5$ 个奇数选择；第二位有 $5$ 个偶数选择；之后剩余不同数字分别有 $8$ 和 $7$ 个选择

First digit: $5$ odd options; second: $5$ even options; then $8$ and $7$ for the distinct remaining digits

解答：

第一位是奇数，有 $5$ 个选择。第二位是偶数，有 $5$ 个选择，而且不会与第一位的奇数重复。

第三位可以是剩下 $8$ 个未用数字中的任意一个，第四位有 $7$ 个选择。总数为 $5 \times 5 \times 8 \times 7 = 1400$ 。

所以正确答案是 B。

The first digit is odd: $5$ choices. The second is even: $5$ choices (none of which repeats the odd first digit).

The third digit is any of the $8$ unused digits, and the fourth is any of the $7$ remaining. In total, $5 \times 5 \times 8 \times 7 = 1400.$

Thus, the correct answer is B .

24.

在平行四边形 $ABCD$ 中， $\overline{DE}$ 是到底边 $\overline{AB}$ 的高， $\overline{DF}$ 是到底边 $\overline{BC}$ 的高。（两幅图表示同一个平行四边形。）如果 $DC = 12$ ， $EB = 4$ ，且 $DE = 6$ ，那么 $DF =$

In parallelogram $ABCD,$ $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}.$ (Both pictures represent the same parallelogram.) If $DC = 12,$ $EB = 4,$ and $DE = 6,$ then $DF =$

$6.4$

$7$

$7.2$

$8$

$10$

知识点：平行四边形勾股定理

难度评级：1150

小提示：

平行四边形对边相等，所以 $AB = 12$ ；先求 $AE$ ，再用直角三角形 $ADE$ 求 $AD$

Opposite sides are equal, so $AB = 12;$ find $AE,$ then $AD$ with the right triangle $ADE$

大提示：

面积可以用任一底边 $\times$ 对应高来表示，列出 $AB \cdot DE = BC \cdot DF$ 即可

The area equals base $\times$ height for either base: $AB \cdot DE = BC \cdot DF$

解答：

因为 $AB = DC = 12$ ，所以 $AE = 12 - 4 = 8$ 。在直角三角形 $ADE$ 中， $AD = \sqrt{8^2 + 6^2} = 10$ ，因此 $BC = AD = 10$ 。

面积为 $AB \cdot DE = 12 \cdot 6 = 72$ ，同时也等于 $BC \cdot DF = 10 \cdot DF$ 。所以 $DF = \dfrac{72}{10} = 7.2$ 。

所以正确答案是 C。

Since $AB = DC = 12,$ we get $AE = 12 - 4 = 8.$ In right triangle $ADE,$ $AD = \sqrt{8^2 + 6^2} = 10,$ so $BC = AD = 10.$

The area is $AB \cdot DE = 12 \cdot 6 = 72,$ and also $BC \cdot DF = 10 \cdot DF.$ So $DF = \dfrac{72}{10} = 7.2.$

Thus, the correct answer is C .

25.

从达拉斯到休斯敦的巴士每小时整点发车。从休斯敦到达拉斯的巴士每小时半点发车。两城之间的行程需要 $5$ 小时。假设巴士都在同一条公路上行驶，一辆开往休斯敦的巴士会在公路上（不在车站内）遇到多少辆开往达拉斯的巴士？

Buses from Dallas to Houston leave every hour on the hour. Buses from Houston to Dallas leave every hour on the half hour. The trip from one city to the other takes $5$ hours. Assuming the buses travel on the same highway, how many Dallas-bound buses does a Houston-bound bus pass on the highway (not in the station)?

$5$

$6$

$9$

$10$

$11$

知识点：路程、速度与时间

难度评级：1260

小提示：

跟踪一辆开往休斯敦的巴士在 $5$ 小时行程中，会与哪些迎面巴士同时在路上

Track one Houston-bound bus over its $5$ -hour trip and see which oncoming buses share the road with it

大提示：

数它出发时已经在公路上的开往达拉斯的巴士，加上它到达前出发的那些

Count the Dallas-bound buses already on the highway when it starts, plus those that leave before it arrives

解答：

考虑一辆 $6\!:\!00$ 从达拉斯出发、 $11\!:\!00$ 到休斯敦的巴士。它会遇到所有在这段时间内与它同在公路上的开往达拉斯的巴士。

开往达拉斯的巴士每半点从休斯敦出发，行程 $5$ 小时。与这辆车在路上相遇且不在车站相遇的，是 $1\!:\!30$ 、 $2\!:\!30$ 、 $\ldots$ 、 $10\!:\!30$ 从休斯敦出发的巴士，共 $10$ 辆。

所以正确答案是 D。

Consider a bus leaving Dallas at $6\!:\!00,$ arriving in Houston at $11\!:\!00.$ It meets every Dallas-bound bus that is on the highway during that window.

Dallas-bound buses leave Houston on the half hour and take $5$ hours. The ones sharing the road (meeting away from a station) are those that left Houston at $1\!:\!30,$ $2\!:\!30,$ $\ldots,$ $10\!:\!30,$ which is $10$ buses.

Thus, the correct answer is D .