1987 AMC 8 真题与在线限时测验

计时

40:00

1.

$0.4 + 0.02 + 0.006$ 等于多少？

What is $0.4 + 0.02 + 0.006?$

$0.012$

$0.066$

$0.12$

$0.24$

$0.426$

答案：E

知识点：小数位值

难度评级：450

小提示：

相加前先把小数点对齐

Line up the decimal points before adding

大提示：

把这些数写成 $0.400$ 、 $0.020$ 和 $0.006$

Write the numbers as $0.400,$ $0.020,$ and $0.006$

解答：

按位相加， $0.4 + 0.02 + 0.006 = 0.426$ 。

所以正确答案是 E。

Adding place by place, $0.4 + 0.02 + 0.006 = 0.426.$

Thus, the correct answer is E .

2.

$\dfrac{2}{25}$ 写成小数是多少？

What is $\dfrac{2}{25}$ written as a decimal?

$0.008$

$0.08$

$0.8$

$1.25$

$12.5$

答案：B

知识点：分数小数

难度评级：560

小提示：

把这个分数改写成分母为 $100$ 的分数

Rewrite the fraction with a denominator of $100$

大提示：

分子和分母同时乘以 $4$

Multiply the numerator and denominator by $4$

解答：

分子和分母同时乘以 $4$ ，得 $\dfrac{2}{25} = \dfrac{8}{100} = 0.08$ 。

所以正确答案是 B。

Multiplying top and bottom by $4$ gives $\dfrac{2}{25} = \dfrac{8}{100} = 0.08.$

Thus, the correct answer is B .

3.

下式的值是多少？

$\begin{gathered} 2(81 + 83 + 85 + 87 + 89 \\ {}+ 91 + 93 + 95 + 97 + 99) \end{gathered}\text{？}$

What is the value of

$\begin{gathered} 2(81 + 83 + 85 + 87 + 89 \\ {}+ 91 + 93 + 95 + 97 + 99)? \end{gathered}$

$1600$

$1650$

$1700$

$1750$

$1800$

答案：E

知识点：等差数列整数运算

难度评级：660

小提示：

这十个数 $81$ 、 $83$ 、 $\ldots$ 、 $99$ 等间隔排列

The ten numbers $81,$ $83,$ $\ldots,$ $99$ are evenly spaced

大提示：

它们的平均数是 $\dfrac{81 + 99}{2} = 90$ ，所以和是 $10 \times 90$

Their average is $\dfrac{81 + 99}{2} = 90,$ so their sum is $10 \times 90$

解答：

这十个等间隔数的平均数是 $90$ ，所以它们的和是 $10 \times 90 = 900$ 。

加倍后得到 $2 \times 900 = 1800$ 。

所以正确答案是 E。

The ten evenly spaced numbers have average $90,$ so their sum is $10 \times 90 = 900.$

Doubling gives $2 \times 900 = 1800.$

Thus, the correct answer is E .

4.

火星人用“克勒特”（clert）衡量角度。一个整圆有 $500$ 克勒特。一个直角有多少克勒特？

Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?

$90$

$100$

$125$

$180$

$250$

答案：C

知识点：比与比例

难度评级：560

小提示：

直角是整圆的 $\dfrac14$

A right angle is $\dfrac14$ of a full circle

大提示：

求 $500$ 克勒特的 $\dfrac14$

Take $\dfrac14$ of $500$ clerts

解答：

直角是整圆的四分之一，所以它包含 $\dfrac14 \times 500 = 125$ 克勒特。

所以正确答案是 C。

A right angle is a quarter of a full circle, so it contains $\dfrac14 \times 500 = 125$ clerts.

Thus, the correct answer is C .

5.

图中矩形区域的面积是多少平方米？

What is the area, in square meters, of the rectangular region shown?

$0.088 \text{ m}^2$

$0.62 \text{ m}^2$

$0.88 \text{ m}^2$

$1.24 \text{ m}^2$

$4.22 \text{ m}^2$

答案：A

知识点：面积小数

难度评级：660

小提示：

矩形面积等于长乘以宽

The area of a rectangle is its length times its width

大提示：

把 $0.4$ 和 $0.22$ 相乘

Multiply $0.4$ by $0.22$

解答：

面积是 $0.4 \times 0.22 = 0.088$ 平方米。

所以正确答案是 A。

The area is $0.4 \times 0.22 = 0.088$ square meters.

Thus, the correct answer is A .

6.

从集合 $\{ -7, -5, -1, 1, 3 \}$ 中选两个数相乘，可能得到的最小乘积是

The smallest product one could obtain by multiplying two numbers in the set $\{ -7, -5, -1, 1, 3 \}$ is

$-35$

$-21$

$-15$

$-1$

$3$

答案：B

知识点：最优化

难度评级：800

小提示：

最小的乘积是最负的乘积，来自一个负因数和一个正因数

The smallest product is the most negative one, which comes from one negative and one positive factor

大提示：

把最负的数和最大的正数配对

Pair the most negative number with the largest positive number

解答：

负乘积来自负数乘正数；要最负，就让两个因数的绝对值尽可能大。

最负的乘积是 $(-7)(3) = -21$ 。

所以正确答案是 B。

A negative product comes from multiplying a negative number by a positive number, and it is most negative when both factors are largest in size.

The most negative product is $(-7)(3) = -21.$

Thus, the correct answer is B .

7.

图中的大立方体由 $27$ 个大小相同的小立方体组成。大立方体每个面的相对面都用同样方式涂色。至少有一个面被涂色的小立方体共有多少个？

The large cube shown is made up of $27$ identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is

$10$

$16$

$20$

$22$

$24$

答案：C

知识点：正方体补集计数

难度评级：1200

小提示：

小立方体只有在某个外露小方格被涂色时才有涂色面；先数完全未涂色的小立方体，包括最中心隐藏的那个

A small cube shows a shaded face only when one of its surface squares is shaded, so count the cubes that stay completely blank, beginning with the single hidden cube at the center of the block

大提示：

除中心小立方体外，找出所有外露方格都落在空白位置的小立方体，再从 $27$ 中减去未涂色数

Besides the hidden center cube, find the cubes whose every exposed square lands on a blank position, then subtract the number of blank cubes from $27$

解答：

先数没有任何涂色面的立方体，再从 $27$ 中减去。一个小立方体未涂色，当且仅当它所有外露的小方格都是空白。

三种面图案如下：顶面和底面只有中心小方格涂色；一对相对侧面是四个角和中心涂色；另一对相对侧面是四条边的中点涂色。

恰有 $7$ 个小立方体避开所有涂色小方格：大立方体最中心隐藏的一个；那对边中点涂色面中心处的两个小立方体，它们唯一外露的小方格是空白中心；以及四个在“角和中心涂色”的面与顶面或底面相交边中点处的小立方体，因为那里每个外露小方格都是空白边格。

因此有 $27 - 7 = 20$ 个小立方体至少有一个涂色面。

所以正确答案是 C。

Count the cubes with no shaded face and subtract from $27.$ A small cube is unshaded exactly when every one of its exposed squares is blank.

The three patterns are: the top and bottom faces show only their center square shaded; one pair of opposite side faces shows the four corners and the center shaded; the remaining pair shows the four edge-midpoints shaded.

Exactly $7$ small cubes avoid every shaded square: the one hidden cube at the very center of the block; the two cubes at the centers of the edge-midpoints faces, whose only exposed square is that blank center; and the four cubes at the midpoints of the edges where a corners-and-center face meets the top or bottom face, since there each exposed square is a blank edge cell.

Hence $27 - 7 = 20$ cubes have at least one shaded face.

Thus, the correct answer is C .

8.

在下面的加法中， $A$ 和 $B$ 是非零数字：

$\begin{array}{r} 9876 \\ A32 \\ +B1 \\ \hline \end{array}$

这三个整数的和有多少位数字（数字不一定互不相同）？

In the addition problem below, $A$ and $B$ are nonzero digits:

$\begin{array}{r} 9876 \\ A32 \\ +B1 \\ \hline \end{array}$

How many digits (not necessarily different) are in the sum of the three whole numbers?

$4$

$5$

$6$

$9$

取决于 $A$ 和 $B$ 的值

depends on the values of $A$ and $B$

答案：B

知识点：位值极限情形界定

难度评级：930

小提示：

$9876$ 已经很接近 $10{,}000$ ，所以再加任何正数都会超过它

The number $9876$ is already close to $10{,}000,$ so adding anything pushes the sum past it

大提示：

也检查最大可能和，取 $A = B = 9$

Check the largest possible sum too, using $A = B = 9$

解答：

因为 $A$ 和 $B$ 至少为 $1$ ，所以和至少是 $9876 + 132 + 11 = 10{,}019$ ，有 $5$ 位。

另一端，当 $A = B = 9$ 时，和为 $9876 + 932 + 91 = 10{,}899$ ，仍然是 $5$ 位。因此和总是恰好有 $5$ 位。

所以正确答案是 B。

Since $A$ and $B$ are at least $1,$ the sum is at least $9876 + 132 + 11 = 10{,}019,$ which has $5$ digits.

At the other extreme, with $A = B = 9$ the sum is $9876 + 932 + 91 = 10{,}899,$ still $5$ digits. So the sum always has exactly $5$ digits.

Thus, the correct answer is B .

9.

求和

$\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 + \dfrac17$

时，使用的最小公分母是多少？

When finding the sum

$\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 + \dfrac17,$

what is the least common denominator used?

$120$

$210$

$420$

$840$

$5040$

答案：C

知识点：最小公倍数

难度评级：800

小提示：

最小公分母是 $2$ 、 $3$ 、 $4$ 、 $5$ 、 $6$ 、 $7$ 的最小公倍数

The least common denominator is the least common multiple of $2,$ $3,$ $4,$ $5,$ $6,$ $7$

大提示：

用 $4$ 覆盖因数 $4$ （也覆盖 $2$ ），再包含 $3$ 、 $5$ 和 $7$

Use $4$ to cover the factor $4$ (and $2$ ), then include $3,$ $5,$ and $7$

解答：

最小公倍数必须包含 $4 = 2^2$ 、 $3$ 、 $5$ 和 $7$ （因数 $6 = 2 \times 3$ 已经被覆盖）。

所以它是 $4 \times 3 \times 5 \times 7 = 420$ 。

所以正确答案是 C。

The least common multiple must include $4 = 2^2,$ $3,$ $5,$ and $7$ (the factor $6 = 2 \times 3$ is already covered).

So it is $4 \times 3 \times 5 \times 7 = 420.$

Thus, the correct answer is C .

10.

下式的值是多少？

$4(299) + 3(299) + 2(299) + 298\text{？}$

What is the value of

$4(299) + 3(299) + 2(299) + 298?$

$2889$

$2989$

$2991$

$2999$

$3009$

答案：B

知识点：分配律

难度评级：800

小提示：

合并前三项： $4(299) + 3(299) + 2(299)$ $= 9 \times 299$

Combine the first three terms: $4(299) + 3(299) + 2(299)$ $= 9 \times 299$

大提示：

计算 $9 \times 299$ ，再加 $298$

Compute $9 \times 299,$ then add $298$

解答：

前三项合并为 $(4 + 3 + 2)(299)$ $= 9 \times 299 = 2691$ 。

再加最后一项，得 $2691 + 298 = 2989$ 。

所以正确答案是 B。

The first three terms combine to $(4 + 3 + 2)(299)$ $= 9 \times 299 = 2691.$

Adding the last term gives $2691 + 298 = 2989.$

Thus, the correct answer is B .

11.

和 $2\dfrac17 + 3\dfrac12 + 5\dfrac{1}{19}$ 位于哪两个值之间？

The sum $2\dfrac17 + 3\dfrac12 + 5\dfrac{1}{19}$ is between which two values?

$10$ 和 $10\dfrac12$

$10$ and $10\dfrac12$

$10\dfrac12$ 和 $11$

$10\dfrac12$ and $11$

$11$ 和 $11\dfrac12$

$11$ and $11\dfrac12$

$11\dfrac12$ 和 $12$

$11\dfrac12$ and $12$

$12$ 和 $12\dfrac12$

$12$ and $12\dfrac12$

答案：B

知识点：分数估算

难度评级：860

小提示：

先加整数部分： $2 + 3 + 5 = 10$

Add the whole-number parts first: $2 + 3 + 5 = 10$

大提示：

分数部分是 $\dfrac12 + \dfrac17 + \dfrac{1}{19}$ ；它大于 $\dfrac12$ ，但小于 $1$

The fractional parts are $\dfrac12 + \dfrac17 + \dfrac{1}{19};$ this is more than $\dfrac12$ but less than $1$

解答：

整数部分的和是 $10$ ，分数部分是 $\dfrac12 + \dfrac17 + \dfrac{1}{19}$ 。

因为 $\dfrac17 + \dfrac{1}{19}$ 是一个较小的正数，所以分数部分总和大于 $\dfrac12$ ，但远小于 $1$ 。因此原和位于 $10\dfrac12$ 和 $11$ 之间。

所以正确答案是 B。

The whole-number parts sum to $10.$ The fractional parts are $\dfrac12 + \dfrac17 + \dfrac{1}{19}.$

Since $\dfrac17 + \dfrac{1}{19}$ is a small positive amount, the fractional total is more than $\dfrac12$ but well under $1.$ So the sum lies between $10\dfrac12$ and $11.$

Thus, the correct answer is B .

12.

这个 $12$ 乘 $18$ 的大矩形区域中，有几分之几被涂色？

What fraction of the large $12$ by $18$ rectangular region is shaded?

$\dfrac{1}{108}$

$\dfrac{1}{18}$

$\dfrac{1}{12}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

答案：C

知识点：面积分数

难度评级：930

小提示：

水平线把矩形分成上半部分和下半部分

The horizontal line splits the rectangle into a top half and a bottom half

大提示：

涂色部分是右下四分之一区域中最右边的三分之一

The shaded part is the rightmost third of the bottom-right quarter of the rectangle

解答：

整个区域面积是 $12 \times 18 = 216$ 。涂色区域宽 $3$ 、高 $6$ ，面积为 $3 \times 6 = 18$ 。

所以涂色比例是 $\dfrac{18}{216} = \dfrac{1}{12}$ 。等价地，它是整个区域的 $\dfrac14$ 的 $\dfrac13$ 。

所以正确答案是 C。

The whole region has area $12 \times 18 = 216.$ The shaded region is $3$ wide and $6$ tall, with area $3 \times 6 = 18.$

So the shaded fraction is $\dfrac{18}{216} = \dfrac{1}{12}.$ Equivalently, it is $\dfrac13$ of $\dfrac14$ of the whole.

Thus, the correct answer is C .

13.

下列哪个分数的值最大？

Which of the following fractions has the largest value?

$\dfrac{3}{7}$

$\dfrac{4}{9}$

$\dfrac{17}{35}$

$\dfrac{100}{201}$

$\dfrac{151}{301}$

答案：E

知识点：分数

难度评级：930

小提示：

把每个分数都和 $\dfrac12$ 比较

Compare each fraction to $\dfrac12$

大提示：

一个分数大于 $\dfrac12$ ，当且仅当它的分子大于分母的一半

A fraction is more than $\dfrac12$ exactly when its numerator is more than half its denominator

解答：

对 $\dfrac37$ 、 $\dfrac49$ 、 $\dfrac{17}{35}$ 和 $\dfrac{100}{201}$ ，每个分子都小于分母的一半，所以它们都小于 $\dfrac12$ 。

对 $\dfrac{151}{301}$ ，分子 $151$ 大于 $301$ 的一半，所以这个分数大于 $\dfrac12$ ，也是最大的。

所以正确答案是 E。

For $\dfrac37,$ $\dfrac49,$ $\dfrac{17}{35},$ and $\dfrac{100}{201},$ each numerator is less than half its denominator, so each is less than $\dfrac12.$

For $\dfrac{151}{301},$ the numerator $151$ is more than half of $301,$ so this fraction exceeds $\dfrac12$ and is the largest.

Thus, the correct answer is E .

14.

一台计算机每秒可以做 $10{,}000$ 次加法。它一小时可以做多少次加法？

A computer can do $10{,}000$ additions per second. How many additions can it do in one hour?

$6$ 百万

$6$ million

$36$ 百万

$36$ million

$60$ 百万

$60$ million

$216$ 百万

$216$ million

$360$ 百万

$360$ million

答案：B

知识点：速率单位换算

难度评级：730

小提示：

一小时有 $60 \times 60 = 3600$ 秒

There are $60 \times 60 = 3600$ seconds in an hour

大提示：

把速率 $10{,}000$ 乘以一小时的秒数

Multiply the rate $10{,}000$ by the number of seconds in an hour

解答：

一小时有 $3600$ 秒，所以计算机可以做 $10{,}000 \times 3600 = 36{,}000{,}000$ 次加法。

也就是 $36$ 百万。

所以正确答案是 B。

One hour has $3600$ seconds, so the computer does $10{,}000 \times 3600 = 36{,}000{,}000$ additions.

That is $36$ million.

Thus, the correct answer is B .

15.

一个促销广告写着：“按原价买三条轮胎，第四条轮胎只要 $\$3$ 。”山姆在促销中花 $\$240$ 买了一套四条轮胎。一条轮胎的原价是多少？

A sale ad read: “Buy three tires at the regular price and get the fourth tire for $\$3.$ ” Sam paid $\$240$ for a set of four tires at the sale. What was the regular price of one tire?

$\$59.25$

$\$60$

$\$70$

$\$79$

$\$80$

答案：D

知识点：逆推法

难度评级：860

小提示：

$\$240$ 包括三条原价轮胎和第四条的 $\$3$

The $\$240$ covers three tires at the regular price plus $\$3$ for the fourth

大提示：

先减去 $\$3$ ，再把剩下的钱平均分给三条原价轮胎

Subtract the $\$3,$ then divide the rest among the three full-price tires

解答：

三条原价轮胎共花 $\$240 - \$3 = \$237$ 。

所以一条轮胎的原价是 $\dfrac{\$237}{3} = \$79$ 。

所以正确答案是 D。

The three regular-price tires cost $\$240 - \$3 = \$237.$

So one tire costs $\dfrac{\$237}{3} = \$79.$

Thus, the correct answer is D .

16.

篮球赛季前三场比赛中，乔伊丝前 $30$ 次投篮命中 $12$ 次，所以她的赛季投篮命中率是 $40\%$ 。在下一场比赛中，她投篮 $10$ 次，并把赛季命中率提高到 $50\%$ 。这 $10$ 次投篮中她命中了多少次？

Joyce made $12$ of her first $30$ shots in the first three games of the basketball season, so her seasonal shooting average was $40\%.$ In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%.$ How many of these $10$ shots did she make?

$2$

$3$

$5$

$6$

$8$

答案：E

知识点：百分数平均数

难度评级：960

小提示：

下一场之后，她一共投篮 $30 + 10 = 40$ 次

After the next game she has attempted $30 + 10 = 40$ shots

大提示：

$40$ 次投篮的 $50\%$ 命中率表示总共命中 $20$ 次；她之前已有 $12$ 次

A $50\%$ average on $40$ shots means $20$ made in all; she already had $12$

解答：

第四场后她共投篮 $40$ 次。 $50\%$ 命中率表示她总共命中 $\tfrac12 \times 40 = 20$ 次。

她之前已经命中 $12$ 次，所以最后 $10$ 次中命中 $20 - 12 = 8$ 次。

所以正确答案是 E。

After the fourth game she has taken $40$ shots. A $50\%$ average means she made $\tfrac12 \times 40 = 20$ shots in total.

She had already made $12,$ so she made $20 - 12 = 8$ of the last $10.$

Thus, the correct answer is E .

17.

艾比、布雷特、卡尔和达娜坐在一排编号为 $1$ 到 $4$ 的四个座位上。乔看着他们说：“布雷特挨着卡尔。”“艾比坐在布雷特和卡尔之间。”然而乔的每一句话都是假的。布雷特实际坐在 $3$ 号座位。谁坐在 $2$ 号座位？

Abby, Bret, Carl, and Dana are seated in a row of four seats numbered $1$ to $4.$ Joe looks at them and says: “Bret is next to Carl.” “Abby is between Bret and Carl.” However, each one of Joe’s statements is false. Bret is actually sitting in seat $3.$ Who is sitting in seat $2?$

艾比

Abby

布雷特

Bret

卡尔

Carl

达娜

Dana

信息不足，无法确定

There is not enough information to be sure

答案：D

知识点：逻辑推理

难度评级：1060

小提示：

“布雷特挨着卡尔”是假的，所以卡尔不在 $2$ 号或 $4$ 号座位；卡尔只能在 $1$ 号座位

“Bret is next to Carl” is false, so Carl is not in seat $2$ or seat $4;$ the only remaining seat for Carl is seat $1$

大提示：

卡尔在 $1$ 号座位后，“艾比在布雷特和卡尔之间”是假的，所以艾比不能在 $2$ 号座位；把艾比放进最后一个可行座位，再看剩下谁

Once Carl is in seat $1,$ “Abby is between Bret and Carl” is false, so Abby cannot be in seat $2;$ place Abby in the last open seat and see who is left

解答：

布雷特在 $3$ 号座位。因为“布雷特挨着卡尔”是假的，卡尔不在 $2$ 号或 $4$ 号座位，所以卡尔必须在 $1$ 号座位。

布雷特（ $3$ 号）和卡尔（ $1$ 号）之间的座位是 $2$ 号。因为“艾比在布雷特和卡尔之间”是假的，艾比不在 $2$ 号，所以艾比必须在 $4$ 号。

因此达娜坐在 $2$ 号座位。

所以正确答案是 D。

Bret is in seat $3.$ Since “Bret is next to Carl” is false, Carl is not in seat $2$ or seat $4,$ so Carl must be in seat $1.$

The seat between Bret (seat $3$ ) and Carl (seat $1$ ) is seat $2.$ Since “Abby is between Bret and Carl” is false, Abby is not in seat $2,$ so Abby must be in seat $4.$

That leaves Dana in seat $2.$

Thus, the correct answer is D .

18.

房间里一半的人离开了。剩下的人中有三分之一开始跳舞。此时有 $12$ 人没有跳舞。房间里原来有多少人？

Half the people in a room left. One third of those remaining started to dance. There were then $12$ people who were not dancing. What was the original number of people in the room?

$24$

$30$

$36$

$42$

$72$

答案：C

知识点：分数逆推法

难度评级：960

小提示：

一半人离开后，剩下的人数才是跳舞比例所作用的总数

After half leave, the number remaining is what the dancing fraction applies to

大提示：

$12$ 个没跳舞的人是剩余人数的 $\dfrac23$

The $12$ non-dancers are $\dfrac23$ of those remaining

解答：

在留下的人中， $\dfrac13$ 跳舞，所以 $\dfrac23$ 没跳舞。因此剩余人数的 $\dfrac23$ 等于 $12$ ，得剩余 $18$ 人。

这 $18$ 人是原来人数的一半，所以房间一开始有 $2 \times 18 = 36$ 人。

所以正确答案是 C。

Of those remaining, $\dfrac13$ danced, so $\dfrac23$ did not. Thus $\dfrac23$ of the remaining people equals $12,$ giving $18$ remaining.

Those $18$ are half the original group, so the room started with $2 \times 18 = 36$ people.

Thus, the correct answer is C .

19.

一个计算器有一个平方键 $\boxed{x^2}$ ，会把当前显示的数替换成它的平方。例如，如果显示 $\boxed{\phantom{00}3}$ ，按下 $\boxed{x^2}$ 键后显示 $\boxed{\phantom{00}9}$ 。如果显示的是 $\boxed{\phantom{00}2}$ ，要按多少次 $\boxed{x^2}$ 键才能得到大于 $500$ 的显示数？

A calculator has a squaring key $\boxed{x^2}$ that replaces the number currently displayed with its square. For example, if the display reads $\boxed{\phantom{00}3}$ and $\boxed{x^2}$ is pressed, the display becomes $\boxed{\phantom{00}9}.$ If the display reads $\boxed{\phantom{00}2},$ how many times must $\boxed{x^2}$ be pressed to produce a displayed number greater than $500?$

$4$

$5$

$8$

$9$

$250$

答案：A

知识点：指数

难度评级：1030

小提示：

从 $2$ 开始，每按一次就平方，得到 $2$ 、 $4$ 、 $16$ 、 $256$ 、 $\ldots$ ；继续观察何时超过目标值

Starting from $2,$ each press squares the number: $2,$ $4,$ $16,$ $256,$ $\ldots$

大提示：

找出这个列表中第一个超过 $500$ 的数

Find the first term in that list that exceeds $500$

解答：

重复按键得到 $2 \to 4 \to 16 \to 256 \to 65536$ 。

因为 $256 \lt 500 \lt 65536$ ，显示数第一次超过 $500$ 是第四次按键。

所以正确答案是 A。

Pressing the key repeatedly gives $2 \to 4 \to 16 \to 256 \to 65536.$

Since $256 \lt 500 \lt 65536,$ the display first exceeds $500$ on the fourth press.

Thus, the correct answer is A .

20.

考虑命题：“如果整数 $n$ 不是质数，那么整数 $n - 2$ 不是质数。”下列哪个 $n$ 的值说明这个命题是假的？

Consider the statement: “If a whole number $n$ is not prime, then the whole number $n - 2$ is not prime.” Which of the following values of $n$ shows this statement to be false?

$9$

$12$

$13$

$16$

$23$

答案：A

知识点：反例质数

难度评级：1010

小提示：

要否定一个“如果……那么……”命题，需要找到“如果”部分成立但“那么”部分不成立的例子

To disprove an “if ... then” statement, find a case where the “if” part holds but the “then” part fails

大提示：

寻找不是质数的 $n$ ，同时 $n - 2$ 是质数

Look for $n$ that is not prime while $n - 2$ is prime

解答：

反例需要 $n$ 不是质数，但 $n - 2$ 是质数。当 $n = 9$ 时， $9$ 不是质数，而 $9 - 2 = 7$ 是质数，所以它推翻了这个命题。

其他选项不能作为反例： $12 - 2 = 10$ 和 $16 - 2 = 14$ 不是质数，而 $13$ 和 $23$ 本身就是质数。

所以正确答案是 A。

A counterexample needs $n$ not prime but $n - 2$ prime. For $n = 9,$ the number $9$ is not prime while $9 - 2 = 7$ is prime, so it breaks the statement.

The others fail to be counterexamples: $12 - 2 = 10$ and $16 - 2 = 14$ are not prime, while $13$ and $23$ are themselves prime.

Thus, the correct answer is A .

21.

假设 $n^\ast$ 表示 $\dfrac1n$ ，即 $n$ 的倒数。例如， $5^\ast = \dfrac15$ 。下面四个命题中有多少个是真的？

$\mathrm{(i)}\quad 3^\ast + 6^\ast = 9^\ast$

$\mathrm{(ii)}\quad 6^\ast - 4^\ast = 2^\ast$

$\mathrm{(iii)}\quad 2^\ast \cdot 6^\ast = 12^\ast$

$\mathrm{(iv)}\quad 10^\ast \div 2^\ast = 5^\ast$

Suppose $n^\ast$ means $\dfrac1n,$ the reciprocal of $n.$ For example, $5^\ast = \dfrac15.$ How many of the following four statements are true?

$\mathrm{(i)}\quad 3^\ast + 6^\ast = 9^\ast$

$\mathrm{(ii)}\quad 6^\ast - 4^\ast = 2^\ast$

$\mathrm{(iii)}\quad 2^\ast \cdot 6^\ast = 12^\ast$

$\mathrm{(iv)}\quad 10^\ast \div 2^\ast = 5^\ast$

$0$

$1$

$2$

$3$

$4$

答案：C

知识点：自定义运算分数

难度评级：1030

小提示：

把每个 $n^\ast$ 替换成 $\dfrac1n$ ，逐个检验等式两边

Replace each $n^\ast$ with $\dfrac1n$ and test the two sides of each statement

大提示：

对 $\mathrm{(iv)}$ ， $10^\ast \div 2^\ast = \dfrac{1}{10} \div \dfrac12 = \dfrac{1}{10} \cdot 2$

For $\mathrm{(iv)},$ $10^\ast \div 2^\ast = \dfrac{1}{10} \div \dfrac12 = \dfrac{1}{10} \cdot 2$

解答：

$\mathrm{(i)}$ $\dfrac13 + \dfrac16 = \dfrac12$ ，但 $9^\ast = \dfrac19$ ，所以为假。 $\mathrm{(ii)}$ $\dfrac16 - \dfrac14 = -\dfrac{1}{12}$ ，但 $2^\ast = \dfrac12$ ，所以为假。

$\mathrm{(iii)}$ $\dfrac12 \cdot \dfrac16 = \dfrac{1}{12} = 12^\ast$ ，为真。 $\mathrm{(iv)}$ $\dfrac{1}{10} \div \dfrac12 = \dfrac15 = 5^\ast$ ，为真。

因此正好有 $2$ 个命题为真。

所以正确答案是 C。

$\mathrm{(i)}$ $\dfrac13 + \dfrac16 = \dfrac12,$ but $9^\ast = \dfrac19,$ so false. $\mathrm{(ii)}$ $\dfrac16 - \dfrac14 = -\dfrac{1}{12},$ but $2^\ast = \dfrac12,$ so false.

$\mathrm{(iii)}$ $\dfrac12 \cdot \dfrac16 = \dfrac{1}{12} = 12^\ast,$ true. $\mathrm{(iv)}$ $\dfrac{1}{10} \div \dfrac12 = \dfrac15 = 5^\ast,$ true.

So exactly $2$ of the statements are true.

Thus, the correct answer is C .

22.

$ABCD$ 是一个矩形， $D$ 是圆心， $B$ 在圆上。如果 $AD = 4$ 且 $CD = 3$ ，那么阴影区域面积位于哪两个值之间？

$ABCD$ is a rectangle, $D$ is the center of the circle, and $B$ is on the circle. If $AD = 4$ and $CD = 3,$ then the area of the shaded region is between which two values?

$4$ 和 $5$

$4$ and $5$

$5$ 和 $6$

$5$ and $6$

$6$ 和 $7$

$6$ and $7$

$7$ 和 $8$

$7$ and $8$

$8$ 和 $9$

$8$ and $9$

答案：D

知识点：勾股定理扇形

难度评级：1120

小提示：

对角线 $DB$ 是半径；用勾股定理求出它

The diagonal $DB$ is the radius; find it with the Pythagorean theorem

大提示：

阴影区域是四分之一圆减去矩形： $\dfrac14 \pi (5)^2 - 3 \cdot 4$

The shaded region is a quarter of the circle with the rectangle removed: $\dfrac14 \pi (5)^2 - 3 \cdot 4$

解答：

因为 $AD = 4$ 且 $CD = 3$ ，所以对角线 $DB = \sqrt{3^2 + 4^2} = 5$ ，这就是半径。

阴影区域是以 $D$ 为圆心的四分之一圆减去矩形： $\begin{gathered} \dfrac14 \pi (5)^2 - (3)(4) = \dfrac{25\pi}{4} - 12 \\ \approx 19.6 - 12 = 7.6 \end{gathered}\text{。}$

这个值位于 $7$ 和 $8$ 之间。

所以正确答案是 D。

Since $AD = 4$ and $CD = 3,$ the diagonal $DB = \sqrt{3^2 + 4^2} = 5,$ which is the radius.

The shaded region is the quarter circle at $D$ minus the rectangle: $\begin{gathered} \dfrac14 \pi (5)^2 - (3)(4) = \dfrac{25\pi}{4} - 12 \\ \approx 19.6 - 12 = 7.6. \end{gathered}$

This lies between $7$ and $8.$

Thus, the correct answer is D .

23.

假设下表列出了美国 $1980$ 年各地区、各族裔的人口数，单位为百万。按最接近的百分数计算，美国黑人总人口中有百分之多少住在南部？

	东北	中西	南部	西部
白人	$42$	$52$	$57$	$35$
黑人	$5$	$5$	$15$	$2$
亚裔	$1$	$1$	$1$	$3$
其他	$1$	$1$	$2$	$4$

Assume the adjoining chart shows the $1980$ U.S. population, in millions, for each region by ethnic group. To the nearest percent, what percent of the U.S. Black population lived in the South?

	NE	MW	South	West
White	$42$	$52$	$57$	$35$
Black	$5$	$5$	$15$	$2$
Asian	$1$	$1$	$1$	$3$
Other	$1$	$1$	$2$	$4$

$20\%$

$25\%$

$40\%$

$56\%$

$80\%$

答案：D

知识点：百分数

难度评级：820

小提示：

把四个地区的黑人人口相加，得到总数

Add the Black population across all four regions to get the total

大提示：

用南部的人口除以总数，再化成百分数

Divide the South’s figure by that total, then convert to a percent

解答：

按百万人计，黑人总人口为 $5 + 5 + 15 + 2 = 27$ 。

南部所占比例是 $\dfrac{15}{27} = \dfrac59 \approx 55.6\%$ ，四舍五入为 $56\%$ 。

所以正确答案是 D。

The total Black population is $5 + 5 + 15 + 2 = 27$ million.

The South’s share is $\dfrac{15}{27} = \dfrac59 \approx 55.6\%,$ which rounds to $56\%.$

Thus, the correct answer is D .

24.

一场选择题考试共有 $20$ 题。计分规则是：每答对一题得 $+5$ 分，每答错一题得 $-2$ 分，未答题得 $0$ 分。约翰的考试成绩是 $48$ 分。他最多可能答对了多少题？

A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John’s score on the examination is $48.$ What is the maximum number of questions he could have answered correctly?

$9$

$10$

$11$

$12$

$16$

答案：D

知识点：一次方程最优化奇偶性

难度评级：1150

小提示：

如果 $c$ 题答对、 $w$ 题答错，则 $5c - 2w = 48$

If $c$ answers are correct and $w$ are wrong, then $5c - 2w = 48$

大提示：

从最大的可能 $c$ 试起： $5c - 48$ 必须是非负偶数 $2w$ ，并且 $c + w \le 20$

Try the largest possible $c$ : $5c - 48$ must be a nonnegative even number $2w,$ with $c + w \le 20$

解答：

设 $c$ 为答对题数， $w$ 为答错题数，则 $5c - 2w = 48$ 。于是 $2w = 5c - 48$ ，所以 $c$ 必须为偶数。

试取 $c = 14$ ，得到 $w = 11$ ，但 $c + w = 25 \gt 20$ ，超过了题目总数；任何更大的偶数 $c$ 都会需要回答更多题。试取 $c = 12$ ，得到 $w = 6$ ，且 $c + w = 18 \le 20$ ，符合条件。因此最大值是 $12$ 。

所以正确答案是 D。

Let $c$ be the number correct and $w$ the number wrong, so $5c - 2w = 48.$ Then $2w = 5c - 48,$ which requires $c$ to be even.

Trying $c = 14$ gives $w = 11,$ but $c + w = 25 \gt 20,$ too many; every larger even $c$ would require still more answered questions. Trying $c = 12$ gives $w = 6$ and $c + w = 18 \le 20,$ which works. So the maximum is $12.$

Thus, the correct answer is D .

25.

罐子里有编号 $1$ 到 $10$ 的十个球。杰克伸手随机取出一个球。然后吉尔伸手随机取出另一个不同的球。取出的两个球上的数字之和为偶数的概率是多少？

Ten balls numbered $1$ to $10$ are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. What is the probability that the sum of the two numbers on the balls removed is even?

$\dfrac{4}{9}$

$\dfrac{9}{19}$

$\dfrac{1}{2}$

$\dfrac{10}{19}$

$\dfrac{5}{9}$

答案：A

知识点：基本概率奇偶性

难度评级：1090

小提示：

两个数的和为偶数，当且仅当两个数同为奇数或同为偶数

The sum of two numbers is even exactly when both are odd or both are even

大提示：

有 $5$ 个奇数球和 $5$ 个偶数球；在全部 $\binom{10}{2}$ 对中数出同奇偶性的对数

There are $5$ odd and $5$ even balls; count the matching-parity pairs out of all $\binom{10}{2}$ pairs

解答：

和为偶数时，两个球同为奇数或同为偶数。全奇的配对有 $\binom{5}{2} = 10$ 对，全偶的配对也有 $\binom{5}{2} = 10$ 对，共 $20$ 对有利情况。

总配对数是 $\binom{10}{2} = 45$ ，所以概率是 $\dfrac{20}{45} = \dfrac49$ 。

所以正确答案是 A。

The sum is even when both balls are odd or both are even. There are $\binom{5}{2} = 10$ all-odd pairs and $\binom{5}{2} = 10$ all-even pairs, for $20$ favorable pairs.

The total number of pairs is $\binom{10}{2} = 45,$ so the probability is $\dfrac{20}{45} = \dfrac49.$

Thus, the correct answer is A .