1989 AMC 8 详解 | LIVE by Po-Shen Loh

1.

下列表达式的值是多少？

$\begin{aligned} &(1 + 11 + 21 + 31 + 41) \\ &\quad {}+ (9 + 19 + 29 + 39 + 49) \end{aligned}$

What is the value of

$\begin{aligned} &(1 + 11 + 21 + 31 + 41) \\ &\quad {}+ (9 + 19 + 29 + 39 + 49)? \end{aligned}$

$150$

$199$

$200$

$249$

$250$

知识点：配对与分组

难度评级：560

小提示：

寻找能配成整十数的项

Look for pairs of terms that combine into round numbers

大提示：

把 $1$ 与 $49$ 、 $11$ 与 $39$ 等配对，每一对的和都是 $50$

Pairing $1$ with $49,$ $11$ with $39,$ and so on, each pair sums to $50$

解答：

把各项配对，使每一对的和都是 $50$ ： $1 + 49$ 、 $11 + 39$ 、 $21 + 29$ 、 $31 + 19$ 、 $41 + 9$ 。

一共有 $5$ 对，所以总和是 $5 \times 50 = 250$ 。

所以正确答案是 E。

Pair the terms so each pair sums to $50$ : $1 + 49,$ $11 + 39,$ $21 + 29,$ $31 + 19,$ and $41 + 9.$

There are $5$ such pairs, so the total is $5 \times 50 = 250.$

Thus, the correct answer is E .

2.

下列表达式的值是多少？

$\frac{2}{10} + \frac{4}{100} + \frac{6}{1000}$

What is the value of

$\frac{2}{10} + \frac{4}{100} + \frac{6}{1000}?$

$0.012$

$0.0246$

$0.12$

$0.246$

$246$

知识点：小数位值

难度评级：450

小提示：

按位值把每个分数写成小数

Write each fraction as a decimal using its place value

大提示：

$\frac{2}{10} = 0.2$ ， $\frac{4}{100} = 0.04$ ， $\frac{6}{1000} = 0.006$ 。

$\frac{2}{10} = 0.2,$ $\frac{4}{100} = 0.04,$ and $\frac{6}{1000} = 0.006$

解答：

这三个分数分别表示不同数位上的小数： $\frac{2}{10} = 0.2$ ， $\frac{4}{100} = 0.04$ ， $\frac{6}{1000} = 0.006$ 。

相加得 $0.2 + 0.04 + 0.006 = 0.246$ 。

所以正确答案是 D。

Each fraction is a decimal in a different place: $\frac{2}{10} = 0.2,$ $\frac{4}{100} = 0.04,$ and $\frac{6}{1000} = 0.006.$

Adding these gives $0.2 + 0.04 + 0.006 = 0.246.$

Thus, the correct answer is D .

3.

下列哪个数最大？

Which of the following numbers is the largest?

$0.99$

$0.9099$

$0.9$

$0.909$

$0.9009$

知识点：小数位值

难度评级：450

小提示：

可以在小数末尾补零，把它们写成相同位数再比较

Give every number the same number of decimal places by appending zeros

大提示：

从左向右逐位比较；百分位上的数字决定这些以 $0.9$ 开头的数的大小

Compare digit by digit from the left; the hundredths place decides between the leading $0.9$ numbers

解答：

把这些数都写成四位小数： $0.9900$ 、 $0.9099$ 、 $0.9000$ 、 $0.9090$ 、 $0.9009$ 。

从左向右比较可知， $0.9900$ 的百分位数字最大，所以 $0.99$ 最大。

所以正确答案是 A。

Write each number with four decimal places: $0.9900,$ $0.9099,$ $0.9000,$ $0.9090,$ $0.9009.$

Comparing from the left, $0.9900$ has the largest hundredths digit, so $0.99$ is the largest.

Thus, the correct answer is A .

4.

下列哪个数最接近

$\frac{401}{0.205}\text{。}$

Estimate to determine which of the following is closest to

$\frac{401}{0.205}.$

$0.2$

$2$

$20$

$200$

$2000$

知识点：估算

难度评级：660

小提示：

把 $401$ 和 $0.205$ 都估成容易计算的数

Round $401$ and $0.205$ to simple numbers

大提示：

$\frac{401}{0.205} \approx \frac{400}{0.2}$

解答：

把分子和分母分别取成便于计算的近似数，得到 $\frac{401}{0.205} \approx \frac{400}{0.2}$ 。

由于 $\frac{400}{0.2} = 2000$ ，原数最接近 $2000$ 。

所以正确答案是 E。

Round the numerator and denominator: $\frac{401}{0.205} \approx \frac{400}{0.2}.$

Since $\frac{400}{0.2} = 2000,$ the value is closest to $2000.$

Thus, the correct answer is E .

5.

下列表达式的值是多少？

$-15 + 9 \times (6 \div 3)$

What is the value of

$-15 + 9 \times (6 \div 3)?$

$-48$

$-12$

$-3$

$3$

$12$

知识点：运算顺序

难度评级：560

小提示：

先算括号里的运算，再做乘法，最后做加法

Follow the order of operations: parentheses first, then multiplication, then addition

大提示：

$6 \div 3 = 2$ ，所以原式变成 $-15 + 9 \times 2$ 。

$6 \div 3 = 2,$ so the expression becomes $-15 + 9 \times 2$

解答：

先算括号： $6 \div 3 = 2$ 。再做乘法， $9 \times 2 = 18$ 。

最后 $-15 + 18 = 3$ 。

所以正确答案是 D。

Inside the parentheses, $6 \div 3 = 2.$ Then multiplication comes before addition, so $9 \times 2 = 18.$

Finally, $-15 + 18 = 3.$

Thus, the correct answer is D .

6.

数轴上各标记点之间的距离相等。 $y$ 是多少？

If the markings on the number line shown are equally spaced, what is the number $y?$

$3$

$10$

$12$

$15$

$16$

知识点：比与比例

难度评级：660

小提示：

数一数从 $0$ 到 $20$ 有多少个相等的间隔

Count how many equal spaces lie between $0$ and $20$

大提示：

从 $0$ 到 $20$ 有 $5$ 个相等间隔，所以每个间隔长 $\frac{20}{5} = 4$ ；再数一数到 $y$ 有几个间隔

There are $5$ spaces from $0$ to $20,$ so each space is $\frac{20}{5} = 4;$ then count the spaces up to $y$

解答：

从 $0$ 到 $20$ 被分成 $5$ 个相等间隔，所以每个间隔的长度是 $\frac{20}{5} = 4$ 。

$y$ 距 $0$ 有 $3$ 个间隔，所以 $y = 3 \times 4 = 12$ 。

所以正确答案是 C。

From $0$ to $20$ there are $5$ equal spaces, so each space is $\frac{20}{5} = 4.$

The mark labeled $y$ is $3$ spaces from $0,$ so $y = 3 \times 4 = 12.$

Thus, the correct answer is C .

7.

$20$ 枚二十五美分硬币和 $10$ 枚十美分硬币的总价值，等于 $10$ 枚二十五美分硬币和 $n$ 枚十美分硬币的总价值。 $n$ 是多少？

The value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes. What is $n?$

$10$

$20$

$30$

$35$

$45$

知识点：钱币一次方程

难度评级：770

小提示：

先求 $20$ 枚二十五美分硬币和 $10$ 枚十美分硬币的总价值

First find the total value of $20$ quarters and $10$ dimes

大提示：

去掉 $10$ 枚二十五美分硬币的价值后，剩下的价值都要由 $10$ 美分硬币补足

After using $10$ quarters, find how much value the remaining dimes must supply, then divide by $10$ ¢

解答：

$20$ 枚二十五美分硬币和 $10$ 枚十美分硬币的总价值是 $\$5.00 + \$1.00 = \$6.00$ 。

十枚二十五美分硬币值 $\$2.50$ ，所以 $n$ 枚十美分硬币必须值 $\$6.00 - \$2.50 = \$3.50$ 。这相当于 $35$ 枚十美分硬币，因此 $n = 35$ 。

所以正确答案是 D。

The value of $20$ quarters and $10$ dimes is $\$5.00 + \$1.00 = \$6.00.$

Ten quarters are worth $\$2.50,$ so the $n$ dimes must supply $\$6.00 - \$2.50 = \$3.50.$ That is $35$ dimes, so $n = 35.$

Thus, the correct answer is D .

8.

下列表达式的值是多少？

$(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)$

What is the value of

$(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)?$

$1$

$3$

$9$

$24$

$26$

知识点：分配律分数

难度评级：820

小提示：

先算 $2 \times 3 \times 4 = 24$ ，再把它乘进括号内的每一项

$2 \times 3 \times 4 = 24;$ distribute it across the three fractions

大提示：

把前面的乘积分别乘进括号各项，得到 $24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}$ ，再相加

$24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}$

解答：

$2 \times 3 \times 4 = 24$ ，所以原式等于 $24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}$ 。

这等于 $12 + 8 + 6 = 26$ 。

所以正确答案是 E。

Since $2 \times 3 \times 4 = 24,$ distribute it over the sum: $24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}.$

This equals $12 + 8 + 6 = 26.$

Thus, the correct answer is E .

9.

约翰逊老师的数学班中，男生与女生的比例是 $2$ 比 $3$ 。班上一共有 $30$ 名学生。男生占全班的百分之几？

There are $2$ boys for every $3$ girls in Ms. Johnson’s math class. If there are $30$ students in her class, what percent of them are boys?

$12\%$

$20\%$

$40\%$

$60\%$

$66\dfrac{2}{3}\%$

知识点：比与比例百分数

难度评级：660

小提示：

每 $2$ 个男生配 $3$ 个女生，所以每 $5$ 名学生中有 $2$ 名男生

For every $2$ boys there are $3$ girls, so boys are $2$ out of every $5$ students

大提示：

把 $\frac{2}{5}$ 化成百分数

Convert the fraction $\frac{2}{5}$ to a percent

解答：

男生占每 $2 + 3 = 5$ 名学生中的 $2$ 名，也就是全班的 $\frac{2}{5}$ 。全班共有 $30$ 人这一信息其实不必使用。

化成百分数， $\frac{2}{5} = 40\%$ 。

所以正确答案是 C。

Boys make up $2$ out of every $2 + 3 = 5$ students, which is $\frac{2}{5}$ of the class. The total of $30$ students is not even needed.

As a percent, $\frac{2}{5} = 40\%.$

Thus, the correct answer is C .

10.

七点整时，时针和分针之间较小的夹角是多少？

How many degrees are in the smaller angle between the hour hand and the minute hand of a clock that reads seven o’clock?

$50^\circ$

$120^\circ$

$135^\circ$

$150^\circ$

$165^\circ$

知识点：时钟

难度评级：730

小提示：

钟面上 $12$ 个小时刻度把一圈分成 $12$ 等份

The $12$ hour marks split the clock face into $12$ equal angles

大提示：

七点整时，两根针沿较短方向相隔 $5$ 个小时刻度，每个刻度对应 $30^\circ$

Each hour mark is $30^\circ;$ at seven o’clock the hands are $5$ marks apart the short way

解答：

钟面上的 $12$ 个数字把一周分成 $12$ 等份，每份是 $\frac{360^\circ}{12} = 30^\circ$ 。

七点整时，分针指向 $12$ ，时针指向 $7$ 。沿较短方向相隔 $5$ 个刻度，所以角度是 $5 \times 30^\circ = 150^\circ$ 。

所以正确答案是 D。

The $12$ numbers divide the clock into $12$ equal sections of $\frac{360^\circ}{12} = 30^\circ$ each.

At seven o’clock the hands point to $12$ and $7,$ which are $5$ sections apart the short way, giving $5 \times 30^\circ = 150^\circ.$

Thus, the correct answer is D .

11.

下列哪个图形与所给图形关于虚线对称？

Which of the five “T-like shapes” would be symmetric to the one shown with respect to the dashed line?

知识点：变换对称性

难度评级：860

小提示：

关于虚线对称表示沿虚线折叠后，两个图形会完全重合

Two figures are symmetric across the dashed line when folding the paper along that line makes one land exactly on the other

大提示：

竖直反射会保持上下位置不变，但会左右互换；注意小方块和斜线方向如何改变

A reflection across a vertical line keeps top and bottom fixed but swaps left and right; track where the small corner square ends up and how the slanted strokes turn

解答：

两个图形关于虚线对称，指的是沿这条线把纸折起来后，一个图形恰好落在另一个图形上。关于竖直虚线反射时，图形的左、右位置互换，上、下位置保持不变。

在这个反射下，原图中左上角的小方块会移到右上角，中间那条横杆的倾斜方向会相反，而原本朝右下方伸出的竖杆必须改为朝左下方伸出。只有一个选项同时具备这三个特征。

所以正确答案是 B。

Two figures are symmetric with respect to the dashed line when folding the paper along that line makes one coincide with the other. Reflecting across the vertical dashed line swaps left and right while leaving top and bottom unchanged.

Under this reflection the small square in the top-left corner moves to the top-right corner, the straight crossbar reverses its slant, and the stem that points down and to the right must instead point down and to the left. Only one choice has all three of these features.

Thus, the correct answer is B .

12.

下列表达式的值是多少？

$\frac{1 - \frac{1}{3}}{1 - \frac{1}{2}}$

What is the value of

$\frac{1 - \frac{1}{3}}{1 - \frac{1}{2}}?$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{3}{2}$

$\dfrac{4}{3}$

知识点：分数

难度评级：770

小提示：

先分别化简分子和分母

Simplify the numerator and denominator separately

大提示：

$1 - \frac{1}{3} = \frac{2}{3}$ ，且 $1 - \frac{1}{2} = \frac{1}{2}$ ；然后做除法

$1 - \frac{1}{3} = \frac{2}{3}$ and $1 - \frac{1}{2} = \frac{1}{2};$ then divide

解答：

分子是 $1 - \frac{1}{3} = \frac{2}{3}$ ，分母是 $1 - \frac{1}{2} = \frac{1}{2}$ 。

因此 $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3}$ 。

所以正确答案是 E。

The numerator is $1 - \frac{1}{3} = \frac{2}{3},$ and the denominator is $1 - \frac{1}{2} = \frac{1}{2}.$

Dividing gives $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3}.$

Thus, the correct answer is E .

13.

下列哪个式子等于

$\frac{9}{7 \times 53}$

Which of the following is equal to

$\frac{9}{7 \times 53}?$

$\dfrac{0.9}{0.7 \times 53}$

$\dfrac{0.9}{0.7 \times 0.53}$

$\dfrac{0.9}{0.7 \times 5.3}$

$\dfrac{0.9}{7 \times 0.53}$

$\dfrac{0.09}{0.07 \times 0.53}$

知识点：分数位值

难度评级：920

小提示：

如果把分子除以十，分母中的某一个因数也必须除以十，分数值才不变

Dividing the numerator and one factor of the denominator by the same number keeps the fraction unchanged

大提示：

要把 $9$ 变成 $0.9$ ，需除以 $10$ ；为保持分数值不变，也把分母中的一个因数除以 $10$

To turn $9$ into $0.9,$ divide by $10;$ to keep the value equal, divide one factor of the denominator by $10$ as well

解答：

将分子 $9$ 除以 $10$ ，得到 $0.9$ 。为了保持分数值不变，分母整体也要除以 $10$ ，即把其中一个因数除以 $10$ 。

把分母中的因数 $7$ 除以 $10$ ，而另一个因数保持不变，便得到 $\frac{0.9}{0.7 \times 53}$ 。其他选项会使原式改变 $10$ 倍或 $100$ 倍。

所以正确答案是 A。

To rewrite the numerator $9$ as $0.9,$ divide it by $10.$ To keep the fraction equal, divide the denominator by $10$ too, which means dividing one of its factors by $10.$

Dividing the factor $7$ by $10$ gives $\frac{0.9}{0.7 \times 53},$ which equals the original. Each other choice changes the value by a factor of $10$ or $100.$

Thus, the correct answer is A .

14.

把数字 $2$ 、 $4$ 、 $5$ 、 $6$ 、 $9$ 分别填入这个减法算式的五个方框中，每个数字恰好用一次，可能得到的最小差是多少？

When placing each of the digits $2,$ $4,$ $5,$ $6,$ $9$ in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?

$58$

$123$

$149$

$171$

$176$

知识点：位值最优化

难度评级：860

小提示：

要让差尽量小，就让三位数尽量小，同时让两位数尽量大

Make the three-digit number as small as possible and the two-digit number as large as possible

大提示：

用这些数字能组成的最大两位数使用 $9$ 和 $6$ ；再用剩下的数字组成最小三位数

The largest two-digit number from these digits uses $9$ and $6;$ build the smallest three-digit number from the digits that remain

解答：

差最小时，上面的三位数应尽量小，下面要减去的两位数应尽量大。

最大的两位数是 $96$ ，使用数字 $9$ 和 $6$ 。剩下的数字 $2$ 、 $4$ 、 $5$ 能组成的最小三位数是 $245$ ，所以最小差是 $245 - 96 = 149$ 。

所以正确答案是 C。

The difference is smallest when the three-digit number (the top) is as small as possible and the two-digit number (subtracted) is as large as possible.

The largest two-digit number is $96,$ using the digits $9$ and $6.$ The smallest three-digit number from the remaining digits $2,$ $4,$ $5$ is $245.$ So the smallest difference is $245 - 96 = 149.$

Thus, the correct answer is C .

15.

如图，在平行四边形 $ABCD$ 中，阴影区域 $BEDC$ 的面积是多少？

In parallelogram $ABCD$ shown, what is the area of the shaded region $BEDC?$

$24$

$48$

$60$

$64$

$80$

知识点：平行四边形面积分割

难度评级：860

小提示：

阴影面积等于整个平行四边形面积减去三角形 $ABE$ 的面积

The shaded region is the whole parallelogram with the triangle $ABE$ removed

大提示：

平行四边形面积是 $10 \times 8$ ；而三角形 $ABE$ 的底 $AE = 10 - 6$ ，高为 $8$ 。

The parallelogram area is $10 \times 8;$ triangle $ABE$ has base $AE = 10 - 6$ and height $8$

解答：

平行四边形的底为 $10$ ，高为 $8$ ，所以面积是 $10 \times 8 = 80$ 。

未涂色三角形 $ABE$ 的底为 $AE = AD - ED = 10 - 6 = 4$ ，高为 $8$ ，所以面积是 $\frac{1}{2} \times 4 \times 8 = 16$ 。阴影面积是 $80 - 16 = 64$ 。

所以正确答案是 D。

The parallelogram has base $10$ and height $8,$ so its area is $10 \times 8 = 80.$

The unshaded triangle $ABE$ has base $AE = AD - ED = 10 - 6 = 4$ and height $8,$ so its area is $\frac{1}{2} \times 4 \times 8 = 16.$ The shaded area is $80 - 16 = 64.$

Thus, the correct answer is D .

16.

$47$ 可以用多少种方式写成两个质数的和？

In how many ways can $47$ be written as the sum of two primes?

$0$

$1$

$2$

$3$

多于 $3$

more than $3$

知识点：质数奇偶性

难度评级：860

小提示：

$47$ 是奇数，所以两个质数中必须有一个是偶数

$47$ is odd, so one of the two primes must be even

大提示：

唯一的偶质数是 $2$ ，这会迫使另一个数是 $45$ 。

The only even prime is $2,$ which would force the other prime to be $45$

解答：

因为 $47$ 是奇数，若两个质数的和等于 $47$ ，其中必有一个偶质数。唯一的偶质数是 $2$ 。

于是另一个数必须是 $47 - 2 = 45$ ，但 $45 = 9 \times 5$ ，不是质数。因此没有这样的表示方式。

所以正确答案是 A。

Since $47$ is odd, a sum of two primes equal to $47$ needs one even prime and one odd prime. The only even prime is $2.$

That would require the other number to be $47 - 2 = 45,$ but $45 = 9 \times 5$ is not prime. So there is no way.

Thus, the correct answer is A .

17.

$N$ 是一个大于 $9$ 且小于 $17$ 的数。下列哪个数可能是 $6$ 、 $10$ 和 $N$ 的平均数？

The number $N$ is between $9$ and $17.$ Which of the following could be the average of $6,$ $10,$ and $N?$

$8$

$10$

$12$

$14$

$16$

知识点：平均数不等式

难度评级：800

小提示：

平均数是 $\frac{6 + 10 + N}{3}$ ；考察 $N$ 取值范围两端时的结果

The average is $\frac{6 + 10 + N}{3};$ find its value at each end of the range for $N$

大提示：

当 $N$ 在 $9$ 和 $17$ 之间时，这个平均数在 $\frac{25}{3}$ 和 $11$ 之间

As $N$ runs from $9$ to $17,$ the average runs from $\frac{25}{3}$ to $11$

解答：

平均数为 $\frac{6 + 10 + N}{3} = \frac{16 + N}{3}$ 。当 $N = 9$ 时，这个式子为 $\frac{25}{3} \approx 8.3$ ；当 $N = 17$ 时，它等于 $11$ 。

所以平均数严格介于约 $8.3$ 和 $11$ 之间。选项中只有 $10$ 落在这个范围内。

所以正确答案是 B。

The average is $\frac{6 + 10 + N}{3} = \frac{16 + N}{3}.$ When $N = 9$ this is $\frac{25}{3} \approx 8.3,$ and when $N = 17$ it is $11.$

So the average lies strictly between about $8.3$ and $11.$ Among the choices, only $10$ falls in this range.

Thus, the correct answer is B .

18.

计算器上有一个倒数键，按下它会把当前显示的数换成它的倒数。例如，若显示屏上是 $4$ ，按一次该键后显示 $0.25$ 。如果显示屏上是 $32$ ，至少按几次倒数键才能再次显示 $32$ ？

A calculator has a reciprocal key that replaces the number currently displayed with its reciprocal. For example, if the display shows $4$ and the key is pressed, the display becomes $0.25.$ If $32$ is currently displayed, what is the fewest number of times the reciprocal key must be pressed so that the display again reads $32?$

$1$

$2$

$3$

$4$

$5$

知识点：函数分数

难度评级：730

小提示：

按一次会把 $32$ 变成它的倒数 $\frac{1}{32}$ 。

Pressing the key once turns $32$ into $\frac{1}{32}$

大提示：

一个数的倒数再取倒数，会回到原来的数

The reciprocal of the reciprocal of a number is the number itself

解答：

从 $32$ 开始，按一次倒数键会得到 $\frac{1}{32}$ 。

再按一次，就是对倒数再取倒数，得到 $\frac{1}{\frac{1}{32}} = 32$ 。所以最少需要按 $2$ 次。

所以正确答案是 B。

Pressing the key once changes $32$ to its reciprocal $\frac{1}{32}.$

Pressing it a second time takes the reciprocal again, returning to $\frac{1}{\frac{1}{32}} = 32.$ So $2$ presses are enough.

Thus, the correct answer is B .

19.

图中曲线表示冲浪城政府在 $1988$ 年的累计花费，单位为百万美元。例如，截至二月初约已花费 $0.5$ 百万美元，截至四月底约已花费 $2$ 百万美元。六月、七月和八月大约一共花费了多少百万美元？

The graph below shows the total accumulated dollars (in millions) spent by the Surf City government during $1988.$ For example, about $0.5$ million had been spent by the beginning of February and approximately $2$ million by the end of April. Approximately how many millions of dollars were spent during the summer months of June, July, and August?

$1.5$

$2.5$

$3.5$

$4.5$

$5.5$

知识点：数据与图表解读估算

难度评级：920

小提示：

这是累计花费图，所以一段时间内的花费等于图上数值的增加量

Because the graph shows an accumulated total, the amount spent over a span of months is the increase in that total across the span

大提示：

读出六月开始附近和八月结束附近的累计值，然后相减

Read the height of the curve at the beginning of June and at the end of August, then subtract the two values

解答：

图显示的是累计花费，所以六月、七月、八月的花费等于这段时间累计值的增加量。

从图上看，六月开始时累计值略多于 $2$ 百万美元，八月结束时略多于 $4.5$ 百万美元，差约为 $4.5 - 2 = 2.5$ 百万美元。

所以正确答案是 B。

The graph gives the total accumulated spending, so the amount spent during June, July, and August equals the accumulated total at the end of August minus the accumulated total at the beginning of June.

The curve is at a bit more than $2$ million at the beginning of June and a bit more than $4.5$ million by the end of August. The difference is about $4.5 - 2 = 2.5$ million.

Thus, the correct answer is B .

20.

图中的展开图折成一个数字立方体。在立方体的每个顶点，都会有三个面相交。相交于同一个顶点的三个面上的数字，可能得到的最大和是多少？

The figure shown may be folded along the lines to form a number cube. Three faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?

$11$

$12$

$13$

$14$

$15$

知识点：展开图（立体几何）最优化

难度评级：920

小提示：

同一个顶点上的三个面必须两两相邻，不能包含一对相对的面

A corner uses three faces that are mutually adjacent, so no two of them can be opposite faces

大提示：

由展开图可知相对面是 $1$ 和 $3$ 、 $2$ 和 $5$ 、 $4$ 和 $6$ ；从每一对中取较大的数

Folding the net, the opposite face pairs are $1$ & $3,$ $2$ & $5,$ and $4$ & $6;$ take the larger of each pair

解答：

折成立方体后，相对面不能在同一个顶点相遇。由展开图可判断相对面为 $1$ 与 $3$ 、 $2$ 与 $5$ 、 $4$ 与 $6$ 。

要让三个相邻面的和最大，就从每对相对面中取较大的数 $3$ 、 $5$ 和 $6$ 。它们可以在同一个顶点相遇，和为 $3 + 5 + 6 = 14$ 。

所以正确答案是 D。

When the net is folded, the pairs of opposite faces are $1$ and $3,$ $2$ and $5,$ and $4$ and $6.$ Three faces meeting at a corner must come one from each opposite pair.

To maximize the sum, take the larger number from each pair: $3,$ $5,$ and $6.$ These three faces do meet at a corner, and their sum is $3 + 5 + 6 = 14.$

Thus, the correct answer is D .

21.

杰克有一袋 $128$ 个苹果。他先把其中的 $25\%$ 卖给吉尔。接着，他又把袋中剩余苹果的 $25\%$ 卖给琼。之后，他把袋中剩下的苹果中最亮的一个送给了老师。最后袋中还剩多少个苹果？

Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?

$7$

$63$

$65$

$71$

$111$

知识点：百分数

难度评级：860

小提示：

卖掉 $25\%$ 后会剩下 $75\% = \frac{3}{4}$

Selling $25\%$ leaves $75\% = \frac{3}{4}$ of the apples

大提示：

把 $128$ 连续乘以 $\frac{3}{4}$ 两次，然后再减去送给老师的那个苹果

Multiply $128$ by $\frac{3}{4}$ twice, then subtract the one given to the teacher

解答：

杰克第一次卖掉 $25\%$ 后，剩下 $\frac{3}{4} \times 128 = 96$ 个苹果。第二次又卖掉余数的 $25\%$ ，剩下 $\frac{3}{4} \times 96 = 72$ 个苹果。

再送给老师 $1$ 个后，剩下 $72 - 1 = 71$ 个。

所以正确答案是 D。

After selling $25\%$ to Jill, Jack keeps $\frac{3}{4} \times 128 = 96$ apples. After selling $25\%$ of those to June, he keeps $\frac{3}{4} \times 96 = 72$ apples.

He then gives $1$ to his teacher, leaving $72 - 1 = 71.$

Thus, the correct answer is D .

22.

字母 $A$ 、 $J$ 、 $H$ 、 $S$ 、 $M$ 、 $E$ 和数字 $1$ 、 $9$ 、 $8$ 、 $9$ 分别循环移动，每次移动一位，用来组成一个编号列表。从 AJHSME $1989$ 开始，列表的第 $1$ 行是 JHSMEA $9891$ ，第 $2$ 行是 HSMEAJ $8919$ ，第 $3$ 行是 SMEAJH $9198$ ，依此类推。AJHSME $1989$ 第一次会出现在第几行？

The letters $A,$ $J,$ $H,$ $S,$ $M,$ $E$ and the digits $1,$ $9,$ $8,$ $9$ are each cycled separately (shifted one place at a time) to build a numbered list. Starting from AJHSME $1989,$ the list begins: line $1$ is JHSMEA $9891,$ line $2$ is HSMEAJ $8919,$ line $3$ is SMEAJH $9198,$ and so on. On what numbered line will AJHSME $1989$ appear for the first time?

$6$

$10$

$12$

$18$

$24$

知识点：最小公倍数

难度评级：1020

小提示：

六个字母每 $6$ 行回到原来的顺序；数字 $1989$ 每 $4$ 行回到原来的顺序

The six letters return to AJHSME every $6$ lines; the four digits return to $1989$ every $4$ lines

大提示：

两者同时回到原样的行号是 $6$ 和 $4$ 的最小公倍数

Both return together on the least common multiple of $6$ and $4$

解答：

字母串 AJHSME 每移动 $6$ 次回到原顺序，数字串 $1989$ 每移动 $4$ 次回到原顺序。

两者第一次同时回到原样是在 $6$ 和 $4$ 的最小公倍数 $12$ 次移动后。因此 AJHSME $1989$ 第一次出现在第 $12$ 行。

所以正确答案是 C。

The six letters cycle back to their original order AJHSME every $6$ lines, and the four digits cycle back to $1989$ every $4$ lines.

Both happen on the same line at the least common multiple of $6$ and $4,$ which is $12.$ So AJHSME $1989$ first reappears on line $12.$

Thus, the correct answer is C .

23.

一位艺术家有 $14$ 个边长为 $1$ 米的正方体。她把它们放在地面上，搭成如图所示的雕塑，然后给雕塑露出的表面上漆。她一共要漆多少平方米？

An artist has $14$ cubes, each with an edge of $1$ meter. She stands them on the ground to form a sculpture as shown. She then paints the exposed surface of the sculpture. How many square meters does she paint?

$21$

$24$

$33$

$37$

$42$

知识点：表面积正方体

难度评级：1050

小提示：

分别数四个竖直侧面的露出面积，再加上从上方看到的面积

Count the exposed faces on the four vertical sides, then the faces seen from directly above

大提示：

每个侧面看到的阶梯面积是 $6$ ，上表面的投影覆盖整个 $3 \times 3$ 底面

Each of the four sides shows $6$ exposed faces, and looking straight down covers the full $3 \times 3$ footprint of top faces

解答：

因为每一层都靠齐在同一个角上，所以从每个竖直方向看，露出的阶梯侧面都是 $3 + 2 + 1 = 6$ 个小正方形面，四个方向共 $4 \times 6 = 24$ 个侧面。

从上方看，露出的顶面覆盖整个 $3 \times 3$ 的底面投影，所以顶面积是 $9$ 平方米。底面不漆，因此总上漆面积是 $24 + 9 = 33$ 平方米。

所以正确答案是 C。

Because each higher layer is flush into one corner, each of the four vertical sides shows a stepped profile of $3 + 2 + 1 = 6$ exposed square faces, for $4 \times 6 = 24$ side faces.

Viewed from directly above, the top faces cover the full $3 \times 3$ footprint, adding $9$ more faces. The bottom rests on the ground and is not painted, so the total is $24 + 9 = 33$ square meters.

Thus, the correct answer is C .

24.

把一张正方形纸沿竖直方向对折，再沿图中的虚线把折好的纸剪成两半。这样会得到三个长方形：一个大长方形和两个小长方形。一个小长方形的周长与大长方形周长之比是多少？

Suppose a square piece of paper is folded in half vertically. The folded paper is then cut in half along the dashed line. Three rectangles are formed—a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

$\dfrac{5}{6}$

知识点：折纸周长

难度评级：950

小提示：

可以假设原正方形边长为 $4$

Give the original square a convenient side length, such as $4,$ and track the pieces

大提示：

含折痕的部分展开成一个 $2 \times 4$ 的大长方形，另外两层各展开成一个 $1 \times 4$ 的小长方形

The piece containing the fold unfolds to a $2 \times 4$ large rectangle, while each other layer is a $1 \times 4$ small rectangle

解答：

设原正方形边长为 $4$ 。对折后得到一个 $2 \times 4$ 的双层长方形，再用一条与折痕平行的线把折好的纸切开，分成含折痕的一条和不含折痕的一条。

含折痕的那块展开后是 $2 \times 4$ 的大长方形，周长为 $2(2 + 4) = 12$ 。另一块展开后成为两个 $1 \times 4$ 的小长方形，每个周长为 $2(1 + 4) = 10$ 。所求比为 $\frac{10}{12} = \frac{5}{6}$ 。

所以正确答案是 E。

Let the square have side $4.$ Folding in half makes a $2 \times 4$ stack of two layers. Cutting parallel to the fold splits it into a strip containing the fold and a strip that does not.

The strip with the fold unfolds into the large rectangle, $2 \times 4,$ with perimeter $2(2 + 4) = 12.$ The other strip is two separate small rectangles, each $1 \times 4,$ with perimeter $2(1 + 4) = 10.$ The ratio is $\frac{10}{12} = \frac{5}{6}.$

Thus, the correct answer is E .

25.

每次转动图中的两个转盘时，两个指针各选出一个数。所选两个数之和为偶数的概率是多少？

Every time the two wheels shown are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?

$\dfrac{1}{6}$

$\dfrac{3}{7}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{7}$

知识点：基本概率奇偶性

难度评级：920

小提示：

两个数之和为偶数，当且仅当两个数同为偶数或同为奇数

A sum is even exactly when both numbers are even or both are odd

大提示：

第一个转盘有 $2$ 个偶数、 $2$ 个奇数；第二个转盘有 $1$ 个偶数、 $2$ 个奇数

The first wheel has $2$ even and $2$ odd numbers; the second has $1$ even and $2$ odd

解答：

和为偶数需要两个数同奇偶。第一个转盘的偶数为 $\{4, 8\}$ ，奇数为 $\{3, 5\}$ ，所以两种奇偶性的概率都是 $\frac{2}{4} = \frac{1}{2}$ 。第二个转盘的偶数为 $\{6\}$ ，概率为 $\frac{1}{3}$ ；奇数为 $\{7, 9\}$ ，概率为 $\frac{2}{3}$ 。

因此和为偶数的概率是 $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}$ 。

所以正确答案是 C。

The sum is even when both numbers are even or both are odd. The first wheel has evens $\{4, 8\}$ and odds $\{3, 5\},$ each with probability $\frac{2}{4} = \frac{1}{2}.$ The second wheel has even $\{6\}$ with probability $\frac{1}{3}$ and odds $\{7, 9\}$ with probability $\frac{2}{3}.$

So the probability of an even sum is $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.$

Thus, the correct answer is C .