2020 年 AMC 8 真题（P版）

1 答案 E

糖是柠檬汁的二倍，所以卢卡需要 \(2\cdot3=6\) 杯糖。水是糖的 \(4\) 倍，所以卢卡需要 \(4\cdot6=24\) 杯水。

正确答案是 E。

English

Since Luka needs twice as much sugar as lemon, he needs \(2\cdot3=6\) cups of sugar. Since Luka also needs \(4\) times as much water as sugar, he needs \(4\cdot6=24\) cups of water.

Thus, the correct answer is E.

2 答案 C

他们总共赚了 \(\$15+\$20+\$25+\$40=\$100\)。

平均分给四个人后，每人应得 \(\frac{100}{4}=25\) 美元。

因此，赚 \(\$40\) 的朋友要给其他人 \(\$40-\$25=\$15\)。

正确答案是 C。

English

First, the total amount of money that they make is \(\$15+\$20+\$25+\$40=\$100.\)

Since they divide this equally, they each get \(\frac{100}{4}=25\) dollars.

The friend who earned \(\$40\) therefore gives away \(\$40-\$25=\$15.\)

Thus, the correct answer is C.

3 答案 D

花园面积为 \(6\cdot8=48\) 平方英尺。

每平方英尺可种 \(4\) 株，所以凯莉总共可种 \(4\cdot 48=192\) 株草莓。

每株平均 \(10\) 个草莓，所以总数为 \(10\cdot 192=1920\)。

正确答案是 D。

English

First, the size of the garden is \(6\cdot8=48\) square feet.

Next, since there are \(4\) plants per square foot, Carrie can plant \(4\cdot 48=192\) plants total.

Finally, since there are \(10\) strawberries per plant, Carrie can harvest \(10\cdot 192=1920\) strawberries total.

Thus, the correct answer is D.

4 答案 B

前三个六边形分别有 \(1\)、\(7\) 和 \(19\) 个点。每增加一圈，比上一圈多加 \(6\) 个点。

第四个六边形会在第三个基础上再增加 \(18\) 个点，所以共有 \(19+18=37\) 个点。

正确答案是 B。

English

The first three hexagons contain \(1,\) \(7,\) and \(19\) dots. Each new band adds \(6\) more dots than the previous band.

The fourth hexagon adds \(18\) dots around the third hexagon, so it contains \(19+18=37\) dots.

Thus, the correct answer is B.

5 答案 C

每个杯子得到水壶容量 \(\dfrac{3}{4}\) 的五分之一，也就是 \[ \dfrac{1}{5}\cdot\dfrac{3}{4}=\dfrac{3}{20}=15\%\text{。}\]

因此正确答案是 C。

English

Each cup receives one fifth of the \(\dfrac{3}{4}\) of a pitcher, or \[ \dfrac{1}{5}\cdot\dfrac{3}{4}=\dfrac{3}{20}=15\%. \]

Thus, the correct answer is C.

6 答案 A

玛伦坐在最后一节。由于亚伦正好在莎伦后面，且达伦在亚伦前方，在玛伦前面的车厢中只需安排达伦、莎伦和亚伦的相对位置。

如果莎伦和亚伦分别在第 \(1\) 和第 \(2\) 节，达伦就无法坐在亚伦前面；如果他们在第 \(3\) 和第 \(4\) 节，达伦和凯伦只能在第 \(1\) 和第 \(2\) 节，会相邻。

因此莎伦和亚伦必须在第 \(2\) 和第 \(3\) 节，达伦在第 \(1\) 节，凯伦在第 \(4\) 节，亚伦坐在中间车厢。

正确答案是 A。

English

Maren is in the last car. Since Aaron is directly behind Sharon and Darren is in front of Aaron, the possible placements of Darren, Sharon, and Aaron before Maren are limited.

If Sharon and Aaron were in cars \(1\) and \(2,\) Darren could not be in front of Aaron. If they were in cars \(3\) and \(4,\) then Darren and Karen would have to occupy cars \(1\) and \(2,\) which are adjacent.

Thus Sharon and Aaron must be in cars \(2\) and \(3.\) Darren is then in car \(1,\) Karen is in car \(4,\) and Aaron is in the middle car.

Thus, the correct answer is A.

7 答案 C

千位必须是 \(2\)。因为数字互不相同且递增，百位必须是 \(3\)。

最后两位必须从 \(4\)、\(5\)、\(6\)、\(7\)、\(8\)、\(9\) 中选择；选定两个数字后，它们的顺序也就确定了。

共有 \(\binom{6}{2}=15\) 个这样的整数。

正确答案是 C。

English

The thousands digit must be \(2.\) Since the digits are distinct and increasing, the hundreds digit must be \(3.\)

The last two digits must be chosen from \(4,\) \(5,\) \(6,\) \(7,\) \(8,\) \(9.\) Once the two digits are chosen, their order is forced.

There are \(\binom{6}{2}=15\) such integers.

Thus, the correct answer is C.

8 答案 C

最大金额对应 \(2019\) 枚五美分硬币和 \(1\) 枚一美分硬币；最小金额对应 \(1\) 枚五美分硬币和 \(2019\) 枚一美分硬币。

从第二种情况变到第一种情况，共把 \(2018\) 枚一美分硬币换成五美分硬币。每次替换增加 \(5-1=4\) 美分，所以差为 \[ 2018\cdot4=8072 \] 美分。

所以正确答案是 C。

English

The greatest value occurs with \(2019\) nickels and \(1\) penny, while the least occurs with \(1\) nickel and \(2019\) pennies.

Between these two cases, \(2018\) pennies have been replaced by nickels. Each replacement adds \(5-1=4\) cents, so the difference is \[ 2018\cdot4=8072 \] cents.

Thus, the correct answer is C.

9 答案 D

顶层中，每条边上非角的小块恰好有两个糖霜面。共有 \(4\) 条边，每条边有 \(2\) 个这样的块，共 \(8\) 块。

在下面 \(3\) 层中，恰好有两个糖霜面的只有每层的 \(4\) 个角块，共 \(3\cdot4=12\) 块。

总数为 \(8+12=20\)。

正确答案是 D。

English

In the top layer, the non-corner edge pieces have icing on exactly two sides. There are \(4\) edges with \(2\) such pieces each, for \(8\) pieces.

In each of the other \(3\) layers, the only pieces with exactly two iced sides are the \(4\) corner pieces. This adds \(3\cdot4=12\) pieces.

The total is \(8+12=20.\)

Thus, the correct answer is D.

10 答案 C

四个弹珠共有 \(4!=24\) 种排列。

若斯蒂利和泰格相邻，把它们看成一个整体，则有 \(3!\) 种放置整体和另外两个弹珠的方法，整体内部有 \(2\) 种顺序，共 \(3!\cdot2=12\) 种。

不相邻的排法为 \(24-12=12\)。

正确答案是 C。

English

There are \(4!=24\) total arrangements of the marbles.

If the Steelie and Tiger are adjacent, treat them as one block. Then there are \(3!\) ways to arrange the block with the other two marbles, and \(2\) orders inside the block, for \(3!\cdot2=12\) adjacent arrangements.

Therefore, \(24-12=12\) arrangements keep the Steelie and Tiger separated.

Thus, the correct answer is C.

11 答案 E

娜奥米用 \(10\) 分钟走完 \(6\) 英里，平均速度为 \[\dfrac{6}{\frac{10}{60}}=36\] 英里每小时。

玛雅用 \(30\) 分钟走完 \(6\) 英里，平均速度为 \[\dfrac{6}{\frac{30}{60}}=12\] 英里每小时。

差为 \(36-12=24\)。

正确答案是 E。

English

Naomi traveled \(6\) miles in \(10\) minutes, so her average speed was \[\dfrac{6}{\frac{10}{60}}=36\] miles per hour.

Maya traveled \(6\) miles in \(30\) minutes, so her average speed was \[\dfrac{6}{\frac{30}{60}}=12\] miles per hour.

This difference is \(36-12=24.\)

Thus, the correct answer is E.

12 答案 A

因为 \(5!=120=12\cdot10\)，原方程可写成 \[ 12\cdot10\cdot9!=12\cdot N!\text{。}\]两边约去 \(12\)，得到 \(N!=10\cdot9!=10!\)，所以 \(N=10\)。

正确答案是 A。

English

Because \(5!=120=12\cdot10,\) the equation becomes \[ 12\cdot10\cdot9!=12\cdot N!. \] Canceling \(12\) gives \(N!=10\cdot9!=10!,\) so \(N=10.\)

Thus, the correct answer is A.

13 答案 B

设贾马尔加入 \(s\) 只紫色袜子，则紫色袜子共有 \(s+18\) 只。

原来共有 \(36\) 只袜子，加入后总数为 \(s+36\) 只。

因为之后抽到紫色袜子的概率是 \(60\%\)，所以 \[\dfrac{s+18}{s+36}=0.6\text{。}\]

解 \(s\) 得 \[\begin{align*}s+18 &= 0.6s+21.6 \\ 0.4s &= 3.6\\ s &= 9\end{align*}\text{。}\]

所以加入了 \(9\) 只袜子。

正确答案是 B。

English

Suppose Jamal adds \(s\) purple socks. Then, there will be \(s+18\) purple socks.

Also, since there are \(36\) total socks to begin with, we have \(s+36\) socks after adding the socks.

Since we have a \(60\%\) chance of choosing a purple sock afterwards, we know \[\dfrac{s+18}{s+36}=0.6.\]

Solving for \(s\) yields: \[\begin{align*}s+18 &= 0.6s+21.6 \\ 0.4s &= 3.6\\ s &= 9.\end{align*}\]

Therefore, \(9\) socks are added.

Thus, the correct answer is B.

14 答案 D

从虚线读出平均人口约为 \(4750\)。

有 \(20\) 个城市，所以总人口约为 \[20\cdot4750=95{,}000\text{。}\]

因此总人口约为 \(95{,}000\)。

正确答案是 D。

English

Looking at the horizontal dashed line, the average population is around \(4750.\)

Since there are \(20\) cities, the total population is approximately \[20\cdot4750=95{,}000.\]

Therefore, the total population is approximately \(95{,}000.\)

Thus, the correct answer is D.

15 答案 C

题目条件给出 \(0.15x=0.20y\)。

解出 \(y\)，得到 \(y=\dfrac{0.15}{0.20}x=\dfrac{3}{4}x\)。

因此 \(y\) 是 \(x\) 的 \(75\%\)。

正确答案是 C。

English

The statement gives \(0.15x=0.20y.\)

Solving for \(y\) gives \(y=\dfrac{0.15}{0.20}x=\dfrac{3}{4}x.\)

Therefore, \(y\) is \(75\%\) of \(x.\)

Thus, the correct answer is C.

16 答案 E

除 \(B\) 外，每个点都在两条线上，所以在五个线和中被计入两次；\(B\) 在三条线上，因此还要多计一次。

六个数字恰为 \(1\)、\(2\)、\(3\)、\(4\)、\(5\)、\(6\)，总和是 \(21\)。因此 \[ 2\cdot21+B=47\text{，}\]所以 \(B=5\)。

正确答案是 E。

English

Every point is counted in two of the line sums except \(B,\) which is counted in three. Thus the total of the five line sums is twice the sum of all six digits, plus one extra \(B.\)

The digits are exactly \(1,\) \(2,\) \(3,\) \(4,\) \(5,\) and \(6,\) whose sum is \(21.\) Therefore \[ 2\cdot21+B=47, \] so \(B=5.\)

Thus, the correct answer is E.

17 答案 B

因为 \(2020=2^2\cdot5\cdot101\)，所以它有 \[ (2+1)(1+1)(1+1)=12 \] 个正因数。

在这些因数中，只有 \(1\) 和质数 \(2\)、\(5\)、\(101\) 的因数少于三个，而 \(4\) 恰有三个因数。因此其余 \(12-5=7\) 个因数本身都拥有超过三个因数。

所以正确答案是 B。

English

Since \(2020=2^2\cdot5\cdot101,\) it has \[ (2+1)(1+1)(1+1)=12 \] positive factors.

Among those factors, only \(1\) and the primes \(2,\) \(5,\) and \(101\) have fewer than three factors, while \(4\) has exactly three. The other \(12-5=7\) factors therefore have more than three factors.

Thus, the correct answer is B.

18 答案 A

因为 \(FE\) 是半圆直径，直径长为 \(34\)，半径为 \(17\)。设 \(O\) 为直径的中点。

因此 \(OF\) 的长度为 \(17\)。

由于 \(D\) 在 \(OF\) 上，\[\begin{align*} OD + FD &= OF\\ OD + 9 &= 17 \\ OD &= 8 \end{align*}\text{。}\]

又因为该点在半圆上，所以 \(OC = 17\)。

矩形 \(ABCD\) 中，\(\angle ODC\) 是直角。用勾股定理求 \(DC\)：\[\begin{align*} OD^2+DC^2&=OC^2 \\ 8^2+DC^2 &= 17^2 \\ DC &= 15 \end{align*}\text{。}\]

因此矩形面积为 \(DC\cdot DA = 15\cdot 16=240\)。

正确答案是 A。

English

Since \(FE\) is the diameter of the semicircle, we know the length of the diameter is \(34,\) and so the radius is \(17.\) Let \(O\) be the center of the diameter.

The length from \(OF\) therefore is \(17.\)

Since \(D\) is on \(OF,\) we know \[\begin{align*} OD + FD &= OF\\ OD + 9 &= 17 \\ OD &= 8. \end{align*}\]

Also, since we have a semicircle, we know \(OC = 17.\)

Finally, since \(ABCD\) is a rectangle, we know \(\angle ODC\) is a right angle. This means we can find \(DC\) by the Pythagorean Theorem. We know \[\begin{align*} OD^2+DC^2&=OC^2 \\ 8^2+DC^2 &= 17^2 \\ DC &= 15. \end{align*}\]

Thus, the area of the rectangle is \(DC\cdot DA = 15\cdot 16=240.\)

Thus, the correct answer is A.

19 答案 B

一个五位交替数形如 \(ABABA\)，其中 \(A\ne0\) 且 \(A\ne B\)。能被 \(5\) 整除要求末位 \(A\) 为 \(0\) 或 \(5\)。因为 \(A\ne0\)，所以必须有 \(A=5\)。

因此这个数形如 \(5B5B5\)。它的各位数字之和为 \(15+2B\)，所以能被 \(3\) 整除要求 \(B\) 是 \(3\) 的倍数。可能值为 \(B=0\)、\(B=3\)、\(B=6\)、\(B=9\)，共得到 \(4\) 个数。

所以正确答案是 B。

English

A five-digit flippy number has the form \(ABABA,\) where \(A\ne0\) and \(A\ne B.\) Divisibility by \(5\) requires the last digit \(A\) to be \(0\) or \(5.\) Since \(A\ne0,\) we must have \(A=5.\)

The number therefore has the form \(5B5B5.\) Its digit sum is \(15+2B,\) so divisibility by \(3\) requires \(B\) to be a multiple of \(3.\) The possibilities are \(B=0,\) \(B=3,\) \(B=6,\) and \(B=9,\) giving \(4\) numbers.

Thus, the correct answer is B.

20 答案 B

第 \(2\) 棵树高 \(11\) 米。由于相邻两棵树高度相差因子 \(2\)，且高度都是整数，第 \(1\) 棵和第 \(3\) 棵都必须高 \(22\) 米。

前三棵树总高为 \(22+11+22=55\) 米。第 \(4\) 棵和第 \(5\) 棵的可能整数高度对为 \((11,22)\)、\((44,22)\)、\((44,88)\)。

这三种情况对应的平均数分别为 \(17.6\)、\(24.2\)、\(37.4\)。笔记表明平均数以 \(0.2\) 结尾，所以只能是 \(24.2\)。

正确答案是 B。

English

Tree \(2\) is \(11\) meters tall. Since all heights are integers and neighboring trees differ by a factor of \(2,\) trees \(1\) and \(3\) must both be \(22\) meters tall.

The first three trees total \(22+11+22=55\) meters. The possible integer pairs for trees \(4\) and \(5\) are \((11,22),\) \((44,22),\) and \((44,88).\)

These give averages \(17.6,\) \(24.2,\) and \(37.4,\) respectively. The notebook shows that the average ends in \(0.2,\) so it must be \(24.2.\)

Thus, the correct answer is B.

21 答案 A

每一步都必须向上一行，并向左或向右移动。从 \(P\) 开始逐行计算，每个可达方格的路径数等于下面两个相邻方格路径数之和。

方格 \(Q\) 处的计数是 \(28\)，所以共有 \(28\) 条路径。

正确答案是 A。

English

Each move must go up one row and either left or right. Counting row by row from \(P,\) each reachable square gets the sum of the counts from the two adjoining squares below it.

The count at \(Q\) is \(28,\) so there are \(28\) paths.

Thus, the correct answer is A.

22 答案 E

从 \(1\) 倒推。一个值 \(m\) 总可以由偶数输入 \(2m\) 产生；当 \(\frac{m-1}{3}\) 是正奇数时，也可以由这个奇数输入产生。

列出每次倒推后所有可能的值： \[ \begin{aligned} \{1\}&\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\\ &\leftarrow\{2,16\}\leftarrow\{4,5,32\}\\ &\leftarrow\{1,8,10,64\}\text{。} \end{aligned} \] 因此可能的起始值为 \(1\)、\(8\)、\(10\) 和 \(64\)，它们的和是 \(83\)。

正确答案是 E。

English

Work backward from \(1.\) A value \(m\) can always come from the even input \(2m.\) It can also come from the odd input \(\frac{m-1}{3}\) when that expression is a positive odd integer.

Listing the possible values after each backward step gives \[ \begin{aligned} \{1\}&\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\\ &\leftarrow\{2,16\}\leftarrow\{4,5,32\}\\ &\leftarrow\{1,8,10,64\}. \end{aligned} \] Thus the possible starting values are \(1,\) \(8,\) \(10,\) \(64,\) whose sum is \(83.\)

Thus, the correct answer is E.

23 答案 B

若没有“每人至少一个奖项”的限制，每个奖项有 \(3\) 个学生可选，\(5\) 个奖项共有 \(3^5=243\) 种分配。

减去至少一名学生没有获奖的情况。指定某名学生没有获奖时，五个奖项只能给另外两人，有 \(2^5\) 种；三名学生都可能是没有获奖者，所以先减去 \(3\cdot2^5\)。不过所有奖项都给同一个人的 \(3\) 种情况被多减了一次。

由容斥，符合条件的分配数为 \(3^5-3\cdot2^5+3=150\)。

正确答案是 B。

English

There are \(3^5=243\) ways to give each of the \(5\) distinct awards to one of the \(3\) students.

Subtract the distributions in which at least one student receives no award. If a particular student receives none, the awards go to the other two students in \(2^5\) ways. This gives \(3\cdot2^5\) counts, but the \(3\) cases in which one student receives all awards have each been subtracted twice.

By inclusion-exclusion, the desired number is \(3^5-3\cdot2^5+3=150.\)

Thus, the correct answer is B.

24 答案 A

当 \(n=24\) 时，阴影瓷砖总面积为 \(24^2s^2\)。大正方形一边有 \(24\) 块瓷砖和 \(25\) 条边框，所以边长为 \(24s+25d\)。

阴影部分占 \(64\%=\dfrac{16}{25}\)，所以 \[ \dfrac{24^2s^2}{(24s+25d)^2}=\dfrac{16}{25}\text{。} \] 取正平方根得 \(\dfrac{24s}{24s+25d}=\dfrac{4}{5}\)。

因此 \(120s=96s+100d\)，所以 \(24s=100d\)，\(\dfrac{d}{s}=\dfrac{6}{25}\)。

正确答案是 A。

English

For \(n=24,\) the shaded tile area is \(24^2s^2.\) Each side of the large square consists of \(24\) tiles and \(25\) borders, so its side length is \(24s+25d.\)

The shaded tiles cover \(64\%=\dfrac{16}{25}\) of the large square, so \[ \dfrac{24^2s^2}{(24s+25d)^2}=\dfrac{16}{25}. \] Taking positive square roots gives \(\dfrac{24s}{24s+25d}=\dfrac{4}{5}.\)

Thus \(120s=96s+100d,\) so \(24s=100d\) and \(\dfrac{d}{s}=\dfrac{6}{25}.\)

Thus, the correct answer is A.

25 答案 A

设三个正方形的边长分别为 \(s_1,s_2,s_3\)。沿着大矩形的宽度可得 \[ s_1+s_2+s_3=3322\text{。} \]

矩形 \(R_2\) 的高度为 \(s_1-s_2\)，所以大矩形的总高度为 \[ s_1-s_2+s_3=2020\text{。} \] 用宽度方程减去高度方程，得 \(2s_2=1302\)，所以 \(s_2=651\)。

正确答案是 A。

English

Let \(s_1,s_2,s_3\) be the side lengths of the three squares. Across the width of the large rectangle, \[ s_1+s_2+s_3=3322. \]

The height of \(R_2\) is \(s_1-s_2,\) so the total height is \[ s_1-s_2+s_3=2020. \] Subtracting the height equation from the width equation gives \(2s_2=1302,\) so \(s_2=651.\)

Thus, the correct answer is A.