2022 年 AMC 8 真题（P版）

1 答案 A

标志由 \(5\) 个全等的倾斜正方形组成。每个倾斜正方形的两条对角线长度均为 \(2\)，所以面积为 \(\frac{2\cdot2}{2}=2\) 平方英寸。

总面积为 \(5\cdot2=10\)。

正确答案是 A。

English

The logo is made from \(5\) congruent tilted squares. Each tilted square has both diagonals of length \(2\), so its area is \(\frac{2\cdot2}{2}=2\) square inches.

The total area is \(5\cdot2=10\).

Thus, the correct answer is A.

2 答案 D

根据定义，\[\begin{align*} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align*}\]

正确答案是 D。

English

Using the definitions, \[\begin{align*} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align*}\]

Thus, the correct answer is D.

3 答案 E

因为 \(a< b< c\) 且 \(abc=100\)，所以 \(a^3<100\)，因此 \(a\le4\)。同时 \(a\) 必须整除 \(100\)，故 \(a\in\{1,2,4\}\)。

若 \(a=1\)，则 \((b,c)\) 可为 \((2,50),(4,25),(5,20)\)。若 \(a=2\)，唯一可行的数对是 \((5,10)\)。若 \(a=4\)，则 \(bc=25\)，不存在整数 \(b\) 满足 \(4< b< c\)。

共有 \(4\) 种可能的三元组。

所以正确答案是 E。

English

Since \(a< b< c\) and \(abc=100\), we have \(a^3<100\), so \(a\le4\). Also \(a\) must divide \(100\), so \(a\in\{1,2,4\}\).

If \(a=1\), the pairs \((b,c)\) are \((2,50),(4,25),(5,20)\). If \(a=2\), the only possible pair is \((5,10)\). If \(a=4\), then \(bc=25\), and there is no integer \(b\) with \(4< b< c\).

There are \(4\) possible triples.

Thus, the correct answer is E.

4 答案 E

先把字母关于斜线 \(q\) 镜像，再把得到的图像关于水平线 \(p\) 镜像。最终图像位于 \(p\) 下方、竖直线右侧，方向与选项 E 相同。

正确答案是 E。

English

Reflecting over the diagonal line \(q\) changes the orientation of the letter as if the diagonal were a mirror. Reflecting that image over the horizontal line \(p\) then places the final letter below \(p\), to the right of the vertical line, with the orientation shown in choice E.

Thus, the correct answer is E.

5 答案 C

Bella 五年前 \(6\) 岁，所以现在 \(11\) 岁。

小猫五年前刚出生，所以现在 \(5\) 岁。

三者共有 \(3\) 个年龄，总和为 \(30\)，所以 Anna 的年龄是 \[30-5-11 = 14\]

Anna \(14\) 岁，Bella \(11\) 岁，所以 Anna 大 \(3\) 岁。

正确答案是 C。

English

If Bella was \(6\) five years ago, then she is \(11\) right now.

If the kitten was a newborn five years ago, then it is \(5\) right now.

Since the sum of all \(3\) ages is \(30,\) Anna’s age is \[30-5-11 = 14\]

Since Anna is \(14\) and Bella is \(11,\) she is \(3\) years older than Bella.

Thus, the correct answer is C.

6 答案 C

设最小数为 \(x\)，则最大数为 \(4x\)。因为中间数是最小数和最大数的平均数，所以 \[ \frac{x+4x}{2}=15\text{。}\]因此 \(5x=30\)，\(x=6\)。

正确答案是 C。

English

Let the smallest number be \(x\). Then the largest is \(4x\). Since the three numbers are equally spaced, the middle number is the average of the smallest and largest: \[ \frac{x+4x}{2}=15. \] Thus \(5x=30\), so \(x=6\).

Thus, the correct answer is C.

7 答案 B

歌曲大小为 \(4.2\cdot8000=33{,}600\) 千比特。以每秒 \(56\) 千比特下载，需要 \[ \frac{33{,}600}{56}=600 \] 秒，即 \(10\) 分钟。

正确答案是 B。

English

The song has \(4.2\cdot8000=33{,}600\) kilobits. At \(56\) kilobits per second, the download takes \[ \frac{33{,}600}{56}=600 \] seconds, which is \(10\) minutes.

Thus, the correct answer is B.

8 答案 B

从 \(3\) 到 \(20\) 的每个整数都既作为分母出现一次，也作为分子出现一次，所以全部相消。

相消后，分子只剩 \(1\) 和 \(2\)，分母只剩 \(21\) 和 \(22\)。

剩下的分数是 \(\dfrac{1 \cdot 2}{21 \cdot 22} \)，化简为 \(\dfrac{2}{462} = \dfrac{1}{231}\)。

正确答案是 B。

English

Since every integer from \(3\) to \(20\) occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only \(1\) and \(2\) left as numerators and \(21\) and \(22\) left as denominators.

The remaining fraction is \( \dfrac{1 \cdot 2}{21 \cdot 22} .\) This simplifies to \( \dfrac{2}{462} = \dfrac{1}{231} \)

Thus, the correct answer is B.

9 答案 B

当前水温和室温的差为 \(212-68 = 144 \)。

\(15\) 分钟包含三个 \(5\) 分钟，所以温差会减半 \(3\) 次。

也就是说温差乘以 \((\dfrac{1}{2})^3 = \dfrac{1}{8} \)，新的温差是 \(\dfrac{1}{8}\cdot144=18\)。水温为 \(68+18=86\)。

正确答案是 B。

English

The current difference is \(212-68 = 144 .\)

Since we have \(15\) minutes while halving every \(5\) minutes, we halve the difference \(3\) times.

This means the difference is multiplied by \((\dfrac{1}{2})^3 = \dfrac{1}{8} ,\) so our new difference is \(\dfrac{1}{8}\cdot144=18.\) This makes our final temperature \(68+18=86.\)

Thus, the correct answer is B.

10 答案 E

她以每小时 \(45\) 英里的速度开 \(2\) 小时，所以步道入口离家 \(90\) 英里。图像应从 \(0\) 英里上升到 \(90\) 英里，对应上午 \(8\) 点到 \(10\) 点，然后在徒步的 \(3\) 小时内保持水平。

她下午 \(1\) 点返程。返程路程为 \(90\) 英里，以每小时 \(60\) 英里的速度行驶需要 \(1.5\) 小时，所以下午 \(2{:}30\) 到家。符合的是选项 E。

正确答案是 E。

English

She drives \(45\) miles per hour for \(2\) hours, so the trail is \(90\) miles from her house. The graph must rise from \(0\) to \(90\) miles between \(8\) AM and \(10\) AM, then stay flat for the \(3\)-hour hike.

She starts home at \(1\) PM. Driving \(90\) miles at \(60\) miles per hour takes \(1.5\) hours, so she gets home at \(2{:}30\) PM. This matches choice E.

Thus, the correct answer is E.

11 答案 D

从一整段变成 \(10\) 段，需要咬 \(9\) 口。咬了 \(9\) 口，每口吃掉 \(3\) 英寸，所以共吃掉 \(27\) 英寸。

已经吃掉 \(27\) 英寸，最后还剩 \(17\) 英寸，因此原长为 \(27+17 = 44\) 英寸。

正确答案是 D。

English

Since there are \(10\) pieces, there were \(9\) locations where a bite was made. Since we have \(9\) bites and \(3\) inches are removed per bite, a total of \(27\) inches were removed.

With \(27\) inches removed and \(17\) inches remaining, we know we started with \(27+17 = 44 \) inches.

Thus, the correct answer is D.

12 答案 B

转盘 A 决定十位数字，转盘 B 决定个位数字。共有 \(4\cdot4=16\) 个等可能的两位数。

在 \(51\) 到 \(84\) 之间，可能的完全平方数是 \(64\) 和 \(81\)，且二者都可以由转盘得到。因此概率为 \[ \frac{2}{16}=\frac{1}{8}\text{。}\]

正确答案是 B。

English

Spinner A gives the tens digit and Spinner B gives the ones digit. There are \(4\cdot4=16\) equally likely two-digit numbers.

The possible perfect squares between \(51\) and \(84\) are \(64\) and \(81\), and both can occur. Thus the probability is \[ \frac{2}{16}=\frac{1}{8}. \]

Thus, the correct answer is B.

13 答案 D

设较小的数为 \(x\)，空格中的数为 \(c\)。那么另一个数是 \(2x+c\)，其中 \(x\) 和 \(c\) 都是正整数。

由两数之和得 \(x+(2x+c)=28\)，所以 \(c=28-3x\)。要使 \(c\) 为正数，需要 \(28-3x>0\)，即 \(x<\frac{28}{3}\)。

因此 \(x\) 可以是从 \(1\) 到 \(9\) 的任意整数，对应 \(9\) 个可能的空格值。

正确答案是 D。

English

Let the smaller number be \(x\), and let the blank be \(c\). Then the other number is \(2x+c\), where both \(x\) and \(c\) are positive integers.

The sum condition gives \(x+(2x+c)=28\), so \(c=28-3x\). For \(c\) to be positive, \(28-3x>0\), so \(x<\frac{28}{3}\).

Thus \(x\) can be any integer from \(1\) through \(9\), giving \(9\) possible values of the blank.

Thus, the correct answer is D.

14 答案 D

有 \(5\) 个 E 和 \(4\) 个非 E 字母 B、K、P、R。为了不让两个 E 相邻，唯一可能的排列形式是 \[ E\_E\_E\_E\_E\text{，}\]四个非 E 字母分别填入四个空隙。

B、K、P、R 可以用 \(4!=24\) 种方式排列。

正确答案是 D。

English

The word has \(5\) E’s and \(4\) other letters: B, K, P, and R. To keep no two E’s adjacent, the only possible pattern is \[ E\_E\_E\_E\_E, \] with the four non-E letters in the four gaps.

The letters B, K, P, and R can be arranged in those gaps in \(4!=24\) ways.

Thus, the correct answer is D.

15 答案 C

对于每种重量，只有价格最低的那个点才可能给出最低的每盎司价格。

在 \(1\) 盎司时，最低价格超过 \(1\) 美元，所以每盎司价格超过 \(1\) 美元。

在 \(2\) 盎司时，最低价格为 \(2\) 美元，所以每盎司价格为 \(1\) 美元。

在 \(3\) 盎司时，最低价格约为 \(2.5\) 美元，所以每盎司价格约为 \(\frac{2.5}{3}\)，小于 \(1\) 美元。

在 \(4\) 盎司时，最低价格接近 \(3.9\) 美元，每盎司约为 \(\frac{3.9}{4}\)，仍高于 \(3\) 盎司选项。

在 \(5\) 盎司时，最低价格接近 \(4.5\) 美元，每盎司约为 \(\frac{4.5}{5}\)，也高于 \(3\) 盎司选项。

所以每盎司价格最低的是 \(3\) 盎司。

正确答案是 C。

English

For each weight, only the lowest-price point at that weight can give the lowest price per ounce.

At \(1\) ounce, the lowest price is more than \(1\) dollar, so the price per ounce is more than \(1\).

At \(2\) ounces, the lowest price is \(2\) dollars, so the price per ounce is \(1\).

At \(3\) ounces, the lowest price is about \(2.5\) dollars, so the price per ounce is about \(\frac{2.5}{3}\), less than \(1\).

At \(4\) ounces, the lowest price is close to \(3.9\) dollars, so the price per ounce is close to \(\frac{3.9}{4}\), still larger than the \(3\)-ounce option.

At \(5\) ounces, the lowest price is close to \(4.5\) dollars, so the price per ounce is close to \(\frac{4.5}{5}\), also larger than the \(3\)-ounce option.

The price per ounce is lowest at \(3\) ounces.

Thus, the correct answer is C.

16 答案 B

设四个数依次为 \(a,b,c,d\)。

因为 \(a\) 和 \(b\) 的平均数是 \(21\)，所以它们的和是 \(21\cdot2 = 42\)。
因为 \(c\) 和 \(d\) 的平均数是 \(30\)，所以它们的和是 \(30\cdot2 = 60\)。
也就是 \(a+b = 42\)，\(c+d = 60 \)，从而 \[\begin{align*} a+b+c+d &= 42 + 60\\ &= 102 \end{align*}\]

现在中间两个数 \(b\) 和 \(c\) 的平均数是 \(26\)，所以它们的和是 \(26\cdot2=52\)。因此 \(b+c = 52\)。

用所有数的总和减去这个结果，得 \(a+d =50\)。
所以第一个和最后一个数的平均数是 \(\dfrac{a+d}{2}=25\)，答案为 \(25\)。

正确答案是 B。

English

Let the numbers be \(a,b,c,d\) in order.

Since the average of \(a\) and \(b\) is \(21,\) we know their sum is \(21\cdot2 = 42.\)
Since the average of \(c\) and \(d\) is \(30,\) we know their sum is \(30\cdot2 = 60.\)
Since \(a+b = 42\) and \(c+d = 60 ,\) we know \[\begin{align*} a+b+c+d &= 42 + 60\\ &= 102 \end{align*}\]

Now, with the average of \(b\) and \(c\) being \(26,\) we know their sum is \(26\cdot2=52.\) This means \(b+c = 52.\)

Subtracting this result from the sum of all the terms yields \(a+d =50.\)
Since \(\dfrac{a+d}{2}=25,\) our answer is \(25.\)

Thus, the correct answer is B.

17 答案 B

对于每个满足 \(n\ge10\) 的偶数，乘积 \(n!!\) 都含有因子 \(10\)，所以该项的个位数字是 \(0\)。只有前四项会影响个位数字。

这四项的和为 \[ \begin{gathered} 2!!+4!!+6!!+8!!\\ =2+8+48+384\\ =442 \end{gathered}\text{。}\]因此整个和的个位数字为 \(2\)。

正确答案是 B。

English

For every even \(n\ge10,\) the product \(n!!\) contains a factor of \(10,\) so that term has units digit \(0.\) Only the first four terms can affect the units digit.

They sum to \[ \begin{gathered} 2!!+4!!+6!!+8!!\\ =2+8+48+384\\ =442. \end{gathered} \] Therefore the full sum has units digit \(2.\)

Thus, the correct answer is B.

18 答案 C

四个中点形成一个平行四边形。它的一条水平底边长度为 \(2-(-3)=5\)，高度为 \(4\)，面积为 \(5\cdot4=20\)。

连接任意矩形四边中点形成的平行四边形面积等于原矩形面积的一半，所以原矩形面积为 \(2\cdot20=40\)。

正确答案是 C。

English

The four given midpoints form a parallelogram. Its horizontal base has length \(2-(-3)=5\), and its height is \(4\), so its area is \(5\cdot4=20\).

For any rectangle, the parallelogram formed by joining the side midpoints has half the area of the rectangle. Therefore the rectangle’s area is \(2\cdot20=40\).

Thus, the correct answer is C.

19 答案 C

所有成绩都是 \(5\) 的倍数，加上 \(5\) 分后仍然如此。如果第 \(10\) 和第 \(11\) 个成绩的平均数是 \(85\)，但两者不都是 \(85\)，那么第 \(11\) 个成绩至少为 \(90\)。这就要求至少有 \(10\) 个成绩不低于 \(90\)。

可是重新评分后，只有原来不低于 \(85\) 的成绩才可能达到 \(90\)。点图中这样的成绩只有 \(7\) 个，所以中间两个成绩必须都是 \(85\)。

点图中原来有 \(7\) 个不低于 \(85\) 的成绩。要使第 \(10\) 和第 \(11\) 个成绩都为 \(85\)，至少要有 \(11\) 个成绩不低于 \(85\)。把四个 \(80\) 分提高到 \(85\) 分既是必要的，也是充分的。

所以正确答案是 C。

English

All scores are multiples of \(5,\) and adding \(5\) preserves that fact. If the \(10\)th and \(11\)th scores average \(85\) but are not both \(85,\) then the \(11\)th score must be at least \(90.\) That would require at least \(10\) scores of \(90\) or more.

However, even after regrading, only the original scores of \(85\) or more can reach \(90.\) The plot has only \(7\) such scores, so the two middle scores must both be \(85.\)

The plot initially has \(7\) scores of at least \(85.\) To make the \(10\)th and \(11\)th scores both \(85,\) there must be at least \(11\) scores of at least \(85.\) Raising four of the \(80\)’s to \(85\) is both necessary and sufficient.

Thus, the correct answer is C.

20 答案 D

由顶行可知每行、每列的和都必须为 \(12\)。

在第一列中，\(x\) 上方的缺失数为 \(14-x\)。在底行中，\(x\) 右边的缺失数为 \(4-x\)。中间行剩下的缺失数为 \(x-1\)。

因为 \(x\) 大于另外三个缺失数，需要 \(x>14-x\)、\(x>4-x\)、\(x> x-1\)。最强条件是 \(x>7\)，所以最小整数为 \(8\)。

正确答案是 D。

English

Adding the numbers in the top row shows that every row and column must have sum \(12\).

In the first column, the missing number above \(x\) is \(14-x\). In the bottom row, the missing number to the right of \(x\) is \(4-x\). In the middle row, the remaining missing number is \(x-1\).

Since \(x\) is larger than the other missing numbers, we need \(x>14-x\), \(x>4-x\), and \(x> x-1\). The strongest condition is \(x>7\), so the smallest possible integer value is \(8\).

Thus, the correct answer is D.

21 答案 C

Steph 总共进了 \(25\) 球，出手 \(30\) 次。Candace 也出手了 \(12+18=30\) 次，且两人的总命中率相同，所以 Candace 也进了 \(25\) 球。

设 Candace 上半场进 \(f\) 球，下半场进 \(s\) 球，则 \(f+s=25\)。

Steph 上半场命中率为 \(\frac{15}{20}=\frac34\)，所以 \(\frac{f}{12}<\frac34\)，得到 \(f<9\)。Steph 下半场命中率为 \(1\)，所以 \(s<18\)。

因此 \(f\le8\) 且 \(s\le17\)。又 \(f+s=25\)，唯一可能是 \(f=8\)、\(s=17\)，所以 \(s-f=9\)。

正确答案是 C。

English

Steph made \(25\) baskets in \(30\) attempts. Candace also took \(12+18=30\) attempts, and their overall percentages were the same, so Candace also made \(25\) baskets.

Let \(f\) be Candace’s first-half baskets and \(s\) be her second-half baskets. Then \(f+s=25\).

Steph’s first-half percentage was \(\frac{15}{20}=\frac34\), so \(\frac{f}{12}<\frac34\), giving \(f<9\). Steph’s second-half percentage was \(1\), so \(s<18\).

Thus \(f\le8\) and \(s\le17\). Since \(f+s=25\), the only possibility is \(f=8\) and \(s=17\), so \(s-f=9\).

Thus, the correct answer is C.

22 答案 A

公交车需要 \(2\) 分钟开到下一站，再等 \(1\) 分钟，所以每 \(3\) 分钟完成一个站距周期。

Zia 只在到达公交站时做决定，也就是每 \(5\) 分钟一次。以公交车出发站为第零站，则 Zia 从第 \(3\) 站出发。

\(5\) 分钟后，Zia 在第 \(4\) 站，公交车在第 \(2\) 站等待，她继续走。\(10\) 分钟后，Zia 在第 \(5\) 站，公交车在第 \(3\) 站和第 \(4\) 站之间，她继续走。\(15\) 分钟后，Zia 在第 \(6\) 站，公交车在第 \(5\) 站，正好是前一站，所以她停下等待。

公交车再用 \(2\) 分钟到达她所在的站，所以她在 \(17\) 分钟后上车。

正确答案是 A。

English

The bus takes \(2\) minutes to drive to the next stop and then waits \(1\) minute, so it moves through one-stop cycles every \(3\) minutes.

Zia makes a decision only when she reaches a stop, every \(5\) minutes. Measure stops from the bus’s starting stop. Zia starts at stop \(3\).

After \(5\) minutes, Zia is at stop \(4\), while the bus is waiting at stop \(2\), so she keeps walking. After \(10\) minutes, Zia is at stop \(5\), while the bus is between stops \(3\) and \(4\), so she keeps walking. After \(15\) minutes, Zia is at stop \(6\), while the bus is at stop \(5\), the previous stop, so she waits.

The bus then takes \(2\) more minutes to reach her stop, so she boards after \(17\) minutes.

Thus, the correct answer is A.

23 答案 D

设三角形个数为 \(k\)。要同时有三角形线和圆形线，必须有 \(3\le k\le6\)。

若 \(k=3\)，三角形必须成一条线。行或列有效，因为剩下的六个圆形中仍有完整的一行或一列；对角线无效，因为其余位置没有完整的圆形线。因此共有 \(6\) 种 \(k=3\) 的配置。由对称性，也有 \(6\) 种 \(k=6\) 的配置。

若 \(k=4\)，先选一条三角形线。若这条线是行或列，额外的三角形有 \(6\) 个位置可选，每一种都可行，因为总有一条平行的行或列全是圆形，共 \(6\cdot6=36\) 种；若这条线是对角线，则无论额外的三角形放在哪里，都不能留下完整的圆形线。由对称性，\(k=5\) 也有 \(36\) 种。

总数为 \(6+36+36+6=84\)。

正确答案是 D。

English

Let \(k\) be the number of triangles. To have both a triangle line and a circle line, we need \(3\le k\le6\).

If \(k=3\), the three triangles must form a line. A row or column works, because the remaining six circles contain a full circle row or column. A diagonal does not work, because its complement contains no full line of circles. Thus there are \(6\) configurations for \(k=3\). By symmetry, there are also \(6\) configurations for \(k=6\).

If \(k=4\), choose the triangle line and then the extra triangle. If the line is a row or column, all \(6\) possible extra positions work, because one parallel row or column remains all circles. This gives \(6\cdot6=36\) configurations. If the triangle line is a diagonal, no extra position leaves a full circle line. By symmetry, \(k=5\) also gives \(36\) configurations.

The total number of configurations is \(6+36+36+6=84\).

Thus, the correct answer is D.

24 答案 C

折叠后匹配边长给出 \(GF=EF=8\)。因为 \(GFCB\) 是矩形，所以 \(BC=8\)。折叠时 \(AB\) 与 \(BC\) 对应，所以 \(AB=8\)。在矩形 \(HJBA\) 中，\(HJ=8\)，且 \(BJ\) 等于 \(AH=8\)。

因为 \(GH=14\)，所以 \(GJ=14-8=6\)。一个三角形底面是直角三角形 \(BJG\)，面积为 \[ \frac{6\cdot8}{2}=24\text{。}\]棱柱长度为 \(GF=8\)，所以体积为 \(24\cdot8=192\)。

正确答案是 C。

English

When the net is folded, the matching side lengths give \(GF=EF=8\). Since \(GFCB\) is a rectangle, \(BC=8\). The fold identifies \(AB\) with \(BC\), so \(AB=8\). In rectangle \(HJBA\), this gives \(HJ=8\), and the opposite side \(BJ\) equals \(AH=8\).

Since \(GH=14\), we have \(GJ=14-8=6\). Thus one triangular base of the prism is right triangle \(BJG\), with area \[ \frac{6\cdot8}{2}=24. \] The prism length is \(GF=8\), so the volume is \(24\cdot8=192\).

Thus, the correct answer is C.

25 答案 E

设 \(p_n\) 为跳 \(n\) 次后在起始叶子上的概率，\(p_0=1\)。

若蟋蟀在起点，下一跳一定离开；若不在起点，下一跳有 \(3\) 个等可能选择，其中一个会回到起点。因此 \[ p_{n+1}=\frac{1-p_n}{3}\text{。}\]

逐次计算得 \[ \begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27} \end{gathered}\text{。}\]

正确答案是 E。

English

Let \(p_n\) be the probability that the cricket is on its starting leaf after \(n\) hops. We have \(p_0=1\).

If the cricket is on the starting leaf, the next hop must leave it. If the cricket is not on the starting leaf, exactly one of the \(3\) possible hops returns to the start. Therefore \[ p_{n+1}=\frac{1-p_n}{3}. \]

Thus \[ \begin{gathered} p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad \\ p_4=\frac{1-\frac29}{3}=\frac{7}{27}. \end{gathered} \]

Thus, the correct answer is E.