2019 年 AMC 8 真题（P版）

1 答案 D

买 \(6\) 个三明治需要 \(6 \times 4.5 = 27\) 美元，已经不能再买一个三明治。

剩下 \(30 - 27 = 3\) 美元，可以买 \(3\) 杯汽水。商品总数为 \(6 + 3 = 9\)。

正确答案是 D。

English

If they buy \(6\) sandwiches, they would spend \(6 \times 4.5 = 27\) dollars. This means that they would not be able to buy any more sandwiches.

They would then have \(30 - 27 = 3\) dollars left. With this, they could buy \(3\) sodas. They would therefore buy a total of \(6 + 3 = 9\) items.

Thus, the correct answer is D.

2 答案 E

从图中可见，小矩形的长边等于两个短边，所以长边为 \(2 \cdot 5 = 10\) 英尺。

大矩形的一边为 \(BC = 10\) 英尺，另一边为 \(DC = 10 + 5 = 15\) 英尺。面积为 \(10 \cdot 15 = 150\) 平方英尺。

正确答案是 E。

English

From the figure, we can see that the longer side has the same length as two of the shorter sides. This makes it \(2 \cdot 5 = 10\) feet long.

This tells us that \(BC = 10\) feet and \(DC = 10 + 5 = 15\) feet. Therefore, the area is \(10 \cdot 15 = 150\) square feet.

Thus, the correct answer is E.

3 答案 E

先把三个分数改写为同整数部分加一个分数：\[\begin{align*} \dfrac{15}{11} &= 1 + \dfrac{4}{11} \\ \dfrac{17}{13} &= 1 + \dfrac{4}{13} \\ \dfrac{19}{15} &= 1 + \dfrac{4}{15} \end{align*}\text{。}\]

分子相同的正分数，分母越大，分数越小。

因此正确顺序是 \[ \dfrac{19}{15} \lt \dfrac{17}{13} \lt \dfrac{15}{11}\text{。}\]

正确答案是 E。

English

We can rewrite the fraction as follows: \[\begin{align*} \dfrac{15}{11} &= 1 + \dfrac{4}{11} \\ \dfrac{17}{13} &= 1 + \dfrac{4}{13} \\ \dfrac{19}{15} &= 1 + \dfrac{4}{15}. \end{align*}\]

Recall that if two fractions have the same numerator, then the fraction with the larger denominator is smaller.

Using this fact, we can see that the correct ordering is \[ \dfrac{19}{15} \lt \dfrac{17}{13} \lt \dfrac{15}{11}. \]

Thus, the correct answer is E.

4 答案 D

可以把菱形分成 \(4\) 个如图所示的直角三角形。

我们知道这是直角三角形，因为菱形的两条对角线互相垂直；斜边长为 \(52 \div 4 = 13\)。

用勾股定理，另一条直角边长为 \[ \sqrt{13^2 - 12^2} = \sqrt{25} = 5\text{。}\]

因此菱形面积为 \[ 4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120\text{。}\]正确答案是 D。

English

We can split the rhombus up into \(4\) triangles, each of which looks like this.

We know that this is a right triangle, since the diagonals of a rhombus are perpendicular, and we know the hypotenuse is \(52 \div 4 = 13.\)

Using the Pythagorean theorem, we get the other leg to be \[ \sqrt{13^2 - 12^2} = \sqrt{25} = 5. \]

This means that the area of the rhombus is \[ 4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120. \] Thus, the correct answer is D.

5 答案 B

兔子对应的那条路线，就是中间有一段水平线的那条，这段水平线代表兔子打盹的那段时间。

我们还知道，兔子跑到终点所用的时间比乌龟长。

这说明兔子那条路线的终点应该有更大的 \(x\) 值。符合这一点的正是图 \(B\)。

正确答案是 B。

English

The hare’s path is the one with the horizontal line in the middle, which represents the period when the hare took a nap.

We also know that the time it takes the hare to reach the finish line is longer than the time it took the tortoise.

This means that the end of the hare’s path should have a larger \(x\)-value. This is exactly graph \(B.\)

Thus, the correct answer is B.

6 答案 C

正方形的对称轴只有两条对角线和两条连接对边中点的直线。

每条对称轴上有 \(8\) 个格点（不包括 \(P\)）。四条对称轴仅在 \(P\) 相交，而该点不能选取，所以共有 \(4 \cdot 8 = 32\) 个不同的点可作为 \(Q\)。

所求概率为 \(\dfrac{32}{80} = \dfrac{2}{5}\)。

正确答案是 C。

English

Note the only lines of symmetry for a square are the two diagonals and the two lines connecting opposite midpoints.

On each symmetry line there are \(8\) grid points other than \(P.\) The four lines intersect only at \(P,\) which is not eligible, so these give \(4 \cdot 8 = 32\) distinct choices for \(Q.\)

The probability is therefore \(\dfrac{32}{80} = \dfrac{2}{5}.\)

Thus, the correct answer is C.

7 答案 A

要让其中一次成绩尽可能低，就让另一次成绩尽可能高。假设 Shauna 第四次测试得 \(100\) 分。

这 \(4\) 次测试的总分为 \[ 76 + 94 + 87 + 100 = 357\text{。}\]若五次平均分为 \(81\)，总分必须为 \(5 \cdot 81 = 405\)，所以最后一次需要 \(405 - 357 = 48\) 分。

正确答案是 A。

English

To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a \(100\) on her fourth test.

The sum of the \(4\) tests is then \[ 76 + 94 + 87 + 100 = 357. \] For an average of \(81,\) Shauna’s test scores must add to \(5 \cdot 81 = 405.\) This means that she needs to get a \(405 - 357 = 48\) on her last test.

Thus, the correct answer is A.

8 答案 E

假设 Gilda 原来有 \(100\) 颗弹珠。给 Pedro \(20\) 颗后，还剩 \(100 - 20 = 80\) 颗。

接着她给 Ebony \(80 \times 0.1 = 8\) 颗，还剩 \(80 - 8 = 72\) 颗。

最后她给 Jimmy \(72 \times 0.25 = 18\) 颗，还剩 \(72 - 18 = 54\) 颗，也就是原来的 \(54\%\)。

正确答案是 E。

English

Assume that Gilda starts off with \(100\) marbles. After giving \(20\) marbles to Pedro, she has \(100 - 20 = 80\) marbles left.

She then gives \(80 \times 0.1 = 8\) marbles to Ebony, with her having \(80 - 8 = 72\) marbles left.

Finally, Gilda gives \(72 \times 0.25 = 18\) marbles to Jimmy. She has a total of \(72 - 18 = 54\) marbles left, which is \(54\%\) of her original total.

Thus, the correct answer is E.

9 答案 B

Felicia 罐头的直径是 Alex 罐头的两倍，这说明她的罐头半径也是两倍。

圆的面积公式是 \(\pi r^2\)。半径加倍时，底面积变为原来的四倍。

因此 Felicia 罐头的底面积是 Alex 罐头的 \(4\) 倍。

另一方面，Alex 罐头的高度是 Felicia 罐头的两倍。圆柱体积等于底面积乘高。

所以 Felicia 罐头的体积是 Alex 的 \(4 \div 2 = 2\) 倍。

正确答案是 B。

English

Felicia’s diameter is twice as much as Alex’s which means that her radius is also twice as much.

Recall that the formula for the area of a circle is \(\pi r^2.\) If the radius is doubled, then the area is quadrupled.

This means that the base of Felicia’s can has \(4\) times the area as the base of Alex’s can.

We also see that Alex’s can is twice as tall as Felicia’s can. The formula for the volume of a can is base times height.

Using this formula, we see that Felicia’s can will have twice the volume of Alex’s can since \(4 \div 2 = 2.\)

Thus, the correct answer is B.

10 答案 B

修正后多出 \(5\) 名参与者，所以平均数增加 \(5 \div 5 = 1\)。

原来的中位数是 \(20\)。把 \(16\) 换成 \(21\) 后，中位数变为 \(21\)。

所以中位数也增加 \(1\)。

正确答案是 B。

English

There are \(5\) more participants, which means the mean is increased by \(5 \div 5 = 1.\)

Right now, we can see that the median is \(20.\) If \(16\) gets replaced with \(21,\) then the median becomes \(21.\)

This shows that the median is also increased by \(1.\)

Thus, the correct answer is B.

11 答案 D

两门课都上的人数为 \[ 70 + 54 - 93 = 31\text{。}\] 用上数学课的人数减去这个数，就得到只上数学课的人数。

所求人数为 \(70 - 31 = 39\)。

正确答案是 D。

English

The number of kids that are taking both classes is \[ 70 + 54 - 93 = 31. \] Subtracting this from the number of kids taking a math class will give us the number of kids taking only a math class.

The desired number is \(70 - 31 = 39.\)

Thus, the correct answer is D.

12 答案 A

比较第一和第三个视图，可知棕色与紫色相对。

比较第一和第二个视图，可知绿色与白色相对。

剩下的颜色是水蓝色和红色，所以水蓝色与红色相对。

正确答案是 A。

English

Using the first and third cubes, we can see that brown is opposite purple.

Similarly, with the first and second cubes, we see that green and white are opposites.

The only color left to pair with aqua is red, so aqua and red are opposite faces.

Thus, the correct answer is A.

13 答案 A

\(2\) 位回文数的十位和个位相同，所以都是 \(11\) 的倍数。

如果一个 \(3\) 位数是 \(3\) 个不同的 \(2\) 位回文数之和，那么它也必须是 \(11\) 的倍数。

最小的 \(3\) 位 \(11\) 的倍数且不是回文数的是 \(110\)。它可以写成 \(11 + 22 + 77\)。

因此 \(N = 110\)，数字和为 \(1 + 1 + 0 = 2\)。

正确答案是 A。

English

Note that \(2\)-digit palindromes have the tens and units digits the same. This means that they are all multiples of \(11.\)

If a \(3\)-digit number is the sum of \(3\) \(2\)-digit palindromes, then it itself must also be a multiple of \(11.\)

The smallest \(3\)-digit multiple of \(11\) that is not a palindrome is \(110.\) This can be achieved by adding \(11 + 22 + 77.\)

Therefore, \(N = 110.\) The sum of its digits is \(1 + 1 + 0 = 2.\)

Thus, the correct answer is A.

14 答案 C

逐个检查选项即可。

每隔 \(10\) 天兑换一次，星期几会前进 \(3\) 天。

若第一次是星期 \(x\)，六次兑换的星期依次为 \(x, x+3, x+6, x+2, x+5\) 和 \(x+1\)（模 \(7\)）。

唯一没出现的是 \(x+4\)，所以星期日必须在第一次兑换日后四天。星期日前四天是星期三。

正确答案是 C。

English

Let us go through the answer choices.

Each redemption is \(10\) days later, which advances the weekday by \(3\) days.

Starting on day \(x,\) the six redemption days are \(x, x+3, x+6, x+2, x+5,\) and \(x+1\) modulo \(7.\)

The only missed weekday is \(x+4,\) so Sunday must be four days after the first redemption day. Four days before Sunday is Wednesday.

Thus, the correct answer is C.

15 答案 B

概率为 \(\dfrac{2}{5}\)，表示戴帽子的人中有 \(\dfrac{2}{5}\) 也戴太阳镜。

所以两者都戴的人数为 \(35 \cdot \dfrac{2}{5} = 14\)。

从戴太阳镜的五十人中随机选，选到也戴帽子的概率为 \(\dfrac{14}{50} = \dfrac{7}{25}\)。

正确答案是 B。

English

If the probability is \(\dfrac{2}{5},\) that means \(\dfrac{2}{5}\) of the people wearing caps are also wearing sunglasses.

This means that \(35 \cdot \dfrac{2}{5} = 14\) people are wearing both caps and sunglasses.

The probability of a person wearing sunglasses also wearing a cap is \(\dfrac{14}{50} = \dfrac{7}{25}.\)

Thus, the correct answer is B.

16 答案 D

前 \(15\) 英里用时 \(\dfrac{15}{30} = \dfrac{1}{2}\) 小时。

设还需开 \(x\) 英里，则这段用时为 \(\dfrac{x}{55}\) 小时。

整趟路程为 \(15 + x\) 英里。平均速度为 \(50\) 英里每小时，所以总用时也等于 \(\dfrac{x + 15}{50}\) 小时。

令两种总用时相等，得到 \[ \dfrac{1}{2} + \dfrac{x}{55} = \dfrac{x + 15}{50}\text{。}\]

交叉相乘并化简得 \[ 15 + x = 25 + \dfrac{10x}{11} \Rightarrow x = 110\text{。}\]

正确答案是 D。

English

For the first \(15\) miles, Qiang drove for \(\dfrac{15}{30} = \dfrac{1}{2}\) an hour.

Let \(x\) be the distance that Qiang must drive to satisfy the condition. It will take him \(\dfrac{x}{55}\) hours to drive this distance.

The total trip is now \(15 + x\) miles. We know the average speed is \(50\) miles per hour, so this will take him \(\dfrac{x + 15}{50}\) hours.

Setting the two times equal to each other, we get \[ \dfrac{1}{2} + \dfrac{x}{55} = \dfrac{x + 15}{50}. \]

Cross-multiplying and simplifying yields \[ 15 + x = 25 + \dfrac{10x}{11} \Rightarrow x = 110. \]

Thus, the correct answer is D.

17 答案 B

可以把所有因子重新组合为 \[\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}\text{。}\]

由这个形式可见，中间各项全部相消，只剩 \[ \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}\text{。}\]

正确答案是 B。

English

We can regroup all the factors as follows. \[\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}\]

From this representation, we can see that all the middle terms cancel leaving only \[ \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}. \]

Thus, the correct answer is B.

18 答案 C

和为偶数有两种情况：两个都是偶数，或两个都是奇数。

单个骰子掷出偶数的概率为 \(\dfrac{2}{6} = \dfrac{1}{3}\)，掷出奇数的概率为 \(\dfrac{4}{6} = \dfrac{2}{3}\)。

因此所求概率为 \[ \dfrac{1}{3} \cdot \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{5}{9}\text{。}\]

正确答案是 C。

English

The only two ways that the sum can be even is if both rolls are even or both are odd.

The probability that a roll is even is \(\dfrac{2}{6} = \dfrac{1}{3},\) and the probability that it is odd is \(\dfrac{4}{6} = \dfrac{2}{3}.\)

Therefore, the desired probability is \[ \dfrac{1}{3} \cdot \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{5}{9}. \]

Thus, the correct answer is C.

19 答案 C

每支前三名队伍与后三名队伍比赛 \(6\) 场，所以从这些比赛中最多得到 \(6 \cdot 3 = 18\) 分。

前三名队伍之间有 \(3\) 对，因此共有 \(6\) 场比赛。每场比赛总共最多产生 \(3\) 分，所以这六场比赛给三支队伍的总分至多为 \(18\) 分。若三队最终同分，每队从中至多得到 \(\frac{18}{3}=6\) 分。

因此每支前三名队伍最多得到 \(18+6=24\) 分。这个上界可以达到：让每支前三名队伍赢下所有对后三名队伍的比赛，并让每对前三名队伍在两场交锋中各赢一场。这样每支前三名队伍都得到 \(18+6=24\) 分。

所以正确答案是 C。

English

Each top team plays \(6\) games against the bottom three teams, so it can earn at most \(6 \cdot 3 = 18\) points from those games.

Among the top three teams, there are \(3\) pairs and hence \(6\) games. Each game awards at most \(3\) points total, so these games award at most \(18\) points among the three teams. If their final scores are equal, each can therefore receive at most \(\frac{18}{3}=6\) of those points.

Thus each top team has at most \(18+6=24\) points. This bound is attainable: let every top team win all its games against the bottom teams, and for each pair of top teams let each team win one of their two games. Then each top team earns \(18+6=24\) points.

Thus, the correct answer is C.

20 答案 D

回顾平方差公式：\[a^2 - b^2 = (a + b)(a - b)\text{。}\]于是可以作如下变形：\[ \begin{aligned} (x^2 - 5)^2 - 4^2 &= 0 \\ (x^2 - 1)(x^2 - 9) &= 0 \end{aligned} \]

因此可以看出，共有 \(4\) 个 \(x\) 的可能取值。

正确答案是 D。

English

Recall that \[a^2 - b^2 = (a + b)(a - b).\] We can do the following rearrangement. \[ \begin{aligned} (x^2 - 5)^2 - 4^2 &= 0 \\ (x^2 - 1)(x^2 - 9) &= 0 \end{aligned} \]

From this we can see that there are \(4\) possible values for \(x.\)

Thus, the correct answer is D.

21 答案 E

先求三角形的顶点。直线 \(y = 5\) 与另外两条直线的交点为 \((4, 5)\) 和 \((-4, 5)\)。

求第三个交点时，令两条斜线相等，得到 \[ 1 + x = 1 - x \Rightarrow x = 0\text{。}\]

所以第三个顶点是 \((0, 1)\)。

前两个交点关于 \(y\) 轴对称，因此这个三角形是等腰三角形。

底边长为 \(2 \cdot 4 = 8\)，高为 \(5 - 1 = 4\)。三角形面积为 \[ \dfrac{1}{2} \cdot 8 \cdot 4 = 16\text{。}\]

正确答案是 E。

English

First, let us find the vertices of the triangle. The intersections between \(y = 5\) and the other two lines are \((4, 5)\) and \((-4, 5)\).

To find the third intersection, we can equate the two equations to get \[ 1 + x = 1 - x \Rightarrow x = 0. \]

This gives us the point \((0, 1).\)

Note that the first two intersection points are mirrored across the \(y\)-axis. This tells us that the triangle is isosceles.

The base is therefore \(2 \cdot 4 = 8,\) and the height is \(5 - 1 = 4.\) The area of the triangle is therefore \[ \dfrac{1}{2} \cdot 8 \cdot 4 = 16. \]

Thus, the correct answer is E.

22 答案 E

设这个百分比写成小数为 \(x\)。涨价会把原价乘以 \(1 + x\)。

之后降价会把新价格乘以 \(1 - x\)。因此最终价格是原价乘以 \[ (1 + x)(1 - x) = 1 - x^2 = 0.84\text{。}\]

于是 \[ x^2 = 0.16 \Rightarrow x = 0.4\text{。}\]

所以这个百分比是 \(40\%\)。

正确答案是 E。

English

Let \(x\) be the percent converted to a real number. Then the percent increase would multiply the original by \(1 + x.\)

The percent decrease would multiply the new price by \(1 - x.\) The final price will then be the original price multiplied by \[ (1 + x)(1 - x) = 1 - x^2 = 0.84. \]

This tells us that \[ x^2 = 0.16 \Rightarrow x = 0.4. \]

This tells us that the percent would be \(40\%.\)

Thus, the correct answer is E.

23 答案 B

设总得分为 \(x\)，所求答案为 \(y\)。

由题意可得 \[ \dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x\text{。}\]

化简得 \[ y + 15 = \dfrac{13x}{28}\text{。}\]

我们知道 \(y \leq 14\)，因为其余 \(7\) 名队员每人的得分不超过 \(2\) 分。

此外，\(x\) 必须是 \(28\) 的倍数。若 \(x = 28\)，则 \[ y + 15 = 13 \Rightarrow y = -2\text{，}\]不合要求。

若 \(x = 2 \cdot 28 = 56\)，则 \[ y + 15 = 26 \Rightarrow y = 11\text{，}\]符合要求。

下一个可能的总分是 \(x = 84\)，这会得到 \(y = 24 > 14\)。更大的 \(28\) 的倍数会使 \(y\) 更大，所以 \(y = 11\) 是唯一可能值。

所以正确答案是 B。

English

Let \(x\) be the total number of points and \(y\) be the desired answer.

Then from the problem statement we get that \[ \dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x. \]

Simplifying yields \[ y + 15 = \dfrac{13x}{28}. \]

We know that \(y \leq 14\) since none of the \(7\) team members scored more than \(2\) points.

We also know that \(x\) must be a multiple of \(28.\) If \(x = 28,\) then we get that \[ y + 15 = 13 \Rightarrow y = -2, \] which is not allowed.

If \(x = 2 \cdot 28 = 56,\) then we have that \[ y + 15 = 26 \Rightarrow y = 11, \] which works.

The next possible total is \(x = 84,\) which would give \(y = 24 > 14.\) Every larger multiple of \(28\) makes \(y\) still larger, so \(y = 11\) is the only possible value.

Thus, the correct answer is B.

24 答案 B

因为 \(AD : DC = 1 : 2\)，三角形 \(ABD\) 与 \(DBC\) 的面积比也是 \(1 : 2\)。所以 \([ABD] = 120\)，\([DBC] = 240\)。

因为 \(E\) 是 \(BD\) 的中点，三角形 \(ABE\) 和 \(AED\) 的面积各为 \(60\)。设 \(x = [EBF]\)。同样有 \([DEF] = x\)，因为 \(BE = ED\)，且两个三角形的第三个顶点都在直线 \(AF\) 上。

线段 \(DF\) 分割 \(\triangle DBC\)，所以 \([DFC] = 240 - 2x\)。另外，\(\triangle ADF\) 和 \(\triangle DFC\) 的底边 \(AD\) 与 \(DC\) 在同一直线上，面积比为 \(1 : 2\)。

因此 \[ \dfrac{60 + x}{240 - 2x} = \dfrac{1}{2}\text{。}\]解得 \(120 + 2x = 240 - 2x\)，所以 \(x = 30\)。

正确答案是 B。

English

Since \(AD : DC = 1 : 2,\) triangles \(ABD\) and \(DBC\) have areas in the ratio \(1 : 2.\) Thus \([ABD] = 120\) and \([DBC] = 240.\)

Because \(E\) is the midpoint of \(BD,\) triangles \(ABE\) and \(AED\) each have area \(60.\) Let \(x = [EBF].\) Then \([DEF] = x\) as well, since \(BE = ED\) and both triangles have their third vertex on line \(AF.\)

Segment \(DF\) splits \(\triangle DBC,\) so \([DFC] = 240 - 2x.\) Also, \(\triangle ADF\) and \(\triangle DFC\) have bases \(AD\) and \(DC\) on the same line, so their areas are in the ratio \(1 : 2.\)

Therefore \[ \dfrac{60 + x}{240 - 2x} = \dfrac{1}{2}. \] Solving gives \(120 + 2x = 240 - 2x,\) so \(x = 30.\)

Thus, the correct answer is B.

25 答案 C

先给三个人每人 \(2\) 个苹果，剩下 \(24 - 3 \cdot 2 = 18\) 个。

接下来我们要解方程 \[ a + b + c = 18\text{，}\] 其中 \(a\)、\(b\) 和 \(c\) 分别是 Alice、Becky 和 Chris 额外得到的苹果数。

因为每人至少两个苹果的条件已经满足，所以这些变量都是非负整数。

由隔板法，方案数为 \[ \binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = 190\text{。}\]

正确答案是 C。

English

Let us assign everybody \(2\) apples. This leaves us with \(24 - 3 \cdot 2 = 18\) apples.

Then, we want to solve \[ a + b + c = 18, \] where \(a,\) \(b,\) and \(c\) are the additional apples given to Alice, Becky, and Chris.

Note that these are all nonnegative, since we already satisfied the only condition we needed to.

We can use stars and bars to get the number of solutions as \[ \binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = 190. \]

Thus, the correct answer is C.