2013 年 AMC 8 真题（P版）

1 答案 A

为了让丹妮卡把模型车排成每行正好六辆，她拥有的模型车数量必须是 \(6\) 的倍数。大于 \(23\) 的最小这样的数是 \(24\)，所以她必须再买 \(1\) 辆车。

所以正确答案是 A。

English

In order for Danica to arrange her model cars in rows of exactly six, the number of model cars she has must be a multiple of six. The smallest multiple of \(6\) greater than \(23\) is \(24,\) so she must buy \(1\) more car to attain this amount.

Thus, A is the correct answer.

2 答案 D

促销时半磅价格为 \(\$3\)，所以一磅促销价为 \(2 \cdot \$3 = \$6\)。

促销价是正常价格的 \(50\%\)，所以正常价格是促销价的两倍。因此一磅鱼的正常价格是 \(2 \cdot \$6 = \$12\)。

所以正确答案是 D。

English

During the sale, the price for half a pound is \(\$3,\) and therefore, the price for a full pound is \(2 \cdot \$3 = \$6.\)

The sale price is \(50\%\) off the regular price, and so the regular price is twice the sale price. Therefore, the regular price for a full pound of fish is \(2 \cdot \$6 = \$12.\)

Thus, D is the correct answer.

3 答案 E

在括号内，将相邻项配对：\[ \begin{aligned} &(-1+2)+(-3+4) \\ &\quad {}+\cdots+(-999+1000) \end{aligned}\text{。}\]

每一对的和都是 \(1\)，共有 \(500\) 对，所以括号内的值是 \(500\)。再乘以 \(4\)，得到 \(2000\)。

所以正确答案是 E。

English

Within the parentheses, pair consecutive terms: \[ \begin{aligned} &(-1+2)+(-3+4) \\ &\quad {}+\cdots+(-999+1000). \end{aligned} \]

Each pair has sum \(1\), and there are \(500\) pairs, so the value inside the parentheses is \(500\). Multiplying by \(4\) gives \(2000\).

Thus, E is the correct answer.

4 答案 C

朱迪应付的那一份由大家额外支付的钱完全覆盖，所以她应付的金额为 \[7 \cdot \$2.50 = \$17.50\text{。}\]

因为每个人原本应付相同金额，所以总账单为 \[8 \cdot \$17.50 = \$140\text{。}\]

所以正确答案是 C。

English

Since Judi’s portion was fully covered by everyone’s contributions, the portion of the bill she was responsible for was equal to \[7 \cdot \$2.50 = \$17.50.\]

As everyone paid the same amount, we can conclude that the total bill is \[8 \cdot \$17.50 = \$140.\]

Thus, C is the correct answer.

5 答案 E

体重的中位数就是排序后正中间的那个值，可以这样找出： \[ \cancel{ 5 },\cancel{ 5 },\underline{6},\cancel{8},\cancel{106} \] 因此中位数是 \(6\) 磅。

平均数可以计算如下： \[ \begin{align*} &\dfrac{106 + 5 + 5 + 6 + 8}{5} \\&= \dfrac{130}{5}\\ &= 26\text{。} \end{align*} \] 这个值比中位数大，具体来说，平均数比中位数大 \(20\) 磅。

所以正确答案是 E。

English

The median weight is the middle value, which we can find as: \[\cancel{ 5 },\cancel{ 5 },\underline{6},\cancel{8},\cancel{106}\] Therefore, the median is \(6\) pounds.

The average (mean) can be calculated to be: \[ \begin{align*} &\dfrac{106 + 5 + 5 + 6 + 8}{5} \\&= \dfrac{130}{5}\\ &= 26. \end{align*} \] This is greater than the median, specifically, the mean is \(20\) pounds greater than the median.

Thus, E is the correct answer.

6 答案 C

第二行两个数的乘积是 \(600\)，所以第二行右侧缺失的数是 \(600 \div 30 = 20\)。

现在知道 \(5\) 与第一行缺失数的乘积为 \(20\)。因此缺失数是 \(20 \div 5 = 4\)。

所以正确答案是 C。

English

The product of the numbers in the second row is \(600,\) so the missing number in the middle right box is \(600 \div 30 = 20.\)

Now we know that the product of \(5\) and the missing number in the first row is \(20.\) Therefore, the missing number is \(20 \div 5 = 4.\)

Thus, C is the correct answer.

7 答案 C

\(2\) 分 \(45\) 秒等于 \(2 \cdot 60 + 45 = 165\) 秒。因为 \(10\) 秒通过 \(6\) 节车厢，所以设 \(165\) 秒通过 \(x\) 节车厢，满足 \[\dfrac{6}{10}=\dfrac{x}{165}\text{。}\]

解得 \[ x=\dfrac{165\cdot6}{10}=99\text{。} \] 最接近的选项是 \(100\)。

所以正确答案是 C。

English

Converting \(2\) minutes and \(45\) seconds into seconds, we get \(2 \cdot 60 + 45 = 165\) seconds. Since \(6\) cars passed in \(10\) seconds, at the same rate the number of cars \(x\) passing in \(165\) seconds satisfies \[\dfrac{6}{10}=\dfrac{x}{165}.\]

Solving gives \[x=\dfrac{165\cdot6}{10}=99.\] The closest answer choice is \(100\).

Thus, C is the correct answer.

8 答案 C

至少有两个连续正面的结果是 \(HHT\)、\(THH\) 和 \(HHH\)。三次公平抛硬币共有 \(2^3=8\) 个等可能结果，所以概率为 \(\dfrac{3}{8}\)。

所以正确答案是 C。

English

The outcomes with at least two consecutive heads are \(HHT\), \(THH\), and \(HHH\). There are \(2^3=8\) equally likely outcomes from three fair coin tosses, so the probability is \(\dfrac{3}{8}\).

Thus, C is the correct answer.

9 答案 C

第 \(n\) 跳时，浩克跳 \(2^{n-1}\) 米。需要找最小的 \(n\)，使 \(2^{n-1}>1000\)。

因为 \(2^9=512\)，而 \(2^{10}=1024\)，所以第一次超过 \(1000\) 米时 \(n-1=10\)，即 \(n=11\)。

所以正确答案是 C。

English

On the \(n\)th jump, the Hulk jumps \(2^{n-1}\) meters. We need the smallest \(n\) for which \(2^{n-1}>1000\).

Since \(2^9=512\) and \(2^{10}=1024\), the first jump longer than \(1000\) meters has \(n-1=10\), so \(n=11\).

Thus, C is the correct answer.

10 答案 C

将两个数质因数分解：\[180=2^2\cdot3^2\cdot5\text{，}\]\[594=2\cdot3^3\cdot11\text{。}\]

最大公因数取较小的幂，所以为 \(2\cdot3^2=18\)。最小公倍数取较大的幂，所以为 \[ 2^2\cdot3^3\cdot5\cdot11=5940\text{。} \] 所求比值为 \(\dfrac{5940}{18}=330\)。

所以正确答案是 C。

English

Prime-factorize the two numbers: \[180=2^2\cdot3^2\cdot5,\] \[594=2\cdot3^3\cdot11.\]

The greatest common factor uses the smaller powers, so it is \(2\cdot3^2=18\). The least common multiple uses the larger powers, so it is \[2^2\cdot3^3\cdot5\cdot11=5940.\] Therefore the desired ratio is \(\dfrac{5940}{18}=330\).

Thus, C is the correct answer.

11 答案 D

祖父星期一慢跑用时 \(\frac{2}{5}\) 小时，星期三步行用时 \(\frac{2}{3}\) 小时，星期五用时 \(\frac{2}{4}=\frac{1}{2}\) 小时。换算为分钟，星期一为 \[ \dfrac{2}{5} \cdot 60 = 24 \] 分钟，星期三为 \[ \dfrac{2}{3} \cdot 60 = 40 \] 分钟，星期五为 \[ \dfrac{1}{2} \cdot 60 = 30 \] 分钟。因此实际总时间为 \(94\) 分钟。

如果他一直以每小时 \(4\) 英里的速度走，每天用时 \(\frac{2}{4}=\frac{1}{2}\) 小时，即 \(\frac{2}{4} \cdot 60 = 30\) 分钟，\(3\) 天共 \(3 \cdot 30 = 90\) 分钟。

因此少用的时间是 \(94-90=4\) 分钟。

所以正确答案是 D。

English

Grandfather spent \(\frac{2}{5}\) hours jogging on Monday. He spent \(\frac{2}{3}\) hours walking on Wednesday and \(\frac{2}{4}=\frac{1}{2}\) hours on Friday. Converting these times to minutes, we get \[\dfrac{2}{5} \cdot 60 = 24\] minutes for Monday, \[\dfrac{2}{3} \cdot 60 = 40\] minutes for Wednesday, and \[\dfrac{1}{2} \cdot 60 = 30\] minutes for Friday. Therefore, Grandfather totaled \(94\) minutes of exercise throughout the week.

If he walked at a pace of \(4\) miles per hour each day, he would have spent \(\frac{2}{4}=\frac{1}{2}\) hours each day walking. This equals \(\frac{2}{4} \cdot 60 = 30\) minutes every day. If he did this for \(3\) days, he would have totaled \(3 \cdot 30 = 90\) minutes of exercise.

Therefore, Grandfather would have spent \(94-90=4\) fewer minutes on the treadmill.

Thus, D is the correct answer.

12 答案 B

哈维尔第二双节省 \(0.40 \cdot 50 = 20\) 美元。第三双节省 \(\frac{50}{2} = 25\) 美元。因此他一共节省 \(45\) 美元。

因为 \[ \frac{45}{150} = \frac{3}{10} = 0.30\text{，} \] 所以与正常价格相比，他节省了 \(30 \%\)。

所以正确答案是 B。

English

Javier saved \(0.40 \cdot 50 = 20\) dollars on the second pair. He also saved \(\frac{50}{2} = 25\) dollars on the third pair. This shows that he saved a total of \(45\) dollars.

As \[\frac{45}{150} = \frac{3}{10} = 0.30,\] we can conclude that Javier saved \(30 \%\) compared to the regular price.

Thus, B is the correct answer.

13 答案 A

如果这个分数的后两位是 \(10a+b\)，交换后得到 \(10b+a\)。差为 \[ \begin{aligned} &(10a+b)-(10b+a) \\ &= 9(a-b)\text{，} \end{aligned} \] 因此错误总和与正确总和之差必须是 \(9\) 的倍数。

选项中只有 \(45\) 能被 \(9\) 整除。

所以正确答案是 A。

English

If the last two digits of the score are \(10a+b\), then reversing those digits gives \(10b+a\). The difference is \[ \begin{aligned} &(10a+b)-(10b+a) \\ &= 9(a-b), \end{aligned} \] so the incorrect sum must differ from the correct sum by a multiple of \(9\).

Among the answer choices, only \(45\) is divisible by \(9\).

Thus, A is the correct answer.

14 答案 C

艾布和比娅选择软糖共有 \(2\cdot4=8\) 种等可能结果。颜色相同可能是两人都选绿色，有 \(1\) 种；也可能两人都选红色，有 \(1\cdot2=2\) 种。

因此 \(8\) 个结果中有 \(3\) 个颜色相同，概率为 \(\dfrac{3}{8}\)。

所以正确答案是 C。

English

There are \(2\cdot4=8\) equally likely ways for Abe and Bea to choose their jelly beans. The colors match if they both choose green, which can happen in \(1\) way, or if they both choose red, which can happen in \(1\cdot2=2\) ways.

Thus \(3\) of the \(8\) outcomes match, for probability \(\dfrac{3}{8}\).

Thus, C is the correct answer.

15 答案 B

从每个方程中减去已知项：\[3^p=90-81=9\text{，}\]\[2^r=76-44=32\text{，}\]\[6^s=1421-125=1296\text{。}\]

因此 \(p=2\)、\(r=5\)、\(s=4\)。它们的乘积是 \(2\cdot5\cdot4=40\)。

所以正确答案是 B。

English

Subtract the known number from each equation: \[3^p=90-81=9,\] \[2^r=76-44=32,\] \[6^s=1421-125=1296.\]

Therefore \(p=2\), \(r=5\), and \(s=4\). Their product is \(2\cdot5\cdot4=40\).

Thus, B is the correct answer.

16 答案 E

\(8\) 年级学生人数必须同时是 \(5\) 和 \(8\) 的倍数，所以最少为 \(40\)。这时 \(6\) 年级人数为 \(40 \cdot \dfrac{3}{5} = 24\)，\(7\) 年级人数为 \(40 \cdot \dfrac{5}{8} = 25\)。

因此学生总数为 \(40 + 24 + 25 = 89\)。

所以正确答案是 E。

English

The number of \(8\)th-graders must be a multiple of \(5\) and \(8,\) which means that it is at least \(40.\) Using this number, we get that the number of \(6\)th-graders is \(40 \cdot \dfrac{3}{5} = 24.\) Similarly, the number of \(7\)th-graders is \(40 \cdot \dfrac{5}{8} = 25.\)

The total number of students is therefore \(40 + 24 + 25 = 89.\)

Thus, E is the correct answer.

17 答案 B

设这六个连续整数为 \(x\)、\(x+1\)、\(x+2\)、\(x+3\)、\(x+4\)、\(x+5\)。它们的和为 \[6x+15=2013\text{。}\]

因此 \(6x=1998\)，所以 \(x=333\)。最大整数为 \(x+5=338\)。

所以正确答案是 B。

English

Let the six consecutive integers be \(x,\) \(x+1,\) \(x+2,\) \(x+3,\) \(x+4,\) \(x+5\). Their sum is \[6x+15=2013.\]

Thus \(6x=1998\), so \(x=333\). The largest integer is \(x+5=338\).

Thus, B is the correct answer.

18 答案 B

外表面确定的长方体体积为 \(12\cdot10\cdot5=600\) 立方英尺。内部空心部分在长上少 \(2\) 英尺、宽上少 \(2\) 英尺、高上少 \(1\) 英尺，所以尺寸为 \(10\times8\times4\)。

空心体积为 \(10\cdot8\cdot4=320\) 立方英尺，所以堡垒含有 \(600-320=280\) 个一英尺见方的立方块。

所以正确答案是 B。

English

The rectangular prism determined by the outside faces has volume \(12\cdot10\cdot5=600\) cubic feet. The empty inside is smaller by \(2\) feet in length, \(2\) feet in width, and \(1\) foot in height, so its dimensions are \(10\times8\times4\).

The empty volume is \(10\cdot8\cdot4=320\) cubic feet, so the fort contains \(600-320=280\) one-foot cubical blocks.

Thus, B is the correct answer.

19 答案 D

因为卡西能推断自己不是最低分，她一定比汉娜分数高。同样，布里奇特的话说明她一定比汉娜分数低。

因此从高到低的排名是卡西、汉娜、布里奇特。

所以正确答案是 D。

English

Since Cassie deduced she didn’t get the lowest score, she must have got a higher score than Hannah. Similarly, Bridget’s statement reveals that she must have got a lower score than Hannah.

Therefore, the ranking from highest to lowest is Cassie, Hannah, and then Bridget.

Thus, D is the correct answer.

20 答案 C

由勾股定理，半圆半径为 \(\sqrt{2}\)。因此半圆面积为 \[ \dfrac{1}{2} \cdot \pi \cdot \sqrt{2}^2 = \pi\text{。} \] 所以正确答案是 C。

English

Using the Pythagorean theorem, we get that the radius of the semicircle is \(\sqrt{2}.\) This means that the area of the semicircle is \[\dfrac{1}{2} \cdot \pi \cdot \sqrt{2}^2 = \pi.\] Thus, C is the correct answer.

21 答案 E

为了最快到达公园西南角，萨曼莎必须向东走 \(2\) 个街区、向北走 \(1\) 个街区，这些步骤可排列成 \(3\) 种方式。经过公园对角线小路后，她必须再向东走 \(2\) 个街区、向北走 \(2\) 个街区到学校，这些步骤可排列成 \(\binom{4}{2}=6\) 种方式。

公园前后的选择相互独立，所以最短路线总数为 \(3\cdot6=18\)。

所以正确答案是 E。

English

To reach the southwest corner of the park as quickly as possible, Samantha must go \(2\) blocks east and \(1\) block north, which can be arranged in \(3\) ways. After the diagonal path through the park, she must go \(2\) blocks east and \(2\) blocks north to reach school, which can be arranged in \(\binom{4}{2}=6\) ways.

The choices before and after the park are independent, so the total number of shortest routes is \(3\cdot6=18\).

Thus, E is the correct answer.

22 答案 E

一个 \(60\) 乘 \(32\) 网格有 \(61\) 条竖直网格线，每条由 \(32\) 根牙签组成；还有 \(33\) 条水平网格线，每条由 \(60\) 根牙签组成。

因此牙签总数为 \[ \begin{aligned} 61\cdot32+33\cdot60 &= 1952+1980 \\ &= 3932 \end{aligned}\text{。}\]

所以正确答案是 E。

English

A \(60\)-by-\(32\) grid has \(61\) vertical grid lines, each made of \(32\) toothpicks, and \(33\) horizontal grid lines, each made of \(60\) toothpicks.

Therefore the total number of toothpicks is \[ \begin{aligned} 61\cdot32+33\cdot60 &= 1952+1980 \\ &= 3932. \end{aligned} \]

Thus, E is the correct answer.

23 答案 B

\(\overline{AB}\) 上的半圆面积为 \(8\pi\)，所以对应整圆面积为 \(16\pi\)。其半径为 \(4\)，因此 \(AB=8\)。

半径为 \(r\) 的半圆弧长为 \(\pi r\)。因为 \(\overline{AC}\) 上的弧长为 \(8.5\pi\)，其半径为 \(8.5\)，所以 \(AC=17\)。

在直角三角形 \(ABC\) 中，由勾股定理， \[ BC=\sqrt{17^2-8^2}=\sqrt{225}=15\text{。} \] 因此 \(\overline{BC}\) 上半圆的半径为 \(\frac{15}{2}=7.5\)。

所以正确答案是 B。

English

The semicircle on \(\overline{AB}\) has area \(8\pi\), so its full circle would have area \(16\pi\). Its radius is therefore \(4\), and \(AB=8\).

A semicircle of radius \(r\) has arc length \(\pi r\). Since the arc on \(\overline{AC}\) has length \(8.5\pi\), its radius is \(8.5\), so \(AC=17\).

Using the Pythagorean theorem in right triangle \(ABC\), \[BC=\sqrt{17^2-8^2}=\sqrt{225}=15.\] The radius of the semicircle on \(\overline{BC}\) is \(\frac{15}{2}=7.5\).

Thus, B is the correct answer.

24 答案 C

设每个正方形边长为 \(1\)，则三个正方形总面积为 \(3\)。从总面积中减去对角线 \(AJ\) 左侧的非阴影区域会更容易。

根据解答图，长方形 \(EDKF\) 面积为 \(1\cdot\dfrac12=\dfrac12\)，三角形 \(AKJ\) 面积为 \[ \dfrac12\cdot\dfrac32\cdot2=\dfrac32\text{。} \] 因此非阴影面积为 \(\dfrac12+\dfrac32=2\)，阴影五边形面积为 \(3-2=1\)。

所求比值为 \(\dfrac{1}{3}\)。

所以正确答案是 C。

English

Let each square have side length \(1\), so the total area of the three squares is \(3\). It is easier to subtract the unshaded region to the left of diagonal \(AJ\) from this total.

Using the solution diagram, rectangle \(EDKF\) has area \(1\cdot\dfrac12=\dfrac12\), and triangle \(AKJ\) has area \[\dfrac12\cdot\dfrac32\cdot2=\dfrac32.\] Thus the unshaded area is \(\dfrac12+\dfrac32=2\), so the shaded pentagon has area \(3-2=1\).

The requested ratio is \(\dfrac{1}{3}\).

Thus, C is the correct answer.

25 答案 A

球的半径为 \(2\)。滚动时，球心沿着与轨道平行的半圆弧运动。第一段和第三段圆弧中球心路径在轨道圆弧内侧，所以半径分别为 \(100-2=98\) 和 \(80-2=78\)。中间圆弧中球心路径在轨道圆弧外侧，所以半径为 \(60+2=62\)。

球心行进的总距离是三段半圆弧长度之和：\[(98+62+78)\pi=238\pi\text{。}\]

所以正确答案是 A。

English

The ball has radius \(2\). As the ball rolls, its center follows semicircular arcs parallel to the track. On the first and third arcs the center path is inside the track arcs, so those radii are \(100-2=98\) and \(80-2=78\). On the middle arc the center path is outside the track arc, so that radius is \(60+2=62\).

The total distance traveled by the center is the sum of the three semicircle lengths: \[(98+62+78)\pi=238\pi.\]

Thus, A is the correct answer.