2014 年 AMC 8 真题（P版）

1 答案 A

Harry 正确计算：\[\begin{align*} 8 - (2 + 5) &= 8 - 7 \\ &= 1 \end{align*}\text{。}\]Terry 错误计算：\[\begin{align*} 8 - 2 + 5 &= 6 + 5 \\ &= 11 \end{align*}\text{。}\]因此 \[\begin{align*} H - T &= 1 - 11 \\ &= -10 \end{align*}\text{。}\]

所以正确答案是 A。

English

Harry calculates it correctly as follows: \[\begin{align*} 8 - (2 + 5) &= 8 - 7 \\ &= 1. \end{align*}\] Terry calculates it incorrectly as follows: \[\begin{align*} 8 - 2 + 5 &= 6 + 5 \\ &= 11. \end{align*}\] Therefore: \[\begin{align*} H - T &= 1 - 11 \\ &= -10. \end{align*}\]

Thus, A is the correct answer.

2 答案 E

要使用最多硬币，Paul 可以全用 \(5\) 美分硬币，得到 \(\frac{35}{5}=7\) 枚。

要使用最少硬币，他可以用一枚 \(25\) 美分和一枚 \(10\) 美分硬币，共 \(2\) 枚。差为 \(7-2=5\)。

所以正确答案是 E。

English

To use the largest number of coins, Paul would use only \(5\)-cent coins, which gives \(\frac{35}{5}=7\) coins.

To use the smallest number of coins, Paul would use a \(25\)-cent coin and a \(10\)-cent coin, for a total of \(2\) coins. The difference is \(7-2=5\).

Thus, E is the correct answer.

3 答案 B

前三天 Isabella 读了 \(36 \cdot 3 = 108\) 页。

接下来三天 Isabella 读了 \(44 \cdot 3 = 132\) 页。

因此她一共读了 \(108 + 132 + 10 = 250\) 页。

所以正确答案是 B。

English

During the first three days, Isabella read \(36 \cdot 3 = 108\) pages.

During the next three days, Isabella read \(44 \cdot 3 = 132\) pages.

Therefore, she read a total of \(108 + 132 + 10 = 250\) pages.

Thus, B is the correct answer.

4 答案 E

两个奇质数之和是偶数。因为 \(85\) 是奇数，所以其中一个质数必须是唯一的偶质数 \(2\)。

另一个质数是 \(85-2=83\)，所以乘积为 \(2\cdot83=166\)。

所以正确答案是 E。

English

The sum of two odd primes is even. Since \(85\) is odd, one of the primes must be the only even prime, \(2\).

The other prime is \(85-2=83\), so the product is \(2\cdot83=166\).

Thus, E is the correct answer.

5 答案 C

用 \(\$20\)，Margie 可以买 \(\frac{20}{4}=5\) 加仑汽油。

每加仑可行驶 \(32\) 英里，所以她可以行驶 \(5\cdot32=160\) 英里。

所以正确答案是 C。

English

With \(\$20\), Margie can buy \(\frac{20}{4}=5\) gallons of gas.

At \(32\) miles per gallon, she can drive \(5\cdot32=160\) miles.

Thus, C is the correct answer.

6 答案 D

每个长方形面积是 \(2\) 乘对应长度。可以先把共同因子 \(2\) 提出来，只需求长度之和，即 \(91\)。

因此面积之和为 \(2 \cdot 91 = 182\)。

所以正确答案是 D。

English

To find the area of each rectangle we multiply \(2\) by their respective lengths. This means that we can factor out the \(2\) and we are left with the sum of the lengths. Adding together the lengths, we get \(91.\)

Therefore, the sum of the areas is \(2 \cdot 91 = 182.\)

Thus, D is the correct answer.

7 答案 B

设班上男生人数为 \(x\)，则女生有 \(x + 4\) 人。因为共有 \(28\) 名学生，\[\begin{align*} x + x + 4 &= 28\\ 2x&=24 \\ x&=12 \end{align*}\text{。}\]因此女生与男生之比为 \(16 : 12 = 4 : 3\)。

所以正确答案是 B。

English

Let \(x\) be the number of boys in the class. This means that there are \(x + 4\) girls. As there are \(28\) students, we know that: \[\begin{align*} x + x + 4 &= 28\\ 2x&=24 \\ x&=12. \end{align*}\] Therefore, the ratio of girls to boys is \(16 : 12 = 4 : 3.\)

Thus, B is the correct answer.

8 答案 D

因为 \(11\) 个人支付相同金额，所以总额能被 \(11\) 整除。因此 \(\underline{1} \underline{A} \underline{2}\) 必须能被 \(11\) 整除。

\(11\) 的整除规则是：交错位数字和之差若能被 \(11\) 整除，则原数能被 \(11\) 整除。

对 \(\underline{1} \underline{A} \underline{2}\)，这个差为 \[1 + 2 - A = 3 - A\text{。}\]要让它能被 \(11\) 整除，只能有 \(A = 3\)。

所以正确答案是 D。

English

Note that since \(11\) people paid the same amount, then the resulting sum is divisible by \(11.\) Therefore, \(\underline{1} \underline{A} \underline{2}\) must be divisible by \(11.\)

Remember the divisibility rule for \(11\): if we take the difference of the sums of alternating digits, and this difference is divisible by \(11,\) the whole number is divisible by \(11.\)

For \(\underline{1} \underline{A} \underline{2},\) the aforementioned difference is \[1 + 2 - A = 3 - A.\] The only way for this to be divisible by \(11\) is if \(A = 3.\)

Thus, D is the correct answer.

9 答案 D

因为 \(BD=DC\)，三角形 \(BDC\) 是等腰三角形，所以 \(\angle DBC=\angle BCD=70^\circ\)。

因此 \[\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ\end{aligned}\text{。}\]又因为 \(A,D,C\) 共线，\(\angle ADB\) 与 \(\angle BDC\) 构成平角，所以 \(\angle ADB=180^\circ-40^\circ=140^\circ\)。

所以正确答案是 D。

English

Since \(BD=DC\), triangle \(BDC\) is isosceles, so \(\angle DBC=\angle BCD=70^\circ\).

Thus \[\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ.\end{aligned}\] Because \(A,D,C\) are collinear, \(\angle ADB\) and \(\angle BDC\) form a straight angle, so \(\angle ADB=180^\circ-40^\circ=140^\circ\).

Thus, D is the correct answer.

10 答案 A

第七届 AMC \(8\) 是第一届之后 \(6\) 年举行的。因此 Samantha 参加的是 \(1985 + 6 = 1991\) 年的竞赛。

她那年 \(12\) 岁，所以出生年份为 \(1991 - 12 = 1979\)。

所以正确答案是 A。

English

The seventh AMC \(8\) would have been administered \(6\) years after the first one. Therefore, Samantha took it in \(1985 + 6 = 1991.\)

This means that Samantha was born \(12\) years prior in \(1991 - 12 = 1979.\)

Thus, A is the correct answer.

11 答案 A

用 E 表示向东一个街区，N 表示向北一个街区。最短路线使用三个 E 和两个 N。为了避开离 Jack 家以东一格、以北一格的危险路口，前两步必须是 EE 或 NN。

可能路线为 \(EEENN\)、\(EENEN\)、\(EENNE\) 和 \(NNEEE\)，共 \(4\) 条路线。

所以正确答案是 A。

English

Let E represent traveling one block east and N represent traveling one block north. A shortest route uses three E moves and two N moves. To avoid the dangerous intersection one block east and one block north of Jack’s house, the first two moves must be either EE or NN.

The possible routes are \(EEENN\), \(EENEN\), \(EENNE\), and \(NNEEE\), for \(4\) routes.

Thus, A is the correct answer.

12 答案 B

读者给三位名人配对共有 \(3! = 6\) 种方式。其中只有一种是正确配对。因此随机猜对全部三组的概率是 \(\frac{1}{6}\)。

所以正确答案是 B。

English

Notice that there are \(3! = 6\) total ways that a reader could match the celebrities. However, only one of these is the correct matching. Therefore, the probability that the reader guesses it correctly is \(\frac{1}{6}.\)

Thus, B is the correct answer.

13 答案 D

因为 \(n^2 + m^2\) 是偶数，所以 \(n^2\) 与 \(m^2\) 要么都是奇数，要么都是偶数。

若它们都是奇数，则 \(n\) 和 \(m\) 都是奇数。若它们都是偶数，则 \(n\) 和 \(m\) 都是偶数。只要 \(n\) 和 \(m\) 同奇偶，它们的和都是偶数。

因此 \(n + m\) 不可能是奇数。

所以正确答案是 D。

English

Since \(n^2 + m^2\) is even, either \(n^2\) and \(m^2\) are both odd, or both even.

If they are both odd, then \(n\) and \(m\) are both odd. If they are both even, then \(n\) and \(m\) are both even. If \(n\) and \(m\) are both odd or even, their sum will always be even.

Therefore, \(n + m\) is never odd.

Thus, D is the correct answer.

14 答案 B

长方形 \(ABCD\) 的面积为 \[5 \cdot 6 = 30\text{。}\]

而这个三角形的面积为 \[\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*}\] 因此 \[\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12 \end{align*}\text{。}\]再由勾股定理，\[\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13 \end{align*}\text{。}\]

所以正确答案是 B。

English

The area of \(ABCD\) is \[5 \cdot 6 = 30.\]

The area of \[\begin{align*} \triangle DCE &= \frac{1}{2}DC \cdot CE \\ &= 30 \end{align*}\] Therefore: \[\begin{align*} \dfrac{1}{2} \cdot 5 \cdot CE &= 30 \\ CE &= 12. \end{align*}\] Then using the Pythagorean theorem we get that \[\begin{align*} DE &= \sqrt{5^2 + 12^2} \\ &= \sqrt{169} \\ &= 13. \end{align*}\]

Thus, B is the correct answer.

15 答案 C

\(12\) 段弧等分圆周，所以每段弧对应的圆心角为 \(\frac{360^{\circ}}{12} = 30^{\circ}\)。

\(\angle AOE\) 跨过 \(4\) 段弧，所以 \[\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}\text{。}\]类似地，\[\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}\text{。}\]我们还知道这两个三角形都是等腰三角形，因为它们各有两条边是半径。因此 \[ x = \dfrac{180 - 120}{2} = 30^{\circ} \] 且 \[ y = \dfrac{180 - 60}{2} = 60^{\circ}\text{。}\]所以 \(x + y = 90^{\circ}\)。

所以正确答案是 C。

English

Note that each of the \(12\) arcs splits the circle evenly, so they each cover \(\frac{360^{\circ}}{12} = 30^{\circ}.\)

\(\angle AOE\) spans \(4\) of these arcs, so \[\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}.\] Similarly, \[\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}.\] We also know that both triangles are isosceles since two of their sides are radii. Therefore, \[ x = \dfrac{180 - 120}{2} = 30^{\circ} \] and \[ y = \dfrac{180 - 60}{2} = 60^{\circ}. \] Therefore, \(x + y = 90^{\circ}.\)

Thus, C is the correct answer.

16 答案 B

每支球队有 \(4\) 场非联盟比赛，所以涉及 Middle School Eight 球队与非联盟对手的比赛有 \(8\cdot4=32\) 场。

联盟内部，这 \(8\) 支球队中的每支都有 \(7\) 场对阵其他球队的主场比赛，所以联盟比赛有 \(8\cdot7=56\) 场。总数为 \(32+56=88\)。

所以正确答案是 B。

English

Each team plays \(4\) non-conference games, for \(8\cdot4=32\) games involving Middle School Eight teams and non-conference opponents.

Within the conference, each of the \(8\) teams has \(7\) home games against the other teams, so there are \(8\cdot7=56\) conference games. The total is \(32+56=88\).

Thus, B is the correct answer.

17 答案 B

George 平常走 \(1\) 英里时速度为每小时 \(3\) 英里，需要 \(\dfrac{1}{3}\) 小时，也就是 \(20\) 分钟。

今天他走前 \(\dfrac{1}{2}\) 英里时，速度为每小时 \(2\) 英里，用时 \(\dfrac{1}{2}\div2=\dfrac{1}{4}\) 小时，也就是 \(15\) 分钟。他还剩 \(5\) 分钟，即 \(\dfrac{1}{12}\) 小时，来走完 \(\dfrac{1}{2}\) 英里。

所需速度为 \(\dfrac{1}{2}\div\dfrac{1}{12}=6\) 英里每小时。

所以正确答案是 B。

English

If George normally walks \(1\) mile at \(3\) miles per hour, it takes him \(\dfrac{1}{3}\) hour, or \(20\) minutes, to get to school.

Today he walked the first \(\dfrac{1}{2}\) mile at \(2\) miles per hour, taking \(\dfrac{1}{2}\div2=\dfrac{1}{4}\) hour, or \(15\) minutes. He has \(5\) minutes, which is \(\dfrac{1}{12}\) hour, left to cover \(\dfrac{1}{2}\) mile.

His required speed is \(\dfrac{1}{2}\div\dfrac{1}{12}=6\) miles per hour.

Thus, B is the correct answer.

18 答案 D

四个孩子按出生顺序共有 \(2^4=16\) 个等可能结果。各类别数量为：全是男孩 \(1\) 种；全是女孩 \(1\) 种；两男两女 \(\binom{4}{2}=6\) 种；三人一种性别、一人另一种性别 \(2\binom{4}{1}=8\) 种。

最大数量是 \(8\)，所以最可能的结果是三个孩子是一种性别，一个孩子是另一种性别。

所以正确答案是 D。

English

There are \(2^4=16\) equally likely birth-order outcomes. The counts by category are: all boys, \(1\); all girls, \(1\); two boys and two girls, \(\binom{4}{2}=6\); and three of one gender and one of the other, \(2\binom{4}{1}=8\).

The largest count is \(8\), so the most likely outcome is three children of one gender and one of the other.

Thus, D is the correct answer.

19 答案 A

为了最小化可见白色面积，把一个白色小立方体放在大立方体中心，这样没有任何面可见。把另外五个白色小立方体放在大立方体的面中心位置，每个只贡献一个可见白色小正方形。

可见白色表面积为 \(5\) 平方英寸。边长 \(3\) 英寸的立方体总表面积为 \(6\cdot3^2=54\) 平方英寸，所以白色部分占 \(\dfrac{5}{54}\)。

所以正确答案是 A。

English

To minimize visible white area, place one white cube in the center of the large cube, where no faces are visible. Place each of the other five white cubes at the center of a face of the large cube, where each contributes only one visible white square.

The visible white surface area is therefore \(5\) square inches. The total surface area of the \(3\)-inch cube is \(6\cdot3^2=54\) square inches, so the fraction that is white is \(\dfrac{5}{54}\).

Thus, A is the correct answer.

20 答案 B

三个圆在长方形内的部分都是四分之一圆，半径分别为 \(1\)、\(2\)、\(3\)。它们的总面积为 \[\dfrac{\pi}{4}+\pi+\dfrac{9\pi}{4}=\dfrac{7\pi}{2}\text{。}\]

用 \(\pi\approx\dfrac{22}{7}\)，这个总面积约为 \(11\)。长方形面积为 \(3\cdot5=15\)，所以所求面积约为 \(15-11=4\)。

所以正确答案是 B。

English

The parts of the three circles inside the rectangle are quarter-circles with radii \(1\), \(2\), and \(3\). Their total area is \[\dfrac{\pi}{4}+\pi+\dfrac{9\pi}{4}=\dfrac{7\pi}{2}.\]

Using \(\pi\approx\dfrac{22}{7}\), this total is about \(11\). The rectangle has area \(3\cdot5=15\), so the desired area is about \(15-11=4\).

Thus, B is the correct answer.

21 答案 A

为使 \(\underline{74A52B1}\) 能被 \(3\) 整除，数字和 \(19+A+B\) 必须是 \(3\) 的倍数。因此 \(A+B\) 比 \(3\) 的某个倍数小 \(1\)。

为使 \(\underline{326AB4C}\) 能被 \(3\) 整除，数字和 \(15+A+B+C\) 必须是 \(3\) 的倍数。因为 \(15\) 已经能被 \(3\) 整除，所以 \(A+B+C\) 必须能被 \(3\) 整除。

因此 \(C\) 必须比 \(3\) 的某个倍数大 \(1\)。选项中只有 \(1\) 符合这个条件。

所以正确答案是 A。

English

For \(\underline{74A52B1}\) to be divisible by \(3\), the digit sum \(19+A+B\) must be a multiple of \(3\). Hence \(A+B\) is \(1\) less than a multiple of \(3\).

For \(\underline{326AB4C}\) to be divisible by \(3\), the digit sum \(15+A+B+C\) must be a multiple of \(3\). Since \(15\) is already divisible by \(3\), \(A+B+C\) must be divisible by \(3\).

Therefore \(C\) must be \(1\) more than a multiple of \(3\). Among the answer choices, only \(1\) has that form.

Thus, A is the correct answer.

22 答案 E

设这个数为 \(10a+b\)，其中 \(a\) 是十位数字，\(b\) 是个位数字。条件给出 \[ab+a+b=10a+b\text{。}\]

两边消去 \(b\)，得 \(ab+a=10a\)，所以 \(ab=9a\)。由于 \(a\neq0\)，得到 \(b=9\)。

所以正确答案是 E。

English

Let the number be \(10a+b\), where \(a\) is the tens digit and \(b\) is the units digit. The condition gives \[ab+a+b=10a+b.\]

Canceling \(b\) from both sides gives \(ab+a=10a\), so \(ab=9a\). Since \(a\neq0\), we get \(b=9\).

Thus, E is the correct answer.

23 答案 A

任意两人的号码和必须是月份中的日期，所以至多为 \(31\)。两个两位质数之和不超过 \(31\) 的可能有 \(11+13=24\)、\(11+17=28\)、\(11+19=30\) 和 \(13+17=30\)。

三个日期分别是早些时候、今天和晚些时候，所以必须互不相同且递增。唯一可能的一组三个不同日期是 \(24,28,30\)，对应号码 \(11\)、\(13\)、\(17\)。

Bethany 的日期是 Ashley 加 Caitlin，Ashley 的日期是 Bethany 加 Caitlin，Caitlin 的晚些时候日期是 Ashley 加 Bethany。因此 Caitlin 的号码是三者中最小的，即 \(11\)。

所以正确答案是 A。

English

Any pair-sum must be a date in the month, so it is at most \(31\). The possible sums of two two-digit primes that are at most \(31\) are \(11+13=24\), \(11+17=28\), \(11+19=30\), and \(13+17=30\).

The three dates are earlier, today, and later, so they must be distinct and increasing. The only possible set of three distinct dates is \(24,28,30\), coming from the numbers \(11\), \(13\), and \(17\).

Bethany’s date is Ashley plus Caitlin, Ashley’s date is Bethany plus Caitlin, and Caitlin’s later date is Ashley plus Bethany. This means Caitlin’s number is the smallest of the three, namely \(11\).

Thus, A is the correct answer.

24 答案 C

将 \(100\) 位顾客的购买数从小到大排列。中位数是第 \(50\) 项和第 \(51\) 项的平均数。为了最大化它，让前 \(49\) 项尽可能小，都等于 \(1\)。

如果第 \(50\) 项至少为 \(4\)，那么总数至少为 \(49+51\cdot4=253\)，超过二百五十二。所以第 \(50\) 项至多为 \(3\)。

这个最大值可以达到：取 \(49\) 项等于 \(1\)，第 \(50\) 项等于 \(3\)，最后 \(50\) 项都等于 \(4\)。总数为 \(49+3+50\cdot4=252\)，中位数为 \(\frac{3+4}{2}=3.5\)。

所以正确答案是 C。

English

Order the \(100\) purchases from least to greatest. The median is the average of the \(50\)th and \(51\)st entries. To maximize it, make the first \(49\) entries as small as possible, so set them all equal to \(1\).

If the \(50\)th entry were at least \(4\), then the total would be at least \(49+51\cdot4=253\), too many cans. So the \(50\)th entry is at most \(3\).

This maximum is attainable: use \(49\) entries equal to \(1\), the \(50\)th entry equal to \(3\), and the final \(50\) entries equal to \(4\). The total is \(49+3+50\cdot4=252\), and the median is \(\frac{3+4}{2}=3.5\).

Thus, C is the correct answer.

25 答案 B

每个半圆让 Robert 沿公路前进 \(40\) 英尺，所以一英里路段需要 \(\frac{5280}{40}=132\) 个半圆。

每个半圆直径为 \(40\) 英尺，因此弧长为 \(20\pi\) 英尺。Robert 骑行总距离为 \(132\cdot20\pi=2640\pi\) 英尺，也就是 \(\dfrac{2640\pi}{5280}=\dfrac{\pi}{2}\) 英里。

速度为每小时 \(5\) 英里，所以时间为 \(\dfrac{\pi}{2}\div5=\dfrac{\pi}{10}\) 小时。

所以正确答案是 B。

English

Each semicircle advances Robert \(40\) feet along the highway, so a one-mile stretch requires \(\frac{5280}{40}=132\) semicircles.

Each semicircle has diameter \(40\) feet, so its arc length is \(20\pi\) feet. The total distance Robert rides is \(132\cdot20\pi=2640\pi\) feet, which is \(\dfrac{2640\pi}{5280}=\dfrac{\pi}{2}\) miles.

At \(5\) miles per hour, the time is \(\dfrac{\pi}{2}\div5=\dfrac{\pi}{10}\) hours.

Thus, B is the correct answer.