2015 年 AMC 8 真题（P版）

1 答案 A

一边长 \(12\) 英尺，即 \(\dfrac{12}{3} = 4\) 码。

另一边长 \(9\) 英尺，即 \(\dfrac{9}{3} = 3\) 码。

因为尺寸是 \(4\) 码乘 \(3\) 码，所以面积为 \[ 4\cdot3=12 \] 平方码。

所以正确答案是 A。

English

Since one side is \(12\) feet, it would be \(\dfrac{12}{3} = 4\) yards.

Since another side is \(9\) feet, it would be \(\dfrac{9}{3} = 3\) yards.

Since the dimensions are \(4\) yards by \(3\) yards, the area is \[ 4\cdot3=12 \] square yards.

Thus, the correct answer is A .

2 答案 D

先注意，八边形可分成 \(8\) 个面积相等的三角形，每个三角形都以 \(O\) 为一个顶点。因此每个三角形占八边形总面积的 \(\dfrac {1}{8}\)。

阴影部分包含三个完整三角形和三角形 \(ABO\) 的一半。因此阴影面积是八边形总面积的 \[\dfrac{3.5}{8} = \dfrac{7}{16}\] 倍。

所以正确答案是 D。

English

First notice that there are \(8\) equally sized triangles that can be created with \(O\) and any two consecutive points. Therefore, they each take up \(\dfrac {1}{8} \) of the total area of the octagon.

The shaded area has three complete triangles and half of the triangle \(ABO.\) Therefore, the shaded area is \[\dfrac{3.5}{8} = \dfrac{7}{16}\] of the total area of the octagon.

Thus, the correct answer is D .

3 答案 D

杰克走 \(4\) 英里需要 \(60\) 分钟，所以他到游泳池需要 \(\dfrac{60}{4} = 15\) 分钟。

吉尔骑行 \(10\) 英里需要 \(60\) 分钟，所以她到游泳池需要 \(\dfrac{60}{10} = 6\) 分钟。

两人的时间差为 \(15-6 = 9\) 分钟。

所以正确答案是 D。

English

Jack travels at a rate of \(4\) miles per \(60\) minutes. Therefore, it takes him \(\dfrac{60}{4} = 15\) minutes to get to the pool.

Jill travels at a rate of \(10\) miles per \(60\) minutes. Therefore it takes her \(\dfrac{60}{10} = 6\) minutes to get to the pool.

Therefore, the difference in their times is \(15-6 = 9\) minutes.

Thus, the correct answer is D .

4 答案 E

两名男生坐在两端有 \(2! = 2\) 种方式。三名女生坐在中间三个座位有 \(3! = 6\) 种方式。

因此总排列数为 \(2\cdot6=12\)。

所以正确答案是 E。

English

There are \(2! = 2\) ways to place the two boys at the two ends. There are \(3! = 6\) ways to arrange the three girls in the middle seats.

Thus the total number of arrangements is \(2\cdot6=12.\)

Thus, E is the correct answer.

5 答案 A

考虑全部 \(12\) 场比赛时，得分 \(40\) 来自第 \(12\) 场，并会成为最低分。因此，与前 \(11\) 场的极差相比，全部 \(12\) 场的极差从 \(73-42 = 31\) 增加到 \(73-40=33\)。

所以正确答案是 A。

English

When considering all \(12\) games, \(40,\) from the \(12\)th game, will be the lowest score. Therefore, compared to the range of just the first \(11\) games, the range of all \(12\) games would increase from \(73-42 = 31\) to \(73-40=33.\)

Thus, the correct answer is A .

6 答案 B

从 \(B\) 向 \(AC\) 作高，交 \(AC\) 于 \(X\)。因为 \(AB=BC\)，点 \(X\) 是 \(AC\) 的中点，所以 \(AX=21\)。

在直角三角形 \(ABX\) 中，高为 \[ \begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20 \end{aligned}\text{。}\]

因此 \(\triangle ABC\) 的面积为 \(\dfrac12\cdot42\cdot20=420\)。

所以正确答案是 B。

English

Drop the altitude from \(B\) to \(AC,\) meeting \(AC\) at \(X.\) Since \(AB=BC,\) point \(X\) is the midpoint of \(AC,\) so \(AX=21.\)

In right triangle \(ABX,\) the altitude is \[ \begin{aligned} BX &= \sqrt{29^2-21^2} \\ &= \sqrt{400} = 20. \end{aligned} \]

The area of \(\triangle ABC\) is \(\dfrac12\cdot42\cdot20=420.\)

Thus, B is the correct answer.

7 答案 E

只有当两枚筹码都是奇数时，乘积才是奇数。每个盒子中有两个奇数筹码 \(1\) 和 \(3\)，共三个筹码，所以乘积为奇数的概率是 \((\dfrac23)^2=\dfrac49\)。

乘积为偶数的概率是补集，等于 \(1-\dfrac49=\dfrac59\)。

所以正确答案是 E。

English

The product is odd only when both chips are odd. Each box has two odd chips, \(1\) and \(3,\) out of three chips, so the probability of an odd product is \((\dfrac23)^2=\dfrac49.\)

The probability of an even product is the complement, \(1-\dfrac49=\dfrac59.\)

Thus, E is the correct answer.

8 答案 D

设第三边长为 \(s\)。三角不等式给出 \(s<5+19=24\)，所以周长满足 \[5+19+s<48\text{。}\]

周长可以从下方任意接近 \(48\)，因此大于所有这类三角形周长的最小整数是 \(48\)。

所以正确答案是 D。

English

Let the third side length be \(s.\) The triangle inequality gives \(s<5+19=24,\) so the perimeter satisfies \[5+19+s<48.\]

Perimeters can be made arbitrarily close to \(48\) from below, so the smallest whole number larger than the perimeter of any such triangle is \(48.\)

Thus, D is the correct answer.

9 答案 D

我们要计算 \[ 1+3 + \cdots + 39\text{。} \] 把两端的项依次配对，可得 \[ \begin{aligned} &(1+39)+(3+37)+\cdots\\ &\qquad +(19+21) \end{aligned}\text{。}\]共有 \(10\) 对，每对和为 \(40\)，所以 \[ 10\cdot40=400\text{。}\]

所以正确答案是 D。

English

We want to find \[1+3 + \cdots + 39.\] Pairing terms from the ends gives \[\begin{aligned} &(1+39)+(3+37)+\cdots\\ &\qquad +(19+21). \end{aligned}\] There are \(10\) pairs, and each pair sums to \(40.\) Therefore, the sum is \[10\cdot40=400.\]

Thus, the correct answer is D .

10 答案 B

首先，千位有 \(9\) 种选择，因为 \(0\) 不能选。

接着，百位有 \(9\) 种选择，十位有 \(8\) 种选择，个位有 \(7\) 种选择。因此这样的整数共有 \(9\cdot9\cdot8\cdot7 = 4536\) 个。

所以正确答案是 B。

English

First, there are \(9\) digits to choose for the thousands digit since \(0\) can’t be chosen.

Then, after that, there are \(9\) ways to choose the hundreds digit, \(8\) ways to choose the tens digit, and \(7\) ways to choose the ones digit. Therefore, we get \(9\cdot9\cdot8\cdot7 = 4536\) ways to choose such an integer.

Thus, the correct answer is B .

11 答案 B

第一个符号有 \(5\) 种选择，第二个有 \(21\) 种选择，第三个必须是不同的非元音字母，所以有 \(20\) 种选择，最后一个数字有 \(10\) 种选择。

因此可能牌照数为 \(5\cdot21\cdot20\cdot10=21{,}000\)。其中正好一个是 AMC8，所以概率为 \(\dfrac{1}{21{,}000}\)。

所以正确答案是 B。

English

There are \(5\) choices for the first symbol, \(21\) choices for the second, \(20\) choices for the third because it must be a different non-vowel, and \(10\) choices for the final digit.

Thus there are \(5\cdot21\cdot20\cdot10=21{,}000\) possible plates. Exactly one of these is AMC8, so the probability is \(\dfrac{1}{21{,}000}.\)

Thus, B is the correct answer.

12 答案 C

立方体有 \(12\) 条棱。对任意一条棱，都有另外 \(3\) 条棱与它平行。

这样每一对平行棱都被数了两次，所以平行棱对数为 \(\dfrac{12\cdot3}{2}=18\)。

所以正确答案是 C。

English

A cube has \(12\) edges. For any edge, there are \(3\) other edges parallel to it.

This counts each pair twice, once from each edge in the pair, so the number of pairs of parallel edges is \(\dfrac{12\cdot3}{2}=18.\)

Thus, C is the correct answer.

13 答案 D

原集合的和为 \(1+2+\cdots+11=66\)。移除两个数后剩下 \(9\) 个数，平均数要为 \(6\)，所以剩余数总和必须是 \(9\cdot6=54\)。

因此被移除的两个数之和必须是 \(66-54=12\)。可能的二元素子集为 \(\{1,11\}\)、\(\{2,10\}\)、\(\{3,9\}\)、\(\{4,8\}\)、\(\{5,7\}\)，共有 \(5\) 种。

所以正确答案是 D。

English

The original set has sum \(1+2+\cdots+11=66.\) After removing two numbers, \(9\) numbers remain and must have mean \(6,\) so their sum must be \(9\cdot6=54.\)

Therefore the two removed numbers must have sum \(66-54=12.\) The possible two-element subsets are \(\{1,11\},\) \(\{2,10\},\) \(\{3,9\},\) \(\{4,8\},\) \(\{5,7\},\) so there are \(5\) choices.

Thus, D is the correct answer.

14 答案 D

设四个连续奇整数为 \(2k+1\)、\(2k+3\)、\(2k+5\) 和 \(2k+7\)。它们的和为 \[8k+16=8(k+2)\text{。}\]

所以这样的和必须是 \(8\) 的倍数。选项中唯一不能被 \(8\) 整除的是 \(100\)。

所以正确答案是 D。

English

Let the four consecutive odd integers be \(2k+1,\) \(2k+3,\) \(2k+5,\) and \(2k+7.\) Their sum is \[8k+16=8(k+2).\]

So any such sum must be a multiple of \(8.\) The only answer choice that is not divisible by \(8\) is \(100.\)

Thus, D is the correct answer.

15 答案 D

因为 \(29\) 名学生两个议题都反对，所以 \(198-29 = 169\) 名学生至少支持一个议题。

已知 \(149\) 人支持第一个议题，\(119\) 人支持第二个议题，且 \(169\) 人至少支持一个议题。由容斥原理，两个都支持的人数为 \[149+119-169 = 99\text{。}\]

所以正确答案是 D。

English

Since \(29\) students voted against both, we know that \(198-29 = 169\) people voted for at least one.

As we know that \(149\) students voted for the first issue, and \(119\) students voted for the second issue, and \(169\) students that voted for at least one issue, we conclude that the number of students that voted for both is \[149+119-169 = 99.\]

Thus, the correct answer is D .

16 答案 B

设六年级学生有 \(s\) 人，九年级学生有 \(n\) 人。配对的九年级学生数等于配对的六年级学生数，所以 \[\dfrac13 n=\dfrac25 s\text{。}\]

得 \(5n=6s\)，所以 \(n:s=6:5\)。学生总数对应 \(11\) 份。

配对的九年级学生占全体学生的 \(\dfrac13\cdot\dfrac6{11}=\dfrac2{11}\)，配对的六年级学生占全体学生的 \(\dfrac25\cdot\dfrac5{11}=\dfrac2{11}\)。合计有 \(\dfrac4{11}\) 的学生有伙伴。

所以正确答案是 B。

English

Let there be \(s\) sixth graders and \(n\) ninth graders. The number of paired ninth graders equals the number of paired sixth graders, so \[\dfrac13 n=\dfrac25 s.\]

This gives \(5n=6s,\) so \(n:s=6:5.\) The total number of students is therefore proportional to \(11\) parts.

The paired ninth graders make up \(\dfrac13\cdot\dfrac6{11}=\dfrac2{11}\) of all students, and the paired sixth graders make up \(\dfrac25\cdot\dfrac5{11}=\dfrac2{11}\) of all students. Altogether, \(\dfrac4{11}\) of the students have a buddy.

Thus, B is the correct answer.

17 答案 D

设高峰期速度为每小时 \(s\) 英里。\(20\) 分钟是 \(\dfrac13\) 小时，所以距离为 \(\dfrac{s}{3}\)。

没有交通拥堵时，速度为每小时 \(s+18\) 英里，路程用时 \(12\) 分钟，即 \(\dfrac15\) 小时。同一距离为 \(\dfrac{s+18}{5}\)。

令两种距离相等：\[ \dfrac{s}{3}=\dfrac{s+18}{5} \] 得 \(5s=3s+54\)，所以 \(s=27\)。距离为 \(\dfrac{27}{3}=9\) 英里。

所以正确答案是 D。

English

Let the rush-hour speed be \(s\) miles per hour. The \(20\)-minute rush-hour trip takes \(\dfrac13\) hour, so the distance is \(\dfrac{s}{3}.\)

Without traffic, the speed is \(s+18\) miles per hour and the trip takes \(12\) minutes, or \(\dfrac15\) hour. The same distance is \(\dfrac{s+18}{5}.\)

Set the distances equal: \[\dfrac{s}{3}=\dfrac{s+18}{5}.\] Then \(5s=3s+54,\) so \(s=27.\) The distance is \(\dfrac{27}{3}=9\) miles.

Thus, D is the correct answer.

18 答案 B

在任意五项等差数列中，中间项是首项和末项的平均数。

顶行的中间项为 \(\dfrac{1+25}{2}=13\)，底行的中间项为 \(\dfrac{17+81}{2}=49\)。

再对中间列使用同样事实，得 \(X=\dfrac{13+49}{2}=31\)。

所以正确答案是 B。

English

In any five-term arithmetic sequence, the middle term is the average of the first and last terms.

The middle entry of the top row is \(\dfrac{1+25}{2}=13,\) and the middle entry of the bottom row is \(\dfrac{17+81}{2}=49.\)

Now apply the same fact to the middle column: \(X=\dfrac{13+49}{2}=31.\)

Thus, B is the correct answer.

19 答案 A

整个网格的面积是 \(6\cdot5=30\)。要求三角形面积，可把它放在下图所示的宽为 \(4\)、高为 \(3\) 的长方形中。

长方形的面积为 \(4\cdot3=12\)。三个角上三角形的面积分别为 \(\dfrac12(4)(2)=4\)、\(\dfrac12(1)(3)=\dfrac32\) 和 \(\dfrac12(3)(1)=\dfrac32\)。因此 \[\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5 \end{aligned}\text{。}\]

因此覆盖比例为 \(\dfrac{5}{30} = \dfrac{1}{6} \)。

所以正确答案是 A。

English

The total area of the grid is \(6\cdot5=30.\) To find the area of the triangle, place it inside the \(4\)-by-\(3\) rectangle shown below.

The rectangle has area \(4\cdot3=12.\) The three corner triangles have areas \(\dfrac12(4)(2)=4,\) \(\dfrac12(1)(3)=\dfrac32,\) and \(\dfrac12(3)(1)=\dfrac32.\) Therefore, \[\begin{aligned} [\triangle ABC] &= 12-4-\dfrac32-\dfrac32\\ &=5. \end{aligned}\]

Therefore, the fraction of the area is \(\dfrac{5}{30} = \dfrac{1}{6} .\)

Thus, the correct answer is A .

20 答案 D

设 \(a\)、\(b\)、\(c\) 分别为 \(\$1\)、\(\$3\)、\(\$4\) 的袜子双数。则 \[a+b+c=12\text{，}\]\[a+3b+4c=24\text{。}\]

两式相减得 \(2b+3c=12\)。因为每种至少买一双，所以 \(b>0\)、\(c>0\)。又 \(3c<12\)，所以 \(c<4\)。从方程模 \(2\) 看，\(c\) 必须是偶数，所以 \(c=2\)。

于是 \(2b+6=12\)，得 \(b=3\)，所以 \(a=12-3-2=7\)。

所以正确答案是 D。

English

Let \(a,\) \(b,\) and \(c\) be the numbers of \(\$1,\) \(\$3,\) and \(\$4\) pairs, respectively. Then \[a+b+c=12,\] \[a+3b+4c=24.\]

Subtracting gives \(2b+3c=12.\) Since at least one pair of each type was bought, \(b>0\) and \(c>0.\) Also \(3c<12,\) so \(c<4.\) Modulo \(2,\) the equation gives \(c\) even, so \(c=2.\)

Then \(2b+6=12,\) so \(b=3,\) and \(a=12-3-2=7.\)

Thus, D is the correct answer.

21 答案 C

面积为 \(18\) 的正方形边长为 \(JB=\sqrt{18}=3\sqrt2\)。因为 \(\triangle JBK\) 是等边三角形，所以 \(BK=3\sqrt2\)。

面积为 \(32\) 的正方形边长为 \(FE=\sqrt{32}=4\sqrt2\)。因为 \(FE=BC\)，所以 \(BC=4\sqrt2\)。

在 \(B\) 点处，等边三角形、正方形和等角六边形的内角分别为 \(60^\circ\)、\(90^\circ\) 和 \(120^\circ\)。因此 \[\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ \end{aligned}\text{。}\]所以 \[[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12\text{。}\]

所以正确答案是 C。

English

The square with area \(18\) has side length \(JB=\sqrt{18}=3\sqrt2.\) Since \(\triangle JBK\) is equilateral, \(BK=3\sqrt2.\)

The square with area \(32\) has side length \(FE=\sqrt{32}=4\sqrt2.\) Since \(FE=BC,\) we have \(BC=4\sqrt2.\)

At \(B,\) the equilateral triangle, square, and equiangular hexagon contribute angles of \(60^\circ,\) \(90^\circ,\) and \(120^\circ,\) respectively. Hence \[\begin{aligned} \angle KBC &= 360^\circ-60^\circ-90^\circ\\ &\qquad-120^\circ\\ &=90^\circ. \end{aligned}\] Therefore \[[\triangle KBC]=\dfrac12(3\sqrt2)(4\sqrt2)=12.\]

Thus, C is the correct answer.

22 答案 C

每行可站的人数正好是学生总数的正因数。因为六月 \(1\) 日至六月 \(12\) 日有不同排列，而六月 \(13\) 日没有新的排列方式，所以总人数必须恰好有 \(12\) 个正因数。

总人数必须同时能被 \(15\) 和 \(6\) 整除，因此能被 \(\operatorname{lcm}(15,6)=30=2\cdot3\cdot5\) 整除。而这个数只有 \(8\) 个因数。

最小的 \(30\) 的倍数且有 \(12\) 个因数的是 \(60=2^2\cdot3\cdot5\)，其因数个数为 \((2+1)(1+1)(1+1)=12\)。

所以正确答案是 C。

English

The possible numbers of students per row are exactly the positive divisors of the total number of students. Since June \(1\) through June \(12\) give different arrangements and June \(13\) gives no new one, the total number of students must have exactly \(12\) positive divisors.

The number must be divisible by both \(15\) and \(6,\) hence by \(\operatorname{lcm}(15,6)=30=2\cdot3\cdot5.\) This number has only \(8\) divisors.

The smallest multiple of \(30\) with \(12\) divisors is \(60=2^2\cdot3\cdot5,\) which has \((2+1)(1+1)(1+1)=12\) divisors.

Thus, C is the correct answer.

23 答案 D

所有纸条的和为 \(35\)，所以五个连续整数杯子和的平均数必须是 \(7\)。因此杯子 \(A\)、\(B\)、\(C\)、\(D\)、\(E\) 的和分别为 \(5\)、\(6\)、\(7\)、\(8\)、\(9\)。

杯子 \(B\) 已经有一张 \(3\)，且总和必须为 \(6\)，所以它还必须有另一张 \(3\)。杯子 \(E\) 已经有一张 \(2\)，所以 \(E\) 中其他纸条之和必须为 \(7\)。

\(3.5\) 不能放在 \(A\)，因为杯子 \(A\) 还需要 \(1.5\)。它不能放在 \(B\)，因为这个杯子已经满了。它不能放在 \(C\) 或 \(E\)，因为那样还需要另一个 \(3.5\)，而剩余纸条不能凑出这个和。杯子 \(D\) 可行，例如 \(3.5+4.5=8\)。

所以正确答案是 D。

English

The sum of all the slips is \(35,\) so the five consecutive integer cup sums must average \(7.\) Therefore cups \(A,\) \(B,\) \(C,\) \(D,\) and \(E\) must have sums \(5,\) \(6,\) \(7,\) \(8,\) and \(9,\) respectively.

Cup \(B\) already contains a \(3\) and must sum to \(6,\) so it must contain another \(3.\) Cup \(E\) already contains a \(2,\) so the other slips in \(E\) must sum to \(7.\)

The \(3.5\) slip cannot go in \(A,\) because cup \(A\) would need another \(1.5.\) It cannot go in \(B,\) which is already full. It cannot go in \(C\) or \(E,\) because either would then need another \(3.5,\) and no remaining slips can make that total. Cup \(D\) works, for example with \(3.5+4.5=8.\)

Thus, D is the correct answer.

24 答案 B

每支球队在本分区内打 \(3N\) 场，对另一个分区打 \(4M\) 场，所以 \[3N+4M=76\text{。}\]

因为 \(N>2M\)，所以 \(3N>6M\)，从而 \(76=3N+4M>10M\)。因此 \(M<7.6\)。又 \(M>4\)，所以 \(M\in\{5,6,7\}\)。

将 \(3N+4M=76\) 模 \(3\)，得 \(M\equiv1\pmod3\)，所以 \(M=7\)。因此每支球队打 \(4M=28\) 场非分区比赛，本分区比赛为 \(76-28=48\) 场。

所以正确答案是 B。

English

Each team plays \(3N\) games within its own division and \(4M\) games against the other division, so \[3N+4M=76.\]

Since \(N>2M,\) we have \(3N>6M,\) and hence \(76=3N+4M>10M.\) Thus \(M<7.6.\) Together with \(M>4,\) this gives \(M\in\{5,6,7\}.\)

Reducing \(3N+4M=76\) modulo \(3\) gives \(M\equiv1\pmod3,\) so \(M=7.\) Therefore the team plays \(4M=28\) non-division games and \(76-28=48\) division games.

Thus, B is the correct answer.

25 答案 C

对于能放入的最大正方形，它的每条边都必须接触一个被挖去的边长为 \(1\) 英寸的小正方形的内角，否则还可以继续放大。因此它围住中央的 \(3\times3\) 正方形，并在四边各增加一个全等直角三角形。

中央正方形面积为 \(3\cdot3=9\)。每个增加的三角形两条直角边为 \(3\) 和 \(1\)，四个三角形总面积为 \[4\left(\dfrac{3\cdot1}{2}\right)=6\text{。}\]

放入的正方形面积为 \(9+6=15\)。

所以正确答案是 C。

English

For a largest fitted square, each side must touch an inner corner of one of the removed \(1\)-inch squares; otherwise the fitted square could be enlarged. Thus it surrounds the central \(3\times3\) square and adds four congruent right triangles, one along each side.

The central square has area \(3\cdot3=9.\) Each added triangle has base \(3\) and height \(1,\) so the four triangles have total area \[4\left(\dfrac{3\cdot1}{2}\right)=6.\]

The fitted square has area \(9+6=15.\)

Thus, C is the correct answer.