2012 年 AMC 8 真题（P版）

1 答案 E

\(8\) 个汉堡是 \(24\) 个汉堡的 \(\dfrac{1}{3}\)，所以 \(3\) 磅也是所需肉量的 \(\dfrac{1}{3}\)。因此总肉量为 \(\dfrac{1}{\dfrac{1}{3}} \cdot 3 = 3\cdot3 = 9\)。

所以正确答案是 E。

English

If we have \(8\) hamburgers, we have \(\dfrac{1}{3}\) of the \(24\) hamburgers. This means we have \(\dfrac{1}{3} \) of the meat when we have \(3\) pounds. The total amount of meat is therefore \(\dfrac{1}{\dfrac{1}{3}} \cdot 3 = 3\cdot3 = 9.\)

Thus, the answer is E .

2 答案 B

每 \(8\) 小时出生 \(1\) 个婴儿，所以每 \(24\) 小时出生 \(3\) 个婴儿。因此每天有 \(3\) 人出生、\(1\) 人死亡，平均每天人口净增加 \(2\) 人。一年有 \(365\) 天，每天净增加 \(2\) 人，所以一年净增加约 \(2\cdot 365 = 730 \) 人，约为 \(700\)。

所以正确答案是 B。

English

Since we have \(1\) birth every \(8\) hours, we have \(3\) births every \(24\) hours. Therefore, we have \(3\) births a day and \(1\) death a day. The net change in population every day should be on average \(2.\) Since we have \(365\) days in a year and \(2\) added to the population every day, the net change in population should be around \(2\cdot 365 = 730 .\) This is approximately \(700.\)

Thus, the answer is B .

3 答案 B

从 \(6:57 \text{AM}\) 起 \(10\) 小时后是 \(4:57 \text{PM} \)。也就是说，日出后 \(10\) 小时 \(3\) 分钟是 \(5:00 \text{PM} \)。距日落还有 \(21\) 分钟，所以实际日落时间是 \(5:21 \text{PM} \)。

所以正确答案是 B。

English

Since \(10\) hours after \(6:57 \text{AM} \) is \(4:57 \text{PM} ,\) we can then say \(10\) hours and \(3\) minutes after sunrise is \(5:00 \text{PM} .\) We then have \(21\) more minutes until sunset, so sunset is \(5:21 \text{PM} .\)

Thus, the answer is B .

4 答案 C

Peter 吃了 \(1\) 整片，又吃了他分得的那片的 \(\dfrac 12\)。

因此他吃了整个披萨的 \(\dfrac{\frac{3}{2}}{12}\)，也就是 \(\dfrac 18\)。

所以正确答案是 C。

English

Peter ate \(1\) full slice, and he ate \(\dfrac 12 \) of the slice that he split.

Therefore, he ate \( \dfrac{\frac{3}{2}}{12} \) of the pizza, which is equivalent to \(\dfrac 18.\)

Thus, the answer is C .

5 答案 E

先从右侧高度相加求图形总高度，得 \(1+2+1+6 = 10\)。

再从左侧高度相加，得 \(1+1+1+2+X = 5+X\)。

两个高度相同，所以 \(10=5+X \)，因此 \(X = 5\)。

所以正确答案是 E。

English

First, we can find the height of the object by getting the sum of the heights on the right. Therefore, the height is \(1+2+1+6 = 10.\)

Next, we can find the height of the object by getting the sum of the heights on the left. Therefore, the height is \(1+1+1+2+X = 5+X.\)

Since the heights are the same, we know \(10=5+X ,\) so \(X = 5.\)

Thus, the answer is E .

6 答案 E

每一边都增加 \(2\) 英寸，所以高度和宽度各总共增加 \(4\) 英寸。外部长方形尺寸为 \(12 \times 14\)，面积为 \(12\cdot 14 = 168\)。

照片本身面积为 \(8 \cdot 10 =80\)。

因此边框面积为 \(168-80=88\)。

所以正确答案是 E。

English

If we add \(2\) inches on each side, we add \(4\) inches total on both sides. This means that the dimensions of the outer part of the frame are \(12 \times 14.\) The area of this is \(12\cdot 14 = 168.\)

However, we must take out the area of the inner part of the frame which has area \(8 \cdot 10 =80.\)

Therefore, the total area is \(168-80=88.\)

Thus, the answer is E .

7 答案 B

若平均分为 \(95\)，四次测试总分为 \(95\cdot 4 = 380 \)。由于已知前两次成绩，后两次测试总分是 \(380\) 减去前两次成绩。

后两次测试总分必须为 \(380-91-97 = 192\)。因此后两次成绩之和为 \(192\)。

要使第三次成绩最低，就让第四次成绩最高，为 \(100\)。于是第三次成绩为 \(192 -100 = 92\)。

所以正确答案是 B。

English

If the average is \(95,\) then the sum of the tests is \(95\cdot 4 = 380 .\) Since we have the first two tests, the sum of the last two tests is \(380\) minus the first two scores.

This makes the sum of the last two scores equal to \(380-91-97 = 192.\) Her last two scores therefore have a sum of \(192.\)

Given the sum of the tests we try to minimize one score, then we must maximize the other test. Therefore, we maximize the fourth test by making it \(100.\) This would make the third test equal to \(192 -100 = 92.\)

Thus, the answer is B .

8 答案 D

设原价为 \(p\)。半价后价格为 \(0.5p\)。

再打 \(20\%\) 折扣表示支付价格的 \(80\%\)，所以最终价格为 \(0.5p\cdot 0.8 =0.4p\)。

因此减少了 \(0.6p\)，折扣率为 \(\dfrac{0.6p}{p} = 0.6\)，即 \(60\%\)。

所以正确答案是 D。

English

Let \(p\) be the original price. If everything is half off, we have the new price as \(0.5p.\)

Having a \(20\%\) discount makes it such that we keep \(80\%\) of the price, so the price is \(0.5p\cdot 0.8 =0.4p.\)

This would have \(0.6p\) off, so we get a discount of \(\dfrac{0.6p}{p} = 0.6,\) which is \(60\%.\)

Thus, the answer is D .

9 答案 C

设 \(f\) 为 \(4\) 条腿动物的数量，\(t\) 为 \(2\) 条腿动物的数量。

数腿得到 \(2t+4f = 522\)，数头得到 \(t + f = 200\)。将第二式乘以四，得 \(4t+4f = 800\)。减去第一式，得到 \(2t = 278 \implies t = 139\)。因此有 \(139\) 只两条腿的鸟。

所以正确答案是 C。

English

Let \(f\) be the number of animals with \(4\) legs and let \(t\) be the number of animals with \(2\) legs.

Counting the number of legs yields \(2t+4f = 522\) and counting the number of heads yields \(t + f = 200.\) This means \(4t+4f = 800,\) and subtracting the first equation from the second yields \(2t = 278 \implies t = 139.\) This means there are \(139\) two-legged birds.

Thus, the answer is C .

10 答案 D

\(0\) 不能放在千位。因此有 \(3\) 个位置可放。放好零后，数字 \(1\) 有 \(3\) 个可用位置，两个 \(2\) 自动占据剩余位置。因此共有 \(3\cdot 3 = 9\) 个数。

所以正确答案是 D。

English

First, we can’t have the \(0\) in the thousands position. Therefore, we have \(3\) spots we can put it. Then, we have \(3\) available positions for the \(1,\) and then the two \(2\)s are placed. This makes it such that we have \(3\cdot 3 = 9\) combinations.

Thus, the answer is D .

11 答案 D

已列出的数中，除 \(6\) 外每个数只出现一次，而 \(6\) 出现两次。若 \(x\) 等于其他已列出的数，就会有两个众数，不再有唯一众数。因此唯一众数必须是 \(6\)，平均数也必须是 \(6\)。共有 \(7\) 个数，所以总和为 \(6\cdot 7 = 42\)。另一方面，总和还等于 \[3+4+5+6+6+7+x\] \[= 31+x \] \[= 42\text{，}\]所以 \(x = 11\)。

所以正确答案是 D。

English

Every listed value except \(6\) appears once while \(6\) appears twice. If \(x\) equals any of the other listed values, then we have two modes, which means we do not have a unique mode. Otherwise, the only value that shows up more than once is \(6\) making that the unique mode. This also means \(6\) is the mean. Since there are \(7\) elements, the sum of the elements is \(6\cdot 7 = 42.\) The sum is also \[3+4+5+6+6+7+x\] \[= 31+x \]\[= 42,\] so \(x = 11.\)

Thus, the answer is D .

12 答案 A

要求个位数字，只需计算 \(13^{2012} \mod 10\)。

\[\begin{align*}13^{2012} &\equiv 3^{2012} \mod 10\\&\equiv 81^{503} \mod 10\\&\equiv 1^{503} \mod 10\\&\equiv 1 \mod 10\end{align*}\]

这说明 \(13^{2012}\) 的个位数字与 \(1\) 相同，所以 \(13^{2012}\) 的个位数字是 \(1\)。

所以正确答案是 A。

English

We have to find \(13^{2012} \mod 10.\) The following is true:

\[\begin{align*}13^{2012} &\equiv 3^{2012} \mod 10\\&\equiv 81^{503} \mod 10\\&\equiv 1^{503} \mod 10\\&\equiv 1 \mod 10\end{align*}\]

This means \(13^{2012}\) has the same units digit as \(1,\) so the units digit of \(13^{2012} \) is \(1.\)

Thus, the answer is A .

13 答案 C

设每支铅笔价格为 \(c\) 美分。则 \(c\) 同时整除 \(143\) 和 \(187\)，且 \(c\gt1\)。

因为 \(143=11\cdot13\)，且 \(187=11\cdot17\)，最大公因数是 \(11\)。所以每支铅笔 \(11\) 美分。

Sharona 多付了 \(187-143=44\) 美分，所以她多买了 \(\frac{44}{11}=4\) 支铅笔。所以正确答案是 C。

English

Let \(c\) be the price of one pencil in cents. Then \(c\) divides both \(143\) and \(187\), and \(c\gt1\).

Since \(143=11\cdot13\) and \(187=11\cdot17\), the greatest common divisor is \(11\). Thus each pencil costs \(11\) cents.

Sharona paid \(187-143=44\) cents more, so she bought \(\frac{44}{11}=4\) more pencils. Thus, the answer is C .

14 答案 B

若有 \(N\) 支球队，每支球队要与 \(N-1\) 支球队比赛。不过，每场比赛涉及 \(2\) 支球队，所以把 \(N\) 与 \(N-1\) 相乘会把每场比赛数两次。因此 \(N(N-1) = 42\)。这给出 \[N^2 -N +0.25 = 42.25\] 从而 \[(N-0.5)^2 = 6.5^2\text{，}\]于是 \[N-0.5 = 6.5\]\[ N = 7\]

所以正确答案是 B。

English

Each of the \(N\) teams plays \(N-1\) games. However, \(2\) teams play each game, so multiplying \(N\) and \(N-1\) would be twice the number of games. Therefore, we know \(N(N-1) = 42.\) This leads to \[N^2 -N +0.25 = 42.25\] which implies \[(N-0.5)^2 = 6.5^2. \] In turn, this suggests: \[N-0.5 = 6.5\]\[ N = 7\]

Thus, the answer is B .

15 答案 D

设这个数为 \(x\)。因为它除以 \(3\)、\(4\)、\(5\) 和 \(6\) 时余数都是 \(2\)，所以 \(x-2\) 是 \(3\)、\(4\)、\(5\) 和 \(6\) 的公倍数。这表示 \(x-2\) 是 \(\operatorname{lcm}(3,4,5,6)\) 的倍数，而这个最小公倍数是 \(60\)。因此 \(x-2\) 必须是 \(60\) 的倍数。下一个满足条件的数出现在 \(x-2 = 60 \implies x = 62 \) 时。

所以正确答案是 D。

English

Let the number be \(x.\) Since it leaves a remainder of \(2\) when divided by \(3\), \(4\), \(5\), and \(6,\) we know \(x-2\) is a multiple of \(3\), \(4\), \(5\), and \(6.\) This means \(x-2 \) is a multiple of \(\operatorname{lcm}(3,4,5,6)\), which is \(60.\) Therefore, \(x-2\) must be a multiple of \(60.\) The next number such that this occurs is when \(x-2 = 60 \implies x = 62 .\)

Thus, the answer is D .

16 答案 C

要让两个五位数的和最大，从左到右的各位数字应尽可能大。因此两个数的第一位必须是 \(9\) 或 \(8\)，第二位是 \(7\) 或 \(6\)，第三位是 \(5\) 或 \(4\)，第四位是 \(3\) 或 \(2\)，最后一位是 \(1\) 或 \(0\)。

给出的数中，唯一满足这种位置要求的是 \(87431\)。

所以正确答案是 C。

English

To construct two five-digit numbers, the digits in the largest place values must be as great as possible. Therefore, the leftmost digits must be the greatest two digits. This means the first digit must be either \(9\) or \(8,\) the second digit must be either \(7\) or \(6,\) the third digit must be either \(5\) or \(4,\) the fourth digit must be either \(3\) or \(2,\) and the last digit must be either \(1\) or \(0.\)

The only one of the given numbers that satisfy this is \(87431.\)

Thus, the answer is C .

17 答案 B

因为这 \(10\) 个小正方形的边长都是正整数，所以每个小正方形的面积至少为 \(1\)，总面积至少为 \(10\)。边长不超过 \(3\) 的整数边长正方形面积至多为 \(3^2=9\)，所以大正方形的边长不能小于 \(4\)。

下图把一个 \(4\times4\) 正方形切成十个整数边长的正方形，其中八个是单位正方形。因此边长 \(4\) 可以达到，而且是最小值。

所以正确答案是 B。

English

Since all \(10\) squares have positive integer side lengths, each has area at least \(1\). Their total area is therefore at least \(10\). A square with integer side length at most \(3\) has area at most \(3^2=9\), so its side length cannot be less than \(4\).

The following configuration cuts a \(4\times4\) square into ten integer-sided squares, eight of which are unit squares. Therefore side length \(4\) is attainable and is the minimum.

Thus, the answer is B .

18 答案 A

这个数是合数，不是质数，并且没有小于 \(50\) 的质因数。因此最小可能的质因数是 \(53\)、\(59\)、\(61\)，依此类推。

像 \(53^2\) 这样的平方数不允许，而 \(53^3\) 远大于 \(53\cdot59\)。最小的非平方合数是 \(53\cdot59=3127\)。

所以正确答案是 A。

English

The number is composite, not prime, and has no prime factor below \(50\). The smallest possible prime factors are therefore \(53\), \(59\), \(61\), and so on.

A square such as \(53^2\) is not allowed, and \(53^3\) is much larger than \(53\cdot59\). The smallest allowed nonsquare composite is \(53\cdot59=3127\).

Thus, the answer is A .

19 答案 C

设红、绿、蓝弹珠数分别为 \(r,g,b\)。由题意可知 \[r+g+b -r=6\text{，}\]\[r+g+b-g = 8\text{，}\]\[ r+g+b-b = 4\text{。}\]把这些方程相加，得到 \(3(r+g+b) -(r+g+b) = 18\)。因此 \(2(r+g+b) = 18\)，所以 \(r+g+b = 9\)。也就是说，弹珠总数为 \(9\)。

所以正确答案是 C。

English

Let \(r,g,b\) be the number of red marbles, green marbles, and blue marbles respectively. We then know \[r+g+b -r=6,\]\[r+g+b-g = 8,\]\[ r+g+b-b = 4\] by the statements given. Adding these equations yields \(3(r+g+b) -(r+g+b) = 18.\) This would mean \(2(r+g+b) = 18,\) so \(r+g+b = 9.\) Therefore, the sum of all of the marbles is \(9.\)

Thus, the answer is C .

20 答案 B

中间的分数为 \(\dfrac{7}{21}=\dfrac13\)。又因为 \(3\cdot5=15\lt19\)，所以 \(\dfrac{5}{19}\lt\dfrac13\)；因为 \(3\cdot9=27\gt23\)，所以 \(\dfrac{9}{23}\gt\dfrac13\)。因此 \[\dfrac{5}{19}\lt\dfrac{7}{21}\lt\dfrac{9}{23}\text{。}\]

所以正确答案是 B。

English

The middle fraction is \(\dfrac{7}{21}=\dfrac13\). Also, \(\dfrac{5}{19}\lt\dfrac13\) because \(3\cdot5=15\lt19\), while \(\dfrac{9}{23}\gt\dfrac13\) because \(3\cdot9=27\gt23\). Therefore \[\dfrac{5}{19}\lt\dfrac{7}{21}\lt\dfrac{9}{23}.\]

Thus, the answer is B .

21 答案 D

立方体总表面积为 \(6\cdot 10\cdot 10 = 600\)。因此未被绿色覆盖的面积为 \(600-300 = 300\) 平方英尺，平均到每个面是 \(\dfrac{300}{6} = 50\) 平方英尺。

所以正确答案是 D。

English

The total surface area is \(6\cdot 10\cdot 10 = 600.\) Therefore, \(600-300 = 300\) square feet aren’t covered. This would be \( \dfrac{300}{6} = 50 \) square feet per face.

Thus, the answer is D .

22 答案 D

九个互不相同整数排序后，中位数是第五个数。

中位数不能小于 \(3\)，否则 \(3\)、\(4\)、\(6\)、\(9\) 和 \(14\) 已经给出至少五个更大的元素。中位数也不能大于 \(9\)，否则 \(2\)、\(3\)、\(4\)、\(6\) 和 \(9\) 已经给出至少五个更小的元素。

从 \(3\) 到 \(9\) 的每个整数都可以通过适当选择另外三个整数成为中位数。因此可能的中位数有 \(7\) 个。所以正确答案是 D。

English

In a sorted set of nine distinct integers, the median is the fifth number.

The median cannot be below \(3\), since then \(3\), \(4\), \(6\), \(9\), and \(14\) would already give at least five larger elements. It cannot be above \(9\), since \(2\), \(3\), \(4\), \(6\), and \(9\) would already give at least five smaller elements.

Each integer from \(3\) through \(9\) can be made the median by choosing the three missing integers appropriately. Therefore there are \(7\) possible medians. Thus, the answer is D .

23 答案 C

设三角形边长为 \(s\)。其周长为 \(3s\)，所以六边形边长为 \(\frac{3s}{6} = \frac s2\)。

如图，正六边形可分成 \(6\) 个边长为 \(\dfrac{s}{2}\) 的等边三角形。每个小三角形是原三角形按比例 \(\dfrac{1}{2}\) 缩小得到的，所以面积缩小为 \((\dfrac{1}{2})^2 = \dfrac{1}{4} \)。因此每个小三角形面积为 \(4 \cdot \dfrac{1}{4} = 1\)。共有 \(6\) 个，所以六边形面积为 \(6\cdot 1 = 6\)。

所以正确答案是 C。

English

Let the side length of the triangle be \(s.\) This means the perimeter is \(3s.\) Therefore, the side length for the hexagon is \( \frac{3s}{6} = \frac s2.\)

A hexagon can be made of \(6\) equilateral triangles with side length \( \dfrac{s}{2} \) as shown above. Each triangle is the original triangle scaled down by \(\dfrac{1}{2},\) so the area is scaled down by \( (\dfrac{1}{2})^2 = \dfrac{1}{4} .\) Therefore, the area of each of these triangles is \(4 \cdot \dfrac{1}{4} = 1.\) Since there are \(6\) of them, the area is \(6\cdot 1 = 6.\)

Thus, the answer is C .

24 答案 A

原圆面积为 \(\pi\cdot2^2=4\pi\)。

连接四个四分之一圆弧端点形成一个正方形。这个正方形的两条对角线分别是 \(4\) 和 \(4\)，所以面积为 \(\frac12\cdot4\cdot4=8\)。

圆内但正方形外的部分面积为 \(4\pi-8\)。这四块与正方形内但星形外的四块全等。

因此星形面积为 \(8-(4\pi-8)=16-4\pi\)。所求比值为 \(\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}\)。所以正确答案是 A。

English

The area of the original circle is \(\pi\cdot2^2=4\pi\).

Join the four quarter-circle endpoints to form a square. The square has diagonals \(4\) and \(4\), so its area is \(\frac12\cdot4\cdot4=8\).

The part inside the circle but outside this square has area \(4\pi-8\). Those four pieces are congruent to the pieces inside the square but outside the star.

Thus the star area is \(8-(4\pi-8)=16-4\pi\). The desired ratio is \(\dfrac{16-4\pi}{4\pi}=\dfrac{4-\pi}{\pi}\). Thus, the answer is A .

25 答案 C

四个三角形都可以通过旋转互相得到，所以它们全等。因此可以如图标出 \(a\)。两个正方形之间的总面积为 \(5-4 = 1\)，所以有 \(4\) 个全等三角形的总面积为 \(1\)。每个三角形面积为 \(\dfrac{1}{4}\)。每个三角形的面积也等于 \(\frac {ab}2\)，所以 \(\frac{ab}2 = \dfrac{1}{4}\)。因此 \(ab = \dfrac{1}{2} \)。

所以正确答案是 C。

English

Since all the triangles can be made from each other by rotating them around, they are all congruent. Therefore, we can place the \(a\) as we have. The total area of the triangles is \(5-4 = 1,\) so we have \( 4 \) congruent triangles with a combined area of \(1.\) This means the area of each triangle is \( \dfrac{1}{4}.\) The area of each triangle is also \( \frac {ab}2,\) so \( \frac{ab}2 = \dfrac{1}{4}.\) This means \(ab = \dfrac{1}{2} .\)

Thus, the correct answer is C .