2012B AMC 10 真题（P版）

1 答案 C

每个教室中学生比兔子多 \(18-2=16\)。

\(4\) 个教室中总共多 \(4\cdot16=64\)。

所以正确答案是 C。

English

Each classroom has \(18-2=16\) more students than rabbits.

Across \(4\) classrooms, the difference is \(4\cdot16=64\).

Thus, C is the correct answer.

2 答案 E

短边等于圆的直径，即 \(2r=2\cdot 5=10\)。

长与宽之比为 \(2:1\)，所以长边为 \(10\cdot 2=20\)。

长方形面积为 \(20\cdot 10=200\)。

所以正确答案是 E。

English

The smaller side of the rectangle is equal to the diameter, which is \(2r=2\cdot 5=10.\)

Due to the \(2:1\) ratio, the long side is \(10\cdot 2=20.\)

Therefore, the area is \(20\cdot 10=200.\)

Thus, the correct answer is E .

3 答案 B

注意到这里的直线是完全水平的。这意味着，只要从该点向这条直线作垂线段，再把沿这条线段的距离加倍，就能得到反射后点的坐标。

点 \((1000,2012)\) 在 \(y=2000\) 上方 \(12\) 个单位，所以对称点在直线另一侧同样相距 \(12\) 个单位，即 \((1000,1988)\)。

所以正确答案是 B。

English

Notice that the line in question is perfectly horizontal. This means that if we were to construct a perpendicular line segment from the point to the line, to find the reflected coordinates of the point, we simply double the distance along that line segment.

The line segment from \((1000,2012)\) to \(y=2000\) is of length \(12,\) so the reflected point is along this same line segment, but a distance of \(12\) on the other side of the horizontal line. This yields \((1000,1988).\)

Thus, the correct answer is B .

4 答案 A

Ringo 的弹珠数除以 \(6\) 余 \(4\)，所以可写成 \(6x+4\)，其中 \(x\) 为整数。

同理，Paul 的弹珠数可写为 \(6y+3\)，其中 \(y\) 为整数。

因此总数为 \[(6x+4)+(6y+3) \] \[= 6(x+y+1)+1\] 这表示合在一起后除以 \(6\) 余 \(1\)。

所以正确答案是 A。

English

As we know that when Ringo’s marbles are divided by \(6,\) we have a remainder of \(4,\) we conclude that he has \(6x+4\) marbles for some \(x.\)

Using the same logic, we can also conclude that Paul has \(6y+3\) marbles for some \(y.\)

Therefore, the total number of marbles is \[(6x+4)+(6y+3) \]\[= 6(x+y+1)+1\] which, when divided by \(6,\) only leaves \(1\) left over.

Thus, the correct answer is A .

5 答案 D

设税前餐费为 \(x\)。税为 \(0.1x\)，小费为 \(0.15x\)。总花费满足 \[\begin{aligned}x+0.1x+0.15x&=1.25x\\&=27.5\end{aligned}\] 所以 \(x=22\)。

所以正确答案是 D。

English

Suppose the original price is \(x.\) Then, the tax is \(0.1x\) and the tip is \(0.15x.\) This makes the total payment equal to: \[\begin{aligned}x+0.1x+0.15x&=1.25x\\&=27.5\end{aligned}\] Therefore, \(x=22.\)

Thus, the correct answer is D .

6 答案 A

设这段很小的舍入量为 \(d>0\)。Xiaoli 用 \(x+d\) 代替 \(x\)，用 \(y-d\) 代替 \(y\)。因此估计值为 \[(x+d)-(y-d)=x-y+2d\text{，}\] 它大于 \(x-y\)。

所以正确答案是 A。

English

Let the small rounding amount be \(d>0.\) Xiaoli replaces \(x\) by \(x+d\) and \(y\) by \(y-d.\) Her estimate is therefore \[(x+d)-(y-d)=x-y+2d,\] which is larger than \(x-y.\)

Thus, the correct answer is A .

7 答案 D

设两者各藏了 \(x\) 颗橡子。花栗鼠用了 \(\dfrac x3\) 个洞，松鼠用了 \(\dfrac x4 \) 个洞。

两者所用洞数之差为四，所以 \[\dfrac x3 - \dfrac x4 = \dfrac{x}{12}= 4\] 解得 \(x=48\)。

所以正确答案是 D。

English

Let the number of acorns they each hid be \(x.\) Then, the number of holes from the chipmunk is \(\dfrac x3\) and the number of holes from the squirrel is \(\dfrac x4 .\)

This means \[\dfrac x3 - \dfrac x4 = \dfrac{x}{12}= 4\] Therefore, \(x=48.\)

Thus, the correct answer is D .

8 答案 B

若 \(x=2+k\) 是一个解，则 \(x=2-k\) 也是解，因为 \[((2+k)-2)^2 = ((2-k)-2)^2\] 这两个解的和为 \(4\)。因此，所有整数解之和等于四乘以可行的正数 \(k\) 的个数。

正数 \(k\) 必须满足 \[1 < k^2 < 25\text{，}\] 共有 \(3\) 个，即 \(k=2,3,4\)。

所以总和为 \(4\cdot3=12\)。

所以正确答案是 B。

English

Suppose we have \(x=2+k\) as a solution. Then, \(x=2-k\) would also be a solution as \[((2+k)-2)^2 = ((2-k)-2)^2\] The sum of these two solutions would be \(4.\) Thus, the sum of all integer solutions to the above equation is four times the number of positive \(k\)’s that work.

To find the number of \(k\)’s, we need to find the number of positive solutions to: \[1 < k^2 < 25,\] which would be \(3,\) as \(k=2,3,4.\)

Therefore, the sum of all the solutions is \(4\cdot3=12.\)

Thus, the correct answer is B .

9 答案 A

前两个整数的和 \(26\) 是偶数，所以它们可以都是奇数。接下来两个数之和为 \(41-26=15\)。

这个和是奇数，所以这一对必须有一个偶数和一个奇数。最后两个数之和为 \(57-41=16\)，所以它们可以都是奇数。因此至少有一个偶数，而且这个下界可以达到。

所以正确答案是 A。

English

The first two integers have even sum \(26\), so they can both be odd. The next two integers have sum \(41-26=15\), which is odd, so one of them must be even and one odd.

The last two integers have sum \(57-41=16\), so they can both be odd. Therefore at least one integer must be even, and one is attainable.

Thus, A is the correct answer.

10 答案 D

交叉相乘得 \(MN = 36\)。因此可让 \(M\) 取 \(36\) 的任一正因数，再唯一确定 \(N\)。

由于 \(36=2^2\cdot 3^2\)，共有 \((2+1)(2+1)=9\) 个可能的 \(M\)，每个也对应唯一的 \(N\)。

所以正确答案是 D。

English

By cross multiplying, we can see that \(MN = 36.\) Thus, we can make \(M\) any factor of \(36\) and then determine \(N\) from it.

Since \(36=2^2\cdot 3^2,\) we have \((2+1)(2+1)=9\) possible choices for \(M,\) each of which also determine a unique \(N.\)

Thus, the correct answer is D .

11 答案 A

星期五固定为蛋糕。星期六不能是蛋糕，所以有 \(3\) 种选择。

从星期五往前，星期四有 \(3\) 种选择；再往前每一天也都只有不能等于后一日的限制，因此各有三种选择。星期日到星期四共 \(5\) 天，所以总数为 \(3^5\cdot 3 = 729\)。

所以正确答案是 A。

English

We know that there must be cake served on Friday, and as such, on Saturday, we cannot have cake. Therefore, we have \(3\) choices for Saturday’s menu.

Furthermore, for each of the \(5\) previous days, we could go backwards and have \(3\) choices on each day, making the total number of choices \(3^5\cdot 3 = 729.\)

Thus, the correct answer is A .

12 答案 B

因为 \(AB\) 与 \(BC\) 垂直，所以 \[AB^2 + BC^2 = (10\sqrt 2)^2 = 200\text{。}\] 又因为 \(\angle BAC = 45^\circ\)，\(\triangle ABC\) 是等腰直角三角形，所以 \(AB = BC \)，从而 \(2AB^2 = 200\)。因此 \(AB = BC = 10\)。

于是 \[BD= BC+CD = 30\text{。}\] 由勾股定理， \[\begin{aligned}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{aligned}\] 因为 \[31^2 \lt AD^2 \lt 32^2\text{，}\] 所以 \[31 \lt AD\lt 32\]

所以正确答案是 B。

English

We know \(AB\) and \(BC\) are perpendicular, so \[AB^2 + BC^2 = (10\sqrt 2)^2 = 200.\] Also, as \(\angle BAC = 45^\circ,\) we know that \(\triangle ABC\) is an isosceles right triangle, so \(AB = BC ,\) making \(2AB^2 = 200.\) Thus, \(AB = BC = 10.\)

As such, we know that \[BD= BC+CD = 30.\] Thus, by the Pythagorean Theorem, we have that \[\begin{aligned}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{aligned}\] Thus, since \[31^2 \lt AD^2 \lt 32^2,\] we have \[31 \lt AD\lt 32\]

Thus, the correct answer is B .

13 答案 B

设 \(d\) 为扶梯的长度，\(c\) 为 Clea 在静止扶梯上步行的速度，\(e\) 为运行扶梯本身的速度。由速度等于路程除以时间，\(c = \dfrac d{60}\)，且 \[c+e = \dfrac d{24}\text{。}\] 因此 \[e = \dfrac d{24}-\dfrac d{60} = \dfrac d{40}\text{，}\] 这表明她站着不动时需要 \(40\) 秒。

所以正确答案是 B。

English

Let \(d\) be the escalator’s length, and let \(c\) represent Clea’s speed walking down the non-operational escalator. Similarly, let \(e\) represent Clea’s speed standing still on the operational escalator. Then, as speed is equal to distance over time, we know that \(c = \dfrac d{60}\) and \[c+e = \dfrac d{24}.\] Therefore, \[e = \dfrac d{24}-\dfrac d{60} = \dfrac d{40},\] meaning that Clea takes \(40\) seconds to descend the escalator by simply standing still.

Thus, the correct answer is B .

14 答案 D

这个菱形可分成两个全等的等边三角形。设每个小等边三角形的边长为 \(s\)，则一个面积为 \(\dfrac{s^2 \sqrt 3}4\)，两个总面积为 \(\dfrac{s^2 \sqrt 3}2\)。

大等边三角形边长为 \(2 \sqrt 3\)，其高为 \(2 \sqrt 3 \sin(60^\circ) \)，也就是 \(3\)。

正方形半边长为 \(\sqrt 3\)，所以较小三角形的高为 \(3 - \sqrt 3 \)。因此 \(s\) 与 \(2 \sqrt 3\) 的比为 \(\dfrac {3-\sqrt 3}3 \)。

因此 \[s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2\text{。}\]

所以菱形面积为 \[\begin{aligned} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{aligned}\]

所以正确答案是 D。

English

This rhombus is created by placing two congruent equilateral triangles. Let the side length of it be \(s.\) Then, the area of one of them is \(\dfrac{s^2 \sqrt 3}4,\) making the total area \(\dfrac{s^2 \sqrt 3}2.\)

The side length of the larger equilateral triangle is \(2 \sqrt 3.\) The height of it is \(3\) since the height is equal to \(2 \sqrt 3 \sin(60^\circ) .\)

Half of the square is \(\sqrt 3,\) so the height of the smaller triangle is \(3 - \sqrt 3 .\) Thus, the ratio between \(s\) and \(2 \sqrt 3\) is \(\dfrac {3-\sqrt 3}3 .\)

As such, \[s = \dfrac{2 \sqrt 3(3-\sqrt 3)}3 = 2\sqrt 3-2 .\]

Therefore, the combined area is \[\begin{aligned} \dfrac{s^2 \sqrt 3}2 &= \dfrac{(2\sqrt 3-2)^2 \sqrt 3 }2 \\&= \dfrac{(16-8\sqrt 3) \sqrt 3 }2 \\&= 8 \sqrt 3 - 12 \end{aligned}\]

Thus, the correct answer is D .

15 答案 D

总比赛数，即总胜场数，为 \(\binom 62 = 15\)。

若 \(6\) 支队伍全都并列，每队应有 \(2.5\) 胜，不可能。

若有 \(5\) 支队伍并列第一，每队可以有 \(3\) 胜；只要一支队伍输掉全部比赛，而这 \(5\) 支获胜队伍在彼此之间平分胜负，就能做到。

把这 \(5\) 支队伍编号为 \(1\) 到 \(5\)，并指定第 \(6\) 支队伍输掉所有比赛。对每个 \(1\le x \le 5\)，让该队击败队伍 \(x+1 \mod 5\)、队伍 \(x+2 \mod 5\)，以及队伍 \(6\)。

所以正确答案是 D。

English

They would have to share \(\binom 62 = 15\) wins.

This means we cannot have a \(6\) way tie as that would be \(2.5 \) wins per team.

If we had a \(5\) way tie, each team could have \(3\) wins, which is possible if one team loses all of its games, and out of the \(5\) winning teams, they each split their games.

If we label the teams from \(1\) to \(5,\) and designate team \(6\) to lose all their games. To get a \(5\) way tie, we could have each team \(1\le x \le 5\) beat team \(x+1 \mod 5\) and team \(x+2 \mod 5,\) as well as team \(6.\)

Thus, the correct answer is D .

16 答案 A

连接三个圆心，得到边长为 \(4\) 的等边三角形，面积为 \(4\sqrt3\)。

每个圆中计入的部分是 \(300^\circ\) 扇形，面积为 \(\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}\)。

所以总面积为 \(3\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3\)。

所以正确答案是 A。

English

Connect the centers of the three circles. This forms an equilateral triangle of side \(4\), with area \(4\sqrt3\).

The included part of each circle is a \(300^\circ\) sector, whose area is \(\dfrac{300}{360}\pi\cdot2^2=\dfrac{10\pi}{3}\).

The total area is \(3\cdot\dfrac{10\pi}{3}+4\sqrt3=10\pi+4\sqrt3\).

Thus, A is the correct answer.

17 答案 C

每个扇形形成的圆锥母线长为 \(12\)。较小扇形的圆心角为 \(120^\circ\)，弧长为 \(\dfrac13\cdot2\pi\cdot12=8\pi\)，所以较小圆锥底面半径为 \(4\)，高为 \(\sqrt{12^2-4^2}=8\sqrt2\)。

较大扇形弧长为 \(16\pi\)，底面半径为 \(8\)，高为 \(\sqrt{12^2-8^2}=4\sqrt5\)。

体积比为 \[\dfrac{\frac13\pi\cdot4^2\cdot8\sqrt2}{\frac13\pi\cdot8^2\cdot4\sqrt5}=\dfrac{\sqrt{10}}{10}\text{。}\]

所以正确答案是 C。

English

Each sector forms a cone with slant height \(12\). The smaller sector has angle \(120^\circ\), so its arc length is \(\dfrac13\cdot2\pi\cdot12=8\pi\), giving base radius \(4\). Its cone height is \(\sqrt{12^2-4^2}=8\sqrt2\).

The larger sector has arc length \(16\pi\), giving base radius \(8\). Its cone height is \(\sqrt{12^2-8^2}=4\sqrt5\).

The volume ratio is \[\dfrac{\frac13\pi\cdot4^2\cdot8\sqrt2}{\frac13\pi\cdot8^2\cdot4\sqrt5}=\dfrac{\sqrt{10}}{10}.\]

Thus, C is the correct answer.

18 答案 C

在 \(500\) 人中，大约 \(1\) 人患病并检测为阳性。其余 \(499\) 人中约有 \(2\%\) 假阳性，也就是约 \(10\) 人。

因此约 \(11\) 个阳性结果中，只有约 \(1\) 个是真阳性，概率最接近 \(\dfrac1{11}\)。

所以正确答案是 C。

English

Among \(500\) people, about \(1\) person has the disease and tests positive. Of the remaining \(499\) people, about \(2\%\) test falsely positive, which is about \(10\) people.

So among about \(11\) positive tests, only about \(1\) is a true positive. The probability is closest to \(\dfrac1{11}\).

Thus, C is the correct answer.

19 答案 C

四边形 \(BFDG\) 是梯形，底边为 \(DG\) 与 \(BF\)，高为 \(6\)。又因为 \(G\) 是 \(A\) 与 \(D\) 的中点，所以 \(GD= 15\)。

由相似三角形 \(EBF \sim EAD \)，有 \[\dfrac{BF}{AD} = \dfrac{EB}{EA} \] \[\dfrac{BF}{30} = \dfrac 28 \] \[BF = 7.5\]

因此 \(BFDG\) 的面积为 \(\dfrac{6(15+7.5)}2 = \dfrac{135}2\)。

所以正确答案是 C。

English

The polygon \(BFDG \) is a trapezoid with bases \(DG\) and \(BF\) and height \(6.\) Also, since \(G\) is the midpoint between \(A\) and \(D,\) we have \(GD= 15.\)

We can see that \(EBF \sim EAD ,\) so \[\dfrac{BF}{AD} = \dfrac{EB}{EA} \] \[\dfrac{BF}{30} = \dfrac 28 \]\[BF = 7.5\]

This makes the area of \(BFDG\) equal to \(\dfrac{6(15+7.5)}2 = \dfrac{135}2.\)

Thus, the correct answer is C .

20 答案 A

设初始数为 \(x\)。Bernardo 依次产生的数为 \[\begin{gathered} 2x,\quad4x+100,\quad8x+300,\\ 16x+700,\quad32x+1500 \end{gathered}\text{。}\] Silvia 在每个数的基础上再加 \(50\)。Bernardo 要在某一轮获胜，他产生的数必须小于 \(1000\)，而 Silvia 随后产生的数必须至少为 \(1000\)。

对 Bernardo 的前四轮，满足这两个不等式的最小整数 \(x\) 依次为 \(475,213,82\) 和 \(16\)。Bernardo 的第五个输出已经至少为 \(1500\)。因此最小的获胜初始数为 \(N=16\)，其各位数字之和为 \(1+6=7\)。

所以正确答案是 A。

English

If the initial number is \(x,\) Bernardo’s successive outputs are \[\begin{gathered} 2x,\quad4x+100,\quad8x+300,\\ 16x+700,\quad32x+1500. \end{gathered}\] Silvia’s output after each of these is \(50\) larger. Bernardo wins on a given turn exactly when his output is below \(1000\) but the following output from Silvia is at least \(1000.\)

For the first four Bernardo turns, the smallest integer \(x\) satisfying those two inequalities is, respectively, \(475,213,82,\) and \(16.\) A fifth Bernardo output is already at least \(1500.\) Thus the smallest winning initial number is \(N=16,\) whose digit sum is \(1+6=7.\)

Thus, the correct answer is A .

21 答案 A

把四条长度为 \(a\) 的线段看成以这四个点为顶点的图的边。若其中没有三角形，它们就会构成一个 \(4\) 环，而长度为 \(2a\) 的线段会是它的一条对角线。另外两个点各自给出一条总长为 \(2a\) 的两边路径，连接该对角线的两个端点。三角不等式取等号会迫使两个中间点都成为这条对角线的中点，与四点互异矛盾。因此有三个点构成边长为 \(a\) 的等边三角形，记为 \(A,B,C\)。

第四个点 \(D\) 到其中一点（设为 \(A\)）的距离为 \(a\)，到另一点（设为 \(B\)）的距离为 \(2a\)。由于 \(BA+AD=BD\)，点 \(B,A,D\) 共线，且 \(A\) 是 \(\overline{BD}\) 的中点。因此 \(BD\) 是过 \(B,C,D\)、圆心为 \(A\) 的圆的直径，所以 \(\triangle BCD\) 是直角三角形。

于是 \(b^2=(2a)^2-a^2=3a^2\)，所以 \(\frac{b}{a}=\sqrt3\)。

所以正确答案是 A。

English

Regard the four length-\(a\) segments as edges of a graph on the four points. If they contained no triangle, they would form a \(4\)-cycle. The length-\(2a\) segment would then be one of its diagonals. Each of the other two points gives a two-edge path of total length \(2a\) between the diagonal’s endpoints. Equality in the triangle inequality would force both intermediate points to be the midpoint of that diagonal, contradicting that the four points are distinct. Therefore three points do form an equilateral triangle of side \(a;\) call them \(A,B,C.\)

The fourth point \(D\) is distance \(a\) from one of these points, say \(A,\) and distance \(2a\) from another, say \(B.\) Because \(BA+AD=BD,\) the points \(B,A,D\) are collinear and \(A\) is the midpoint of \(\overline{BD}.\) Thus \(BD\) is a diameter of the circle through \(B,C,D\) centered at \(A,\) so \(\triangle BCD\) is right.

Thus \(b^2=(2a)^2-a^2=3a^2\), so \(\frac{b}{a}=\sqrt3\).

Thus, A is the correct answer.

22 答案 B

每个前缀 \(a_1,a_2,\ldots,a_i\) 都必须构成一个连续整数区间，因为每个新加入的数必须比已出现的某个数大一或小一。特别地，完整集合就是区间 \([1,10]\)。

反向阅读这个排列。每一步删去的项必须是当前区间的最小值或最大值；若删去内部的数，那么正向排列中该数前面就没有与它相邻的整数。反之，每一种依次删去端点的方式都能产生一个合格排列。对 \(a_{10},a_9,\ldots,a_2\) 中的每一项都有两种选择，之后 \(a_1\) 被唯一确定，所以总数为 \(2^9=512\)。

所以正确答案是 B。

English

Every prefix \(a_1,a_2,\ldots,a_i\) must form an interval of consecutive integers: each new term is required to be one more or one less than a term already present. In particular, the full set is the interval \([1,10].\)

Read the list backward. At each step, the term removed must be either the smallest or the largest integer in the current interval; removing an interior term would leave that term with no adjacent predecessor in the forward list. Conversely, every sequence of endpoint removals produces a valid list. There are two choices for each of \(a_{10},a_9,\ldots,a_2,\) after which \(a_1\) is forced. Hence there are \(2^9=512\) lists.

Thus, the correct answer is B .

23 答案 D

被切下的四面体可以看作底面为直角等腰三角形、两直角边为 \(1\)，高为 \(1\)，所以体积为 \(\dfrac13\cdot\dfrac12\cdot1=\dfrac16\)。

切面是边长为 \(\sqrt2\) 的等边三角形，面积为 \(\dfrac{\sqrt3}{4}(\sqrt2)^2=\dfrac{\sqrt3}{2}\)。设从相对顶点到切面的高为 \(h\)，由体积公式 \(\dfrac13\cdot\dfrac{\sqrt3}{2}\cdot h=\dfrac16\)，得 \(h=\dfrac{\sqrt3}{3}\)。

单位立方体的体对角线长为 \(\sqrt3\)。切面朝下放置后，高为 \(\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{2\sqrt3}{3}\)。

所以正确答案是 D。

English

The discarded tetrahedron has a right isosceles triangle of leg \(1\) as one base and height \(1\), so its volume is \(\dfrac13\cdot\dfrac12\cdot1=\dfrac16\).

The cut face is an equilateral triangle of side \(\sqrt2\), so its area is \(\dfrac{\sqrt3}{4}(\sqrt2)^2=\dfrac{\sqrt3}{2}\). If \(h\) is the height from the opposite vertex to this cut face, then \(\dfrac13\cdot\dfrac{\sqrt3}{2}\cdot h=\dfrac16\), so \(h=\dfrac{\sqrt3}{3}\).

The full cube diagonal has length \(\sqrt3\). After the tetrahedron is removed and the cut face is placed down, the height is \(\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{2\sqrt3}{3}\).

Thus, D is the correct answer.

24 答案 B

分两种情况：每一对恰好有 \(1\) 首共同喜欢的歌，或某一对女孩共同喜欢 \(2\) 首歌。这是因为每一对都必须至少有一首共同喜欢的歌，而比这两种情况更多的共同歌曲会需要 \(5\) 首歌。

情况 \(1\)：三对女孩各自恰好有一首只被这对喜欢的歌。

给第一对选歌有 \(4\) 种，第二对有 \(3\) 种，第三对有 \(2\) 种；剩下的歌可以被 \(3\) 人中的恰好一人喜欢或无人喜欢。因此这一类有 \[4\cdot 3\cdot 2\cdot (3+1)=96\text{。}\]

情况 \(2\)：某一对女孩共同喜欢 \(2\) 首歌。

选共同喜欢两首歌的那一对有 \(3\) 种；从四首歌中选出这对喜欢的 \(2\) 首有 \(\binom 42 = 6\) 种；再把剩下两首分配给另外两对，有 \(2! =2\) 种。因此这一类有 \[3\cdot 6\cdot 2=36\]

总数为 \(96+36 = 132\)。

所以正确答案是 B。

English

There are two cases: Each pair has exactly one liked song in common, or some pair has \(2\) liked songs in common. This is because each pair must have at least \(1\) liked song in common, and any more pairs than in the cases would result in \(5\) songs.

Case \(1:\) Each pair has exactly one liked song in common

There are \(4\) ways to choose the song that one pair likes, \(3\) ways to choose the song that the second pair likes, and \(2\) ways to choose the song the third pair likes if we choose some order for them. Then, for the last song, one of them could like it which has \(3\) cases or none of them likes it which is another case. Thus, the number of solutions in this case is \[4\cdot 3\cdot 2\cdot (3+1)=96.\]

Case \(2:\) Some pair has \(2\) liked songs in common

There are \(3\) ways to choose the pair that has \(2\) liked songs in common. Then, there are \(\binom 42 = 6\) ways to choose which two songs they like. Finally, there are \(2! =2 \) ways to assign the two remaining songs to the other two pairs. Thus, the number of solutions in this case is \[3\cdot 6\cdot 2=36\]

The total amount is then \(96+36 = 132.\)

Thus, the correct answer is B .

25 答案 E

按路径使用的反向箭头集合 \(S\) 分类。若 \(S=\varnothing\)，路径由每一列选择一条前进边确定，共 \(2\cdot2\cdot4\cdot4\cdot4\cdot2\cdot2=2^{10}\) 条。

若 \(S\) 只含一个外侧反向箭头，每种外侧选择有 \(2^8\) 条；若含两个外侧反向箭头但不含中间箭头，有 \(2^6\) 条。

若 \(S\) 只含中间反向箭头，有 \(2^9\) 条；若含中间箭头和恰好一个外侧反向箭头，对每一种外侧箭头的选择都有 \(2^7\) 条；若三个反向箭头都使用，有 \(2^5\) 条。

总数为 \(2^{10}\) \(+2\cdot2^8\) \(+2^6\) \(+2^9\) \(+2\cdot2^7\) \(+2^5=2400\)。

所以正确答案是 E。

English

Classify a path by the set \(S\) of backward arrows it uses. If \(S=\varnothing\), the path is determined by choosing one forward arrow in each column, giving \(2\cdot2\cdot4\cdot4\cdot4\cdot2\cdot2=2^{10}\) paths.

If \(S\) uses only the left backward arrow, there are \(2^8\) paths, and by symmetry the same for only the right backward arrow. If it uses both outer backward arrows but not the middle one, there are \(2^6\) paths.

If \(S\) uses only the middle backward arrow, there are \(2^9\) paths. If it uses the middle arrow and exactly one outer backward arrow, there are \(2^7\) paths for each choice of outer arrow. If it uses all three backward arrows, there are \(2^5\) paths.

The total is \(2^{10}\) \(+2\cdot2^8\) \(+2^6\) \(+2^9\) \(+2\cdot2^7\) \(+2^5=2400\).

Thus, E is the correct answer.