1978 AMC 12 真题（P版）

1 答案 B

令 \(u=\frac{2}{x}\)，原方程化为 \(1-2u+u^2=0\)，即 \((u-1)^2=0\)。因此 \(\frac{2}{x}=u=1\)。

因此，正确答案是 B。

English

With \(u=\frac{2}{x},\) the equation becomes \(1-2u+u^2=0,\) or \((u-1)^2=0.\) Therefore \(\frac{2}{x}=u=1.\)

Therefore, the correct answer is B.

2 答案 C

题设条件为 \[ 4\left(\frac1{2\pi r}\right)=2r\text{。} \] 两边乘以 \(\pi r\)，得到 \(2=2\pi r^2\)，所以面积 \(\pi r^2\) 等于 \(1\)。

因此，正确答案是 C。

English

The condition is \[ 4\left(\frac1{2\pi r}\right)=2r. \] Multiplying by \(\pi r\) gives \(2=2\pi r^2,\) so the area \(\pi r^2\) equals \(1.\)

Therefore, the correct answer is C.

3 答案 D

由于 \(x=\frac{1}{y}\)，也有 \(y=\frac{1}{x}\)。因此该乘积为 \[ (x-y)(y+x)=x^2-y^2\text{。} \]

因此，正确答案是 D。

English

Since \(x=\frac{1}{y},\) we also have \(y=\frac{1}{x}.\) Thus the product is \[ (x-y)(y+x)=x^2-y^2. \]

Therefore, the correct answer is D.

4 答案 B

每个变量以正号出现三次、以负号出现一次，所以该式为 \[ \begin{gathered} 2(a+b+c+d)\\ =2a+2b+2c+2d\\ =2+20+200+2000\\ =2222\text{。} \end{gathered} \]

因此，正确答案是 B。

English

Each variable appears positively three times and negatively once, so the expression is \[ \begin{gathered} 2(a+b+c+d)\\ =2a+2b+2c+2d\\ =2+20+200+2000\\ =2222. \end{gathered} \]

Therefore, the correct answer is B.

5 答案 C

设前三人的付款额分别为 \(w\)、\(x\) 和 \(y\)。则 \[ \begin{aligned} w&=\frac12(60-w),\\ x&=\frac13(60-x),\\ y&=\frac14(60-y)\text{。} \end{aligned} \] 因此 \(w=20\)、\(x=15\)、\(y=12\)。第四个男孩支付了 \(60-20-15-12=13\) 美元。

因此，正确答案是 C。

English

Let the first three payments be \(w,\) \(x,\) and \(y.\) Then \[ \begin{aligned} w&=\frac12(60-w),\\ x&=\frac13(60-x),\\ y&=\frac14(60-y). \end{aligned} \] Hence \(w=20,\) \(x=15,\) \(y=12.\) The fourth payment is \(60-20-15-12=13\) dollars.

Therefore, the correct answer is C.

6 答案 E

若 \(y=0\)，第一个方程给出 \(x=x^2\)，所以 \((x,y)=(0,0)\) 或 \((1,0)\)。若 \(y\ne0\)，第二个方程给出 \(x=\frac{1}{2}\)。代入第一个方程得到 \(y^2=\frac{1}{4}\)，从而还有 \((\frac{1}{2},\frac{1}{2})\) 和 \((\frac{1}{2},-\frac{1}{2})\)。共有四个数对。

因此，正确答案是 E。

English

If \(y=0,\) the first equation gives \(x=x^2,\) so \((x,y)=(0,0)\) or \((1,0).\) If \(y\ne0,\) the second equation gives \(x=\frac{1}{2}.\) The first then gives \(y^2=\frac{1}{4},\) producing \((\frac{1}{2},\frac{1}{2})\) and \((\frac{1}{2},-\frac{1}{2}).\) There are four pairs.

Therefore, the correct answer is E.

7 答案 E

两条相对边之间的距离是边心距的两倍，因此为 \(s\sqrt3\)。于是 \(s\sqrt3=12\)，所以 \(s=\frac{12}{\sqrt3}=4\sqrt3\)。

因此，正确答案是 E。

English

The distance between opposite sides is twice the apothem, hence \(s\sqrt3.\) Therefore \(s\sqrt3=12,\) so \(s=\frac{12}{\sqrt3}=4\sqrt3.\)

Therefore, the correct answer is E.

8 答案 D

第一个公差是 \(\frac{y-x}{3}\)，第二个公差是 \(\frac{y-x}{4}\)。由于 \(x\ne y\)，它们的比为 \[ \frac{\frac{y-x}{3}}{\frac{y-x}{4}}=\frac43\text{。} \]

因此，正确答案是 D。

English

The first common difference is \(\frac{y-x}{3},\) and the second is \(\frac{y-x}{4}.\) Since \(x\ne y,\) their ratio is \[ \frac{\frac{y-x}{3}}{\frac{y-x}{4}}=\frac43. \]

Therefore, the correct answer is D.

9 答案 B

由于 \(x-1\lt0\)，所以 \(\sqrt{(x-1)^2}=|x-1|=1-x\)。因此 \[ \begin{aligned} \left|x-(1-x)\right|&=|2x-1|\\ &=1-2x\text{，} \end{aligned} \] 因为 \(2x-1\lt0\)。

因此，正确答案是 B。

English

Because \(x-1\lt0,\) \(\sqrt{(x-1)^2}=|x-1|=1-x.\) Therefore \[ \begin{aligned} \left|x-(1-x)\right|&=|2x-1|\\ &=1-2x, \end{aligned} \] since \(2x-1\lt0.\)

Therefore, the correct answer is B.

10 答案 B

对于任意 \(A\ne P\)，圆上离 \(A\) 最近的点是从 \(P\) 经过 \(A\) 的射线与圆的交点。仅当 \(A\) 位于从 \(P\) 经过 \(B\) 的射线上时，该交点才是 \(B\)。圆心 \(P\) 也符合条件，因为它到圆上各点的距离相等。因此所求轨迹就是这条射线。

因此，正确答案是 B。

English

For any \(A\ne P,\) the closest point of the circle to \(A\) is where the ray from \(P\) through \(A\) meets the circle. This point is \(B\) exactly when \(A\) lies on the ray from \(P\) through \(B.\) The center \(P\) also qualifies because every point on the circle is equally distant from it. Thus the locus is that ray.

Therefore, the correct answer is B.

11 答案 C

原点到直线 \(x+y-r=0\) 的距离为 \(\frac{r}{\sqrt2}\)。相切要求该距离等于半径 \(\sqrt r\)，所以 \[ \frac r{\sqrt2}=\sqrt r\text{。} \] 由于 \(r\gt0\)，两边平方再除以 \(r\)，得到 \(r=2\)。

因此，正确答案是 C。

English

The distance from the origin to \(x+y-r=0\) is \(\frac{r}{\sqrt2}.\) Tangency requires this to equal the radius \(\sqrt r,\) so \[ \frac r{\sqrt2}=\sqrt r. \] Since \(r\gt0,\) squaring and dividing by \(r\) gives \(r=2.\)

Therefore, the correct answer is C.

12 答案 E

令 \(x=\angle BAC=\angle BCA\)、\(y=\angle CBD=\angle CDB\)、\(z=\angle DCE=\angle DEC\)。依次应用外角定理可得 \[ y=2x,\qquad z=x+y=3x\text{。} \] \(\triangle ADE\) 的三个角依次为 \(x\)、\(140^\circ\) 和 \(z\)，所以 \(x+140^\circ+3x=180^\circ\)。故 \(x=10^\circ\)。

因此，正确答案是 E。

English

Let \(x=\angle BAC=\angle BCA,\) \(y=\angle CBD=\angle CDB,\) and \(z=\angle DCE=\angle DEC.\) The exterior-angle theorem applied successively gives \[ y=2x,\qquad z=x+y=3x. \] The angles of \(\triangle ADE\) are \(x,\) \(140^\circ,\) and \(z,\) so \(x+140^\circ+3x=180^\circ.\) Hence \(x=10^\circ.\)

Therefore, the correct answer is E.

13 答案 B

由韦达定理， \[ \begin{aligned} c+d&=-a,& cd&=b,\\ a+b&=-c,& ab&=d\text{。} \end{aligned} \] 两个关于根的和的方程推出 \(b=d\)。由于 \(b=d\ne0\)，两个关于根的积的方程给出 \(a=c=1\)。再由 \(a+b=-c\) 得 \(b=d=-2\)，因此 \(a+b+c+d=-2\)。

因此，正确答案是 B。

English

Vieta’s formulas give \[ \begin{aligned} c+d&=-a,& cd&=b,\\ a+b&=-c,& ab&=d. \end{aligned} \] The two sum equations imply \(b=d.\) Since \(b=d\ne0,\) the product equations give \(a=c=1.\) Then \(a+b=-c\) gives \(b=d=-2,\) and therefore \(a+b+c+d=-2.\)

Therefore, the correct answer is B.

14 答案 C

用通常的记法，\(a=(18)_n=n+8\)。由于一个根是 \(n\)，另一个根必为 \(8\)。两根之积为 \(b=8n\)，它在 \(n\) 进制中的表示为 \(80\)。

因此，正确答案是 C。

English

In ordinary notation, \(a=(18)_n=n+8.\) Since one root is \(n,\) the other root must be \(8.\) Their product is \(b=8n,\) whose base-\(n\) representation is \(80.\)

Therefore, the correct answer is C.

15 答案 A

两边平方得到 \(1+2\sin x\cos x=\frac{1}{25}\)，所以 \(\sin x\cos x=-\frac{12}{25}\)。因此 \(\sin x\) 和 \(\cos x\) 是下面这个方程的两根： \[ t^2-\frac15t-\frac{12}{25}=0\text{，} \] 即 \(\frac{4}{5}\) 和 \(-\frac{3}{5}\)。在给定区间内 \(\sin x\ge0\)，故 \(\sin x=\frac{4}{5}\)，\(\cos x=-\frac{3}{5}\)。于是 \(\tan x=-\frac{4}{3}\)。

因此，正确答案是 A。

English

Squaring gives \(1+2\sin x\cos x=\frac{1}{25},\) so \(\sin x\cos x=-\frac{12}{25}.\) Thus \(\sin x\) and \(\cos x\) are the roots of \[ t^2-\frac15t-\frac{12}{25}=0, \] namely \(\frac{4}{5}\) and \(-\frac{3}{5}.\) Because \(\sin x\ge0\) on the given interval, \(\sin x=\frac{4}{5}\) and \(\cos x=-\frac{3}{5}.\) Hence \(\tan x=-\frac{4}{3}.\)

Therefore, the correct answer is A.

16 答案 E

若一人少握了一次手，那么这次未握手所涉及的另一人也没有与所有人握过手。因此，至多有 \(N-2\) 人与所有其他人都握过手。让恰好两个人彼此不握手，而其余握手全部发生，就能达到这个上界。由于选项中没有 \(N-2\)，答案是“以上都不是”。

因此，正确答案是 E。

English

If one person has missed a handshake, the other person in that missed pair also has not shaken hands with everyone. Thus at most \(N-2\) people can have shaken hands with everyone. This is attainable when exactly two people fail to shake hands with each other and every other handshake occurs. Since \(N-2\) is not listed, the answer is “none of these.”

Therefore, the correct answer is E.

17 答案 D

令 \(x=\frac{3}{y}\)，它是正数。于是 \[ \begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x\text{。} \end{aligned} \] 因此所求表达式为 \[ \left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2\text{。} \]

因此，正确答案是 D。

English

Set \(x=\frac{3}{y},\) which is positive. Then \[ \begin{aligned} x^2+1&=\frac{9+y^2}{y^2},\\ \sqrt{\frac{12}{y}}&=2\sqrt x. \end{aligned} \] Therefore the requested expression is \[ \left(\left[f(x^2+1)\right]^{\sqrt x}\right)^2=k^2. \]

Therefore, the correct answer is D.

18 答案 C

有理化可得 \[ \sqrt n-\sqrt{n-1} =\frac1{\sqrt n+\sqrt{n-1}}\text{。} \] 当 \(n=2500\) 时，分母为 \(50+\sqrt{2499}\lt100\)，所以该差大于 \(0.01\)。当 \(n=2501\) 时，分母为 \(\sqrt{2501}+50\gt100\)，所以该差小于 \(0.01\)。分母随 \(n\) 增大而增大，因此 \(2501\) 是满足条件的最小整数。

因此，正确答案是 C。

English

Rationalizing gives \[ \sqrt n-\sqrt{n-1} =\frac1{\sqrt n+\sqrt{n-1}}. \] For \(n=2500,\) the denominator is \(50+\sqrt{2499}\lt100,\) so the difference exceeds \(0.01.\) For \(n=2501,\) the denominator is \(\sqrt{2501}+50\gt100,\) so the difference is less than \(0.01.\) The denominator increases with \(n,\) making \(2501\) the least such integer.

Therefore, the correct answer is C.

19 答案 C

总概率为 \(50p+50(3p)=200p=1\)，所以 \(p=0.005\)。不超过 \(50\) 的完全平方数有七个，大于 \(50\) 的还有三个：\(64\)、\(81\)、\(100\)。因此所求概率为 \[ 7p+3(3p)=16p=0.08\text{。} \]

因此，正确答案是 C。

English

The total probability is \(50p+50(3p)=200p=1,\) so \(p=0.005.\) There are seven perfect squares at most \(50\) and three more, \(64,\) \(81,\) \(100,\) above \(50.\) Hence the desired probability is \[ 7p+3(3p)=16p=0.08. \]

Therefore, the correct answer is C.

20 答案 A

令它们的公共值为 \(t\)。由前两个方程可得 \[ (b-c)(t+2)=0\text{，} \] 循环比较可得另外两个类似关系。因此要么 \(a=b=c\)，此时 \(x=8\)；要么 \(t=-2\)。在后一种情形下，\(a+b=-c\)、\(b+c=-a\)、\(c+a=-b\)，所以 \[ x=\frac{(-c)(-a)(-b)}{abc}=-1\text{。} \] 条件 \(x\lt0\) 选出后一数值。

因此，正确答案是 A。

English

Let the common value be \(t.\) The first two resulting equations imply \[ (b-c)(t+2)=0, \] and cyclic comparisons give the analogous relations. Thus either \(a=b=c,\) which gives \(x=8,\) or \(t=-2.\) In the latter case \(a+b=-c,\) \(b+c=-a,\) and \(c+a=-b,\) so \[ x=\frac{(-c)(-a)(-b)}{abc}=-1. \] The condition \(x\lt0\) selects the latter value.

Therefore, the correct answer is A.

21 答案 A

令 \(S\) 表示题给和式。换底并合并对数可得 \[ \begin{aligned} S&=\frac1{\log_3x}+\frac1{\log_4x}\\ &\quad+\frac1{\log_5x}\\ &=\log_x(3\cdot4\cdot5)\\ &=\log_x60\\ &=\frac1{\log_{60}x}\text{。} \end{aligned} \]

因此，正确答案是 A。

English

Let \(S\) denote the given sum. Changing bases and combining logarithms gives \[ \begin{aligned} S&=\frac1{\log_3x}+\frac1{\log_4x}\\ &\quad+\frac1{\log_5x}\\ &=\log_x(3\cdot4\cdot5)\\ &=\log_x60\\ &=\frac1{\log_{60}x}. \end{aligned} \]

Therefore, the correct answer is A.

22 答案 D

若假话的实际数量 \(m\) 是 \(1\)、\(2\)、\(3\)、\(4\) 中的一个，则上面的四句话中恰有一句为真，也就是声称假话数量为 \(m\) 的那一句。因此恰有三句话是假话，从而必须有 \(m=3\)。这确实自洽：第三句话为真，其余三句为假。

因此，正确答案是 D。

English

If the actual number \(m\) of false statements is one of \(1,\) \(2,\) \(3,\) \(4,\) exactly one displayed statement—the one naming \(m\)—is true. Therefore exactly three statements are false, forcing \(m=3.\) This is consistent: the third statement is true and the other three are false.

Therefore, the correct answer is D.

23 答案 C

令 \(A=(0,0)\)、\(B=(s,0)\)、\(D=(0,s)\)，其中 \(s=\sqrt{1+\sqrt3}\)。经过 \(F\) 的两条直线的方程为 \[ \begin{aligned} AE:\quad y&=\sqrt3x,\\ BD:\quad y&=s-x\text{。} \end{aligned} \] 因此 \(F\) 到 \(AB\) 的高为 \(\frac{s\sqrt3}{1+\sqrt3}\)。于是 \[ [\triangle ABF] =\frac{s^2\sqrt3}{2(1+\sqrt3)} =\frac{\sqrt3}{2}\text{。} \]

因此，正确答案是 C。

English

Let \(A=(0,0),\) \(B=(s,0),\) and \(D=(0,s),\) where \(s=\sqrt{1+\sqrt3}.\) The two lines containing \(F\) have equations \[ \begin{aligned} AE:\quad y&=\sqrt3x,\\ BD:\quad y&=s-x. \end{aligned} \] Hence the altitude of \(F\) above \(AB\) is \(\frac{s\sqrt3}{1+\sqrt3}.\) Therefore \[ [\triangle ABF] =\frac{s^2\sqrt3}{2(1+\sqrt3)} =\frac{\sqrt3}{2}. \]

Therefore, the correct answer is C.

24 答案 A

三个表达式之和为 \[ \begin{aligned} &x(y-z)+y(z-x)\\ &\qquad+z(x-y)=0\text{。} \end{aligned} \] 将这个非零等比数列写成 \(u\)、\(ur\)、\(ur^2\)，可得 \(u(1+r+r^2)=0\)。由于 \(u\ne0\)，所以 \(r^2+r+1=0\)。

因此，正确答案是 A。

English

The three expressions have sum \[ \begin{aligned} &x(y-z)+y(z-x)\\ &\qquad+z(x-y)=0. \end{aligned} \] Writing the nonzero geometric progression as \(u,\) \(ur,\) \(ur^2\) gives \(u(1+r+r^2)=0.\) Since \(u\ne0,\) \(r^2+r+1=0.\)

Therefore, the correct answer is A.

25 答案 D

前两个条件构成正方形 \([\frac{a}{2},2a]\times[\frac{a}{2},2a]\)。在这个正方形内，\(x+y\ge a\) 自动成立。最后两个条件等价于 \(|x-y|\le a\)；它们的边界直线切去顶点 \((2a,\frac{a}{2})\) 和 \((\frac{a}{2},2a)\)。从正方形切去相对的两个角会得到六边形，所以边界有六条边。

因此，正确答案是 D。

English

The first two conditions form the square \([\frac{a}{2},2a]\times[\frac{a}{2},2a].\) Within this square, \(x+y\ge a\) is automatic. The last two conditions are equivalent to \(|x-y|\le a;\) their boundary lines cut off the corners \((2a,\frac{a}{2})\) and \((\frac{a}{2},2a).\) Cutting two opposite corners from a square produces a hexagon, so the boundary has six sides.

Therefore, the correct answer is D.

26 答案 B

设 \(H\) 为从 \(C\) 向 \(AB\) 所作高的垂足。在所有经过 \(C\) 且与 \(AB\) 相切的圆中，当切点为 \(H\) 时半径最小，所以 \(CH\) 是直径。由直角三角形的面积可得 \[ CH=\frac{AC\cdot BC}{AB}=\frac{8\cdot6}{10}=4.8\text{。} \] 此外，\(\angle QCR=90^\circ\)，所以 \(QR\) 是圆 \(P\) 的直径。因此 \(QR=CH=4.8\)。

因此，正确答案是 B。

English

Let \(H\) be the foot of the altitude from \(C\) to \(AB.\) Among circles through \(C\) tangent to \(AB,\) the least radius occurs when the tangency point is \(H,\) so \(CH\) is a diameter. The area of the right triangle gives \[ CH=\frac{AC\cdot BC}{AB}=\frac{8\cdot6}{10}=4.8. \] Also \(\angle QCR=90^\circ,\) so \(QR\) is a diameter of circle \(P.\) Therefore \(QR=CH=4.8.\)

Therefore, the correct answer is B.

27 答案 C

符合条件的整数对从 \(2\) 到 \(11\) 的每个整数取模都与 \(1\) 同余，因此也可将模数取为 \[ \begin{aligned} L&=\operatorname{lcm}(2,\ldots,11)\\ &=2^3\cdot3^2\cdot5\cdot7\cdot11\\ &=27720\text{。} \end{aligned} \] 大于 \(1\) 的最小两个符合条件的整数是 \(L+1\) 和 \(2L+1\)，它们的差为 \(L=27720\)。

因此，正确答案是 C。

English

A qualifying integer is congruent to \(1\) modulo every integer from \(2\) through \(11,\) hence modulo \[ \begin{aligned} L&=\operatorname{lcm}(2,\ldots,11)\\ &=2^3\cdot3^2\cdot5\cdot7\cdot11\\ &=27720. \end{aligned} \] The two smallest qualifying integers greater than \(1\) are \(L+1\) and \(2L+1,\) whose difference is \(L=27720.\)

Therefore, the correct answer is C.

28 答案 E

令 \(\mathbf d_n=\overrightarrow{A_nA_{n+1}}\)。由中点关系可得 \(\mathbf d_{n+3}=\frac12(\mathbf d_n+\mathbf d_{n+1})\)，并且 \(\mathbf d_n+\mathbf d_{n+1}+\mathbf d_{n+2}=\frac12\mathbf d_n\)。因此， \[ \begin{aligned} \mathbf d_{n+4} &=\frac12(\mathbf d_{n+1}+\mathbf d_{n+2})\\ &=-\frac14\mathbf d_n\text{。} \end{aligned} \] 所以 \(\mathbf d_{43}\) 和 \(\mathbf d_{44}\) 分别是 \(\mathbf d_3\) 和 \(\mathbf d_4\) 的同一正数倍，故 \(\angle A_{44}A_{45}A_{43}=\angle A_4A_5A_3\)。由于 \(A_4\) 和 \(A_5\) 分别是 \(A_1A_2\) 和 \(A_2A_3\) 的中点，所以 \(A_4A_5\parallel A_1A_3\)。再利用等边三角形的角，可得 \(\angle A_4A_5A_3=120^\circ\)。

因此，正确答案是 E。

English

Let \(\mathbf d_n=\overrightarrow{A_nA_{n+1}}.\) The midpoint rule gives \(\mathbf d_{n+3}=\frac12(\mathbf d_n+\mathbf d_{n+1})\) and also \(\mathbf d_n+\mathbf d_{n+1}+\mathbf d_{n+2}=\frac12\mathbf d_n.\) Consequently, \[ \begin{aligned} \mathbf d_{n+4} &=\frac12(\mathbf d_{n+1}+\mathbf d_{n+2})\\ &=-\frac14\mathbf d_n. \end{aligned} \] Thus \(\mathbf d_{43}\) and \(\mathbf d_{44}\) are the same positive scalar multiple of \(\mathbf d_3\) and \(\mathbf d_4,\) respectively, so \(\angle A_{44}A_{45}A_{43}=\angle A_4A_5A_3.\) Since \(A_4\) and \(A_5\) are the midpoints of \(A_1A_2\) and \(A_2A_3,\) \(A_4A_5\parallel A_1A_3.\) The equilateral-triangle angles then give \(\angle A_4A_5A_3=120^\circ.\)

Therefore, the correct answer is E.

29 答案 D

利用位置向量，相等的延长线段给出 \[ \begin{aligned} A'&=2A-D,& B'&=2B-A,\\ C'&=2C-B,& D'&=2D-C\text{。} \end{aligned} \] 将这些式子代入有向多边形面积和式，可见对角的叉积项相互抵消： \[ \begin{aligned} &A'\mathbin{\times}B' +B'\mathbin{\times}C'\\ &\quad+C'\mathbin{\times}D' +D'\mathbin{\times}A'\\ &=5A\mathbin{\times}B +5B\mathbin{\times}C\\ &\quad+5C\mathbin{\times}D +5D\mathbin{\times}A\text{。} \end{aligned} \] 因此外部四边形的面积是原四边形面积的五倍，即 \(5\cdot10=50\)。

因此，正确答案是 D。

English

Using position vectors, the equal extensions give \[ \begin{aligned} A'&=2A-D,& B'&=2B-A,\\ C'&=2C-B,& D'&=2D-C. \end{aligned} \] Substitution into the oriented polygon-area sum shows that the diagonal cross terms cancel: \[ \begin{aligned} &A'\mathbin{\times}B' +B'\mathbin{\times}C'\\ &\quad+C'\mathbin{\times}D' +D'\mathbin{\times}A'\\ &=5A\mathbin{\times}B +5B\mathbin{\times}C\\ &\quad+5C\mathbin{\times}D +5D\mathbin{\times}A. \end{aligned} \] Hence the outer area is five times the original area, or \(5\cdot10=50.\)

Therefore, the correct answer is D.

30 答案 E

比赛总数为 \(\frac{3n(3n-1)}{2}\)，所以女子必须赢得 \(\frac{7n(3n-1)}{8}\) 场。若女子在 \(2n^2\) 场男女对阵中赢了 \(k\) 场，则 \[ \frac{n(n-1)}2+k=\frac{7n(3n-1)}8\text{，} \] 从而 \(k=\frac{n(17n-3)}{8}\)。由上界 \(k\le2n^2\) 可得 \(n\le3\)。当 \(n=1\)、\(2\) 时，该公式给出的值不是整数；当 \(n=3\) 时，得到 \(k=18\)，这是可能的。因此 \(n=3\)，不在列出的数值选项中。

因此，正确答案是 E。

English

There are \(\frac{3n(3n-1)}{2}\) matches, so women must win \(\frac{7n(3n-1)}{8}\) matches. If women win \(k\) of the \(2n^2\) mixed matches, then \[ \frac{n(n-1)}2+k=\frac{7n(3n-1)}8, \] giving \(k=\frac{n(17n-3)}{8}.\) The bound \(k\le2n^2\) forces \(n\le3.\) For \(n=1,\) \(2,\) this formula is not an integer, while \(n=3\) gives \(k=18,\) which is possible. Thus \(n=3,\) which is not among the listed numerical choices.

Therefore, the correct answer is E.