2007A AMC 10 真题（P版）

1 答案 C

Susan 支付 \(4 \cdot 0.75 \cdot 20 = 60\) 美元。

Pam 支付 \(5 \cdot 0.70 \cdot 20 = 70\) 美元。

差为 \(70 - 60 = 10\) 美元。

所以正确答案是 C。

English

Susan pays \(4 \cdot 0.75 \cdot 20 = 60\) dollars.

Pam pays \(5 \cdot 0.70 \cdot 20 = 70\) dollars.

The difference is \(70 - 60 = 10\) dollars.

Thus, the correct answer is C.

2 答案 A

分子为 \(6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8\)。

分母为 \(6\#2 = 6 + 2 - 6 \cdot 2^2\) \(= 8 - 24\) \(= -16\)。

商为 \[ \dfrac{8}{-16} = -\dfrac{1}{2}\text{。}\]

所以正确答案是 A。

English

The numerator is \(6@2 = 6 \cdot 2 - 2^2 = 12 - 4 = 8.\)

The denominator is \(6\#2 = 6 + 2 - 6 \cdot 2^2\) \(= 8 - 24\) \(= -16.\)

The quotient is \[ \dfrac{8}{-16} = -\dfrac{1}{2}. \]

Thus, the correct answer is A.

3 答案 D

砖的体积为 \(40 \cdot 20 \cdot 10 = 8000\) 立方厘米。

若水面上升 \(h\) 厘米，增加的水体积为 \(100 \cdot 40 \cdot h = 4000h\) 立方厘米。

令二者相等，得 \(4000h = 8000\)，所以 \(h = 2\)。

所以正确答案是 D。

English

The brick has volume \(40 \cdot 20 \cdot 10 = 8000\) cubic centimeters.

If the water rises by \(h\) centimeters, the added volume is \(100 \cdot 40 \cdot h = 4000h\) cubic centimeters.

Setting these equal gives \(4000h = 8000,\) so \(h = 2.\)

Thus, the correct answer is D.

4 答案 A

设较小整数为 \(x\)，则较大整数为 \(x + 2\)，并且 \[ x + 2 = 3x \] 因此 \(x = 1\)。

两个整数为 \(1\) 和 \(3\)，和为 \(4\)。

所以正确答案是 A。

English

Let the smaller integer be \(x.\) Then the larger is \(x + 2,\) and \[ x + 2 = 3x, \] so \(x = 1.\)

The two integers are \(1\) and \(3,\) and their sum is \(4.\)

Thus, the correct answer is A.

5 答案 B

设 \(p\) 和 \(n\) 分别为一支铅笔和一本笔记本的价格，单位为美分。\[ 7p + 8n = 415 \] \[ 5p + 3n = 177\text{。}\]

解这个方程组，得到 \(p = 9\) 和 \(n = 44\)。

所以 \(16\) 支铅笔和 \(10\) 本笔记本共需 \(16(9) + 10(44) = 584\) 美分，即 \(\$5.84\)。

所以正确答案是 B。

English

Let \(p\) and \(n\) be the prices in cents of a pencil and a notebook. Then \[ 7p + 8n = 415 \]\[ 5p + 3n = 177. \]

Solving this system gives \(p = 9\) and \(n = 44.\)

So \(16\) pencils and \(10\) notebooks cost \(16(9) + 10(44) = 584\) cents, or \(\$5.84.\)

Thus, the correct answer is B.

6 答案 A

从 \(2002\) 到 \(2003\)，增长为 \[ \dfrac{6}{60} = \dfrac{1}{10} = 10\%\text{。}\]

其他增长率为 \(\dfrac{4}{66}\)、\(\dfrac{6}{70}\)、\(\dfrac{2}{76}\)、\(\dfrac{7}{78}\)，都小于 \(\dfrac{1}{10}\)。

所以最大百分比增长发生在 \(2002\) 和 \(2003\) 之间。

所以正确答案是 A。

English

From \(2002\) to \(2003,\) the increase is \[ \dfrac{6}{60} = \dfrac{1}{10} = 10\%. \]

The other increases are \(\dfrac{4}{66},\) \(\dfrac{6}{70},\) \(\dfrac{2}{76},\) and \(\dfrac{7}{78},\) each less than \(\dfrac{1}{10}.\)

So the largest percentage increase was between \(2002\) and \(2003.\)

Thus, the correct answer is A.

7 答案 D

联邦税后，Public 先生保留遗产的 \(80\%\)。

州税取这部分的 \(10\%\)，即遗产的 \(8\%\)。

总税额为遗产的 \(20\% + 8\% = 28\%\)，所以遗产为 \[ \dfrac{10{,}500}{0.28} = 37{,}500\text{。}\]

所以正确答案是 D。

English

After federal taxes, Mr. Public keeps \(80\%\) of his inheritance.

State taxes take \(10\%\) of that, which is \(8\%\) of the inheritance.

The total tax is \(20\% + 8\% = 28\%\) of the inheritance, so the inheritance is \[ \dfrac{10{,}500}{0.28} = 37{,}500. \]

Thus, the correct answer is D.

8 答案 D

因为 \(\triangle ABC\) 是等腰三角形，\(\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ\)。

因为 \(\triangle ADC\) 是等腰三角形，\(\angle DAC = \tfrac12(180^\circ - 140^\circ) = 20^\circ\)。

因此 \(\angle BAD = \angle BAC - \angle DAC\) \(= 70^\circ - 20^\circ\) \(= 50^\circ\)。

所以正确答案是 D。

English

Since \(\triangle ABC\) is isosceles, \(\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ.\)

Since \(\triangle ADC\) is isosceles, \(\angle DAC = \tfrac12(180^\circ - 140^\circ) = 20^\circ.\)

Therefore \(\angle BAD = \angle BAC - \angle DAC\) \(= 70^\circ - 20^\circ\) \(= 50^\circ.\)

Thus, the correct answer is D.

9 答案 E

两个方程变为 \(3^a = 3^{4(b+2)}\) 和 \(5^{3b} = 5^{a-3}\)。

所以 \(a = 4(b + 2)\)，且 \(3b = a - 3\)。

解得 \(a = -12\)、\(b = -5\)，因此 \(ab = 60\)。

所以正确答案是 E。

English

The equations become \(3^a = 3^{4(b+2)}\) and \(5^{3b} = 5^{a-3}.\)

So \(a = 4(b + 2)\) and \(3b = a - 3.\)

Solving gives \(a = -12\) and \(b = -5,\) so \(ab = 60.\)

Thus, the correct answer is E.

10 答案 E

设孩子数为 \(N\)，全家总年龄为 \(T\)。

则 \(20 = \dfrac{T}{N + 2}\)，且 \(16 = \dfrac{T - 48}{N + 1}\)。

这给出 \(20N + 40 = T\) 和 \(16N + 64 = T\)，所以 \(20N + 40 = 16N + 64\)。

因此 \(4N = 24\)，所以 \(N = 6\)。

所以正确答案是 E。

English

Let \(N\) be the number of children and \(T\) the total age of the family.

Then \(20 = \dfrac{T}{N + 2}\) and \(16 = \dfrac{T - 48}{N + 1}.\)

These give \(20N + 40 = T\) and \(16N + 64 = T,\) so \(20N + 40 = 16N + 64.\)

Hence \(4N = 24\) and \(N = 6.\)

Thus, the correct answer is E.

11 答案 C

每个顶点恰好属于三个面，所以把六个面的数字和加起来得到 \[ \begin{aligned} 3(1 + 2 + \cdots + 8) &= 3 \cdot 36 \\ &= 108 \end{aligned}\text{。}\]

共有六个面，所以共同和为 \(108 \div 6 = 18\)。

所以正确答案是 C。

English

Each vertex belongs to exactly three faces, so summing the numbers over all six faces gives \[ \begin{aligned} 3(1 + 2 + \cdots + 8) &= 3 \cdot 36 \\ &= 108. \end{aligned} \]

There are six faces, so the common sum is \(108 \div 6 = 18.\)

Thus, the correct answer is C.

12 答案 D

每名游客独立选择两名导游之一，共有 \(2^6 = 64\) 种安排。

其中恰好有两种安排会让某名导游没有游客，所以答案为 \(64 - 2 = 62\)。

所以正确答案是 D。

English

Each tourist independently chooses one of the two guides, giving \(2^6 = 64\) arrangements.

Exactly two of these leave a guide with no tourists, so the answer is \(64 - 2 = 62.\)

Thus, the correct answer is D.

13 答案 B

设 Yan 到家的距离和到体育场的距离分别为 \(x\) 和 \(y\)，步行速度为 \(w\)。

直接步行到体育场需要 \(\dfrac{y}{w}\)。走回家再骑车需要 \(\dfrac{x}{w} + \dfrac{x + y}{7w}\)。

令二者相等，得 \(7y = 8x + y\)，所以 \(8x = 6y\)，从而 \(\dfrac{x}{y} = \dfrac{3}{4}\)。

所以正确答案是 B。

English

Let \(x\) and \(y\) be Yan’s distances from home and from the stadium, and let \(w\) be his walking speed.

Walking to the stadium takes \(\dfrac{y}{w}.\) Walking home and biking takes \(\dfrac{x}{w} + \dfrac{x + y}{7w}.\)

Setting these equal gives \(7y = 8x + y,\) so \(8x = 6y\) and \(\dfrac{x}{y} = \dfrac{3}{4}.\)

Thus, the correct answer is B.

14 答案 A

设三边为 \(3x, 4x, 5x\)。

这个三角形是直角三角形，所以斜边为圆的直径。因此 \(5x = 2 \cdot 3 = 6\)，得到 \(x = \tfrac65\)。

面积为 \[ \begin{aligned} \tfrac12 (3x)(4x) &= 6x^2 = 6 \cdot \tfrac{36}{25} \\ &= 8.64 \end{aligned}\text{。}\]

所以正确答案是 A。

English

Let the sides be \(3x, 4x, 5x.\) The triangle is right-angled, so its hypotenuse is a diameter.

Thus \(5x = 2 \cdot 3 = 6,\) giving \(x = \tfrac65.\)

The area is \[ \begin{aligned} \tfrac12 (3x)(4x) &= 6x^2 = 6 \cdot \tfrac{36}{25} \\ &= 8.64. \end{aligned} \]

Thus, the correct answer is A.

15 答案 B

考虑连接半径 \(2\) 的圆心和两个相邻小圆圆心形成的等腰直角三角形。它的两条直角边长为 \(2 + 1 = 3\)，所以斜边为 \(3\sqrt2\)。

正方形边长比这条斜边多 \(2\)（两端各一个小圆半径），所以 \(s = 2 + 3\sqrt2\)。

面积为 \[ \begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Consider the isosceles right triangle joining the center of the radius-\(2\) circle to the centers of two adjacent small circles. Its legs have length \(2 + 1 = 3,\) so its hypotenuse is \(3\sqrt2.\)

The side of the square exceeds this hypotenuse by \(2\) (one radius on each end), so \(s = 2 + 3\sqrt2.\)

The area is \[ \begin{aligned} (2 + 3\sqrt2)^2 &= 4 + 12\sqrt2 + 18 \\ &= 22 + 12\sqrt2. \end{aligned} \]

Thus, the correct answer is B.

16 答案 E

从 \(0\) 到 \(2007\) 的整数中一半为奇数，所以 \(ad\) 为奇数的概率为 \(\tfrac12 \cdot \tfrac12 = \tfrac14\)，为偶数的概率为 \(\tfrac34\)。\(bc\) 也是如此。

\(ad - bc\) 为偶数当两个乘积奇偶性相同：\[ \tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58\text{。}\]

所以正确答案是 E。

English

Half the integers from \(0\) to \(2007\) are odd, so each of \(ad\) and \(bc\) is odd with probability \(\tfrac12 \cdot \tfrac12 = \tfrac14\) and even with probability \(\tfrac34.\)

The difference \(ad - bc\) is even when both products have the same parity: \[ \tfrac14 \cdot \tfrac14 + \tfrac34 \cdot \tfrac34 = \tfrac{1}{16} + \tfrac{9}{16} = \tfrac58. \]

Thus, the correct answer is E.

17 答案 D

因为 \(n^3 = 75m = 3 \cdot 5^2 \cdot m\)，每个质因数的指数都必须是三的倍数。

最小这样的 \(m\) 是 \(3^2 \cdot 5 = 45\)，此时 \(n^3 = 3^3 \cdot 5^3\)，所以 \(n = 15\)。

因此 \(m + n = 45 + 15 = 60\)。

所以正确答案是 D。

English

Since \(n^3 = 75m = 3 \cdot 5^2 \cdot m,\) every prime factor must occur a multiple of three times.

The smallest such \(m\) is \(3^2 \cdot 5 = 45,\) giving \(n^3 = 3^3 \cdot 5^3\) and \(n = 15.\)

Then \(m + n = 45 + 15 = 60.\)

Thus, the correct answer is D.

18 答案 C

将图形放在坐标系中，取 \(A = (-2, 6)\)、\(B = (2, 6)\)、\(C = (2, 2)\)、\(G = (2, -6)\)、\(H = (-2, -6)\)。

直线 \(AG\) 为 \(y = -3x\)，直线 \(CH\) 为 \(y = 2x - 2\)。

它们的交点为 \(M = \left(\tfrac25, -\tfrac65\right)\)。

对 \(A, B, C, M\) 使用鞋带公式，面积为 \[ \begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Put the figure on coordinates with \(A = (-2, 6),\) \(B = (2, 6),\) \(C = (2, 2),\) \(G = (2, -6),\) and \(H = (-2, -6).\)

Line \(AG\) is \(y = -3x,\) and line \(CH\) is \(y = 2x - 2.\)

Their intersection is \(M = \left(\tfrac25, -\tfrac65\right).\)

Applying the shoelace formula to \(A, B, C, M\) gives \[ \begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5}. \end{aligned} \]

Thus, the correct answer is C.

19 答案 C

设正方形边长为 \(s\)，画刷宽度为 \(w\)，一个未涂色等腰直角三角形的直角边为 \(x\)。每个三角形面积为 \(\tfrac18 s^2\)，所以 \(\tfrac12 x^2 = \tfrac18 s^2\)，得到 \(x = \tfrac{s}{2}\)。

直角边加上画刷宽度等于半条对角线：\(x + w = \tfrac{\sqrt2}{2} s\)。因此 \(w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}\)。

所以 \[ \dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2\text{。}\]

所以正确答案是 C。

English

Let \(s\) be the side, \(w\) the brush width, and \(x\) the leg of one unpainted isosceles right triangle. Each triangle has area \(\tfrac18 s^2,\) so \(\tfrac12 x^2 = \tfrac18 s^2\) and \(x = \tfrac{s}{2}.\)

The leg plus the brush width is half the diagonal: \(x + w = \tfrac{\sqrt2}{2} s.\) Thus \(w = \tfrac{\sqrt2}{2} s - \tfrac{s}{2}.\)

Therefore \[ \dfrac{s}{w} = \dfrac{2}{\sqrt2 - 1} = 2\sqrt2 + 2. \]

Thus, the correct answer is C.

20 答案 D

将 \(a + a^{-1} = 4\) 平方，得到 \(a^2 + 2 + a^{-2} = 16\)，所以 \(a^2 + a^{-2} = 14\)。

再平方得到 \(a^4 + 2 + a^{-4} = 196\)，所以 \(a^4 + a^{-4} = 194\)。

所以正确答案是 D。

English

Squaring \(a + a^{-1} = 4\) gives \(a^2 + 2 + a^{-2} = 16,\) so \(a^2 + a^{-2} = 14.\)

Squaring again gives \(a^4 + 2 + a^{-4} = 196,\) so \(a^4 + a^{-4} = 194.\)

Thus, the correct answer is D.

21 答案 C

外部立方体每个面的面积为 \(24 \div 6 = 4\)，所以边长为 \(2\)，球的直径为 \(2\)。

这个直径是内部立方体的空间对角线，所以 \(l\sqrt3 = 2\)，得到 \(l^2 = \tfrac43\)。

内部立方体表面积为 \(6 l^2 = 6 \cdot \tfrac43 = 8\)。

所以正确答案是 C。

English

Each face of the outer cube has area \(24 \div 6 = 4,\) so its side is \(2,\) and the sphere has diameter \(2.\)

This diameter is the space diagonal of the inner cube, so \(l\sqrt3 = 2,\) giving \(l^2 = \tfrac43.\)

The inner cube’s surface area is \(6 l^2 = 6 \cdot \tfrac43 = 8.\)

Thus, the correct answer is C.

22 答案 D

在整个数列中，每个数字作为百位、十位、个位出现的次数相同。

若 \(k\) 是所有项个位数字之和，则 \(S = 111k = 3 \cdot 37 \cdot k\)，所以 \(S\) 总能被 \(37\) 整除。

数列 \(123\)、\(231\)、\(312\) 给出 \(S = 666 = 2 \cdot 3^2 \cdot 37\)，没有更大的质因数被强制整除，所以答案是 \(37\)。

所以正确答案是 D。

English

Each digit appears as a hundreds digit, a tens digit, and a units digit the same number of times across the sequence.

If \(k\) is the sum of the units digits of all terms, then \(S = 111k = 3 \cdot 37 \cdot k,\) so \(S\) is always divisible by \(37.\)

The sequence \(123,\) \(231,\) \(312\) gives \(S = 666 = 2 \cdot 3^2 \cdot 37,\) which has no larger prime factor forced, so \(37\) is the answer.

Thus, the correct answer is D.

23 答案 B

因数 \(m + n\) 和 \(m - n\) 的奇偶性相同。因为它们的乘积是 \(96\)，所以不可能都是奇数，只能都是偶数。

偶数因数对为 \((48, 2)\)、\((24, 4)\)、\((16, 6)\) 和 \((12, 8)\)，对应 \((m, n) = (25, 23)\)、\((14, 10)\)、\((11, 5)\) 和 \((10, 2)\)。

所以共有 \(4\) 个有序对。

所以正确答案是 B。

English

The factors \(m + n\) and \(m - n\) have the same parity. Since their product is \(96,\) they cannot both be odd, so both must be even.

The even factor pairs are \((48, 2),\) \((24, 4),\) \((16, 6),\) and \((12, 8),\) giving \((m, n) = (25, 23),\) \((14, 10),\) \((11, 5),\) and \((10, 2).\)

So there are \(4\) ordered pairs.

Thus, the correct answer is B.

24 答案 B

矩形 \(ABFE\) 面积为 \(AE \cdot AB = 2 \cdot 4\sqrt2 = 8\sqrt2\)。

直角三角形 \(ACO\) 和 \(BDO\) 的斜边均为 \(2\sqrt2\)，且一条直角边为 \(2\)，所以每个都是面积为 \(2\) 的等腰直角三角形。

角 \(CAE\) 和 \(DBF\) 都是 \(45^\circ\)，所以扇形 \(CAE\) 和 \(DBF\) 的面积各为 \(\tfrac18 \pi \cdot 2^2 = \tfrac{\pi}{2}\)。

阴影面积为 \[ \begin{gathered} 8\sqrt2 - 2 \cdot 2 - 2 \cdot \tfrac{\pi}{2} \\ = 8\sqrt2 - 4 - \pi \end{gathered}\text{。}\]

所以正确答案是 B。

English

Rectangle \(ABFE\) has area \(AE \cdot AB = 2 \cdot 4\sqrt2 = 8\sqrt2.\)

Right triangles \(ACO\) and \(BDO\) each have hypotenuse \(2\sqrt2\) and a leg of \(2,\) so each is isosceles right with area \(2.\)

Angles \(CAE\) and \(DBF\) are each \(45^\circ,\) so sectors \(CAE\) and \(DBF\) each have area \(\tfrac18 \pi \cdot 2^2 = \tfrac{\pi}{2}.\)

The shaded area is \[ \begin{gathered} 8\sqrt2 - 2 \cdot 2 - 2 \cdot \tfrac{\pi}{2} \\ = 8\sqrt2 - 4 - \pi. \end{gathered} \]

Thus, the correct answer is B.

25 答案 D

若 \(n \le 2007\)，则 \(S(n) \le 28\)，且 \(S(S(n)) \le 10\)，所以 \(n \ge 2007 - 28 - 10 = 1969\)。

因为 \(n\)、\(S(n)\)、\(S(S(n))\) 模 \(9\) 都同余，且 \(2007\) 是 \(9\) 的倍数，所以它们都必须是 \(3\) 的倍数。

检查 \(1969\) 到 \(2007\) 之间的 \(3\) 的倍数，满足条件的是 \(1977\)、\(1980\)、\(1983\)、\(2001\)。

所以共有 \(4\) 个 \(n\)。

所以正确答案是 D。

English

If \(n \le 2007,\) then \(S(n) \le 28\) and \(S(S(n)) \le 10,\) so \(n \ge 2007 - 28 - 10 = 1969.\)

Since \(n,\) \(S(n),\) and \(S(S(n))\) all leave the same remainder modulo \(9\) and \(2007\) is a multiple of \(9,\) each must be a multiple of \(3.\)

Checking the multiples of \(3\) between \(1969\) and \(2007,\) the condition holds for \(1977,\) \(1980,\) \(1983,\) and \(2001.\)

So there are \(4\) values of \(n.\)

Thus, the correct answer is D.