2007B AMC 10 真题（P版）

1 答案 E

一间卧室的墙面面积为 \(2(12+10)\cdot 8 = 44\cdot 8 = 352\) 平方英尺。减去门洞和窗户的 \(60\) 平方英尺，每间卧室需粉刷 \(352-60=292\) 平方英尺。

有 \(3\) 间卧室，所以共需粉刷 \(3\cdot 292 = 876\) 平方英尺。

所以正确答案是 E。

English

The walls of one bedroom have area \(2(12+10)\cdot 8 = 44\cdot 8 = 352\) square feet. Subtracting the \(60\) square feet of doorways and windows leaves \(352-60=292\) square feet per bedroom.

With \(3\) bedrooms, the total is \(3\cdot 292 = 876\) square feet.

Thus, the correct answer is E.

2 答案 E

因为 \(3\star 5=(3+5)\cdot 5=40\)，且 \(5\star 3=(5+3)\cdot 3=24\)，所以差为 \(40-24=16\)。

所以正确答案是 E。

English

Since \(3\star 5=(3+5)\cdot 5=40\) and \(5\star 3=(5+3)\cdot 3=24,\) the difference is \(40-24=16.\)

Thus, the correct answer is E.

3 答案 B

回家用了 \(\frac{120}{30}=4\) 加仑，返程用了 \(\frac{120}{20}=6\) 加仑，共用 \(10\) 加仑，行驶 \(240\) 英里。

平均为 \(\frac{240}{10} = 24\) 英里/加仑。

所以正确答案是 B。

English

The student used \(\frac{120}{30}=4\) gallons driving home and \(\frac{120}{20}=6\) gallons returning, for \(10\) gallons over \(240\) miles.

The average is \(\frac{240}{10} = 24\) miles per gallon.

Thus, the correct answer is B.

4 答案 D

因为 \(OA=OB=OC\)，三角形 \(AOB, BOC\) 和 \(COA\) 都是等腰三角形。底角给出 \(\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ\)，\(\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ\)。

因此 \(\angle ABC=20^\circ+30^\circ=50^\circ\)。

所以正确答案是 D。

English

Since \(OA=OB=OC,\) triangles \(AOB, BOC,\) and \(COA\) are isosceles. The base angles give \(\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ\) and \(\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.\)

Therefore \(\angle ABC=20^\circ+30^\circ=50^\circ.\)

Thus, the correct answer is D.

5 答案 D

将陈述写成蕴含关系：Crup 蕴含 Dramp，Dramp 蕴含 Arog，Arog 蕴含 Braf，即 \(C\Rightarrow D\Rightarrow A\Rightarrow B\)。

所以每个 Crup 都是 Dramp、Arog 和 Braf；选项中唯一必然成立的是：所有 Crups 都是 Arogs 和 Brafs。

所以正确答案是 D。

English

Writing the statements as implications, being a Crup implies being a Dramp, a Dramp implies being an Arog, and an Arog implies being a Braf: \(C\Rightarrow D\Rightarrow A\Rightarrow B.\)

So every Crup is a Dramp, an Arog, and a Braf. The only listed statement guaranteed true is that all Crups are Arogs and Brafs.

Thus, the correct answer is D.

6 答案 D

三道空题给 \(3\cdot 1.5 = 4.5\) 分，所以 Sarah 需要从前 \(22\) 题得到 \(100-4.5=95.5\) 分。

因为 \(\frac{95.5}{6}\) 介于 \(15\) 和 \(16\) 之间，她至少要答对 \(16\) 题，此时总分为 \(100.5\)。

所以正确答案是 D。

English

The three blank problems give \(3\cdot 1.5 = 4.5\) points, so Sarah needs \(100-4.5=95.5\) points from the first \(22.\)

Since \(\frac{95.5}{6}\) lies between \(15\) and \(16,\) she must answer at least \(16\) correctly, which would give a score of \(100.5.\)

Thus, the correct answer is D.

7 答案 E

因为 \(AB=BC=EA\) 且 \(\angle A=\angle B=90^\circ\)，四边形 \(ABCE\) 是正方形，所以 \(\angle AEC=90^\circ\)。

余下的边满足 \(CD=DE=EC\)，所以 \(\triangle CDE\) 是等边三角形，\(\angle CED=60^\circ\)。

因此 \(\angle E=\angle AEC\) \(+\angle CED=90^\circ\) \(+60^\circ=150^\circ\)。

所以正确答案是 E。

English

Because \(AB=BC=EA\) and \(\angle A=\angle B=90^\circ,\) quadrilateral \(ABCE\) is a square, so \(\angle AEC=90^\circ.\)

The remaining sides satisfy \(CD=DE=EC,\) so \(\triangle CDE\) is equilateral and \(\angle CED=60^\circ.\)

Therefore \(\angle E=\angle AEC\) \(+\angle CED=90^\circ\) \(+60^\circ=150^\circ.\)

Thus, the correct answer is E.

8 答案 D

一旦选定 \(a\) 和 \(c\)，\(b=\dfrac{a+c}{2}\) 就确定，且 \(a\lt b\lt c\) 自动成立。为了使 \(b\) 为整数，\(a\) 和 \(c\) 必须同奇偶。

从 \(\{0,2,4,6,8\}\) 中选两个，有 \(\binom{5}{2}=10\) 对；从 \(\{1,3,5,7,9\}\) 中选两个，又有 \(10\) 对。

因此共有 \(20\) 个符合条件的五位数。

所以正确答案是 D。

English

Once \(a\) and \(c\) are chosen, \(b=\dfrac{a+c}{2}\) is determined, and \(a\lt b\lt c\) holds automatically. For \(b\) to be an integer, \(a\) and \(c\) must share parity.

Choosing two even digits from \(\{0,2,4,6,8\}\) gives \(\binom{5}{2}=10\) pairs, and choosing two odd digits from \(\{1,3,5,7,9\}\) gives another \(10.\)

This yields \(20\) valid numbers.

Thus, the correct answer is D.

9 答案 D

最后一个 s 是消息中字母 s 的第 \(12\) 次出现，所以它向右移动 \(1+2+\cdots+12=\dfrac{12\cdot 13}{2}=78\) 位。

因为 \(78=3\cdot 26\) 是字母表长度 \(26\) 的倍数，所以移动后回到同一个字母 s。

所以正确答案是 D。

English

The final s is the \(12\)th appearance of the letter s in the message, so it is shifted \(1+2+\cdots+12=\dfrac{12\cdot 13}{2}=78\) places to the right.

Since \(78=3\cdot 26\) is a multiple of the alphabet length \(26,\) the shift returns to the same letter, s.

Thus, the correct answer is D.

10 答案 A

以 \(BC\) 为底，面积为 \(\dfrac12(BC)d\)，其中 \(d\) 是 \(A\) 到直线 \(BC\) 的距离。面积等于 \(1\) 当且仅当 \(d=\dfrac{2}{BC}\)。

到直线 \(BC\) 距离固定的点形成两条平行于 \(BC\) 的直线，分居两侧。

所以正确答案是 A。

English

Taking \(BC\) as the base, the area is \(\dfrac12(BC)d,\) where \(d\) is the distance from \(A\) to line \(BC.\) The area equals \(1\) exactly when \(d=\dfrac{2}{BC}.\)

The points at this fixed distance from line \(BC\) form two lines parallel to \(BC,\) one on each side.

Thus, the correct answer is A.

11 答案 C

三角形边长为 \(3\)、\(3\)、\(2\)。它的面积为 \(\dfrac12\cdot 2\cdot\sqrt{3^2-1^2}=2\sqrt2\)。

外接圆半径为 \(R=\dfrac{abc}{4K}\) \(=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2}\) \(=\dfrac{9}{4\sqrt2}\) \(=\dfrac{9\sqrt2}{8}\)。

圆面积为 \(\pi R^2=\pi\cdot\dfrac{81\cdot 2}{64}=\dfrac{81}{32}\pi\)。

所以正确答案是 C。

English

The triangle has sides \(3,\) \(3,\) and \(2.\) Its area is \(\dfrac12\cdot 2\cdot\sqrt{3^2-1^2}=2\sqrt2.\)

The circumradius is \(R=\dfrac{abc}{4K}\) \(=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2}\) \(=\dfrac{9}{4\sqrt2}\) \(=\dfrac{9\sqrt2}{8}.\)

The area of the circle is \(\pi R^2=\pi\cdot\dfrac{81\cdot 2}{64}=\dfrac{81}{32}\pi.\)

Thus, the correct answer is C.

12 答案 D

\(N\) 年前 Tom 的年龄为 \(T-N\)，三个孩子年龄之和为 \(T-3N\)。

条件给出 \(T-N=2(T-3N)\)，所以 \(T-N=2T-6N\)，化简为 \(5N=T\)。

因此 \(\frac{T}{N}=5\)。

所以正确答案是 D。

English

\(N\) years ago Tom’s age was \(T-N,\) and the sum of his three children’s ages was \(T-3N.\)

The condition gives \(T-N=2(T-3N),\) so \(T-N=2T-6N,\) which simplifies to \(5N=T.\)

Therefore \(\frac{T}{N}=5.\)

Thus, the correct answer is D.

13 答案 D

两个圆交于 \((0,0)\) 和 \((2,2)\)。

由对称性，交集的一半是从一个圆的四分之一圆中去掉直角边为 \(2\) 的等腰直角三角形。四分之一圆面积为 \(\dfrac14\pi(2)^2=\pi\)，三角形面积为 \(\dfrac12(2)^2=2\)。

所以交集面积为 \(2(\pi-2)\)。

所以正确答案是 D。

English

The two circles intersect at \((0,0)\) and \((2,2).\)

By symmetry, half the intersection is formed by removing an isosceles right triangle of leg length \(2\) from a quarter of one circle. The quarter-circle has area \(\dfrac14\pi(2)^2=\pi\) and the triangle has area \(\dfrac12(2)^2=2.\)

Therefore the whole region has area \(2(\pi-2).\)

Thus, the correct answer is D.

14 答案 C

两名女生离开，两名男生加入，所以小组总人数不变。离开的两名女生代表全组的 \(40\%-30\%=10\%\)。

所以小组共有 \(2\div 0.10=20\) 人，起初女生人数是 \(20\) 人的 \(40\%\)，即 \(8\)。

所以正确答案是 C。

English

Two girls leave and two boys arrive, so the group size is unchanged. The two girls who left therefore represent \(40\%-30\%=10\%\) of the group.

Thus the group has \(2\div 0.10=20\) people, and the original number of girls was \(40\%\) of \(20,\) or \(8.\)

Thus, the correct answer is C.

15 答案 D

设 \(x=\angle A\)。则 \(\angle B=\dfrac{x}{2}\)，\(\angle C=\dfrac{x}{3}\)，\(\angle D=\dfrac{x}{4}\)。

四边形内角和为 \(360^\circ\)，所以 \(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}=\dfrac{25x}{12}=360\)。

因此 \(x=\dfrac{12\cdot 360}{25}=172.8\approx 173\)。

所以正确答案是 D。

English

Let \(x=\angle A.\) Then \(\angle B=\dfrac{x}{2},\) \(\angle C=\dfrac{x}{3},\) and \(\angle D=\dfrac{x}{4}.\)

The angles sum to \(360^\circ,\) so \(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}=\dfrac{25x}{12}=360.\)

Thus \(x=\dfrac{12\cdot 360}{25}=172.8\approx 173.\)

Thus, the correct answer is D.

16 答案 C

假设班上有 \(10\) 名学生：一名低年级生和九名高年级生。总分为 \(10\cdot 84=840\)。

九名高年级生总分为 \(9\cdot 83=747\)，所以低年级生的分数为 \(s=840-747=93\)。

所以正确答案是 C。

English

Suppose the class has \(10\) students: one junior and nine seniors. The total of all scores is \(10\cdot 84=840.\)

The nine seniors total \(9\cdot 83=747,\) so the junior’s score is \(s=840-747=93.\)

Thus, the correct answer is C.

17 答案 D

设边长为 \(s\)。从 \(P\) 作出的垂线是三角形 \(APB\)、\(BPC\)、\(CPA\) 的高，所以它们的面积分别为 \(\dfrac{s}{2}\)、\(s\)、\(\dfrac{3s}{2}\)。

面积和等于 \(\triangle ABC\) 的面积 \(\dfrac{\sqrt3}{4}s^2\)，因此 \(3s=\dfrac{\sqrt3}{4}s^2\)。

正解为 \(s=4\sqrt3\)。

所以正确答案是 D。

English

Let the side length be \(s.\) The perpendiculars from \(P\) are the heights of triangles \(APB,\) \(BPC,\) and \(CPA,\) so their areas are \(\dfrac{s}{2},\) \(s,\) and \(\dfrac{3s}{2}.\)

Their sum equals the area of \(\triangle ABC,\) which is also \(\dfrac{\sqrt3}{4}s^2.\) Hence \(3s=\dfrac{\sqrt3}{4}s^2.\)

The positive solution is \(s=4\sqrt3.\)

Thus, the correct answer is D.

18 答案 B

连接四个外圆的圆心形成正方形。相邻外圆相切，所以每条边长为 \(2r\)。

正方形对角线经过中心圆的圆心，长度为 \(1+r+r+1=2+2r\)。边长为 \(2r\) 的正方形对角线为 \(2r\sqrt2\)，所以 \(2(2r)^2=(2+2r)^2\)。

展开得 \(1+2r+r^2=2r^2\)，即 \(r^2-2r-1=0\)。正根为 \(r=1+\sqrt2\)。

所以正确答案是 B。

English

Connect the centers of the four outer circles to form a square. Adjacent outer circles are tangent, so each side has length \(2r.\)

The diagonal of the square passes through the center of the central circle, giving length \(1+r+r+1=2+2r.\) Since a square with side \(2r\) has diagonal \(2r\sqrt2,\) we get \(2(2r)^2=(2+2r)^2.\)

Expanding gives \(1+2r+r^2=2r^2,\) so \(r^2-2r-1=0.\) The positive root is \(r=1+\sqrt2.\)

Thus, the correct answer is B.

19 答案 C

阴影方格对应两个余数同奇或同偶。第一个余数为偶数（来自数字 \(2\) 和 \(6\)）的概率为 \(\dfrac13\)，为奇数的概率为 \(\dfrac23\)。

第二个余数为偶数的概率为 \(\dfrac12\)，为奇数的概率为 \(\dfrac12\)。

它们奇偶性相同的概率为 \(\dfrac13\cdot\dfrac12+\dfrac23\cdot\dfrac12=\dfrac12\)。

所以正确答案是 C。

English

The shaded squares are those where the two remainders are both odd or both even. The first remainder is even (from the numbers \(2\) and \(6\)) with probability \(\dfrac13\) and odd with probability \(\dfrac23.\)

The second remainder is even with probability \(\dfrac12\) and odd with probability \(\dfrac12.\)

The probability that they share parity is \(\dfrac13\cdot\dfrac12+\dfrac23\cdot\dfrac12=\dfrac12.\)

Thus, the correct answer is C.

20 答案 C

选择 \(5\) 行中的 \(3\) 行有 \(\binom{5}{3}=10\) 种，选择 \(5\) 列中的 \(3\) 列也有 \(\binom{5}{3}=10\) 种。

三个选出的方块必须占据不同的行和列，所以它们是在三行与三列之间建立一个匹配，共有 \(3!=6\) 种。

总数为 \(10\cdot 10\cdot 6=600\)。

所以正确答案是 C。

English

Choose \(3\) of the \(5\) rows in \(\binom{5}{3}=10\) ways and \(3\) of the \(5\) columns in \(\binom{5}{3}=10\) ways.

The three chosen blocks must occupy distinct rows and columns, so they form a matching between the three rows and three columns, which can be done in \(3!=6\) ways.

The total is \(10\cdot 10\cdot 6=600.\)

Thus, the correct answer is C.

21 答案 B

设正方形边长为 \(s\)，从 \(B\) 到 \(AC\) 的高为 \(h\)。则 \(h=\dfrac{AB\cdot BC}{AC}=\dfrac{3\cdot 4}{5}=\dfrac{12}{5}\)。

正方形上方的小三角形与 \(\triangle ABC\) 相似，并以正方形上边为底，因此 \(\dfrac{h-s}{h}=\dfrac{s}{AC}\)，所以 \(s=\dfrac{AC\cdot h}{AC+h}\)。

代入可得 \(s=\dfrac{5\cdot\frac{12}{5}}{5+\frac{12}{5}}=\dfrac{12}{\frac{37}{5}}=\dfrac{60}{37}\)。

所以正确答案是 B。

English

Let \(s\) be the side of the square and \(h\) the altitude from \(B\) to \(AC.\) Then \(h=\dfrac{AB\cdot BC}{AC}=\dfrac{3\cdot 4}{5}=\dfrac{12}{5}.\)

The small triangle above the square is similar to \(\triangle ABC\) with the square’s top side as its base, giving \(\dfrac{h-s}{h}=\dfrac{s}{AC},\) so \(s=\dfrac{AC\cdot h}{AC+h}.\)

Substituting, \(s=\dfrac{5\cdot\frac{12}{5}}{5+\frac{12}{5}}=\dfrac{12}{\frac{37}{5}}=\dfrac{60}{37}.\)

Thus, the correct answer is B.

22 答案 B

每个骰子底面出现所选数字的概率为 \(\dfrac14\)。所以该数字出现 \(0\)、\(1\)、\(2\) 次的概率分别为 \(P(0)=\dfrac{9}{16},\ \) \(P(1)=\dfrac{6}{16},\ \) \(P(2)=\dfrac{1}{16}\)。

期望收益为 \((-1)\cdot\dfrac{9}{16}\) \(+(1)\cdot\dfrac{6}{16}\) \(+(2)\cdot\dfrac{1}{16}\) \(=\dfrac{-9+6+2}{16}\) \(=-\dfrac{1}{16}\)。

所以正确答案是 B。

English

Each die shows the chosen number on the bottom with probability \(\dfrac14.\) So the number appears \(0,\) \(1,\) or \(2\) times with probabilities \(P(0)=\dfrac{9}{16},\ \) \(P(1)=\dfrac{6}{16},\ \) \(P(2)=\dfrac{1}{16}.\)

The expected return is \((-1)\cdot\dfrac{9}{16}\) \(+(1)\cdot\dfrac{6}{16}\) \(+(2)\cdot\dfrac{1}{16}\) \(=\dfrac{-9+6+2}{16}\) \(=-\dfrac{1}{16}.\)

Thus, the correct answer is B.

23 答案 E

设原棱锥高为 \(h\)，小棱锥高为 \(h-2\)。

两个棱锥相似，所以表面积之比等于高之比的平方。小棱锥表面积为原来的一半，因此 \(\left(\dfrac{h-2}{h}\right)^2=\dfrac12\)，从而 \(\dfrac{h}{h-2}=\sqrt2\)。

于是 \(h=\sqrt2(h-2)\)，所以 \(h(\sqrt2-1)=2\sqrt2\)，\(h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2\)。

所以正确答案是 E。

English

Let \(h\) be the altitude of the original pyramid; the smaller pyramid has altitude \(h-2.\) The two pyramids are similar, so the ratio of their surface areas is the square of the ratio of their altitudes.

The smaller surface area is half the original, so \(\left(\dfrac{h-2}{h}\right)^2=\dfrac12,\) giving \(\dfrac{h}{h-2}=\sqrt2.\)

Then \(h=\sqrt2(h-2),\) so \(h(\sqrt2-1)=2\sqrt2\) and \(h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2.\)

Thus, the correct answer is E.

24 答案 C

因为 \(n\) 能被 \(9\) 整除，其数字和是 \(9\) 的倍数。若四这个数字出现 \(k\) 次，九这个数字出现 \(m\) 次，则数字和为 \(4k+9m\)，所以 \(4k\) 能被 \(9\) 整除，迫使 \(k\) 也能被 \(9\) 整除。因此 \(k\ge 9\)，并且至少有一个 \(9\)，这个数至少有十位。

要被 \(4\) 整除，最后两位必须组成 \(4\) 的倍数。在 \(44\)、\(49\)、\(94\)、\(99\) 中，只有 \(44\) 可行，所以 \(n\) 以 \(44\) 结尾。

最小的这样十位数把唯一的 \(9\) 放在最低的可用数位上，得到 \(4{,}444{,}444{,}944\)，最后四位是 \(4944\)。

所以正确答案是 C。

English

Since \(n\) is divisible by \(9,\) its digit sum is a multiple of \(9.\) With \(k\) fours and \(m\) nines, the digit sum is \(4k+9m,\) so \(4k\) is divisible by \(9,\) forcing \(k\) to be divisible by \(9.\) Thus \(k\ge 9,\) and with at least one \(9,\) the number has at least ten digits.

For divisibility by \(4,\) the last two digits must form a multiple of \(4,\) and among \(44,\) \(49,\) \(94,\) and \(99\) only \(44\) works, so \(n\) ends in \(44.\)

The smallest such ten-digit number places the single \(9\) in the lowest available position, giving \(4{,}444{,}444{,}944.\) Its last four digits are \(4944.\)

Thus, the correct answer is C.

25 答案 A

合并后，表达式为 \(\dfrac{9a^2+14b^2}{9ab}\)。若它是整数，则 \(a\) 必须整除 \(9a^2+14b^2\)，因此 \(14b^2\) 能被 \(a\) 整除。由于 \(\gcd(a,b)=1\)，得 \(14\) 能被 \(a\) 整除。同理 \(9a^2\) 能被 \(b\) 整除，所以 \(9\) 能被 \(b\) 整除。

所以 \(a\in\{1,2,7,14\}\)，\(b\in\{1,3,9\}\)。检查这些情况，只有 \(b=3\) 时对每个允许的 \(a\) 都得到整数。

有效数对为 \((1,3)\)、\((2,3)\)、\((7,3)\)、\((14,3)\)，共 \(4\) 对。

所以正确答案是 A。

English

Combining, the expression is \(\dfrac{9a^2+14b^2}{9ab}.\) For this to be an integer, \(a\) must divide \(9a^2+14b^2,\) hence \(14b^2\) is divisible by \(a.\) Since \(\gcd(a,b)=1,\) we get that \(14\) is divisible by \(a.\) Similarly, \(9a^2\) is divisible by \(b,\) so \(9\) is divisible by \(b.\)

So \(a\in\{1,2,7,14\}\) and \(b\in\{1,3,9\}.\) Checking these, only \(b=3\) makes the expression an integer for each allowed \(a.\)

The valid pairs are \((1,3),\) \((2,3),\) \((7,3),\) and \((14,3),\) for a total of \(4.\)

Thus, the correct answer is A.