2008B AMC 10 真题（P版）

1 答案 E

如果 \(k\) 个进球值 \(3\) 分，其余值 \(2\) 分，总分为 \(2(5-k)+3k=10+k\)。

当 \(k\) 取 \(0,1,\ldots,5\) 时，总分取从 \(10\) 到 \(15\) 的每个整数，共 \(6\) 种可能。

所以正确答案是 E。

English

If \(k\) of the baskets are worth \(3\) points and the rest worth \(2,\) the total is \(2(5-k)+3k=10+k.\)

As \(k\) ranges over \(0,1,\ldots,5,\) the total takes every integer value from \(10\) to \(15,\) giving \(6\) possibilities.

Thus, the correct answer is E.

2 答案 B

将第二行反为 \(11,10,9,8\)，第四行反为 \(25,24,23,22\) 后，两条对角线分别为 \(1,10,17,22\) 和 \(4,9,16,25\)。

它们的和为 \(1+10+17+22=50\) 和 \(4+9+16+25=54\)，正差为 \(54-50=4\)。

所以正确答案是 B。

English

After reversing the second row to \(11,10,9,8\) and the fourth row to \(25,24,23,22,\) the two diagonals are \(1,10,17,22\) and \(4,9,16,25.\)

Their sums are \(1+10+17+22=50\) and \(4+9+16+25=54,\) so the positive difference is \(54-50=4.\)

Thus, the correct answer is B.

3 答案 D

因为 \(\sqrt{x}=x^{\frac{1}{2}}\)，所以 \(x\sqrt{x}=x^{1}\cdot x^{\frac{1}{2}}=x^{\frac{3}{2}}\)。

取立方根相当于指数乘以 \(\tfrac13\)，得到 \(\left(x^{\frac{3}{2}}\right)^{\frac{1}{3}}=x^{\frac{1}{2}}\)。

所以正确答案是 D。

English

Since \(\sqrt{x}=x^{\frac{1}{2}},\) we have \(x\sqrt{x}=x^{1}\cdot x^{\frac{1}{2}}=x^{\frac{3}{2}}.\)

Taking the cube root multiplies the exponent by \(\tfrac13,\) giving \(\left(x^{\frac{3}{2}}\right)^{\frac{1}{3}}=x^{\frac{1}{2}}.\)

Thus, the correct answer is D.

4 答案 C

当其他 \(20\) 名球员各拿最低工资 \(\$15{,}000\) 时，某一名球员工资最大。

剩下给这名球员的是 \(\$700{,}000-20\cdot\$15{,}000\) \(=\$700{,}000-\$300{,}000\) \(=\$400{,}000\)。

所以正确答案是 C。

English

One player’s salary is largest when the other \(20\) players each earn the minimum \(\$15{,}000.\)

That leaves \(\$700{,}000-20\cdot\$15{,}000\) \(=\$700{,}000-\$300{,}000\) \(=\$400{,}000\) for the single player.

Thus, the correct answer is C.

5 答案 A

因为 \((y-x)^2=(x-y)^2\)，两个输入完全相同。

因此 \((x-y)^2\,\$\,(y-x)^2\) \(=\left((x-y)^2-(x-y)^2\right)^2\) \(=0^2=0\)。

所以正确答案是 A。

English

Since \((y-x)^2=(x-y)^2,\) the two inputs are identical.

Therefore \((x-y)^2\,\$\,(y-x)^2\) \(=\left((x-y)^2-(x-y)^2\right)^2\) \(=0^2=0.\)

Thus, the correct answer is A.

6 答案 C

因为 \(AB=4\,BD\) 且 \(AB+BD=AD\)，可得 \(5\,BD=AD\)，所以 \(BD=\tfrac15 AD\)。

因为 \(AC=9\,CD\) 且 \(AC+CD=AD\)，可得 \(10\,CD=AD\)，所以 \(CD=\tfrac{1}{10}AD\)。

因此 \(BC=BD-CD\) \(=\tfrac15 AD-\tfrac{1}{10}AD\) \(=\tfrac{1}{10}AD\)。

所以正确答案是 C。

English

Since \(AB=4\,BD\) and \(AB+BD=AD,\) we get \(5\,BD=AD,\) so \(BD=\tfrac15 AD.\)

Since \(AC=9\,CD\) and \(AC+CD=AD,\) we get \(10\,CD=AD,\) so \(CD=\tfrac{1}{10}AD.\)

Then \(BC=BD-CD\) \(=\tfrac15 AD-\tfrac{1}{10}AD\) \(=\tfrac{1}{10}AD.\)

Thus, the correct answer is C.

7 答案 C

大三角形边长是小三角形的 \(10\) 倍，所以面积是小三角形的 \(10^2=100\) 倍。

因为小三角形无重叠地铺满它，正好需要 \(100\) 个。

所以正确答案是 C。

English

The large triangle has side length \(10\) times that of a small triangle, so its area is \(10^2=100\) times as large.

Since the small triangles tile it without overlap, exactly \(100\) of them are required.

Thus, the correct answer is C.

8 答案 C

若买 \(r\) 朵玫瑰和 \(c\) 朵康乃馨，则 \(3r+2c=50\)。因为 \(2c\) 和 \(50\) 都是偶数，\(3r\) 必须是偶数，所以 \(r\) 是偶数。

又 \(3r\le 50\)，所以 \(r\le 16\)。偶数值 \(r=0,2,4,\ldots,16\) 各自给出一个有效的 \(c\)，共有 \(9\) 种花束。

所以正确答案是 C。

English

If \(r\) roses and \(c\) carnations are bought, then \(3r+2c=50.\) Because \(2c\) and \(50\) are even, \(3r\) must be even, so \(r\) is even.

Also \(3r\le 50,\) so \(r\le 16.\) The even values \(r=0,2,4,\ldots,16\) each give a valid \(c,\) which is \(9\) bouquets.

Thus, the correct answer is C.

9 答案 A

由韦达定理，方程 \(ax^2-2ax+b=0\) 的两根之和为 \(\dfrac{-(-2a)}{a}=2\)。

平均数是它的一半，即 \(1\)。

所以正确答案是 A。

English

By Vieta’s formulas, the sum of the roots of \(ax^2-2ax+b=0\) is \(\dfrac{-(-2a)}{a}=2.\)

The average is half of this, namely \(1.\)

Thus, the correct answer is A.

10 答案 A

设 \(O\) 为圆心，\(D\) 为 \(AB\) 的中点。于是 \(OD\perp AB\)，且 \(AD=3\)，所以 \(OD=\sqrt{5^2-3^2}=4\)。

因为 \(C\) 是小弧中点，所以 \(O\)、\(D\)、\(C\) 共线，且 \(DC=OC-OD=5-4=1\)。

因此 \(AC=\sqrt{AD^2+DC^2}\) \(=\sqrt{3^2+1^2}\) \(=\sqrt{10}\)。

所以正确答案是 A。

English

Let \(O\) be the center and \(D\) the midpoint of \(AB.\) Then \(OD\perp AB\) with \(AD=3,\) so \(OD=\sqrt{5^2-3^2}=4.\)

Since \(C\) is the midpoint of the minor arc, \(O,\) \(D,\) \(C\) are collinear and \(DC=OC-OD=5-4=1.\)

Then \(AC=\sqrt{AD^2+DC^2}\) \(=\sqrt{3^2+1^2}\) \(=\sqrt{10}.\)

Thus, the correct answer is A.

11 答案 B

由递推式，\(u_5=2u_4+u_3=2u_4+9\)，且 \(u_6=2u_5+u_4\) \(=2(2u_4+9)+u_4\) \(=5u_4+18\)。

令 \(5u_4+18=128\)，得 \(u_4=22\)，所以 \(u_5=2\cdot 22+9=53\)。

所以正确答案是 B。

English

Using the recurrence, \(u_5=2u_4+u_3=2u_4+9\) and \(u_6=2u_5+u_4\) \(=2(2u_4+9)+u_4\) \(=5u_4+18.\)

Setting \(5u_4+18=128\) gives \(u_4=22,\) so \(u_5=2\cdot 22+9=53.\)

Thus, the correct answer is B.

12 答案 A

每次翻转计 \(100000\) 步，所以 Pete 共走了 \(44\cdot 100000+50000=4{,}450{,}000\) 步。

除以 \(1800\)，约为 \(2472\) 英里，最接近 \(2500\)。

所以正确答案是 A。

English

Each flip counts \(100000\) steps, so Pete took \(44\cdot 100000+50000=4{,}450{,}000\) steps.

Dividing by \(1800\) gives about \(2472\) miles, which is closest to \(2500.\)

Thus, the correct answer is A.

13 答案 B

因为前 \(n\) 项平均数为 \(n\)，所以这些项之和为 \(n^2\)。

第 \(n\) 项为 \(n^2-(n-1)^2=2n-1\)，所以第 \(2008\) 项为 \(2\cdot 2008-1=4015\)。

所以正确答案是 B。

English

Since the mean of the first \(n\) terms is \(n,\) their sum is \(n^2.\)

The \(n\)th term is \(n^2-(n-1)^2=2n-1,\) so the \(2008\)th term is \(2\cdot 2008-1=4015.\)

Thus, the correct answer is B.

14 答案 B

因为 \(\angle ABO=90^\circ\)，线段 \(AB\) 竖直，所以 \(A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right)\)。

绕原点逆时针旋转 \(90^\circ\) 将 \((x,y)\) 变为 \((-y,x)\)，所以 \(A\) 的像为 \(\left(-\tfrac{5\sqrt3}{3},\,5\right)\)。

所以正确答案是 B。

English

Because \(\angle ABO=90^\circ,\) segment \(AB\) is vertical, so \(A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right).\)

A \(90^\circ\) counterclockwise rotation about the origin sends \((x,y)\) to \((-y,x),\) so the image of \(A\) is \(\left(-\tfrac{5\sqrt3}{3},\,5\right).\)

Thus, the correct answer is B.

15 答案 A

由 \(a^2+b^2=(b+1)^2\) 得 \(a^2=2b+1\)，所以 \(a\) 是奇数，且 \(a^2\) 是奇完全平方数。

因为 \(b\lt 100\)，需要 \(a^2=2b+1\lt 201\)，且当 \(b\ge 4\) 时有 \(a^2\ge 9\)。奇平方数 \(9,25,49,81,121,169\) 给出 \(a=3,5,7,9,11,13\)，共有 \(6\) 个三角形。

所以正确答案是 A。

English

From \(a^2+b^2=(b+1)^2\) we get \(a^2=2b+1,\) so \(a\) is odd and \(a^2\) is an odd perfect square.

Since \(b\lt 100,\) we need \(a^2=2b+1\lt 201,\) and \(a^2\ge 9\) for \(b\ge 4.\) The odd squares \(9,25,49,81,121,169\) give \(a=3,5,7,9,11,13,\) which is \(6\) triangles.

Thus, the correct answer is A.

16 答案 A

只要至少掷一个骰子，由对称性，点数和为奇数的概率为 \(\tfrac12\)。

不掷骰子只在两枚硬币都为反面时发生，概率为 \(\tfrac14\)；此时点数和 \(0\) 是偶数。因此所求概率为 \(\left(1-\tfrac14\right)\cdot\tfrac12=\tfrac38\)。

所以正确答案是 A。

English

Whenever at least one die is rolled, by symmetry the sum is odd with probability \(\tfrac12.\)

No die is rolled only when both coins are tails, with probability \(\tfrac14;\) that sum \(0\) is even. So the answer is \(\left(1-\tfrac14\right)\cdot\tfrac12=\tfrac38.\)

Thus, the correct answer is A.

17 答案 B

三次中恰好一次认可有 \(\binom{3}{1}=3\) 种方式，每种概率为 \((0.7)(0.3)(0.3)=0.063\)。

总概率为 \(3\cdot 0.063=0.189\)。

所以正确答案是 B。

English

Exactly one approval among three occasions arises in \(\binom{3}{1}=3\) ways, each with probability \((0.7)(0.3)(0.3)=0.063.\)

The total is \(3\cdot 0.063=0.189.\)

Thus, the correct answer is B.

18 答案 B

设砖块数为 \(n\)。Brenda 每小时砌 \(\tfrac{n}{9}\) 块，Brandon 每小时砌 \(\tfrac{n}{10}\) 块，所以一起每小时砌 \(\tfrac{n}{9}+\tfrac{n}{10}-10\) 块。

他们工作 \(5\) 小时共完成 \(n\) 块：\[5\left(\tfrac{n}{9}+\tfrac{n}{10}-10\right)=n\text{。}\]解得 \(\tfrac{5n}{9}+\tfrac{n}{2}-50=n\)，所以 \(n=900\)。

所以正确答案是 B。

English

Let \(n\) be the number of bricks. Brenda lays \(\tfrac{n}{9}\) per hour and Brandon \(\tfrac{n}{10},\) so together they lay \(\tfrac{n}{9}+\tfrac{n}{10}-10\) per hour.

Over \(5\) hours this equals \(n:\) \[5\left(\tfrac{n}{9}+\tfrac{n}{10}-10\right)=n.\] Solving, \(\tfrac{5n}{9}+\tfrac{n}{2}-50=n,\) which gives \(n=900.\)

Thus, the correct answer is B.

19 答案 E

被水浸没的横截面是圆弓形。弦在圆心下方 \(4-2=2\) 英尺，且 \(\cos\theta=\tfrac{2}{4}=\tfrac12\)，所以半角为 \(60^\circ\)，圆心角为 \(120^\circ\)。

扇形面积为 \(\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3}\)，两条半径形成的三角形面积为 \(\tfrac12(4)^2\sin 120^\circ=4\sqrt3\)，所以圆弓形面积为 \(\tfrac{16\pi}{3}-4\sqrt3\)。

乘以长度 \(9\)，体积为 \(9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3\)。

所以正确答案是 E。

English

The submerged cross-section is a circular segment. The chord is \(4-2=2\) feet below the center, and \(\cos\theta=\tfrac{2}{4}=\tfrac12,\) so the half-angle is \(60^\circ\) and the central angle is \(120^\circ.\)

The sector area is \(\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3},\) and the triangle formed by the two radii has area \(\tfrac12(4)^2\sin 120^\circ=4\sqrt3.\) The segment area is \(\tfrac{16\pi}{3}-4\sqrt3.\)

Multiplying by the length \(9\) gives \(9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3.\)

Thus, the correct answer is E.

20 答案 B

在 \(36\) 个等可能结果中，和为 \(5\) 的有 \((1,4),(2,3),(2,3),(4,1)\)，共 \(4\) 个。

和为 \(7\) 的有 \((1,6),(2,5),(2,5)\)、\((3,4),(3,4),(4,3)\)，共 \(6\) 个；和为 \(9\) 的有 \((1,8),(3,6),(3,6),(4,5)\)，共 \(4\) 个。

因此概率为 \(\tfrac{4+6+4}{36}=\tfrac{14}{36}=\tfrac{7}{18}\)。

所以正确答案是 B。

English

Of the \(36\) equally likely outcomes, the pairs giving sum \(5\) are \((1,4),(2,3),(2,3),(4,1),\) which is \(4\) outcomes.

Sum \(7\) comes from \((1,6),(2,5),(2,5),\) \((3,4),(3,4),(4,3),\) which is \(6,\) and sum \(9\) from \((1,8),(3,6),(3,6),(4,5),\) which is \(4.\)

The probability is \(\tfrac{4+6+4}{36}=\tfrac{14}{36}=\tfrac{7}{18}.\)

Thus, the correct answer is B.

21 答案 C

先安排女性。第一位女性可坐任意 \(10\) 把椅子，而男女必须交替，所以其余女性在剩下四个同类座位中有 \(4!\) 种排法，共 \(10\cdot 4!=240\) 种安排。

固定一位女性在椅子 \(1\)。她的配偶必须坐在椅子 \(4\) 或椅子 \(8\)；每个选择都会一致地迫使其他男性的位置。因此每种女性安排恰有 \(2\) 种有效男性安排。

总数为 \(2\cdot 240=480\)。

所以正确答案是 C。

English

Seat the women first. The first woman may take any of the \(10\) chairs, and since seats alternate, the remaining women fill their four seats in \(4!\) ways, giving \(10\cdot 4!=240\) arrangements.

Fix a woman in chair \(1.\) Her spouse must sit in chair \(4\) or chair \(8;\) each choice then forces the placement of every other man consistently. So each seating of the women yields exactly \(2\) valid seatings of the men.

The total is \(2\cdot 240=480.\)

Thus, the correct answer is C.

22 答案 C

可区分的排列共有 \(\tfrac{6!}{3!\,2!}=60\) 种。三颗红珠必须占据不相邻的位置，可能的红珠位置为 \(\{1,3,5\},\{2,4,6\},\{1,3,6\}\) 和 \(\{1,4,6\}\)。

对于 \(\{1,3,5\}\) 和 \(\{2,4,6\}\)，剩余位置彼此不相邻，所以蓝珠可在 \(3\) 个位置中任意选择，共 \(3+3=6\) 种。对于 \(\{1,3,6\}\) 和 \(\{1,4,6\}\)，剩余位置中有两个相邻，所以蓝珠必须隔开两颗白珠，共 \(2+2=4\) 种。

有效排列共 \(10\) 种，所以概率为 \(\tfrac{10}{60}=\tfrac16\)。

所以正确答案是 C。

English

There are \(\tfrac{6!}{3!\,2!}=60\) distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are \(\{1,3,5\},\{2,4,6\},\{1,3,6\},\) and \(\{1,4,6\}.\)

For \(\{1,3,5\}\) and \(\{2,4,6\},\) the remaining seats are mutually non-adjacent, so the blue bead can go in any of the \(3,\) giving \(3+3=6.\) For \(\{1,3,6\}\) and \(\{1,4,6\},\) two remaining seats are adjacent, so the blue must separate the whites, giving \(2+2=4.\)

That is \(10\) valid orderings, so the probability is \(\tfrac{10}{60}=\tfrac16.\)

Thus, the correct answer is C.

23 答案 B

被涂色矩形为 \((a-2)\times(b-2)\)，且是地板面积的一半，所以 \(ab=2(a-2)(b-2)\)。

展开得 \(0=ab-4a-4b+8\)，两边加 \(8\) 得 \((a-4)(b-4)=8\)。

在 \(b\gt a\gt 0\) 下，分解 \(8=1\cdot 8=2\cdot 4\)，得到 \((a,b)=(5,12)\) 和 \((6,8)\)，所以共有 \(2\) 种可能。

所以正确答案是 B。

English

The painted rectangle is \((a-2)\times(b-2),\) and it is half the floor, so \(ab=2(a-2)(b-2).\)

Expanding gives \(0=ab-4a-4b+8,\) and adding \(8\) yields \((a-4)(b-4)=8.\)

With \(b\gt a\gt 0,\) the factorizations \(8=1\cdot 8=2\cdot 4\) give \((a,b)=(5,12)\) and \((6,8).\) So there are \(2\) possibilities.

Thus, the correct answer is B.

24 答案 C

令 \(M\) 为使 \(\triangle BMC\) 为等边三角形的点，且与 \(A\) 在 \(BC\) 的同侧。于是 \(\angle ABM=70^\circ-60^\circ=10^\circ\)，且 \(\angle MCD=170^\circ-60^\circ=110^\circ\)。

因为 \(AB=BM\) 且 \(MC=CD\)，三角形 \(ABM\) 和 \(MCD\) 是等腰三角形，所以 \(\angle AMB=85^\circ\)，且 \(\angle CMD=35^\circ\)。

于是 \(\angle AMD=360^\circ-85^\circ-60^\circ\) \(-35^\circ\) \(=180^\circ\)，所以 \(M\) 在 \(\overline{AD}\) 上，且 \(\angle BAD=\angle BAM=85^\circ\)。

所以正确答案是 C。

English

Let \(M\) be the point with \(\triangle BMC\) equilateral, on the same side of \(BC\) as \(A.\) Then \(\angle ABM=70^\circ-60^\circ=10^\circ\) and \(\angle MCD=170^\circ-60^\circ=110^\circ.\)

Since \(AB=BM\) and \(MC=CD,\) triangles \(ABM\) and \(MCD\) are isosceles, giving \(\angle AMB=85^\circ\) and \(\angle CMD=35^\circ.\)

Then \(\angle AMD=360^\circ-85^\circ-60^\circ\) \(-35^\circ\) \(=180^\circ,\) so \(M\) lies on \(\overline{AD}\) and \(\angle BAD=\angle BAM=85^\circ.\)

Thus, the correct answer is C.

25 答案 B

给垃圾桶编号，使 Michael 在 \(0\) 号桶，垃圾车在 \(1\) 号桶。Michael 在 \(n\) 号桶的到达时间为 \(40n\) 秒；垃圾车在 \(n\) 号桶的离开时间为 \(50(n-1)\) 秒，到达时间为 \(50(n-1)-30\) 秒。

Michael 和垃圾车在 \(n\) 号桶相遇当且仅当 \(50(n-1)-30\le 40n\) \(\le 50(n-1)\)，化简得 \(5\le n\le 8\)。

在 \(5\) 号桶他们在垃圾车离开时相遇，在 \(6\) 和 \(7\) 号桶 Michael 经过停着的车，在 \(8\) 号桶他们在垃圾车到达时相遇。此外，在 \(6\) 号和 \(7\) 号桶之间，垃圾车还必须超过 Michael 一次，所以总共相遇 \(5\) 次。

所以正确答案是 B。

English

Number the pails so Michael is at pail \(0\) and the truck at pail \(1.\) Michael reaches pail \(n\) at \(40n\) seconds. The truck leaves pail \(n\) at \(50(n-1)\) seconds and arrives there at \(50(n-1)-30\) seconds.

Michael and the truck are together at pail \(n\) when \(50(n-1)-30\le 40n\) \(\le 50(n-1),\) which simplifies to \(5\le n\le 8.\)

At pail \(5\) they meet as the truck departs, at pails \(6\) and \(7\) Michael passes it, and at pail \(8\) they meet as the truck arrives. Between pails \(6\) and \(7\) the truck must overtake Michael once more, so in total they meet \(5\) times.

Thus, the correct answer is B.