2001 AMC 10 真题（P版）

1 答案 E

这 \(9\) 个数已经按递增顺序排列，中位数是第 \(5\) 项 \(n+6\)，所以 \(n+6=10\)，得 \(n=4\)。

代入后，各项之和的计算过程为 \(9n\) \(\small {}+(3+4+5+6+8+10+12+15)\) \(=9n+63\) \(=99\)，所以平均数是 \(\frac{99}{9}=11\)。

所以正确答案是 E。

English

The list has \(9\) numbers in increasing order, so the median is the \(5\)th term, \(n+6.\) Setting \(n+6=10\) gives \(n=4.\)

The sum of the terms is \(9n\) \(\small {}+(3+4+5+6+8+10+12+15)\) \(=9n+63\) \(=99,\) so the mean is \(\frac{99}{9}=11.\)

Thus, the correct answer is E.

2 答案 C

\(x\) 的倒数是 \(\dfrac1x\)，加法逆元是 \(-x\)，它们的乘积为 \(\dfrac1x\cdot(-x)=-1\)。

因此 \(x=2+(-1)=1\)，位于区间 \(0\lt x\le2\)。

所以正确答案是 C。

English

The reciprocal of \(x\) is \(\dfrac1x\) and its additive inverse is \(-x.\) Their product is \(\dfrac1x\cdot(-x)=-1.\)

So \(x=2+(-1)=1,\) which lies in the interval \(0\lt x\le2.\)

Thus, the correct answer is C.

3 答案 E

设两个数为 \(a\) 和 \(b\)，且 \(a+b=S\)。每个数加上 \(3\) 再加倍后，所得两数之和为 \(2(a+3)\) \(+2(b+3)\) \(=2(a+b)+12\) \(=2S+12\)。

所以正确答案是 E。

English

Let the numbers be \(a\) and \(b,\) so \(a+b=S.\) After adding \(3\) to each and doubling, the sum is \(2(a+3)\) \(+2(b+3)\) \(=2(a+b)+12\) \(=2S+12.\)

Thus, the correct answer is E.

4 答案 E

三角形的每条边都是线段，最多与圆相交于 \(2\) 个点；三角形有 \(3\) 条边，所以最多有 \(3\cdot2=6\) 个交点，而且这个上界可以达到。

所以正确答案是 E。

English

Each side of the triangle is a segment, which can intersect a circle in at most \(2\) points. With \(3\) sides, the maximum is \(3\cdot2=6\) points, and this is achievable.

Thus, the correct answer is E.

5 答案 D

逐一检查这些五连方的反射对称性，恰好有六个至少有一条对称轴：直条形、十字形、\(T\) 形、\(U\) 形、\(V\) 形和 \(W\) 形。其余六个都没有对称轴。

所以正确答案是 D。

English

Checking each pentomino for a reflection line, exactly six have at least one: the straight bar, the plus, the \(T\)-shape, the \(U\)-shape, the \(V\)-shape, and the \(W\)-shape. The remaining six have no line of symmetry.

Thus, the correct answer is D.

6 答案 E

设 \(N=10a+b\)，则 \(P(N)=ab\)，\(S(N)=a+b\)，所以 \(10a+b=ab+a+b\)。化简得 \(9a=ab\)，因 \(a\ne0\)，故 \(b=9\)。

因此该数的个位数字是 \(9\)。所以正确答案是 E。

English

Write \(N=10a+b.\) Then \(P(N)=ab\) and \(S(N)=a+b,\) so \(10a+b=ab+a+b.\) This reduces to \(9a=ab,\) and since \(a\ne0,\) we get \(b=9.\)

The units digit is \(9.\) Thus, the correct answer is E.

7 答案 C

设原数为 \(x\)。小数点向右移动四位相当于乘以 \(10000\)，所以 \(10000x=4\cdot\dfrac1x\)，即 \(x^2=\dfrac{4}{10000}\)。

因为 \(x\gt0\)，所以 \(x=\dfrac{2}{100}=0.02\)，这就是原数。

所以正确答案是 C。

English

Moving the decimal four places right multiplies \(x\) by \(10000.\) So \(10000x=4\cdot\dfrac1x,\) giving \(x^2=\dfrac{4}{10000}.\)

Since \(x\gt0,\) \(x=\dfrac{2}{100}=0.02.\)

Thus, the correct answer is C.

8 答案 B

他们再次同时工作需要经过 \(\text{lcm}(3,4,6,7)\) 个上课日。因为 \(4=2^2\)，\(6=2\cdot3\)，所以最小公倍数为 \(2^2\cdot3\cdot7=84\)。

所以正确答案是 B。

English

They meet again after \(\text{lcm}(3,4,6,7)\) days. Since \(4=2^2\) and \(6=2\cdot3,\) the least common multiple is \(2^2\cdot3\cdot7=84.\)

Thus, the correct answer is B.

9 答案 B

设年收入为 \(x\ge28000\)。

\[ \begin{aligned} &\tfrac{p}{100}(28000) \\ &\quad {}+\tfrac{p+2}{100}(x-28000) \\ &\quad =\tfrac{p+0.25}{100}x \end{aligned}\text{。}\]

两边乘以 \(100\) 并展开，所有含 \(p\) 的项抵消，得到 \(2x-56000=0.25x\)。因此 \(1.75x=56000\)，所以 \(x=32000\)。

所以正确答案是 B。

English

Let her income be \(x\ge28000.\) Then

\[ \begin{aligned} &\tfrac{p}{100}(28000) \\ &\quad {}+\tfrac{p+2}{100}(x-28000) \\ &\quad =\tfrac{p+0.25}{100}x. \end{aligned} \]

Multiplying by \(100\) and expanding, all the \(p\) terms cancel, leaving \(2x-56000=0.25x.\) So \(1.75x=56000\) and \(x=32000.\)

Thus, the correct answer is B.

10 答案 D

用 \(xz=48\) 除以 \(xy=24\)，得 \(z=2y\)。再由 \(yz=2y^2=72\)，可得 \(y=6\)，于是 \(z=12\)，\(x=\frac{24}{y}=4\)。

因此 \(x+y+z=22\)。所以正确答案是 D。

English

Dividing \(xz=48\) by \(xy=24\) gives \(z=2y.\) Then \(yz=2y^2=72,\) so \(y=6,\) \(z=12,\) and \(x=\frac{24}{y}=4.\)

Hence \(x+y+z=22.\) Thus, the correct answer is D.

11 答案 C

第 \(n\) 圈是一个 \((2n+1)\times(2n+1)\) 正方形的外边框，内部去掉一个 \((2n-1)\times(2n-1)\) 正方形，因此含有 \((2n+1)^2-(2n-1)^2=8n\) 个单位正方形。

当 \(n=100\) 时，个数为 \(800\)。所以正确答案是 C。

English

The \(n\)th ring is the border of a \((2n+1)\times(2n+1)\) square surrounding a \((2n-1)\times(2n-1)\) square, so it contains \((2n+1)^2-(2n-1)^2=8n\) unit squares.

For \(n=100,\) that is \(800.\) Thus, the correct answer is C.

12 答案 D

三个连续整数中，一定有一个是 \(3\) 的倍数，且乘积 \(n\) 也有一个偶数因子，因此能被 \(6\) 整除。再加上题设的因数 \(7\)，它一定能被 \(6\)、\(14\)、\(21\) 和 \(42\) 整除。

但 \(28=2^2\cdot7\) 需要两个因数 \(2\)，这不一定保证；例如 \(5\cdot6\cdot7=210\) 能被 \(7\) 整除，却不能被 \(28\) 整除。

所以正确答案是 D。

English

Among three consecutive integers, at least one is even and one is a multiple of \(3,\) so \(n\) is divisible by \(6.\) With the given factor of \(7,\) it is divisible by \(6,\) \(14,\) \(21,\) and \(42.\)

But \(28=2^2\cdot7\) requires two factors of \(2,\) which is not guaranteed: \(5\cdot6\cdot7=210\) is divisible by \(7\) but not by \(28.\)

Thus, the correct answer is D.

13 答案 E

连续奇数数字 \(GHIJ\) 是 \(9753\) 或 \(7531\)，因此留给 \(A\)、\(B\)、\(C\) 的奇数数字是 \(1\) 或 \(9\)。由于 \(A+B+C=9\)，这个奇数数字只能是 \(1\)，所以 \(ABC\) 中两个偶数数字的和为 \(8\)。

连续偶数数字组成的 \(DEF\) 可能是 \(864\)、\(642\) 或 \(420\)，于是留给 \(ABC\) 的偶数数字对分别为 \(\{2,0\}\)、\(\{8,0\}\) 或 \(\{8,6\}\)。只有 \(\{8,0\}\) 的和为 \(8\)，所以 \(ABC=810\)，从而 \(A=8\)。

所以正确答案是 E。

English

The consecutive odd digits \(GHIJ\) are \(9753\) or \(7531,\) leaving one odd digit (\(1\) or \(9\)) for \(A,\) \(B,\) \(C.\) Since \(A+B+C=9,\) the odd digit there must be \(1,\) so the two even digits in \(ABC\) sum to \(8.\)

The consecutive even digits \(DEF\) are \(864,\) \(642,\) or \(420,\) leaving even-digit pairs \(\{2,0\},\) \(\{8,0\},\) or \(\{8,6\}\) for \(ABC.\) Only \(\{8,0\}\) sums to \(8,\) so \(ABC=810\) and \(A=8.\)

Thus, the correct answer is E.

14 答案 A

设 \(n\) 张全价票每张 \(p\) 美元。\(np+(140-n)\dfrac p2=2001\)，\(p(n+140)\) \(=4002\) \(=2\cdot3\cdot23\cdot29\)。

则 \(140\le n+140\le280\)。在 \(4002\) 的因数中，落在这个范围内的只有 \(174=2\cdot3\cdot29\)。所以 \(n+140=174\)，得 \(n=34\)、\(p=23\)，全价票收入为 \(34\cdot23=782\) 美元。

所以正确答案是 A。

English

Let \(n\) full-price tickets sell at \(p\) dollars each. Then \(np+(140-n)\dfrac p2=2001,\) so \(p(n+140)\) \(=4002\) \(=2\cdot3\cdot23\cdot29.\)

Since \(140\le n+140\le280,\) the only factor of \(4002\) in range is \(174=2\cdot3\cdot29.\) So \(n+140=174,\) giving \(n=34\) and \(p=23.\) The full-price tickets raise \(34\cdot23=782\) dollars.

Thus, the correct answer is A.

15 答案 C

人行横道是一个平行四边形。以路缘上的 \(15\) 英尺为底、街道宽 \(40\) 英尺为高，其面积为 \(15\cdot40=600\)。

以一条 \(50\) 英尺的条纹为底，面积等于 \(50\) 乘以两条条纹之间的距离 \(d\)，所以 \(d=\frac{600}{50}=12\)。

所以正确答案是 C。

English

The crosswalk is a parallelogram. Using the curb (\(15\) ft) as base and the street width (\(40\) ft) as height, its area is \(15\cdot40=600\) square feet.

Using a stripe (\(50\) ft) as base, the area equals \(50\) times the distance \(d\) between the stripes, so \(d=\frac{600}{50}=12.\)

Thus, the correct answer is C.

16 答案 D

设平均数为 \(m\)。最小数为 \(m-10\)，最大数为 \(m+15\)，中位数为 \(5\)。

于是 \(m\) 满足 \[ \begin{aligned} &\tfrac13\big((m-10)+5+(m+15)\big) \\ &\quad =m \end{aligned} \] 化简得 \(m=10\)。

因此三个数的和为 \(3m=30\)。所以正确答案是 D。

English

Let \(m\) be the mean. The least number is \(m-10\) and the greatest is \(m+15,\) with median \(5.\)

Since the mean of the three is \(m,\) \[ \begin{aligned} &\tfrac13\big((m-10)+5+(m+15)\big) \\ &\quad =m, \end{aligned} \] which gives \(m=10.\)

The sum is \(3m=30.\) Thus, the correct answer is D.

17 答案 C

卷成圆锥时，扇形半径 \(10\) 成为圆锥的母线长，扇形弧长成为底面周长。

弧长为 \(\dfrac{252}{360}\cdot2\pi\cdot10=14\pi\)，所以 \(2\pi r=14\pi\)，底面半径 \(r=7\)。圆锥母线长为 \(10\)，底面半径为 \(7\)。

所以正确答案是 C。

English

When rolled into a cone, the sector’s radius \(10\) becomes the slant height, and the arc length becomes the base circumference.

The arc length is \(\dfrac{252}{360}\cdot2\pi\cdot10=14\pi,\) so \(2\pi r=14\pi\) gives base radius \(r=7.\) The cone has slant height \(10\) and base radius \(7.\)

Thus, the correct answer is C.

18 答案 D

考虑一个重复的 \(3\times3\) 方块，共九个小正方形；其中四个不属于五边形，所以五边形覆盖的比例为 \(1-\dfrac49=\dfrac59\approx55.6\%\)。

这个百分比最接近 \(56\)。所以正确答案是 D。

English

Consider a repeating \(3\times3\) block of nine small squares. Four of these nine squares are not part of the pentagons, so the pentagons cover \(1-\dfrac49=\dfrac59\approx55.6\%\) of the area.

This is closest to \(56.\) Thus, the correct answer is D.

19 答案 D

不同选择的数量等于方程 \(g+c+p=4\) 的非负整数解个数。由插板法，解的个数为 \(\dbinom{4+2}{2}=\dbinom62=15\)。

所以正确答案是 D。

English

The number of selections is the number of nonnegative integer solutions of \(g+c+p=4.\) By stars and bars, this is \(\dbinom{4+2}{2}=\dbinom62=15.\)

Thus, the correct answer is D.

20 答案 B

设八边形边长为 \(x\)。每个被切去的等腰直角三角形的斜边也是八边形的一条边，所以它的直角边为 \(\dfrac{x}{\sqrt2}\)。

沿正方形的一条边，有两段这样的直角边和一条八边形边，因此 \(2\cdot\dfrac{x}{\sqrt2}+x=2000\)，即 \(x(\sqrt2+1)=2000\)，所以 \(x=\dfrac{2000}{\sqrt2+1}=2000(\sqrt2-1)\)。

所以正确答案是 B。

English

Let each octagon side be \(x.\) It is the hypotenuse of each cut isosceles right triangle, whose legs are \(\dfrac{x}{\sqrt2}.\)

Along one side of the square, two legs and one octagon side give \(2\cdot\dfrac{x}{\sqrt2}+x=2000,\) so \(x(\sqrt2+1)=2000\) and \(x=\dfrac{2000}{\sqrt2+1}=2000(\sqrt2-1).\)

Thus, the correct answer is B.

21 答案 B

取轴截面。圆锥底面半径为 \(5\)、高为 \(12\)；圆柱截面的宽为 \(2r\)、高为 \(2r\)。

由相似三角形，\(\dfrac{12-2r}{r}=\dfrac{12}{5}\)，所以 \(5(12-2r)=12r\)，即 \(60=22r\)，从而 \(r=\dfrac{30}{11}\)。

所以正确答案是 B。

English

Take an axial cross-section. The cone has base radius \(5\) and height \(12;\) the cylinder appears as a rectangle of width \(2r\) and height \(2r.\)

By similar triangles, \(\dfrac{12-2r}{r}=\dfrac{12}{5},\) so \(5(12-2r)=12r,\) giving \(60=22r\) and \(r=\dfrac{30}{11}.\)

Thus, the correct answer is B.

22 答案 D

因为 \(v\) 位于第一行、第一列和主对角线上，这三条线中另外两个数的和相等：\(25+18=24+w=21+x\)。所以 \(w=19\)，\(x=22\)。

反对角线上的 \(25\)、\(22\)、\(19\) 之和为 \(66\)，所以幻方和为 \(66\)。于是 \(v=66-24-19=23\)，\(y=66-18-22=26\)，\(z=66-25-21=20\)。

因此 \(y+z=46\)。所以正确答案是 D。

English

Since \(v\) sits in the first row, first column, and main diagonal, the remaining two entries of each of those lines have equal sums: \(25+18=24+w=21+x.\) So \(w=19\) and \(x=22.\)

The anti-diagonal \(25,\) \(22,\) \(19\) sums to \(66,\) so the magic sum is \(66.\) Then \(v=66-24-19=23,\) \(y=66-18-22=26,\) and \(z=66-25-21=20.\)

Hence \(y+z=46.\) Thus, the correct answer is D.

23 答案 D

想象先给五个筹码排一个随机顺序。游戏停在白色筹码上，恰好表示两个白色筹码都在所有三个红色筹码取完之前出现，也就是完整顺序中的最后一个筹码是红色。

最后一个筹码为红色的概率为 \(\dfrac{3}{5}\)。所以正确答案是 D。

English

Imagine drawing all five chips in a random order. The drawing stops on a white chip exactly when both white chips come out before all three reds, which happens precisely when the very last chip in the full ordering is red.

That probability is \(\dfrac{3}{5}.\) Thus, the correct answer is D.

24 答案 B

从 \(B\) 向 \(CD\) 作垂线，交于 \(E\)。于是 \(BE=AD=7\)，\(CE=CD-AB\)，并且 \(BC^2=BE^2+CE^2\)。

又因为 \(BC=CD+AB\)，所以 \[ \begin{aligned} &(CD+AB)^2 \\ &\quad {}-(CD-AB)^2 \\ &\quad =BE^2=49 \end{aligned}\text{。}\]

左边等于 \(4\cdot AB\cdot CD\)，因此 \(AB\cdot CD=\dfrac{49}{4}=12.25\)。

所以正确答案是 B。

English

Drop a perpendicular from \(B\) to \(CD,\) meeting it at \(E.\) Then \(BE=AD=7\) and \(CE=CD-AB.\) By the Pythagorean theorem, \(BC^2=BE^2+CE^2.\)

Since \(BC=CD+AB,\) \[ \begin{aligned} &(CD+AB)^2 \\ &\quad {}-(CD-AB)^2 \\ &\quad =BE^2=49. \end{aligned} \]

The left side equals \(4\cdot AB\cdot CD,\) so \(AB\cdot CD=\dfrac{49}{4}=12.25.\)

Thus, the correct answer is B.

25 答案 B

不超过 \(2001\) 的 \(3\) 或 \(4\) 的倍数共有 \(\left\lfloor\tfrac{2001}{3}\right\rfloor\) \(+\left\lfloor\tfrac{2001}{4}\right\rfloor\) \(-\left\lfloor\tfrac{2001}{12}\right\rfloor\) \(=667+500-166\) \(=1001\)。

其中是 \(5\) 的倍数者，也就是 \(15\) 或 \(20\) 的倍数，共有 \(133\) 加 \(100\) 个，但要减去重复计算的 \(60\) 的倍数，共 \(33\) 个，因此共有 \(133+100-33=200\) 个。

所以所求个数为 \(1001-200=801\)。正确答案是 B。

English

Multiples of \(3\) or \(4\) up to \(2001:\) \(\left\lfloor\tfrac{2001}{3}\right\rfloor\) \(+\left\lfloor\tfrac{2001}{4}\right\rfloor\) \(-\left\lfloor\tfrac{2001}{12}\right\rfloor\) \(=667+500-166\) \(=1001.\)

Among these, remove the multiples of \(5:\) there are \(133\) multiples of \(15\) and \(100\) multiples of \(20,\) re-adding the \(33\) multiples of \(60:\) \(133+100-33=200.\)

So the count is \(1001-200=801.\) Thus, the correct answer is B.