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2006 AMC amc10b 真题 答案 详解

分类:AMC 数学竞赛 标签:amc amc10b 发布:2006-11-02

2006 年 amc10b 真题,中英双语,答案,详解和视频讲解

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2006-amc10b-paper-eng.pdf 英文真题 209.48KB 5 页
2006-amc10b-solution-eng.pdf 真题文字详解(英文) 1020.22KB 17 页
2006-amc10b-solution-eng-zh.pdf 真题文字详解(中英双语) 1.16MB 26 页

英文真题

2006 AMC 10B Problems

Problem 1

What is $(-1)^{1} + (-1)^{2} + \ldots + (-1)^{2006}$? (A) -2006
(B) -1
(C) 0
(D) 1
(E) 2006

Problem 2

For real numbers (x) and (y), define (x \spadesuit y = (x+y)(x-y)). What is (3 \spadesuit (4 \spadesuit 5))? (A) -72
(B) -27
(C) -24
(D) 24
(E) 72

Problem 3

A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
(A) 10
(B) 14
(C) 17
(D) 20
(E) 24

Problem 4

Circles of diameter 1 inch and 3 inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
(A) 2
(B) 3
(C) 6
(D) 8
(E) 9

Problem 5

A (2\times 3) rectangle and a (3\times 4) rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
(A) 16
(B) 25
(C) 36
(D) 49
(E) 64

Problem 6

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure (\frac{2}{\pi}), as shown. What is the perimeter of this region?
(A) (\frac{4}{\pi})
(B) 2
(C) (\frac{8}{\pi})
(D) 4
(E) (\frac{16}{\pi})

Problem 7

Which of the following is equivalent to (\sqrt{\frac{x}{1-\frac{x-1}{x}}}) when (x < 0)?

英文真题

真题文字详解(英文)

Problem 1 What is $(-1)^{1}+(-1)^{2}+\ldots+(-1)^{2006}$?
(A) -2006 (B) -1 (C) 0 (D) 1 (E) 2006

Solution 1 Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$:
$$(-1)^{1}+(-1)^{2}+\ldots+(-1)^{2006}=(-1)+(1)+\ldots+(-1)+(1)=\boxed{(C)\ 0}.$$

Solution 2 The Using the formula for the first n terms of a geometric series, with n=2006, a=$(-1)^{1}=-1$, and r=-1, the sum can also be obtained:
$$\frac{a(1-r^{n})}{1-r}=\frac{-1(1-(-1)^{2006})}{1+1}=\frac{-1(1-1)}{2}=\frac{0}{2}=\boxed{(C)\ 0}$$

Problem 2 For real numbers x and y, define $x \spadesuit y=(x+y)(x-y)$. What is $3\spadesuit(4\spadesuit5)$?
(A) -72 (B) -27 (C) -24 (D) 24 (E) 72

Solution 1 Since $x \spadesuit y=(x+y)(x-y)$:
$$3\spadesuit(4\spadesuit5)=3\spadesuit((4+5)(4-5))=3\spadesuit(-9)=(3+(-9))(3-(-9))=\boxed{(A)\ -72}$$

Solution 2 From difference of squares, we have $x \spadesuit y=(x+y)(x-y)=x^2-y^2$
So:
$$3\spadesuit(4\spadesuit5)=3\spadesuit(4^2-5^2)=3\spadesuit(16-25)=3\spadesuit(-9)=(3^2)-(-9)^2=9-81=\boxed{(A)\ -72}$$

Problem 3 A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
(A) 10 (B) 14 (C) 17 (D) 20 (E) 24

真题文字详解(英文)

真题文字详解(中英双语)

Problem 1 What is $(-1)^{1}+(-1)^{2}+\ldots+(-1)^{2006}$?
(A) -2006 (B) -1 (C) 0 (D) 1 (E) 2006

Solution 1 Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$:
由于 $-1$ 乘以奇数次方等于 $-1$,而 $-1$ 乘以偶数次方等于 $1$:
$$(-1)^{1}+(-1)^{2}+\ldots+(-1)^{2006}=(-1)+(1)+\ldots+(-1)+(1)=\boxed{\text{(C)}\ 0}.$$

Solution 2 The Using the formula for the first n terms of a geometric series, with n = 2006, a = (-1)^1 = -1, and r = -1, the sum can also be obtained:
利用求前n项的几何级数公式,其中n=2006,a=(-1)^1=-1,和r=-1,也可以得到和:
$$\frac{a(1-r^n)}{1-r}=\frac{-1(1-(-1)^{2006})}{1+1}=\frac{-1(1-1)}{2}=\frac{0}{2}=\boxed{\text{(C)}\ 0}$$

Problem 2 For real numbers x and y, define x♠y=(x+y)(x-y). What is 3♠(4♠5)?
对于实数x和y,定义x♠y=(x+y)(x-y)。3♠(4♠5)是什么?
(A) -72 (B) -27 (C) -24 (D) 24 (E) 72

Solution 1 Since x♠y=(x+y)(x-y):
Since x♠y=(x+y)(x-y):
$$3♠(4♠5)=3♠((4+5)(4-5))=3♠(-9)=(3+(-9))(3-(-9))=\boxed{\text{(A)}\ -72}$$

Solution 2 From difference of squares, we have x♠y=(x+y)(x-y)=x^2-y^2
From difference of squares, we have x♠y=(x+y)(x-y)=x^2-y^2
So:
So:
$$3♠(4♠5)=3♠(4^2-5^2)=3♠(16-25)=3♠(-9)=(3^2)-(-9)^2=9-81=\boxed{\text{(A)}\ -72}$$

Problem 3 A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
举行了一场足球比赛,比赛双方是美洲狮队和黑豹队。两队的总得分为34分,美洲狮队以14分的优势获胜。黑豹队得了多少分?
(A) 10 (B) 14 (C) 17 (D) 20 (E) 24

真题文字详解(中英双语)