2017 AMC amc12b 真题 答案 详解

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中英双语真题

2017 AMC12B

Problem 1

Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?

Kymbrea 的漫画书集目前有 30 本漫画书，并且她现在还在以每个月 2 本漫画书的速度向她的漫画书集中增添新书。Lashawn 的漫画书集目前有 10 本漫画书，并且目前在以每个月 6 本漫画书的速度向他的漫画书集中增添新书。问经过多少个月 Lashawn 的漫画书是 Kymbrea 书的 2 倍？

(A) 1 (B) 4 (C) 5 (D) 20 (E) 25

Problem 2

Real numbers x, y, and z satisfy the inequalities 0 < x < 1, -1 < y < 0, and 1 < z < 2. Which of the following numbers is necessarily positive?

实数 x, y, z 满足不等式 0 < x < 1, -1 < y < 0, 1 < z < 2，下面哪个数字是正数？

(A) y + x^2 (B) y + xz (C) y + y^2 (D) y + 2y^2 (E) y + z

Problem 3

Supposed that x and y are nonzero real numbers such that (\frac{3x+y}{x-3y} = -2). What is the value of (\frac{x+3y}{3x-y})?

假设 x 和 y 是非零实数，满足 (\frac{3x+y}{x-3y} = -2)，则 (\frac{x+3y}{3x-y}) 的值为多少？

(A) -3 (B) -1 (C) 1 (D) 2 (E) 3

英文真题

2017 AMC 12B Problems

Problem 1

Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?

(A) 1 (B) 4 (C) 5 (D) 20 (E) 25

Problem 2

Real numbers x, y, and z satisfy the inequalities 0 < x < 1, -1 < y < 0, and 1 < z < 2. Which of the following numbers is necessarily positive?

(A) y + x^2 (B) y + xz (C) y + y^2 (D) y + 2y^2 (E) y + z

Problem 3

Supposed that x and y are nonzero real numbers such that (\frac{3x+y}{x-3y} = -2). What is the value of (\frac{x+3y}{3x-y})?

(A) -3 (B) -1 (C) 1 (D) 2 (E) 3

Problem 4

Samia set off on her bicycle to visit her friend, traveling at an average speed of 17 kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at 5 kilometers per hour. In all it took her 44 minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?

(A) 2.0 (B) 2.2 (C) 2.8 (D) 3.4 (E) 4.4

Problem 5

The data set [6, 19, 33, 33, 39, 41, 41, 43, 51, 57] has median (Q_2 = 40), first quartile (Q_1 = 33), and third quartile (Q_3 = 43). An outlier in a data set is a value that is more than 1.5 times the interquartile range below the first quartile ((Q_1)) or more than 1.5 times the interquartile range above the third quartile ((Q_3)), where the interquartile range is defined as (Q_3 - Q_1). How many outliers does this data set have?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

Problem 6

The circle having (0, 0) and (8, 6) as the endpoints of a diameter intersects the x-axis at a second point. What is the x-coordinate of this point?

(A) (4\sqrt{2}) (B) 6 (C) (5\sqrt{2}) (D) 8 (E) (6\sqrt{2})

Problem 7

The functions sin(x) and cos(x) are periodic with least period (2\pi). What is the least period of the function cos(sin(x))?

(A) (\frac{\pi}{2}) (B) (\pi) (C) (2\pi) (D) (4\pi) (E) It's not periodic.

Problem 8

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

(A) (\frac{\sqrt{3}-1}{2}) (B) (\frac{1}{2}) (C) (\frac{\sqrt{5}-1}{2}) (D) (\frac{\sqrt{2}}{2}) (E) (\frac{\sqrt{6}-1}{2})

真题文字详解(英文)

Problem 1
Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea?
(A) 1 (B) 4 (C) 5 (D) 20 (E) 25

Solution
Kymbrea has 30 comic books initially and every month, she adds two. This can be represented as (30 + 2x) where (x) is the number of months elapsed. LaShawn's collection, similarly, is (10 + 6x). To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for (x) with the equation (2(2x + 30) = 6x + 10) and get (x = \boxed{25}).

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Problem 2
Real numbers (x), (y), and (z) satisfy the inequalities (0 < x < 1), (-1 < y < 0), and (1 < z < 2). Which of the following numbers is necessarily positive?
(A) (y + x^2) (B) (y + xz) (C) (y + y^2) (D) (y + 2y^2) (E) (y + z)

Notice that (y + z) must be positive because (|z| > |y|). Therefore the answer is (E) (y + z).
The other choices:
(A) As (x) grows closer to 0, (x^2) decreases and thus becomes less than (y).
(B) (x) can be as small as possible ((x > 0)), so (xz) grows close to 0 as (x) approaches 0.
(C) For all (-1 < y < 0), (|y| > |y^2|) and thus it is always negative.
(D) The same logic as above, but when (-\frac{1}{2} < y < 0) this time.

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Problem 3
Supposed that (x) and (y) are nonzero real numbers such that (\frac{3x+y}{x-3y}=-2). What is the value of (\frac{x+3y}{3x-y})?
(A) -3 (B) -1 (C) 1 (D) 2 (E) 3

Solution 1
Rearranging, we find (3x + y = -2(x - 3y)), or (5x = 5y \implies x = y). Substituting, we can convert the second equation into (\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{(D) 2}).
More step-by-step explanation:
[ \begin{align} \frac{3x+y}{x-3y}&=-2\ 3x+y&=-2(x-3y)\ 3x+y&=-2x+6y\ 5x&=5y\ x&=y\ \frac{x+3y}{3x-y}&=\frac{1+3(1)}{3(1)-1}=\frac{4}{2}=2. \end{align} ] We choose (D) 2.

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Solution 2
Substituting each (x) and (y) with 1, we see that the given equation holds true, as (\frac{3(1)+1}{1-3(1)}=-2). Thus, (\frac{x+3y}{3x-y}=\boxed{(D) 2}).

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Solution 3
Let (y=ax). The first equation converts into (\frac{(3+a)x}{(1-3a)x}=-2) which simplifies to (3+a=-2(1-3a)). After a bit of algebra we found out (a=1), which means that (x=y). Substituting (y=x) into the second equation it becomes (\frac{4x}{2x}=\boxed{(D) 2}).

真题文字详解(中英双语)

Problem 1

Kymbrea's comic book collection currently has 30 comic books in it, and she is adding to her collection at the rate of 2 comic books per month. LaShawn's collection currently has 10 comic books in it, and he is adding to his collection at the rate of 6 comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea?

(A) 1 (B) 4 (C) 5 (D) 20 (E) 25

Solution

Kymbrea has 30 comic books initially and every month, she adds two. This can be represented as (30 + 2x) where x is the number of months elapsed. LaShawn's collection, similarly, is (10 + 6x). To find when LaShawn will have twice the number of comic books as Kymbrea, we solve for x with the equation (2(2x + 30) = 6x + 10) and get (x = \boxed{25}).

Kymbrea最初有30本漫画书，并且每个月都在增加两本。这可以表示为(30 + 2x)，其中x是经过的月数。LaShawn的收藏类似地是(10 + 6x)。为了找到LaShawn何时会有Kymbrea两倍的漫画书，我们用方程(2(2x + 30) = 6x + 10)求解x，得到(x = \boxed{25})。

Problem 2

Real numbers (x), (y), and (z) satisfy the inequalities (0 < x < 1), (-1 < y < 0), and (1 < z < 2). Which of the following numbers is necessarily positive?

(A) (y + x^2) (B) (y + xz) (C) (y + y^2) (D) (y + 2y^2) (E) (y + z)

Solution

Notice that (y + z) must be positive because (|z| > |y|). Therefore the answer is (\boxed{(E)\ y + z}).

注意到(y + z)必须是正数，因为(|z| > |y|)。因此答案是(\boxed{(E)\ y + z})。

The other choices:

(A) As (x) grows closer to 0, (x^2) decreases and thus becomes less than (y). (B) (x) can be as small as possible ((x > 0)), so (xz) grows close to 0 as (x) approaches 0. (C) For all (-1 < y < 0), (|y| > |y^2|) and thus it is always negative. (D) The same logic as above, but when (-\frac{1}{2} < y < 0) this time.

(A) 当(x)接近0时，(x^2)减少，因此小于(y)。 (B) (x)可以尽可能小（(x>0)），所以当(x)接近0时，(xz)接近于0。 (C) 对于所有(-1<y<0)，(|y|>|y^2|)，因此它总是负数。 (D) 与上述相同的逻辑，但这次当(-\frac{1}{2}<y<0)。

Problem 3

Supposed that (x) and (y) are nonzero real numbers such that (\frac{3x+y}{x-3y}=-2). What is the value of (\frac{x+3y}{3x-y})?

假设(x)和(y)是非零实数，并且满足(\frac{3x+y}{x-3y}=-2)。(\frac{x+3y}{3x-y})的值是多少？

(A) -3 (B) -1 (C) 1 (D) 2 (E) 3

Solution 1

Rearranging, we find (3x + y = -2(x - 3y)) or (5x = 5y \Rightarrow x = y). Substituting, we can convert the second equation into (\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{(D)\ 2}).

重排后，我们得到(3x + y = -2(x - 3y))，或者(5x = 5y \Rightarrow x = y)。代入后，我们可以将第二个方程转换为(\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{(D)\ 2}\