2021 AMC amc12b 真题 答案 详解

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中英双语真题

2021Spring AMC12B

Problem 1
How many integer values of ( x ) satisfy ( |x| < 3\pi )?

有多少个整数值满足 ( |x| < 3\pi )?

(A) 9 (B) 10 (C) 18 (D) 19 (E) 20

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Problem 2
At a math contest, 57 students are wearing blue shirts, and another 75 students are wearing yellow shirts. The 132 students are assigned into 66 pairs. In exactly 23 of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

在一次数学竞赛中，57名学生穿着蓝色衬衫，另外75名学生穿着黄色衬衫。132名学生被分成了66对。其中恰好有23对，每对的两名学生都穿着蓝色衬衫。问两名学生都穿着黄色衬衫的对有多少个？

(A) 23 (B) 32 (C) 37 (D) 41 (E) 64

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Problem 3
Suppose [ 2 + \frac{1}{1 + \frac{1}{2 + \frac{2}{3 + \frac{x}{2}}}} = \frac{144}{53} ] What is the value of ( x )?

求( x )的值。

(A) (\frac{3}{4}) (B) (\frac{7}{8}) (C) (\frac{14}{15}) (D) (\frac{37}{38}) (E) (\frac{52}{53})

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英文真题

2021 AMC 12B Problems

Problem 1

How many integer values of ( x ) satisfy ( |x| < 3\pi )?

(A) 9 (B) 10 (C) 18 (D) 19 (E) 20

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Problem 2

At a math contest, 57 students are wearing blue shirts, and another 75 students are wearing yellow shirts. The 132 students are assigned into 66 pairs. In exactly 23 of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?

(A) 23 (B) 32 (C) 37 (D) 41 (E) 64

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Problem 3

Suppose

[ 2 + \frac{1}{1 + \frac{1}{2 + \frac{2}{3 + x}}} = \frac{144}{53}. ]

What is the value of ( x )?

(A) ( \frac{3}{4} ) (B) ( \frac{7}{8} ) (C) ( \frac{14}{15} ) (D) ( \frac{37}{38} ) (E) ( \frac{52}{53} )

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Problem 4

Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is 84, and the afternoon class's mean score is 70. The ratio of the number of students in the morning class to the number of students in the afternoon class is ( \frac{3}{4} ). What is the mean of the score of all the students?

(A) 74 (B) 75 (C) 76 (D) 77 (E) 78

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Problem 5

The point ( P(a,b) ) in the ( xy )-plane is first rotated counterclockwise by ( 90^\circ ) around the point ( (1,5) ) and then reflected about the line ( y=-x ). The image of ( P ) after these two transformations is at ( (-6,3) ). What is ( b-a )?

(A) 1 (B) 3 (C) 5 (D) 7 (E) 9

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Problem 6

An inverted cone with base radius 12 cm and height 18 cm is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of 24 cm. What is the height in centimeters of the water in the cylinder?

(A) 1.5 (B) 3 (C) 4 (D) 4.5 (E) 6

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Problem 7

Let ( N=34\cdot34\cdot63\cdot270 ). What is the ratio of the sum of the odd divisors of ( N ) to the sum of the even divisors of ( N )?

(A) 1:16 (B) 1:15 (C) 1:14 (D) 1:8 (E) 1:3

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Problem 8

Three equally spaced parallel lines intersect a circle, creating three chords of lengths 38, 38, and 34. What is the distance between two adjacent parallel lines?

(A) ( 5\frac{1}{2} ) (B) 6 (C) ( 6\frac{1}{2} ) (D) 7 (E) ( 7\frac{1}{2} )

真题文字详解(英文)

Problem1 How many integer values of ( x ) satisfy ( |x| < 3\pi )?
(A) 9 (B) 10 (C) 18 (D) 19 (E) 20

Solution 1 Since ( 3\pi \approx 9.42 ), we multiply 9 by 2 for the integers from 1 to 9 and the integers from -1 to -9 and add 1 to account for 0 to get (D) 19.

Solution 2 ( |x| < 3\pi \Leftrightarrow -3\pi < x < 3\pi ). Since ( \pi ) is approximately 3.14, ( 3\pi ) is approximately 9.42. We are trying to solve for ( -9.42 < x < 9.42 ), where ( x \in \mathbb{Z} ). Hence, ( -9 \leq x \leq 9 ), for ( x \in \mathbb{Z} ). The number of integer values of ( x ) is ( 9 - (-9) + 1 = 19 ). Therefore, the answer is (D) 19.

Solution 3 ( 3\pi \approx 9.4 ). There are two cases here. When ( x > 0 ), ( |x| > 0 ), and ( x = |x| ). So then ( x < 9.4 ). When ( x < 0 ), ( |x| > 0 ), and ( x = -|x| ). So then ( -x < 9.4 ). Dividing by -1 and flipping the sign, we get ( x > -9.4 ). From case 1 and 2, we know that ( -9.4 < x < 9.4 ). Since ( x ) is an integer, we must have ( x ) between -9 and 9. There are a total of ( 9 - (-9) + 1 = (D) 19 ) integers.

~PureSwag

Solution 4 Looking at the problem, we see that instead of directly saying ( x ), we see that it is ( |x| ). That means all the possible values of ( x ) in this case are positive and negative. Rounding ( \pi ) to 3 we get ( 3(3) = 9 ). There are 9 positive solutions and 9 negative solutions: ( 9 + 9 = 18 ). But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is ( 9 + 9 + 1 = 18 + 1 = (D) 19 ).

Solution 5 There are an odd number of integer solutions ( x ) to this inequality since if any non-zero integer ( x ) satisfies this inequality, then so does ( -x ), and we must also account for 0, which gives us the desired. Then, the answer is either (A) or (D), and since ( 3\pi > 3 \cdot 3 > 9 ), the answer is at least ( 9 \cdot 2 + 1 = 19 ), yielding (D) 19.

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2021 AMC 10B #1 and 2021 AMC 12B #1, so both problems redirect to this page. 以下问题来自2021年AMC 10B第1题和2021年AMC 12B第1题，因此这两个问题都指向这一页。

How many integer values of ( x ) satisfy ( |x| < 3\pi )?

有多少个整数值满足条件？

(A) 9 (B) 10 (C) 18 (D) 19 (E) 20

Solution 1 Since ( 3\pi \approx 9.42 ), we multiply 9 by 2 for the integers from 1 to 9 and the integers from -1 to -9 and add 1 to account for 0 to get (D) 19.

由于，我们将乘以对于从到的整数和从到的整数，并加上以计算从到的值，得到。

Solution 2 ( |x| < 3\pi \Leftrightarrow -3\pi < x < 3\pi ). Since ( \pi ) is approximately 3.14, ( 3\pi ) is approximately 9.42. We are trying to solve for ( -9.42 < x < 9.42 ), where ( x \in \mathbb{Z} ). Hence, ( -9 \leq x \leq 9 ), for ( x \in \mathbb{Z} ). The number of integer values of ( x ) is ( 9 - (-9) + 1 = 19 ). Therefore, the answer is (D) 19.

由于大约等于，大约等于。我们试图求解，其中。因此，，满足条件的整数值的数量是。所以答案是。

Solution 3 ( 3\pi \approx 9.4 ). There are two cases here. 这里有两种情况。

When ( x > 0 ), ( |x| > 0 ), and ( x = |x| ). So then ( x < 9.4 )

当( x > 0 )，( |x| > 0 )，和( x = |x| )。那么( x < 9.4 )

When ( x < 0 ), ( |x| > 0 ), and ( x = -|x| ). So then ( -x < 9.4 ). Dividing by -1 and flipping the sign, we get ( x > -9.4 ).

当( x < 0 )，( |x| > 0 )，和( x = -|x| )。那么( -x < 9.4 )。除以-1并改变符号，我们得到( x > -9.4 )。

From case 1 and 2, we know that ( -9.4 < x < 9.4 ). Since ( x ) is an integer, we must have ( x ) between -9 and 9. There are a total of ( 9 - (-9) + 1 = (D) 19 ) integers.

从情况1和2，我们知道 ( -9.4 < x < 9.4 )。因为( x )是一个整数，我们必须有( x )在-9和9之间。总共有~PureSwag

Solution 4 Looking at the problem, we see that instead of directly saying ( x ), we see that it is ( |x| ). That means all the possible values of ( x ) in this case are positive and negative. Rounding ( \pi ) to 3 we get ( 3(3) = 9 ). There are 9 positive solutions and 9 negative solutions: ( 9 + 9 = 18 ). But what about zero? Even though zero is neither negative nor positive, but we still need to add it into the solution. Hence, the answer is ( 9 + 9 + 1 = 18 + 1 = (D) 19 ).

观察这个问题，我们看到不是直接说它，而是看到它是 ( |x| )，这意味着在这种情况下 ( x ) 的所有可能值都是正数和负数。将π四舍五入到3我们得到3(3)=9。有9个正解和9个负解：9+9=18。但是呢？即使零既不是负数也不是正数，但仍然需要将其加入解决方案中。因此，答案是9+9+1=18+1=(D) 19。

Solution 5