2021 AMC amc10b 真题 答案 详解

---

中英双语真题

2021Spring AMC10B

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

---

Problem 2
What is the value of ( \sqrt{(3 - 2\sqrt{3})^2} + \sqrt{(3 + 2\sqrt{3})^2} )?

算式 ( \sqrt{(3 - 2\sqrt{3})^2} + \sqrt{(3 + 2\sqrt{3})^2} ) 的值是多少？

(A) 0 (B) (4\sqrt{3} - 6) (C) 6 (D) (4\sqrt{3}) (E) (4\sqrt{3} + 6)

---

Problem 3
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the 28 students in the program, 25% of the juniors and 10% of the seniors are on the debate team. How many juniors are in the program?

在面向高二和高三学生的某课外课程中，有一个辩论队，每个年级参加辩论队的人数相同。已知在这个课程的28名同学中，有25%的高二学生和10%的高三学生参加了辩论队。问这个课程中有多少名高二的学生？

(A) 5 (B) 6 (C) 8 (D) 11 (E) 20

英文真题

2021 AMC 10B Problems
Problem 1
How many integer values of ( x ) satisfy ( |x| < 3\pi )?
(A) 9 (B) 10 (C) 18 (D) 19 (E) 20

Problem 2
What is the value of ( \sqrt{(3 - 2\sqrt{3})^2} + \sqrt{(3 + 2\sqrt{3})^2} )?
(A) 0 (B) (4\sqrt{3} - 6) (C) 6 (D) (4\sqrt{3}) (E) (4\sqrt{3} + 6)

Problem 3
In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the 28 students in the program, 25% of the juniors and 10% of the seniors are on the debate team. How many juniors are in the program?
(A) 5 (B) 6 (C) 8 (D) 11 (E) 20

Problem 4
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

Problem 5
The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give 24, while the other two multiply to 30. What is the sum of the ages of Jonie's four cousins?
(A) 21 (B) 22 (C) 23 (D) 24 (E) 25

Problem 6
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 scores of all the students?
(A) 74 (B) 75 (C) 76 (D) 77 (E) 78

Problem 7
In a plane, four circles with radii 1, 3, 5, and 7 are tangent to line ( \ell ) at the same point A, but they may be on either side of ( \ell ). Region S consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region S?
(A) (24\pi) (B) (32\pi) (C) (64\pi) (D) (65\pi) (E) (84\pi)

Problem 8
Mr. Zhou places all the integers from 1 to 225 into a 15 by 15 grid. He places 1 in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

真题文字详解(英文)

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.

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, -9.42 < x < 9.42 ( \Rightarrow ) -9 ≤ x ≤ 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 · 2 + 1 = 19, yielding (D) 19.

Problem 2 What is the value of ( \sqrt{(3-2\sqrt{3})^2} + \sqrt{(3+2\sqrt{3})^2} )?

(A) 0 (B) ( 4\sqrt{3}-6 ) (C) 6 (D) ( 4\sqrt{3} ) (E) ( 4\sqrt{3}+6 )

Solution 1

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

Solution 1 Since 3π ≈ 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| < 3π ⇔ −3π < x < 3π. Since π is approximately 3.14, 3π is approximately 9.42. We are trying to solve for −9.42 < x < 9.42, where x ∈ Z. Hence, −9.42 < x < 9.42 ⇒ −9 ≤ x ≤ 9, for x ∈ Z. The number of integer values of x is 9 − (−9) + 1 = 19. Therefore, the answer is (D) 19.

Solution 3 3π ≈ 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 π 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 3π > 3 · 3 > 9, the answer is at least 9 · 2 + 1 = 19, yielding (D) 19.

Problem 2 What is the value of √((3-2√3)^2) + √((3+2√3)^2)?
(A) 0 (B) 4√3 - 6 (C) 6 (D) 4√3 (E) 4√3 + 6

Solution 1