2022 AMC amc10a 真题 答案 详解

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中英双语真题

2022 AMC 10A

1. What is the value of the below expression? 问下式的值是多少?

3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}}

(A) \frac{31}{10} (B) \frac{49}{15} (C) \frac{33}{10} (D) \frac{109}{33} (E) \frac{15}{4}

1. Mike cycled 15 laps in 57 minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first 27 minutes? Mike 用 57 分钟骑了 15 圈. 假设他一直以恒定的速度骑行. 那么他在前 27 分钟大约完成了多少圈?

(A) 5 (B) 7 (C) 9 (D) 11 (E) 13

1. The sum of three numbers is 96. The first number is 6 times the third number, and the third number is 40 less than the second number. What is the absolute value of the difference between the first and second numbers? 三个数之和为 96. 第一个数是第三个数的 6 倍, 第三个数比第二个数少 40. 问第一个数与第二个数之差的绝对值是多少?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

英文真题

2022 AMC 10A Problems Problem 1 What is the value of (3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{3}}})?
(A) (\frac{31}{10}) (B) (\frac{49}{15}) (C) (\frac{33}{10}) (D) (\frac{109}{33}) (E) (\frac{15}{4})

Problem 2 Mike cycled 15 laps in 57 minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first 27 minutes?
(A) 5 (B) 7 (C) 9 (D) 11 (E) 13

Problem 3 The sum of three numbers is 96. The first number is 6 times the third number, and the third number is 40 less than the second number. What is the absolute value of the difference between the first and second numbers?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Problem 4 In some countries, automobile fuel efficiency is measured in liters per 100 kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals (m) miles, and 1 gallon equals (l) liters. Which of the following gives the fuel efficiency in liters per 100 kilometers for a car that gets (x) miles per gallon?
(A) (\frac{x}{100lm}) (B) (\frac{xlm}{100}) (C) (\frac{lm}{100x}) (D) (\frac{100}{xlm}) (E) (\frac{100lm}{x})

Problem 5 Square ABCD has side length 1. Points P, Q, R, and S each lie on a side of ABCD such that APQCRS is an equilateral convex hexagon with side length (s). What is (s)?
(A) (\frac{\sqrt{2}}{3}) (B) (\frac{1}{2}) (C) (2 - \sqrt{2}) (D) (1 - \frac{\sqrt{2}}{4}) (E) (\frac{2}{3})

Problem 6 Which expression is equal to
for (a < 0?)
[ \left| a - 2 - \sqrt{(a-1)^2} \right| ]
(A) (3 - 2a) (B) (1-a) (C) 1 (D) (a+1) (E) 3

Problem 7 The least common multiple of a positive integer (n) and 18 is 180, and the greatest common divisor of (n) and 45 is 15. What is the sum of the digits of (n)?
(A) 3 (B) 6 (C) 8 (D) 9 (E) 12

真题文字详解(英文)

Problem 1 The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page. What is the value of $$3+ \frac{1}{3+\frac{1}{3+\frac{1}{\ddots}}}$$?
(A) ( \frac{31}{10} )
(B) ( \frac{49}{15} )
(C) ( \frac{33}{10} )
(D) ( \frac{109}{33} )
(E) ( \frac{15}{4} )

Solution 1 We have
$$3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{\ddots}}} = 3 + \frac{1}{3 + \frac{1}{3 + \frac{1}{(\frac{10}{3})}}}$$
$$= 3 + \frac{1}{3 + \frac{3}{10}}$$
$$= 3 + \frac{1}{\left( \frac{33}{10} \right)}$$
$$= 3 + \frac{10}{33}$$
$$= \boxed{(D)\ \frac{109}{33}}.$$

~MRENTHUSIASM

Solution 2 Continued fractions with integer parts ( q_i ) and numerators all 1 can be calculated as
$$\frac{\left[ q_0, q_1, q_2, \ldots, q_n \right]}{\left[ q_1, q_2, \ldots, q_n \right]}$$ where
$$\left[ q_0, q_1, q_2, \ldots, q_n \right] = q_0 \left[ q_1, q_2, \ldots, q_n \right] + \left[ q_2, \ldots, q_n \right]$$

Solution 3 Finite continued fractions of form ( n + \frac{1}{n + \frac{1}{n + \cdots}} = \frac{x}{y} ) have linear combinations of ( x, y ) that solve Pell's Equation. Specifically, the denominator ( y ) and numerator ( x ) are solutions to the Diophantine equation ( \left(n^2 + 4\right) \left(\frac{y}{2}\right)^2 - \left(x - \frac{ny}{2}\right)^2 = \pm 1 ). So for this problem in particular, the denominator ( y ) and numerator ( x ) are solutions to the Diophantine equation ( 13 \left(\frac{y}{2}\right)^2 - \left(x - \frac{3y}{2}\right)^2 = \pm 1 ). That leaves two answers. Since the number of 1's in the continued fraction is odd, we further narrow it down to ( 13 \left(\frac{y}{2}\right)^2 - \left(x - \frac{3y}{2}\right)^2 = -1 ), which only leaves us with 1 answer and that is ( (x, y) = (109, 33) ) which means ( \boxed{(D)\ \frac{109}{33}} ).

(Note: Integer solutions increase exponentially, so our next solution will have a numerator greater than ( 3^2(109) ). Therefore, when you don't see numerators greater than ( 3^2(109) ) in the answer choices, this method should be fine.)

真题文字详解(中英双语)

Problem 1 The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page. 以下问题既来自2022年AMC 10A第1题，也来自2022年AMC 12A第1题，因此这两个问题都指向这一页。

What is the value of $$3+ \frac{1}{3+\frac{1}{\frac{1}{3}}}$$?
(A) ( \frac{31}{10} )
(B) ( \frac{49}{15} )
(C) ( \frac{33}{10} )
(D) ( \frac{109}{33} )
(E) ( \frac{15}{4} )

Solution 1 We have
$$3 + \frac{1}{3 + \frac{1}{\frac{1}{3}}} = 3 + \frac{1}{3 + \left(\frac{1}{\frac{1}{3}}\right)} = 3 + \frac{1}{3 + \frac{3}{10}} = 3 + \frac{1}{\frac{39}{10}} = 3 + \frac{10}{33} = \boxed{(D)\frac{109}{33}}.$$

~MRENTHUSIASM 我们有~MRENTHUSIASM

Solution 2 Continued fractions with integer parts ( q_i ) and numerators all 1 can be calculated as
$$\frac{\left[ q_0, q_1, q_2, \ldots, q_n \right]}{\left[ q_1, q_2, \ldots, q_n \right]}$$ where 分数部分为( q_i )，分子全为1的连分数可以计算为，其中
$$\left[ q_0, q_1, q_2, \ldots, q_n \right] = q_0 \left[ q_1, q_2, \ldots, q_n \right] + \left[ q_2, \ldots, q_n \right]$$