2020 AMC amc10a 真题 答案 详解

---

中英双语真题

2020 AMC10A 2020 AMC10A Problem 1 What value of x satisfies $$x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?$$ x为何值时满足方程 $$x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3} ?$$ (A) $-\frac{2}{3}$ (B) $\frac{7}{36}$ (C) $\frac{7}{12}$ (D) $\frac{2}{3}$ (E) $\frac{5}{6}$ Problem 2 The numbers 3, 5, 7, a, and b have an average (arithmetic mean) of 15. What is the average of a and b? 数字3, 5, 7, a和b的平均值（算数平均值）是15。则a和b的平均值是多少？ (A) 0 (B) 15 (C) 30 (D) 45 (E) 60 Problem 3 Assuming (a \neq 3), (b \neq 4), and (c \neq 5), what is the value in simplest form of the following expression? 假设(a \neq 3)，(b \neq 4)且(c \neq 5)，那么下式化成最简形式后的值为多少？ $$\frac{a-3}{5-c}\cdot\frac{b-4}{3-a}\cdot\frac{c-5}{4-b}$$ (A) -1 (B) 1 (C) $\frac{abc}{60}$ (D) $\frac{1}{abc}-\frac{1}{60}$ (E) $\frac{1}{60}-\frac{1}{abc}$

英文真题

2020 AMC 10A Problems Problem 1 What value of x satisfies $$x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?$$ (A) $-\frac{2}{3}$ (B) $\frac{7}{36}$ (C) $\frac{7}{12}$ (D) $\frac{2}{3}$ (E) $\frac{5}{6}$ Problem 2 The numbers 3, 5, 7, a, and b have an average (arithmetic mean) of 15. What is the average of a and b? (A) 0 (B) 15 (C) 30 (D) 45 (E) 60 Problem 3 Assuming a ≠ 3, b ≠ 4, and c ≠ 5, what is the value in simplest form of the following expression? $$\frac{a-3}{5-c}\cdot\frac{b-4}{3-a}\cdot\frac{c-5}{4-b}$$ (A) -1 (B) 1 (C) $\frac{abc}{60}$ (D) $\frac{1}{abc}-\frac{1}{60}$ (E) $\frac{1}{60}-\frac{1}{abc}$ Problem 4 A driver travels for 2 hours at 60 miles per hour, during which her car gets 30 miles per gallon of gasoline. She is paid $0.50 per mile, and her only expense is gasoline at $2.00 per gallon. What is her net rate of pay, in dollars per hour, after this expense? (A) 20 (B) 22 (C) 24 (D) 25 (E) 26 Problem 5 What is the sum of all real numbers x for which |x^2 - 12x + 34| = 2? (A) 12 (B) 15 (C) 18 (D) 21 (E) 25 Problem 6 How many 4-digit positive integers (that is, integers between 1000 and 9999, inclusive) having only even digits are divisible by 5? (A) 80 (B) 100 (C) 125 (D) 200 (E) 500 Problem 7 The 25 integers from -10 to 14, inclusive, can be arranged to form a 5-by-5 square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum? (A) 2 (B) 5 (C) 10 (D) 25 (E) 50 Problem 8 What is the value of 1+2+3−4+5+6+7−8+…+197+198+199−200? (A) 9,800 (B) 9,900 (C) 10,000 (D) 10,100 (E) 10,200

真题文字详解(英文)

Problem 1 What value of ( x ) satisfies ( x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3} )?

[ \begin{array}{ll} (A) & -\frac{2}{3} \ (B) & \frac{7}{36} \ (C) & \frac{7}{12} \ (D) & \frac{2}{3} \ (E) & \frac{5}{6} \ \end{array} ]

Solution 1 Adding ( \frac{3}{4} ) to both sides, ( x = \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12} = \left( E \right) \frac{5}{6} ).

Solution 2 Multiplying 12 on both sides gets us ( 12x - 9 = 1 \Rightarrow 12x = 10 ), therefore ( x = \left( E \right) \frac{5}{6} ).

---

Problem 2 The numbers 3, 5, 7, ( a ), and ( b ) have an average (arithmetic mean) of 15. What is the average of ( a ) and ( b )?

[ \begin{array}{ll} (A) & 0 \ (B) & 15 \ (C) & 30 \ (D) & 45 \ (E) & 60 \ \end{array} ]

Solution The arithmetic mean of the numbers 3, 5, 7, ( a ), and ( b ) is equal to ( \frac{3 + 5 + 7 + a + b}{5} = \frac{15 + a + b}{5} = 15 ). Solving for ( a + b ), we get ( a + b = 60 ). Dividing by 2 to find the average of the two numbers ( a ) and ( b ) gives ( \frac{a + b}{2} = \left( C \right) 30 ).

---

Solution (two solutions - balancing and summing)

We know the average is 15. 3, 5, and 7 are respectively, 12, 10, and 8 below the average. So far then we are -12 - 10 - 8 or 30 below the average. We have to make this up with ( a ) and ( b ) so on average each is ( \frac{30}{2} = 15 ) above the average. The average is 15 so each is on average 15+15 = 30, or answer C.

Incidentally we can immediately rule out answers A and B as being too small - with these we'd never catch up to get our average of 15. An alternative solution is that to get an average of 15 we need a sum of 15*5 = 75. We have 3, 5, and 7 which adds up to 15, so ( a ) and ( b ) must make up the remaining 75-15 = 60. 60/2 = 30 so the average of ( a ) and ( b ) is 30.

~Dilip

---

Problem 3 Assuming ( a \neq 3 ), ( b \neq 4 ), and ( c \neq 5 ), what is the value in simplest form of the following expression?

[ \frac{a - 3}{5 - c} \cdot \frac{b - 4}{3 - a} \cdot \frac{c - 5}{4 - b} ]

[ \begin{array}{ll} (A) & -1 \ (B) & 1 \ (C) & \frac{abc}{60} \ (D) & \frac{1}{abc} - \frac{1}{60} \ (E) & \frac{1}{60} - \frac{1}{abc} \ \end{array} ]

Solution 1 (Negatives)

真题文字详解(中英双语)

Problem 1 What value of x satisfies (x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3})?
(A) (-\frac{2}{3}) (B) (\frac{7}{36}) (C) (\frac{7}{12}) (D) (\frac{2}{3}) (E) (\frac{5}{6})

Solution 1
Adding (\frac{3}{4}) to both sides, (x = \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12} = \left( E \right)\frac{5}{6}).
在两边加上 (\frac{3}{4})，得到 (x = \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12} = \left( E \right)\frac{5}{6}).

Solution 2
Multiplying 12 on both sides gets us (12x - 9 = 1 \Rightarrow 12x = 10), therefore (x = \left( E \right)\frac{5}{6}).
在两边乘以 12 ，我们得到 (12x - 9 = 1 \Rightarrow 12x = 10)，因此 (x = \left( E \right)\frac{5}{6}).

Problem 2 The numbers 3, 5, 7, a, and b have an average (arithmetic mean) of 15. What is the average of a and b?
数字 3, 5, 7, a 和 b 的平均数（算术平均值）是 15 。a 和 b 的平均数是多少？
(A) 0 (B) 15 (C) 30 (D) 45 (E) 60

Solution
The arithmetic mean of the numbers 3, 5, 7, a, and b is equal to (\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15). Solving for a + b, we get (a + b = 60). Dividing by 2 to find the average of the two numbers a and b gives (\frac{60}{2}=\left( C \right)30).
数字 3, 5, 7, a 和 b 的算术平均值等于 (\frac{3+5+7+a+b}{5}=\frac{15+a+b}{5}=15)。求解 a + b，我们得到 (a + b = 60)。除以 2 以找到数字 a 和 b 的平均值得到 (\frac{60}{2}=\left( C \right)30)。

Solution (two solutions - balancing and summing)
解答（两种解法-平衡和求和）

We know the average is 15. 3, 5, and 7 are respectively, 12, 10, and 8 below the average. So far then we are -12 - 10 - 8 or 30 below the average. We have to make this up with a and b so that each is 30/2 = 15 above the average. The average is 15 so each is on average 15+15 = 30, or answer C. 我们知道平均数是 15。3、5 和 7 分别比平均数低 12、10 和 8。到目前为止，我们已经低于平均数 -12 - 10 - 8 或 30。我们需要用 a 和 b 来弥补这个差距，所以每个都是 30/2 = 15 高于平均数。平均数是 15，所以每个的平均值是 15+15 = 30，或答案 C。