2020 澳洲 AMC E-Senior 真题 答案 详解

中英双语真题

2020 Australian Mathematics Competition – Senior 1-10 题，每题3分 Questions 1 to 10, 3 marks each 1. 请问算式 2020 ÷ 20 等于多少？ What is the value of 2020 ÷ 20? (A) 2000 (B) 2040 (C) 11 (D) 101 (E) 1001 2. 在右图中，请问 x 与 y 之和等于多少？ In the diagram provided, find the sum of x and y. (A) 30 (B) 75 (C) 95 (D) 105 (E) 180 3. 请问根式 √(7 + 18 ÷ (10 - 15)) 的值等于多少？ Evaluate √(7 + 18 ÷ (10 - 15)). (A) 5/3 (B) 9 (C) 3 (D) 5 (E) 1/27 4. 小莎心里想着两个数，它们的和为 26、差为 14。请问小莎想的这两个数的乘积是多少？ Sebastien is thinking of two numbers whose sum is 26 and whose difference is 14. The product of Sebastien’s two numbers is (A) 80 (B) 96 (C) 105 (D) 120 (E) 132 5. 已知 7/8 的 * 的 5/6 的 4/5 等于 1，请问 * 的值为多少？ If 4/5 of 5/6 of * of 7/8 is equal to 1, then the value of * is (A) 6 (B) 8 (C) 10 (D) 12 (E) 14 6. 将一个面积为 10000 m² 正方形花园的长与宽各增加 10%，请问这个花园的面积会增加多少 m²？ A square garden of area 10 000 m² is to be enlarged by increasing both its length and width by 10%. The increase in area, in square metres, is (A) 1000 (B) 2000 (C) 2100 (D) 2400 (E) 4000 7. 已知 f(x) = 2x² - 3x + c 且 f(2) = 6，请问 c 的值是多少？ Given that f(x) = 2x² - 3x + c and f(2) = 6, then c is equal to (A) 4 (B) 3 (C) 6 (D) 8 (E) 12

英文真题

Senior Division Questions 1 to 10, 3 marks each

1. What is the value of (2020 \div 20)?
2. (A) 2000
3. (B) 2040
4. (C) 11
5. (D) 101

6. (E) 1001

7. In the diagram provided, find the sum of (x) and (y).

8. (A) 30
9. (B) 75
10. (C) 95
11. (D) 105

12. (E) 180

13. Evaluate (\sqrt{7 + 18 \div (10 - 15)}).

14. (A) (\frac{5}{3})
15. (B) 9
16. (C) 3
17. (D) 5

18. (E) (\frac{1}{27})

19. Sebastien is thinking of two numbers whose sum is 26 and whose difference is 14. The product of Sebastien's two numbers is:

20. (A) 80
21. (B) 96
22. (C) 105
23. (D) 120

24. (E) 132

25. If (\frac{4}{5} \times \frac{5}{6} \times \frac{7}{7} \times \frac{7}{8}) is equal to 1, then the value of * is:

26. (A) 6
27. (B) 8
28. (C) 10
29. (D) 12

30. (E) 14

31. A square garden of area 10,000 m² is to be enlarged by increasing both its length and width by 10%. The increase in area, in square metres, is:

32. (A) 1000
33. (B) 2000
34. (C) 2100
35. (D) 2400

36. (E) 4000

37. Given that (f(x) = 2x^2 - 3x + c) and (f(2) = 6), then (c) is equal to:

38. (A) 4
39. (B) 3
40. (C) 6
41. (D) 8
42. (E) 12

真题文字详解(英文)

72 2020 AMC Senior Solutions Senior Solutions 1. \frac{2020}{20} = \frac{2000}{20} + \frac{20}{20} = 100 + 1 = 101, hence (D). 2. (Also I3) The supplementary angle to 105° is x = 75. Since the triangle is isosceles, the other base angle is 75° and the apex angle is y = 180 - 2 × 75 = 30. Then x + y = 75 + 30 = 105, hence (D). 3. \sqrt{7+18÷(10-15)} = \sqrt{7+18÷(10-1)} = \sqrt{7+18÷9} = \sqrt{7+2} = \sqrt{9} = 3, hence (C). 4. (Also J10, I5) Alternative 1 The average of the two numbers is 26 ÷ 2 = 13, so that they are equal distances above and below 13. Since they differ by 14, the distance above and below 13 is 14 ÷ 2 = 7. That is, the two numbers are 13 - 7 = 6 and 13 + 7 = 20, and their product is 6 × 20 = 120, hence (D). Alternative 2 Let the smaller number be x, so that the larger is 14 + x. Then x + (14 + x) = 26, giving 2x = 26 - 14 = 12, and so x = 6. Therefore, the numbers are 6 and 20, with product 6 × 20 = 120, hence (D). 5. (Also I8) Let x be the value of *. Then 1 = \frac{4}{5} \times \frac{5}{6} \times \frac{x}{7} \times \frac{7}{8} = \frac{4 \times 5 \times x \times 7}{5 \times 6 \times 7 \times 8} = \frac{4 \times x}{6 \times 8} = \frac{x}{12} so that x = 12, hence (D). 6. The garden is 100 m × 100 m. Increasing the length by 10% gives a new length of 110 m. Similarly, the new width is 110 m. The new area is 110 m × 110 m = 12100 m^2, which is an increase of 2100 m^2, hence (C). 7. f(2) = 2 \times 2^2 - 3 \times 2 + c = 2 + c. But f(2) = 6, so c = 4, hence (A). www.amt.edu.au