2016 澳洲 AMC E-Senior 真题 答案 详解

🗓 2016-10-03 📁 auamc 🏷️ auamc E-Senior


序号 文件列表 说明
1 2016-E-Senior-paper-eng-zh.pdf 11 页 6.04MB 中英双语真题
2 2016-E-Senior-paper-eng.pdf 6 页 134.59KB 英文真题
3 2016-E-Senior-key.pdf 1 页 482.59KB 真题答案
4 2016-E-Senior-solution-eng.pdf 11 页 222.91KB 真题文字详解(英文)

中英双语真题

1-10 Questions, 3 marks each

1. 请问百位数码为7、个位数码为8的三位数共有多少个? How many 3-digit numbers have the hundreds digit equal to 7 and the units digit equal to 8? (A) 10 (B) 100 (C) 20 (D) 19 (E) 90

2. 若 p = 7、q = -4,则 p^2 - 3q^2 等于 If p = 7 and q = -4, then p^2 - 3q^2 = (A) 49 (B) 48 (C) 0 (D) 97 (E) 1

3. 在下面的这些正方形中,标记的长度都是 1 单位长。请问哪一个正方形的周长最长? In each of these squares, the marked length is 1 unit. Which of the squares would have the greatest perimeter? (A) P (B) Q (C) R (D) S (E) all are the same

中英双语真题

英文真题

Senior Division
Questions 1 to 10, 3 marks each

  1. How many 3-digit numbers have the hundreds digit equal to 7 and the units digit equal to 8?
    (A) 10
    (B) 100
    (C) 20
    (D) 19
    (E) 90

  2. If ( p = 7 ) and ( q = -4 ), then ( p^2 - 3q^2 = )
    (A) 49
    (B) 48
    (C) 0
    (D) 97
    (E) 1

  3. In each of these squares, the marked length is 1 unit. Which of the squares would have the greatest perimeter?
    (A) P
    (B) Q
    (C) R
    (D) S
    (E) all are the same

  4. Given that ( n ) is an integer and ( 7n + 6 \geq 200 ), then ( n ) must be
    (A) even
    (B) odd
    (C) 28 or more
    (D) either 27 or 28
    (E) 27 or less

  5. King Arthur's round table has a radius of three metres. The area of the table top in square metres is closest to which of the following?
    (A) 20
    (B) 30
    (C) 40
    (D) 50
    (E) 60

  6. In the diagram, the value of ( x ) is
    (A) 120
    (B) 108
    (C) 105
    (D) 135
    (E) 112.5

英文真题

真题文字详解(英文)

Solutions – Senior Division
1. They are 708, 718, 728, 738, 748, 758, 768, 778, 788, 798, hence (A).
2. ( p^2 - 3q^2 = 7^2 - 3 \times (-4)^2 = 49 - 3 \times 16 = 1 ), hence (E).
3. (Also I8) The perimeters of P, Q, R and S are ( 4\sqrt{2} ), 8, ( 2\sqrt{2} ) and 4 respectively, hence (B).
4. Solving, ( 7n \geq 194 ), and so ( n \geq \frac{194}{7} = 27\frac{5}{7} ). So the smallest ( n ) can be is 28, and any value ( n \geq 28 ) is also a solution, hence (C).
5. The table top is a circle with radius 3 metres, so its area in square metres is ( \pi(3)^2 = 9\pi ). Because ( \pi ) is slightly greater than 3, ( 9\pi ) is slightly greater than 27. Of the given choices, 30 is closest, hence (B).
6. (Also II1)
Alternative 1
The sum of the exterior angles of the pentagon is ( 360 = 90 + 4 \times (180 - x) ), so that ( 180 - x = \frac{270}{4} = 67.5 ) and ( x = 180 - 67.5 = 112.5 ), hence (E).
Alternative 2
The sum of the interior angles of the pentagon is ( 3 \times 180 = 90 + 4x ), so that ( x = \frac{450}{4} = 112.5 ), hence (E).
7. (Also II2) If C is the point directly below A and to the left of B, then the right triangle ABC has sides 16 and 12, and hypotenuse ( x ). Then ( x^2 = 16^2 + 12^2 = 400 ) and ( x = 20 ), hence (A).
8. Square both sides of the equation to obtain:
( \sqrt{x^2 + 1} = x + 2 \quad \Rightarrow \quad x^2 + 1 = x^2 + 4x + 4 \quad \Rightarrow \quad 4x = -3 ).
Hence, the only possible solution is ( x = -\frac{3}{4} ). We substitute this value into the original equation and verify that both sides are indeed equal to ( \frac{5}{4} ), hence (B).

2016 AMC – Senior Solutions

真题文字详解(英文)

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