2016 澳洲 AMC E-Senior 真题 答案 详解
2016-10-03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
英文真题
Senior Division
Questions 1 to 10, 3 marks each
-
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 -
If ( p = 7 ) and ( q = -4 ), then ( p^2 - 3q^2 = )
(A) 49
(B) 48
(C) 0
(D) 97
(E) 1 -
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 -
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 -
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 -
In the diagram, the value of ( x ) is
(A) 120
(B) 108
(C) 105
(D) 135
(E) 112.5
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