2013 澳洲 AMC E-Senior 真题 答案 详解

中英双语真题

Advanced Test 1-10 questions, each worth 3 points
1. The expression (\frac{0.6}{12}) equals
(A) 5 (B) 0.5 (C) 0.05 (D) 0.005 (E) 0.0005

1. Which one of the following is a right triangle?
   (A)
   (B)
   (C)
   (D)
   (E)

2. A plot of farmland increased its yield by 20% this season to reach 114 tons. What was the yield last season?
   (A) 90 (B) 91 (C) 93 (D) 94 (E) 95

3. Which number is closest to the cube root of 201320132013?
   (A) 600 (B) 5000 (C) 6000 (D) 50000 (E) 60000

4. If (p = 4b + 26) and (b) is an integer, then (p) cannot be evenly divided by which of the following numbers?
   (A) 2 (B) 4 (C) 5 (D) 6 (E) 7

5. The first five terms of a sequence are: 1, 2, 1, -1, -2,... It is known that from the second term onwards, each term is obtained by subtracting the previous term from the term before that. What is the sum of the first 42 terms of this sequence?
   (A) 0 (B) 4 (C) 12 (D) 24 (E) 30

英文真题

Senior Division Questions 1 to 10, 3 marks each 1. The expression (\frac{0.6}{12}) is equal to (A) 5 (B) 0.5 (C) 0.05 (D) 0.005 (E) 0.0005 2. Which of the following is a right-angled triangle? (A) (B) (C) (D) (E) 3. The yield from a farm increased by 20% this season to 114 tonnes. Last season the number of tonnes yielded was (A) 90 (B) 91 (C) 93 (D) 94 (E) 95 4. The cube root of 201320132013 is closest to (A) 600 (B) 5000 (C) 6000 (D) 50000 (E) 60000 5. If (p = 4b + 26) and (b) is a positive integer, then (p) could not be divisible by (A) 2 (B) 4 (C) 5 (D) 6 (E) 7 6. The first five terms of a sequence are shown. 1, 2, 1, -1, -2,... After the second term, each term can be obtained by subtracting from the previous term the term before that. What is the sum of the first 42 terms of this sequence? (A) 0 (B) 4 (C) 12 (D) 24 (E) 30

真题文字详解(英文)

SOLUTIONS - SENIOR DIVISION

1. ( \frac{0.6}{12} ) is the same as ( 0.1 \times \frac{6}{12} = 0.05 ), hence (C).

2. (A) ( 3^2 + 4^2 = 25 \neq (5.2)^2 ) (B) ( 1^2 + 4^2 = 17 \neq (\sqrt{5})^2 ) (C) ( 6^2 + 8^2 = 100 = 10^2 ) (D) ( 1^2 + 2.1^2 = 5.41 \neq 3^2 ) (E) ( 2^2 + 3^2 = 13 \neq 7^2 ) hence (C).

3. If ( y ) were the previous yield, then ( 1.2y = 114 ) tonnes so ( y = \frac{114}{1.2} = \frac{1140}{12} = 95 ) tonnes, hence (E).

4. First consider that ( 5000^3 = 125000000000 ) and ( 6000^3 = 216000000000 ). The number is between these two, so its cube root is between 5000 and 6000 but obviously closer to 6000, hence (C).

5. (Also I7) When ( b = 1, p = 30 ) (which is divisible by 2, 5 and 6). When ( b = 4, p = 42 ) (which is divisible by 7). This eliminates alternatives A, C, D and E. However, ( 4p + 26 = 4(p + 6) + 2 ) so ( p ) cannot be divisible by 4, hence (B).

6. There is a repeating cycle of six terms in this sequence 1, 2, 1, -1, -2, -1; 1, 2, 1, -1, -2, -1; ... and so 42 terms cover seven such cycles. The terms in each cycle add to 0, hence (A).

7. Intercepts for the equations are (A) ( x = \frac{12}{5}, y = 1 ) (B) ( x = -\frac{1}{4}, y = -1 ) (C) ( x = -\frac{7}{3}, y = 1 ) (D) ( x = -\frac{2}{3}, y = 1 ) (E) ( x = \frac{3}{2}, y = 1 ) Only A fits the given line, hence (A).

8. If all three were running, when Cody wins, Chris has only run 95 m. Then Joseph has run ( \frac{9}{10} ) of this, or 85.5 metres. Cody wins by 14.5 metres, hence (C).