2017 澳洲 AMC E-Senior 真题 答案 详解

---

中英双语真题

1-10 Questions, 3 marks each
Questions 1 to 10, 3 marks each

1. A、B、C、D、E、F、G、H 这八位小孩在游泳池进行游泳比赛。将他们随意地分配到1号至8号水道。请问是 F、G 或 H 被分配到1号水道的机率是什么？
   Eight children in a swimming race are John, Iain, Hans, Ivan, Giovanni, Beth, Liz and Elisa. They are put in lanes 1 to 8 randomly. What is the probability that Beth, Liz or Elisa is in lane 1?
   (A) ( \frac{3}{8} ) (B) ( \frac{1}{2} ) (C) ( \frac{3}{5} ) (D) ( \frac{5}{3} ) (E) ( \frac{1}{3} )

2. 已知 ( a = 20 ) 且 ( b = 17 )，请问 ( 17a + 20b ) 的值是多少？
   What is the value of ( 17a + 20b ) when ( a = 20 ) and ( b = 17 )?
   (A) 680 (B) 689 (C) 1720 (D) 2017 (E) 3737

3. 1 的 1000% 等于
   1000% of 1 is
   (A) 0.1 (B) 1 (C) 10 (D) 100 (E) 1000

4. 算式 ( 4^2 + 3^3 + 2^4 ) 之值等于
   ( 4^2 + 3^3 + 2^4 = )
   (A) 29 (B) 33 (C) 43 (D) 59 (E) 73

5. 图中的风车星形是将一个直角三角形围绕它的一个顶点旋转而构成的。请问图中标记有点上的角之度数为何？
   This pinwheel star is formed by rotating a right-angled triangle around one of its corners. What is the angle at each of the nine tips that are marked with dots?
   (A) 30° (B) 40° (C) 45° (D) 50° (E) 60°

英文真题

Senior Division

Questions 1 to 10, 3 marks each

1. Eight children in a swimming race are John, Iain, Hans, Ivan, Giovanni, Beth, Liz and Elisa. They are put in lanes 1 to 8 randomly. What is the probability that Beth, Liz or Elisa is in lane 1?
(A) ( \frac{3}{8} ) (B) ( \frac{1}{2} ) (C) ( \frac{3}{5} ) (D) ( \frac{5}{3} ) (E) ( \frac{1}{3} )

2. What is the value of (17a + 20b) when (a = 20) and (b = 17)?
(A) 680 (B) 689 (C) 1720 (D) 2017 (E) 3737

3. 1000% of 1 is
(A) 0.1 (B) 1 (C) 10 (D) 100 (E) 1000

4. (4^2 + 3^3 + 2^4 = )
(A) 29 (B) 33 (C) 43 (D) 59 (E) 73

5. This pinwheel star is formed by rotating a right-angled triangle around one of its corners. What is the angle at each of the nine tips that are marked with dots?
(A) 30° (B) 40° (C) 45° (D) 50° (E) 60°

真题文字详解(英文)

Solutions – Senior Division
1. (Also I2) Each of the 8 children is equally likely to be in lane 1, so the probability that one of the three identified students is in lane 1 is (\frac{3}{8}); hence (A).
2. (17 \times 20 + 20 \times 17 = 340 + 340 = 680); hence (A).
3. (Also J9, I4) 100% of 1 is 1, and 1000% is 10 times greater, which is 10; hence (C).
4. (4^2 + 3^3 + 2^4 = 16 + 27 + 16 = 59); hence (D).
5. (Also J12) The nine equal angles meeting at the centre of the pinwheel must each be (\frac{360}{9} = 40^\circ). Within each right-angled triangle, the angles add to (180^\circ), so they are (40^\circ), (90^\circ) and (50^\circ), where (50^\circ) is the angle marked with a dot; hence (D).
6. (Also J11, I7) There are many ways to solve this with 4 swaps. For example,
[ \begin{aligned} &\text{WORDS} \rightarrow \text{WORSD} \rightarrow \text{WOSRD} \rightarrow \text{WSORD} \rightarrow \text{SWORD}\ \end{aligned} ]
This suggests that the answer is 4. Suppose this isn’t the smallest: some solution with three or fewer swaps changes WORDS to SWORD. Each swap involves two letters, and each letter must be swapped at least once. When swapping five objects with these conditions, the only pattern of swaps possible is shown in the diagram: there are three swaps, one object is swapped twice and all other objects are swapped only once. In particular, there is a pair of objects that are swapped only with each other. Such a pattern of swaps can't transform WORDS into SWORD, since there is no pair of letters which have only been swapped with each other. Consequently the smallest possible number of swaps is 4; hence (B).