2019 澳洲 AMC E-Senior 真题 答案 详解

2019 年 E-Senior 真题,中英双语,答案,详解

中英双语真题

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

1. What is the value of (201 \times 9)?
2. (A) 189
3. (B) 1809
4. (C) 1818
5. (D) 2001

6. (E) 2019

7. What is the area of the shaded triangle?

8. (A) (8 m^2)
9. (B) (12 m^2)
10. (C) (14 m^2)
11. (D) (20 m^2)

12. (E) (24 m^2)

13. What is 19% of $20?

14. (A) $20.19
15. (B) $1.90
16. (C) $0.19
17. (D) $3.80

18. (E) $0.38

19. What is the value of (z)?

20. (A) 30
21. (B) 35
22. (C) 45
23. (D) 50

24. (E) 55

25. The value of (2^0 + 1^9) is

26. (A) 1
27. (B) 2
28. (C) 3
29. (D) 10

30. (E) 11

31. Let (f(x) = 3x^2 - 2x). Then (f(-2)) equals

32. (A) -32
33. (B) -8
34. (C) 16
35. (D) 32
36. (E) 40

英文真题

Senior Division
Questions 1 to 10, 3 marks each

1. What is the value of (201 \times 9)?
2. (A) 189
3. (B) 1809
4. (C) 1818
5. (D) 2001

6. (E) 2019

7. What is the area of the shaded triangle?

8. (A) (8m^2)
9. (B) (12m^2)
10. (C) (14m^2)
11. (D) (20m^2)

12. (E) (24m^2)

13. What is 19% of $20?

14. (A) $20.19
15. (B) $1.90
16. (C) $0.19
17. (D) $3.80

18. (E) $0.38

19. What is the value of (z)?

20. (A) 30
21. (B) 35
22. (C) 45
23. (D) 50

24. (E) 55

25. The value of (2^0 + 1^9) is

26. (A) 1
27. (B) 2
28. (C) 3
29. (D) 10

30. (E) 11

31. Let (f(x) = 3x^2 - 2x). Then (f(-2)) =

32. (A) -32
33. (B) -8
34. (C) 16
35. (D) 32
36. (E) 40

真题文字详解(英文)

2019 AMC Senior Solutions 1. $201 \times 9 = 1809$, hence (B). 2. (Also I4) The shaded triangle has base $b=6\mathrm{~m}$ and perpendicular height $h=4\mathrm{~m}$. Its area is $A=\frac{1}{2}bh=\frac{1}{2} \times 6 \times 4=12$ square metres, hence (B). 3. $\frac{19}{100} \times 20=\frac{380}{100}=3.8$, hence (D). 4. (Also J13) In the right-hand triangle, $a=180-45-50=85$. Since $a$ and $b$ are vertically opposite, $b=85$. Finally, $z=180-85-60=35$, hence (B). 5. $2^{0}+1^{9}=1+1=2$, hence (B). 6. $f(-2)=3(-2)^{2}-2(-2)=3 \times 4+4=16$, hence (C). 7. The angle sum of a quadrilateral is $360^{\circ}$. So $\frac{10}{3} \theta=360$, and $\theta=\frac{3}{10} \times 360=108$, hence (E). 8. (Also I9) The pattern repeats 1, 4, 7, 4. Since $1+4+7+4=16$, each full cycle contributes 16, and then there is an additional amount that is 0, 1, 1 + 4 = 5 or 1 + 4 + 7 = 12. That is, the sum is of the form $16n, 16 n+1, 16 n+5$ or $16 n+12$. Of the numbers given, only $65=16 \times 4+1$ can be written this way, hence (E). 9. They meet after $60 \times 20=1200$ seconds. In this time, Mia has walked $1200 \times 1.5=1800$ metres, or $1.8 \mathrm{~km}$, and Crystal has walked $1200 \times 2=2400$ metres, or $2.4 \mathrm{~km}$. So the track is $1.8+2.4=4.2 \mathrm{~km}$ long, hence (B).