Page 20 - 1202 Question Bank Mathematics Form 4

Given minimum point is (−2.5, −2.25), Side length of triangle = 5 – 1 Paper 2
Axis of symmetry, x = –2.5. = 4 cm
2
(c) When f(x) = x + 5x + 4 is reflected 19. Area of square – Area of triangle = 40 Section A
in the x-axis, the quadratic function 1 1. (a) 4 2 (b) 6 2
]
2
change to f(x) = –x – 5x – 4 where the 2y(2y) – [ 2 (y + 2)(2y) = 40 (c) 8 1 (d) 2 5
value of a will change to −a.
4y – y – 2y – 40 = 0 2. (a) 3 × 7 = 21 (b) 4 × 5 = 500
2
2
3
1
14. Price = RM300 3y – 2y – 40 = 0 (c) 7 × 9 = 63 (d) 2 × 3 = 162
2
4
1
(8x + 14)(5x) = 300 (3y + 10)(y – 4) = 0 0 1
4
2
40x + 70x – 300 = 0 10 3. (a) 32 = 2 × 4 + 3 × 4
= 2 + 12

(x – 2)(4x + 15) = 0 x = – 3 (Rejected) or y = 4
15 = 14
x = 2, x = – (Rejected) Perimeter of the combination objects
4 (b) 11101 2
1
2
3
0
Number of text books bought by Sofea = (4 + 6) + (4 + 2) + [3 × 2(4)] = 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 +
©PAN ASIA PUBLICATIONS
= 10 + 6 + 8 + 8 + 8
= 8(2) + 14 = 40 cm 1 × 2 4
= 30 books = 1 + 4 + 8 + 16
1
20. (a) L = (y)(y + 4) = 29
15. Volume of the box = 4 500 2 (c) 413 = 3 × 5 + 1 × 5 + 4 × 5 2
1
0
1
(x + 5)(x)(30) = 4 500 L = (y + 4y) 5 = 3 + 5 + 100
2
30x + 150x – 4 500 = 0 2 = 108
2
1
2
x + 5x – 150 = 0 L = y + 2y (d) 624 = 4 × 7 + 2 × 7 + 6 × 7 2
2
1
0
(x – 10)(x + 15) = 0 2 7 = 4 + 14 + 294
x = 10, x = –15 (b) Area of triangle = 48 = 312
(Rejected) 1 2 y + 2y = 48
2
Thus, the value of x is 10 cm. 1 4. (a) 534 to base two
10
2
16. Area of the box = 432 2 y + 2y – 48 = 0 2 534 Remainder
2
(2x + 4)(3x + 3) = 432 y + 4y – 96 = 0
2
6x + 12x + 6x + 12 = 432 (y + 12)(y – 8) = 0 2 267 0
CHAPTER 1 – CHAPTER 2
2
6x + 18x – 420 = 0 y = 8, y = –12 2 133 1
2
x + 3x – 70 = 0 (Rejected)
(x + 10)(x – 7) = 0 Thus, y = 8 cm. 2 66 1
x = 7, x = –10 (c) Area of the polygon = 6 × 48 2 33 0
(Rejected) = 288 cm 2 2 16 1
Substitute x = 7, Thus, the area of polygon is 288 cm 2
Diameter 4 balls = 3(7) + 3 and its name is hexagon. 2 8 0
= 21 + 3 21. (a) L = (35 + y)(65 + y) 2 4 0
2
= 24 cm L = 2 275 + 65y + 35y + y
2
So, diameter 1 ball = 24 ÷ 4 L = y + 100y + 2 275 2 2 0
= 6 cm (b) Area of the tile = 6 175 2 1 0
2
Substitute x = 7 into 2x + 4, y + 100y + 2 275 = 6 175 0 1
2
Diameter 3 balls = 2(7) + 4 y + 100y – 3 900 = 0 534 = 1000010110
= 14 + 4 (y – 30)(y + 130) = 0 10 2
= 18 cm y = 30, y = –130 (Rejected) (b) 534 to base eight
10
So, diameter 1 ball = 18 ÷ 3 Thus, y = 30 cm. 8 534 Remainder
= 6 cm (c) The smallest part of tile represented
by region DEFI. 8 66 6
Section C Area of DEFI = 0.3 m × 0.3 m
17. (a) L = (x + 6)(x + 3) = 0.09 m 2 8 8 2
2
L = x + 6x + 3x + 18 Number of tiles needed 8 1 0
2
L = x + 9x + 18 = 1.08 ÷ 0.09
(b) Area of the cake = 270 = 12 tiles 0 1
x + 9x + 18 = 270 22. (a) Area A – Area B = 12 cm 2 534 = 1026 8
2
10
2
x + 9x + 18 – 270 = 0 2x(x + 3) – x(x + 5) = 12 (c) 534 to base five
10
2
2
x + 9x – 252 = 0 2x + 6x – x – 5x – 12 = 0
2
(x – 12)(x + 21) = 0 x + x – 12 = 0 5 534 Remainder
2
x = 12 cm (x – 3)(x + 4) = 0 5 106 4
(c) Number of small cakes x = 3, x = – 4 (Rejected)
= 270 cm ÷ 9 cm 2 Thus, x = 3 cm. 5 21 1
2
= 30 cakes (b) Perimeter of new arrangement 5 4 1
Thus, the small cakes is enough to = 6 + 6 + 6 + 8 + 3 + 8 + 3
give to all Safi’s friend. = 40 cm 0 4
534 = 4114
18. (a) Area of triangle, L 10 5
1
= [2(x + 3)](x – 1) CHAPTER 2 (d) 534 to base four
2 10
= (x + 3)(x – 1) 4 534 Remainder
= x + 2x – 3 Paper 1
2
(b) Area of triangle = 32 1. A 2. D 3. D 4. A 5. C 4 133 2
2
x + 2x – 3 = 32 6. C 7. A 8. B 9. B 10. D 4 33 1
2
x + 2x – 35 = 0 11. C 12. D 13. B 14. D 15. C
(x – 5)(x + 7) = 0 16. A 17. C 18. C 19. B 20. A 4 8 1
x = 5, x = –7 21. C 22. B 23. B 24. D 25. A 4 2 0
(Rejected) 26. C 27. D 28. A 29. B 30. D
When x = 5, 31. C 32. C 33. A 34. B 35. D 0 2
Height of triangle = 5 + 3 36. D 37. B 38. C 39. D 40. C 534 = 20112 4
10
= 8 cm 41. B 42. B
116
Answers 1202QB Maths Form 4.indd 116 21/02/2022 6:27 PM