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Mathematics Form 3 PT3 Model Paper Mathematics Form 3 PT3 Model Paper
7
(c) Amount to be paid by instalment c = (ii) New probability of green pen: n() = 60
2
= RM84.62 18 2 = 1 – = 3 17 – x + x + 8 + x + 2 + 23 – x + 6 + y = 60
= RM1 523.16 ∴ Equation of the straight line is 5 5 17 – 3 + 3 + 8 + 3 + 2 + 23 – 3 + 6 + y = 60
1
Amount of interest to be paid y = – x + 7 (c) 2 – p  2p – 4 56 + y = 60
= RM1 523.16 – RM1 360 2 2 2 + 4  2p + p y = 4
= RM163.16 3p  6
28. (a) x + y = 12 ......................(1)
4y – x = 18 .....................(2) p  2 30. (a) (i) (a) Transformation A is reflection
k + 5 + 6 + 6 + 8 + 12 + 12 2p – 4  26 – 3p at line y = 1.
27. (a) (i) + 12 = 8 From (1) 2p + 3p  26 + 4 (b) Transformation B is an
8 y = –x + 12 ...................(3) 5p  30
k + 61 = 64 enlargement with scale factor
k = 3 From (2) p  6 of 2 at centre (–4, 0).
(ii) mode = 12 x + 18
y = ......................(4) p 6
2
(iii) 3, 5, 6, 6, 8, 12, 12, 12 4 p 2 (ii) ∆STU = 2 × ∆JKL
6 + 8 (3) = (4) = 4 × 25 cm 2
median, m = 2 6
2 x + 18 2 p 6 = 100 cm 2
= 7 –x + 12 = 4
k – m = 3 – 7 –4x + 48 = x + 18 Therefore, the possible values of p are

= – 4 –5x = –30 3, 4 and 5. (b) (i) z = 1 x – 2
x = 6 3 y
(b) x – 2x – 3 = x(3 – x) Substitute x = 6 into (1) 8 x – 2
2
x – 2x – 3 = 3x – x 2 6 + y = 12 29. (a) sin x = 17 3z = 
2
y
2
2x – 5x – 3 = 0 x – 2
y = 6 AG = 8 9z =
2
y
2x 1 x ∴ coordinates T is (6, 6) BG 17 9yz = x – 2
2
x –3 –6x 16 8 2
= x = 9yz + 2
2x 2 –3 –5x 3 4 BG 17
(b) (i) P(Green pen) = 1 – = BG = 34 cm
2
(2x + 1)(x – 3) = 0 7 7 (ii) x = 9yz + 2
1 Total number of pens: DG = BG – BD
1
x = – , x = 3 = 34 – 29   2
2 4 x = 9(–1) + 2
× Total number of pens = 24 = 5 cm x = 1 3
7
1 1 7 5
(c) x + y = 5 Total number of pens = 24 × cos y =
4 2 4 13 31. (a) Volume of small cubes
1
1 y = – x + 5 = 42 y = cos –1 5
2 4 13 = Volume of sphere
1 Number of pink pens = 42 – 24 y = 67.4° 4 851
y = – x + 10 = 18 4 22
2   (21) 3
1 New probability of pink pen: (b) Price of juice Kavita has to pay = 3 7
∴ Gradient of the line, m = – 18 + 2 4 851
2 = = (150 × RM0.10) + (200 × RM0.08)
42 + 2 + 6 = 8 cm 3
Equation of straight line 2 + [(1 000 – 150 – 200) × RM0.03]
–1 = = RM15 + RM16 + RM19.50 3
8
y = x + c 5 Length of the side of small cube = 
2 = RM50.50 = 2 cm
Substitute x = 3, y = 2
1 (c) n(P  R  Q) = 16 Area of one of the faces of the small
2 = – (3) + c cube
2 x + 8 + x + 2 = 16
2x = 6 = 2 × 2 2
x = 3 = 4 cm
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BOOKLET ANS MATH F3.indd 31 03/01/2020 10:21 AM