Top One (Maths F3) Penerbitan Pelangi Sdn Bhd

Mathematics Form 3 Chapter 8 Loci in Two Dimensions (c)

(c) P Q

SQ

4. (a) R P
P Q

5. (a)

S

(b) Q P
P
Q R

(b)

R
P

S
Q

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(c) Mathematics Form 3 Chapter 8 Loci in Two Dimensions

R (c) S B R

S

P Q
Q

P

C DA

6. (a) D P C

7. (a) A
QQ
R

RS

A Q B B DC
Q P
(b)
(b) B
A
P A

D B Z C
R YX

C DE

S © Penerbitan Pelangi Sdn. Bhd.

43 A2

Mathematics Form 3 Chapter 8 Loci in Two Dimensions PT3 Standard Practice 8

(c) Section A

A 1. Answer: C

BQ E 2. Locus of X is a circle of radius 6 cm with
P centre N. From the rectangle KLMN, locus
R X is arc KQR.
C D Answer: D

3. Answer: D

4. Answer: C

5. Locus X is arc STU and locus Y is line OB.
Hence, the point of intersection is at A.
Answer: A

6. Locus X is line BD and locus Y is arc ADE at
centre B. Hence the point of intersection is
at D.
Answer: D

Section B

1. (a) An ant moves on a
straight horizontal
wire.

Seekor semut bergerak
pada wayar mendatar
yang lurus.

(b) The tip of a fan blade
that is spinning.

Hujung bilah kipas yang
sedang berpusing.

(c) The tip of hour hand
in a clock from 4
o’clock to 8 o’clock.

Hujung jarum jam dari
pukul 4 ke pukul 8.

(d) A lift moving in tall
building.

Sebuah lif bergerak
dalam bangunan tinggi.

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Mathematics Form 3 Chapter 8 Loci in Two Dimensions

2. (b) P Q
A moving point that its
distance from the line KL and Line UR
NM is always constant. POR Locus L

Suatu titik bergerak yang jaraknya Arc T Locus M S
sentiasa sama dari garis KL dan KVM
NM. (c)
Line
A moving point that its LON S 5 cm T
distance is the same as LM
from point L. Line
QOS
Suatu titik bergerak yang jaraknya
adalah sama dengan LM dari titik L.

A moving point that its
distance from the line KN and
NM is always constant.

Suatu garis bergerak yang jaraknya
sentiasa sama dari garis KN dan
NM.

A moving point that is
equidistant from the points P
and R.

Suatu titik bergerak yang berjarak
sama dari titik P dan R.

Section C

3. (a) (i) The locus of the screw is a parallel
line to the horizontal line.

(ii) The locus of the apple is a vertical
line going down to the ground.

(iii) The locus of the cow is a circle and
the pole as its centre.

(iv) The locus of the ship is a
perpendicular line between Johor
and Singapore.

45 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 8 Loci in Two Dimensions

HOTS Challenge

13 m Pool

Tree Kolam

Pokok 7.6 m

X

Stool

Bangku

Y

The possible locations of the treasure are marked X or Y.

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Top One (MATHS F3) Penerbitan Pelangi Sdn BhdCHAP
TER

9 Straight Lines
Garis Lurus

1. (a) m = undefined (c) Gradient m = –7 5 = –7
(b) m < 0 11 – 6
(c) m = 0 –7
y = 6 x + c,

substitute (5, 0)

2. (a) y = –5x + 2 0 = –7 (5) + c
6
(b) y = 5
35
(c) y = 3 c = 6

(d) y = 1 x y = – 7 x + 35
2 6 6

(e) y = 3 x + 2
4 5
5. (a) 3y = –5x + 2

3. (a) m = –5, c = –6 y = –5 x + 2
3 3
1
(b) m = 2 , c = 7

(c) m = 0, c = 8 (b) 6y = 2x +3

y = 2 x + 3
6 6

4. (a) Gradient m = 3–5 = 1 y = 1 x + 1
c, –4 – 0 2 3 2
1
y = 2 x +

substitute (0, 5) (c) 4y = –x – 1

5= 0 + c y = – x – 1
4 4

c=5 (d) –8y = –4x +12

y = 1 x + 5 –4 12
2 –8 8
y = x –

(b) Gradient m = –2 – 1 y = 1 x – 3
10 – 9 2 2

= –3 y x
1 8 4
6. (a) = – + 1
= –3

y = –3x + c, y = – 8 x + 8
4
substitute (9, 1)
y = –2x + 8
1 = –3(9) + c

c = 28

y = –3x + 28

47 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 9 Straight Lines

(b) y = – x + 1 9. (a) LHS: 3(2) = 6
5 10 RHS: 2(0) + 6 = 6
Conclusion: Yes
y = – 5x + 5
10 (b) LHS: 4(–2) = –8
RHS: 3(–1) + 1 = –2
y = – 1 x + 5 Conclusion: No
2
(c) LHS: 0
(c) – y = – x + 1 RHS: 5(–3) – 1 = –16
4 3 Conclusion: No

y = 4x – 4
3

(d) – y = 4x + 1
9 9
10. (a) L1: 3y = –6x + 7
y = –4x – 9
7
y = –2x + 3 , gradient m1 = –2

7. (a) y = 1 x – 9 L2: –4y = 8x – 5
6
y = –2x + 5 , gradient m2 = –2
6y = x – 54 4

x – 6y – 54 = 0 m1 = m2, therefore L1 is parallel to L2.

(b) x + y = 1 (b) L1: 2y + 1=– 1 x + 6
2 8 2

 x + y = 1 × 8 y = – 1 x + 5 , gradient m1 = – 1
2 8 4 2 4

4x + y = 8 L2: 3y = 8x – 3

4x + y – 8 = 0 8 8
3 3
y= x – 1, gradient m2 =

8. (a) 5y = 2x + 10 m1 ≠ m2, therefore L1 is not parallel to
L2.
–2x + 5y = 10

– 2x + 5y = 10
10 10 10
x
– 5 + y = 1 (c) L1: 3y + 2x = –6
2
2 2
(b) 6y – 2x – 8 = 0 y = – 3 x – 2, gradient m1 = – 3

(–2x + 6y = 8) ÷ 8 L2: 6 – 5y = 8x – 1

–x + 3y = 1 y = – 8 x + 7 , gradient m2 = – 8
4 4 5 5 5

(c) y = 1 x – 6 m1 ≠ m2, therefore L1 is not parallel to
2 L2.

 1 x – y = 6 ÷ 6
2

112x – y = 1
6

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11. (a) 2y = 6 – 3x Mathematics Form 3 Chapter 9 Straight Lines

y = –3 x +3 12. (a) y = 2x – 4
2 xy
0 –4
m = – 3 20
2
y = 6 – 2x
new equation xy
06
y = – 3 x + c 30
2
From the graph,
pada titik (5, 0) Point of intersection
= (2.5, 1)
0 = – 3 (5) + c
2 y = 6 – 2x y

c = 15 6
2
4 y = 2x – 4
Maka, y = – 3 x + 15
2 2

(b) 6x – 3y = 8

y = 6 x – 8
3 3

= 2x – 8
3

m=2

new equation 2 (2.5, 1) x
y = 2x + c 24
–4 –2 0
at the point (8, –1) –2
–1 = 2(8) + c

c = –17

Therefore, y = 2x – 17

(c) 2y = 3x + 10 –4

y = 3 x + 5
2

m= 3
2

new equation

y = 3 x + c
2

at the point(–2, –1)

–1 = 3 (–2) + c
2

c=2

Therefore, y = 3 x + 2
2

49 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 9 Straight Lines From the graph,
Point of intersection = (2, 0)
(b) y = –x – 2
y
xy
0 –2 y = –x + 2 4
–2 0
2 y = –21 x – 1
y = 2x – 2
–4 –2 0 24 x
xy –2
0 –2
10

From the graph,
Point of intersection
= (0, –2)

y

4 y = 2x – 2 13. (a) y = 2x – 1 ................(1)
2 y = 3 – x ..................(2)

(1) – (2)

y = –x – 2 3x – 4 = 0

x = 1 1
3

–4 –2 0 24 x Substitute x = 1 1 into (2)
–2 3

y = 3 – 1 1 = 1 2
3 3

Therefore the point of intersection is

= 1 1 , 1 2 
3 3
1
(c) y = 2 x – 1

xy (b) 2y = 3x – 2 ................(1)
y = 2x + 1 ................(2)
0 –1
(2) × 2
20 2y = 4x + 2 ................(3)

y = –x + 2 (1) – (3)
0 = –x – 4
xy x = –4
02
20 Substitute x = –4 into (2)
y = 2(–4) + 1
y = –7

Therefore the point of intersection is
= (–4, –7)

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Mathematics Form 3 Chapter 9 Straight Lines

(c) y = 1 x – 3 ..................(1) substitute x= 7 into (3)
2 2

4y = 6 – x ...................(2) y = 8 7  – 26
2
(1) × 4
4y = 2x – 12 ...............(3) y=2

(2) – (3) Therefore, the point of intersection is
0 = –3x + 18
x =6 =  7 , 2
2

Substitute x = 6 into (1) (c) y = 1 x – 16 ....................(1)
5 5
y = 1 (6) – 3 = 0
2
–3x + 7y = –8 ..................(2)
Therefore, the point of intersection is
= (6 , 0) Substitute (1) into (2)

14. (a) 7x – 4y = –7 ..........(1)  –3x+ 7 1 x – 16 = –8
5x + y = 22 ............(2) 5 5

From (2) –3x + 7 x – 112 = –8
y = 22 – 5x ............(3) 5 5

Substitute (3) into (1) – 8 x = 72
7x – 4(22 – 5x) = –7 5 5

7x – 88 + 20x = –7 x = –9
27x = 81
x=3 substitute x = –9 into (2)
–3(–9) + 7y = –8
Substitute x = 3 into (3)
y = 22 – 5(3) y = –5
y=7
Therefore the point of intersection is
Therefore, the point of intersection is = (–9, –5)
= (3, 7)
15. (a) y = –2x + 5,
(b) 4x – 7y = 0 .......................(1) Gradient PQ = –2
8x – y – 26 = 0 .................(2) Equation PQ, y = –2x + c
Substitute x = 2 and y = –1,
From (2) –1 = –2(2) + c
y = 8x – 26 ........................(3) c=3

substitute (3) into (1) Therefore the equation is y = –2x + 3

4x – 7(8x – 26) = 0

4x – 56x +182 = 0

52x = 182

x= 7
2

51 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 9 Straight Lines

(b) Gradient CD = 2– (–2) = 4 Substitute x = 0, y = 4
4– (–1) 5
4 = 2 (0) + c
Gradient AB = y–0 = 4 3
3 – (–3) 5 c =4

y = 4 y = 2 x + 4
6 5 3
4 2
y = 4 4 mQR = – 6 = – 3
5
4
y = 5 x + c Equation of QR is :

At the point (–3, 0 ) y = – 2 x + c
3
substitute x = –3 and y = 0 2
at the point (0, 4), c = 4, y = – 3 x + 4
4
0 = 5 (–3) + c

c = 12 (e) R(–3, 3)
5 mRP = 0
Equation of RP is y = 3
Therefore equation of AB is

y = 4 x + 12 (f) Q(4, 0)
5 5
8
and the equation of AB in general form mPQ = 0 – 4 = –2

is mRS = mPQ = –2

 y = 4 x + 12 ×5 Equaton of RS:
5 5

5y = 4x + 12 y = –2x + c

4x – 5y + 12 = 0 Substitute x = –3, y = 8
8 = –2(–3) + c
(c) Gradient mPQ = 6– 1 = 5 c =8–6=2
7– 6 y = –2x + 2

mOR = mPQ = 5

Equation of OR: y = 5x + c

Substitute x = 0, y = 0

c=0

y = 5x

(d) mPR = undefined
Equation of PR is x = 6

Q (0,4)

mPQ = 8–4 = 2
6 3

Equation of PQ is y = 2 x + c
3

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Mathematics Form 3 Chapter 9 Straight Lines

PT3 Standard Practice 9 Section B

Section A 1. Diagram Equation
(a)
6 – (–4) Rajah Persamaan
8 – (–2)
1. Gradient of KL = y
K
=1 y = – 2 + 2
2 5
Equation of KL 5 Lx
y=x+c O
at L(8, 6)
6=8+c (b) y K(8, 4) y= 1 x
c = –2 x 2
LO
∴y=x–2

Answer: D

2. 4x + 3y = 15 2. Equation

3y = –4x + 15 Persamaan Point

y = – 4 x + 5 (a) y = 3x – 7 Titik
3
(6, 11) ✓
4
∴ m = – 3 , c = 5

Answer: B (b) –5x + y = 2 (–3, –13) ✓
(2, –3) ✗
3. x-intercept, y = 0 (c) 3y = 1 x + 8 (–10, 6) ✗
2x – 3y = 6 2

2x – 3(0) = 6 (d) 1 y + 1 x = 1
2x = 6 3 5
x=3
Section C
∴ x-intercept = 3
1. (a) (i) Rhombus has equal length of sides.
Answer: B OK = KL = LM = MO
OK = (–5)2 + 122
4. mKL = –1, c = –5
= 13 units
∴ equation of the straight line KL is ∴ OM = 13 units
y = –x – 5
M(–13, 0)
Answer: C

5. – 2 = 5 (ii) mKO = 12 – 0
3 –5 – (0)
y x
12
3y – 2x = 15 = – 5

2x – 3y + 15 = 0 y = mx + c

Answer: C y = – 152x + c

Substitute x = 0, y = 0,

c=0

Equation of LM
y = – 152x

(iii) y = 12

53 © Penerbitan Pelangi Sdn. Bhd.

A2

Mathematics Form 3 Chapter 9 Straight Lines

(b) Equation of line PQ 2. (a) (i) k located at line y = 3
Therefore k(a, 3)
y = – 5 x + 5 ..............(1) y = 2x – 4
3 3 = 2(a) – 4
a = 3.5
Equation of line RS

y = x + 3 ....................(2)

(1) = (2), ∴ k(3.5, 3)

 – 5 x + 5 = x + 3 ×3 (ii) x = –4
3
–5x + 15= 3x + 9
(iii) mML = mKN
x= 3 =2
4
y = 2x + c
Substitute x = 3 into (2) Substitute x = –4, y = 0,
4 0 = 2(–4) + c
3 c =8
y= 4 +3 ∴ y = 2x + 8

y = 15
4

∴ k 3 , 15  (iv) Coordinates P = (p, 0)
4 4 Substitute x = p, y = 0 into y = 2x – 4,
0 = 2p – 4
(c) Substitute x = 2, y = 8, p=2
Length of MP = 2 – (–4)
y = k x + 5m = 6 units
3 Area of KLMP = (3 × 6)
= 18 units2
8 = 2k + 5m
3

24 = 2k + 15m ...................(1)

Substitute x = –5, y = –13

y = k x + 5m (b) 2x + 3y – 4 = 0
3
3y = –2x + 4
–13 = –5k + 5m
3 y = – 2 x + 4
3 3
–39 = –5k + 15m ...............(2)
2
(1) – (2) gradient, m = – 3

24 – (–39)= 2k – (–5k)

63 = 7k (c) 5x + 2y = 8

k=9 Divide both sides by 8,

Substitute k = 9 into (1) 5x + 2y = 8
8 8 8
24 = 2(9) + 15m y
x + 4 =1
15m = 6 8

m = 2 5
5
Therefore, x-intercept = 8
∴m = 2 , k =9 5
5 y-intercept = 4

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Mathematics Form 3 Chapter 9 Straight Lines

HOTS Challenge

Types AB

Packaging 10 15
Packet (x) 8 20
Boxes (y)
740 1 350

10x + 8y = 740 ....................(1)
15x + 20y =1 350 ................(2)

From (1)
10x + 8y = 740

x = (740 – 8y) ÷ 10
x = 74 – 0.8y

Substitute (3) into (2)
15(74 – 0.8y) + 20y = 1 350

1 110 – 12y + 20y = 1 350
8y = 240
y = 30

x = 74 – 0.8 × 30
x = 50

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A2

PT3 Model Paper

Section A 5. 4.3 × 10–5 + 0.00174
1. = 4.3 × 10–5 + 1.74 × 10–3
= (0.043 + 1.74) × 10–3
1 +2 2 +2 = 1.783 × 10–3

3 +2 4 +2 Answer: C
=5
    6.
∴ 5n + 2, n = 1, 2, 3, 4, … 10 –3 = 17 3
Answer: A
17 10
2.
8 8, 16, 64  17 3 1
1, 2, 8 10
Answer: B = 2

3. Scale 3
= length of drawing : length of object
= 0.16 cm : 5 m  17 2
= 0.16 : 500
= 1 : 3 125 = 10
Answer: C
Answer: C
4. Answer: D
7. x + 90° + 0.5x + 69° + 2x + 68° = 360°
3.5x + 227° = 360°
3.5x = 133°
x = 38°

Answer: A

8. Answer: B

x1 + x2 y1 + y2
2 2
 9. Midpoint = ,

–6 + (–2) –9 + 3
2 2
 S = ,

= (–4, –3)

Answer: A

10. Coordinates of image
= (–1 + 7, –5 + –5)
= (6, –10)

Answer: C

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Mathematics Form 3 PT3 Model Paper

11. Total time = 2345 – 2015 16. Mean
= 3.5 hours
(12 × 5) + (15 × 3) + (18 × 7) + (21 × 2)

Average speed = 250 = + (24 × 6)
3.5 5+3+7+2+6

= 71.43 km/h     = 417
23
Answer: B
= 18 .13

12. (Cos 60° + sin 45°)(Cos 60° – sin 45°) Answer: A

  1+  2 1 –  2 17. Percentage of students weight more than
2 2 2 60 kg.
2

= – 1 25 + 5
4 15 + 30 + 10 + 15 + 25 + 5
= × 100%
Answer: D

= 30%

13. Price after discount Answer: D

P = RM60 × 80  = RM48 18. The interest Vadivoo has to pay
100
5.85
Q = RM78 × 75  = RM58.50 = RM55 000 × 100  × 2
100
= RM6 435

R = RM53 × 95  = RM50.35 Answer: B
100

S = RM94 × 60  = RM56.40 19. Answer: A
100

∴ The cheapest type of diaper is diaper P 20. K(0, y) J(–2, 6)
Answer: A
Gradient JK = 3

14. ∠UQP = ∠TSQ = 65° 0 y–6   = 3
∠QUP = 180° – 65° – 72° – (–2)
= 43°
y–6=6
Answer: C
y = 12

∴ OK = 12 units

15. Probability of green dresses = 1 – 2  – 1 OL = 132 – 122
5 5 = 5 units

2 ∴ L(5, 0)
5
= x-intercept of KL is 5
Answer: C
2  × Total dresses = 10
5 dresses = 10 × 
5
Total 2

    = 25 pieces

Answer: A

57 © Penerbitan Pelangi Sdn. Bhd.

A3

Mathematics Form 3 PT3 Model Paper 0.5 cm
0.5 cm
Section B

21. (i) K = –2.4 = –2 2
5
4
L = –1.8 = –1 5

(ii) KL – 1.203
= –2.4 (–1.8) – 1.203
= 4.32 – 1.203
= 3.117

22. (a) (i) Line KM / Garis KM 25. (i) ✓
(ii) Line AC / Garis AC (ii) ✗

(b) 6 – 3y  4y – 8
6 + 8  4y + 3y
14  7y

y2

23. (a) Centre (b) (i) FALSE
(b) Chord (ii) TRUE
(c) Major sector
(d) Circumference Section C

24. (a) Scale = 2 cm : 10 cm 26. (a) (i) 1 = 1
=1:5 3 64 4

(b) AB = 36 × 1 = 6 cm    (ii) 2 – 9 2 2 – 2
6 3 3
= 3

BC = EF = 21 × 1 = 3.5 cm = – 7 2
6
3

DE = 9 × 1 = 1.5 cm = 49
6 9

AH = 42 × 1 = 7 cm
6
x2 x –3 1
HE = 2 AB × 1 (b) (i) 8y2 × x –5 = 8y2 = 8x3y2
3 6

= 2 (36) × 1 = 4 cm 1
3 6
(ii) 93k – 1 = 729 × 81 2
3–2

36k – 2 × 3–2 = 36 × 32

36k – 4 = 38

∴ 6k – 4 = 8
6k = 12

k=2

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Mathematics Form 3 PT3 Model Paper

(c) Amount to be paid by instalment c = 7
= RM84.62 18 2
= RM1 523.16
∴ Equation of the straight line is
Amount of interest to be paid 1 7
= RM1 523.16 – RM1 360 y = – 2 x + 2
= RM163.16
28. (a) x + y = 12 ......................(1)
k + 5 + 6 + 6 + 8 + 12 + 12 4y – x = 18 .....................(2)

27. (a) (i) + 12 =8 From (1)
8 k + 61 = 64 y = –x + 12 ...................(3)

k=3 From (2)

(ii) mode = 12 y = x + 18 ......................(4)
4
(iii) 3, 5, 6, 6, 8, 12, 12, 12
6 + 8 (3) = (4)
median, m = 7 2
= x + 18
–x + 12 = 4

k–m =3–7 –4x + 48 = x + 18
= –4
–5x = –30

x=6

(b) x2 – 2x – 3 = x(3 – x) Substitute x = 6 into (1)
x2 – 2x – 3 = 3x – x2 6 + y = 12

2x2 – 5x – 3 = 0 y= 6

2x 1 x ∴ coordinates T is (6, 6)

x –3 –6x

2x2 –3 –5x (b) (i) P(Green pen) =1 – 3 = 4
7 7
(2x + 1)(x – 3) = 0
1
x = – 2 , x = 3 Total number of pens:

4 × Total number of pens = 24
7

(c) 1 x + 1 y = 5 Total number of pens = 24 × 7
4 2 4

1 y = – 1 x + 5 = 42
2 4
Number of pink pens = 42 – 24
y = – 1 x + 10 = 18
2
New probability of pink pen:
∴ Gradient of the line, m = – 1 18 + 2
2 = 42 + 2 + 6

Equation of straight line = 2
–1 5
y = 2 x + c

Substitute x = 3, y = 2

2 = – 1 (3) + c
2

59 © Penerbitan Pelangi Sdn. Bhd.

A3

Mathematics Form 3 PT3 Model Paper

(ii) New probability of green pen: n() = 60
2 3 17 – x + x + 8 + x + 2 + 23 – x + 6 + y = 60
= 1 – 5 = 5 17 – 3 + 3 + 8 + 3 + 2 + 23 – 3 + 6 + y = 60

(c) 2 – p  2p – 4 56 + y = 60
2 + 4  2p + p y=4
3p  6
p2 30. (a) (i) (a) Transformation A is reflection
2p – 4  26 – 3p at line y = 1.

2p + 3p  26 + 4 (b) Transformation B is an
5p  30 enlargement with scale factor
p6 of 2 at centre (– 4, 0).

pϽ6 pϾ2 (ii) ∆STU = 22 × ∆JKL
= 4 × 25 cm2
2 6 = 100 cm2
2ϽpϽ6

Therefore, the possible values of p are (b) (i) z = 1 x–2
3, 4 and 5. 3 y

29. (a) sin x = 8 3z = x–2
17 y

AG = 8 9z2 = x – 2
BG 17 y

16 8 9yz2 = x – 2
BG 17
= x = 9yz2 + 2

BG = 34 cm (ii) x = 9yz2 + 2
DG = BG – BD
 x 1 2
= 34 – 29 3
= 5 cm x
= 9(–1) + 2
= 1

cos y= 5
13
5 31. (a) Volume of small cubes
y = cos–1 13
Volume of sphere
y = 67.4° = 4 851

 = 4 22 (21)3
3 7
(b) Price of juice Kavita has to pay
= (150 × RM0.10) + (200 × RM0.08) 4 851
+ [(1 000 – 150 – 200) × RM0.03]
= RM15 + RM16 + RM19.50 = 8 cm3

= RM50.50 Length of the side of small cube = 38
= 2 cm

(c) n(P  R  Q) = 16 Area of one of the faces of the small
x + 8 + x + 2 = 16 cube
2x = 6
x=3 =2×2
= 4 cm2

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Top One (MATHS F3) Penerbitan Pelangi Sdn Bhd

(b) Average speed = Total distance Mathematics Form 3 PT3 Model Paper
Time
(ii) ∠FBC = 180° – 150°
[1.5 × (10 × 60) + 5 000 + x] = 7 000 = 30°
10 + 32.5 + 17.5 60
∴ x + 130° + 30° = 180°
x = 1 100m x = 20°

∴ Ken walked for 1 100 m. ∠DEB + x = 180°
∠DEB = 180° – 20°
(c) (i) tan ∠KJL = 10.24 = 160°
8
y = 360 – 160°
10.24 = 200°
8
 ∠KJL
= tan–1

∠KJL = 52°

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A3

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