nota 2 matematik tingkatan 4 dan 5 spm

Fb: 2011 SPM Tips/Ramalan/Soalan Bocor (Public page)

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Form 4 – Chapter 1 Standard Form Passport To Success

(Fully–worked Solutions)

Paper 1 1 27.035 = 27.0 (3 sig. fig.)

3 5

Answer: B

2 4.23 10–4 = 0.0004.23

= 0.000423

Answer: B

429 000 4.29 105 3 ————– = —————

1.5 10–2 1.5 10–2

4.29 105 = —— —— 10–2 1.5

= 2.86 105 – (–2) = 2.86 107

Answer: B

4 2.35 108 – 2.48 107 = 2.35 108 – 0.248 101 107 = 2.35 108 – 0.248 108 = (2.35 – 0.248) 108 = 2.102 108

Answer: D

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Suc Math SPM (Passport).indd 4 10/7/2008 3:17:36 PM

https://www.facebook.com/pages/2011-SPM-TipsRamalanSoalan-Bocor/150247588399481


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Form 4 – Chapter 2 Quadratic Expressions and Equations Passport To Success


Paper 1 1 3h(1 – h) + (h – 1)2

= 3h – 3h2 + h2 – 2h + 1 = –2h2 + h + 1

Answer: D

Paper 2

2 8 3m = — – 10

m 8 – 10m

3m = ————– m

2 3m = 8 – 10m 2 3m + 10m – 8 = 0

(3m – 2)(m + 4) = 0 2

m = — or –4 3

3 (9p – 1)2 81p2 – 18p + 1 72p2 – 18p + 1

(12p – 1)(6p – 1)

9p2 9p2 0 0

1 1 p = —– or —

12 6

= = = =

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Form 4 – Chapter 3 Sets Passport To Success


Paper 1 1 A = {4, 9}

Set A has 2n = 22 = 4 subsets. The subsets are {4}, {9}, {4, 9}, { }.

Answer: D

2

4 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,

14, 15, 16, 17, 18, 19, 20} P = {6, 11, 16} Q = {1, 2, 3, 4, 6, 8, 12} P’ Q = {1, 2, 3, 4, 8, 12} n(P ’ Q) = 6

Answer: B

Paper 2 5 (a)

K T

15 5 30 P

Q 20

R It is given that n(K T ) = 5.

n(K ) – n(K = 20 – 5 = 15

T) n(T ) – n(K = 35 – 5 = 30

T) (P

(b) P

Q) R

Q

R n(ξ) – n(K T) = 70 – (15 + 5 + 30) = 20

Answer: A

3

P R

Set (P

Union

P R

Q)’

with

Q

Set R

Q

In the above diagram,


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represents the set (Q R)’, and

(b) the shaded region represents the set P.

The intersection of (a) and (b) is the set that is required by the shaded region of the question i.e. (Q R)’ P.

Answer: C

Uniting (P Q)’ and R, we have (P Q)’ R, as shown in the following Venn diagram.

P R

Q

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Form 4 – Chapter 4 Mathematical Reasoning Passport To Success


Paper 2 1 (a) 3m2 + 5m – 2 = 0 is not a statement.

This is because we cannot determine its truth value.

3 (a) (i) 15 3 = 5 and 72 = 14 is false.

‘15 ÷ 3 = 5’ is true. ‘72 = 14’ is false. ‘true’ and ‘false’ is ‘false’.

(b) Premise 1: All sets which contain n elements have 2n subsets.

Premise 2: Set A contains 3 elements. Conclusion: Set A has 23 subsets.

The given argument is a type 1 argument. Premise 1: All P is Q. Premise 2: R is P. Conclusion: R is Q. where P : ‘3 elements’ Q : ‘have 23 subsets’ R : ‘Set A’

1 1 (ii) 24 is a multiple of 6 or — > — is

7 5 true.

‘24 is a multiple of 6’ is true. 1 1

‘— — ’ is false. 7 5

‘true’ or ‘false’ is ‘true’.

(c) 1 = 2(1)3 – 1 15 = 2(2)3 – 1 53 = 2(3)3 – 1 127 = 2(4)3 – 1 The nth term is 2n3 – 1, n = 1, 2, 3, 4, …

2 (a) ‘Some quadratic equations have two distinct roots.’

A quadratic equation may have ‘two distinct roots’, ‘two equal roots’ or ‘no roots’.

(b) Premise 1: If the side a rhombus is 5 cm, then its perimeter is 20 cm.

Premise 2: The side of rhombus P is 5 cm.

Conclusion: The perimeter of rhombus P is 20 cm.

The given argument is a type 2 argument. Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. where p : ‘The side of rhombus P is 5 cm.’ q : ‘The perimeter of rhombus P is 20 cm.’

(b) If x 3, then x < 8.

The converse of the above statement is ‘If x 8, then x 3.’

The converse is false.

When x 8, x = 7, 6, 5, 4,… but x = 7, 6, 5 and 4 is not less than 3.

(c) 5x 10 if and only if x 2. Implication 1: If 5x 10, then x Implication 2: If x 2, then 5x

2. 10.

(c) Premise 1: If set M is a subset of set N, then M N = M.

Premise 2: M NM


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Conclusion: Set M is not a subset of set N.

The given argument is a type 3 argument. Premise 1: If p, then q. Premise 2: Not q. Conclusion: Not p. where p : ‘set M is a subset of set N ’ q : ‘M N = M’

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Form 4 – Chapter 5 The Straight Line Passport To Success


Paper 1 1 2x + 5y = 7

5y = –2x + 7 2 7

y = – —x + — 5 5 2 m = –— 5

Answer: A

2 3x + 6y + 5 = 0 6y = –3x – 5

1 5 y = – —x – —

2 6 5 c = –— 6

Answer: B

Paper 2 3

D(–2, 3)

E G(–2, 0) O

Since DE passes through point D(–2, 3), x = –2 and y = 3.

1 3 = – — (–2) + c

3 2 7

c=3–—=— 3 3

Hence, the equation of DE is

1 7 y = – —x + —.

3 3

At the x-axis, y = 0. 1 7

0 = – —x + — 3 3

0 = –x + 7 x =7 x-intercept = 7

(b) G is point (–2, 0). The equation of GF is y = mx + c, i.e.

1 y = – —x + c

3

x

y

F

1 (a) mDE = mGF = – —

3

The equation of DE is y = mx + c, i.e. 1

y = – —x + c 3

Since GF passes through point G(–2, 0), x = –2 and y = 3.

1 0 = – — (–2) + c

3 2

c = –— 3

Hence, the equation of GF is 1 2

y = –—x – — 3 3

3y = –x – 2

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Form 4 – Chapter 6 Statistics III Passport To Success


Paper 2

1 (a) Distance (km)

21 – 30

31 – 40

41 – 50

51 – 60

61 – 70

71 – 80

81 – 90

Midpoint (x)

25.5

35.5

45.5

55.5

65.5

75.5

85.5

Tally f

2

4

11

10

8

4

1

= 40

2160 (b) x = —–– = —––– = 54 km

40

(c) (i), (ii) 40

35

fx

51.0

142.0

500.5

555.0

524.0

302.0

85.5

= 2160

Class boundaries

20.5 – 30.5

30.5 – 40.5

40.5 – 50.5

50.5 – 60.5

60.5 – 70.5

70.5 – 80.5

80.5 – 90.5

12 Cumulative

frequency 30

25

20

15

10

10

Frequency 8

6

4

2 5

0 29.5 0

20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5 39.5 49.5 59.5 69.5

75 79.5 89.5 99.5

Marks Distance (km)

2 Marks

20 – 29

30 – 39

40 – 49

50 – 59

60 – 69

70 – 79

80 – 89

90 – 99

Upper boundary

29.5

39.5

49.5

59.5

69.5

79.5

89.5


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99.5

Tally f

0

4

5

7

10

7

5

2

Cumulative frequency

0

4

9

16

26

33

38

40

(c) (i) Q3 = 75 3

(ii) The third quartile means — of the 4

students (i.e. 30 students) have marks of 75 and below.

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Bab 7 tidak ada


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Form 4 – Chapter 8 Circles III Passport To Success


Paper 1 1

126° B

54° 50°

27° D

t°

A

E

54° A

27°

2 C

P y°

20° 30°

O 120°

30° x°

B C

Angle in the alternate segment

Q

= = 50°


Angles on a straight line

= = 54°

= 180° –

= 180° – 54° = 126°

180° – = ——————–

2 180° – 126°

= —————– 2

= 27°

t° = = 27°

Answer: A

+ = 50° 20° + = 50°

= 30°

= = 30° OB = OQ

BD = BC

= 180° – –

= 180° – 30° – 30° Angles in a triangle = 120°

y° = ———– = —— = 60°

2 2


The angle subtended by an arc at the centre of a circle is twice the angle at the circumference.

x° = = 60° Angle in the alternate segment

x° + y° = 60° + 60° = 120°

Answer: D

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Form 4 – Chapter 9 Trigonometry II Passport To Success


Paper 1 1

8 4 cos = –cos = – —– = – — 10 5

x°

13 cm

12 cm

T 5 cm

R 10 cm Q y°

P y

13 cm

U

is the basic angle which corresponds to . cos is negative because is an angle in the third quadrant.

V

S

Answer: A

3 The graph of y = cos x for 0° as shown below.

x 180° is

5 sin x° = —

9 5 sin = — 9

RQ —— = SQ 10 —— = SQ

5SQ = SQ =

5 — 9 5 — 9

90 18

1

is the basic angle which corresponds to the obtuse (x °). sin x ° is positive because x ° is an angle in the second quadrant.

180° 0

90°

–1

x

Answer: B

4 cos = –0.4226 Basic = 65°

TQ = SQ – ST = 18 – 13 = 5 cm

In UTQ, based on the Pythagorean triples, TU = 12 cm.

12 tan y° = –tan = – —– 5

is the basic angle which corresponds to the obtuse (y °). tan y ° is negative because y ° is an angle in the second quadrant.

S A

65° 65°

2 1

T C

Answer: D


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y

P(8, 6)

10 6

x

1 = 180° – 65° = 115° 2 = 180° + 65° = 245°

cos is negative in the second and third quadrants.

Answer: A

O T

8 R

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Form 4 – Chapter 10 Angles of Elevation and Depression Passport To Success


Paper 1 1

2

Angle of depression

40°

Angle of elevation

P 40° 15 m

xm P

T 40°

3m 1m R

R

Q

xm

18 m

Q 15 m S In PQR,

In RTP, x

tan 40° = —– 15

x = 15 tan 40°

= 12.586 m RS = 2x = 2 12.586 = 25.17 m

Answer: C

x tan 40° = —–

18 x = 18 tan 40°

= 15.10 m

Height of tree = 15.10 + (3 – 1)

= 17.10 m

Answer: B

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Form 4 – Chapter 11 Lines and Planes in 3–Dimensions Passport To Success


Paper 1 1

Orthogonal projection

J

Normal

M

A N D

G H

12 cm

F 5 cm M 5 cm E

N

Paper 2 3

B 10 cm C

8 cm

K L

The line is KN. The normal is KJ. The orthogonal projection is JN.

The angle between the line KN and the plane NMJ is the angle between the line KN and its orthogonal projection (JN), i.e.

Answer: A

2 P N Q

The line of intersection of the planes ABM and ABCD is AB. is a right angle on the plane ABM. is a right angle on the plane ABCD.

Hence, the angle between the planes ABM and ABCD is

A 5 cm N

D

M

A B

C

8 cm

The line of intersection of the planes NCM and QBC is MC. is a right angle on the plane NCM. is a right angle on the plane QBC.

Hence, the angle between the planes NCM and QBC is

Answer: A

M

Let N be the midpoint of AD. In ANM,

8 tan = — 5


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= 57.99° (or = 57.99°)

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Form 5 – Chapter 1 Number Bases Passport To Success


Paper 1 1

23405 = 2 53 + 3 52 + 4 51 + 0 50

But it is given that: 23405 = 2 53 + 3 52 + y 51 + 0 50

Hence, by comparison, y = 4.

Answer: D

4

2 100

421

4

Answer: D

3 53

2

52

3

51

4

50

05

000 1112

421

0

421

78

8 157

8 19 –5

8 2

0

–3

–2

1 1 1 1 0 0 12

+ 1 1 1 0 12

1 1 0 1 1 02

12 + 12 = 102 12 + 12 + 12 = 112

Answer: A

15710 = 2358

But it is given that 15710 = 2k58. Hence, by comparison, k = 3.

Answer: C

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Form 5 – Chapter 2 Graphs of Functions II Passport To Success


Paper 1 1 The general shape of the cubic graph

y = –2x3 – 9 is

3 (a) When x = –1, y = 8 – (–1)3 = 9 When x = 1.5, y = 8 – (1.5)3 = 4.625

(b)

y

10

8.4 8

The y–intercept of y = –2x3 – 9 is –9. Hence, the graph of y = –2x3 – 9 is as shown below.

y

x –1.0 –0.5 –0.75

6 y = 8 – x3

4

2 1.25

0

–2

–4

0.5 1.0 y = –x

1.5 2.0 2.15

2.5 x

O

–9 –6

–8

Answer: D

Paper 2 2 For y –2x + 4, shade above the straight

line y = –2x + 4 and it should be a solid line.

For y x + 1, shade below the straight line


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y = x + 1 and it should be a solid line.

For x 4, shade to the left of the straight line x = 4 and it should be a dashed line.

The region which satisfies all the given inequalities is as follows.

y

6

4

(c) From the graph, (i) when x = 1.25, y = 6

(ii) when y = 8.4, x = –0.75 Graph drawn (d) y = 8 – x3

3 +0=x –x–8 Given equation y = –x

This is the equation of the straight line which has to be drawn.

From the graph, the value of x which satisfies the equation x3 – x – 8 = 0 is the x-coordinate of the point of intersection of the curve y = 8 – x3 and the straight line y = –x, i.e. x = 2.15.

y 2

= x + 1

–2 O

–2

–4

x 2 4

+4 –2x

y=

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Form 5 – Chapter 3 Transformations III Passport To Success


Paper 2 1

6

(b)

y Centre of rotation R

G H (i) A(1, 3) A’(5, 0) A’’(0, 5)

H G (ii) A(1, 3) A’(3, 1) A’’(7, –2)

4 A

2

B O 2

C

4 6 x

P Q

(c)

8

6

4

y K Q L

y=6 R


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A and P. Construct a perpendicular bisector of the line segment AP. …➀

Draw a line segment to join the points BQ. Construct a perpendicular bisector of the line segment BQ. …➁

(i) The centre of rotation is the point of intersection of the perpendicular bisectors of ➀ and ➁, i.e. (3, 5).

(ii) The angle of rotation is 90° (anticlockwise).

(i)

P 2

Q M

O 2 4 6 8 x

PQR is transformed to KQ’R under transformation V, i.e. reflection in the straight line y = 6.

(ii) KQ’R is transformed to KLM under transformation W, i.e. enlargement with centre (5, 9) and a scale factor of 2.

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Form 5 – Chapter 4 Matrices Passport To Success


Paper 1

1 –2 h – 2 0 2 = –2 3 5 –3 k 9 –2 h – 0 4 = –2 3 5 –6 2k 9

–2 h – 4 = –2 9 5 – 2k 9

h–4=0 h =4

h+k=4–1=3

Answer: C

2 2A – 1 –1

0 = 5 2 4 9 8 2A = 5 2 + 1

9 8 –1 2A = 6 2

8 12 1 A =— 6 2 2 8 12

A = 3 1 4 6

5 – 2k 5–7

–2 k

Hence, by comparison, k = 17 and h = 4.

(b) 3x – 4y = –5 2x + 3y = 8

The matrix equation is

3 –4 x = –5 2 3 y 8

1 x = —– 3 4 –5 17 –2 3 y 8 1 x = —– 17

x 17 34 P y = y

x x = 1 P –1P y

y 2 x I

0 7 0 7 0 7

= = = =

7 2k 2k –1

–5 8 = P –1 –5

8

x = 1, y = 2

0 4

5 (a) Let A =

Answer: D

3 (k 4) 2 0 = –k 7

k(0) + 4(7)) = (–2k 28) =

–2k = k=

(14

(14 (14 14 –7

28)

28) 28)

(k(2) + 4(–k)

2 –1 –6 4

1 A–1 = —————–—– 4 1 2(4) – (–1)(–6) 6 2

1 =— 4 1 2 6 2

1 2 — 2 =

3 1 But it is given that A–1 = 2 h .

3 1 1

Hence, by comparison, h = — . 2

(b) 2m – n = 6 –6m + 4n = –20

The matrix equation is

2 –1 m = 6 –6 4 n –20

1 m = 2 — 6 2

n 3 1 –20

1 m = 2 6 – — (20) 2

n 3 6 – 1(20)

m = 2 n –2

m = 2, n = –2

m n m –1

A A n m

I n m n

A =

–5 = P –1

y 8 x –5 –1

=P y 8

Answer: D

Paper 2

4 (a) PQ = 1 0

PP –1 = 1 0

Q = P –1

1 = —————–— 3 3(3) – (–4)(2) –2

1 = —– 3 17 –2

4 3

h . 3

4 3

0 1 0 1

6 –20

6 –20

6 –20

6 –20

= A–1

= A–1

= A–1

1 But it is given that Q = — 3 k –2

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Form 5 – Chapter 5 Variations Passport To Success


Paper 1 1 y x3

y = kx3, where k is a constant When x = 3, y = 9, 9 = k(3)3

9 k = —–

27 1

k=— 3

1 y = — x3 3

8 When x = k and y = — ,

3 8 = — k3 1 — 3 3

3 k =8 k =2

Answer: A

1 2 Q —— 3 R k

Q = ——, where k is a constant 1 —

R3

Q = kR 1

–— 3

m 3 s— n km

s = —– , where k is a constant n

1 When m = 2 and n = 8, s = —

2 k(2) 1 = —––

— 8 2

4k = 8 k =2

2m s = —– n

When s = 25 and m = 50, 2(50)

25 = —––– n

100 n = —––

25 n =4

Answer: C

Answer: B

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Chapter 6 Gradient and Area Under a Graph Passport To Success


Paper 2 1

Speed (m s–1) 16

400

10 8

P

O

Q

Time (s)

O 5 11 22

Time (min)

2 Distance (m)

d

4 7 n 18

(a) Rate of change of speed from nth s to 5

18th s = – — m s–2 2

10 – 0 5 – ———– = – —

18 – n 2

5(18 – n) 90 – 5n

5n 5n

20 20 90 – 20 70 70

n = —– 5

n = 14

= = = =

(a) The length of time Normala stops for a rest = 11 – 5

Horizontal part of = 6 minutes the graph

(b) Speed in the first 5 minutes 400

Gradient = —–– 5

= 80 m min–1

(c) Average speed

Total distance ——————–

Total time d

—– 22

d

= 30 m min–1

= 30

= 30

= 660

Negative gradient

(b) (i) Length of time the particle travels at a uniform speed =n–7

Horizontal part of = 14 – 7 the graph

=7s

(ii) Average speed in the first 7 s Total distance

= ——————– Total time

Area P + Area Q = ——————–—

Total time 1 1 — (8 + 16)(4) + — (16 + 10)(3) 2 2

= —————————————— 7

87 = —–

7 3

= 12 — m s–1 7

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Form 5 – Chapter 7 Probability I & II Passport To Success


Paper 1 1 Let S – Sample space

A – Event that the card drawn is a factor of 48

A = {6, 12, 24, 16, 3, 4, 8} = {3, 4, 6, 8, 12, 16, 24}

n(A) = 7 7 n(A) P(A) = —––– = —–

12 n(S)

Answer: D

2 Let M – Event that a male fish is chosen F – Event that a female fish is chosen S – Sample space

5 7 P(F) = 1 – —– = —–

12 12 n(F) 7 —––– = —–

12 n(S) 35 7

—––– = —– 12 n(S) 12 n(S) = —– 35 7

= 60

Answer: A

3 Let R B H S

P(B) =

n(H) = = =

– – – –

Event of drawing a red pen Event of drawing a blue pen Event of drawing a black pen Sample space

2 — 50 = 20 5 n(S) – n(R) – n(B) 50 – 18 – 20 12

4 Let M – Event that a male student is chosen F – Event that a female student is chosen H – Event that a student carrying a

handphone is chosen S – Sample space

n(S) = n(M) + n(F) = 24 + 16 = 40 5 n(H) = P(H) n(S) = — 40 = 25 8

Hence, the number of male students who carry handphones = Total number of students who carry

handphones – Number of female students who carry handphones

= 25 – 7 = 18

Answer: D

Paper 2 5 (a) Let R – Event that a red cube is drawn

Y – Event that a yellow cube is drawn P(RR or YY)

2 4 3 6 = — —– + — —– 5 10 5 10

13 Outcomes = —– 5 — 25 R RR 11

5R 2 (b) — 5 R

2R 3Y Jar 3

— 5

6 —

Bowl 11 4 — 11

Y 4R 7Y

7 — Bowl 11

6Y Y

R

RY

YR

Y YY

Answer: A

P(RR or YY) = P(RR) + P(YY) 2 5 3 7 = — —– + — —– 5 11 5 11

31 = —–

55

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Form 5 – Chapter 8 Bearing Passport To Success


Paper 1 1

Bearing of R from T N2

T 42°

N1 S

42° 42°

R P

TS = TR

RS //VT and alternate angles are equal.

V Q

2 N1

50° 20°

130° R

N2

Bearing of P from R

= = 42°

= = 42°

Bearing of R from T

= 180° + 42° = 222°

Answer: C

1QR = 50° It is given that the bearing of R from Q is 050°.

2RQ = 180° – 50° = 130°

QN1//RN2 and the sum of interior angles is 180°.

Bearing of P from R

= 360° – (130° + 20°)


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Answer: C

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Form 5 – Chapter 9 Earth as a Sphere Passport To Success


Paper 1 1

Paper 2 3

Q N N

P

F 65°

G

0°

O 35°

H

35°E

J

M

55°W

42° O

T

S

42°S 10°E

S

Since the longitude of point H is 35°E, = 35°.

Since the difference in longitude between point F and point H is 100°, = 100°. = 100° – 35° = 65°

Therefore, the longitude of point F is 65°W.

Hence, the longitude of point J is (180 – 65)°E = 115°E

Answer: B

2

D H

A 40° O 40° B

F 40°S

S

N

50°N

(a) Longitude of point P = (180 – 55)°E = 125°E

(b) Distance of MT = (55 + 10) 60 cos 42°

= 2898.3 n.m.

(c) Distance of MQ = 4740 n.m.


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4740 = ——— = 79° 60

Hence, the latitude of point Q = (79 – 42)°N = 37°N

Distance of MNP (d) Time = ——————–—–

Speed

180 60 = ————–

660 = 16.36 hours = 16 hours 22 minutes

= 180° because MP is the diameter of the earth.

0.36 hours = 0.36 60 = 22 minutes

Since the difference in latitude between point D and point F is 90°, then = 90° – 50° = 40°

Therefore, = 40° because FOH is the

diameter of the earth.

Hence, the latitude of point H is 40°N.

Answer: A

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Form 5 – Chapter 10 Plans and Elevations Passport To Success


Paper 2 1 (a)

(ii) M/N 3 cm J/R

T/S 2 cm V/U

L 3 cm B/A 4 cm J/M 1 cm N/T

7 cm 5 cm

Q/P

U/P L/V Plan

K/Q

C Elevation as viewed from Y

R/S

(b) (i) L/A/M 3 cm B/J U/T 2 cm 1 cm

V/N

6 cm 5 cm

P/S Q/C/R


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Elevation as viewed from X

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nota 2 matematik tingkatan 4 dan 5 spm

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