Www.ketam.pja.My -Nota Matematik Tingkatan 4 Dan 5 SPM

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SPM ZOOM-IN(Fully-worked Solutions)

Form 4: Chapter 1 Standard Form

Paper 1 1 0.009495 = 0.00950 (3 sig. fig.)

Answer: D

2 709 000 = 709 000 = 7.09 × 105

Answer: B

3 0.049 + 3 × 10–4

= 4.9 × 10–2 + 0.03 × 102 × 10–4

= 4.9 × 10–2 + 0.03 × 10–2

= (4.9 + 0.03) × 10–2

= 4.93 × 10–2

Answer: A

4 196 × 1010————–25 × 104

= 196 × 106————–

25

= 196 × 106

————–— 25

= 14 × 103———–

5 = 2.8 × 103

Answer: A

5

Suc Math SPM (Zoom IN).indd 1Suc Math SPM (Zoom IN).indd 1 10/9/2008 1:56:47 PM10/9/2008 1:56:47 PM

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Form 4: Chapter 2 Quadratic Expressions and Equations

Paper 1 1 6p2 – p(3 – p) = 6p2 – 3p + p2

= 7p2 – 3p

Answer: C

2 If p = –2 is a root of the equation p2 – kp – 6 = 0, then we substitute p = –2 into p2 – kp – 6 = 0

(–2)2 – k(–2) – 6 = 0 4 + 2k – 6 = 0 2k – 2 = 0 2k = 2

k = 2—2

k = 1

Answer: A

Paper 2

1 3p2 + 10p————

p + 2 = 3

3p2 + 10p = 3(p + 2) 3p2 + 10p = 3p + 6 3p2 + 10p – 3p – 6 = 0 3p2 + 7p – 6 = 0 (3p – 2)(p + 3) = 03p – 2 = 0 or p + 3 = 0 3p = 2 p = –3

p = 2—3

∴ p = 2—3

or –3

2

3(x2 + 9)————

2x = 9

3(x2 + 9) = 9(2x) 3x2 + 27 = 18x 3x2 – 18x + 27 = 0 3(x2 – 6x + 9) = 0 3(x – 3)(x – 3) = 0 x – 3 = 0 ∴ x = 3


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SPM ZOOM-IN(Fully-worked Solutions)Form 4: Chapter 3 Sets

Paper 1 1

A : Regions I and II B : Region I only A’ : Region III only B’ : Regions II and III

Shaded region: Region II only

∴ Shaded region is A � B’ = B’ � A ↓ ↓ I and II II and III

Answer: B

2

A’ : II, III, IVB’ : III, IVC’ : IV

A’ � B’ : III, IV ∴ A’ � B’ = B’A’ � C’ : IV ∴ A’ � C’ = C’

Answer: A

3

n(Q’ � R) = y n(Q’ � P) = 4

n(Q’ � R) – 3 = n(Q’ � P) y – 3 = 4 y = 4 + 3 y = 7

∴ n(R) = 2 + y = 2 + 7 = 9

Answer: B

ξ

Paper 2 1

(a) C : II, III, IV A’ : III, IV, V B’ : I, II, III

C � A’ : II, III, IV, V C � A’ � B’ : II, III

Answer:

(b)

A’ : II, III C’ : I, II A’ � C’ : II (A’ � C’)’ : I, III

Answer:

2 ξ = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}(a) P = {19}(b) Q = { }(c) n(Q) = 0(d) Q’ = {11, 12, 13, 14, 15, 16, 17, 18, 19,

20} P � Q’ = {19} ∴ n(P � Q’) = 1

AB

I IIIII

A � B � C

A BC

III III

IV

P Q R

4 1 2 y

A B

C

III

III

IVV

A B C

I II III

A B C

Since the universal set is less than 22

A B

C

II

III


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Form 4: Chapter 4 Mathematical Reasoning

Paper 2 1 (a) (i) Only common multiples of 6 and 7

are divisible by 7. All other multiples of 6 are not divisible by 7.

∴ Some multiples of 6 are divisible by 7.

(ii) Hexagon means a six-sided polygon. ∴ All hexagons have 6 sides.

(b) The converse of ‘If p, then q’ is ‘If q, then p’. p: k � 4, q: k � 12 ∴ Converse: If k � 12, then k � 4. If k � 12, then k = 13, 14, 15, … All values greater than 12 are greater

than 4 (e.g. 13 � 4). ∴ The converse is true.(c) This is a form III type of argument. Premise 1: If p, then q. Premise 2: Not q is true. Conclusion: Not p is true. p: Set A is a subset of set B. q: A � B = A ∴ Premise 2: A � B ≠ A

2 (a)

Since Q � P, all elements of Q are also elements of P.

∴ Some elements of set Q are elements of set P.

(b) 8 – 15 = –4 is false but 6 × 6 = 65 × 6–3 is true. ↓ ↓

62 = 65 – 3 = 62

To make a compound statement true from one true and one false statement, the word ‘or’ must be used.

∴ 8 – 15 = –4 or 6 × 6 = 65 × 6–3.(c) The argument is a form III type of argument. Premise 1: If p, then q. Premise 2: Not q is true. Conclusion: Not p is true. ∴ q: n = 0, p: 5n = 0 ∴ Premise 1: If 5n = 0, then n = 0.

3 (a) –8 × (–5) = 40 and –9 � –3 ↓ ↓

‘True’ and ‘False’ is ‘False’. ∴ The statement is false.(b) Implication 1: If p, then q. Implication 2: If q, then p. Statement: p if and only if q.

p: x—y

is an improper fraction, q: x � y.

The required statement is x—y

is an

improper fraction if and only if x > y.(c) Argument form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. ∴ p: x � 7, q: x � 2 ∴ Premise 1: If x � 7, then x � 2.

4 (a) If ‘antecedent’, then ‘consequent’.

∴ If 1% = 1—–

100, then 20% of 200 = 40.

(b) Argument form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. p: cos θ = 0.5, q: θ = 60° Premise 2: cos θθ = 0.5

(c) 34

—–32 = 34 – 2

Let 4 = a and 2 = b.

∴ 3a

—–3b = 3a – b

P

Q

Generalisation

A � B is not A.

False statement


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Form 4: Chapter 5 The Straight Line

Paper 1 1 4x + 3y = 12 3y = –4x + 12

y = – 4—3

x + 4

At y-intercept, x = 0 ∴ y = 4 ∴ y-intercept = 4

Answer: D 2 7x + 4y = 5

4y = –7x + 5

y = – 7—4

x + 5—4

y = mx + c

∴ m = – 7—4

∴ Gradient = – 7—4

Answer: B

3 P(–5, –6), Q(–3, 2), R(1, k) m

PQ = m

PR

2 – (–6)————––3 – (–5)

= k – (–6)

————–1 – (–5)

8—2

= k + 6——–

6

k + 6 = 24 k = 18

Answer: D

Paper 2 1 (a) 4x – 9y + 36 = 0

At G, x = 0, ∴ 4(0) – 9y + 36 = 0 9y = 36 y = 4 ∴ G(0, 4)(b) Let the equation of the straight line JK

be y = mx + c. m

JK = m

GH = 4—

9

y = 4—9

x + c

At J�–4 1—2

, 0�, 0 = 4—9 �– 9—

2 � + c

0 = –2 + c ∴ c = 2

∴ y = 4—9

x + 2

or 9y = 4x + 18 or 4x – 9y + 18 = 0

2 (a) O(0, 0), P(2, 6)

mOP

= 6 – 0——–2 – 0

= 3

The gradient of OP is 3.

(b) RQ//OP ∴ m

RQ = m

OP = 3

Let the equation of the straight line QR be y = 3x + c.

At point R(7, 3), y = 3 and x = 7. ∴ 3 = 3(7) + c 3 = 21 + c c = –18 The equation of the straight line QR is

y = 3x – 18.

(c) PQ//OR

∴ mPQ

= mOR

= 3—7

Let the equation of the straight line PQ

be y = 3—7

x + c.

At point P(2, 6), y = 6 and x = 2.

∴ 6 = 3—7

(2) + c

6 = 6—7

+ c

c = 36—–7

The y-intercept of the straight line

PQ is 36—–7

.

P, Q, R are points on a straight line.


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Form 4: Chapter 6 Statistics III

Paper 1

1 Number of guidebooks 2 3 4 5 6

Frequency 3 7 8 10 8

Cumulative frequency 3 10 18 28 36

The mode is 5. Answer: C

2 Score Frequency

1 5

2 4

3 6

4 3

5 2

Mean, x = Σfx

——Σf

=

(1 × 5) + (2 × 4) + (3 × 6) + (4 × 3) + (5 × 2)

———————————— 5 + 4 + 6 + 3 + 2

= 53—–20

= 2.65

The scores higher than the mean (2.65) are 3, 4 and 5 with the frequencies 6, 3 and 2 participants respectively.

Hence, the number of participants getting scores higher than the mean score is 6 + 3 + 2 = 11

Answer: C

Paper 2

1 (a)

Mass (g) Upper boundary Tally Frequency Cumulative

frequency

600 – 619 619.5 2 2

620 – 639 639.5 3 5

640 – 659 659.5 10 15

660 – 679 679.5 12 27

680 – 699 699.5 7 34

700 – 719 719.5 4 38

720 – 739 739.5 2 40

(b) Mass (g) Upper boundary

Cumulative frequency

580 – 599 599.5 0

600 – 619 619.5 2

620 – 639 639.5 5

640 – 659 659.5 15

660 – 679 679.5 27

680 – 699 699.5 34

700 – 719 719.5 38

720 – 739 739.5 40

The ogive is as shown below.

Mode is the value of data with the highest frequency.

Cumulative frequency

O

40

35

30

25

20

15

10

5

Mass (g)599.5 619.5 639.5 659.5 679.5

667.5

699.5 719.5 739.5

(c) From the ogive,

(i) 1—2

× 40 fish = 20 fish

Hence, the median mass = 667.5 g (ii) The median mass means that 50%

(20) of the fish have masses of less than or equal to 667.5 g.

2 (a)

Average marks

Midpoint (x)

TallyFrequency

(f)fx

5 – 9 7 4 28

10 – 14 12 7 84


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Frequency

9

Averagemarks

8

7

6

5

4

3

2

1

09.54.5 19.514.5 24.5 29.5 34.5 39.5 44.5

Average marks

Midpoint (x)

TallyFrequency

(f)fx

15 – 19 17 9 153

20 – 24 22 8 176

25 – 29 27 5 135

30 – 34 32 4 128

35 – 39 37 5 185

40 – 44 42 3 126

Σf = 45 Σfx = 1015

(b) Mean = Σfx

——Σf

= 1015——–

45 = 22 5—

9

(c)

(d) Percentage of students who need to attend extra classes

= 9 + 7 + 4——–——

45 × 100

= 44 4—9

%


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Form 4: Chapter 7 Probability I

Paper 1 1 S = {15, 16, 17, 18, 19, 20, 21, 22, 23, 24,

25, 26, 27, 28, 29, 30} n(S) = 16

A = Event that the sum of digits of the number on the chosen card is even

A = {15, 17, 19, 20, 22, 24, 26, 28} n(A) = 8

P(A) = 8—–16

= 1—2

Answer: A

2 Graduate Non-graduate Total

Male 18

Female 28 4 32

Total 50

The information in the above table is given.

Number of graduate teachers = P(graduate teacher) × Total number of teachers

= 4—5

× 50

= 40

Thus, the table can now be completed, as shown below:

Graduate Non-graduate Total

Male 12 6 18

Female 28 4 32

Total 40 10 50

Hence, the number of male non-graduate teachers is 6.

Answer: A

3 Marks Number of students

1 – 40 h

41 – 70 88

71 – 100 8

P(marks not more than 70) = 14—–15

h + 88——–——–h + 88 + 8

= 14—–15

h + 88——–—h + 96

= 14—–15

15(h + 88) = 14(h + 96) 15h + 1320 = 14h + 1344 15h – 14h = 1344 – 1320 h = 24

Answer: A


Bab 8 tidak ada

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Form 4: Chapter 9 Trigonometry II

Paper 1 1 tan θ = –1.7321 Basic ∠ = 60° θ = 360° – 60° θ = 300°

Answer: C

2 tan y° = 24—–7

QS—–QR

= 24—–7

QS—–14

= 24—–7

QS = 24—–7

× 14 QS = 48 cm

QT = 1—4

QS

QT = 1—4

(48)

QT = 12 cm

PT = PQ2 + QT 2

PT = 52 + 122

PT = 13 cm

300°

60°x

cos x° = –cos ∠PTQ

= –QT—–PT

= – 12—–13

Answer: B

3 The information on special angles of the unit circle is used to draw the graph of y = tan x°. Therefore, the graph of y = tan x° is D.

Answer: D

tan 90° = ∞

tan 180° = 0 tan 0° = 0tan 360° = 0

tan 270° = –∞

180°

90°

0°360°

270°

O

y

O 90° 180° 270° 360°x


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Form 4: Chapter 10 Angles of Elevation and Depression

Paper 1 1

Answer: A 2 CB—–

AB = tan 16°

CB = AB × tan 16° = 35 × tan 16° = 10.0361

∴ The height of the pole, CB, is 10 m, correct to the nearest integer.

Answer: A

3 XW—–4.2

= tan 53°

XW = 4.2 × tan 53°

YW—–4.2

= tan 19°

YW = 4.2 × tan 19°

XY = XW – YW = 4.2(tan 53°) – 4.2(tan 19°) = 4.2(tan 53° – tan 19°) = 4.127 m = 4.1 m (correct to one decimal place)

Answer: B

R T

US

Bird

Cat

Angle of depression = ∠TRS

AB

C

35 m16°

X

Y

W V4.2 m

53°

19°


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Form 4: Chapter 11 Lines and Planes in 3-Dimensions

Paper 1 1

The angle between the line SM and the plane PTWS is ∠MSN,

where MN – Normal to the plane PTWS SN – Orthogonal projection on the

plane PTWS

Answer: B

2

The angle between the plane HGB and the plane DHGC is ∠BGC.

Answer: D

Paper 2 1

The angle between the line PM and the plane PSTU is ∠NPM.

In �NUP, using the Pythagoras’ Theorem,

NP = 42 + 62 = 52 = 7.2111 cm

In �NMP, tan ∠NPM = NM—––NP

tan ∠NPM = 8—–––7.2111

tan ∠NPM = 1.1094 ∠NPM = 47°58’

2

The angle between the plane SABM and the plane SDCR is ∠ASD.

Based on �SDA,

tan ∠ASD = AD—––SD

tan ∠ASD = 4—–20

∠ASD = 11°19’

P

Q R

S

T

U V

WN

M

A B

CDE F

GHJ

• The line of intersection of the planes HGB and DHGC is HG.

• The line that lies on the plane DHGC and is perpendicular to the line of intersection HG is GC.

• The line that lies on the plane HGB and is perpendicular to the line of intersection HG is GB.

• Hence, the angle between the plane HGB and the plane DHGC is the angle between the lines GC and GB, i.e. ∠BGC.

S

P Q

RS

T

U V

W

N M

6 cm

4 cm

8 cm

8 cm

A B

CD

P Q

RS M

20 cm

4 cm

• The line of intersection of the planes SABM and SDCRM is SM.

• The line that lies on the plane SABM and is perpendicular to the line of intersection (SM) is SA.

• The line that lies on the plane SDCR and is perpendicular to the line of intersection (SM) is SD.

• The angle between the plane SABM and the plane SDCR is the angle between the lines SA and SD, i.e. ∠ASD.

A D4 cm

20 cm

The angle between the line SM and its orthogonal projection (SN) is ∠MSN.


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Form 5: Chapter 1 Number Bases

Paper 1 1

Answer: B

2 1 1 0 1 0 02

– 1 1 12

0 10

1 1 0 1 0 02

– 1 1 12

0 10 10

1 1 0 1 0 0 – 1 1 12

0 12

0 10 10 10 10

1 1 0 1 0 02

– 1 1 12

1 0 1 1 0 12

Answer: C

1 1 1 1

1 1 1 1 12

+ 1 1 1 1 12

1 1 1 1 1 02

12 + 12 = 102

12 + 12 + 12 = 112

102 – 12 = 12

102 – 12 = 12

102 – 12 = 12

12 – 12 = 0

52 51 50

1 1 3

3 52 + 5 + 3 = 1 × 52 + 1 × 51 + 3 × 5°

∴ 52 + 5 + 3 = 1135

Answer: C

4 83 + 5 = 1 × 83 + 0 × 82 + 0 × 81 + 5 × 80

∴ 83 + 5 = 10058

Answer: A

5 1 110 111 011 0002

421 421 421 421 421

1 6 7 3 0

∴ 11101110110002 = 167308

Answer: D

83 82 81 80

1 0 0 5

⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩

⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩ ⎧ ⎨ ⎩1 1

1

12 – 12 = 0


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16y = – ––– x


Form 5: Chapter 2 Graphs of Functions II

Paper 1 1 y = ax2

The greater the value of ‘a’, the graph will be closer to the y-axis.

∴ When a = 5, it is graph I,

a = 1, it is graph II and

a = 1—2

, it is graph III.

∴ I: a = 5, II: a = 1, III: a = 1—2

Answer: D

2 y = – 18—–x

is a reciprocal graph.

B is a quadratic graph.C and D are cubic graphs.

Answer: A

Paper 2 1 (a) Substitute x = –2, y = k

into y = 2x2 – 3x – 5. k = 2(–2)2 – 3(–2) – 5 k = 8 + 6 – 5 k = 9

Substitute x = 3, y = m into y = 2x2 – 3x – 5. m = 2(3)2 – 3(3) – 5 m = 18 – 9 – 5 m = 4

(b)

(c) From the graph, (i) when x = –1.5, y = 4, (ii) when y = 20, x = 4.35.

(d) To find the equation of the suitable straight line to be drawn, do the following:

y = 2x2 – 3x – 5 ......➀ 0 = 2x2 + 3x – 17......➁

➀ – ➁: y = –6x + 12 Draw the straight line y = –6x + 12 by

plotting the following points: When x = 0, y = –6(0) + 12 = 12. ∴ Plot (0, 12). When x = 1, y = –6(1) + 12 = 6. ∴ Plot (1, 6). When x = 2, y = –6(2) + 12 = 0. ∴ Plot (2, 0).

The solution from the graph is x = 2.25.

2 (a) Substitute x = –2 into y = – 16—–x

, then

y = –16——–2

= 8

Substitute x = 3 into y = – 16—–x

, then

y = –16——3

= –5.3

(b), (d)

(c) y = – 16—–x

......➀

0 = – 16—–x

+ 2x ......➁

➀ – ➁: y = –2x Draw the straight line y = –2x by

plotting the following points: When x = 0, y = –2(0) = 0. ∴ Plot (0, 0). When x = 1, y = –2(1) = –2. ∴ Plot (1, –2). When x = –1, y = –2(–1) = 2. ∴ Plot (–1, 2). From the graph, the solutions are

x = 2.85 and x = –2.85.

30

25

20

15

10

5

x

y

4

–2 –1

–1.5

O 1 2

2.25

3 4

4.35

5

–5

y = –6x + 12

y = 2x2 – 3x – 5

Graph drawn.

Equation that has to be solved such that 2x 2 + 3x – 17 = 0 is rearranged.

x

y

20

15

10

5

–4 –3

–2.85

–2 –1 O 1 22.85

3 4

–5

–10

–15 16y = – ––– x

x = 1y = 5x + 5

y = 5

y = –2x

Graph drawn.

Equation that has to be solved such that –

16—–x + 2x = 0

is rearranged.


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Form 5: Chapter 3 Transformations III

Paper 2

1 (a) (i) H(4, 4) ⎯→ H’(6, 1) ⎯→ H’’(0, 1)

(ii) H(4, 4) ⎯→ H’(2, 4) ⎯→ H’’(4, 1)

(b) X – Translation 5� � 3

Y – Anticlockwise rotation of 90° about the point N(7, 10)

(c) (i) Scale factor = 2, Centre = (–1, 8) (ii) Area of �EFG = 22 × Area of �ABC 52 = 4 × Area of �ABC Area of �ABC = 13 units2

∴ Area of �LMN = Area of �ABC = 13 units2

W V

V W

2 (a) (i) P(–2, 2) ⎯→ P’(0, 2) ⎯→ P’’(2, 1)

(ii) P(–2, 2) ⎯→ P’(0, 1) ⎯→ P’’(–1, 0)

(b) (i) V – Reflection in the straight line y = x

W – Enlargement with centre (4, –1) and a scale factor of 3

(ii) Area of �DEF = 32 × Area of �LMN 54 = 9 × Area of �LMN Area of �LMN = 6 units2

∴ Area of the shaded region = Area of �DEF – Area of �LMN = 54 – 6 = 48 units2

R T

T R


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Form 5: Chapter 4 Matrices

Paper 1 1 3(6 p) + q(3 –3) = (15 12) (18 3p) + (3q –3q) = (15 12)

∴ 18 + 3q = 15 ......➀ 3q = 15 – 18 3q = –3 q = – 3—

3 q = –1

3p + (–3q) = 12 3p – 3q = 12 ......➁

Substitute q = –1 into ➁:∴ 3p – 3(–1) = 12 3p + 3 = 12 3p = 12 – 3 3p = 9

p = 9—

3 p = 3

∴ p + q = 3 + (–1) = 3 – 1 = 2

Answer: A

2 1� � 7 + p� � 3

= 4� � q

1 + p = 4 7 + 3 = q ∴ p = 4 – 1 ∴ q = 10 p = 3 p × q = 3 × 10 = 30

Answer: C

Paper 2

1 (a) Inverse of 4 –5� � 6 –7

= 1

—————————(4 × –7) – (–5 × 6)

–7 5� � –6 4

= 1

——————–(–28) – (–30)

–7 5� � –6 4

= 1

————–28 + 30

–7 5� � –6 4

= 1—2

–7 5� � –6 4 = k –7 q� � –6 4

∴ k = 1—2

and q = 5

(b) 4 –5� � 6 –7 x� � y

= –2� � 41—2

–7 5� � –6 4 4 –5� � 6 –7

x� � y = 1—

2 –7 5� � –6 4

–2� � 4

1 0� � 0 1 x� � y

= 1—2

(–7 × –2) + (5 × 4)� � (–6 × –2) + (4 × 4)

x� � y = 1—

2 14 + 20� � 12 + 16

= 1—2

34� � 28

= 34—–2� �

28—–2

= 17� � 14

∴ x = 17 and y = 14

2 (a) If no inverse, ad – bc = 0. ∴ (2 × –4) – (4 × d) = 0 –8 – 4d = 0 –4d = 8

d = – 8—4

d = –2

(b) Q = 2 –3� � 4 –4 if d = –3

Q–1 = 1

——————–——–(2 × –4) – (–3 × 4)

–4 3� � –4 2

= 1

———––8 + 12

–4 3� � –4 2

= 1—4

–4 3� � –4 2

= –1 3—4� �

–1 1—2

(c) QP = 2� � 6

2 –3� � 4 –4 a� � b

= 2� � 6

1—4

–4 3� � –4 2 2 –3� � 4 –4

a� � b = 1—

4 –4 3� � –4 2

2� � 6

1 0� � 0 1 a� � b

= 1—4

(–4 × 2) + (3 × 6)� � (–4 × 2) + (2 × 6)

a� � b = 1—

4 –8 + 18� � –8 + 12


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a� � b = 1—

4 10� � 4

= 10—–

4� � 1

= 5—

2� � 1

∴ a = 5—2 or 2 1—

2 , b = 1


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Form 5: Chapter 5 Variations

Paper 1 1 Given s ∝ 1—–

r2,

∴ s = k—–r2

When r = 2 and s = 5,

5 = k—–22

k = 20

∴ s = 20—–r2

Answer: D

2 s ∝ r3

s = kr3, where k is a constant

When s = 192 and r = 4, 192 = k(4)3

k = 3 ∴ s = 3r3

When s = –24, –24 = 3r3

r3 = –8 r = –2

Answer: B

3 r ∝ s2—–

t

r = ks2—–

t, where k is a constant

When r = 8, s = 2 and t = 3,

8 = k(2)2

—–—3

24 = 4k k = 6

∴ r = 6s2

—–t

When r = 27, s = 6 and t = u,

27 = 6(6)2

—–—u

u = 216—––27

u = 8

Answer: C

k is a constant.


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Form 5: Chapter 6 Gradient and Area Under a Graph

Paper 2 1 (a) Average speed of the lorry for the whole

journey from point P to point Q

= Total distance travelled

———————————Total time taken

= 300——16

= 18 3—4

m s–1

(b) Speed of the car for the whole journey = Gradient of the straight line ABC

= –� Vertical axis———————Horizontal axis �

= –� 300 – 0————10 – 0 �

= –30 m s–1

Hence, the speed of the car for the whole journey from point Q to point P is 30 m s–1.

(c) The point on the distance–time graph when the lorry and the car meet is the intersection point of the graph OBD and the graph ABC, i.e. point B.

Hence, the distance from point Q when the lorry and the car meet is 300 – 60 = 240 m

2 (a) Total distance travelled by the particle for the whole journey is 310 m.

Total area under the graph = 310

(4 × 25) + 1—2

(25 + 40)(4) + 1—2

(t – 8)(40) = 310

100 + 130 + 20(t – 8) = 310 20(t – 8) = 80 t – 8 = 4 t = 12

(b) Rate of speed of the particle from the 4th second to the 8th second

= Gradient of the graph from the 4th second to the 8th second

= 40 – 25————8 – 4

= 3 3—4

m s–2

(c) Average speed of the particle in the first 8 seconds

= Total distance——————–

Total time

= Area under the graph in the first 8 s

——————–——————————8

= 100 + 130—————8

= 28 3—4

m s–1

From (a)


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New value of P(R)


Form 5: Chapter 7 Probability II

Paper 1 1 Let

R = Event of obtaining a round biscuitSq = Event of obtaining a square biscuitT = Event of obtaining a triangular biscuitS = Sample space

P(T) = 1 – P(R) – P(Sq)

P(T) = 1 – 3—7

– 1—4

P(T) = 9—–28

n(T)—–—n(S)

= 9—–28

36—–—n(S)

= 9—–28

9 × n(S) = 36 × 28 n(S) = 36 × 28————

9 n(S) = 112

n(R) + n(Sq) + n(T) = 112 n(R) + n(Sq) + 36 = 112 n(R) + n(Sq) = 112 – 36 n(R) + n(Sq) = 76

Answer: C

2 LetG = Event of obtaining a green discB = Event of obtaining a blue discS = Sample space

P(B) = 1 – P(G) = 1 – 6—–11

= 5—–11

n(B)—–—n(S)

= 5—–11

30—–—n(S)

= 5—–11

5 × n(S) = 30 × 11

n(S) = 30 × 11————5

n(S) = 66

∴ n(G) = n(S) – n(B) = 66 – 30 = 36

Answer: A

3 Let B = Event of drawing a blue ball R = Event of drawing a red ball S = Sample space

Given P(R) = 5—8

,

n(R)—–—n(S)

= 5—8

n(R)—–—

32 = 5—

8

n(R) = 5—8

× 32

n(R) = 20

Let the number of blue balls added = hTherefore, n(S) = 32 + h

P(R) = 5—9

n(R)—–—n(S)

= 5—9

20———32 + h

= 5—9

5(32 + h) = 180 160 + 5h = 180 5h = 20 h = 4

Hence, the number of new blue balls that have to be added to the bag is 4.

Answer: D

Paper 2 1 (a) P(letter M)

= n(M)—–—n(S)

= 2 + 5——————2 + 5 + 3 + 4

= 7—–14

= 1—2


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(b) P(both the cards drawn are cards with the letter N)

= 7—–14

× 6—–13

= 3—–13

(c) P(both the cards drawn are of different colours) = P(GY or YG) = P(GY) + P(YG)

= � 5—–14

× 9—–13 � + � 9—–

14 × 5—–

13 �

= 45—–91

G – GreenY – Yellow

Initially, there are 5 green cards out of 14 cards.

After 1 green card is taken out, it is left with 13 cards and so there are 9 yellow cards out of the 13 cards.

After 1 card with the letter N is taken out, it is left with 6 cards with the letter N out of the balance of 13 cards.

Initially, there are 7 cards with the letter N out of 14 cards.

2 (a) P(Z) = 1—5

Number of male students from school Z

——————————————Total number of male students

from all the three schools

= 1—5

10—————–k + 22 + 10

= 1—5

k + 32 = 50 k = 18

(b) P(Two students from school Y are of the same gender) = P(MM or FF) = P(MM) + P(FF)

= � 22—–40

× 21—–39 � + � 18—–

40 × 17—–

39 � = 32—–

65


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Form 5: Chapter 8 Bearing

Paper 1 1

Bearing of A from B= 180° + 60°= 240°

Answer: C

2 Label the north direction and write down all the provided information onto the diagram.

Let the bearing of point K from point H be θ.

θ = 360° – 35° – 40° = 285°

Answer: C

3 Label the east and north direction and write down all the provided information onto the diagram.

From the above diagram, the bearing of point Q from point P is 030°.

Answer: A

North

North

A

B

60°

120°

60°

Bearing of Afrom B

North

North

North

K

F

H

65°65°

35°40°

40°

35°

θ

Bearing of F from K = 065°

Alternate angles are equal.

180° – 100°∠FHK = —————– 2

North

North

P

Q

R

East

30°

60°

30°

30°60°

Alternate angles are equal.


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Form 5: Chapter 9 Earth as a Sphere

Paper 1 1

Based on the above diagram:

The latitude of point D = (90 – 50)°N = 40°N

The longitude of point D = (180° – 32°)E = 148°E

Hence, the location of point D is (40°N, 148°E).

Answer: A

2

The latitude of point R = (90 – 50)°S = 40°SThe longitude of point R = (180 – 80)°W

= 100°W

Hence, the position of point R is (40°S, 100°W).

Answer: B

Paper 2 1

148°

N

O

S

32°

50°40°

0° 148°E

D

N

O

S

PR

0°

40°

80°100°

50°

(a) (i) The latitude of point R is 35°S. (ii) The longitude of point L is

(180 – 70)°E = 110°E(b) Distance of ML (along the parallel of

latitude) = 180 × 60 × cos 35° = 8846.8 n.m.(c) Distance of LMR (via the North Pole) = 180 × 60 = 10 800 n.m.

Average speed

= Distance————Speed

= 10 800———600

= 18 hours

Hence, the time the aeroplane reached point R is 0500 + 1800 = 2300.

2

(a) The position of point Q is (30°S, (180 – 35)°W) = (30°S, 145°W)

(b) (i) JK = 3300 n.m. ∠JOK × 60 = 3300

∠JOK = 3300——–60

∠JOK = 55°

∴ x = 55 – 30 = 25

N

O

S

M L

R

180°

35° N

35°35°

70°W110°E

N

O

S

Q

J

K

L

M

C30°N

35°E

30°30°

30°N

x°S

Equator 0°

J

K

55°


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(ii) JL = 4936 n.m. ∠JCL × 60 × cos 30° = 4936

∠JCL = 4936—————–60 × cos 30°

∠JCL = 95°

∴ y = 95 – 35 = 60

y°W 0° 35°E

L G J

95°

0.23 hours= 0.23 × 60= 13.8= 14 (correct to the nearest minute)

(c) Time taken = Total distance travelled

———————————Speed

= JL + LM————600

=

4936 + [(30 + 40) × 60]———————————

600 =

4936 + 4200——————

600

= 9136———600

= 15.23 hours = 15 hours 14 minutes


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Form 5: Chapter 10 Plans and Elevations

Paper 2 1 (a)

(b) (i)

(ii)

2 (a)

(b) (i)

(ii)

K/Q L/P

A/D B/C

J/G M/N

4 cm

Plan

4 cm

A B

H/J/K M/L

D/G/Q N/D/P E/F/C

4 cm

2 cm4 cm

4 cm

4 cm 4 cm

6 cm

Elevation as viewed from X

5 cmH M/J

B/A

L/K

P/QC/DF/N/GE/D

4 cm

5 cm

4 cm

4 cm

2 cm

Elevation as viewed from Y Elevation as viewed from X

M N W P

QR

S T

I J

D/A C/B

K/F/GL/E/H

2 cm

2 cm

3 cm

1 cm 6 cm 1 cm

C/D B/A

G/HF/EK/L

J/I

2 cm

5 cm

3 cm

4 cm 3 cm

Elevation as viewed from Y

M/A/HN/R/S W/Q/T

P/B/G

C/J/F

KL

D/I/E

1 cm 1 cm

3 cm

4 cm

8 cmPlan


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