Focus SPM Mathematics samplebook by Pelangi Publishing Thailand

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CC038231

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FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books with special features to enhance students’ learning as a whole. This series covers the latest Kurikulum Standard Sekolah Menengah (KSSM) and integrates Sijil Pelajaran Malaysia (SPM) requirements. A great resource for every student indeed!

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MATHEMATICS SPM

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CC038231 ISBN: 978-967-2930-41-9

CVR_Focus_SPM_Maths_titles.indd 4-6

Ng Seng How • Ooi Soo Huat Samantha Neo • Yong Kuan Yeoh

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REVISION REVISI

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SPM

MATHEMATICS

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Format: 190mm X 260mm TP Focus SPM Maths BI pgi

MATHEMATICS Ng Seng How Ooi Soo Huat Samantha Neo Yong Kuan Yeoh

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All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2930-41-9 eISBN: 978-967-431-630-3 (eBook) First Published 2021

Lot 8, Jalan P10/10, Kawasan Perusahaan Bangi, Bandar Baru Bangi, 43650 Bangi, Selangor Darul Ehsan, Malaysia. Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: pelangi@pelangibooks.com Enquiry: customerservice@pelangibooks.com Printed in Malaysia by QPS Press Sdn. Bhd. Lot 52, Jalan 6/2, Seri Kembangan Industrial Area, 43300 Seri Kembangan, Selangor Darul Ehsan. Please log on to www.ePelangi.com/errata for up-to-date adjustments to the contents of the book (where applicable).

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CONTENTS Mathematical Formulae Form r

pte

Cha

Quadratic Functions and Equations in One Variable

2 13

pte

Number Bases

2.1 Number Bases SPM Practice 2

17 24

ter

p Cha

Logical Reasoning

3.1 Statements 3.2 Argument SPM Practice 3

Operations on Sets

4.1 Intersection of Sets 4.2 Union of Sets 4.3 Combined Operation on Sets SPM Practice 4

27 36 45

49 53 59 64

5.1 Network SPM Practice 5

70 81

Linear Inequalities in Two Variables 86 System of Linear Inequalities in Two Variables 91 97 SPM Practice 6 r

pte

Cha

Graphs of Motion

101

7.1 Distance-Time Graphs 7.2 Speed-Time Graphs SPM Practice 7

102 106 113

pte

Cha

Measures of Dispersion for Ungrouped Data

8.1 Dispersion 8.2 Measures of Dispersion SPM Practice 8

118

119 121 135

pte

Cha

Probability of Combined Events

9.1 9.2

Combined Events Dependent Events and Independent Events 9.3 Mutually Exclusive Events and Non-Mutually Exclusive Events 9.4 Application of Probability of Combined Events SPM Practice 9

139

140 141 147 154 157

pte

Network in Graph Theory

Linear Inequalities in Two Variables

6.1 6.2

Cha

te hap

pte

Cha

1.1 Quadratic Functions and Equations SPM Practice 1 Cha

pte

Cha

Consumer Mathematics: Financial Management

10.1 Financial Planning and Management SPM Practice 10

161

162 172

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Form

pte

Cha

pt Cha

Variation

175

1.1 Direct Variation 1.2 Inverse Variation 1.3 Combined Variation SPM Practice 1

176 181 185 188

The Value of Sine, Cosine and Tangent for Angle q, 0° < q < 360° 261 6.2 The Graphs of Sine, Cosine and Tangent Functions 270 SPM Practice 6 277 r

pte

ter

Matrices

2.1 Matrices 2.2 Basic Operation on Matrices SPM Practice 2

191

192 194 206

ter

Consumer Mathematics: Insurance

209

3.1 Risk and Insurance Coverage SPM Practice 3

210 219

pte

Cha

Consumer Mathematics: Taxation

4.1 Taxation SPM Practice 4

Dispersion Measures of Dispersion SPM Practice 7

283

284 294 301

pte

Mathematical Modeling

305

8.1 Mathematical Modeling SPM Practice 8

306 316

SPM Model Paper

318

Answers

335

223 232

pte

Cha

222

Measures of Dispersion for Grouped Data

7.1 7.2

Cha p Cha

260

6.1

Cha p Cha

Ratios and Graphs of Trigonometric Functions

Congruency, Enlargement and Combined Transformations 234

5.1 Congruency 5.2 Enlargement 5.3 Combined Transformation 5.4 Tessellation SPM Practice 5

235 239 247 254 257

iii

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Mathematical Formulae

_ +

– 2 Σfx2 Σf(x – x) = – x–2 Σf Σf

1. n(A  B) = n(A) + n(B) – n(A  B)

21. Variance, σ 2 =

n(A) = n() – n(A)

n(A  B) = n(A  B)

22. Standard deviation, –2 Σx2 –2 Σ(x – x) = –x σ = N N

n(A  B) = n(A  B)

P(A) =

Complement of event A, P(A) = 1 – P(A)

P(A and B) = P(A  B) P(A  B) = P(A) × P(B)

P(A or B) = P(A  B)

10.

P(A  B) = P(A) + P(B) – P(A  B)

11. Σd(v) = 2E; v  V Gradient, m =

13. m = 14. 15. 16. 17. 18. 19. 20.

Vertical distance Horizontal distance

y2 – y1 x2 – x1

y-intercept x-intercept Distance Speed = Time Total distance travelled Average speed = Total time taken Change of speed Acceleration = Change in time Σx x– = N Σfx x– = Σf – 2 Σx2 Σ(x – x) Variance, σ 2 = = – x–2 N N m=–

23. Standard deviation, –2 Σfx2 –2 Σf(x – x) = –x σ = Σf Σf

 

n(A) n(S)

12.

 

24. A–1 =

1 d –b ad – bc –c a

25. Premium =

Face value of policy Premium rate × RMx per RMx

26. Amount of required insurance Percentage of Insurable value = × co-insurance of property 27. Chargeable income Tax = Total annual – – Tax relief exemption income 28. Property assessment tax = Property assessment tax rate × Annual value 29. Quit rent = Quit rent rate per unit area × Total land area 30. Scale factor, k Distance of corresponding point of image from O = Distance of point of object from O Length of corresponding side of image = Length of side of object 31. Area of image = k2 × Area of object 32. Range = Midpoint of the – Midpoint of the highest class lowest class 33. Interquartile range = Third quartile, Q3 − First quartile, Q1

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Learning Area: Relationship and Algebra

r te

p ha

Form 4

Linear Inequalities in Two Variables

You need more exercise.

My weight is 80 kg and my height is 1.7 m. Where is my category in body mass index?

Body Mass Index (BMI) 2.00

eigh

se be O

1.60

rm No

1.70

erw

1.80

Und

Height (m)

1.90

ini

b yo

1.50 1.40 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 Weight (kg)

• Boundary line – Garis sempadan • Half-plane – Separuh satah • Linear inequality – Ketaksamaan linear • Linear inequalities in two variables – Ketaksamaan linear dalam dua pemboleh ubah • Region – Rantau • System of linear inequalities – Sistem ketaksamaan linear • Variable – Pemboleh ubah

Concept map

f linear system o a r o f inventory lution d the so ny makes a in p f als m o o t c d a w materi r of n use tribute ra e manage are ofte is h t d s , le le rs b e p a e m ri a gin exa in two v occur, en ints. For equalities lve some constra ould not w e g ra o Linear in o t v s . es that in l in stock imized and so on inequaliti of capita in m te s is a t w s e o h c so that t way so that the st in the be

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Inequalities in Two 6.1 Linear Variables A

Representing situations in the form of linear inequalities

Linear inequality in two variables is an unequal relation between two variables where the highest power of the variables is one. For example, x + y  1, 2x – y  5.

Form 4

1 Represent the following situations in the form of linear inequalities. (a) Encik Zaki has two children who are studying in Indonesia and Australia respectively. The call rate to Indonesia is RM0.28 per minute while the call rate to Australia is RM0.66 per minute. Encik Zaki wants to make a phone call to both of his children and he has prepaid of RM30 in his handphone. (b) In a written quiz, the correct answer for a multiple choice question will be given 2 marks while the correct answer for a structural question will be given 5 marks. Participants who earn a total score of more than 50 are eligible to the second round. (c) Pak Abu needs 3 days to produce a rattan basket and 5 days to produce a rattan chair. He receives an order to produce x rattan baskets and y rattan chairs that need to be completed in less than 45 days. Solution (a) Let e = the duration of call, in minute made to Indonesia and f = the duration of call, in minute made to Australia. The call rate to Indonesia is RM0.28 per minute while to Australia is RM0.66 per minute. Total prepaid is RM30. Therefore, 0.28e + 0.66f  30. (b) Let x = the number of correct answer of multiple choice questions and y = the number of correct answer of structural questions. The marks for the correct answer for a multiple choice question is 2 while for a structural question is 5. Participants with the total marks of more than 50 is eligible to the second round. Therefore, 2x + 5y  50.

(c) Let x = the number of rattan basket ordered and y = the number of rattan chair ordered. The number of days needed to produce a rattan basket is 3 days while to produce a rattan chair is 5 days. Total number of days to complete the order of rattan basket and rattan chairs is less than 45 days. Therefore, 3x + 5y  45. Try Question 1 in Try This! 6.1

Verifying the conjecture about the points in the region and the solution of certain linear inequalities

1. The straight line drawn on a Cartesian plane divides the plane into two regions (half-plane). The straight line is called boundary line. y Region above

y = mx + c

Region below

y Region above y=h O Region below

Region below

Region above y = mx + c

Region on the left O

Region on the right

x=k

2. Observe the following diagram. The straight line y = x – 1 is a straight line that divides the Cartesian plane into two regions.. (a) All points that lie on the straight line, for example, E(–3, –4), satisfy the equation y = x – 1. (b) All points that lie in the region above the straight line, for example, A(–4, 0), satisfy the equation y . x – 1. (c) All points that lie in the region below the straight line, for example, C(5, –2), satisfy the equation y , x – 1.

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables Point A (–4, 0)

Point B (2, 3)

y-coordinate p 0 . –4 – 1 q x-coordinate

y-coordinate p 3.2–1 q x-coordinate

A –4

–2

–2 E

–4

2(–6) + 5 =

–7

∴ y = 2x + 5 Therefore, point (–6, –7) lies on the straight line y = 2x + 5.

y=x–1

–6

–7

C D Half-plane

Point C (5, –2)

Point D (0, –3)

y-coordinate p –2 , 5 – 1 q x-coordinate

y-coordinate p –3 , 0 – 1 q x-coordinate

Determine whether each of the following point is the solution for the linear inequalities given. (a) (4, 5); y  –x + 2 (b) (–1, 2); y  3x – 4 (c) (–3, –7); 2x + y  –5 (d) (–3, 3); y  x + 6 Solution (a)



-x + 2

Left side

3. Points that lie on the straight line, for example E(–3, –4) are the solutions for the straight line y = x – 1.

Right side –4 + 2

5 5

-2



2 Determine whether each of the following point lies on the straight line, in the region above or below the straight line y = 2x + 5. (a) (–4, 2) (b) (1, 3) (c) (–6, –7)

Point satisfies the inequality.

Therefore, point (4, 5) is the solution of linear inequality y  –x + 2. (b)

Solution (a) For point (–4, 2) y

2x + 5

2(– 4) + 5





Right side

3(-1) – 4

Point does not satisfy the inequality.

Therefore, point (–1, 2) is not the solution of linear inequality y  3x – 4.

Therefore, point (–4, 2) lies in the region above the straight line y = 2x + 5. 2x + 5

2(1) + 5



∴ y  2x + 5 Therefore, point (1, 3) lies in the region below the straight line y = 2x + 5.

-7



∴ y  2x + 5

3x – 4

Left side

-3

(b) For point (1, 3)

Form 4

2x + 5

–7

y Half-plane

(c)

2x + y



Left side

Right side –5

2(-3) + (–7) -13

–5



–5

Therefore, point (–3, –7) is not the solution of linear inequality 2x + y  –5. 87

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

(d)



x+6

Left side 3 3

Right side -3 + 6 =

Therefore, point (–3, 3) is the solution of linear inequality y  x + 6.

y mx + c

O y mx + c

Form 4

REMEMBER! The symbol  means less than or equal to.

Try Questions 2 – 4 in Try This! 6.1

Example of HOTS Question In a Cartesian plane, there are two points, P and Q, and a straight line R. P(3, –5) lies on the straight line, y = 4 – 3x and point Q(1, 2) lies in the region above the straight line. Both points P and Q are not the solutions of inequality R. Determine the linear inequality R. Solution: Point P lies on the straight line y = 4 – 3x and point Q lies in the region above the straight line y = 4 – 3x. Points P and Q are not the solution of linear inequality R. Therefore, linear inequality R is y  4 – 3x. Try this HOTS Question

On a Certesian plane, there is a straigh line E, y = 2x + 5 and three points K, L and M. Given that point K lies in the region below the straight line E, point L lies on the straight line E and point M lies in the region above the straight line E. Determine the linear inequality E if only point K is the solution.

2. For the graph of straight line y = h, (a) the points that lie on the straight line satisfy the equation y = h. (b) the points that lie in the region above the straight line satisfy the equation y . h. (c) the points that lie in the region below the straight line satisfy the equation y , h. y y h

y h x

O y h

3. For the graph of straight line x = k, (a) the points that lie on the straight line satisfy the equation x = k. (b) the points that lie in the region on the right of the straight line satisfy the equation x . k. (c) the points that lie in the region on the left of the straight line satisfy the equation x , k. y

Answer: y  2x + 5

x k

Determining the region that satisfies a linear inequality

1. For the graph of straight line y = mx + c, (a) the points that lie on the straight line satisfy the equation y = mx + c. (b) the points that lie in the region above the straight line satisfy the equation y . mx + c.

x k

4. If the boundary of the shaded region is a dashed line, then the points on the boundary line are not included in the region that satisfies the inequality given.

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

5. If the boundary of the shaded region is a solid line, then the points on the boundary line are included in the region that satisfies the inequality given. Type of straight line

y x+3

Points on the straight line Not included in the solution region.

Dashed line

 or 

Solution (a)

Solid line Included in the solution region.

 or 

x 2 x

6. The following diagrams shows the region that satisfies the given inequality.

(b)

y x+3 y mx + c

y mx + c

Form 4

Inequality symbol

(b) y > x + 3

O y mx + c

The regions do not include the points lie on the straight line y = mx + c.

5 y y mx + c

y mx + c

O y mx + c

The regions include the points lie on the straight line y = mx + c.

State the inequality that defines each of the following shaded regions. (a) y 1 y = —x – 1 3 0

4 Shade the region that satisfies the given linear inequality.

(b)

x y=2

(a) x , 2 0

y x 2 O

Solution x

1 x – 1 is a solid line and the shaded 3 region lies above the line. Therefore, the 1 inequality is y  x – 1. 3

(a) Line y =

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

(b) Line x + y = 2 is a dashed line and the shaded region lies below the line. Therefore, the inequality is x + y  2. Try Questions 5 – 7 in Try This! 6.1

Form 4

Draw and shade the region that satisfy the following linear inequalities. (a) y  3x + 1 (b) y  –2x + 3 1 (c) y > x + 2 2 Solution (a) y = 3x + 1 • Convert the given x

–1

–2

linear inequality to linear equation form to draw the straight line.

y 6 4

• Check the inequality symbol and draw the straight line. • Shade the region that satisfies the inequality.

y 3x + 1

2 –4

–2

–1

4 y = –2x 3

–2

1. Represent the following situations in the form of linear inequalities. (a) A patient is at risk of having a heart problem if the blood pressure in the ankle, k mm Hg, is less than 90% of the blood pressure in the arm, p mm Hg. (b) Aisyah’s father and Haida’s father gives pocket money several times each month respectively. Each time, Aisyah will get RM5 while Haida gets RM8. Let’s say Aisyah’s father gave her pocket money for a times while Haida’s father gave her pocket money for h times last month. The total amount of cash they received last month was more than RM50. (c) Ananda has a 2.4 m of wood to make a rectangular photo frame for the project of the Living Skills subject. 2. Determine whether each of the following point lies on the straight line, in the region above or in the region below the straight line y = –2x + 7. (a) (1, 5) (b) (2, 4) (c) (0, –3)

4. Determine whether each of the following point is the solution for the linear inequalities given. (a) (3, – 5); y  –3x + 8 (b) (– 6, 10); y  x + 5 (c) (–1, –3); x – 2y  4 (d) (2, 2); 4 + x  3y

6.1

3. Determine whether each of the following points satisfies y = 4 – 5x, y  4 – 5x or y  4 – 5x. (a) (–1, 7) (b) (2, – 6) (c) (3, 0)

(b) y = –2x + 3 x

Try This!

5. Shade the region that satisfies the given inequalities. (a) y  3x + 2

1 x+2 2

–2

y = 3x 2 x

y 4

(b) y  –1

2 y = 1x + 2 2 –2

0 –2

Try Question 8 in Try This! 6.1

y x

O y –1

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

1 x+1 2

(b)

y = 3x – 1

(c)

(d) y  2x + 7

y y = 2x 7 0

1 y = – —x 1 2 x

Form 4

1 y = —x 1 2 x

8. Draw and shade the region that satisfy the following linear inequalities.

(e) y  –x + 1

(a) y  –x – 1

(b) y  2x y = –x 1

1 (c) y  – x – 1 2 1 (d) x–y2 3

6. For each of the following graph, determine whether the shaded region satisfies the inequality given. (a) y  –2x – 5 y 0

y = –2x – 5

(b) x – y  3 y x

0 x–y=3

of Linear Inequalities 6.2 Systems in Two Variables A

Representing situations in the form of system of linear inequalities

1. System of linear inequalities in two variables is a set of two or more linear inequalities with similar variables. For example, x + 2y  3 and y  x – 4 is a system of linear inequalities in two variables that consists of two linear inequalities. 2. The table below shows the linear inequalities used to represent certain situations. Example of situation

7. State the inequality that defines each of the following shaded regions. (a) y 3 y = –—x 2 0

Linear inequality

y is greater than x

y.x

y is less than x

y,x

y is not less than x

y>x

y is not more than x

y<x

y is at least k times x

y > kx

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Example of situation

(ii) Kim does not want to spend more than RM20 000 for the laptop inventory every time.

Linear inequality

y is at most k times x

y < kx

Minimum of y is k

y>k

Maximum of y is k

y<k

Form 4

Sum of x and y is greater than k

x+y.k

Difference between y and x is less than k

y–x,k

Sum of x and y is at most k

x+y<k

y is more than x by at least k

y–x>k

7 Represent each of the following situation in the form of system of linear inequalities.

Given the cost of laptop A is RM2 000 and laptop B is RM2 300. Solution (a) Let x represents the number of notebook and y represents the number of drawing book. 3x + 7y  30

xy

Process

Pineapple jam

Strawberry jam

Cooking (hour)

Cooling (hour)

1 4

5 6

3x +

(c) Kim sells two types of laptop in his shop. He needs to make inventory for the laptops every month. The following is his consideration when preparing the inventory. (i) Inventory of laptop A is at most two times the inventory of laptop B.

11 y  11 4

5 5 x+ y8 4 6

First inequality: Total operation time for cooking process. Second inequality: Total operation time for cooling process.

First inequality: The number of laptop A is at most 2 times the number of laptop B.

2 000x + 2 300y  20 000

3 4

The cooking and cooling section operate for 11 hours and 8 hours daily respectively. In a day, the factory is able to produce x bottles of pineapple jam and y bottles of strawberry jam.

Second inequality: Number of notebook purchased is less than drawing book.

(b) Let x represents the number of bottles for pineapple jam and y represents the number of bottles for strawberry jam.

(a) Alia has a RM30 voucher that can be used at a stationery store. A notebook and a drawing book cost RM3 and RM7 respectively. Alia plans to spend not more than the voucher value. The number of notebooks to be purchased is less than the number of drawing books. (b) A factory produces pineapple and strawberry jam. The table shows the process and time used to produce the jam.

First inequality: Spend not more than the voucher value of RM30.

Second inequality: Inventory of laptop A and laptop B is not more than RM20 000. Try Question 1 in Try This! 6.2

Verifying the conjecture about the points in the region and the solution of linear inequalities system

1. Points that satisfy all linear inequalities lie in the common region of the inequalities. Therefore, the points are known as the solutions of the system of linear inequalities.

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

SPM

An art centre offers two classes, handicraft and drawing. The number of pupils in the handicraft class is x and the number of pupils in the drawing class is y. The maximum number of pupils for both classes is 30 pupils. The number of pupils in handicraft class is at most twice the number of pupils in drawing class.

On the graph below, shade the region that satisfies all the three inequalities y  –x + 7, y  1 x – 2 and 2 x  2. y

(c) From the graph, (i) if the number of pupils in handicraft class is 15, find the number of pupils in drawing class. (ii) can this art centre receive 25 pupils in handicraft class? Give your justification.

2 O

–2

(b)

x + y 30

2y x

10 O

y x=2

y = –x 7

4 2 0

Try This!

4 6 1 y = —x – 2 2

6.2

1. Represent each of the following situations in the form of system of linear inequalities.

y(drawing) 30

4 6 1 y = —x – 2 2

Solution

–2

Solution (a) x + y < 30 ⇒ y < −x + 30 1 x < 2y ⇒ y > x 2

y = –x 7

(a) Write two linear inequalities other than x > 0 and y > 0 that represent the situation above. (b) Draw and shade the region that satisfies above system of linear inequalities.

Highlights

Form 4

x (handicraft)

If the number of pupils in handicraft class is 15, the maximum number of pupils in drawing class = 15.

(ii) No. This is because the value x = 25 is outside the shaded region. Try Questions 6 – 11 in Try This! 6.2

(a) Sekolah Setia holds a charity concert in a hall. Two types of concert ticket, VIP and Special, will be sold among parents. The number of VIP ticket sold is x and the number of Special ticket sold is y. The hall can provide not more than 1 000 seats. The number of Special tickets must exceed twice the number of VIP tickets. (b) Lily uses flour and sugar to bake cakes. Given the mass of sugar, s kg, being used is less than half the mass of flour, t kg. The price of sugar is RM2 per kilogram while the price of flour is RM2.50 per kilogram. The total cost of making a cake is less than RM50. (c) Hassan likes to eat burgers and chicken nuggets. However, Hassan has been advised by doctor that his fat and salt intake for each meal should not exceed 18 g and 1.6 g respectively. The following table shows the fat and salt contain in a burger and chicken nugget.

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

Food

Fat (g)

Salt (g)

Burger

1.2

Chicken nugget

0.2

6. State three inequalities that define the shaded region in the diagram below. y

Let the number of burgers taken is x and the number of nuggets taken is y.

y=x–2 x

0 x 2y = 4

2. Determine whether each of the following points is the solution of the system of linear inequalities. (a) (3, –2); y  –2x + 1, x – 2y  4

Form 4

(b) (–1, 7) ; y  x + 9, y  1 x 3 (c) (7, – 4) ; y  4x , y  –2x + 10

7. State four inequalities that define the shaded region in the diagram below. y y = 2x 2 1 y = —x 2

3. The diagram below shows three straight lines. y

y = 2x 3 y = –2x 3 0

y = –x 5

1 y = – —x – 1 2

8. State three inequalities that define the shaded region in the diagram below. y

Shade the region that satisfies the system of linear 1 inequalities y  – x – 1, y  2x + 3 and y  –2x + 3. 2

y = –2x 4 x 2y = 4

4. On the diagram below, shade the region that satisfies the system of linear inequalities y  2x, y  –x + 2 and y  0.

9. On the diagram below, shade the region that satisfies the system of linear inequalities y  –x + 4, y  2x + 4 and y  –1.

y = 2x

y = –x 2

y = 2x 4

5. On the diagram below, shade the region that 1 satisfies the system of linear inequalities y  x, 3 2y  x – 2, y  –x + 4 and y  0.

–2 –10

y = –x 4 2

6 y = –1

y 1 y = —x 3 0

2y = x – 2 x y = –x 4

10. Mr Goh wants to plant two types of flower trees, C and D in his house area. The table below shows the prices of a tree C and of a tree D. Tree

Price per tree (RM)

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Mathematics SPM Chapter 6 Linear Inequalities in Two Variables

11. A patient needs a minimum of 100 units nutrient A and less than 120 units nutrient B per day. The table below shows the content of both nutrients in two types of food. Food

Nutrient A

Nutrient B

(b) Draw and shade the region that satisfies the above system of linear inequalities.

0.5 unit

0.2 unit

0.3 unit

(c) From the graph, (i) determine the maximum number of tree C that can be planted if Mr Goh planted 5 tree D. (ii) can Mr Goh plant 20 tree D? Justify your answer.

Given that the intake of food P is not more than twice the intake of food Q. Let x be the number of intake of food P and y be the number of intake of food Q. (a) Write a system which consists of three linear inequalities that represent the above situation. (b) Using a scale of 2 cm to 100 grams on both axes, construct and shade the region that satisfies the system of linear inequalities. (c) From the graph, if the patient takes 270 grams of food P, find the maximum number of grams of food Q that he can take on that day.

SPM Practice PAPER 1 1.

Nutrient content in every 1 gram

(a) Write two linear inequalities other than x > 0 and y > 0 which represent the above situation.

Nutritionists recommend that fat calories, y kcal, which consumed daily, must be at most 30% of the total calories, x kcal, taken daily. Choose linear inequality that represent the situation above. A y  0.3x B y  30x C y  0.3x D y  30x

2. Which of the points lies below the line y = 4x + 3? A (2, 15) B (–3, 0) C (–1, –2) D (1, 7) 3. Point P(2, –3) satisfies inequality Q. Which of the following is inequality Q? A y7+x B y  4x – 6 C y  1 – 3x D y  5 – 2x

Form 4

Given the number of tree C planted should not more than the number of tree D by 5 and the cost of planting trees must less than RM120. Let the number of tree C planted is x and the number of tree D planted is y.

4. Which of the points is the solution of the linear inequality x – 3y  4? A (1, 3) C (5, –1) B (2, –1) D (–3, –7) 5. Choose the inequality that represents the graph below. y 4 2 –4

–2

–2 –4

A B

y  1 – 2x y  1 – 2x

C D

y  1 – 2x y  1 – 2x

6. Which of the following is the correct statement for the region of inequality y  2x + 7? A Line y = 2x + 7 is a dashed line and the region above the line is shaded. B Line y = 2x + 7 is a solid line and the region below the line is shaded. C Line y = 2x + 7 is a dashed line and the region below the line is shaded. D Line y = 2x + 7 is a solid line and the region above the line is shaded.

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