Practice paper, not an official KPM paper

Ujian Akhir Sesi AkademikEnd of Academic Session Test

MatematikMathematics

Tingkatan 3, 2 jam / 2 hours, 100 marks

Nama / Name

Kelas / Class

Tarikh / Date

Markah / Marks: / 100

Instructions

Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C.This question paper consists of three sections: Section A, Section B and Section C.
Jawab semua soalan.Answer all questions.
Tunjukkan langkah-langkah penting dalam kerja mengira anda.Show your working. It may help you to get marks.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams in this paper are not drawn to scale unless stated.
Anda dibenarkan menggunakan kalkulator saintifik.You may use a scientific calculator.

Answer schemes are in English (DLP).

Practice shows answers as you go. Exam times you and hides answers until you submit.

Bahagian ASection A

20 marks

Jawab semua soalan. Setiap soalan diikuti oleh empat pilihan jawapan, A, B, C dan D. Pilih satu jawapan yang terbaik.Answer all questions. Each question is followed by four options, A, B, C and D. Choose the best answer.

Question 1

Nyatakan 7⁻² sebagai satu pecahan.Express 7⁻² as a fraction.

−49−49
114114
149149
4949

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A negative index means the reciprocal: 7⁻² = 1/7² = 1/49. The negative sign of the index does not make the value negative, and 7² is 49, not 14.

For the teacher: Understanding, low difficulty, Chapter 1, standard 1.2.5

Textbook p. 14

Question 2

Hitung nilai bagi 27^2/3 × 2⁻².Calculate the value of 27^2/3 × 2⁻².

9494
9292
3636
136136

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27^2/3 = (∛27)² = 3² = 9 and 2⁻² = 1/4, so the product is 9 × 1/4 = 9/4.

For the teacher: Applying, medium difficulty, Chapter 1, standard 1.2.6

Textbook p. 17

Question 3

Diberi 2^{x + 3} − 2^x = 112. Cari nilai x.Given 2^{x + 3} − 2^x = 112. Find the value of x.

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2^x+3 = 2^x × 2³ = 8 × 2^x, so the left side is 8 × 2^x − 2^x = 7 × 2^x. Then 7 × 2^x = 112, 2^x = 16 = 2⁴, and x = 4.

For the teacher: Applying, high difficulty, Chapter 1, standard 1.2.9

Textbook p. 26

Question 4

Bundarkan 0.007 185 2 kepada 3 angka bererti.Round off 0.007 185 2 to 3 significant figures.

0.0070.007
0.00720.0072
0.007180.00718
0.007190.00719

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The significant figures start at the first non-zero digit: 7, 1, 8. The next digit is 5, so the third figure rounds up from 8 to 9, giving 0.00719. 0.00718 is truncated, not rounded, and 0.0072 has only 2 significant figures.

For the teacher: Remembering, low difficulty, Chapter 2, standard 2.1.2

Textbook p. 35

Question 5

Nyatakan 0.000 060 5 dalam bentuk piawai.Write 0.000 060 5 in standard form.

6.05 × 10⁵6.05 × 10⁵
6.05 × 10⁻⁵6.05 × 10⁻⁵
6.5 × 10⁻⁵6.5 × 10⁻⁵
60.5 × 10⁻⁶60.5 × 10⁻⁶

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The decimal point moves 5 places to the right to give 6.05, so the index is −5: 6.05 × 10⁻⁵. 60.5 × 10⁻⁶ has the right value but 60.5 is not between 1 and 10, so it is not in standard form.

For the teacher: Remembering, low difficulty, Chapter 2, standard 2.2.1

Textbook p. 37

Question 6

Panjang sejenis virus ialah 1.2 × 10⁻⁷ m. Berapakah bilangan virus itu yang apabila disusun hujung ke hujung akan menjadi 6 × 10⁻⁴ m?A virus is 1.2 × 10⁻⁷ m long. How many of these viruses placed end to end would measure 6 × 10⁻⁴ m?

7.2 × 10⁻¹¹7.2 × 10⁻¹¹
5 × 10⁻³5 × 10⁻³
5 × 10³5 × 10³
5 × 10⁴5 × 10⁴

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(6 × 10⁻⁴) ÷ (1.2 × 10⁻⁷) = 5 × 10^{−4−(−7)} = 5 × 10³. 7.2 × 10⁻¹¹ comes from multiplying instead of dividing.

For the teacher: Applying, medium difficulty, Chapter 2, standard 2.2.3

Textbook p. 43

Question 7

Mei Ling menyimpan RM2 500 dalam sebuah akaun yang membayar faedah mudah 4% setahun. Berapakah faedah yang diperoleh selepas 3 tahun?Mei Ling saves RM2 500 in an account that pays simple interest of 4% per annum. How much interest does she earn after 3 years?

RM30RM30
RM100RM100
RM200RM200
RM300RM300

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I = Prt = 2 500 × 0.04 × 3 = RM300. RM100 is the interest for one year only.

For the teacher: Applying, low difficulty, Chapter 3, standard 3.1.2

Textbook p. 55

Question 8

Rohaya menyimpan RM8 000 dalam akaun yang membayar faedah 3% setahun dikompaun setiap setengah tahun. Hitung nilai simpanannya selepas 2 tahun.Rohaya deposits RM8 000 in an account that pays interest of 3% per annum compounded half-yearly. Calculate the value of her savings after 2 years.

RM8 490.91RM8 490.91
RM8 487.20RM8 487.20
RM8 480.00RM8 480.00
RM8 960.00RM8 960.00

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MV = 8 000(1 + 0.03/2)^{2 × 2} = 8 000(1.015)⁴ = RM8 490.91. RM8 487.20 is yearly compounding (8 000 × 1.03²), RM8 480.00 is simple interest and RM8 960.00 adds 12% of RM8 000.

For the teacher: Applying, medium difficulty, Chapter 3, standard 3.1.2

Textbook p. 58

Question 9

Pelan sebuah dewan dilukis dengan skala 1 : 500. Panjang dewan itu pada pelan ialah 6 cm. Berapakah panjang sebenar dewan itu?The plan of a hall is drawn to a scale of 1 : 500. The length of the hall on the plan is 6 cm. What is the actual length of the hall?

0.3 m0.3 m
3 m3 m
30 m30 m
300 m300 m

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Actual length = 6 × 500 = 3 000 cm = 30 m. Dividing 3 000 cm by 10 instead of 100 gives 300 m.

For the teacher: Applying, low difficulty, Chapter 4, standard 4.1.3

Textbook p. 93

Question 10

Sebuah peta dilukis dengan skala 1 : 40 000. Luas sebuah taman pada peta itu ialah 6 cm². Berapakah luas sebenar taman itu, dalam km²?A map is drawn to a scale of 1 : 40 000. The area of a park on the map is 6 cm². What is the actual area of the park, in km²?

0.240.24
0.960.96
2.42.4
9.69.6

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1 cm on the map = 40 000 cm = 0.4 km, so 1 cm² = 0.4 × 0.4 = 0.16 km². Area = 6 × 0.16 = 0.96 km². (2.4 uses the linear scale 0.4 without squaring.)

For the teacher: Applying, high difficulty, Chapter 4, standard 4.1.5

Textbook p. 99

Question 11

Dalam segi tiga bersudut tegak PQR, ∠Q = 90°, PQ = 8 cm dan QR = 15 cm. Nyatakan nilai tan ∠P.In the right-angled triangle PQR, ∠Q = 90°, PQ = 8 cm and QR = 15 cm. State the value of tan ∠P.

158158
815815
15171517
817817

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PR = √(8² + 15²) = 17 cm. For angle P, the opposite side is QR = 15 and the adjacent side is PQ = 8, so tan P = 15/8. 8/15 is tan R.

For the teacher: Understanding, low difficulty, Chapter 5, standard 5.1.4

Textbook p. 114

Question 12

Diberi sin θ = 513, dengan keadaan θ ialah sudut tirus. Cari nilai kos θ.Given sin θ = 513, where θ is an acute angle. Find the value of cos θ.

512512
813813
13121312
12131213

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Draw a right-angled triangle with opposite 5 and hypotenuse 13. The adjacent side is √(13² − 5²) = 12, so cos θ = 12/13.

For the teacher: Applying, medium difficulty, Chapter 5, standard 5.1.4

Textbook p. 116

Question 13

Sebatang dawai penyokong sepanjang 6 m diikat dari puncak sebuah tiang tegak ke tanah mengufuk. Dawai itu membuat sudut 65° dengan tanah. Hitung tinggi tiang itu, betul kepada dua tempat perpuluhan.A support wire 6 m long is tied from the top of a vertical pole to the horizontal ground. The wire makes an angle of 65° with the ground. Calculate the height of the pole, correct to two decimal places.

2.54 m2.54 m
5.44 m5.44 m
6.62 m6.62 m
12.87 m12.87 m

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The height is opposite the 65° angle and the wire is the hypotenuse: height = 6 sin 65° = 5.44 m. 2.54 m is 6 cos 65°.

For the teacher: Applying, medium difficulty, Chapter 5, standard 5.1.7

Textbook p. 122

Question 14

Dalam Rajah 1, PQRS ialah sisi empat kitaran. Cari nilai x.In Diagram 1, PQRS is a cyclic quadrilateral. Find the value of x.

Cyclic quadrilateral PQRS in a circle; angle PQR is 108 degrees and angle PSR is x — Rajah 1Diagram 1

36°36°
54°54°
72°72°
108°108°

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Opposite angles of a cyclic quadrilateral add up to 180°: x = 180° − 108° = 72°. 108° is the angle at Q itself.

For the teacher: Understanding, low difficulty, Chapter 6, standard 6.2.2

Textbook p. 147

Question 15

Dalam Rajah 2, PA dan PB ialah tangen kepada bulatan berpusat O di A dan B. ∠APB = 48°. Cari nilai x.In Diagram 2, PA and PB are tangents to the circle with centre O at A and B. ∠APB = 48°. Find the value of x.

Circle with centre O; tangents PA and PB meet at P with angle APB 48 degrees; x is angle OAB — Rajah 2Diagram 2

24°24°
42°42°
48°48°
66°66°

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PA = PB, so triangle PAB is isosceles: ∠PAB = (180° − 48°) ÷ 2 = 66°. OA ⟂ PA, so ∠OAP = 90° and x = ∠OAB = 90° − 66° = 24°. (66° is only ∠PAB.)

For the teacher: Analysing, high difficulty, Chapter 6, standard 6.3.3

Textbook p. 154

Question 16

Rajah 3 menunjukkan sebuah pepejal berbentuk prisma segi tiga yang terletak pada satah mengufuk. Antara berikut, yang manakah pelan pepejal itu?Diagram 3 shows a triangular prism on a horizontal plane. Which of the following is the plan of the solid?

A triangular prism with a triangular front face 6 cm wide and 4 cm high, and length 5 cm, lying on a horizontal plane — Rajah 3Diagram 3

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The plan is the view from above. The prism has a rectangular outline 6 cm by 5 cm, and the ridge (the top edge, in the middle of the triangle) is seen from above as a line down the middle, parallel to the 5 cm edges. The triangle is the front elevation.

For the teacher: Understanding, low difficulty, Chapter 7, standard 7.2.1

Textbook p. 182

Question 17

Sebuah titik X bergerak supaya jaraknya sentiasa 4 cm dari sebuah tembereng garis lurus AB yang panjangnya 6 cm. Apakah lokus bagi titik X?A point X moves so that it is always 4 cm from a straight line segment AB of length 6 cm. What is the locus of X?

Satu bulatan berjejari 4 cm berpusat di tengah-tengah ABA circle of radius 4 cm with its centre at the midpoint of AB
Dua garis lurus yang selari dengan AB, masing-masing 4 cm dari ABTwo straight lines parallel to AB, each 4 cm from AB
Dua garis selari yang dihubungkan oleh dua semibulatan berjejari 4 cm, mengelilingi ABTwo parallel lines joined by two semicircles of radius 4 cm, going around AB
Pembahagi dua sama serenjang bagi ABThe perpendicular bisector of AB

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Points 4 cm from a segment lie on two parallel lines 4 cm from AB (along its length) and on a semicircle of radius 4 cm around each end point A and B, forming a closed curve around AB. Two lines alone would be the locus for a long straight line, not a segment.

For the teacher: Understanding, low difficulty, Chapter 8, standard 8.2.3

Textbook p. 208

Question 18

Apakah kecerunan garis lurus 3x + 2y = 12?What is the gradient of the straight line 3x + 2y = 12?

−3−3
−32−32
3232
66

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Rewrite as y = −(3/2)x + 6. The gradient is the coefficient of x, −3/2. The number 6 is the y-intercept.

For the teacher: Understanding, low difficulty, Chapter 9, standard 9.1.2

Textbook p. 231

Question 19

Sebuah garis lurus melalui titik (2, −1) dan titik (6, 7). Apakah persamaan garis lurus itu?A straight line passes through the points (2, −1) and (6, 7). What is the equation of the line?

y = 2x + 5y = 2x + 5
y = 12x − 2y = 12x − 2
y = −2x + 3y = −2x + 3
y = 2x − 5y = 2x − 5

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m = (7 − (−1)) ÷ (6 − 2) = 2. Using (2, −1): −1 = 2(2) + c, so c = −5, and the equation is y = 2x − 5. Check with (6, 7): 2(6) − 5 = 7.

For the teacher: Applying, medium difficulty, Chapter 9, standard 9.1.5

Textbook p. 239

Question 20

Garis lurus hx + 3y = 8 adalah selari dengan garis lurus 2x − y = 5. Cari nilai h.The straight line hx + 3y = 8 is parallel to the straight line 2x − y = 5. Find the value of h.

−6−6
−23−23
2323
66

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2x − y = 5 gives y = 2x − 5, gradient 2. hx + 3y = 8 gives y = −(h/3)x + 8/3, gradient −h/3. Parallel lines have equal gradients: −h/3 = 2, so h = −6.

For the teacher: Applying, high difficulty, Chapter 9, standard 9.1.4

Textbook p. 238

Bahagian BSection B

20 marks

Jawab semua soalan.Answer all questions.

Question 21

Hadi meminjam RM18 000 daripada sebuah bank untuk membeli sebuah motosikal. Bank mengenakan kadar faedah rata 4.5% setahun dan pinjaman itu perlu dijelaskan dalam masa 5 tahun.Hadi borrows RM18 000 from a bank to buy a motorcycle. The bank charges a flat interest rate of 4.5% per annum and the loan must be settled in 5 years.

[2 marks] (a)Hitung jumlah faedah yang perlu dibayar.Calculate the total interest to be paid.

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18 000 × 0.045 × 5 [1]1
RM4 050 [1]1

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[2 marks] (b)Hitung ansuran bulanan yang perlu dibayar oleh Hadi.Calculate Hadi's monthly instalment.

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(18 000 + 4 050) ÷ 60 [1]1
RM367.50 [1]1

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For the teacher: Applying, low difficulty, Chapter 3, standard 3.2.5

Textbook p. 79

Question 22

Sebuah peta dilukis dengan skala 1 : 50 000. Jarak antara Pekan Rimba dengan Bandar Indah pada peta itu ialah 6.4 cm.A map is drawn to a scale of 1 : 50 000. The distance between Pekan Rimba and Bandar Indah on the map is 6.4 cm.

[2 marks] (a)Hitung jarak sebenar, dalam km, antara kedua-dua tempat itu.Calculate the actual distance, in km, between the two places.

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6.4 × 50 000 = 320 000 cm [1]1
3.2 km [1]1

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[2 marks] (b)Sebuah kereta bergerak dari Pekan Rimba ke Bandar Indah dengan purata laju 48 km/j. Hitung masa perjalanan itu, dalam minit.A car travels from Pekan Rimba to Bandar Indah at an average speed of 48 km/h. Calculate the journey time, in minutes.

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3.2 ÷ 48 = 1/15 hour [1]1
4 minutes [1]1

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For the teacher: Applying, medium difficulty, Chapter 4, standard 4.1.5

Textbook p. 101

Question 23

Sebuah papan segi empat tepat PQRS berukuran 12 cm × 5 cm diletakkan supaya sisi PQ terletak pada satah mengufuk dan sisi SR berada 3 cm di atas satah mengufuk itu. S′ dan R′ ialah unjuran ortogon bagi S dan R pada satah mengufuk. Rajah 4 menunjukkan pandangan sisi papan itu, dengan PS = 5 cm.A rectangular board PQRS measuring 12 cm × 5 cm is placed so that side PQ lies on the horizontal plane and side SR is 3 cm above the horizontal plane. S′ and R′ are the orthogonal projections of S and R on the horizontal plane. Diagram 4 shows the side view of the board, with PS = 5 cm.

Side view of a board PS leaning with S 3 cm above the ground, PS 5 cm, S prime on the ground directly below S — Rajah 4Diagram 4

[1 mark] (a)Namakan unjuran ortogon bagi garis PS pada satah mengufuk.Name the orthogonal projection of the line PS on the horizontal plane.

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PS′1

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[2 marks] (b)Hitung panjang PS′.Calculate the length of PS′.

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PS′ = √(5² − 3²) [1]1
4 cm [1]1

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[1 mark] (c)Hitung luas unjuran ortogon bagi PQRS pada satah mengufuk.Calculate the area of the orthogonal projection of PQRS on the horizontal plane.

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12 × 4 = 48 cm² [1]1

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For the teacher: Applying, medium difficulty, Chapter 7, standard 7.1.2

Textbook p. 180

Question 24

Titik A dan titik B terletak 8 cm antara satu sama lain. Titik P bergerak supaya jaraknya sentiasa sama dari A dan dari B. Titik P juga berada 5 cm dari A.Points A and B are 8 cm apart. A point P moves so that it is always the same distance from A and from B. P is also 5 cm from A.

[1 mark] (a)Nyatakan lokus bagi titik yang sama jarak dari A dan B.State the locus of points that are equidistant from A and B.

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The perpendicular bisector of AB [1]1

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[1 mark] (b)Nyatakan lokus bagi titik yang berjarak 5 cm dari A.State the locus of points that are 5 cm from A.

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A circle with centre A and radius 5 cm [1]1

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[1 mark] (c)Berapakah bilangan kedudukan yang mungkin bagi P?How many possible positions of P are there?

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[1 mark] (d)Hitung jarak, dalam cm, setiap kedudukan P dari garis AB.Calculate the distance, in cm, of each position of P from the line AB.

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√(5² − 4²) = 3 cm [1] (P is above the midpoint of AB, which is 4 cm from A)1

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For the teacher: Applying, low difficulty, Chapter 8, standard 8.2.6

Textbook p. 214

Question 25

Rajah 5 menunjukkan sebuah bangunan TB yang tingginya 40 m. Sudut tunduk sebuah kereta di C dari T ialah 32°. Kereta itu kemudian bergerak ke D, yang lebih dekat dengan bangunan itu, dan sudut tunduk dari T menjadi 50°.Diagram 5 shows a building TB that is 40 m tall. The angle of depression of a car at C from T is 32°. The car then moves to D, nearer to the building, and the angle of depression from T becomes 50°.

Building TB 40 m tall with a car at C and nearer at D; angles of depression 32 degrees and 50 degrees from T — Rajah 5Diagram 5

[2 marks] (a)Hitung BC, betul kepada dua tempat perpuluhan.Calculate BC, correct to two decimal places.

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∠TCB = 32° (alternate angles), tan 32° = 40 / BC [1]1
BC = 40 ÷ tan 32° = 64.01 m [1]1

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[2 marks] (b)Hitung jarak CD, dalam m.Calculate the distance CD, in m.

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BD = 40 ÷ tan 50° = 33.56 m [1]1
CD = 64.01 − 33.56 = 30.45 m [1]1

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For the teacher: Applying, high difficulty, Chapter 5, standard 5.1.7

Textbook p. 123

Bahagian CSection C

60 marks

Jawab semua soalan.Answer all questions.

Question 26

Rajah 6 menunjukkan sebuah bulatan yang melalui titik P, Q, R dan S. TPU ialah tangen kepada bulatan itu di P. Diberi ∠TPS = 50°, ∠SPQ = 92° dan ∠SQR = 32°.Diagram 6 shows a circle that passes through the points P, Q, R and S. TPU is a tangent to the circle at P. Given ∠TPS = 50°, ∠SPQ = 92° and ∠SQR = 32°.

Circle through P, Q, R, S with tangent TPU at P; angle TPS 50 degrees, angle SPQ 92 degrees, angle SQR 32 degrees — Rajah 6Diagram 6

[2 marks] (a)Cari ∠PQS. Berikan satu sebab bagi jawapan anda.Find ∠PQS. Give a reason for your answer.

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∠PQS = 50° [1]1
Angle in the alternate segment (equal to the angle between the tangent and the chord PS) [1]1

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[2 marks] (b)Cari ∠QRS.Find ∠QRS.

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Opposite angles of a cyclic quadrilateral add up to 180° [1]1
∠QRS = 180° − 92° = 88° [1]1

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[2 marks] (c)Cari ∠UPQ.Find ∠UPQ.

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Angles on the straight line TPU: 50° + 92° + ∠UPQ = 180° [1]1
∠UPQ = 38° [1]1

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[1 mark] (d)Cari ∠PQR.Find ∠PQR.

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∠PQR = 50° + 32° = 82° [1]1

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[3 marks] (e)Aiman berkata bahawa PR ialah diameter bulatan itu. Adakah Aiman betul? Berikan alasan anda.Aiman says that PR is a diameter of the circle. Is Aiman correct? Give your reason.

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If PR were a diameter, ∠PQR would be 90° (angle in a semicircle) [1]1
∠PQR = 82°, which is not 90° [1]1
Aiman is not correct [1]1

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.3.5

Textbook p. 158

Question 27

Jawab semua bahagian soalan ini.Answer all parts of this question.

[2 marks] (a)Hitung nilai bagi 8^−2/3.Calculate the value of 8^−2/3.

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8^−2/3 = 1 ÷ (∛8)² [1]1
1/4 [1]1

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[3 marks] (b)Ringkaskan (4x⁶y⁻²)^1/2 × (x⁻¹y³)². Berikan jawapan dalam indeks positif.Simplify (4x⁶y⁻²)^1/2 × (x⁻¹y³)². Give your answer in positive indices.

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(4x⁶y⁻²)^1/2 = 2x³y⁻¹ [1]1
(x⁻¹y³)² = x⁻²y⁶ [1]1
2xy⁵ [1]1

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[3 marks] (c)Selesaikan persamaan 3^{x + 1} × 9^{x − 2} = 27.Solve the equation 3^{x + 1} × 9^{x − 2} = 27.

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3^x+1 × 3^2x−4 = 3³ [1]1
3x − 3 = 3 [1]1
x = 2 [1]1

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[2 marks] (d)Hana mendakwa bahawa 2^{n + 2} − 2ⁿ = 2² untuk semua nilai n. Dengan menggunakan n = 3, tunjukkan bahawa dakwaan Hana tidak betul. Kemudian, nyatakan 2^{n + 2} − 2ⁿ sebagai gandaan 2ⁿ.Hana claims that 2^{n + 2} − 2ⁿ = 2² for all values of n. Using n = 3, show that Hana's claim is not correct. Then express 2^{n + 2} − 2ⁿ as a multiple of 2ⁿ.

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n = 3: 2⁵ − 2³ = 24, but 2² = 4, so the claim is not correct [1]1
2ⁿ(2² − 1) = 3 × 2ⁿ [1]1

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For the teacher: Evaluating, low difficulty, Chapter 1, standard 1.2.9

Textbook p. 26

Question 28

Jawab semua bahagian soalan ini.Answer all parts of this question.

[2 marks] (a)Bundarkan 0.004 037 2 kepada 3 angka bererti. Kemudian, nyatakan jawapan itu dalam bentuk piawai.Round off 0.004 037 2 to 3 significant figures. Then state the answer in standard form.

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0.00404 [1]1
4.04 × 10⁻³ [1]1

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[3 marks] (b)Jisim satu atom emas ialah 3.27 × 10⁻²² g. Hitung bilangan atom emas dalam sebentuk cincin emas yang berjisim 6.54 g. Berikan jawapan dalam bentuk piawai.The mass of one gold atom is 3.27 × 10⁻²² g. Calculate the number of gold atoms in a gold ring of mass 6.54 g. Give your answer in standard form.

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6.54 ÷ (3.27 × 10⁻²²) [1]1
= 2 ÷ 10⁻²² [1]1
2 × 10²² [1]1

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[2 marks] (c)Cahaya bergerak dengan laju 3 × 10⁵ km/s. Jarak Bumi dari Matahari ialah 1.5 × 10⁸ km. Hitung masa, dalam saat, yang diambil oleh cahaya Matahari untuk sampai ke Bumi. Berikan jawapan dalam bentuk piawai.Light travels at a speed of 3 × 10⁵ km/s. The distance from the Earth to the Sun is 1.5 × 10⁸ km. Calculate the time, in seconds, light from the Sun takes to reach the Earth. Give your answer in standard form.

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(1.5 × 10⁸) ÷ (3 × 10⁵) [1]1
5 × 10² s [1]1

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[3 marks] (d)Neptun terletak 4.5 × 10⁹ km dari Matahari. Zara berkata cahaya Matahari mengambil masa lebih daripada 20 kali ganda lebih lama untuk sampai ke Neptun berbanding ke Bumi. Adakah Zara betul? Tunjukkan pengiraan anda.Neptune is 4.5 × 10⁹ km from the Sun. Zara says that sunlight takes more than 20 times as long to reach Neptune as it takes to reach the Earth. Is Zara correct? Show your calculation.

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(4.5 × 10⁹) ÷ (1.5 × 10⁸) [1]1
= 3 × 10¹ = 30 times [1]1
30 > 20, so Zara is correct [1]1

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For the teacher: Evaluating, low difficulty, Chapter 2, standard 2.2.3

Textbook p. 43

Question 29

Rajah 7 menunjukkan segi empat selari PQRS yang dilukis pada suatu satah Cartes.Diagram 7 shows a parallelogram PQRS drawn on a Cartesian plane.

Parallelogram PQRS on a Cartesian plane with P(−3, 0), Q(3, 3), R(5, 7), S(−1, 4) — Rajah 7Diagram 7

[3 marks] (a)Cari persamaan garis lurus PQ.Find the equation of the straight line PQ.

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Gradient = (3 − 0) ÷ (3 − (−3)) = 1/2 [1]1
0 = 1/2(−3) + c, c = 3/2 [1]1
y = 1/2 x + 3/2 [1]1

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[3 marks] (b)Cari persamaan garis lurus SR.Find the equation of the straight line SR.

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SR is parallel to PQ, so its gradient is 1/2 [1]1
4 = 1/2(−1) + c, c = 9/2 [1]1
y = 1/2 x + 9/2 [1]1

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[2 marks] (c)Tunjukkan bahawa PS adalah selari dengan QR.Show that PS is parallel to QR.

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Gradient PS = (4 − 0) ÷ (−1 + 3) = 2 [1]1
Gradient QR = (7 − 3) ÷ (5 − 3) = 2, equal gradients, so PS ∥ QR [1]1

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[2 marks] (d)Garis lurus x + y = 3 bersilang dengan PQ di titik M. Cari koordinat M.The straight line x + y = 3 intersects PQ at the point M. Find the coordinates of M.

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3 − x = 1/2 x + 3/2 (or elimination) [1]1
x = 1, y = 2: M = (1, 2) [1]1

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For the teacher: Analysing, medium difficulty, Chapter 9, standard 9.1.7

Textbook p. 245

Question 30

Rajah 8 menunjukkan sebuah pepejal yang terdiri daripada sebuah kuboid dan sebuah prisma segi tiga yang dilekatkan pada permukaan atas kuboid itu. Tapak kuboid ialah sebuah segi empat tepat 8 cm × 5 cm dan tingginya 3 cm. Keratan rentas prisma ialah sebuah segi tiga sama kaki dengan tapak 8 cm dan tinggi 3 cm.Diagram 8 shows a solid made up of a cuboid and a triangular prism attached to the top surface of the cuboid. The base of the cuboid is a rectangle 8 cm × 5 cm and its height is 3 cm. The cross-section of the prism is an isosceles triangle with base 8 cm and height 3 cm.

A solid: a cuboid 8 cm long, 5 cm deep and 3 cm high with a triangular prism, 3 cm high, on top — Rajah 8Diagram 8

[2 marks] (a)Hitung luas dongakan depan pepejal itu, dalam cm².Calculate the area of the front elevation of the solid, in cm².

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Rectangle 8 × 3 = 24 [1]1
Triangle 1/2 × 8 × 3 = 12, total 36 cm² [1]1

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[2 marks] (b)Pelan pepejal itu ialah sebuah segi empat tepat dengan satu garis di dalamnya. Nyatakan ukuran segi empat tepat itu dan kedudukan garis itu.The plan of the solid is a rectangle with a line inside it. State the measurements of the rectangle and the position of the line.

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Rectangle 8 cm × 5 cm [1]1
The ridge is a line parallel to the 5 cm sides, halfway along the 8 cm side (4 cm from each end) [1]1

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[3 marks] (c)Hitung isi padu pepejal itu, dalam cm³.Calculate the volume of the solid, in cm³.

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Cuboid: 8 × 5 × 3 = 120 [1]1
Prism: 1/2 × 8 × 3 × 5 = 60 [1]1
180 cm³ [1]1

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[3 marks] (d)Rizal berkata bahawa dongakan sisi pepejal itu (dilihat dari sebelah kanan) mempunyai luas yang sama dengan dongakan depan. Adakah Rizal betul? Tunjukkan pengiraan anda.Rizal says that the side elevation of the solid (viewed from the right) has the same area as the front elevation. Is Rizal correct? Show your calculation.

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Side elevation is a rectangle 5 cm wide and (3 + 3) = 6 cm high [1]1
Area = 30 cm², front elevation 36 cm² [1]1
The areas are not equal, so Rizal is not correct [1]1

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For the teacher: Analysing, medium difficulty, Chapter 7, standard 7.2.3

Textbook p. 190

Question 31

Jawab semua bahagian soalan ini.Answer all parts of this question.

[3 marks] (a)Aisyah menyimpan RM12 000 dalam sebuah akaun deposit tetap yang membayar faedah 4% setahun dikompaun setiap suku tahun. Hitung jumlah simpanannya pada akhir 3 tahun.Aisyah deposits RM12 000 in a fixed deposit account that pays interest of 4% per annum compounded quarterly. Calculate the total value of her savings at the end of 3 years.

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MV = 12 000(1 + 0.04/4)^{4 × 3} [1]1
= 12 000(1.01)¹² [1]1
RM13 521.90 [1]1

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[2 marks] (b)Raju membeli 2 000 unit amanah saham pada harga RM0.80 seunit. Selepas setahun, dia menjual semua unit itu pada harga RM0.95 seunit dan menerima agihan pendapatan RM60. Hitung pulangan pelaburan (ROI), dalam peratus.Raju buys 2 000 unit trust units at RM0.80 each. After a year, he sells all the units at RM0.95 each and receives an income distribution of RM60. Calculate the return on investment (ROI), as a percentage.

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Gain = 2 000 × (0.95 − 0.80) + 60 = RM360 [1]1
ROI = 360 ÷ 1 600 × 100% = 22.5% [1]1

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[2 marks] (c)Pak Salleh ingin membeli sebuah kereta berharga RM30 000 dengan membuat pinjaman. Tawaran A: kadar faedah rata 3.5% setahun selama 5 tahun. Tawaran B: kadar faedah rata 3.2% setahun selama 6 tahun. Hitung ansuran bulanan bagi setiap tawaran.Pak Salleh wants to buy a car costing RM30 000 with a loan. Offer A: flat interest rate of 3.5% per annum for 5 years. Offer B: flat interest rate of 3.2% per annum for 6 years. Calculate the monthly instalment for each offer.

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A: 30 000 × 1.175 = 35 250; 35 250 ÷ 60 = RM587.50 [1]1
B: 30 000 × 1.192 = 35 760; 35 760 ÷ 72 = RM496.67 [1]1

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[3 marks] (d)Pak Salleh mahu jumlah bayaran balik yang paling rendah, manakala anaknya mahu ansuran bulanan yang paling kecil. Tawaran manakah yang sesuai bagi setiap orang? Berikan sebab anda.Pak Salleh wants the lowest total repayment, while his son wants the smallest monthly instalment. Which offer suits each of them? Give your reasons.

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Total repayment: A = RM35 250, B = RM35 760 [1]1
Pak Salleh: Offer A, because the total repayment is lower [1]1
His son: Offer B, because the monthly instalment (RM496.67) is smaller than A (RM587.50) [1]1

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For the teacher: Evaluating, low difficulty, Chapter 3, standard 3.1.2

Textbook p. 58

Test specification (JSU)

Chapter	Remembering	Understanding	Applying	Analysing	Evaluating	Creating	Items	marks
1. Indices	–	1 (1)	2 (2)	–	1 (10)	–	4	13
2. Standard Form	2 (2)	–	1 (1)	–	1 (10)	–	4	13
3. Consumer Mathematics: Savings and Investments, Credit and Debt	–	–	3 (6)	–	1 (10)	–	4	16
4. Scale Drawings	–	–	3 (6)	–	–	–	3	6
5. Trigonometric Ratios	–	1 (1)	3 (6)	–	–	–	4	7
6. Angles and Tangents of Circles	–	1 (1)	–	2 (11)	–	–	3	12
7. Plans and Elevations	–	1 (1)	1 (4)	1 (10)	–	–	3	15
8. Loci in Two Dimensions	–	1 (1)	1 (4)	–	–	–	2	5
9. Straight Lines	–	1 (1)	2 (2)	1 (10)	–	–	4	13
Total	2 (2)	6 (6)	16 (31)	4 (31)	3 (30)	0 (0)	31	100

Each cell shows items (marks).