Matematik Tingkatan 4 Kertas 2Mathematics Form 4 Paper 2

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Question 1

Skrin sebuah tablet berbentuk segi empat tepat dengan panjang (x + 6) cm dan lebar x cm. Luas skrin itu ialah 91 cm².The screen of a tablet is a rectangle with length (x + 6) cm and width x cm. The area of the screen is 91 cm².

[1 mark] (a)Bentukkan persamaan kuadratik dalam bentuk am.Form a quadratic equation in general form.

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• x² + 6x − 91 = 01

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[3 marks] (b)Selesaikan persamaan itu dengan pemfaktoran dan nyatakan lebar skrin itu.Solve the equation by factorisation and state the width of the screen.

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• (x + 13)(x − 7) = 01
• x = −13 or x = 71
• Reject x = −13 because a length cannot be negative; width = 7 cm1

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For the teacher: Applying, medium difficulty, Chapter 1, standard 1.1.8

Textbook p. 26

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Question 2

Sebuah peranti menyimpan data dalam sistem nombor yang berlainan.A device stores data in different number systems.

[2 marks] (a)Tukarkan 1011010₂ kepada nombor dalam asas lapan.Convert 1011010₂ to a number in base eight.

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• Group in threes from the right: 001 011 0101
• 132₈1

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[2 marks] (b)Hitung 11011₂ + 1101₂. Beri jawapan dalam asas dua.Calculate 11011₂ + 1101₂. Give the answer in base two.

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• Correct vertical addition with carrying (or 27 + 13 = 40)1
• 101000₂1

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[1 mark] (c)Nyatakan nilai digit bagi digit 3 dalam 2306₇, dalam asas sepuluh.State the digit value of the digit 3 in 2306₇, in base ten.

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• 3 × 7² = 1471

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

Textbook p. 41

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Question 3

[1 mark] (a)Nyatakan sama ada pernyataan majmuk ini benar atau palsu: “Sebuah heksagon mempunyai 6 sisi atau 12 ialah nombor ganjil.”State whether this compound statement is true or false: “A hexagon has 6 sides or 12 is an odd number.”

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• True (one part is true, so “or” is true)1

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[2 marks] (b)Tulis akas bagi implikasi “Jika suatu nombor boleh dibahagi tepat dengan 6, maka nombor itu boleh dibahagi tepat dengan 3.” Nyatakan sama ada akas itu benar atau palsu, dan beri satu contoh penyangkal jika palsu.Write the converse of the implication “If a number is divisible by 6, then the number is divisible by 3.” State whether the converse is true or false, and give one counter-example if it is false.

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• Converse: If a number is divisible by 3, then the number is divisible by 6. It is false1
• Counter-example: 9 (or 15, 21)1

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For the teacher: Understanding, low difficulty, Chapter 3, standard 3.1.4

Textbook p. 64

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Question 4

Diberi set semesta ξ = {x : 1 ≤ x ≤ 15, x ialah integer}, set P = {faktor bagi 12} dan set Q = {nombor ganjil}.Given the universal set ξ = {x : 1 ≤ x ≤ 15, x is an integer}, set P = {factors of 12} and set Q = {odd numbers}.

[1 mark] (a)Senaraikan unsur bagi P ∩ Q.List the elements of P ∩ Q.

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• {1, 3}1

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[2 marks] (b)Cari n(P ∪ Q).Find n(P ∪ Q).

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• n(P) = 6, n(Q) = 8, n(P ∩ Q) = 21
• 6 + 8 − 2 = 121

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[1 mark] (c)Senaraikan unsur bagi (P ∪ Q)′.List the elements of (P ∪ Q)′.

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• {8, 10, 14}1

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For the teacher: Applying, low difficulty, Chapter 4, standard 4.2.3

Textbook p. 113

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Question 5

Rajah 1 menunjukkan satu rangkaian yang menghubungkan lima stesen pengecasan, A, B, C, D dan E, bagi kenderaan elektrik.Diagram 1 shows a network that connects five charging stations, A, B, C, D and E, for electric vehicles.

[1 mark] (a)Nyatakan bilangan bucu dan bilangan tepi bagi rangkaian itu.State the number of vertices and the number of edges of the network.

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• 5 vertices and 7 edges (the loop counts as one edge)1

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[2 marks] (b)Cari darjah bucu D dan jumlah darjah semua bucu.Find the degree of vertex D and the sum of the degrees of all vertices.

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• Degree of D = 31
• Sum = 2 + 3 + 3 + 3 + 3 = 141

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[2 marks] (c)Satu pokok dibentuk menggunakan kesemua lima bucu. Nyatakan bilangan tepi pokok itu dan senaraikan satu set tepi yang boleh membentuk pokok itu.A tree is formed using all five vertices. State the number of edges of the tree and list one set of edges that can form it.

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• 4 edges1
• Any valid tree, e.g. AB, BC, CD, DE (no cycle, all vertices joined)1

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For the teacher: Understanding, low difficulty, Chapter 5, standard 5.1.1

Textbook p. 132

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Question 6

[1 mark] (a)Adakah titik (1, −2) memenuhi ketaksamaan y ≤ 2x − 3? Tunjukkan pengiraan anda.Does the point (1, −2) satisfy the inequality y ≤ 2x − 3? Show your working.

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• 2(1) − 3 = −1 and −2 ≤ −1, so it satisfies the inequality1

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[2 marks] (b)Suatu garis lurus 3x + 2y = 12 dilukis sebagai garis sempang. Rantau yang mengandungi asalan berada di sebelah bawah garis itu. Tulis ketaksamaan bagi rantau itu.The straight line 3x + 2y = 12 is drawn as a dashed line. The region that contains the origin is below the line. Write the inequality for the region.

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• Dashed line, so the inequality is strict (< )1
• 3x + 2y < 12 (origin: 0 < 12 is true)1

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For the teacher: Applying, medium difficulty, Chapter 6, standard 6.1.2

Textbook p. 160

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Question 7

Rajah 2 menunjukkan graf laju-masa bagi sebuah kereta dalam tempoh 25 saat.Diagram 2 shows the speed-time graph of a car over 25 seconds.

[1 mark] (a)Hitung pecutan kereta itu dalam 8 saat pertama.Calculate the acceleration of the car in the first 8 seconds.

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• 20 ÷ 8 = 2.5 m s⁻²1

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[2 marks] (b)Hitung jumlah jarak yang dilalui oleh kereta itu.Calculate the total distance travelled by the car.

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• ½ × (12 + 25) × 201
• 370 m1

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[1 mark] (c)Hitung nyahpecutan kereta itu dalam 5 saat terakhir.Calculate the deceleration of the car in the last 5 seconds.

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• 20 ÷ 5 = 4 m s⁻²1

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For the teacher: Applying, medium difficulty, Chapter 7, standard 7.2.4

Textbook p. 203

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Question 8

Bilangan aplikasi yang dimuat turun oleh lapan orang murid dalam seminggu ialah 3, 12, 6, 9, 5, 16, 9 dan 8.The numbers of apps downloaded by eight pupils in a week are 3, 12, 6, 9, 5, 16, 9 and 8.

[1 mark] (a)Cari julat data itu.Find the range of the data.

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• 16 − 3 = 131

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[2 marks] (b)Cari julat antara kuartil.Find the interquartile range.

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• Ordered: 3, 5, 6, 8, 9, 9, 12, 16; Q1 = 5.5 and Q3 = 10.51
• IQR = 10.5 − 5.5 = 51

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[1 mark] (c)Setiap nilai ditambah 4. Nyatakan julat antara kuartil yang baharu.Each value is increased by 4. State the new interquartile range.

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• 5 (unchanged when a constant is added)1

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For the teacher: Applying, low difficulty, Chapter 8, standard 8.2.1

Textbook p. 220

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Question 9

[2 marks] (a)Sekeping syiling adil dilambung dan sebiji dadu adil dilambung serentak. Cari kebarangkalian mendapat kepala dan nombor perdana.A fair coin is tossed and a fair die is rolled at the same time. Find the probability of getting a head and a prime number.

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• P(head) = ½ and P(prime) = 3/6 (2, 3, 5)1
• ½ × ½ = ¼1

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[2 marks] (b)Peristiwa M dan peristiwa N ialah peristiwa saling eksklusif. Diberi P(M) = 0.35 dan P(N) = 0.45. Cari (i) P(M atau N), (ii) kebarangkalian bahawa tiada satu pun peristiwa berlaku.Event M and event N are mutually exclusive events. Given P(M) = 0.35 and P(N) = 0.45. Find (i) P(M or N), (ii) the probability that neither event occurs.

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• (i) 0.35 + 0.45 = 0.81
• (ii) 1 − 0.8 = 0.21

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For the teacher: Applying, medium difficulty, Chapter 9, standard 9.3.3

Textbook p. 259

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Question 10

Pendapatan bulanan Encik Rahim ialah RM4 200. Perbelanjaan bulanannya ialah: sewa rumah RM1 100, ansuran kereta RM650, makanan RM900, hiburan RM350 dan bil RM420.Encik Rahim's monthly income is RM4 200. His monthly expenses are: house rent RM1 100, car instalment RM650, food RM900, entertainment RM350 and bills RM420.

[1 mark] (a)Hitung jumlah perbelanjaan bulanan.Calculate the total monthly expenses.

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• RM3 4201

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[2 marks] (b)Hitung aliran tunai bersih dan nyatakan sama ada ia lebihan atau defisit.Calculate the net cash flow and state whether it is a surplus or a deficit.

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• 4 200 − 3 420 = RM7801
• Surplus (positive cash flow)1

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[1 mark] (c)Dia ingin membeli komputer riba berharga RM3 900 hanya dengan menggunakan lebihan itu. Berapa bulan yang diperlukan?He wants to buy a laptop costing RM3 900 using only this surplus. How many months are needed?

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• 3 900 ÷ 780 = 5 months1

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For the teacher: Applying, low difficulty, Chapter 10, standard 10.1.1

Textbook p. 274

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Question 11

Rajah 3 menunjukkan rangkaian yang menghubungkan enam bilik pelayan, P, Q, R, S, T dan U. Nombor pada setiap tepi ialah panjang kabel gentian optik, dalam km, yang boleh dipasang.Diagram 3 shows a network that connects six server rooms, P, Q, R, S, T and U. The number on each edge is the length of fibre-optic cable, in km, that can be laid.

[2 marks] (a)Nyatakan bilangan tepi rangkaian itu dan darjah bucu S.State the number of edges of the network and the degree of vertex S.

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• 9 edges1
• Degree of S = 41

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[2 marks] (b)Rangkaian dalam Rajah 3 bukan sebuah pokok. Beri satu sebab.The network in Diagram 3 is not a tree. Give one reason.

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• It contains a cycle, e.g. P-Q-R-P (or: a tree on 6 vertices has only 5 edges)1
• A tree has no cycles [or: 9 edges ≠ 5]1

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[3 marks] (c)Cari pokok yang mempunyai jumlah berat minimum bagi rangkaian itu. Senaraikan tepi yang dipilih dan hitung jumlah panjang kabel.Find the tree with the minimum total weight for the network. List the edges chosen and calculate the total length of cable.

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• Choose edges in increasing weight without forming a cycle: ST 2, QR 3, PQ 4, TU 41
• Add RS 5 to join the two parts (5 edges for 6 vertices)1
• Total = 2 + 3 + 4 + 4 + 5 = 18 km1

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[2 marks] (d)Seorang juruteknik mencadangkan memasang kesemua sembilan kabel untuk keandalan. Kos pemasangan ialah RM12 000 bagi setiap km. Hitung kos tambahan berbanding pokok yang mempunyai jumlah berat minimum.A technician suggests laying all nine cables for reliability. The cost of laying is RM12 000 per km. Calculate the extra cost compared with the tree of minimum total weight.

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• All nine cables: 4 + 7 + 3 + 6 + 5 + 8 + 2 + 9 + 4 = 48 km, extra = 48 − 18 = 30 km1
• 30 × 12 000 = RM360 0001

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For the teacher: Applying, medium difficulty, Chapter 5, standard 5.1.3

Textbook p. 140

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Question 12

Sebuah kedai menjual x buah tablet dan y buah telefon pintar dalam sehari. Rajah 4 menunjukkan rantau berlorek R yang memenuhi tiga syarat jualan.A shop sells x tablets and y smartphones in a day. Diagram 4 shows the shaded region R that satisfies three sales conditions.

[3 marks] (a)Tulis tiga ketaksamaan yang mentakrifkan rantau R.Write the three inequalities that define region R.

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• y ≥ 11
• y < 2x1
• x + y ≤ 91

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[2 marks] (b)Adakah titik (2, 4) berada di dalam rantau R? Terangkan jawapan anda.Is the point (2, 4) inside region R? Explain your answer.

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• No1
• Substituting: 4 < 2(2) = 4 is false (the point lies on the dashed line, which is not included)1

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[1 mark] (c)Terangkan maksud ketaksamaan y < 2x dalam konteks jualan itu.Explain the meaning of the inequality y < 2x in the context of the sales.

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• The number of smartphones sold is less than twice the number of tablets sold1

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[3 marks] (d)Keuntungan ialah RM60 bagi setiap tablet dan RM40 bagi setiap telefon pintar. Dengan menggunakan nilai integer x dan y dalam rantau R, cari keuntungan maksimum dalam sehari.The profit is RM60 for each tablet and RM40 for each smartphone. Using integer values of x and y in region R, find the maximum profit in a day.

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• Profit = 60x + 40y; each tablet earns more, so make x as large as possible1
• x = 8, y = 1 (check: 1 ≥ 1, 1 < 16, 8 + 1 ≤ 9)1
• 60(8) + 40(1) = RM5201

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For the teacher: Analysing, high difficulty, Chapter 6, standard 6.2.3

Textbook p. 169

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Question 13

Rajah 5 menunjukkan graf jarak-masa bagi sebuah van penghantaran yang bergerak dari depot dan kembali ke depot itu.Diagram 5 shows the distance-time graph of a delivery van that travels from a depot and returns to the depot.

[2 marks] (a)Nyatakan tempoh masa, dalam minit, van itu berhenti dan jaraknya dari depot semasa berhenti.State the length of time, in minutes, that the van is stationary and its distance from the depot while stationary.

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• 30 minutes (0.75 h to 1.25 h)1
• 45 km1

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[2 marks] (b)Hitung laju van itu semasa perjalanan pulang, dalam km j⁻¹.Calculate the speed of the van on the return journey, in km h⁻¹.

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• 120 ÷ (3.75 − 2.25)1
• 80 km h⁻¹1

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[2 marks] (c)Hitung laju purata van itu bagi seluruh perjalanan, dalam km j⁻¹.Calculate the average speed of the van for the whole journey, in km h⁻¹.

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• Total distance = 120 + 120 = 240 km1
• 240 ÷ 3.75 = 64 km h⁻¹1

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[3 marks] (d)Pada masa berapakah, dalam jam selepas van meninggalkan depot, van itu berada 60 km dari depot semasa perjalanan pulang?At what time, in hours after the van leaves the depot, is the van 60 km from the depot on the return journey?

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• Distance left to travel 60 km at 80 km h⁻¹1
• 60 ÷ 80 = 0.75 h after leaving the farthest point1
• 2.25 + 0.75 = 3.0 hours1

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

Textbook p. 188

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Question 14

Rajah 6 menunjukkan graf bagi fungsi kuadratik f(x) = x² + 2x − 8. Graf itu memotong paksi-x di titik P dan Q. M ialah titik minimum.Diagram 6 shows the graph of the quadratic function f(x) = x² + 2x − 8. The graph meets the x-axis at points P and Q. M is the minimum point.

[2 marks] (a)Berdasarkan Rajah 6, nyatakan punca-punca bagi persamaan x² + 2x − 8 = 0.Based on Diagram 6, state the roots of the equation x² + 2x − 8 = 0.

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• x = −41
• x = 21

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[2 marks] (b)Nyatakan persamaan paksi simetri dan koordinat titik M.State the equation of the axis of symmetry and the coordinates of point M.

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• x = −11
• M = (−1, −9)1

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[2 marks] (c)Selesaikan x² + 2x − 8 = 7 dengan pemfaktoran.Solve x² + 2x − 8 = 7 by factorisation.

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• x² + 2x − 15 = 0, (x + 5)(x − 3) = 01
• x = −5 or x = 31

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[3 marks] (d)Tinggi, h m, sebuah dron pada masa t saat diberi oleh h = 20t − 5t². Cari masa apabila dron itu berada pada ketinggian 15 m.The height, h m, of a drone at time t seconds is given by h = 20t − 5t². Find the times when the drone is at a height of 15 m.

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• 20t − 5t² = 15, so t² − 4t + 3 = 01
• (t − 1)(t − 3) = 01
• t = 1 s and t = 3 s (going up and coming down)1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.1.6

Textbook p. 21

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Question 15

Ain bekerja sebagai jurutera perisian. Pendapatannya sebulan ialah gaji RM3 600 dan sewa bilik yang diterimanya RM450. Perbelanjaan tetapnya ialah sewa rumah RM900, ansuran pinjaman RM450 dan insurans RM120. Perbelanjaan tidak tetapnya ialah makanan RM780, pengangkutan RM320 dan pembelian dalam talian RM260. Dia bercadang mengumpul RM9 600 dalam 8 bulan sebagai wang pendahuluan.Ain works as a software engineer. Her monthly income is a salary of RM3 600 and room rent of RM450. Her fixed expenses are house rent RM900, loan instalment RM450 and insurance RM120. Her variable expenses are food RM780, transport RM320 and online shopping RM260. She plans to save RM9 600 in 8 months as a deposit.

[2 marks] (a)Hitung jumlah pendapatan bulanan Ain dan nyatakan jumlah pendapatan pasifnya.Calculate Ain's total monthly income and state her passive income.

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• 3 600 + 450 = RM4 0501
• Passive income = RM450 (room rent)1

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[3 marks] (b)Hitung jumlah perbelanjaan bulanan dan aliran tunai bersih.Calculate the total monthly expenses and the net cash flow.

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• Fixed = 900 + 450 + 120 = RM1 470; variable = 780 + 320 + 260 = RM1 3601
• Total expenses = RM2 8301
• Net cash flow = 4 050 − 2 830 = RM1 220 (surplus)1

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[2 marks] (c)Hitung simpanan bulanan yang diperlukan dan nyatakan sama ada rancangan itu boleh dilaksanakan.Calculate the monthly saving needed and state whether the plan is feasible.

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• 9 600 ÷ 8 = RM1 200 per month1
• Feasible, because the surplus RM1 220 ≥ RM1 2001

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[2 marks] (d)Perbelanjaan tidak tetap Ain meningkat 25%. Adakah rancangan itu masih boleh dilaksanakan? Tunjukkan pengiraan anda.Ain's variable expenses increase by 25%. Is the plan still feasible? Show your working.

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• Increase = 25% × 1 360 = RM340, so surplus = 1 220 − 340 = RM8801
• RM880 < RM1 200, so not feasible (she must cut spending or extend the time)1

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For the teacher: Evaluating, high difficulty, Chapter 10, standard 10.1.2

Textbook p. 285

16Choose 1

Question 16

Rajah 7 menunjukkan graf laju-masa bagi sebuah kereta api elektrik yang bergerak dari satu stesen ke stesen lain dalam tempoh 34 saat.Diagram 7 shows the speed-time graph of an electric train travelling from one station to the next in 34 seconds.

[2 marks] (a)Nyatakan laju awal dan laju seragam kereta api itu.State the initial speed and the uniform speed of the train.

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• 8 m s⁻¹1
• 20 m s⁻¹1

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[2 marks] (b)Hitung pecutan kereta api itu dalam 6 saat pertama.Calculate the acceleration of the train in the first 6 seconds.

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• (20 − 8) ÷ 61
• 2 m s⁻²1

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[2 marks] (c)Hitung nyahpecutan kereta api itu dalam 10 saat terakhir.Calculate the deceleration of the train in the last 10 seconds.

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• 20 ÷ 101
• 2 m s⁻²1

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[3 marks] (d)Hitung jumlah jarak antara kedua-dua stesen itu.Calculate the total distance between the two stations.

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• First 6 s: ½ × (8 + 20) × 6 = 84 m1
• Uniform: 20 × 18 = 360 m (last 10 s: ½ × 20 × 10 = 100 m)1
• Total = 84 + 360 + 100 = 544 m1

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[2 marks] (e)Hitung laju purata kereta api itu bagi keseluruhan perjalanan.Calculate the average speed of the train for the whole journey.

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• 544 ÷ 341
• 16 m s⁻¹1

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[2 marks] (f)Seorang penumpang berkata bahawa kereta api itu bergerak pada laju seragam untuk lebih daripada separuh jarak perjalanan. Adakah pernyataan itu betul? Sokong jawapan anda dengan pengiraan.A passenger says that the train travelled at uniform speed for more than half of the distance of the journey. Is the statement correct? Support your answer with calculation.

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• Half of the distance = 272 m1
• 360 m > 272 m, so the statement is correct1

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[2 marks] (g)Cari laju kereta api itu pada t = 3 s dan masa apabila lajunya 10 m s⁻¹ semasa nyahpecutan.Find the speed of the train at t = 3 s and the time when its speed is 10 m s⁻¹ during deceleration.

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• 8 + 2 × 3 = 14 m s⁻¹1
• Speed drops 2 m s⁻¹ every second: 10 m s⁻¹ is 5 s after 24 s, so t = 29 s1

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For the teacher: Analysing, medium difficulty, Chapter 7, standard 7.2.4

Textbook p. 203

17Choose 1

Question 17

Encik Hafiz ialah seorang juruteknik. Pendapatan bulanannya ialah gaji RM4 500, upah kerja sambilan RM600 dan dividen saham RM150. Perbelanjaan tetapnya ialah ansuran rumah RM1 250, ansuran kereta RM680, telefon dan internet RM180 serta insurans RM220. Perbelanjaan tidak tetapnya ialah makanan RM1 050, petrol RM400, utiliti RM260 dan hiburan RM300. Matlamatnya ialah “Menyimpan RM21 840 dalam 24 bulan untuk wang pendahuluan kereta baharu”.Encik Hafiz is a technician. His monthly income is a salary of RM4 500, part-time wages of RM600 and share dividends of RM150. His fixed expenses are house instalment RM1 250, car instalment RM680, phone and internet RM180 and insurance RM220. His variable expenses are food RM1 050, petrol RM400, utilities RM260 and entertainment RM300. His goal is “Save RM21 840 in 24 months as a deposit for a new car”.

[3 marks] (a)Hitung jumlah pendapatan, jumlah perbelanjaan dan aliran tunai bersih Encik Hafiz setiap bulan.Calculate Encik Hafiz's total income, total expenses and net cash flow each month.

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• Income = RM5 2501
• Expenses = 2 330 + 2 010 = RM4 3401
• Net cash flow = RM910 (surplus)1

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[2 marks] (b)Matlamat itu mematuhi konsep SMART. Nyatakan ciri SMART yang ditunjukkan oleh (i) RM21 840 dan (ii) 24 bulan.The goal follows the SMART concept. State the SMART characteristic shown by (i) RM21 840 and (ii) 24 months.

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• (i) Measurable (or Specific)1
• (ii) Time-bound1

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[2 marks] (c)Hitung simpanan bulanan yang diperlukan dan bandingkan dengan lebihan bulanannya.Calculate the monthly saving needed and compare it with his monthly surplus.

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• 21 840 ÷ 24 = RM9101
• Equal to the surplus: feasible but with no spare money1

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[4 marks] (d)Selepas 10 bulan, Encik Hafiz menggunakan RM2 800 daripada simpanannya untuk membayar bil perubatan. Hitung simpanan bulanan baharu yang diperlukan untuk 14 bulan yang tinggal supaya matlamat itu tercapai.After 10 months, Encik Hafiz uses RM2 800 of his savings to pay a medical bill. Calculate the new monthly saving needed for the remaining 14 months for the goal to be reached.

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• Savings after 10 months = 910 × 10 = RM9 1001
• After the bill: 9 100 − 2 800 = RM6 3001
• Still needed = 21 840 − 6 300 = RM15 5401
• 15 540 ÷ 14 = RM1 110 per month1

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[4 marks] (e)Adakah lebihan bulanannya mencukupi untuk simpanan baharu itu? Cadangkan satu cara, dengan pengiraan, untuk mencapai matlamat itu.Is his monthly surplus enough for the new saving? Suggest one way, with calculation, to reach the goal.

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• Not enough: 910 < 1 1101
• Shortfall = 1 110 − 910 = RM200 per month1
• Any sensible suggestion, e.g. reduce entertainment from RM300 to RM100 (or RM200 more from variable expenses)1
• New surplus = 910 + 200 = RM1 110, which equals the saving needed1

Tick each point you got.

For the teacher: Evaluating, high difficulty, Chapter 10, standard 10.1.2

Textbook p. 287

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