Matematik Tingkatan 4 Kertas 2Mathematics Form 4 Paper 2

Print questions onlyCetak soalan sahaja Print with answersCetak dengan skema Print answer key onlyCetak kunci jawapan sahaja Practice shows answers as you go. Exam times you and hides answers until you submit.Latihan menunjukkan jawapan semasa menjawab. Peperiksaan menetapkan masa dan menyembunyikan jawapan sehingga dihantar.

1

QuestionSoalan 1

Seorang petani mempunyai sebuah plot sayur berbentuk segi empat tepat dengan keliling 34 m dan luas 60 m².A farmer has a rectangular vegetable plot with a perimeter of 34 m and an area of 60 m².

[1 markmarkah] (a)Anggapkan panjang plot ialah x m. Nyatakan lebar plot itu dalam sebutan x, kemudian bentukkan persamaan kuadratik dalam bentuk am dengan menggunakan luas plot.Taking the length as x m and the perimeter as 34 m, express the width of the plot in terms of x, then form a quadratic equation in general form using the area of the plot.

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• Width = (17 − x) m, so x(17 − x) = 60 and x² − 17x + 60 = 01

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[2 marksmarkah] (b)Selesaikan persamaan itu dengan pemfaktoran dan nyatakan ukuran plot itu.Solve the equation by factorisation and state the dimensions of the plot.

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• (x − 5)(x − 12) = 0, so x = 5 or x = 121
• Dimensions: 12 m by 5 m (both roots give the same plot)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1.8

TextbookBuku teks p.ms. 26

2

QuestionSoalan 2

Sebuah mesin kira stok di sebuah kedai tani menyimpan kuantiti barang dalam asas yang berbeza.A stock counter at a farm-supply shop stores item quantities in different bases.

[2 marksmarkah] (a)Tukarkan 215₆ kepada nombor dalam asas sepuluh.Convert 215₆ to a number in base ten.

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• 2 × 6² + 1 × 6 + 51
• 831

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[2 marksmarkah] (b)Hitung 11101₂ − 1110₂. Beri jawapan dalam asas dua.Calculate 11101₂ − 1110₂. Give the answer in base two.

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• Correct vertical subtraction with borrowing (or 29 − 14 = 15)1
• 1111₂1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.1.3

TextbookBuku teks p.ms. 47

3

QuestionSoalan 3

[1 markmarkah] (a)Lengkapkan kesimpulan: Premis 1: Semua gandaan 5 berakhir dengan digit 0 atau 5. Premis 2: 135 ialah gandaan 5. Kesimpulan: …Complete the conclusion. Premise 1: All multiples of 5 end with the digit 0 or 5. Premise 2: 135 is a multiple of 5. Conclusion: …

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• 135 ends with the digit 0 or 5 (Argument Form I)1

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[1 markmarkah] (b)Lengkapkan kesimpulan: Premis 1: Jika sebuah sisi empat ialah segi empat sama, maka semua sisinya sama panjang. Premis 2: Sisi empat PQRS tidak mempunyai semua sisi yang sama panjang. Kesimpulan: …Complete the conclusion. Premise 1: If a quadrilateral is a square, then all its sides are equal in length. Premise 2: Quadrilateral PQRS does not have all sides equal in length. Conclusion: …

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• Quadrilateral PQRS is not a square (Argument Form II)1

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[2 marksmarkah] (c)Perhatikan corak 3 = 2² − 1, 8 = 3² − 1, 15 = 4² − 1, 24 = 5² − 1. Buat satu kesimpulan umum secara aruhan dan gunakannya untuk mencari nombor ke-8 dalam corak itu.Look at the pattern 3 = 2² − 1, 8 = 3² − 1, 15 = 4² − 1, 24 = 5² − 1. Make a general conclusion by induction and use it to find the 8th number in the pattern.

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• The nth number = (n + 1)² − 11
• 8th number = 9² − 1 = 801

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For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.2.2

TextbookBuku teks p.ms. 76

4

QuestionSoalan 4

Diberi set semesta ξ = {x : 1 ≤ x ≤ 20, x ialah integer}, set A = {nombor genap} dan set B = {gandaan 3}.Given the universal set ξ = {x : 1 ≤ x ≤ 20, x is an integer}, set A = {even numbers} and set B = {multiples of 3}.

[1 markmarkah] (a)Senaraikan unsur bagi A ∩ B.List the elements of A ∩ B.

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• {6, 12, 18}1

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[2 marksmarkah] (b)Cari n(A ∪ B)′.Find n(A ∪ B)′.

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• n(A ∪ B) = 10 + 6 − 3 = 131
• n(A ∪ B)′ = 20 − 13 = 71

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[1 markmarkah] (c)Senaraikan unsur bagi A′ ∩ B.List the elements of A′ ∩ B.

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• {3, 9, 15} (multiples of 3 that are not even)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.2.3

TextbookBuku teks p.ms. 113

5

QuestionSoalan 5

Rajah 1 menunjukkan lima pusat pengumpulan hasil ladang, S, T, U, V dan W. Nombor pada setiap tepi ialah jarak jalan, dalam km.Diagram 1 shows five farm-produce collection points, S, T, U, V and W. The number on each edge is the length of the road, in km.

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

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• 7 edges1
• Degree of U = 41

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[2 marksmarkah] (b)Cari laluan terpendek dari S ke W dan nyatakan jaraknya.Find the shortest path from S to W and state its length.

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• S − T − U − V − W (5 + 3 + 4 + 6)1
• 18 km (every other path gives 19 km or more)1

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For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1.2

TextbookBuku teks p.ms. 137

6

QuestionSoalan 6

[2 marksmarkah] (a)Garis lurus y = 3x − 6 dilukis sebagai garis sempang. Asalan berada dalam rantau berlorek. Tulis ketaksamaan bagi rantau itu dan tunjukkan bagaimana anda menentukannya.The straight line y = 3x − 6 is drawn as a dashed line. The origin is in the shaded region. Write the inequality for the region and show how you decided.

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• Dashed line, so the inequality is strict1
• y > 3x − 6 (origin: 0 > −6 is true)1

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[2 marksmarkah] (b)Seorang penternak menternak x ekor ayam dan y ekor itik. Jumlah haiwan tidak melebihi 60 ekor dan bilangan ayam sekurang-kurangnya tiga kali bilangan itik. Tulis dua ketaksamaan bagi syarat ini.A farmer keeps x chickens and y ducks. The total number of animals is not more than 60 and the number of chickens is at least three times the number of ducks. Write two inequalities for these conditions.

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• x + y ≤ 601
• x ≥ 3y1

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[1 markmarkah] (c)Adakah titik (45, 15) memenuhi kedua-dua ketaksamaan dalam (b)? Tunjukkan pengiraan anda.Does the point (45, 15) satisfy both inequalities in (b)? Show your working.

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• 45 + 15 = 60 ≤ 60 and 45 ≥ 3(15) = 45, so it satisfies both (it lies on both boundary lines)1

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.1.2

TextbookBuku teks p.ms. 160

7

QuestionSoalan 7

Sebuah traktor meningkatkan lajunya secara seragam daripada 3 m s⁻¹ kepada 15 m s⁻¹ dalam masa 4 saat.A tractor increases its speed uniformly from 3 m s⁻¹ to 15 m s⁻¹ in 4 seconds.

[1 markmarkah] (a)Hitung pecutan traktor itu.Calculate the acceleration of the tractor.

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• (15 − 3) ÷ 4 = 3 m s⁻²1

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[2 marksmarkah] (b)Hitung jarak yang dilalui oleh traktor itu dalam tempoh 4 saat itu.Calculate the distance travelled by the tractor in the 4 seconds.

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• ½ × (3 + 15) × 41
• 36 m1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 7, standard 7.2.3

TextbookBuku teks p.ms. 200

8

QuestionSoalan 8

Bilangan bakul durian yang dituai oleh sebuah kebun dalam lima hari ialah 5, 7, 8, 9 dan 11.The numbers of baskets of durian harvested by an orchard in five days are 5, 7, 8, 9 and 11.

[1 markmarkah] (a)Cari min bagi data itu.Find the mean of the data.

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• (5 + 7 + 8 + 9 + 11) ÷ 5 = 81

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[2 marksmarkah] (b)Hitung varians dan sisihan piawai.Calculate the variance and the standard deviation.

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• Variance = (9 + 1 + 0 + 1 + 9) ÷ 5 = 4 (or 340 ÷ 5 − 8² = 4)1
• Standard deviation = 21

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[2 marksmarkah] (c)Setiap nilai didarabkan dengan 3. Nyatakan min dan sisihan piawai yang baharu.Each value is multiplied by 3. State the new mean and the new standard deviation.

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• New mean = 241
• New standard deviation = 61

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.2.2

TextbookBuku teks p.ms. 224

9

QuestionSoalan 9

[2 marksmarkah] (a)Peluang biji benih jenis X bercambah ialah 0.8 dan peluang biji benih jenis Y bercambah ialah 0.7. Kedua-dua peristiwa itu tidak bersandar. Cari kebarangkalian bahawa kedua-dua biji benih bercambah.The probability that a seed of type X germinates is 0.8 and the probability that a seed of type Y germinates is 0.7. The two events are independent. Find the probability that both seeds germinate.

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• 0.8 × 0.71
• 0.561

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[2 marksmarkah] (b)Peristiwa R dan peristiwa S ialah peristiwa saling eksklusif. Diberi P(R) = 2/5 dan P(S) = 1/3. Cari (i) P(R atau S), (ii) kebarangkalian bahawa tiada satu pun peristiwa berlaku.Events R and S are mutually exclusive events. Given P(R) = 2/5 and P(S) = 1/3. Find (i) P(R or S), (ii) the probability that neither event occurs.

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• (i) 2/5 + 1/3 = 11/151
• (ii) 1 − 11/15 = 4/151

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 9, standard 9.3.3

TextbookBuku teks p.ms. 259

10

QuestionSoalan 10

Pendapatan bulanan Encik Ng ialah gaji RM2 900, komisen RM350 dan faedah simpanan RM50. Perbelanjaan tetapnya RM1 450 dan perbelanjaan tidak tetapnya RM1 310.Encik Ng's monthly income is a salary of RM2 900, commission of RM350 and savings interest of RM50. His fixed expenses are RM1 450 and his variable expenses are RM1 310.

[1 markmarkah] (a)Hitung jumlah pendapatan bulanan Encik Ng.Calculate Encik Ng's total monthly income.

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• 2 900 + 350 + 50 = RM3 3001

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[1 markmarkah] (b)Hitung aliran tunai bersih bulanannya.Calculate his monthly net cash flow.

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Any 1 pointsMana-mana 1 isi

• Expenses = 1 450 + 1 310 = RM2 7601
• 3 300 − 2 760 = RM540 (positive)1

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[2 marksmarkah] (c)Perbelanjaannya meningkat 15%. Hitung aliran tunai bersih yang baharu dan nyatakan sama ada ia positif atau negatif.His expenses increase by 15%. Calculate the new net cash flow and state whether it is positive or negative.

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• New expenses = 115% × 2 760 = RM3 1741
• 3 300 − 3 174 = RM126, positive1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 10, standard 10.1.1

TextbookBuku teks p.ms. 276

11

QuestionSoalan 11

Rajah 2 menunjukkan graf fungsi keuntungan P(x) = −x² + 8x − 7 bagi sebuah kedai jus. P ialah keuntungan dalam ratus ringgit dan x ialah bilangan unit jus yang dijual dalam ratus. Graf itu memotong paksi-x di A dan B, dan paksi-y di C. M ialah titik maksimum.Diagram 2 shows the graph of the profit function P(x) = −x² + 8x − 7 of a juice stall. P is the profit in hundreds of ringgit and x is the number of units of juice sold in hundreds. The graph meets the x-axis at A and B, and the y-axis at C. M is the maximum point.

[2 marksmarkah] (a)Berdasarkan Rajah 2, nyatakan punca-punca bagi persamaan −x² + 8x − 7 = 0.Based on Diagram 2, state the roots of the equation −x² + 8x − 7 = 0.

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• x = 11
• x = 71

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[2 marksmarkah] (b)Nyatakan persamaan paksi simetri dan keuntungan maksimum, dalam RM.State the equation of the axis of symmetry and the maximum profit, in RM.

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• x = 41
• Maximum P = 9, so RM9001

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[2 marksmarkah] (c)Nyatakan koordinat titik C dan terangkan maksudnya dalam konteks perniagaan itu.State the coordinates of point C and explain its meaning in the context of the business.

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• C = (0, −7)1
• When no units are sold there is a loss of RM700 (the fixed cost)1

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[3 marksmarkah] (d)Pengurus mahukan keuntungan RM500. Selesaikan −x² + 8x − 7 = 5 dan nyatakan bilangan unit yang perlu dijual jika dia mahu mencapai keuntungan RM500 dengan menjual unit yang paling sedikit.The manager wants a profit of RM500. Solve −x² + 8x − 7 = 5 and state the number of units that must be sold if he wants to reach a profit of RM500 by selling the fewest units.

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• x² − 8x + 12 = 0, so (x − 2)(x − 6) = 01
• x = 2 or x = 61
• Fewest units: x = 2, which is 200 units1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1.6

TextbookBuku teks p.ms. 21

12

QuestionSoalan 12

Rajah 3 menunjukkan graf laju-masa bagi sebuah trak ladang yang membawa hasil dari kebun ke pasar dalam tempoh 24 saat.Diagram 3 shows the speed-time graph of a farm truck that carries produce from an orchard to a market in 24 seconds.

[2 marksmarkah] (a)Nyatakan laju seragam trak itu dan tempoh masa trak itu bergerak pada laju seragam.State the uniform speed of the truck and the length of time it travels at that speed.

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• 18 m s⁻¹1
• 14 − 6 = 8 s1

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[2 marksmarkah] (b)Hitung pecutan trak itu dalam 6 saat pertama.Calculate the acceleration of the truck in the first 6 seconds.

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• 18 ÷ 61
• 3 m s⁻²1

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[3 marksmarkah] (c)Hitung jumlah jarak yang dilalui oleh trak itu.Calculate the total distance travelled by the truck.

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• Speeding up: ½ × 6 × 18 = 54 m1
• Uniform: 8 × 18 = 144 m; slowing down: ½ × 10 × 18 = 90 m1
• Total = 54 + 144 + 90 = 288 m (or trapezium ½ × (8 + 24) × 18)1

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[2 marksmarkah] (d)Cari semua nilai masa apabila laju trak itu ialah 9 m s⁻¹.Find all the times at which the speed of the truck is 9 m s⁻¹.

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• Accelerating: 9 ÷ 3 = 3, so t = 3 s1
• Decelerating at 18 ÷ 10 = 1.8 m s⁻²: 9 ÷ 1.8 = 5 s after 14 s, so t = 19 s1

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For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.2.4

TextbookBuku teks p.ms. 203

13

QuestionSoalan 13

Sebuah kotak mengandungi 9 bolt, 3 daripadanya rosak. Dua bolt dikeluarkan secara rawak satu demi satu tanpa pengembalian.A box contains 9 bolts, 3 of which are defective. Two bolts are taken out at random one after the other without replacement.

[1 markmarkah] (a)Cari kebarangkalian bahawa bolt pertama yang dikeluarkan tidak rosak.Find the probability that the first bolt taken out is not defective.

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• 6/9 = 2/31

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[2 marksmarkah] (b)Cari kebarangkalian bahawa kedua-dua bolt rosak.Find the probability that both bolts are defective.

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• 3/9 × 2/81
• 1/121

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[3 marksmarkah] (c)Cari kebarangkalian bahawa tepat satu bolt rosak.Find the probability that exactly one bolt is defective.

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• Defective then good: 3/9 × 6/8 = 1/41
• Good then defective: 6/9 × 3/8 = 1/41
• 1/4 + 1/4 = 1/21

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[3 marksmarkah] (d)Penyelia menolak sebuah kotak jika kedua-dua bolt yang diperiksa rosak. Dia memeriksa dua kotak yang sama kandungannya, secara berasingan. Cari kebarangkalian bahawa kedua-dua kotak itu tidak ditolak.The supervisor rejects a box if both bolts checked are defective. He checks two boxes with the same contents, separately. Find the probability that neither box is rejected.

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• P(a box is not rejected) = 1 − 1/12 = 11/121
• The two boxes are independent events1
• 11/12 × 11/12 = 121/1441

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.3.3

TextbookBuku teks p.ms. 259

14

QuestionSoalan 14

Seorang kontraktor mengupah x orang pekerja mahir dan y orang pekerja am untuk sebuah tapak pembinaan. Gaji ialah RM200 sehari bagi seorang pekerja mahir dan RM100 sehari bagi seorang pekerja am. Rajah 4 menunjukkan rantau berlorek R yang memenuhi tiga syarat pengambilan.A contractor hires x skilled workers and y general workers for a construction site. The wage is RM200 a day for a skilled worker and RM100 a day for a general worker. Diagram 4 shows the shaded region R that satisfies three hiring conditions.

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

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• x ≥ 21
• y > 31
• 2x + y ≤ 151

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[2 marksmarkah] (b)Adakah titik (5, 4) berada di dalam rantau R? Terangkan jawapan anda.Is the point (5, 4) inside region R? Explain your answer.

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• Yes1
• 5 ≥ 2, 4 > 3 and 2(5) + 4 = 14 ≤ 15, so all three inequalities are satisfied1

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[1 markmarkah] (c)Terangkan maksud ketaksamaan y > 3 dalam konteks pekerja itu.Explain the meaning of the inequality y > 3 in the context of the workers.

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• The contractor hires more than 3 general workers1

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[3 marksmarkah] (d)Kontraktor itu mahu menggunakan keseluruhan bajet gaji RM1 500 dengan bilangan pekerja mahir yang paling banyak. Dengan menggunakan nilai integer x dan y dalam rantau R, cari bilangan setiap jenis pekerja.The contractor wants to use the whole wage budget of RM1 500 with as many skilled workers as possible. Using integer values of x and y in region R, find the number of each type of worker.

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• Wage condition: 200x + 100y = 1 500, so y = 15 − 2x1
• y > 3 gives x ≤ 5 (x = 6 gives y = 3, which is not allowed)1
• x = 5 skilled workers and y = 5 general workers1

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For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.2.4

TextbookBuku teks p.ms. 172

15

QuestionSoalan 15

Rajah 5 menunjukkan rangkaian yang boleh menghubungkan enam plot ladang, A, B, C, D, E dan F, dengan paip pengairan. Nombor pada setiap tepi ialah panjang paip, dalam km, yang boleh dipasang.Diagram 5 shows a network that can connect six farm plots, A, B, C, D, E and F, with irrigation pipes. The number on each edge is the length of pipe, in km, that can be laid.

[2 marksmarkah] (a)Nyatakan darjah bucu D dan jumlah darjah semua bucu.State the degree of vertex D and the sum of the degrees of all the vertices.

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• Degree of D = 41
• Sum = 2 × 9 = 181

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[2 marksmarkah] (b)Namakan satu kitaran dalam rangkaian itu dan nyatakan bilangan tepi dalam pokok yang menghubungkan kesemua enam bucu.Name one cycle in the network and state the number of edges of a tree that joins all six vertices.

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• Any cycle, e.g. A−B−C−A1
• 5 edges1

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[3 marksmarkah] (c)Cari pokok dengan jumlah berat minimum. Senaraikan tepi yang dipilih dan hitung jumlah panjang paip.Find the tree with the minimum total weight. List the edges chosen and calculate the total length of pipe.

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• Increasing weights without a cycle: EF 2, DE 3, BC 4, AB 51
• CD 6 joins the two parts (5 edges)1
• Total = 2 + 3 + 4 + 5 + 6 = 20 km1

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[2 marksmarkah] (d)Kos memasang paip ialah RM8 500 setiap km dan peruntukan ialah RM180 000. Adakah peruntukan itu mencukupi untuk pokok itu? Tunjukkan pengiraan anda.The cost of laying pipe is RM8 500 per km and the allocation is RM180 000. Is the allocation enough for the tree? Show your working.

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• 20 × 8 500 = RM170 0001
• 170 000 < 180 000, so it is enough (RM10 000 left)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1.3

TextbookBuku teks p.ms. 141

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QuestionSoalan 16

Encik Daud ialah seorang petani sayur. Pendapatan bulanannya ialah hasil jualan sayur RM6 750 dan sewa stor RM600. Perbelanjaan tetapnya ialah sewa tanah RM900, ansuran traktor RM750 dan insurans RM180. Perbelanjaan tidak tetapnya ialah baja dan benih RM1 600, upah pekerja RM1 250, perbelanjaan keluarga RM1 150 dan pengangkutan RM470. Matlamatnya ialah membeli sebuah pam air berharga RM6 000 dalam masa 8 bulan.Encik Daud is a vegetable farmer. His monthly income is vegetable sales of RM6 750 and storeroom rent of RM600. His fixed expenses are land rent RM900, tractor instalment RM750 and insurance RM180. His variable expenses are fertiliser and seeds RM1 600, workers' wages RM1 250, family expenses RM1 150 and transport RM470. His goal is to buy a water pump costing RM6 000 in 8 months.

[3 marksmarkah] (a)Hitung jumlah pendapatan, jumlah perbelanjaan dan aliran tunai bersih bulanan Encik Daud.Calculate Encik Daud's total monthly income, total expenses and net cash flow.

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• Income = 6 750 + 600 = RM7 3501
• Expenses = (900 + 750 + 180) + (1 600 + 1 250 + 1 150 + 470) = 1 830 + 4 470 = RM6 3001
• Net cash flow = RM1 050 (surplus)1

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[2 marksmarkah] (b)Hitung simpanan bulanan yang diperlukan untuk matlamat itu dan nyatakan sama ada ia boleh dilaksanakan. Berapakah lebihan yang tinggal selepas menyimpan?Calculate the monthly saving needed for the goal and state whether it is feasible. How much of his surplus is left after saving?

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• 6 000 ÷ 8 = RM7501
• Feasible, because 750 ≤ 1 050; RM300 is left1

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[2 marksmarkah] (c)Satu panduan perancangan kewangan ialah menyimpan 10% daripada jumlah pendapatan terlebih dahulu. Adakah simpanan RM750 memenuhi panduan itu? Tunjukkan pengiraan anda.One financial planning guide is to save 10% of total income first. Does the saving of RM750 meet this guide? Show your working.

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• 10% × 7 350 = RM7351
• 750 ≥ 735, so the guide is met1

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[3 marksmarkah] (d)Selepas 3 bulan, hasil jualan sayur jatuh 10% kerana kemarau, manakala perbelanjaan tidak berubah. Hitung aliran tunai bersih yang baharu dan jumlah yang masih diperlukan untuk membeli pam itu.After 3 months, the vegetable sales fall by 10% because of a drought, while expenses stay the same. Calculate the new net cash flow and the amount still needed to buy the pump.

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• Fall = 10% × 6 750 = RM675, so net cash flow = 1 050 − 675 = RM3751
• Saved in 3 months = 3 × 750 = RM2 2501
• Still needed = 6 000 − 2 250 = RM3 7501

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[3 marksmarkah] (e)Encik Daud mahu terus menyimpan RM750 sebulan untuk 5 bulan yang tinggal. Hitung jumlah perbelanjaan tidak tetap yang perlu dikurangkan sebulan dan nyatakan sebagai peratusan perbelanjaan tidak tetapnya, betul kepada satu tempat perpuluhan.Encik Daud wants to keep saving RM750 a month for the remaining 5 months. Calculate the amount of variable expenses that must be cut each month and express it as a percentage of his variable expenses, correct to one decimal place.

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• 3 750 ÷ 5 = RM750 needed, but net cash flow is RM375, so shortfall = RM3751
• Cut RM375 from variable expenses1
• 375 ÷ 4 470 × 100 = 8.4%1

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[2 marksmarkah] (f)Jika Encik Daud hanya menyimpan aliran tunai bersih baharunya sebulan, berapa bulan lebih lambat daripada 8 bulan asal dia akan memperoleh pam itu? Nyatakan satu kelebihan memotong perbelanjaan tidak tetap berbanding pilihan ini.If Encik Daud saves only his new net cash flow each month, how many months later than the original 8 months will he get the pump? State one advantage of cutting variable expenses compared with this option.

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• 3 750 ÷ 375 = 10 more months, so 3 + 10 = 13 months, 5 months later1
• Advantage: the pump is bought on time, as planned (any sensible reason)1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 10, standard 10.1.2

TextbookBuku teks p.ms. 284

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QuestionSoalan 17

Rajah 6 menunjukkan gambar rajah Venn bagi 50 orang petani dalam sebuah koperasi. Set P = {petani yang menanam padi}, set V = {petani yang menanam sayur} dan set F = {petani yang menanam buah-buahan}. Nombor dalam rajah ialah bilangan petani.Diagram 6 shows a Venn diagram of 50 farmers in a cooperative. Set P = {farmers who grow paddy}, set V = {farmers who grow vegetables} and set F = {farmers who grow fruit}. The numbers in the diagram are numbers of farmers.

[2 marksmarkah] (a)Diberi n(ξ) = 50. Cari nilai x.Given n(ξ) = 50. Find the value of x.

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• 12 + 9 + 7 + 6 + 5 + 4 + x + 4 = 501
• x = 31

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[3 marksmarkah] (b)Cari (i) n(P), (ii) n(P ∩ V), (iii) n(V ∪ F).Find (i) n(P), (ii) n(P ∩ V), (iii) n(V ∪ F).

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• (i) 12 + 6 + 4 + 3 = 251
• (ii) 6 + 4 = 101
• (iii) 50 − 12 − 4 = 34 (or 9 + 6 + 4 + 5 + 7 + 3)1

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[2 marksmarkah] (c)Cari bilangan petani yang menanam tepat dua jenis tanaman.Find the number of farmers who grow exactly two types of crop.

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• Regions in exactly two sets: 6, 5 and 3 (not the 4 in all three)1
• 6 + 5 + 3 = 141

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[2 marksmarkah] (d)Nyatakan kawasan “hanya buah-buahan” (7 orang petani) menggunakan tatatanda set.Express the region “fruits only” (7 farmers) using set notation.

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• F ∩ (P ∪ V)′1
• (or P′ ∩ V′ ∩ F)1

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[3 marksmarkah] (e)Koperasi memberi subsidi RM250 kepada setiap petani yang menanam sekurang-kurangnya dua jenis tanaman. Jumlah subsidi ialah RM4 500. Adakah ini betul? Tunjukkan pengiraan anda.The cooperative gives a subsidy of RM250 to every farmer who grows at least two types of crop. The total subsidy is RM4 500. Is this correct? Show your working.

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• At least two crops = 14 + 4 = 18 farmers1
• 18 × 250 = RM4 5001
• Correct1

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[3 marksmarkah] (f)Pengerusi berkata bahawa lebih separuh daripada petani menanam sayur. Adakah pernyataan itu betul? Berapakah bilangan paling kurang petani yang perlu mula menanam sayur supaya pernyataan itu betul?The chairman says that more than half of the farmers grow vegetables. Is the statement correct? What is the least number of farmers who must start growing vegetables for the statement to be correct?

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• n(V) = 9 + 6 + 4 + 5 = 241
• Half of 50 = 25, and 24 < 25, so the statement is not correct1
• Need n(V) ≥ 26, so at least 2 more farmers1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.3.3

TextbookBuku teks p.ms. 121

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