MatematikMathematics

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

Bilangan hari, d, sekarung makanan ayam boleh bertahan berubah secara songsang dengan bilangan ayam, c. Sekarung makanan itu bertahan 24 hari untuk 50 ekor ayam. Berapa harikah sekarung makanan itu bertahan untuk 80 ekor ayam?The number of days, d, that a sack of chicken feed lasts varies inversely as the number of chickens, c. A sack lasts 24 days for 50 chickens. For how many days will a sack last for 80 chickens?

• A38.4 hari38.4 days
• B15 hari15 days
• C30 hari30 days
• D54 hari54 days

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B

d = k/c, so k = dc = 24 × 50 = 1 200. For c = 80, d = 1 200 ÷ 80 = 15 days. 38.4 days comes from treating it as direct variation (24 × 80 ÷ 50).

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.2

TextbookBuku teks p.ms. 21

2

QuestionSoalan 2

Masa, T jam, untuk menuai sebuah ladang berubah secara langsung dengan luas ladang itu, A hektar, dan secara songsang dengan bilangan pekerja, w. Diberi T = 6 apabila A = 3 dan w = 4. Berapakah bilangan pekerja yang diperlukan untuk menuai 5 hektar dalam masa 5 jam?The time, T hours, to harvest a field varies directly as the area of the field, A hectares, and inversely as the number of workers, w. Given T = 6 when A = 3 and w = 4. How many workers are needed to harvest 5 hectares in 5 hours?

• A44
• B66
• C88
• D1010

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C

T = kA/w. 6 = k(3)/4, so k = 8. For T = 5 and A = 5: 5 = 8(5)/w, so w = 40 ÷ 5 = 8 workers.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.3.2

TextbookBuku teks p.ms. 28

3

QuestionSoalan 3

Diberi P = 2−134 dan Q = 52. Cari PQ.Given P = 2−134 and Q = 52. Find PQ.

• A823823
• B8787
• C12231223
• D238238

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A

Row 1: 2 × 5 + (−1) × 2 = 8. Row 2: 3 × 5 + 4 × 2 = 23. The product is a 2 × 1 matrix. 12 comes from treating −1 × 2 as +2.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2.3

TextbookBuku teks p.ms. 50

4

QuestionSoalan 4

Matriks m632 tidak mempunyai matriks songsang. Cari nilai m.The matrix m632 has no inverse. Find the value of m.

• A33
• B4.54.5
• C66
• D99

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D

A matrix has no inverse when its determinant is 0: 2m − 6 × 3 = 0, so 2m = 18 and m = 9.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2.4

TextbookBuku teks p.ms. 60

5

QuestionSoalan 5

Antara berikut, yang manakah contoh insurans am?Which of the following is an example of general insurance?

• AInsurans hayat seumur hidupWhole life insurance
• BInsurans kebakaran untuk bangsal ladangFire insurance for a farm shed
• CInsurans hayat berjangkaTerm life insurance
• DInsurans endowmenEndowment insurance

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B

General insurance covers loss or damage other than the risks to life, such as fire, motor, medical and travel. The other three are life insurance policies.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1.1

TextbookBuku teks p.ms. 76

6

QuestionSoalan 6

Encik Salim, 36 tahun, seorang petani yang tidak merokok, membeli polisi insurans hayat dengan nilai muka RM140 000. Kadar premium bagi umurnya ialah RM2.45 bagi setiap RM1 000. Hitung premium tahunan yang perlu dibayar.Encik Salim, 36, a non-smoking farmer, buys a life insurance policy with a face value of RM140 000. The premium rate for his age is RM2.45 per RM1 000. Calculate the annual premium he has to pay.

• ARM34.30RM34.30
• BRM57.14RM57.14
• CRM3 430.00RM3 430.00
• DRM343.00RM343.00

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D

Premium = 140 000 ÷ 1 000 × 2.45 = 140 × 2.45 = RM343.00. Dividing by 100 gives RM3 430, and dividing the face value by the rate gives RM57.14.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1.2

TextbookBuku teks p.ms. 81

7

QuestionSoalan 7

Nilai boleh insurans sebuah gudang ladang ialah RM300 000. Polisi insurans kebakarannya mempunyai syarat insurans bersama 80% dan tiada deduktibel. Gudang itu diinsuranskan dengan RM180 000. Berapakah pampasan yang akan diterima jika kerugian sebahagian akibat kebakaran ialah RM48 000?The insurable value of a farm warehouse is RM300 000. Its fire insurance policy has an 80% co-insurance clause and no deductible. The warehouse is insured for RM180 000. How much compensation will be received if a partial loss from a fire is RM48 000?

• ARM36 000RM36 000
• BRM48 000RM48 000
• CRM28 800RM28 800
• DRM38 400RM38 400

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A

Amount required = 80% × 300 000 = RM240 000. Compensation = (180 000 ÷ 240 000) × 48 000 = RM36 000. RM28 800 uses 100% of the value instead of the 80% clause; RM48 000 ignores the penalty.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.1.3

TextbookBuku teks p.ms. 88

8

QuestionSoalan 8

Jumlah pendapatan tahunan Puan Hajar ialah RM69 500. Dia mendapat pengecualian cukai sebanyak RM1 500 dan jumlah pelepasan cukai RM17 000. Hitung pendapatan bercukainya.Puan Hajar's total annual income is RM69 500. She has a tax exemption of RM1 500 and a total tax relief of RM17 000. Calculate her chargeable income.

• ARM68 000RM68 000
• BRM52 500RM52 500
• CRM51 000RM51 000
• DRM88 000RM88 000

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C

Chargeable income = total income − exemption − relief = 69 500 − 1 500 − 17 000 = RM51 000. RM52 500 forgets the exemption and RM68 000 forgets the relief.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 104

9

QuestionSoalan 9

Pendapatan bercukai Encik Mazlan ialah RM38 600. Dia membayar zakat RM300. Menggunakan jadual: RM35 001 hingga RM50 000, cukai asas RM600 dan 8% bagi setiap ringgit selepas RM35 000, hitung cukai pendapatan yang perlu dibayar.Encik Mazlan's chargeable income is RM38 600. He paid zakat of RM300. Using the table: RM35 001 to RM50 000, base tax RM600 plus 8% of every ringgit above RM35 000, calculate the income tax he has to pay.

• ARM888RM888
• BRM588RM588
• CRM188RM188
• DRM2 788RM2 788

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B

Tax = 600 + (38 600 − 35 000) × 8% = 600 + 288 = RM888. Subtract the zakat rebate: 888 − 300 = RM588. The RM400 individual rebate applies only when chargeable income is RM35 000 or less, so it is not used here (RM188 wrongly subtracts it as well as the zakat).

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 105

10

QuestionSoalan 10

Puan Aminah membayar cukai pendapatan sebanyak RM2 100 selepas rebat zakat RM400. Menggunakan jadual: RM50 001 hingga RM70 000, cukai asas RM1 800 dan 14% bagi setiap ringgit selepas RM50 000, cari pendapatan bercukainya.Puan Aminah paid income tax of RM2 100 after a zakat rebate of RM400. Using the table: RM50 001 to RM70 000, base tax RM1 800 plus 14% of every ringgit above RM50 000, find her chargeable income.

• ARM65 000RM65 000
• BRM52 143RM52 143
• CRM50 700RM50 700
• DRM55 000RM55 000

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D

Tax before the rebate = 2 100 + 400 = RM2 500. Then 2 500 = 1 800 + (x − 50 000) × 0.14, so (x − 50 000) = 700 ÷ 0.14 = 5 000 and x = RM55 000. RM52 143 forgets the zakat; RM65 000 forgets both the base tax and the zakat.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 109

11

QuestionSoalan 11

Panjang sebuah petak sayur segi empat tepat pada pelan ialah 6 cm. Pelan itu dibesarkan dan panjang imej petak itu ialah 15 cm. Cari faktor skala pembesaran itu.The length of a rectangular vegetable plot on a plan is 6 cm. The plan is enlarged and the length of the image of the plot is 15 cm. Find the scale factor of the enlargement.

• A0.40.4
• B22
• C2.52.5
• D99

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C

Scale factor = length of image ÷ length of object = 15 ÷ 6 = 2.5. 0.4 is the inverse (6 ÷ 15), and 9 is the difference 15 − 6.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.2.2

TextbookBuku teks p.ms. 139

12

QuestionSoalan 12

Luas sebuah pelan ladang ialah 64 cm². Pelan itu dibesarkan supaya luas imejnya ialah 196 cm². Sebuah sisi pada pelan asal ialah 6 cm. Cari panjang, dalam cm, sisi yang sepadan pada imej.The area of a plan of a farm is 64 cm². The plan is enlarged so that the area of its image is 196 cm². One side of the original plan is 6 cm. Find the length, in cm, of the corresponding side on the image.

• A10.510.5
• B18.418.4
• C1414
• D88

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A

Area scale factor = 196 ÷ 64 = k², so k = 14 ÷ 8 = 1.75. Length of side = 6 × 1.75 = 10.5 cm. 18.4 uses the area ratio instead of the length ratio.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.2.4

TextbookBuku teks p.ms. 146

13

QuestionSoalan 13

Rajah 1 menunjukkan segi tiga JKL dan imejnya J′K′L′ di bawah satu pembesaran. Antara berikut, yang manakah pusat dan faktor skala pembesaran itu?Diagram 1 shows triangle JKL and its image J′K′L′ under an enlargement. Which of the following is the centre and the scale factor of the enlargement?

• APusat (3, 4), faktor skala 2Centre (3, 4), scale factor 2
• BPusat (3, 4), faktor skala −2Centre (3, 4), scale factor −2
• CPusat (4, 3), faktor skala −2Centre (4, 3), scale factor −2
• DPusat (3, 4), faktor skala −1/2Centre (3, 4), scale factor −1/2

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B

Join J to J′ and K to K′. The lines meet at (3, 4), the centre. From the centre C(3, 4), J is at (−2, +1) and J′ is at (+4, −2), which is −2 times as far in the opposite direction. So k = −2.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 5, standard 5.2.2

TextbookBuku teks p.ms. 139

14

QuestionSoalan 14

Antara berikut, yang manakah sama dengan tan 300°?Which of the following is equal to tan 300°?

• Atan 60°tan 60°
• Btan 30°tan 30°
• C−tan 30°−tan 30°
• D−tan 60°−tan 60°

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D

300° is in quadrant IV, where tangent is negative. The reference angle is 360° − 300° = 60°, so tan 300° = −tan 60°.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 6, standard 6.1.2

TextbookBuku teks p.ms. 179

15

QuestionSoalan 15

Antara fungsi berikut, yang manakah mempunyai nilai maksimum 5, nilai minimum −1 dan tempoh 180° bagi 0° ≤ x ≤ 360°?Which of the following functions has a maximum value of 5, a minimum value of −1 and a period of 180° for 0° ≤ x ≤ 360°?

• Ay = 3 cos 2x + 2y = 3 cos 2x + 2
• By = 2 cos 3x + 3y = 2 cos 3x + 3
• Cy = 3 cos 180x + 2y = 3 cos 180x + 2
• Dy = 5 sin 2x − 1y = 5 sin 2x − 1

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A

Amplitude = (5 − (−1)) ÷ 2 = 3 and the graph is shifted up by (5 + (−1)) ÷ 2 = 2. Period 180° means b = 360° ÷ 180° = 2. So y = 3 cos 2x + 2. Option B has period 120° and minimum 1.

For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.2.2

TextbookBuku teks p.ms. 188

16

QuestionSoalan 16

Cari titik tengah bagi selang kelas 45 – 54.Find the midpoint of the class interval 45 – 54.

• A4949
• B5050
• C49.549.5
• D9999

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C

Midpoint = (45 + 54) ÷ 2 = 99 ÷ 2 = 49.5. Taking the mean of the lower limit and the next lower limit gives 50, and 99 is the sum.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 7, standard 7.1.1

TextbookBuku teks p.ms. 199

17

QuestionSoalan 17

Jadual di bawah menunjukkan hujan harian, dalam mm, selama 20 hari.

Hujan / Rainfall (mm)	1 – 10	11 – 20	21 – 30	31 – 40
Bilangan hari / Days	4	6	8	2

Cari min hujan harian itu.The table above shows the daily rainfall, in mm, over 20 days. Find the mean daily rainfall.

• A20.5 mm20.5 mm
• B19.5 mm19.5 mm
• C25.5 mm25.5 mm
• D22.5 mm22.5 mm

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B

Midpoints 5.5, 15.5, 25.5, 35.5. Σfx = 4(5.5) + 6(15.5) + 8(25.5) + 2(35.5) = 22 + 93 + 204 + 71 = 390. Mean = 390 ÷ 20 = 19.5 mm. 25.5 mm is only the midpoint of the modal class.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.1.3

TextbookBuku teks p.ms. 208

18

QuestionSoalan 18

Varians bagi satu set data ialah 9. Setiap nilai dalam set itu didarab dengan 2 dan kemudian ditambah dengan 6. Cari varians bagi set data yang baharu.The variance of a set of data is 9. Each value in the set is multiplied by 2 and then 6 is added. Find the variance of the new set of data.

• A3636
• B2424
• C1818
• D1515

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A

Multiplying every value by k multiplies the variance by k², so 9 × 2² = 36. Adding 6 shifts all values equally and does not change the variance. (Standard deviation would become 3 × 2 = 6.)

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

TextbookBuku teks p.ms. 215

19

QuestionSoalan 19

Apakah langkah pertama dalam proses pemodelan matematik?What is the first step in the mathematical modelling process?

• AMentafsir penyelesaianInterpreting the solution
• BMenguji dan memurnikan modelTesting and refining the model
• CMengenal pasti dan mentakrifkan masalahIdentifying and defining the problem
• DMenulis laporan penyelidikanWriting the report

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C

Modelling begins by identifying and defining the problem. Next come the variables and assumptions, then the formulation and solution of the model, the interpretation and testing, and finally the report.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.1.1

TextbookBuku teks p.ms. 228

20

QuestionSoalan 20

Hasil, Y kg, sebidang kebun cili dimodelkan oleh fungsi Y = 1 200 + 30x − x², dengan x kg ialah jisim baja yang digunakan. Cari hasil maksimum, dalam kg, kebun itu.The yield, Y kg, of a chilli plot is modelled by the function Y = 1 200 + 30x − x², where x kg is the mass of fertiliser used. Find the maximum yield, in kg, of the plot.

• A1 2001 200
• B1 6501 650
• C1 4001 400
• D1 4251 425

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D

Complete the square: Y = 1 425 − (x − 15)². The maximum is 1 425 kg when x = 15. (1 400 is the yield at x = 10; 1 200 is the yield with no fertiliser.)

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.1.2

TextbookBuku teks p.ms. 240

21

QuestionSoalan 21

Jumlah baja, F kg, yang diperlukan oleh sebuah batas sayur segi empat sama berubah secara langsung dengan kuasa dua panjang sisinya, s m. Sebuah batas dengan sisi 3 m memerlukan 0.9 kg baja.The mass of fertiliser, F kg, needed by a square vegetable bed varies directly as the square of the length of its side, s m. A bed with a side of 3 m needs 0.9 kg of fertiliser.

[1 markmarkah] (a)Pilih hubungan antara F dengan s.Choose the relation between F and s.

AF = 0.1s²BF = 0.3sCF = 0.3s²

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F = 0.1s²

F = ks². 0.9 = k(3²) = 9k, so k = 0.1 and F = 0.1s².

[1 markmarkah] (b)Pilih jisim baja yang diperlukan oleh sebuah batas dengan sisi 8 m.Choose the mass of fertiliser needed by a bed with a side of 8 m.

A0.8 kgB6.4 kgC2.4 kg

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6.4 kg

F = 0.1 × 8² = 0.1 × 64 = 6.4 kg. (2.4 kg comes from forgetting to square.)

[1 markmarkah] (c)Pilih panjang sisi sebuah batas yang memerlukan 14.4 kg baja.Choose the length of the side of a bed that needs 14.4 kg of fertiliser.

A12 mB144 mC24 m

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12 m

14.4 = 0.1s², so s² = 144 and s = 12 m.

[1 markmarkah] (d)Tandakan BENAR atau PALSU: Jika panjang sisi sebuah batas digandakan, jumlah baja yang diperlukan juga digandakan.Mark TRUE or FALSE: If the side of a bed is doubled, the mass of fertiliser needed is also doubled.

ABenarTrueBPalsuFalse

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False

F is proportional to s². Doubling s multiplies F by 2² = 4, so the fertiliser is four times as much.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1.2

TextbookBuku teks p.ms. 8

22

QuestionSoalan 22

Rajah di bawah ialah ringkasan borang cukai pendapatan Encik Zulkifli, seorang petani. Gunakan jadual: RM50 001 hingga RM70 000, cukai asas RM1 800 dan 14% bagi setiap ringgit selepas RM50 000.The table shows the summary of the income tax form of Encik Zulkifli, a farmer. Use the table: RM50 001 to RM70 000, base tax RM1 800 plus 14% of every ringgit above RM50 000.

Butiran / Item	Jumlah / Amount (RM)
Jumlah pendapatan / Total income	72 000
Pengecualian cukai / Tax exemption	2 000
Pelepasan cukai: individu 9 000, KWSP 4 000, perubatan 3 000 / Tax relief: individual 9 000, EPF 4 000, medical 3 000	16 000
Zakat dibayar / Zakat paid	360
Potongan cukai bulanan (PCB) / Monthly tax deduction (PCB)	150 × 12 bulan / months

[1 markmarkah] (a)Pilih pendapatan bercukai Encik Zulkifli.Choose Encik Zulkifli's chargeable income.

ARM54 000BRM56 000CRM70 000

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RM54 000

72 000 − 2 000 − 16 000 = RM54 000. (RM56 000 forgets the exemption.)

[1 markmarkah] (b)Pilih cukai pendapatan sebelum rebat zakat.Choose the income tax before the zakat rebate.

ARM2 360BRM2 600CRM7 560

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RM2 360

Tax = 1 800 + (54 000 − 50 000) × 14% = 1 800 + 560 = RM2 360.

[1 markmarkah] (c)Pilih cukai pendapatan yang perlu dibayar selepas rebat zakat.Choose the income tax payable after the zakat rebate.

ARM2 000BRM2 720CRM1 640

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RM2 000

2 360 − 360 = RM2 000. Zakat is a rebate, so it is subtracted from the tax.

[1 markmarkah] (d)Tandakan BENAR atau PALSU: Jumlah PCB kurang daripada cukai yang perlu dibayar, maka Encik Zulkifli perlu membayar baki RM200 kepada LHDN.Mark TRUE or FALSE: The total PCB is less than the tax payable, so Encik Zulkifli has to pay a balance of RM200 to the IRB.

ABenarTrueBPalsuFalse

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True

Total PCB = 150 × 12 = RM1 800. The balance = 2 000 − 1 800 = RM200 still to be paid.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 106

23

QuestionSoalan 23

Rajah 2 ialah pelan sebuah kebun sayur segi empat tepat yang dilukis dengan skala 1 cm : 4 m.Diagram 2 is the plan of a rectangular vegetable garden drawn to a scale of 1 cm : 4 m.

[1 markmarkah] (a)Pilih panjang sebenar kebun itu.Choose the actual length of the garden.

A18 mB4.5 mC9 m

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18 m

4.5 cm × 4 = 18 m.

[1 markmarkah] (b)Pilih lebar sebenar kebun itu.Choose the actual width of the garden.

A3 mB12 mC7 m

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12 m

3 cm × 4 = 12 m.

[1 markmarkah] (c)Pilih luas sebenar kebun itu.Choose the actual area of the garden.

A216 m²B54 m²C13.5 m²

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216 m²

18 × 12 = 216 m². (54 m² multiplies the plan area of 13.5 cm² by 4 instead of by 4² = 16 in m².)

[1 markmarkah] (d)Tandakan BENAR atau PALSU: Faktor skala luas bagi pembesaran daripada pelan kepada kebun sebenar ialah 400.Mark TRUE or FALSE: The area scale factor of the enlargement from the plan to the real garden is 400.

ABenarTrueBPalsuFalse

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False

The length scale factor is 400 (1 cm : 4 m = 1 : 400). The area scale factor is 400² = 160 000, not 400.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.2.2

TextbookBuku teks p.ms. 139

24

QuestionSoalan 24

Jadual 1 menunjukkan hasil susu harian, dalam liter, bagi 30 ekor lembu di sebuah ladang.Table 1 shows the daily milk yield, in litres, of 30 cows on a farm.

Hasil susu / Milk yield (l)	5 – 9	10 – 14	15 – 19	20 – 24	25 – 29
Bilangan lembu / Number of cows	2	6	10	k	4

[1 markmarkah] (a)Pilih nilai k.Choose the value of k.

A6B8C10

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8

2 + 6 + 10 + k + 4 = 30, so 22 + k = 30 and k = 8.

[1 markmarkah] (b)Pilih selang kelas yang mengandungi median.Choose the class interval that contains the median.

A10 – 14B15 – 19C20 – 24

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15 – 19

The median is between the 15th and 16th values. Cumulative frequencies: 2, 8, 18, so the 15th and 16th values lie in 15 – 19.

[1 markmarkah] (c)Pilih min hasil susu itu.Choose the mean milk yield.

A17.5 lB18.0 lC19.5 l

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18.0 l

Midpoints 7, 12, 17, 22, 27. Σfx = 14 + 72 + 170 + 176 + 108 = 540. Mean = 540 ÷ 30 = 18.0 l.

[1 markmarkah] (d)Tandakan BENAR atau PALSU: Sempadan bawah dan sempadan atas bagi kelas 20 – 24 ialah 19.5 dan 24.5.Mark TRUE or FALSE: The lower boundary and the upper boundary of the class 20 – 24 are 19.5 and 24.5.

ABenarTrueBPalsuFalse

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True

The data are whole numbers, so the boundaries are half a unit either side: 19.5 and 24.5.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.1.3

TextbookBuku teks p.ms. 208

25

QuestionSoalan 25

Isi padu air, V liter, dalam sebuah tangki air hujan dimodelkan oleh fungsi linear V = at + b, dengan t hari ialah masa selepas permulaan bulan. Pada t = 2, V = 740 dan pada t = 5, V = 500.The volume of water, V litres, in a rainwater tank is modelled by the linear function V = at + b, where t days is the time after the start of the month. At t = 2, V = 740 and at t = 5, V = 500.

[1 markmarkah] (a)Pilih nilai a.Choose the value of a.

A−80B80C−240

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−80

a = (500 − 740) ÷ (5 − 2) = −240 ÷ 3 = −80.

[1 markmarkah] (b)Pilih persamaan model itu.Choose the equation of the model.

AV = −80t + 900BV = −80t + 740CV = 80t + 900

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V = −80t + 900

740 = −80(2) + b, so b = 740 + 160 = 900.

[1 markmarkah] (c)Pilih nilai t, betul kepada satu tempat perpuluhan, apabila isi padu air ialah 150 liter.Choose the value of t, correct to one decimal place, when the volume of water is 150 litres.

A9.4B11.3C7.5

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9.4

150 = −80t + 900, so 80t = 750 and t = 9.375 ≈ 9.4 days. (11.3 is when the tank is empty.)

[1 markmarkah] (d)Tandakan BENAR atau PALSU: Tangki itu akan kosong selepas 11 hari.Mark TRUE or FALSE: The tank will be empty after 11 days.

ABenarTrueBPalsuFalse

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False

V = 0 when t = 900 ÷ 80 = 11.25 days. At t = 11, V = 900 − 880 = 20 litres, so water is still left.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.1.2

TextbookBuku teks p.ms. 240

26

QuestionSoalan 26

Encik Borhan membeli anak benih betik dan anak benih cili untuk kebunnya. Lima anak benih betik dan dua anak benih cili berharga RM100. Tiga anak benih betik dan empat anak benih cili berharga RM116. Harga sepokok anak benih betik ialah RMp dan harga sepokok anak benih cili ialah RMq.Encik Borhan buys papaya seedlings and chilli seedlings for his garden. Five papaya seedlings and two chilli seedlings cost RM100. Three papaya seedlings and four chilli seedlings cost RM116. The price of a papaya seedling is RMp and the price of a chilli seedling is RMq.

[2 marksmarkah] (a)Tulis maklumat di atas dalam bentuk persamaan matriks.Write the information above as a matrix equation.

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• 5p + 2q = 100 and 3p + 4q = 1161
• 5234 pq = 1001161

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[2 marksmarkah] (b)Cari matriks songsang bagi 5234.Find the inverse matrix of 5234.

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• Determinant = 5(4) − 2(3) = 141
• 114 4−2−351

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[3 marksmarkah] (c)Seterusnya, hitung nilai p dan nilai q.Hence, calculate the value of p and the value of q.

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• pq = 114 4−2−35 1001161
• p = (400 − 232) ÷ 14 = 121
• q = (−300 + 580) ÷ 14 = 201

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[1 markmarkah] (d)Hitung jumlah harga 8 anak benih betik dan 6 anak benih cili.Calculate the total price of 8 papaya seedlings and 6 chilli seedlings.

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• 8(12) + 6(20) = RM2161

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2.6

TextbookBuku teks p.ms. 62

27

QuestionSoalan 27

Suhu, T °C, di dalam sebuah rumah hijau pada t jam selepas pukul 6 pagi dimodelkan oleh fungsi T = 28 + 5 sin (15t)°, dengan 0 ≤ t ≤ 24.The temperature, T °C, inside a greenhouse t hours after 6 a.m. is modelled by the function T = 28 + 5 sin (15t)°, where 0 ≤ t ≤ 24.

[2 marksmarkah] (a)Nyatakan suhu maksimum dan suhu minimum rumah hijau itu.State the maximum temperature and the minimum temperature of the greenhouse.

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• Maximum = 28 + 5 = 33 °C1
• Minimum = 28 − 5 = 23 °C1

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[2 marksmarkah] (b)Hitung suhu, betul kepada satu tempat perpuluhan, pada t = 4.Calculate the temperature, correct to one decimal place, when t = 4.

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• T = 28 + 5 sin 60°1
• = 28 + 4.33 = 32.3 °C1

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[2 marksmarkah] (c)Cari semua nilai t, bagi 0 ≤ t ≤ 12, apabila suhu ialah 30.5 °C.Find all the values of t, for 0 ≤ t ≤ 12, when the temperature is 30.5 °C.

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• 28 + 5 sin 15t = 30.5, so sin 15t = 0.5 and 15t = 30° or 150°1
• t = 2 and t = 101

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[2 marksmarkah] (d)Nyatakan tempoh fungsi itu, dalam jam, dan nilai t apabila suhu paling rendah.State the period of the function, in hours, and the value of t when the temperature is lowest.

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• Period = 360° ÷ 15° = 24 hours1
• Lowest when sin 15t = −1: 15t = 270°, so t = 18 (midnight)1

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.2.3

TextbookBuku teks p.ms. 190

28

QuestionSoalan 28

Kos elektrik, C ringgit, bagi sebuah perkakas berubah secara langsung dengan kadar kuasa, P kW, dan masa penggunaan, t jam. Sebuah pemanas air 2 kW yang digunakan selama 3 jam berkos RM2.64.The cost of electricity, RMC, of an appliance varies directly as its power rating, P kW, and the time it is used, t hours. A 2 kW water heater used for 3 hours costs RM2.64.

[2 marksmarkah] (a)Ungkapkan C dalam sebutan P dan t.Express C in terms of P and t.

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• C = kPt, so 2.64 = k(2)(3) and k = 0.441
• C = 0.44Pt1

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[2 marksmarkah] (b)Hitung kos menggunakan sebuah perkakas 1.5 kW selama 8 jam.Calculate the cost of using a 1.5 kW appliance for 8 hours.

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• C = 0.44 × 1.5 × 81
• RM5.281

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[3 marksmarkah] (c)Bilangan hari, d, sebuah tong gas LPG bertahan berubah secara langsung dengan jisim gas, m kg, dan secara songsang dengan bilangan orang, n, dalam rumah itu. Sebuah tong 14 kg bertahan 28 hari bagi 4 orang. Cari d dalam sebutan m dan n, dan hitung bilangan hari sebuah tong 12 kg bertahan bagi 6 orang.The number of days, d, that an LPG gas cylinder lasts varies directly as the mass of gas, m kg, and inversely as the number of people, n, in the house. A 14 kg cylinder lasts 28 days for 4 people. Find d in terms of m and n, and calculate the number of days a 12 kg cylinder lasts for 6 people.

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• 28 = k(14)/4, so k = 81
• d = 8m/n1
• d = 8(12)/6 = 16 days1

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[3 marksmarkah] (d)Sebuah keluarga seramai 6 orang mahu sebuah tong gas bertahan sekurang-kurangnya 20 hari. Tong yang ada ialah 12 kg, 14 kg dan 18 kg. Tentukan tong yang paling kecil yang sesuai.A family of 6 wants a gas cylinder to last at least 20 days. The cylinders available are 12 kg, 14 kg and 18 kg. Determine the smallest suitable cylinder.

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• 20 = 8m/61
• m = 15 kg is needed at least1
• 18 kg cylinder (12 kg gives 16 days and 14 kg gives 18.7 days, both less than 20)1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.3.2

TextbookBuku teks p.ms. 28

29

QuestionSoalan 29

Encik Idris, seorang petani di Semenanjung Malaysia, mempunyai sebuah pikap 1 998 cc dengan jumlah diinsuranskan RM48 000. Dia berhak mendapat diskaun tanpa tuntutan (NCD) 30%. Jadual 2 menunjukkan petikan Tarif Motor. Kadar komprehensif ialah bagi RM1 000 pertama jumlah diinsuranskan, dan RM26 ditambah bagi setiap RM1 000 berikutnya atau sebahagiannya.Encik Idris, a farmer in Peninsular Malaysia, has a 1 998 cc pickup with a sum insured of RM48 000. He is entitled to a 30% no claim discount (NCD). Table 2 shows an extract of the Motor Tariff. The comprehensive rate is for the first RM1 000 of the sum insured, and RM26 is added for each further RM1 000 or part of it.

Kapasiti enjin tidak melebihi (cc) Engine capacity not exceeding (cc)	Komprehensif (RM) Comprehensive (RM)	Pihak ketiga (RM) Third party (RM)
1 400	273.80	120.60
1 650	305.50	135.00
2 200	339.10	151.20
3 050	372.60	167.40

[3 marksmarkah] (a)Hitung premium asas bagi polisi komprehensif.Calculate the basic premium for a comprehensive policy.

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• 2 200 cc row: RM339.10 for the first RM1 0001
• Extra = 26 × 47 = RM1 2221
• Basic premium = 339.10 + 1 222 = RM1 561.101

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[2 marksmarkah] (b)Hitung premium kasar yang perlu dibayar selepas NCD.Calculate the gross premium he has to pay after the NCD.

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• NCD = 30% × 1 561.10 = RM468.331
• Gross premium = 1 561.10 − 468.33 = RM1 092.771

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[2 marksmarkah] (c)Hitung premium asas jika dia memilih polisi pihak ketiga, kebakaran dan kecurian.Calculate the basic premium if he chooses a third party, fire and theft policy.

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• 75% × 1 561.101
• RM1 170.831

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[3 marksmarkah] (d)Isteri Encik Idris, 34 tahun, seorang bukan perokok, membeli polisi insurans hayat dengan nilai muka RM200 000 dengan kadar RM1.38 bagi setiap RM1 000. Dia menambah perlindungan penyakit kritikal sebanyak 25% daripada nilai muka dengan kadar RM1.64 bagi setiap RM1 000. Hitung jumlah premium tahunannya.Encik Idris's wife, 34, a non-smoker, buys a life insurance policy with a face value of RM200 000 at a rate of RM1.38 per RM1 000. She adds critical illness cover of 25% of the face value at a rate of RM1.64 per RM1 000. Calculate her total annual premium.

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• Critical illness cover = 25% × 200 000 = RM50 0001
• 200 × 1.38 + 50 × 1.641
• RM276 + RM82 = RM358.001

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1.2

TextbookBuku teks p.ms. 83

30

QuestionSoalan 30

Puan Rohani ialah seorang guru yang juga menyewakan tanah ladangnya. Maklumat cukai pendapatannya bagi satu tahun taksiran ditunjukkan dalam Jadual 3, bersama sebahagian jadual kadar cukai pendapatan.Puan Rohani is a teacher who also rents out her farmland. Her income tax information for one year of assessment is shown in Table 3, together with part of the income tax rate table.

Penyata cukai Puan Rohani / Tax statement of Puan Rohani
Gaji / Salary	RM6 500 × 12 bulan / months
Bonus tahunan / Annual bonus	RM8 000
Sewa tanah ladang / Rent from farmland	RM6 000
Pengecualian cukai / Tax exemption	RM1 500
Pelepasan cukai: individu RM9 000, pasangan RM4 000, dua orang anak RM2 000 setiap seorang, KWSP dan insurans hayat RM7 000 / Tax relief: individual RM9 000, spouse RM4 000, two children RM2 000 each, EPF and life insurance RM7 000	—
Zakat dibayar / Zakat paid	RM650
PCB / Monthly tax deduction	RM300 × 12 bulan / months

Pendapatan bercukai (RM) Chargeable income (RM)	Cukai asas (RM) Base tax (RM)	Kadar bagi baki Rate on the balance
50 001 – 70 000	1 800	14%
70 001 – 100 000	4 600	21%

[2 marksmarkah] (a)Hitung jumlah pendapatan tahunan Puan Rohani.Calculate Puan Rohani's total annual income.

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• 6 500 × 12 = RM78 0001
• 78 000 + 8 000 + 6 000 = RM92 0001

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[2 marksmarkah] (b)Hitung pendapatan bercukainya.Calculate her chargeable income.

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• Total relief = 9 000 + 4 000 + 2(2 000) + 7 000 = RM24 0001
• 92 000 − 1 500 − 24 000 = RM66 5001

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[3 marksmarkah] (c)Hitung cukai pendapatan sebelum rebat zakat.Calculate the income tax before the zakat rebate.

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• Base tax RM1 800 (range 50 001 – 70 000)1
• (66 500 − 50 000) × 14% = RM2 3101
• Tax = 1 800 + 2 310 = RM4 1101

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[2 marksmarkah] (d)Hitung cukai yang perlu dibayar selepas rebat zakat, dan nyatakan sama ada PCB mencukupi.Calculate the tax payable after the zakat rebate, and state whether the PCB is enough.

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• 4 110 − 650 = RM3 4601
• PCB = 300 × 12 = RM3 600, which is more than RM3 460, so it is enough; RM140 is refunded1

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[3 marksmarkah] (e)Puan Rohani berkata, "Jika saya menyumbang RM3 000 kepada skim persaraan swasta (PRS) yang layak mendapat pelepasan cukai tambahan, cukai saya akan berkurang sebanyak RM3 000." Adakah dia betul? Tunjukkan pengiraan anda.Puan Rohani says, "If I contribute RM3 000 to a private retirement scheme (PRS) that qualifies for extra tax relief, my tax will fall by RM3 000." Is she correct? Show your calculation.

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• New chargeable income = 66 500 − 3 000 = RM63 5001
• New tax = 1 800 + (63 500 − 50 000) × 14% = RM3 690; saving = 4 110 − 3 690 = RM4201
• She is not correct: relief lowers the income that is taxed, so the tax falls only by 14% of RM3 000 = RM4201

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For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 105

31

QuestionSoalan 31

Rajah 3 menunjukkan segi tiga A, B dan C pada satah Cartes. Rajah itu ialah pelan tiga petak tanaman. Segi tiga B ialah imej bagi segi tiga A di bawah pembesaran V. Segi tiga C ialah imej bagi segi tiga B di bawah transformasi W.Diagram 3 shows triangles A, B and C on a Cartesian plane. The diagram is the plan of three planting plots. Triangle B is the image of triangle A under the enlargement V. Triangle C is the image of triangle B under the transformation W.

[3 marksmarkah] (a)Huraikan selengkapnya pembesaran V.Describe in full the enlargement V.

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• Enlargement1
• Centre (−4, 0)1
• Scale factor 21

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[2 marksmarkah] (b)Huraikan selengkapnya transformasi W.Describe in full the transformation W.

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• Reflection1
• In the line x = 21

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[2 marksmarkah] (c)Luas bagi segi tiga C mewakili 3 600 m². Hitung luas sebenar, dalam m², yang diwakili oleh segi tiga A.The area of triangle C represents 3 600 m². Calculate the actual area, in m², represented by triangle A.

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• Area C = area B = 2² × area A = 4 × area A1
• Area A = 3 600 ÷ 4 = 900 m²1

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[2 marksmarkah] (d)Cari koordinat imej bagi titik P(−3, 3) di bawah gabungan transformasi WV.Find the coordinates of the image of point P(−3, 3) under the combined transformation WV.

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• V: P(−3, 3) → (−2, 6)1
• W: (−2, 6) → (6, 6)1

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[3 marksmarkah] (e)Cari koordinat imej bagi P di bawah VW. Seterusnya, nyatakan sama ada VW sama dengan WV, dengan memberi satu sebab.Find the coordinates of the image of P under VW. Hence, state whether VW is the same as WV, giving a reason.

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• W first: (−3, 3) → (7, 3)1
• V: (7, 3) → (18, 6)1
• Not the same: the image of P is (18, 6) under VW but (6, 6) under WV, so the order matters1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 5, standard 5.3.1

TextbookBuku teks p.ms. 152

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