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

Nilaikan 216¹³.Evaluate 216¹³.

• A7272
• B66
• C14.714.7
• D3636

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B

A fractional index 1/3 means the cube root: 216^1/3 = ∛216 = 6, because 6 × 6 × 6 = 216. Dividing 216 by 3 gives 72, 14.7 is the square root, and 36 is 6 × 6.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.7

TextbookBuku teks p.ms. 20

2

QuestionSoalan 2

Ringkaskan (2p³q⁻¹)³ × p⁻⁴q⁵.Simplify (2p³q⁻¹)³ × p⁻⁴q⁵.

• A6p⁵q²6p⁵q²
• B8p¹³q²8p¹³q²
• C8p⁵q⁻⁸8p⁵q⁻⁸
• D8p⁵q²8p⁵q²

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D

(2p³q⁻¹)³ = 2³p⁹q⁻³ = 8p⁹q⁻³. Then 8p⁹q⁻³ × p⁻⁴q⁵ = 8p⁹⁻⁴q⁻³⁺⁵ = 8p⁵q². Multiplying 2 by 3 gives the wrong coefficient 6.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.2.3

TextbookBuku teks p.ms. 11

3

QuestionSoalan 3

Diberi 3^x+2 − 3^x = 216. Cari nilai x.Given 3^x+2 − 3^x = 216. Find the value of x.

• A33
• B2727
• C22
• D55

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A

3^x+2 = 3^x × 3² = 9 × 3^x. So 9 × 3^x − 3^x = 8 × 3^x = 216, giving 3^x = 27 = 3³ and x = 3. Check: 3⁵ − 3³ = 243 − 27 = 216. The value 27 is 3^x, not x.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 1, standard 1.2.9

TextbookBuku teks p.ms. 23

4

QuestionSoalan 4

Jisim sebutir vitamin ialah 0.0503 g. Berapakah bilangan angka bererti bagi 0.0503?The mass of a vitamin tablet is 0.0503 g. How many significant figures does 0.0503 have?

• A22
• B33
• C44
• D55

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B

The zeros before the 5 are only place holders and are not significant. The significant figures are 5, 0 and 3, so there are 3. The zero between 5 and 3 is significant because it lies between non-zero digits.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.1.1

TextbookBuku teks p.ms. 34

5

QuestionSoalan 5

Cahaya bergerak pada laju 3.0 × 10⁵ km/s. Jarak Musytari dari Matahari ialah 7.78 × 10⁸ km. Hitung masa, dalam s, yang diambil oleh cahaya Matahari untuk sampai ke Musytari. Ungkapkan jawapan dalam bentuk piawai, betul kepada tiga angka bererti.Light travels at a speed of 3.0 × 10⁵ km/s. The distance of Jupiter from the Sun is 7.78 × 10⁸ km. Calculate the time, in s, that sunlight takes to reach Jupiter. Give the answer in standard form, correct to three significant figures.

• A2.59 × 10³2.59 × 10³
• B2.59 × 10⁴2.59 × 10⁴
• C2.33 × 10¹⁴2.33 × 10¹⁴
• D3.86 × 10⁻⁴3.86 × 10⁻⁴

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A

Time = distance ÷ speed = 7.78 × 10⁸ ÷ 3.0 × 10⁵ = 2.593 × 10³ = 2.59 × 10³ s. The value 3.86 × 10⁻⁴ comes from dividing speed by distance, and 2.33 × 10¹⁴ from multiplying.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.2.2

TextbookBuku teks p.ms. 42

6

QuestionSoalan 6

Jisim satu atom suatu unsur ialah 2.5 × 10⁻²³ g. Hitung bilangan atom unsur itu dalam 1 kg, dalam bentuk piawai.The mass of one atom of an element is 2.5 × 10⁻²³ g. Calculate the number of atoms of the element in 1 kg, in standard form.

• A4 × 10²²4 × 10²²
• B4 × 10²⁵4 × 10²⁵
• C2.5 × 10⁻²⁰2.5 × 10⁻²⁰
• D4 × 10¹⁹4 × 10¹⁹

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B

1 kg = 1 000 g = 10³ g. Number of atoms = 10³ ÷ (2.5 × 10⁻²³) = 0.4 × 10²⁶ = 4 × 10²⁵. The answer 4 × 10²² forgets to change kg to g.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 2, standard 2.2.3

TextbookBuku teks p.ms. 43

7

QuestionSoalan 7

Puan Sarah membeli sebidang tanah pada harga RM80 000 dan menjualnya pada harga RM92 000 selepas beberapa tahun. Tiada pendapatan lain diterima. Hitung pulangan atas pelaburan (ROI) beliau.Puan Sarah buys a piece of land for RM80 000 and sells it for RM92 000 after some years. No other income is received. Calculate her return on investment (ROI).

• A13.04%13.04%
• B12%12%
• C15%15%
• D87%87%

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C

Capital gain = 92 000 − 80 000 = RM12 000. ROI = 12 000 ÷ 80 000 × 100% = 15%. The value 13.04% divides by the selling price instead of the amount invested.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1.6

TextbookBuku teks p.ms. 71

8

QuestionSoalan 8

Seorang pelanggan mengambil pinjaman RM8 000 dengan faedah 9% setahun ke atas baki pinjaman. Ansuran bulanan ialah RM300. Selepas ansuran bulan pertama dibayar, berapakah baki pinjaman itu?A customer takes a loan of RM8 000 with interest of 9% per annum on the outstanding balance. The monthly instalment is RM300. After the first instalment is paid, what is the outstanding balance of the loan?

• ARM7 700RM7 700
• BRM7 820RM7 820
• CRM7 740RM7 740
• DRM7 760RM7 760

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D

Interest for the first month = 8 000 × 0.09 ÷ 12 = RM60. Balance = 8 000 + 60 − 300 = RM7 760. RM7 700 ignores the interest, and RM7 820 adds the interest to the wrong side of the instalment.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.2.5

TextbookBuku teks p.ms. 78

9

QuestionSoalan 9

Panjang sebuah dewan ialah 18 m. Dewan itu dilukis pada sebuah pelan lantai dengan skala 1 : 400. Berapakah panjang dewan itu pada pelan?The length of a hall is 18 m. The hall is drawn on a floor plan to the scale 1 : 400. What is the length of the hall on the plan?

• A45 cm45 cm
• B4.5 cm4.5 cm
• C0.45 cm0.45 cm
• D720 000 cm720 000 cm

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B

18 m = 1 800 cm. Length on plan = 1 800 ÷ 400 = 4.5 cm. The values 45 cm and 0.45 cm come from a slip with the units, and 720 000 cm multiplies by the scale.

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

TextbookBuku teks p.ms. 93

10

QuestionSoalan 10

Satu kawasan mempunyai luas 12 cm² pada sebuah pelan. Luas sebenar kawasan itu ialah 108 m². Cari skala pelan itu.A region has an area of 12 cm² on a plan. The actual area of the region is 108 m². Find the scale of the plan.

• A1 : 3001 : 300
• B1 : 9001 : 900
• C1 : 90 0001 : 90 000
• D1 : 3 0001 : 3 000

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A

108 m² = 1 080 000 cm². The area scale is 1 080 000 ÷ 12 = 90 000 = 300², so the length scale is 1 : 300. The value 1 : 90 000 is the area ratio, which has not been square-rooted.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.1.5

TextbookBuku teks p.ms. 101

11

QuestionSoalan 11

Tanpa menggunakan kalkulator, nilaikan tan 45° + kos 60°.Without using a calculator, evaluate tan 45° + cos 60°.

• A0.50.5
• B11
• C22
• D1.51.5

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D

tan 45° = 1 and cos 60° = 0.5, so the sum is 1 + 0.5 = 1.5. The value 1 leaves out cos 60°, and 2 treats cos 60° as 1.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1.5

TextbookBuku teks p.ms. 119

12

QuestionSoalan 12

Dari puncak sebuah tebing setinggi 40 m, sudut tunduk sebuah bot di laut ialah 35°. Hitung jarak mengufuk bot itu dari kaki tebing, betul kepada 0.1 m.From the top of a cliff 40 m high, the angle of depression of a boat at sea is 35°. Calculate the horizontal distance of the boat from the foot of the cliff, correct to 0.1 m.

• A28.0 m28.0 m
• B69.7 m69.7 m
• C57.1 m57.1 m
• D48.8 m48.8 m

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C

The angle of depression equals the angle of elevation of the cliff top from the boat. The height 40 m is opposite and the distance is adjacent: distance = 40 ÷ tan 35° = 57.1 m. The value 28.0 m is 40 tan 35°, 69.7 m is 40 ÷ sin 35° and 48.8 m is 40 ÷ cos 35°.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.1.7

TextbookBuku teks p.ms. 123

13

QuestionSoalan 13

Dalam segi tiga bersudut tegak ABC, ∠B = 90°, sin ∠A = 513 dan BC = 10 cm. Hitung luas, dalam cm², segi tiga itu.In the right-angled triangle ABC, ∠B = 90°, sin ∠A = 513 and BC = 10 cm. Calculate the area, in cm², of the triangle.

• A6060
• B120120
• C130130
• D240240

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B

sin A = BC ÷ AC = 5/13, so AC = 10 × 13 ÷ 5 = 26 cm. AB = √(26² − 10²) = √576 = 24 cm. Area = 1/2 × 24 × 10 = 120 cm². The value 240 forgets the 1/2.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 5, standard 5.1.4

TextbookBuku teks p.ms. 116

14

QuestionSoalan 14

Dalam Rajah 1, AB ialah diameter bulatan berpusat O. Cari nilai x.In Diagram 1, AB is the diameter of the circle with centre O. Find the value of x.

• A38°38°
• B76°76°
• C52°52°
• D90°90°

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C

The angle in a semicircle is 90°, so ∠ACB = 90°. In triangle ABC, x = 180° − 90° − 38° = 52°. The value 76° doubles the given angle.

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

TextbookBuku teks p.ms. 139

15

QuestionSoalan 15

Dalam Rajah 2, PQRS ialah sisi empat kitaran dan QRT ialah garis lurus. Cari nilai x.In Diagram 2, PQRS is a cyclic quadrilateral and QRT is a straight line. Find the value of x.

• A74°74°
• B106°106°
• C37°37°
• D148°148°

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A

The exterior angle of a cyclic quadrilateral equals the opposite interior angle, so x = ∠SRT = 74°. The value 106° is the interior angle ∠SRQ = 180° − 74°.

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

TextbookBuku teks p.ms. 148

16

QuestionSoalan 16

Sebuah piramid tegak mempunyai tapak segi empat sama dan puncaknya terletak tepat di atas pusat tapak. Yang manakah menerangkan pelan piramid itu?A right pyramid has a square base and its apex is directly above the centre of the base. Which describes the plan of the pyramid?

• ASegi empat sama tanpa sebarang garis di dalamnyaA square with no lines inside it
• BSegi tiga sama kakiAn isosceles triangle
• CSegi empat tepat dengan satu garis menegak di tengahA rectangle with one vertical line through the middle
• DSegi empat sama dengan dua pepenjuru yang bertemu di tengahA square with two diagonals that meet at the centre

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D

Looking from above, the base is a square and the apex is at its centre. The edges joining the apex to the four corners appear as the two diagonals, which meet at the centre. An isosceles triangle is a front elevation, not the plan.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 7, standard 7.2.1

TextbookBuku teks p.ms. 183

17

QuestionSoalan 17

Titik B dan titik C terletak 8 cm antara satu sama lain. Berapakah bilangan titik yang berjarak 5 cm daripada B dan juga 5 cm daripada C?Points B and C are 8 cm apart. How many points are 5 cm from B and also 5 cm from C?

• A00
• B22
• C11
• D44

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B

The circle of radius 5 cm about B and the circle of radius 5 cm about C intersect, because 5 + 5 = 10 is more than 8. They meet at two points on the perpendicular bisector of BC, each √(25 − 16) = 3 cm from the midpoint of BC.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 8, standard 8.2.6

TextbookBuku teks p.ms. 214

18

QuestionSoalan 18

Nyatakan kecerunan garis lurus y = 5 − 3x.State the gradient of the straight line y = 5 − 3x.

• A55
• B33
• C−3−3
• D−5−5

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C

Write the equation as y = −3x + 5. Comparing with y = mx + c gives m = −3. The value 5 is the y-intercept.

For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 9, standard 9.1.1

TextbookBuku teks p.ms. 229

19

QuestionSoalan 19

Yang manakah antara titik berikut terletak pada garis lurus 2x + 3y = 12?Which of the following points lies on the straight line 2x + 3y = 12?

• A(3, 2)(3, 2)
• B(6, 2)(6, 2)
• C(0, 3)(0, 3)
• D(2, 3)(2, 3)

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A

Substitute each point. For (3, 2): 2(3) + 3(2) = 6 + 6 = 12. The other points give 18, 9 and 13.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 9, standard 9.1.3

TextbookBuku teks p.ms. 234

20

QuestionSoalan 20

Garis lurus yang melalui titik P(−2, 1) dan Q(4, k) adalah selari dengan garis 2x − 3y = 6. Cari nilai k.The straight line through the points P(−2, 1) and Q(4, k) is parallel to the line 2x − 3y = 6. Find the value of k.

• A44
• B−3−3
• C1010
• D55

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D

2x − 3y = 6 gives y = (2/3)x − 2, so the gradient is 2/3. Parallel lines have equal gradients: (k − 1) ÷ (4 − (−2)) = 2/3, so k − 1 = 4 and k = 5. The value 10 uses the gradient 3/2.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.1.4

TextbookBuku teks p.ms. 238

21

QuestionSoalan 21

Soalan ini tentang hukum indeks.This question is about the laws of indices.

[2 marksmarkah] (a)Ringkaskan 12m⁵n⁻² ÷ 4m²n⁻⁵.Simplify 12m⁵n⁻² ÷ 4m²n⁻⁵.

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• 12 ÷ 4 = 3 [1]1
• 3m⁵⁻²n^{−2−(−5)} = 3m³n³ [1]1

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[2 marksmarkah] (b)Selesaikan 9^x = 27^x−2.Solve 9^x = 27^x−2.

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• 3^2x = 3^3(x−2) = 3^3x−6 [1]1
• 2x = 3x − 6, so x = 6 [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.9

TextbookBuku teks p.ms. 22

22

QuestionSoalan 22

Sebuah peta dilukis dengan skala 1 : 25 000. Jarak antara stesen bas dengan sebuah sekolah pada peta ialah 6.0 cm.A map is drawn to the scale 1 : 25 000. The distance between a bus station and a school on the map is 6.0 cm.

[1 markmarkah] (a)Hitung jarak sebenar, dalam km, antara stesen bas dengan sekolah itu.Calculate the actual distance, in km, between the bus station and the school.

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• 6.0 × 25 000 = 150 000 cm = 1.5 km1

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[3 marksmarkah] (b)Sebuah bas bergerak dari stesen bas ke sekolah dengan laju purata 30 km/j. Hitung masa perjalanan, dalam minit.A bus travels from the bus station to the school at an average speed of 30 km/h. Calculate the journey time, in minutes.

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• Time = 1.5 ÷ 30 [1]1
• = 0.05 hour [1]1
• 0.05 × 60 = 3 minutes [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1.5

TextbookBuku teks p.ms. 99

23

QuestionSoalan 23

Rajah 3 menunjukkan segi tiga bersudut tegak ABC dengan ∠C = 90°, ∠A = 30° dan AB = 12 cm.Diagram 3 shows a right-angled triangle ABC with ∠C = 90°, ∠A = 30° and AB = 12 cm.

[2 marksmarkah] (a)Tanpa menggunakan kalkulator, hitung panjang BC, dalam cm.Without using a calculator, calculate the length of BC, in cm.

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• BC = 12 sin 30° = 12 × 1/2 [1]1
• 6 cm [1]1

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[2 marksmarkah] (b)Tanpa menggunakan kalkulator, hitung panjang AC, dalam cm. Biarkan jawapan dalam bentuk surd.Without using a calculator, calculate the length of AC, in cm. Leave the answer as a surd.

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• AC = 12 cos 30° = 12 × √3/2 [1]1
• 6√3 cm [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1.5

TextbookBuku teks p.ms. 119

24

QuestionSoalan 24

Rajah 4 menunjukkan dongakan depan dan pelan sebuah pondok yang berbentuk prisma. Dongakan depan ialah muka hujung pondok itu.Diagram 4 shows the front elevation and the plan of a shelter in the shape of a prism. The front elevation is the end face of the shelter.

[1 markmarkah] (a)Nyatakan ukuran dongakan sisi, dalam m, apabila pondok itu dilihat dari sebelah kanan.State the measurements, in m, of the side elevation when the shelter is viewed from the right.

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• A rectangle 10 m by 5 m (3 m + 2 m high)1

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[1 markmarkah] (b)Hitung luas dongakan depan, dalam m².Calculate the area of the front elevation, in m².

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• 6 × 3 + 1/2 × 6 × 2 = 24 m²1

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[2 marksmarkah] (c)Hitung isi padu pondok itu, dalam m³.Calculate the volume of the shelter, in m³.

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• Volume = area of cross-section × length [1]1
• 24 × 10 = 240 m³ [1]1

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

TextbookBuku teks p.ms. 190

25

QuestionSoalan 25

Garis lurus L mempunyai persamaan y = 3x − 4. Titik P(k, 2k + 1) terletak pada L.The straight line L has the equation y = 3x − 4. The point P(k, 2k + 1) lies on L.

[1 markmarkah] (a)Cari nilai k.Find the value of k.

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• 2k + 1 = 3k − 4, so k = 51

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[1 markmarkah] (b)Cari pintasan-x bagi L.Find the x-intercept of L.

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• 0 = 3x − 4, so x = 4/31

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[2 marksmarkah] (c)Garis lurus M selari dengan L dan melalui pintasan-y bagi garis x + 2y = 10. Cari persamaan bagi M.The straight line M is parallel to L and passes through the y-intercept of the line x + 2y = 10. Find the equation of M.

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• y-intercept of x + 2y = 10: x = 0, y = 5 [1]1
• M is parallel to L, so m = 3 and the equation is y = 3x + 5 [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.1.5

TextbookBuku teks p.ms. 240

26

QuestionSoalan 26

Soalan ini tentang indeks dan pereputan bahan radioaktif.This question is about indices and the decay of a radioactive substance.

[2 marksmarkah] (a)Ringkaskan (3x⁻²y³)² × 4x⁵y⁻¹.Simplify (3x⁻²y³)² × 4x⁵y⁻¹.

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• 9x⁻⁴y⁶ × 4x⁵y⁻¹ [1]1
• 36xy⁵ [1]1

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[2 marksmarkah] (b)Nilaikan 16^−3/4 × 32^2/5.Evaluate 16^−3/4 × 32^2/5.

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• 16^−3/4 = (2⁴)^−3/4 = 2⁻³ = 1/8 and 32^2/5 = 2² = 4 [1]1
• 1/8 × 4 = 1/2 [1]1

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[3 marksmarkah] (c)Jisim, M gram, suatu bahan radioaktif selepas t separuh hayat diberi oleh M = 80 × (1/2)^t. (i) Hitung jisim bahan itu selepas 3 separuh hayat. (ii) Cari bilangan separuh hayat yang diambil supaya jisim bahan itu menjadi 0.625 g.The mass, M grams, of a radioactive substance after t half-lives is given by M = 80 × (1/2)^t. (i) Calculate the mass of the substance after 3 half-lives. (ii) Find the number of half-lives taken for the mass to become 0.625 g.

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• (i) M = 80 × (1/2)³ = 80 × 1/8 = 10 g [1]1
• (ii) (1/2)^t = 0.625 ÷ 80 = 1/128 = (1/2)⁷ [1]1
• t = 7 half-lives [1]1

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[3 marksmarkah] (d)Lina berkata bahawa 2ⁿ⁺¹ + 2ⁿ⁺¹ = 4ⁿ⁺¹ bagi semua nilai n. Adakah Lina betul? Berikan satu bukti.Lina says that 2ⁿ⁺¹ + 2ⁿ⁺¹ = 4ⁿ⁺¹ for every value of n. Is Lina correct? Give evidence.

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• No, Lina is not correct [1]1
• 2ⁿ⁺¹ + 2ⁿ⁺¹ = 2 × 2ⁿ⁺¹ = 2ⁿ⁺², which is not 4ⁿ⁺¹ = 2²ⁿ⁺² [1]1
• Counter-example: n = 1 gives 4 + 4 = 8, but 4² = 16 [1]1

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For the teacherUntuk guru: EvaluatingMenilai, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.9

TextbookBuku teks p.ms. 26

27

QuestionSoalan 27

Jadual menunjukkan beberapa data tentang Bumi dan Marikh. Laju cahaya ialah 3.0 × 10⁵ km/s.

Data	Nilai
Jarak Bumi dari Matahari	1.5 × 108 km
Jisim Bumi	5.97 × 1024 kg
Jisim Marikh	6.42 × 1023 kg
Jarak Bumi ke Marikh pada satu ketika	7.8 × 107 km

The table shows some data about the Earth and Mars. The speed of light is 3.0 × 10⁵ km/s.

Data	Value
Distance of the Earth from the Sun	1.5 × 108 km
Mass of the Earth	5.97 × 1024 kg
Mass of Mars	6.42 × 1023 kg
Distance from the Earth to Mars at one time	7.8 × 107 km

[2 marksmarkah] (a)Hitung masa, dalam s, yang diambil oleh cahaya Matahari untuk sampai ke Bumi. Ungkapkan jawapan dalam bentuk piawai.Calculate the time, in s, that sunlight takes to reach the Earth. Give the answer in standard form.

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• 1.5 × 10⁸ ÷ 3.0 × 10⁵ [1]1
• 5.0 × 10² s [1]1

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[3 marksmarkah] (b)Hitung jumlah jisim Bumi dan Marikh, dalam kg, dalam bentuk piawai.Calculate the total mass of the Earth and Mars, in kg, in standard form.

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• 6.42 × 10²³ = 0.642 × 10²⁴ [1]1
• 5.97 × 10²⁴ + 0.642 × 10²⁴ = 6.612 × 10²⁴ [1]1
• 6.612 × 10²⁴ kg [1]1

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[2 marksmarkah] (c)Berapa kali ganda jisim Bumi berbanding jisim Marikh? Berikan jawapan betul kepada tiga angka bererti.How many times is the mass of the Earth the mass of Mars? Give the answer correct to three significant figures.

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• 5.97 × 10²⁴ ÷ 6.42 × 10²³ [1]1
• 9.30 [1]1

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[3 marksmarkah] (d)Sebuah kapal angkasa bergerak dengan laju 2.5 × 10⁴ km/j. Siti berkata kapal itu sampai ke Marikh, pada jarak yang diberi dalam jadual, dalam masa kurang daripada 100 hari. Adakah Siti betul? Tunjukkan pengiraan anda.A spacecraft travels at a speed of 2.5 × 10⁴ km/h. Siti says that the spacecraft reaches Mars, at the distance given in the table, in less than 100 days. Is Siti correct? Show your calculation.

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• Time = 7.8 × 10⁷ ÷ 2.5 × 10⁴ = 3.12 × 10³ hours [1]1
• 3 120 ÷ 24 = 130 days [1]1
• 130 days is more than 100 days, so Siti is not correct [1]1

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

TextbookBuku teks p.ms. 44

28

QuestionSoalan 28

Rajah 5 ialah pelan lantai sebuah bilik darjah dan sebuah koridor, dilukis dengan skala 1 : 200.Diagram 5 is a floor plan of a classroom and a corridor, drawn to the scale 1 : 200.

[2 marksmarkah] (a)Cari panjang dan lebar sebenar, dalam m, bilik darjah itu.Find the actual length and width, in m, of the classroom.

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• 7.2 × 200 = 1 440 cm = 14.4 m [1]1
• 4.5 × 200 = 900 cm = 9 m [1]1

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[1 markmarkah] (b)Sebuah pintu berlebar 1.0 m. Berapakah lebarnya, dalam cm, pada pelan itu?A door is 1.0 m wide. What is its width, in cm, on the plan?

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• 100 ÷ 200 = 0.5 cm1

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[4 marksmarkah] (c)Lantai bilik darjah itu akan dipasang jubin segi empat sama bersisi 0.6 m. Hitung bilangan jubin yang diperlukan.The floor of the classroom is to be covered with square tiles of side 0.6 m. Calculate the number of tiles needed.

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• Actual area = 14.4 × 9 = 129.6 m² (or 32.4 × 200² ÷ 10 000) [1]1
• Area of one tile = 0.6 × 0.6 = 0.36 m² [1]1
• 129.6 ÷ 0.36 [1]1
• 360 tiles [1]1

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[3 marksmarkah] (d)Syukri melukis semula pelan itu dengan skala 1 : 100 dan berkata bahawa luas bilik darjah pada pelan baharu itu dua kali ganda luas pada pelan 1 : 200. Adakah Syukri betul? Terangkan dengan pengiraan.Syukri redraws the plan to the scale 1 : 100 and says that the area of the classroom on the new plan is twice the area on the 1 : 200 plan. Is Syukri correct? Explain with a calculation.

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• Not correct [1]1
• On the new plan the room is 14.4 cm by 9 cm, so the area is 129.6 cm², compared with 32.4 cm² [1]1
• Each length is doubled, so the area is multiplied by 2² = 4, not 2 [1]1

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

TextbookBuku teks p.ms. 92

29

QuestionSoalan 29

Rajah 6 menunjukkan sebuah bulatan berpusat O. Garis lurus PA dan PB ialah tangen kepada bulatan itu di A dan B. Titik C dan D terletak pada bulatan itu. Diberi ∠APB = 50° dan PA = 12 cm.Diagram 6 shows a circle with centre O. The straight lines PA and PB are tangents to the circle at A and B. The points C and D lie on the circle. Given ∠APB = 50° and PA = 12 cm.

[2 marksmarkah] (a)Hitung ∠AOB.Calculate ∠AOB.

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• ∠OAP = ∠OBP = 90° (tangent is perpendicular to the radius) [1]1
• 360° − 90° − 90° − 50° = 130° [1]1

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[2 marksmarkah] (b)Hitung ∠ACB dan ∠ADB.Calculate ∠ACB and ∠ADB.

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• ∠ACB = 130° ÷ 2 = 65° [1]1
• ∠ADB = 180° − 65° = 115° (opposite angles of a cyclic quadrilateral) [1]1

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[3 marksmarkah] (c)Zaki berkata bahawa ∠PAB = ∠ADB. Adakah Zaki betul? Terangkan dengan pengiraan.Zaki says that ∠PAB = ∠ADB. Is Zaki correct? Explain with a calculation.

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• PA = PB (tangents from a point), so ∠PAB = (180° − 50°) ÷ 2 = 65° [1]1
• ∠ADB = 115°, so the angles are not equal; Zaki is not correct [1]1
• ∠PAB equals ∠ACB (angle in the alternate segment), which is 65° [1]1

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[3 marksmarkah] (d)Hitung jejari bulatan itu, dalam cm, betul kepada dua tempat perpuluhan.Calculate the radius of the circle, in cm, correct to two decimal places.

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• OP bisects ∠APB, so ∠OPA = 25° and ∠OAP = 90° [1]1
• tan 25° = OA ÷ 12, so OA = 12 tan 25° [1]1
• 5.60 cm [1]1

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For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.3.4

TextbookBuku teks p.ms. 155

30

QuestionSoalan 30

Rajah 7 menunjukkan dua batang jalan lurus, OP dan OQ, yang bertemu di O dengan keadaan ∠POQ = 80°. Sebatang telaga akan digali pada titik W yang sama jarak dari kedua-dua jalan itu dan berjarak 120 m dari O.Diagram 7 shows two straight roads, OP and OQ, which meet at O such that ∠POQ = 80°. A well is to be dug at a point W that is equidistant from the two roads and is 120 m from O.

[2 marksmarkah] (a)Huraikan lokus bagi titik yang sama jarak dari OP dan OQ.Describe the locus of points that are equidistant from OP and OQ.

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• The bisector of angle POQ [1]1
• a straight line from O making an angle of 40° with each road [1]1

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[1 markmarkah] (b)Huraikan lokus bagi titik yang berjarak 120 m dari O.Describe the locus of points that are 120 m from O.

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• A circle (arc) with centre O and radius 120 m1

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[3 marksmarkah] (c)Hitung jarak serenjang, dalam m, W dari jalan OP, betul kepada 0.1 m.Calculate the perpendicular distance, in m, of W from the road OP, correct to 0.1 m.

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• W is on the bisector, so the angle between OW and OP is 40° [1]1
• Distance = 120 sin 40° [1]1
• 77.1 m [1]1

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[2 marksmarkah] (d)Peraturan daerah menetapkan bahawa telaga mesti sekurang-kurangnya 80 m dari mana-mana jalan. Adakah kedudukan W mematuhi peraturan itu? Berikan alasan anda.A district rule says that a well must be at least 80 m from any road. Does the position of W follow the rule? Give your reason.

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• No [1]1
• 77.1 m is less than 80 m [1]1

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[2 marksmarkah] (e)Cari jarak minimum, dalam m (nombor bulat), W dari O supaya peraturan itu dipatuhi.Find the least distance, as a whole number of m, of W from O so that the rule is followed.

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• OW sin 40° ≥ 80, so OW ≥ 80 ÷ sin 40° = 124.5 m [1]1
• 125 m [1]1

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For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 8, standard 8.2.6

TextbookBuku teks p.ms. 214

31

QuestionSoalan 31

Rajah 8 menunjukkan segi empat selari ABCD pada satah Cartesan dengan A(1, 1), B(5, 3), C(7, 7) dan D(3, 5).Diagram 8 shows a parallelogram ABCD on a Cartesian plane with A(1, 1), B(5, 3), C(7, 7) and D(3, 5).

[2 marksmarkah] (a)Cari persamaan garis lurus AB.Find the equation of the straight line AB.

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• m = (3 − 1) ÷ (5 − 1) = 1/2 [1]1
• 1 = (1/2)(1) + c, so y = (1/2)x + 1/2 [1]1

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[2 marksmarkah] (b)Cari persamaan garis lurus DC.Find the equation of the straight line DC.

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• DC is parallel to AB, so m = 1/2 [1]1
• 5 = (1/2)(3) + c, so y = (1/2)x + 7/2 [1]1

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[2 marksmarkah] (c)Hafiz berkata bahawa titik E(10, 8) terletak pada garis lurus DC. Adakah Hafiz betul? Tunjukkan pengiraan anda.Hafiz says that the point E(10, 8) lies on the straight line DC. Is Hafiz correct? Show your calculation.

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• x = 10: y = (1/2)(10) + 7/2 = 8.5 [1]1
• 8.5 ≠ 8, so E does not lie on DC; Hafiz is not correct [1]1

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[4 marksmarkah] (d)Cari koordinat titik persilangan pepenjuru AC dan BD.Find the coordinates of the point of intersection of the diagonals AC and BD.

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• AC: m = (7 − 1) ÷ (7 − 1) = 1, so y = x [1]1
• BD: m = (5 − 3) ÷ (3 − 5) = −1, so y = −x + 8 [1]1
• x = −x + 8, so x = 4 [1]1
• (4, 4) [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.1.7

TextbookBuku teks p.ms. 245

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