MatematikMathematics

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1

QuestionSoalan 1

Isi padu satu larutan dikecilkan kepada 181 daripada isi padu asalnya. Ungkapkan 181 sebagai nombor dengan asas 3.The volume of a solution is reduced to 181 of its original volume. Express 181 as a number with base 3.

• A3⁻⁴3⁻⁴
• B3⁴3⁴
• C3⁻³3⁻³
• D3⁻²⁷3⁻²⁷

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A

81 = 3⁴, and a negative index gives the reciprocal: 1/81 = 1 ÷ 3⁴ = 3⁻⁴. Option B has the right base and power but the wrong sign, and 3⁻³ is 1/27.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.5

TextbookBuku teks p.ms. 14

2

QuestionSoalan 2

Ringkaskan (p⁵q⁻²)³ ÷ p⁴q⁻⁷.Simplify (p⁵q⁻²)³ ÷ p⁴q⁻⁷.

• Apq⁵pq⁵
• Bp¹¹q⁻¹³p¹¹q⁻¹³
• Cp¹⁹q⁻¹³p¹⁹q⁻¹³
• Dp¹¹qp¹¹q

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D

(p⁵q⁻²)³ = p¹⁵q⁻⁶. Then p¹⁵q⁻⁶ ÷ p⁴q⁻⁷ = p¹⁵⁻⁴q^{−6−(−7)} = p¹¹q. Option B adds the powers of q instead of subtracting.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.2.4

TextbookBuku teks p.ms. 11

3

QuestionSoalan 3

Diberi 4^x+1 = 8^x−1. Cari nilai x.Given 4^x+1 = 8^x−1. Find the value of x.

• A33
• B55
• C77
• D−5−5

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B

Write both sides with base 2: 4^x+1 = 2^2x+2 and 8^x−1 = 2^3x−3. Equate the indices: 2x + 2 = 3x − 3, so x = 5. The value 3 comes from wrongly solving 4(x + 1) = 8(x − 1).

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

TextbookBuku teks p.ms. 23

4

QuestionSoalan 4

Jarak purata Bulan dari Bumi ialah 384 400 km. Ungkapkan jarak itu dalam bentuk piawai.The mean distance of the Moon from the Earth is 384 400 km. Express the distance in standard form.

• A3.844 × 10⁵3.844 × 10⁵
• B38.44 × 10⁴38.44 × 10⁴
• C3.844 × 10⁻⁵3.844 × 10⁻⁵
• D3.844 × 10⁶3.844 × 10⁶

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A

Move the decimal point 5 places to the left to get 3.844, so the index is +5: 3.844 × 10⁵. Option B is not in standard form because 38.44 is not between 1 and 10.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2.1

TextbookBuku teks p.ms. 37

5

QuestionSoalan 5

Sebutir pasir berjisim 2.5 × 10⁻⁵ kg. Hitung bilangan butir pasir dalam 6 × 10² kg pasir dan ungkapkan jawapan dalam bentuk piawai.A grain of sand has a mass of 2.5 × 10⁻⁵ kg. Calculate the number of grains in 6 × 10² kg of sand and give the answer in standard form.

• A1.5 × 10⁻²1.5 × 10⁻²
• B2.4 × 10⁻³2.4 × 10⁻³
• C2.4 × 10⁷2.4 × 10⁷
• D2.4 × 10⁸2.4 × 10⁸

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C

Number of grains = 6 × 10² ÷ 2.5 × 10⁻⁵ = (6 ÷ 2.5) × 10²⁻⁽⁻⁵⁾ = 2.4 × 10⁷. When dividing, subtract the indices: 2 − (−5) = 7.

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

TextbookBuku teks p.ms. 40

6

QuestionSoalan 6

Seramai 482 693 orang mengunjungi satu pameran sains. Bundarkan 482 693 kepada tiga angka bererti.A total of 482 693 people visited a science exhibition. Round off 482 693 to three significant figures.

• A48 30048 300
• B482 000482 000
• C482 700482 700
• D483 000483 000

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D

The three significant figures are 4, 8, 2 and the next digit (6) rounds the 2 up to 3. The number keeps its size, so 482 693 ≈ 483 000. Option A has lost a place value and 482 700 is four significant figures.

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

TextbookBuku teks p.ms. 34

7

QuestionSoalan 7

Cikgu Rahim menyimpan RM7 500 dalam akaun yang membayar faedah mudah 3.2% setahun. Hitung jumlah simpanannya selepas 4 tahun.Cikgu Rahim saves RM7 500 in an account that pays simple interest of 3.2% per annum. Calculate his total savings after 4 years.

• ARM960RM1 120
• BRM7 740RM8 224
• CRM8 460RM9 120
• DRM9 600RM10 240

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C

I = Prt = 7 500 × 0.032 × 4 = RM960. Total savings = 7 500 + 960 = RM8 460. RM960 is only the interest, and RM7 740 is the savings after 1 year.

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

TextbookBuku teks p.ms. 55

8

QuestionSoalan 8

Harga tunai sebuah komputer riba ialah RM3 200. Zahar membayar deposit 15% dan baki dibayar secara ansuran selama 2 tahun dengan kadar faedah rata 4% setahun ke atas baki. Hitung ansuran bulanannya.The cash price of a laptop is RM3 200. Zahar pays a 15% deposit and pays the balance by instalments over 2 years at a flat rate of 4% per annum on the balance. Calculate his monthly instalment.

• ARM113.33RM113.33
• BRM122.40RM122.40
• CRM133.33RM133.33
• DRM144.00RM144.00

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B

Balance = 85% × 3 200 = RM2 720. Interest = 2 720 × 0.04 × 2 = RM217.60. Total to repay = RM2 937.60, so the monthly instalment = 2 937.60 ÷ 24 = RM122.40. RM113.33 leaves out the interest and RM144.00 charges interest on the full price.

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

TextbookBuku teks p.ms. 78

9

QuestionSoalan 9

Skala sebuah peta ialah 1 : 40 000. Jarak sebenar antara dua stesen ialah 6 km. Berapakah jarak antara dua stesen itu pada peta?The scale of a map is 1 : 40 000. The actual distance between two stations is 6 km. What is the distance between the two stations on the map?

• A1.5 cm1.5 cm
• B15 cm15 cm
• C150 cm150 cm
• D1 500 cm1 500 cm

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B

6 km = 600 000 cm. Distance on the map = 600 000 ÷ 40 000 = 15 cm. Dividing converts the actual distance to the map distance; multiplying by 40 000 would give the opposite.

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

TextbookBuku teks p.ms. 93

10

QuestionSoalan 10

Pelan sebuah dewan dilukis pada skala 1 : 200 dan luas dewan pada pelan itu ialah 18 cm². Dewan yang sama dilukis semula pada skala 1 : 500. Berapakah luas dewan pada lukisan baharu itu?The plan of a hall is drawn to the scale 1 : 200 and the area of the hall on the plan is 18 cm². The same hall is redrawn to the scale 1 : 500. What is the area of the hall on the new drawing?

• A2.88 cm²2.88 cm²
• B7.2 cm²7.2 cm²
• C45 cm²45 cm²
• D112.5 cm²112.5 cm²

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A

Lengths shrink by 200/500 = 2/5, so areas shrink by (2/5)² = 4/25: 18 × 4/25 = 2.88 cm². Option B uses the ratio of the lengths without squaring.

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

TextbookBuku teks p.ms. 101

11

QuestionSoalan 11

Dalam segi tiga bersudut tegak KLM, ∠L = 90°, KL = 9 cm dan LM = 40 cm. Cari nilai sin ∠LKM.In the right-angled triangle KLM, ∠L = 90°, KL = 9 cm and LM = 40 cm. Find the value of sin ∠LKM.

• A40414041
• B941941
• C940940
• D409409

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A

KM = √(9² + 40²) = √1 681 = 41 cm. The side opposite ∠LKM is LM = 40 cm, so sin ∠LKM = 40/41. The ratio 40/9 is tan ∠LKM.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1.4

TextbookBuku teks p.ms. 114

12

QuestionSoalan 12

Dari puncak sebuah rumah api setinggi 45 m, sudut tunduk sebuah bot ialah 28°. Hitung jarak mengufuk, betul kepada 0.1 m, bot itu dari kaki rumah api.From the top of a lighthouse 45 m high, the angle of depression of a boat is 28°. Calculate the horizontal distance of the boat from the foot of the lighthouse, correct to 0.1 m.

• A23.9 m23.9 m
• B39.7 m39.7 m
• C84.6 m84.6 m
• D95.9 m95.9 m

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C

The angle of depression equals the angle of elevation of the top from the boat. tan 28° = 45 ÷ d, so d = 45 ÷ tan 28° = 84.6 m. The value 23.9 m is 45 tan 28° and 95.9 m is the distance along the line of sight, 45 ÷ sin 28°.

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

TextbookBuku teks p.ms. 122

13

QuestionSoalan 13

Dalam sebuah segi tiga bersudut tegak, tan θ = 512 dengan θ ialah satu sudut tirus. Hipotenus segi tiga itu ialah 39 cm. Hitung luas segi tiga itu.In a right-angled triangle, tan θ = 512 where θ is an acute angle. The hypotenuse of the triangle is 39 cm. Calculate the area of the triangle.

• A90 cm²90 cm²
• B195 cm²195 cm²
• C270 cm²270 cm²
• D540 cm²540 cm²

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C

The sides are in the ratio 5 : 12 : 13 and 39 = 13 × 3, so the sides are 15 cm and 36 cm. Area = ½ × 15 × 36 = 270 cm². Option D forgets the ½.

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

TextbookBuku teks p.ms. 115

14

QuestionSoalan 14

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

• A37°37°
• B53°53°
• C90°90°
• D143°143°

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B

The angle in a semicircle is 90°, so ∠ACB = 90°. In triangle ABC, x = 180° − 90° − 37° = 53°. The value 37° is just the given angle.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 6, standard 6.1.1

TextbookBuku teks p.ms. 139

15

QuestionSoalan 15

Dalam Rajah 2, PQRS ialah sebuah sisi empat kitaran dan QR dipanjangkan ke T. Cari nilai ∠SRT.In Diagram 2, PQRS is a cyclic quadrilateral and QR is extended to T. Find the value of ∠SRT.

• A34°34°
• B51°51°
• C59°59°
• D85°85°

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C

The exterior angle of a cyclic quadrilateral equals the opposite interior angle: 3x + 8 = 5x − 26, so x = 17. Then ∠SRT = 3(17) + 8 = 59°. The values 34, 51 and 85 are 2x, 3x and 5x.

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

TextbookBuku teks p.ms. 147

16

QuestionSoalan 16

Sebatang rod pepejal silinder berjejari 2.5 cm dan panjang 12 cm diletakkan terbaring pada satah mengufuk. Apakah bentuk pelan rod itu?A solid cylindrical rod of radius 2.5 cm and length 12 cm lies on its curved surface on a horizontal plane. What is the shape of the plan of the rod?

• ABulatan berdiameter 5 cmA circle of diameter 5 cm
• BSegi empat tepat 12 cm × 2.5 cmA rectangle 12 cm × 2.5 cm
• CBulatan berjejari 12 cmA circle of radius 12 cm
• DSegi empat tepat 12 cm × 5 cmA rectangle 12 cm × 5 cm

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D

Seen from above, the rod shows its length (12 cm) and its full width, which is the diameter, 2 × 2.5 = 5 cm. So the plan is a rectangle 12 cm × 5 cm. The circle is the view from one end.

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

TextbookBuku teks p.ms. 182

17

QuestionSoalan 17

Dua garis lurus AB dan CD bersilang di O. Titik X bergerak supaya sentiasa sama jarak dari AB dan CD, dan sentiasa berjarak 5 cm dari O. Berapakah bilangan kedudukan X?Two straight lines AB and CD intersect at O. A point X moves so that it is always the same distance from AB and CD, and always 5 cm from O. How many positions of X are there?

• A11
• B22
• C33
• D44

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D

Points equidistant from two intersecting lines lie on both angle bisectors, which are two lines through O. The circle of radius 5 cm about O cuts each bisector twice, giving 2 × 2 = 4 positions.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.2.6

TextbookBuku teks p.ms. 214

18

QuestionSoalan 18

Titik yang manakah terletak pada garis lurus 3x − 2y = 12?Which point lies on the straight line 3x − 2y = 12?

• A(0, 6)(0, 6)
• B(4, 1)(4, 1)
• C(6, −3)(6, −3)
• D(2, −3)(2, −3)

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D

Substitute each point: (2, −3) gives 3(2) − 2(−3) = 6 + 6 = 12, which matches. (4, 1) gives 10, (0, 6) gives −12 and (6, −3) gives 24.

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

TextbookBuku teks p.ms. 233

19

QuestionSoalan 19

Cari kecerunan garis lurus x/4 − y/3 = 1.Find the gradient of the straight line x/4 − y/3 = 1.

• A−34−34
• B3434
• C4343
• D−3−3

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B

Multiply by 12: 3x − 4y = 12, so 4y = 3x − 12 and y = (3/4)x − 3. The gradient is 3/4. The value −3 is the y-intercept, and 4/3 turns the fraction upside down.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 9, standard 9.1.2

TextbookBuku teks p.ms. 231

20

QuestionSoalan 20

Garis y = 2x + 1 dan 3x + y = 16 bersilang di P. Satu garis lurus melalui P dan selari dengan x + 2y = 6. Cari persamaan garis lurus itu.The lines y = 2x + 1 and 3x + y = 16 intersect at P. A straight line passes through P and is parallel to x + 2y = 6. Find the equation of the line.

• Ax + 2y = 17x + 2y = 17
• Bx + 2y = 6x + 2y = 6
• Cx − 2y = −11x − 2y = −11
• D2x + y = 132x + y = 13

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A

Substitute y = 2x + 1 into 3x + y = 16: 5x + 1 = 16, so x = 3 and y = 7, giving P(3, 7). The line x + 2y = 6 has gradient −1/2, so the parallel line is x + 2y = 3 + 14 = 17. Option B is parallel but does not pass through P.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 9, standard 9.1.6

TextbookBuku teks p.ms. 242

21

QuestionSoalan 21

Satu nanometer (nm) ialah 10⁻⁹ m. Sebuah cip komputer menggunakan transistor yang berlebar 7 nm.One nanometre (nm) is 10⁻⁹ m. A computer chip uses transistors that are 7 nm wide.

[1 markmarkah] (a)Ungkapkan 7 nm dalam m, dalam bentuk piawai.Express 7 nm in m, in standard form.

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• 7 × 10⁻⁹ m1

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[2 marksmarkah] (b)Sebanyak 2 500 transistor disusun bersebelahan dalam satu baris tanpa jurang. Hitung jumlah panjang baris itu, dalam m, dan ungkapkan jawapan dalam bentuk piawai.2 500 transistors are placed side by side in one row with no gaps. Calculate the total length of the row, in m, and express the answer in standard form.

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• 2.5 × 10³ × 7 × 10⁻⁹ = 17.5 × 10⁻⁶ [1]1
• 1.75 × 10⁻⁵ m [1]1

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[1 markmarkah] (c)Satu mikrometer (µm) ialah 10⁻⁶ m. Ungkapkan jawapan di (b) dalam µm.One micrometre (µm) is 10⁻⁶ m. Express the answer to (b) in µm.

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• 1.75 × 10⁻⁵ ÷ 10⁻⁶ = 17.5 µm1

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

TextbookBuku teks p.ms. 43

22

QuestionSoalan 22

Encik Zaki membeli sebuah lot kedai dengan harga RM360 000. Dia menyewakan lot kedai itu pada kadar RM1 800 sebulan selama 2 tahun, kemudian menjualnya dengan harga RM396 000.Encik Zaki buys a shop lot for RM360 000. He rents it out at RM1 800 a month for 2 years and then sells it for RM396 000.

[1 markmarkah] (a)Hitung jumlah sewa yang diterima dalam 2 tahun.Calculate the total rent received in 2 years.

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• 1 800 × 24 = RM43 2001

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[1 markmarkah] (b)Hitung keuntungan modal apabila lot kedai itu dijual.Calculate the capital gain when the shop lot is sold.

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• 396 000 − 360 000 = RM36 0001

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[2 marksmarkah] (c)Hitung jumlah pulangan dan kadar pulangan atas pelaburan (ROI) selepas 2 tahun.Calculate the total return and the return on investment (ROI) after 2 years.

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• Total return = 43 200 + 36 000 = RM79 200 [1]1
• ROI = 79 200/360 000 × 100% = 22% [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.1.4

TextbookBuku teks p.ms. 66

23

QuestionSoalan 23

Skala sebuah peta ialah 1 : 250 000. Panjang sebatang lebuh raya pada peta itu ialah 7.2 cm. Sebuah kereta bergerak di sepanjang lebuh raya itu dengan laju malar 60 km/j.The scale of a map is 1 : 250 000. The length of a highway on the map is 7.2 cm. A car travels along the highway at a constant speed of 60 km/h.

[1 markmarkah] (a)Berapa kilometerkah yang diwakili oleh 1 cm pada peta itu?How many kilometres are represented by 1 cm on the map?

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• 250 000 cm = 2.5 km1

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[1 markmarkah] (b)Hitung panjang sebenar lebuh raya itu, dalam km.Calculate the actual length of the highway, in km.

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• 7.2 × 2.5 = 18 km1

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[2 marksmarkah] (c)Hitung masa yang diambil oleh kereta itu, dalam minit, untuk melalui seluruh lebuh raya itu.Calculate the time taken by the car, in minutes, to travel the whole highway.

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• Time = 18 ÷ 60 = 0.3 hours [1]1
• 0.3 × 60 = 18 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. 101

24

QuestionSoalan 24

Rajah 3 menunjukkan dua bulatan berpusat O dan P yang berjejari 10 cm dan 2 cm. AB ialah tangen sepunya yang menyentuh bulatan itu di A dan B. OP = 17 cm.Diagram 3 shows two circles with centres O and P and radii 10 cm and 2 cm. AB is a common tangent that touches the circles at A and B. OP = 17 cm.

[1 markmarkah] (a)Nyatakan nilai ∠OAB.State the value of ∠OAB.

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• 90° (the radius is perpendicular to the tangent at the point of contact)1

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[1 markmarkah] (b)Garis putus-putus PC dilukis selari dengan AB dan bertemu OA di C. Cari panjang OC.The dashed line PC is drawn parallel to AB and meets OA at C. Find the length of OC.

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• ABPC is a rectangle, so CA = BP = 2 cm; OC = 10 − 2 = 8 cm1

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[2 marksmarkah] (c)Hitung panjang AB.Calculate the length of AB.

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• ∠OCP = 90°, so PC² = 17² − 8² = 225 [1]1
• AB = PC = 15 cm [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 6, standard 6.3.5

TextbookBuku teks p.ms. 158

25

QuestionSoalan 25

Rajah 4 menunjukkan dua batang jalan lurus, QP dan QR, yang bertemu di Q dengan ∠PQR = 64°. Sebuah lampu jalan L dipasang supaya sama jarak dari kedua-dua jalan itu dan berjarak 50 m dari Q.Diagram 4 shows two straight roads, QP and QR, that meet at Q with ∠PQR = 64°. A street lamp L is fixed so that it is the same distance from both roads and 50 m from Q.

[1 markmarkah] (a)Huraikan lokus bagi L yang sama jarak dari QP dan QR.Describe the locus of L that is equidistant from QP and QR.

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• The bisector of ∠PQR (a straight line through Q that divides the angle into two equal parts)1

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[1 markmarkah] (b)Hitung sudut antara QL dengan QP.Calculate the angle between QL and QP.

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• 64° ÷ 2 = 32°1

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[2 marksmarkah] (c)Hitung jarak serenjang, dalam m, lampu itu dari jalan QP, betul kepada 0.1 m.Calculate the perpendicular distance, in m, of the lamp from the road QP, correct to 0.1 m.

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• Distance = 50 sin 32° [1]1
• 26.5 m [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.2.5

TextbookBuku teks p.ms. 211

26

QuestionSoalan 26

Rajah 5 menunjukkan sebuah menara tegak TF di atas tanah mengufuk. Dua orang pemerhati berdiri di A dan B, di kedua-dua belah menara itu, dengan A, F dan B berada dalam satu garis lurus. AB = 120 m. Sudut dongakan T dari A ialah 42° dan dari B ialah 31°. Tinggi menara ialah h m.Diagram 5 shows a vertical tower TF on horizontal ground. Two observers stand at A and B, on opposite sides of the tower, with A, F and B in a straight line. AB = 120 m. The angle of elevation of T from A is 42° and from B is 31°. The height of the tower is h m.

[1 markmarkah] (a)Ungkapkan AF dalam sebutan h.Express AF in terms of h.

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• AF = h ÷ tan 42°1

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[1 markmarkah] (b)Ungkapkan BF dalam sebutan h.Express BF in terms of h.

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• BF = h ÷ tan 31°1

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[3 marksmarkah] (c)Hitung h, betul kepada 0.1 m.Calculate h, correct to 0.1 m.

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• h ÷ tan 42° + h ÷ tan 31° = 120 [1]1
• h(1.1106 + 1.6643) = 120, so h = 120 ÷ 2.7749 [1]1
• 43.2 m [1]1

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[2 marksmarkah] (d)Pemerhati di A memandang ke atas ke arah T. Nyatakan sudut tunduk A dari T dan hitung jarak TA, betul kepada 0.1 m.Observer A looks up at T. State the angle of depression of A from T and calculate the distance TA, correct to 0.1 m.

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• 42° (equal to the angle of elevation, alternate angles) [1]1
• TA = 43.24 ÷ sin 42° = 64.6 m [1]1

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[3 marksmarkah] (e)Aini berkata bahawa kaki menara F lebih dekat dengan B kerana B mempunyai sudut dongakan yang lebih kecil. Adakah Aini betul? Tunjukkan pengiraan anda.Aini says that the foot of the tower F is closer to B because B has the smaller angle of elevation. Is Aini correct? Show your calculation.

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• Not correct [1]1
• AF = 43.24 ÷ tan 42° = 48.0 m and BF = 43.24 ÷ tan 31° = 72.0 m [1]1
• F is closer to A: for the same height, the smaller angle (at B) means a longer horizontal distance [1]1

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

TextbookBuku teks p.ms. 122

27

QuestionSoalan 27

Rajah 6 menunjukkan dongakan sisi, dongakan hadapan dan pelan bagi sebuah pepejal. Pepejal itu terletak pada satah mengufuk dan terdiri daripada sebuah kuboid dengan sebuah piramid tapak segi empat sama di atasnya. Tapak piramid itu terletak penuh di atas muka atas kuboid.Diagram 6 shows the side elevation, the front elevation and the plan of a solid. The solid rests on a horizontal plane and is made up of a cuboid with a square-based pyramid on top. The base of the pyramid lies entirely on the top face of the cuboid.

[2 marksmarkah] (a)Nyatakan panjang, lebar dan tinggi kuboid itu.State the length, the width and the height of the cuboid.

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• Length 10 cm and width 6 cm [1]1
• Height 4 cm [1]1

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[2 marksmarkah] (b)Hitung tinggi condong bagi satu muka segi tiga piramid itu, dalam cm.Calculate the slant height of one triangular face of the pyramid, in cm.

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• Half the base = 3 cm; height = 4 cm [1]1
• √(3² + 4²) = 5 cm [1]1

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[3 marksmarkah] (c)Hitung isi padu pepejal itu, dalam cm³. [Isi padu piramid = 13 × luas tapak × tinggi]Calculate the volume of the solid, in cm³. [Volume of pyramid = 13 × base area × height]

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• Cuboid = 10 × 6 × 4 = 240 cm³ [1]1
• Pyramid = 1/3 × 6 × 6 × 4 = 48 cm³ [1]1
• Total = 288 cm³ [1]1

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[3 marksmarkah] (d)Piramid itu digelongsor 2 cm ke kanan di sepanjang muka atas kuboid. Hakim berkata bahawa dongakan sisi akan berubah kerana piramid itu telah bergerak. Adakah Hakim betul? Nyatakan juga pandangan yang berubah.The pyramid is slid 2 cm to the right along the top face of the cuboid. Hakim says that the side elevation will change because the pyramid has moved. Is Hakim correct? State also which views change.

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• Not correct [1]1
• The side elevation looks along the 10 cm length, so a slide along the length does not change what is seen (still a 6 cm by 4 cm rectangle with a triangle of height 4 cm) [1]1
• The plan and the front elevation change, because the pyramid appears 2 cm further to the right in both [1]1

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

TextbookBuku teks p.ms. 190

28

QuestionSoalan 28

Rajah 7 menunjukkan pelan sebuah padang segi empat tepat PQRS yang dilukis pada skala 1 cm : 10 m. PQ = 12 cm dan QR = 8 cm. Sebuah penyiram air X dan sebuah paip Z akan dipasang di dalam padang itu.Diagram 7 shows the plan of a rectangular field PQRS drawn to the scale 1 cm : 10 m. PQ = 12 cm and QR = 8 cm. A water sprinkler X and a tap Z are to be fixed inside the field.

[2 marksmarkah] (a)X sama jarak dari P dan Q. Huraikan lokus bagi X dan nyatakan jaraknya dari PS pada pelan.X is the same distance from P and Q. Describe the locus of X and state its distance from PS on the plan.

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• The perpendicular bisector of PQ (a line parallel to PS and QR) [1]1
• 6 cm from PS [1]1

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[2 marksmarkah] (b)X juga sama jarak dari PQ dan PS. Nyatakan lokus bagi syarat ini dan hitung jarak sebenar, dalam m, X dari PQ.X is also the same distance from PQ and PS. State the locus for this condition and calculate the actual distance, in m, of X from PQ.

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• The bisector of ∠SPQ (the line at 45° to PQ through P) [1]1
• X is 6 cm from PS, so also 6 cm from PQ: 6 × 10 = 60 m [1]1

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[3 marksmarkah] (c)Z terletak pada lokus di (a) dan berjarak 7.5 cm dari R pada pelan. Hitung jarak sebenar, dalam m, Z dari PQ.Z lies on the locus in (a) and is 7.5 cm from R on the plan. Calculate the actual distance, in m, of Z from PQ.

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• Z is 12 − 6 = 6 cm from QR, so the vertical distance from R is √(7.5² − 6²) = 4.5 cm [1]1
• Distance from PQ = 8 − 4.5 = 3.5 cm [1]1
• 3.5 × 10 = 35 m [1]1

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[3 marksmarkah] (d)Rani berkata bahawa sebatang pokok boleh ditanam di dalam padang itu pada kedudukan yang sama jarak dari PS dan QR, dan berjarak 4 cm dari R. Adakah Rani betul? Terangkan.Rani says that a tree can be planted inside the field at a position that is the same distance from PS and QR and 4 cm from R. Is Rani correct? Explain.

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• Not correct [1]1
• The points equidistant from PS and QR lie on the line 6 cm from QR [1]1
• The circle of radius 4 cm about R does not reach this line (6 cm > 4 cm), so no position exists [1]1

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For the teacherUntuk guru: CreatingMencipta, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.2.7

TextbookBuku teks p.ms. 216

29

QuestionSoalan 29

Cik Nora ingin membeli sebuah motosikal berharga RM9 000. Dia mempunyai simpanan RM900 dan mampu membayar paling banyak RM270 sebulan. Dua tawaran diberi.
Tawaran A (kedai): deposit 10%, baki dibayar selama 3 tahun dengan kadar faedah rata 5% setahun ke atas baki.
Tawaran B (bank): pinjaman penuh RM9 000 selama 3 tahun dengan kadar faedah rata 4.2% setahun, tanpa deposit.Cik Nora wants to buy a motorcycle that costs RM9 000. She has savings of RM900 and can pay at most RM270 a month. Two offers are made.
Offer A (shop): a 10% deposit, the balance repaid over 3 years at a flat rate of 5% per annum on the balance.
Offer B (bank): a loan of the full RM9 000 over 3 years at a flat rate of 4.2% per annum, with no deposit.

[2 marksmarkah] (a)Hitung faedah dan ansuran bulanan bagi Tawaran A.Calculate the interest and the monthly instalment for Offer A.

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• Balance = 90% × 9 000 = RM8 100; interest = 8 100 × 0.05 × 3 = RM1 215 [1]1
• Instalment = (8 100 + 1 215) ÷ 36 = RM258.75 [1]1

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[2 marksmarkah] (b)Hitung jumlah bayaran balik dan ansuran bulanan bagi Tawaran B.Calculate the total repayment and the monthly instalment for Offer B.

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• Interest = 9 000 × 0.042 × 3 = RM1 134, total repayment = RM10 134 [1]1
• Instalment = 10 134 ÷ 36 = RM281.50 [1]1

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[2 marksmarkah] (c)Hitung jumlah kos motosikal itu di bawah Tawaran A (termasuk deposit). Tawaran yang manakah lebih murah dan berapakah bezanya?Calculate the total cost of the motorcycle under Offer A (including the deposit). Which offer is cheaper overall and by how much?

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• Total cost A = 900 + 9 315 = RM10 215 [1]1
• Offer B is cheaper by 10 215 − 10 134 = RM81 [1]1

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[4 marksmarkah] (d)Tawaran yang manakah patut dipilih oleh Cik Nora? Berikan alasan anda.Which offer should Cik Nora choose? Give your reasons.

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• Offer A [1]1
• Its instalment RM258.75 is within her limit of RM270 [1]1
• Offer B costs RM281.50 a month, which is more than RM270, so she cannot afford it [1]1
• Offer A costs RM81 more overall, but only A fits her budget; she has RM900 for the deposit [1]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: EvaluatingMenilai, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.2.5

TextbookBuku teks p.ms. 78

30

QuestionSoalan 30

Rajah 8 menunjukkan sebuah bulatan berpusat O. PA dan PB ialah tangen kepada bulatan itu di A dan B. AC ialah diameter bulatan itu. Diberi ∠APB = 50°.Diagram 8 shows a circle with centre O. PA and PB are tangents to the circle at A and B. AC is a diameter of the circle. Given ∠APB = 50°.

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

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• ∠OAP = ∠OBP = 90° [1]1
• 360° − 90° − 90° − 50° = 130° [1]1

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

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• Angle at the circumference is half the angle at the centre on the same arc AB [1]1
• 130° ÷ 2 = 65° [1]1

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[2 marksmarkah] (c)Hitung ∠BAC.Calculate ∠BAC.

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• ∠ABC = 90° (angle in a semicircle) [1]1
• 180° − 90° − 65° = 25° [1]1

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[4 marksmarkah] (d)Hadi berkata bahawa BC selari dengan OP. Adakah Hadi betul? Terangkan.Hadi says that BC is parallel to OP. Is Hadi correct? Explain.

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• Correct [1]1
• PA = PB (tangents from P) and OA = OB (radii), so OP is the perpendicular bisector of AB; OP ⟂ AB [1]1
• ∠ABC = 90° (angle in a semicircle), so BC ⟂ AB [1]1
• OP and BC are both perpendicular to AB, so they are parallel [1]1

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For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.3.3

TextbookBuku teks p.ms. 154

31

QuestionSoalan 31

Soalan ini tentang indeks dan peluruhan sejenis ubat dalam darah.This question is about indices and the decay of a drug in the blood.

[2 marksmarkah] (a)Ringkaskan (3a²b⁻¹)³ × a⁻⁵b⁴.Simplify (3a²b⁻¹)³ × a⁻⁵b⁴.

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• 27a⁶b⁻³ × a⁻⁵b⁴ [1]1
• 27ab [1]1

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

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• 16^3/4 = 8 and 4⁻² = 1/16 [1]1
• 8 ÷ 1/16 = 128 [1]1

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[3 marksmarkah] (c)Jisim ubat, A mg, dalam darah seorang pesakit selepas t jam diberi oleh A = 640 × 2^−t/4. (i) Hitung jisim ubat selepas 12 jam. (ii) Cari bilangan jam untuk jisim ubat menjadi 5 mg.The mass of a drug, A mg, in a patient's blood after t hours is given by A = 640 × 2^−t/4. (i) Calculate the mass of the drug after 12 hours. (ii) Find the number of hours for the mass of the drug to become 5 mg.

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• (i) 640 × 2⁻³ = 80 mg [1]1
• (ii) 2^−t/4 = 5 ÷ 640 = 1/128 = 2⁻⁷ [1]1
• t/4 = 7, so t = 28 hours [1]1

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[3 marksmarkah] (d)Maya berkata bahawa ubat itu akan habis sepenuhnya (0 mg) selepas 32 jam. Adakah Maya betul? Tunjukkan pengiraan anda.Maya says that the drug will be completely gone (0 mg) after 32 hours. Is Maya correct? Show your calculation.

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• Not correct [1]1
• A = 640 × 2^−32/4 = 640 × 2⁻⁸ = 2.5 mg [1]1
• 2^−t/4 is always greater than 0, so the mass becomes very small but never exactly 0 [1]1

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

TextbookBuku teks p.ms. 22

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