Practice paper, not an official KPM paperKertas latihan, bukan kertas rasmi KPM
50
Ujian Akhir Sesi AkademikEnd of Academic Session Test
MatematikMathematics
Tingkatan 5, 2 jam / 2 hours, 100 marksmarkah
Nama / NameNama
Kelas / ClassKelas
Tarikh / DateTarikh
Markah / marksmarkah: / 100
InstructionsArahan
- Kertas soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C.This question paper consists of three sections: Section A, Section B and Section C.
- Jawab semua soalan.Answer all questions.
- Tunjukkan langkah-langkah penting dalam kerja mengira anda.Show your working. It may help you to get marks.
- Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams in this paper are not drawn to scale unless stated.
- Anda dibenarkan menggunakan kalkulator saintifik.You may use a scientific calculator.
Answer schemes are in English (DLP).Skema jawapan dalam Bahasa Inggeris (DLP).
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.
Bahagian ASection A
20 marksmarkahJawab semua soalan. Setiap soalan diikuti oleh empat pilihan jawapan, A, B, C dan D. Pilih satu jawapan yang terbaik.Answer all questions. Each question is followed by four options, A, B, C and D. Choose the best answer.
1
QuestionSoalan 1
Jarak, d km, yang dilalui oleh sebuah kereta api ekspres berubah secara langsung dengan masa, t jam. Kereta api itu bergerak sejauh 200 km dalam 2.5 jam. Berapakah jarak yang dilalui dalam 4 jam?The distance, d km, travelled by an express train varies directly as the time, t hours. The train travels 200 km in 2.5 hours. What distance does it travel in 4 hours?
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C
d = kt, so k = 200 ÷ 2.5 = 80. For t = 4, d = 80 × 4 = 320 km. (500 km comes from multiplying 200 by 2.5.)
2
QuestionSoalan 2
Masa, t minit, untuk memuat turun sebuah fail berubah secara songsang dengan kelajuan sambungan, s Mbps. Pada kelajuan 8 Mbps, fail itu dimuat turun dalam 15 minit. Cari kelajuan yang diperlukan untuk memuat turun fail itu dalam 6 minit.The time, t minutes, to download a file varies inversely as the connection speed, s Mbps. At 8 Mbps the file downloads in 15 minutes. Find the speed needed to download the file in 6 minutes.
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A
t = k/s, so k = ts = 15 × 8 = 120. For t = 6, s = 120 ÷ 6 = 20 Mbps. (3.2 Mbps comes from treating it as direct variation: 8 × 6 ÷ 15.)
3
QuestionSoalan 3
Kuasa, P watt, sebuah motor skuter elektrik berubah secara bersama dengan voltan, V volt, dan arus, I ampere. Diberi P = 150 apabila V = 10 dan I = 6. Hitung nilai P apabila V = 16 dan I = 4.5.The power, P watts, of an electric scooter motor varies jointly as the voltage, V volts, and the current, I amperes. Given P = 150 when V = 10 and I = 6, calculate P when V = 16 and I = 4.5.
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D
P = kVI. 150 = k(10)(6) = 60k, so k = 2.5. P = 2.5 × 16 × 4.5 = 180 W. (144 comes from using k = 2.)
4
QuestionSoalan 4
Diberi P = 2−134 dan Q = 5−2. Cari PQ.Given P = 2−134 and Q = 5−2. Find PQ.
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B
Row 1: 2 × 5 + (−1)(−2) = 12. Row 2: 3 × 5 + 4(−2) = 7. Options A and C each have one row wrong because the sign of −2 was ignored.
5
QuestionSoalan 5
Matriks M = 4−32k tidak mempunyai matriks songsang. Cari nilai k.The matrix M = 4−32k has no inverse. Find the value of k.
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C
A matrix has no inverse when its determinant is 0. Determinant = 4k − (−3)(2) = 4k + 6 = 0, so k = −1.5. (1.5 comes from 4k − 6 = 0.)
6
QuestionSoalan 6
Matriks A berperingkat 3 × 2 dan matriks B berperingkat 2 × 4. Apakah peringkat bagi matriks AB?Matrix A has order 3 × 2 and matrix B has order 2 × 4. What is the order of matrix AB?
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A
The inner numbers match (2 and 2), so AB can be found. Its order is given by the outer numbers: 3 × 4.
7
QuestionSoalan 7
Premium asas bagi polisi insurans motor komprehensif sebuah kereta ialah RM1 250. Pemilik kereta itu layak mendapat Diskaun Tanpa Tuntutan (NCD) sebanyak 25%. Berapakah premium kasar yang perlu dibayar?The basic premium for the comprehensive motor insurance policy of a car is RM1 250. The owner is entitled to a No Claim Discount (NCD) of 25%. What is the gross premium to be paid?
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D
NCD = 25% × 1 250 = RM312.50. Gross premium = 1 250 − 312.50 = RM937.50. (RM312.50 is only the discount; RM1 562.50 adds it instead of subtracting.)
8
QuestionSoalan 8
Premium tahunan sebuah polisi insurans hayat ialah RM696. Kadar premium ialah RM5.80 bagi setiap RM1 000 nilai muka. Hitung nilai muka polisi itu.The annual premium of a life insurance policy is RM696. The premium rate is RM5.80 per RM1 000 of face value. Calculate the face value of the policy.
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B
Premium = (face value ÷ 1 000) × rate, so face value ÷ 1 000 = 696 ÷ 5.80 = 120. Face value = 120 × 1 000 = RM120 000.
9
QuestionSoalan 9
Encik Karim mempunyai perniagaan sendiri. Bilakah tarikh akhir dia perlu menghantar Borang Nyata Pendapatan (ITRF) bagi tahun sebelumnya?Encik Karim has his own business. By which date must he submit the Income Tax Return Form (ITRF) for the previous year?
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B
A taxpayer with a business files Form B by 30 June. A taxpayer with no business files Form BE by 30 April.
10
QuestionSoalan 10
Encik Jamal menyewakan sebuah kedai dengan sewa RM1 500 sebulan. Pihak berkuasa tempatan mengenakan cukai taksiran sebanyak 6% daripada nilai tahunan. Berapakah cukai taksiran yang perlu dibayar bagi setiap setengah tahun?Encik Jamal rents out a shop for RM1 500 a month. The local authority charges property assessment tax of 6% of the annual value. How much property assessment tax must be paid for each half-year?
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D
Annual value = 1 500 × 12 = RM18 000. Tax for the year = 6% × 18 000 = RM1 080. For a half-year: 1 080 ÷ 2 = RM540. (RM1 080 is the tax for a whole year.)
11
QuestionSoalan 11
Sebuah model pesawat dibuat dengan skala 1 : 50. Luas sayap model itu ialah 120 cm². Hitung luas sayap pesawat sebenar, dalam m².A model aeroplane is made to a scale of 1 : 50. The area of the wing of the model is 120 cm². Calculate the area of the wing of the actual aeroplane, in m².
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A
The linear scale factor is 50, so the area scale factor is 50² = 2 500. Actual area = 120 × 2 500 = 300 000 cm² = 30 m² (÷ 10 000). 0.6 forgets to square the scale factor (120 × 50 = 6 000 cm²).
12
QuestionSoalan 12
Titik P(5, 3) dipetakan kepada titik P′ di bawah gabungan transformasi RT. Transformasi T ialah translasi −24 dan transformasi R ialah pantulan pada paksi-x. Cari koordinat P′.Point P(5, 3) is mapped onto point P′ under the combined transformation RT. Transformation T is the translation −24 and transformation R is a reflection in the x-axis. Find the coordinates of P′.
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C
RT means T first, then R. T: (5, 3) → (3, 7). R: reflection in the x-axis changes the sign of y, so (3, 7) → (3, −7). (3, 1) comes from doing R first (the order TR).
13
QuestionSoalan 13
Tanpa menggunakan kalkulator, cari nilai kos 330°.Without using a calculator, find the value of cos 330°.
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C
330° is in quadrant IV, where cosine is positive. Reference angle = 360° − 330° = 30°, so cos 330° = cos 30° = √3/2.
14
QuestionSoalan 14
Diberi sin θ = 0.8 bagi 0° ≤ θ ≤ 360°. Cari nilai-nilai θ, betul kepada satu tempat perpuluhan.Given sin θ = 0.8 for 0° ≤ θ ≤ 360°. Find the values of θ, correct to one decimal place.
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A
Sine is positive in quadrants I and II. Reference angle α = sin⁻¹(0.8) = 53.1°. θ = 53.1° or θ = 180° − 53.1° = 126.9°.
15
QuestionSoalan 15
Antara berikut, yang manakah BUKAN ukuran serakan?Which of the following is NOT a measure of dispersion?
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D
Variance, interquartile range and standard deviation all describe how spread out the data are. The median is a measure of central tendency.
16
QuestionSoalan 16
Jadual di bawah menunjukkan jarak larian 20 orang peserta dalam sebuah kejohanan.
Hitung min jarak larian itu, dalam km.The table below shows the distance run by 20 participants in a tournament.
Calculate the mean distance run, in km.
| Jarak larian (km) / Distance run (km) | 1 – 3 | 4 – 6 | 7 – 9 |
|---|---|---|---|
| Bilangan peserta / Number of runners | 6 | 10 | 4 |
| Jarak larian (km) / Distance run (km) | 1 – 3 | 4 – 6 | 7 – 9 |
|---|---|---|---|
| Bilangan peserta / Number of runners | 6 | 10 | 4 |
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B
Midpoints are 2, 5 and 8. Σfx = 2(6) + 5(10) + 8(4) = 94, Σf = 20, so mean = 94 ÷ 20 = 4.7 km. (5.0 ignores the frequencies; 94 forgets to divide by Σf.)
17
QuestionSoalan 17
Bagi 20 nilai data, min ialah 10 dan varians ialah 16. Cari nilai Σfx².For 20 values of data, the mean is 10 and the variance is 16. Find the value of Σfx².
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A
Variance = Σfx² ÷ Σf − mean², so 16 = Σfx² ÷ 20 − 100 and Σfx² ÷ 20 = 116. Σfx² = 116 × 20 = 2 320.
18
QuestionSoalan 18
Sebuah syarikat sewa kereta mengenakan yuran tetap RM60 ditambah RM0.40 bagi setiap kilometer yang dipandu. Fungsi manakah yang memodelkan jumlah bayaran, RMC, bagi d km?A car rental company charges a fixed fee of RM60 plus RM0.40 for every kilometre driven. Which function models the total charge, RMC, for d km?
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C
A fixed amount plus a constant rate per kilometre is a linear model: C = 60 + 0.4d. Check: d = 100 gives 60 + 40 = RM100.
19
QuestionSoalan 19
Seorang pembangun permainan video mencatat bilangan mata, y, yang diperoleh seorang pemain pada tahap x.
Fungsi manakah yang paling sesuai memodelkan data itu?A video game developer records the number of points, y, that a player earns at level x.
Which function best models the data?
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y | 3 | 4 | 7 | 12 | 19 |
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y | 3 | 4 | 7 | 12 | 19 |
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D
The first differences are 1, 3, 5, 7 (not constant), but the second differences are all 2, so the model is quadratic with a = 1. At x = 0, y = 3, so c = 3 and y = x² + 3. Check x = 4: 16 + 3 = 19.
20
QuestionSoalan 20
Hadi ingin mempunyai RM10 000 selepas 5 tahun untuk membeli sebuah basikal. Sebuah akaun simpanan memberi faedah 3% setahun dikompaunkan setiap setengah tahun. Berapakah jumlah yang perlu didepositkan sekarang, betul kepada sen terdekat?Hadi wants to have RM10 000 after 5 years to buy a bicycle. A savings account pays interest of 3% per annum compounded half-yearly. How much must be deposited now, correct to the nearest sen?
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B
A = P(1 + r/n)^(nt): 10 000 = P(1.015)^10, so P = 10 000 ÷ 1.160541 = RM8 616.67. RM8 626.09 compounds yearly (10 000 ÷ 1.03^5).
Bahagian BSection B
20 marksmarkahJawab semua soalan. Setiap soalan mempunyai empat bahagian; pilih jawapan yang betul bagi setiap bahagian (1 markah setiap bahagian).Answer all questions. Each question has four parts; choose the correct answer for each part (1 mark each).
21
QuestionSoalan 21
Harga tiket bas untuk seorang dewasa ialah RMx dan untuk seorang kanak-kanak ialah RMy. Satu kumpulan membeli 3 tiket dewasa dan 5 tiket kanak-kanak dengan harga RM110. Kumpulan lain membeli 1 tiket dewasa dan 2 tiket kanak-kanak dengan harga RM40. Maklumat itu ditulis sebagai 3512xy = 11040.The price of a bus ticket for an adult is RMx and for a child is RMy. One group buys 3 adult tickets and 5 child tickets for RM110. Another group buys 1 adult ticket and 2 child tickets for RM40. The information is written as 3512xy = 11040.
[1 markmarkah]
(a)Pilih nilai penentu bagi matriks 3512.Choose the determinant of the matrix 3512.
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1
Determinant = 3(2) − 5(1) = 6 − 5 = 1.
[1 markmarkah]
(b)Pilih nilai x.Choose the value of x.
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20
The inverse is (1/1)[2 −5; −1 3]. x = 2 × 110 − 5 × 40 = 220 − 200 = 20.
[1 markmarkah]
(c)Pilih nilai y.Choose the value of y.
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10
y = −1 × 110 + 3 × 40 = −110 + 120 = 10. Check: 3(20) + 5(10) = 110 and 20 + 2(10) = 40.
[1 markmarkah]
(d)Tandakan BENAR atau PALSU: Matriks 6432 mempunyai matriks songsang.Mark TRUE or FALSE: The matrix 6432 has an inverse.
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False
Determinant = 6(2) − 4(3) = 12 − 12 = 0. A matrix with determinant 0 has no inverse.
22
QuestionSoalan 22
Jadual 1 menunjukkan kadar premium tahunan, dalam RM, bagi setiap RM1 000 nilai muka polisi insurans hayat. Encik Razak, 35 tahun, seorang perokok, membeli polisi insurans hayat dengan nilai muka RM250 000.Table 1 shows the annual premium rates, in RM, per RM1 000 of face value of a life insurance policy. Encik Razak, 35, is a smoker and buys a life insurance policy with a face value of RM250 000.
| Umur (tahun) Age (years) | Bukan perokok Non-smoker | Perokok Smoker |
|---|---|---|
| 25 | 1.80 | 3.10 |
| 35 | 2.40 | 4.20 |
| 45 | 4.10 | 7.50 |
[1 markmarkah]
(a)Pilih kadar premium bagi Encik Razak.Choose Encik Razak's premium rate.
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4.20
Age 35 and a smoker: RM4.20 per RM1 000.
[1 markmarkah]
(b)Pilih premium tahunan yang perlu dibayar.Choose the annual premium he has to pay.
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RM1 050
Premium = 250 000 ÷ 1 000 × 4.20 = 250 × 4.20 = RM1 050. (RM600 uses the non-smoker rate.)
[1 markmarkah]
(c)Pilih jumlah yang dibayar sebulan jika premium dibayar secara bulanan.Choose the amount paid each month if the premium is paid monthly.
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RM87.50
1 050 ÷ 12 = RM87.50.
[1 markmarkah]
(d)Tandakan BENAR atau PALSU: Jika Encik Razak berhenti merokok sebelum polisi diperbaharui, premium tahunannya menjadi RM450 lebih rendah.Mark TRUE or FALSE: If Encik Razak stops smoking before the policy is renewed, his annual premium becomes RM450 lower.
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True
Non-smoker premium = 250 × 2.40 = RM600. The saving is 1 050 − 600 = RM450.
23
QuestionSoalan 23
Sebuah logo kelab sukan dibesarkan pada satah Cartes dengan pusat asalan, O, dan faktor skala 1.5. Titik A(4, −2) terletak pada logo itu dan luas logo itu ialah 20 cm².A sports club logo is enlarged on a Cartesian plane with the origin, O, as the centre and a scale factor of 1.5. Point A(4, −2) lies on the logo and the area of the logo is 20 cm².
[1 markmarkah]
(a)Pilih koordinat imej bagi titik A.Choose the coordinates of the image of point A.
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(6, −3)
Image = 1.5 × (4, −2) = (6, −3), since the centre is the origin.
[1 markmarkah]
(b)Pilih luas imej logo itu.Choose the area of the image of the logo.
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45 cm²
Area of image = k² × area of object = 1.5² × 20 = 2.25 × 20 = 45 cm². (30 cm² multiplies by k, not k².)
[1 markmarkah]
(c)Satu lagi pembesaran dengan pusat O memetakan logo asal kepada imej berluas 80 cm². Pilih faktor skala positifnya.Another enlargement with centre O maps the original logo onto an image of area 80 cm². Choose its positive scale factor.
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2
k² = 80 ÷ 20 = 4, so k = √4 = 2.
[1 markmarkah]
(d)Tandakan BENAR atau PALSU: Pembesaran dengan faktor skala −1.5 menghasilkan imej yang lebih kecil daripada objek.Mark TRUE or FALSE: An enlargement with a scale factor of −1.5 gives an image smaller than the object.
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False
The size depends on |k| = 1.5, which is greater than 1, so the image is larger. The negative sign only puts the image on the opposite side of the centre.
24
QuestionSoalan 24
Kekuatan isyarat, y unit, daripada sebuah penderia pada jam kecergasan di sudut x dimodelkan oleh y = 3 kos 4x + 2 bagi 0° ≤ x ≤ 360°.The signal strength, y units, from a sensor in a fitness watch at angle x is modelled by y = 3 cos 4x + 2 for 0° ≤ x ≤ 360°.
[1 markmarkah]
(a)Pilih nilai maksimum bagi y.Choose the maximum value of y.
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5
Maximum = a + c = 3 + 2 = 5.
[1 markmarkah]
(b)Pilih kala bagi graf itu.Choose the period of the graph.
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90°
Period = 360° ÷ b = 360° ÷ 4 = 90°.
[1 markmarkah]
(c)Pilih nilai y apabila x = 30°.Choose the value of y when x = 30°.
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0.5
y = 3 cos 120° + 2 = 3(−0.5) + 2 = 0.5. (cos 120° is negative, in quadrant II.)
[1 markmarkah]
(d)Tandakan BENAR atau PALSU: Graf itu mempunyai 4 kitaran lengkap bagi 0° ≤ x ≤ 360°.Mark TRUE or FALSE: The graph has 4 complete cycles for 0° ≤ x ≤ 360°.
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True
Each cycle is 90° long, and 360° ÷ 90° = 4 cycles.
25
QuestionSoalan 25
Bilangan tontonan, V, sebuah video sukan pada hari ke-t selepas dimuat naik ditunjukkan dalam jadual.The number of views, V, of a sports video on day t after it is uploaded is shown in the table.
| t (hari / day) | 0 | 1 | 2 |
|---|---|---|---|
| V | 500 | 600 | 720 |
[1 markmarkah]
(a)Pilih nisbah sepunya bagi bilangan tontonan dari satu hari ke hari berikutnya.Choose the common ratio of the number of views from one day to the next.
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1.2
600 ÷ 500 = 1.2 and 720 ÷ 600 = 1.2. The ratio is constant, but the difference (100, 120) is not.
[1 markmarkah]
(b)Pilih fungsi yang memodelkan data itu.Choose the function that models the data.
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V = 500(1.2)^t
A constant ratio means an exponential model, V = Ca^t, with C = 500 (the value at t = 0) and a = 1.2.
[1 markmarkah]
(c)Pilih bilangan tontonan pada hari ke-5, betul kepada nilai bulat terdekat.Choose the number of views on day 5, to the nearest whole number.
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1 244
V = 500(1.2)^5 = 500 × 2.48832 = 1 244.
[1 markmarkah]
(d)Tandakan BENAR atau PALSU: Model itu meramalkan bahawa bilangan tontonan mencapai 5 000 dalam masa 10 hari.Mark TRUE or FALSE: The model predicts that the number of views reaches 5 000 within 10 days.
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False
At t = 10, V = 500(1.2)^10 = 500 × 6.19 = 3 096, which is less than 5 000. Views reach 5 000 only after about 12.6 days.
Bahagian CSection C
60 marksmarkahJawab semua soalan. Tunjukkan langkah-langkah penting dalam kerja mengira anda.Answer all questions. Show the important steps in your working.
26
QuestionSoalan 26
Kecerahan, B unit, sebuah lampu limpah stadium berubah secara langsung dengan kuasanya, P W, dan secara songsang dengan kuasa dua jaraknya, d m, dari lampu itu. Diberi B = 40 apabila P = 500 dan d = 5.The brightness, B units, of a stadium floodlight varies directly as its power, P W, and inversely as the square of the distance, d m, from the lamp. Given B = 40 when P = 500 and d = 5.
[2 marksmarkah]
(a)Tulis hubungan antara B, P dan d, dan cari pemalar variasi, k. Seterusnya, tulis persamaan yang menghubungkan B, P dan d.Write the relation between B, P and d, and find the constant of variation, k. Hence, write the equation that connects B, P and d.
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Tick each point you got.Tandakan setiap isi yang betul.
[2 marksmarkah]
(b)Hitung nilai B apabila P = 800 dan d = 8.Calculate the value of B when P = 800 and d = 8.
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Tick each point you got.Tandakan setiap isi yang betul.
[2 marksmarkah]
(c)Hitung jarak, d, dari lampu itu apabila P = 450 dan B = 18, betul kepada dua tempat perpuluhan.Calculate the distance, d, from the lamp when P = 450 and B = 18, correct to two decimal places.
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Tick each point you got.Tandakan setiap isi yang betul.
[2 marksmarkah]
(d)Apakah yang berlaku kepada B jika kuasa lampu itu digandakan dan jaraknya juga digandakan? Jelaskan.What happens to B if the power of the lamp is doubled and the distance is also doubled? Explain.
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Tick each point you got.Tandakan setiap isi yang betul.
27
QuestionSoalan 27
Encik Zul dan isterinya, Puan Nadia, bekerja sebagai pemandu pelancong dan jurulatih sukan. Pendapatan bercukai Encik Zul ialah RM62 000 dan pendapatan bercukai Puan Nadia ialah RM38 000. Mereka tidak membayar zakat. Sebahagian daripada jadual kadar cukai pendapatan ditunjukkan dalam Jadual 2.Encik Zul and his wife, Puan Nadia, work as a tour guide and a sports coach. The chargeable income of Encik Zul is RM62 000 and the chargeable income of Puan Nadia is RM38 000. They do not pay zakat. Part of the income tax rate table is shown in Table 2.
| Pendapatan bercukai (RM) Chargeable income (RM) | Cukai asas (RM) Base tax (RM) | Kadar bagi baki Rate on the balance |
|---|---|---|
| 35 001 – 50 000 | 600 | 8% |
| 50 001 – 70 000 | 1 800 | 14% |
| 70 001 – 100 000 | 4 600 | 21% |
[2 marksmarkah]
(a)Hitung cukai pendapatan yang perlu dibayar oleh Encik Zul.Calculate the income tax Encik Zul has to pay.
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Tick each point you got.Tandakan setiap isi yang betul.
[2 marksmarkah]
(b)Hitung cukai pendapatan yang perlu dibayar oleh Puan Nadia.Calculate the income tax Puan Nadia has to pay.
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Tick each point you got.Tandakan setiap isi yang betul.
[2 marksmarkah]
(c)Jika mereka memilih taksiran bersama, hitung cukai pendapatan yang perlu dibayar.If they choose joint assessment, calculate the income tax payable.
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[2 marksmarkah]
(d)Bandingkan taksiran berasingan dengan taksiran bersama. Nyatakan kaedah yang lebih menjimatkan dan berapakah penjimatannya.Compare separate assessment with joint assessment. State which method saves more and by how much.
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28
QuestionSoalan 28
Rajah 1 ialah satu ogif yang menunjukkan masa, dalam minit, yang diambil oleh 60 orang peserta dalam sebuah larian santai 10 km.Diagram 1 is an ogive that shows the times, in minutes, taken by 60 participants in a 10 km fun run.
[3 marksmarkah]
(a)Berdasarkan ogif itu, nyatakan kuartil pertama, median dan kuartil ketiga bagi masa itu.Based on the ogive, state the first quartile, the median and the third quartile of the times.
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[2 marksmarkah]
(b)Hitung julat antara kuartil. Nyatakan maksud nilai itu.Calculate the interquartile range. State what the value means.
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[1 markmarkah]
(c)Berapakah bilangan peserta yang mengambil masa lebih daripada 54.5 minit?How many participants took more than 54.5 minutes?
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[1 markmarkah]
(d)Peserta yang tergolong dalam 25% terpantas menerima pingat. Nyatakan masa paling lama yang masih layak untuk mendapat pingat.Participants in the fastest 25% receive a medal. State the longest time that still qualifies for a medal.
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[3 marksmarkah]
(e)Pada tahun lepas, median masa ialah 58.0 minit dan julat antara kuartil ialah 11 minit. Bandingkan larian tahun ini dengan larian tahun lepas, dan nyatakan kesimpulan anda.Last year, the median time was 58.0 minutes and the interquartile range was 11 minutes. Compare this year's run with last year's run and state your conclusion.
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29
QuestionSoalan 29
Sebuah kedai komputer menjual pemacu USB dan kad memori. Harga sebuah pemacu USB ialah RMx dan harga sebuah kad memori ialah RMy. Lima pemacu USB dan dua kad memori berharga RM114. Tiga pemacu USB dan empat kad memori berharga RM102.A computer shop sells USB drives and memory cards. The price of a USB drive is RMx and the price of a memory card is RMy. Five USB drives and two memory cards cost RM114. Three USB drives and four memory cards cost RM102.
[2 marksmarkah]
(a)Tulis maklumat di atas dalam bentuk persamaan matriks.Write the information above as a matrix equation.
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[2 marksmarkah]
(b)Cari matriks songsang bagi matriks pekali dalam (a).Find the inverse of the coefficient matrix in (a).
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[4 marksmarkah]
(c)Dengan menggunakan kaedah matriks, hitung nilai x dan nilai y.By using the matrix method, calculate the value of x and the value of y.
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[2 marksmarkah]
(d)Cawangan P memesan 20 pemacu USB dan 15 kad memori. Cawangan Q memesan 30 pemacu USB dan 10 kad memori. Dengan menggunakan pendaraban matriks, cari jumlah kos pesanan bagi setiap cawangan.Branch P orders 20 USB drives and 15 memory cards. Branch Q orders 30 USB drives and 10 memory cards. By using matrix multiplication, find the total cost of the order of each branch.
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30
QuestionSoalan 30
Puan Aini tinggal di Semenanjung Malaysia dan memandu sebuah kereta 1 590 cc dengan nilai diinsuranskan RM48 000. Dia berhak mendapat NCD sebanyak 30%. Jadual 3 menunjukkan sebahagian daripada Jadual Tarif Motor. Kadar komprehensif ialah untuk RM1 000 pertama nilai diinsuranskan dan RM26 dikenakan bagi setiap RM1 000 berikutnya atau sebahagian daripadanya. Premium asas bagi polisi pihak ketiga, kebakaran dan kecurian ialah 75% daripada premium asas komprehensif.Puan Aini lives in Peninsular Malaysia and drives a 1 590 cc car with a sum insured of RM48 000. She is entitled to an NCD of 30%. Table 3 shows part of the Motor Tariff Schedule. The comprehensive rate is for the first RM1 000 of the sum insured and RM26 is charged for each further RM1 000 or part of it. The basic premium for a third party, fire and theft policy is 75% of the comprehensive basic premium.
| 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 |
[1 markmarkah]
(a)Nyatakan kadar komprehensif bagi RM1 000 pertama untuk kereta Puan Aini.State the comprehensive rate for the first RM1 000 for the car of Puan Aini.
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[3 marksmarkah]
(b)Hitung premium asas komprehensif.Calculate the comprehensive basic premium.
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[2 marksmarkah]
(c)Hitung premium kasar yang perlu dibayar bagi polisi komprehensif selepas NCD.Calculate the gross premium to be paid for the comprehensive policy after the NCD.
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[2 marksmarkah]
(d)Hitung premium kasar bagi polisi pihak ketiga, kebakaran dan kecurian selepas NCD.Calculate the gross premium for the third party, fire and theft policy after the NCD.
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[2 marksmarkah]
(e)Suami Puan Aini, Encik Rahim, 38 tahun dan bukan perokok, membeli polisi insurans hayat dengan nilai muka RM300 000 pada kadar RM2.60 bagi setiap RM1 000. Dia menambah perlindungan penyakit kritikal sebanyak 40% daripada nilai muka pada kadar RM1.95 bagi setiap RM1 000. Hitung jumlah premium tahunan Encik Rahim.The husband of Puan Aini, Encik Rahim, is 38 and a non-smoker. He buys a life insurance policy with a face value of RM300 000 at a rate of RM2.60 per RM1 000. He adds critical illness cover of 40% of the face value at a rate of RM1.95 per RM1 000. Calculate the total annual premium of Encik Rahim.
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[2 marksmarkah]
(f)Pada tahun lepas, Puan Aini terundur ke dinding dan keretanya memerlukan pembaikan berharga RM3 500. Polisi yang manakah dalam (c) dan (d) akan membayar kos itu? Adakah premium tambahan bagi polisi itu berbaloi? Berikan justifikasi anda.Last year, Puan Aini reversed into a wall and her car needed repairs costing RM3 500. Which policy in (c) and (d) would have paid for the repairs? Is the extra premium for that policy worth it? Justify your answer.
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31
QuestionSoalan 31
Rajah 2 menunjukkan graf bagi fungsi y = a tan bx bagi 0° ≤ x ≤ 120°, dengan a dan b ialah pemalar positif. Fungsi ini memodelkan bacaan sebuah antena radar yang berputar. Garis putus-putus ialah asimptot, pada x = 30° dan x = 90°. Graf itu melalui titik (15°, 2).Diagram 2 shows the graph of the function y = a tan bx for 0° ≤ x ≤ 120°, where a and b are positive constants. The function models the reading of a rotating radar antenna. The dashed lines are asymptotes, at x = 30° and x = 90°. The graph passes through the point (15°, 2).
[2 marksmarkah]
(a)Nyatakan kala bagi graf itu dan seterusnya cari nilai b.State the period of the graph and hence find the value of b.
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[2 marksmarkah]
(b)Dengan menggantikan titik (15°, 2), cari nilai a.By substituting the point (15°, 2), find the value of a.
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[2 marksmarkah]
(c)Hitung nilai y apabila x = 40°, betul kepada dua tempat perpuluhan.Calculate the value of y when x = 40°, correct to two decimal places.
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[3 marksmarkah]
(d)Selesaikan persamaan 2 tan 3x = −2 bagi 0° ≤ x ≤ 120°.Solve the equation 2 tan 3x = −2 for 0° ≤ x ≤ 120°.
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[1 markmarkah]
(e)Hafiz berkata nilai maksimum bagi y dalam julat itu ialah 2. Adakah Hafiz betul? Jelaskan.Hafiz says that the maximum value of y in this range is 2. Is Hafiz correct? Explain.
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[2 marksmarkah]
(f)Graf yang diperoleh apabila graf y = a tan bx dipantulkan pada paksi-x digunakan untuk mereka bentuk antena kedua. Tulis persamaan graf itu dan nyatakan kalanya.The graph obtained when the graph of y = a tan bx is reflected in the x-axis is used to design a second antenna. Write the equation of that graph and state its period.
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Test specification (JSU)Jadual Spesifikasi Ujian (JSU)
| ChapterBab | RememberingMengingat | UnderstandingMemahami | ApplyingMengaplikasi | AnalysingMenganalisis | EvaluatingMenilai | CreatingMencipta | Items | marksmarkah |
|---|---|---|---|---|---|---|---|---|
| 1. Variation | – | – | 4 (11) | – | – | – | 4 | 11 |
| 2. Matrices | – | 1 (1) | 4 (16) | – | – | – | 5 | 17 |
| 3. Consumer Mathematics: Insurance | – | – | 3 (6) | – | 1 (12) | – | 4 | 18 |
| 4. Consumer Mathematics: Taxation | 1 (1) | – | 1 (1) | – | 1 (8) | – | 3 | 10 |
| 5. Congruency, Enlargement and Combined Transformations | – | – | 3 (6) | – | – | – | 3 | 6 |
| 6. Ratios and Graphs of Trigonometric Functions | – | – | 3 (6) | – | 1 (12) | – | 4 | 18 |
| 7. Measures of Dispersion for Grouped Data | 1 (1) | – | 1 (1) | 2 (11) | – | – | 4 | 13 |
| 8. Mathematical Modeling | – | – | 2 (2) | 2 (5) | – | – | 4 | 7 |
| TotalJumlah | 2 (2) | 1 (1) | 21 (49) | 4 (16) | 3 (32) | 0 (0) | 31 | 100 |
Each cell shows items (marks).Setiap petak menunjukkan item (markah).