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A1 Kertas Percubaan SPM 2019: Mathematics – Answers

Kertas Percubaan 1Kertas 1 1 B 2 C 3 B 4 D 5 B 6 A 7 D 8 C 9 D 10 A11 A 12 D 13 C 14 B 15 A16 A 17 C 18 D 19 D 20 C21 C 22 B 23 A 24 B 25 A26 B 27 A 28 D 29 B 30 B31 B 32 D 33 A 34 A 35 D36 B 37 B 38 D 39 C 40 C

Kertas2Section A 1 (a) P = {1, 3, 9, 27}

P Q

• 1• 27

• 3• 6

(b) A B C

2 m – n3 = 1 ...................... ①

m2 + n = 4 (× 2)

m + 2n = 8 ...................... ② ② – ① : m + 2n = 8 (–) m – n

3 = 1

7n3 = 7

7n = 21 n = 3 Substitute n = 3 into ① Gantikan n = 3 ke dalam ① m – 3

3 = 1

m – 1 = 1 m = 2 3 (a) ∠VNQ

(b) tan ∠VNQ = 34

∠VNQ = 36.87° or/atau ∠VNQ = 36°52'

4 Volume of the solid/Isi padu pepejal= volume of the half cylinder + volume of the prism= isi padu separuh silinder + isi padu prisma

= 12

× 227

× 72 2

2 × 6 + 1

2 (5 + 12) × 6 × 7

= 115 12

+ 357

= 472 12

cm3 or/atau 472.5 cm3

5 (a) (i) True/Benar (ii) True/Benar

(b) m3 –64(c) 4n + 6, n = 1, 2, 3, ...

6 (a) y-intercept/pintasan-y = 7 mAB = mCD = – 1

2 y = – 1

2x + 7

(b) C = (x, 0) y = – 1

2x + 7, y = 0

0 = –x + 14 x = 14

7 (a) Area of the shaded region/Luas kawasan berlorek = area of sector OPQ – area of sector ORS = luas sektor OPQ – luas sektor ORS

= 120°360°

× 227

× 2122 – 30°360°

× 227

× 1422 = 462 – 511

3 = 4102

3 cm2

(b) Perimeter of the whole diagram/Perimeter seluruh rajah = arc/lengkok PQ + QS + arc/lengkok ST + POT

= 120°360°

× 2 × 227

× 212 + 7 + 60°360°

× 2 × 227

× 142 + 35

= 44 + 7 + 14 23

+ 35 = 100 2

3 cm

8 (a) Duration of time/Tempoh masa = 5 – 2 = 3 hours/jam(b) (i) 350 km

(ii) 540 – 350 = 190 km(c) From P to Q/Dari P ke Q = 540 – 350

9 – 5

= 190 km4 h/j

= 47.5 km h–1/47.5 km j–1 (d) Average speed/Laju purata = 540

90 = 60 km h–1/60 km j–1

9 Sample space/Ruang sampel= {(K, 5), (K, 6), (K, 8), (E, 5), (E, 6), (E, 8), (N, 5), (N, 6),

(N, 8), (M, 5), (M, 6), (M, 8)}(a) {(N, 6), (N, 8)} P(letter N and even number) P(huruf N dan nombor genap) = 2

12 = 16

(b) {(K, 6), (K, 8), (E, 6), (E, 8), (N, 5), (N, 6), (N, 8), (M, 6), (M, 8)}

P(letter N or even number) P(huruf N atau nombor genap) = 9

12

= 34

10 102 – 162 2

2 = 100 – 64

= 36 = 6tan–1 12

6 2 = 63.43° = 63° 26’

11 (a) M = 53

– 2

23

– 1 2

Answers

Kertas Percubaan SPM Maths Jaw 4th.indd 1 23-Apr-19 9:07:21 AM

A2© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2019

(b) 6 –35 –2 2m

n 2 = 9–3 2

mn 2 = 9

–353

– 2

23

– 1 2 2

=

9 × – 23 + –3 × 1 29 × – 53 + –3 × 2 = –6 – 3

–15 – 6 2 = –9

–212 m = –9, n = –21

Section B12 (a) y = –x3 + 7x – 15

p = –(–2)3 + 7(–2) – 15 = 8 – 14 – 15 = –21 q = –(1)3 + 7(1) – 15 = –1 + 7 – 15 = –9(b)

–3–3.5 –2 –1

–5

–10

–15

–20

–25

–30

–35

0 1

5

2 3 3.5

y

x

y = – x – 10

y = – x³ + 7x – 15

(c) (i) y = –21.5 (ii) x = –3.4

(d) y = –x3 + 7x – 15 ... ① 0 = 8x – x3 – 5 ... ② ① – ②: y – 0 = –x3 + 7x – 15 – 8x + x3 + 5 y = –x – 10 By drawing the straight line y = –x – 10 on the graph,

Dengan melukis garis lurus y = –x – 10 pada graf, x = –3.1, 0.65, 2.4

13 (a) (i) P = Translation –40 2.

P = Translasi –40 2.

(ii) Q = Clockwise rotation of 90° about the centre (1, 9).Q = Putaran 90° ikut arah jam pada pusat (1, 9).

\ Clockwise rotation of 90° about the centre (3, 11).Putaran 90° ikut arah jam pada pusat (3, 11).

(b) V = Translation –30 2.

V = Translasi –30 2.

W = Enlargement about the centre (7, 6) with a scale factor of 3.

W = Pembesaran pada pusat (7, 6) dengan faktor skala 3.

14 (a) Time (minutes)Masa (minit)

FrequencyKekerapan

MidpointTitik tengah

30 – 34 2 3235 – 39 4 3740 – 44 5 4245 – 49 8 4750 – 54 4 5255 – 59 7 5760 – 64 5 62

(b) 5(c) Estimated mean of time

Min anggaran masa

= 2(32) + 4(37) + 5(42) + 8(47) + 4(52) + 7(57) + 5(62)35

= 1 71535

= 49(d)

FrequencyKekerapan

Time (minutes)Masa (minit)

8

6

4

2

7

5

3

1

0 29.5 34.5 39.5 44.5 49.5 54.5 59.5 64.5

(e) Modal class = 45 – 49Kelas mod = 45 – 49

15 (a)

M/Q

N/R

C/E/H

B/JA/K

D/F/G

L/P

3 cm4 cm

2 cm

3 cm



(b) (i)

U/V W/Z

Q

M

R/H/J

N/E

C/BD/AT/S

P/G/K

L/F

2 cm

5 cm

1 cm3 cm3 cm

7 cm

(ii) Q M

L/N

T/D/C

S/A/BK/JV/Z

U/WG/H

P/R

F/E

4 cm1 cm

1 cm

3 cm

2 cm

2 cm

16 (a) 180° – 50° = 130° J(30° N, 130° E) J(30° U, 130° T)

(b) (i) 6 00060

= 100

100° – 30° = 70° \x = 70

(ii) θ × 60 × kos 30 = 5 196 51.96θ = 5 196 θ = 100 100° – 50° = 50° \y = 50

(c) Distance RM/Jarak RM (30 + 20) × 60 = 3 000 n.m Time/ Masa = Distance/Jarak

Speed/Laju = 5 196 + 3 000

750 = 8 196

750 = 10.93 hours/jam

Kertas Percubaan 2Kertas 1 1 D 2 B 3 B 4 C 5 B 6 C 7 D 8 B 9 D 10 A11 B 12 A 13 C 14 C 15 C16 A 17 D 18 C 19 A 20 A21 B 22 A 23 A 24 C 25 A26 C 27 C 28 D 29 C 30 D31 D 32 C 33 D 34 D 35 C36 D 37 A 38 B 39 A 40 C

Kertas2Section A 1

y = x + 2y = 2x – 2

2

4

6

-2-4 0 2 4 6

y

x

–2 y = –2

2 (a) ∠EPM(b) tan θ = 7

10 θ = 34.99° ≈ 35°

3 (a) Volume of the pyramid/Isi padu piramid = 1

3 × 14 × 8 × 12 = 448 cm3

(b) Volume of the prism/Isi padu prisma = 1 064 – 448 = 616 cm3

Let the height of BQ = h cm Andaikan tinggi BQ = h cm 12 (14 + 8) × h × 8 = 616 88h = 616 h = 7 \ Height of BQ = 7 cm

\ Tinggi BQ = 7 cm 4 Each animal has 2 wings/Setiap haiwan ada 2 sayap

402 = 20 animals/ekor haiwan

Assume that/Andaikans = swan/angsaf = fly/lalat s + f = 20 ... 2s + 6f = 76 ... ÷ 2 s + 3f = 38 ... – s + 3f = 38 (–)s + f = 20 2f = 18 f = 9Substitute f = 9 into /Gantikan f = 9 ke dalam s + 9 = 20 s = 11\ Swan/angsa = 11 Fly/lalat = 9

5 (a) False/Palsu 3√ –64 = –4 and/dan 1

3–2 = 9(b) Implication 1: If √ 12.25 = 3.5, then 3.52 = 12.25.

Implikasi 1:Jika √ 12.25 = 3.5, maka 3.52 = 12.25.

Implication 2: If 3.52 = 12.25 then √ 12.25 = 3.5.

Implikasi 2:Jika 3.52 = 12.25, maka √ 12.25 = 3.5.

(c) The size of the interior angle of a regular decagon is 144°.Saiz sudut pedalaman bagi sebuah dekagon sekata ialah 144°.

6 (x)(x + 11)(25) cm3 = 11 900 cm3

x2 + 11x = 476 x2 + 11x – 476 = 0 (x – 17)(x + 28) = 0x = 17 or/atau x = –28\ x = 17 cm



7 (a) mSR = –4 – (–2)3 – (–1)

= – 12

y – 4 = – 12

(x – 7)

y = – 12

x + 152

(b) y = – 12

x + 152

, y = 0

0 = –x + 15 x = 15

8 (a) P(boys/lelaki) = 1228

= 37

(b) 14T = 1

6 T = 84 x = 84 – (15 + 14 + 9 + 16) = 30

9 (a) 12 m s–1

(b) (i) 12(t – 4) = 72 12t – 48 = 72 12t = 120 t = 10

(ii) Total distance/Jumlah jarak = 1

2 (4)(12 + 16) + 12 × (20 – 10) × 12 + 72

= 56 + 60 + 72 = 188 m Average speed/Purata laju = 188

20 = 9.4 m s–1

10 (a) Inverse matrix of 4 7 3 62

Matriks songsang bagi 4 7 3 62

= 14(6) – 3(7) 6 –7

–3 4 2 = 1

3 6 –7 –3 4 2

Compare with/Bandingkan dengan 1h

6 –7 k 4 2.

\ h = 3, k = –3

(b) 4 7 3 6 2 x

y 2 = 1312 2

xy 2 = 1

3 6 –7 –3 4 2 13

12 2 = 1

3 6(13) – 7(12)–3(13) + 4(12)2

= 13 78 – 84

–39 + 48 2 = 1

3 –69 2 = –2

3 2 \ x = –2, y = 3

11 (a) Perimeter of the whole diagramPerimeter seluruh rajah

= 5823 cm

(b) Area of the shaded region/Luas kawasan berlorek = 2051

3 cm2

Section B12 (a) x 2.5 6

y 12 5

(b)

x

y

10

5

10

15

20

25

30

2 3 4 5 6 7 8

y = 30 — x

y = −4x + 29

(c) (i) y = 9 (ii) x = 1.5

(d) y = 30x ...

0 = 30x + 4x – 29 ...

– : y = –4x + 29 By drawing the straight line y = –4x + 29 on the

graph,Dengan melukis garis lurus y = –4x + 29 pada graf,

x = 1.2, 6 13 (a) (i) (9, –3) (ii) (7, 2)

(b) (i) G = Anticlockwise rotation 90° about the centre (0, 1).

= Putaran 90° lawan arah jam pada pusat (0, 1). H = Enlargement about the centre P with a scale

factor of 2. = Pembesaran pada pusat P dengan faktor skala 2. (ii) Area of object × k2 = Area of image

Luas objek × k2 = Luas imej x × 22 – x = 36 cm2

4x – x = 36 cm2

3x = 36 cm2

x = 12 cm2

14 (a) Class intervalSelang kelas


FrequencyKekerapan

11 – 15 13 416 – 20 18 521 – 25 23 926 – 30 28 1031 – 35 33 636 – 40 38 441 – 45 43 2

(b) Estimated mean mass Min anggaran jisim

= 4(13) + 5(18) + 9(23) + 10(28) + 6(33) + 4(38) + 2(43)40

= 1 06540

= 26.63 kg (correct to 2 decimal places)= 26.63 kg (betul kepada 2 tempat perpuluhan)



(c)FrequencyKekerapan

Mass/Jisim (kg)

10

9

8

7

6

5

4

3

2

1

08 13 18 23 28 33 38 43 48

(d) 1215 (The diagrams are not drawn to full scale).

(Rajah tidak dilukis dengan skala penuh). (a)

E/F D/G

C/H

A/K B/J

3 cm

4 cm

6 cm

6 cm

(b) (i) 6 cm

3 cm

3 cm

3 cm 3 cm

H/J C/B

G M/SD/R

F/K N/P Q/A

(ii)

3 cm 3 cm

3 cm

2 cm

4 cm

6 cm

1 cm

2 cm

F/G N/M D

C

Q/R

A/BK/J

H

P/S

16 (a) 180° – 50° = 130° E/130°T

(b) 4 50060 = 75°

75 – 60 = 15° Latitude of M/Latitud M = 15°S(c) 180° × 60 × cos 60° = 5 400 nautical miles/batu nautika(d) (60° × 60) + 4 500 = 8 100 n.m Distance = time taken × speed

Jarak = masa yang diambil × laju 8 100 = time taken/masa diambil × 720 time/masa = 8 100 720 = 11.25 hours/jam

Kertas Percubaan 3Kertas 1 1 D 2 A 3 D 4 B 5 D 6 A 7 D 8 C 9 C 10 B11 A 12 A 13 C 14 C 15 A16 D 17 C 18 D 19 D 20 A21 B 22 D 23 A 24 A 25 B26 C 27 A 28 D 29 D 30 C31 A 32 B 33 C 34 B 35 B36 D 37 D 38 C 39 B 40 A

Kertas2Section A 1

10

6

2

8

4

0 2 4 6 8 10

y

x

y = –4x + 8

y = 12 x + 1

y = 5

y –4x + 8

y 12

x + 1

y 5 2 (a) PQU/RST

(b) sin PQU = 120°360°

PQU = 19.47°

3 2x – 1 = 5x

3 + x 2(3 + x) = 5x(x – 1) 6 + 2x = 5x2 – 5x 5x2 – 7x – 6 = 0 (5x + 3)(x – 2) = 0 x = – 3

5 , 2



4 Volume of jar6 = Volume of container

Isi padu jar6 = Isi padu bekas

π3d

2 22

(12)

6 = π d

2 22

h

π9d2

4 22 = π d2

4 2h 18d2

4 = d2h4

\ h = 18 cm

5 (a) mOB = 1 – 03 – 0 = 1

3 mCBD × 1

3 = –1 mCBD = –3 y = –3x + c 1 = –3(3) + c c = 10 y = –3x + 10

(b) x = 3 13

6 2x + 2(y + 5) = 75 ... y + 5 = 4x ... Substitute into Gantikan ke dalam 2x + 2(4x) = 75 10x = 75 x = 7.5 y + 5 = 4x = 4(7.5) = 30\Length/Panjang = 30 m

7 (a) (i) True/Benar (ii) True/Benar

(b) (i) If mn is a proper fraction, then m and n are integers with 0 m n.Jika mn ialah pecahan wajar, maka m dan n ialah integer dengan 0 m n.

(ii) If m and n are integers with 0 m n, then mn is a proper fraction.Jika m dan n ialah integer dengan 0 m n, maka mn ialah pecahan wajar.

(c) Number of vertices of regular polygon = Number of sides of regular polygonBilangan bucu sebuah poligon sekata = Bilangan sisi sebuah poligon sekata

8 (a) x + y = 13 9x + 7y = 107

(b) 1 1 9 7 2 x

y 2 = 131072

xy 2 = 1

1(7) – 1(9) 7 –1

–9 1 2 131072

= – 12

91 – 107–117 + 1072

= – 12

–16–102

x = 8, y = 5

9 (a) OPQR = 210360

π(12)2

= 712

227 2(144)

= 264 cm2

AOB = 60360

π(7)2

= 16

227 2(49)

= 25.67 cm2

Shaded region/Kawasan berlorek = 264 – 25.67 = 238.33 cm2

(b) AB = 60360

× 2π(7)

= 16

× 2227 2(7)

= 7.33 cm

PQR = 210360

× 2π(12)

= 712

× 2 227 2(12) = 44 cm

Perimeter = 44 + 5 + 7.33 + 7 + 12 = 75.33 cm10 Shampoo/Syampu : S

Soap bar/Sabun buku : BToothpaste/Ubat gigi : T (a) Sample space/Ruang sampel = { (S, 5), (S, 10), (S, 20), (S, 50), (B, 5), (B, 10), (B, 20),

(B, 50), (T, 5), (T, 10), (T, 20), (T, 50)}(b) (i) {(S, 20), (B, 5), (B, 10), (B, 20), (B, 50), (T, 20)}

612 = 1

2 (ii) { (S, 5), (S, 10), (S, 20), (S, 50), (B, 5), (B, 10),

(B, 20), (B, 50), (T, 5), (T, 10), (T, 20)}

= 1112

11 (a) 18 m s–1

(b) 18 – 38.0 = 1.875 m s–2

(c) 12 (18 + 3)(8) = 12 (t – 14) 18(4)

21(8) = (t – 14) 18(4) 168 = (t – 14) (72)

t – 14 = 16872

t = 16 13 s

Section B12 (a) y = –2x2 + 3x + 9

= –2(–2)2 + 3(–2) + 9 = –5 \ x = –2, y = 5(b) y = –2(1)2 + 3(1) + 9 y = 10 \ x = 1, y = 10(c) (i) 0

(ii) –2.8(d) –2x2 + 3x + 9 = y 2x2 – 2x – 14 = 0 x – 5 = y

x 0 1 2y –5 –4 –3

x = –2.2, 3.2



(b), (d)

15

5

10

0–1–2–3–4 1 2 3 4

–5

–10

–15

–20

–25

–30

–35

y

x3.2–2.8 –2.2 4.3

y = x – 5

y = –2x2 + 3x + 9

13 (a) (i) A(3, 9) T A’(5, 13) T A’’(7, 17) (ii) A(3, 9) R A’(9, –3) T A’’(11, 1)

(b) (i) (a) V : A 90° anticlockwise rotation about the centre (4, 3).

V : Putaran 90° lawan arah jam pada pusat (4, 3). (b) L : An enlargement of scale factor 3 about the

centre (3, 3). L : Pembesaran dengan faktor skala 3 pada pusat

(3, 3). (ii) Area of object × k2 = Area of image

Luas objek × k2 = Luas imej 30 × 32 = 270 m2

270 – 30 = 240 m2

14 (a)

ExpensesPerbelanjaan


FrequencyKekerapan

Upper boundary Sempadan

atas

Cumulative frequencyKekerapan longgokan

5 – 9 7 0 9.5 010 – 14 12 2 14.5 215 – 19 17 4 19.5 620 – 24 22 6 24.5 1225 – 29 27 8 29.5 2030 – 34 32 12 34.5 3235 – 39 37 10 39.5 4240 – 44 42 6 44.5 4845 – 49 47 2 49.5 50

(b) 12(2) + 17(4) + 22(6) + 27(8) + 32(12) + 37(10) + 42(6) + 47(2)50

= RM1 54050

= RM30.80

(c)Number of studentsBilangan pelajar

Expenses (RM)Perbelanjaan (RM)

50

4037.5

30

20

10

45

35

25

15

5

0 9.5 19.5 29.5 39.5 44.5 49.514.5 24.5 34.537

(d) Third quartile/Kuartil ketiga

= 34

× 50

= 37.5 \ RM37

15 (a) Y/S W/R

T/P V/Q

3 cm 3 cm

2 cm

X

U(b)

B/A

E/F

U/T X/Y

W

R/SQ/C/D/P

V

1 cm

3 cm

2 cm

2 cm

4 cm

(c) T/Y U/X

V/W

Q/RB/CA/DP/S

F E

3 cm

2 cm

2 cm3 cm1 cm

5 cm4 cm

16 (a) 30 × 60 = 1 800 n.m./batu nautika 120 × 60 cos x/kos x = 1 800 x = 75.52°(b) 1 800 + (75.52 × 60) + 1 800 = 8 131.2 n.m./batu nautika



(c)

W(0°, 100° W)

N/U

P(0°, 70° W)

H(x°S, 70° W)M(x°S, 50° E)

Q(x°S, 80° E)

180° – 75.52° = 104.48° 104.48 × 60 = 6 268.8 n.m./batu nautika

(d) A : 8 131.2t

= 800 t = 10.164 hrs/jam

B : 6 268.8t

= 800 t = 7.836 hrs/jam 10.164 – 7.836 = 2.328 hours/jam = 2 hours/jam 19.68 minutes/minit ≈ 2 hours 20 mins ≈ 2 jam 20 minit

Kertas Percubaan 4Kertas 1 1 C 2 D 3 C 4 C 5 A 6 A 7 A 8 A 9 B 10 A11 D 12 A 13 B 14 D 15 C16 B 17 C 18 A 19 A 20 C21 A 22 D 23 A 24 D 25 D26 D 27 D 28 A 29 C 30 C31 C 32 C 33 A 34 C 35 A36 B 37 B 38 C 39 B 40 C

Kertas2Section A 1 (a) M = {2, 3, 5, 7,}

M N

• 2• 5

• 3• 7

(b) P Q R’ 2 (a)

T

QP

MW

VN

U R

S16 cm

10 cm

NQM(b) tan NQM = 16

10 NQM = 58.0°

3 Assume the radius of the cone = j,Katakan jejari tapak kon = j,

13 × π × j22 × 7.5 = 40π

j2 = 16 j = 4 cmThe volume of hemisphere/Isi padu hemisfera= 2

3 × 227 × (4)3

= 134 221 cm3

Q

N M16 cm

58°10 cm

4 x = price of one food coupon/harga satu kupon makanany = price of one drink coupon/harga satu kupon minuman

35

42 2x

y 2 = 2634

2 x

y 2 = 13(2) – 4(5) 2

–5 –43 2 26

34 2

= – 114 –84

–282 = 62 2\ x = RM6, y = RM2

5 (a) False/Palsu(b) If x = 4, then 2x – 3 = 5.

Jika x = 4, maka 2x – 3 = 5.(c) 14 is divisible by 2.

14 boleh dibahagi tepat dengan 2.(d) The surface area of sphere with radius 8 cm is

256π cm2.Luas permukaan sfera dengan jejari 8 cm ialah 256π cm2.

6 The area of the land/Luas tanah = 7 × 2 = 14 m2

x(3x – 19) = 14 3x2 – 19x – 14 = 0 (3x + 2)(x – 7) = 0 x = – 2

3, 7

x 0, \x = 7 7 (a) Sample space/Ruang sampel

{(4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9)}(b) {(4, 8), (6, 8)} = 2

9(c) {(4, 7), (4, 8), (4, 9), (5, 7), (5, 9), (6, 7), (6, 8), (6, 9)} = 8

9 8 (a) 2 – p

4 – 0 = – 1

4 p = 3(b) Gradient/Kecerunan = – y-intercept

x-intercept 2 / pintasan-ypintasan-x 2

= – 3–6 2

= 12

y = mx + c

y = 12

x + 3

9 (a) 150°360°

π(14)2 + 12

(21 + 26)(6) – 50°360°

π(14)2

= 100°360° 22

7 2 (14)2 + 12

(47)(6)

= 171.11 + 141 = 312.11 cm2

(b) 150 + 50360

2π(14) + 4(14) + 21 + 6 + 12

= 48.89 + 56 + 39 = 143.89 cm

10 (a) 30 – 20 = 10 minutes/minit

(b) 24 + 1430

= 1.27 km/minute / km/minit

(c) 24 + 24t

= 8060

48t

= 43

t = 36 minutes/minit



11 (a) 1k

p –3 –2 8 2 8 3

2 1 2 = 2 0 0 2 2

12 1

k 2 1 –3 –2 8 2 8 3

2 1 2 = 1 0 0 1 2

p = 1

12 1

k 2 = 18(1) – 3(2)

12k

= 18 – 6

12k

= 12

k = 1

(b) 8 3 2 1 2x

y2 = 722

xy2 = 1

2 1 –3–2 8 27

22 = 1

2 7 – 6–14 + 16 2

xy2 =

1212

x = 12

, y = 1

Section B12 (a) y = 17x2 – 120x + 280

x = 1, y = 177 x = 6, y = 172(b)

10

20

40

606980

100

120

140

160170180

200

220

2 3 4 5 6 7

y

x1.1 1.7 3.5 4.2 5.95

y = –20x + 160

y = 17x2 – 120x + 280

(c) x = 3.5, y = 69 y = 170, x = 1.1, 5.95(d) y = 17x2 – 120x + 280 0 = 17x2 – 100x + 120 y = –20x + 160 x = 1.7, 4.2

13 (a) (i) Q(8, –4) V Q’(10, 0) V Q’’(12, 4)

(ii) Q(8, –4) W Q’(0, 4) V Q’’(2, 8)(b) (i) Y : A clockwise 90° rotation about the center (6, 2).

Y : Putaran 90° mengikut arah jam pada pusat (6, 2).

x 1 2 3y 140 120 100

(ii) X : An enlargement with scale factor 52

about the center (0, 0).

X : Pembesaran dengan faktor skala 52

pada pusat (0, 0).

(c) OAB × 52 2

2

= 52.4

OAB = 8.384 unit2

52.4 – 8.384 = 44.016 unit2

14 (a) 5(1) + 15(2) + 25(4) + 35(6) + 45(13) + 55(22) + 65(8)

56

= 2 66056

= 47.5 cm(b)

Height (cm)Tinggi (cm)

Upper boundary

Sempadan atas

Cumulative frequencyKekerapan longgokan

0 0.5 0

0 x 10 10.5 1

10 x 20 20.5 3

20 x 30 30.5 7

30 x 40 40.5 13

40 x 50 50.5 26

50 x 60 60.5 48

60 x 70 70.5 56

(c)

55

50

4542

40

3532

30

25

20

1514

10

5

0 0.5 20.5 40.5 60.510.5 30.5 50.5 70.5

Number of plantsBilangan tumbuhan

41.5 57.553

Height (cm)Tinggi (cm)

60

(d) (i) 1st quartile/Kuartil pertama = 56 × 14

= 14 3rd quartile/Kuartil ketiga = 56 × 3

4 = 42 Interquartile range = 57.5 – 41.5 = 16

Julat antara kuartil = 57.5 – 41.5 = 16 (ii) 56 – 32 = 24 plants/24 tumbuhan



15 (a) Q/N U/R V/W J/M

P/O T/S Y/XK/L

3 cm

3 cm

3 cm

1 cm

1 cm

(b)

O/N

P/Q

A

S/R

T/U Y/V

X/W K/J

L/M

5 cm

3 cm

1 cm

4 cm

1 cm

5 cm

3 cm

(c)

Y/T

A

V/U

J/W/R/Q

M/NL/O

K/X/S/P

4 cm5 cm

3 cm

3 cm

16 (a) 180° – 80° = 100°100° E/100° T

(b) (i) y × 60 cos/kos 47° = 6 138 y = 150 150 – 80 = 70°

70° E/70° T (ii) y × 60 = 4 560 y = 76° 76° – 47° = 29°

29° N/29° U(c) 60 × 180 = 10 800 n.m./batu nautika

(d) 6 138 + 10 800t = 900

t = 18.82 hours/jam


answers - penerbitilmubakti.com · a1 ker er 2019 answers kertas percubaan 1 kertas 1 1 b 2 c 3 b 4...

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