MATH CATCH

sekeping kad berlabel dengan A = {36}
huruf vokal B = {30, 35, 40}
A = Event of picking a card C = {29, 38, 39}
labelled with 74 and a card A ∪ B ∪ C = {29, 30, 35, 36, 38, 39,
labelled with a vowel 40}
= {(74, I)} (A ∪ B ∪ C)' = {27, 28, 31, 32, 33,
P(A) 34, 37, 41, 42, 43, 44}
n[(A ∪ B ∪ C)'] = 11
n(A)
= n(S) 32 (a) Sebilangan persamaan kuadratik
mempunyai punca yang negatif.
1 Some quadratic equations have
= 25 negative roots.

(ii) B = Peristiwa memilih sekeping (b) Premis 2:
kad berlabel dengan 49 atau Premise 2:
sekeping kad berlabel dengan 97 tidak boleh dibahagi dengan 3.
huruf konsonan 97 is not divisible by 3.
B = Event of picking a card
labelled with 49 or a card (c) 3(2)n + n
labelled with a consonant (d) Implikasi 1:
= {(49, G), (49, H), (49, R), (49,
X)} Implication 1:
P(B) mm
n(B)
= n(S) Jika 8 > 9 , maka m > 0.
4 mm
= 25
If 8 > 9 , then m > 0.
30 (a) Sebilangan prisma tegak mempunyai
Implikasi 2:
keratan rentas dalam bentuk Implication 2:

trapezium. mm
Jika m > 0, maka 8 > 9 .
Some right prisms have cross-
mm
sections in the form of trapezium. If m > 0, then 8 > 9 .

(b) Luas segi tiga ABC adalah 1 × 15 × 33 (a) Palsu
2 False

48 iaitu 360 cm2. (b) Jika m adalah gandaan 4, maka m
adalah gandaan 16.
1 If m is a multiple of 4, then m is a
The area of triangle ABC is 2 × 15 × multiple of 16.
48 that is 360 cm2. ∴ Akas itu adalah palsu.
∴ The converse is false.
(c) Premis 1:
(c) Premis 2:
Premise 1: Premise 2:
Luas bulatan itu bukan 64π cm2.
Jika xy  2x + 5 ialah suatu The area of the circle is not 64π cm2.

persamaan kuadratik dalam x, maka

nilai y adalah 2.

If xy  2x + 5 is a quadratic equation

in x, then the value of y is 2.

31 (a) ξ = {27, 28, 29, 30, 31, 32, 33, 34, 34 (a) {(1, P), (1, Q), (1, R), (1, S), (2, P),
(2, Q), (2, R), (2, S), (3, P), (3, Q),
35, 36, 37, 38, 39, 40, 41, 42, 43, 44} (3, R), (3, S), (4, P), (4, Q), (4, R),
(4, S), (5, P), (5, Q), (5, R), (5, S), (6,
B = {30, 35, 40} P), (6, Q), (6, R), (6, S)}
(b) ξ = {27, 28, 29, 30, 31, 32, 33, 34,
(b) (i) A = Peristiwa bagi penunjuk itu
35, 36, 37, 38, 39, 40, 41, 42, 43, 44} menunjukkan sektor R
A = Event of the pointer points
A = {36} at sector R
= {(1, R), (2, R), (3, R), (4, R),
C = {29, 38, 39}
A∩C=∅
(c) ξ = {27, 28, 29, 30, 31, 32, 33, 34,

35, 36, 37, 38, 39, 40, 41, 42, 43, 44}

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(5, R), (6, R)} (ii) Jarak
P(A) Distance
1
n(A) = 2 × (176 +128) × 12 + 128 × 4
= n(S)
1
1 + 2 × 128 × 16
=4
= 3360 m
(ii) A = Peristiwa bagi dadu Laju purata
menunjukkan nombor ganjil atau Average speed
penunjuk menunjukkan sektor S
A = Event of the dice shows an 3360
odd number or the pointer points = 32
at sector S = 105 m s−1
= {(1, P), (1, Q), (1, R), (1, S),
(3, P), (3, Q), (3, R), (3, S), (5, 37 (a) (i) Palsu
P), (5, Q), (5, R), (5, S), (2, S), False
(4, S), (6, S)}
P(A) (ii) Benar
n(A) True
= n(S)
5 (b) Premis 2:
=8 Premise 2:
16 adalah suatu nombor kuasa dua
35 (a) (i) 5 ialah faktor bagi 19 dan 25 sempurna.
ialah nombor perdana. 16 is a perfect square.
5 is a factor of 19 and 25 is a
prime number. (c) Implikasi 1:
Implication 1:
(ii) Palsu Jika 3k2 > 12, maka k > 2.
False If 3k2 > 12, then k > 2.

(b) s ≠ 2 Implikasi 2:
Implication 2:
(c) Implikasi 1: Jika dua garis lurus Jika k > 2, maka 3k2 > 12.
Implication adalah serenjang, maka If k > 2, then 3k2 > 12.
1: hasil darab kecerunan
dua garis itu adalah −1 38 (a) (i) Benar
If two straight lines are
perpendicular, then the True
product of the
gradients of the two (ii) Benar
lines is −1
True
Implikasi 2: Jika hasil darab
Implication kecerunan dua garis (b) Implikasi 1: Jika ΔABC ialah segi
2: lurus adalah −1, maka
Implication tiga bersudut tegak,
dua garis itu adalah 1:
serenjang maka AC2 = AB2 +
If the product of the
two straight lines is BC2.
−1, then the two lines If ΔABC is a right-
are perpendicular
angled triangle, then
36 (a) Laju seragam AC2 = AB2 + BC2.
Uniform speed
= 128 m s−1 Implikasi 2: Jika AC2 = AB2 + BC2,
Implication maka ΔABC ialah segi
(b) (i) 512 = 128 × (t − 12)
512 2: tiga bersudut tegak.

t = 128 + 12 If AC2 = AB2 + BC2,
= 16 then ΔABC is a right-

angled triangle.

(c) ∠ABC ≤ 90°

39 (a) (4x + 26 + 10 + 19) − (10 + 19 + 22

+ 9) = 15
4x + 55 − 60 = 15

4x  5 = 15

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4x = 20 Jika 3p = 5q  4, maka q = 3p + 4
x=5 5 .

20 + 26 + 10 + 19 + y + 22 + 9 = 113 If 3p = 5q  4, then q = 3p + 4
y + 106 = 113 5 .
y=7
(b) 10 Implikasi 2:
(c) 26 + 19 + 22
= 67 Implication 2:

40 Ruang sampel, S Jika q = 3p + 4 maka 3p = 5q  4.
Sample space, S 5 ,
= {(H, N), (H, 3), (H, 4), (J, N), (J, 3), (J,
4), (L, N), (L, 3), (L, 4), (1, N), (1, 3), (1, If q = 3p + 4 then 3p = 5q  4.
4), (2, N), (2, 3), (2, 4)} 5 ,

(a) A = Peristiwa kedua-dua keping kad (b) (i) Kesimpulan:
adalah dilabel dengan satu nombor Conclusion:
A = Event of both cards are labelled q mempunyai dua garis simetri.
with a number q has two lines of symmetry.
= {(1, 3), (1, 4), (2, 3), (2, 4)}
P(A) (ii) Premis 2:
n(A) Premise 2:
= n(S) 3 adalah satu faktor bagi 24.
4 3 is a factor of 24.
= 15
43 (a) 12 − 4 = 8 s
(b) B = Peristiwa sekeping kad dilabel
dengan satu huruf dan kad yang lain (b) Kadar perubahan laju
dilabel dengan satu nombor
B = Event of one card is labelled Rate of change of speed
with a letter and the other card is 35 − 15
labelled with a number
= {(H, 3), (H, 4), (J, 3), (J, 4), (L, 3), =4
(L, 4), (1, N), (2, N)}
P(B) = 5 m s-2
n(B)
= n(S) (c) 1 1
8 662 = 2 × (15 + 35) × 4 + 35 × 8 + 2
= 15
× (35 + 59) × (t − 12)
41 (a) Palsu
False 662 − 100 − 280 = 1 × 94 × (t − 12)
2
(b) Sebilangan
Some 282 × 2
t = 94 + 12
(c) Implikasi 1:
Jika 2x  5 = 1, maka x = 3 = 18
Implication 1:
If 2x  5 = 1, then x = 3 44 (a) Laju purata
Implikasi 2:
Jika x = 3, maka 2x  5 = 1 Average speed
Implication 2:
If x = 3, then 2x  5 = 1 2812
= 37
42 (a) Implikasi 1: = 76 m s−1
Implication 1:
(b) Kadar perubahan laju

Rate of change of speed
−18

=6
= −3 m s−2

(c) 1 (v + 18) × 20 + 18 × 11 +
2 812 = 2 ×

1
2 × 18 × 6

2 812 = (v + 18) × 10 + 198 + 54

v = 2 812 − 198 − 54 − 18
10

= 238

53

45 (a) (i) 2 1
(ii) 8 = 10
(iii) 7 + 11 + 2
= 20 (b) B = Peristiwa memilih sekeping kad
dengan nombor gandaan 10 atau
(b) 2x + 8 + 7 + 2 + 11 + 2 = 36 sekeping card dilabel dengan huruf Q
2x + 30 = 36 B = Event of picking a card with a
2x = 6 number which is a multiple of 10 or a
x=3 card labelled with letter Q
= {(40, A), (40, N), (40, P), (40, Q),
46 (a) A = Peristiwa sehelai baju putih (40, S), (43, Q)}
dipilih. P(B)
A = Event that a white shirt is n(B)
drawn. = n(S)
P(A) 6
9 = 10
= 17 3
=5
(b) B = Peristiwa sehelai baju putih dan
sehelai skirt hijau dipilih. 48 ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
B = Event that a white shirt and a P = {8}
green skirt are drawn. P ∪ Q = {4, 6, 8, 10, 12, 14}
P(B)
93 (a) (i) P ⊂ Q, P ∪ Q = Q
= 17 × 15 = {4, 6, 8, 10, 12, 14}
9
= 85 (ii) Q ' ∪ P = {5, 7, 8, 9, 11, 13}

(c) C = Peristiwa sehelai baju dan (b) P ' = {4, 5, 6, 7, 9, 10, 11, 12, 13, 14}
sehelai skirt yang sama warna R = {5, 7, 9, 11, 13}
dipilih. P ' ∩ R = {5, 7, 9, 11, 13}
C = Event that a shirt and a skirt of n(P ' ∩ R) = 5
the same colour are drawn.
P(C) 49 (a) 70 − 40 = 30 minit
97 25 63 70 − 40 = 30 minutes
= (17 × 15 ) + (17 × 15 ) + (17 × 15 )
21 2 6 (b) Average speed = 70
= 85 + 51 + 85 105
91
= 255 = 2 km minit−1
3
47 Ruang sampel, S
Sample space, S 2
= {(40, A), (40, N), (40, P), (40, Q), (40, = 3 × 60
S), (43, A), (43, N), (43, P), (43, Q), (43, = 40 km j-1
S)}
70
(a) A = Peristiwa memilih sekeping kad Laju putara = 105
dengan nombor genap dan sekeping
kad dilabel dengan huruf N = 2 km minute−1
A = Event of picking a card with an 3
even number and a card labelled
with letter N 2
= {(40, N)} = 3 × 60
P(A) = 40 km h-1
n(A)
= n(S) (c) (i) 28 km

(ii) 70 minit

70 minutes

50 (a) (i) Sebilangan nombor ganjil adalah
nombor perdana.
Some odd numbers are prime
numbers.

54

(ii) Semua oktagon mempunyai a letter card X.
lapan sisi. = {(J, X)}
All octagon have eight sides. n(B) = 1
∴ P(B)
(b) Jika a > 5, maka a > 7.
If a > 5, then a > 7. n(B)
∴ Akas itu adalah palsu. = n(S)
∴ The converse is false.
1
(c) Premis 2: = 42
Premise 2:
P ∩ Q ≠ P. (c) C = Peristiwa kedua-dua keping kad
P ∩ Q ≠ P. yang dipilih adalah daripada huruf
yang sama.
51 Ruang sampel, S C = Event that the two cards drawn
Sample space, S are of the same letter.
= {(V, J), (V, S), (V, V), (V, X), (V, G), = {(V, V), (V, V)}
(V, H), (J, V), (J, S), (J, V), (J, X), (J, G), n(C) = 2
(J, H), (S, V), (S, J), (S, V), (S, X), (S, ∴ P(C)
G), (S, H), (V, V), (V, J), (V, S), (V, X), n(C)
(V, G), (V, H), (X, V), (X, J), (X, S), (X, = n(S)
V), (X, G), (X, H), (G, V), (G, J), (G, S), 2
(G, V), (G, X), (G, H), (H, V), (H, J), (H, = 42
S), (H, V), (H, X), (H, G)} 1
= 21

52 (a) (i) Palsu
False

(ii) Benar
True

(b) Premis 2: 9 lebih besar daripada
Premise 2: sifar.
9 is greater than zero.

(a) A = Peristiwa kad pertama yang (c) Implikasi 1: Jika 2n > 20, maka n >
dipilih bukan kad huruf G. Implication 10
A = Event that the first card drawn is 1: If 2n > 20, then n > 10
not a letter card G.
= {(V, J), (V, S), (V, V), (V, X), (V, Implikasi 2: Jika n > 10, maka 2n >
G), (V, H), (J, V), (J, S), (J, V), (J, Implication 20
X), (J, G), (J, H), (S, V), (S, J), (S, 2: If n > 10, then 2n > 20
V), (S, X), (S, G), (S, H), (V, V), (V,
J), (V, S), (V, X), (V, G), (V, H), (X, 53 (a) 1 1 atau 9−2 = 1
V), (X, J), (X, S), (X, V), (X, G), (X, 5 >2 81
H), (H, V), (H, J), (H, S), (H, V), (H,
X), (H, G)} 1 > 1 or 9−2 = 1
n(A) = 36 5 2 81
∴ P(A)
n(A) (b) Kesimpulan:
= n(S)
36 Conclusion:
= 42
6 Prisma tegak itu tidak mempunyai
=7 luas keratan rentas 90 cm2 dan

(b) B = Peristiwa kad pertama yang ketinggian 15 cm.
dipilih ialah kad huruf J dan kad
kedua ialah kad huruf X. The right prism does not have a
B = Event that the first card drawn is cross-sectional area of 90 cm2 and a
a letter card J and the second card is
height of 15 cm.

(c) Semua nombor 5, 12, 31, 68, ... boleh
ditulis dalam bentuk n3 + 4, n = 1, 2,

3, 4, ...

All the numbers of 5, 12, 31, 68, ...
can be written in the form n3 + 4, n =

1, 2, 3, 4, ...

55

54 (a) (i) Ruang sampel, S (b) B = Peristiwa Lukman dan Yahya
Sample space, S akan memasuki kelas tuisyen Bahasa
= {(Ganesh, Jeffri), (Ganesh, Inggeris.
Kamaruddin), (Ganesh, B = Event that Lukman and Yahya
Raihani), (Jeffri, Kamaruddin), will join English tuition class.
(Jeffri, Raihani), (Kamaruddin, P(B)
Raihani)} 31
= 8 × 16
(ii) {(Ganesh, Raihani), (Jeffri, 3
Raihani), (Kamaruddin, = 128
Raihani)}
P(a boy and a girl) (c) C = Peristiwa salah satu daripada
3 mereka akan memasuki kelas tuisyen
=6 Matematik dan yang lain satu akan
1 memasuki kelas tuisyen Bahasa
=2 Inggeris.
C = Event that one of them will join
(b) (i) Ruang sampel, S Mathematics tuition class and the
Sample space, S other will join English tuition class.
= {(Musa, Kristy), (Musa, P(C)
Hazila), (Musa, Nuraizah), 3 1 31
(Musa, Raihani), (Ganesh, = (14 × 16 ) + (8 × 20 )
Kristy), (Ganesh, Hazila), 33
(Ganesh, Nuraizah), (Ganesh, = 224 + 160
Raihani), (Jeffri, Kristy), (Jeffri, 9
Hazila), (Jeffri, Nuraizah), = 280
(Jeffri, Raihani), (Kamaruddin,
Kristy), (Kamaruddin, Hazila), 56 ξ = {37, 38, 39, 40, 41, 42, 43, 44, 45, 46,
(Kamaruddin, Nuraizah), 47, 48, 49, 50, 51, 52}
(Kamaruddin, Raihani)} A = {40, 45, 50}
B = {50}
(ii) {(Musa, Kristy), (Musa, Hazila), C = {37, 38, 39, 45, 46, 47, 48, 49}
(Musa, Nuraizah)} (a) (i) B = {50}
P(both from the Group P)
3 (ii) C = {37, 38, 39, 45, 46, 47, 48,
= 16 49}

55 (a) A = Peristiwa Yahya akan memasuki (b) (i) A ∩ C = {45}
n(A ∩ C) = 1
kelas tuisyen Matematik dan kelas
(ii) B ∪ C = {37, 38, 39, 45, 46, 47,
tuisyen Bahasa Inggeris. 48, 49, 50}
(B ∪ C)' = {40, 41, 42, 43, 44,
A = Event that Yahya will join 51, 52}
n[(B ∪ C)'] = 7
Mathematics tuition class and

English tuition class.

A' = Peristiwa Yahya tidak akan

memasuki kelas tuisyen Matematik

atau kelas tuisyen Bahasa Inggeris.

A' = Event that Yahya will not join

Mathematics tuition class or English

tuition class.

P(A')
= 1 − P(A)

= 1 − (210 × 1 )
16

= 1 − 1
320

319
= 320

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