KOLEKSI SOALAN BAHAGIAN A

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Diagram 4 shows two packages of stationeries sold in a bookstore.

Rajah 4 menunjukkan dua pakej alat tulis yang dijual di sebuah kedai buku.

For the two packages of the stationeries, the price of each bottle of
glue is the same, so as the price of each pen.
Calculate the price, in RM, of a pen and a bottle of glue.

Bagi dua pakej alat tulis itu, harga setiap botol gam adalah sama, begitu juga
dengan harga setiap pen.
Hitung harga, dalam RM, bagi sebatang pen dan sebotol gam.

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Solution by matrix method is not allowed to answer this question.

Penyelesaian dengan kaedah matriks tidak dibenarkan untuk menjawab soalan
ini.

Diagram 4 shows two identical scales, and . A few boxes, and
are arranged on scales.

Rajah 4 menunjukkan dua penimbang yang serupa, dan . Beberapa kotak
dan disusun di atas penimbang itu.

It is given that the difference of mass on scale is 864 .
Calculate the mass, in kg, of a box and of a box .

Diberi bahawa beza jisim pada penimbang ialah 864 kg.
Hitung jisim, dalam kg, bagi satu kotak dan satu kotak .

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Solution using matrix method is not allowed in this question.

Penyelesaian menggunakan kaedah matriks tidak dibenarkan untuk soalan ini.

A hotel has 500 rooms which consists of standard rooms and deluxe
rooms. The prices of a standard room and a deluxe room per day are
RM 120 and RM 220 respectively. On a public holiday, all the rooms
in the hotel are rent out. The total income received in RM 90000.
Find the number of each type of room in the hotel.

Sebuah hotel mempunyai 500 bilik yang terdiri daripada bilik standard dan
bilik deluxe. Harga sebuah bilik standard dan sebuah bilik deluxe sehari masing
– masing ialah RM 120 dan RM 220. Pada hari cuti kelepasan am, semua bilik
hotel telah disewa. Jumlah pendapatan yang diperoleh ialah RM 90000.
Cari bilangan bagi setiap jenis bilik di hotel itu.

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PAHANG GERAK GEMPUR TERENGGANU MODUL 2
+ = 40 + 2 = 80
+ 3 = 64 3 − 2 = 20 @ setara
2 = 24 , = 12 4 = 100 or −8 = −220 orsetara
+ 12 = 40 , = 28 27.50
2 + 2 = 2(28) + 2(12)
= 80 JOHOR SET 2
SPM ULANGAN 4 + 5 = 9700 @ 2 + 3 =
5 + 2 = 3 + 8 @ 2 = 6 5100
12 − 2 = 852 @ setara 4 + 6 = 10200 @ setara
6 = 852 = 1800 , = 500
= 142 , = 426
PAHANG JUJ SET 2
JOHOR MUAR − = 10 @ 4 + 6 = 78
6 − 2 = 18 @ 2 − 4 = 16 6 − 6 = 60 @ 4 + 4 =40
10 = 138 @ 10 = 38
10 = 13.8
5 = 10 @ − = 10 SELANGOR 2
2 + 2 = 44 @ 3 + 4 = 72
3 2 + 2 = 44 × 2

MRSM 4 + 4 = 88
= + 6.10 or + 5 = 10.90 3 + 4 = 72
6 = 4.80 @ setara
Pencil = 142
Geometry set = 426

PULAU PINANG = 16 , = 6

16 − 6 = 40 @ = −6 − 23 PERLIS
17 = 17 @ − 51 = 51 = 4 @ − 3.75 = + 3.75
4 − 3.75 =
4
+ 3.75 @
= 1 , = −4 3 = 7.50 @ equivalent
= 2.50 = 2 500 or
= 4(2500) = 10000
12 500

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PAHANG JUJ SET 1 TERENGGANU MODUL 1 (1)
2 + 2 = 11 @ + 3 = 12.5
2 + 6 = 2 @ = 12.5 − 3 = 9 − @ = 9 − @
= 2 , = 3.50 = 2 − 12 @ = 12+

2

3 = 6 @ 3 = 21
= 7 , = 2

NEGERI SEMBILAN SELANGOR 1
= 2+ or = 3 − 2 @ 2 + 3 = 20 @ + 2 = 11
3 + 2 = 11 × 2
= − +5
= −4 + 5 or 4 2 + 4 = 22

4 + (2+ ) = 5 or equivalent = 7, = 2

3 SELANGOR 3
or + = 500
13 = 13
120 + 220 = 90000
= 1 , = 1 = 500 −
120(500 − ) + 220 = 90000
TERENGGANU MODUL 3 = 200 , = 300

+ 3 = −3 or = 9 + 3

equivalent

−6 = 12 @ equivalent

= 3 , = −2

SBP KEDAH SET 1
4 + 3 = 2 + 6 or 2 = 3 or 6 + 9 = 54 @ 6 − 8 = −14
10 − 3 = 864 @ setara @ 17 = 68 @ 17 = 51
8 = 864 @ setara = 3 , = 4
= 108 , = 72

KEDAH SET 2 PAHANG TRIAL
2 − 6 = 30 @ 4 − 12 = 60
@ 6 + 12 = −30 @ 10 = 30
= 3 , = −4

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TERENGGANU MODUL 1 (2) BEBAS

2 + 5 = 31
3 + = 27
= 31−5 @ = 27 − 3

2

13 = 39 @ 13 = 104
= 8 , = 3

KELANTAN

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QUESTION 5 / SOALAN 5
(4M)

- ISIPADU
(VOLUME)

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Answer / Jawapan :6

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SELANGOR 2 PAHANG GERAK GEMPUR
1 22 1 22
3 × 7 × 3.52 × ℎ = 308
ℎ = 24 × × 92 × 21
Height of the container = 24 × 14.6 37
7 1 22
= 1752 @ 50 2 @ 50.06
× × 92 × 21 − 6 × 6 × 6
35 35 37
= 1566
BEBAS
PAHANG TRIAL

TERENGGANU MODUL 2 PERLIS
1 1

12 (8 + 11)(4) × 21 (6 + 8)(7) × ℓ
22 22
14 × 7 × 52 × × × 42 × 7
27
1 22 1 1 22
2 (8 + 11)(4) × + 4 × 7 × 52 × (6 + 8)(7) × ℓ − × × 42 × 7
2 27
= 403.5
= 12
=7

PULAU PINANG JOHOR SET 2

1 Volume of cake mixture / Isipadu
2 (5 + 3)(5)(5)
(5)(3)(7) adunan 22
1 2 22
= ( × × 283) − 2 × × 212 × 14
(5 + 3)(5)(5) + (5)(3)(7) 37 7
2
= 205 3 Isipadu adunan kek yang tinggal = 7186.67

PAHANG JUJ SET 2 TERENGGANU MODUL 1

1 × 22 × 22 × 7 or equivalent 1 1 22
( × × × 3.52 × 7)
27 23 7
7×7×7
7 × 4 × ℎ 28ℎ 1 1 22
( × × × 3.52 × 7) + (7 × 7 × 7)
28ℎ + 44 = 212 or equivalent 23 7
ℎ=6
11 4655
387.92 @ 387 12 @ 12

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SPM ULANGAN JOHOR (MUAR)
Baki isipadu = Isipadu jag – 5 1 22
(Isipadu cawan) = × × 52 × 9

22 37
= ( × 6.52 × 21) 1 4 22
= 2 × 3 × 7 × 23
7 31 272 × 52 × 9
3 1 4 22 = 1 × 22
−5 (4× [2 ×3 ×7 43]) × 4× × 23
= (2788.50) − (502.86) 23 7
= 2285.64 @ 2285 16 235.71
= 16.76
25

= 14.06 ≈ 14

PAHANG JUJ SET 1 KEDAH SET 2
24 16 22
× 82 × 4
9 = , = 6
1 22 7
× × 92 × 24 22
37
1 22 × 72 × 4 4
7

× × 62 × 16 12000 ÷ 188 7
37
1 22 1 22
× × 92 × 24 − × × 62 × 16 63

37 37
= 1433.14

KEDAH SET 1 NEGERI SEMBILAN
1
8×8×8×2
22 21 × 14 × × 16
22
50 × × 2 × 0.5 2 × 7 × 82 ×
7 1 1 22
22 [×2 14 × × 16] + [ ×2 ×7 82 × ]
= 2763 3
8 × 8 × 8 × 2 = 50 × × 2 × 0.5
7 7

3.61

= 13

SELANGOR 1 KELANTAN
(a) 1 × (10 + 8) × 4 × 8

2

= 288
(b) 288 + 1 × 8 × 8 × = 420

3

= 6.19

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SELANGOR 3 SBP
Isipadu prisma 11

1 2×3×6×8×5
= [(30 + 15)18] × 15 1 4 22
4 × 3 × 7 × 53
2
= 6075 11 1 4 22
Isipadu separuh silinder × × 6 × 8 × 5 + × × × 53
= 6075 − 4440 23 43 7
= 1635 20 3590
170.95 @ 170 @
1 22 21 21
( × 2 × 15) = 1635
MRSM
27
= 8.33 1
TERENGGANU MODUL 3 (7 + 13) × 7 ×

1 222 3
× (6 + 10) × 7 × 7 × 72 × 2 ×

2 74
2 × 22 × 3.53 or equivalent 1 (7 + 13) × 7 × + 22 × 72 × 2 × 3
2 74
37 11

1 2 22
× (6 + 10) × 7 × 7 − × × 3.53

2 37
1 1813

302.17 @ 302 6 @ 6

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QUESTION 6 / SOALAN 6
(6M)

- PENAKULAN MATEMATIK
(MATHEMATICAL
REASONING)

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(a) It is given that is a negative number. Complete the
mathematical sentences in the answer space by using the
symbol “>” or “<” to form a true statement.

Diberi bahawa ialah satu nombor negative. Lengkapkan ayat
matematik di ruang jawapan dengan menggunakan symbol “>” atau
“<” untuk membentuk satu penyataan benar.

(b) Write down premise 2 to complete the following arguments :

Tulis premis 2 untuk melengkapkan hujah berikut :

Premise 1 : If the interior angle of a polygon is 120°, then the
polygon is a regular hexagon.

Premis 1 : Jika sudut pedalaman bagi sebuah polygon ialah 120°, maka
polygon itu ialah heksagon sekata.

Premise 2 : ………………………………………………………………………

Premis 2 ...........................................................................

Conclusion : Polygon is a regular hexagon.

Kesimpulan : Poligon ialah sebuah heksagon sekata.

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(a) …………………………………………………………………………………………………………
(b) 102 = 20 ………………… 9 × 0 = 0
(c) …………………………………………………………………………………………………………
(d) ………………………………………………………………………………………………………….

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