04/12/2023FAUZIAH AHMAD

1

Teknik Menjawab

Matematik Kertas 2

OLEH :CIKGU FAUZIAH AHMAD

04/12/2023 FAUZIAH AHMAD 2

6 4 7ribu sa

6 4 0 0 7

04/12/2023 FAUZIAH AHMAD 3

32 745

30 000

04/12/2023 FAUZIAH AHMAD 4

2 x 2

= 4

5 x 2 10 = 0.4

04/12/2023 FAUZIAH AHMAD 5

2.3 kg /2.35 kg

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5 0· 32 ÷ 4 = ( 1 markah )

0. 0 8 4

0. 3 2

- 03

- 03 2

- 3 20

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1 4+

1 x 3

= 1 4+

3

9 3 x 3 9 9

= 1 79

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silinder

2

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1

1. 4 3x 41 7 2

2 910 10

2. 3 0 0− 1 7 2

1 2 8

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Rajah 3

Jam Minit8 15

+ + 408 55

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RM125x 3

RM375

RM 3 7 . 5 010 RM 3 7 5 . 0 0

- 3 07 5

- 7 05 0

- 5 00

- 00

30 x RM125100

= 3 x RM12510

= RM37510

= RM37.50

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4.05 m = 4.05 x 100 = 405.00 cm

3 10

405 cm − 155 cm 250 cm

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RM5 X 2 = RM10.00RM2 X 2 = RM 4.00RM1 X 2 = RM 2.00 + RM16.00

50 + 20 + 20 = 90 sen

RM16.00+ RM 0.90 RM16.90

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Purata = 750 g + 3.9 kg + 2.07 kg 3

750 ÷ 1000 = 0.750 kg

0.750 3.900+ 2.070 6.720

2 . 2 43 6 . 7 2

6761 21 2

Purata = 750 g + 3.9 kg + 2.07 kg 3 = 6.720 kg 3 = 2.24 kg

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5

5

5 7

1

Perimeter = ( 1 + 7 + 6 + 7 + 5 + 5 + 5 ) cm = 36 cm

1 cm 7 cm 6 cm 7 cm 5 cm 5 cm+ 5 cm 36 cm

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1 minggu = 7 hari

6 minggu7 4 5 hari

- 4 23 hari

45 hari = 6 minggu 3 hari

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2 2 cm3 6 cm

6 0

2

Luas 1 petak = 2 cm x 2 cm = 4 cm²

Luas kawasan yang tidak berlorek = 4 cm² x 6 = 24 cm²

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RM2.50 ÷ 10 x 6 =

RM 0 . 2 510 RM 2 . 5 0

- 02 5

- 2 05 0

- 5 00

1 3

RM 0.25x 6 RM 1.50

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2.6 l = 2.6 x 1000 = 2600.0 ml

2600 ml− 850 ml 1750 ml

850 ml = 850 ÷ 1000 = 0.850 l

2.600 l − 0.850 l 1.750 l

CARA 1 CARA 2

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2 2

RM 0.65x 4 RM 2.60+ RM 2.00 RM 4.60

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Kawasan berlorek = 5

Kawasan tidak berlorek = 5

Seluruh rajah = 10

Peratus kawasan berlorek = 5 x 100 % 10

= 50 %