Bengkel Teknik Menjawab Soalan Math UPSR 2010 Jidin

1

Strategi dan teknik menjawab soalan

Matematik UPSR 2011

FOKUS

2

Tetapkan Matlamat

CIPTA ANGAN-ANGAN KEJAYAAN

MATEMATIK = 100% GRED ‘A’ DALAM UPSR

{Walaupun kamu tak mampu membaca tetapi kamu mampu berangan

(membayangkan) apa yang kamu hajati. Itulah benih yang paling baik

sebelum menghadapi UPSR}

3

UPSR - MathematicsPaper 1 (Kertas 1)– 015/1Paper 2 (Kertas 2)– 015/2

Mulai 2003 soalan Matematik UPSR dikemukakan dalam dwibahasa (Bahasa Malaysia dan Bahasa Inggeris). Bagaimanapun, calon hanya perlu memberi tumpuan penuh untuk menjawab dalam satu bahasa yang dipilih.

4

Topic Type of question

Duration Number of

questions

Marks

• Whole numbers• Mixed numbers• Fractions• Decimals• Percentage• Money• Shape & space• Measurement: mass, length & volume of liquid• Data handling

Multiple choice

Answer all

questions

1 hour 40 40

Paper 1 015/1

5

1. UTAMAKAN SOALAN YANG MUDAH

Examples:

(1)40.1 – 29.04 – 7 = A 4.06 B 4.14 C 10.36 D 10.99

(2)RM350 – RM192.05 + 15 sen = A RM157.80 B RM158.10 C RM162.20 D RM162.80

(3) 0.5 million + 400 000 = A 0.9 million B 0.09 million C 0.45 million D 0.405 million

PENTING

6

PENTING

2. MEMAHAMI KEHENDAK SOALAN

3. MERANCANG STRATEGI UNTUK PENYELESAIAN MASALAH •Baca dan fahami kehendak soalan •Baca sekali lagi dan gariskan maklumat penting•Lakarkan maklumat supaya mudah difahami•Pastikan kehendak soalan•Tentukan cara penyelesaian (Operasi)•Selesaikan•Semak

7

4. GUNAKAN KAEDAH YANG PALING RINGKAS

Contoh :1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11+ 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 =

PENTING

Jawapan :

210

‘Biar Lambat Asal Selamat’Kira Mesti Cepat Jawapan Mesti Tepat

8

2 356 709647 935768 171

1 452 355+ 84 953

Contoh :

3

Jawapan :2

2

2

1

3

0

3

1

3

3

2

5

4

0

2

8

2

6

4

2

3

2

9

5. WASPADA TERHADAP PENGGANGGU

PENTING

Example:

A shop sells 630 red and yellow roses altogether. The shop sells 124 more red roses than yellow roses.How many yellow roses does the shop sell?

A. 568 B. 516 C. 382 D. 253

Sebuah kedai menjual 630 kuntum bunga ros merah dan bunga ros kuning. Kedai itu telah menjual lebih 124 kuntum bunga ros merah daripada bunga ros kuning. Berapakah bunga ros kuning yang dijual oleh kedai itu?

A. 568 B. 516 C. 382 D. 253

Sebuah kedai menjual 630 kuntum bunga ros merah dan bunga ros kuning. Kedai itu telah menjual lebih 124 kuntum bunga ros merah daripada bunga ros kuning. Berapakah bunga ros kuning yang dijual oleh kedai itu?

A. 568 B. 516 C. 382 D. 253

10

6. MEMASTIKAN SEMUA SOALAN TELAH DIJAWAB.

Semua 40 soalan telah dijawab Jangan conteng kertas jawapan. Hitamkan jawapan dengan penuh. Gunakan pensel 2B

Contohnya: (A) (B) (C) (D) Padamkan dengan bersih jika hendak menukar jawapan.

PENTING

11

Langkah-langkah menjawab soalan:

1) Soalan 1 markah = 1 minit (max) atau kurang 1 minit.

2) Soalan 2 markah dan 3 markah = gunakan masa seminimum mungkin iaitu kurang daripada 2 minit.

3) Masa yang lebih digunakan untuk menyemak jawapan (sekiranya ada)

4) Pastikan tulisan anda KEMAS dan JELAS.5) Setiap soalan perlu ditunjukkan langkah

pengiraan yang TERATUR dan LENGKAP. Pastikan jawapan akhir mengikut kehendak soalan.Langkah pengiraan mempunyai taburan markah yang tertentu. Walaupun jawapan akhir salah tetapi masih berpeluang mendapat markah untuk langkah pengiraan.

12

LULUS?????• Pass at least C : 18 questions – Paper 1

18 marks – Paper 2∞Lulus sekurang-kurangnya Gred C ∞: 18 soalan - kertas 1∞: 18 markah - kertas 2

• Gred A : perform both papers BUT Paper 2

determines candidate to get Gred A.∞ Gred A: Lulus kedua-dua kertas dengan

cemerlang TETAPI Kertas 2 menentukan calon untuk mendapatkan Gred A.

13

Topic analysis

PAPER 2: KOD 015/2

14

PENTING!!!!!

Salin soalan dengan JELAS dan TEPAT.

Jawapan yang dibawa ke bawah mesti JELAS dan TEPAT.

Pastikan unit/jawapan ditulis berdasarkan kehendak soalan.

15

PENTING!!!!!

Penukaran unit mesti ditulis dengan JELAS.

Sekiranya soalan menggunakan rumus/formula, gantikan dengan nilai yang sepatutnya.

16

PENTING!!!!!

Gunakan Sistem 1, 2, 3 dalam penyelesaian masalah. 1: Ayat matematik 2: Langkah pengiraan 3: Jawapan akhir

Jangan tinggalkan tempat kosong.

17

QUESTION 1 TO 5 (1 MARK)

18

TOPIC ANALYSIS

WHOLE NUMBERSFRACTIONS, DECIMALS &

PERCENTAGEMEASUREMENT : MASS,

LENGTH & VOLUME OF LIQUIDTIMESHAPE & SPACE

19

WHOLE NUMBERS

Write number in numeral or in words

Determine place value and digit value

Round off

20

State the digit value of 4 in the number 45 090. (UPSR 2007)

45 09040 000

Underlined digit 4

Replace other digits with 0

21

5 7 3 4 9

State the place value of the numeral 7 in 57 349. (UPSR 2004)

Determine place value for each digit

OTHThTTh

Underlined digit 7

ThousandsAnswer

22

5 7 . 3 4 9

State the place value of the digit 4 in 57.349

Underlined digit 4

OT


HundredthsAnswer

23

5 7 . 3 4 9

State the digit value of the numeral 3 in 57.349

Underlined digit 3


0 0 . 3 0 00 0 . 0 4 00 0 . 0 0 9

0.3Answer

24

Write ‘fifty thousand and three’ in numerals. (UPSR 2006)

TTh Th H T O

Draw place value chart

Insert digit in the place value column

5 0 30 0

Empty column: insert 0

Tuliskan 320 105 dalam perkataan.

Kumpulkan 3 digit dan garis.

320 105RIBU

Tiga ratus dua puluh

seratus lima

Jawapan :

Tiga ratus dua puluh

ribu

Seratus lima

25

Round off 38 572 to the nearest ten thousand. (UPSR 2003)

3 8 5 7 2Underlined digit in ten thousands

Circle the digit on the right

+1

4 0 0 0 0

Jika digit yang dibulatkan adalah 0, 1, 2, 3, 4, maka kekalkan digit yang

digaris.

Jika digit yang dibulatkan adalah 5, 6, 7, 8, 9, maka tambah ‘1’ pada

digit yang digaris.

Gantikan dengan 0 semua digit selepas digit yangDibundarkan / digaris.

26

FRACTIONS, DECIMALS & PERCENTAGE

Write fractions in words, numeral, simplest form and equivalent fraction

Decimals: place value and digit value

Conversion

27

Convert to a decimal. (UPSR 2007)100

78

100

78 = 8 x 100 + 7 = 807

8 0 71 0 0

= 8.07

28

Convert 0.13 to a fraction.

(UPSR 2004)

Answer:

0 . 1 3 =13

tenths

hundredths

001

29

Convert 3.09 to a mixed number. (UPSR 2005)

Answer:

3 . 0 9 =9

ones

3

tenths

hundredths

001

30

Convert 93% to decimals. (UPSR 2006)

93% =93

100

93 ÷ 100 = 9 31 0 00

= 0.93

31

Measurement: mass, length and volume of liquid

Read the measurement

i) Mass – kg and gii) Length – mm, cm, m and

kmiii) Volume of liquid – l and ml

32

Diagram 1 shows the weight of the apple. (UPSR 2004)

What is the weight, in g, of the apple?

0

250750

1kg

500

Diagram 1

33

1 kg

250

500

750

01 kg

250

500

750

01 kg

250

500

750

0

250 ÷ 5 = 50g

50

50

5050

50

250 + 100 = 350gAnswer:

34

TIME

Read a clock face12-hour system24-hour system

35

Diagram 2 shows Azwan’s dinner time. (UPSR 2007)

Diagram 2

Write the time in the 24-hour system

36

1

2

3

4

56

7

8

9

10

1112

8.00 p.m.

5 mins

10 mins

15 mins

20 mins

Answer 2020 hours0800 + 1200 = 2000

24-hour systemAfter 12.00 p.m.+ 1200

37

SHAPE AND SPACE

Name the shape 2 D shapes : square, rectangle, triangle, circle etc.

3 D shapes: cube, cuboid, cone, pyramid, cylinder etc.

Characteristics of the shape Edge, vertex, side, flat surface, curve surface etc

38

Diagram 3 shows a cuboid. (UPSR 2007)

State the number of vertices.

vertex

8 vertices

39

Table 1 shows the characteristics of a three dimensional shape. (UPSR 2006)

Number of vertices 8

Number of equal edges 12

Name three dimensional shape.

edge

vertex

cube

Lorekkan daripada rajah di bawah.

25

Caranya mudah…

Lorekkan 2 daripada 5 petak. ATAU

Loreklan 2 daripada setiap 5 petak.

40

QUESTION 6 TO 15

(2 MARKS)

41

TOPIC ANALYSIS:

WHOLE NUMBERS, FRACTIONS, DECIMALS, PERCENTAGE, TIME, MONEY, MASS, LENGTH, VOLUME OF LIQUID, SHAPE & SPACE.

BASIC OPERATIONS (+, -, X, ÷)MIXED OPERATIONSCONVERSION OF UNITPERIMETER & AREA VOLUME OF 3D SHAPESAVERAGE

42

(UPSR 2002)

39 + 13 x 4 =

BODMAS ruleB: Bracket

O: operations

M: Multiplication

D: Division

A: Addition

S: Subtraction

13 x 4 =

52

39 + 52 =

91

43

UPSR 2002

5

21

5

4+ =

16

5

655

1

1

1 + 115 2

1

5

44

UPSR 2002

45.3 – 7.09 =

45.3 - 7.09

10 102

1283 .

03

45

0.72 ÷ 9 = ____ (UPSR 2002)

0 . 7 290

0707 2

08

7 2

-

-

. 9876543210

01234567890

x

0 . 7 2 9

0 .0

0

078

72

Jawapan : 0.08

SIFIR 9

46

100 ÷ 5= 2020 ÷ 5= 4

96 ÷ 4 =

100

25X 96

Answer:

UPSR 2002

Calculate 25% of 96

5

20

25 ÷ 5= 55 ÷ 5= 1

24

1

4

47

(UPSR 2003)

Express as a percentage. 4

3

100%x43 =

2575%

1

100%x43 = 4

300= 75%

48

(UPSR 2003) Find the value of of RM270.

5

2

RM270x52 =RM108

2 X RM270 = RM 540

RM540 ÷ 5 = RM108

49

35% of 3000g. (UPSR 2005)

43.5 – 2.76 =___(UPSR 2005)

504 – 396 ÷ 12 = __ (UPSR 2006)

Try these out:

(252 + 84 ) ÷ 14 = __ (UPSR 2007)

50

UPSR 2007

6 days x 9 = __ weeks __ days

Notes: 1 week = 7 days6 x 9 = 54

54 ÷ 7 =7 remainder 5

Answer: 7 weeks 5 days

51

(UPSR 2006) Rajah 6 menunjukkan masa Sue Lan tiba di sekolah pada suatu pagi.

Diagram 6

Dia bertolak dari rumah pada jam 6:45 a.m. Berapa lamakah masa yang diambil oleh Sue Lan untuk tiba di sekolah?

Lakarkan Jadual berikut :

Tiba / Tamat (End)

Mula / bertolak(Start)

Tempoh masa(Duration)

7:15 6:45 ?END – START = DURATION

END – DURATION = START

Tiba / Tamat (End)



7:15 ? 30 minit

END = DURATION + START

Tiba / Tamat (End)



? 6:45 30 minit

52

Hour Minute

7 1545-

Anwer: 30 minutes

6

6 +60

30

53

(UPSR 2004) Calculate 3 hours 35 minutes + hour.

Give the answer, in hours and minutes.

2

1

Important!4

115 minuteshour=

4

345 minuteshour=

2

130 minuteshour=

54

Hour Minute

3 3530+

3 65601+ -

4 5Answer: 4 hours 5 minutes

55

(UPSR 2006) 3 years 7 months + 5 years 8 months = ____years __________months

(UPSR 2005) 4 hours 15 minutes + 1 hour 50 minutes

Try these out:

56

.00

(UPSR 2004) RM9 – RM0.75 + RM2.35 =___

RM9RM0.75

81910

-

RM8.25RM2.35+

RM10.60

1

57

Table 1 shows the quantities of notes and coins. ( UPSR 2005)

Money Quantity

RM5 4

50 sen 3

Calculate the total value of the money.

TABLE 1

58

Answer:

RM 5 x 4 = RM 20

RM 5 + RM 5 + RM 5 + RM 5 = RM 20

@

50 sen + 50 sen + 50 sen = 150 sen

50 sen x 3 = 150 sen @

150 sen ÷ 100 = RM1.50

59

Continued…

RM 20 + RM 1.50 =

RM 20.00+ RM 1.50 RM 21.50

60

(UPSR 2007)Calculate 4.03 km – 750 m. Give your answer, in m.

CONVERSION OF UNIT

4.03 km = ______ m km m

x 1000

÷ 10004. 0 3 x 1000 =4030 m

61

1 km = 1000 m

4

4.03 km - 750 m

0 37 5 0

0

-2 8 03

Answer:

3 280 m31

9 10

62

(UPSR 2006)Calculate 2.05 kg x 12. Give your answer, in g.

(UPSR 2007) Calculate 1.27 l ÷ 10.Give your answer, in ml.

Try these out:

63

Diagram 4 consists of three triangles, PQR, RSU and STU of the same size.(UPSR 2005)

8 cm

6 cm

10 cm

P

Q

RS

U T

Calculate the perimeter, in cm, of the whole diagram

64

Perimeter?????

6 + 8 + 10 +8 + 8 + 6 + 6 =

8 cm

6 cm

10 cm

P

Q

RS

U T

52 cm

65

Jangan takut menghadapi kegagalan kerana dengan kegagalan kita beroleh pengetahuan tentang kekurangan dan kelemahan diri kita , yang ditakuti ialah gagal untuk kali kedua dalam perkara yang sama - Hamka

66

QUESTION 16 TO 20 (3 MARKS)

67

TOPIC ANALYSIS: PROBLEM SOLVINGINCLUDE WHOLE NUMBERS,

PERCENTAGE , MONEY, MEASUREMENT ( MASS, LENGTH & VOLUME OF LIQUID ), TIME, SHAPE & SPACE AND AVERAGE.

THE PROBLEM SOLVING WILL COMBINE 2 OR MORE SKILLS IN A QUESTION.

68

TIPS…Problem solving:Penyelesaian masalah (cara-caranya)

a) Read the questionBaca soalan yang diberi

b) Extract the informationKumpul maklumat:

What is given? Apa yang diberi? What is asked for? Apa kehendak soalan? What opeartion is needed?

Apa operasi yang diperlukan?c) Calculate step by step.

Kira (langkah pengiraan yang TERATUR dan LENGKAP)

d) Check the answerSemak semula

69

Ahmad, Bala and Chong collected used tins. The average number of tins collected by the 3 boys was 16. Ahmad and Chong collected 9 and 17 tins, respectively.Calculate the number of tins collected by Bala. (UPSR 2003)

Untuk mencari PURATA, Tambah lepas tu bahagi.Ahmad 3

=Bala Chon

g+ +

16 Purata diberi

9 17

3 x 16 = 48 tin9 + BALA + 17 = 48 tin48 – 17 – 9 = 22 tin.

70

Average : Total value of unitNumber of unit

Purata: Hasil tambah semua kumpulan

Bilangan kumpulan

71

Diagram 6 shows a collection of three types of shapes. (UPSR 2007)

Calculate the percentage of of the whole collection.

72

=100%x205

= 5 piecesTotal shapes = 20

x100%

1

4

=

20%

ANSWER:

20

5

Calculate the percentage of of the whole collection.

5 x 100 ÷ 20

73

Musa has RM100. He buys a pair of shoes for RM72.90 and 3 pens for RM4.50 each.How much money does he have left? (UPSR 2004)

RM4.50 X 3 = RM13.50

RM72.90 + RM13.50 =RM100.00 – RM86.40 =

pen

Total

RM13.60

Left

Answer

74

Lina’s mother used of 2 kg of sugar to bake a cake.How much sugar did she use? Give your answer, in g. (UPSR 2003)

1

4

14

x 2000 g=

500 g2 kg = 2000 g

1 x 2000 ÷ 4

75

Notes to remember!

Topic/

Fraction

Mass

(kg=g)

Length

(km=m)

Volume of liquid

(l=ml)

1

2

0.5 kg

500 g

0.5 km

500 m

0.5 l500 ml

1

4

0.25 kg

250 g

0.25 km

250 m

0.25 l250 ml

3

4

0.75 kg

750 g

0.75 km

750 m

0.75 l750 ml

76

A bottle contains 1.5 l of water. of the water is split.

How much is the volume, in ml, of water is left in the bottle? (UPSR 2004)

5

2

TOPIC/ FRACTION

MASS LENGTH VOLUME OF LIQUID

0.2kg = 200g 0.2km = 200m 0.2l = 200ml

0.4kg = 400g 0.4km = 400m 0.4l = 400ml

0.6kg = 600g 0.6km = 600m 0.6l = 600ml

0.8kg =800g 0.8km =800m 0.8l =800ml

15

25

35

45

77

Answer:

1 500 ml – 400 ml =

1.5 l x 1000= 1 500 ml

1 100 ml

1 500400

1 100

-

78

Table 1 shows the time taken by Mohan to revise his lesson. (UPSR 2007)

4

3

Subject Time taken

Mathematics 1 hour 30 minutes

Science hour

Mohan starts his revision at 2.15 p.m.At what time does Mohan finish his revision?

79

TIME RELATIONSHIP (HOUR & MINUTE)

Topic/

Fraction

TIME

(hour to minute)

1

2

30 minutes

1

4

15 minutes

3

4

45 minutes

80

Answer:

2 151 303 45

45903

1 60- +

+

+

304at

4.30p.m.

MinuteHour

81

Diagram 5 shows a rectangle PQRS and a square PKLM. (UPSR 2005)

P

3 cm

M

K

L

S12 cm

7 cm

Q

R

Calculate the area, in cm2, of the shaded region.

82

Area: Length x Breadth

12 cm x 7 cm = 84cm2PQRS

PKLM3 cm x 3 cm = 9cm2

Shaded region

84cm2 – 9cm2 = 75cm2

Answer: 75cm2

83

(UPSR 2007) Diagram 4 shows a rectangle JKLM and a right angled triangle LKN.

J

M L

K N6 cm

8 cm

JKN is straight line. The length of KN is equal to the length of KL. Calculate the area, in cm2, of the coloured region.

84

Formula for AREA of Triangle:

12

x Length

x Breadth

12

x (6 cm + 8 cm)

x 6 cm

14 x 6 =2 1

3

=

42 cm2

J

M L

K N6 cm

8 cm

JKN is straight line. The length of KN is equal to the length of KL. Calculate the area, in cm2, of the coloured region.

85

(UPSR 2006) Table 2 shows the number of classes and the average number of pupils in Year One and Year Two.

YearNumber of

classesAverage number of

pupils

One 4 31

Two 3 35

Find the difference between the number of pupils in Year One and Year Two.

86


Manipulate the formulaTotal value of unit : average x number of unit

Year 1: 31 x 4 = 124

Year 2: 35 x 3 = 105

Difference: 124 - 10515 pupils


Manipulate the formulaTotal value of unit : average x number of unit


Manipulate the formula

87

Table 2 shows the length of 4 ribbons bought by Karina. (UPSR 2005)

Ribbon Length

Blue 4.2 m

Green 1.25 m

Yellow 0.5 m

Red 2.25 m

Calculate the average length, in m, of the ribbons.

TABLE 2

88


Average:

4.2 m + 1.25 m + 0.5 m + 2.25 m4

= 8.20 m = 2.05 m 4


89

~Tiada kejayaan semanis madu tanpa usaha sepahit hempedu ~

Sekian……..

90

Salin soalan :JELAS & TEPAT

5900 + 789 – 1405 =

5900789+

6689

66891405-

UPSR 2006

7986698

6698

52935284

91

Jawapan :JELAS & TEPAT

5900 + 789 – 1405 =

5900789+

6689

66981405-5293

UPSR 2006

Bengkel Teknik Menjawab Soalan Math UPSR 2010 Jidin

Documents

Transcript of Bengkel Teknik Menjawab Soalan Math UPSR 2010 Jidin