bengkel teknik menjawab soalan math upsr 2010 jidin
TRANSCRIPT
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Strategi dan teknik menjawab soalan
Matematik UPSR 2011
FOKUS
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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}
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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.
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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Topic analysis
PAPER 2: KOD 015/2
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PENTING!!!!!
Salin soalan dengan JELAS dan TEPAT.
Jawapan yang dibawa ke bawah mesti JELAS dan TEPAT.
Pastikan unit/jawapan ditulis berdasarkan kehendak soalan.
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PENTING!!!!!
Penukaran unit mesti ditulis dengan JELAS.
Sekiranya soalan menggunakan rumus/formula, gantikan dengan nilai yang sepatutnya.
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PENTING!!!!!
Gunakan Sistem 1, 2, 3 dalam penyelesaian masalah. 1: Ayat matematik 2: Langkah pengiraan 3: Jawapan akhir
Jangan tinggalkan tempat kosong.
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QUESTION 1 TO 5 (1 MARK)
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TOPIC ANALYSIS
WHOLE NUMBERSFRACTIONS, DECIMALS &
PERCENTAGEMEASUREMENT : MASS,
LENGTH & VOLUME OF LIQUIDTIMESHAPE & SPACE
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WHOLE NUMBERS
Write number in numeral or in words
Determine place value and digit value
Round off
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State the digit value of 4 in the number 45 090. (UPSR 2007)
45 09040 000
Underlined digit 4
Replace other digits with 0
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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
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5 7 . 3 4 9
State the place value of the digit 4 in 57.349
Underlined digit 4
OT
Determine place value for each digit
HundredthsAnswer
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5 7 . 3 4 9
State the digit value of the numeral 3 in 57.349
Underlined digit 3
Determine place value for each digit
0 0 . 3 0 00 0 . 0 4 00 0 . 0 0 9
0.3Answer
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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
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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.
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FRACTIONS, DECIMALS & PERCENTAGE
Write fractions in words, numeral, simplest form and equivalent fraction
Decimals: place value and digit value
Conversion
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Convert to a decimal. (UPSR 2007)100
78
100
78 = 8 x 100 + 7 = 807
8 0 71 0 0
= 8.07
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Convert 0.13 to a fraction.
(UPSR 2004)
Answer:
0 . 1 3 =13
tenths
hundredths
001
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Convert 3.09 to a mixed number. (UPSR 2005)
Answer:
3 . 0 9 =9
ones
3
tenths
hundredths
001
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Convert 93% to decimals. (UPSR 2006)
93% =93
100
93 ÷ 100 = 9 31 0 00
= 0.93
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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
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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
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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:
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TIME
Read a clock face12-hour system24-hour system
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Diagram 2 shows Azwan’s dinner time. (UPSR 2007)
Diagram 2
Write the time in the 24-hour system
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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
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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
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Diagram 3 shows a cuboid. (UPSR 2007)
State the number of vertices.
vertex
8 vertices
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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.
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QUESTION 6 TO 15
(2 MARKS)
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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
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(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
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UPSR 2002
5
21
5
4+ =
16
5
655
1
1
1 + 115 2
1
5
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UPSR 2002
45.3 – 7.09 =
45.3 - 7.09
10 102
1283 .
03
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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
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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
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(UPSR 2003)
Express as a percentage. 4
3
100%x43 =
2575%
1
100%x43 = 4
300= 75%
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(UPSR 2003) Find the value of of RM270.
5
2
RM270x52 =RM108
2 X RM270 = RM 540
RM540 ÷ 5 = RM108
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35% of 3000g. (UPSR 2005)
43.5 – 2.76 =___(UPSR 2005)
504 – 396 ÷ 12 = __ (UPSR 2006)
Try these out:
(252 + 84 ) ÷ 14 = __ (UPSR 2007)
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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
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(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)
Mula / bertolak(Start)
Tempoh masa(Duration)
7:15 ? 30 minit
END = DURATION + START
Tiba / Tamat (End)
Mula / bertolak(Start)
Tempoh masa(Duration)
? 6:45 30 minit
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Hour Minute
7 1545-
Anwer: 30 minutes
6
6 +60
30
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(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=
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Hour Minute
3 3530+
3 65601+ -
4 5Answer: 4 hours 5 minutes
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(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:
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.00
(UPSR 2004) RM9 – RM0.75 + RM2.35 =___
RM9RM0.75
81910
-
RM8.25RM2.35+
RM10.60
1
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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
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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
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Continued…
RM 20 + RM 1.50 =
RM 20.00+ RM 1.50 RM 21.50
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(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
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1 km = 1000 m
4
4.03 km - 750 m
0 37 5 0
0
-2 8 03
Answer:
3 280 m31
9 10
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(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:
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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
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Perimeter?????
6 + 8 + 10 +8 + 8 + 6 + 6 =
8 cm
6 cm
10 cm
P
Q
RS
U T
52 cm
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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
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QUESTION 16 TO 20 (3 MARKS)
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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.
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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
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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.
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Average : Total value of unitNumber of unit
Purata: Hasil tambah semua kumpulan
Bilangan kumpulan
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Diagram 6 shows a collection of three types of shapes. (UPSR 2007)
Calculate the percentage of of the whole collection.
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=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
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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
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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
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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
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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
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Answer:
1 500 ml – 400 ml =
1.5 l x 1000= 1 500 ml
1 100 ml
1 500400
1 100
-
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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?
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TIME RELATIONSHIP (HOUR & MINUTE)
Topic/
Fraction
TIME
(hour to minute)
1
2
30 minutes
1
4
15 minutes
3
4
45 minutes
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Answer:
2 151 303 45
45903
1 60- +
+
+
304at
4.30p.m.
MinuteHour
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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.
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Area: Length x Breadth
12 cm x 7 cm = 84cm2PQRS
PKLM3 cm x 3 cm = 9cm2
Shaded region
84cm2 – 9cm2 = 75cm2
Answer: 75cm2
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(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.
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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.
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(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.
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Average : Total value of unitNumber of unit
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
Average : Total value of unitNumber of unit
Manipulate the formulaTotal value of unit : average x number of unit
Average : Total value of unitNumber of unit
Manipulate the formula
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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
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Average : Total value of unitNumber of unit
Average:
4.2 m + 1.25 m + 0.5 m + 2.25 m4
= 8.20 m = 2.05 m 4
Average : Total value of unitNumber of unit
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~Tiada kejayaan semanis madu tanpa usaha sepahit hempedu ~
Sekian……..
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Salin soalan :JELAS & TEPAT
5900 + 789 – 1405 =
5900789+
6689
66891405-
UPSR 2006
7986698
6698
52935284
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Jawapan :JELAS & TEPAT
5900 + 789 – 1405 =
5900789+
6689
66981405-5293
UPSR 2006
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