kelemahan calon matematik upsr
DESCRIPTION
MathematikTRANSCRIPT
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SALAM………SEJAHTERA
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KELEMAHAN CALON MATEMATIK UPSR
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A. Tidak Menguasai Fakta Asas Matematik
Example 1:
1995 (No. 1)/Paper 2
37 + 865 =Mistakes of the candidates :
37
+ 865
892
37
+ 865
802
37
+ 865
992
Answer : 37
+ 865
902
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B. Tidak Menguasai Kemahiran Matematik
Example 1 :1995 (No. 1)/Paper 2
37 + 865 =
Jawapan : 37
+ 865
902Mistakes of candidates :
37
+ 865
1235
37
+ 865
892
37 + 865 = 37865
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Example 2 :2005 (No. 8)/Paper 2
43.5 – 2.76 =
Answer : 43.50
- 2.76
40.74
Mistakes of candidates :
a. 43.5
- 2.76
15.9
b. 4350
- 2760
1590
c. 43.5
- 2.76
40.86
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Example 2:
1997 (No. 15)/Paper 2
Draw a isosceles triangle PQR in the diagram below. (2 marks)
Answer
Mistakes of candidates
Q
Q
P
R Q
P
R Q
P
R
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Example 3:
1999 (No. 9)/Paper 2
120 + 15 ÷ 3 = (2 marks)
Answer : 15 ÷ 3 = 5 + 120
= 125
Mistakes of candidates : 120
+ 15
135 ÷ 3 = 45
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Answer : 4 years 6 months
4 ) 18 years
-16
2 years = 24 months
24 months
Example 4:2003 (No. 15)/Paper 1
1 of 18 years = ___ years ___ months 4
4 yrs 2 mths
4 ) 18 years
16
2 years
4 years 5 months
4 ) 18 years
16
2 0
2 0
Mistakes of candidates
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Example 5 :2002 (No. 4)/Paper 1
3432 x 6 ÷ 12 =
Mistakes of candidates : 12 ÷ 6 = 2
3432
x 2
6864
Answer (i)
3432
x 6
20592 ÷ 12
= 1716
Answer (ii)
3432 x 6
12
= 3432 ÷ 2
= 1716
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C. Tidak Menguasai Rumusan MatematikExample 1 :1998 (No. 16)/Paper 2
Answer : (i) 4+4+4+4+4+4 = 24 cm (ii) 4 x 6 = 24 cm
2+2+2+2 = + 8 cm 2 x 4 = + 8 cm
32 cm 32 cm
Diagram beside is form from squares
of same size. Calculate the perimetre
of the shaded region. (3 marks)
2 cm
4 cm
Mistakes of candidates : 4 x 2 = 8 cm²
x 4
32 cm²
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Contoh 2 :1999 (No. 40)/Paper 1In Diagram 14, PQS is a right angle triangle and PRS is isosceles, PR = RS. Calculate the area of triangle PRS, in cm²
S
4 cm 5 cm
3 cm
P
Q RRajah 14
Answer : Area = Base x Height ÷ 2
= RS x PQ ÷ 2
= 5 cm x 4 cm ÷ 2
= 20 cm² ÷ 2
= 10 cm²
(i) 3 cm x 4cm ÷ 2
= 12 cm² ÷ 2
= 6 cm² Mistakes of candidates
(ii) 5 cm x 5cm ÷ 2
= 25 cm² ÷ 2
= 12.5 cm²
(iii) 8 cm x 4cm ÷ 2
= 32 cm² ÷ 2
= 16 cm²
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Example 3 :1996 (No. 5)/Paper 2Write a equivalent fraction of 3/5
Answer : (i) 3 x 2 = 6 (ii) 3 x 3 = 9 dll
5 x 2 10 5 x 3 15
Mistakes of candidates : (i) 3 x 3 = 9 dll
5 x 5 25
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D.Tidak Memahami Maklumat/Kata kunci/Kehendak soalanExample 1 :1995 (No. 4)/Paper 2580 manggis shared same amount among Ang, Beng, Dol, Eng and Fan. How many manggis each one can get?
Answer : 580 ÷ 5 = 116
Mistakes of candidates : (i) 580 ÷ 4
(ii) 580 + 5
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Example 2:2002 (No. 19)/Paper 2
A pail contains 2 liters 30 ml water. Azmi pours 450ml of water from another container to the pail. Calculate the amount of water, in ml, in the pail now?
Answer : 2030 ml
+ 450 ml
2480 ml
Mistakes of
candidates : 2030 ml
- 450 ml
1580 ml
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Example 3:2002 (No. 15)/Paper 2
Azlan woke up and looked at the watch same as the watch below. Bus will takes him at 6.50 a.m. How many minutes did Azlan has before the bus arrived?
Answer : 6.50
- 6.10
0.40 = 40 minutes or
= 0 hour 40 minutes
Mistakes of candidates : 6.50
6.10 -
0.40 minutes
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E. Lemah Dalam Membaca Rajah Pengukur/Penyukat/PenimbangExample 1:1994 (No. 42)/Paper 1WEIGHT MEASUREMENT
Example 2:1995 (No. 11)/Paper 2TimeWhat is the time ¼ hour after the time shown by watch beside?
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Contoh 3:1997 (No. 32)/Paper 1 LENGTH MEASUREMENT
Diagram 12 shows the length of two pencils. The difference of length, in mm, between the two pencils is?
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Example 4:2002 (No. 14)/Paper 1MASS
Between these measurement, which shows the right sum of 1.5kg + 250 g?
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F. Tidak Cekap Dalam KBKKExample 1:1994 (No. 41)/Paper 1
Gaji Puan Tan ialah RM850. Gaji suaminya 20% lebih daripada gaji Puan Tan. Berapakah jumlah gaji mereka?
Kesilapan calon : (i) 20 x RM850 (ii) RM850
100 + 20
= 2 x RM85 RM870
= RM170
Jawapan :
(i) Pn. Tan = RM850
lebih = 20/100 x 850 = RM170
suami = 850 + 170 = 1020
JUMLAH = 850 + 1020 = RM1870
(ii) 120 x 850
100
=10200/100
= RM1020
J = 850+1020
= RM1870
(ii) 120 x 850
100
=12 x 85
= RM1020J = 850+1020 = RM1870
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Contoh 2:1995 (No. 18)/Kertas 1
1800 biji manik dibahagi antara Karen dan Mona. Karen mendapat 200 biji manik lebih daripada Mona. Bilangan manik yang Mona dapat ialah
Jawapan : 1800
- 200
1600 ÷ 2 = 800
Kesilapan calon:(i)1800 ÷ 2 = 900
- 200
700
(ii) 1800 ÷ 2 = 900
+ 200
1100
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Contoh 3:1996 (No. 14)/Kertas 1
Rajah 4 menunjukkan bekas X yang mengandungi 2.4 liter dan bekas Y yang mengandungi 600 ml air. Berapa mlkah air yang perlu dituang daripada bekas X ke dalam bekas Y supaya kandungan air dalam kedua-dua bekas itu sama banyak?
Jawapan : (i) 2400 ml
- 600 ml
1800 ml ÷ 2 = 900 ml
(i) 2400 ml
+ 600 ml
3000 ml ÷ 2 = 1500 – 600 = 900ml
Kesilapan calon : (i) 2400 ÷ 2 = 1200 (ii) 2400
- 600 - 600
600 ml 1800 ml
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Example 4:2003 (No. 4)/Paper 1
2 x = 4805 Answer : (i) 2 = 480
5
1 = 480 ÷ 2
5
= 240 x 5 = 1200
(ii) 2 = 480
5
= 5 x 480
2
= 2400 ÷ 2
= 1200
(ii) 2 = 480
5
= 480 x 5
2
= 1200
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G. Lemah Dalam Pertukaran Unit
Contoh 1:2008 (No. 8)/Kertas 2
Give the answer in cm.
Calculate 2m 7cm ÷ 9, Answer : 207cm ÷ 9
= 23cm
a.Wrong 27cm ÷ 9 Answer : = 3 cm
b.Wrong 2007 ÷ 9 Answer : = 223 cm
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Example 2:2001 (No. 11)/Paper 2
Convert 1 3/5 hour to minutes.
Answer :
(i) 1 hour x 60 = 60 minutes
3/5 x 60 = 36 minutes
96 minutes
(ii) 8/5 x 60
= 480 ÷ 5
= 96 minutes
(ii) 8/5 x 60
= 8 x 12
= 96 minutes
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Example 3:2003 (No. 8)/Paper 1
7.4 kg ÷ 20 = _______ gram
Answer :
7400 ÷ 20 = 370 gram
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H. Terpengaruh Dengan Soalan Lain/Terdahulu
Contoh 1:1999 (No. 15)/Kertas 2
Rajah di atas menunjukkan sebuah piramid bertapak segiempat tepat. Nyatakan,(a) bilangan bucu(b) bentuk permukaan-permukaan condong bagi piramid itu.
Answer : (a) 5
(b) segitiga sama kaki
Mistakes of candidates : (a) 5
(b) 4
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Example 2:2001 (No. 11)/Paper 2
Hour minute(x 60)
Contoh 3:2001 (No. 14)/Kertas 2
Day Hour(x 24)
Mistakes of candidates :
( x 60 )
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I. Fobia Dengan Soalan Penyelesaian Masalah
1. Lemah 3M
2. Lemah imaginasi
Ingat a. Fahami kehendak soalan.
4 langkah : b. Tentukan operasi.
c. Kira dengan tepat.
d. Semak semula.
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J. Bentuk Tulisan Mengelirukan
CONTOH
4 9 dan sebaliknya
0 6 dan sebaliknya
8 6 dan sebaliknya
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K. Pengiraan Caca Marba
CONTOH
SOALAN……………………………………………
……………………………………………................
Langkah 1……………………………………………..
Langkah 2……………………………………………..
Langkah 3……………………………………………..
BUATLAH BEGINI
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RM8 40 sen + RM3.24 =
Jawapan = RM10.84
Tidak diterima RM1084 @ 10.84 sen30 % of 150
30 × 150 100
Tidak diterima 45%
= 45
L. Jawapan Tidak Konsisten
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