skema matematik ppt tingkatan 3 perak 2010
DESCRIPTION
matematikTRANSCRIPT
SULIT
SULIT
JABATAN PELAJARAN PERAK
PEPERIKSAAN PERTENGAHAN TAHUN 2010
TINGKATAN 3
MATHEMATICS 50
Mei / Jun2010
MATHEMATICS
PERATURAN PEMARKAHAN
ANSWER
MATHEMATICS PAPER 1
1D21B
2C22D
3D23B
4B24C
5B25C
6A26D
7B27A
8D28C
9B29A
10C30B
11C31A
12B32B
13C33C
14A34B
15D35D
16A36A
17D37A
18B38D
19A39D
20A40B
MARKING SCHEME
MATHEMATICS PAPER 2
NoSolution and Mark SchemeSub MarkFull Mark
1 or
K1N12
2(a)
(b) 10 + 27 or 10 or 27
37
N1
K1N13
3(a) 13.59 + 10.59 or 6.48 + 3.48 3
(b) 7 ( 3 ) or 3
10
K1N1
K1
N14
NoSolution and Mark SchemeSub MarkFull Mark
4
All points are plotted correctly and image is drawn completely
Note : 1. 2 points plotted correctly and image drawn completely P1 2. Only plotted all points correctly but the image is not drawn
completely P1 3. Dont accept sketch the image
4. Ignore labelP22
5
All sides are drawn correctlyNote: 1. 4 or 5 sides are drawn correctly P1 2. Dont accept sketch the image
3. Ignore labelP22
NoSolution and Mark SchemeSub MarkFull Mark
6
All points are plotted correctly and image is drawn completely.
Note : 1. 4 or 5 points plotted correctly and image drawn completely P1 2. Dont accept sketch the image
3. Ignore labelP22
7(a) (0, 1)
(b) Reflection // Pantulan y-axis as a axis of reflection // paksi-y sebagai paksi pantulanP1
P1
P13
8a2 = b + 25
K1
N12
9(a) y = 5 (b)
k = 4 N1K1
N13
10(a) 4b(a 2 )
(b) m2 2m m + 2 6
(m + 1)(m 4)N1
K1
N13
NoSolution and Mark SchemeSub MarkFull Mark
112(m 3) = 5n
or 2m = 5n + 6
or
K1
K1
N13
12(a) 3m + mn(b) 25x2 10xy 10xy + 4y2 25x2 20xy + 4y2
N1K1
N13
13 or equivalent
or equivalent
K1
K1
N13
145(u2 16)
5(u + 4)(u 4)
K1
N12
15(a) Uniform scale
All points are plotted correctly
Note: 3 or 4 points are plotted correctly P1
Line is drawn to join the points
Note: Dont accept sketch the line(b) Monday
K1P2
N1
N15
NoSolution and Mark SchemeSub MarkFull Mark
16(a) 100o 0.1(b)
Construct line segment AD
Construct line segment CD
Completely parallelogram ABCD
Construct arc to cut line AB
Construct perpendicular DE to ABN1
K1
K1
K1
K1
K16
17(a) Line POR or Line PR
Note : Only line drawn in the diagram P1 (b)
Draw locus L Draw locus M
( correctly markedP2
K1
K1
N15
NoSolution and Mark SchemeSub MarkFull Mark
1832x -1 = 33x = 2K1
N12
19(a) 4a10b2(b) or
6N1
K1
N13
2012 8 or SQ = 12
4K1
N12
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