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Modul ENRICHMENT Matematik SPM 2014
Tajuk : Persamaan Linear/Linear Equations
Masa : 30 minit
1. Diberi xx  3
2
1
, hitung nilai x .
Given xx  3
2
1
, calculate the value of x .
A. 6
B. 3
C. 3
D. 6
2. Diberi vv 243
3
1
 , hitung nilai v.
Given vv 243
3
1
 , calculate the value of v.
A. 3
B. 2
C. 2
D. 3
3. Diberi 12
2
1
2  uu , hitung nilai u.
Given 12
2
1
2  uu , calculate the value of u.
A. 2
B. 1
C. 1
D. 2
4. Diberi
4
5
1
3
)2(3 pp


, hitung nilai p.
Given
4
5
1
3
)2(3 pp


, calculate the value of p.
A. 4
B. 3
C. 3
D. 4
5. Diberi 185)2(9  xx , cari nilai x .
Given 185)2(9  xx , find the value of x .
A. 36
B. 18
C. 9
D. 5
6. Diberi 142)2(3  hh , cari nilai h.
Given 142)2(3  hh , find the value of h.
A. 2
B. 3
C. 4
D. 5
7. Diberi 4
2
1
3
1



 xx
, hitung nilai x .
Given 4
2
1
3
1



 xx
, calculate the value of x .
A. 2
B. 3
C. 5
D. 6
8. Diberi
xx 

 1
2
53
7
, hitung nilai x .
Given that
xx 

 1
2
53
7
, calculate the value of x .
A. 1
B.
3
1

C.
3
1
D. 2
9. Diberi
3
2
1
2
)5(3 

 pp
, hitung nilai p
Given that
3
2
1
2
)5(3 

 pp
, calculate the value of p
A.
7
55

B.
7
52

C.
7
50

D.
7
47

10. Diberi x
x


2
5
4
1
, hitung nilai x
Given that x
x


2
5
4
1
, calculate the value of x
A. 4
B. 3
C. 3
D. 4
JAWAPAN
1. xx  3
2
1
6
2
1
3
2
1
3



x
x
xx
JAWAPAN : A
2. vv 243
3
1

3
7
3
7
34
3
1
2



v
v
vv
JAWAPAN : D
3. 12
2
1
2  uu
2
2
3
3
2
1
212



u
u
uu
JAWAPAN : D
4.
4
5
1
3
)2(3 pp


4
312
31224
12324
12152412
1
12
152412
13
4
5
4
3
63


















 
p
p
p
p
pp
pp
pp
JAWAPAN : A
5. 185)2(9  xx
9
4
36
364
181859
185189





x
x
x
xx
xx
JAWAPAN : C
6. 142)2(3  hh
4
5
20
205
61423
14263





h
h
h
hh
hh
JAWAPAN : C
7. 4
2
1
3
1



 xx
5
255
2415
243322
4
6
33
6
22
4
23
)1(3
23
)1(2














x
x
x
xx
xx
xx
JAWAPAN : C
8.
xx 

 1
2
53
7
3
1
13
76107
10677
)3(2)1(7





x
x
xx
xx
xx
JAWAPAN : B
9.
3
2
1
2
)5(3 

 pp
7
55
4967
6497
1
6
42459
12
3
2
3
2
153










 





 
p
p
p
pp
pp
JAWAPAN : A
10. x
x


2
5
4
1
3
3
9
39
49
4101
4
101







x
x
x
xx
xx
x
x
JAWAPAN : C

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Linear equations

  • 1. Modul ENRICHMENT Matematik SPM 2014 Tajuk : Persamaan Linear/Linear Equations Masa : 30 minit 1. Diberi xx  3 2 1 , hitung nilai x . Given xx  3 2 1 , calculate the value of x . A. 6 B. 3 C. 3 D. 6 2. Diberi vv 243 3 1  , hitung nilai v. Given vv 243 3 1  , calculate the value of v. A. 3 B. 2 C. 2 D. 3 3. Diberi 12 2 1 2  uu , hitung nilai u. Given 12 2 1 2  uu , calculate the value of u. A. 2 B. 1 C. 1 D. 2 4. Diberi 4 5 1 3 )2(3 pp   , hitung nilai p. Given 4 5 1 3 )2(3 pp   , calculate the value of p. A. 4 B. 3 C. 3 D. 4
  • 2. 5. Diberi 185)2(9  xx , cari nilai x . Given 185)2(9  xx , find the value of x . A. 36 B. 18 C. 9 D. 5 6. Diberi 142)2(3  hh , cari nilai h. Given 142)2(3  hh , find the value of h. A. 2 B. 3 C. 4 D. 5 7. Diberi 4 2 1 3 1     xx , hitung nilai x . Given 4 2 1 3 1     xx , calculate the value of x . A. 2 B. 3 C. 5 D. 6 8. Diberi xx    1 2 53 7 , hitung nilai x . Given that xx    1 2 53 7 , calculate the value of x . A. 1 B. 3 1  C. 3 1 D. 2
  • 3. 9. Diberi 3 2 1 2 )5(3    pp , hitung nilai p Given that 3 2 1 2 )5(3    pp , calculate the value of p A. 7 55  B. 7 52  C. 7 50  D. 7 47  10. Diberi x x   2 5 4 1 , hitung nilai x Given that x x   2 5 4 1 , calculate the value of x A. 4 B. 3 C. 3 D. 4
  • 4. JAWAPAN 1. xx  3 2 1 6 2 1 3 2 1 3    x x xx JAWAPAN : A 2. vv 243 3 1  3 7 3 7 34 3 1 2    v v vv JAWAPAN : D 3. 12 2 1 2  uu 2 2 3 3 2 1 212    u u uu JAWAPAN : D 4. 4 5 1 3 )2(3 pp   4 312 31224 12324 12152412 1 12 152412 13 4 5 4 3 63                     p p p p pp pp pp JAWAPAN : A
  • 5. 5. 185)2(9  xx 9 4 36 364 181859 185189      x x x xx xx JAWAPAN : C 6. 142)2(3  hh 4 5 20 205 61423 14263      h h h hh hh JAWAPAN : C 7. 4 2 1 3 1     xx 5 255 2415 243322 4 6 33 6 22 4 23 )1(3 23 )1(2               x x x xx xx xx JAWAPAN : C 8. xx    1 2 53 7 3 1 13 76107 10677 )3(2)1(7      x x xx xx xx JAWAPAN : B
  • 6. 9. 3 2 1 2 )5(3    pp 7 55 4967 6497 1 6 42459 12 3 2 3 2 153                    p p p pp pp JAWAPAN : A 10. x x   2 5 4 1 3 3 9 39 49 4101 4 101        x x x xx xx x x JAWAPAN : C