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Published by SUZHANA SIPON, 2020-12-22 23:16:14

KBAT ADDMATH

SOALAN KBAT ADDMATHS

MODUL KBAT SPM – KERTAS 2

QUESTION 8:
CHAPTER: __________________________

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MODUL KBAT SPM – KERTAS 2

QUESTION 9:
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QUESTION 10:
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MODUL KBAT SPM – KERTAS 2

QUESTION 11:
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below
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MODUL KBAT SPM – KERTAS 2
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MODUL KBAT SPM – KERTAS 2

QUESTION 12:
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MODUL KBAT SPM – KERTAS 2

QUESTION 13:
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MODUL KBAT SPM – KERTAS 2

QUESTION 14:
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QUESTION 15:
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MODUL KBAT SPM – KERTAS 2

QUESTION 16:
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MODUL KBAT SPM – KERTAS 2

QUESTION 17:
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MODUL KBAT SPM – KERTAS 2

QUESTION 18:
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MODUL KBAT SPM – KERTAS 2

QUESTION 19:
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MODUL KBAT SPM – KERTAS 2

QUESTION 20:
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MODUL KBAT SPM – KERTAS 2

QUESTION 21:
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QUESTION 22:
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MODUL KBAT SPM – KERTAS 2

QUESTION 23:
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MODUL KBAT SPM – KERTAS 2

QUESTION 24:
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MODUL KBAT SPM – KERTAS 2

QUESTION 25:
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MODUL KBAT SPM – KERTAS 2
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MODUL KBAT SPM – KERTAS 2

QUESTION 26:
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MODUL KBAT SPM – KERTAS 2

QUESTION 27:
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MODUL KBAT SPM – KERTAS 2

QUESTION 28:
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MODUL KBAT SPM – KERTAS 2

QUESTION 29:
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B
A

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MODUL KBAT SPM – KERTAS 2

QUESTION 30:
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MODUL KBAT SPM – KERTAS 2

QUESTION 31:
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MODUL KBAT SPM – KERTAS 2

QUESTION 32:
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QUESTION 33:
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MODUL KBAT SPM – KERTAS 2

QUESTION 34:
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MODUL KBAT SPM – KERTAS 2

QUESTION 35:
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MODUL KBAT SPM – KERTAS 2

QUESTION 36:
CHAPTER: __________________________

Diagram 1 shows a rectangular room. The shaded region is a rectangularcarpet which covered the
room and placed 1 m from each of the walls of the room.
Rajah 1 menunjukkansebuahbilik yang berbentuk segiempattepat. Rantau berlorek itu dilitupi oleh
permaidani segiempat tepat yang diletakkan 1 m daripada dinding-dinding bilik itu.

1m
1m 1m

1m
Diagram 1 / Rajah 1
If the area and the perimeter of the carpet are 8.75 m2 and 12 m respectively, find the measurement
of the room.
Jikaluasdan perimeter permaidani itu masing-masing ialah 8.75 m2 dan 12 m, cari ukuran bilik
itu.

[7 marks/7markah]

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MODUL KBAT SPM – KERTAS 2

QUESTION 37:
CHAPTER: __________________________

Blue and yellow cards in equilateral triangle in a box need to be arrange as in Diagram 2. The blue
cards are arranged in odd row while the yellow cards are in even row.
Kad-kad berbentuk segitiga sama berwarna biru dan kuning di dalam kotak ingin disusun seperti
Rajah 2 di bawah. Kad-kad disusun di mana barisan ganjil berwarna biru manakala barisan
genap berwarna kuning.

Diagram 2 /Rajah 2
Find
Cari
(a) the number of cards use to form 15th row.

Bilangan kad yang digunakan untuk membentuk barisan yang ke -15
[2 marks/markah]

(b) the total number of yellow cards are used if there are twenty rows.
jumlah bilangan kad berwarna kuning yang digunakan jika susunan yang dibuat adalah
sebanyak dua puluh barisan.
[2 marks/markah]

(c) the maximum row will be formed if in the box have 800 yellow cards and 435 blue cards.
Bilangan baris maksimum yang boleh dibentuk jika di dalam kotak dibekalkan 800 kad kuning
dan 435 kad biru.
[3marks/markah]

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MODUL KBAT SPM

ANSWER

Q1: Answer Marks ANSWER PAPER 1
(−1.3542 , 0) 4
B3 Marks
 8  82  4(1)(9) B2 4
x B1 B3
B2
2(1) 4 B1
2 + 8 + 2 − 6 + 9 = 0 B3
B2
(−4, 3) B1

Alternative way:
(−1.3542 , 0)
h   8  82  4(1)(9)

2(1)
√(ℎ − (−4))2 + (0 − 3)2 = 4

(-4, 3)

No. Marking Scheme
Q2: Elite 5 and Honest 4

x  14210
x 10000  0.842

5000

z  0.842

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ANSWER PAPER 1

Q3: Marking Scheme MARK
  16 4
15
B3
r 5 B2

11  12  2r B1
9 8  2r

Hayunan ketiga = 8m

Q5:

Q6:
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ANSWER PAPER 1

Q7:
Q8:
Q9:
Q10:

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ANSWER PAPER 1

Q11:

Q12:
(2 × 1 × 3 × 2) × (2 × 1 × 1) = 24
24 × 6 = 144
144 + 144 = 288
Q13:
ii) ⃗ ⃗ ⃗⃗ ⃗ = −3 + 6
b) 3354 km

Q14:
a) y = 2
b) Cicak yang berada di Q dan 13.42

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ANSWER PAPER 1

Q15:
Q16:
Q17:
Q18:
Q19:
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ANSWER PAPER 1

Q20:
Q21:
Q22:

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ANSWER PAPER 1

Q23:
Q24:

Q25:
Q26:
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ANSWER PAPER 1

Q27:
Q28:
Q29:
Q30:

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ANSWER PAPER 1

Q31:
Q32:
Q33:

Q34:
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ANSWER PAPER 1

Q35:

Q36:
Q37:
Q38:
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ANSWER PAPER 1

Q39:
Q40:
Q41:

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ANSWER PAPER 1

Q42:
Q43:
Q44:

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ANSWER PAPER 1

Q45:
Q46:

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Q1: ANSWER PAPER 2

1a. Method 1 : Use
completing
y  x2  2x the square

   x 12 1 K1 method

 x 12 1 N1
Height = 1m
Height = 1 m
Method 2 : Use x   b
2a
x b
2a to find x and y.
 2 K1
2(1)
1 N1
Height = 1m
y  12  2(1) 1
Use dy  0 to
Height = 1m dx
Method 3 :
K1 find x and y
y  x2  2x
N1
dy  2x  2  0 Height = 1m
dx
Use x=0 or x=2
x 1 K1 to find y

y  (1)2  2(1) 1 Use ( 0,0)
K1 Or(2,0) to
Height = 1 m
find a.
1b. x  0, y  0 or x  2, y  (2)2  2(2)  0 N1 a=-2

(0,0) and (2,0) N1 y  2(x 1)2  2

y  a(x 1)2  2 at (0,0) or (2,0)

0  a(0 1)2  2 , a = -2
y  2(x 1)2  2

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ANSWER PAPER 2

Q2:

0.7855 rad. P1

Area  1 r2  1 (rh)
28

1 (12)2  ) K1 K1 1 (2)( )(12)(8)
( 8
24

56.556 cm2 75.408 cm2

Area = 56.556* + 75.408*
K1

131.964 cm2

N1

Total cost = 131.964 x 0.03 N1
RM 3.959

3x  y  320

x2 100x  5y  xy  3025  0

y  320  3x

x2 100x  5320  3x x320  3x 3025  0

Q3:

x  952x 15  0

x  7.5, x  95
y  729.5, y  35
x  95, y  35

600  650  1250
1250 RM15  RM18750 , Yes

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ANSWER PAPER 2
Q4:

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ANSWER PAPER 2
Q5:

Q6:

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