Answers

MODULE 1- ANSWERS TOPIC: SIMULTENOUS LINEAR EQUATIONS 1. Calculate the values of m and n that satisfy the simultaneous linear equations: 4m + n = 2 and 2m – 3n = 8 [ 4 marks] 12m + 3n = 6 2m – 3n = 8 14m = 14 m=1

1 1 1

n = -2

1

2. Calculate the values of m and n that satisfy the simultaneous linear equations:

1 m - 3n = 10 2 5m + 6n = -8 m - 6n = 20 5m + 6n = -8 6m = 12 m=2 n =-2

[ 4 marks] 1 1 1 1

3. Calculate the values of m and n that satisfy the simultaneous linear equations: 2m – n = 7 m – 2n = 5 [ 4 marks] 2m – n = 7 2m – 4n = 10 3n = -3 n = -1 m =3

1 1 1 1

4. Calculate the values of p and q that satisfy the simultaneous linear equations: p + 2q = 6

3 p – q = -7 2 [ 4 marks] p + 2q = 6 3p – 2q = -14 4p= -8 p = -2 q =4

1 1 1 1

5. Calculate the values of m and n that satisfy the simultaneous linear equations: 4m - 3n = 7 and m + 6n = 4 [ 4 marks] 4m - 3n = 7 4m + 24n = 16 -27n = -9

1 1

1 3

1

m =2

1

n=

6. Calculate the values of p and q that satisfy the simultaneous linear equations: 2p - 3q = 13 4p + q = 5 [ 4 marks] 4p - 6q = 26 4p + q = 5 -7q = 21 q = -3

1

p =2

1

1 1

7. Calculate the values of p and q that satisfy the simultaneous linear equations:

1 p – 2q = 13 2 3p + 4q = -2

[ 4 marks]

p – 4q = 26 3p + 4q = -2 4p = 24 p=6

1

q = -5

1

1 1

8. Calculate the values of k and w that satisfy the simultaneous linear equations: 2k – 3w = 10 and 4k + w = -1 [ 4 marks] 4k – 6w = 20 4k + w = -1 -7w = 21 w = -3 k =

1 2

1 1 1 1

9. Calculate the values of m and n that satisfy the simultaneous linear equations: 2m - 5n = -12 and 3m + n = -1 [ 4 marks] 2m - 5n = -12 15m + 5n = -5 17m = 17 m = -1 n =2

1 1 1 1

10. Calculate the values of x and y that satisfy the simultaneous linear equations: x – 3y = 5 3x – y = 3 [ 4 marks] 3x – 9y = 15 3x – y = 3 -8y = 12

3 y=2 x=

1 2

1 1 1

1

11. Calculate the values of s and t that satisfy the simultaneous linear equations: 8s + 3t = 12 6s – 9t = 24 [ 4 marks] 24s + 9t = 36 6s – 9t = 24 30s = 60 s=2 t=-

4 3

1 1 1 1

12. Calculate the values of m and n that satisfy the simultaneous linear equations: m - 3n = 6 and 2m – n = 7 [ 4 marks] 2m - 6n = 12 2m - n = 7 -5n = 5 n = -1

1 1 1

m =3

1

13. Calculate the values of k and h that satisfy the simultaneous linear equations: 4k - 3h = 10 2k – 5h = 12 [ 4 marks] 4k - 3h = 10 4k – 10h = 24 -7h = -14 h=2 k=4

1 1 1 1

14. Calculate the values of m and n that satisfy the simultaneous linear equations: 2m + 3n = 9

1 m–n=2 3

[ 4 marks]

2m + 3n = 9 m – 3n = 6 3m = 15 m= 5 n =

1 1 1

1 3

1

15. Calculate the values of m and n that satisfy the simultaneous linear equations: 2m – n = 2 4m – 3n = 5 [ 4 marks] 6m – 3n = 6 4m – 3n = 5 2m = 1

1 1

1 2

1

n =1

1

m=

MODULE 2 – ANSWERS TOPIC : SOLID GEOMETRY

1. Volume of solid = Volume of cuboid + volume of half-cylinder 1 22 = 14 x 6 x 4 + x x 32 x 14 2 7 336 cm3

=

≈

+

197.92cm3

= 533.92 cm3 534 cm3

2. Volume of remaining solid = volume of big pyramid = = = =

1 ( x 152 x 18 ) 3 1 ( x 225 x 18 ) 3

-

(1350 - 48) 1302 cm3

3. Volume of solid = Volume of cone + Volume of hemisphere

1 2 1 4 πr h + × × πr 3 3 2 3  1 22  21  2   1 4 22  21  3  4042.5 =  × ×   × h +  × × ×     3 7  2    2 3 7  2   =

4042.5 = 115.5 h + 2425.5 h = 14 cm 4. Volume of pyramid = volume of solid – volume of cuboid = 1100 cm3 − (10 ×10 × 8) = 1100 - 800 = 300 cm3 Volume of pyramid =

1 x Area of base x h 3

=

1 x (10 x 10) x h 3

=

100 xh 3

100 h = 300 3

volume of small pyramid

1 x 62 x 4 ) 3 1 ( x 36 x 4 ) 3 (

= 300 x

3 100

= 9 cm 5. Volume of cylinder = πr2h =

22 x 3.5 2 x 4 7

= 154 cm3 Volume of cone

Volume of solid 154 +

=

1 2 πr h 3

=

1 22 x x 3.5 2 x t 3 7

=

269.5 t cm3 21

= 231 cm3

269.5 t = 231 21 269.5 t = 1617 t=

1617 269.5

= 6 cm

6. Volume of remaining solid = Volume of cylinder – volume of cone = πr2h -

=[

1 2 πr h 3

22 1 22 x 7 x 7 x 20] – [ x x 7 x 7 x 9] 7 3 7

= 3080 – 462 = 2618 cm3

7. Volume of solid = Volume of half – cylinder + volume of prism Volume of half – cylinder =

1 x πr2h 2

Volume of prism = area of base x height Volume of solid = Volume of half – cylinder + volume of prism =[

1 22 14 2 1 x x( ) x 6] + [ x 14 x 8 x 6] 2 7 2 2

= 462 + 336 = 798 cm2

MODULE 3 - ANSWERS TOPIC: CIRCLE, AREA AND PERIMETER 1 (a)

90 22 120 22 × 2 × ×12 @ × 2× ×7 360 7 360 7 90 22 120 22 × 2 × × 12 + × 2 × × 7 + 12 + 5 360 7 360 7 57.53

K1 K1 N1

(b)

90 22 120 22 2 × × 12 2 @ × ×7 360 7 360 7

K1

90 22 120 22 1 × × 12 2 + × × 7 2 − × 7 × 12 360 7 360 7 2 122.48

K1 N1

2 (a)

∠FEC = 135

K2

135 22 × 2× ×7 360 7 16.5

K1 N1

(b)

135 22 × ×7 ×7 360 7

K1

1  Shaded area = (21 × 14) −  × 14 × 14  − L3 2  = 138.25

K1

L3 =

N1 3 a)

b)

270 22 90 22 × × 2 × 21 atau × ×7×2 360 7 360 7

K1

270 22 90 22 × × 2 × 21 + × × 7 × 2 + 14 + 14 360 7 360 7

K1

= 138

N1

270 22 × × 21 × 21 atau 360 7

2×

90 22 × ×7×7 360 7

K1

270 22 90 22 × × 21 × 21 - 2 × × ×7×7 360 7 360 7

K1

= 962.5 cm2

N1

4 a)

60 22 × 2 × × 28 360 7

K1

60 22 × 2 × × 28 + 14 + 14 + 14 + 14 + 28 360 7 1 113 atau 113⋅33 3 b)

60 22 60 22 × × 28 × 28 atau × ×14 ×14 360 7 360 7 60 22 60 22 × × 28 × 28 − × ×14 ×14 + 14 × 14 360 7 360 7 504

K1 N1

K1 K1 N1

5 a)

120 22 240 22 × × 14 × 14 atau × ×7×7 360 7 360 7 120 22 240 22 × × 14 × 14 + × ×7×7 360 7 360 7 308

b)

K1 K1 N1

120 22 240 22 × 2 × × 14 atau × 2× × 7 360 7 360 7 120 22 240 22 × 2 × × 14 + × 2× × 7 + 7 + 7 360 7 360 7 2 72 3

K1 K1 N1

6 (a)

45 22 ×2× × 14 360 7

K1

22  45  ×2× × 14  + 14 + 14 + 14 + 14  360 7  

K1

70

(b)

2 3

45 22 × × 14 × 14 360 7

N1 or

90 22 × × 14 × 14 360 7

K1

90 22  45 22    × × 14 × 14  + 2 14 × 14 − × × 14 × 14   360 7 360 7    

K1

161

N1

7 (i)

(ii)

(iii)

210 22 × 2 × × 35 360 7 1 128 @ 128.33 3

K1 N1

210 22 210 22 × 2× ×7 : × 2 × × 35 360 7 360 7

K1

1: 5

N1

210 22 210 22 × × 35 2 or × × 72 360 7 360 7 210 22 210 22 × × 35 2 − × × 72 360 7 360 7 2156

K1 K1 N1

8 (a)

45 22 ×2× × 14 360 7

or

14 2 + 14 2 − 14

11 + 14 + 14 + 14 + 5.799 58.80 (2 d. p) (b)

45 22 × × 14 x 14 360 7 45 22 1 90 22 × 14 × 14 − × ×7 × 7 + × × 14 × 14 2 360 7 360 7

90 22 × ×7 × 7 360 7

or

136.5

K1 K1 N1 K1 K1 N1

9 (a)

(b)

90 22 60 22 × × 14 × 14 and A2 = × ×7×7 360 7 360 7 A1 – A2 1 128 3 90 22 180 22 × 2 × × 14 or P2 = ×2× ×7 P1 = 360 7 360 7 A1 =

P1 + P2 + 14 58

K1 K1 N1 K1 K1 N1

10 (a)

AB =

14 2 + 14 2 = 392 = 19.80 150 22 60 22 90 22 × 2 × × 14 atau × 2 × × 14 atau × 2 × × 14 360 7 360 7 360 7

Lengkok AC + 14 + 14 + 19.80 atau Lengkok AB + lengkok BC + 14 + 14 + 19.80 84.47 (b)

150 22 1 × × 14 2 atau × 14 × 14 360 7 2 150 22 1 × × 14 2 - × 14 × 14 atau 360 7 2 770 – 98 3 2 476 158 atau atau 158.67 3 3

K1 K1 K1 N1 K1 K1

N1

MODULE 4 - ANSWERS TOPIC: QUADRATIC EXPRESSIONS AND EQUATIONS. Answer:

(a)

3x(x− 1) 2

=x+6

3x2 –x – 12 = 0 ………..1 (3x+ 4)(x – 3) =0 ……….1

4 3

x=

x= 3 …………….1,1

(b) (w – 1)2 – 32 = 0 w2– 2w – 8= 0 …………..1 (w+2)(w–4)=0 …………..1 w=4

w= –2 ……………1,1

(c) 2a2 = 3(1 + a) + 2 2a2– 3a – 5 = 0 …………1 (a+1)(2a – 5) = 0………..1 a= –1 x =

5 …………..1,1 2

2

(d)

5p + 3p 1+ p

=4

5p2– p – 4 = 0 …………1 (5p– 4)(p+1) = 0…….....1 p=

4 5

p = – 1………..1,1

2

(e)

3t

2

= 7t – 4

3t2+ 14t – 8 = 0 ………..1 (3t – 2)(t – 4)=0 ………..1 t=

2 3

t = 4 …………1,1

(f) x2 – 2 =

11x− 5 4

4x2– 11x – 5 = 0 ………1 (4x+1)(x - 3) = 0 ………1

1 4

x=

(g)

x=3

2 m −6 5

=m

m2– 5m – 6 = 0 ………..1 (m - 6)(m + 1) = 0………1 m=6

m= -1 …………1,1

(h) (2x + 1)(x – 2)=7 2x2– 3x – 9 = 0 …………1 (x+3)(x - 3) = 0 …………1 x = -3 x = 3 ……………. 1,1 (i) p + 2 =

p+ 2 p− 3

p2 – 2p – 8 = 0 …………1 (p+2)(p– 4)=0 ………….1 p= –2

p = 4……………1,1

(j) 3x(2x – 1)+ 8x = 1 6x2+ 5x – 1 = 0 …………1 (x+1)(6x - 1) = 0 ………..1 x = -1 x =

1 …………..1,1 6

2 (k) 3y −2 = y

5

3y2 – 5y – 2 = 0………………1 (y+1)(y– 2)=0 ………………..1 y= -1

y = 2 …………………1,1

(l) (r –1)(r + 3) = 5(r + 3) r2 – 3yr– 18 = 0 ……………..1

(r+6)(r– 3)=0 ………………..1 y= -6

y = 3 ……………….1,1

(m) 7p – 2p2 = 2(1 + p) 2p2– 5p + 2 = 0 …………..1 (2p– 1)(2p+5) = 0 ………..1

1 2

p=

−5 …………1,1 2

p=

x 2 + 25 2

(n) 5x =

x2 – 10x + 25 = 0 …………1 (x-5)(x-5)=0……………….1 x=5 ………………………..1 2

(o)

p +5 6

=p

p2– 6p + 5 = 0 ………….1 (p– 1)(p+5) = 0 …………..1 p=1

p = -5……………..1,1

(p) 4(5x – 1) =

− 3(5x− 1) x

20x2– 11x -3 = 0 ……….1 (5x+ 1)(4x-3) = 0 ………..1 x=-

(q) d =

1 5

x=

1 ………….1,1 4

7 − 6d d

d2+ 6d -7 = 0 ………..1 (d– 1)(d+7) = 0 ………1 x=1 (r)

x = -7 ……….1,1

2 2m + 5m m+ 1

=2

2m2+ 3m -2 = 0 ……….1

(2m– 1)(m+1) = 0………1 x = -2

(s)

x=

3m(m− 1) m+ 1

1 ………….1,1 2

=2

3m2- 5m -2 = 0 …………..1 (3m+1)(m-2) = 0 ………….1 x=2

x=-

1 …………1,1 3

(t) y2 + 9y – 1 = 3(y – 2) y2+ 6y+5 = 0 ……………1 (y+1)(y+5) = 0 ………….1 x = -1

x = -5 ………….1,1

MODULE 5 – ANSWERS TOPIC: SETS 1 (a)

(b)

2 (a)

(b)

ξ

ξ

1

PQ

R

PQ

R

J ∩ L′

2

1 J

K

( K ∪ L) ∩ J ′ J

2

K

3

L

3 (a)

1

Q R

P

(b)

R

P

2 3

Q

4 (a)

5

1

(b)

30 –(5 + 6 + 2 + 1 ) 16

1 1

(c)

9

1 4

5 (a) (b)

{10,11,12,13 }

1

A = { 9,12,15 } B = { 7,9,11,13,15} A ∪ B = { 7,9,11,12,13,15}

1 1

N(AUB)=6

6 (a)

3

2

ξ

G

H

24 12

(b)

7

6

8

1

24

(a)

{4 , 8} Note:

(b)

6

3

1 Accept without bracket. 2

3

8

1

ξ P

R

3

Q

P Q

2

R

1.

(c)

MODULE 6 TOPIC : MATHEMATICAL REASONING (i) some 1m (ii) all 1m If x > 4 then x > 7 1m False. 1m P∩Q≠P 1m

(a) (b) (c)

Non-statement 1m 3 is a factor of 24 2m 3(2n) – n and n = 1,2,3,……

(a)

(i) True 1m (ii) False 1m The determinant of matrix a = 0 If A ⊂ B then A ∩ B = A If A ∩ B = 0 then A ⊂ B

(a) (b)

2.

3.

(b) (c)

4.

(a) (b) (c)

5.

(a) (b) (c)

6.

(a) (b) (c)

7.

(a)

(b) (c) 8.

9.

(a) (b)

True If p3= 8 then p = 2 If p = 2 then p3 = 8 If x = 9 then x = 3

2m

1m 1m 1m

1m 1m 1m 2m

(i) True 1m (ii) False 1m Premise 1 : If p = 7, then 6 x p = 42 1m 4n2 + 7, n = 1,2,3,4…. 2m (i) Statement. 1m (ii) Non-statement 1m Some 1m Premise 2 : p is a positive integer (i) False (ii) False t≠0 If x ± 3, then x2 = 9 True

2m

1m 1m 1m 1m 1m

(c)

Statement 1m If x is an even number then x can be divided by 2. If x can be divided by 2 then x is an even number. n2 + 2 , n = 0, 1, 2,……… 2m

(a)

(i)

True

1m

1m 1m

(ii) False 1m 2 If n = 2, then x + x is a quadratic expressions. If 1 – m > 2 then m < -1 If m < -1 then 1 – m > 2

(b) (c) 10.

(a) (b)

1m 1m 1m

It is a statement because it has a truth value which is false. 2m (i) True 1m (ii) False 1m y>3 1m

(c)

MODULE 7- ANSWERS TOPIC: THE STRAIGHT LINES

12 − 2 3 − (−2) 10 = 2

1. a) m =

b) y = 2x + c Point (5, 5),

= 2

5 = 2(5) + c

= 10 + c c = -5 y = 2x - 5 Equation of SRT is y = 2x

–5 c) x – intercept = - ﴾ =

5 2 2 − (−5) 4 − ( −1)

2. a) Gradient = =

−5 ﴿ 2

b) y – intercept = 5

7 5

Point E = (-5, -2), gradient =

5 7/5 − 25 = 7

x –intercept = -

7 5

y= mx + c -2 = mx + c -2 =

7 (−5) + c 5

c=5 y =

7 x+5 5

2− p 6−0 2− p = 6

3. a) The gradient = -3 -18

= 2–p p= 20 Coordinates of point T = (0 , 20)

b) y = mx + c m = -3, c = 20 y = -3x + 20

1 x + 18 3 1 x+9 y= 6

c) 2y =

The value of p = 9, gradient =

6−3 −1− 4 3 = 5

4. a) m =

1 6 b) m = -3 (4 – q) -12 + 3q

3 5

3 5

3−0 4−q

=

= 3(5) = 15 3q = 27 q = 9

c) D = (-1 , 0) , F = (4 , 3)

3−0 4 − (−1)

m =

3 5

=

0 − (−8) 0 − (−4) 8 = 4

5. a) Gradient =

= 6 = 6 = c= y =

b) x-intercept = -

10 2

= -5

2

y = mx + c 2(-2) + c -4 + c 10 2x + 10

3−0 5−0 3 = 5 3 = 5

5. a) Gradient =

z−4 10 5z - 20

= 30

5z

= 50

b) gradient =

3 , E = (-2 , 4) 5 y

= mx + c

4

=

-

c

=

26 5

6 + c 5

Equation of line EF is y =

26 3 x + 5 5

z

= 10

c) x – intercept of line EF = -

26 5 3 5

= −

26 3 b) x-intercept of HIJ = − (

7. a) F = (0,4) , G = (-4 , 0)

4−0 0 − (−4) 4 = 4

Gradient

=

=

m=

9 4

y = 5x - 3

9 ) 4

- 0

Q = (-5 ,

9 3 ), gradient M = 4 4

y = mx + c

9 4 3

c=6

9 1 x 4 3 3 = 4

=

c) R = (0, 6) , S = (6, 0)

0−6 6−0 −6 = 6

m=

= -1 9. a) i) Equation of LK is x = 7 ii)

b)

9 3 = (−5) + c 4 4 15 9 + c = 4 4

-5 – (- 8) =

3 5

c) y = mx + c

= 1

8. a) P = (-8, 0) , Q = (-5 ,

−3 ) 5

y = mx + c 8 = 2(-2) + c 8 = -4 + c 12 = c Equation of EFG is y = 2x + 12

y-intercept = 6

8 − (−4) −2−7 12 = − 9 4 − = 3

b) m =

y = mx + c 8= −

4 (−2) + c 3

16 =c 3 4 16 y= − x + 3 3 Coordinates of H = (0,

16 ), 3

Coordinates of J is (x, 0) ,

−

y= −

4 16 x+ 3 3

4 16 x+ = 0 3 3

-4x + 16 = 0 -4x = -16 X = 4 Therefore coordinates of J = (4, 0) 10 a). 3y – 6x = 3 -----------------(1) 4x = y – 7 y = 4x + 7 _________(2) Substitute (2) into (1) 3(4x + 7) - 6x = 3 12x + 21 - 6x = 3 6x = 3 – 21 6x = -18 x = -3 y = 4(-3) + 7 = -12 + 7 = -5 Point of intersection is (-3, -5)

2 x + 3 ---------------------(1) 3 4 y = x + 1 ---------------------(2) 3

b) y =

(1) to (2)

2 4 x + 3 = x +1 3 3 4 2 x − x = 3 −1 3 3

2 x=2 3 x=3 y =

2 (3) + 3 3

=2+ 3 =5 Point of intersection is (3, 5) MODULE 8 – ANSWERS TOPIC: STATISTICS 1) a) Body mass(kg) 15-19

Midpoint 17

Lower boundary 14.5

Upper boundary 19.5

Frequency 2

20-24

22

19.5

24.5

4

25-29

27

24.5

29.5

9

30-34

32

29.5

34.5

13

35-39

37

34.5

39.5

9

40-44

42

39.5

44.5

3

b)

i) Modal class 30-34 kg ii)9+3=12 iii)Mean= (17x2)+(22x4)+(27x9)+(32x13)+(37x9)+(42x3) 40 = 1240/40 = 31 kg c) graph d) graph 2) a) Marks 11-15

freq 1

Cumulative freq 1

16-20

3

4

21-25

6

10

26-30

10

20

31-35

11

31

36-40

7

38

41-45

2

40

b) c) 3) a)

graph i) upper quartile ¾ x 40 = 30 = 35 i) 30 students scored less than 35

Age 40-44 45-49 50-54 55-59 60-64 b) c) c)

Frequency 6 12 18 9 5

Midpoint 42 47 52 57 62

Mean age = 51.5 The modal age is 50-54 years old graph

4) a) Donation (RM) 10-19

Frequency 5

Cumulative frequency 5

20-29

8

13

30-39

11

24

40-49

13

37

50-59

3

40

c)

i) ii)

RM43.50 There are 10 persons who donated RM43,50 or more

b) graph 5) a)

b)

Mass(kg) 30-34

Frequency 2

Midpoint 32

35-39

10

37

40-44

12

42

45-49

8

47

50-54

5

52

55-59

3

57

i) ii)

40-44 kg (32x2)+(37x10)+(42x12)+(47x8)+(52x5)+(57x3) 40 = 1745/40 = 43.625 kg

c) graph

6) a) Pocket money(RM) 31-35

Frequency 3

Midpoint 33

36-40

5

38

41-45

6

43

46-50

9

48

51-55

8

53

56-60

6

58

61-65

3

63

b) i) 46-50 ii) mean = (33x3)+(38x5)+(43x6)+(48x9)+(53x8)+(58x6)+(63x3) 40 = 1940/40 = 48.5 c) graph

MODULE 9 - ANSWERS TOPIC: LINES AND PLANES IN 3-DIMENSIONS 1

Identify ∠PUQ @ ∠SCR tan ∠PUQ =

9 12

∠PUQ = 360 52’ @ 36.870

2

3

 PRQ 12 tan PRQ  5 PRQ = 67.380 or 670 22’

Identify

Identify ∠VST tan ∠VST =

5 12

22.60 or 220 37’

4

Identify ∠VTU tan ∠VTU =

6 4

∠VTU = 56⋅31° or 56°18’ 5

Identify ∠ GSJ tan ∠ GSJ =

6 8

36.87° or 36° 52’ 6

Identify ∠PWS tan ∠PWS =

13 10 2 − 6 2

∠PWS = 58⋅39° or 58⋅4° or 58°24’ 7

Identify ∠ TRM Tan ∠ TRM =

5 82 + 62

∠ TRM = 26.57° or 8

26 ° 34 ′

Identify ∠PTQ or ∠QTP tan ∠PTQ =

9 12

∠PTQ = 36052’ or 36.90

9

a) b)

10 Identify ∠EPF 7 10

Tan ∠EPF =

10

c)

340 59’ ∠UQV

a)

13 cm

b)

Identify ∠SMW tan ∠SMW =

7 13

28.30 or 28018’

c)

∠RUQ or ∠QUR

MODULE 10 – ANSWERS TOPIC: GRAPHS OF FUNCTIONS 1.

a) x=-2 y=15

x=0.5 y= -5

x=3 y=0

b) graph

40

35

30

25

y(x) = 2⋅x2-5⋅x-3 20

20.5

15

y(x) = 3 ⋅x+4 10

5

-10

-5

-2.4

5

-0.75 -5

-10

c) i) ii)

x=-2.4 y= 20.5 when 2x2 – 5x – 3 = 0 y=0 Then the values of x is -0.5 and 3

b) y=2x2-5x-3 0=2x2-8x-7 (-) ----------------------Y= 3x + 4 X Y

0 4

3 13

From the graph x= -0.75 and 4.75

4.75

10

2.

a)x=2.5 y=-2.25

x=4 y=0

b)graph 2 c)

i) ii)

y=1.75 x=7.4

b) straight line y=2x+1 x=0.45 and 6.55 3.

a) x=-1 y=-5

x=2.5 y=2

b)graph 16

C 14

12

10

y(x) = 3⋅x

8

6

4

2

A -5

-4

-3

-2

-1

B

1 -2

y(x) =

5 x

-4

-6

-8

-10

-12

-14

c)

i) ii)

x=1.8 y= 2.8 y=-6 x=-0.8

b) The graph is y=3x X Y

0 2 0 6

2

3

4

5

The values of x= -1.3 and 1.3

4.

a)

x=-2 y=-1

x=2.5 y=0.8

b) graph c)

i) ii)

y=-1.3 x=1.7

b) The straight line is y=

3 x–2 4

The values of x = -0.75 and 3.45 5.

a) x=-3 y=30 b) graph

x=2 y=0 45

40

35

34

30

y=x^3-1 3x+18 25

20

y=-2x+20 15

10

5

3.85

-3.2 -6

c)

i)

-0.25

-4

x=-1.5 y=34 ii) y=25

-2

-5

x=3.85

b) y=x3-13x+18 0=x3-11x-2 (-) -------------------------Y= -2x+20 X 0 4 Y 20 12

2

3.35 4

6

X= -3.2, -0.25 and 3.35

6.

a)

x=-2 y=15

x=3 y=-5

b) graph 6 c)

i) ii)

y=10.75 x=-1.5

b)

y=-12x-5 x=-3.6, 0 and 2.75

JAWAPAN MODUL11 TOPIC : TRANSFORMATIONS 1 (a) (i) (ii)(a) (b) (b)(i)(a) (b) (ii)

2 (a)(i) (ii) (b)(i)(a) (b) (ii)

3 (a)(i) (ii) (b)(i)(a) (b) (ii)

4 (a)(i) (ii) (b)(i)(a) (b) (ii)

5 (a)(i) (ii) (b)(i)(a)

(0, -1) (-3, -4) (-1, -2)

1 1 1

M is a rotation of 90o clockwise about point (1,3) N is an enlargement with centre at (2,0) and a scale factor of 2 Area EFGH = k2(Area ABCD) = 22(25) = 100 unit2

3 3

(4, -2) (1, 0)

1 2

U is a rotation of 90o clockwise about the point (1, 1) V is an enlargement with centre at (4, 0) and scale factor of 2 Area OFJK = k2(Area ABCD) 120 + Area ABCD = 22(Area ABCD) Area ABCD = 40 unit2

3 3

(12, 7) (6, 7)

1 2

V is a rotation of 90o clockwise about point (7, 0) W is an enlargement with centre at (7, 3) and scale factor of 3 Area EFG = k2(Area PQR) 72 = 32(Area PQR) Area PQR = 8 unit2

3 3

(-3, 6) (6, 11)

1 2

 − 8   3 

U is a translation 

3

3

3

1

V is an enlargement with centre at (-3, 8) and scale factor of 2. Area WXYS = k2(Area RSTU) 150 + RSTU = 22(Area RSTU) Area RSTU = 50 cm2

3

(-3, 0) (4, 4)

2 2

S is a reflection in the line x =1

2

4

Q is an enlargement with centre at (-11, 2) and scale factor of 2. Area ABCD + Area of shaded region= k2(Area ABCD) 64 + Area of shaded region = 22(64) Area of shaded region = (256 – 64) cm2 Area of the shaded region = 192 cm2

(ii)

MODULE 12 - ANSWERS TOPIC : MATRICES 1.

m= −

(a)

1 2

1m

n =2 (b)  3

 5

1m

− 2  x  8    =   − 4   y  13   x  1 − 4   =   y 2 − 5 x=3 y= −

2.

(a)

 2

1 2

1m 1m

3  p    = 4  q 

 1     − 8

p=2 q = -3 (a)

(b)

5 3 

A=

1m

1m

 p  1  4 − 3  1    =     q  14  − 2 5  − 8 

3.

2  8    3 13 

1m

n =4 m=5

(b)  5

1m

1m 1m 1m 1m

− 2  − 1 

2m

 −1 2  r   − 4     =   − 3 5  s   − 9

1m

 r  1  5 − 2  − 4    =     s  1  3 − 1  − 9 

1m

3 3

4.

(a)

5.

(a)

(b)

s = -3

1m 1m

1  8 − 5   2  − 6 4 

1m

 4 5  m   7     =    6 8  n  10 

1m

 m  1  8 − 5  7    =     n  2  − 6 4 10 

1m

m=3

1m

n = -1

1m

 − 5 3  1m 1   − 20 − (−24)  8 4  1  − 5 3 1m  =  4  8 4 

P =

 4 − 3  h   − 7     =    8 − 5  k   − 11  h  1  − 5 3  − 7    =    k 8 4 − 11 2      =

6.

1m

1  8 − 5   32 − 30  − 6 4 

P=

=

(b)

r = -2

1 2    2  − 100 

1m 1m

h=1 k = -50

1m 1m

(a)

k = -12

1m

(b)

(i)

h = 26

1m

 x 1  8 6  − 4    =     y  26  5 7  1  1  − 26    = 26  − 13 

(ii)

1m 1m

x = -1 y= − 7.

(a)

1m

1 2

- 4 – 5k = 0

1m 1m

5k = -4 k= − (b)

8.

4 5

1m

 2 5  x   13     =    1 − 2  y   − 7 

1m

 x 1  − 2 − 5  13    = −    9  − 1 2  − 7   y

1m

x = -1 y=3

1m 1m

1 0  0 1

(a)

M=  

2m

(b)

 x 1  3 − 4  6    =     y  6 − 4  − 1 2  5 

1m

1  3 − 4  6     2  − 1 2  5  1  − 2 =   2 4  =

1m

x = -1 y=2

9.

(a)

1  − 2 1   − 6 + 5  − 5 3 

1m 1m

1m

=

1  − 2 1   − 1  − 5 3 

1m

− 1  d   7     =   5 − 3   e  16   d  1  − 3 1  7    =     e  − 1  − 5 2 16 

(b) 2

=

1m 1m

1  − 5   − 1  − 3 

5 =    3 d=5 e=3 10.

(a)

 5 2 1   5 − (−4)  − 2 1  1  5 2  =  9  − 2 1  − 2  u   8   2 5  v  =  7        u  1  5 2  8    =     v  9  − 2 1  7 

(b)  1

1m 1m

1m 1m

1m 1m

1  54  =   9  − 9 6 =    − 1 u=6 v = −1

1m 1m

MODULE 13 - ANSWERS TOPIC : GRADIENT AND AREA UNDER A GRAPH 1 (a) (b)

20

1

23 − 5 0−6

1 1

−3 atau nyahpecutan 3 atau awapecutan 3 (c)

1 1 × 6( 23 + 5) + 4 × 5 + × 5( 5 + k ) = 139 2 2

2 1

k=9

2(a) (b)

(c)

2 saat

1

8−0 6−0 4 atau setara 3 1 × 8( 2 + 8) 2 1 × 8( 2 + 8) − 4 × 8 2

1 1 1 1 1

8 3 (a)

105 km

1

(b)

0800 a.m

1

(c)

105 – 60 = 45km

1 1

(d)

105 = 42km / j 2.5

1 1

6

6

4

(a)

0 − 20 30 − 25

1 1

− 4 ms −2

(b)

(c)

5s

1

1 1 × 10 × ( v + 20) + × 20 × (15 + 20 ) = 525 2 2 v = 12 ms

2

−1

1 6

. 5

(b)

15

1

(b)

20 − 0 25 − 30

1 1

−4 (c)

1 1 × (u + 20) × 10 + × 5 × 20 = 190 2 2

2

u=8

1 6

6 (a) (b)

20

23 − 5 0−6

−3 atau nyahpecutan 3 atau awapecutan 3 (c)

1 1 × 6( 23 + 5) + 4 × 5 + × 5( 5 + k ) = 139 2 2 k=9

1 1 1 2 1

7

(c)

(b)

(c)

45 20

= 135

1-1

km jam

60

3 9 minit @ jam @ 0.15 jam 20

120 95

= 75.79 km jam

1

2- 1

60 6

MODULE 8: ANSWERS TOPIC : PROBABILIty Answer: 1.

2.

(b)

5 1 = 15 3 5 5 4 4 3 6 = ×  +  ×  +  +   15 14   15 14   15 14  31 = 105

(a)

=

(a)

(b)

3.

(a) (b)

4.

(a) (b)

5.

(a)

(b)

=

17 16 136 × = 26 25 325  9 8   15 14  = ×  +  ×   24 23   24 23  47 = 92 5 4  4 5 5 = ×  +  ×  = 9 8 9 8 9 7 6 5 4 = ×  +  ×   12 11   12 11  31 = 66 5 3 1 = ×  =  15 14  14 5 4 3 2 7 6 = ×  +  ×  +  ×   15 14   15 14   15 14  34 = 105 5 4 = × 9 8 5 = 18  3 2  4 3 = ×  +  ×  7 6 7 6 3 = 7

6.

(a)

(b)

7.

(a)

(b)

8.

(a) (b)

9.

(a)

(b)

10.

(a)

(b)

8 7 × 12 11 14 = 33  3 7  7 3  3 2 = ×  +  ×  +  ×   10 9   10 9   10 9  8 = 15 =

3 2 + 10 9 1 = 15  5 3  3 5 = ×  +  +   10 9   10 9  1 = 3 =

1 12 1 2 = + 12 12 3 = 12 =

1 3 =1-  +  6 4 1 = 12 1 2 3 1   1 1 = ×  +  ×  +  ×   6 5   4 10   12 2  11 = 60 9 8 × 16 15 3 = 10 =

5 4 2 1 = ×  +  ×   16 15   16 15  11 = 120

MODULE 15 - ANSWERS TOPIC : PLAN AND ELEVATION

1.

8 cm

V/W

(a)

P/Q 2 cm

U

T 3 cm

Y/X

1.

S/R

V

(b) (i)

P

3 cm 6 cm

A/F/E

3 cm Y/U

S/T

2 cm 4 cm

C/D 1.

8 cm

B/X/W

R/Q P/V

(b) (ii)

3 cm A

F

5 cm

5 cm S/Y 2 cm

B/C

E

3 cm

R/X

3 cm T/U 5 cm

Q/W/D

2.

K/L

(a)

5 cm

B/A

4 cm

J/I

C/D 4 cm

H/G

2

E/F

P/Q

(b) (i)

J/K

N

C/B 6 cm

2 cm E/D

H/I 2 cm G/L

8 cm

P 2

F/A

5 cm

4 cm

M

N/Q

(b) (ii) 2 cm C/J

4 cm

4 cm

E/H

4 cm

D/I

B/K

4 cm

2 cm F/G

M/A/L

3.

C

(a)

H 3 cm

B

G

4 cm

4 cm

A/D 3.

F/E

(b) (i) C/K/D

4 cm

H/E

6 cm

M/B/A

G/F H/C

3.

1 cm

(b) (ii)

M

L

K

2 cm G/B 6 cm 4 cm

F/A

6 cm

E/D

4.

E

(a)

A

8 cm

F/D 4.

B/C

7 cm

(b) (i)

K/L

6 cm

N/M

4 cm 7 cm E/D

A/C 3 cm

6 cm J/P/G

F 4.

B

7 cm

(b) (ii)

I/H

I/J

7 cm

N/K

A/E 11 cm P

B/F

3 cm

8 cm

H/G 3 cm C/D 4 cm

M/L

5.

(a)

F/E

A/D 2 cm

G

B

5 cm

6 cm

H 5.

C

(b) (i)

L/R

M/N 3 cm G/F

B/S/A

9 cm

6 cm

H/E 5.

6 cm

(b) (ii) L/M

J/C/P/D

7 cm

6 cm

K/Q

R/N 3 cm 1 cm B/G S

A/F

6 cm

1 cm C/H K/J

D/E 6 cm

Q/P 1 cm

F/G 6.

A/H

(a)

4 cm

6 cm

E/M 3 cm D/L 3 cm 6 cm

B/J 6.

C/K

(b) (i)

A/P/B

H/J 3 cm

G/M

F/E

3 cm

10 cm

D/Q/C 6.

L/K

(b) (ii)

F

G 1 cm H

A

3 cm D/E

L/M Q/P

2 cm 3 cm

C/B

3 cm

10 cm

K/J

MODULE 16 - ANSWERS TOPIC: EARTH AS A SPHERE 1.

(a)

110 °W

(b)(i)

∠ HOK =

2m

=

4800 60

1m

80°

1m

Latitude of K = 80 – 50 = 30°N

1m

(ii) Distance F to H = ( 70 + 45 ) 60 cos 50° = 4435.23 n.m

2m 1m

(iii) Time = 4435.23 + 48003m

680

= 13.58 hrs

2.

1m

(a)

60°S

2m

(b)

Point H Point L L ( 60°N, 160°E )

1m 1m 1m

(c)

Distance H to south pole = 30 X 60 = 1800 n.m.

(d)

1m 1m

(i)

Distance = 450 x 8 1m = 3600 n.m. 1m

(ii)

∠ GOP =

3600 60 cos 60°

2m

= 120° Longitude = 120° - 20° = 100°E

1m

3.

(a)

Longitude of R = 95°E

2m

(b)

Distance PR = 40 x 60 = 2400 n.m.

1m 1m

(c)

Distance PQ = ( 85 – 30 ) 60 cos 70° = 1128.67 n.m.

3m 1m

(d)

Time =

160 × 60 600

3m

= 16 hrs

4.

1m

(a)

Longitude of S = 100° E

2m

(b)

∠ ROT = 3600

1m

60

= 60°

(c)

(e)

1m

Latitude of T = 60 – 40 = 20°S

1m 1m

Distance R to S = 100 x 60 = 6000 n.m.

1m 1m

Average speed = 180 × 60 × cos 40 + 3600

3m

20 8273.30 + 3600 = 20 = 593.66 n.m. 5.

1m

(a)

θ =170°E

2m

(b)

Point G Point H

2m 2m

(c)

Distance G to H = 50 x 60 = 3000 n.m

(d)

2m 1m

Time =

3000 375

2m

=

8 hrs

1m

6.

(a)

(i) Longitude of R = 140°E

2m

(ii) Distance PR via S : Distance PR via N 80x 60 : 280 x 60 2m 2 : 7 1m (iii) Distance P to Q = ( 40 + 90 ) 60 x cos 50° = 5013.74 n.m

(b)

∠ QOT =

3900 60

= 65° Latitude of T = 65º – 50º = 15°N

3m 1m

1m 1m 1m