```Solution
Method 1
1(ai)Kristian’s Amount
corresponds to
3
=
× \$   ℎ
3+2
3
\$72 = × \$   ℎ
5
5 × \$72
ℎ =
3
ℎ = \$120
Stephanie=\$120 − \$72 = \$48
Method 2
Kristian:Stephanie=3:2
\$72: 3
\$ℎ: 2
\$72 × 2
ℎ =
3
= \$24 × 2
= \$48
aii) Price of computer game=45%\$72
45
=
× \$72
100
162
=
5
= \$32.4
iii) Total money spent=\$8.40 + \$32.4
= \$40.8
Money Left=\$72 − \$40.8
= \$31.2
\$31.2     \$72
31.2
=
72
312
=
720
104
=
240
13
=
30
iv) Original price:100%
Reduced sale price :(100-20)%=80%
Original price:100%
\$19.20: 80%
Original price=
\$19.20×100%
\$1920
=
80
80%
= \$24
Method 2
Let  be the original price
Then 20% of  will be 0.2
− 0.2 = \$19.20
0.8 = \$19.20
\$19.20
=
0.8
\$192
=
8
= \$24
b) Amount = Principal +Simple Interest
Simple interest =
××
100
\$550 × 10 × 2
=
100
= \$110
Amount=\$550 + \$110
= \$660
c) Amount = (1 + %)
= \$550(1 + 1.9%)10
= \$550(1.019%)10
= \$550 × 1.207096
= \$663.9
d) Amount=\$638.30
(1 + %) = \$638.30
\$550(1 + %)10 = \$638.30
\$638.30
10
(1 + %) =
\$550
(1 + %)10 = 1.160545454
1
(1.160545454)10
1 + % =
1 + % = 1.015
% = 1.015 − 1

= 0.015
100
= 0.015 × 100
= 1.5
2a)
(−1, −2)-----------(4, −4)
(−3, −2)------------(2,-4)
(−3, −3)---------(2,-5)
.
. .
T’’
. .
.
T’
(x,2k-y)
ii. (−1, −2)-----------(−1,4)
(−3, −2)------------(-3,4)
(−3, −3)---------(-3,5)
iii. Enlargement, scale factor 3,
centre of enlargement (-6,-5)
b)i
1 2
1 3
+ =(
)+(
)
3 4
0 6
To find the sum of two matrices just add
the corresponding entries
1+1 2+3
+ =(
)
3+0 4+6
2 5
=(
)
3 10
b)ii
1 3 1 2
= (
)(
)
0 6 3 4
Multiply the rows in the first matrix by
the columns in the second
1×1+3×3 1×2+3×4
= (
)
0×1+6×3 0×2+6×4
1+9
= (
0 + 18
10
= (
18
2 + 12
)
0 + 24
14
)
24
b) iii
|| = ||
The determinant of a 2 × 2 matrix say

(
) is given by  − .

1×4−2×3=4×−3×1
4 − 6 = 4 − 3
1 = 4
1
=
4
c) If we transform any arbitrary point
0 −1
2
(, ) in  by the matrix (
)
1 0
we obtain ′( ′ ,  ′ ).
′
0 −1
i.e (
) () = ( )
′
1 0
0 −
′
(
)=( )
− 0
′
−
′
( )=( )

′
Hence the given matrix maps
(, ) onto ′(−, ).
This is an anti-clockwise rotation
through angle of 90° about the origin.
Ciii
If the point (, ) is reflected in the
line  = , we obtain
′ ( ′ ,  ′ ) = ′(, )
i.e
′ =
′ =
This is the same as;
′ = 0 + --------(1)
′ =  + 0-------(2)
Equations (1) and (2) are systems of
linear equations in  and .
We can transform this into a
2 × 2 matrix.
0 +
′
( )=(
)
+ 0
′
′
0 1
( )=(
) ()
′
1 0
0 1
Hence the matrix is (
)
1 0
3a)
i.
1
Median = ∑  ℎ occurrence
2
1
= (200) = 100th occurrence
2
Tracing from graph,this corresponds to
= 4003
ii.
1
The lower quartile is ∑ th
4
occurrence
1
= (200) = 50th occurrence
4
Tracing from graph,this corresponds to
= 3503
iii. The inter-quartile range = upper
quartile – lower quartile
= 420 − 350
= 703
iv.
From the graph 3003
corresponds to 30 students. The
number of students who estimate
that the volume is greater than
300 3 = 200 − 30 = 170
3b
Area(2 )
Mid-Values()
Frequency()

20 <  ≤ 60
60 <  ≤ 100
100 <  ≤ 150
150 <  ≤ 250
40
80
125
200
32
64
80
24
1280
5120
10,000
4800
∑  = 445
∑  = 200
∑
= 21,200
∑
̅ =
∑
21200
=
200
= 106
ii.
Frequency
density
0.8
1.6
1.6
0.24
iii.
(    > 1002 )
80
24
=
+
200 200
104
=
200
If the students are two then event is
mutually inclusive i.e the occurrence of
the first event will affect the second.
P(both student estimate
2
104
A>100 )= ×
200
10712
=
39800
103
99
4. a. The volume, V, of a sphere with
4
3
Substitute  = 15
4
= (15)3
3
= 45003
= 14137.173
= 141403 correct 4 s.f
bi)
Volume of cylinder,
=  2 ℎ
= (25)2 60
= 375003
= 117809.723
volume of water required to fill the
tank=    −
ℎ
= 117809.72 − 14140
= 103669.723
= 1037003 to 4 s.f
ii. There is no change in volume of water,
(25)2  = 1037003
1037003
=
6252
= 52.81cm
ci)
The volume, V, of a cone with radius r
1
and height h is  =  2 ℎ
3
The volume of the sphere is now equal
to the volume of the cone. But this cone
has height, ℎ = 54.
1 2
× 54 = 14140
3
14140
2
=
18
2 = 250.05
= √250.05
= 15.81
ii.
Total surface area =base area + curved
surface area
. .  =  2 +
.

54
15.81
Using Pythagoras’ theorem we can find
2 = 542 + 15.812
= √3165.9561
= 56.2668
. .
= ( × 15.812
+  × 15.81 × 56.2668)2
= 1,139.532
= 3,579.952
= 35802 to 4 s.f.
5a.
20
( ) =
+ ,  ≠ 0

−10 −8 −5 −2 −1.6
() −12 −10.5 −9 −12 −14.1
1.6 2 5 8 10
14.1 12 9 10.5 12
.
C)
from the graph  = 2.4  8.5
d) for values of k within: −8.94 <  < 8.94, () is
has no solution.
The prime numbers within this range are:
K=2,3,5,7
e)
We first determine the gradient function:

20
=− 2 +1

For gradient to be −4, then:

= −4

20
− 2 + 1 = −4

2 = 4
= −2  2
Hence the other coordinate is:
(−2, −12)
f)
( ) =  2
20
+  = 2

Multiplying through by ,we obtain:
20 +  2 =  3
3 −  2 − 20 = 0
Comparing coefficients to:
3 +  2 +  = 0
= −1   = −20
f)  =  2

−4
16
−2
4
0
0
2
4
4
16
ii)We plot and draw the graph as follows
iii. from the graph the solution
to  3 −  2 − 20 = 0 is  = 3.2
iv From iii,  = 3.2 since that is solution of
3 −  2 − 20 = 0
6. ai.The perimeter of the rectangle is given
by,
= 2 + 2
Let’s substitute  = 80,  = ,  =  ⇒

= .

⇒ 80 = 2 + 2 ( )

Multiplying through by , we have,
2 2 − 80 + 2 = 0
Divide through by 2,
⇒  2 − 40 +  = 0
ii. When  = 300, the equation becomes;
2 − 40 + 300 = 0
2 − 10 − 30 + 300 = 0
( − 10) − 30( − 10) = 0
( − 10)( − 30) = 0
− 10 = 0   − 30 = 0
= 10 0  = 30
iii. When  = 200, the equation becomes
2 − 40 + 200 = 0
= 1,  = −40,  = 200
− ± √ 2 − 4
=
2
−(−40) ± √(−40)2 − 4 × 1 × 200
=
2×1
40 ± √1600 − 800
=
2
40 ± √800
=
2
40 ± 20√2
=
2
= 20 ± 10√2
= 34.14   = 5.86
b) i
Average speed =

200
=

200
=

Also;
200
( + 10) =

200
=
( + 10)
200
200
− =
−
( + 10)

− =
− =
− =
200−200(+10)
(+10)
200−200+2000)
(+10)
2000)
(+10)
ii.
When  = 80,
We have;
2000)
− =
80(80 + 10)
2000
=
7200
5
=
ℎ
18
5
=
× 60
18
50
=

3
2
= 16 +
3
2
= 16 + × 60
3
= 16 40
7ai.
1
⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗

2
1
= ⃗⃗⃗⃗⃗

2
1
=
2
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
. ⃗⃗⃗⃗⃗⃗
=

1
⃗⃗⃗⃗⃗⃗
=  + ⃗⃗⃗⃗⃗

3
1
⃗⃗⃗⃗⃗⃗
=  + ⃗⃗⃗⃗⃗

3
1
⃗⃗⃗⃗⃗⃗
=  − ⃗⃗⃗⃗⃗

3
1
1
= −
2
3
.
⃗⃗⃗⃗⃗
= ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗

2
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
=  + ⃗⃗⃗⃗⃗

3
2
=+
3

From triangle ORU,
⃗⃗⃗⃗⃗⃗
= ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗

= ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗⃗

= ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗⃗

Since ∆ ≡ ∆
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
=
⃗⃗⃗⃗⃗⃗
= ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗
+ ⃗⃗⃗⃗⃗⃗

1
=++
2
3
=+
2
2
⃗⃗⃗⃗⃗⃗ | = √180
c) ⃗⃗⃗⃗⃗⃗
= ( ) and |
−
⃗⃗⃗⃗⃗⃗ |,
Using the components of |
⃗⃗⃗⃗⃗⃗ | = √(2)2 + (−)2
|
⃗⃗⃗⃗⃗⃗ | = √5 2
|
On substitution,
√180 = √5 2
Squaring both sides,
180 = 5 2
2 = 36
Taking positive square root of both sides,
= √36
=6
8. a
( ) = (1)
2 + 1 = 12 + 4
2 + 1 = 5
2 = 4
=2
b) ℎ(3) = 2(23 ) + 1 = 16 + 1 = 17
c)
( ) = 2 + 1
We know that:
( −1 ( )) =
2 −1 ( ) + 1 =
2 −1 ( ) =  − 1
−1
−1 ( )
=
2
d)  ( ) = (2 + 1)2 + 4
= (2 + 1)(2 + 1) + 4
= 4 2 + 2(2 ) + 1 + 4
= 4 2 + 4 + 5
c) ℎ( ) = 2
ℎ(ℎ−1 ( )) =
ℎ−1 ( )
2
=
Taking logarithm of both sides to base ;
ℎ−1 ( ) ln 2 = ln
ln
−1 ( )
ℎ  =
ln 2
Now back to the question,
ℎ−1 ( ) = 0.5
ln
= 0.5
ln 2
ln  = 0.5 ln 2
ln  = ln √2
= √2
f)
1
ℎ()
= 2
1
⟹  = 2
2
⟹ 2− = 2
Since the base is the same,
− =
= −1
9) ai. Given (4,0)  (0,2)
We first of all find the gradient;
∆
=
∆
2−0
=
0−4
1
=−
2
We know -intercept to be 2
1
=− +2
2
ii)
2 2
+
=1
2  2
The point (4,0) lies on this curve,
42 02
⟹ 2+ 2=1

16 0
⟹ 2+ 2=1

16
⟹ 2 +0=1

16
⟹ 2=1

16
⟹
= 2
1
2 = 16
The point (0,2) lies on this curve,
02 22
⟹ 2+ 2=1

0
4
⟹ 2+ 2=1

4
⟹0+ 2 =1

4
⟹ 2=1

4
⟹ = 2
1
2 = 4
b) If (2, )or (2, −) lies on
2 2
+
=1
16 4
Then it must satisfy it.
(2)2 (−)2
+
=1
16
4
4 2
+
=1
16 4
4 + 4 2 = 16
4 2 = 12
2 = 3
∴  = √3
ii. The point (2, √3) can be used to
determine angle ,
√3
2
1
√3
tan ( < ) =
2
2
1
3
−1 √
<  = tan ( )
2
2
<  = 2 tan
−1
= 81.78
c)
=
Comparing:
2 2
+ 2=1
2

to
√3
( )
2
2 2
+
=1
16 4
2 = 16, ⟹  = 4
2 = 4, ⟹  = 2
∴  =  × 4 × 2
= 8
ii. If the curve intersects the x-axis at
(12,0)  (−12,0) , then either point
must satisfy the equation of the curve;
122 02
+ 2=1
2

2
12
+0=1
2
2 = 122
= 12
By comparison, the length of  = 12 is a
magnification of  = 4, by a scale factor of
3.
The corresponding scale factor for the area
is 32 = 9
Hence the new area= 9 × 8 = 72
```