# 425119-june-2016-detailed-solution-41 ```Solution
Method 1
1(ai)Kristian’s Amount
corresponds to
3
=
&times; \$ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑠ℎ𝑎𝑟𝑒𝑑
3+2
3
\$72 = &times; \$ 𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑠ℎ𝑎𝑟𝑒𝑑
5
5 &times; \$72
𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑠ℎ𝑎𝑟𝑒𝑑 =
3
𝑇𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑠ℎ𝑎𝑟𝑒𝑑 = \$120
Stephanie=\$120 − \$72 = \$48
Method 2
Kristian:Stephanie=3:2
\$72: 3
\$𝑆𝑡𝑒𝑝ℎ𝑎𝑛𝑖𝑒: 2
\$72 &times; 2
𝑆𝑡𝑒𝑝ℎ𝑎𝑛𝑖𝑒 =
3
= \$24 &times; 2
= \$48
aii) Price of computer game=45%𝑜𝑓\$72
45
=
&times; \$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 𝑎𝑠 𝑎 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 \$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&times;100%
\$1920
=
80
80%
= \$24
Method 2
Let 𝑥 be the original price
Then 20% of 𝑥 will be 0.2𝑥
𝑥 − 0.2𝑥 = \$19.20
0.8𝑥 = \$19.20
\$19.20
𝑥=
0.8
\$192
=
8
= \$24
b) Amount = Principal +Simple Interest
Simple interest =
𝑃&times;𝑇&times;𝑅
100
\$550 &times; 10 &times; 2
=
100
= \$110
Amount=\$550 + \$110
= \$660
c) Amount = 𝑃(1 + 𝑟%)𝑡
= \$550(1 + 1.9%)10
= \$550(1.019%)10
= \$550 &times; 1.207096
= \$663.9
d) Amount=\$638.30
𝑃(1 + 𝑟%)𝑡 = \$638.30
\$550(1 + 𝑥%)10 = \$638.30
\$638.30
10
(1 + 𝑥%) =
\$550
(1 + 𝑥%)10 = 1.160545454
1
(1.160545454)10
1 + 𝑥% =
1 + 𝑥% = 1.015
𝑥% = 1.015 − 1
𝑥
= 0.015
100
𝑥 = 0.015 &times; 100
𝑥 = 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
𝑀+𝑃 =(
)+(
)
3 4
0 6
To find the sum of two matrices just add
the corresponding entries
1+1 2+3
𝑀+𝑃 =(
)
3+0 4+6
2 5
=(
)
3 10
b)ii
1 3 1 2
𝑃𝑀 = (
)(
)
0 6 3 4
Multiply the rows in the first matrix by
the columns in the second
1&times;1+3&times;3 1&times;2+3&times;4
𝑃𝑀 = (
)
0&times;1+6&times;3 0&times;2+6&times;4
1+9
𝑃𝑀 = (
0 + 18
10
𝑃𝑀 = (
18
2 + 12
)
0 + 24
14
)
24
b) iii
|𝑀| = |𝑁|
The determinant of a 2 &times; 2 matrix say
𝑎 𝑏
(
) is given by 𝑎𝑑 − 𝑏𝑐.
𝑐 𝑑
1&times;4−2&times;3=4&times;𝑘−3&times;1
4 − 6 = 4𝑘 − 3
1 = 4𝑘
1
𝑘=
4
c) If we transform any arbitrary point
0 −1
2
𝑃(𝑥, 𝑦) in 𝑅 by the matrix (
)
1 0
we obtain 𝑃′(𝑥 ′ , 𝑦 ′ ).
𝑥′
0 −1 𝑥
i.e (
) (𝑦) = ( )
𝑦′
1 0
0𝑥 − 𝑦
𝑥′
(
)=( )
𝑥 − 0𝑦
𝑦′
−𝑦
𝑥′
( )=( )
𝑥
𝑦′
Hence the given matrix maps
𝑃(𝑥, 𝑦) onto 𝑃′(−𝑦, 𝑥).
This is an anti-clockwise rotation
through angle of 90&deg; about the origin.
Ciii
If the point 𝑃(𝑥, 𝑦) is reflected in the
line 𝑦 = 𝑥, we obtain
𝑃′ (𝑥 ′ , 𝑦 ′ ) = 𝑃′(𝑦, 𝑥)
i.e
𝑥′ = 𝑦
𝑦′ = 𝑥
This is the same as;
𝑥 ′ = 0𝑥 + 𝑦--------(1)
𝑦 ′ = 𝑥 + 0𝑦-------(2)
Equations (1) and (2) are systems of
linear equations in 𝑥 and 𝑦.
We can transform this into a
2 &times; 2 matrix.
0𝑥 + 𝑦
𝑥′
( )=(
)
𝑥 + 0𝑦
𝑦′
𝑥′
0 1 𝑥
( )=(
) (𝑦)
𝑦′
1 0
0 1
Hence the matrix is (
)
1 0
3a)
i.
1
Median = ∑ 𝑓 𝑡ℎ occurrence
2
1
= (200) = 100th occurrence
2
Tracing from graph,this corresponds to
= 400𝑚3
ii.
1
The lower quartile is ∑ 𝑓th
4
occurrence
1
= (200) = 50th occurrence
4
Tracing from graph,this corresponds to
= 350𝑚3
iii. The inter-quartile range = upper
quartile – lower quartile
= 420 − 350
= 70𝑚3
iv.
From the graph 300𝑚3
corresponds to 30 students. The
number of students who estimate
that the volume is greater than
300 𝑚3 = 200 − 30 = 170
3b
Area(𝐴𝑚2 )
Mid-Values(𝑥)
Frequency(𝑓)
𝑓𝑥
20 &lt; 𝐴 ≤ 60
60 &lt; 𝐴 ≤ 100
100 &lt; 𝐴 ≤ 150
150 &lt; 𝐴 ≤ 250
40
80
125
200
32
64
80
24
1280
5120
10,000
4800
∑ 𝑥 = 445
∑ 𝑓 = 200
∑ 𝑓𝑥
= 21,200
∑ 𝑓𝑥
𝑋̅ =
∑𝑥
21200
=
200
= 106
ii.
Frequency
density
0.8
1.6
1.6
0.24
iii.
𝑃(𝑎 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝐴 &gt; 100𝑚2 )
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&gt;100𝑚 )= &times;
200
10712
=
39800
103
99
4. a. The volume, V, of a sphere with
4
radius r is 𝑉 = 𝜋𝑟 3
3
Substitute 𝑟 = 15𝑐𝑚
4
𝑉 = 𝜋(15𝑐𝑚)3
3
= 4500𝜋𝑐𝑚3
= 14137.17𝑐𝑚3
= 14140𝑐𝑚3 correct 4 s.f
bi)
Volume of cylinder,
𝑉 = 𝜋𝑟 2 ℎ
= 𝜋(25𝑐𝑚)2 60𝑐𝑚
= 37500𝜋𝑐𝑚3
= 117809.72𝑐𝑚3
volume of water required to fill the
tank= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 −
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑝ℎ𝑒𝑟𝑒
= 117809.72 − 14140
= 103669.72𝑐𝑚3
= 103700𝑐𝑚3 to 4 s.f
ii. There is no change in volume of water,
𝜋(25𝑐𝑚)2 𝑑 = 103700𝑐𝑚3
103700𝑐𝑚3
𝑑=
625𝜋𝑐𝑚2
𝑑 = 52.81cm
ci)
The volume, V, of a cone with radius r
1
and height h is 𝑉 = 𝜋𝑟 2 ℎ
3
The volume of the sphere is now equal
to the volume of the cone. But this cone
has height, ℎ = 54𝑐𝑚.
1 2
𝜋𝑟 &times; 54 = 14140
3
14140
2
𝑟 =
18𝜋
𝑟 2 = 250.05
𝑟 = √250.05
= 15.81𝑐𝑚
ii.
Total surface area =base area + curved
surface area
𝑇. 𝑆. 𝐴 = 𝜋𝑟 2 + 𝜋𝑟𝑙
.
𝑙𝑐𝑚
54𝑐𝑚
15.81𝑐𝑚
Using Pythagoras’ theorem we can find 𝑙
𝑙 2 = 542 + 15.812
𝑙 = √3165.9561
= 56.2668𝑐𝑚
𝑇. 𝑆. 𝐴
= (𝜋 &times; 15.812
+ 𝜋 &times; 15.81 &times; 56.2668)𝑐𝑚2
= 1,139.53𝜋𝑐𝑚2
= 3,579.95𝑐𝑚2
= 3580𝑐𝑚2 to 4 s.f.
5a.
20
𝑓 (𝑥 ) =
+ 𝑥, 𝑥 ≠ 0
𝑥
𝑥
−10 −8 −5 −2 −1.6
𝑓(𝑥) −12 −10.5 −9 −12 −14.1
1.6 2 5 8 10
14.1 12 9 10.5 12
.
C)
from the graph 𝑥 = 2.4 𝑜𝑟 8.5
d) for values of k within: −8.94 &lt; 𝑘 &lt; 8.94, 𝑓(𝑥) is
has no solution.
The prime numbers within this range are:
K=2,3,5,7
e)
We first determine the gradient function:
𝑑𝑦
20
=− 2 +1
𝑑𝑥
𝑥
For gradient to be −4, then:
𝑑𝑦
= −4
𝑑𝑥
20
− 2 + 1 = −4
𝑥
𝑥2 = 4
𝑥 = −2 𝑜𝑟 2
Hence the other coordinate is:
(−2, −12)
f)
𝑓 (𝑥 ) = 𝑥 2
20
+ 𝑥 = 𝑥2
𝑥
Multiplying through by 𝑥,we obtain:
20 + 𝑥 2 = 𝑥 3
𝑥 3 − 𝑥 2 − 20 = 0
Comparing coefficients to:
𝑥 3 + 𝑝𝑥 2 + 𝑞 = 0
𝑝 = −1 𝑎𝑛𝑑 𝑞 = −20
f) 𝑦 = 𝑥 2
𝑥
𝑦
−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 𝑥 3 − 𝑥 2 − 20 = 0 is 𝑥 = 3.2
iv From iii, 𝑛 = 3.2 since that is solution of
𝑥 3 − 𝑥 2 − 20 = 0
6. ai.The perimeter of the rectangle is given
by,
𝑝 = 2𝑤 + 2𝑙
Let’s substitute 𝑝 = 80, 𝑤 = 𝑥, 𝐴 = 𝑙𝑥 ⇒
𝐴
𝑙= .
𝑥
𝐴
⇒ 80 = 2𝑥 + 2 ( )
𝑥
Multiplying through by 𝑥, we have,
2𝑥 2 − 80𝑥 + 2𝐴 = 0
Divide through by 2,
⇒ 𝑥 2 − 40𝑥 + 𝐴 = 0
ii. When 𝐴 = 300, the equation becomes;
𝑥 2 − 40𝑥 + 300 = 0
𝑥 2 − 10𝑥 − 30𝑥 + 300 = 0
𝑥 (𝑥 − 10) − 30(𝑥 − 10) = 0
(𝑥 − 10)(𝑥 − 30) = 0
𝑥 − 10 = 0 𝑜𝑟 𝑥 − 30 = 0
𝑥 = 10 0𝑟 𝑥 = 30
iii. When 𝐴 = 200, the equation becomes
𝑥 2 − 40𝑥 + 200 = 0
𝑎 = 1, 𝑏 = −40, 𝑐 = 200
−𝑏 &plusmn; √𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
−(−40) &plusmn; √(−40)2 − 4 &times; 1 &times; 200
𝑥=
2&times;1
40 &plusmn; √1600 − 800
𝑥=
2
40 &plusmn; √800
𝑥=
2
40 &plusmn; 20√2
𝑥=
2
𝑥 = 20 &plusmn; 10√2
𝑥 = 34.14 𝑜𝑟 𝑥 = 5.86
b) i
Average speed =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
200
𝑥=
𝑡
200
𝑡=
𝑥
Also;
200
(𝑥 + 10) =
𝑇
200
𝑇=
(𝑥 + 10)
200
200
𝑇−𝑡 =
−
(𝑥 + 10)
𝑥
𝑇−𝑡 =
𝑇−𝑡 =
𝑇−𝑡 =
200𝑥−200(𝑥+10)
𝑥(𝑥+10)
200𝑥−200𝑥+2000)
𝑥(𝑥+10)
2000)
𝑥(𝑥+10)
ii.
When 𝑥 = 80,
We have;
2000)
𝑇−𝑡 =
80(80 + 10)
2000
=
7200
5
=
ℎ𝑜𝑢𝑟𝑠
18
5
=
&times; 60𝑚𝑖𝑛𝑠
18
50
=
𝑚𝑖𝑛𝑠
3
2
= 16𝑚𝑖𝑛𝑠 + 𝑚𝑖𝑛𝑠
3
2
= 16𝑚𝑖𝑛𝑠 + &times; 60𝑠𝑒𝑐𝑠
3
= 16𝑚𝑖𝑛𝑠 40𝑠𝑒𝑐𝑠
7ai.
1
⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
𝑀𝑄
𝑅𝑄
2
1
= ⃗⃗⃗⃗⃗
𝑂𝑃
2
1
= 𝒑
2
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
𝑖𝑖. ⃗⃗⃗⃗⃗⃗
𝑀𝑇 = 𝑀𝑄
𝑄𝑇
1
⃗⃗⃗⃗⃗⃗
= 𝑀𝑄 + ⃗⃗⃗⃗⃗
𝑄𝑃
3
1
⃗⃗⃗⃗⃗⃗
= 𝑀𝑄 + ⃗⃗⃗⃗⃗
𝑅𝑂
3
1
⃗⃗⃗⃗⃗⃗
= 𝑀𝑄 − ⃗⃗⃗⃗⃗
𝑂𝑅
3
1
1
= 𝒑− 𝒓
2
3
𝑖𝑖𝑖.
⃗⃗⃗⃗⃗
𝑂𝑇 = ⃗⃗⃗⃗⃗
𝑂𝑃 + ⃗⃗⃗⃗⃗
𝑃𝑇
2
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝑂𝑇 = 𝑂𝑃 + ⃗⃗⃗⃗⃗
𝑃𝑄
3
2
=𝒑+ 𝒓
3
𝑈
From triangle ORU,
⃗⃗⃗⃗⃗⃗
𝑂𝑈 = ⃗⃗⃗⃗⃗
𝑂𝑅 + ⃗⃗⃗⃗⃗
𝑅𝑈
= ⃗⃗⃗⃗⃗
𝑂𝑅 + ⃗⃗⃗⃗⃗
𝑅𝑄 + ⃗⃗⃗⃗⃗⃗
𝑄𝑈
= ⃗⃗⃗⃗⃗
𝑂𝑅 + ⃗⃗⃗⃗⃗
𝑅𝑄 + ⃗⃗⃗⃗⃗⃗
𝑄𝑈
Since ∆𝑀𝑇𝑄 ≡ ∆𝑈𝑇𝑄
⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
𝑄𝑈 = 𝑀𝑄
⃗⃗⃗⃗⃗⃗
𝑂𝑈 = ⃗⃗⃗⃗⃗
𝑂𝑅 + ⃗⃗⃗⃗⃗
𝑅𝑄 + ⃗⃗⃗⃗⃗⃗
𝑄𝑈
1
=𝒓+𝒑+ 𝒑
2
3
=𝒓+ 𝒑
2
2𝑘
⃗⃗⃗⃗⃗⃗ | = √180
c) ⃗⃗⃗⃗⃗⃗
𝑀𝑇 = ( ) and |𝑀𝑇
−𝑘
⃗⃗⃗⃗⃗⃗ |,
Using the components of |𝑀𝑇
⃗⃗⃗⃗⃗⃗ | = √(2𝑘)2 + (−𝑘)2
|𝑀𝑇
⃗⃗⃗⃗⃗⃗ | = √5𝑘 2
|𝑀𝑇
On substitution,
√180 = √5𝑘 2
Squaring both sides,
180 = 5𝑘 2
𝑘 2 = 36
Taking positive square root of both sides,
𝑘 = √36
𝑘=6
8. a
𝑓 (𝑥 ) = 𝑔(1)
2𝑥 + 1 = 12 + 4
2𝑥 + 1 = 5
2𝑥 = 4
𝑥=2
b) 𝑓ℎ(3) = 2(23 ) + 1 = 16 + 1 = 17
c)
𝑓(𝑥 ) = 2𝑥 + 1
We know that:
𝑓(𝑓 −1 (𝑥 )) = 𝑥
2𝑓 −1 (𝑥 ) + 1 = 𝑥
2𝑓 −1 (𝑥 ) = 𝑥 − 1
𝑥−1
−1 ( )
𝑓 𝑥 =
2
d) 𝑔𝑓 (𝑥 ) = (2𝑥 + 1)2 + 4
= (2𝑥 + 1)(2𝑥 + 1) + 4
= 4𝑥 2 + 2(2𝑥 ) + 1 + 4
= 4𝑥 2 + 4𝑥 + 5
c) ℎ(𝑥 ) = 2𝑥
ℎ(ℎ−1 (𝑥 )) = 𝑥
ℎ−1 (𝑥 )
2
=𝑥
Taking logarithm of both sides to base 𝑒;
ℎ−1 (𝑥 ) ln 2 = ln 𝑥
ln 𝑥
−1 ( )
ℎ 𝑥 =
ln 2
Now back to the question,
ℎ−1 (𝑥 ) = 0.5
ln 𝑥
= 0.5
ln 2
ln 𝑥 = 0.5 ln 2
ln 𝑥 = ln √2
𝑥 = √2
f)
1
ℎ(𝑥)
= 2𝑘𝑥
1
⟹ 𝑥 = 2𝑘𝑥
2
⟹ 2−𝑥 = 2𝑘𝑥
Since the base is the same,
−𝑥 = 𝑘𝑥
𝑘 = −1
9) ai. Given 𝐴(4,0) 𝑎𝑛𝑑 𝐵(0,2)
We first of all find the gradient;
∆𝑦
𝑚=
∆𝑥
2−0
=
0−4
1
=−
2
We know 𝑦-intercept to be 2
1
𝑦 =− 𝑥+2
2
ii)
𝑥2 𝑦2
+
=1
𝑎2 𝑏 2
The point 𝐴(4,0) lies on this curve,
42 02
⟹ 2+ 2=1
𝑎
𝑏
16 0
⟹ 2+ 2=1
𝑎
𝑏
16
⟹ 2 +0=1
𝑎
16
⟹ 2=1
𝑎
16
⟹
= 𝑎2
1
𝑎2 = 16
The point 𝐵(0,2) lies on this curve,
02 22
⟹ 2+ 2=1
𝑎
𝑏
0
4
⟹ 2+ 2=1
𝑎
𝑏
4
⟹0+ 2 =1
𝑏
4
⟹ 2=1
𝑏
4
⟹ = 𝑏2
1
𝑏2 = 4
b) If 𝑃(2, 𝑘)or 𝑄(2, −𝑘) lies on
𝑥2 𝑦2
+
=1
16 4
Then it must satisfy it.
(2)2 (−𝑘)2
+
=1
16
4
4 𝑘2
+
=1
16 4
4 + 4𝑘 2 = 16
4𝑘 2 = 12
𝑘2 = 3
∴ 𝑘 = √3
ii. The point 𝑃(2, √3) can be used to
determine angle 𝑃𝑂𝑄,
√3
2
1
√3
tan ( &lt; 𝑃𝑂𝑄) =
2
2
1
3
−1 √
&lt; 𝑃𝑂𝑄 = tan ( )
2
2
&lt; 𝑃𝑂𝑄 = 2 tan
−1
= 81.78
c)
𝐴𝑟𝑒𝑎 = 𝜋𝑎𝑏
Comparing:
𝑥2 𝑦2
+ 2=1
2
𝑎
𝑏
to
√3
( )
2
𝑥2 𝑦2
+
=1
16 4
𝑎2 = 16, ⟹ 𝑎 = 4
𝑏 2 = 4, ⟹ 𝑏 = 2
∴ 𝐴𝑟𝑒𝑎 = 𝜋 &times; 4 &times; 2
= 8𝜋
ii. If the curve intersects the x-axis at
(12,0) 𝑎𝑛𝑑 (−12,0) , then either point
must satisfy the equation of the curve;
122 02
+ 2=1
2
𝐴
𝐵
2
12
+0=1
𝐴2
𝐴2 = 122
𝐴 = 12
By comparison, the length of 𝐴 = 12 is a
magnification of 𝑎 = 4, by a scale factor of
3.
The corresponding scale factor for the area
is 32 = 9
Hence the new area= 9 &times; 8𝜋 = 72𝜋
```