QUESTIONS
&
SOLUTIONS
LOOBWE, A. N.
nloobwe@gmail.com
0978739107
PREPARED BY LOOBWE, A. N.
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SOLUTIONS 2019 INTERNAL PAPER 2 (4024)
6 3 3 36
Q1
(a)
18 2 36
(i)
(ii)
2 36 18
A45 |7| Adj A
5
2 18
1 3 9
36 2 6
:
9 :;
:
:=
6 9
A
2 3
:
<?
:
>
(b) (i)
1-. Pick
4
9
5
9
1
223 Pick
3
8
R
G
4
8
5
8
4
8
R
POSSIBLE OUTCOMES
RR
RG
G
GR
R
GG
G
(ii) POne Red and one Green PRG & GR
PRG PGR
4 5
5 4
" "
9 8
9 8
5
5
18 18
#
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Page 11 of 139
MSc. Loobwe, A. N.
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Q2
@A B 4CD
A4C
@A B 4D
A4C
(a)
@A4CAEC
A4C
@AEC
5
> ;
(b)
(i)
b √ac
√256 1024
S2 (iii)
S55 √262 144
#:;
(ii)
TUV 45
U45
T2 ar 245
TC
r
T5
8
r 2
4
T55 425545
425I
DCWW 45
C45
42 048 1
21
42 047
1
= :==
41024
<J >
Q3.
(a)
a 1 b 1
2
K L √K C 4MN
2M
1 L O1C 414
21
1 L √17
2
Either
c 4
1 L √1 16
4
1 √17
2
:. #>
or
RS
1 √17
2
;. #>
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Page 12 of 139
MSc. Loobwe, A. N.
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(b)
nB nB ∪ CY x 4 8 5 7 10
(i)
4 8 5 7 10
9 17
17 9
=
nA nB
(ii)
7 8 5 2[
15 5 2[
15 5 2[
10 2[
\#
nE 7 8 10 8 4 5 25 <<
(iii)
(iv)
Q 4.
(a)
nBY 7 10 25 ;]
^
D
_AE^
CA4D
53 5 42 4
2 43 5
15 25 8 16
2 43 5
15 8 25 16
2 43 5
] <:
; <` #
3
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Page 13 of 139
MSc. Loobwe, A. N.
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(b)
Begin
Enter l
Is
b c 0?
Yes
"error message"
re-enter positive length
No
Enter h
Is
ℎ c 0?
Yes
"error message"
re-enter positive length
No
h
1
∗j∗b∗ℎ
3
Display volume
End
4
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Page 14 of 139
MSc. Loobwe, A. N.
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Q5
C
a ii BC 13.3 cm
bi
3cm
A
110°
biii
Note: The zigzag line on the boundary of the shaded
region symbolise that the line is not part of the solution.
(a) (i)
(a)
Q6
(b)
(C)
(ii)
5
8cm
bii
35°
B
lllll⃗ 3OA
lllll⃗
AB
`n
lllll⃗ BC
lllll⃗
lllll⃗ AB
AC
`n ;o
lllll⃗ BM
llllll⃗
llllll⃗
OM OB
<n o
lllll⃗ AX
lllll⃗
lllll⃗ OA
OX
lllll⃗ 5AX
lllll⃗
AC
lllll⃗ 3a 2b
5AX
1
lllll⃗
AX 3a 2b
5
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lllll⃗
OX lllll⃗
OA lllll⃗
AX
1
M 3a 2b
5
3
2
M a b
5
5
8
2
a b
5
5
# <n o A.R
;
(b)
rs
rA
15 C 12 2
151C 121 2
15 12 2
5
t5 tC 1
5tC 1
tC 1
5
1
[ N
5
1
2 1 N
5
2
\
u
:
#
1
N
5
11
5
::
#
Or
#\ ::
6
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Page 16 of 139
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Q7
7
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MSc. Loobwe, A. N.
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Q8
(a) SD wx
∑ z {4 {| B
∑z
̅ ∑ z{
}
2
^^I E 5@5^I E 5C^I E 5_^IE 5DD^IE 5C^^I E @^I E C^I
^ E 5@ E 5 E 5 E 5D E 5C E  E C
C^I E CDII E DC^I E ^€^I E @_II E @@II E D^^I E 5^II
€I
31800
90
̅ 353. 3
SD wx
w
∑ z {4 {| B
∑z
}
^^I4_^_._B E 5@ 5^I4_^_._B E 5C^I4_^_._B E 5_^I4_^_._B E5DD^I4_^_._B E 5C^^I4_^_._B E @^I4_^_._B E C^I4_^_._B
€I
5_I€5C.D@ E D@DC€I.@E @5@C5@.C_ E _5DD5.
wD^€€^D.D^ E @@5C€D.CD E 55DI^.5_ E 5^.5_ E€I
w
CC€III.5
€I
√31433.33444
=177.2944851
∴
SD= 177 .3 (1 d.p)
8
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MSc. Loobwe, A. N.
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(b) Remaining answers are on the graph
9
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MSc. Loobwe, A. N.
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Q9
_ 2 6
Q 10
(a)
(i)
(a)
_ 2 1 5
[ 5
:, #
[ 10
J. : ƒ„
(b)
(iii)
(b)
10
rs
rA
(a)
A
5
C
3 C 2
;.
= 32C 2
14
∴ †‡ˆ ‰SnŠ‹ˆŒ† Š :<
M Kℎ
1
1
2 10 1 10 32 1
2
2
1
1
12 42
2
2
6 21
;] ŽŒŠ† ;
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Page 20 of 139
MSc. Loobwe, A. N.
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(b)
5 3 C 4
_
3 _ 4 C 3
‘
’
3
2 1
“ _ 2 C ”
3
1
“3_ 23C ” “1_ 21C ”
27 18 1 2
45 3
<; •ŒŠ†;
11
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MSc. Loobwe, A. N.
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(a)
90° 45° :`#°
(i)
PQ Q 11.
(ii)
(a)
2πRCosα
135
2 3.142 3437 Cos65
360
3,422.90872
`<;J Œš
(3 s.f)
PQ (b)
—
_@I
PT _@I 2πR
—
5CI
3.142 5I
3437
7199.369333
];JJ Œš (3 s.f)
(b)
(i)
MC K C N C
AB C BC C AC C
(ii)
8C 8C ›u C
u ¡¢ £r¤
¥s¦
u ¡¢ 64 64 ›u C
5.66
10
u ¡¢ 0.566
›u C 128
¢ u ¡ 45 0.566
√128
¢ 55.52823806
11.3137085
¢ 55.53°
1
CP AC
2
∴ †‡ˆ nŒ‰§ˆ ¨ž Š ##. #° (1 d.p)
1
CP 11.31
2
CP 5.66
OP C œu C COC
C 5.66C 10C
C 100 32.0356
12
√67.9644
8.244052402
∴ †‡ˆ ‡Šˆ‰‡† ž Š =. ; Ÿš (1 d.p)
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Page 22 of 139
MSc. Loobwe, A. N.
(a)
mC k C nC 2knCosM
mC 10C 8C 2108cos92°
mC 100 64 160Cos92°
mC 164 5.5839195
mC 164 5.5839195
(i)
Q 12
PREPARED BY LOOBWE, A. N.
√mC √169.5839195
m 13.02243908
©ª :`. J «š
A C k n SinM
(ii)
5
C 10 8 Sin92°
5
39.97563308
¬ <J «š;
A C bh
5
(iii)
40 C 13 h
DI
@.^
5
@.^®
@.^
DI
h @.^
‡ >. ; «š
(b)
h 6.153846154
2tan θ 3
3
2
45 _
θ tan C
tan θ 56.3°
tan is negative in the 2±r
θ 180° α
θ 180° 56.3° 123.7°
∴ ² :;`. ]°
(c)
13
C^³´
µB
¶ C5µ´ ^³·
³
5^µ
25pD 21qD
p
C
@
7q
5p
15q
21 q q q q
p
º
25 p p p p
5 p p p p p p 15q »
7qq
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Page 32 of 139
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SOLUTIONS 2019 G.C.E. PAPER 2 (4024)
Q1
(a)
(b) (i) 16 ----------- equ. 1
(ii)
4 ----------- equ. 2
$
%
"# 1
(
64
512
1
8
∴ the tenth term is
1
2
∴ the first term is 64 and the common ratio
(iii) "# 1
64 2
1
64 512
1
4
1
16 2
16 4
64
1
64 2
Divide equation 1 into equation 2
4 16 1
41
1
4
*
64
1
1&2
64
1
2
128
∴ the sum to infinity is 128
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Page 33 of 139
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Q2
8 ,8 -+ & ,12,- & 4++
(a) (i+
8 8- – 12- / 48
8 & 48 &4&40 &41 2
(ii) Q |5| Adjoint of Q
2
>
* &?
B
D
&
&
@ A CD
E
*
&
C
(i) - & 3 / - / 2 / - & 2 12
1B
(ii) 114
(iii) (a) 109
(b) 5
G
Q3.
H
(a)
&
H
G,H + ,H+
,H+,H +
GH H I
,H+,H +
JID
,JD+,J+
(b) (i)
(b)
(c)
2
(a)
LLLLL⃗ / OB
LLLLL⃗
LLLLL⃗
AB AO
(ii)
&O / P
LLLLLL⃗
AM G , W & 4b+
LLLLLL⃗ AO
LLLLL⃗ / OM
LLLLLL⃗
AM
1
& / LLLLL⃗
OC
3
LLLLL⃗ LLLLL⃗
AB
AC
D ,&O / P+
LLLLL⃗ / AC
LLLLL⃗
LLLLL⃗ OA
OC
1
/ R&a / bU
3
1
1
& a/ b
3
3
DO / P
D
1 2
1
& / a / b
3 3
3
2
1
& / a / b
6
6
4
1
& a/ b
6
6
1
4
b& a
6
6
? , P & CP+ A.R
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Q 4.
,a+ ,ii+
J
,a+ ,i+
,a+ ,i+
6 cm
Angle JLK 36°
10 cm
,b+ ,ii+
,b+ ,i+
K
8 cm
,b+ ,iii+
L
Note: The zigzag line on the boundary of the shaded region symbolise that
the line is not part of the solution.
Q5
(a) a &5 b &9 c 13
-
&,&9+ Z ],&9+ & 4,&5+,13+
2,&5+
9 Z √81 / 260
&10
9 & √341
&10
Either
3
&W Z √W & 4\
2
9 Z √341
&10
- 0.946618531
1 2. aB
or
or
bc
- 9 / √341
&10
- &2.746618531
1 &. dB
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(b)
1~ Pick
7
13
6
12
O
6
13
2H Pick
6
12
7
12
E
5
12
(i) P,Both Old+ P,OO+
7
6
13 12
O
POSSIBLE OUTCOMES
OO
OE
E
O
EO
E
EE
?
d
(ii) P,Only one Even+ P,EO & OE+
7
6
6
7
/ 13 12
13 12
7
7
/
26 26
D
d
Q6
(a)
s
s
6- / 8
t,6- / 8+dx
v
4
G / 8x / C
v 3- / 8x / C
2 3,1+ / 8,1+ / C
23/8/C
C 2 & 11
C &9
∴ the equation of the curve is D1 / *1 & a
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Page 36 of 139
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(b)
Start
Enter base Area (A)
If base Area ‚ 0
Then display “error message” Area must be positive
Else enter height
If height ‚ 0
Then display “error message” height must be positive
Else volume A ∗ h
End if
Display volume
Stop
Q7
5
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Page 37 of 139
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$
†
(a) (i) use sine rule (AAS)
~„H
Q8
~„H ‡
275
sin 125°
sin v°
.
125° / 40° / v° 180°
v 180° & 165° 15°
275
sin 125°
sin 15°
- ˆ‰Š 15° 275 ˆ‰Š 125°
-
‹Œ
‹Œ °
°
870.3641268
∴the distance BC is *d2 Ž
(ii)
A A ab sinC
870 275 sin40°
d?, *aD ‘ (iii)
bh,
A 76,893
870 h 76 893
435h 76 893
h 176.7655172
∴ the shortest distance is 177 km
(b)
13cos θ = 5
5
cosθ 13
θ cos > @
= 67.4°
Cos is positive in the 1~ and 4“ quadrant.
θ 360° & α
θ 360° & 67.4° = 292.6°
∴ • ?d. C°
bc
• a. ?°
(c)
6
–
I
, (+
I
,+,I+
I
,1 & D+
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Q9
(a) SD ˜™
xŸ ∑ ›, œ œ +
∑›
∑ ›œ
ž
H
, +I , +I, +I , +I ,–+I ,G +I, G+
I I I I–IGI
I
I G I ( I I ( I 407
13.56666667
30
SD ˜
˜
˜
,.G+ I , .G+ I , .G+ I ,.G+ I –,.G+ IG, .G+ I ,G.G+
I .
.
I .– I .
I . – I . G I .
√1.78
1.334166406
∴SD= 1.3 Correct to one decimal place.
7
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(b) Remaining answers are on the graph
8
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Page 40 of 139
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Q 10
(a) V π,R / r / R r+ h
1
3.142 ,8 / 3 / 8 3+ 10
3
31.42
,64 / 9 / 24+
3
31.42
,97+
3
1015.913333
£ 22 ¤D
r h
R H
3
h
8 h / 10
8ℎ 3ℎ / 30
8ℎ & 3ℎ 30
5ℎ 30
ℎ6
§ 16
1
1
V πR H & πr h
3
3
1
V π,R H & h+
3
1
1
V πR H & πr h
3
3
1
V π,8 16 & 3 6+
3
3.142
,1024 & 54+
3
3.142
,970+
3
1015.913333
D
£ 22 ¤
(3 s.f)
Or
(b) (i) 30° / 60° 90°
(ii)
(a)
¨
LM G 2πR
45° / 50° 95°
95°
3.142 6370
180
10,563.22944
2, ?22 ‘ (3 s.f)
9
¨
(b) KL G 2πRCosα
90
3.142 6370cos50
180
6432.549163
(3 s.f.)
?CD2 ‘
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Q 11.
(a) (i) the centre of enlargement is the origin (0, 0)
(ii) scale factor
ª«
4
©
&
ª«
2
(b) the transformation that took place is a rotation of 90 clockwise direction with centre (0,0)
(c)
(i) Let the matrix be
W
>
@, then pick any two coordinates of the object which corresponds to the image
\ ¬
and
&4 0
W 2 0
>
@>
@>
@
0 1
\ ¬ 0 1
2 / 0W &4 ----------------------- i
0 / W 0
2\ / 0¬ 0
0\ / ¬ 1
----------------------- ii
------------------- iii
------------------- iv
Solve the equations formed and you have
Equation I and ii
Equation iii and iv
2a &4
a &2
b0
2\ 0
∴ ­he required matrix is >
\
& 2
W
@>
@
2 ¬
\0
¬1
©
(ii) to find the area scale factor, we compare the standard matrix >
0
&2 0
the invariant line and >
@ . we see that  &
0 1
0
@ for a stretch with y axis as
1
(d) To find the coordinates of ∆S, multiply the given matrix with the coordinates of ∆P
1 0 2 2
>
@>
2 1 2 0
12/02
>
22/12
2/0 2/0
>
4/2 4/0
2 2 0
>
@
6 4 1
10
0
@
1
12/00
22/10
0/0
@
0/1
10/01
@
20/11
∴The coordinates are ¯C ,, ?+, °C ,, C+, ±C ,2, +
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Q 12
11
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Page 50 of 139
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Q1
SOLUTIONS 2018 INTERNAL PAPER 2 (4024)
1
(a) (i) Det A 4 2
5
8– 5
85
Det 13
Det B 8
13 40 – 3y
13 – 40 27 y=9
8
(ii) B = 3
5
8
B=
3
(b)
1AB Pick
6
15
9
15
3
5
3y
3 dividing both sides by 3 we have,
9
5
2CD Pick
B
G
y
5
14
9
14
6
14
8
14
(i) PBoth Black PBB
6
5
15 14
B
!
"
#
!
POSSIBLE OUTCOMES
BB
BG
G
B
GB
G
GG
,
1
(ii) PBoth different colours PBG & GB
6
9
9
6
6
76
7
15 14
15 14
9
9
35 35
!
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Page 51 of 139
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Q2
(a)
a 3 b 1
1 G H 1 I
23
1 G √1 60
F 6
1 G √61
F 6
F Either
c F K 1
√61
6
. M
5
43 5
1 √61
6
K . M,
or
F NO
(b) (i)
E
Physic
Maths
Chemistry
2
4
3
5
6
(ii) (a) 6
(b) 2+5 = 7
(c) 2+4+6 = 12
Q3.
(a)
PR2 x
I
x I dx
2x ST
I
22 RW
X
V
Y
I T
I
I U
V
2 1 R T
I
R U
V
Z
I
2
SU I
V R
M
[
\]^_`[
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(b) y x
y4
y 3
(4, 3)
Db
DS
Db
DS
a
S
a
a
1
Normal equation
mR mI a
a
1 RY
V
a
a
a
ST
1 a
V
Q 4.
y 3 3 gradient of the tangent
c mI mI T
M
d
[
Ik
1
V
a
V
a
V
RY
V
c
a
V
1
xc
4 C
c
Z
a i
b ii
7 cm
P
X
a ii XZ 5.5 cm
b i
9 cm
gradient of the normal
Note: The zigzag line
on the boundary of the
shaded region
symbolise that the line
is not part of the
solution.
38°
Y
c See on the diagram
d See on the diagram
3
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lm
Q5
mT lT
(a)
Rmnl
ml mnl
op
q
r = qT
(b)
r
sna
s
IsRk
s
k I 2k k 4
15 k 4
k I 2k I 8k
kI
2k I kI 7k
7k
60
15k
60
60
k I 7k 60 0
kI
5k 12k 60 0
k k 5 12k 5 0
Either
k 12
k 5 0
k 12 0 or k 5 0
k 12 or k 5
∴ positive constant k = 12
(ii) first three terms of the progression 12 +4 , 12, 2(12)- 15
u, [,
(iii)
4
Sw Sw Sw m
Rx
RY
U
R
y
RY
r
y
Sw 16 z a
{w uM
R
Mathematics a school of problem solving where reasoing is the main meal | MSc. Loobwe, A. N.
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|||||⃗
AE Q6
(a) (i)
(a)
|||||⃗
AC
R
V
|||||⃗
|||||⃗ BC
|||||⃗
AC AB
=a 2 b
|||||⃗
AE R
(a
V
2 b)
o
[
|||||⃗ + AE
|||||⃗
|||||⃗ BA
BE
(b)
R
a a
V
[
o
[
I
V
p
b
|||||⃗ + AD
|||||⃗
|||||⃗
BD BA
(c)
(ii)
p
o p
|||||⃗
|||||⃗
BE h ED
|||||⃗
~
||||||⃗
~€
h
|||||⃗ AD
|||||⃗
|||||⃗ BA
BD
=b
|||||⃗
~
||||||⃗
~€
a b
|||||⃗
BE 5
a
R
Vl m
R lm
I
V
|||||⃗
BD
This shows that B, E, and d are collinear points with the length of BE being
BD
[
the
length of
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(b)
Start
Enter x, y
M Sqrt x ∗ x ∗ y ∗ y
Is
M ‡ 0?
Yes
Error " M must be positive "
No
Display M
Stop
6
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Q7
(a)
(b)
V
a
SnR SR
VSR aSnR
SnR ‰
VSVaSa
SnR SnR
7
VSaSVa
SR SnR
d,
dn d
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Q8
(a)
KR
Š
A‹C Œ
’W
A‹C kI

Ž‘

ŽYW
b sin 52° 80° sin 60°
b=
’W A‹C YW°
A‹C kI°
b = 87.9m
∴ KR = 87.9 m
“ (b) (i)
R
I
R
I
”
•
•ℎ
–—˜ 60
“ 3260
80 ℎ 3260
aW œ
aW
–—˜ ™
1
80 50
2
,[. š[
(ii) “ R
I
VIYW
aW
h = 81.5m,
∴ the shortest distance is 81.5m
(b)
™– ž, 0° Ÿ ž Ÿ 360°
ž
™– ž
0W
1
1
0
8
1
90W
0
180W
-1
270W
0
360W
1
™– ž
90°
180°
270°
360°
F
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RIDCU
(c)
Rk
DU
Rk
DU
RIDCU
z RW
Z UC
T DT
RW T DT
Z UC
RI D C C C
Rk
!][
Q9
RW
D D D
D D
Z
C
¡[
(a) SD ¢£
∑ ¥ S S¦ T
∑¥
∑ ¥S
ẍ SD ¢
C
Rk n RW
aRaW
46
ZW
Ik nRk Vk n IV ak n VW
InRWnRknIVnVWnRW
kk n RW
Yk
IRkaY T n RW IkaY T n RkVkaY T n IVakaY T n VWkkaY T nRWYkaY T
SD ¢
SD ¢
I
§
ZW
RZIInaaRWnR’RknIVnIaVWnVYRW
RaIRW
ZW
ZW
SD √157.8888889
SD=12 .56538455
∴SD= 12 .6 Correct to one decimal place.
9
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(b) Remaining answers are on the graph
10
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Q 10
(a) It is a clockwise rotation of 90° about the origin.
(b) It is an enlargement, centre (0, 0) and scale factor 2
(c)
”
©
(b)
Let the matrix be
•
, then pick any two coordinates of P which corresponds to V and
ª
4
4
a b 2 2
1
4
c d 1 1
2” • 2” 4• 4 ----------------------- i
4
2© ª 1
2© 4ª 4
------------------- ii
------------------- iii
------------------- iv
Solve the equations formed using simultaneous method
Equation I and ii
2” • 2” 4• 3b 0
p«
2a 0 2a o
4
[
Equation iii and iv
4
2© ª 1
4
2© 4ª 4
3ª ®
4
a
∴ ¬he required matrix is c
3
2© 1 1
[ «
b
« d
2© 0
¯«
(d) To find the coordinates of ∆S, multiply the given matrix with the coordinates of ∆P
1 0 2 2 4
2 1 1 4 1
1 20 1
1 20 4
2 21 1
2 21 4
20
20
40
41
44
81
2 2 4
3 0
7
1 40 1
2 41 1
∴The coordinates are [, , [, « , M, ,
11
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Q 11.
12
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Q 12
(a) The question was not properly crafted. The shape was not a frustum for its sides were not in
the same ratio. The question was producing two answers which should not be the case. Therefore,
the question was invalid.
(b)
N
15° N
70° E
A
B
C
35° S
40° E
S
(i)
AC °
VYW
2πR
θ 15 35 50
kW
VYW
2
3.142
kW
40029.08
VYW
I,WWRaka
VYW
6370
5559.594444
AC = 5560 km (3 S.f)
°
(ii) (a) BQ = VYW 2 πR
°
900 = VYW 2
900 =
3.142
VIX’Z.ZWIXR °
VYW
6370 cos 35
324,000 = 32789.90271 θ
θ = 9.9°
∴the difference in longitude is 9.9°
(b) longitude of Q = 70 – 9.9 = 60.1
13
´ u«. °
∴ Position of µ°, u«. °
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Page 64 of 139
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Page 65 of 139
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Page 66 of 139
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Page 67 of 139
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Page 68 of 139
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Page 69 of 139
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Page 70 of 139
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Q1
SOLUTIONS 2018 G.C.E. PAPER 2 (4024)
(a) (i) Det A 2 2 3
23 2
6 2
(ii) A = :
;=:
;
2 6 12
3
3
3 3
1
2 18
A1( Adj A
|A|
9
= √9
(b) a 1 b 4
Q2
c 2
√ 4
2
4 4 412
21
4 √16 8
2
4 √24
2
Either
4 √24
4 √24
or
2
2
. !
"#
. !
$%&'
()*+ ,'
(a)
)*' ,
-%+ &'
$%&&
()***,,
(b)
&
( ,%'
3./ ar 21(
…… equ 1.
4./ ar 01(
4./ ar 41(
$
1
ar 2 ……… equ 2.
'
5
'
'6
(
2
7.'
(
B
3 2
:
;
3 6
9
9
9
C
9D
=
)**,
-%%%&&
(i) 3./ ar 01(
ar
3
(
(
7.+
3
3
ar
a :2;
2 9 9
(
9 18
< = (first term)
.
3r = 1
∴ r = (common ratio)
9
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(ii)
S0 S) 7 (1.F H K
; L
'IJ
M
(1
+
G(1:
2
(iii) S` (1.
( (
4
-(
-(
4
-(
(
(
Q3.
H
+
(1
2 N! = O9 2.99 . 1 dp
O
(1.
2 a
(
(
7
PREPARED BY LOOBWE, A. N.
2
2
N` (a) (i) 4 x 3 7 22
14 x 22
x 22 – 14
T O
(ii) (a) 14 + 7 + 7 = 28
(b)
4+2+3=9
2
2
5
(b) DL = :
;, OM = : ;, ON = : ;
10
2
11
YYYYY⃗ OM
YYYYYY⃗
YYYYYY⃗
LM LO
2
2
:
; : ;
10
2
4
: ;
12
1
4: ;
3
YYYYYY⃗ + ON
YYYYY⃗
YYYYYY⃗ = MO
MN
2
5
: ; : ;
2
11
3
1
=: ; 3 : ;
9
3
2
Since YYYYYY⃗
]^ collinear
9 YYYYYYY⃗
9
: ;, ^_
=3: ; and M is a common point , hence the points L, M and H are
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Q 4.
(a)
(i)
R
Note: the zigzag lines on the
boundaries of the shaded region
symbolise that the lines are not part of
the solution
(ii) 7.8cm
b (ii)
b (iii)
8 cm
50°
T
P
Q5
10 cm
(a)
2ij Pick
1%& Pick
(i)
9f
!
9=
!
B
=
C
!
9=
C
=
!
R
!
C
=
!
W
!
3
(ii) PWW 2) 2d
4
2
b (i)
Q
Possible outcomes
B
BB
BR
R
BW
99
!
9=
!
W
B
R
RB
RR
RW
W
B
WB
R
WR
W
WW
9
9!
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(b)
Start
Enter a, r, n
R 1r
Is
R= 0?
S0 "the value of rnot valid"
a1 r
R
Display S0
Stop
Q6
.
V 2 lA( A A( A mh
(
2 l10 4 √10 4 m 9
(
2 l100 16 √100 16m 9
(
2 l116 √1600m 9
(
116 40 9
(
2
156 3
4
o COpq
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2πr

Q7
2d€
(a) (i) BC =
θ 60 60
(ii) S =
( €
2d€
2 3.142 3437
)3(,$$ .3d
2d€
BC = 7,200 nm
/†
/‡
4x 3
43 3
9
m( m 1
m 3
(
normal
2)33.d(
S = 300 knots
(3 S.f)
y 2x 3x 2
9m 1
2 3.142 3437 cos60
D = 3599.68
7199.369333
(b) (i)
( €
2d€
2d€ 21,598.108
( €

Distance CD = 2d€ 2 π R cos∅
=120
BC ’
“
(ii)
(
Š€ x 2x 3dx
‡+
: 2 x 3x;
(+
€+
: 2 1 31; : 0 30;
2
(
2130
gradient of the
1
0
99
‹Œ<#Ž ‘‹
y 3x c
(
7 3 3 C
7 2
(
2
c
(
c
‰ fT 9
==
5
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Q8
(a) (i) use sine rule
•
%–0 —
()
%–0 4€°
˜
%–0 ™
(iii) A .
˜
%–0 $3°
b sin40° 15 x sin79°
b
(
(
bh,
x 23h 150.9
A 150.9
h 13.1km
∴ the shortest distance is 13.1km
() š›i3$°
š›i4€°
= 22.9°
(ii) A =
(
(
œžŸ
x 15 x 23 x sin61°
=150.9 ¡=
(b)
Cos ¢ = 0.937
¢ £œ 1((0.937)
= 20.4°
Cos is positive in the 1š¤ and 4¤¥ quadrant.
¢ 360° ¢ 360° 20.4° = 339.6°
(c)
∴ ¦ =. °
§¨
¦ f. C°
y = sin¢, 0° © ¢ © 360°
¢
0€
90€ 180€ 270€
0
1
0
-1
ªž ¢
360€
0
6
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Q9
(a)
7
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Q 10
Q 11.
(a) SD = √{
𝑥̅ =
𝑥̅ =
𝑥̅ =
∑ f( x− 𝑥̅ )2
∑f
}
∑ fx
n
(2 × 1 )+ (3 × 5 )+(4 × 4)+ (5 × 6)+ (6 × 10)+ (7×16)+(8 ×18)
1+5+4+6+10+16+18
379
60
𝑥̅ = 6.3
1(2−6.3)2 +5(3−6.3)2 + 4(4−6.3)2 +6(5−6.3)2 + 10(6−6.3)2 +16(7−6.3)2 + 18(8−6.3)2
SD = √
60
165
8
SD = √ 60
= √2.75 = 1.658312395
SD=1.7
(1dp)
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(b) (i)
Q 12
(a)
2
¯1)
4
- ¯12
2¯12 1 4 ¯1)
¯1)¯12
2¯131-¯´ €
¯1)¯12
2¯1-¯13´ €
¯1)¯12
9
1!´99
=1!1
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(b) (i)
(a) T "# T "# T (b) 2 12 2 10
µ 2 10
x -2 -1 0
1
y 6 8 10 12
2
14
T . ¶ "# T . ¶ "# T .
(ii)
(a) 4.6, 0 3, 18
m
m (-1€
12114.d
m
(12´4.d
((.d
q !
(b)
A A (
30 28 1 1
1
58 52
2
2
(
30 22 1
A 29 26
· !! ‘‹ =
10
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Page 81 of 139
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Page 83 of 139
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Page 84 of 139
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Page 86 of 139
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Page 87 of 139
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Page 88 of 139
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Q1
SOLUTIONS 2017 INTERNAL PAPER 2 (4024)
(a) (i) Det M = (3 5 2
22
3 – 10
22
3 10
22 10
3
12
3
(ii)
M ||
Adj M
3 2
M=
5 4
(b) (i)
Maize
E
! "
(ii) (a) 19 +2 +3 +11 + 5 +15 +9 + =70
S. Potatoes
(b) 19
19
5
9
11
6
3
(c) 11 + 15 + 5 +9 = 40
15
2
Cassava
Q2
(a) (i)
1=> Pick
4
12
3
12
1
4
11
3
W
5
12
2@A Pick
ii PSame colour
11
11
5
11
B
Y
4
5
2
11
3
11
11
3
4
11
W
B
W
Y
B
Y
W
B
Y
11
PWW PBB PYY
1
4
22
33
POSSIBLE OUTCOMES
WW
WB
WY
BW
BR
BY
YW
YB
YY
55
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(b) (i) (a)
(b)
(c)
DDDDD⃗
PQ
DDDDD⃗ OQ
DDDDDD⃗
PO
2p 4q
DDDD⃗
PX
DDDDD⃗
PQ
2
2p 2
4q
I J
DDDDD⃗
OX
DDDDD⃗ PX
DDDD⃗
OP
3
K
2p p q
2
2
3
K
2p 2 p q
2
I J
(iii)
DDDDD⃗
LM
DDDDD⃗
MS
PREPARED BY LOOBWE, A. N.
DDDDD⃗
ℎLO
K
K
ℎ P Q
2
2
KR
P
2
KR
Q
2
DDDDD⃗ LS
DDDDDD⃗
ML
KR
P
2
4Q KR
Q
2
KR
Q
2
T
4Q
KR
P
2
T
U
A.R
2
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Page 90 of 139
D
(b) 8.5 ab
Q3.
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(C) e
C
11ab
P
(C) ee
A
7ab
60°
120°
10ab
B
Note: The zigzag line on the boundary of the shaded region symbolise
that the line is not part of the solution.
(a)
Q 4.
a
2 b
Either
3
6 c
3
6 X Y63 423
22
6 X √36 24
4
6 X √60
4
6 √60
4
[ . or
]^
6 √60
4
. Develop love for mathematics, solve it, talk about it gossip it and you will pass it!! | MSc. Loobwe, A. N.
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(b)
2
V=
2.K3
2
=
=
(a)
2.K3
2
2.K3
2
637 20
12740
V = 13 343.03cl
(i)
r
r
mn
mo
1
3i
^
(ii)
49 441 147 20
Kii3j.ik
2
=
Q5
3.142 73 213 7 21 20
2
=
=
πR3 r 3 R rh
the n>q term
n>q
ar @
r
rst
(iii)
s@
[ ry1
v wx w
Sk
o |
{
o
{
3i z }
o
3i | {
3i ~
{
oo
€~€
~
{
20 ‚112‚ 2
‚112‚
4
ƒ„
26
4
[45
K
5. 5
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Page 92 of 139
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ˆ
K†~
j‡n
(b)
K†~
j‡n
‰†{
k‡~
k‡~
‰†{
K ŠŠŠ
j‹‹
Œ
Q6
k‹‹‹
‰ŠŠŠŠ
Start
Enter r
If r y 0 Ž
PRINT “Error r must be positive”
ELSE
Area = 3 square radius sinθ
Display area
END IF
Stop
Q7
(a)
(i)
A = 3 ’“ ”e•M
3
1.9 1.3 –e•130°
0.95
(ii)
1.0 —l
h3
s 3 t 3 2st Cos H
ℎ3
1.93 1.33 21.91.3 Cos 130°
h3
3.61 1.69 4.94 cos 130°
h3
√h3
šƒ
(iii)
5
A
3
3
4.85 3.175370792
√8.025370792
. „ —l
bh,
2.8ℎ
ℎ
A
0.95
0.95
0.7œb
∴ the shortest distance is 0.7 km
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Page 93 of 139
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Cos θ
(b)
0.666666666
θ
Cos 0.666666666
„°
3†n k
†ž3
(c)
3†n K
†ž3
3†n ž 3n †ž3
3†ž3†3
†ž3
2(x -2)
Q8
(a)
SD = Ÿ
∑ ¡ŠŠ¢ n
∑¡
̅ =
̅
̅
̅
∑ ¡Š
∑¡
23.1ž3‰‰.1ž213.1ž‚‰.1ž‰33.1ž33‰.1
‚1
ii
2ž3‰ž21ž‚ž‰ž3
11.7
SD = Ÿ
23.1.‰n ž3‰‰.1.‰n ž213.1.‰n ž‚‰.1.‰n ž‰33.1.‰n ž33‰.1.‰n
ii
SD = Ÿ ii
2K12
SD = √34.53
SD = 5.9
6
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Page 94 of 139
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(b) the remaining answers on the graph
7
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Q9
(a)
WY
(i)
(ii)
80 30
[[
¥
2‚i
XZ
110
2 3.142 3437
360
110
21598,108
360
2375,791.88
360
XZ
XZ
XZ
XZ
(b)
¦§
2 πR
6599.42..
55[[ rl (3 s.f)
¥
2‚i
YZ
θ
2 π Rcos∅
15 105
120©
120i
2 3.142 3437cos30
360i
120i
18,704.5102
360i
2244541.224
360i
6234.84
ª§
5 [ rl
( 3 s.f)
Q 10
(a) (i)
(a)
y
0
¬
(b)
]^ ¬
x 2 3x 3 x
]^ ¬
5
x 2 3x 3 x 3
x 2 3x 3 x 3
8
y
¬
2
. 5 ]^ ¬
53
2
. ]^ ¬
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Page 96 of 139
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(ii)
m
(a)
‡n ‡o
†n †o
3,0 0,24
24 0
0 3
m
24
3
m
(b)
l
A
A
2.2 3
3
A
†K
1.7 10 3 1.7 1 10
58 10 19.5 13.5
®
(b)
„
3
52 10
¯r°s± 3
1†
15x 1 2x 4
x 45x 1
5x 1 2x 8
x 45x 1
¬
¬ ¬ 9
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Page 97 of 139
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Q 11.
10
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Q 12
11
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Page 99 of 139
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Page 100 of 139
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Page 101 of 139
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Page 102 of 139
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Page 103 of 139
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Page 104 of 139
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Page 105 of 139
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Page 106 of 139
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Page 107 of 139
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SOLUTIONS 2017 G.C.E PAPER 2 (4024)
Q1
(a)
(i)
|k| = (10 2 11 2
20 – 22
20
22
Det k = 2
(ii)
A
|| Adj A
a 3 b 7 c 1
(b)
7 " #7$ 431
23
! ! ! 7 " √49 12
6
7 " √37
6
Either
7 √37
6
! + . Q2
or
-.
7
! √60
6
+ . /
01 (a)
01 0
01 1
00 020 0 0 32
3
(b)
(i)
425
625
n$
1
725
425
20n
n $ n$
100 n$
21n
90
20n 21n 90 100
n 10
9 Solve it, “Obvious is the dangerous word in mathematics” | MSc. Loobwe, A. N.
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Page 108 of 139
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(ii)
6 + 10, 10 + 10, 15 + 10, ...
16, 20, 25, …
r
$4
6
. :
< => S5 (iii)
= @ B
A
@
A
6 ? S6 C
11529
16 4096 1
4
11529
256
1
4
E
/ F /. :
(a)
Q3.
1NO Pick
3
10
7
10
25R Pick
2
9
F
F
FF
7
9
3
9
G
6
9
2
POSSIBLE OUTCOMES
G
FG
F
GF
GG
G
I
6
K
(i)
PGG 4 J
(ii)
PFG
M
I
GF 4 J
I
M
I
4 J
M4
I
M4
K
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(b) (i)
2y + 1= 7
2y = 7 – 1
2y = 6
y=3
(ii)
Q 4.
(a)
(a)
6+2= 8
(b)
4+1+2=7
(c)
6 + 8 + 7 = 21
M
7S $
$
S2M
MS2M $ 7S $
7S $S2M
MS 4S2J2T
7S $S2M
KU2V
U U2V
(b)
7
W$ 3x $
x M
2dx
2x
5
2
Y5M
25Z Y2M
125
10 8
22Z
4
135 12
V \9]^_ 3
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Page 110 of 139
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Q5
(a) (i) the diagram
(ii) 9 cm
Note: The zigzag line on the boundary of
the shaded region symbolise that the line
is not part of the solution.
R
(b) (iii)
10cm
(b) (i)
3cm
(b) (ii)
7cm
60°
P
Q6
(a)
Q
9cm
(i)
bbbbb⃗ de
bbbbb⃗
OB
(ii)
bbbbb⃗
bbbbb⃗
OE OD
bbbbb⃗
ef
g
h
bbbbb⃗
DE
1
bbbbb⃗
OB
2
bbbbb⃗
bbbbb⃗
AC AO
1
bbbbb⃗
AC
4
a
2b
V
n
1
bbbbb⃗ la
OE
2
bbbbb⃗
OC
1
la 2bm
4
1
1
1
a b a
b
2
4
2
1
1
1
a a b
b
2
4
2
2bm
:g
(iii)
4
bbbbb⃗
oe
CD $ bbbbb⃗
1
la 2bm
2
pn
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Page 111 of 139
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(b)
Start
Enter a, r
IF |r| r 1 stuv
Sw <
=
ELSE Display “sum to infinity”
END IF
Stop
Q7
5
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Page 112 of 139
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Q8
(a) SD = x
∑ zS S{ 1
∑z
x| =
∑ zS
∑z
7$I.7 2 TM$.7 2 IMI.7 2 T$.7 2 $TI.7 2 }7$.7 2 7I.7
$4T7
T}
7 2 T 2 I 2 2 $ 2 } 2 !̅ 42.6
7$I.7 T$.61 2TM$.7 T$.61 2IMI.7 T$.61 2T$.7 T$.61 2$TI.7 T$.61 2}7$.7 T$.61 27I.7 T$.61
x
T}
=x
M4$T.T}
T}
SD = √63.01
SD = 7.9
6
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Page 113 of 139
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(b)
7
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Page 114 of 139
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Q9
(a)
(i)
0.6, 10 2,5
M=
€1 €
S1 S
724
4.6
=$
7
.T
=
M = 10.7
(ii)
U V
(a)
U V . /
(b)
(iii)
A = $ 9
$
8 1
= $ 17
8.5
$
-.
8
U -. U . V
-.
U . :
3 1
11
5.5
: \9]^_ (b)
R€
RS
2x 3
22 3
43
m 1
y x
c
y 2$ 32 4
y 4 6 4
y 6
2, 6
y x
6 2
8
c
c
6 2 c
C 8
„ U /
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Page 115 of 139
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Q 10
(a)
(i)
†‡5 ˆ
=
†‡5 ‰
46°
36°
x° 98°
x° 180°
36.5
r
Sin 46° Sin 98°
r Sin 46° 36.5 Sin 98°
36.5 Sin 98°
Sin 46°
r
r 50.2 m
∴ ‹Œ . 3
A abSinθ
$
(ii)
$ 36.5 50.2 Ž36
V/. ‘
bh
$
(iii)
538.5
1
50.2 ℎ 538.5
2
21.5ℎ 538.5
” . ‘
(b)
Ž • 0.6792
• Ž
0.6792
• 42.8°
Sine is positive in the first and second quadrants
180° • –
180° 42.8° –
– 137.2°
∴ — :. /°
(c)
9
1 ˜™
T
}
˜
-.
š 2p$ q
˜˜˜
T

ž
}
˜
— VK. °
$
˜
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Page 116 of 139
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Q 11.
10
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Page 117 of 139
Q 12
S
T
(a) (i)
PREPARED BY LOOBWE, A. N.
S27
$
12x 4x
60
12x – 4x 60
8x
60
8
8
∴ ^Ÿ Ÿ ]¡Ÿ^ -¢ £¤ K. ¥3
x = 7.5 cm
(ii)
V=
M
πR$
r$
R rh
= M 3.142 4$
=
M.T$
M
16
144
12$
4 12 15
48 15
= 3 .142 208 5
= 3 .142 x 1040
= 3267.68
V = 3270 c3V
(b) (i)
N
Q
P
R
S
80° N
85° S
70° E
10° E
N
(ii) (a)
QR =
¨
M64
2πR
θ 80°
= 165°
85°
67
QR = M64 2 3.142 3437
67
QR = M64 21598.108
QR =
M76M,6}I.}$
M64
QR = 9,899.13
QR = 9,900 nm
(b)
C = 2 πR cos ∅
= 2 x 3.142 x 3437 x Cos 85
=21, 598.108 x Cos 85
11
C = 1,882 . 40
C = 1,880 nm
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Page 118 of 139
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Page 119 of 139
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Page 120 of 139
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Page 121 of 139
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Page 122 of 139
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Page 123 of 139
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Page 124 of 139
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Page 125 of 139
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Page 126 of 139
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SOLUTIONS 2016 INTERNAL PAPER 2 (4024)
Q1
(a)
Det Q = 3 4 2
(i)
2
12 – 2
2
12
2 – 12
2
10
2
5
A
(ii)
2
||
Adj A
3 2
Q=
5 4
4 2
5 3
2 1
Q
a
(b)
1 b
2
x
x
c
2 % '2 417
21
x
Either
x
+
2 % √4
2
28
2 % √32
2
2 √32
2
or
, . .,
/0
2 √32
2
x
+
1. .,
(a) (i)
E
Sports
News
10
8
1
7
b % √b 4ac
2a
x
Q2
2
7
4
9
5
5
(ii)
(a) 5
(b) 10
(c) 8
9
5
5
7
34
35
Music
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Page 127 of 139
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(b)
1JK Pick
2MN Pick
P
0.6
0.6
PP
P
0.4
0.4
POSSIBLE OUTCOMES
N
0.6
0.4
N
P
N
(i) P(one negative other positive)
ii Pboth negative
NN
5. 1?
PN
NP
NN
PN
0.24
5. 4.
NP
0.24
2
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Page 128 of 139
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(a)
Q3.
Start
Enter r
Is
“Error message”
re-enter r
U V 0?
Enter h
Is
ℎ V 0?
“Error message”
re-enter h
Display V
Stop
(b) (i)
(a)
(b)
OOOOO⃗
AB
OOOOO⃗
AO
,Q
OOOOO⃗
OD
OOOOO⃗
OD
3
OOOOO⃗
OB
?R
OOOOO⃗
OA
OOOOO⃗
OA
OOOOO⃗
AD
1
OOOOO⃗
AB
3
3a
1
3a
3
3Q
3R
3a a
2b
6b Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
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Page 129 of 139
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OOOOO⃗
BC
(c)
OOOOO⃗
BO
6b
OOOOO⃗
OC
2
OOOOO⃗
OA
5
6b
OOOOO⃗
BE
(ii)
2Y3aZ
?R
OOOOO⃗
hBC
h [6b
6hb
?Q
6a
\
5
6
ha
5
1
? [ Q R\ ]
C
Q 4.
(a) (i) refer to the diagram
(b) (i)
(b) (iii)
7cm
7cm
A
(a)
7cm
(ii) ⟨CAB
B
(b) (ii)
60°
Note: The zigzag lines on the boundaries of the shaded region symbolise that the
lines are not part of the solution.
Q5
(a)
^
^ _ ^
^ _ _
4
^
^^`
1
`1
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Page 130 of 139
(b)
(i)
^ PREPARED BY LOOBWE, A. N
^
^ ^ – 3 3 – 3
3 3
6
6
6
6
10
6
since x
1
1 1 1
1
5
3
3
1,
e
term a
f
Q1
3
3
,
h
Sg
r
1
a
(iii)
9
9 1
∴
(ii)
9
i
T
T
4 8
j
3 3
4 3
3 8
1
2
8
3
1
1 2
8
3 1
2
8 2
3 3
16
9
1
5
kg
1
l
m
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Page 131 of 139
1
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Q6
(a)
Nn
3x 3x
No
= 3(2 32
= 12 – 6
m
6
y
2 8– 6
y
2
y
2
2
Nn
No
q
q
r
C
(b)
2
x
2
C
C
C
1
,∴ t
+
?
l
,
3x 3x
Nn
No
0
0
3xx – 1
0 = 3x 3x
0
x– 1
y
x y
0 0
x
y
6
1
or
or
0
y
1 y
1
y
,
x
x
0
0 values of x
x
3
1
2
1
∴ 5, 5 Quv 1, 3 Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
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Page 132 of 139
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Q7
(a) SD = w
∑ yooz _
∑y
x{ =
∑ yo
∑y
re ` e ` fe ` e ` e|e ` eee
r``f``e`e
f}
}}
+{
3.. 3
SD =
w
w
ref._ `ef._ `fef._ `ef._ `e|ef._ `eeef._
}}
qrrq
}}
√167.76
SD = 13
7
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Page 133 of 139
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(b)
8
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Page 134 of 139
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Q8
(a) (i) to find the centre of enlargement, join any corresponding two points between the object and the image.
The point of intersection between the drawn lines is the centre of enlargement.
∴ the center of enlargement is 1, 2
(ii) Scale factor

(b)
3.2
1.6
‚
„
3
ƒ
1 3
4 4
‚
3‚
„
3„
1 3
1 5
4ƒ
4ƒ
4
1 ----------------------- i
3
1
4
5
------------------- ii
------------------- iii
------------------- iv
Solve the equations formed using simultaneous method
Equation I and ii
‚
3‚
2a
a
1
1
4b
b
0
4b
0
4ƒ
4ƒ
2
Equation I and ii
1
3
2„
„
1
‚
∴ †he required matrix is „
„
2
ƒ
1
3
5
1
4
4
3„
2
4
4
4
1
4
1
5
1
|
(c) ∆ ABC is mapped onto ∆ A B C by anticlockwise rotation of 90° about C(0,0) or by clockwise
rotation of 270° about C(0,0).
3 0

(d) (i) comparing two matrices ‚ˆ 0 1
0
transformation is ,.
(ii)
9
3 0 1 3 1
0 1 4 4 5
3 1 0 4 3 3 0 4
01 14
03 14
3 0 9 0 3 0
0 4
0 4
0 5
3 9 3
4
4
5
0
, we see that the scale factor of this
1
3 1 0 5
01 15
∴The coordinates are ‰4 4, 4, Š4 m, 4, ‹4 ,, Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
PREPARED BY LOOBWE, A. N.
Page 135 of 139
PREPARED BY LOOBWE, A. N
Q9
(a)
70°
(i)
C
(ii)
2πRCosθ
1,.55 u
AD

f}°
‰‘
V
(b)
120°
2 π 3437 Cos50°
AD
(iii)
50°
(3 S.F)
πR
120°
π 3437
180°
1.55 u
(3 S.F)
blh
1.2 0.9 10
10.8m
1 1000 litres
10.8 x
x
10,800 litres
∴ “]” u•R”0 /– —˜“0”™ –/0 –•”— ˜u “]” “Quš ˜™ 15, .55 —˜“0”™.
(c)
a
h
12
b
r
144
' a
x 5
radius
S. A
S. A
x
c
x
l
13
13cm
12cm
169
√125
5cm
πrl
r
3.142 513
15.7118
5
3.3. l.›3
10
Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
PREPARED BY LOOBWE, A. N.
Page 136 of 139
PREPARED BY LOOBWE, A. N
Q 10
(a)
(i)
a
b
c 2bcCosA
a
25
9 30Cos110°
a
5
a
34 10.2606043
a
'a
‚
44.2606043
√44.2606043
6.7 km
∴ Š
(ii)
A
A
3 253Cos110°
?. lš
abSinθ
1
5 3 Sin110°
2
A
7.05 km
1
bh
2
A
1
6.7km h
2
7.05km
3.35 km h
6.7kmh
6.7km
h
7.05km
7.05km
6.7km
2.1km
∴ “]” ™]/0“”™“ v˜™“Qu›” ˜™ 3. 1 š
(b)
tanθ
0.7
θ
tan 0.7
θ
34.9920202
θ
35°
∴ ™˜u›” “Qu ˜™ /u—t ž/™˜“˜Ÿ” ˜u “]” •v0Qu“, ¡ ˜™ ,°
(c)
r¢_
}h_
j
e¢_
eh
17k 5a
20a 51k 11
r ¢ ¢
}hh
eh
e¢¢
1
13Q
Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
PREPARED BY LOOBWE, A. N.
Page 137 of 139
PREPARED BY LOOBWE, A. N
Q 11.
12
Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
PREPARED BY LOOBWE, A. N.
Page 138 of 139
PREPARED BY LOOBWE, A. N
Q 12
(b)
o
o`
2 3x 1 1 2x 1
3x 12x 1
6x 2x 2 1
3x 12x 1
13
4+ ,
3+ 1,+
1
Maths the Brain cleanser, so let’s detox our brains | MSC. Loobwe, A. N.
PREPARED BY LOOBWE, A. N.
Page 139 of 139