Suggested Solutions to Past CXC
Examination Papers
2005-2010
Compiled By Experienced CXC Instructors
Solution
Page
Paper 2 - 2005
1
Paper 2 - 2006
15
Paper 2 - 2007
31
Paper 2 - 2008
47
Paper 2 - 2009
63
Paper 2 - 2010
80
Questions? Comments? Email mr~norbert@live.com
Mathematics
General Proficiency
May 2005
1.
a)
21
= 5
=
10
3
63-50
15
-- 13/15
b)
i)
ii)
A:
12.50 x 3 = $37.50 'v
B:
$33.90
=$li.25
--
c:
--
=5
D:
15
X
100
108.28 = $16.24
2
$31.00
$6.20
6 x 0.75 = $4.50
6 x 0.40 = $2.40
Total made = $6.90
Total spent = $5.88
*NB:The exact profit was not
necessary
in this case.
(Difference: $6.90 - $5.88 = $1.02)
Amanda made a Rmfit
IIPage
a)
2
ab(5a
ii)
iii)
(3k-l)
(3k+ 1)
2yL4y-y+
2
2y(y - 2) -l(y - 2)
(2y -1) (y - 2)
b)
6xz - 8x + lSx -20
6xz + 7x - 20
c)
i)
2(x - 3)
= 2x-6
ii)
x + x - 3 + 2x - 6 = 39
4 x - 9 = 39
4x =48
x = 12
i)
3x + x = 24
l
3.
+ b)
i)
a)
·Solving the equation was not necessary
in this case.
4x=24
x=6
~ students
b)
take both Music and Drama.
=
ii)
Student who take drama only
7 + x
:. 7 + 6 = 1.3. students take only Drama.
i)
y- 5
2
2
= -(x
+ 3)
3
y='3x+2+S
2
y='3x+7
y=mx+c
or
2
5=-(-3)+c
3
5
= -2 + c
c=7
ii)
21Page
2 x - 3y = 0
2
Y = -x
3
+7
-3y = -2x
y=Y
-2x
-3
= -X23
m =!. .:for the line 2x - 3y = 0 therefore
3
it is parallel to y = !.x + 7
3
(Parallel lines have the same gradient)
4.
a)
Area of small pizza
= rr(7.5)2
= 176.79 em-
Area of medium pizza = rr(152)
= 707.14em2
707.14
176.79
=4
The medium pizza is more than twice as large as the small pizza. It is four
times as large.
b)
·
f1-0 f me diIUmplzza=--=
.
707.14
SIzeo
3
3
23571
. em 2
ratio of area to cost of medium pizza = 235.71 = 14.78 em2/$
15.95
ratio of area to cost of small pizza
= 176.79
= 13.65 em2/$
12.95
Therefore, it is better to buy a slice of medium pizza, rather than a small
pizza.
31Page
5.
The graph below shows the answer for 5(a), 5(b) (i) 5(b) (ii) - lern: 1 unit on both
axis.
_J __
c.
..
,
1; ;
. ,
~,
~~"
"
;\
,
.
"
'
" ..
,
.; ....,
"
..
.
.
.'
;',
.
.
"
,
.'
.
,
'.
:
..
.,.. ..
.1.· ..
'
.
:
! ; ..~. ; ,
'
.'
.'F';'
"
....
·
I'
.: ' I·
~
.'
./
"
. "
..
'\'
s :
.
. "r:,
F,
'.
I,·:'
.i
',:
I'
,"
. -: ,
,
', ..
'"
;
':
,:.. ',',y-,
. ,:
"
,
.., "Et','
. ,;.:' ':~.
:ii
,n
·c.
,
,
.
.'
1
~.
,l
J' I: .;
:" ';', I,
I :
.
"
-
....
",
,,>
"'"
.• ,.
"
'.
:,';'i, i,':'
.
'I',
.
",:
....
'.,
4\Page
'
".,
cT,
',: ..
, ;t.
"
,;
! ,
:'
"
Hi) Composite transformation
c)
1.8
tan a =2
a = tan"! 0.9
a = 42°
6.
a)
i)
EAD
= 57°
Reason: Alternate angles of parallel lines
Then x = 180 - (57 + 80)
= 43°
b)
ii)
DAB=108-57
= 51°
Since angle of a quadrilateral add up to 360°.
Then y = 360 - (51 + 57 + 90)
= 360 -198
= 162°
i)
32 + (-3)2 = 9 + 9
=18
ii)
=
Ifx
Yzy + 5
x-5=Yzy
y=2(x-5)
r1(x) = 2x -10
rl(6) = 2(6) -10
12 -10
=~
=
iii)
fg(x) = 1fzX2 + 5
f g(2) = 1fz(22) + 5
=Yz(4)+5
=2+5
=Z
SIPage
7.
a)
Cumulative Frequency Table showing the height of applicants
Height/ern
Cumulative Frequency
$155
10
$160
65
$165
170
$170
280
$175
360
$180
390
$185
400
Cumulative Frequency Curve of Heights of Applicants
y
400
3!lO
300
J
250
~
Q
200
',.~"
150
100
50
!SS
16S
110
H.ipt
61Page
(cmj
,17,S;
1&0
lSS
x
b)
8.
a)
i)
275 applicants
ii)
167 em
iii)
162.5 em
iv)
95
400
i)
ii)
I
103
(4 X 72) + (3 x 6) + 2
216
(8 X 112) + (3 x 10) + 2
1000
~------~------------------------
iii)
b)
1,--_n_3 --l.1_C(_n _-2_)x_C_n_+1_)2_+_(3_X_n~_2_1,--
(a - b) (a - b) (a + b) + ab(a + b)
= (a - b) (a- - b2) + a-b + ab= a3 - ab- - a-b + b3 + a2b + ab= a3 - ab- + ab- - a-b + a2b + b3
= a3 + b3
__
n3__
Difference
of squares
• (a - b)(a
+ b) = aL b'
71Page
9.
a)
5xz + 2x-7
5(xZ + ~x) - 7
5
5 [Xl + ~X + (!)Z ~ - 7 - 5(!)2
5
5
1
Sex +_)25
b)
c)
5
7-1
5
1
i)
- 7-
ii)
When X + 5 = 0, then x = - '5
5
1
1
S(x + !)2 = 36
5
5
(x + !)2 = 36
5
X + ~=
1
25
../36/25
6
x=--+5 - -5
E"Ith er x = - -1 + -6
5
5
1
6
5
7
5
x=----
or
-
=1
5
d)
v
x
81Page
10.
a)
i)
40-0_'J£.77/2
£".,J,LL
m
15-0
--
s
== gradient
Acceleration
of first portion
of graph
0
Distance travelled = 22 (40 + 50)
ii)
= 10(90)
= 9.illlm
20
b)
c)
= 6km/h
i)
12km
2h
ii)
He took a rest or stopped.
iii)
~=8km/h
i)
x=6
ii)
x~6
1.5
5
y>-x +5
- 8
"evaluate ::-point in the region for
each line.
1
y<-x+5
-6
li.
a)
i)
x
7.5
"Sine rule:
triangle.
pL-
"h
ab sin C
= Area of
~ Q
91Page
(4.5)Sin 0' = 13.5
1h (7.5)
.
13.5
S InO' =--
16.875
b)
0"'
= sino! 0.8
0"'
= 53°
ii)
1h (15)
(9) sin 53 = 67.5 sin 53
= 54 cm2
i)
a)
180 - 136 44°
SJM = 44-
=
2
= 22° (base angles of an isosceles triangle)
b)
=
SJM = JMK
22° (alternate
:. JKM 180 - (124 + 22)
180-146
34°
=
=
=
ii)
a)
MJ
angles)
50
--
sin 136
sin22
MJ =9.2..Zm
b)
JK
92.7
sin 22
sin 34
--=--
JK =Qllm
G
12.
a)
17°N
10 I P age
A-Antigua
B - Belize
G - Greenwich
E - Equator
Meridian
b)
C = 2 x 3.14 x 6400
= 40192
= 40192 cos17
= 38423.6
1
= 38423.6 (~)
360
::::2776km
c)
1 = 40192 (~)
360
::::6140km
13.
a)
i)
AB=DC
Therefore. AB = 3K
ii)
BD=BA+/iD
= - 3K - 3y
1 ---->
---.-.
Hi)
DP=--BD
3
1
:::: -"3 (-3x - 3y)
=x+y
b)
AP=AD+DP
= -3y + x + y
=x 2y
c)
PH=PD+DE
and fiE =
3
liP+ PH
1
= -x - y + - X
= X - 2y + -x - Y
::::Yzx-y
::::-x -
2
2
3
2
3y
=> AP=2PH
Therefore. A, P and E are collinear.
lllPage
A
d)
DA == 3 G)
=G)
IDAI = .../3 + 3 = ..JI8
2
2
AB=AP+PE
= X - 2y + lh X - Y
3
= 2"X - 3y
= ~(~)- 3G)
=
G) - (D
== (~3)
IABI = J 02 + (-3)2 = ~ = 3
DE =~ (~)
= (~)
IDEI == .../3 + 0
2
2
=~
=3
A
Therefore; triangle AED is isosceles.
o '-----t+----" E
12 I P age
14.
a)
i)
(2) (15) - (7) (5)
= 30 - 35
= -5
=> M is a non-singular matrix
ii)
M-I = 2. (15 -5)
-5
-7 2
iii)
iv)
(X) =
0)
(10 1
Y
1 (15
-5 -7
- 85)
(X)Y = ~-5 (-45
21 + 34
(X)Y = ~-5 (-130)
55
y =:.11
X = 26
b)
i)
R= (-1
ii)
N= (-10 - ~)
iii)
T= (~3)
0
1°)
• The general matrix of
rotation for iT' in the
anuclockwlse direction
about the
origine:,:! -::::)
13IPage
iv)
RN=(~1
~)(-~
= (~
-~)
pI = (~
_~)
(161)
= (-1~)
pl = (6. -11)
(161) + (-;3) = (136)
P" = (-g _~) (136)
=
(-=-136)
P" = (-3.
141 P age
-16)
_°1)