Suggested Solutions to Past CXC Examination Papers
2005-2010
Compiled By Experienced CXC Instructors
Page
Solution 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
[email protected]
Mathematics General Proficiency May 2005
1.
a) = =
21
10
5
3
63-50 15
-- 13/15
b)
i)
ii)
A:
12.50 x 3 = $37.50 'v
B:
$33.90 --
c:
--
D:
15 X 100
=$li.25
2
$31.00 $6.20
=5 108.28 = $16.24
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
(Difference: $6.90 - $5.88
in this case.
= $1.02)
Amanda made a Rmfit
IIPage
a)
2
ab(5a
ii) iii)
(3k-l) (3k+ 1) 2yL4y-y+ 2 2y(y - 2) -l(y - 2) (2y -1) (y - 2)
b)
6xz - 8x + lSx -20 6xz + 7x - 20
c)
i)
2(x - 3) = 2x-6
ii)
x + x - 3 + 2x - 6 = 39 4 x - 9 = 39 4x =48 x = 12
i)
3x
l
3.
+ b)
i)
a)
+
x = 24 4x=24 x=6
~ students
b)
·Solving the equation was not necessary in this case.
take both Music and Drama.
=
ii)
Student who take drama only 7 + x :. 7 + 6 = 1.3. students take only Drama.
i)
y- 5 2
2 = -(x + 3) 3
y='3x+2+S 2
y='3x+7
y=mx+c or
2
5=-(-3)+c 3
5
= -2 + c
c=7 ii) 21Page
2 x - 3y = 0
2
Y = -x 3
+7
-3y = -2x y=Y m
=
-2x -3 2 -X 3
=!.3 .:for the line 2x - 3y = 0 therefore it is parallel to y = !. x +7 3
(Parallel lines have the same gradient)
4.
a)
Area of small pizza
= rr(7.5)2 = 176.79 em-
Area of medium pizza = rr(152) = 707.14em2 707.14 176.79
=4
The medium pizza is more than twice as large as the small pizza. It is four times as large.
b)
· SIzeo
f1-0
3
f me diIUmplzza=--= . 707.14 3
23571 . em 2
ratio of area to cost of medium pizza =
ratio of area to cost of small pizza
235.71 15.95
= 14.78 em2/$
= 176.79 = 13.65 12.95
em2/$
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-,
. ,:
"
, . ,;.:'
:ii
,n
.., "Et','
':~. ·c.
,
,
.
.'
1
~. ,l
I: .;
J'
:" ';', I,
I : .
"
-
....
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,,>
"'"
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:,';'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) (4
ii)
I
103
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
71Page
b'
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
25
+ ~ = ../36/25 1
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
Distance travelled = ii)
0
22
of graph
(40 + 50)
= 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
1h
(7.5) (4.5)Sin
.
= 13.5
0'
13.5
S InO' =--
16.875
0"' 0"'
b)
ii)
1h
i)
a)
= sino! 0.8 = 53°
(15) (9) sin 53 = 67.5 sin 53 = 54 cm2
=
180 - 136 44° SJM = 442
= 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.
ii)
AB = 3K
BD=BA+/iD = - 3K - 3y 1 ---->
---.-.
Hi)
DP=--BD 3 1
:::: -"3 (-3x - 3y) =x+y b)
AP=AD+DP +x +y =x 2y
= -3y
c)
PH=PD+DE = -x - y
and 3
+ -2
X
::::Yzx-y
=>
fiE
=
liP+ PH
=X 3
-
::::-x -
2
2y
+
1
-x - Y 2
3y
AP=2PH
Therefore. A, P and E are collinear. lllPage
A
d)
DA == 3 G)
=G)
IDAI = .../3
2
+ 32 = ..JI8
AB=AP+PE =X
-
2y
+ lh X
-
Y
3
= 2"X - 3y
= ~(~)- 3G) =
G) - (D
==
(~3)
IABI = J 02 + (-3)2 DE
=~ =3
=~ (~)
= (~)
IDEI == .../3
2
+ 02
=~ =3
A
Therefore; triangle AED is isosceles.
o '-----t+----"
12
I P age
E
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 =~
(X)Y
= ~ (-130)
-5
-5
(15
-7
(-45 - 85)
(X)Y
X
b)
1 -5
21
+
34
55
= 26
y =:.11
(-1 1°) 0
i)
R=
ii)
N= (-10 - ~)
iii)
T= (~3)
• 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)
