Suggested Solutions to Past CXC Examination Papers · Suggested Solutions to Past CXC Examination...
Transcript of Suggested Solutions to Past CXC Examination Papers · Suggested Solutions to Past CXC Examination...
Suggested Solutions to Past CXC
Examination Papers
2005-2010
Compiled By Experienced CXCInstructors
a)
b)
I"c))
MathematicsGeneral Proficiency
May 2006
i) 151.29 - 0.082 = 151.208
ii) 1.5..<L(2 s.t)
i) p = $40,000 - $(E.. x 40,000)100
= $40,000 - $4,800= $35.200
Total depreciation of car:25,000 - 21,250 = $3,750
= 3750 x100q 25000
= 15%
ii) (Depreciation of taxi) in year 2 = E.. x $35200 = $4,224100
Value of Taxi after 2 years = $3,5200 - $4,224 = $30.976
i) GUY$:US$1 : 0.01
60000: x GUY$60,000 x 0.01 = US$600
ii) US$: EC$0.37: 1925: x
$925 = EC$2.5000.37
15 I P age
2. a)
b)
c)
16 I P age
x-3 _ x-2 = 5 (x-3)- 3 (x-2)3 5 15
5x-15-3x+ 6=
152x-9
= 15
i) a) x(x - 5)
b) (x -9)(x+9)
ii) a2 + 4a factorized= a (a + 4)
Simplifying a2 + 3a - 4 = a2+ 4a - a - 4= a(a + 4) -l(a + 4)= (a - 1) (a + 4)
a (a+4)~ -"-----'---(a-l)(a+4)a=-a-I
i) 2x + 3y = 175 (1)4x + y = 125 (2)
ii) 4x + 6y= 350-4x+ y=125
5y= 225225y=-
5Y= 45 One CD cost ill
Equation (1) x 2:Equation (2):
Therefore,4x + 45 = 1254x = 125 -454x=80
80x=-4
X = 20 One cassette costs.$2.Q
3 a)
b) i)
i)
ii)
iii)
ii)
iii)
LNK= 40°Angles of an isosceles triangle*
angles of a triangle sum up to 180°therefore, <KLN= 180 - (40 + 40)
= 100°¢ <NLM= 140° -100°
= 40°
KNM = KNL + LNM= 40° + 70°= 110°
• Angles of an isosceles mangle.
u
18 - X + X + 15 - X + 3x = 39.or 33 + 2x = 39
2x = 39 - 332x= 6
6X =
2X =~
171 Page
4. a)
1lCft\
b) 7em
e) 7 + 5 + 8 = 20 em
d) - See Diagram -
e) 4.3 em
f) Area = Yzbh= 1fz (8) (4.3)= 17.3 em-
s. a) a =-2
b) x =-1
e) (I, -4)
18 I P age
b=4
x=3
6.
d) { -1, 0, 1, 2, 3}
e) using (3, -1) and (2, -3)
. -3+1gradient = --2-3
-2=
-1
a) N ex + 7)km
H
13 km
b) X2 + (x + 7)2 = 132X2 + X2 + 14x + 49 = 1692X2 + 14x -120:= 0X2 + 7x - 60 = 0
c) X2 + 12x - 5x - 60 = 0x(x + 12) - 5(x+ 12)= 0ex - 5) ex + 12) = 0Therefore, x - 5 = 0
"Choose two points on the tangent.
F
• Divide the equation by 2.
• x + 12 = 0is an invalid solution
x=5km
19 I P age
7. a)
20 I Prig P
d) 5COS£} =-13£} = C05-12.
13£} = 67.4 :. Bearing ofF from G = 067.4°
Frequency Mid-Internal Values (kg)
2 7
29 12
37 17
16 22.*
14 U'
2 32'
b) x = -,2 (:.....;7):.....;+_2_9(.0..1.....:2):....+_3_7.0..( 1.....:7):....+_1_6.0..(2.....:2 ):....+_1_4(.0..2.....:7):....+_2..::..(3--<;.2)100
i)= 17.85 kg
ii) Frequency Polygon for the Gain in Mass of the Cows
5 3$
Mass {I<g"j
c) P b bility 16+14+2ro a II =----100
= .0.32.
21 I P age
8. i) Drawing showing a sequence of too ~ .
" II ~~ .•
--+-r t i' . i
. i J
: l > i I
• l', .. ; ;' +1++-1..•.. • i'
It' I'. .., i.
i.• i. J
, ~ !
.+
ii)4 4x5x2
7 7x8x2
22 I P age
. i!1
111 ..
H-f+H-H.
40
112
! •
. .II .., .; ;
~.H+i. , .-t~~+'! J'
..: :
.• 1
:~
b) i) n I r = n X en + 1) x 2
·~::::;100,n' +n-110 =0n' + 11n -10n-110 =n(n + 11) -10(n + 11):
I s= 10 IlOX llx2 220 (n-10)(n+11):O
ii) ;n=10
.; f
! •
J. a) X2 = X + 2X2 - X - 2 = 0X2 - 2x + x - 2 = 0x(x - 2) + 1 (x - 2) = 0(x + 1) (x - 2) = 0
x =:.l or x=~
b) i) Length of wire = Perimeter of square + Perimeter of Rectangle= 4x + 6 + 2l
ii) 4x + 21+ 6 = 324x + 2l = 262x + I = 13
"Divide by 2.
I = 13 - 2x
Hi) S = X2 + 31S = X2 + 3(13 - 2x) -Area of square e Xl
S = X2 + 39 - 6xArea of rectangle ~ 31
S = X2 - 6x + 39
iv) X2 - 6x + 39 = 30.25X2 - 6x + 8.75 = 0
x = 6±.J (-6)2 -4(1)(8.75)
2(1)6±"'36-35
x=2
6±1X=-
2
Either x =? 5or X=-2 2
~i) X + y:5 60
ii) y;;:: 10iii) y:5 2x
iv) y = 60 - x y = 2x
23 I P age
(0,0)(10,20)
(0,60)(60,0)
1 ' ~
1. i ': ~.~B:Broken lines must beised for inequalities..olld lines were only usediere for clarity.
;;", l
,
r.\,' j.";j: ',: ..
;~II, :
, : :i
." ; 1
.., ;!
' .
.;~[XIXrxIXlXrxlV:,\:'IAIA ,,:'0 :
1/ . ., r . r'\:!, ,: :..,
i , ~.. !
'.; , .. ' ~. :.
v) Total fees = 6x + Sy
vi) (20,40)(50,10)(5,10)
24 I P age
vii) AT(20, 40) Total Fees = 6(20) + 5(40)= 120 + 200=$320
At (SO,10) Total Fees = 6(50) + 5(10)= 300 + SO= $350
At (5,10) Total Fees = 6(5) +5(10)= 30 + 50=$80
Maximumfees charged = .$.3.5.fi
11. a) i) w
T
~-~~--~--~----------~F28m
ii) TFtan 40 = 28
TF= 28tan 40TF= 23.5m
WFtan 54 =-28
WF = 28 tan 54
25 I P age
WF = 38.5mTW = 38.5 - 23.5
= 15J:n
~ i) aCE = 90°(angle between radius and tange ::a tact)
ii) BAc = 70° (angle between a ord = angle in thealternate segrnen )
iii) BOC = 140° (angle at the centre = (7 e at circumference)
iv) BVC = 180 - (70 + 70)= 40° (base angles of an isosceles e are equal and angles
of a triangle add up to 180=
12. a) i) HF2= 4.22 + 62- 2(4.2) (6) cos 70HF2= 17.64 + 36 -17.24HF2= 36.4HF =ji.cm
ii) A = ab sin rr= (4.2) (6) sin 70 'Z['h absin a}
=absina
= 23.68 em-
26 I P age
b) i)
s
ii) a) see diagram
b) r = R cos (J
= 6370 cos 41= 4807.5 Ian
C = 2nr= 2(3.14) (4807.5)= 3ill!l1 km
c) S = (30191) (:680)
=.6..Sil km
Greenwich Meridian
41°N
Equator
27 I P age
13. a)
281Page
<. J.,
.,' ;
•..,~.~~~~~'t'. _~n" j ••.•• >-;
~~ t ..."t-1 ~~ - ,~.-6,,~~T·:--J I • \
J 1. \;, ~ •
f 1t si
; .
:t: .t t
,I .ti ';
b) -+ _ (6)OA 2
~=(~)
i)
ii)
iii)
c) i) ~=-+--+OB OA AB
= (~)+ (~)= (~)
--+=~--+OG 08
=~(~)=G)
G(3,3)
ii) ~::::--++--+AG AO OG
::::(=~)+ (~)::::(~3)
~=-+--+GC GO oc
= (=~)+ (~)
= (~3)--+::::--+::::> A, G and C lie on a straight line.AG GC
14. a) i) 2x + 3::::92x::::6x::::~
ii) M·l ::::!e -;)9 1
Hi) ~e - 3) (2 3) ::::~(6 + 3 9 - 9)9 1 2 -1 3 9 2 - 2 3+6
29 I P age
b)
30 I P age
= ~(~ ~)
= (~ ~)=1
i) a)
ii)
b) (:4 _53)
(~~)(! ~)= (~4 -~)
(2P + 4q Sp + 3q) _ ( 2 5 )2r + 4s Sr + 3s - -4 - 3
2p + 4q = 2Sp + 3q = 5-lOp+ 20q = 1010 P + 69 = 10
14q= 0q=O
Sp + 3(0) = SSp = S
p=l
(p q) _ (1 0)r s - 0-1
2x2 z x i
C :) must be written
2r + 4s = -4Sr + 3s =-3lOr + 20s = -2010 r + 6s = -6
14s = -14s =-1
2r + 4(-1) =-42r- 4 =-4
2r=0r =0