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