Suggested Solutions to Past CXC

Examination Papers

2005-2010

Compiled By Expertenced CXCInstructors

MathematicsGeneral Proficiency

May 2007

1. a) 13.69 - 4.8 = B...B.9.

b) i) teachers: students1: 30x: 1200

1200- = 1.0teachers30 -

ii) 3- x 1200 = 720 students5

iii) Students who own computers = 1200 - 720=480

39" x 4-80"=144 students have play stations:wtr 1

144 3- -1200 25

2. a) i) 8(4) (8)-- 4= 32 - 2=3..Q

ii) 2 * 30 = (2) (30) 30-2

= 60 -15=1.5

b) 5JYX J!::. = _5_

¥! 4p 12p

31 I P age

c)

3 a)

32 I P age

i) Sa + 3b = 1054a + b = 63

ii) Sa+3b=10S12a +3b = 1897a = -84a = -84a =12

4(12) + b = 63b = 63 -48b =15.

i) a) Hockey and Tennis

b) Hockey, Tennis and Squash

c) Hockey

ii) Members who do not play Hockey bet play squash.

b)

4. a)

i) a) b)

G

Triangle PQR in which QR = 8.5 em. PQ = 6 em and PR = 7.5 em

ii) a) 60° (found by measuring with a protractor)

b) 5.2 em

i) 3 em rep. 3 x 4000 em= 12000 em= 12.llJ:n

33 I P age

ii) 5.8 x 4000em = 23200 em = 2..32..m

iii) 1 em- = 1 em x 1 em= 4000 em x 4000 em= 40mx 40m= 1600 m-

iv) 27 ern- = 27 x 1600= 43200 m2

b) i) Vol. = Area x Length=>960 = AB2 x lS

AB2 = 96015

AB2 = 64AB =.8....rul

ii) 4(15 x 8) + 2 (64)= 480 + 128= 608 em-

5. a) 1y ex.-X2

IeY= X2

/'

"-

b) i) ~=232

k=9x2=18

ii) 18r=-1.82

r =.5...6

341 Page

· Hi)

f2=~8

f=..J2.2S=1.5

c) Y _ 7 == 2ex - 4)y-7 ==2x - 8Y = 2x - 8 + 7Y ==2x-1

6. a) i) a) K==2

b) (0,0)

35 I p age

ii)

b)

36\ P age

i) PR2 = 52 + 102 - 2(5)(10) cos 20°PR2 = 125 - 94PR = illPR =5...6km

ii) 142 - 70 = 72180 - 72 = 108°Bearing of R from P = 108°

7. a)TimejS

Frequency

50-54 3

55-59 4

60-64 6

65-69 ~

70-74 Z

75-79 ~

80-84 ~

b) 83 - 51 = 32

37 I P age

c) Frequenq Polygon showing time taken by Om race

.,-I., t

c,

. '.';' .1.,

: ..

,· :.I : ', .' .

'. ~

· ,· J'", .. ';"

. ,'I ; "

u. ii' ,"'. '. I'~.. r ,; -

d)

381Page

8. a)Part Fraction

A 14

1B -

6

1C -

24

5D -

24

1E -

9

1F -

6

1G -

18

b) A, D,F, B,E,C,G

c) i) 12square units

ii)

39 I P age

9. a) i) g(-2) = 2(-2)+ 15

-4+1-5

-3= -

5

ii) gf(x) = 2(x+4)+ 1

5

2x+8+1--5

2x+9= -

5

Hi) 2y+lx=-

55x = 2y + 15x -1 = 2y

5x-ly=-

25x-lTheretore.qt (x) =-

2

b) i) (2x - 1) (x + 3)2X2 + 6x - x-32X2 + 5x - 3

ii) 2X2 + 5x - 3 = 2942X2 + 5x - 297 = 02X2 + 27x - 22x - 297= 0x(2x + 27) - 11(2x + 27) = 0(x - 1l)(2x + 27) = 0

x-ll=O 2x + 27 = 0Therefore, x = ll.crn 2x = -27 (not valid)

40lPage

i!i) Length= 2(11) - 1 width = 11 + 3= 22-1 = .14...crn= 2.1J:m

10. a) (2) y~ 15(3) x + y ~ 60

b) The number of gold stars must be less than twice the number of silver stars.

c) y = 60 - x(0,60)(60,0)

y = lhx(0,0)(40,20)

41\ P age

NB: Broken lines must beused for inequalities.Solid lines were only usedhere for clarity.

d)

42 I P age

Graphshowing the conditions for packaiPne eo d and silver stars.

~; -1:,::",1 ,: .-,

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It"

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I....•. ~." .• ,.,' 'I... ... ,'; .' \ .j, ... : . p ,. , ...:

II

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I,\ ;:.~ ". ..

A

.. .. ,'

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i V"''./ .

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AandC

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11.

12.

a) i) ,

ii)

b) i)

ii)

a)

b)

a)

b)

.,f3- Y3 1

_2 =-x-Y3 2 Y3

=¥Z=ll.5.

r = R cos (J

= 6370 cos 37= 5087.3

C = 2rrr= 2rr(5087.3)=31!M.8km

5390 = (31948) (50 + X)360

5390 50+x----31948 360

60.7 = 50 +xx = 10.7°

i) 360=~8 -

ii) A = ¥.z ab sin C= ¥.z (6) (6) sin 45= 12.7 cm2

iii) 12.7 x 8 = 1ill..& em-

a) TPQ = 90° (angle in a semi circle)

43 I P age

13. a)

b)

44IPage

b) MTQ = 90° (radius is perpendicular to a tangent)

c) STQ = 180 - (90 + 23)= 67° (angles <TO, MTS and STQ are complementary)

TQS = 180 - (67 + 90)= 23° (or angle between a chord tangent is equal to the angle M

the alternate segment)

d) SRQ = SRT + TRQ

SRT = 23° «between tangent and chord = < in alternate segment)

TRQ = 90° « in a semi circle is a right angle)

:. SRQ = 23° + 90°= 113°( opp. Angles of cyclic quadrilateral are supplementary)

K

R K

o -,-, M

i) ~=~+--tMK MO OK

=-m+kii) --t=_+--7

RM RO OM= -lhk + m

iii) ~=-+~KS KO OS

1= -k +-m3-

iv) -+=-+~RS RO OS

1= -lhk+-m- 3-

c) ~=-+-+KL KR RL

= -lhk + lh ( -lhk + m)

= lhk - 14k+ lhm

= -%k+ lhm--+ 3--+

=>KL =-RS ..2

RS is parallel to KL.

14. a) i) 3A= 3 (~ ~)=ea 3b)

3c 3d

ii) B-1- J. (2 -;)- 10-9 -3= (2 - 3)

-3 5

iii) 3A+ B-1 = ea + 2 3b - 3)3c - 3 3d + 5

iv) (3a+2 3b-3)=(14 05)3c - 3 3d + 5 -9

3a + 2 = 143a = 12a=i

45 I P age

b)

461Page

3b - 3 = 0

3b = 3b=l

3c- 3 =-93c =-6c =.:.Z

3d + 5 = 53d=0d=Q

i) a)b)

ii) a)

b)

A= (4 1)-2 0

A translation of -4 units parallel to the y-axisA rotation of 180· about the origin OR andenlargement of scale factor -1, about the origin.

iii) pi = (~) + (~4)

= (~2)= pi (6,-2)

iv)