Suggested Solutions to Past CXC

Examination Papers

2005-2010

Compiled By Experienced CXCInstructors

Paper 2 - 2010 80

Solution

Paper 2 - 2005

Page

1

Paper 2 - 2006 15

Paper 2 - 2007 31

Paper 2 - 2008 47

Paper 2 - 2009 63

Questions? Comments? Email mr~norbert@live.com

MathematicsGeneral Proficiency

May 2005

1. a)

21 10= 5 3

63-50= 15

- 13/- 15

b) i) A: 12.50 x 3 = $37.50 'v

B:$33.90 =$li.25--

2

c: $31.00 =5--$6.20

D:15

- X 108.28 = $16.24100

ii) 6 x 0.75 = $4.506 x 0.40 = $2.40Total made = $6.90Total 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

2 a) i) ab(5a + b)

ii) (3k-l) (3k+ 1)iii) 2yL4y-y+ 2

2y(y - 2) -l(y - 2)(2y -1) (y - 2)

b) 6xz - 8x + lSx -206xz + 7x - 20

c) i) 2(x - 3)= 2x-6

l ii) x + x - 3 + 2x - 6 = 394 x - 9 = 39

4x =48x = 12

·Solving the equation was not necessaryin this case.

3. a) i) 3x + x = 244x=24

x=6~ students take both Music and Drama.

ii) Student who take drama only = 7 + x:. 7 + 6 = 1.3.students take only Drama.

b) i) 2y- 5 = -(x + 3) y=mx+c3

2 2y='3x+2+S or 5=-(-3)+c

32

5 = -2 + cy='3x+7c=7

ii) 2 x - 3y = 02

Y = -x + 73

21Page

-3y = -2x-2xy=- -32

Y = -X3

m =!. .:for the line 2x - 3y = 0 therefore it is parallel to y = !.x + 73 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 = 4176.79

The medium pizza is more than twice as large as the small pizza. It is fourtimes as large.

b) S· f1 f di . 707.14 23571 2Izeo -0 me IUmplzza=--= . em3 3

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 smallpizza.

31Page

5. The graphbelow shows the answer for 5(a), 5(b) (i) 5(b) (ii) - lern: 1 unit on bothaxis.

_J __ ,

c. 1; ;;\

. , ;

.

F'. ,

;', .

" , '.

. ,

.; ....,

.' .

.' ,

~~"

..~,

" ,

" ' " ... ,

.. ".

...'..,.... . 'F';'

: .1.· .. · ~ I'.: ' I·

.'

! ; ..~. ; , :.'

" "

' .... ./. "

',: I' ,"

. "r:,I,·:'

.i . -: ,

, ', .. ':

s :

.. '\' .

",,:.. ',',y-,. ,;.:' ':~.:ii

'" ; " . ,:

,.., "Et','

,n

,·c.

, :" ';', I,

1 ,l

.' .~ .

I :

J' I: .;

.... "

. "

- ,,> "'" .• ,.

'.

:,'; i,':' 'I','i, ",:

.

4\Page

.... "., 'cT,'., ',: .. ", ;t.

! ,:' "

Hi) Composite transformation

c)1.8

tan a=-2

a = tan"! 0.9a = 42°

6. a) i) EAD = 57° Reason: Alternate angles of parallel lines

Then x = 180 - (57 + 80)= 43°

ii) DAB=108-57= 51°

Since angle of a quadrilateral add up to 360°.Then y = 360 - (51 + 57 + 90)

= 360 -198= 162°

b) i) 32 + (-3)2 = 9 + 9=18

ii) Ifx = Yzy+ 5x-5=Yzyy=2(x-5)

r1(x) = 2x -10rl(6) = 2(6) -10

= 12 -10=~

iii) fg(x) = 1fzX2 + 5f g(2) = 1fz(22) + 5

=Yz(4)+5=2+5=Z

SIPage

7.

61Page

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

400

3!lO

300

J 250'"~,.~ 200Q

150

100

50

Cumulative Frequency Curve of Heights of Applicantsy

!SS 16S 110 ,17,S; 1&0 lSS x

H.ipt (cmj

b) i) 275 applicants

ii) 167 em

iii) 162.5 em

iv) 95-400

8. a) i)(4 X 72) + (3 x 6) + 2 216

ii) I~------~------------------------

103 (8 X 112) + (3 x 10) + 2 1000

iii) 1,--_n_3 --l.1_C(_n_-2_)x_C_n_+1_)2_+_(3_X_n~_2_1,-- __ n3__

b) (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

Difference of squares• (a - b)(a + b) =aL b'

71Page

9. a) 5xz + 2x-75(xZ + ~x) - 7

5

5 [Xl + ~X + (!)Z ~ - 7 - 5(!)25 5 5

1 1Sex +_)2- 7-5 5

b) i) 1- 7-5

ii) 1 1WhenX + 5 = 0,then x = - '5

c) S(x + !)2 = 365 5

(x + !)2 = 365 25

X + ~ = ../36/251 6x=--+ -5 - 5

E" h 1 6 1 6It er x = - - + - or x=----

5 5 5 5

=17-5

d)v

x

81Page

10. a) i) 40-0_'J£.77/2- £".,J,LL m s15-0 --

Acceleration == gradientof first portion of graph

ii)Distance travelled = 22

0 (40 + 50)= 10(90)= 9.illlm

20

b) i) 12km = 6km/h2h

ii) He took a rest or stopped.

iii) ~=8km/h1.5

c) i) x=6

ii) x~6

5 +5y>-- x- 8

1y<-x+5-6

li. a) i) x

7.5

"evaluate ::-point in the region foreach line.

"Sine rule: "h ab sin C = Area oftriangle.

pL- ~ Q

91Page

1h (7.5) (4.5)Sin 0' = 13.5S

. 13.5InO' =--

16.8750"' = sino! 0.8

0"' = 53°

ii) 1h (15) (9) sin 53 = 67.5 sin 53= 54 cm2

b) i) a) 180 - 136 = 44°SJM = 44-

2= 22° (base angles of an isosceles triangle)

b) SJM = JMK = 22° (alternate angles):. JKM = 180 - (124 + 22)

= 180-146= 34°

ii) a) MJ 50--sin 136 sin22

MJ =9.2..Zm

b) --=--JK 92.7

sin 22 sin 34JK =Qllm

12. a)

17°N

A-AntiguaB - BelizeG - Greenwich MeridianE - Equator

G

10 I P age

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=DCTherefore. AB = 3K

ii) BD=BA+/iD= - 3K - 3y

Hi)---.-. 1 ---->DP=--BD

31:::: -"3 (-3x - 3y)

=x+y

b) AP=AD+DP= -3y + x + y=x 2y

c) PH=PD+DE3= -x - y +- X2

::::Yzx-y

and fiE = liP+PH1= X - 2y + -x - Y2

3::::-x - 3y

2

=> AP=2PHTherefore. A, P and E are collinear.

lllPage

d) DA == 3G)=G)

IDAI = .../32 + 32 = ..JI8

AB=AP+PE= X - 2y + lh X - Y

3= 2"X - 3y= ~(~)-3G)= G) - (D== (~3)

IABI = J 02+ (-3)2 = ~ = 3

DE =~ (~)= (~)

IDEI == .../32 + 02=~=3

A

o '-----t+----" E

12 I P age

A

Therefore; triangle AED is isosceles.

14. a) i)

ii)

iii)

(2) (15) - (7) (5)= 30-35= -5

=> M is a non-singular matrix

M-I = 2. (15 -5)-5 -7 2

iv) (10 0) (X) 1 (151 Y = -5 -7

(X) = ~ (-45 - 85)Y -5 21 + 34

(X) = ~ (-130)Y -5 55

X = 26 y =:.11

b) i) (-1 1°) • The general matrix ofR= 0 rotation for iT' in theanuclockwlse directionabout the

ii) (-1-~)

origine:,:! -::::)N= 0

iii) T= (~3)

13IPage

141 P age

iv) RN=(~1 ~)(-~ _°1)

= (~ -~)

pI = (~ _~) (161)= (-1~)

pl = (6. -11)

(161)+ (-;3) = (136)

P" = (-g _~) (136)= (-=-136)

P" = (-3. -16)