JUNE 2005 CXC MATHEMATICS GENERAL PROFICIENCY (PAPER 2)

Section I

1. a. Required To Calculate:

Calculation:

b. Data: Table showing Amanda’s shopping bill

(i) Required To Calculate: The values of A, B, C and D

Calculation: 3 T-shirts at $12.50 each cost a total of

2 CD’s cost a total of $33.90

The unit price is

÷øö

çèæ ´- 3911

514

( ) ( )

form)exactin(1513151315506315

105213310

521

313

514

310

514

3910

514

3911

514

=

=

-=

-=

-=

-=

-=

÷øö

çèæ ´-=

÷øö

çèæ ´-

50.37$50.37$50.12$3

=\=´

A

\290.33$

95.16$=95.16$=\B

C posters at $6.20 each cost $31.00

The total bill is $108.28

15% VAT

(ii) Required To Determine: Whether Amanda made a profit or a loss Solution: Price paid for 6 stickers at $0.75 each and 6 stickers at $0.40 each

The cost of 12 stickers to Amanda = $5.88 Since the selling price > Cost price, then Amanda acquired a profit of

2. a. Required To Factorise: (i) , (ii) , (iii) Factorising: (i)

(ii)

This is the difference of two squares

(iii)

500.6$00.31$

=\

=\

C

C

\ 28.108$10015

´=

centnearesttheto24.16$224.16$

==

24.16$=\D

( ) ( )

90.6$40.2$50.4$

40.0675.06

=+=

´+´=

( )02.1$

88.5$90.6$=

-

225 abba + 19 2 -k 252 2 +- yy

225 abba +

( )baabbbabaa

+=+=

5.....5

19 2 -k( ) 22 )1(3 -= k

( )( )1313 +- kk

252 2 +- yy( )( )212 -- yy

b. Required To Simplify: Solution: Simplifying

c. Data: Card game played among 3 people. Solution: Score by Adam points Imran’s score is 3 less than Adam’s score (data) (i) Required To Find: an expression in terms of x for the number of points

scored by Shakeel. Solution:

Shakeel’s score is 2 times Imran’s score points (ii) Required To Find: an equation which may be used to find the value of x. Solution:

Total score = 39 points

3. a. Data: Venn diagram illustrating the students in a class who study Music and /or Dance. Solution:

(i) Required To Calculate: the number of students who take both Music and

Drama. Calculation:

(data)

( )( )4352 -+ xx

( )( )4352 -+ xx

2076208156

2

2

-+=

--+=

xxxxx

x=

( )3-= x

( )32 -= x

( ) ( )

12and48439943962339323

===-=-+-+=-+-+\

xx

xxxxxxx

( ) 24=Mn

And , that is number of students who take both Music and Dance = 6

(ii) Required To Calculate: the number of students who take Drama only. Calculation:

Hence

That is, the number of students who take Dance only = 13

b. Data: Line with gradient passes through P (-3, 5)

(i) Required To Find: the equation of the line through P (-3, 5) and with

gradient .

Solution:

Equation of line is

is of the form , where and .

6and244243

===+\

xxxx

( )DMn Ç

( ) 13only =Dn

32

32

( ) 32

35

=---

xy

732

212362153

+=

+=+=-

xy

xyxy

cmxy +=32

=m 7=c

(ii) Required To Prove: the above line is parallel to the line Solution:

is of the form , where is the gradient.

Hence and are parallel since they both have the

same gradient and parallel lines have the same gradient.

4. Data: Diagrams of

a. Required To Determine: Whether a medium pizza is twice as large as a

small pizza. Solution:

The pizzas are 3-dimensional, hence a comparison of sizes must be made by comparing their volumes. Both have the same height (thickness).

Volume of small pizza Volume of medium pizza

So we see that the medium pizza has 4 times the volume of a small pizza. So that statement – A medium pizza is twice as large as a small pizza is INCORRECT.

032 =- yx

032 =- yx

xy

xy

3223

=

=

cmxy +=32

=m

732

+= xy 032 =- yx

÷øö

çèæ=32

( )3

2

2

cm25.565.7hh

hrVs

p

p

p

=

=

=( )

( )hhh

hrVm

pp

p

p

25.564cm225

153

2

2

==

=

=

b. Data: The prices for each slice of a medium pizza and for one small pizza.

Required To Find: Whether it is better to buy 1 medium pizza or 4 small pizzas. Solution:

Cost of medium pizza = $15.95

Cost of an entire medium pizza Cost of 1 small pizza = $12.95 Since 4 small pizzas 1 medium pizza Then the equivalent cost of 4 small pizzas

4 small pizzas is equivalent in volume to 1 medium pizza which costs $47.95

The ‘better buy’ (which supposedly means more pizza at a lesser price) is obtained by buying a medium pizza.

5. a. Data: The coordinates of the vertices of a triangle, D, E and F. Required To Draw: . Solution:

b. (i) Required to draw: Solution:

31

\ 395.15$ ´=85.47$=

º

80.51$495.12$

=´=

\

DEFD

FED ¢¢¢D

( )1,7=¢D ( )1,5=¢E ( )4,7=¢F

(ii) Required To Draw: Solution:

(iii) Required to identify: the type of transformation that maps onto

Solution:

Hence

FED ¢¢¢¢¢¢D

( )4,7 -=¢¢D ( )4,5 -=¢¢E ( )1,7 -=¢¢F

DEFDFED ¢¢¢¢¢¢D

FEDFEDDEF x ¢¢¢¢¢¢D¾¾¾¾¾ ®¾¢¢¢D¾¾¾¾¾ ®¾D÷÷ø

öççè

æ-

== 5

0nTranslatio

4inReflection

FEDDEF ¢¢¢¢¢¢D¾¾¾¾ ®¾D reflectionGlide

c. Data: Stick 1.8 m casts a shadow 2 m long.

Required To Calculate: Angle of elevation of the sun Calculation: Let the angle of elevation be

6. a. Data: Diagram of a pentagon ABCDE Required To Calculate: Calculation:

(i) (alternate angles)

a

( )

degree)nearesttheto(429.41

9.0tan28.1tan

1

°=°=

=

=

-

aaa

a

°° yx ,

°= 57ˆDAE( )

°=°+°-°=°

435780180x

(Sum of angles in a triangle = 180°) (ii)

(Sum of angles in a quadrilateral = 360°)

b. Data:

(i) Required To Evaluate: Solution:

(ii) Required To Evaluate: Solution:

Let

Replace y by x

(iii) Required To Evaluate:

Solution:

°-°= 57108ˆDAB°= 51

( )°=

°+°+°-°=\162

578051360y

( ) 521

+= xxf ( ) 2xxg =

( ) ( )33 -+ gg

( ) ( )33 -+ gg( ) ( )

1899

33 22

=+=

-+=

( )61-f

521

+= xy

xy

xy

=-

=-

102215

( )( ) ( )

210626102

1

1

=-=\

-=-

-

fxxf

( )2fg

( ) ( )422 2

==g

( ) ( )

( )

752

542142

=+=

+=

= ffg

7. Data: Table showing the height of 400 applicants for the police service. a. Required To Draw: the cumulative frequency curve of heights given in the table. Solution:

The data shows a Continuous variable and we create the table as:

Height in cm, x L.C.B. U.C.B No. of applicants

Cumulative frequency

Points to be plotted

(U.C.B, CF) 151 – 155 10 10 (150.5, 0)

(155.5, 10) 156 – 160 55 65 (160.5, 65) 161 – 165 105 170 (165.5, 170) 166 – 170 110 280 (170.5, 280) 171 – 175 80 360 (175.5, 360) 176 – 180 30 390 (180.5, 390) 181 – 185 10 400 (185.5, 400)

The point (150.5, 0) is obtained by extrapolation, so as to start the curve on the horizontal axis.

5.1555.150 <£ x

5.1605.155 <£ x5.1655.160 <£ x5.1705.165 <£ x5.1755.170 <£ x5.1805.175 <£ x5.1855.180 <£ x

400=å f

b. (i) Required To Estimate: the number of applicants whose heights are less than 170 cm. Solution: From the graph, 265 applicants are less than 170 cm (read off).

(ii) Required to estimate: the median height of applicants. Solution:

The median height of applicants 167 cm (read off). (iii) Required to estimate: the height that 25% of the applicants are less than

Solution:

25% of the applicants

100 applicants are less than 162 cm (read off).

»

»

40010025

´=

100=

(iv) Required To Estimate: the probability that a randomly selected applicant has a height no more than 162 cm. Solution: P( applicant’s height is no more than 162 cm)

8. a. Data: Table showing a number pattern Required To Complete: the table given. Solution:

8

27

64

(i)

216

(ii)

1000

(iii)

41400100

applicantsofNo.cm162applicantsofNo.

=

=

£=

32 ( ) ( ) 22330 2 +´+´33 ( ) ( ) 23341 2 +´+´34 ( ) ( ) 24352 2 +´+´

35

36 ( ) ( ) ( )( ) ( ) 26374

26316262

7

+´+´=

+´++´-

! ! !310 ( ) ( ) ( )

( ) ( ) 21031182103110210

2

2

+´+´=

+´++´-

! ! !3n ( ) ( ) ( ) 2312 2 +´++´- nnn 3n

b. Required to prove: Proof: L.H.S.

Q.E.D.

Section II

9. a. Required to express: in the form , Solution:

Equating the coefficient of

Equating the coefficient of x

Equating constants

OR

( ) ( ) ( ) 332 babaabbaba +=+++-

( ) ( ) ( )( )( ) ( )

R.H.S.

222

33

22322223

22

2

=+=

+++-++-=

++++-=

+++-

baabbababbaabbaa

baabbabababaabbaba

725 2 -+ xx ( ) cbxa ++ 2 ÂÎcba ,,

( )( )

cababxaxcbbxxa

cbxa

+++=

+++=

++

22

22

2

22

2xÂÎ= 5a

( )

ÂÎ=

=

51252

b

b

517

515725

517

751

7515

22

2

-÷øö

çèæ +º-+\

ÂÎ-=

-=+

-=+÷øö

çèæ

xxx

c

c

c

7525

725

2

2

-÷øö

çèæ +=

-+

xx

xx

Half the coefficient of x is

=

is of the form , where

b. (i) Required To Determine: the minimum value of Solution:

and

51

52

21

=÷øö

çèæ

7517

5125

251

525

2

2

-

-

++=

÷øö

çèæ ++=

xx

xx

517

515

2

-÷øö

çèæ +x

( ) cbxa ++ 2

ÂÎ-=

ÂÎ=

ÂÎ=

517

515

c

b

a

725 2 -+= xxy

725 2 -+= xxy

517

515

2

-÷øö

çèæ += xy

517

5170

0515

min

2

-=

-=\

"³÷÷ø

öççè

æ+

y

xx

(ii) Required To Determine: the value of x at which the minimum point occurs.

Solution: When

OR has an axis of symmetry at

At minimum point and

at

0515

2

=÷øö

çèæ +x

51

051

051 2

-=

=÷øö

çèæ +

=÷øö

çèæ +

x

x

x

725 2 -+= xxy( )( )

51522

-=

-=x

51

-=x 7512

515

2

-÷øö

çèæ-+÷

øö

çèæ-=y

517-=

517min -=y

51

-=x

c. Required To Solve: Solution:

OR

OR

Find the square root

0725 2 =-+ xx

0725 2 =-+ xx( )( )

57or1

0175

-=\

=-+

x

xx

( ) ( ) ( )( )( )

521or1

1010or

10141012210144210

14042

5275422 2

-=

-=

±-=

±-=

+±-=

--±-=x

536

515

0517

515

0725

2

2

2

=÷øö

çèæ +

=-÷øö

çèæ +

=-+

x

x

xx

2536

51 2

=÷øö

çèæ +x

1or5756156

5156

51

-=

±-=

±-=

±=÷øö

çèæ +

x

x

c. Required To Sketch: the graph of , showing the coordinates of the minimum point, the value of the y – intercept and the points where the graph cuts the x – axis. Solution: When

Curve cuts the y – axis at (0, -7) and the x – axis at 1 and .

Minimum point =

10. a. Data: Speed – time graph for the movement of a cyclist. (i) Required To Calculate: The acceleration of the cyclist during the first 15

seconds. Calculation:

Since the branch for the first 15 seconds of the journey is a straight line, then the acceleration is constant.

Gradient Acceleration

725 2 -+= xxy

( ) ( ) 7702050 2 -=-+== yx

\57

-

÷øö

çèæ --

517,

51

015040

--

=322= \ 2ms

322 -=

(ii) Required To Calculate: The distance travelled by the cyclist between and .

Calculation:

The distance covered between and is the area of the region, A, shown in the diagram which describes a trapezium.

b. Data: Diagram showing the distance – time journey of an athlete (i) Required To Calculate: The average speed during the first 2 hours.

Calculation:

The average speed during the first 2 hours

(ii) Required To Determine: What the athlete did between 2 and 3 hours after the start of the journey. Solution:

15=t 35=t

15=t 35=t

( ) ( )

m900

1535504021

=

-´+=

1kmh6h2km12

takentimeTotalcovereddistanceTotal

-=

=

=

At ,distance = 12 km and at , distance = 12 km.

This is a indicated by a horizontal branch in the graph. Hence, between 2 and 3 hours after the start, the cyclist did NOT travel OR the cyclist stopped cycling for that 1 hour interval.

(iii) Required To Calculate: the average speed on the return journey.

Solution:

The return journey took

hours.

Average speed

(c) Data: Diagram of a triangle bounded by lines GH, GK and HK. (i) Required To Find: the equation of the line HK

Solution:

2=t 3=t

2135 -

211=

\takentimeTotalcovereddistanceTotal

=

1kmh8

h211

km12

-=

=

HK is a vertical line that cuts the x – axis at 6. Therefore, the equation of HK is .

(ii) Required to find: the set of 3 inequalities which define the shaded region in the diagram. Solution: Region shaded is on the left of . Hence, (and including line).

The region shaded is on the side with the smaller angle. Hence,

(including line).

The side shaded is that with the larger angle. Therefore, region is (including

line). Hence the three inequalities that define the shaded region are:

6=x

6=x 6£x

561

+£ xy

585

+-³ xy

6£x

561

+£ xy

585

+-³ xy

11. a. Data: P, Q are midpoints of with XP = 7.5 cm, XQ = 4.5 cm and area of

(i) Required To Calculate: the size of Calculation:

Let

(ii) Required To Calculate: Area of Calculation:

OR

XYZD2cm5.13=DXPQ

PXQÐ

q=QXP ˆ

( )( )

degree)nearestthe(to531.538.0

5.45.725.13sin

5.13sin5.45.721

°=°=

=´´

=\

=\

q

q

q

YXZD

( )

( )

( )( )

unitssquare5454915

21ofArea

8.054sin

95.42

155.72

=

´=D\

=

====

YXZ

or

XZ

XY

q

and are equivalent or similar.

b. Data: Diagram of trapezium SJKM with SJ parallel to MK, m,

and Required To Calculate: (i)(a) (b) (ii)(a) MJ (b) JK Calculation:

(i) (a) (base angles of isosceles triangle)

(b) (alternate angles)

(ii) (a) (sine rule)

OR

XPQD XYZD

2

22

cm54413.5ofArea

4:12:1ofArea:ofArea

2:1:

=

´=\==\

=

ΔXYZ

ΔXYZΔXPQXYXP

50== SJSM°= 124ˆKJM °= 136ˆTSM

MJS ˆ

MKJ ˆ

JMSMJS ˆˆ =

)180inanglesof(sum222136180ˆ

°=D°=

°-°=\ MJS

°= 22ˆKMJ( )

)180inanglesof(sum3422124180ˆ

°=D°=°+°-°=\ MKJ

°=

° 22sin50

136sinMJ

placedecimal1tom92.7m17.9222sin136sin50

==

°°´

=\MJ

(Cosine Rule)

(b) (Sine Rule)

12. This question is not done since it involves latitude and longitude (Earth Geometry) which has been removed from the syllabus. 13. a. Data: ABCD is a parallelogram with , and P on DB such that

.

(i) Required To Express: in terms of and Solution:

(Equal in magnitude and parallel, as expected for opposite sides of a parallelogram).

(ii) Required To Express: in terms of and

Solution: Similarly as (a)

( ) ( ) ( )( ) °-+= 136cos505025050 222MJ

placedecimal1tom7.9217.92

==MJ

°=

° 34sin71.92

22sinJK

placedecimal1tom1.62m01.6234sin22sin71.92

==

°°´

=JK

xDC 3= yDA 3=2:1: =PBDP

AB x y

DCAB º

xAB 3=\

BD x y

yDACB 3=º

( ) ( )yx

xyCDBCBD

33

33

--=

-+-=+=

(iii) Required To Express: in terms of and Solution:

Since , then and

b. Required To Prove: Proof:

c. Data: E is the midpoint of DC

Required To Prove: A, P and E are collinear. Solution:

DP x y

( )yxDB 33 ---=

yx 33 +=

2:1: =PBDP DBDP31

=

( )yx

yxDP

+=

+= 3331

yxAP 2-=

( ) ( )

Q.E.D.

2

3

yx

yxyDPADAP

-=

++-=+=

( )

( )

yx

xy

DEADAE

xDE

3211

2113

321

-=

+-=

+=

=

( )

AP

yxyx

211

22113

211

=

-=-

is a scalar multiple of . Therefore, is parallel to . Since A is a common point, P lies on AE and A, P and E are collinear.

d. Data: and

Required To Prove: is isosceles. Proof:

In only 2 sides, AE and DE, are equal, therefore the triangle is isosceles. Q.E.D

AE AP AE AP

÷÷ø

öççè

æ=02

x ÷÷ø

öççè

æ=11

y

AEDD

( ) ( )

( ) ( )

÷÷ø

öççè

æ-÷÷ø

öççè

æ=

÷÷ø

öççè

æ-÷÷ø

öççè

æ=

-=

=

+=

÷÷ø

öççè

æ=

÷÷ø

öççè

æ=

=

=

+=\

÷÷ø

öççè

æ=

÷÷ø

öççè

æ=

=

33

03

11

302

211

3211

3

03

03

02

211

211

18

33

33

11

3

3

22

22

yxAE

DE

xDE

DA

yDA

( ) ( )3

30

30

22

=

-+=

÷÷ø

öççè

æ-

=

AE

AEDD

12. a. Data:

(i) Required To Prove: M is a non-singular matrix Solution:

Det

Hence, and so M is non-singular. (ii) Required To Find:

Solution:

(iii) Required To Find: Solution:

where I is the identity matrix .

÷÷ø

öççè

æ=

15752

M

( ) ( )75152 ´-´=M

053530

¹-=-=

1-$ M

1-M

( )( ) ÷÷

ø

öççè

æ-

--=-

27515

511M

÷÷

ø

ö

çç

è

æ

-

-=

52

57

13

1-´MM

IMM =´ -1 22´ ÷÷ø

öççè

æ=

1001

( )

176

57532

52

57

13

15752

11

2221

1211

2221

121122

122

=+-=

÷øö

çèæ ´+-´=

÷÷ø

öççè

æ=÷

÷

ø

ö

çç

è

æ

-

-´÷÷ø

öççè

æ

÷÷ø

öççè

æ=´ ´

-´

e

eeee

eeee

MM

(iv) Required To Solve:

Solution:

Equating corresponding entries and

( )

( )

( )

IMM

MM

e

e

e

=´

÷÷ø

öççè

æ=´\

=-=

÷øö

çèæ -´+´=

=+-=

÷øö

çèæ ´+-´=

=-=

÷øö

çèæ -´+´=

-

-

1

1

22

21

12

1001

167

521517

02121

571537

022

52512

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ÷÷ø

öççè

æ173

15752

yx

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ÷÷ø

öççè

æ173

15752

yx

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ´÷÷ø

öççè

æ

´

--

-

173

15752 11

1

Myx

M

M

( ) ( )

÷÷ø

öççè

æ-

=

÷÷÷

ø

ö

ççç

è

æ

÷øö

çèæ ´-+÷

øö

çèæ -´

´+-´-=

÷÷ø

öççè

æ -÷÷

ø

ö

çç

è

æ

-

-=÷÷

ø

öççè

æ

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ´ -

1126

17523

57

17133

173

52

57

13

1731

yx

Myx

I

26=x 11-=y

b. (i) Required To Find: matrix R, which represents a reflection in the y – axis. Solution:

The matrix, R, which represents a reflection in the y – axis is .

(ii) Required To Find: matrix N, which represents a clockwise rotation of

180° about the origin. Solution: The matrix, N, which represents a clockwise rotation of 180° about O is

.

(iii) Required To Find: matrix T, which represents a translation of – 3 units

parallel to the x – axis and 5 units parallel to the y – axis. Solution: A translation of – 3 units parallel to the x – axis (3 units horizontally to the left) and 5 units parallel to the y – axis (5 units vertically upwards) may be represented by

(iv) Data: , that is N first, then R second. Required To Find: the coordinates of and . Solution:

÷÷ø

öççè

æ -1001

÷÷ø

öççè

æ-

-1001

÷÷ø

öççè

æ -=

53

T

PP RN ¢¾®¾P¢ P ¢¢

( ) ( )( ) ( )

( ) ( )( ) ( )

( )11,6116

1116011061

116

1001

116

1116011061

116

1001

-=¢\

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-´+-´-´+-´-

=÷÷ø

öççè

æ--

÷÷ø

öççè

æ-

÷÷ø

öççè

æ--

=

÷÷ø

öççè

æ´-+´

´+´-=÷÷

ø

öççè

æ÷÷ø

öççè

æ-

-

P

( ) ( )( ) ( )

( )16,3163

1613016031

163

1001

163

163

51136

116 5

3

--=¢¢\

÷÷ø

öççè

æ--

=

÷÷ø

öççè

æ´-+´´+´-

=÷÷ø

öççè

æ÷÷ø

öççè

æ-

-

¢¢¾®¾÷÷ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

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