CXC JUNE 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER … · Calculation: Midpoint of b. Data: The...

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CXC JUNE 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER 2) (REST OF THE CARIBBEAN BESIDES TRINIDAD & TOBAGO)

Section I

1. a. (i) Required To Calculate: Calculation:

(ii) Required To Calculate:

Calculation: Numerator:

Hence,

b. (i) Data:

Required To Calculate: CAN$250 in JA$. Calculation:

Cost of camera = CAN$250 Hence cost of camera in JA$

( ) 6724.027.09.3 +´

( )form)exactin(873.1

)calculatorofuseby(82.0053.16724.027.09.3=

+=+´

4354

212 -

10171082554

25

54

212

=

-=

-=-

terms)lowestitsinfractionsingleaas(1534

34

1017431017

4354

212

=

´=

=-

50.72$JA00.1$CAN º

50.72250´=12518$JA=

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 28

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(ii) Required To Calculate: Remaining money on credit card in CAN$.

Calculation: Limit on credit card = JA$30 000 Available remainder after buying the camera

The equivalent in CAN$

2. a. Data: (i) Required To Calculate: Calculation:

(ii) Required To Calculate:

Calculation:

b. (i) Required To Find: Algebraic expression for the statement given. Solution: Four times the sum of x and 5. ( )

(ii) Required To Find: Algebraic expression for the statement given. Solution: 16 larger than the product of a and b. 16 + ( )

( )12518$00030$JA -=

50.721251800030 -

=

3and1,2 =-== cba( )cba +

( ) ( )( )422

312

==

+-=+ cba

cbaacb++

- 24 2

( ) ( )( )( )

( )

24841214

3123221424 22

-=

-=

-=

+-+--

=++

-cbaacb

´4 5+x( )54 += x

abba =´ab+= 16


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c. Data: Required To Calculate: x Calculation:

d. Required To Factorise: (i)

Solution: (i)

(ii)

3. Data: Results of 1 080 students’ choices in a career guidance seminar.

Career Lawyer Teacher Doctor Artist Salesperson Number of

students 240 189 t 216 330

a. Required To Calculate: t

Calculation:

( )132415 +=- xx

( )

1031

10131310

4621526415132415

=

=

=+=-+=-+=-

x

xxx

xxxx

592)(,126 2432 -++ mmiibaba

baabbababa ´´´´+´´´=+ 2222432 626126( )( )222

222

2626babaabba

+=

+=

( )( )512592 2 +-=-+ mmmm

( )105

3302161892400801)data(0801330216189240

=+++-=

=++++t

t


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b. (i) Required To Calculate: Size of the angles of the sectors in the pie chart. Calculation:

Sector illustrating Angle of sector Lawyer

Teacher

Doctor

Artist

Salesperson

(ii) Required To Draw: Pie chart to represent the information given, using a circle of radius = 4 cm. Solution:

°=°´ 803600801240

°=°´ 633600801189

°=°´ 353600801105

°=°´ 723600801216

°=°´ 1103600801330


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4. a. Data: Given the universal set U and sets M and N are defined. (i) Required To Draw: Venn diagram for the information given. Solution:

(ii) Required To List: Elements of the set . Solution:

(as illustrated on the Venn diagram)

b. (i) Required To Construct: Parallelogram ABCD, in which AB = AD = 7 cm and . Solution:

( )¢È NM

( ) { }25,21,15=¢È NM

°= 60BAD


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(ii) Required To Find: Length of diagonal AC. Solution: AC = 12.1 cm (by measurement).

5. Data: Plan of a floor with given dimensions.

a. (i) Required To Calculate: Length RS. Calculation:

(ii) Required To Calculate: x

Calculation:

b. Required To Calculate: Perimeter of the entire floor.

Calculation: Using R as the starting point and checking the total distance of all the edges, to find the perimeter of the floor. Perimeter

m6ofLength28

)diagramFrom(82

=-=

=+

RSRS

RS

m4106

)diagramFrom(m10

==+=+

xxRSx

( )m383251054 +++++++=m40=


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c. Required To Calculate: Area of the entire floor. Calculation: Area of the entire floor = Area of region A + Area of region B

d. Data: Part A is to covered with flooring boards measuring 1 m by 20 cm.

Required To Calculate: Number of flooring boards needed to cover A. Calculation: Area of 1 flooring board

Number of flooring boards needed to cover A

No. of boards = 250

6. a. Data: Diagram showing a vertical pole and H, J and K, points on the horizontal ground. (i) Required To Complete: Diagram by inserting the angles of elevation. Solution:

( ) ( ){ }

2m742450

83105

=

+=´+´=

cm20m1 ´=

2

2

m51

m10020

m10020m1

=

=

´=

board1ofAreaAofArea

=

25051105

=

´=


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(ii) (a) Required To Calculate: Length of HJ. Calculation:

(b) Required To Calculate: The length of JK.

Calculation:

Length of JK

b. Data: Diagram on axes illustrating figure P and congruent figure Q.

place)decimal1to(m2.19m02.1932tan12

1232tan

==

°=\

=°

HJ

HJ

m55.2327tan12

1227tan

=°

=\

=°

HK

HK

2.1955.23 -=

place)decimal1to(m4.453.4

==


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(i) Required To Describe: The transformation that P undergoes to produce Q. Solution:

P is mapped onto Q by a horizontal shift of 2 units to the right and a vertical shift of 7 units downwards.

This is a translation, T, where .

(ii) (a) Required To Draw: The line on the answer sheet.

(b) Required To Draw: S, the image of P under a reflection in the line .

Solution: P onto S by a reflection in the line .

÷÷ø

öççè

æ-

=72

T

xy =

xy =

xy =

SP

SP

ofverticestheofverticestheof

sCoordinatesCoordinate47642244

22444764

0110

0110

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ

¾¾ ®¾\÷÷ø

öççè

æ


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7. Data: Diagram of a straight line cutting the axes at A and B.

a. (i) Required To Calculate: c Calculation: From the diagram, the straight line cuts the vertical axis at A (0, 7), Hence, . 7=c


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(ii) Required To Calculate: m Calculation: From the diagram B = (2, 0)

Gradient of AB = m

(iii) Required To Calculate: Midpoint of AB.

Calculation:

Midpoint of

b. Data: The point (-2, k) lies on the line.

Required To Calculate: k Calculation:

Equation of AB is , where and , that is

(-2, k) lies on the graph.

c. Required To Calculate: Point of intersection of and the given line.

Calculation:

2007

--

=

27

-=m

÷øö

çèæ ++

=207,

220AB

÷øö

çèæ=

213,1

cmxy +=213-=m 7=c

7213 +-= xy

( )

14

72213

=

+-÷øö

çèæ-=k

2-= xy


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To determine the point of intersection of and AB, we solve the equations simultaneously. Let

…(1)

…(2) Equating (1) and (2)

When

The point of intersection of the two lines is (2, 0).

8. Data: Annie bought 6 each of four different kind of stamps. a. Required To Calculate: Total cost of the stamps. Calculation: Cost of 6 stamps at $4.00 each = $24.00 Cost of 6 stamps at $2.50 each = $15.00 Cost of 6 stamps at $1.20 each = $ 7.20 Cost of 6 stamps at $1.00 each = $ 6.00 Total Hence, the total cost of the stamps that Annie bought = $52.20 b. Data: Cost of Annie posting a parcel was $25.70

(i) Required To Calculate: Stamps that she can select using as many $4.00 stamps as possible. Calculation: Cost of posting a parcel is $25.70 Using 5 stamps at $4.00 each = $20.00 1 stamp at $2.50 each = $ 2.50 1 stamp at $1.20 each = $ 1.20 2 stamps at $1.00 each = $ 2.00 Total is exactly = $ 25.70

2-= xy

7213 +-= xy

2-= xy

2214

92149

21327

27213

=

=

=

+=+

-=+-

x

x

xx

xx

2=x 22 -=y0=

\

20.52$=


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(ii) Required To Calculate: Stamps that she can select using all her $1.00 stamps. Calculate: Using All 6 stamps at $1.00 each = $ 6.00 4 stamps at $4.00 each = $16.00 1 stamp at $2.50 each = $ 2.50 1 stamp at $1.20 each = $ 1.20 Total is exactly = $25.70

c. (i), (ii) Required To Calculate: The largest number of stamps she can use and list the selection of stamps she can use. Calculation: To use the most number of stamps, Annie must use 6 stamps at $1.00 = $ 6.00 6 stamps at $1.20 = $ 7.20 5 stamps at $2.50 = $12.50 Total is exactly = $25.70 And the maximum number of stamps

9. a. (i) Required To Simplify: Solution:

(ii) Required To Simplify: Solution:

b. Data:

(i) Required To Calculate: Calculation:

566 ++= 17=

432 xxx ÷´

xxxxxx

==

=÷´ -+

1

432432

325

23

abba ´

( )

42

23

25

21

23

23

21

25

23

21

325

23

325

23

baba

baba

abbaabba

=

=

=

´=´

++

( ) 32 -= xxf( )2f

( ) ( )

1343222

=-=

-=f


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(ii) Required To Calculate: Calculation: Let

Replace y by x

(iii) Required To Calculate: Calculation:

c. Data: Table of values of temperature vs time for a liquid as its cools.

(i) Required To Draw: The curve to represent the information given. Solution:

( )01-f

23

2332

+=

=+-=

yx

xyxy

( )

( )

211

2300

23

1

1

=

+=

+=

-

-

f

xxf

( )21 ff -

( )( ) ( )

223112

1211

=

+=

=\

=-- fff

f


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Note: Scale is Horizontal axis, 2 cm 10 minutes, and not 10 seconds as printed.

(ii) (a) Required To Find: Temperature of the liquid after 15 minutes. Solution:

The temperature of the liquid after 15 minutes is approximately 49.5°C. (Read off).

(b) Required To Find: Rate of cooling of the liquid at . Solution: The tangent to the curve at cuts the axes at A and B as

shown. A is (0, 50) and B is ( 70, 0)

Gradient AB

(The negative sign temperature is decreasing.) The rate of cooling of the liquid a minutes is 0.7 °C min-1.

10. a. Data: and Required To Calculate: x and y Calculation: Let …(1) …(2) From (1) Substitute in (2)

When When Hence, and OR and .

º

30=t

30=t

7050

-=

17.0-=1min7.0 -°-= C

Þ30=t

274 =+ xy 40=+ xxy

274 =+ xy40=+ xxy

xy 427 -=

( )

( )( )5or2

0520107

40402844042740427

2

2

2

==--=+-

÷=+-

=+-

=+-

xxxxx

xxxxxxxx

2=x ( )2427 -=y 19=5=x ( )5427 -=y 7=2=x 19=y 5=x 7=y


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b. (i) Data: A shaded region illustrating a set of inequalities representing boys and girls in a cricket club .

(a) Required To State: Whether the cricket club can have 10 boys and 5 girls.

Solution: x = no. of boys and y = no. of girls

If and , we note that the point (10, 5) does not lie in the shaded region. Hence, the club cannot have 10 boys and 5 girls.

(b) Required To State: Whether the cricket club can have 6 boys and

6 girls. Solution:

If and we note that the point (6, 6) lies within the shaded region. Hence, the club can have 6 boys and 6 girls.

(ii) Required To Find: The inequalities that defines the shaded region. Solution:

The region with the shaded is on the side with the smaller angle. Hence, the region is . (It may be if the line is included).

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

is . (It may be if the line is included).

The region shaded is above the horizontal . Hence, the region is

. (It may be if the line is included).

10=x 5=y

6=x 6=y

xy 2< xy 2£

1254

+-< xy 1254

+-£ xy

2=y2>y 2³y


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(iii) (a) Data: Profit of $3.00 made on boy’s uniform and profit of $5.00 made of a girl’s uniform.

Required To Find: An expression in x and y for the total profit made. Solution: The profit on x boy’s uniforms at $3 each and y girls uniforms at $5 each Let the total profit be P.

(b) Required To Calculate: The minimum profit the company can

make. Calculation:

The minimum profit will be made when and .

11. a. Data: Diagram, shown below, with circle centre O, and OS parallel to PR.

( ) ( )yx ´+´= 53

yxP 53 +=

1=x 2=y( ) ( )13$

2513min

=+=\P

°= 26ˆRPS


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(i) Required To Calculate: Calculation:

(alternate angles).

(radii)

(Base angles of an isosceles triangle are equal). Hence,

(Sum of angles in a triangle = 180°).

(The angle subtended by a chord at the centre of the circle is twice the angle subtended at the circumference, standing on the same arc).

(ii) Required To Calculate: Calculation:

(The angle made by a tangent to a circle with radius, at the point of contact = 90°).

b. Data: Diagram showing a circle, centre O and chord AB = 14.5 cm.

STP ˆ

°= 26ˆPSO

OPOS =°= 26ˆRPO

( )°=

°+°-°=128

2626180ˆPOS

( )

°=

°=

64

12821ˆSTP

QPR ˆ

°= 90ˆQPO

( )°=

°+°-°=38

262690ˆQPR


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(i) Required To Calculate: The value of . Calculation:

OR

(ii) Required To Calculate: Area of triangle AOB.

Calculation:

Area of

OR

q

( ) ( ) ( ) ( )( )

degree)nearesttheto(11706.117)454.0(cos

)(coscos5.85.825.85.85.141

222

°==-=

-+=-q

q law

degree)nearesttheto(11760.1172533.58

533.582

5.825.7

2sin

°==

´=

=

=

q

q

q

( )( ) °=D 117sin5.85.821AOB

2cm2.32=


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Area

OR

Area (Heron’s Formula)

(iii) Required To Calculate: Area of the shaded region. Calculation: Area of the shaded segment = Area of sector AOB – Area of

(iv) Required To Calculate: Length of major arc AB.

Calculation: Angle of major sector

Length of major arc

12. Data: Ship sails from point R to S and then to T. a. Required To Draw: Diagram showing the journey of the ship from R to S to T. Solution:

( ) ( )

437.46875.19

25.75.8 22

==

-=h

2437.45.14 ´

=

2cm2.32=

75.152

5.145.85.8

=

++=s

( )( )( )5.1475.155.875.155.875.1575.15 ---=2cm2.32=

AOBD

( )2

2

cm5.41

2.325.8360117

=

-´°°

= p

°-°= 117360

( )5.82360117360 p´÷

øö

çèæ

°°-°

=

cm05.36=


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om b. (i) Required To Calculate: Calculation:

(ii) Required To Calculate: Calculation:

𝐿𝑒𝑡𝑅𝑇'𝑆 = 𝜃+,-./0

= 1+-./232°

(sin 𝑟𝑢𝑙𝑒)

sin𝜃 = +,-./232°1+

= 0.733 𝜃 = sinA20.733

= 47.13_

= 47.1° (𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 0.1 𝑜𝑓 𝑎 deg𝑟𝑒𝑒)

(iii) Required To Calculate: Bearing from R to T. Calculation:

TSR ˆ

°=°+°=

1013368ˆTSR

STR ˆ


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om The bearing from R to T c. (ii) Required To Calculate: Distance TX. Calculation:

13. Data: 𝑂𝐴PPPPP⃗ = 𝑎, 𝑂𝐵PPPPP⃗ = 𝑏, P is on OA such that and M is on BA such that BM = MA and OB is produced to N such that OB = BN. a. Required To Draw: The diagram showing the points P and M. Solution:

\ °+°+°= 1.4733180°= 1.260

( )°=

°+°-°=9.9

331.4790a

km88.739.9cos75

759.9cos

=°=

=°

TX

TX

PAOP 2=


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om b. Required To Express: and in terms of a and b. Solution:

𝐴𝐵PPPPP⃗ = 𝐴𝑂PPPPP⃗ + 𝑂𝐵PPPPP⃗ = −(𝑎) + 𝑏 = −𝑎 + 𝑏

If , then

𝑂𝑃PPPPP⃗ = X

Y𝑂𝐴PPPPP⃗

= XY𝑎

and 𝑃𝐴PPPPP⃗ = 2

Y𝑎

𝑃𝑀PPPPPP⃗ = 𝑃𝐴PPPPP⃗ + 𝐴𝑀PPPPPP⃗ = 2

Y𝑎 + 𝐴𝑀PPPPPP⃗

𝐴𝑀PPPPPP⃗ = 2X𝐴𝐵PPPPP⃗

𝑃𝑀PPPPPP⃗ = 2Y𝑎 + 2

X(−𝑎 + 𝑏)

= 2Y𝑎 − 2

X𝑎 + 2

X𝑏

= − 2,𝑎 + 2

X𝑏

c. Required To Prove: P, M and N are collinear.

Proof:

PAAB, PM

PAOP 2=


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𝑂𝐵 = 𝐵𝑁𝑂𝑁PPPPPP⃗ = 2𝑏

𝑃𝑁PPPPPP⃗ = 𝑃𝑂PPPPP⃗ + 𝑂𝑁PPPPPP⃗

= −23𝑎 + 2𝑏

= 4(−16𝑎 +

12𝑏)

, that is a scalar multiple. Hence, is parallel to , P is a common point, M must lie on PN and P, M and N lie on the same straight line, that is they are collinear.

Q.E.D

d. Data: and

Required To Calculate: Length of AN. Calculation: 𝐴𝑁PPPPPP⃗ = 𝐴𝑂PPPPP⃗ + 𝑂𝑁PPPPPP⃗ = −𝑎 + 2𝑏

= −(62) + 2(12)

= (−4 2)

Length of

PMPN 4=\ PN PM

÷÷ø

öççè

æ=26

a ÷÷ø

öççè

æ=21

b

( ) ( )22 24 +-=AN

units52

20

=

=


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14. a. Data: and .

Required To Calculate: Calculation:

and

b. Data: Required To Find: matrix that maps back to Q. Solution:

Let

If M maps Q onto , then maps onto Q. det

÷÷ø

öççè

æ-=

1502

X ÷÷ø

öççè

æ -=

7314

Y

YX +2

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

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ´+´´+-´´+´´+-´-

=

÷÷ø

öççè

æ=

÷÷ø

öççè

æ-÷÷ø

öççè

æ-=

1504

1105512510025022

1502

1502

2221

1211

2

eeee

X

÷÷ø

öççè

æ-

-=

÷÷ø

öççè

æ -+÷÷ø

öççè

æ-

=+

8218

7314

15042 YX

QQ ¢¾¾ ®¾÷÷ø

öççè

æ3121

22´ Q¢

÷÷ø

öççè

æ¾¾ ®¾÷÷

ø

öççè

æ

¢¾¾ ®¾

÷÷ø

öççè

æ

÷÷ø

öççè

æ

75

21 31

21

3121

QQ

÷÷ø

öççè

æ=

3121

M

Q¢ 1-M Q¢( ) ( )1231 ´-´=M

123

=-=


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by

Note: The coordinates of Q and are totally irrelevant in this question.

c. Data: (i) (a) Required To Calculate: Scale factor k. Calculation:

Let the enlargement be L. L has a centre O and scale factor k. L

may be represented by .

Equating corresponding entries.

OR

(b) Required To Find: Coordinates of and Solution: The translation matrix for L is

and similarly

( )( )

÷÷ø

öççè

æ-

-=

÷÷ø

öççè

æ-

-=-

1123

1123

111M

QQ ®¢\ ÷÷ø

öççè

æ-

-1123

Q¢

FEDDEF ¢¢¢¾¾¾ ®¾ tEnlargemen

÷÷ø

öççè

æk

k00

÷÷

ø

ö

çç

è

æ=÷÷

ø

öççè

æ

÷÷

ø

ö

çç

è

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ

18217

125

18217

125

00

kk

kk

211

2175

=

=

k

k

211

1812

=

=

k

k

E¢ F ¢

÷÷÷÷

ø

ö

çççç

è

æ

=

2110

0211

L


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and

and

(ii) Data: undergoes a clockwise rotation of 90° about the origin.

(a) Required To Find: matrix that represents the transformation.

Solution: The matrix that represents a 90° clockwise rotation about O

is .

(b) Required To Find: Coordinates of , and Solution:

Hence , and .

(c) Required To Find: matrix that maps DEF onto . Solution:

÷÷

ø

ö

çç

è

æ=÷÷

ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ

62110

123

4782

2110

0211

÷øö

çèæ=¢

2110,3E ( )6,12=¢F

FED ¢¢¢22´

22´

÷÷ø

öççè

æ- 01

10

D ¢¢ E ¢¢ F ¢¢

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

÷÷÷÷

ø

ö

çççç

è

æ

---=

÷÷÷÷

ø

ö

çççç

è

æ

´+´-÷øö

çèæ -´+´-´+÷

øö

çèæ ´-

´+´÷øö

çèæ ´+´´+÷

øö

çèæ ´

=

÷÷÷÷

ø

ö

çççç

è

æ

÷÷ø

öççè

æ-

123217

6211018

601212110031180

2171

611202110130181

2170

6211018

123217

0110

÷øö

çèæ -=¢¢

217,18D ÷

øö

çèæ -=¢¢ 3,

2110E ( )12,6 -=¢¢F

22´ FED ¢¢¢¢¢¢


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Hence, the single matrix that maps onto is

.

( ) ( )

( ) ( )

÷÷÷÷

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Date post:	01-Jun-2020
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CXC JUNE 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER … · Calculation: Midpoint of b. Data: The...

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