june 2009 cxc mathematics general proficiency (paper … · added to the square of the number in...

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JUNE 2009 CXC MATHEMATICS GENERAL PROFICIENCY (PAPER 2)

Section I

1. a. (i) Required To Calculate:

Calculation: Numerator:

(ii) Required To Calculate:

Calculation:

526

511

322 +

( ) ( )

)(48299658

325

15585321558526

15133

rDenominatoNumerator

15133

1513253

511

322

exact=

=

´=

=

=

=+

+

÷øö

çèæ250256.0

fromstandardin102.3

032.0250256.0

2-´=

=÷øö

çèæ

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 29

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b. Data: Truck driver earns $560 for a 40 hour work week.

(i) Required To Calculate: His hourly rate. Calculation:

Hourly rate of the driver

(ii) Data: Overtime is paid at one and a half times the basic rate. Required To Calculate: Overtime wage for 10 hours of overtime. Calculation:

Overtime rate

Overtime wage for 10 hours of overtime

(iii) Required To Calculate: Total wages earned by driver for a 55 hour

week. Calculation: For a 55 hour work week, total wages = $560 for the first 40 hours + for the next 15 hours of overtime

2. a. Required To Factorise: (i).

Solution: (i) (ii) (This is a difference of 2 squares) (iii)

40560$

=

hourper14$=

hourper14$211 ´=

hourper21$=

1021´=210$=

( )1521$ ´( )

875$315$560$

1521$560$

=+=

´+=

1543)(and205)(,3232 22 -+---+ xxiiixiibybxayax

( ) ( )yxbyxabybxayax 32323232 +-+=--+( )( )bayx -+= 32

( )45205 22 -=- xx( ) ( ){ }22 25 -= x

( )( )225 +-= xx

( )( )3531543 2 +-=-+ xxxx


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b. Data: Cost of biscuits at $x per pack and ice cream at $y per cup. (i) Required To Find: Pair of simultaneous equations in x and y to represent

the information given. Solution: 1 pack of biscuits and 2 cups of ice cream cost Hence, Similarly, 3 packs of biscuits and 1 cup of ice cream cost

. Hence, (ii) Required To Solve: Equations to find x and y. Solution: Let …(1) …(2) From equation (1) Substitute in (2)

Substitute 1 pack of biscuits costs $2 and 1 cup of ice cream costs $3.

3. a. Data: Survey results of 50 students who owned cellular phones and digital cameras. (i) Required To Complete: Venn diagram to represent the information

given. Solution:

( ) ( ){ }yx ´+´ 21$)data(82 =+ yx

( ) ( ){ }yx ´+´ 13$93 =+ yx

82 =+ yx93 =+ yx

yx 28 -=

( )

31555924952496249283

===-=-=+-=+-

yy

yyyyyy

3=y ( )328 -=x2=

\


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(ii) Required To Find: Expression in terms of x for the total number of students in the survey.

Solution: Total number of students in the survey

(iii) Required To Calculate: x Calculation:

b. (i) Required To Construct: Rhombus PQRS, with diagonal PR = 6 cm and

. Solution:

Since , then . Arc PS = Arc RS = Arc PQ.

( ) ( )x

xxxx+=

+-++-=41

21823

94150

)data(5041

=-=

=+xx

°= 60ˆQPR

°= 60ˆRPQ °= 60ˆPRQ


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(ii) Required To Find: Length of QS. Solution: PS = 10.4 cm (by measurement).

4. a. Data: Table showing the readings of an aircraft flight record. (i) Required To Calculate: Distance travelled in km. Calculation: The distance travelled in km = Final reading – Previous reading

(ii) Required To Calculate: Average speed of the aircraft in kmh-1

Calculation: The time of flight _

Hence, average speed

b. Data: Map with a scale of 1 : 50 000 showing two points L and M.

(i) Required To Measure: Distance on the map from L to M. Solution: Distance LM on the map = 7.0 cm (by measurement) (ii) Required To Calculate: Actual distance from L to M. Calculation:

Actual distance from L to M

(iii) Data: Actual distance on the map between 2 points is 4.5 km. Required To Calculate: Distance on the map between the 2 points.

km1269571083

=-=

07:09=

hours6012

minutes1255:08

=

takentimeTotalcovereddistanceTotal

=

1kmh6306012126

-=

=

km1000001000507´

´=

km5.3=


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Calculation: Actual distance between 2 point = 4.5 km Distance between 2 points on map

5. a. Data: and (i) Required To Calculate: . Calculation:

(ii) Required To Calculate: . Calculation:

(iii) Required To Calculate: . Calculation: Let

Replace y by x

Hence, f-1 (3) = (3 + 5) / 2 = 4

b. Data: and an incomplete table of x and y values.

(i) Required To Complete: Table of x and y values. Solution:

cm10000015.4 ´´=

( ) 0005010000015.4 ÷´´=cm9=

( ) 52 -= xxf ( ) 312 -= xxg( )2-f

( ) ( )

9545222

-=--=

--=-f

( )1gf

( ) ( )

( ) ( )( )

22319

313

313525121

2

-=-=

--=

-=-=-=-=

ggf

f

( )31-f

25

2552

+=

=+-=

yx

xyxy

( )251 +

=- xxf

322 -+= xxy


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x -4 -3 -2 -1 0 1 2 y 5 (0) -3 -4 -3 (0) 5 When

When (ii) Required To Draw: Graph of for . Solution:

6. Data: Diagram showing triangles A, B and D. a. (i) Required To Find: Transformation which maps triangle A onto triangle D. Solution:

3-=x ( ) ( ) 3323 2 --+-=y

0369

=--=

1=x ( ) ( ) 3121 2 -+=y0=

322 -+= xxy 24 ££- x


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Triangle A is mapped onto triangle D by a horizontal shift of 3 units to the right and 4 units vertically upwards. This may be represented by the

translation, T, where .

(ii) Required To Find: Transformation which maps triangle A onto triangle

B. Solution: Since orientation changes, B is a rotation of A.

The above diagram shows how to obtain the centre of rotation. Join two pairs of corresponding image and object points. Then construct the perpendicular bisector of each line. The point at which the two bisectors meet is the center of rotation. The origin is therefore the center of rotation.

÷÷ø

öççè

æ=43

T


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To obtain the angle of rotation measure the angle AOB. Triangle A is mapped onto triangle B by a rotation of 90° clockwise about O.

(iii) Required To Find: Coordinates of the vertices of triangle C. Solution: Triangle A is mapped onto triangle C after a reflection in the line .

The vertices of C are (1, 2), (1, 5) and (3, 5).

BA ¾¾¾ ®¾÷÷ø

öççè

æ- 01

10

xy =

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ

¾¾¾¾¾ ®¾ =

552311

311552

0110

inReflection CA xy


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7. Data: Table showing the waiting time of students at a school canteen. a. Required To Complete: Table showing cumulative frequency. Solution: The data is that of a continuous variable.

Waiting time in minutes, t

L.C.B U.C.B. No. of students, f

Cumulative frequency

Points to be plotted

(U.C.B., C.F.) (0.5, 0)

1 – 5 4 4 (5.5, 4) 6 – 10 7 11 (10.5, 11) 11 – 15 11 22 (15.5, 22) 16 – 20 18 40 (20.5, 40) 21 – 25 22 62 (25.5, 62) 26 – 30 10 72 (30.5, 72) 31 – 35 5 77 (35.5, 77) 36 – 40 3 80 (40.5, 80)

The point (0.5, 0) is obtained by extrapolation so that the curve starts from the horizontal axis.

5.55.0 <£ t5.105.5 <£ t5.155.10 <£ t5.205.15 <£ t5.255.20 <£ t5.305.25 <£ t5.355.30 <£ t5.405.35 <£ t


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b. Required To Complete: Cumulative frequency curve.

Solution:

c. (i) Required To Find: Median for the data: Solution:

From the graph, or the table of values, the median for the data = 20.5 minutes.

(ii) Required To Find: Number of students who waited for no more than 29

minutes. Solution: From the graph, 70 students waited no longer than 29 minutes. (iii) Required To Calculate: Probability a randomly chosen student waited no

more than 17 minutes. Solution:


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8. Data: Sequence of diagrams. Required To Draw: Next diagram in the sequence. Solution: Answer Sheet for Question 8

Diagram Number

Number of Unit Squares

Pattern for Calculating Number of Unit Squares

1 1 2 5 3 13 4 25

10 181 15 421

We notice that the expression in column 3 is the square of the number in column 1 added to the square of the number in column 1 of the previous row. Column 2 is the result of column 3. (i) Required To Complete: The table given for diagram 4.

Solution: Column 3 should be

( )

8027

studentsofno.Total17minswaitedwhostudentsofNo.minutes17waitedStudent

=

£=£P

22 01 +22 12 +22 23 +22 34 +22 910 +22 1415 +

22 34 +


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Column 2

(ii) Required To Complete: The table given for diagram 10.

Solution: Column 3 should be Column 2

(iii) Required To Complete: The table given for diagram with 421 unit

squares.

Solution: Let column 1 be n. Therefore, column 3 will be .

Hence, in (iii), column 1 is 15, and column 3 is .

22 34 +=

25916

=+=

22 910 +22 910 +=

18181100

=+=

( )22 1-+ nn

( )

0210042022421124211

2

2

22

22

=--

=--

=+-+

=-+

nnnnnnnnn

( )( )

15ve

14or1501415

=\-¹

-==+-

nnn

nn

22 1415 +


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Section II

9. a. Data: and . Required To Calculate: x and y. Calculation: Let …(1) and …(2) Equating (1) and (2)

When

When

Hence, and OR and .

b. Required To Express: in the form .

Solution:

Equating coefficient of .

Equating coefficient of x.

Equating constants.

xy 24 -= 132 2 +-= xxy

xy 24 -=132 2 +-= xxy

( )( )

211or1

0132032024132

13224

2

2

2

-=

=+-=--

=+-+-

+-=-

x

xxxxxxx

xxx

1-=x ( )124 --=y6=

211=x ÷

øö

çèæ-=21124y

1=

1-=x 6=y211=x 1=y

132 2 +- xx ( ) khxa ++ 2

( ) ( )kahahxaxkhhxxakhxa+++=

+++=++22

222

22

2xÂÎ= 2a

( )

ÂÎ-=

=-

43223

h

h


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And .

OR

Half the coefficient of x is .

= *

Hence, is of the form , where

, and .

c. (i) Required To Find: The minimum value of . Solution: Let

ÂÎ-=

=+

=+÷øö

çèæ-

81

189

1432

2

k

k

k

81

432132

22 -÷

øö

çèæ -º+- xxx

1232132 22 +÷

øö

çèæ -=+- xxxx

43

23

21

-=÷øö

çèæ-

*432

2

+÷øö

çèæ -= x

18181132

169

232 22

-

+-=÷øö

çèæ +- xxxx

81

432132

22 -÷

øö

çèæ -º+- xxx ( ) khxa ++ 2

ÂÎ= 2a ÂÎ-=43h ÂÎ-=

81k

132 2 +- xx

81810

0432

81

432

132

min

2

2

2

-=

-=\

"³÷øö

çèæ -

-÷øö

çèæ -=

+-=

y

xx

x

xxy


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(ii) Required To Find: x at which minimum value occurs. Solution: At

Minimum value of is at .

OR The minimum value of occurs at

When the minimum value is

d. Required To Sketch: The graph of showing coordinates of the

minimum point, the y intercept and the values of x where the graph cuts the x – axis.

Solution:

Let

43

043

0432

2

2

=

=÷øö

çèæ -

=÷øö

çèæ -

x

x

x

132 2 +- xx81

-43

=x

132 2 +- xx( )( )

43223

=

--=x

43

=x

81

412

812

1412

811

1412

16921

433

432

2

-=

-=

+-=

+-÷øö

çèæ=+÷

øö

çèæ-÷

øö

çèæ

132 2 +- xx

81

432

1322

2

-÷øö

çèæ -=

+-=

x

xxy


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OR

and the curve cuts the x – axis at 1 and .

When y cuts the y – axis at 1.

at .

is the minimum point of .

21or1

41

4341

43

161

43

81

432

081

432

2

2

2

=

±=

±=-

=÷øö

çèæ -

=÷øö

çèæ -

=-÷øö

çèæ -

x

x

x

x

x

( )( )

1or2101120132 2

=

=--=+-

x

xxxx

21

0=x ( ) ( ) 10302 2 +-=y1=

81

min -=y43

=x

÷øö

çèæ -\

81,

43 132 2 +-= xxy


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10. Data: Conditions involving a shop’s demands when buying x guitars and y violins. a. (i) Required To Find: The inequality to represent the information given. Solution: No. of violins must be less or equal to the no. of guitars. …(1) (ii) Data: Cost of one guitar is $150 and the cost of one violin is $300. Required To Find: The inequality to represent the information given. Solution: The cost of x guitars at $150 each and y violins at $300 each is . Maximum amount to spend is $4 500.

(iii) Data: Owner of the shop must buy at least 5 violins. Required To Find: The inequality to represent the information given. Solution: The owner must buy at least 5 violins. …(3)

xy £

( ) ( )yx ´+´ 300$150$

)2...(302150

5004300150

£+÷

£+\

yx

yx

5³\ y


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b. (i)-(ii) Required To Draw: The graphs of the lines associated with the inequalities. Solution: The line is a straight horizontal line. The region which satisfies is

Obtaining two points on the line .

x y 0 0 30 30

The region with the smaller angle represents the region.

The region which satisfies is

Obtaining 2 points on the line .

5=y5³y

xy =

£

xy £

302 =+ yx


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x y 0 15 30 0

The region with the smaller angle represents the region.

The region which satisfies is

The feasible region is the area in which all three previously shaded regions overlap.

£

302 £+ yx


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(iii) Required To State: The coordinates of the vertices of the shaded region. Solution:

The vertices of the region that satisfies all the inequalities in ABC (the feasible region). A (5, 5) B (10, 10) C (20, 5)

c. Data: Profit of $60 made on each guitar and a profit of $100 made on each violin.

(i) Required To Express: The total profit in terms of x and y. Solution: Let the profit on x guitars at $60 each and y violins at $100 each be P.

(ii) Required To Calculate: Maximum profit. Calculation: Testing (10, 10)

( ) ( )yxP

yxP10060

10060+=

´+´=\

( ) ( )6001$

101001060=

+=P


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Testing (20, 5)

Maximum profit = $1 700 when shop sells 20 guitars and 5 violins.

11. a. Data: Diagram showing a circle with centre O. Tangent DCE touches the circle at C. and .

(i) Required To Calculate: Calculation:

(The angle made by a tangent to a circle and chord, at the point of contact is equal to the angle in the alternate segment).

(ii) Required To Calculate: Calculation:

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

(iii) Required To Calculate: Calculation:

(The base angles of an isosceles triangle are equal).

( ) ( )7001$

51002060=

+=P

°= 48ˆECA °= 26ˆBCO

CBA ˆ

°= 48ˆCBA

COA ˆ

( )°=°=

96482ˆCOA

DCB ˆ

ACOCAO

OAOCˆˆ

)radii(

=

=


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(The sum of the angles in a triangle = 180°).

(Sum of angles in a triangle = 180°). (iv) Required To Calculate: Calculation:

(The angle made by a tangent to a circle and a chord, at the point of contact equal the angle in the alternate segment).

(v) Required To Calculate: Calculation: (from iii) (iv) Required To Calculate: Calculation:

b. Data: Diagram showing two hurricane tracking stations relative to a fixed point.

(i) Required To Complete: Diagram labelling the bearings 025° and 080°. Solution:

°=

°-°=

42296180ˆACO

( )°=

°+°+°-°=64

264248180ˆDCB

CAB ˆ

°= 64ˆCAB

CAO ˆ

°= 42ˆCAO

BAO ˆ

°=°-°=

°=

224264ˆ

64ˆ

BAO

CAB

P

N


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(ii) (a) Required To Calculate: Calculation:

(b) Required To Calculate: Length of OQ. Calculation:

(c) Required To Calculate: Bearing of Q from O. Calculation: Let

The bearing of Q from O

12. NO SOLUTION HAS BEEN OFFERED FOR THIS QUESTION SINCE IT IS BASED ON EARTH GEOMETRY WHICH IS NO LONGER ON THE SYLLABUS

13. a. Data: and .

QPO ˆ

°=°+°=

°=°-°=

°=

12525100ˆ

)linestraightainAngles(10080180ˆ

)anglesAlternate(25ˆ

QPO

SQP

SPO

( ) ( ) ( )( )

km)nearesttheto(km985km496.9858.971202

)(cos125cos7004002700400 222

===

°-+=

OQ

lawOQ

°= qQOP ˆ

°==

°=

°=

6.355818.0

496.985125sin700sin

)(sin125sin496.985

sin700

q

q

qlaw

°+°= 6.3525°= 6.060

÷÷ø

öççè

æ-=

52

OA ÷÷ø

öççè

æ=24

OB


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(i) Required To Express: and in the form .

Solution:

is of the form , where x = 6 and y = - 3

is of the form , where and .

(ii) Required To Express: in the form .

Solution:

AB AG ÷÷ø

öççè

æyx

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ+÷÷ø

öççè

æ--=

+=

36

24

52OBAOAB

÷÷ø

öççè

æyx

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-

=

=

12

36

3131 ABAG

÷÷ø

öççè

æyx

2=x 1-=y

OG ÷÷ø

öççè

æyx


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is of the form , where and .

b. Data: A vector diagram with B, the midpoint of OX, C the midpoint of AB and

OD = 2DA, 𝑂𝐴#####⃗ = 3𝑎 and 𝑂𝐵#####⃗ = 𝑏.

𝑂𝐵#####⃗ = 𝑏𝐵𝑋#####⃗ = 𝑏

𝑂𝐷 = 2𝐷𝐴𝑂𝐷######⃗ = -

.𝑂𝐴#####⃗

= -.(3𝑎)

= 2𝑎𝐷𝐴#####⃗ = 𝑎

(i) (a) Required To Find: in terms of a and b. Solution:

𝐴𝐵#####⃗ = 𝐴𝑂#####⃗ + 𝑂𝐵#####⃗ = −(3𝑎) + 𝑏 = −3𝑎 + 𝑏

(b) Required To Find: in terms of a and b.

Solution:

÷÷ø

öççè

æ=

÷÷ø

öççè

æ-

+÷÷ø

öççè

æ-=

+=

40

12

52AGOAOG

÷÷ø

öççè

æyx

0=x 4=y

AB

AC


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𝐴𝐶#####⃗ =12𝐴𝐵#####⃗

=12 (−3𝑎 + 𝑏)

= −112𝑎 +

12𝑏

(c) Required To Find: in terms of a and b. Solution:

𝐷𝐶#####⃗ = 𝐷𝐴#####⃗ + 𝐴𝐶#####⃗ = 𝑎 + (−1 5

-𝑎 + 5

-𝑏)

= − 5-𝑎 + 5

-𝑏

(d) Required To Find: in terms of a and b. Solution:

𝐷𝑋#####⃗ = 𝐷𝑂######⃗ + 𝑂𝑋#####⃗ = −(2𝑎) + 2𝑏 = −2𝑎 + 2𝑏

(ii) Required To Find: Two geometrical relationships between DX and DC. Solution:

𝐷𝑋#####⃗ = −2𝑎 + 2𝑏

= 4(−5-𝑎 + 5

-𝑏)

= 4𝐷𝐶#####⃗

and is parallel to .

(iii) Required To Find: One geometrical relationship between points D, C and

X. Solution: Since and are parallel and D is a common point, then C must lie

on DX. Therefore, D, C and X lie on the same straight line OR are collinear.

DC

DX

DCDX ´=\ 4 DX DC

DC DX


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14. a. Data: .

Required To Calculate: x Calculation:

Hence,

b. Data: and .

R represents a reflection in the y – axis. S represents a 90° clockwise rotation about O. (i) Required To Find: Matrix to represent combined transformation S

followed by R. Solution: The combined transformation S followed by R means S first and R second.

This is written as RS and is

(ii) Required To Calculate: Image of the point (5, -2) under the combined

transformation: Calculation: (5, -2) under RS is

The image is (2, -5).

c. Data:

(i) Required To Calculate: Calculation:

13detand34

=÷÷ø

öççè

æ= M

xx

M

( ) ( )12

34det2 -=

´-´=

xxxM

5251312

2

2

±==\

=-

xx

x

÷÷ø

öççè

æ -=

1001

R ÷÷ø

öççè

æ-

=0110

S

÷÷ø

öççè

æ-

-=÷÷

ø

öççè

æ-÷÷

ø

öççè

æ -0110

0110

1001

( ) ( )( ) ( )

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-´+´--´-+´

=÷÷ø

öççè

æ-÷÷

ø

öççè

æ-

-

52

20512150

25

0110

÷÷ø

öççè

æ -=

5213

N

1-N


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(ii) Required To Solve: For x and y in and . Solution:

Equating corresponding entries. and

( ) ( )

( )( )

÷÷÷÷

ø

ö

çççç

è

æ

-=

÷÷ø

öççè

æ-

--=\

=+=

´--´=

-

173

172

171

175

3215

171

17215

2153det

1N

N

53 =- yx 952 =+ yx

53 =- yx952 =+ yx

3 −12 5

⎛

⎝⎜

⎞

⎠⎟

xy

⎛

⎝⎜

⎞

⎠⎟ =

59

⎛

⎝⎜⎞

⎠⎟

×N −1

3 −12 5

⎛

⎝⎜

⎞

⎠⎟ × N

−1 × xy

⎛

⎝⎜

⎞

⎠⎟ = N

−1 59

⎛

⎝⎜⎞

⎠⎟

I xy

⎛

⎝⎜

⎞

⎠⎟ = N

−1 59

⎛

⎝⎜⎞

⎠⎟

xy

⎛

⎝⎜

⎞

⎠⎟ =

517

117

− 217

317

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

59

⎛

⎝⎜⎞

⎠⎟

=

2517

+ 917

−1017

+ 2717

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

= 21

⎛

⎝⎜⎞

⎠⎟

2=x 1=y


june 2009 cxc mathematics general proficiency (paper … · added to the square of the number in...

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