07. CSEC Maths JANUARY 2008 · (ii) Required To Determine: Smallest value of x that satisfies the above inequality. Solution: ... The formula for the sum is a constant, , example

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CSEC JANUARY 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER 2)

Section I

1. a. (i) Required To Calculate:

Calculation: Numerator:

Denominator:

Hence,

(ii) Required To Calculate:

Calculation:

51

212

43

711

´

-

( ) ( )

281128213228

378443

78

43

711

=

-=

-=

-=-

21

51

25

51

212

=

´=´

form)exactin(1411

12

2811212811

51

212

43

711

=

´=

=´

-

15.024.02 -

form)exactin(4.0

6.1215.024.02

=

-=-

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 26

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b. Data: Bicycle with prices for cash or hire purchase (H.P.) payments. (i) Required To Calculate: Total hire purchase price for the bicycle. Calculation:

Hire purchase price of the bicycle = Deposit + Total for 10 installments

(ii) Required To Calculate: Difference between hire purchase price and cash

price. Calculation: Difference between the hire purchase price and the cash price

(iii) Required To Express: Difference between 2 prices as a percentage of the

cash price. Solution: The difference in price as a percentage of the cash price

2. a. (i) Required To Solve: . Solution:

(ii) Required To Determine: Smallest value of x that satisfies the above

inequality. Solution:

If , then the smallest x is 0.

( )00.354$

50.28$1000.69$=

+=

05.34$95.319$00.354$

=-=

places)decimal2to(%64.10%642.10

%10095.31905.34

==

÷øö

çèæ ´=

723 <- x

22

421

42372

723

->÷

->-´

<--<-

<-

x

x

xxx

{ },...3,2,1,0=WWxÎ


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b. Required To Factorise: (i) Solution: (i) (ii) This is a difference of 2 squares

And (iii)

c. Data: Table showing types of cakes, cost and the number sold. (i) Required To Find: An expression, in terms of k, for the amount of money

collected for the sale of sponge cakes. Solution: The cost of 2 sponge cakes at each (ii) Required To Find: An expression in terms of k, for the total amount of

collected. Solution: Cost of sponge cakes Cost of 10 chocolate cakes at $k each Cost of 4 fruit cakes at $2k each Total amount of money collected (iii) Data: Total amount of money collected = $140.00

Required To Calculate: k Calculation: Total amount of money collected = $140.00

pqpqpiiiaiixyx +---- 222 22)(,1)(,

yxxxxyx ..2 -=-( )yxx -=

( ) ( )222 11 -=- aa

( )( )1112 +-=- aaa

( ) ( )qppqppqpqp ---=+-- 222 2

( )( )pqp --= 2

( )5$ +k ( )52 +´= k( )102$ += k

( )102$ += k

10´= kk10$=

42 ´= kk8$=

( ) kkk 810102 +++=( )1020$ += k

10140201401020

-==+\

kk

5.62013013020

=

=

=

k

k


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3. a. Data: Given the elements of universal set, U, S and T. (i) Required To Draw: Venn diagram to represent the information given. Solution:

(ii) (a) Required To List: Members of . Solution: as shown.

(b) Required To List: Members of . Solution:

as shown.

b. Data: Quadrilateral ABCD with AB = AD, , and AB parallel to DC.

(i) Required To Calculate: Calculation:

(Co-interior angles are supplementary).

{ }{ }{ }pkTS

qpkTpmlkS

,,,,,,

=Ç==

TS È

{ }qpkmlTS ,,,,=È

S ¢

{ }nqrS ,,=¢

°= 90ˆDCB °= 42ˆCBD

CBA ˆ

°=°-°=

9090180ˆCBA


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(ii) Required To Calculate: Calculation:

(iii) Required To Calculate: Calculation: (data) (Base angles of an isosceles triangle are equal).

(Sum of angles in a triangle = 180°).

4. a. Data: John left Port A at 07:30 hours and travelled to Port B in the same time

zone. He arrives at Port B at 14:20 hours. (i) Required To Find: Duration of the journey. Solution: Departure time from A is 07:30 hours. Arrival time at B is 14:20 hours. Duration of the journey

(ii) Data: John travelled 410 km.

Required To Calculate: Average speed. Calculation:

b. Data: Diagram showing a circle of radius 3.5 cm, centre O and square OPQR.

DBA ˆ

°=°-°=

484290ˆDBA

DAB ˆ

ADAB =°=\ 48ˆBDA

( )°=

°+°-°=84

4848180ˆDAB

20:14=

hours656

minutes50hours6

50:6

30:7

=

=

-

1kmh60

hours656

km410takentimeTotalcovereddistanceTotalspeedAverage

-=

=

=


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(i) Required To Calculate: Area of the circle. Calculation:

Area of circle

(ii) Required To Calculate: Area of square OPQR.

Calculation: Area of square OPQR

(iii) Required To Calculate: Area of the shaded region. Calculation: Area of the shaded region Area of square OPQR – Area of sector OPR.

5. Data: Bar graph illustrating the number of books read by the boys of a book club. a. Required To Draw: The frequency data representing the information given. Solution: From the bar graph

No. of books read, x No. of boys, f 0 2 1 6 2 17 3 8 4 3

b. Required To Find: The number of boys in the book club.

( )25.3722

=

2cm5.38=

5.35.3 ´=2cm25.12=

=

( )

places)decimal2to(cm63.2cm562.2

5.384125.12

2

2

=

=

-=

å = 36f


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Solution: No. of boys in the book club

c. Required To Find: Modal number of books read.

Solution: Modal number of books read = 2, as corresponds to the maximum frequency, 17.

d. Required To Calculate: Total number of books read.

Calculation: Total number of books read

e. Required To Calculate: The mean number of books read.

Calculation: The mean number of books read, .

This value is a whole number and may be represented by the integer 2.

f. Required To Calculate: Probability a randomly chose boy reads at least books.

Calculation:

6. a. Data: Diagram showing a pattern of congruent right angled triangles as

å= f36=

2=x

( ) ( ) ( ) ( ) ( )43382171602 ´+´+´+´+´=

761224346

=+++=

x

1.23676

=

=

=ååffx

x

( )

3611

3638

boysofno.Totalbooks3readingboysofNo.books3readsboychosenRandomly

=

=+

=

³=³

å f

P


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(i) Required To Describe: Transformation that maps onto . Solution:

L is a common point. Image is re-oriented and congruent. BLF is a straight line.

is mapped onto HFL by a rotation of 180° (clockwise or anti-clockwise) about L.

(ii) Required To Describe: Transformation that maps onto .

Solution:

and have the same ‘orientation’.

is mapped onto by a horizontal shift 4 units to the right and

1 unit vertically downwards. This may be represented by the translation, T

where .

b. (i) Required To Construct: Parallelogram WXYZ, in which, WX = 7.0 cm, WZ = 5.5 cm and . Solution:

BCLD FHLD

HFLBCL D®D

BCL\

BCLD HFGD

BCLD HFGD

BCLD HFGD

T =4-1

æ

èç

ö

ø÷

°= 60ˆZWX


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(ii) Required To Find: The length of WY. Solution: WY = 10.9 cm (by measurement.)

7. Data: Incomplete table for . a. Required To Complete: The table of values for . Solution:

x -1 0 1 2 3 4 5 y 5 0 -3 -4 -3 0 5

When When When

xxy 42 -=xxy 42 -=

0=x ( ) ( )040 2 -=y0=

3=x ( ) ( )343 2 -=y3-=

5=x ( ) ( )545 2 -=y5=


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b. Required To Draw: The graph of . Solution:

c. (ii) Required To Find: Points at which the curve meets the line. Solution: The line and the curve meet at and .

(iii) Required To Find: The equation whose roots are the coordinates found

above. Solution:

and are the roots of the equation or .

xxy 42 -=

2=y xxy 42 -= 5.0-=x 5.4=x

5.0-=x 5.4=x 242 =- xx0242 =-- xx


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8. a. Data: Table showing the sum of rational numbers. Required To Complete: The table given.

n SERIES SUM FORMULA 1 1 1

2 1 + 2 3

3 1 + 2 + 3 6

4 1 + 2 + 3 + 4 10

5 1 + 2 + 3 + 4 + 5 15

(i) 6 1 + 2 + 3 + 4 + 5 + 6 21

8 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 36

(ii) n 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …+

n

The formula for the sum is a constant, , example

When Sum is

When Sum

For (ii), the formula is .

b. Data:

1 + 2 + 3 = 6 and = 36

So too, and

( )( )11121

+

( )( )12221

+

( )( )13321

+

( )( )14421

+

( )( )15521

+

( )( )16621

+

! ! ! !

( )( )18821

+

! ! ! !

( )121

+nn

( )121

+nn

5=n ( )( )15521

+= 15=

15Sum =\

6=n ( )( )16621

+= 21=

21Sum =\

( )121

+nn

2333 6321 =++

104321 =+++ 23333 104321 =+++100=


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(i) Required To Find: Solution: From the table

(ii) Required To Find:

Solution:

c. Required To Find:

Solution:

9. a. Data: Volume, V of a gas varies inversely as the pressure, P, with temperature constant. (i) Required To Find: Equation relating V to P. Solution:

(k is constant of proportion or variation) (ii) Required To Find: k when and . Solution: when

33333333 87654321 +++++++

3687654321 =+++++++

12963687654321 233333333

==+++++++

3333 ...321 n++++

( )

( )2

3333 121...321

121...321

þýü

îíì +=++++\

+=++++

nnn

nnn

33333 12...4321 +++++

( )( )

( )6084784321

78

112122112...4321

23333

==+++\

=

+=+++++

PkV

PV

1

1

´=

µ

8.12=V 500=P

8.12=V 500=P

64005008.12

50018.12

=´=

´=

k

k


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(iii) Required To Calculate: V when . Calculation: When

b. Data: Right-angled triangle of sides .

(i) Required To Find: An equation in terms of a to relate the three sides,

using Pythagoras’ Theorem. Solution:

(ii) Required To Calculate: a

Calculation:

If , the side would compute to be a negative value. Hence, only.

(iii) Required To State: The lengths of the 3 sides of the triangle.

Solution: The lengths of the 3 sides are

480=P

3113

48016400

480

=

´=

=

V

P

( ) ( )1and7, +- aaa

( ) ( ) ( )

04816124914

Theorem'Pythagoras17

2

222

222

=+-

++=+-+

+=-+

aaaaaaa

aaa

( )( )12or4

0412048162

==--=+-

aaaaa

4=a ( )7-a12=a


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The sides are 5, 12 and 13 units.

10. a. Data: School buys x balls and y bats. (i) Required To Find: Inequality for the information given.

Solution: Total number of balls and bats is no more than 30. Hence, …(1)

(ii) Data: School allows no more than $360 to be spent on bats and balls. The cost of a ball is $6 and the cost of a bat is $24. Required To Find: Inequality to represent the information given. Solution: The cost of x balls at $6 each and y bats at $24 each is

Budget allows no more than $360. Similarly,

b. Required To Draw: The graphs of the inequalities shown above, shade the

region that satisfies the inequalities and state the vertices of the feasible region. Solution: and identifies the 1st quadrant. Obtaining 2 points on the line When The line passes through the point (0, 30). When The line passes through the point (30, 0).

yx + 30£30£+ yx

( ) ( ) yxyx 246246 +=´+´

)1...(6046

360246

£+÷

£+

yx

yx

0³x 0³y

30=+ yx0=x 300 =+ y

30=y30=+ yx

0=y 300 =+x30=x

30=+ yx


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The region with the smaller angle corresponds to the region. The region which satisfies is

Obtaining 2 points on the line . When

The line passes through the point (0, 15). When The line passes through the point (60, 0).

The region with the smaller angle corresponds to the region. The region which satisfies is

£30£+ yx

604 =+ yx0=x 6040 =+ y

15460604

=

=

=

y

y

604 =+ yx0=y ( ) 6004 =+x

60=x604 =+ yx

£604 £+ yx


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aths.c

om The feasible region is the area in which both previously shaded regions overlap.


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The vertices of the feasible region are O (0, 0), A (0, 15), B (20, 10) and C (30, 0).

c. Data: Profit made of each ball is $1 and profit made on each bat is $3.

(i) Required To Find: The profit for reach of the combinations above. Solution:

Testing the point (0, 15), (30, 0) and (20, 10). When and

yxP 3+=

0=x 15=y( )

45$1530

=+=P


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When and

When and

As a point of interest, the only point to be considered should be (20, 10), where and since the question specifically indicates – a school buys x balls AND y balls. (They cannot buy 0 bats or 0 balls, that is x and y .)

(ii) Required To Find: Maximum profit that may be made.

Solution: The maximum profit, , which occurs when and .

11. a. Data: O is the centre of the circle WXY and

(i) Required To Calculate:

Calculation:

(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).

30=x 0=y( )

30$0330

=+=P

20=x 10=y( )

50$10320

=+=P

20=x 10=y

+Î Z

50$max =P 20=x 10=y

°= 50ˆYXW

YOW ˆ

( )°=°=

100502ˆYOW


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

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

b. Data: Three buoys A, B and C, their relative distances apart and their positions. (i) Required To Sketch: Diagram showing the information given.

Solution:

(ii) Required To Calculate: Distance AC. Calculation:

(cos law)

YWO ˆ

OYOW =

°=

°-°=

=

402100180

)(ˆˆ equalaretriangleisoscelesanofanglesbasetheWYOYWO

( ) ( ) ( )( )

m1.134m14.134

80cos75125275125

801090ˆ

222

==

°-+=

°=°-°=

ACACAC

CBA


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(iii) Required To Calculate: Bearing of C from A. Calculation: Let

The bearing of C from A

12. a. Data: Diagram illustrating a vertical pole and tower standing on horizontal ground.

(i) Required To Calculate: Horizontal distance AB. Calculation: (alternate angles).

°=

°=

°=

°=

4.33134.175sin80sin

Law)Sine(80sin1.134

sin75

ˆ

a

a

a

aCAB

°+°= 4.3390degreenearesttheto123°=

°= 5ˆABD

tan5° = 2.5AB

AB = 2.5tan5°

= 28.57m= 28.6m (to 1 decimal place)


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(ii) Required To Calculate: Height of the tower BC. Calculation:

(Opposite sides of a rectangle).

Height of tower

b. No solution has been offered for this question as it is based on latitude and longitude (Earth Geometry) which has been removed from the syllabus)

13. a. Data: P and Q are the midpoints of AB and BC of vector triangle ABC. (i) Required To Sketch: Diagram to show the information given. Solution:

57.28=DE

m39.1020tan57.28

57.2820tan

=°=

=°

CE

CE

39.105.2 +=BC

m9.12m98.12

==

( )

y

y

y

QCBQ

yBC

xPBAP

xAB

211

23

321

3

2

=

=

=

=\

=

==\

=


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(ii) (a) Required To Find: Expression in terms of x and y for . Solution:

(b) Required To Find: Expression in terms of x and y for .

Solution:

(iii) Required To Prove:

Proof:

Q.E.D.

b. Data: and

AC

yxBCABAC32 +=+=

PQ

yx

BQPBPQ

211+=

+=

ACPQ21

=

ACPQ

PQ

yx

yxAC

yxPQ

212

2112

32211

=

=

÷øö

çèæ +=

+=

+=

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ=

61

,43OSOR ÷÷

ø

öççè

æ-

=25

OT


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

Solution:

(b) Required To Express: in the form .

Solution:

(ii) (a) Required To Find: The position vector of F. Solution: If , then F is the midpoint of .

OR

RT ÷÷ø

öççè

æba

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-

+÷÷ø

öççè

æ-=

+=

62

25

43OTRORT

SR ÷÷ø

öççè

æba

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ+÷÷ø

öççè

æ --=

+=

24

43

61ORSOSR

FTRF = RT

÷÷ø

öççè

æ=

÷÷÷÷

ø

ö

çççç

è

æ

-

+

=

14224253

OF


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(b) Required To State: Coordinates of F.

Solution:

If , then F = (4, 1).

14. a. Data: and

. Required To Calculate: x and y. Calculation:

If then

Equating corresponding entries.

÷÷ø

öççè

æ=

÷÷ø

öççè

æ-

+÷÷ø

öççè

æ=

+=

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-

=

=

14

31

43

31

62

2121

RFOROF

RTRF

÷÷ø

öççè

æ=14

OF

( ) ÷÷ø

öççè

æ-

==2

1,12

yx

BA ( )65=C

CAB =

( )

( ) ( ) ( ) ( )( )( )222

2121122

112

-+=-´+´´+´=

÷÷ø

öççè

æ-

=

xyxy

yx

AB

CAB =

( ) ( )65222 =-+ xy

352

==+yy

482622

===-

xx

x

3and4 ==\ yx


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b. Data: .

(i) Required To Show: R is a non-singular matrix. Proof: det = 6 + 1 = 7 Since , then R is non-singular.

(ii) Required To Find:

Solution:

(iii) Required To Prove:

Proof:

Q.E.D.

÷÷ø

öççè

æ -=

3112

R

( ) ( )1132 ´--´=R0¹R

1-R

( )( )

÷÷÷÷

ø

ö

çççç

è

æ

-=

÷÷ø

öççè

æ-

--=-

72

71

71

73

2113

711R

IRR =-1

÷÷ø

öççè

æ=

÷÷÷÷

ø

ö

çççç

è

æ

-÷÷ø

öççè

æ -=´ -

2221

1211

1

72

71

71

73

3112

eeee

RR

I

RR

e

e

e

e

=

÷÷ø

öççè

æ=\

=

÷øö

çèæ ´+÷

øö

çèæ ´=

=

÷øö

çèæ -´+÷

øö

çèæ ´=

=

÷øö

çèæ ´-+÷

øö

çèæ ´=

=

÷øö

çèæ -´-+÷

øö

çèæ ´=

-

1001

1723

711

0713

731

0721

712

1711

732

1

22

21

12

11


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(iv) Data: and

Required To Calculate: x and y by matrix method. Calculation:

…(1) …(2)

Equating corresponding entries. and

02 =- yx 73 =+ yx

02 =- yx73 =+ yx

÷÷ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ

-=÷÷

ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´

´

÷÷ø

öççè

æ=÷÷

ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ -

-

--

-

70

72

71

71

73

70

70

70

70

3112

1

11

1

yx

Ryx

I

Ryx

RR

R

yx

R

yx

÷÷ø

öççè

æ=

÷÷÷÷

ø

ö

çççç

è

æ

÷øö

çèæ ´+÷

øö

çèæ ´-

÷øö

çèæ ´+÷

øö

çèæ ´

=

21

7720

71

7710

73

1=x 2=y


Documents

07. CSEC Maths JANUARY 2008 · (ii) Required To Determine: Smallest value of x that satisfies the above inequality. Solution: ... The formula for the sum is a constant, , example