january 2005 cxc mathematics general ......(v) required to find: the value of x if 30 students...

www.fasp

assm

aths.c

om

JANUARY 2005 CXC MATHEMATICS GENERAL PROFICIENY (PAPER 2)

Section I

1. a. Required To Calculate: to 3 decimal places.

Calculation:

(using a calculator)

to 3 decimal places ( as required). b. Data: Table illustrating telephone rates.

Required To Prove: The cost of using the land line is less than the cost for using the cellular phone. Proof: (i) Duration of calls in the month is 1 hour 5 minutes = (60 + 5)

= 65 minutes Cost of using the land line phone = Rental fee + Charges per min.

Cost using the cellular phone

Hence the cost of using the land line telephone is less ($54.75) than using the cellular phone ($55.25).

Required To Calculate: Duration of the calls for the month of March

(ii) In March, the cost of using the land line phone accounted for a bill of $54.60.

Cost of only the calls = $54.60 – Rental fee = $54.60 – $45.00 = $9.60

Duration of the land line calls minutes.

33.02.13

5324.6

4033.02.13

=

=

325.6=

( )75.54$

15.06500.45$=

´+=

6585.0$ ´=25.55$=

\ 6415.060.9

==

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 30

www.fasp

assm

aths.c

om

2. a. Data:

Required To Calculate: The value of r when and Solution: (i) When and ,

Required To : Make q the subject

(ii)

OR

b. Required To Factorise Completely:

(i) (ii) (iii)

Solution: (i)

(ii) (iii)

This is a difference of 2 squares and

32 2

-=qpr

6=p 12=q

6=p 12=q( ) ( ) 8

972

9362

31262 2

===-

=r

32 2

-=qpr

( )( )

÷÷ø

öççè

æ+=

+=

+=

=-

´=-

-=

rr

rp

rrpq

rprqprrqpqr

qpr

3232

3223

1233

21

22

2

2

2

2

32 2

+rp

mtmgtg 2233 -+-823 2 -+ xx

273 2 -x

mtmgtg 2233 -+-( ) ( )( )( )mtg

tgmtg2323+-=

-+-=

823 2 -+ xx( )( )243 +- xx

273 2 -x( )( ) ( ){ }22

2

33

93

-=

-=

x

x

( )( )333 +-= xx


www.fasp

assm

aths.c

om

c. Data: Table of values of x and y and that y varies inversely as x. Required To Calculate: The value of a

Solution:

(k = the constant of proportionality)

And

From the table when

and

When

3. a. Data: Information to complete the given sketch of a Venn diagram. Required To Complete: The Venn diagram to represent the information given

Solution: (i)

(ii) Required To Find: an equation in x for the number of candidates in U. Solution:

xy 1µ

÷øö

çèæ´=\x

ky 1

xky =

2=x 8=y

162

8

=\

=\

k

k

xy 16=

32=x

21213216

=\

=

=

a

y

y

( ) ( ) ( ) 3218911 =+-++-= xxxUn3238 =-\ x


www.fasp

assm

aths.c

om

(iii) Required to Calculate: the value of x. Solution:

(iv) Required: To shade the region

Solution:

b. Data: Diagram showing the cross-section of a shed.

(i) Required To Find: an expression in terms of y for the area of the figure.

Solution: The cross section is divided into 2 regions, A and B, as shown on the diagram.

3238 =- x

63238

==-\

xx

SF Ç¢


www.fasp

assm

aths.c

om

Area of rectangle A

m2

Area of triangle B

m2 Cross sectional area of the figure m2. (ii) Required To Calculate: the value of y Solution:

Total area m2

4. a. Constructing a rectangle PQRS with PQ = 8 cm and PS = 6 cm

Diagonal PR = 10.0 cm (by measurement)

b. Data: Diagram with Y due east of W and V north of W.

y28´=y16=

28 y´

=

y4=\ yy 416 +=

y20=

28=

521

2081

2820

=

=\

=\

y

y

y


www.fasp

assm

aths.c

om

(i) Required to Calculate:

Solution: ( in a straight line)

(ii) Required To Calculate: Solution:

(to the nearest 0.10)

(iii) Required To Calculate: The length VZ Solution:

(Pythagoras’ Theorem)

(iii) Required To Calculate: the bearing from V to X

Solution: The bearing of V from X = 270° + 40° = 310°

VXZ ˆ

( )°+°-°= 5040180ˆVXZ Ð°= 90

XVZ ˆ

815ˆtan =XVZ

°=

÷øö

çèæ=\ -

9.61815tanˆ 1XVZ

( ) ( )222 158 +=VZ

m17289

22564

==

+=\VZ


www.fasp

assm

aths.c

om

5. a. Data: and

(i) Required To Calculate: The value of Solution:

(ii) Required To Calculate: The value of Solution:

(iii) Required To Calculate:

Solution:

Let

Replace y by x

and

and

b. Required To Express: as a single fraction in its simplest form

Solution:

( )452

-+

=xxxf ( ) 32 -= xxg

( )4g

( ) ( ) 3424 -=g

538

=-=

( )2fg

( ) ( ) 3222 -=g

( ) ( )( )

373741512

12134

-=

-=

-+

=

=\=

-=

ffg

( )71-g

( ) 32 -= xxg32 -= xy

23

23+

=

=+yx

xy

( )231 +

=- xxg

( )23771 +

=-g

5210

=

=

143+

+xx


www.fasp

assm

aths.c

om

as a single fraction

c. Required To Find: The value of Solution:

6. a. Data: Straight line drawn through A (1, 1) and B (5, -2).

(i) Required To Calculate: The gradient of the line AB Solution:

Gradient of AB

(ii) Required To Find: The gradient of the line perpendicular to AB

Solution:

Gradient of ANY line perpendicular to AB

(Product of the gradients of perpendicular lines = - 1) (iii) Required To Find: the equation of the line passing through D (3, 2)

which is perpendicular to AB. Solution:

143+

+xx

( ) ( )( ) ( )1

4331413

+++

=+++

xxxx

xxxx

( )137

++

=xxx

032

21

489 ´´

189489 3 2032

21

´´=´´

121431643 3

=´´=

´´=

( )5121

---

=

4343

-=

-=

34

=


www.fasp

assm

aths.c

om

Equation of the line through D and perpendicular to AB is

is of the form , where and .

b. Data: Coordinates of the 3 vertices of a triangle named A. (i) Required To Draw: Triangle A with coordinates (2, 1), (3, 3) and (4, 3). Solution:

(ii) Required To Draw: Triangle B, which is the reflection of Triangle A in

the line . Solution: Coordinates of the vertices of B are (2, -3), (3, -5) and (4, -5)

234

4342

34

32

-=

-=-

=--

xy

xy

xy

cmxy +=34

=m 2-=c

1-=y


www.fasp

assm

aths.c

om

(iii) Required To Draw: Triangle C which is the translation of Triangle A by

the vector .

Solution:

Coordinates of the vertices of C are (-2, 3), (-1, 5) and (0, 5)

÷÷ø

öççè

æ -24

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ+-

¾¾ ®¾÷÷ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ+-

¾¾ ®¾÷÷ø

öççè

æ

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-

¾¾ ®¾÷÷ø

öççè

æ

¾¾ ®¾

÷÷ø

öççè

æ -

÷÷ø

öççè

æ -

÷÷ø

öççè

æ -

÷÷ø

öççè

æ -=

51

2343

33

50

2344

34

32

2142

12

CA

24

24

24

24

T

\


www.fasp

assm

aths.c

om

7. Data: Diagram of a cumulative frequency curve for the marks on a test by 80 students.

(i) Required to Find: The number of students who scored less than 23 marks Solution:

The vertical at 23 corresponds to the horizontal at 52. Hence 52 candidates scored less than 23 marks.

(ii) Required To Find: the number of students who scored more than 17 marks. Solution:

The vertical at 17 corresponds to the horizontal at 26. Number of candidates who scored more than 17 marks = 80 – 26 = 54.

(iii) Required To Find: the inter-quartile range of the marks scored. Solution:

of 80 = 20

of 80 = 60

The cumulative frequency value of 20 corresponds to a mark of 15 (lower quartile, Q1).

\

41

43


www.fasp

assm

aths.c

om

The cumulative frequency value of 60 corresponds to a mark of 25 (upper Quartile, Q3). Interquartile range = Q3 – Q1

(Upper quartile – Lower quartile) = 25 – 15 = 10 (iv) Required To Calculate: the probability that a randomly chosen student

scored between 17 and 23 marks. Solution:

(v) Required to Find: the value of x if 30 students scored more than x marks Solution:

If 30 students scored more than x, then x is the horizontal value that

\

( )

40138026802652

studentsofno.Total23and17betweenscoringstudentsofNo.23and17betweenscoredstudent

=

=

-=

=P


www.fasp

assm

aths.c

om

corresponds to a vertical (cumulative frequency) value of (80 – 50) = 30

8. Data: Diagrams showing the link from a chain.

a. Required To Calculate: the volume of a single link of chain Solution:

Volume of a single link = External volume – Internal volume mm3 = 165 mm3 to 3 significant figures b. Required To Prove: the length of chain is 16 mm + 14 mm.

Solution:

Length of link = PQRST (as shown on the diagram) mm (radius of inner circle of link) mm (radius of outer circle of link) Length of link mm = 16 mm + 14 mm

c. Required To Complete: a table showing the length of the chain formed when rings are linked in a straight line. Solution:

No. of rings, n Length of chain, L 1 16 2 30 3 44

Trying to obtain a pattern or sequence between L and n.

2122»x

9.164=

7== RSQR8== STPQ

\ ( ) ( )7282 +=


www.fasp

assm

aths.c

om

n = 1 2, 3,… L = 16 30, 44,… L = 16+ 14(0) 16 + 14(1), 16 + 14(2),… When When

No. of rings, n Length of chain, L

1 16 2 30 3 44

6 (86)

(12) 170

Section II 9. a. Data:

Required to Calculate: x and y Solution: Let …(1) and …(2) Substitute in (2)

And or 2 When = 4 − 9 = 5 When = 4− 4 = 0 Hence, and OR and

\ 6=n ( ) 86161416 =-+=L170=L ( )11416170 -+= n

( )

12111

141541

154114

=+=

=-

=-

nn

n

n

! !

! !

yx -= 42

2+= yx

yx -= 42 2+= yx24 xy -=\

( )( )

( )( ) 02306024024

24

2

2

2

2

=-+=-+

=--+

=---

+-=

xxxxxxxxxx

3-=x3-=x ( )234 --=y2=x ( )224 -=y3-=x 5-=y 2=x 0=y


www.fasp

assm

aths.c

om

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

Q.E.D c. (i) Required to Express: in the form Solution:

is of the form where

and

(ii) Required To Find: the minimum value of Solution:

( )( ) ( ) 258343232 22 -+º--+- xxxxx

( )( ) ( )( ) ( )

RHS2583

1689416449664

43232

2

22

22

2

=-+=

-+--=

+----+-=

--+-

xxxxx

xxxxxxxxx

2583 2 -+ xx ( ) khxa ++ 2

2583 2 -+ xx

3130

343

253130

31583

?916

383

?343

25383

2

2

2

2

2

-÷øö

çèæ +=

-

-

+++=

+÷øö

çèæ ++=

+÷øö

çèæ +

-÷øö

çèæ +

x

xx

xx

x

xx

( ) khxa ++ 2

ÂÎ= 3a

ÂÎ=34h

ÂÎ-=3130k

2583 2 -+ xx

3130

3432583

22 -÷

øö

çèæ +º-+ xxx


www.fasp

assm

aths.c

om

Minimum value of the function is .

ALTERNATIVE METHOD

For a quadratic , a maximum or minimum value of the

function occurs at .

The minimum value of the function occurs at

When , the minimum value is

d.

OR

xx "³÷øö

çèæ + 0

343

2

\3130-

cbxax ++2

abx2-

=

( )( ) 3

4328

-=-

=x

34

-=x313025

348

343

2

-=-÷øö

çèæ-+÷

øö

çèæ-

02583 2 =-+ xx( ) ( )( )

( )

placedecimal1to5.4or8.115.4or48.1

636486

300648

32253488 2

-=-=

±-=

+±-=

--±-=

x

x

991

34

3130

343

03130

343

02583

2

2

2

2

=÷øö

çèæ +

=÷øö

çèæ +

=-÷øö

çèæ +\

=-+

x

x

x

xx


www.fasp

assm

aths.c

om

10. Data: Information about the numbers, prices and special conditions involving calculators and folders bought for a school. (i) Required To Find: An inequality to represent the information given. Solution:

No. of calculators bought = x No. of calculators bought is at least 5 …(1) (ii) Required To Find: An inequality to represent the information given. Solution:

The number of folders must be at least twice the number of calculators. y and …(2) (iii) Required To Find: An inequality to represent the information given. Solution:

Cost of x calculators at $20 each and y folders at $5 each Amount available for spending is not more than $300

(iii) Required To Draw: the lines of the three equalities and hence shade the

region that satisfies all three inequalities, stated above. Solution: The line is a vertical straight line. The region which satisfies is

placedecimal1to5.4or8.115.4or48.1

3914391

34991

34

-=-=

±-=

±-=

±=+

x

x

x

x

5³\ x

³ x´2xy 2³

( ) ( ) yxyx 520520 +=´+´=

)3...(6045300520

£+÷

£+\

yx

yx

5=x5³x


www.fasp

assm

aths.c

om

Finding two points on the line so as to draw the line. The line passes through the origin (0, 0). When The line passes through (5, 10).

The side that makes the larger angle satisfies the region.

Therefore, the region which satisfies is

Finding two points on the line . When The line passes through the point (0, 60). When

xy 2=xy 2=

5=x ( )52=y10=

xy 2=

³

xy 2³

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

60=y604 =+ yx

0=y 6004 =+x


www.fasp

assm

aths.c

om

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

The side that makes the smaller angle satisfies the region. Therefore, the region which satisfies is

Therefore, the region which satisfies all three inequalities is the area where all three shaded regions, shown above, overlap.

15460

=

=x

604 =+ yx

£604 £+ yx


www.fasp

assm

aths.c

om

This is identified as the feasible region is ABC. (v) Required To Find: an expression in terms of x and y for the total profit, P. Solution:

Total profit = P Profit of x calculators at $6 each and y folders at $2 each is

(vi) Required To Find: the coordinates of the vertices of the shaded region. Solution: Coordinates of the vertices of the shaded region are A (5, 10), B (5, 40), C (10, 20) (vii) Required To Calculate: maximum profit.

( ) ( )yxP

yxyx26

2626+=\

+=´+´


www.fasp

assm

aths.c

om

Solution: is to be found by testing the coordinates of the vertices of the feasible

region. The point A obviously need not be tested as it has both lower x and y values than the other points.

Point B When and

Point C When and

Maximum profit is $110, when 10 calculators and 20 folders are sold.

11. a. Data: Diagram illustrating a vertical pole standing on horizontal ground.

(i) Required To Calculate: the angle of elevation from U to S. Solution: (angles in a straight line) Angle of elevation of U from S is 60° (as illustrated)

(ii) Required To Calculate: the length to UT Solution:

(sum of angles in a triangle total 180o)

(sine rule)

maxP

5=x 40=y( ) ( )110$

40256=

+=P

10=x 20=y( ) ( )100$

202106=

+=P

\

°-°= 120180ˆRSU°= 60

\

( )°=

°+°-°=20

40120180ˆTUS

°=

° 120sin20sin15 UT


www.fasp

assm

aths.c

om

(iii) Required To Calculate: the length of RU Solution:

b. Data: Diagram showing a circle, centre O. LMNR is a tangent. LSOP, NOQ and

MSQ are straight lines.

(i) Required to Calculate: Angle

Solution:

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

(ii) Required To Calculate: Angle

Solution:

(The angle made by the tangent to a circle and radius, at the point of contact is 90°)

placedecimal1tom0.38m89.3720sin120sin15

==

°°´

=\UT

3840sin RU

=°

placedecimal1tom4.24m14.2440sin98.37

==

°=\RU

°= 35ˆNPS

SON

( )°= 352ˆNOS°= 70

NMQ

°= 90ˆMNO

°=\ 90ˆMNO°= 35ˆNQS


www.fasp

assm

aths.c

om

(The angles subtended by chord SN, at the circumference of a circle, standing on the same arc are equal.)

(Sum of the angles in a triangle is 180°)

(iii) Required To Calculate: Solution:

In

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

(iv) Required To Calculate:

Solution: (The angle made by the tangent to a circle and a chord, at the point of contact, is equal to the angle in the alternate segment.) 12.a. This part is not done since it involves latitude and longitude (Earth Geometry) which has been removed from the syllabus.

12. b. (i) Required To Complete: the table of values for Solution:

x 0° 30° 60° 90° 120° 150° 180° y (1) 1.1 1.5 (2) 2.5 (2.9) 3

When

When

When

( )°=

°+°-°=\55

3590180ˆQMN

PLNÐ

OLND( )

°=°+°-°=

207090180ˆNLO

SNMÐ

°= 35ˆMNS

xy cos2 -=xy cos2 -=

0=x ( )0cos2 -=y

112

=-=

°= 90x ( )°-= 90cos2y

202

=-=

°= 150x ( )°-= 150cos2y( )9.2

87.02=

--=


www.fasp

assm

aths.c

om

(iii) Required To Draw: the graph of Solution:

(ii) Required To Find: the value of x for which . Solution:

Draw the horizontal,

xy cos2 -=

8.1cos2 =- x

8.1=y


www.fasp

assm

aths.c

om

13. a. Data: and

(i) Required To Calculate: giving the answer in the form

Solution:

is of the form

where and

°==- 5.78at8.1cos2 xx

÷÷ø

öççè

æ-

=÷÷ø

öççè

æ=÷÷

ø

öççè

æ=

31

,32

,11

nmOP nmPQ 2+=

PQ ÷÷ø

öççè

æyx

÷÷ø

öççè

æ-

+÷÷ø

öççè

æ=

+=

31

2322nmPQ

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ-

+÷÷ø

öççè

æ=

34

62

32

÷÷ø

öççè

æyx

4=x 3-=y


www.fasp

assm

aths.c

om

(ii) Required To Calculate:

Solution:

b. Data: Vector diagram with M the midpoint of CE, and

If , then and

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

(ii) Required To Express: in terms of and

Solution:

PQ

( ) ( )

units525

34 22

==

-+=PQ

OFFE 2=

OFFE 2= aFE 2= aaaOE 32 =+=

CF a b

( )baabOFCOCF

-=+-=+=

CE a b

baabOECOCE

-=+-=+=

33


www.fasp

assm

aths.c

om

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

(iv) Required To Express: in terms of and and k

Solution:

(v) Required To Calculate: the value of k for which

Solution:

Equating components

OR

CM a b

( )

ba

ba

CECM

21

211

32121

-=

-=

=

MG a b

CFkCG =

( )

( )

bkak

bkakba

bakba

CGMCMG

bakCG

÷øö

çèæ -+÷

øö

çèæ -=

-++-=

-+÷øö

çèæ --=

+=

-=\

21

211

21

211

21

211

COMG =

COMG =

( )ba

bbkak

-+º

-=÷øö

çèæ -+÷

øö

çèæ -\

021

211

211

0211

=

=-

k

k

211

121

=

-=-

k

k


www.fasp

assm

aths.c

om

14. a. (i) Data:

Required To Find: Calculation: Det

(ii) Required To Calculate: the values of x and y for which

Solution:

Equating corresponding entries, and

b. Data: Transformation matrix,

A and B are mapped onto and

÷÷ø

öççè

æ-

=3112

M

1-M

( ) ( )1132 -´-´=M716 =+=

( )( )

÷÷÷÷

ø

ö

çççç

è

æ -=

÷÷ø

öççè

æ--

-=-

72

71

71

73

2113

711M

÷÷ø

öççè

æ=÷÷

ø

öççè

æ112

yx

M

÷÷ø

öççè

æ=÷÷

ø

öççè

æ112

yx

M

÷÷ø

öççè

æ=

÷÷÷÷

ø

ö

çççç

è

æ

÷øö

çèæ ´÷

øö

çèæ ´

÷øö

çèæ ´-÷

øö

çèæ ´

=

÷÷ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ -=÷÷

ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´

´

-

--

-

25

17212

71

17112

73

112

72

71

71

73

112

112

1

11

1

yx

Myx

I

Myx

MM

M

5=x 2=y

÷÷ø

öççè

æ=

srqp

T

A¢ B¢ AA T ¢¾®¾


www.fasp

assm

aths.c

om

(i) Required To Calculate: the values of p, q, r and s. Solution:

Equating corresponding entries

Similarly

Equating corresponding entries

(1) + (3)

Subsituting in (1)

(2) + (4)

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ-÷÷ø

öççè

æ42

24

srqp

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ+-+-

42

2424srqp

)2...(222

424)1...(12

2224

-=--÷

=+-=-

-÷-=+-

sr

srqp

qp

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ+-+-

÷÷ø

öççè

æ -=÷÷

ø

öççè

æ-÷÷ø

öççè

æ

¢¾®¾

25

5252

25

52

srqp

srqpBB T

)4...(252)3...(552

=+--=+-

srqp

1

44

55212

-=

-=

-=+-=-

q

q

qpqp

004224

==--=+-

pp

p

0

04

25222

=\

=

=+--=-

s

s

srsr


www.fasp

assm

aths.c

om

Substitute into (2)

(ii) Required To Calculate: The coordinates of the point C.

Solution:

0=s

0and1,1,01

202

=-=-==\-=

-=-

srqprr

CC T ¢¾®¾( ) ( )

( ) ( )

( )2,222

20212120

22

0110

-=¢\

÷÷ø

öççè

æ -=

÷÷ø

öççè

æ´+-´-´-+-´

=÷÷ø

öççè

æ -÷÷ø

öççè

æ-

-\

C


january 2005 cxc mathematics general ......(v) required to find: the value of x if 30 students...

Documents