JUNE 2008 CXC MATHEMATICS GENERAL PROFICIENY (PAPER … · JUNE 2008 CXC MATHEMATICS GENERAL...

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JUNE 2008 CXC MATHEMATICS GENERAL PROFICIENY (PAPER 2) (TRINIDAD AND TOBAGO PAPER ONLY)

Section I

1. a. Required To Calculate:

Calculation: Numerator:

Hence,

b. Data: Table showing the tax allowances for Mr. Allen.

(i) Required To Calculate: His annual salary. Calculation: For 2007 Mr. Allen’s monthly salary is $7 500. Hence, annual salary

851

311

512 -

( ) ( )

151315203315

4511334

511

311

512

=

-=

-=

-=-

form)exactin(158

138

15138131513851

1513

851

311

512

=

´=

=

=-

125007$ ´=00090$=

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

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(ii) Required To Calculate: Total allowances for 2007. Calculation: Total tax allowances for Mr. Allen according to the given table: For self $10 000 For wife $ 5 000 For 3 children $ 7 500 Total $22 500 (iii) Required To Calculate: Mr. Allen’s income tax for 2007. Calculation: Taxable income = Gross salary – Tax allowances

Mr. Allen’s income tax

(iv) Required To Calculate: Percentage of Mr. Allen’s annual salary paid in

income tax. Calculation: Percentage of Mr. Allen’s annual salary paid in income tax

2. a. Required To Simplify: Solution:

( )25003´

50067$50022$00090$

=-=

50067$10022

´=

85014$=

%5.16

1000009085014

100SalaryGrossTaxIncome

=

´=

´=

( )213 -a

( ) ( )( )

1691339

131313

2

2

2

+-=

+--=

--=-

aaaaa

aaa


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b. Data:

Required To Make: p the subject of the formula. Solution:

c. Required To Factorise (i)

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

d. Data: and

Required To Calculate: x and y Calculation: Let

…(1) …(2)

From equation (1)

Substituting in equation (2)

pq

+=35

( )( )

qqp

qpqpqqpq

pq

pq

3535

5315335

1

35

-=

-==+

´=++

=

+=

222 25)(,63 qpiinmn --

nnmnnmn ´´´-=- 23363 2

( )nmn 23 -=

( ) ( )2222 525 qpqp -=-

( )( )qpqp +-= 55

1923 =- yx 432 =+ yx

1923 =- yx432 =+ yx

31921923

+=

+=yx

yx

( ) ( ) ( )12938443331922

3

4331922

=++=++

´

=+÷øö

çèæ +

yyyy

yy


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When

Hence, and . OR Let

…(1) …(2)

Equation (1) …(3)

Equation (2) …(4)

Substitute in equation (1).

Hence, and . OR

213262613381213

-=

-=

-=-=

y

yy

2-=y ( ) 19223 =--x

5153

4193

==

-=

xxx

5=x 2-=y

1923 =- yx432 =+ yx

3´5769 =- yx

2´864 =+ yx

51365

6513

8645769

=

=

=

+=+=-

x

x

yxyx

5=x( )

22442

1519219253

-=-

=

=--=-

=-

y

yyy

5=x 2-=y


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Finding the coordinates of 2 points on the line. x y 3 - 5 7 1

Find the coordinates of 2 points on the line.

x y -1 2 8 -4

If we draw the two straight lines on the same axes,

The point of intersection is (5, -2) from which we deduce and . OR Let

…(1) …(2)

Expressing in matrix form

Let

19321923

-==-xy

yx

423432

+-==+

xyyx

5=x 2-=y

1923 =- yx432 =+ yx

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ -419

3223

yx

÷÷ø

öççè

æ -=

3223

A


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Det

Matrix equation

Equating corresponding entries. and .

3. a. Data: Venn diagram.

( ) ( )2233 ´--´=A

( )( )

÷÷÷÷

ø

ö

çççç

è

æ

-=

÷÷ø

öççè

æ-

--=

=

-

133

132

132

133

3233

131

13

1A

1-A

÷÷÷÷

ø

ö

çççç

è

æ

-=

÷÷÷÷

ø

ö

çççç

è

æ

+-

+=÷÷

ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ

÷øö

çèæ ´+÷

øö

çèæ ´-

÷øö

çèæ ´+÷

øö

çèæ ´

=÷÷ø

öççè

æ´

÷÷ø

öççè

æ

÷÷÷÷

ø

ö

çççç

è

æ

-=÷÷

ø

öççè

æ´´ -

13261365

1312

1338

138

1357

413319

132

413219

133

419

133

132

132

133

1

yx

yx

I

yx

AA

÷÷ø

öççè

æ-

=25

5=x 2-=y


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(i) (a) Required To List: Members of . Solution: as seen on the given diagram.

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

(ii) Required To Find: . Solution:

(iii) (a) Required To Describe: U in words. Solution: U = {Even numbers from 2 to 30 inclusive}

(b) Required To Describe: H in words. Solution: H = {multiples of 6} (Since the universal set is from 2 to 30 we need not say ‘multiples of 6 from 2 to 30’).

b. (i) Required To Construct: Quadrilateral PQRS, in which PQ = 8 cm, QR = 4 cm, PS = 6.5 cm, and . NOTE: The angles 1250 and 700 cannot be constructed and are drawn using the

protractor.

HGÇ

{ }24,12=Ç HG

HG ¢Ç

{ }{ }{ }20,16,8,4

2,20,16,12,8,428,26,22,20,16,14,10,8,4,2

=¢Ç\==¢

HGGH

( )HGn È

( ) { }( ) 9

30,18,6,24,12,20,16,8,4=È\=È

HGnHG

°= 125ˆRQP °= 70ˆSPQ


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Solution:

(ii) Required To Find: Length of RS. Solution: RS = 8.6 cm (by measurement)

4. Data: Given a triangular prism with right-angled isosceles triangles at both ends.

a. Required To Calculate: Area of Calculation:

Area of

ABCD

244´

=DABC2cm8=


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b. Required To Calculate: Length of edge CD.

Calculation: Volume of prism = Cross-sectional area Length CD

c. Required To Calculate: Length of edge AC. Calculation:

d. Required To State: Number of faces, edges and vertices of the prism.

Solution: No. of faces: 2 triangular

1 rectangular side 1 rectangular base 1 rectangular sloping

Total 5 No. of edges: AB, BC, AC, EF, FD, ED, AE, BE, CF Total = 9 No. of vertices: A, B, C, D, E, F Total = 6

5. Data: drawn on labeled axes. a. Required To Draw: Line on the diagram. Solution: Line is shown on the diagram.

´

cm9728

==´

CDCD

( ) ( ) ( )

place)decimal1to(cm7.566.532

32)Theorem'Pythagoras(44

2

222

===

=

+=

AC

ACAC

LMND4=x

4=x


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b. Required To Draw: , the image when is reflected in the line

. Solution:

as shown on the diagram. c. Data:

(i) Required To Draw: . Solution: , and plotted and drawn, as shown on the diagram. (ii) Required To Describe: The transformation that maps onto

. Solution: and pass through O.

by an enlargement, centre O and scale factor 2.

d. Required To Calculate:

Calculation: Since, Then,

NML ¢¢¢D LMND4=x

( ) ( ) ( )2,5and6,5,1,7

4xinReflection

=¢=¢=¢¢¢¢D¾¾¾¾¾ ®¾D =

NMLNMLLMN

( ) ( ) ( )4,6,12,6,2,2 --=¢¢--=¢¢--=¢¢ NMLNML ¢¢¢¢¢¢D

L ¢¢ M ¢¢ N ¢¢ NML ¢¢¢¢¢¢D

LMNDNML ¢¢¢¢¢¢D

MMLL ¢¢¢¢ , NN ¢¢

MLLM

NLLN

MNNM

¢¢¢¢=

¢¢¢¢=

=

=¢¢¢¢

248

NMLLMN ¢¢¢¢¢¢D®D\

ΔLMNofAreaNMLΔofArea ¢¢¢¢¢¢

1:2: =¢¢¢¢ LMML

( )( )

1412

ΔLMNofAreaNMLΔofArea

2

2

=

=¢¢¢¢¢¢


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6. Data: Table showing the population of country in five year periods from 1980 to 2005. a. Required To Draw: Line graph to represent the information given. Solution:


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b. Required To Find: Population in 1998. Solution: From the graph, the population in 1998 is 1.8 million. c. (i) Required To Find: The five year period with the greatest increase in population. Solution: The greatest increase is 1990 to 1995 (line has the highest positive grad.) (ii) Required To Find: The five year period with the smallest increase in population. Solution: The smallest increase is 1995 to 2000. (line has the smallest gradient) d. Required To Explain: How the answers above are shown on the graph. Solution: The greatest increase would be shown by the steepest line segment or the line

segment with the greatest positive gradient, that is 1990 to 1995. The smallest increase would be shown by the least steep line segment or the line

with the smallest positive gradient, that is 1995 to 2000.


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7. a. Data: M (4, 5) is the midpoint of AB, with A (1, 8) and B (j, k). Required To Calculate: j and k. Calculation:

Using the midpoint formula.

Equating

and

b. Data: and

(i) Required To Calculate: . Calculation:

(ii) Required To Calculate: Calculation:

( ) ÷øö

çèæ ++

=28,

215,4 kj

781

421

==+

=+

jj

j

2108

528

==+

=+

kk

k

2and7 ==\ kj

25: -® xxf14:+

®x

xg

( )0f

( ) ( )2

2050-=

-=f

( )2g

( )

34

1242

=

+=g


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(iii) Required To Calculate: Calculation: Let

Replace y by x

(iv) Required To Calculate: x when Calculation:

𝑓𝑔(𝑥) = 5( ()*+

) − 2

1 = /0)*+

− 2 (𝑓𝑔(𝑥) = 1)/0)*+

= 3 20 = 3𝑥 + 3 17 = 3𝑥 𝑥 = 5 /

5

8. Data: Geometrical pattern of grey and white triangles. a. Required To Sketch: Geometrical pattern whose side is 5 cm long. Solution:

b. (i) Required To Complete: Table for a pattern whose side is 6 cm. Solution:

( )xf 1-

52

5225

+=

=+-=

yx

xyxy

( )521 +

=- xxf

( ) 1=xfg


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Length of one side of the pattern (cm)

Total number of triangular shapes

used to make pattern

Number of white triangular shapes

used

Number of grey triangular shapes

used

2 4 3 1 3 9 6 3 4 16 10 6

So far it is obvious that from the figures and the tables, Total number of triangular shapes used (column 2) = No. of white triangular shapes (column 3 ) + No. of grey triangular shapes used (column 4). Example:

Also, column 2 is the square of the column 1 values. 2 4 3 9 4 16 When (column 1) Column 2

Column 4

Column 3 = Column 2 – Column 4

The 4th row is 6 36 21 15

(ii) Required To Complete: Table for a pattern whose side is 20 cm. Solution:

61016369134

+=+=+=

( )22=( )23=( )24=

6=n( )26=36=

( )21-

=nn

( )

152166

=

-=

211536

=-=


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When Column 2

Column 4

The 5th row is 20 400 210 190

c. Required To Complete: Table for a pattern whose side is n cm long. Solution: The last row of the table:

n *

The last row may now be completed as

n

And the completed table is

20=n( )220=400=( )21-

=nn

( )

190212020

=

-=

2n ( )21-nn

( )

( )

( )21

121

21

21*

21

21*

2*

21*

2

22

22

2

+=

+=

+=

-+=

-+=

-+=

nn

nn

nn

nnn

nnn

nnn

2n ( )21+nn ( )

21-nn


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Length of one side of pattern (cm)

Total number of triangular shapes

used to make pattern

Number of white triangular shapes

used

Number of grey triangular shapes

used

2 4 3 1 3 9 6 3 4 16 10 6

6 36 21 15

20 400 210 190

n

9. a. Data: and Required To Calculate: x and y. Calculation: Let

…(1) …(2)

Equating (1) and (2)

When When

and , and b. Data: (i) Required To Express: In the form Solution:

Half the coefficient of

! ! ! !

! ! ! !

! ! ! !2n ( )1

21

+nn ( )121

-nn

1232 --= xxy 162 -= xy

1232 --= xxy162 -= xy

( )( )4or1

041045

1621232

2

==--=+-

-=--

xxxxx

xxx

1=x ( ) 1612 -=y14-=

4=x ( ) 1642 -=y8-=

1=x 14-=y 4=x 8-=y

1232 --= xxy( ) khxy +-= 2

*23123

22 +÷

øö

çèæ -=-- xxx

( )3213 -=- x

23

-=


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( = * )

is of the form , where and .

OR

Equating coefficient of x.

Equating constants.

(ii) Required To Calculate: Minimum value of . Calculation:

124114

41232

-

-

+- xx

( ) khx +- 2

23

=h4114-=k

( )khhxx

khxxxy++-=

+-=--=22

22

2123

2323

=

-=-

h

h

4114

41212

2312

2

-=

+=-

+÷øö

çèæ-=-

k

k

k

4114

23123

22 -÷

øö

çèæ -=--\ xxx

1232 --= xxy

4114

41140

0)2/3()2/3(4114

23123

min

22

-=

-=\

"³--

-÷øö

çèæ -=--

y

xforisxx

xxx


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(iii) Required To Calculate: The roots of . Calculation: If

OR

c. Required To Sketch: The graph of . Solution: When Curve cuts y – axis at – 12.

01232 =-- xx

( ) ( ) ( )( )( )

place)decimal1to(3.2or3.572.2or72.5

257324893

12121433

01232

2

-=-=

±=

+±=

---±--=

=--

x

xx

placedecimal1to3.2or3.572.272.5

2573257

23257457

23

457

23

04114

23

0123

2

2

2

-=-=

±=

±=

±=

±=-

=÷øö

çèæ -

=-÷øö

çèæ -

=--

or

x

x

x

x

xx

1232 --= xxy

0=x ( ) ( ) 12030 2 --=y12-=


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at

The minimum point is .

Curve cuts the x – axis at and .

10. Data: Curve with equation for .

a. Required To Complete: Table of values given. Solution:

x 2 3 4 5 6 7 y 5.0 2.2 1.3 0.8 0.6 0.4

When

When

When

4114min -=y

211=x

÷øö

çèæ -

4114,

211

2573 +

2573 -

220x

y = 72 ££ x

2=x( )2220

=y

5=

5=x( )2520

=y

8.0=

7=x( )2720

=y

4.0=


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b. Required To Plot: The points above and draw a smooth curve. Solution:

c. (i) Required To Find: y when x = 4.5 Solution: When , (Read off from graph). (ii) Required To Find: x when Solution: When , (Read off from graph). d. Required To Estimate: Gradient of the curve at (3, 2.2). Solution: and

5.4=x 1=y

5.3=y

5.3=y 4.2=x

6=P 7.4=Q


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Gradient of the curve at (3, 2.2)

11. Data: Given R, S and T lying on a straight line. The bearing of Q from R is 033° and the bearing of S from Q is 118°. a, b Required To Draw: Diagram showing the information given. Solution:

c. (i) Required To Calculate: . Calculation:

(ii) Required To Calculate: Calculation:

(iii) Required To Calculate: Calculation:

(Exterior angles of a triangle = sum of the interior opposite angles).

7.46

-=

28.1727.1

-=-=

SRQ ˆ

°=°-°=

573390ˆSRQ

SQR ˆ

( )

°=°+°=

°-°+°=

956233

289033ˆSQR

TSQ ˆ

°=°+°=

1529557ˆTSQ

Q


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d. (i) Required To Calculate: Distance QS. Calculation:

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

𝑄𝑇/ = (6.7)/ + (3)/ − 2(6.7) (3)cos152° (Cosine Rule)𝑄𝑇/ = 89.384 𝑄𝑇 = √89.384 = 9.4 km

= 9 km to the nearest km

e. Required To Calculate: Bearing of Q from S. Calculation:

Bearing of Q from S

kmnearestthetokm7km7.695sin57sin8

)ruleSine(95sin8

57sin

==

°°

=\

°=

°

QS

QS

°+°= 28270 °= 298


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12. a. Data: Diagram, as shown below, with , PNQ is the tangent at N and KN = KL.

(i) Required To Calculate: Calculation:

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

(ii) Required To Calculate: Calculation: (The base angles of an isosceles triangle are equal).

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

(Opposite angles of a cyclic quadrilateral are equal). b. This part of the question has not been solved as it involves Earth Geometry which has since been removed from the syllabus.

°= 24ˆKNP

NLK ˆ

°= 24ˆNLK

LKN ˆ

°= 24ˆLNK

( )°=

°+°-°=132

2424180ˆLKN

NML ˆ

°=°-°=

48132180ˆNML


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13. Data: Points O, A, B and C and their coordinates.

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

Solution:

is of the form where and .

(ii) Required To Express: in the form .

Solution:


(iii) Required Express: in the form .

Solution:


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

Solution:

is of the form , where and .

(ii) Required To Express: in terms of p, in the form

Solution:

OA ÷÷ø

öççè

æyx

( )3,1-=A

÷÷ø

öççè

æ -=\

31

OA ÷÷ø

öççè

æyx

1-=x 3=y

OB ÷÷ø

öççè

æyx

( )4,5-=B

÷÷ø

öççè

æ -=\

45

OB ÷÷ø

öççè

æyx

5-=x 4=y

OC ÷÷ø

öççè

æyx

( )pC ,7=

÷÷ø

öççè

æ=\p

OC7

÷÷ø

öççè

æyx

7=x py =

BA ÷÷ø

öççè

æyx

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ -+÷÷ø

öççè

æ --=

+=

14

31

45OABOBA

÷÷ø

öççè

æyx

4=x 1-=y

AC ÷÷ø

öççè

æyx


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c. Data:

Required To Calculate: p Calculation: If ,

then,

d. Data: D is such that

Required To Prove: A, B and D are collinear. Proof:

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ+÷÷ø

öççè

æ--=

+=

38

731

p

p

OCAOAC

÷÷ø

öççè

æyx

8=x 3-= py

10=AC

10=AC

( ) ( )( ) ( )

( )( )3or9

0390276100966410038

1038

2

2

22

22

-==+-=--

=+-+

=-+

=-+

pppppppp

p

÷÷ø

öççè

æ=

011

OD

÷÷ø

öççè

æ-

=

÷÷ø

öççè

æ+÷÷ø

öççè

æ--=

+=

÷÷ø

öççè

æ-

=

416

011

45

14

ODBOBD

BA

÷÷ø

öççè

æ-

=14

4


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is a scalar multiple of and hence they are parallel.

B is a common point to both line segments, therefore, B, A and D are collinear.

BD\ BA


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14. a. Data:

Required To Prove: M is singular. Proof: Det

Hence, M is singular.

b. Data:

Required To Calculate: a and b. Calculation:

Equating corresponding entries.

and

Hence, and .

c. (i) Data: is mapped onto under where

.

Required To Calculate: x and y. Calculation:

÷÷ø

öççè

æ--

=101256

M

( ) ( )125106 -´-´-=M

06060

=+-=

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ785

412

ba

( ) ( )( ) ( )

÷÷ø

öççè

æ=÷÷

ø

öççè

æ++

÷÷ø

öççè

æ=÷÷

ø

öççè

æ´+´´+´

÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ

78

4510

78

45152

785

412

bab

bab

ba

2810-=

=+bb

( )

31557245

===-+

aa

a

3=a 2-=b

( )2,5E ( )5,2 -¢E EE S ¢¾®¾

÷÷ø

öççè

æ=

00yx

S


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Therefore, and .

(ii) Data: G (7, 1) is mapped onto under the translation, T, where

and H is mapped onto .

(a) Required To Calculate: Coordinates of . Calculation:

(b) Required To Calculate: p and q Calculation:


Hence, and .

( ) ( )( ) ( )

÷÷ø

öççè

æ-

=÷÷ø

öççè

æ\

÷÷ø

öççè

æ-

=÷÷ø

öççè

æ´+´´+´

÷÷ø

öççè

æ-

=÷÷ø

öççè

æ÷÷ø

öççè

æ

52

52

52

205250

52

25

00

yx

yx

yx

122

==

xx

155-=-=

yy

1=x 1-=y

G¢

÷÷ø

öççè

æ-=

43

T H ¢

G¢

( )

( )5,454

17

4137

17

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(iii) Data: P (10, 12) undergoes the combined translation S followed by T. Required To Calculate: Image of P. Calculation: Let

and

The image of P under the combined transformation is (9, -6).

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69

410312

1012 4

3T


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