CXC JUNE 2008 MATHEMATICS GENERAL PROFICIENCY (PAPER 2 ...

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

Section I

1. a. (i) Required To Calculate: (3.9 ? 0.27) + 0.6724

Calculation:

(3.9 ? 0.27) + 0.6724 = 1.053 + 0.82

(by use of calculator)

b.

= 1.873 ( in exact form)

m 2 1 - 4 o (ii) Required To Calculate: 2 5

.c 3

4

s Calculation:

Numerator:

th 2 1 - 4 = 5 - 4 a 2 5 2 5

= 25 - 8

m 10

s = 17 s 10

Hence,

a 2 1 - 4 17

p 2 5 = 10

s 3

3

.fa4

4

= 17 ? 4 10 3

w = 34 ( as a single fraction in its lowest terms) 15

ww(i) Data: CAN$1.00 ? JA$72.50

Required To Calculate: CAN$250 in JA$.

Calculation:

Cost of camera = CAN$250

Hence cost of camera in JA$ = 250 ? 72.50

= JA$18 125

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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 = JA$(30 000 - $18125)

The equivalent in CAN$ = 30 000 - 18125 72.50

2. a. b.

m Data: a = 2, b = -1 and c = 3 o (i) Required To Calculate: a(b + c)

Calculation:

.c a(b + c) = 2(-1+ 3) s = 2(2) th = 4

a 4b2 - 2ac

(ii) Required To Calculate: a+b+c

m Calculation:

s 4b2 s a +

b

2ac +c

=

4(-1)2 - 2(2)(3) 2 + (-1) + 3

a = 4(1)-12 p4

s= -8 .fa 4

= -2

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

4?

( x + 5)

= 4(x + 5)

(ii) Required To Find: Algebraic expression for the statement given.

Solution:

16 larger than the product of a and b.

16 +

( a ? b = ab )

= 16 + ab

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c. Data: 15 - 4x = 2(3x + 1)

Required To Calculate: x

Calculation:

15 - 4x = 2(3x + 1)

15 - 4x = 6x + 2

15 - 2 = 6x + 4x

10x = 13

x = 13

10 =1 3

m 10 o d. Required To Factorise: (i) 6a2b3 + 12a4b, (ii) 2m2 + 9m - 5

.c Solution: s (i) 6a2b3 + 12a4b = 6 ? a2 ? b ? b2 + 2 ? 6 ? a2 ? a2 ? b

th ( ) = 6a2b b2 + 2a2

( ) = 6a2b 2a2 + b2

a (ii) 2m2 + 9m - 5 = (2m -1)(m + 5)

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

pa Career s Number of .fa students

Lawyer 240

Teacher 189

Doctor t

Artist 216

a. Required To Calculate: t Calculation:

w 240 + 189 + t + 216 + 330 = 1080 (data) w t = 1080 - (240 + 189 + 216 + 330) w =105

Salesperson 330

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

Sector illustrating Lawyer

Angle of sector 240 ? 360? = 80? 1080

Teacher

189 ? 360? = 63? 1080

Doctor Artist

.com Salesperson

105 ? 360? = 35? 1080 216 ? 360? = 72? 1080 330 ? 360? = 110? 1080

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

faspassma Solution:

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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:

b.

m (ii) Required To List: Elements of the set (M ? N )?. o Solution: .c (M ? N )? = {15, 21, 25} (as illustrated on the Venn diagram)

ths (i) Required To Construct: Parallelogram ABCD, in which AB = AD = 7 cm and BAD = 60? .

faspassma Solution:

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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. b.

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

RS + 2 = 8 (From diagram)

s RS = 8 - 2 a Length of RS = 6 m

sp (ii) Required To Calculate: x

.fa Calculation:

x + RS = 10 m

(From diagram)

x + 6 = 10

w x = 4m

wRequired To Calculate: Perimeter of the entire floor. wCalculation:

Using R as the starting point and checking the total distance of all the edges, to

find the perimeter of the floor.

Perimeter = (4 + 5 + 10 + 5 + 2 + 3 + 8 + 3)m

= 40 m

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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

= {(5 ?10) + (3? 8)}

= 50 + 24

= 74 m2

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.

6. a.

Calculation: Area of 1 flooring board = 1m ? 20 cm

m = 1m ? 20 m o 100

.c = 20 m2 100

s = 1 m2 th 5 a Number of flooring boards needed to cover A = Area of A

Area of 1 board

m = 5?10 s1 s5 a = 250

No. of boards = 250

.fasp 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.

wwwSolution:

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(ii) (a) Required To Calculate: Length of HJ. Calculation: tan 32? = 12 HJ \ HJ = 12 tan 32? = 19.20 m

= 19.2 m (to1decimal place)

b.

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

m tan 27? = 12 o HK

.c \HK = 12 tan 27? s = 23.55m

th Length of JK = 23.55 -19.2 a = 4.35

= 4.4 m (to 1 decimal place)

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

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