Mathematics model exam_vol-2_ppt_design

CSEC Model Exam 1 Paper 2










CSEC MODEL EXAMINATIONS

Caribbean Educational Publishers Ltd.

CSEC MODEL EXAMINATION 1MATHEMATICS

Paper 190 minutes

Answer ALL the questions

NEXT

1. The decimal fraction 0.625 written as a common fraction, in its lowest terms, is

ANSWER

(C)(A) (D)(B)4

5

3

45

88

9

Model Exam 1 Paper 1

1. Multiply the decimal fraction with three

decimal places by , which is 1, to make

the decimal fraction a common fraction.

625 25

1 000 2525

4025 5

405

8

5

1 0000.625 0.625

1 000625

1 000

1000

1000

Divide both the numerator and the denominator by their common factor 25.


This is the common fraction written in its lowest terms.


2. The number 8.150 46 written correct to 3 decimal places is

ANSWER

(C) 8.151(A) 8.149 (D) 8.152(B) 8.150


2. 8.150 46 = 8.150 4 6 = 8.150

The digit in the 4th decimal place is 4, which is less than 5, so we do not add 1 to the digit in the 3rd decimal place. The digit in the 3rd decimal place remains unchanged.


3. The exact value of 0.615 × 0.07 is

ANSWER

(C) 4.305

(A) 0.043 05

(D) 43.05

(B) 0.430 5


3. 0.615 οr

0.07

0.043 05

0.615 × 0.07 = 0.043 05 3 dp + 2 dp = 5 dp

615

7

4 305


4. The exact value of 7 ÷ (0.01)2 is

ANSWER

(C) 7 000

(A) 0.000 7

(D) 70 000

(B) 0.000 07


4.

Invert the product of fractions which is the divisor and multiply instead of divide.


22

2

77 (0.01)

(0.01)

7(0.01) 0.01 0.01

0.01 0.017 1

0.011 1 100

100 1007 100 100

1 1 170 000

5. If $450 is divided into two portions in the ratio 4:5, then the smaller portion is

ANSWER

(C) $200(A) $50 (D) $250(B) $150


5. The number of equal parts = 4 + 5 = 9

the smaller portion= of $450

= × $450

= 4 × $50 = $200

4

9

4

9


6. If 40% of a number is $70, what is the number?

ANSWER

(C) $175(A) $110 (D) $200(B) $150


6.


3 2

40% of the number $70

100100% of the number $70

40

$

$ 5

70

7

0

14

7. What is the least number of cherries that can be shared equally among 5, 10 or 15 children?

ANSWER

(C) 60(A) 30 (D) 75(B) 45


7. 2 5, 10, 15 3 5, 5, 15 5 5, 5, 5

1, 1, 1

LCM = 2 × 3 × 5 = 30The answer to this question is the LCM of 5, 10 and 15.


8. What is the greatest number that can divide exactly into 12, 20 and 60?

ANSWER

(C) 6(A) 2 (D) 8(B) 4


8. 2 12, 20, 60 2 6, 10, 30 3, 5, 15

HCF = 2 × 2 = 4The answer to this question is the HCF of 12, 20 and 60.

2 is a common factor of thethree numbers.2 is a common factor of thethree numbers.


9. The exact value of 85 × 104 is

ANSWER

(C) (85 × 100) (85 × 4)

(A) (85 × 100) + 4

(D) (85 × 100) + (85 × 4)

(B) (85 × 100) – (85 × 4)


9. 85 × 104 = 85 × (100 + 4)= 85 × 100 + 85

× 4= (85 × 100) +

(85 × 4)

Using the distributivelaw.


10. The value of the digit 3 in 736.2 is

ANSWER

(C) 3 tens

(A) 3 tenths

(D) 3 hundreds

(B) 3 ones


10. Hundreds Tens Ones Tenths7 3 6 2

The value of the digit 3 in 736.2 is 3 tens.


11. The simple interest earned on $600 at 5% perannum for 3 years is given by

ANSWER

(C)

(A)

(D)

(B)600 5

$100 3

600 3$

100 5

600 100$

3 5

600 5 3$

100


11. The simple interest,

P = $600R = 5 %T = 3 years$60

1000 5 3

100

PRTI


12. A woman bought a sheep for $800 and sold it for $1200. Her gain as a percentage of the cost price is

ANSWER

(C) 45%(A) (D) 50%(B) 40%1

33 %3


12.


The gain $(1 200 800) $400

$ 400The percentage gain

1

$ 8002

100%

50%

13. An insurance salesman is paid 4% of his sales as commission. His sales for March were $5 025. How much commission was he paid?

ANSWER

(C) $201.00

(A) $50.25

(D) $402.00

(B) $100.50


13. The commission 4% of $5 025

4$5 025

1004 $50.25

$201.00


14. If the simple interest on $700 for 4 years is $168, then the rate of interest per annum is

ANSWER

(C) 7%(A) 5% (D) 8%(B) 6%


14. The rate of interest per annum,

100

100

IR

PT

168

42

7 00 41

%

6%

I = $168P = $700T = 4 years


15. The exchange rate for one United States dollar (US $1.00) is six dollars and thirty-four cents in Trinidad and Tobago currency (TT $6.34). What is the value of US $50 in TT currency?

ANSWER

(C) $264(A) $634 (D) $134(B) $317


15. US $1.00 TT $6.34

US $50 TT $6.34 50

TT $317


16. A sales tax of 10% is charged on an article. How much does a customer pay for an article marked as $75?

ANSWER

(C) $80.00

(A) $82.50

(D) $79.50

(B) $82.00


16. The amount the

customer paid (100 10)% of $75

110% of $75

1.10 $75

$82.50


17. Samuel invested $800 for 3 years at 5% per annum. Marina invested $600 at the same rate.If they both earned the same sum as simpleinterest, how many years did Marina invest hermoney?

ANSWER

(C) 4(A) 6 (D) 3(B) 5


17. Samuel’s interest,

P = $800R = 5%T = 3 years

The time,

I = $120P = $600R = 5%


18. A discount of of the marked price is offered

for cash. What is the discount on a dress with amarked price of $150?

ANSWER

(C) $37.50

(A) $25.00

(D) $40.50

(B) $30.00

1

4


18.1

The discount of $15041

$1504$37.50


19. If P = {2, 3, 5, 7, 9}, Q = {3, 7, 8} andS = {7, 8, 9}, then

ANSWER

(C) {7}

(A) { }

(D) {2, 3, 5, 7, 8, 9}

(B) {2}


19.


20. U = {integers} N = {natural numbers}Zn = {negative integers}

Which of the Venn diagrams given below illustrates the statement:


“No natural numbers are negative integers”?

ANSWER

(C)

(A)

(D)

(B)


20. No natural numbers are negative integers:


21.

ANSWER

(C) 6(A) 2 (D) 8(B) 4

In the Venn diagram shown above, n(L) = 8, n(M) = 10 and What is


21.


22.

ANSWER

The two circles above represent set X and set Y.If X = {factor of 8} and Y = {factor of 12}, thenthe shaded region represents

(C) {2, 4, 6, 8}

(A) { }

(D) {4, 6, 8, 12}

(B) {1, 2, 4}


22. 8 1 8

2 4

12 1 12

2 6

3 4


23. The scale on a map is stated as 1:500 000. Thedistance between two towns as measured onthe map is 1.8 cm. What is the actual distancebetween the two towns?

ANSWER

(C) 9.0 km

(A) 0.9 km

(D) 50 km

(B) 1.8 km


23. 1:500000 cm 500000 cm

5 00 000

100 000

km

5 km

1.8 cm 5 km×1.8

9.0 km


24. The number of kilometres travelled by a vehicle in t hours at a rate of s km per hour is

ANSWER

(C) (A) (D) (B) stst

60ts

ts


24. Speed,

Distance, d = st

ds

t Formula

Multiply both sides by t.d is the subject of the formula.


25. A cuboid with dimensions 12 cm, 10 cm and5 cm occupies space of volume

ANSWER

(C) 81 cm3

(A) 27 cm3

(D) 600 cm3

(B) 54 cm3


25. The volume of the cuboid,

The formula for the volume of acuboid.

3

V

12cm 10cm 5cm

600cm

lbh


26. A cylindrical block of cheese 8 cm thick has a volume of 500 cm3. A student cuts a uniform slice of 2 cm thickness. What volume of the cheese did the student take?

ANSWER

(C) 100 cm3

(A) 50 cm3

(D) 125 cm3

(B) 75 cm3


26.


3

3

3

The volume of

2the cheese taken 500 cm

81

500 cm4

125 cm

27.

ANSWER

The figure above, not drawn to scale, shows the sector of a circle with centre O. The length of the minor arc PQ is 7 cm. The length of the circumference of the circle is

(C) 56 cm

(A) 21 cm

(D) 63 cm

(B) 28 cm


27.


360The number of 45 in 360 8

45The length of the circumference 7 cm 8

56 cm

28. The distance around the edge of a circular table top is 352 cm. The radius of the table top, in centimetres, is

ANSWER

(C) (A) 88 (D) (B) 352176

352


28.


The circumference, 2

So 352 2

352 i.e.

1762

C r

r

r

29. A plane left Guyana at 21:00 h. The next day, the plane arrived at its destination in the same time zone at 02:30 h. How many hours did the flight take?

ANSWER

(C) (A) (D) (B) 31

22

15

2

123

2


29.


The number of hours to

midnight (24:00 21:00) h

3 hours

The number of hours after

1midnight 2 hours

2The number of hours taken for

1the flight 3

15 hours

2

2 hours2

30. An aircraft leaves airport A at 07:30 h and arrives at airport B at 12:30 h, the same day, in the same time zone. The distance between the two airports is 3 600 kilometres. What was the average speed of the aircraft for the flight?

ANSWER

(C) 480 km/h

(A) 180 km/h

(D) 720 km/h

(B) 288 km/h


30. The time taken, (12:30 07:30) hours

5 hours

The average speed,

3 600 km

5 h720 km/h

t

ds

t


31. Each of the letters of the word ‘PERFORM’ is written on a piece of paper. One piece of paper is drawn at random. What is the probability that a letter ‘R’ is drawn?

ANSWER

(C) (A) (D) (B) 1

6

1

3

1

7

2

7


31.


The number of Rs 2

The total number of letters 7

(2

7 )P R

32. A die is tossed twice. What is the probability that a ‘2’ followed by an odd number turns up?

ANSWER

(C) (A) (D) (B) 1

12

1

4

1

36

2

3


32.


(2 followed by

an odd number) (2) (odd number)

1 3

6 61 1

61

12

2

P

P P

33.

ANSWER

The bar chart shows the number of students who liked one of five stated colours. How many students took part in the survey?

(C) 80(A) 5 (D) 125(B) 45


33. The number of students = 20 + 10 + 5 + 20 + 25

= 80


34.

ANSWER

The pie-chart above represents the fruit a group of students ate. If 16 students ate mandarin, then the total number of students in the group is

(C) 128(A) 125 (D) 135(B) 45


34. The angle representing the number of students

who ate mandarin 180° 135°

45°

360The total number of student

12

s 1645

6 8

8

1


35. The lowest weekly wage of a group of employees is $520.60. What is the wage of the highest paid employee, if the range of the wages is $63.20?

ANSWER

(C) $520.60

(A) $63.20

(D) $583.80

(B) $457.40


35. The range = The highest weekly wage − The lowest weekly wage $520.60 = The highest weekly wage − $63.20The highest weekly wage = $520.60 + $63.20

= $583.80


36.

ANSWER(C) 8.5 and 14.5

(A) 0 and 2.5

(D) 14.5 and 20.5

(B) 2.5 and 8.5

The lengths of the pencils of 40 students were measured, to the nearest cm, and the information collected is shown in the frequency table above.

The least and greatest length of the class interval 15–20 are

Frequency 9 17 14

Length of pencil (cm) 3–8 9–14 15–20


36.


3 8 9 14 15 20

2.5 8.5 14.5 20.5

class interval

class boundary

37. (5a)2 =

ANSWER

(C) 10a2(A) 10a (D) 25a2(B) 25a


37. Meaning of a square.

Expanding the term.

Grouping like values.

Multiplying like values.

Simplifying.2

2

2

(5 ) 5 5

5 5

5 5

25

25

a a a

a a

a a

a

a


38. 2x3 × 3x2 =

ANSWER

(C) 6x6(A) 5x5 (D) 36x5(B) 6x5


38.

Expanding the term.

Grouping like values.

Multiplying like values.

Adding the indices.

Simplifying.

3 2

3 2

3 2

3+2

5

5

2 3

2 3

2

6

6

6

3

x x

x x

x x

x

x

x


39. (6a) × (3b) =

ANSWER

(C) 9ab(A) 9ab (D) 18ab(B) 18ab


39.

Expanding theterm.

Grouping likevalues.

Multiplying likevalues.

Simplifying.

( 6 ) ( 3 )

( 6) ( 3)

( 6)

18

( 3)

18

a b

a b

a

b

b

a

a

b


40. 4(3x y) − 2(5y 3x) =

ANSWER

(C) 6x 14y

(A) 18x 14y

(D) 18x 6y

(B) 8x 4y


40. 4(3x y) 2(5y 3x) = 4 × 3x + 4 × (y) 2 × 5y 2 × (3x) = 12x 4y 10y 6x = 12x 6x 4y 10y = 18x 14y


Simplifying each term.Grouping like terms.Adding like terms.


41. If

ANSWER

(C) 10(A) (D) 3(B)

2 , then 5 2p q p pq

15 6


41. Substituting the value forp and for q in the formula.Simplifying the twoterms.Subtracting.

25 2 5 5 2

25 1

15

0


42. If 40 − 3x = x + 8, then x =

ANSWER

(C) 8(A) 4 (D) 29(B) 8


42.

Grouping like terms.

Adding like terms.

Dividing both sides by −4.

Simplifying.


40 3 8

3 8 40

So 4 32

2

48

3

x x

x x

x

x

43.

ANSWER

(C)

(A)

(D)

(B)

2 ( 3 ) (3 2 )a a b b a b 2 22 3a ab b 2 22 3 2a ab b

2 22 9 2a ab b 2 22 9 2a ab b


43. Use the distributive

law to remove the

brackets.

Adding the middle

terms.

2

2

2

2

2 ( 3 ) (3 2 )

2 6

2

3 2

3 2

a a b b a b

a ab a

a b

b b

ab


44. If

ANSWER

(C) (A) 5 (D) (B) 5

2

, when 5, then1

vK v K

v

1

64

16

4


44.

Substituting the value for v in theformula.

Squaring and subtracting values.

Dividing.

2

2

1

( 5)

5 125

41

64

vK

v


45. Yuri’s age is ten years less than twice that of Christine’s age. If Christine’s age is x years, then Yuri’s age, in years, is

ANSWER

(C) x 10

(A) 2(x 5)

(D) 2x 5

(B) 2(x 10)


45. Yuri’s age (2 10) years

(2 10) year

2( 5 ears

s

) yx

x

x


46. Which of the equations stated below represents the equation of a straight line?

ANSWER

(C) y = 5x2

(A) y = 3x

(D) y = 4x3

(B)2

yx


46. The equation of a straight line is y = mx + cIf c = 0, then y = mxSo y = −3x is the equation of a straight line.


47. The gradient of the straight line 2y = 4 5x is

ANSWER

(C) 4(A) 5 (D) 2(B) 5

2


47.

Writing the terms on the RHS

in the form mx + c.

Dividing each term by 2.

2 4 5

2 5 4

52

2

y x

y x

y x

It is in the form y = mx + c.

So the gradient,5

2.m


48. If

ANSWER

(C) 11(A) 1 (D) 17(B) 7

2( ) 5, then ( 3)g x x x g


48. Substitute −3 for x.

Simplifying.

Adding.

2( 3) ( 3) ( 3) 5

9 3 5

17

g


49.

ANSWER

The relation diagram shown above represents a function. Which of the following equations best describes the function?

(C) f (x) = x 2

(A) f (x) = x

(D) f (x) = 2(x 1)

(B) f (x) = y


49. (2) 2(2 1) 2(1) 2 2 2(3) 2(3 1) 2(2) 4 3 4(5) 2(5 1) 2(4) 8 5 8(7) 2(7 1) 2

( ) 2( 1)

(6) 12 7 12

f

f

f

f

ff xff

f x


50. Which of the following diagrams is not the graph of a function?

(A)


50. (B)


50. (C)


50. (D)

ANSWER


50.

Using the vertical line test for a function, it can be seen that The graph represents a one-to-many relation and it is therefore not a function.


51.


In the graph above, when y = 2, the values of x are:

ANSWER

(C) ±1.4(A) ±1.2 (D) ±1.5(B) ±1.3


51.

From the construction on the graph, when y = 2, then x = 1.4 and x = 1.4, that is x = ±1.4.


52.

ANSWER

The half-lines BA and CD are parallel. If angle BCD is 65°, then angle ABC is

(C) 130°(A) 65° (D) 145°(B) 115°


52. Interior angles are

supplementary.

Substitute the

value of angle

BCD.

Subtract 65° from

both sides.

Subtracting.

ˆˆ 180°

ˆ 65° 180°

ˆ 180° 65

115°

°

ABC BCD

ABC

ABC


53.

ANSWER

AC and DE are straight lines that intersects at B.Angle ABE = 127°The size of angle ABD is

(C) 127°(A) 53° (D) 233°(B) 74°


53. The sum ofangles on astraight line.Substitute thevalue for angleABE.Subtract 127°from both sides.Subtracting.

ˆ ˆ 180°

ˆ 127° 180°

ˆ 180° 127°

53°

ABD ABE

ABD

ABD


54.

ANSWER

The line segment PQ is mapped onto the line segment P′ Q′ by a translation. The matrix that represents this translation is

(C) (A) (D)(B)1

2

2

1

2

3

3

5


54.

5 3

2

3 2

5 3

3 2

1

T P P

T P P

or

8 6

6

1

5

2

8 6

6 5

T Q Q

T Q Q


or


55.


The shaded triangle is rotated through an angle of 90° in a counter-clockwise direction about thepoint P. Which of the four triangles represent theimage of the shaded triangle?

ANSWER

(C) C(A) A (D) D(B) B


55.


56.


In the diagram above, the line segment PQ is the image of LM after

ANSWER

(C) a reflection in the x-axis

(A) an enlargement of scale factor 1

(D) a rotation through with centre O

(B) a translation by vector2

5

90


Mx means a reflection in the x-axis.

56.


57.


The point P shown in the graph above is reflected in the x-axis. What are the co-ordinates of the image of P?

ANSWER

(C) (2, 3)

(A) (3, 2)

(D) (2, 3)

(B) (3, 2)


57.


58. In a triangle ABC, if angle A = 2x° and angle B = 3x°, then angle C =

ANSWER

(C) (180 5x)°

(A) 36°

(D)

(B) 72°

180

5x


58. The sum ofthe angles of atriangle.Substitute thevalue for angle Aand for angle B.Add the xs.

Subtract 5x fromboth sides.

Simplifying.


ˆ ˆˆ 180°

ˆ2 ° 3 ° 180°

ˆSo 5 ° 180°

ˆ

(18

180°

0 5 )°

5 °

A B C

x x C

x

C

x

C

x

59.

ANSWER

In the right-angled triangle, tan θ =

(C) (A) (D)(B)5

13

12

13

5

12

12

5


59. Definition of the tangent of anangle.

Using the capital lettersnotation.

Substituting the length foreach side.

tanθ

12 cm

Opp

Adj

AB

BC

5 cm

12

5


60.


The diagram above, not drawn to scale, shows that the angle of depression of a point A on the ground from T, the top of a tower, is 40°. A is 25 m from B, the base of the tower. The height, TB, of the tower, in metres, is

ANSWER

(C) 25 tan 40°

(A) 25 sin 40°

(D) 25 sin 60°

(B) 25 cos 40°


60.


Alternateangles.Definition ofthe tangent ofan angle.Substitute thelength of AB.Multiply bothsides by 25 m.


ˆ 40°

tan 40°

25 m25 m tan 40°

or 25 m tan 4 0°

TAB

Opp

Adj

TB

ABTB

TB

TB

CSEC MODEL EXAMINATION 1

MATHEMATICS

Paper 2

2 hours 40 minutes

SECTION I

Answer ALL the questions in this section

All working must be clearly shown

NEXT

1. (a) Using a calculator, or otherwise, calculate

the EXACT value of

ANSWER

(i)

(3 marks)

73

151 3

2 43 5

giving your answer as a common fraction


1. (a) (i)


(ii)

ANSWER

(3 marks)giving your answer in standard form.

0.0225

36


1. (a) Using a calculator, or otherwise, calculate

the EXACT value of

1. (a)(ii)

2

4

2

0.0225 225 10

36 36

15 10

6

2.5 10

Standard form


The basic wage earned by a factory worker for a 40-hour week is $640.00.

(i) Calculate her basic hourly rate.

For overtime work, the factory worker is paid one and a half times the basic hourly rate.

1. (b)

ANSWER

(1 mark)


1. (b) (i) The basic hourly rateThe basic wage

The basic week$640

4

$1

0

6


The basic wage earned by a factory worker for a 40-hour week is $640.00.

(ii) Calculate her overtime wage for 15 hours of overtime.

1. (b)

ANSWER

(2 marks)


1. (b) (ii) The overtime hourly rate = The overtime rate The basic hourly rate

The overtime wage = The overtime hourly rate The number of hours worked overtime= $24 15= $360


1. (b) The basic wage earned by a factory worker for a 40-hour week is $640.00.

(iii) Calculate the total wages earned by the factory worker for a 60-hour week.

ANSWER

(3 marks)Total 12 marks


1. (b)

(iii)

The number of hours worked overtime

= (60 40) hours = 20 hours

The overtime wage = $24 20= $480

The total wages earned

= The basic wage The overtime wage= $(640 480)= $1120


Factorise completely:2. (a)

(i) 8px 5py 8qx 5qy

ANSWER

(2 marks)


2. (a)(i) 8px 5py 8qx 5qy

= p(8x 5y) q(8x 5y)= (8x 5y) (p q)

Factorise pairwise

Factoriseusing 8x 5yas a commonfactor.


(ii) 4x2 36

ANSWER



(2 marks)

2. (a)(ii) 4x2 36

= 4(x2 9)= 4(x2 32)= 4(x 3)(x 3)

Factorise using 4 as the HCF.

Write as the differenceof two squares.Factorise as the difference of two squares.


(iii) 5x2 6x 8

ANSWER



(2 marks)

2

2

610

5( 8)

( 2)(5 4

440

5 6 8

5 10 4 8

5 ( 2 ( 2)

)

) 4

x

x x

x x x

x x

m nm

m

x

n

x

n

2. (a)(iii)

Factorise pairwiseFactorise usingx 2 as a commonfactor.


2. (b)One cup of yogurt costs $x and one granolabar costs $y.

One cup of yogurt and three granola bars cost$32.00, while two cups of yogurt and two granola bars cost $30.00.

(i) Write a pair of simultaneous equations in x and y to represent the given information above.

ANSWER(2 marks)


2. (b) (i) The cost of one cup of yogurt = $x

The cost of one granola bar = $y

The first equation is:x 3y = 32 (in dollars)

The second equation is:2x 2y = 30 (in dollars)

The pair of simultaneous equations in x and y:

x 3y = 32 2x 2y = 30


2. (b)

ANSWER

One cup of yogurt costs $x and one granola bar costs $y.

One cup of yogurt and three granola bars cost $32.00, while two cups of yogurt and two granola bars cost $30.00.

Solve the equations to find the cost of one cup of yogurt and the cost of one granola bar. (4 marks)

Total 12 marks


(ii)

Hence, the cost of a yogurt is $6.50 andthe cost of a granola bar is $8.50.

2. (b) (ii)


So

In a survey of 85 students,

25 played drums

20 played tassa

x played drums and tassa

3x played neither.

Let D represent the set of students in the survey who played drums, and T the set of students who played tassa.

Copy and complete the Venn diagram

below to represent the information obtained from the survey.

(i) ANSWER

3. (a)

(2 marks)

3. (a)

(i)

The Venn diagram is shown above.

The students who played drums only,

The students who played tassa only,


( ) 25 xn D T ( ) 20 xn T D


25 played drums

20 played tassa


3x played neither.

Let D represent the set of students in

the survey who played drums, and T

the set of students who played tassa.

Write an expression in x for the totalnumber of students in the survey.

(ii) ANSWER

(1 mark)


3. (a)

The total number of students in thesurvey, n(U ) = 25 x + x + 20 x + 3x

= 2x + 45

3. (a)

(ii)


3. (a)

ANSWER(2 marks)Calculate the value of x.



25 played drums

20 played tassa


3x played neither.

Let D represent the set of students in the survey who played drums, and T the set of students who played tassa.

(iii)

3. (a)

(iii) n(U ) = 85 and n(U ) = 2x + 45

so we have the following equation:

Hence, the value of x is 20.

Subtract 45 fromboth sides.

Divide both sidesby 2.


2 45 85So 2 85 45i.e. 2 4

20

040

2

xxx

x

(i) Using a ruler, a pencil, and a pair of compasses, construct the kite PQRS accurately. ANSWER(4 marks)


The diagram below, not

drawn to scale, shows a

kite, PQRS, with the

diagonal PR = 6 cm,

3. (b)

3. (b) (i)

Draw a horizontal line greater than 6 cm. Mark a point P to the left of the line. Set your compasses to a separation of 6 cm using a ruler. Place the steel point of the compasses at point P and construct an arc to intersect the horizontal line at point R. PR = 6 cm.


Set your compasses to a separation of

that is, 3.25 cm. With point P as centre, construct an arc above PR. With point R as centre and the same compasses separation, construct another arc to intersect the previous arc at point Q.

13 cm,

4

13 cm.

4PQ RQ

Set your compasses to a separation of 5 cm. With centres P and R, construct two arcs below PR to intersect at point S.

PS = RS = 5 cm.

Use a ruler and pencil to draw the four sides of the kite PQRS.


(ii) Join QS. Measure and state, in centimetres, the length of QS.

ANSWER(2 marks)

Total 11 marks


The diagram below, not drawn to scale,

shows a kite, PQRS, with the diagonal

PR = 6 cm,

3. (b)

Draw a straight line from Q to S. Take a divider and open it from point Q to point S. Measure the separation of the divider using a ruler.

3. (b) (ii)

15 cm 5.

425 cmQS The length of


The table below shows two readings from the records of a train.

4. (a)

Town Time Distance travelled (km)

X 07:20 538

Y 09:50 773

Calculate(i) the number of hours taken for the

journey from town X to town YANSWER

(1 mark)


4. (a) (i) The number of hours taken for the journey from town X to town Y,

(09 : 50 07 : 20) h

2h 30min

12 h

2

t =

=


4. (a)

Calculate (ii) the distance travelled, in kilometres, between the two towns ANSWER(1 mark)




X 07:20 538

Y 09:50 773

4. (a) (ii) The distance travelled between the two towns, (773 538) km

235 kmd


4. (a)

ANSWER(2 marks)


Calculate (iii) the average speed of the train in km/h



X 07:20 538

Y 09:50 773

4. (a)

235 km1

2 h2

235km/h

52

2235 km/h

54794 km/h

2 km/h

ds

t

The average speed of the train,(iii)


4. (b) The map shown below is drawn to a scale of 1:500 000.

ANSWER(2 marks)

(i) Measure along a straight line and state, in centimetres, the distance on the map from P to Q.


Open your divider from P to Q, then measure the separation using a ruler.

4. (b) (i)

The distance on the map from P to Q = 5.8 cm


(ii) Calculate the actual distance, in kilometres, from P to Q ANSWER

(2 marks)



The scale is 1: 500 0004. (b) (ii)

The actual distancefrom P to Q = 5.8 5 km = 29.0 km


That is 1cm 500 000 cm

5 00 000So 1cm

100 000

km

1cm 5 km

(iii) The actual distance between two places is 8.5 km. Calculate the number of centimetres that represent this distance on the map

ANSWER(3 marks)Total 11 marks



4. (b) (iii) 5 km is represented by 1 cm

1 km is represented by1

cm5

8.5 km is represented by 8.5

= 1.7 cm

1cm

5


ANSWER

(1 mark)

5. (a) Given that f (x) = 4x − 7 and g(x) = x2 − 15, calculate the value of

(i) f (−3)


5. (a) (i) ( ) 4 7

( 3) 4( 3) 7

12 7

19

f x x

f

Substitute −3 for x.


ANSWER

(2 marks)



(ii) gf (2)

5. (a) (ii)

Substitute 2 for x.( ) 4 7

(2) 4(2) 7

8 7

1

f x x

f

Substitute 1 for x.

2

2

( ) 15

(1) (1) 15

1 15

(2) 14

g x x

g

gf


Substitute f (x)into g(x) for x.

Or


2

2

2

2

2

( ) 4 7

( ) 15

( ) (4 7) 15

(2) (4 2 7) 15

(8 7) 15

1 15

1 15

14

f x x

g x x

gf x x

gf

ANSWER

(2 marks)



(iii) f −1(−1)

5. (a) (iii)

Defining equation for f(x).

Interchanging x and y.

Adding 7 to both sides.

Dividing both sides by 4.

Defining equation for f −1(x).1

( ) 4 7

4 7

4 7

4 7

7

4

( )7

4

f x x

y x

x y

y x

y

fx

x

x


ANSWER

(2 marks)

5. (b) (i) Given that y = x2 + x − 6, copy and complete the table below.

x −4 −3 −2 −1 0 1 2 3y 6 −4 −6 −6 −4 6


5. (b) (i) 2

2

2

6

when 3, then

( 3) ( 3) 6

9 3 6

when 2, then

2 2

0

4 4

0

6

y x x

x

y

x

y


The completed table is shown below.

x −4 −3 −2 −1 0 1 2 3y 6 0 −4 −6 −6 −4 0 6


ANSWER


(ii) Using a scale of 2 cm to represent 1 unit

on the x-axis and 1 cm to represent

1 unit on the y-axis, draw the graph of

y = x2 + x − 6 for −4 ≤ x ≤ 3.


5. (b) Given that y = x2 + x − 6, copy and complete the table below.

x −4 −3 −2 −1 0 1 2 3y 6 −4 −6 −6 −4 6

5. (b) (ii)

Using the given scales, the graph of y = x2 + x − 6 for −4 ≤ x ≤ 3 was drawn on graph paper as shown above.


ANSWER

6. The diagram below shows trapeziums A, B and C. The line y = −x is also shown.


ANSWER

(3 marks)

6. (a) Describe, fully, the single transformation which maps trapezium A onto

(i) trapezium B


6. (a) (i) The single transformation which mapstrapezium A onto trapezium B is a

translation

with vector

3.

6

Each point on trapezium A is moved 3 units horizontally to the right, then 6units vertically downwards.


ANSWER

(3 marks)6. (a) (ii) trapezium C


6. (a) (ii) The single transformation which maps

trapezium A onto trapezium C is a

reflection in the line y = −x .


ANSWER


6. (b) State the coordinates of the vertices of trapezium D, the image of trapezium B after a reflection in the line y = −x.


6. (b) The coordinates of the vertices of trapezium D are:

(3, 0), (1, 2), (1, 4) and (3, 4).


7. The waiting time, to the nearest minute, experienced by 100 people to catch a bus is shown in the table below.

Waiting Time (minutes)

Number of Students

Cumulative Frequency

1 – 5 9 9 6 – 10 12 2111 – 15 15 3616 – 20 1921 – 25 2226 – 30 1631 – 35 436 – 40 3


ANSWER

(2 marks)7. (a) Use the table given above to construct a

cumulative frequency table.


7. (a) Interval(minutes)

CumulativeFrequency

< 5.5 9 < 10.5 9 + 12 = 21< 15.5 21 + 15 = 36< 20.5 36 + 19 = 55< 25.5 55 + 22 = 77< 30.5 77 + 16 = 93< 35.5 93 + 4 = 97< 40.5 97 + 3 = 100

The cumulative frequency table is shown above.


ANSWER

(4 marks)7. (b) Use the values from your table to draw a

cumulative frequency curve.


7. (b)

The completed cumulative frequency curve is shown above.


ANSWER

(2 marks)

7. (c) Use your graph to estimate

(i) the median for the data


7. (c) (i) Half of the total frequency,

From the graph, the waitingtime corresponding to a total

frequency of 50, Q2 = 19 minutes

Hence, the median for the data is

19 minutes.

1 1100

2 250

n


ANSWER

(2 marks)7. (c) (ii) the number of people who waited less

than 23 minutes


7. (c) (ii) From the graph, the number of people whowaited less than 23 minutes =

65


ANSWER


7. (c) (iii) the probability that a person, chosen at

random from the group, waited for at

least 18 minutes


7. (c) (iii) From the graph, the number of people whowaited less than 18 minutes =

45 The number of people whowaited for at least 18 minutes

P(x ≥ 18 minutes)

100 45

55

55

1011

20

0


8. The first three diagrams in a sequence are shown below. Diagram 1 has a single circle, which can be considered as a square pattern formed by a single circle.

Diagram 2 consists of a square of side two circles with two triangles formed at the ends as shown.

Diagram 3 consists of a square of side three circles with two triangles formed at the ends as shown.

Diagram 1 Diagram 2 Diagram 3


Diagram Number

Number of Circles Forming

the Square

Number of Additional Circles in Two

Triangles

Pattern for Calculating the Total Number of

Circles in the Diagram1 12 1(0) 12 + 1(0)2 22 2(1) 22 + 2(1)3 32 3(2) 32 + 3(2)

(i) 4 42 — —

(ii) — — 8(7) —

(iii) n — — —


ANSWER

(2 marks)8. (a) Draw Diagram 4 in the sequence.


8. (a)

Diagram 4

Diagram 4 in the sequence is shown above.


ANSWER


8. (b) Complete the table by inserting the appropriate values at the rows

marked (i), (ii) and (iii).


(b) Diagram Number

Number of Circles Forming

the Square

Number of Additional Circles in Two

Triangles

Pattern for Calculating the Total Number of

Circles in the Diagram1 12 1(0) 12 + 1(0)2 22 2(1) 22 + 2(1)3 32 3(2) 32 + 3(2)

(i) 4 42 4(3) 42 + 4(3)

(ii) 8 82 8(7) 82 + 8(7)

(iii) n n2 n(n 1) n2 + n(n 1)

8.

The completed table is shown above.


SECTION II

Answer TWO questions in this section


NEXT

ANSWER

(4 marks)

9. (a) Solve the pair of simultaneous equations

y = 1 − 2x

y = 2x2 + 5x − 3


9. (a)


2

2

2

2

2

1 2

2 5 3

: 2 5 3 1 2

Transfer all terms to the LHS:

2 5 2 3 1 0

Add like terms:

2 7 4 0

Factorise the expression on the LHS:

(2 1)( 4) 0 2 2

4 1 4

7 8

y x

y x x

x x x

x x x

x x

x x x x x

x x x

‚

‚

Either 2x – 1 = 0i.e. 2x = 1

Or x + 4 = 0 x = – 4

1

2x

1When , then

21 2

11 2

2

1 1

0

x

y x



When 4, then

1

9

2( 4)

1 8

x

y

10 Hence, , and .4

2 9,x y x y

ANSWER

(3 marks)

9. (b) Express in the form where a, h and k are

realnumbers

22 5 3x x 2( ) ,a x h k


9. (b) Factorise out thecoefficient of x2

i.e. 2.

Write as a perfect square.

The LCM of 2 and 16 is 16.

Adding the fractions.

2 2 21 1 5 5

coefficient of 2 2 2 4

x


2

2

2 22

2

2

2

2 5 3

5 32

2 2

5 5 3 52

2 4 2 4

5 3 252

4 2 16

5 3(8) 25(1)2

4 16

5 24 252

4 16

x x

x x

xx

x

x

x

22 5 5 3 5

22 4 2 4

x x

Simplifying the fraction.

Multiplying the fraction by 2.

It is in the form where a, h and k are 2, respectively.

2

2

5 492

4 16

5 492

4 8

x

x

2( )a x h k


and5

4

49

8

ANSWER

(1 mark)

9. (c) Using your answer from (b) above, or otherwise, calculate.

(i) the minimum value of22 5 3x x


9. (c)

(i) The minimum value of

2

2

2 5 3

5 492

4 8

x x

x

22 5 3

4

16

8

9

8

x x


ANSWER

(1 mark)

9. (c) Using your answer from (b) above, or otherwise, calculate.


(ii) the value of x where the minimum

occurs

9. (c) (ii) The minimum occurs where the value of x

5

41

14


4

50

45

x

x

ANSWER

(4 marks)

9. (d) Sketch the graph of y = 2x2 + 5x − 3, clearly showing

the coordinates of the minimum point.

the value of the y-intercept.

the values of x where the graph cuts the

x-axis.


9. (d) The coordinates of the minimumpoint are

y = 2x2 + 5x − 3 the value of the y intercept, c = 3

y = 2x2 + 5x − 3 and y = 0 on the x-axis,

so 0 = (2x − 1)(x + 3) by factorising the

expression.

1 11 , 6 .

4 8


Hence, x = and x = −3 are the values

of x where the graph cuts the x-axis.

A sketch of the graph of y = 2x2 + 5x − 3 is

shown below.

1

2


ANSWER


9. (e) Sketch on your graph of y = 2x2 + 5x − 3,

the line which intersects the curve at the values of x and y as calculated in (a) above.


9. (e) A sketch of the line y = 1 − 2x which intersects the curve

y = 2x2 + 5x − 3 at the points (−4, 9) and is shown below.

1,0

2


ANSWER(1 mark)(i) ABC


10. (a) The diagram following, not drawn to scale, shows a circle, centre O. The line DCE is a tangent to the circle. Angle ACE = 46° and angle OCB = 25°.

Calculate:

10. (a) (i) 46ABC ACE in alternate segment


ANSWER(1 mark) (ii) AOC



Calculate:

10. (a) (ii) at centre =2· at circumference

2

9

46

2

2

AOC ABC


ANSWER(1 mark)


(iii) BCD


Calculate:

10. (a) (iii) betweenradius andtangent atpoint oftangency.


90

So 2

6

5 90

90

5

25

BCD OCB

BCD

BCD

ANSWER(1 mark)


(iv) BAC


Calculate:

10. (a) (iv) in alternate segment.

65BAC BCD


ANSWER(1 mark)


(v) OAC


Calculate:

10. (a) (v)ΔOAC isisosceles,since OC =OA = r.

180

218

4

28

24

0 92

8

AOCOAC


ANSWER(1 mark) (vi) OAB



Calculate:

10. (a) (vi)

2

65 4

1

4

OAB BAC OAC


10. (b) The diagram below, not drawn to scale, shows the positions of two ships, P and Q, relative to a point O. P is on a bearing of 045° from O and the distance OP = 500 km.

Q is on a bearing of 080° from P and the distance PQ = 800 km.


ANSWER

(2 marks)

10. (b) (i) Copy the diagram above. On you diagram indicate the angles that represent the

bearings of 045° and 080°.


10. (b) (i)

A copy of the diagram is shown above. The angles that represent the bearings of 045° and 080° are indicated. The

distances are also indicated.


10. (b) The diagram below, not drawn to scale, shows the positions of two ships, P and Q, relative to a point O. P is on a bearing of 045° from O and the distance OP = 500 km.

Q is on a bearing of 080° from P and the distance PQ = 800 km.


ANSWER


10. (b) (ii) Calculate

a) OPQ

b) the distance OQ, to the nearest kilometre

c) the bearing of Q from O


10. (b) (ii) a)


Interior angles are supplementary,NO //NP.


180

45 180

180 45

135

NOP NPO

NPO

NPO

Sat apoint.

Hence, OPQ is 145°.


360

135 80 360

215 360

360 215

145

OPQ NPO NPQ

OPQ

OPQ

OPQ

10. (b) (ii) b)


Hence, the distance OQ is 1 243 km, to the

nearest kilometre.

Considering ΔOPQ and using the cosine rule:


2 2 2

2 2

ˆ2 cos

500 800 2 500 800 cos145

250 000 640 000 800 000 ( 0.819)

890 000 655 200

1545 200

1545 200 km

(to1 243km the nearest km)

P OPQOQ OP PQ OP Q

OQ

10. (b) (ii) c)


Hence, the bearing of Q from O is 066.7°.

Considering ΔOPQ and using the sine rule:

1

1 243 800ˆsin145 sin

800 sin145ˆsin1 243

0.369

ˆ sin 0.369

21.7

POQ

POQ

POQ

45 21.7

66.7

NOQ NOP POQ


ANSWER

(4 marks)

11. (a) The value of the determinant of

is −36.

Calculate the values of x.

4

5

xM

x


11. (a)

2

2

2

4

5

The determinant of , 4( 5)

20

And 36.

So 36 20

i.e. 36 20 16

16 4

xM

x

M M x x

x

M

x

x

x

Hence, the values of x are +4 and −4.


ANSWER(2 marks)

11. (b) The transformation R is represented by the

matrix1 0

.0 1

The transformation S is represented by the

matrix1 0

.0 1

(i) Write a single matrix, in the form

to represent the combined

transformation S followed by R.

a b

c d


11. (b) (i) The combined transformation S followed by R,

Hence, the single matrix that represents the combined transformation S followed by R is .

1 0

0 1


ANSWER

(3 marks)


(ii) Calculate the image of the point P(−7, 4) under the combined transformation S followed by R.

11. (b) The transformation R is represented by the

matrix1 0

.0 1

The transformation S is represented by the

matrix1 0

.0 1

11. (b) (ii) RS P P′

P′ (7, 4)

Hence, the image of the point P (−7, 4) under the combined transformation S followed by

R is P′ (7, 4).

1 0 7 1 ( 7) 0 4

0 1 4 0 ( 7) 1 4

7

4

P


(ii)Or


ANSWER

(2 marks)

11. (c)5 2

The matrix .3 1

N

(i) Determine the inverse matrix of N.


11. (c) (i) 5 2The matrix

3 1N

The determinant of , 5( 1) 2(3)

5 6

11

N N

The adjoint matrix of 1 2, adjoint

3 5N N



(ii) Hence, calculate the value of x and the

value of y for which5 2 4

3 9

x y

x y


11. (c)5 2

The matrix .3 1

N

ANSWER

11. (c) (ii)Given 5 2 4 then

3 9

5 2 4

3 1 9

x y

x y

x

y

1

It is in the form

So

NX B

X N B


Hence, x = 2 and y = −3.



Paper 190 minutes


NEXT

1. (1)3 + (3)2 =

ANSWER

(A) 4

(B) 9

(C) 8

(D) 10


1. (1)3 + (3)2 =

= (1) (1) (1) + (3) (3)

= 1 (1) + 9

Use the meaning of a square and a cube.

The product of two negative signs is a positive sign.

The product of a positive sign and a negative sign is negative.

= 1 + 9

= 8 Subtracting.


2. Express as a decimal correct to 3 significant figures.

ANSWER

(A) 5.27

(B) 5.28

(C) 5.29

(D) 5.30

25

7


20.285 . . .

7

25 5.28 5 . . .

5.297

(3 sf )

2. 0.2857 20

1460

56

40

355

The digit after the 3rd significant figure is 5, so we add 1 to the digit 8.


3. The decimal fraction 0.016 expressed as a commonfraction in its lowest terms is

ANSWER

(A)

(B)

(C)

(D)

16

100

4

2502

125

16

1000


Write the decimal fraction as an equivalent common fraction.


This is the common fraction written in its lowest terms.

3. 160.016

1000

16 8

1000 8

2

125


4. In standard form, 8 504 is

ANSWER

(A) 8.504 102

(B) 8.504 103

(C) 8.504 102

(D) 8.504 103


4. 8 504 = 8.504 1 000

= 8.504 103

The first number must have a value between 1 and 10.

That is, 1 < first number < 10


5.

ANSWER

(A)

(B) 12

(C)

(D)

41

3

1

12

1

81

4

3


4

1

1 1 1 1 1

3 3 3 3 3

1 1 1 1

3 3 3 3

81

5.


6. If 70% of a number is 80, then the number is

ANSWER

(A) 10

(B) 56

(C) 80

(D) 2114

7


6. The number 8 0100

7 0

800

72

1147


7. The multiplicative inverse of –5 is

ANSWER

(A) 5

(B) 5

(C)

(D)

1

51

5


5 1

1

5

51

x

x

7. Definition

Divide both sides by –5.

A positive value divided by a negative value is a negative value.


8. The HCF of 15, 30 and 60 is

ANSWER

(A) 3

(B) 5

(C) 15

(D) 45


8. 3 15, 30, 60

5 5, 10, 20

1, 2, 4

Each of the numbers 15, 30 and 60 is divisible by 15.

The HCF = 3 5 = 15


9. If 2n is an even number, which of the following is an odd number?

ANSWER

(A) 2n 1

(B) 2(n + 1)

(C) 2n 2

(D) 2(n + 3)


9. Even number = 2n

Odd number = 2n 1


10. The next term in the sequence 5, 2, 1, 4 is

ANSWER

(A) 5

(B) 6

(C) 7

(D) 8


10. 5, 5 3 = 2,

2 3 = 1,

1 3 = 4,

4 3 = 7

A term in the sequence is obtained by subtracting 3 from the term just to its left (the preceding term).


11. A butcher bought a car for $2 500 and sold it for $3 000. His profit as a percentage of the cost price is

ANSWER

(A) 5%

(B) 10%

(C) 15%

(D) 20%


11. The profit = $(3 000 2 500)

= $500

The percentage profit =

1

$ 500

5

$ 2500100%

= 20%


12. A boutique gives 10% discount for cash. What is the cash price of a dress with a marked price of $350?

ANSWER

(A) $35

(B) $315

(C) $340

(D) $360


12. (100 10)% of $350 = 90% of $350

90$350

100

= $315


13. If J $90.00 is equivalent to US $1.00, then J $5 400.00 equivalent to

ANSWER

(A) US $6.00

(B) US $60.00

(C) US $600.00

(D) US $540


13. J $90.00 US $1.00

1.00J $1.00 US $

90.001.00

J $5 400.00 US $ 5 400.0090.005 400

US $90

US $60


14. The freight charges on a parcel is $150 plus custom duties of 20%. What amount of money was paid to collect the parcel?

ANSWER

(A) $160

(B) $170

(C) $180

(D) $190


14. (100 20)% of $150 120% of $150

120$150

1

$ 0

0

18

0


15. A man pays $0.25 for each unit of electricity used up to 400 units and $0.31 for each unit of electricity used in excess of 400 units. How much does he pay for consuming 1 200 units of electricity?

ANSWER

(A) $56

(B) $324

(C) $348

(D) $672


15. The cost for the first 400 units = $0.25 400

= $25 4 = $100

The cost for the remaining = $0.31 800

800 units = $31 8 = $248

The electricity bill = $(100 + 248)

= $348


16. The table below shows the rates charged by an insurance company for home insurance.

ANSWER

(A) $2 100 (B) $4 500

(C) $4 020 (D) $6 600

House $4.50 per $1 000

Contents $2.10 per $1 000

A house is valued at $800 000 and the contents at $200 000. How much will the owner pay for home insurance?


16. The cost for insuring the house = $4.50

= $4.50 800

= $450 8

= $3 600

$800 000

$1 000

The cost for insuring the contents = $2.10

= $2.10 200

= $210 2

= $420

$200 000

$1 000


The cost for the home insurance = $(3 600 + 420)= $4 020


17. A student bought 12 blue pens at $15 each and 13 green pens at $10 each. What is the mean cost per pen?

ANSWER

(A) $12.40

(B) $12.50

(C) $12.60

(D) $12.70


17. The cost for the 12 blue pens = $15 12

= $180

The cost for the 13 green pens = $10 13

= $130

The total cost for the 25 pens = $(180 + 130)

= $310

The mean cost per pen =

= $12.40

$310

25


18. A woman invested a sum of money at 6% per annum for 2 years. If she collected $ 300 as simple interest, what was the sum of money that she invested?

ANSWER

(A) $2 500

(B) $2 700

(C) $2 800

(D) $10 000


18. I = $300

R = 6%

T = 2 years

100The principal,

100 300$

IP

RT

50

6 2

100 50$

25

12

$100

50

25

$2 0


19.

ANSWER

(A) (P Q)

(B) (P Q)

(C) P Q

(D) P Q

In the Venn diagram above, the shaded region represents


19. The unshaded region represents

P Q P or Q

The shaded region represents

(P Q) Not P or Q


20. If U = {2, 3, 5, 7, 11, 13} and A = {5, 11}, then n(A) =

ANSWER

(A) 2

(B) 4

(C) 6

(D) 8


20. A = {2, 3, 7, 13}

n(A) = 4


21.

ANSWER

(A) {3, 9, 15, 18, 24}

(B) {3, 6, 9, 12, 15}(C) {6, 12, 18, 24}(D) {6, 12}

In the Venn diagram, set L and set M are represented by two intersecting circles. If L = {multiples of 3 less than 16} and M = {multiples of 6 less than 25}, then the shaded region represents


21.


{3, 6, 9, 12, 15}

{6, 12, 18, 24}

which are common elements.{6, 12}

L

M

L M

22. Which of the following pairs of sets are equivalent?

ANSWER

(A) {2, 3} and {a, b, c}

(B) { } and {1, 2, 3}

(C) {a, b, c} and {2, 4}

(D) {1, 2, 3} and {a, b, c}


22. {1, 2, 3} {a, b, c}

n{1, 2, 3} = 3 n{a, b, c} = 3

The number of elements in each of the sets is 3, therefore the sets are equivalent.

Or

1 a

2 b

3 c

There is a 1 1 correspondence between the elements of the two sets, therefore the sets are equivalent.


23. The volume of a cube with edges of length 1 cm is

ANSWER

(A) 1 cm3

(B) 12 cm3

(C) 16 cm3

(D) 24 cm3


23. The volume of the cube,

V = l3

= (1 cm)3

= 1 cm3

The formula for thevolume of a cube.


24. Expressed in millimetres, 470 centimetres is

ANSWER

(A) 4.7

(B) 47

(C) 4 700

(D) 47 000


24. 1 cm = 10 mm

470 cm = 10 470 mm

= 4 700 mm


25. The lengths of the sides of a triangle are x, 2x and 3x centimetres. The perimeter of the triangle is 30 centimetres. What is the value of x?

ANSWER

(A)

(B) 5

(C) 10

(D) 15

3 5


25. The perimeter = (x + 2x + 3x) cm

= 6x cm

Equating the values for the perimeter:

6x = 30

= 5

30

6x


26. If Usain Bolt runs the 100 metres race in 9.6 seconds, what was his average speed in metres per second?

ANSWER

(A)

(B)

(C)

(D) 96

24

254

45

510

12


26. The average speed,d = 100 m and t = 9.6 s

100 m

9.6 s

1000m/s

96250

m/s24

510 m/s

12

ds

t


27. Forty students each drank 2 bottles of sweet drink. Each bottle held 250 millilitres of sweet drink. How many litres of sweet drink were used?

ANSWER

(A) 20

(B) 80

(C) 500

(D) 20 000


27. The number of bottles used = 40 2

= 80

The number of millilitres used = 250 80

= 20 000

The number of litres used =

= 20

20 000

1000


28. The length of a rectangle is three times that of its width. If the area of the rectangle is 108 cm2, then its width, in cm, is

ANSWER

(A) 6

(B) 26

(C) 27

(D) 36


l = 3w cm28.

A = 108 cm2 b = w cm

The area of the rectangle,A = lb

= (3w w) cm2 Substitute 3w for l and w for b.

= 3w2 cm2

Equating the values for the area:2

2

3 108

108

6 cm

363

36

w

w

w


29. A student leaves home at 06:25 h and arrives at school at 07:45 h. The student travels non-stop at an average speed of 60 km/h. What distance, in kilometres, is the student’s home from school?

ANSWER

(A) 40

(B) 50

(C) 70

(D) 80


29. The time taken,

t = (07:45 – 06:25) h

= 1 h 20 min 1

1 h3

The distance,

160 1 km

34

60 km3

80 k20 4 km

m

d st

s = 60 km/h

t 1

1 h3


30.

ANSWER

(A) (B)

(C) (D)

The diagram above shows a sector POQ withsector angle POQ = 45° and radius OQ = r units.The area of the sector POQ is

21π

2r 21

π4

r

21π

6r 21

π8

r


30. 2

2

2

2

θThe area of the sector, π

36045

π3601

π8

1π

8

A r

r

r

r


Items 31 – 34 refer to the following frequency distribution.The distribution shows the mass of parcels, in kilograms, sent to a skybox by an individual.

Mass of parcel (kg) Number of parcel

2 3

3 7

4 2

5 1


31. The mode, in kilograms, of the distribution is.

ANSWER

(A) 2

(B) 3

(C) 4

(D) 5


31. Mode = 3 kg 7 (highest frequency)


32. What is the median, in kilograms, of the distribution?

ANSWER

(A) 4

(B) 3.5

(C) 3

(D) 2


32. The total frequency = 3 + 7 + 2 + 1

= 13

So the middle value is in the 7th ordered position.

The 7th parcel in ascending or descending order has a

mass of 3 kg.

So the median of the distribution has a mass of 3 kg.


33. The total mass, in kilograms, of all the parcels sent to the skybox by the individual is

ANSWER

(A) 13

(B) 14

(C) 40

(D) 182


x(kg) f fx

2 3 63 7 214 2 85 1 5

fx = 40

33.


34. The mean, in kilograms, of the distribution is

ANSWER

(A)

(B)

(C)

(D)

13

131

321

4131

42


34.

13 kg

The mean,

40 kg

13

13

fxx

f


35.

ANSWER

(A) 400

(B) 300

(C) 200

(D) 100

The pie chart shown above represents the ways in which a school of 600 children watched a movie. The number of children who watched the movie at a cinema is approximately


35. 1 1The number of children of 600

3 21

3003100


36. The volume, in millilitres, of five sizes of bottled orange juice are 500, 250, 2 000, 750, 1 000. The range, in millilitres, is

ANSWER

(A) 250

(B) 500

(C) 1 000

(D) 1 750


36. The range

= The greatest volume The least volume

= (2 000 250) ml

= 1 750 ml


37. 5(x 2) =

ANSWER

(A) 5x 2

(B) 5x + 2

(C) 5x 10

(D) 5x + 10


37. 5(x 2) = 5 x5 (2)

= 5x + 10

Use the distributivelaw to remove thebrackets.

The product of apositive and a negativesign is a negativesign. The product oftwo negative signs is apositive sign.


38. 4(2x 1) 3(x 5) =

(A) 5x 11

(B) 5x +11

(C) 5x – 6

(D) 5x + 6

ANSWER


38. 4(2x 1) 3(x 5)

= 8x 4 3x + 15

= 8x 3x + 15 4

= 5x + 11

Use the distributive law on the terms in each pair of brackets.Group like terms.Add like terms.


39. For all x, 4x(x + 3) 2x(5x 1) =

ANSWER

(A) 6x2 + 14x

(B) 6x2 14x

(C) 4x2 10x + 4

(D) 4x2 10x 4


39. 4x(x + 3) –2x(5x 1)

≡ 4x2 + 12x 10x2 + 2x

≡ 4x2 10x2 + 12x + 2x

≡ 6x2 + 14x

Use the distributive law twice toremove the two pairs of brackets.

Group like terms.

Add like terms.


40.

ANSWER

(A) 1

(B) 1

(C)

(D)

If 1 , then 10*25*p

p qq

251

10

251

10


40. 1*

1010 25 1*

2510

15

1 2

1

pp q

q

State the given formula.

Substitute the value for p and for q.

Use the meaning of a square root.

Dividing.

Subtracting.


41. If a = 2 and ab = 10, then (a + b)2 (a2 + b2) =

ANSWER

(A) 20

(B) 20

(C) 78

(D)


2 2 2

2 2 2

2

So

2 5 2 5

7 4 25

49 29

20

a b a b

41. If 2 and 10

10then

10

25

a ab

ba

Substitute 2 for a and 5 for b.


42.

ANSWER

(A)

(B)

(C) 3x

(D)

8 5

9 9x x

13

9x13

6x

1

3x


42.

The common denominator is 9x.

Simplify the values in the numerator by subtracting.

Divide both the numerator and denominator by their common factor 3.


43. The statement “When 2 is added to five times a number n, the result is 40.” May be represented by the equation

ANSWER

(A) 2(5n) = 40

(B) 2 5n = 40

(C) 5n + 2 = 40

(D) 5n 40 = 2


43. Five times a number n = 5n

2 added to five times a number n = 5n + 2

The equation is: 5n + 2 = 40


44. If x and y are numbers with x greater than y, then the statement. “The square of the difference of two numbers is always positive.” May be represented as

ANSWER

(A) (x y)2 > 0

(B) x2 y2 > 0

(C) 2(x y) > 0

(D) (x + y)2 > 0


44. The difference of the two numbers = x y

The square of the difference of the two numbers = (x y)2

The statement is:

(x y)2 > 0

A positive number is greater than zero.


45. Given that 3x + 8 29, then the range of values of x is

ANSWER

(A) x 7

(B) x > 7

(C)

(D)

37

3x

37

3x


45. Subtract 8 from both sides.

Subtracting.

Divide both sides by 3.

Dividing.


3 8 29

3 29 8

3 21

2

7

1

3

x

x

x

x

x

So

46.

ANSWER

(A) y is greater than x

(B) x is a factor of y

(C) x is less than y

(D) x is a multiple of y

The arrow diagram above describes the relation


46. 2 4 = 8

4 2 = 8

3 3 = 9

2 5 = 10

Hence, x is a multiple of y.


47. Which of the following relation diagrams illustrates a function?

ANSWER

(A) (B)

(C) (D)


47.

Each element in the domain is mapped onto one

and only one element in the range.

This relation diagram represents a function.


48. If f(x) = x2 + x 1, then f(3) =

ANSWER

(A) 5

(B) 5

(C) 7

(D) 13


2

2

( ) 1

( 3) ( 3) ( 3) 1

9 3 1

13

f x x x

f

48. Substitute 3 for x.

Simplify each term.

Subtracting.


49. Which of the following sets is represented by the relation f: x x2 3?

ANSWER

(A) {(0, 3), (1, 2), (2, 1), (3, 6)}

(B) {(0, 3), (1, 2), (2, 1), (3, 0)}

(C) {(0, 3), (1, 6), (2, 9), (3, 12)}

(D) {(0, 3), (1, 3), (2, 3), (3, 4)}


49. f(x) = x2 3

f(0) = 02 3 = 3 (0, 3)

f(1) = 12 3 = 1 3 = 2 (1, 2)

f(2) = 22 3 = 4 3 = 1 (2, 1)

f(3) = 32 3 = 9 3 = 6 (3, 6)

The set is {(0, –3), (1, 2), (2, 1), (3, 6)}


50.

ANSWER

(A) y = ax2 + bx

(B) y = bx ax2

(C) y = ax2 + bx + c

(D) y = c + bx ax2

If a, b and c are constants with a > 0, then theequation of the graph could be


a A maximum turning point

c 0 (y-intercept)

Equation is: y = c + bx ax2

50.


51. Which of the following diagrams is the graph of a function?

ANSWER

(A) (B)

(C) (D)


51.

Using the vertical line test for a function:x1 y1

x2 y2

The graph represents a 1 1 relation which is a function.


52.

ANSWER

(A) x = y

(B) x < y

(C) x + y = 180

(D) x + y > 180

In the figure above, AB and CD are parallel. Therelation between x and y is


52. x + y = 180 The interior angles are supplementary.


53. Which of the following plane shapes has no line of symmetry?

ANSWER

(A) (B)

(C) (D)


53. Each of these three plane figures has a line of symmetry.

This figure has no line of symmetry.


54.

ANSWER

(A) 6 8 (B) 6 10

(C) 8 10 (D) 6 16

The area of PQR, in cm2, is given by

1

2

1

2

1

2

1

2


54.

The area of PQR, A = bh

= 6 cm 8 cm

= 6 8 cm2

1

21

21

2


55.

ANSWER

(A) 28

(B) 56

(C) 102

(D) 124

In ABC, angle ABC = x and angle BAC = 28.What is the value of x?


55.

Δ ABC is isosceles since AB = CB.

Also angle BCA = angle BAC = 28°

So x + 28 + 28 = 180i.e. x + 56 = 180 x = 180 – 56 = 124 x = 124

Sum of the angles of a triangle.


56.

ANSWER

(A) 640 m(B) 160 m

(C)

(D)

In the diagram above, not drawn to scale, TB represents a hill which is 320 m high, and S is the position of a ship. The angle of elevation of S from T is 30°. The distance of the ship from the top of the hill is

160 3 m320

m2


56.

320 msin 30

1 320 m

22 320

6 0 m

m

4

ST

STST


57.

ANSWER

(A) DAB = 90(B) ADB = ACB

(C) CAB = ACB

(D) ACB + ABD = 90

In the diagram above, not drawn to scale, BODis a diameter of the circle centre O. Which of thefour statements below is false?


57. DAB = 90ADB = ACB

CAB + ADB = 90CAB = ACB

DAB = 90

ADB = ACB

Each of these three statements is true.

ADB + ABD = 90ACB + ABD = 90 since ADB = ACB

This statement is false.

The angle in a semicircle is 90º

Angles at the circumference standing on the same arc.


58.

ANSWER

(A) (B)

(C) (D)

In the triangle shown above, tan M is

3

5

4

3

3

4

4

5


58.

Opptan

A

4

3

dj

4 cm

3 cm

M

KL

ML

Definition of the tangent of an angle.

Using the capital letters notation.

Substitute the length of each side.


59. A ship sailed 75 km due east from A to B. It then sailed 50 km due south to C. Which of the diagrams below best represents the path of the ship?

ANSWER

(A) (B)

(C) (D)


59. This diagram bestdescribes the pathof the ship.


60.

ANSWER

(A) x = 0

(B) y = 0

(C) y = x

(D) x = –y

In the diagram shown, if the line y = –x is rotated about 0 through a clockwise angle of 90°, then its image is


60.

The image is the line y = x.



MATHEMATICS

Paper 2

2 hours 40 minutes

SECTION I



NEXT

1. (a) (i) Using a calculator, or otherwise, determine the exact value of

(2 marks)

ANSWER

5.8 0.39 0.5625


(ii) Express as a single fraction

(3 marks)

1 35

2 51

32

1. (a) (i) 5.8 0.39 0.5625

2.262 0.75

(exact value3 ).012


1. (a) (ii)1 35

2 51

32

94

101

32

57

5

21

Use the mixed number

Function, , to simplify

the numbers in the numerator. Use the mixed

number function, , to

divide the mixed number in the numerator by the mixed number in the denominator. (single fraction)

ba

c

ba

c


1. (b) In this question, use CAN $1.00 = GUY

$164.00.

(2 marks) ANSWER

(i) While vacationing in Canada, Robert used his credit card to buy a camcorderfor CAN $450.00.

How many Guyanese dollars is Robert owing on his credit card for this transaction?


1. (b) (i) CAN $1.00 = GUY $164.00CAN $450.00 = GUY $164.00 × 450

= GUY $73 800.00

Hence, Robert is owing GUY $73 800.00 on his credit card for the transaction.



ANSWER

(ii) Robert’s credit card balance is GUY $102 500.00. After buying the camcorder, how many canadian dollars does he have left on his credit card for spending?


1. (b) In this question, use CAN $1.00 = GUY

$164.00.

1. (b) (ii) The credit card balance after the transaction

GUY$ 102 500.00 73 800.00

GUY$ 28 700.00

Now GUY $164.00 = CAN $1.00

So GUY $1.00 = CAN

GUY $28 700.00 = CAN × 28 700.00

= CAN $175.00

Hence, Robert has CAN $175.00 left onhis credit card for spending.

1.00$

164.001.00

$164.00


2. (a) Find the value of each of the following algebraic expressions when a 3, b

1and c 2

(1 mark)

ANSWER

(i) a (b c)


2. (a) (i) ( ) 3( 1 2

3( 3)

9

)a b c

Substitute the values for a,b and c intothe algebraicexpression, thensimplify.


(2 marks)

ANSWER

(ii)25 3b ac

a b c


2. (a) Find the value of each of the following algebraic expressions when a 3, b

1and c 2

2. (a) (ii) 2

2

5 3

5( 1) 3( 3)2

3 ( 1) 2

5(1) 18

3 1 25 18

6

53

3

6

6

2

b ac

a b c

Substitute thevalues for a,b and c intothe algebraicexpression,then simplifyaccording tothe arithmeticrules


2. (b) Change the following statements into algebraic expressions:

(1 mark)

ANSWER

(i) Seven times the sum of x and 3.


2. (b) (i) The sum of and 3 3

Seven times the

sum of and 3 7 ( 3)7( 3)

x x

x xx


(2 marks)

ANSWER

(ii) Fifteen more than the product of p and q.


2. (b) Change the following statements into algebraic expressions:

2. (b) (ii) The product of p and q p × q pqFifteen more than theproduct of p and q pq+15


2. (c) Solve the equation3(2x + 1) 4x 1 (2 marks)

ANSWER


2. (c) 3(2 1) 4 1

6 3 4 1

6 4 3 1

2 4

4

22

x x

x x

x x

x

x

Use the distributive law Group like terms

Add like termsDivide both sides by 2

Simplify


So

2. (d) Factorise completely

(2 marks)

ANSWER

(i) 8a3b4 − 16a6b2


2. (d) (i) Factorise using8a3b2 as the HCF

6

3

3

2

4

3 2

2

8 ( 2

6

)

8 1

a b

a

b a

a b b


ANSWER

(ii) 3m2 + 11m − 4 (2 marks)Total 12 marks


2. (d) Factorise completely

2. (d) (ii)

Factorise pairwise.

Factorise using thecommon factor (3m – 1).

2

2

3 11 4

3 12 4

(3 1) 4(3

(3 1)( 4)

1)

m m

m m m

m m m

m m

11

3( 4) 12

1 12 11

( 1) 12 12

p q

pq


3. Students taking part in a community project were surveyed to find out the type of movies that they were most likely to view. Each student choose only one type of movie and 1 260 students were surveyed. The results are shown in the table below.

MovieNumber of Students

Horror 168

Detective 210

Romance r

War 182

Musical 462


3. (a) Calculate the value of r, the number of students who were most likely to view romance movies.

ANSWER

(2 marks)


3. (a) 168 210 182 462 1 260

1022 1 260

1 26

2 8

1

3

0 022

r

r

r

Hence, 238 students were most likely to view romance movies.


So

(i) The data collected in the table are to be represented on a pie chart. Calculate the size of the angle in each of the five sectors of the pie chart.

ANSWER(4 marks)



Horror 168

Detective 210

Romance r

War 182

Musical 462

3. (b)

3. (b) (i)

The sector angle

representing168

horror movies 3601 260

2168

7

The sector angle

representing2

detec

360

1

tive

260

6

212

48

movies 2107

7

60


The sector anglerepresenting romance movies

The sector angle representingromance movies

The sector angle representingmusical movies

238

768

2

282

752

1

2462

7132


(ii) Using a circle of radius 4.5 cm, construct a pie chart to represent the data.

ANSWER




Horror 168

Detective 210

Romance r

War 182

Musical 462

3. (b)

3. (b) (ii)

The constructed pie chart with radius 4.5 cm is shown above.


4. (a) A universal set, U, is defined asU = {25, 26, 27, 28, 29, 30, 31, 32, 33,

34, 35, 36, 37, 38}.Sets P and E are subsets of U such thatP = {Prime Numbers} and E = {EvenNumbers}.

(5 marks)

ANSWER

(i) Draw a Venn diagram to represent the sets P, E and U.


4. (a) (i) U = { 25, 26, 27, 28, 29, 30, 31, 32, 33,

34, 35, 36, 37, 38}, P = {29, 31, 37} and

E = {26, 28, 30, 32, 34, 36, 38}.

The Venn diagram representing the sets P, E and U is shown above.


(ii) List the elements of the set (1 mark)

ANSWER


4. (a) A universal set, U, is defined asU = {25, 26, 27, 28, 29, 30, 31, 32, 33,

34, 35, 36, 37, 38}.Sets P and E are subsets of U such thatP = {Prime Numbers} and E = {EvenNumbers}.

4. (a) (ii) The elements of the set = {25, 27, 33, 35}.


4. (b) (i) Using only a pair of compasses and a

pencil, construct parallelogram ABCD in

which AB = 5 cm, AD = 8 cm and the

angle BAD is 60º.

(5 marks)

ANSWER


4. (b) (i)

The constructed parallelogram ABCD

with AD = BC = 8 cm, AB = DC =

5 cm and BAD = 60°.


4. (b) (ii) Measure and state the length of the

diagonal AC.

ANSWER

(1 mark)Total 12 marks


4. (b) (ii) The length of the diagonal AC = 11.4 cm.


5. The diagram below, not drawn to scale, represents the floor plan of a house. The broken line PS, divides the floor plan into a semi-circle, A, and a rectangle, B.

Use as 22

.7


(a) Calculate the radius of the semi-circle PST. ANSWER(1 mark)

5. (a) The radius of the semi-cir

7m

cle2

27m

2

2

PSPST

QR

r


(b) Calculate the perimeter of the entire floor plan.ANSWER

(3 marks)



Use as 22

.7

5. (b) The length of the arc PTS 2

2

22 7m

7

11 m

2

r

r

The perimeter of the entire floor plan PQRST = (12 + 7 + 12 +11) m

= 42 m


(c) Evaluate the area of the entire floor plan.ANSWER

(4 marks)



Use as 22

.7

5. (c) The area of thesemi-circle PST

2

11

2

1 22

2

r

2

2

27 7m

7 2 2

77

119

4

4

m

m

The area of the rectangle PQRS = lb= 12 × 7 m2

= 84 m2


The area of the entire floor plan 2

21103 m

4

119 84 m

4PQRST


(d) Section B of the floor is to be covered with floor tiles measuring 1 m by 50 cm. How many floor tiles are needed to just completely cover Section B?

ANSWER




Use as 22

.7

5. (d) The area of a floor tile

2

1m 50cm

11m m

11 m

2

2

lb

The number of floor tiles needed to just

completely cover Section B 2

2

12 7 m1

1 m2

12

6

7 2

1 8


6. (a) In the diagram below, not drawn to scale, TB is a vertical lantern post standing on a horizontal plane. B, P and Q are points on the horizontal plane.

TB = 10 metres and the angles of depression from the top of the pole T to P and Q are 35º and 29º respectively.


(i) Copy the diagram and insert the angles of depression. (1 mark)

ANSWER

6. (a) (i)

The angles can be seen inserted in the diagram above.

TBP = TBQ = 90°Vertical post standing on horizontal ground.


(ii) Calculate to one decimal placea) the length of BPb) the length of PQ

ANSWER(5 marks)


6. (a) In the diagram below, not drawn to scale, TB is a vertical lantern post standing on a horizontal plane. B, P and Q are points on the horizontal plane.

TB = 10 metres and the angles of depression from the top of the pole T to P and Q are 35º and 29º respectively.

6. (a) (ii)

In the diagram above:

of depression =

of elevation

35

29

PTU BPT

QTU BQT


(a)


Considering ΔTBP:

tan35

10m

10m

tan3510m

0. 047

1 .3m0

TB

BP

BP

BP

(to one decimal place)


(b)


Considering ΔTBQ:

(to one decimal place)

tan 29

10m

10m

tan 2910m

0.55418.1m

TB

BQ

BQ

BP

The length of

(18.1 14

3 m

.3

8

)m

.

PQ BQ BP


6. (b)

ANSWER


(i) The figure labelled P undergoes a transformation, such that its image is Q. Completely describe this

transformation. (2 marks)

6. (b)


(i) The transformation is a translation represented by

the column vector .

3

8T

(ii) On graph paper, draw and labela) the line y = −xb) S, the image of P under a

reflection in the line y = −x.



6. (b)

(ii) a) The line y = −x can be seen drawn and labelled on graph paper.

b) S, the image of P under a reflection in the line y = −x can be seen drawn and labelled on graph paper.


6. (b)

(a) The equation of the line above is y = mx + c.ANSWER(1 mark) (i) State the value of c.


7. The diagram below shows the graph of a straight line passing through the points A and B.

7. (a)

(i) From the graph, theintercept on

the y-axis = 4.The value of

c = 4.


(ii) Determine the value of m. ANSWER(2 marks)


7. (a) The diagram below shows the graph of a straight line passing through the points A and B.

7. (a) (ii)

The gradient of theline segment AB

The vertical rise

The horizontal shift4

34

34

So the value of 3

m


(ii) From the graph, the slope of AB indicates a negative gradient.

The gradient of theline segment AB

4

3

So the value of 4

3

AO

BO

m

Or


(a) (iii) Determine the coordinates of the midpoint of the line

segment AB.

ANSWER(2 marks)


7. The diagram below shows the graph of a straight line passing through the points A and B.

7. (a) (iii) Let the mid-point of the line segmentAB be M (x, y).

The x-coordinate of M2

3

21

12

BO


The -coordinate of 2

2

4

2

AOy M

So the coordinates of the mid-point of

the line segment AB is .

11 ,2

2M


(iii) From the construction on the graph, thecoordinates of the mid-point of the line

Or

11se ,2

2gment is .MAB


(iii) Using A(0, 4) andB(3, 0), the midpointof the linesegment AB,

Or

1 2 1 2,2 2

0 3 4 0,

2 2

11 , 2

3 4,

2

2

2

x x y yM


7.

ANSWER(3 marks)

14 ,

2p


(b) The point lies on the line. State

the value of p.

7. (b) From the construction on the graph, when

, then y = p = −2.

So the value of p is –2.

14

2x


(b) The equation of AB is

When and y = p, then

Hence, the value of p is –2.

44

3y x

14

2x

4 14 4

3 2

4 94

3 2

2(3) 4

6

2

4

p

Or


7.

ANSWER



(c) Determine the coordinates of the point of intersection of the line y = x − 3 and the line shown previously.

7. (c) 44

33

:=

y x

y x

‚

‚


43 4

34

4 334

1 737

73

7

3

3

7

x x

x x

x

x

x

x

Group like terms


When x = 3, then

y = x − 3

= 3 − 3

= 0

Hence, the coordinates of the point of

intersection of the line y = x − 3 and the line

shown is (3, 0) 4

43

y x


(c) Given y = x − 3, then m = 1 and c = −3.

Using c = −3 and , the graph of the

line y = x − 3 was drawn on the same graph paper as shown above.

The graph of the lines and

y = x − 3 intersect at B (3, 0).

1

1m

44

3y x

Or


8. The first three diagrams in a sequence are shown below. Diagram 1 has a single dot, which can be considered as a triangular pattern formed by a single dot. Diagram 2 consists of a triangle formed by three dots.Diagram 3 consists of a triangle formed by six dots.


8. (a) Draw Diagram 4 in the sequence.

ANSWER

(2 marks)


8. (a)

Diagram 4Diagram 4 in the sequence can be seen above.


8. (b) Complete the table by inserting the appropriate values at the row 2 marked (i), (ii) and (iii). (6 marks)


ANSWER

Diagram Number

Number of Dots Forming

the triangle

Pattern for Calculating the Total Number of Dots in the Diagram

1 1 1 (1 + 1) ÷ 22 3 2 (2 + 1) ÷ 23 6 3 (3 + 1) ÷ 2

(i) 4 — —

(ii) — 21 —

(iii) n — —


Diagram Number

Number of Dots Forming the

triangle

Pattern for Calculating the Total Number of Dots

in the Diagram1 1 1 (1 + 1) ÷ 22 3 2 (2 + 1) ÷ 23 6 3 (3 + 1) ÷ 24 10 4 (4 + 1) ÷ 26 21 6 (6 + 1) ÷ 2n n (n + 1) ÷ 2

8. (b)

The completed table can be seen above.

2

2

n n


8. (c) How many dots will be needed to form the triangle in Diagram 100?

ANSWER



8. (c) The total number ofdots in the diagram 100(100 1) 2

100(101) 2

50(101)

5 050


(c) The total number ofdots in the diagram

Or

2100 100

210 000 100

210 1

25 050

00


SECTION II



9. (a) Simplify

ANSWER

(1 mark)(i) x3 × x4 ÷ x6


9. (a) (i) 3 4 6

3 4 6

7 6

7 6

1

x x x

x x

x x

x

x

x


ANSWER

(2 marks)(ii)5 7

2 2a b ab


9. (a) Simplify

9. (a) (ii)5 7

2 2

5 7 1

2 2 2

5 7 1 1

2 2 2 2

5 1 7 1

2 2 2 2

5 1 7 1

2 2 2 2

6 8

2 2

3 4

3 4

( )

a b ab

a b ab

a b a b

a a b b

a b

a b

a

a b

b


ANSWER

(1 mark)

9. (b) If f(x) = 4x − 1, find the value of

(i) f (3)


9. (b) (i) ( ) 4 1

(3) 4(3) 1

11

12 1

f x x

f


(ii) f –1(0)

ANSWER

(2 marks)



9. (b) (ii) Given

then

So

i.e.

1

1

( ) 4 1

4 1

4 1

1 4

1

41

41

( )4

0 1 (0)

41

4

f x x

y x

x y

x y

xy

xy

xf x

f

is the defining equation for f (x)

Interchanging x and y

Adding 1 to both sides

Dividing both sides by 4

is the defining equation for f–1(x)


(iii) f –1 f (3)

ANSWER

(2 marks)



9. (b) (iii) 1

1

1(3) 11 and ( )

411 1

S o (34

12

3

)

4

xf f x

f f


Or

1 1( ) 4 1 and ( )

4

xf x x f x

1

1

4 1 1So ( )

44

4

(3) 3

xf f x

x

x

f f


(i) Using a scale of 8 cm to represent 100 years on the horizontal axis and a scale of 4 cm to represent 100 kg on the vertical axis, construct a mass-time graph to show

how the solid decays in the 168 years interval.

ANSWER(4 marks)

Draw a smooth curve through all the plotted points.


9. (c) The mass, in kg, of strontium, a radioactivematerial, after a number of years is given inthe table below.

t (time in years)

0 28 56 84 112 140 168

m (mass in kg)

400 200 100 50 25 12.5 6.25

9. (c) (i)

The points were plotted on graph paper and a smooth curve drawn as shown above.


(ii) Use your graph to estimatea) the mass of the solid after 50

yearsb) the rate of decay of the solid at t

= 75 years.ANSWER



9. (c) The mass, in kg, of strontium, a radioactivematerial, after a number of years is given inthe table below.

t (time in years)

0 28 56 84 112 140 168

m (mass in kg)

400 200 100 50 25 12.5 6.25

9. (c) (ii) (a) From the construction on the graph:

The mass of the solid after 50 years

= 116 years


(b) Draw a tangent to the curve at t = 75 years.

Using two points on the tangent, (0, 180) and (112.5, 0), the gradient of the tangent

2 1

2 1

(0 180)kg

(112.5 0) years

180kg year

112.51.6 kg year

m m

t t

Hence, the rate of decay of the solid at t = 75 years is −1.6 kg/year.


10. (a) In the diagram below, not drawn to scale,

PQ is a tangent to the circle, centre O.

PS is parallel to OR and angle RPS = 32º.

ANSWER(2 marks)

Calculate, giving reasons for your answer, the size of

(i) angle PQR


10. (a)


(i) 32

32

180 (32 32 )

180 64

116

258

2116

ORP RPS

OPR ORP

POR

PORPQR

Alternate S. ΔOPR is isosceles sinceOP = OR = r (radius)

Sum of theangles of a Δ

at centre = 2. at circumference

Hence, the size of angle PQR is 58º.


ANSWER(2 marks)


10. (a) In the diagram below, not drawn to scale,

PQ is a tangent to the circle, centre O.

PS is parallel to OR and angle RPS = 32º.

Calculate, giving reasons for your answer, the size of

(ii) angle SPT

10. (a) (ii)

Hence, the size of angle SPT is 26º.

90

( )

90 (32 32 )

90 64

26

OPT

SPT OPT OPR RPS

between radius and tangentat point of tangency.


10. (b) In the diagram below, not drawn to scale, O is centre of the circle of radius 9.5 cm and AB is a chord of length 16.5 cm.

ANSWER(3 marks)

(i) Calculate the value of θ to the nearest degree.


10. (b) (i)

Considering Δ AOB and using the cosine rule:


(b) (i) Considering Δ AOB and using the cosine

rule:

Or

2 2 2

2 2 2

1

2 2 cos

16.5 2(9.5) 2(9.5) cos

272.25 180.5 180.5cos

So 180.5cos 180.5 272.25

180.5cos 91.75

91.75i.e. cos

180.50.5083

cos ( 0.5083)

121 (to the nearest

degree)

AB r r


10. (b) (i)

Δ AOB is isosceles since AO = BO = r (radius)


Considering AOB:

ˆsin

8.25 cm

ADAOB

AO

9.5 cm

1ˆSo sin 0.868 4

60.27

And 2

2(60.27

121 (to the nearest

)

1

deg

20.5

r

4

ee)

AOB

AOB AÔD

16.5cmSo 8.25cm

2AD BD


ANSWER(2 marks)

(ii) Calculate the area of triangle AOB.



10. (b) (ii)


2

1

2

2

The area of ,

1sin

21

sin21

9.5 9.5 sin cm2

0.5 90.25 0.8572 cm

(correct to 3 s.f.)38.7 cm

AOB A

ab C

AO BO

(ii)

Or

1

2

2 2

2

2

The area of , :

1sin

21

(9.5) sin cm2

45.125 0.8572 cm

(correct 38.7 to 3 s.f.)cm

AOB A

r


(ii) The semi-perimeter of ΔAOB,

Or

2(9.5 9.5 16.5)cm

235.5

cm2

17.75cm

a b cs


The area of ΔAOB, A1

2

2

2

( )( )( )

17.75(17.75 9.5)(17.75 9.5)

(17.75 16.5) cm

17.75(8.25)(8.25)(1

38.7c

.25) cm

(correct to 3 s )m .f.

s s a s b s c


ANSWER(3 marks) (iii)Hence, calculate the area of the

shaded region. [Use = 3.14]



10. (b) (iii)

2

2 2

295

3

.

60

2 c

1213.14 9.5 cm

360

(correct to 3 s.f.m )

r

The area of the minor sector AOB, A2


The area of the shaded region,

1

2

2

2(95.2 38.7)cm

56.5cm

A A A


ANSWER (iv) Calculate the length of the major arc AB.




10. (b) (iv)


The major sector angle, reflex

360 121

239

AOB

2360

2392 3.14 9.5 cm360

(correct to 3 s.39 ..6 cm f )

l r

The length of the major arc AB,


ANSWER(2 marks)

(a) Copy the diagram and complete it to show the points of P and M.


11.

In the diagram above, the position vectors of A and B relative to the origin are a and b respectively.

The point P is on OA such that OP = 3 PA.

The point M is on BA such that BM = MA.

11. (a)

The diagram was copied and completed as shown above. The points P and M are learly shown.


ANSWER

(1 mark)(b) OB is produced to N such that OB = 2 BN

(i) Show the position of N on your diagram.


11.

11. (b) (i)

The position of N is shown in the diagram.


ANSWER

(5 marks)

(ii) Express in terms of a and b the vectors.


(b) OB is produced to N such that OB = 2 BN

, and .AB PA PM��

11.

11. (b) (ii)

AB AO OB

OA OB

��

��

a b

b a


3

1

3

PA OP

PA OP

given

3

43

4

OP OA

��

a

1

3

4

1

1 3

3

4

PA OP

��

a

a


given

1

21

21

( )2

1( )

2

BM MA

BM MA BA

AB

��

��

b a

a b


1 1( )

4 21 1 1

4 2

1 1

2

4

1 1

4

2

2

PM PA AM

PA MA

��

��

a a b

a a b

a b

b a

Hence,1 1 1

4, an

2 4dAB PA PM

��b a a b a


ANSWER

(4 marks)11. (c) Use a vector method to prove that P, M

and N are collinear.


11. (c) 2

1

21

21

2

BN OB

BN OB

BN OB

��

b


1 1( )

2 21 1 1

2

2

2 21

21

21 1

22 4

MN M

PM

B BN

BM BN

MN

��

��

a b b

a b b

a b

b a

b a

Since , then the vectors are either parallel or coincident. Since the vectorshave a common point M, then P, M and N are collinear.

2MN PM��


ANSWER


11. (d) Calculate the length of AN if.

8 2 and

4 4

a b


11. (d)


1

23

22 83

4 42

32 82

3 44

2

OA AN ON

AN ON OA

OB BN OA

=

��

��

��

b b a

b a

Considering ΔNOA:


1

23

22 83

4 42

32 82

3 44

2

OA AN ON

AN ON OA

OB BN OA

=

��

��

��

b b a

b a


3 8

6 4

3 8

6 4

5

2


Or


Considering ΔMAN:

1 1( )

2 21 1 1

2 2 21 1 1

2 2 23

28 23

4 42

AN AM MN

��

b a b a

b a b a

a a b b

a b


328 2

4 34

2

8 3

4 6

8 3

4

2

6

5


2 2

(exact value)

5.39 (c

( 5) 2

orrect to 3 s.f

25

29

.)

4

AN

The length of

Hence, the length of AN is 5.39 units.



Paper 190 minutes


NEXT

1. The decimal fraction 0.85 written as a common fraction, in its simplest form, is

ANSWER

(A)

(B)

(C)

(D) 10

17

17

2020

1717

10


1000.85 0.85

10085

10085 5

100 517

20

1. Multiply the decimal fraction with two

decimal places by , which is 1, to make

the decimal fraction a common fraction.

100

100

Divide both the numerator and thedenominator by their common factor 5.

This is the common fraction written in its simplest form.


2. The number 75 836 written correct to 4 significant figures is

ANSWER

(A) 80 000

(B) 76 000

(C) 75 800

(D) 75 840


2. 75 836

= 75 840 (4 s.f.)

The digit after the 4th significant figure is 6, so we add 1 to the digit 3. 0 is needed as a place holder.


3. Given that 768 51.2 = 39 321.6, then 76.8 0.512 =

ANSWER

(A) 3 932.16

(B) 393.216

(C) 39.321 6

(D) 3.932 16


3.

4 dp

768 51.2 39 321.6

1dp 1dp

76.8 0.512

1dp 3 dp 4dp

39.3216


4.

ANSWER

(A) 0.018

(B) 0.18

(C) 1.8

(D) 18

If 144 225 180, then 1.44 2.25


4.

2 dp 2 dp 2 dp 2 dp

144 225 12 12 15 15

12 15

180

1.44 2.25 1.2 1.2 1.5 1.5

1.2 1.5

1.80

1.8


5. y is inversely proportional to the square root of 7 may be expressed as

ANSWER

(A)

(B)

(C) y 72

(D)

7y 1

7y

2

1

7y


5. y is inversely proportional to means 71

7y


6. One hundred thousand written as a power of 10 is

ANSWER

(A) 104

(B) 105

(C) 106

(D) 107


6. One hundred thousand = 100 000

= 105


7. By the distributive law, 74 13 + 74 12 =

ANSWER

(A) 86 87

(B) 74 25

(C) 86 + 87

(D) 74 + 25


7. 74 13 + 74 12

= 74 (13 + 12)

= 74 25

The common

factor is 74.

Adding.


8. The highest common factor of 12, 24 and 30 is

ANSWER

(A) 2

(B) 4

(C) 5

(D) 6


8.

The HCF = 2 3 = 6

2 is a common factor of thethree numbers.3 is a common factor of thethree numbers.

2 12, 24, 30

3 6, 12, 15

2, 4, 5


9. The lowest common multiple of 5, 8 and 20 is

ANSWER

(A) 1

(B) 10

(C) 20

(D) 40


9. 2 5, 8, 20

2 5, 4, 10

2 5, 2, 5

5 5, 1, 5

1, 1, 1

The LCM = 2 2 2 5 = 40


10. The next two terms in the sequence 7, 6, 8 . . . is

ANSWER

(A) 7, 9

(B) 7, 7

(C) 7, 8

(D) 7, 6


10. 7, 6, 8, 7, 9, . . . 1 + 2 –1 + 2


11. A man’s annual income is $60 000. His non-taxable allowances is $15 000. If he pays a tax of 25% on his taxable income, then the tax payable is

ANSWER

(A) $3 750

(B) $11 250

(C) $15 000

(D) $33 750


11.


The taxable income $(60 000 15 000)

$45 000

the tax payable 25% of $45 000

25

100

$45 000

$11 250

12. The basic rate of pay is $28.00 per hour. What is the overtime rate of pay if it is one-and-a-half times the basic rate?

ANSWER

(A) $32.00

(B) $35.00

(C) $36.00

(D) $42.00


12. 1The overtime rate 1 $28

23

$282

4

1

2

$

$

3 4


13. Alfred saved $74 when he bought a cell phone at a sale which gave a discount of 20% on the marked price. What was the marked price of the cellphone?

ANSWER

(A) $370

(B) $296

(C) $222

(D) $148


13. The discount of 20% $74

100the marked price $74

20(which is 100%) $74 5

$370


14. A store offers a discount of 10% off the marked price for cash. If the cash price of a calculator is $135, what is the marked price?

ANSWER

(A) $13.50

(B) $121.50

(C) $148.50

(D) $150.00


14. 90% of the marked price $135

100the marked price $135

90

(which is 100%) $15

$150

10


15. The charge per kWh of electricity used is 35 cents. There is also a fixed charge of $27.00. What amount is the electricity bill if 80 kWh of electricity is consumed?

ANSWER

(A) $55

(B) $62

(C) $142

(D) $307


15. The cost for the electricity = 35¢ 80

= 2 800¢

= $28.00

The fixed charge = $27.00

\ the amount of the bill = $(28.00 + 27.00)

= $55.00


16. The exchange rate for US $1.00 is GUY $200. What amount of Guyanese dollars will a tourist receive for changing US $75.00?

ANSWER

(A) $150

(B) $1 500

(C) $15 000

(D) $150 000


16. US $1.00 = GUY $200

US $75.00 = GUY $200 75

= GUY $15 000


17. Calculate the book value of a computer valued at $3 000, after two years, if it depreciates by 10% each year.

ANSWER

(A) $300

(B) $2 400

(C) $2 430

(D) $2 920


17. The book value after 1 year 90% of $3 000

90$3 000

100

$2 700

The book value after 2 years 90% of $2 700

90$2 700

100

$2 430


18. A man pays $540 as income tax. If income tax is charged at 20% of the taxable income, what was his taxable income?

ANSWER

(A) $1 800

(B) $2 160

(C) $2 700

(D) $3 100


18. 100The taxable income $540

20

(which is 100%) $270 1

$2 70

0

0


19. X = {a, p, e}. How many subsets has the set X?

ANSWER

(A) 3

(B) 6

(C) 8

(D) 10


19. { }, {a}, {p}, {e}

{a, p}, {a, e}, {p, e}

{a, p, e}

The number of subsets = 8.

or

The number of subsets, N = 2n X = {a, p, e}

= 23 n(X) = 3

= 8 n = 3


20. A school has 200 students. 108 students play both soccer and basketball, 52 students play soccer only, and 15 students play neither sport. How many students play basketball only?

ANSWER

(A) 25

(B) 40

(C) 50

(D) 77


20.

Hence, 25 students play basketball only.


108 52 15 200

So 175 200

200 175

25

x

x

x

21. All students in a class play chess or scrabble or both. 15% of the students play chess only, and 37% of the students play scrabble only. What percentage of students play both games?

ANSWER

(A) 22

(B) 48

(C) 52

(D) 78


21.

Hence, 48% of the students play both games.


15 37 100

So 52 100

100 5

8

2

4

x

x

x

22.

ANSWER

(A) X Y

(B) Y X

(C) X Y = { }

(D) X Y { }

The Venn diagram above is best represented by the statement


22. Sets X and Y have no common elements, so X Y = { }.


23. 5:30 p.m. may be represented as.

ANSWER

(A) 05:30 h

(B) 17:30 h

(C) 15:30 h

(D) 18:30 h


23. 5 : 30 p.m. = (12 + 5) : 30 h

= 17 : 30 h


24.

ANSWER

(A) 8 (B) 16

(C) 24 (D) 32

The diagram above shows a circle with centre O and diameter 8 cm. The area of the circle, in cm2, is


24. The diameter of the circle, 8 cm

8 cmt he radius of the circle,

24 cm

d

r

2

2

2

2

The area of the circle, A π

π(4 cm)

π(16 cm )

c6 m1 π

r

Formula for the area of a circle.Substitute r = 4 cm.

Squaring.


25.

ANSWER

(A) (B)

(C) (D)

In the diagram above, POQ is a minor sector of a circle with angle POQ = 60° and OQ = r cm.The area, in cm2, of the minor sector POQ is

1π

3r

21π

6r 21

π3

r

1π

6r


25.


26. Mark takes 35 minutes to drive to university which is 45 km away from his apartment. His speed in km per hour is

ANSWER

(A)

(B)

(C)

(D) 45 35

60

35 60

45

35 45

60

45 60

35


26.


27.

ANSWER

(A) (B)

(C) (D)

The diagram above, not drawn to scale, shows a cone of radius r cm and height r cm. The volume of the cone, in cm3, is

21π

3r

34π

3r 24

π3

r

31π

3r


27.


28. The length of the edge of a cube is 20 cm. Thevolume of the cube is

ANSWER

(A) 8 000 cm3

(B) 400 cm3

(C) 240 cm3

(D) 200 cm3


28. 3

3

3

38 00

The volume of the cube,

(20 cm)

20 20 20 cm

cm0

V l


29. The mass of one tonne of sugar in kilograms is

ANSWER

(A) 100

(B) 1 000

(C) 10 000

(D) 100 000


29. 1 tonne = 1 000 kg


30. Robert has 0.75 kg of sweets. He has bags which can each hold 15 g of sweets. How many bags of sweets can he fill?

ANSWER

(A) 0.5

(B) 5

(C) 50

(D) 500


30.


0.75kg 0.75×1000g

750gThe number of bags that can 750gbe filled with s

5

we

0

ets15g

31. A bowl contains 6 green marbles and 7 yellow marbles. A marble is picked at random from the bowl. The marble is found to be green and it is not replaced. What is the probability that the next ball picked at random from the bowl will be yellow?

ANSWER

(A) (B)

(C) (D) 7

13

1

2

7

12

6

13


The number of green marblesremaining in the bowl = 6 1 = 5

The number of yellow marblesin the bowl = 7

The total number of marblesremaining in the bowl = 5 + 7 = 12

P(second marble is yellow) =

31.

7

12


32. The mode of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is

ANSWER

(A) 4

(B) 5

(C) 6

(D) 7


32. The mode is 5, since it occurs the most number of times.


33. The median of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is

ANSWER

(A) 6

(B) 5.5

(C) 5

(D) 4


1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8

Two middle values

The median of the numbers,2

5 5 10

2 25Q

Q2 = 533.


34. The mean of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is

ANSWER

(A) 4

(B)

(C) 5

(D)

34

4

15

2


34. The sum of the numbers, x = 1 + 2 + 3 + 4 +4 + 5 + 5 + 5 + 6 + 7 + 7 + 8

= 57The total frequency of the number, n = 12

The mean of the numbers,

57

1

4

23

4

xx

n


35. The scores of 100 students who took part in a shooting competition at a May Fair is recorded in the table shown below.

ANSWER

(A) (B)

(C) (D)

Score 0 1 2 3 4 5 6 7 8 9 10

Frequency 2 4 5 7 10 31 20 12 5 3 1

The probability that a student chosen at randomfrom these students scored exactly 6 is

1

1003

101

53

50


35. The frequency of the score 6 20

The total frequency, 100

20P(score 6)

1

1

0

5

0

n


36. The mean of the five numbers 7, p, 5, 9 and 18 is 12. The number p is

ANSWER

(A) 15

(B) 17

(C) 19

(D) 21


The sum of the numbers, p = 7 + p + 5 + 9 + 18

= p + 39

The total frequency, f = 5

The mean of the numbers, 12x

36.


37. (7a) (+2b) =

ANSWER

(A) 14ab

(B) –14ab

(C)

(D) 2

7ab

2

7ab


37. (7a) (+2b)

= 7 a 2 b

= 7 2 a b

= 14ab

Expand each expression.

Group like values.

The product of a negative

sign times a positive sign

is a negative sign.


38. a(a 3b) + b(a 3b) =

ANSWER

(A) (a 3b) (a + b)

(B) (a + 3b) (a b)

(C) (a 3b) (a b)

(D) (a + 3b) (a + b)


38. a(a 3b) + b(a 3b)

= (a 3b)(a + b)

Factorize using (a 3b)

as a common factor.


39.

ANSWER

(A) 2.6

(B) 26

(C) 260

(D) 2 600

5If 13, then

100

xx


39.5

13100

1320

13

260

20

x

x

x

Divide both the numerator and the denominator of the fraction on the LHS by the common factor 5.

Multiply both sides by 20.

Multiply the numbers on the RHS.Multiplying.


40. Given that p q means (p q)2, the value of 5 3 is

ANSWER

(A) 2

(B) 8

(C) 2

(D) 8

1

2


p q = (p q)2

5 3 = (5 3)2

= (2)2

= (4)

= 2

Substitute 5 for p and 3 for q in the formula.

Simplify the numbers in the brackets by subtracting.

Square the number in the brackets.

Multiply.

40.1

2

1

2

1

2

1

2


41. The statement ‘9 is subtracted from four times a certain number and the result is 15’ is represented by the equation

ANSWER

(A) 4x 15 = 9

(B) 9x 4 = 15

(C) 4x 15 = 9

(D) 4x 9 = 15


Four times a certain number = 4x

9 subtracted from four timesa certain number = 4x 9

The equation is:4x 9 = 15

41.


42. If 40 5x = x 14, then x =

ANSWER

(A) 6

(B) 6

(C) 9

(D) 9


42. 40 5 14

5 14 40

6 54

5

9

4

6

x x

x x

x

x

-

Subtract x and 40 from each side.

Add like terms.


A negative sign divided by anegative sign is a positive sign.

or


So

Add 14 and 5x to each side.

Add like terms.

Divide both sides by 6.A positive sign divided by a positive sign is a positive sign.

40 5 14

40 14 5

54 6

5

9

9

4

6

x x

x x

x

x

x

x


So

43. 2x(4x + 5) 4x(3x + 2) =

ANSWER

(A) 4x2 2x

(B) 2x 4x2

(C) 20x2 + 18

(D) 20x 4x2


43.2 2

2 2

2

2

2 (4 5) 4 (3 2)

8 10 1

2 4

2 8

8 12 10 8

4 2

x x x x

x x x x

x x x x

x x

x x

Use the distributivelaw twice.

Group like terms.

Add like terms.

Rearrange terms.


44.

ANSWER

(A) 2k2

(B) 4k2

(C) 6k2

(D) 8k2

The figure above, consists of a triangle resting on a square of length 2k cm. The height of the triangle is 2k cm. The area, in cm2, of the figure is


44.

1

2

2 2

1The area of the triangle,

21

( 2 ) 2 cm2

2 cm

A bh

k k

k


45. The width of a rectangular glass block is w centimeters. Its height is four-fifths its width and its length is 5 times its height. The volume of the rectangular glass block, in centimeters, is

14

13w

316

5w

316

25w

14

3w(A)

(B)

(C)

(D)

ANSWER


45.


46.

ANSWER

(A) x is a multiple of y (B) x is a factor of y

(C) x is greater than y (D) x is divisible by y

The arrow diagram above represents the relation


46. 8 = 1 8 = 2 4 2 and 4 are factor of 8.10 = 1 10 = 2 5 2 and 5 are factor of 10.12 = 1 15 = 2 6 = 3 4 2, 3, 4 and 6 are factors of 12.Hence, the diagram represents the relation, ‘x is a factor of y’.


47.

ANSWER

(A) 3 x < 2

(B) 3 < x 2

(C) 3 < x < 2

(D) 3 x 2

The diagram above is the number line of the inequality


47.

(3 is included) (2 is not included)

3 x < 2


48.

ANSWER

(A) {(x, y) : 2 y 1} (B) {(x, y) : 2 < y < 1}

(C) {(x, y) : 2 < y 1} (D) {(x, y) : 2 < y 1}

The shaded area in the graph above can be represented by


48.

(–2 is not included)

(1 is included)


49.

(A)

(B)

(C) 5

(D) 5 ANSWER

4 1If ( ) , then ( 9)

7

xg x g

37

7-

37

7


49.


4 1If ( )

74( 9) 1

then ( 9)7

36 1

735

75

xg x

g

50. Which of the following sets is represented by the relation f : x → x3 + 1?

(A) {(0, 0), (1, 1), (2, 8), (3, 27)}

(B) {(0, 1), (1, 2), (2, 9), (3, 28)}

(C) {(0, 0), (1, 1), (2, 4), (3, 9)}

(D) {(0, 1), (1, 2), (2, 5), (3, 10)}

ANSWER


50.

3

3

3

3

3

( ) 1

(0) 0 1 0 1 1 (0,1)

(1) 1 1 1 1 2 (1, 2)

(2) 2 1 8 1

(0,1), (1, 2),

9 (2, 9)

(3) 3 1 27 1 28 (

(2, 9), (3, 2

3

8

, )

)

28

f x x

f

f

f

f


51. Which of the following diagrams is the graph of a function?

(A) (B)

(C) (D)

ANSWER


51. Using the vertical line test for a function, the graph below is the graph of a function.

x1 y1

x2 y2

1 1 relation


52.

ANSWER

(A) x + y = 180 (B) x + y < 180

(C) x = y (D) x > y

In the figure above AB and CD are parallel. The relation between x and y is


52. x = y because they are alternate angles.


53.

ANSWER

(A) (B) 50 sin

(C) 50 cos (D) 50 tan

The triangle ABC is right-angled at B. AC = 50 cm and angle ACB = degrees. An expression for the length of AB, in cm, is


53.


54.

ANSWER

(A) (B)

(C) (D)

In the right-angled triangle PQR, angle Q = 90, PR = 90 cm, PQ = 50 cm and RQ = y cm. cos PRQ =

9

5

50

y

90

y

5

9


54.


55.

ANSWER

(A) 10 (B) 12

(C) 14 (D) 16

How many triangles congruent to BCE are needed to completely cover the rectangle ABCD?


55.

The number of triangles needed = 14


56.

ANSWER

(A) 40 (B) 50

(C) 80 (D) 90

The diagram shows ΔABC with AD = BD = CD.The magnitude of angle ABC is


56.


57.

ANSWER

(A) an enlargement about the origin of scale factor 2

(B) a rotation through 180° about the origin

(C) an enlargement about the origin of scale factor 2

(D) a rotation through 180° about the point

The transformation that maps KLM onto PQR is


57.

22

1

PQ QR PR

KL LM KM

The image is on the other side of the centre of enlargement, which is the origin, so the scale factor is 2.

The transformation is an enlargement about the origin of scale factor 2.


58.

(A) (3, 2) (B) (2, 3)

(C) (3, 2) (D) (3, 2)

The point P(−2, 3) is rotated about the origin through an angle of 270° in an anticlockwise direction. The coordinates of the image of P is

ANSWER


58.

The image of P(2, 3) under the rotation is P1(3, 2)


59. PQR is an isosceles triangle with angle P = 70°. The possible values of angle Q are

ANSWER

(A) 50 or 70 (B) 40 or 70

(C) 40 or 55 or 70 (D) 50 or 60 or 70


59.

or

or


60. The image of the point P(5, 8) under the translation is

ANSWER

(A) (3, 5)

(B) (3, 5)

(C) (5, 3)

(D) (5, 3)

23


60.

2 5 2 5

3 83 8

3

5

T: 5, 8 3, 5

T P P

P



Paper 22 hours 40 minutes

SECTION IAnswer ALL the questions in this section


NEXT

1. (a) Using a calculator, or otherwise, calculate the exact value of

ANSWER

(4 marks)

1 23 1

5 33

25

giving your answer as a fraction in its lowest terms.


1. (a)


(i) Calculate his annual salary.

ANSWER

(1 mark)


1. (b) Mr. John has a wife and two children. His monthly salary during the year 2008 was $9 500.

Information on calculating income tax for 2008 is shown in the table below.

Allowances Tax RateWorker $12 000

25% of the taxable incomeSpouse $7 000Each child $3 500Taxable Income = Annual Salary Total Allowances

1. (b) (i) Mr. John’s annual salary = $9 500 12

= $114 000


(ii) Calculate Mr. John’s total allowances for 2008.

ANSWER

(2 marks)






1. (b) (ii) The worker allowance $12 000

The spouse allowance $7 000

The children allowances $3 500 2

$7 000

Mr. John's totalallowances for 2008 $(12 000 7 000 7 000)

$26 000


(iii) Calculate Mr. John’s Income tax for 2008.

ANSWER

(3 marks)






1. (b) (iii)


Mr. John’s taxable

income $(114 000 26 000)

$88 000

Mr. John’s income

tax for 2008 25% of $88 000

25

1

10014

$ 8822

$2

0

0

00

2 00

(iv) What percentage of Mr. John’s annual salary was paid in income tax? ANSWER







1. (b) (iv) The percentage of Mr. John’s annual salary that was paid in income tax

22 000

114 000

19.3%

100%

11 100%

57(correct to 3 s.f.)


2. (a) Simplify completely: (5a 1)2

ANSWER

(2 marks)


2. (a) (5a 1)2 = (5a 1)(5a 1)

= 5a(5a 1) 1(5a 1)

= 25a2 5a 5a + 1

= 25a2 10a + 1


2. (b) Make p the subject of the formula

ANSWER

(3 marks)8

.5

qp


2. (b)


2. (c) Factorize completely

ANSWER

(1 mark)(i) 8mn 6n2


(ii) 49p2 q2 (2 marks)

2. (c) (i) 8mn 6n2

= 2n(4m 3n)

Factorize using theHCF of the two terms.


(ii) 49p2 q2

= (7p)2 q2

= (7p + q)(7p q)

2. (d) Solve the following pair of simultaneous

equations:

ANSWER


5x 2y = 164x + 3y = 1


2. (d) 5x 2y = 16 4x + 3y = 1

3 and 2:15x 6y = 48 8x + 6y = 2

+ :15 8 48 2

23 46

4

2

6

23

x x

x

x


Substitute x = 2 in :

4(2) 3 1

8 3 1

3 1 8

3 9

9

33

y

y

y

y

y

Hence, x = 2, y = 3.


(i) List the elements of the set

ANSWER(2 marks)a) J Kb) J K.


3. (a) The figure below is a Venn diagram showing a Universal set, U, and two subsets, J and K. The numerals in the diagram represent elements of the sets.

3. (a) (i) a) J K = {10, 20, 30}

b) J K = {5, 15, 25, 35}


(ii) Determine the value of n (J K).ANSWER

(1 mark)



3. (a) (ii) J K = {5, 10, 15, 20, 25, 30, 35, 40, 50, 60, 70, 80}

n (J K) = 12


(iii) Describe in words

ANSWER(3 marks)a) the Universal set Ub) the set J



3. (a) (iii) a) The Universal set U = {5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}

The Universal set U is the set of multiples of 5 less than 101 (or no more than 100, or between 1 and 100 inclusive).

b) The set K = {10, 20, 30, 40, 50, 60, 70, 80}

The set K is the set of multiples of 10 less than 81(or no more than 80, or between 1 and 80 inclusive).


3. (b) (i)

ANSWER(5 marks)

Use a ruler and a protractor to draw accurately the quadrilateral PQRS shown below. PQ = 9 cm, QR = 5 cm, PS = 7.5 cm, angle PQR = 135° and angle QPS = 60°.


3. (b) (i)


(ii) Measure and state the length of RS.

ANSWER



3. (b)

3. (b) (ii) By measurement, the length of RS = 9.2 cm


(a) Calculate the area of triangle ABC. ANSWER(2 marks)


4. The diagram below, not drawn to scale, shows a triangular prism with identical right angled isosceles triangles at both ends. Angle ABC = 90° and AB = BC = 6 cm.

4. (a)

2

2

1The area of triangle ,

21

6 cm

18 cm

6 cm2

3 6 cm

ABC A bh


(b) Calculate the length of the edge CD.ANSWER

(3 marks)


The volume of the prism is 270 cm3.

4.

4. (b) The volume of the prism, V = 270 cm3

The volume of the prism, V = Al18

18 270So

270i.e. cm

1830

cm215 cm

l

l

l

CD

Hence, the length of the edge CD is 15 cm.


(c) Calculate, to one decimal place, the length of the edge AC.

ANSWER

(2 marks)


4.

4. (c)

Considering the right-angled ABC and using Pythagoras’ theorem:2 2 2

2 2

8

6 6

36 36

72

5 c

7

.

2 c

m

m

AC AB BC

AC

Hence, the length of the edge AC is 8.5 cm.


4. (d) State the number of faces, edges, and vertices of the prism by completing the table

below.

ANSWER


Face Edge Vertices


4. (d)

Face Edge Vertices

5 9 6



(a) Draw the line x = 1 ANSWER(1 mark)


5. Triangle JKL has coordinates J(2, 1), K(4, 3) and L(2, 6) respectively.

5. (a) The vertical line x = 1 was drawn on the graph paper.


5. (b)

ANSWER

(3 marks)Draw the image of triangle JKL after a reflectionin the line x = 1. Label the image JKL.


The image triangle JKL was drawn or the graph paper. The vertices of triangle JK L are J(0, 1), K(2, 3), and L(0, 6).

5. (b)


(c)

ANSWER(2 marks)

A second transformation maps triangle JKL onto triangle JKL which has vertices J(4, 2), K(8, 6) and L(4, 12).(i) Draw the triangle JKL.


5.

5. (c) (i) The triangle JKL was drawn on the graph paper.The vertices of triangle JKL are J(4, 2), K(8, 6), and L(4, 12) respectively.


ANSWER(4 marks)

(ii) Describe completely the transformation which maps triangle JKL onto triangle JKL.


(c) A second transformation maps triangle JKL onto triangle JKL which has vertices J(4, 2), K(8, 6) and L(4, 12).

5.

5. (c) (ii) 2 ( 12)

6 12 12

510

52

J L

JL

The object and the image are on opposite sides of the centre of enlargement, which is the origin, 0.


∴ the scale factor, k = 2.

So E(0, 2) : JKL J K L.

Hence, the transformation which maps triangle JKLonto triangle JKL is an enlargement with the origin as the centre of enlargement and scale factor 2.


5. (d) Calculate the value of

ANSWER


Area of triangle

Area of triangle

J K L

JKL


5. (d)


6. The table below gives the birth rate per 1 000 people in Trinidad and Tobago during a six-year period.

ANSWER(6 marks)

Year 2003 2004 2005 2006 2007 2008Birth Rate 12.74 12.81 12.72 12.90 13.07 13.22

(a) Using a scale of 10 cm to represent a birth rate of 1 per thousand on the vertical axis and 2 cm to represent 1 year on the horizontal axis, draw a line-graph to represent the birth rate in Trinidad and Tobago from 2003 to 2008.


6. Using the given scales and data the graph was drawn as shown below.

(a)


6. (b) In which year was there

ANSWER

(1 mark)(i) the lowest birth rate?


(ii) the highest birth rate? (1 mark)

6. (b) (i) The lowest birth rate = 12.72 per 1 000 people


(ii) The highest birth rate = 13.22 per 1 000 people

6. (c) During which period was there

ANSWER

(1 mark)(i) the greatest increase in the birth rate?


(ii) the smallest increase in the birth rate?

(1 mark)

6. (c) (i) The period when there was the greatest increase in the birth rate = 2005 to 2006


(ii) The period when there was the smallest increase in the birth rate = 2003 to 2004

6. (d) During which period was there a decrease in the birth rate?

ANSWER



6. (d) The period when there was a decrease in the birth rate = 2004 to 2005


7. (a) A line segment connects the points A(3, 10) and B(f, g). If the mid-point of AB is (1, 2), calculate the values of f and g.

ANSWER

(4 marks)


7. (a) Calculate the x-coordinate of B:

1 2

23

12

3 1(2) 2

2 53

x xx

f

f

f


Hence, f = 5 and g = 6.

21

210

22

6

10 2(2) 4

4 10

y yy

g

g

g


Calculate the y-coordinate of B:

7. (b) Two functions, f and g, are defined as follows:

ANSWER

(1 mark)

: 7 4

5:

1

f x x

g xx

Calculate

(i) f(0)


7. (b) (i) ( ) 7 4

(0) 7(0) 4

0

4

4

f x x

f


(ii) g(2)

ANSWER

(1 mark)



: 7 4

5:

1

f x x

g xx

Calculate

7. (b) (ii)5

( )1

5(2)

2

21

3

15

3

g xx

g


(iii) f1(x)

ANSWER

(2 marks)



: 7 4

5:

1

f x x

g xx

Calculate

7. (b) (iii)

1

( ) 7 4

7 4

7 4

4 7

4

74

7

( )4

7

f x x

y x

x y

x y

xy

xy

xf x

Defining equation for f(x).

Interchanging x and y.

Adding 4 to both sides.


Defining equation for f1(x)


(iv) the value of x if fg(x) = 1

ANSWER




: 7 4

5:

1

f x x

g xx

Calculate

7. (b) (iv) 5( ) 7 4

1

354

1( ) 1

354 1

1

fg xx

xfg x

x

Equating


351 4 5

135

157 1

7 1

6

6

x

x

x

x

x

x

Adding 4 to both sides

Dividing both sides by 5, and multiplying by x + 1.

Subtracting 1 from both sides.



The first three diagrams in a sequence are shownbelow. Diagram 1 has three coins, which can beconsidered as a square pattern with a missing coin.

8.

Diagram 2 consists of a square pattern with amissing coin formed by eight coins.Diagram 3 consists of a square pattern with amissing coin formed by fifteen coins.

Diagram 1 Diagram 2 Diagram 3


ANSWER

(2 marks)Draw Diagram 4 in the sequence.8. (a)


8. (a)

Diagram 4

Diagram 4 in the sequence can be seen above.


ANSWER

(6 marks)

Complete the table by inserting the appropriatevalues at the rows marked (i), (ii) and (iii).

8. (b)

Diagram Number

Number of Coins Forming the Square

Pattern for calculating the Total Number of Coins in the Diagram

1 3 22 12 8 32 13 15 42 14 — —

8 80 —

n — —


8. (b) Diagram Number

Number of Coins Forming the Square

Pattern for calculating the Total Number of Coins in the Diagram

1 3 22 12 8 32 13 15 42 14 24 52 18 80 92 1n n2 + 2n (n + 1)2 1



ANSWER


Hence, determine a formula consisting of asingle term in n for calculating the total number of coins, N, in a diagram.

8. (c)


8. (c) The total number of coins in a diagram,

N = n2 + 2n

N = n(n + 2)

Hence, the formula is N = n(n + 2).


SECTION II



ANSWER

(3 marks)

Copy and complete the table below for the function9. (a)

2

10for 2 7.y x

x

x 2 3 4 5 6 7y 1.1 0.6 0.3


9. (a) When x = 2, then

When x = 5, then

2

10 10 5

2 4 22.5y

2

10 10 2

5 25 50.4y

When x = 7, then

2

10 10(correct to 1 d. p.)

7 40.2

9y

x 2 3 4 5 6 7y 2.5 1.1 0.6 0.4 0.3 0.2



ANSWER(4 marks)

Using a scale of 2 cm to represent 1 uniton both axes, plot the points whose x andy values are given in the table above.

(i)


Copy and complete the table below for the function9. (b)

2

10for 2 7.y x

x

x 2 3 4 5 6 7y 1.1 0.6 0.3

9. (b) (i) Using the given scale the points were plotted as shown.


ANSWER

(1 mark)Draw a smooth curve through the points.(ii)


Copy and complete the table below for the function9. (b)

2

10for 2 7.y x

x

x 2 3 4 5 6 7y 1.1 0.6 0.3

9. (b) (ii) A smooth curve was drawn through the points as shown.


ANSWER

(1 mark)

Use your graph to estimate9. (c)

(i) the value of y when x = 3.5


9. (c) From the construction on the graph:

(i) when x = 3.5, the value of y = 0.8


ANSWER

(2 marks)the value of x when y = 4.09. (c) (ii)


9. (c) (ii) when y = 4.0, the value of x = 1.6


ANSWER


Draw a tangent to the curve at the point (3, 1.1) as accurately as possible. Hence, estimate the gradient of the curve at the point (3, 1.1).

9. (d)


The tangent to the curve at the point (3, 1.1) was drawn as shown:9. (d)Choose two points on the tangent, for example, A(0, 3.3) and B(4.4, 0).The gradient of the tangent AB,

Hence, the gradient of the curve at the point (3, 1.1) is 0.75.

1

2

2 11

2 1

2

0 3.3

4.4 00.

(0,3.3) ( , )

(4.43.3

4.4

7

,

5

0)

0.75 ( , )

A A xy y

mx

y

B

B x y

x

m


The diagram below, not drawn to scale, shows three points P, Q and R on a straight line. Q and R are both due East of P such that PQ = 5 km, and QR = 12 km.

10.


ANSWER

(1 mark)Copy and label the diagram.10. (a)


The diagram was copied and labelled as shown above.10. (a)


ANSWER

(4 marks)

On your diagram, show the point, S, such that– the bearing of S from R is 035°– the bearing of Q from S is 121°

10. (b)


The bearing of S from R is 035° and the bearing of Q from S is 121° are shown in the diagram.

10. (b)


ANSWER

(1 mark)

State the size of

(i) SRQ

10. (c)


(ii) RSQ(iii) SQP

(1 mark)(1 mark)

(i) SRQ = 90° 35° = 55°10. (c)


(ii) NSR = 180 35 = 145

RSQ = 360 (145 + 121)

= 360 266

= 94

(iii) SQP = 55° + 94° = 149°

ANSWER

(2 marks)

Calculate, correct to the nearest kilometre, the distance.

(i) SQ

10. (d)


10. (d) (i)

Considering RSQ and using the sine rule:12 km

sin 55 sin 9412 km sin 55

sin 9412 km 0.819

0.9989.8

0

k

1 km

m

SQ

SQ

Hence, the distance SQ is 10 km.

(correct to the nearest kilometre).


ANSWER

(2 marks)

Calculate, correct to the nearest kilometre, the distance.

(ii) SP

10. (d)


10. (d) (ii)

Considering PSQ and using the cosine rule:

Hence, the distance SP is 14 km.

(correct to the nearest kilometre).

205.026 km14.3 km14 km

SP

2 2 29.8 5 2 9.8 5 cos14996.04 25 98 ( 0.857)121.04 83.986205.026

SP


ANSWER


Calculate the bearing of S from Q.10. (e)


10. (e)

SQR = 180 149 = 31

NQS reflex = 270 + 31 = 301

Hence, the bearing of S from Q is 301.


The points O(0, 0), A(4, 2), B(5, 7) andC(3, r) are plotted on graph paper.

11.

ANSWER


(a)

(i) OA��

(1 mark)

Write the following position vectors in the

form :x

y

11.

The points O, A and B can be seen plotted on graph paper.


(a) (i)4

2OA

��

ANSWER

(1 mark)(ii) OB��



11.

(a) Write the following position vectors in the

form :x

y

(a) (ii)5

7OB

��



11.

ANSWER

(1 mark)(iii)OC��



11.

(a) Write the following position vectors in the

form :x

y

(a) (iii) 3O

rC

��



11.

ANSWER

(2 marks)

Write as a column vector, in the form (b) :x

y

(i) BA��



11.

11. (b) (i)


ANSWER

(2 marks)(ii) in terms of r. AC��


Write as a column vector, in the form (b) :x

y


11.

11. (b) (ii)


ANSWER

(3 marks)

11. (c) Calculate the values ofr for which 37

The point D is such that7

.17

OD

��

AC��


Squaring both sides:

11. (c)


Factorizing as the difference of two squares:


ANSWER


11. (d) Using a vector method, prove that the pointsA, B and D are collinear.


11. (d)

DB and BA have the same direction and a common point B, so A, B, and D are collinear.



Paper 190 minutes


NEXT

1. (−4)2 + (−3)2 =

ANSWER

(C) −25(A) −7 (D) 25(B) 7


1. Expand usingthe definition ofa square.

The product oftwo negativesigns is apositive sign.

2 2( 4) ( 3)

( 4) ( 4) ( 3) ( 3)

16 9

25


2. 25.3 ÷ 0.01 =

ANSWER

(C) 253(A) 0.253 (D) 2 530(B) 2.53


2.Multiply both thenumerator andthe denominatorby 100, to makethe number in thedenominator awhole number.

25.325.3 0.01

0.0125.3 100

0.01 100

2530

1 2 3 5 0


3. The number 8 705 written in standard form is

ANSWER

(C) 8.705 × 103

(A) 8.705 × 10−3

(D) 8.705 × 102

(B) 8.705 × 10−2


3. 8 705 = 8.705 × 1000 = 8.705 × 103


4. The number 0.057 04 written in standard form is

ANSWER

(C) 5.704 × 103

(A) 5.704 × 10−3

(D) 5.704 × 102

(B) 5.704 × 10−2


4.

25.704 10

The first number must be between 1 and 10.

10.057 04 5.704

100


5. In a school, the ratio of the number of boys to the number of girls is 3:5. If the school has 864 students, how many are girls?

ANSWER

(C) 324(A) 172 (D) 540(B) 288


5.


6. A class has 40 students. 75% of the class are girls. 40% of the girls have a calculator. How many girls in the class have a calculator?

ANSWER

(C) 10(A) 30 (D) 4(B) 12


6.


7. Using the distributive law

ANSWER

(C) 70 × 20

(A) 39 × 41

(D) 35 × 10

(B) 35 × 20

35 16 4 35× + × =


7. Using thedistributivelaw.

Adding thenumbers inthe brackets.

35 16 4 35 35 (16 4)

35 20

× + × = × +

= ×


8. 8 05110 =

ANSWER

(C) 8 × 102 + 5 × 10 + 1

(A) 8 × 102 + 5 × 1

(D) 8 × 103 + 5 × 10 + 1

(B) 8 × 103 + 5 × 1


8. 8 05110 = 8 × 103 + 0 × 102 + 5 × 101 + 1 × 100

= 8 × 103 + 5 × 10 + 1 × 1

= 8 × 103 + 5 × 10 + 1

Expandingas a baseten number.


9. The next term in the sequence 1, 5, 25, 125, . . . is

ANSWER

(C) 500(A) 250 (D) 625(B) 450


9.


0 1 2 3

4

1, 5, 25, 125, . . . 5 , 5 , 5 , 5 , . . .

The next term 5

62 5

=

=

=

10. A stick is used to measure cloth of lengths 2.8 m, 4.2 m and 5.6 m. If the rod fitted each length of cloth an exact number of times, what is its greatest length?

ANSWER

(C) 2.1 m(A) 0.7 m (D) 2.8 m(B) 1.4 m


10. 2 28, 42, 56 7 14, 21, 28 2, 3, 4

The HCF of 28, 42 and 56 = 2 × 7 = 14

So the HCF of 2.8, 4.2 and 5.6 = = 1.4

Hence, the greatest length of the rod is 1.4 m.

14

10


11. The rateable value of a house is $1 850. If the rates charged for that area are 25¢ in the $1, then the amount payable per annum for rates is

ANSWER

(C) $1 825

(A) $462.50

(D) $1 875

(B) $925.00


11. The rates 25 ¢ in the $1

25%

0.25

=

=

=

The annual amount payable = 0.25 × $1 850= $462.50


12. The basic hourly rate of pay is $9.00. Overtime is paid for at double the basic rate. How much will Rita receive for overtime if she worked 20 hours overtime?

ANSWER

(C) $180.00

(A) $58.00

(D) $360.00

(B) $90.00


12.

$18.00

The overtime rate of pay $9.00 2

the overtime pay $18.00 20

$360.00


13. If $9 000 is borrowed for 3 years at the rate of 4% per annum, the simple interest is

ANSWER

(C) $750.00

(A) $120.00

(D) $1 080.00

(B) $167.50


13.


14. A woman’s taxable income is stated as $20 500. She pays tax at the marginal rate of 25%. The amount of income tax payable is

ANSWER

(C) $1 640

(A) $5 125

(D) $820

(B) $2 050


14. The amount of income tax payable = 25% of $20 500


1 2

25= $20 500

1001

$20 5004$5125

15. The cash price of a cellphone is $2 500. On a hire-purchase plan, a deposit of $250 is required, followed by 18 monthly payments of $137.50. How much is saved by paying cash?

ANSWER

(C) $200(A) $250 (D) $175(B) $225


15. The deposit = $250The amount of the monthlypayments = $137.50 × 18

= $2 475The hire-purchase price = $(250 + 2 475)

= $2 725The cash price = $2 500The difference between thehire-purchase price andthe cash price = $(2 725 − 2 500)

= $225

Hence, the amount saved by paying cash is $225.


16. After a discount of 40% is given on a stove, Angela saved $720. What was the marked price of the stove?

ANSWER

(C) $960(A) $2 160 (D) $760(B) $1 800


16.


17. A house costing $250 000 can be bought by making a 5% deposit and taking a mortgage loan for the remaining amount. What is the amount of the deposit?

ANSWER

(C) $12 500

(A) $2 500

(D) $25 000

(B) $5 000


17.


18. Orange juice is sold in packets of 250 ml, 500 ml and 750 ml. The size of the smallest container that can be completely filled by a whole number of packets of juice of either size is

ANSWER

(C) 1 500(A) 150 (D) 2 000(B) 750


18. 2 25, 50, 75

3 25, 25, 75

5 25, 25, 25

5 5, 5, 5

1, 1, 1

Hence, the smallest container that can be completely filled by a whole number of packets of juice of either size is 1 500 ml.

The LCM of 25,50 and 75 = 2 × 3 × 5 × 5 = 150So the LCM of 250,500 and 750 = 150 × 10 = 1 500


19.

ANSWER

In the Venn diagram above, the shaded regionrepresents

(C) Q P′

(A) Q′

(D) P Q′

(B) P Q′


19. The shaded region represents P only = P and not Q

= P Q′


20. If A = {1, 2, 3, 4, 5, 6}, then the number ofsubsets of A is

ANSWER

(C) 32(A) 5 (D) 64(B) 25


20. The number ofsubsets of A, N = 2n n(A) = 6

= 26 n = 6= 2 × 2 × 2 × 2 × 2 × 2= 64


21.

ANSWER

The two circles above represent P = {factors of 8} and Q = {factors of 12}. The shaded region represents

(C) {1, 2, 4, 8}

(A) {1, 2, 3, 4, 6, 8, 12}

(D) {1, 2, 3, 4, 6, 12}

(B) {3, 6, 8, 12}


21. 1, 2, 4, 8

1, 2, 3, 4, 6, 12

P

Q

Factors of 8 1 8

2 4

Factors of 12 1 12

2 6

3 4

P Q = P or Q These are the = {1, 2, 3, 4, 6, 8, 12} elements in the

shaded region.


22. If U = {2, 3, 5, 7, 11, 15, 17, 19}, A = {2, 3, 5, 7, 11} and B = {3, 7, 11, 15}, then (A B)′ =

ANSWER

(C) {3, 7, 11, 15}

(A) {2}

(D) {2, 3, 17, 19}

(B) {17, 19}


22.

15,17,19

2, 5,17,19

17,19

or

2, 3, 5, 7,11

3, 7,11,15

2, 3, 5, 7,11,15

2, 3, 5, 7,11,15,17,19

17,19

A

B

A B A B

A

B

A B

U

A B

Ç È

È

È


or

The shaded region in the Venn diagram represents (A B)′.

(A B)′ = {17, 19}


23. The distance around the boundary of a circular lake is 132 m. The radius of the lake is

ANSWER

(C)

(A) 66π

(D) 264π

(B) 132π m

66m

π


23.


The Circumference of the lake,

132 m

2π

2π 132

132

F

2π66

π

Hence, the radius of the lake

ormula for the circumference

of a circle.

66m .is

π

C

C r

r

r

24.

ANSWER

The diagram shows the sector of a circle with centre O, radius 6 cm and angle AOB = 150°. The length of the minor arc AB, in cm, is

(C) 5π(A) (D) 10π(B) 2

π5

5π

2


24.


25.

ANSWER

The area of the trapezium shown above is

(C) 44 cm2

(A) 31.5 cm2

(D) 45 cm2

(B) 34 cm2


25.


26. The volume of a cuboid with edges of lengths 10 cm, 100 cm and 1 000 cm is

ANSWER

(C) 1 110 cm3

(A) 1 000 000 cm3

(D) 60 cm3

(B) 2 000 cm3


26.

The volume of the cuboid, V = lbh Formula= 1 000 × 100 × 10 cm3 = 1 000 000

cm3


27. The length of a rectangle is tripled. By what value must the width of the rectangle be multiplied for its area to remain the same.

ANSWER

(C) (A) 3 (D) (B) 1

2

1

3

1

6


27.


28. The area of a triangle is 31.5 cm2. If the base of the triangle is multiplied by four and the altitude is halved, then the area would be

ANSWER

(C) 63 cm2

(A) 252 cm2

(D) 15.75 cm2

(B) 126 cm2


28.


29.

ANSWER

The diagram shows a cylinder with radius 4 cm and height 25 cm. The volume of the cylinder is

(C) 300π cm3

(A) 100π cm3

(D) 400π cm3

(B) 200π cm3


29. The volume of thecylinder, V = π2h Formula

= π(4)2 × 25 cm3 r = 4 cm and

= π(16) × 25 cm3 h = 25 cm

= 400π cm3


30. If a sailing boat travels a distance of 2 040 km in24 hours, what was its average speed?

ANSWER

(C) 255 km/h

(A) 85 km/h

(D) 340 km/h

(B) 170 km/h


30.


31.

ANSWER

The pie-chart shows the preference in fruits of the students in a school. If 78 students prefer melon, then the total number of students in the school is

(C) 540(A) 312 (D) 624(B) 468


31. 45° represents 78 students

360° represent 78 students ×

= 78 students × 8

= 624 students

Hence, the total number of students in the school

is 624 students.

360

45


32. If the mode of the numbers 2, 3, 4, 5, 5, 6, 6, x, 7, 7, 8 is 6, then x =

ANSWER

(C) 7(A) 5 (D) 8(B) 6


32. The mode is the number that occurs the most,so x = 6.The number 6 occurs three times.


33. An urn contains 5 blue balls and 6 green balls. A ball is picked at random from the urn and it is found to be blue. It is not replaced. What is the probability that the next ball to be taken randomly from the urn will also be blue?

ANSWER

(C) (A) (D) (B) 4

11

2

5

3

5

6

11


33.

P (second ball is blue) 24

10 5

B G Total 5 6 11 4 6 10


34. The ages of five students are: 15, 14, q, 16, 12 (years)If the mean age is 15 years, then q is

ANSWER

(C) 17(A) 19 (D) 16(B) 18


34.


35. If the lower quartile of the distribution of the masses of a class of students is 28 kg and the upper quartile is 65 kg, then the semi-interquartile range is

ANSWER

(C) 55.5 kg

(A) 18.5 kg

(D) 74 kg

(B) 37 kg


35.


36. In a survey to determine the number of children per household, the following table was obtained.

ANSWER

If a house is visited at random, the probability that it contains exactly 2 children is

(C) (A) (D)(B)

Number of children

0 1 2 3 4 5

Frequency 8 7 6 4 2 3

2

15

1

5

1

3

2

3


36.6

(exactly 2 children)8 7 6 4 2 36

5

301

P


37. 4x − 3(x + 6) =

ANSWER

(C) 7x − 18

(A) x − 18

(D) −7x + 18

(B) x + 18


37.

1

4 3( 6 3 1

8

) 4 8x x x x

x



38. 3(x + y) − 4(x − y) =

ANSWER

(C) −7x + 7y

(A) − x

(D) −7x − 7y

(B) − x + 7y


38. Use the distributive law twice.

3( ) 4( )

3 3 4 4

3 4 3

7

4

x y x y

x y x

x y

y

x x y y

Group like terms.Add like terms.


39. For all a and b,

ANSWER

(C)

(A)

(D)

(B)

( ) ( )a a b b a b 2( )ab a b 2 2a b

2 2a b 2 22a ab b


39. Factorise using (a − b) as a common factor.

This is the form of thedifference of two squares.2 2

2 2

2 2

( ) ( )

( )( )

or

( ) ( )

a a b b a b

a b a b

a b

a a b b a b

a ab ab b

a b

Use the distributive lawtwice.

Add like terms.


40. 5x × 5y =

ANSWER

(C) 25xy(A) 25x+y (D) 5xy(B) 5x+y


40. 5x × 5y = 5x+y The bases are the same, so weadd the powers.


41.

ANSWER

(C)

(A)

(D)

(B)

2 3

5 4

x x

y y

8 15

20

xy xy

xy

8 15

20

x y

xy

223

20

x

y

23

20

x

y


41. 2 3

5 4

2 (4) 3 (5) 20 204 5

20 5 4

8 15

20

23

20

x x

y y

x x y y

y y y

x x

y

x

y

The LCM of the denominators 5yand 4y is 20y.

Add the like terms in thenumerator.


42.

ANSWER

(C) −20(A) −10 (D) 10(B)

If 5 , then 30 2k

k ll

112

2


42. * 5

3030*2 5

230,

5

2

15

10

k

k

k

l

l

l


43.

(C) (A) (D) (B)

If 24 16 8(5 ), thenx x x 7

4

7

2

3

4

4

3

ANSWER


43. Use the distributive law.

Group like terms.

Add like terms.


Reduce the fraction to lowest terms.

24 16 8(5 )

24 16 40 8

24 8 40 16

32 24

24

323

4

x x

x x

x x

x

x


So

Use the distributive law.


Group like terms.

Add like terms.


or

24 16 8(5 )

8(3 2) 8(5 )

3 2 5

3 5 2

3

4

4

3

x x

x x

x x

x x

x

x


So

44. If x is an integer which satisfies the inequalities5 < x − 3 < 7, then the value of x is

ANSWER

(C) 8(A) 5 (D) 9(B) 7


44. Add 3 throughout the

inequality.

Add the numbers.

The only possible integeris 9.

5 3 7

5 3 7 3

8

9

10

x

x

x

x


45. Three times the square of the sum of two numbers, x and y, is 21. Which equation below describes the given statement?

ANSWER

(C)

(A)

(D)

(B)23( ) 21x y 2(3 3 ) 21x y

2[3( )] 21x y 2( ) 3(21)x y


45. The sum of two numbers, x and y = x + y

The square of the sum of two

numbers, x and y = (x + y)2

Three times the square of the

sum of two numbers, x and y = 3(x + y)2

The equation is:

3(x + y)2 = 21


46.

ANSWER

The graph of the inequality shown is defined by

(C) −3 < x ≤ 4

(A) −3 < x < 4

(D) −3 ≤ x < 4

(B) −3 ≤ x ≤ 4


46.

All values from −3 to 4 are included.


47.

ANSWER

In the graph above, the shaded region is represented by

(C) {(x, y) : 2 ≤ y ≤ 4}

(A) {(x, y) : 2 ≤ x ≤ 4}

(D) {(x, y) : 2 < y < 4}

(B) {(x, y) : 2 < x < 4}


47.

{(x, y) : 2 ≤ y ≤ 4} represents the shaded region.


48. If the real value of x which

cannot be in the domain of x is

ANSWER

(C) (A) −4 (D) 3 (B)

4 1( ) ,

3 1

xh x

x

1

4 1

3


48.4 1

If ( ) , th

1

en3 1

3 1 0

1

3

3

xh x

xx

x

x

since division by 0 is undefined.


So

49.

ANSWER

(C)

(A) −19

(D)

(B)

5 4If ( ) , then ( 9)

3

xf x f

116

3

116

3

213

3


49. 5 4( )

35( 9) 4

( 9)3

45 4

349

31

163

xf x

f


50. Which of the following sets is represented by therelation g: x x2 − 1?

ANSWER

(C)

(A)

(D)

(B)

{(0, 1), (1, 0), (2, 3), (3, 8)}{(0, 1), (1, 0), (2, 5), (3, 10)}{(0, 1), (1, 0), (2, 4), (3, 9)}{(0, 1), (1, 0), (2, 3), (3, 8)}


50. 2

2

2

2

2

( ) 1

(0) 0 1 0 1 1 (0, 1)

(1) 1 1 1 1 0 (1, 0)

(2) 2 1 4 1 3 (2, 3)

(3) 3 1 9 1 8 (3, 8)

g x x

g

g

g

g

The required set = {(0, 1), (1,0), (2,3), (3,8)}


51.

ANSWER

The relation diagram shown represents a

(C) many–1relation

(A) 1–1 relation

(D) many-to-many relation

(B) 1–many relation


51.

The relation diagram represents a 1– many relation


52.

ANSWER

In the figure above, AB and CD are parallel. The relation between x and y is

(C) x + y < 180

(A) x + y = 180

(D) x > y

(B) x = y


52.

x + y = 180


53.

ANSWER

The triangle ABC is right-angled at C. Angle CAB = 25° and AC = 18 m. If BC represents the height of a tower, then its height, in m, is

(C)

(A) 18 cos 25°

(D) 18 tan 25°

(B) 18 sin 25°

18

cos25


53.


54.

ANSWER

In the right-angled triangle above, not drawn to scale, angle R = 90°, PQ = 100 cm, PR = 90 cm and QR = h cm. tan

(C) (A) (D) (B) 9

10

10

9

90

h

h

90


54.


55.

ANSWER

In the circle ABCD above, O is the centre. Angle CAD = 37° and angle CDE = 105°. Angle ABD =

(C) 71° (A) 37° (D) 105° (B) 68°


55.


So

And

56. Which of the following shapes does not have a line of symmetry?

ANSWER

(C)

(A)

(D)

(B)


56.

These shapes each have a line of symmetry


57.


In the diagram above, the image of B was obtained by a

ANSWER

(C) translation parallel to the y-axis

(A) reflection in the x-axis

(D) translation parallel to the x-axis

(B) reflection in the y-axis


57.

The image of B is:

• congruent to its object• laterally inverted• the same distance away from the y-axis.

Hence, the image of B was obtained by a reflectionin the y-axis.


58. The image of the point P(5, −8) under the

translation is

ANSWER

(C) (−3, 4)

(A) (7, −12)

(D) (3, −4)

(B) (−7, 12)

2

4


58.


59. P1 (−7, 10) is the image of P(x, y) after a reflection in the line y = x. P (x, y) is

ANSWER

(C) P(−10, −7)

(A) P(−10, 7)

(D) P(10, 7)

(B) P(10, −7)


59. 1

1

1

: ( , ) ( , )

: ( , ) ( , )

: ( 7, 1

Reflection

Inverse reflection

(10, 7)0)

y x

y x

y x

M P x y P y x

M P y x P x y

M P P

- -


60.

ANSWER

In the diagram above, not drawn to scale, angle CAB = 29° and AOC is a diameter of the circle with centre O. Angle ACB =

(C) 45° (A) 61° (D) 29° (B) 58°


60.

in a semi-circle.ˆ 90

ˆ ˆSo 9

6

90 9

1

0

2

ABC

ACB CAB



MATHEMATICS

Paper 2

2 hours 40 minutes

SECTION I



NEXT

1. (a) Using a calculator, or otherwise, determine the exact value of (4.7)2 − (7.65 ÷ 1.7). (3 marks)

ANSWER


1. (a) 2(4.7) (7.65 1.7)

22.09

17.59 (exa

4

ct value

.5

)


1. (b) A total of 1 600 students attend Central High School. The ratio of teachers to students

is 1:40

(2 marks)

ANSWER

(i) How many teachers are there at the school?

Three-eights of the students own a laptop

computer.


1. (b) (i) The number of teachers at the school


(2 marks)

ANSWER


is 1:40


(ii) How many students do not own a laptop

computer?

Thirty-six percent of the students who own a laptop computer also own a video game.

1. (b) (ii)


The number of students who do not own

a laptop computer

Hence, 1 000 students do not own a laptop computer.

Or


1. (b) (ii) The fraction of students who do not

own a laptop computer

The number of students who do not own a laptop computer = 1 600 − 600

= 1 000



ANSWER


is 1:40


(iii) What fraction of the students in theschool own a video game?

Express your answer in its lowest terms.

1. (b) (iii) The number of students who own a

video game = 36% of 600

Hence, 216 students own a video game.


2. (a) Given that evaluate

(4 marks)

ANSWER

(i) 2 * 8


(ii) 4 * (2 * 8)

2. (a) (i)


(ii)


2. (b) Simplify, expressing your answer in its

simplest form(2 marks)

ANSWER


2. (b)


2. (c) A theatre has two sections, balcony (B) andhouse (H)

Tickets for section B cost $b each.Tickets for section H cost $h each.

Paula paid $515 for 4 section B tickets and3 section H tickets.

Zuri paid $520 for 5 section B tickets and1 section H ticket.



(i) Write two equations in b and h to represent the information above.

ANSWER(ii) Calculate the values of b and h.


2. (c) (i) The equation for the tickets bought byPaula:4b + 3h = 515 (in dollars)

The equation for the tickets bought byZuri:5b + h = 520 (in dollars)

(ii)

Hence, b = 95 and h = 45.



3. (a) The Venn diagram below represents information on the type of games played by members of a youth club. All members of the youth club play at least one game.

(5 marks)

F represents the set of members who play football.

C represents the set of members who play cricket.

S represents the set of members who play squash.

Ada, Obi and Roy are three members of the youth club.

(i) State what game(s) is/are played bya) Adab) Obic) Roy

ANSWER(ii) Describe in words the members of the setF′ S.


3. (a) (i) a) Ada played all three games. or

Ada played football, cricket and squash. b) Obi played cricket and squash. c) Roy played football.

or Roy played only football.

(ii) F′ S represents the set of members who play squash (and cricket) but

not football.

or

S C means that all members who play squash also play cricket.

So F′ S represents the set of members who play squash and cricket but not football.

3. (b) (i) Use only a pencil, a ruler and a pair of

compasses to perform the followingconstruction.a) Construct a triangle PQR in

which QR = 9.5 cm, QP = 8 cm and PQR = 60°.


b) Construct a line PT such that PT is perpendicular to QR and meets QR at T.


(ii) a) Measure and state the size of angle PRQ.

b) Measure and state the length of PT.

ANSWER

3. (b) (i)


a) Draw a horizontal line greater than 9.5 cm. Mark off a point Q. Set your compasses to a separation of 9.5 cm. With Q as centre mark off the point R to the right of the horizontal line. QR = 9.5 cm.

With Q as centre and a suitable separation, drawn an arc to the right of Q, to intersect the horizontal line at A. With A as centre and the same separation, drawn an arc to intersect the previous arc at B. Draw a straight line passing through the points Q and B. Angle PQR = 60°.

With a compasses separation of 8 cm and centre Q, mark off point P on QB produced. QP = 8 cm.

Draw a straight line from P to R to complete the triangle PQR.


b) With P as centre and a suitable compasses separation, draw an arc to intersect QR at C and D. With C and D as centres, construct two arcs to intersect at E. Draw a straight line through the points P and E to intersect QR at T. PT is perpendicular to QR.

(ii) a) By measurement, the size of angle PRQ = 52°

b) By measurement, the length of PT = 7 cm.


4. (a) The diagram below shows a map of a recreation park drawn on a grid of 1 cm squares. The scale of the map is 1 : 5 000.

(6 marks)


Use the map of the recreation park, find

(i) the distance, to the nearest metre, from St. Maria to St. Albert on the ground.

(ii) the distance, to the nearest metre, from St. Albert to St. Raymond on the ground.

(iii) the area on the ground represented by 1 cm2 on the map.

(iv) the approximate area of the recreation park on the ground, stating the answer in square metres.

ANSWER


4. (a) The diagram below shows a map of a recreation park drawn on a grid of 1

cm squares. The scale of the map is 1 : 5 000.

(6 marks)

4. (a)


The scale of 1 : 5 000 means1 cm represents 5 000 cm

So 1 cm represents 50 m

(i) The distance from St. Maria to St. Albert on the map = 5 cm

The distance from St. Maria to St. Albert on the ground = 5 × 50 m = 250 m


(ii) The distance from St. Albert to St. Raymond on the map = 8.6 cm

The distance from St. Albert to St. Raymond on the ground = 8.6 × 50 m

= 430 m


(iii) The area on the ground represented by 1 cm2 on the map = (50 m)2 = 2 500 m2


(iv) The approximate area of the recreation park on the map = (61 + 14) cm2

= 75 cm2

The approximate area of the recreation park on he ground = 75 × 2 500 m2 = 187 500 m2


4. (b) The diagram below, not drawn to scale, shows a prism of volume 450 cm3. The cross-section ABCD is a

square. The length of the prism is 18 cm.



Calculate

(i) the length of the edge AB, in cm

(ii) the total surface area of the prism, in cm2.ANSWER

4. (b) (i)


The volume of the prism:

Hence, the length of the edge AB is 5 cm.

(ii) The area of one rectangular surface A = lb

= 18 cm × 5 cm = 90 cm2

The total surface area of the prism, TSA = 2 × 25 cm2 + 4 × 90 cm2

= 50 cm2 × 360 cm2 = 410 cm2


5. Two variables x and y are related such that‘y varies inversely as the square root of x’.

(2 marks)

ANSWER

(a) Write an equation in x, y and k to describe the inverse variation, where k is the

constant of variation.


5. (a)


5. (b)

(6 marks) ANSWER

Using the information in the table of values above, calculate the value of

x 1 f 9

y r 12

3

(i) k, the constant of variation

(ii) r

(iii) r


5. (b) (i)

Hence, the value of k is 2.


(ii)

Hence, the value of r is 2.


(iii)

Hence, the value of f is 4.


5. (c) Determine the equation of the line which is parallel to the line y = 4x − 1 and passes

through the point (5, 2).

ANSWER



5. (c) y = 4x − 1 ⇒ m = 4

Using m = 4 and the point (5, 2), then

Hence, the equation of the line is y = 4x − 18.


(i) State

a) the scale factor for the enlargement.

b) the coordinates of the centre of enlargement.

L″M″N″ is the image of LMN under a

reflection in the line y = x.

(ii) Draw and label the triangle L″M″N″ on graph paper.

ANSWER


6. (a) L′M′N′ is the image of LMN under an enlargement.

(5 marks)

(i) a)

The image and the object are on the same side of the centre of enlargement.

the scale factor, k = 2.

Hence, the scale factor for the enlargement is 2.

b) The origin O(0, 0) is the centre of enlargement.

(ii) Triangle L″M″N″ was drawn on graphpaper as shown above.


6. (a)

6. (b)

ANSWER

Three parishes, P, Q, and R are such that the bearing of P from Q is O60°. R is 20 km due east of Q, and PQ = 8 km.

(i) Calculate, correct to one decimal place,

the distance PR.(ii) Given that ∠QPR = 125°, state the bearing of R from P.



6. (b) (i)


Considering ΔPQR and using the cosine rule:

Hence, the distance PR is 13.7 km.

6. (b) (ii)


Hence, the bearing of R from P is 115°.

7. The raw data for the number of minutes per day a class of 40 high school students spent watching TV is given below.

138 146 168 146 161 164 158 126173 145 150 140 138 142 135 132147 176 147 142 144 136 163 135150 125 148 119 153 156 149 152154 140 145 157 144 165 135 128


7. (a) Copy and complete the frequency table to represent this data.

ANSWER

Viewing Time in minutes Frequency118–126 3

127–135 5

136–144 9

145–153

154–162

163–171

172–180 (2 marks)


7. (a)Viewing Time in

minutes Frequency118–126 3

127–135 5

136–144 9

145–153 12

154–162 5

163–171 4

172–180 2


The frequency table way copied and completed as shown above.


7. (b) Using the raw scores, determine the range for the data. (2 marks)

ANSWER


7. (b) The range for the data = The largest observation

− The smallestobservation

= (176 − 119)minutes

= 57 minutes


7. (c) Using a scale of 1 cm to represent 5 minutes on the horizontal axis and a scale of 1cm to represent 1 student on the vertical axis, draw a frequency polygon to represent the data.

(6 marks)

ANSWER


7. (c) Class mid-point (minutes) Frequency

122 3

131 5

140 9

149 12

158 5

167 4

176 2

The frequency polygon was drawn using the table with class mid-points above and the given scales.


7. (d) What is the probability that a student chosen from this class watched TV for less than 145 minutes?

ANSWER



7. (b) The number of studentswho watched TV for = 3 + 5 + 9less than 145 minutes = 17

The total number of students = 40

P (student watched TV < 145 minutes)


8. Rectangle WXYZ below represents one whole plane figure which has been divided into seven smaller parts. These parts are labelled A, B, C, D, E, F and G.


8. (a) Copy and complete the following table, stating what fraction of the rectangle each part

represents.

ANSWER(5 marks)


8. (a)

The table was copied and completed as shown above.


8. (b) Write the parts in order of the size of their perimeters.

ANSWER

(2 marks)


8. (b)


The perimeter of F = (2 + 3 + 4+ 3.6) units = 12.6 units

The perimeter of G = 2 (3 + 2) units= 2 (5) units= 10 units

The parts in order of the size of their perimeters: G, C, B, F, D, E, A.


8. (c) The area of G is 6 square units. C, G and E are rearranged to form a trapezium.

ANSWER

(i) What is the area of the trapezium in

square units?(ii) Sketch the trapezium clearly showing the

outline of each of the three parts.(3 marks)

Total 10 marks


8. (c) (i) The area of the trapezium = The area of (C + G + E)

= (4 + 6 + 9) square units

= 19 square units

or


The area of the trapezium,


(ii)

A sketch of the trapezium clearly showing the outline of each of the three parts is shown above.


SECTION II



9. (a) Given that and f (x) = x + 5

ANSWER

(i) calculate the value of g(−3)(ii) write an expression for g f (x) in

its simplest form

(iii) find the inverse function g−1(x).(7 marks)


9. (a) (i)


(ii)


(iii)


9. (b) The length of the rectangle below is (2x + 1) cm and its width is (x − 3) cm.

ANSWER

(i) Write an expression in the form ax2 + bx + c for the area of the rectangle.

(ii) Given that the area of the rectangle is 225 cm2, determine the value of x.

(iii) Hence, state the dimensions of the rectangle, in centimetres.



9. (b) (i)

Hence, an expression in the form ax2 + bx + c for the area of the rectangle is 2x2 − 5x − 3.


(ii) The area of the rectangle, A = (2x2 − 5x − 3) cm2 and the area of the rectangle = 225 cm2

We can form the equation of equal areas:


(iii)

Hence the dimensions of the rectangle are 25 cm and 9 cm.


10. (a) Given that

ANSWER

(i) express in fractional or surd form

a) cos θ

b) tan θ

(ii) Hence, determine the exact value of

(6 marks)


10. (a) Given that


Considering a right-angled Δ and using Pythagoras’ theorem:


(ii)

Calculate the size of each of the followingangles, giving reasons for your answer

ANSWER

(i) angle TPQ

(ii) angle MTQ

(iii) angle TQS

(iv) angle SRQ


10. (b)

In the diagram above, not drawn to scale, LM is a tangent to the circle at the point, T. O is the centre of the circle and angle MTS = 25°.


10. (b)


(a) Sketch the diagram above. Show the approximate positions of P and Q such that

ANSWER

P is the mid-point of OL

Q in a point on OM such that (2 marks)1

.4

OQ OM��


OL and OM are position vectors such thatand .

OL l��

OM m��

11.

11. (a)

A sketch of the diagram is shown above.

.1

and 4

OP PL OQ OM ��


11. (b) Write down, in terms of l and m the vectors

ANSWER

(i)

(ii)

(iii)

(iv) (8 marks)

ML��

PM��

LQ��

PQ��


11. (b)


11. (c) R is the mid-point of PM. Using a vector method, prove that PQ is not parallel to

LR.

ANSWER



11. (c)


is not a multiple of , they have different directions, so PQ is not parallel to LR.PQ��

LR��



Paper 190 minutes


NEXT

1. The decimal fraction 0.35 written as a common fraction is

ANSWER

(A)

(B)

(C)

(D)7

2

1

47

2013

20


1.100

0.35 0.35100

35

10035 5

1007

5

20

Multiply by which is 1, to write

the decimal as a common fraction.

Divide both the numerator and the

denominator by the common factor 5.

Common fraction reduced to its

lowest terms.

100

100


2. is the same as

ANSWER

(A)

(B)

(C)

(D)

23

4

6

8

8

99

162

4

3


2.2

2

2

2

3 1

4 34

41

3

4

3

A negative power is the reciprocal of the positive power.

Inverting the fraction which is the divisor and multiplying instead.


3. The number 3.142 857 143... written correct to 3 significant figures is

ANSWER

(A) 3.14

(B) 3.15

(C) 3.142

(D) 3.143


3. 3.14 2 857 143= 3.14 correct to 3 sf

The 4th significant figure which is the digit 2 is less than the digit 5, so we do not add 1 to the 3rd significant figure.


4. If 60% of a number is 312, what is the number?

ANSWER

(A) 208

(B) 520

(C) 832

(D) 1 280


4.


5. If $940 is shared in the ratio 3:5:12, then the difference between the largest and the smallest shares is

ANSWER

(A) $141

(B) $235

(C) $423

(D) $564


The total number of equal parts = 3 + 5 + 12 = 20

The difference between the largest andsmallest shares in terms of equal parts = 12 3 = 9

So 20 equal parts = $940

9 equal parts =9

$94020

= $47 9

= $423Hence, the difference between the largest and smallest shares is $423.

5.


6. Written in scientific notation 0.0057 104 is

ANSWER

(A) 5.7 103

(B) 5.7 101

(C) 5.7 103

(D) 5.7 107


The first number must be, between 1 and 10.A negative power is the reciprocal of a positive power.The product of numbers with the same base is simplifiedby adding the powers.

6.


7. What is the face value of the digit 5 in the number 45 368?

ANSWER

(A) 5

(B) 50

(C) 500

(D) 5 000


7. Place value 104 103 102 101 100

Face value 4 5 3 6 8

The face value is 5.


8. The next term in the sequence 1, 3, 7, 13, 21 . . . is

ANSWER

(A) 27

(B) 29

(C) 31

(D) 33


8. 1, 1 + 2, 3 + 4, 7 + 6, 13 + 8, 21 + 10,The next number in the sequence is 31.


9. What is the value of the digit 4 in the number 57.843?

ANSWER

(A)

(B)

(C) 4

(D) 400

4

100

4

10


9. Place value 101 100 101 102 103

Face value 5 7 8 4 3

The value ofthe digit 4 = Face value Place value

2

2

4 10

14

10

4

100

14

100

A negative power isthe reciprocal of thepositive power


10. What is the least number of carambolas that can be shared equally among 8, 12 or 16 children?

ANSWER

(A) 24

(B) 36

(C) 48

(D) 64


10. 2 8,12,16

2 4, 6, 8

2 2, 3, 4

2 1, 3, 2

3 1, 3, 1

1, 1, 1

The LCM of 8,12 and 16 = 2 2 2 2 3= 48

The least number of carambolas is 48.


11. The simple interest on $600 at 5% per annum for 3 years is given by

ANSWER

(A)

(B)

(C)

(D)

600 5$

3 100

600 3$

100 5

600 100$

3 5

600 5 3$

100


11. The simple interest,

600 5 3$

100

100

PRTI

FormulaP = $600, R = 5% and T = 3 years


12. A salesman is paid 4% of his sales as commission. His sales for last month were $8 700. How much commission was he paid?

ANSWER

(A) $34.80

(B) $174.00

(C) $348.00

(D) $435.00


The commission paid 4% of $8 700

4$8 700

100$348

12.


13. A man bought a goat for $700 and sold it for $525. What was his loss as a percentage of the cost price?

ANSWER

(A) 14.3%

(B) 20%

(C) 25%

(D) 33.3%


The loss = $(700 525) = $175

The cost price = $700The loss as a percentage of the cost price =

= 25%

$175100%

$700

13.


14. A sales tax of 8% is charged on an article marked at $140.00. How much does a customer pay for the article?

ANSWER

(A) $252.00

(B) $151.20

(C) $128.80

(D) $11.20


14. The sales tax = 8% of $140

= $11.20

The amount the customer pays = $(140 + 11.20)

= $151.20

8$140

100

$112

10


15. If the simple interest on $900 for 3 years is $108, what is the rate of interest per annum?

ANSWER

(A) 0.25%

(B) 2.8%

(C) 3.24%

(D) 4%


FormulaP = $900T = 3 years I = $108

15.


16. A lot of land is valued at $50 000. Land tax is charged at the rate of $0.60 per $100 value. What is the amount payable for land tax?

ANSWER

(A) $200

(B) $300

(C) $400

(D) $500


The number of $ 100 value =

The amount payable for land tax = $0.60 500

= $60 5

= $300

$50 000500

$10016.


17. A customer buys a computer on hire purchase. He makes a deposit of $350 and pays 30-monthly instalments of $115.50 each. The hire purchase price of the computer is

ANSWER

(A) $495.50

(B) $3 465

(C) $3 495

(D) $3 815


The deposit = $350

The amount of themonthly instalments = $115.50 30

= $3 465.00

The hire-purchase price = $(350 3 465.00)

= $3 815

17.


18. The exchange rate for one United States dollar (US $1.00) is two dollars and seventy cents in Eastern Caribbean currency (EC $2.70). What is the value of US $2 000 in EC currency?

ANSWER

(A) $1 993.70

(B) $2 002.70

(C) $2 700

(D) $5 400


US $1.00 = EC $2.70

US $2 000 = EC $2.70 2 000

= EC $270 20

= EC $5 400

18.


19.

ANSWER

(A) (P Q)

(B) P Q

(C) Q

(D) Q P

In the Venn diagram, the shaded region represents


19.

The unshaded region = Q

∴ the shaded region = Q


20. Which of the following sets is equivalent to {2, 3, 5, 7}?

ANSWER

(A) {p, q}

(B) {a, b, c}

(C) {k, l, m, n}

(D) {p, q, r, s, t}


20.

1 1 correspondence

The two sets are equivalent, since they have the same number of elements there exists a one-to-one correspondence.


21.

ANSWER

(A) 5

(B) 12

(C) 23

(D) 35

In the Venn diagram, if n(X) = 28, n(Y) = 17 and n(X Y) = 40, then n(X Y) =


21.

n(X Y) = n(X) + n(Y) n(X Y) Formula

40 = 28 + 17 n(X Y)

40 = 45 n(X Y)

n(X Y) = 45 40

= 5


22. If P = {2, 3, 5, 7, 11, 13}, Q = {3, 7, 13} and R = {2, 3, 7}, then P Q R =

ANSWER

(A) { }

(B) {3, 7 }

(C) {2, 5, 11}

(D) {5, 11, 13}


22. P Q = {2, 3, 5, 7, 11, 13} {3, 7, 13}

= {3, 7, 13} Common elements

(P Q) R = {3, 7, 13} {2, 3, 7} = {3, 7}

= P Q R Common elements


23. The circumference of a circular table-top is 110 cm.

Given that , then the radius of the circular

table-top, in centimeters, is

ANSWER

(A)

(B) 35

(C) 70

(D) 172

22π

7

117

2


23.


24. A man leaves piarco Airport at 21:45 h and reaches Miami International Airport at 03:15 h the next day. How many hours did the journey take?

ANSWER

(A) 5

(B)

(C)

(D) 25

15

21

182


The number of hoursto midnight = (24:00 21:45) h

= 2 h 15 min

The number of hoursfor the journey = 2 h 15 min + 3 h 15 min

= 5 h 30 min

= 1

5 h2

24.


25. A woman takes 30 minutes to drive to Cheddi Jagan International Airport which is 26 km away from her home. Her speed in km per hour is

ANSWER

(A)

(B)

(C)

(D)

26 60

1 30

30 1

26 60

30 60

26 1

26 30

60 1


25.

Substitute the value for d and for t.


26.

ANSWER

(A) 42 cm

(B) 44 cm

(C) 46 cm

(D) 48 cm

The perimeter of the shape is


26.

The perimeter, P = (14 + 2 + 5 + 7 + 4 + 7 + 5 + 2) cm

= 46 cm


27.

ANSWER

(A) (B)

(C) (D)

The figure above shows a sector of a circle centre O. The radius of the circle is 7 cm and the sector angle is 60°. The perimeter of the sector OPQ is

22Take π

7

228 cm

3

121 cm

3

122 cm

32

27 cm3


27. The length of the arc PQ,

θ2π

36022 60

2 7 cm7 360

12 22 cm

622

cm

7 c3

31

m

l r

FormulaSubstitute each valuefor , r and .


The perimeter of the sector OPQ

121 cm

3

2

17 2 7 cm

3

17 14 cm

3

l r

Substitute the valuefor l and for r.


28. Which of the following drawings represents a uniform solid?

ANSWER

(A) (B)

(C) (D)


28. The wedge is a uniform solid, since its ends or cross-section are identical.


29.

ANSWER

(A) 36 cm2

(B)

(C) 45 cm2

(D) 48 cm2

The area of the parallelogram PQRS is

2140 cm

2


29.

The area of parallelogramPQRS, A = bh

= 9 4 cm2

= 36 cm2

Formula

Substitute the valuefor b and for h.


30. If a square has the same area as a rectangle with sides 20 cm and 5 cm, then the length of a side of the square is

ANSWER

(A) 17.5 cm

(B) 15 cm

(C) 12.5 cm

(D) 10 cm


30.

The area of the rectangle,

A = lb

= 20 5 cm2

= 100 cm2

Substitute thevalue for l and for b.


31. Each of the letters of the word ‘STATISTICS’ is written on a slip of paper. One slip is randomly drawn. What is the probability of drawing a letter ‘S’?

ANSWER

(A)

(B)

(C)

(D)3

5

1

103

101

5


31. The number of letter ‘S’ = 3

The total number of letters = 10

P(letter is an S) = 3

10


32. A man throws a die twice. What is the probability that he will throw a ‘6’ followed by an odd number?

ANSWER

(A)

(B)

(C)

(D)

1

10

7

12

1

125

12


32. The two eventsare independent.


33.

ANSWER

(A) 55 and 59

(B) 54 and 60

(C) 54.5 and 59.5

(D) 54.5 and 59

The masses of 30 children were measured, to the nearest kg. The information is shown in the grouped frequency table above. The class boundaries of the class interval 55–59 are

Mass (kg) 40–44 45–49 50–54 55–59Frequency 5 8 13 4


33. 50 54 55 59 60 64

54.5 59.5

54 55The lower class boundary, LCB

254.5

59 60The upper class boundary, UCB

259.5

+=

=

+=

=

Hence, the class boundaries are 54.5 and 59.5.


34. The mean of a number of plums is 7. If x = 63, how many plums were used in the calculation of the mean?

ANSWER

(A) 3

(B) 6

(C) 9

(D) 12


34.

Formula

Substitute the value for and for x.Multiplying both sides by n.


x

The mean, 7

and 63

Now

63So 7

i.e. 7 63

9

63

7

x

x

xx

n

nn

n


Hence, 9 plums were used in the calculation of the mean.

Items 35–36 refer to the scores in the following box.

12 13 5 88 10 7 4


35. The mode of the scores presented in the box is

ANSWER

(A) 7

(B) 8

(C) 10

(D) 13


The score 8 occurs two times.

The mode = 8.

35.


36. The median of the scores presented in the box is

ANSWER

(A) 5

(B) 7

(C) 8

(D) 12


36. The scores in ascending order:

4, 5, 7, 8, 8, 10,12,13

Middle values

2

8 8The median scor ,

28

e Q

+=


37. The expression ‘y is equal to the square root of x’ can be written as

ANSWER

(A) y = 2x

(B)

(C) y = x2

(D) y2 = x

y x


37. The square root of

The equation is:

x x

y x


38. The expression 5(x 3) =

ANSWER

(A) 5x + 15

(B) 5x 15

(C) 5x + 15

(D) 5x 15


38. 5(x 3) = 5 x 5 (3)

= 5x + 15

Negative times positive is negative. Negative times negative is positive.


39. The expression (2a)4 =

ANSWER

(A) 16a4

(B) 2a4

(C) 16a

(D) 8a4


39. 4

4

4

(2 ) 2 2 2 2

2 2 2 2

1

6

6

1

a a a a a

a a a

a

a

a

Expanding using themeaning of the 4th power.Grouping like terms.Multiplying like terms.Simplifying.


40. If then 6 o (2) =

ANSWER

(A)

(B) 9

(C) 3

(D)

2

ο ,p

p qp q

9

2

1

2


40.2

2

Substitue

ο

66 ο ( 2)

6 ( 2)

36

6 23

6 and

9

2

6

8

2

pp

p q

qp q


41. If F = mv2; when m = 80 and v = 12, then F =

ANSWER

(A) 480

(B) 960

(C) 2 850

(D) 5 760

1

2


41. 2

2 Substitute the valuefor a

1

21

(80) (12)2

nd

40 (14 )

.

5 0

4

76

m v

F mv


42. If 9x 40 = 10 x, then x =

ANSWER

(A) 5

(B) 5

(C)

(D)

33

4

33

4


42.


43. Robert buys $x worth of gas each month. In December, he bought $8 less than twice the regular worth. The worth of gas he bought in December is

ANSWER

(A) $(8 2x)

(B) $(2x 8)

(C) $8x

(D) $16x


43. Twice the regular worth of gas = $x 2= $2x

$8 less than twice the regular worth of gas = $(2x 8)


44. If V = 4R2, then R is

ANSWER

(A)

(B)

(C)

(D)4π

V

1

2 π

V

4π

VR

21

2 π

V


44.


So

45. Using the distributive law, (p r) (q r) is

ANSWER

(A) p(q r)

(B) pq r2

(C) (p q)r

(D) pqr2


45. (p r) (q r)

= (p q)r

r is a common factor.Factorising using thedistributive law.


46. Which of the following equations represents a straight line?

ANSWER

(A) y = x2 1

(B) y = 5 2x

(C)

(D) y = 2 + 3x 5x2

2y

x


46. y = 5 2x

So y = 2x + 5

It is in the form y = mx + c, which is the equation of a straight line.


47. If g(x) = x2 x + 3, the g(4) =

ANSWER

(A) 17

(B) 9

(C) 15

(D) 23


2

2

( ) 3

( 4) ( 4) ( 4) 3

16 4 3

23

g x x x

g

47. Substitute –4 for x.The square of a numberis always positive.Negative timesnegative is positive.


48.

ANSWER

(A) one-to-one

(B) one-to-many

(C) many-to-one

(D) many-to-many

The relationship that best describes the arrow diagram is


48.

The relationship that best describes the arrowdiagram is one-to-many.


ANSWER

49. Which of the following graphs is that of a function?


49. Using the vertical line test for a function, graph (A) is the graph of a function.


50. The range of f: x x3 for the domain {3, 2, 1, 0, 1, 2}

ANSWER

(A) {3, 2, 1, 0, 1, 2}

(B) {9, 6, 3, 0, 3, 6}

(C) {0, 1, 8, 27}

(D) {27, 8, 1, 0, 1, 8}


50. 3

3

3

3

3

3

3

:

: 3 ( 3) 27

: 2 ( 2) 8

: 1 ( 1) 1

: 0 0 0

:1 1 1

: 2 2 8

f x x

f

f

f

f

f

f

The range of ƒ = {27, 8, 1, 0, 1, 8}


51. What is the gradient of the straight line 3y = 5 6x?

ANSWER

(A) 2

(B)

(C)

(D)5

3

1

2

3

5


51.

is in the

3 5 6

3 6 5

6 5

3 35

23

The gradient,

form

2.

y x

y x

y x

y x

m

y mx c


52. P(3, 8) is the image of P(9, 5) under a translation T. T is represented by the column matrix

ANSWER

(A)

(B)

(C)

(D) 16

13

6

13

6

13

1

16

3


52.


3 9

8 5

3 9

8 5

6

13

T P P

T P P

53. A ship was travelling on a bearing of 180º. In what direction was it travelling?

ANSWER

(A) North

(B) South

(C) East

(D) West


53.

The ship was travelling due south.


54.

ANSWER

(A) (1, 2) (B) (1, 2)

(C) (1, 2) (D) (3, 2)

ABC is rotated through 180º about the origin as centre. The coordinates of the image of A under the transformation is


54.

Hence, the coordinates of the image of A are (1, 2).


55.

ANSWER

(A) 45

(B) 120

(C) 135

(D) 150

In the figure above, O is the centre of the circle.The magnitude of angle PQR is


55.


Sum of in a

circle.

at centre 2.

135 at aircumferenc

ˆ (reflex) 360 90

270

e

270ˆ

.2

POR

PQ

s

R

56.

ANSWER

(A) 45 cos 48°

(B) 5 sin 48

(C) 9 sin 48

(D) 45 sin 48

In the parallelogram ABCD, AB = 9 cm, BC = 5 cm and angle ABC = 48º. The area of the parallelogram ABCD, in cm2, is


56.

The area of the parallelogram ABCD,

A = ab sin C Formula

= 9 5 sin 48° cm2

= 45 sin 48° cm2

Two sides and the included angle.


57.

ANSWER

(A) (1.5 + 15 cos 30°) m

(B) (1.5 + 15 sin 30°) m

(C) (1.5 + 15 tan 30°) m

(D) 15 tan 30° m

TAB represents a tower. The height of the tower, TB, is


57.


1.5 m

tan 3015 m

So 15 m tan 30

15 tan 30 m

And

15 tan 30 1

1.5 15 tan 3

.5 m

0 m

AB

TA

TA

TB TA AB

58.

ANSWER

(A) 311°

(B) 131°

(C) 98°

(D) 49°

AC and DE are straight lines intersecting at B.Angle CBE = 49º. The size of angle CBD is


58.

Sum of s on a st line.

Subtracting 49° from both sides.


ˆ 49 180

ˆ 180 49

131

CBD

CBD

Items 59–60 refer to the diagram below. OAB isa right-angled triangle.


59. OA B is the image of OAB under an enlargement by a scale factor of 3. The coordinates of the points A and B are

ANSWER

(A) (0, 2) and (3, 0)

(B) (0, 3) and (6, 0)

(C) (0, 3) and (4, 0)

(D) (0, 4) and (8, 0)


59.


60. Area of OAB : Area of OA B =

ANSWER

(A) 1:9

(B) 1:6

(C) 1:3

(D) 1:1.5


Area of OAB: Area of OAB = 1:9

60.



MATHEMATICS

Paper 2

2 hours 40 minutes

SECTION I



NEXT

1. (a) Using a calculator, or otherwise, determine

the value of (15.4)2 − (0.217 ÷ 7) and write

the answer(2 marks)

ANSWER

(i) exactly

(ii) correct to two significant figures.


1. (a) (i) (15.4)2 − (0.217 ÷ 7)

= 237.16 − 0.031

= 237.129

Squaring anddividingSubtracting(exactly)

(ii) 23

7 .129

= 240

The third significant figure,7, is greater than 5, so weadd 1 to the 3.(correct to two significantfigures)


Calculate

ANSWER

(i) the values of k and l(ii) the book value of the Private Car after 2 years.


(6 marks)

1. (b) The table below gives information on the book values and the rates of depreciation for two motor vehicles.

Motor Vehicle

Initial Book Value

Annual Rate of

Depreciation

Book Value after One Year

Taxi $75 000 15% $kPrivate car $90 000 l% $81 000

1. (b) (i) The book valueof the Taxi afterone year, $k = (100 − 15)% of

$75 000= 85% of $75 000= 0.85 × $75 000= $63 750

the value of k is 63 750


The amount of thedepreciation ofthe Private Carafter one year = $(90 000 – 81

000)= $9 000

The rate of depreciation of

the Private Car, l%

the value of l is 10.


(ii) The book value of thePrivate Car after one year = $81

000

The book valueof the PrivateCar aftertwo years = (100 − 10)% of

$81 000 = 90% of $81 000 = 0.9 × $81 000 = $72 900


1. (c) GUY $1.00 = US $0.005 andJAM $1.00 = US $0.011Calculate the value of

(2 marks)

ANSWER

(i) GUY $80 000 in US $


(ii) US $385 in JAM $. (2 marks)Total 12 marks

1. (c) (i) GUY $1.00 = US $0.005GUY $80 000 = US $0.005× $80 000

= US $400


(ii)

2. (a) Simplify

(3 marks)

ANSWER


2. (a)


2. (b) (i) Factorise

(1 mark)

ANSWER

a) p2 − 7p


b) p2 − 640 (1 mark)

(ii) Simplify (3 marks)

2. (b) (i) a)


b)

2. (b) (ii)


2. (c) Three DVDs and two Blue-ray discs cost $421, while two DVDs and one Blue-ray

disccost $242.

(2 marks)

ANSWER

(i) Given that one DVD costs $d and one

Blue-ray disc costs $b, write two equations

in b and d to represent the information.


(ii) Calculate the cost of one Blue-ray disc. (2 marks)Total 12 marks

2. (c) (i) The cost of one DVD = $dThe cost of one Blue-ray disc = $b

The two equations in b and d are:

3d + 2b= 421 (in dollars)

2d + b= 242 (in dollars)


Hence, the cost of one Blue-ray disc is $116.

2. (c) (ii)


3. (a)


Giving the reason for each step of your

answer, calculate the size of

ANSWER

(2 marks)(i) LNK


3. (a) In the quadrilateral KLMN, not drawn to scale, KL = LN = LM, NKL = 35° and KLM = 140°.

(ii) NLM(iii) KNM

(2 marks)

(2 marks)

(i) LNK = NKL = 35°Base s ofisoscelesΔ KLN


3. (a)

3. (a) (ii) KLN = 180° − (35° + 35°) = 180° − 70°= 110°

NLM = 140° − 110°= 30°

Sum of the s of a Δ


3. (a) (iii)

= 75°


(i) Copy and complete the Venn diagram torepresent the information.


(ii) Write an expression in x for the number of students in the survey.

(iii) Calculate the value of x.


3. (b) In a survey of a class of 38 students, it was found that

15 like Karate12 like Judox like Karate and Judo2x like neither Karate nor Judo.

K is the set of students in the survey who like Karate, and J is the set of students who like Judo.

3. (b) (i)

The Venn diagram represent the information given.


(ii) The number of students in the survey

= 15 − x + x + 12 – x + 2x= 2x + x – x – x + 15 + 12 = x + 27

(iii) The number ofstudents in the survey = 38So x + 27 = 38i.e. x = 38 − 27

= 11



4. (a) Using a ruler, a pencil and a pair of compasses, construct the triangle ABC

in which AB = 10 cm, BAC = 60°, and AC = 7 cm.

ANSWER

(4 marks)


4. (a)

Construct AB = 10 cm, then BAC = 60°.Construct AC = 7 cm, then complete the ΔABC.


4. (b) Measure and state the length of BC.

ANSWER

(1 mark)


4. (b) By measurement, the length of BC = 8.9 cm


4. (c) Find the perimeter of Δ ABC.

ANSWER

(1 mark)


4. (c) The perimeter of ΔABC = (10 + 7 + 8.9) cm

= 25.9 cm


4. (d) Construct on your diagram the line CD which is perpendicular to AB, and

meets AB at D.

ANSWER

(2 marks)


4. (d)

The line CD which is perpendicular to AB was constructed in the preceding diagram.


4. (e) Determine the length of CD.

ANSWER

(2 marks)


4. (e) Considering the right-angled ΔCAD:

Hence, the length of CD is 6.1 cm.


4. (f) Calculate the area of ΔABC, giving your

answer correct to one decimal place.

ANSWER



4. (f)


Or


Use the graph to determine the(a) values of a and b which define the domain of

the graph

ANSWER

(2 marks)


5. The diagram below shows the graph of the function f (x) = 2 − x − x2 for a ≤ x ≤ b. The tangent to the graph at (1, 0) is also drawn.

5. (a) From the graph, the domain is −3 ≤ x ≤ 2.

The value of a is −3; this value is obtained from the left-hand side of the graph.

The value of b is 2; this value is obtained from the right-hand side

of the graph.


(b) values of x for which x2 + x − 2 = 0

ANSWER

(2 marks)



5. (b) Since f (x) = 2 − x − x2,if f (x) = 0, then

0 = 2 − x − x2

So x2 + x – 2 = 0

The solution to the quadratic equation are the

intercepts on the x-axis of the graph.

Hence, x = −2 or x = 1.


(c) coordinates of the maximum point on thegraph

ANSWER(2 marks)



5. (c) The coordinates of the maximum point on the

graph are


(d) gradient of f (x) = 2 − x − x2 at x = 1

ANSWER

(3 marks)



5. (d) Using the points (0, 3) and (3, −6) on the tangent to the graph, then the gradient of

the tangent, m

Hence, the gradient of f (x) = 2 − x − x2 at x = 1 is –3.


(e) values of x for which 2 − x − x2 > −2, where x is a whole number. ANSWER

(2 marks)



5. (e) Draw the line f (x) = −2, then the values of x for which 2 − x

− x2 > −2 is given by the inequality −2.6 ≤ x ≤ 1.6.

Since x is a whole number, then the possible values of x are 0, 1.


(a) Copy the diagram and indicate on the diagram, the distances x km, (x + 2) km and 10 km. ANSWER


6. A man walks x km due north, from point K to point L. He then walks (x + 2) km due west from point L to point M. The distance along a straight line from K to M is 10 km. The diagram below, not drawn to scale, shows the relative positions of K, L and M. The direction of north and west are also shown.

(2 marks)

6. (a)

The diagram was copied and the distancesindicated as shown above.


(b) Apply Pythagoras’ theorem to the diagram to obtain an equation in x. Show that the equation can be

simplified to x2 + 2x – 48 = 0.ANSWER

(3 marks)



6. (b) Using Pythagoras’ theorem:


(c) Solve the equation and state the distance KL. ANSWER(2 marks)



6. (c)

Hence, the distance KL is 6 km.


(d) Determine the bearing of K from M. ANSWER(4 marks)Total 11 marks



6. (d)


Hence, the bearing of K from M is 126.9°.


(a) Copy and complete the mid-interval valuescolumn.

ANSWER


7. In an experiment, the mass of 100 children were recorded in kilograms as shown in the table below.

Mass in kilograms Frequency

Mid-Interval Values (kg)

5–9 5 7

10–14 23 12

15–19 29

20–24 14

25–29 13

30–34 11

35–39 5

(1 mark)

Mid-IntervalValues (kg)

7

12

17

22

27

32

37

7. (a)The mid-intervalvalues column wascopied and completedas shown.


(b) (i) Calculate an estimate of the mean mass

of the 100 children.

ANSWER(3 marks)





5–9 5 7

10–14 23 12

15–19 29

20–24 14

25–29 13

30–34 11

35–39 5

7. (b) (i)

Frequency f


x fx 5 7 35

23 12 276

29 17 493

14 22 308

13 27 351

11 32 352

5 37 185 n = f = 100 fx = 2 000


An estimate of themean mass of the100 children,


(b) (ii) Draw a frequency polygon to represent

the mass of the children.

ANSWER(5 marks)





5–9 5 7

10–14 23 12

15–19 29

20–24 14

25–29 13

30–34 11

35–39 5

The frequency polygon above represents the mass of the children.

7. (b) (ii)


(c) Calculate the probability that a student chosen at random from the experimental group had a mass of 25 kg or more.






5–9 5 7

10–14 23 12

15–19 29

20–24 14

25–29 13

30–34 11

35–39 5

7. (c) The number of children witha mass of 25 kg or more =

13 + 11 + 5= 29

The total number of children = 100

P (student’s mass $ 25 kg)


8. The first three diagrams in a sequence are shown below.Diagram 1 has four disks forming a square pattern.

Diagram 2 is a square pattern formed by using five additional disks.

Diagram 3 is a square pattern formed by using seven additional disks.

(2 marks)


(a) Draw Diagram 4 in the sequence.ANSWER

8. (a)

Diagram 4 in the sequence can be seen above.


8. (b) Complete the table by inserting the appropriate values at

the rows marked (i), (ii) and (iii).

ANSWER

(6 marks)

Diagram Number

Number of Disks Forming

the Square

Number of Additional

Disks Forming the

Square

Pattern for Calculating the

Number of Additional Disks

1 2 × 2 ~ ~2 3 × 3 5 2 × 3 − 13 4 × 4 7 2 × 4 − 14 5 × 5 — —— 8 × 8 15 —n (n + 1) (n + 1) — —


8. (b)


Diagram Number

Number of Disks Forming

the Square

Number of Additional

Disks Forming the

Square

Pattern for Calculating the

Number of Additional Disks

1 2 × 2 ~ ~2 3 × 3 5 2 × 3 − 13 4 × 4 7 2 × 4 − 14 5 × 5 9 2 × 5 − 17 8 × 8 15 2 × 8 − 1n (n + 1) (n + 1) (n + 1) + n 2(n + 1) − 1


8. (c) Hence, determine a formula consisting of a

single term in n for calculating the total number of disks, N, in a diagram.

ANSWER



8. (c) The total number of disks in a diagram,

Hence, the formula is N = (n + 1)2.


SECTION II



9. (a) Solve the pair of simultaneous equations y = x + 3

y = x2

ANSWER

(5 marks)


9. (a)

where a = 1, b = −1 and c = −3.


Using the quadratic formula:


Hence, x = 2.30 or x = −1.30


(i) Write an expression, in terms of l and x, for the length of the strip of wire.

ANSWER(2 marks)


9. (b) A strip of wire of length 58 cm is cut into two pieces. One piece is bent to form a square of

side x cm. The other piece is bent to form arectangle of length l cm and width 5 cm.

The diagrams below, not drawn to scale, show the square and the rectangle.

9. (b) (i) The length of thestrip of wire = [4x + 2l +

2(5)] cm= (4x + 2l + 10) cm

An expression, in terms of l and x, for the

length of the strip of wire is 4x + 2l + 10.


(ii) Show that l = 24 − 2x

ANSWER(2 marks)

The sum of the areas of the square and the rectangle is represented by A.





9. (b) (ii)


(iii) Show that A = x2 − 10x + 120.

ANSWER

(2 marks)





9. (b) (iii) The sum of the areasof the square andthe rectangle


(iv) Calculate the values of x for which A = 96. ANSWER(4 marks)

Total 15 marks





9. (b) (iv)

Hence, the values of x are 4 and 6.


10. (a) The diagram below, not drawn to scale, shows a vertical cellphone tower, TF, and a vertical antenna, WT, mounted on the

top of the tower.

A point P is on the same horizontal ground as F, such that PF = 30 m, and the angles of elevation of T and W from P are 42° and 55° respectively.


10. (a) (i) Copy and label the diagram clearly showing

a) the distance 30 m

b) the angles of 42° and 55°

c) any right angles.

ANSWER

(7 marks) (ii) Calculate the length of the antenna WT.


10. (a) (i)


The diagram was copied and labelledabove, clearly showinga) the distance 30 mb) the angles of 42° and 55°c) the right angle TFP


10. (a) (ii)


Considering the right-angled ΔPWF:


Considering the right-angled ΔPTF:

The length of the antenna WT = WF − TF= (42.8 − 27) m= 15.8 m


10. (b) The diagram below, not drawn to scale,shows a circle with centre O. The lines

BDand DCE are tangents to the circle. AngleBCD = 65°


Calculate, giving reasons for each step ofyour answer.

ANSWER


(i) OCE

(ii) BAC

(iii) BOC

(iv) BDC


10. (b)


10. (b) (i)


10. (b) (ii)


10. (b) (iii)


10. (b) (iv)


11. (a) The value of the

determinant of M = is 22.

ANSWER

(3 marks)(i) Calculate the value of x.


11. (a) (i) The value of the determinant of M,



(ii) For this value of x, find M–1.

ANSWER

(2 marks)




11. (a) (ii)


(iii) Show that M–1 M = I.

ANSWER

(2 marks)




11. (a) (iii)

Hence M –1 M = I.


11. (b) The graph shows the line segment AC andits image A′C′ after a transformation by the

matrix


(i) Write in the form of a single 2 × 2 matrix,the coordinates ofa) A and C ANSWER

(2 marks)



matrix

11. (b) (i) a)


(i) b) A′ and C′.

ANSWER

(2 marks)



matrix

11. (b) (i) b)


(ii) Using matrixes only, write an equation torepresent the transformation of AC intoA′C′.

ANSWER(2 marks)



matrix

11. (b) (ii) The matrix equation that represents thetransformation of AC onto A′C′ is:


(iii) Determine the values of p, q, r and s.




matrix

11. (b) (iii)


Equating corresponding elements in the first row:


Substituting −1 for p in :


Equating corresponding elements in the second row:


Substituting 0 for r in :

Hence, p = −1, q = 0, r = 0 and s = 1.



MATHEMATICS: A COMPLETE COURSE WITH CXC QUESTIONSText © Raymond ToolsieFirst Published in 1996Reprinted in 1997, 1998, 1999, 2000, 2001, 2002, 2003Second Edition November 2004Third Edition 2009

ISBN: 976-8014-13-0

Reprinted in 2004, 2005, 2006, 2007, 2008by Eniath’s Printing Company Limited6 Gaston Street, Lange Park, Chaguanas,Trinidad, West IndiesDesigned by: diacriTech, www.diacritech.com

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Mathematics model exam_vol-2_ppt_design

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