Mathematics_revision_exam_vol-1_ppt_design Kw.pdf

REVISION EXAMINATIONS

Revision Exam 1 Paper 1

Revision Exam 3 Paper 2

Revision Exam 1 Paper 2

Revision Exam 4 Paper 1

Revision Exam 2 Paper 1

Revision Exam 4 Paper 2

Revision Exam 2 Paper 2

Revision Exam 5 Paper 1

Revision Exam 3 Paper 1

Revision Exam 5 Paper 2

REVISION EXAMINATION 1 MATHEMATICS Paper 1 90 minutes Answer ALL the questions

NEXT

Revision Exam 1 Paper 1

1. 35.4 × 0.2 is approximately equal to (A) 0.71

(B) 7.1

(C) 71.0 (D) 710

ANSWER

Revision Exam 1 Paper 1

1.

Revision Exam 1 Paper 1

2. (A) 6.0

(B) 9.2

(C) 21.16 (D) 27.6

ANSWER

Revision Exam 1 Paper 1

2.

Revision Exam 1 Paper 1

Alternative Method:

Revision Exam 1 Paper 1

3. $180.00 is divided among 3 friends in the ratio 3 : 4 : 8. What amount is the largest share? (A) $96

(B) $84

(C) $48

(D) $36

ANSWER

Revision Exam 1 Paper 1

3.

The total number of equal parts = 3 + 4 + 8 = 15 12 8  the largest share = $180 × 15 1 = $96

Revision Exam 1 Paper 1

4. 1 274 in standard form is (A) 1.274 × 10–3

(B) 1.274 × 10–2

(C) 1.274 × 103

(D) 1.274 × 102

ANSWER

Revision Exam 1 Paper 1

4. 1 274 = 1.274 × 1 000 = 1.274 × 103

Revision Exam 1 Paper 1

5. If a vehicle travels 220 kilometres on 16 litres of petrol, then on a full 48 litre-tank it should travel (A) 268 km

(B) 330 km

(C) 660 km

(D) 10 560 km

ANSWER

Revision Exam 1 Paper 1

5.

Revision Exam 1 Paper 1

6. (A) 1.9 × 10

(B) 0.19 × 10

(C) 6.0 × 10

(D) 0.6 × 10

ANSWER

Revision Exam 1 Paper 1

6.

Revision Exam 1 Paper 1

7. The next two terms in the sequence 2, –1, –4, –7, . . . are (A) –10, –13

(B) –3, 0

(C) –9, –11

(D) –9, –11

ANSWER

Revision Exam 1 Paper 1

7. So

Revision Exam 1 Paper 1

8. Using the distributive law –3(4 – 7) = (A) –3×4 – 3×7

(B) –3×[–4] + 3×7

(C) –3×4 – 3×[–7]

(D) 3×[−4] + 3×[−7]

ANSWER

Revision Exam 1 Paper 1

8.

Revision Exam 1 Paper 1

9. 6158 = (A) 6×83 + 1×82 + 5×81 (B) 6×82 + 1×81 + 5×80 (C) 6×81 + 1×80 + 5×8−1 (D) 6×80 + 1×81 + 5×80

ANSWER

Revision Exam 1 Paper 1

9.

Revision Exam 1 Paper 1

10. The HCF of the numbers 18, 27 and 36 is (A) 3

(B) 6

(C) 9

(D) 12

ANSWER

Revision Exam 1 Paper 1

10.

3 18, 27, 36 3 11 6, 9, 12 2, 3,14 The HCF = 3 × 3 = 9

Revision Exam 1 Paper 1

11. When an article is sold for $132.00, a profit of 10% is made. The cost of the article is (A) $118.80

(B) $120.00

(C) $122.00

(D) $142.00

ANSWER

Revision Exam 1 Paper 1

11. 110% of the cost price

= $132.00

100% of the cost price The cost of the article

= $120.00

Revision Exam 1 Paper 1

12. The sum of $6 500 was borrowed from a bank at the rate of 8% per annum for 5 years. The simple interest payable is (A) $40.63

(B) $104

(C) $520

(D) $2 600

ANSWER

Revision Exam 1 Paper 1

12. The simple interest payable,

= $65 × 40 = $2 600

Revision Exam 1 Paper 1

13. The sum of $800 invested at simple interest for 6 years earns $336. The rate of interest per annum is (A) 6%

(B) 6.5%

(C) 7%

(D) 7.5%

ANSWER

Revision Exam 1 Paper 1

13. The rate of interest per annum;

I = $336 P = $800 T = 6 years

= 7%

Revision Exam 1 Paper 1

14. A woman bought a car for $80 500. After one year, it was worth $70 840. Its depreciation, as a percentage of the selling price was (A) 10%

(B) 11%

(C) 12% (D) 13%

ANSWER

Revision Exam 1 Paper 1

14. The depreciation as a percentage of the selling price

=12%

Revision Exam 1 Paper 1

15. If 15% value-added tax is paid on articles purchased, then the ratio of the price of an article inclusive of value-added tax to the price of the article exclusive of the value-added tax is (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

15. The required ratio

Revision Exam 1 Paper 1

16. After a 5% discount the selling price of a pair of pants was $342. The selling price of the pair of pants if a 10% discount was given instead is (A) $307.80

(B) $324.00

(C) $324.90

(D) $360.00

ANSWER

Revision Exam 1 Paper 1

16.

The selling price after a 5% discount = $342 18 The selling price after a 10% 18 90 × discount = $342 95 19 1

= $18 × 18 = $324

Revision Exam 1 Paper 1

17. If TT how much approximately in TT $ would one get for EC $50? (A) $12.00

(B) $20.83

(C) $120.00

(D) $208.30

ANSWER

Revision Exam 1 Paper 1

17.

Revision Exam 1 Paper 1

18. A plane left Norman Manley International Airport at 09:35 h and arrived at Cheddi Jagan International Airport at 11:20 h the same day. The time for the journey was (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

18.

Revision Exam 1 Paper 1

19. If P = {7, 8, 9, 10}, Q = {4, 6, 8, 10, 12} and R = {8, 9, 10, 11, 12}, then = (A) {8, 10}

(B) {8, 9, 10}

(C) {8, 10, 12}

(D) { }

ANSWER

Revision Exam 1 Paper 1

19.

= {8, 10} = {8, 10}

Revision Exam 1 Paper 1

20.

In the Venn diagram above, the shaded portion represents (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

20.

Revision Exam 1 Paper 1

21. The number of possible subsets that can be obtained from the set {1, p, ø} is (A) 3

(B) 6

(C) 8

(D) 10

ANSWER

Revision Exam 1 Paper 1

21.

The subsets = { }, {1}, {p}, {ø}, {1, p}, {1, ø}{p, ø}, {1, p, ø}

The number of subsets = 8.

Revision Exam 1 Paper 1

22.

In the Venn diagram above, the shaded region represents (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

22.

The shaded region =

Revision Exam 1 Paper 1

23. A square has the same area as a parallelogram with altitude 4 cm and base 16 cm. What is the length of the side of the square? (A) 4 cm

(B) 6 cm

(C) 8 cm (D) 12 cm

ANSWER

Revision Exam 1 Paper 1

23.

Revision Exam 1 Paper 1

23.

Revision Exam 1 Paper 1

24.

The area of the trapezium above (not drawn to scale) is (A) 21 cm2

(B) 42 cm2

(C) 84 cm2

(D) 168 cm2

ANSWER

Revision Exam 1 Paper 1

24.

Revision Exam 1 Paper 1

25. The flat surface area of a closed cylinder of height h cm and radius r cm is (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

25.

Revision Exam 1 Paper 1

26. If a sphere has radius r cm and volume V cm3, then V = (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

26.

Revision Exam 1 Paper 1

27. The height of a waterfall is 80 m. The waterfall is represented on a scale drawing by a length of 2 cm. The scale used was (A) 1: 40 000

(B) 1: 4 000

(C) 1: 400

(D) 1: 40

ANSWER

Revision Exam 1 Paper 1

27. Since 2 cm represent then

2 cm represent



1 cm represents



Scale:

1: 4 000

80 m 8 000 cm 4 000 cm

Revision Exam 1 Paper 1

28. A bullet takes 2s to travel a distance of 800 m. The average speed of the bullet is (A)

(B)

(C) 400 m/s

(D) 1 600 m/s

ANSWER

Revision Exam 1 Paper 1

28.

d = 800 m t = 2s

Revision Exam 1 Paper 1

29. A plane travels for 5 h at an average speed of 720 km/h. What was the distance covered? (A) 60 km

(B) 144 km

(C) 360 km

(D) 3 600 km

ANSWER

Revision Exam 1 Paper 1

29.

s = 720 km/h t=5h

Revision Exam 1 Paper 1

30. The length of a rod was measured as 38.7 cm accurate to the nearest cm. The greatest possible length of the rod is

(A) 38.75 cm

(B) 38.71 cm

(C) 38.65 cm

(D) 38.60 cm

ANSWER

Revision Exam 1 Paper 1

30.

Revision Exam 1 Paper 1

31. Each of the letters of the word ‘BELIEVER’ are written on separate pieces of paper. The pieces of paper are then placed in a jug. What is the probability of drawing a letter ‘E’? (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

31.

Revision Exam 1 Paper 1

32. The mean of nine numbers is 15.6. The number 4.4 is removed from the set. The new mean is (A) 20.0

(B) 17.0

(C) 11.2 (D) 8.0

ANSWER

Revision Exam 1 Paper 1

32. The sum of the nine numbers

= 15.6 × 9 = 140.4

The sum of the eight numbers = 140.4 − 4.4 = 136

The mean of the eight numbers = = 17

Revision Exam 1 Paper 1

33.

The pie chart above shows how 240 students spend their physical education period. The number of students who spend it playing Jiu-Jitsu is equal to (A) 50

(B) 60

(C) 75

(D) 80 ANSWER

Revision Exam 1 Paper 1

33. The number of students who spend it playing Jiu-Jitsu

Revision Exam 1 Paper 1

34. The median of the numbers 7, 8, 9, 9, 10, 11, 12, 13 is (A) 9

(B) 9.5

(C) 9.875

(D) 10

ANSWER

Revision Exam 1 Paper 1

34.

7, 8, 9, 9, 10 , 11, 12, 13 Q2 The median,

Revision Exam 1 Paper 1

Items 35–36 refer to the following histogram which shows the points scored in a competition by a group of students.

Revision Exam 1 Paper 1

35. The total number of students who took part in the competition was (A) 30

(B) 49

(C) 100

(D) 105

ANSWER

Revision Exam 1 Paper 1

35. The total number of students = 5 + 10 + 15 + 30 + 20 + 15 + 10 = 105

Revision Exam 1 Paper 1

36. The modal score was (A) 5

(B) 6

(C) 7

(D) 8

ANSWER

Revision Exam 1 Paper 1

36. The highest bar has a frequency of 30 and a score of 7.  the modal score = 7

Revision Exam 1 Paper 1

37. If 4n is an even number, which of the following is an odd number? (A) 4n – 1

(B) 4n – 2

(C) 4n + 2

(D) 4(n + 1)

ANSWER

Revision Exam 1 Paper 1

37. If 4n is an even number, then 4n – 1 is an odd number.

Revision Exam 1 Paper 1

38. 6x – 3(x + 4) = (A) 3x + 12

(B) 3x – 12

(C) –3x + 12

(D) –3x – 12

ANSWER

Revision Exam 1 Paper 1

38. = 3x – 12

Using the distributive law. Subtracting like terms.

Revision Exam 1 Paper 1

39. If p * q = 4p + q, then 1* 5 = (A) 24

(B) 21

(C) 12

(D) 9

ANSWER

Revision Exam 1 Paper 1

39. p * q = 4p + q 1* 5 = 4(1) + 5 =4+5

=9

Revision Exam 1 Paper 1

40. If 20 – 3x = 3x – 10, then x = (A) −5

(B) 0

(C) 5

(D) 10

ANSWER

Revision Exam 1 Paper 1

40. 20 – 3x = 3x – 10 20 + 10= 3x + 3x 30 = 6x =x

5= x 

x=5

Add 10 and 3x to both sides. Add like terms. Divide both sides by 6.

Revision Exam 1 Paper 1

41. 3a2 + (–4a)2 = (A) −13a2

(B) −a2

(C) 7a2

(D) 19a2

ANSWER

Revision Exam 1 Paper 1

41. 3a2 + (–4a)2 = 3a2 + (–4a) (–4a) = 3a2 + 16a2 = 19a2

Squaring Adding like terms

Revision Exam 1 Paper 1

42. If (A)

and

, then

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

42.

Substitute

for x and

for y

Add the numerators, since 3 is the common denominator

Revision Exam 1 Paper 1

43. p(x – y) + q(x – y) = (A) px + qx

(B) (p + q)(x – y)

(C) −px − qy

(D) (px – qy)2

ANSWER

Revision Exam 1 Paper 1

43. p(x – y) + q(x – y) = (x – y)(p + q) = (p + q)(x – y)

Revision Exam 1 Paper 1

44. If v2 = u2 + 2as and u > 0, then (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

44.

v2 = u2 + 2as v2 − 2as = u2

Subtract 2as from both sides Take square roots.



since u > 0

Revision Exam 1 Paper 1

45. x4 y2 ÷ x2 y3 = (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

45.

Alternative Method:

Revision Exam 1 Paper 1

46. The straight lines y = 3 and x = 4 intersect at (A) (0, 0)

(B) (6, 8)

(C) (3, 4)

(D) (4, 3)

ANSWER

Revision Exam 1 Paper 1

46.

Revision Exam 1 Paper 1

47. Which of the following relation diagrams represents a function? (A)

Revision Exam 1 Paper 1

47. (B)

Revision Exam 1 Paper 1

47. (C)

Revision Exam 1 Paper 1

47. (D)

ANSWER

Revision Exam 1 Paper 1

47. 1–1 relation

This relation diagram represents a function, since it is a one-to-one relation.

Revision Exam 1 Paper 1

48. The graph of the straight line 2y + 8 = 6x intersects the y-axis at the point (A) (0, –8)

(B) (0, 8)

(C) (0, –4)

(D) (0, 4)

ANSWER

Revision Exam 1 Paper 1

48. 2y + 8 = 6x

Subtract 8 from both sides. Divide both sides by 2.

So 2y = 6x − 8 

y = 3x − 4

When x = 0,

y = –4

(0, –4)

Revision Exam 1 Paper 1

X

Y

49.

The diagram above represents the mapping (A) x → 3x

(B) x → x + 3

(C) x → x3

(D) x → 3x

ANSWER

Revision Exam 1 Paper 1

49. If

then

x → 3x

1 → 31 = 3 2 → 32 = 9 3 → 33 = 27

and

4 → 34 = 81

Revision Exam 1 Paper 1

50. A straight line passing through the point (–3, 5) and with gradient –2 can be represented by the particular equation (A) y = −2x + 1

(B) y = −2x − 1

(C) y = 2x − 11

(D) y = 2x + 11

ANSWER

Revision Exam 1 Paper 1

50. Using the point (–3, 5) = (x1, y1) and m = –2, then the equation of a straight line y − y1 = m(x − x1) becomes y − 5 = −2(x − [−3]) = −2(x + 3) = −2x − 6  y = −2x − 6 + 5 = −2x − 1

Revision Exam 1 Paper 1

51. The gradient of the straight line 4x − 3y = 6 is (A)

(B)

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

51.

4x − 3y = 6 4x − 6 = 3y



Revision Exam 1 Paper 1

52.

In the triangle PQR above, the ratio expressing tan θ =

(A)

(B)

(C)

(D) ANSWER

Revision Exam 1 Paper 1

52.

Revision Exam 1 Paper 1

53.

In the figure above, AB and CD are parallel. Which of the following statements best describes the relation between x and y? (A) x ≠ y

(B) x − y = 0

(C)

(D) x > y

ANSWER

Revision Exam 1 Paper 1

53.

x =y 

x−y =0

corresponding ∠s since AB || CD

Revision Exam 1 Paper 1

54. If tan θ = cos θ =

(A)

, and

(B)

is acute, then the value of

(C)

(D)

ANSWER

Revision Exam 1 Paper 1

54.

The hypotenuse = 5, since 3, 4 and 5 form a Pythagorean triple.

Revision Exam 1 Paper 1

55.

In the diagram above, if O is the centre of the circle, then y = (A)

(B) x

(C) 2x

(D) 180 − x ANSWER

Revision Exam 1 Paper 1

55.

y° = 2x°



y = 2x

The angle at the centre of a circle is twice the angle at the circumference standing on the same arc (or chord). ∠ at centre 2. ∠ at circumference.

Revision Exam 1 Paper 1

56.

In the triangle PQR above, angle PRS = 135° and angle PQR = 65°. The angle RPQ = (A) 45°

(B) 65° (C) 70°

(D) 100° ANSWER

Revision Exam 1 Paper 1

56.

exterior ∠= sum of two opposite co-interior ∠s Subtract 65° from both sides.

Revision Exam 1 Paper 1

57.

The triangle KLM above is right-angled at L. Angle LKM = 25° and ML = 14 m. The length of KM, in m, is (A)

(B)

(C) 14 cos 25°

(D) 14 tan 25°

ANSWER

Revision Exam 1 Paper 1

57.

Revision Exam 1 Paper 1

58.

In the figure above, AOC is a diameter of a circle centre O. If angle BAO = 52°, what is the size of angle BCA? (A) 38°

(B) 52° (C) 76°

(D) 104° ANSWER

Revision Exam 1 Paper 1

58.



∠ in a semi-circle complementary ∠s Subtract 52° from both sides

Revision Exam 1 Paper 1

59.

In the diagram above, BC is a vertical tower of height 12 m, standing on level ground. The angle of elevation of the top of the tower from a point A is θ degrees. If AB = 24 m, then θ° = (A) 15°

(B) 30° (C) 45°

(D) 60°

ANSWER

Revision Exam 1 Paper 1

59.



Revision Exam 1 Paper 1

60.

In the diagram above, the bearing of the point A from the point B is (A) 085° (B) 095° (C) 265° (D) 275° ANSWER

Revision Exam 1 Paper 1

60.

The bearing of the point A from B is 085°.

REVISION EXAMINATION 1 MATHEMATICS Paper 2 2 hours 40 minutes Answer ALL the questions

NEXT

Revision Exam 1 Paper 2

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

the value of

and state the answer

(i) exactly (ii) correct to one decimal place (iii) correct to one significant figure.

(3 marks)

ANSWER

Revision Exam 1 Paper 2

1. (a)

(i)

= 5.62 + 1.47 = 7.09 (exactly) (ii) 7.0 | 9 = 7.1 (correct to 1 decimal place) The digit 9, which is in the second decimal place, is greater than 5, so we add 1 to the digit 0, which is in the first decimal place. (iii) 7 . | 09 = 7 (correct to 1 significant figure) The digit 0, which is the second significant figure, is less than 5, so we do not add 1 to the digit 7, which is the first significant figure.

Revision Exam 1 Paper 2

1. (b) Divide $500 in the ratio 7:3.

(2 marks)

ANSWER

Revision Exam 1 Paper 2

1. (b) The total number of proportional parts = 7 + 3 = 10

Hence, the larger share is $350 and the smaller share is $150.

Revision Exam 1 Paper 2

1. (c) A computer virus damaged 70% of the computers in an office. If 18 computers were not damaged, calculate how many computers were in the office altogether. (2 marks)

ANSWER

Revision Exam 1 Paper 2

1.

Hence, 60 computers were in the office.

Revision Exam 1 Paper 2

1. (d) Calculate

(3 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

1. 8  7 + 4 = 56 + 4 = 60 5  14 + 1 = 70 + 1 = 71 2  20 + 7 = 40 + 7 = 47

Invert the fraction that is the divisor and multiply instead of divide.

Revision Exam 1 Paper 2

2. (a) If p = 4 and q = – 3, calculate the value of 5p + 2q.

(2 marks)

ANSWER

Revision Exam 1 Paper 2

2. (a)

5p + 2q = 5(4) + 2(–3) = 20 – 6 = 14

Substitute 4 for p and –3 for q in the expression.

Revision Exam 1 Paper 2

2. (b) Write as a single fraction in its simplest

form

(3 marks)

.

ANSWER

Revision Exam 1 Paper 2

2. (b)

The LCM of the denominators 2 and 3 is 6. Use 6 as the common denominator.

Remove the brackets using the distributive law. Group like terms.

Add like terms.

Revision Exam 1 Paper 2

2. (c) Solve the inequality

(3 marks)

ANSWER

Revision Exam 1 Paper 2

2. Subtract 2 from both sides. Multiply both sides by −1.

Divide both sides by 3. So

Revision Exam 1 Paper 2

2. (d) Simplify

(2 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

2. (d)

Remove the brackets using the distributive law Add like terms.

Revision Exam 1 Paper 2

3. (a) A company offers loans of $8 000, with monthly repayments of $426 each for 20 months. Calculate: (i) the interest, in dollars, paid on the loan (ii) the interest as a percentage of the loan amount.

(3 marks)

ANSWER

Revision Exam 1 Paper 2

3. (a)

(i)

The total amount of the monthly repayments

= $426 × 20 = $8 520

The interest paid on the loan (ii)

The interest as a percentage of the loan amount

= $(8 520 − 8 000) = $520

Revision Exam 1 Paper 2

3. (b) The cash price of a computer game is $628. The hire purchase price is 10% deposit and 12 monthly payments of $56.52. Calculate how much more the hire purchase price is than the cash price. (4 marks)

ANSWER

Revision Exam 1 Paper 2

3. (b) The deposit = 10% of $628

The total amount of the monthly payments

= $56.52 × 12 = $678.24

 the hire purchase price

= $(62.80 + 678.24)

= $741.04

Revision Exam 1 Paper 2

The difference between the hire purchase price and the cash price

= $(741.04 − 628.00) = $113.04

Hence, the hire purchase price is $113.04 more than the cash price.

Revision Exam 1 Paper 2

3. (c) Mrs Frank has a job for which the basic rate of pay is $5.00 per hour, and the overtime rate of pay is $7.50 per hour. During a certain week she earned $263.75. She worked

hours overtime

Calculate (i) the amount she earned at the basic rate

(3 marks)

ANSWER

Revision Exam 1 Paper 2

3. (c) (i) Her overtime payment

= $63.75 Her basic payment

= $(263.75 – 63.75)

= $200.00 Hence, she earned $200.00 at the basic rate.

Revision Exam 1 Paper 2

3. (c) Mrs Frank has a job for which the basic rate of pay is $5.00 per hour, and the overtime rate of pay is $7.50 per hour. During a certain week she earned $263.75. She worked

hours overtime

Calculate (ii) the number of hours she worked at the basic rate. (3 marks) ANSWER

Revision Exam 1 Paper 2

3. (c) (ii) The number of hours she earned at the basic rate

= 40

Revision Exam 1 Paper 2

4. (a)

The diagram below, not drawn to scale, shows parallel lines AB and CD intersected by the line EF.

Calculate the size of the angle marked (i) w (ii) z

(2 marks)

ANSWER

Revision Exam 1 Paper 2

4. (a) (i) w + 124 = 180 

w = 180 – 124

Interior s are supplementary.

= 56

(ii)

z = 124

Corresponding s.

Revision Exam 1 Paper 2

4. (b) The diagram below shows a circle ABC with centre O. BC is a diameter and ACB = 32. Calculate the size of OAB. (3 marks)

ANSWER

Revision Exam 1 Paper 2

4. (b)

CAB = 90 s in a semi-circle. AOC is isosceles, since AO = CO (radii) So CAO = ACB = 32  OAB = CAB – CAO = 90 – 32 = 58

Revision Exam 1 Paper 2

4. (c)

The figure below, not drawn to scale, is a sector OAB of a circle with centre O.

Use (i) Calculate the length of the arc AB of the sector of the circle. (5 marks)

ANSWER

Revision Exam 1 Paper 2

4. (c) (i)

The length of the arc AB of the sector of the circle,

Revision Exam 1 Paper 2

4. (c)

The figure below, not drawn to scale, is a sector OAB of a circle with centre O.

Use (ii) Using a protractor, a ruler and a pair of compasses, make an accurate (5 marks) full-scale drawing of the figure. Total 10 marks

ANSWER

Revision Exam 1 Paper 2

4. (c) (ii)

An accurate full-scale diagram of the figure is shown above.

Revision Exam 1 Paper 2

5. (a) An American tourist converted US $700 to TT currency. The exchange rate on that day was TT $6.38 = US $1.00. In addition, 1% government tax is charged on all foreign currency transactions. Calculate, in Trinidad and Tobogo currency: (i) the tax paid (ii) the amount of money that the tourist actually received.

(6 marks)

ANSWER

Revision Exam 1 Paper 2

5. (a) (i) US $1.00 US $700

The tax paid

= TT $6.38 = TT $6.38 × 700 = TT $4 466 = 1% of TT $4 466

(ii) The amount of money that the tourist actually received = TT $(4 466 – 44.66) = TT $4 421.34

Revision Exam 1 Paper 2

5. (b) An antique stool is valued at $500. Its value increases by 7% each year. (i) Find the value of the stool after two years. (ii) Show that the value of the stool after three years is $612.52 to the nearest cent. (4 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

5. (b) (i) The value after one year

= 107% of $500 = 1.07 × $500 = $535

The value after two years = 1.07 × $535

= $572.45 (ii) The value after three years = 1.07 × $572.45 = $612.5215 = $612.52 (to the nearest cent)

Revision Exam 1 Paper 2

6. The diagram below, not drawn to scale, shows a rectangle ABDE joined along the edges AE and BD by two triangles AEF and BDC, so that CDEF is a straight line. AB = 20 m, BC = 18 m, CD = 12.5 m, DE = 20 m, EF = 7.5 m and AF = 15 m.

(a) Calculate the length of AE

(3 marks)

ANSWER

Revision Exam 1 Paper 2

6. (a)

Using Pythagoras’ theorem:

Hence, the length of AE is 13.0 m

Revision Exam 1 Paper 2

6. The diagram below, not drawn to scale, shows a rectangle ABDE joined along the edges AE and BD by two triangles AEF and BDC, so that CDEF is a straight line. AB = 20 m, BC = 18 m, CD = 12.5 m, DE = 20 m, EF = 7.5 m and AF = 15 m.

(b) Determine (i) the perimeter of the trapezium ABCF

ANSWER (5 marks)

Revision Exam 1 Paper 2

6. (b)

(i)

The perimeter of the trapezium ABCF, P = (20 + 18 + 12.5 + 20 + 7.5 + 15) m = 93 m

Revision Exam 1 Paper 2

6. The diagram below, not drawn to scale, shows a rectangle ABDE joined along the edges AE and BD by two triangles AEF and BDC, so that CDEF is a straight line. AB = 20 m, BC = 18 m, CD = 12.5 m, DE = 20 m, EF = 7.5 m and AF = 15 m.

(b) Determine

(ii) the area of the trapezium ABCF.

ANSWER

Revision Exam 1 Paper 2

6. (b) (ii)

The area of the trapezium ABCF,

Revision Exam 1 Paper 2

6. (c) A scale of 1:100 is used to draw an accurate diagram of the trapezium ABCF. Find the area of the scale diagram in square centimetres. (2 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

6. (c)

1:100 means

1 cm: 100 cm

which is

1 cm: 1 m

So

1 cm2 :1 m2



390 cm2: 390 m2

Hence, the area of the scale diagram is 390 cm2.

Revision Exam 1 Paper 2

7. (a) Solve the following simultaneous equations for x and y.

(5 marks)

ANSWER

Revision Exam 1 Paper 2

7. (a)

3: + :

Revision Exam 1 Paper 2

7. (b) Ira had $16 and Rameez had $4. Each of the children earned $x for doing a chore. Write an expression in terms of x for the amount of money (i) Ira now has (ii) Rameez now has After they were paid for the chore, Ira’s amount of money was twice that of Rameez’s.

(iii) Write an equation in terms of x to represent the information given. (iv) Solve the equation. (v) What amount of money was Ira paid for (5 marks) the chore? Total 10 Marks

ANSWER

Revision Exam 1 Paper 2

7. (b) (i) The amount of money Ira has = $ (16 + x) (ii) The amount of money Rameez has = $(4 + x) (iii) The equation in terms of x is: (iv)

Hence x = 8 (v) Ira was paid $8 for the chore.

Revision Exam 1 Paper 2

8. From a point (P) on level ground, which is 100 m from the foot (F) of a church tower, the angle of elevation of the top (T) of the tower is 40. (a) Draw a diagram to represent the information given. Label the diagram showing the church tower and the ground. Insert all measurements given. (2 marks)

ANSWER

Revision Exam 1 Paper 2

8. (a)

The diagram representing the information is shown above.

Revision Exam 1 Paper 2

8. From a point (P) on level ground, which is 100 m from the foot (F) of a church tower, the angle of elevation of the top (T) of the tower is 40.

(b) Calculate, to two significant figures, (i) the height of the tower

(4 marks)

ANSWER

Revision Exam 1 Paper 2

8. (b) (i)

So

Hence, the height of the tower is 84 m.

Revision Exam 1 Paper 2

8. From a point (P) on level ground, which is 100 m from the foot (F) of a church tower, the angle of elevation of the top (T) of the tower is 40.

(b) Calculate, to two significant figures, (ii) the distance from the point (P) to the top of the tower (T).

ANSWER

Revision Exam 1 Paper 2

8. (b)

(ii)



Hence, the distance PT is 130 m.

Revision Exam 1 Paper 2

8. (c) If the point (P) is now 75 m from the foot (F) of the tower, calculate the angle of elevation of the top (T) of the tower.

(4 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

8. (c)

Revision Exam 1 Paper 2

9.

The marks obtained by 100 pupils on a test are shown below. (a) Copy and complete the frequency table below to represent the information given. Marks 1 2 3 4 5 6 7 8 9 10

Frequency 3 5 7 12 28 — 11 8 6 4

ANSWER (2 marks)

Revision Exam 1 Paper 2

9. (a) The completed frequency table is shown, below.

50th and 51st marks

Marks 1 2 3 4 5 6 7 8 9 10

The frequency for the mark

Frequency 3 5 7 12 28 Highest frequency — 11 8 6 4

Revision Exam 1 Paper 2

9.

The marks obtained by 100 pupils on a test are shown below. (b) Using the frequency distribution, state (i) the modal mark (ii) the median mark

(iii) the range.

Marks 1 2 3 4 5 6 7 8 9 10

Frequency 3 5 7 12 28 — 11 8 6 4

(3 marks)

ANSWER

Revision Exam 1 Paper 2

9. (b)

(i) The highest frequency = 28  the modal mark = 5 (ii) The position of the median,

 the median mark, (iii) The range = The largest observation – The smallest observation = (10 – 1) marks = 9 marks

Revision Exam 1 Paper 2

9.

The marks obtained by 100 pupils on a test are shown below. (c) On graph paper, draw a histogram to illustrate the frequency distribution.

Marks 1 2 3 4 5 6 7 8 9 10

Frequency 3 5 7 12 28 — 11 8 6 4

(3 marks)

ANSWER

Revision Exam 1 Paper 2

9. (c) The histogram which illustrates the frequency distribution is shown below.

Revision Exam 1 Paper 2

9. (d) A pupil is chosen at random from the group of pupils. What is the probability that the pupil’s mark is greater than 5? (2 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

9. (d) The number of pupils with a mark greater than 

Revision Exam 1 Paper 2

10.

The cost of renting a sailing boat consists of a basic charge plus a charge per km travelled. The graph below shows the total cost in dollars (y) for the number of km travelled (x).

ANSWER

Revision Exam 1 Paper 2

10.

Revision Exam 1 Paper 2

10. (a)

What is the cost of renting a sailing boat to travel a distance of (i) 150 km

(ii) 260 km?

(2 marks)

ANSWER

Revision Exam 1 Paper 2

From the graph: 10. (a)

(i) The cost of renting a sailing boat to travel a distance of 150 km= $40

(ii) The cost of renting a sailing boat to travel a distance of 260 km = $62

Revision Exam 1 Paper 2

10. (b)

What distance in km was travelled when (1 mark) the cost was $50?

ANSWER

Revision Exam 1 Paper 2

10. (b)

The distance travelled when the cost was $50 = 200 km

Revision Exam 1 Paper 2

10. (c)

What is the amount of the basic charge?

(1 mark)

ANSWER

Revision Exam 1 Paper 2

10. (c)

The amount of the basic charge, c = $10

Revision Exam 1 Paper 2

10. (d) Calculate the gradient of the line.

(2 marks)

ANSWER

Revision Exam 1 Paper 2

10. (d) The gradient of the line,

Revision Exam 1 Paper 2

10. (e)

Write down the equation of the line in (2 marks) the form y = mx + c.

ANSWER

Revision Exam 1 Paper 2

10. (e)

The equation of the line is:

Revision Exam 1 Paper 2

10. (f)

Calculate the cost of renting a sailing boat to travel a distance of 420 km.

(2 marks) Total 10 marks

ANSWER

Revision Exam 1 Paper 2

10. (f)

The cost of renting a sailing boat to travel a distance of 420 km,

REVISION EXAMINATION 2 MATHEMATICS Paper 1 90 minutes Answer ALL the questions

NEXT

Revision Exam 2 Paper 1

1.

expressed as a percentage = (A) 5% (B) 20% (C) 25% (D) 40% ANSWER

Revision Exam 2 Paper 1

1.

expressed as a percentage

Revision Exam 2 Paper 1

2. The decimal equivalent of

(A) 0.062 5 (B) 0.187 5

(C) 0.312 5 (D) 0.437 5

ANSWER

Revision Exam 2 Paper 1

2.

Revision Exam 2 Paper 1

3. 0.004 573 in standard form correct to 3 significant figures is (A) 4.57  10–2

(B) 4.57  10–3 (C) 4.58  10–2 (D) 4.58  10–3 ANSWER

Revision Exam 2 Paper 1

3. 3 is less than 5; we do not add 1 to 7. In standard form correct to 3 significant figures

Revision Exam 2 Paper 1

4. (–2)2 + (–1)3 = (A) –5 (B) –3 (C) 3 (D) 5 ANSWER

Revision Exam 2 Paper 1

4.

Revision Exam 2 Paper 1

5. The square root of 390 lies between (A) 17 and 18 (B) 18 and 19 (C) 19 and 20 (D) 20 and 21

ANSWER

Revision Exam 2 Paper 1

5.

Revision Exam 2 Paper 1

6. (32  43)6 = (A) 336  422

(B) 312  418 (C) 312  43 (D) 32  418 ANSWER

Revision Exam 2 Paper 1

6. (32  43)6 = 32  6  43  6 = 312  418

Revision Exam 2 Paper 1

7. 47  103 is equivalent to (A) (47  100) + 3 (B) (47  100) + (47  3) (C) (47  100) – (47  3)

(D) (47  100)(47  3)

ANSWER

Revision Exam 2 Paper 1

7. 47  103 = 47  (103) = 47  (100 + 3) = (47  100) + (47  3)

Revision Exam 2 Paper 1

8. 2410 can be written in base two as (A) 11 000 (B) 10 100 (C) 10 010 (D) 10 001 ANSWER

Revision Exam 2 Paper 1

8.

Revision Exam 2 Paper 1

9. In the numeral 5238, the value of the digit 2 is (A) 2 (B) 8

(C) 16 (D) 128 ANSWER

Revision Exam 2 Paper 1

9. 5238 = 5  82 + 2  81 + 3  80 The value of the digit 2 = 2  81 = 16 Face Place Value value value

Revision Exam 2 Paper 1

10. What is the least number of plums that can be shared equally among 14, 28 or 35 students? (A) 70 (B) 140

(C) 210 (D) 280 ANSWER

Revision Exam 2 Paper 1

10. 2 14, 28, 35

2

7, 14, 35

5

7, 7, 35

7

7, 7, 7

1, 1, 1 The LCM of 14, 28 and 35 = 2  2  5  7 = 140

Revision Exam 2 Paper 1

11. The marked price of a refrigerator was $5 600. A woman bought the refrigerator on terms by paying $560 down and $308 for 18 months. What amount of money would have been saved if the refrigerator had been bought cash? (A) $180

(B) $308 (C) $504 (D) $560

ANSWER

Revision Exam 2 Paper 1

11.

The down payment

= $560

The sum of the monthly instalments = $308  18 = $5 544 The hire-purchase price = $(560 + 5 544) = $6 100 The difference between the hire-purchase price and the marked price = $(6 100 – 5 600) = $504

Revision Exam 2 Paper 1

12. A vendor bought eggs at $8.00 a dozen and sold them at $13.20 a dozen. The percentage profit was

(A) 65% (B) 35% (C) 25% (D) 20% ANSWER

Revision Exam 2 Paper 1

12. The profit

= $(13.20 – 8.00) = $5.20

Revision Exam 2 Paper 1

13. If

of a sum of money is $75, then the total

sum of money is (A) $750 (B) $900

(C) $975 (D) $1 125 ANSWER

Revision Exam 2 Paper 1

13.

Revision Exam 2 Paper 1

14. The simple interest or $875 invested for 6 years at 5.7% per annum is

ANSWER

Revision Exam 2 Paper 1

14. The simple interest, P = $875 R = 5.7% T = 6 years

Revision Exam 2 Paper 1

15. The cost price of an article is $96. In selling the article a shopkeeper made a profit of 25%. The selling price was (A) $121 (B) $120

(C) $115.20 (D) $72 ANSWER

Revision Exam 2 Paper 1

15.

Revision Exam 2 Paper 1

16. The marked price of a pair of pants is $150. If VAT of 15% is payable, then the price paid by the customer is (A) $127.50 (B) $135.00

(C) $165.00 (D) $172.50 ANSWER

Revision Exam 2 Paper 1

16.

Revision Exam 2 Paper 1

17. In a bank, the interest rate on investments

increased from

% to 6% per annum. The

increase in annual interest on a fixed deposit of $8 000 is

(A) $40 (B) $80 (C) $440

(D) $480

ANSWER

Revision Exam 2 Paper 1

17.

Revision Exam 2 Paper 1

18. By selling a blouse for $75, a businesswoman made a profit of $15. The profit as a percentage of the cost price is (A) 60% (B) 30%

(C) 25% (D) 20% ANSWER

Revision Exam 2 Paper 1

18.

Revision Exam 2 Paper 1

19. If P = {factors of 30} and Q = {factors of 20}, then P  Q =

(A) {5, 10, 15, 20} (B) {1, 2, 5, 10} (C) {4, 20} (D) { } ANSWER

Revision Exam 2 Paper 1

19. 

P = {1, 2, 3, 5, 6, 10, 15, 30} Q = {1, 2, 4, 5, 10, 20} P  Q = {1, 2, 5, 10}

Revision Exam 2 Paper 1

20.

The shaded region in the Venn diagram represents (A) (A  C)  B (B) (A  B)  C (C) (B  C)  A (D) A  B  C

ANSWER

Revision Exam 2 Paper 1

20.

A  C is shaded

B is shaded

(A  C)  B is shaded

Revision Exam 2 Paper 1

21. The set of fractions

written in

ascending order of magnitude is

ANSWER

Revision Exam 2 Paper 1

21.

Revision Exam 2 Paper 1

22. The set of numbers greater than or equal to –4, but less than 9 can be written as

(A) {x: –4 < x  9} (B) {x: –4  x < 9} (C) {x: –4  x  9} (D) {x: –4 < x < 9} ANSWER

Revision Exam 2 Paper 1

22. The set of numbers greater than or equal to –4 is {x: x  –4} = {x: –4  x}

The set of numbers less than 9 is {x: x < 9} The set of numbers greater than or equal to –4, but less than 9 is {x: –4  x, x < 9}

= {x: –4  x < 9}

Revision Exam 2 Paper 1

23. If a circle has a diameter d cm, and area A cm2, then A = (A) d2

(B) (C) 2d (D) 4d2 ANSWER

Revision Exam 2 Paper 1

23. The area of a circle, A = r2

Revision Exam 2 Paper 1

24. If the diagonals of a kite measures 18 cm and 24 cm, then its area is (A) 21 cm2

(B) 108 cm2 (C) 216 cm2

(D) 432 cm2 ANSWER

Revision Exam 2 Paper 1

24. The area of the kite,

d1 = 18 cm

d2 = 24 cm

Revision Exam 2 Paper 1

25. If a rectangle with perimeter 76 centimetres has sides of lengths 13 cm and 5x centimetres, then x = (A) 38 (B) 19

(C) 12 (D) 5 ANSWER

Revision Exam 2 Paper 1

25.

5x cm P = 76 cm

13 cm

Revision Exam 2 Paper 1

26. Which of the following figures best describes a polygon with all its sides equal? (A) convex polygon

(B) re-entrant polygon (C) regular polygon (D) quadrilateral ANSWER

Revision Exam 2 Paper 1

26. A regular polygon has equal sides.

Revision Exam 2 Paper 1

27.

In the circle above, the radius is 24 cm. The length of the arc PQ, in centimetres, is

(A) 6 (B) 12 (C) 18

(D) 24

ANSWER

Revision Exam 2 Paper 1

27. The length of the arc PQ,

Revision Exam 2 Paper 1

28. The volume of a cuboid whose edges are 5 cm, 6 cm and 8 cm is (A) 120

(B) 180 (C) 190 (D) 240 ANSWER

Revision Exam 2 Paper 1

28. The volume of the cuboid, V = lbh

l = 8 cm

= 8 cm  6 cm  5 cm 3

= 240 cm

b = 6 cm h = 5 cm

Revision Exam 2 Paper 1

29. The circumference of a circle of radius

(A) (B) 6 cm (C) (D) 13 cm ANSWER

Revision Exam 2 Paper 1

29.

Revision Exam 2 Paper 1

30. A vehicle travels with a speed of 52 km/h for 15 minutes. The distance travelled in kilometers is

(A) 3.47 (B) 13 (C) 208 (D) 780 ANSWER

Revision Exam 2 Paper 1

30.

Revision Exam 2 Paper 1

31. A bowl has 4 red balls, 6 blue balls and 10 green balls. If one of the balls is chosen at random, what is the probability that it is not blue.

ANSWER

Revision Exam 2 Paper 1

31.

Revision Exam 2 Paper 1

32. The mean of nine numbers is 17.5. The number 12.5 is added to the set. The new mean is

(A) 17 (B) 14.5 (C) 15 (D) 13 ANSWER

Revision Exam 2 Paper 1

32.

The sum of the nine numbers = 17.5  9 = 157.5 The sum of the ten numbers = 157.5 + 12.5 = 170

The mean of the ten numbers = = 17

Revision Exam 2 Paper 1

33. The histogram below shows the number of points scored by participants in a competition. The number of participants is

(A) 125 (B) 126 (C) 127 (D) 128 ANSWER

Revision Exam 2 Paper 1

33. The number of participants = 5 + 12 + 17 + 20 + 25 +

19 + 14 + 8 + 5 + 2 = 127

Revision Exam 2 Paper 1

34. From the histogram above, the mean number of points scored by a participant is

ANSWER

Revision Exam 2 Paper 1

34.

x 1 2 3 4 5 6 7 8 9 10

f 5 12 17 20 25 19 14 8 5 2

fx 5 24 51 80 125 114 98 64 45 20

n = f = 127 fx = 626

Revision Exam 2 Paper 1

35. From the histogram above, the median number of points scored by a participant is

(A) 4 (B) 5 (C) 6 (D) 7 ANSWER

Revision Exam 2 Paper 1

35.

Revision Exam 2 Paper 1

x 1 2 3 4 Q2 = 5 6 7 8 9 10

f 5 12 17 20 25 19 14 8 5 2

64th position

The median number of points scored by a participant, Q2 = 5

Revision Exam 2 Paper 1

36. From the histogram above, the modal number of points scored by a participant is

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

Revision Exam 2 Paper 1

36.

The highest frequency

= 25

The point corresponding to the highest frequency

=5

 the modal number of points scored by a participant =5

Revision Exam 2 Paper 1

37. If x is an odd number, which of the following is also odd?

(A) x – 1 (B) x + 1 (C) x + 3 (D) 2x – 1 ANSWER

Revision Exam 2 Paper 1

37.

2x – 1 is an odd number. For example, if x = 3, then 2x – 1 = 2(3) – 1 = 6–1

=5

Revision Exam 2 Paper 1

38. Raymond is 52 years of age and Yuri is 26 years of age. The ratio of Yuri’s age to Raymond’s age is

(A) 2:1 (B) 1:2 (C) 1:3 (D) 3:1 ANSWER

Revision Exam 2 Paper 1

38. The ratio of Yuri’s age to Raymond’s age = 26:52

= 1:2

Revision Exam 2 Paper 1

39.

ANSWER

Revision Exam 2 Paper 1

39.

The LCM of 9r and 5s is 45rs,

Revision Exam 2 Paper 1

40. 5x + 2(x – 3) = (A) 7x + 6 (B) 7x – 6 (C) 3x + 6 (D) 3x – 6 ANSWER

Revision Exam 2 Paper 1

40. 5x + 2(x – 3) = 5x + 2x – 6 Using the distributive law. = 7x – 6 Adding like terms.

Revision Exam 2 Paper 1

41. 5x  5y = (A) 1x – y (B) 5x – y (C) 5xy (D) 5x  y ANSWER

Revision Exam 2 Paper 1

41.

Revision Exam 2 Paper 1

42. If x and y are both integers, then 3(x + y)2 means (A) three times the sum of their squares (B) six times their sum (C) the square of three times their sum (D) three times the square of their sum ANSWER

Revision Exam 2 Paper 1

42. Their sum The square of their sum

=x+y

= (x + y)2

Three times the square of their sum = 3(x + y)2

Revision Exam 2 Paper 1

43. If x = –5 and y = –8, then x + y is (A) –3 (B) 3 (C) –13 (D) 13 ANSWER

Revision Exam 2 Paper 1

43. x + y = –5 + (–8) = –5 –8 = –13

Revision Exam 2 Paper 1

44. If 0.08x = 40, then x = (A) 500 (B) 50 (C) 5 (D) 0.5 ANSWER

Revision Exam 2 Paper 1

44.

Divide both sides by 0.08. Multiply both the numerator and denominator by 100, to make the denominator a whole number. Dividing by 8.

Revision Exam 2 Paper 1

45. In a certain village, adults are served x grams of meat per meal. What mass of meat, in kilograms, is required for 7 meals to feed 200 adults? (A) 1400x

(B) 140x (C) 14x (D) 1.4x ANSWER

Revision Exam 2 Paper 1

45.

Revision Exam 2 Paper 1

46.

In the figure on the previous page, the point for which the x-coordinate is negative and the y-coordinate is negative is (A) A

(B) B

(C) C

(D) D

ANSWER

Revision Exam 2 Paper 1

46. In the third quadrant, both the x and y-coordinates are negative.

e.g.

Revision Exam 2 Paper 1

47. Which of the inequalities in x best describes the number line above?

(A) {x: > –2} (B) {x: x  –2}

(C) {x: x < –2} (D) {x: x  –2} ANSWER

Revision Exam 2 Paper 1

47. The number line represents {x: x  –2}.

Revision Exam 2 Paper 1

48.

The diagram given represents the mapping

(A) x → 3x + 1 (B) x → 3x – 1 (C) x → –3x + 1 (D) x → –3x – 1

ANSWER

Revision Exam 2 Paper 1

48.

x → 3x – 1 –2 → 3(–2) – 1 = –6 – 1 = –7 –1 → 3(–1) – 1 = –3 – 1 = –4 0 → 3(0) – 1 = 0 – 1 = –1 1 → 3(1) – 1 = 3 – 1 = 2 2 → 3(2) – 1 = 6 – 1 = 5

Revision Exam 2 Paper 1

49. If g: x → x2 + 1, then g(–3) = (A) 10 (B) 8

(C) –10 (D) –8

ANSWER

Revision Exam 2 Paper 1

49.

g(x) = x2 + 1 g(–3) = (–3)2 + 1 =9+1

= 10

Substitute –3 for x Square Add the numbers

Revision Exam 2 Paper 1

50. The gradient of the straight line passing through the points (–2, 5) and (8, –3) is

ANSWER

Revision Exam 2 Paper 1

50. Using the points (–2, 5) and (8, –3), then the gradient of the straight line (x1, y1) = (–2, 5) (x2, y2) = (8, –3)

Revision Exam 2 Paper 1

51. The equation of a straight line with gradient –4 and y-intercept 7 is

(A) y = 4x + 7 (B) y = 4x – 7 (C) y = –4x – 7 (D) y = –4x + 7 ANSWER

Revision Exam 2 Paper 1

51. The gradient,

The y-intercept,

m = –4 c= 7

The equation of the straight line is y = mx + c = –4x + 7

Revision Exam 2 Paper 1

52. A plane is heading on a bearing of 045°, then changes its course to a bearing of 150°. The angle through which the plane turns is

(A) 315 (B) 225 (C) 195 (D) 105 ANSWER

Revision Exam 2 Paper 1

52.

Angle through which the plane turns

The difference between the two bearings = 150 – 45 = 105

Revision Exam 2 Paper 1

53.

In the right-angled ABC above, angle A =

(A) 15

(B) 30 (C) 45 (D) 60

ANSWER

Revision Exam 2 Paper 1

53.

Revision Exam 2 Paper 1

54.

In the triangle ABC, the exterior angle ACD = 125 and angle CAB = 60. Angle ABC = (A) 65

(B) 60 (C) 50 (D) 45

ANSWER

Revision Exam 2 Paper 1

54.

exterior  = sum of two co-interior s

Revision Exam 2 Paper 1

55.

The translation in which PQ is mapped into P′Q′ can be represented by the column matrix

ANSWER

Revision Exam 2 Paper 1

55.

T

Revision Exam 2 Paper 1

56. A plane was flying on a bearing of 045°. In which direction was it flying? (A) north-east

(B) north-west (C) south-east (D) south-west ANSWER

Revision Exam 2 Paper 1

56.

Revision Exam 2 Paper 1

57. The interior angle of a regular hexagon is (A) 45 (B) 60 (C) 72 (D) 120 ANSWER

Revision Exam 2 Paper 1

57. The sum of the interior angles of a regular hexagon (n = 6) = 180  6 – 360

= 1 080 – 360 = 720

The interior angles of a regular hexagon = 120 Alternative Method:

Revision Exam 2 Paper 1

Revision Exam 2 Paper 1

58. When rotated through 180° about the origin, the image of the point (–3, –8) is (A) (–3, 8) (B) (3, –8) (C) (3, 8) (D) (–8, –3) ANSWER

Revision Exam 2 Paper 1

R180: P(x, y) → P(–x, –y)

58.



R180: (–3, –8) → (3, 8)

Revision Exam 2 Paper 1

59.

In the rhombus above, if PQT = 38° then QPT = (A) 19

(B) 38

(C) 52

(D) 90 ANSWER

Revision Exam 2 Paper 1

59.

The diagonals of a rhombus intersect at right angles.

Revision Exam 2 Paper 1

60. The point P(5, 8) is reflected in the line y = x. What are the co-ordinates of the image P ? (A) (–5, –8) (B) (–8, 5)

(C) (8, 5) (D) (–8, –5) ANSWER

Revision Exam 2 Paper 1

60.

My = x: P(x, y) → P(y, x) My = x: P(5, 8) → P(8, 5)

REVISION EXAMINATION 2 MATHEMATICS Paper 2 2 hours 40 minutes Answer All the questions

NEXT

Revision Exam 2 Paper 2

1. (a) Calculate the value of

, expressing

your answer (i) exactly (ii) in standard form

(3 marks)

ANSWER

Revision Exam 2 Paper 2

1. (a)

0

Revision Exam 2 Paper 2

1. (b) In a certain school, 57% of the students are boys. If there are 513 boys in the school, calculate the total number of students in the school. (2 marks)

ANSWER

Revision Exam 2 Paper 2

1. (b) The total number of students in the school

Revision Exam 2 Paper 2

1. (c) The table below shows the points scored by two teams in a computation competition. First Round Second Round

Team X

87

Team Y

65

94

(i) In the second round, the ratio of points scored by Team X to Team Y was 5:3. How many points did Team Y (5 marks) score in the second round? Total 10 marks

ANSWER

Revision Exam 2 Paper 2

1. (c) (i) Let the score of Team Y be x points.

So i.e.

Hence, the score of Team Y must be 39 points.

Revision Exam 2 Paper 2

1. (c) The table below shows the points scored by two teams in a computation competition. First Round Second Round Team X

87

Team Y

65

94

(ii) What is the least number of points Team Y should have scored in the second round so that the total score of Team Y would be more than the total score of Team X? ANSWER

Revision Exam 2 Paper 2

1. (c) (ii) The total score of Team X = (87 + 94) points = 181 points The difference of the teams score = (181– 65) points = 116 points Hence, the least number of points Team Y should have scored is 117 points.

Revision Exam 2 Paper 2

2. (a) How much interest is earned when $900 is invested at 10% per annum compound interest for two years? (3 marks)

ANSWER

Revision Exam 2 Paper 2

2. (a) The principal after one year

The principal after two years

The interest earned for two years = $(1089 – 900) = $189

Revision Exam 2 Paper 2

2. (b) Mr Aaron works a basic week of 40 hours at a rate of $18 an hour. His overtime rate is $9 per hour more than his basic rate.

Calculate (i) his total wage for a basic week

(7 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

2. (b) (i) His basic wage = $18 × 40 = $720

Revision Exam 2 Paper 2

2. (b) Mr Aaron works a basic week of 40 hours at a rate of $18 an hour. His overtime rate is $9 per hour more than his basic rate. (ii) his wage for a week in which he worked 57 hours

ANSWER

Revision Exam 2 Paper 2

2. (b) (ii) The number of hours worked overtime = 57 – 40 = 17 His overtime wage = $(18 + 9) × 17 = $27 × 17

= $459 His wage for the 57 hour week

= $(720 + 459) = $1 179

Revision Exam 2 Paper 2

2. (b) Mr Aaron works a basic week of 40 hours at a rate of $18 an hour. His overtime rate is $9 per hour more than his basic rate. (iii) the number of hours he worked during one week if he was paid a wage of $1 260

ANSWER

Revision Exam 2 Paper 2

2. (b) (iii) His overtime wage = $(1 260 – 720) = $540 The number of hours he worked overtime = 20 The number of hours he worked during the week = 40 + 20 = 60

Revision Exam 2 Paper 2

3. (a) A man who has a wife and two children earns $42 300 annually. The annual tax-free allowances are shown in Table 1. Table 1 Allowance Adult

$1 200 each

Child

$500 each

House $3 600 per family Calculate

(i) his total annual tax-free allowances

(5 marks)

ANSWER

Revision Exam 2 Paper 2

3. (a) (i) The allowance for the adults = $1 200 × 2 = $2 400 The allowance for the two children = $500 × 2 = $1 000

The house allowance = $3 600  his total annual tax-free allowances

= $(2 400 + 1000 + 3 600) = $7 000

Revision Exam 2 Paper 2

3. (a) A man who has a wife and two children earns $42 300 annually. The annual tax-free allowances are shown in Table 1. Table 1 Allowance

Adult

$1 200 each

Child

$500 each

House $3 600 per family (ii) his annual taxable income. Table 2 shows the taxes that are due annually. Table 2 Taxable Income Taxes Due First $25 000

$1 250

Remainder

25% of the remainder

ANSWER

Revision Exam 2 Paper 2

3. (a) (ii) His annual taxable income = $(42 300 – 7 000) = $35 300

Revision Exam 2 Paper 2

3. (a) A man who has a wife and two children earns $42 300 annually. The annual tax-free allowances are shown in Table 1. Table 1 Allowance

Adult

$1 200 each

Child

$500 each

House $3 600 per family (iii) Calculate the taxes he should pay annually.

ANSWER

Revision Exam 2 Paper 2

3. (a) (iii) Tax payable on the first $25 000 = $1 250 The remaining amount to be taxed = $(35 300 – 25 000) = $10 300 Tax payable on the remainder

= 25% of $10 300

The taxes he should pay annually = $(1 250 + 2 575) = $3 825

Revision Exam 2 Paper 2

3. (b) A used car is advertised for sale at $25 600. A discount of 8% is given if it is bought for cash. It can also be bought on hire purchase by paying a deposit of $2 304 followed by 32 monthly payments of $837.20 each. Calculate (i) the cash price

(5 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

3. (b) (i) The cash price = (100 – 8)% of $25 600

Revision Exam 2 Paper 2

3. (b) A used car is advertised for sale at $25 600. A discount of 8% is given if it is bought for cash. It can also be bought on hire purchase by paying a deposit of $2 304 followed by 32 monthly payments of $837.20 each.

(ii) the hire purchase price

ANSWER

Revision Exam 2 Paper 2

3. (b) (ii) The deposit = $2 304 The total monthly payments = $837.20 × 32 = $26 790.40

The hire purchase price = $(2 304 + 26 790.40) = $29 094.40

Revision Exam 2 Paper 2

3. (b) A used car is advertised for sale at $25 600. A discount of 8% is given if it is bought for cash. It can also be bought on hire purchase by paying a deposit of $2 304 followed by 32 monthly payments of $837.20 each. (iii) the amount saved by buying the car for cash rather than on hire purchase.

ANSWER

Revision Exam 2 Paper 2

3. (b) (iii) The amount saved by paying cash = $(29 094.40 – 23 552) = $5 542.40

Revision Exam 2 Paper 2

4. (a) Solve the simultaneous equations. (5 marks)

ANSWER

Revision Exam 2 Paper 2

4. (a)

× 2: – : Substitute y = 1 in : x – 2(1) = 2 x–2=2 x=2+2  x=4 Hence x = 4, y = 1.

Revision Exam 2 Paper 2

4. (b) In the figure below, not drawn to scale, ABC is an isosceles triangle with  ACB = r° and  ABC = (r + 6)°

(i) Write an expression in terms of r for the value of the angle at A. (5 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

4. (b) (i) The expression for the angel at A = (r + 6)°

Revision Exam 2 Paper 2

4. (b) In the figure below, not drawn to scale, ABC is an isosceles triangle with  ACB = r° and  ABC = (r + 6)°

(ii) Determine the size of each angle in the triangle. ANSWER

Revision Exam 2 Paper 2

4. (b) (ii)

r° + (r + 6)° + (r + 6)° = 180°

So r° + r° + r° + 6° + 6° = 180° i.e. 

(Sum of the s of a ∆)

3r° + 12° = 180° 3r° = 180° – 12° = 168°

 

(r + 6)° = (56 + 6)° = 62°

Hence, the size of each angle in the triangle are 56°, 62° and 62°.

Revision Exam 2 Paper 2

5. (a) Given that x = 4 and y = – 3, calculate the value of xy2.

(2 marks)

ANSWER

Revision Exam 2 Paper 2

5. (a) xy2 = 4(–3)2 = 4(9) = 36

Revision Exam 2 Paper 2

5. (b) Simplify

(4 marks)

ANSWER

Revision Exam 2 Paper 2

5. (b)

The LCM of 8 and 12 is 24.

Add like terms

Revision Exam 2 Paper 2

5. (c) (i) Solve the inequality:

8x – 7 – 10x > 5, where x is a real number.

(4 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

5. (c) (i) 8x – 7 – 10x > 5

8x –10x > 5 + 7 –2x > 12 2x < – 12

Add 7 to both sides.

Add like terms. Multiply by –1. Divide both sides by 2. Simplify.

x<–6

Revision Exam 2 Paper 2

5. (c) (ii) Show the solution set for the inequality on a number live.

ANSWER

Revision Exam 2 Paper 2

5. (c) (ii) x < – 6

Revision Exam 2 Paper 2

6. (a) The diagram below shows the line l, which passes through the points R(0, –2) and S(4, 2) (i) Determine the gradient of the line, l.

(4 marks)

ANSWER

Revision Exam 2 Paper 2

6. (a)

(i) Complete a right-angled triangle RST. The gradient of the line,

Revision Exam 2 Paper 2

6. (a) The diagram below shows the line l, which passes through the points R(0, –2) and S(4, 2) (ii) Write down the equation of the line, l.

ANSWER

Revision Exam 2 Paper 2

6. (a) (ii) The y-intercept, c = –2 The equation of the line, l is:

i.e.

y = mx + c Substitute the value for m and for c. y = 1x – 2



y=x–2

Revision Exam 2 Paper 2

6. (b) The table below shows three of the values

of f(x) = x2 – 3x + 1 for values of x from 0 to

. x

0

f(x)

1

1

(i) Copy the table and insert the missing values of f(x).

2

3

–1

(6 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

6. (b) (i) f(x) = x2 – 3x + 1 f(1) = 12 – 3(1) + 1 =1–3+1 = –1 f(3) = 32 – 3(3) + 1 =9–9+1 =1 The completed table for f(x) is shown below. x

0

1

2

3

f(x)

1

–1 –1 1

Revision Exam 2 Paper 2

6. (b) The table below shows three of the values of f(x) = x2 – 3x + 1 for values of x from 0 to.

x

0

f(x)

1

1

2

3

–1

(ii) Using the same graph paper, scales and axes as 6(a) above, draw the graph of f(x) = x2 – 3x + 1. (iii) Using the graphs, write down the coordinates of the points of intersection of the line, l, and the graph of f(x).

ANSWER

Revision Exam 2 Paper 2

6. (b) (ii) The graph of f(x) = x2 – 3x + 1 was drawn on the graph paper as shown above. (iii) The coordinates of the points of intersection of the line, l, and the graph of f(x) are: (1, –1) and (3, 1).

Revision Exam 2 Paper 2

7. (a) A car left Town X at 08:30 am to travel to Town Y which is 36 km away. The car arrived at Town Y at 09:15 am. (i) How long did the car take to travel from Town X to Town Y? (ii) Calculate the average speed, in km h–1, of the car for the journey.

(4 marks)

ANSWER

Revision Exam 2 Paper 2

7. (a) (i) The time taken to travel from Town X to Town Y

45 minutes

(ii) The distance between the two towns = 36 km The average speed of the car,

8

Revision Exam 2 Paper 2

7. (b) The diagram below, not drawn to scale, shows a container in the shape of a rectangular prism.

The base of the container has a length of 95 cm and a width of 50 cm.

(i) Calculate the area, in cm2, of the base of the container. Water is poured into the container, reaching a height of 20 cm. (6 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

7. (b) (i) The area of the base of the container, A = lb = 95 cm × 50 cm = 4 750 cm2

Revision Exam 2 Paper 2

7. (b) The diagram below, not drawn to scale, shows a container in the shape of a rectangular prism.

The base of the container has a length of 95 cm and a width of 50 cm. (ii) Calculate, in cm3, the volume of water in the container. (iii) If the container holds 114 litres when full, calculate the height, h, in cm, of the water when the container is full. ANSWER

Revision Exam 2 Paper 2

7. (b) (ii) The volume of water in the rectangular prism, V = Ah = 4 750 cm2 × 20 cm = 95 000 cm3 = 95 l (iii)



Hence, the height of the water when the container is full is 24 cm.

Revision Exam 2 Paper 2

8. (a) (i) Using a pencil, a ruler and a pair of compasses, construct rectangle KLMN, in which LM = 8 cm, angle LMN = 90° and MN = 6.5 cm. (ii) Measure and write down the length of the diagonal LN, in cm.

(5 marks)

ANSWER

Revision Exam 2 Paper 2

8. (a) (i)

The rectangle KLMN was constructed as shown above. (ii) The length of the diagonal LN = 10.3 cm

Revision Exam 2 Paper 2

8. (b) In the figure below, not drawn to scale, triangles BCD and ACE are right-angled triangles. ACE is an enlargement of BCD with centre C. BC = 8 cm, CD = 6 cm and AE = 50 cm.

(i) Calculate, in cm, the length of the side BD. (ii) Determine the scale factor of the (5 marks) enlargement. Total 10 marks

ANSWER

Revision Exam 2 Paper 2

8. (b) (i) Using Pythagoras’ theorem: BD2 = BC2 + CD2 = 82 + 62 = 64 + 36 = 100 = 10 cm

Hence, the length of the side BD is 10 cm. (ii) The scale factor of the enlargement,

Revision Exam 2 Paper 2

9. (a) The point P(–3, 8) in translated by the column vector

to give the image P' (7, 20).

Calculate the values of x and y.

(2 marks)

ANSWER

Revision Exam 2 Paper 2

9. (a)

Hence, x = 10 and y =12.

P

Revision Exam 2 Paper 2

9. (b) The diagram below, not drawn to scale, shows a vertical flagpole, GF, which is kept in place by two ropes, GM and GN, fixed at M and N on level horizontal ground. MFN is a straight line.

The length of GM is 12 m and the length of GN is 18 m. The angle of elevation of G from N is 35°. Calculate, correct to three significant figures. (4 marks) (i) the length, in metres, of the flagpole

ANSWER

Revision Exam 2 Paper 2

9. (b) (i)

sin 35° =  GF = 18 m × sin 35° = 18 m × 0.576 = 10.368 m = 10.4 m (correct to 3 s.f.)

Revision Exam 2 Paper 2

9. (b) The diagram below, not drawn to scale, shows a vertical flagpole, GF, which is kept in place by two ropes, GM and GN, fixed at M and N on level horizontal ground. MFN is a straight line.

The length of GM is 12 m and the length of GN is 18 m. The angle of elevation of G from N is 35. (ii) the angle of elevation of G from M (4 marks) Total 10 marks

ANSWER

Revision Exam 2 Paper 2

9. (ii)



Revision Exam 2 Paper 2

10. The bar chart shows the shirt sizes of a group of 80 children.

Revision Exam 2 Paper 2

10. (a) How many children wear a size twelve shirt?

(1 mark)

(b) How many children wear a shirt size smaller than size twelve?

(1 mark)

(c) Which shirt size is the modal shirt size?

(1 mark)

(d) Which shirt size is the median shirt size?

(2 marks)

ANSWER

Revision Exam 2 Paper 2

10. (a) The number of children who wear a size twelve shirt = 15 (b) The number of children who wear a shirt size smaller than size twelve = 12 + 25 = 37 (c) The modal shirt size = 11

(d) The position of the median shirt size = 

the median shirt size, Q2

(80 + 1)

= 40.5th = 12

Revision Exam 2 Paper 2

10. The bar chart shows the shirt sizes of a group of 80 children. (e) What is the probability that a child selected at random wears: (i) a shirt size of 11? (ii) a shirt size larger than 12?

(3 marks)

ANSWER

Revision Exam 2 Paper 2

10. (e) (i) The number of children who wear a shirt size of 11 = 25  P (x wears a shirt size of 11) (ii) The number of children who wear a shirt size larger than 12

 P (x wears a shirt size larger than 12)

= 20 + 8 = 28

Revision Exam 2 Paper 2

10. The bar chart shows the shirt sizes of a group of 80 children. (f) Which of these two averages, the mode and the median, would be of greater interest to the owner of a variety store who wishes to stock up on children’s shirts? Give a reason for your answer. (2 marks) (Total 10 marks)

ANSWER

Revision Exam 2 Paper 2

10. (f) The mode would be of greater interest to the owner of a variety store who wishes to stock up on children’s shirts. The mode occurs the most frequent.

It represents the shirt size that is sold the most.

REVISION EXAMINATION 3 MATHEMATICS Paper 1 90 minutes Answer ALL the questions

NEXT

Revision Exam 3 Paper 1

1. Which of the following numbers is the closest approximation for 5 150  3.95? (A) 15 000

(B) 18 000 (C) 20 000 (D) 23 000 ANSWER

Revision Exam 3 Paper 1

1. 5 150  3.95  5 150  4

= 20 600  20 000

Revision Exam 3 Paper 1

2. What is 10 per cent of 25 per cent of $3 500? (A) $87.50 (B) $262.50

(C) $875.00 (D) $5 250.00 ANSWER

Revision Exam 3 Paper 1

2.

Revision Exam 3 Paper 1

3. When expressed to the nearest hundred, 7 584 becomes (A) 7 400 (B) 7 500

(C) 7 600 (D) 7 700 ANSWER

Revision Exam 3 Paper 1

3.

The tens digit 8 is greater than 5, so we add 1 to the hundreds digit 5.

Revision Exam 3 Paper 1

4. What is the value of

in its simplest form?

ANSWER

Revision Exam 3 Paper 1

4.

Add the numbers in the brackets. Simplify as the difference of two squares. Cancel common factors.

Revision Exam 3 Paper 1

5. (0.2 + 0.05) (0.2 – 0.05) = (A) 0.375 (B) 0.0375

(C) 0.625 (D) 0.0625 ANSWER

Revision Exam 3 Paper 1

5. (0.2 + 0.05) (0.2 – 0.05) = 0.25  0.15 = 0.0375 2 d.p. + 2 d.p. = 4 d.p.

Revision Exam 3 Paper 1

6. Which of the following is between

and

?

ANSWER

Revision Exam 3 Paper 1

6.

Revision Exam 3 Paper 1

7. If 99 is added to each of the primes 31 and 37, which of the following is true of the resulting numbers? (A) Both are prime (B) Both are composite (C) The smaller is composite and the larger is prime (D) The smaller is prime and the larger is composite ANSWER

Revision Exam 3 Paper 1

7. 31 + 99 = 130 which is composite. 37 + 99 = 136 which is composite.

Revision Exam 3 Paper 1

8. In the numeral 846p, the place value of the digit 8 is (A) p0 (B) p1

(C) p2 (D) p3 ANSWER

Revision Exam 3 Paper 1

8. 846p = 8  p2 + 4  p1 + 6  p0

Place values Or Place value Digit

p2 8

p1 4

p2 is the place value of the digit 8.

p0 6

Revision Exam 3 Paper 1

9. Which of the following is not a rational number?

ANSWER

Revision Exam 3 Paper 1

9.

is not a rational number, since its denominator is 0.

Revision Exam 3 Paper 1

10. Which of the following sets has an infinite number of members? (A) {prime numbers between 20 and 30} (B) {odd numbers between 20 and 30}

(C) {multiples of 30} (D) {factors of 30} ANSWER

Revision Exam 3 Paper 1

10. {multiples of 30} = 30  1, 30  2, 30  3,

Revision Exam 3 Paper 1

11. The chart below gives details of a special offer for a carnival weekend holiday getaway. Air Fare Accomodation Adults $975 each $200 each Children under or 12 years $540 each $500 for a family of 4

Revision Exam 3 Paper 1

11. A man, his wife, and their two children, aged 13 years and 11 years wish to go on the carnival weekend holiday getaway. What is the least cost to the family for the visit? (A) $3 530

(B) $3 830 (C) $3 965 (D) $4 265

ANSWER

Revision Exam 3 Paper 1

11. The cost for three adult air fares = $975  3 = $2 925 The cost for one child air fare = $540 The cost for the family accomodation = $500 the least total cost = $(2 925 + 540 + 500) = $3 965

Revision Exam 3 Paper 1

12. The cost price of a watermelon is $50 and the profit is 25 per cent of the cost price. What is the selling price of the watermelon? (A) $75.00

(B) $62.50 (C) $60.00 (D) $37.50

ANSWER

Revision Exam 3 Paper 1

12.

Revision Exam 3 Paper 1

13. How much money is due as simple interest on a loan of $580 for one year if the annual rate of interest is per cent? (A) $1.05

(B) $15.95 (C) $31.90 (D) $62.80

ANSWER

Revision Exam 3 Paper 1

13.

Revision Exam 3 Paper 1

14.

Blue-Ray Player Cash $3 700 Hire Purchase Plan: Pay down $300 $212.50 monthly for 20 months

Mr Ganga purchases the Blue-Ray Player using the hire purchase plan instead of paying cash. How much more than $3 700 does Mr Ganga pay by using the hire purchase plan? (A) $550

(B) $850

(C) $3 400

(D) $4 250

ANSWER

Revision Exam 3 Paper 1

14. The sum of the monthly installments = $212.50  20 = $4 250.00 The hire purchase plan = $(300 + 4 250) = $4 550 The amount greater than $3 700 = $(4 550 – 3 700) = $850

Revision Exam 3 Paper 1

15. If a 12 per cent purchase tax is charged, what is the total cost of a skirt marked at $140? (A) $123.20 (B) $128.00 (C) $152.00 (D) $156.80

ANSWER

Revision Exam 3 Paper 1

15. The total cost of the skirt = (100 + 12)% of $140

= 112% of $140

= $156.80

Revision Exam 3 Paper 1

16. If a customer spends more than $40, then a 10 per cent discount is given on their bill. How much is paid by a customer whose bill is $90? (A) $99 (B) $81 (C) $45 (D) $40 ANSWER

Revision Exam 3 Paper 1

16. The amount paid by the customer = (100 – 10)% of $90 = 90% of $90

= $81

Revision Exam 3 Paper 1

17. The interest rate on investments in a bank decreased from 10% per annum to

per annum. What is the decrease in the sum of the annual interest on a deposit of $5 000? (A) $175 (B) $200 (C) $325 (D) $500

ANSWER

Revision Exam 3 Paper 1

17.

= $50  3.5 = $175

Revision Exam 3 Paper 1

18. A desalination plant charges $25.00 per month for the meter rental, $30.00 for the first 1 000 litres, and $2.50 for each additional 100 litres. What is the bill for 3 500 litres of water used in one month?

(A) $57.50 (B) $62.50 (C) $90.50 (D) $117.50

ANSWER

Revision Exam 3 Paper 1

18.

The water rental

= $25.00

The cost for the first 1 000 litres = $30.00 The cost for the remaining 2 500 litres =

= $62.50 The bill for the month

= $(25.00 + 30.00 + 62.50) = $117.50

Revision Exam 3 Paper 1

19.

In the Venn diagram above, the shaded portion represents

(A) (B)

(C) (D)

ANSWER

Revision Exam 3 Paper 1

19.

The shaded region represents P′

The shaded region represents

Revision Exam 3 Paper 1

20.

In the Venn diagram above, R and S denotes the sets of elements in the oval and circle respectively. n(R) = 30 and n(S) = 25. What is the value of x? (A) 1.5 (B) 2 (C) 3

(D) 4.5

ANSWER

Revision Exam 3 Paper 1

20.

n(R) = 4x + 3x + 2x + x = 10x and n(R) = 30 Form an equation: 10x = 30  x=3 Or n(S) = 3x + 2x + 10 = 5x + 10 and n(S) = 25

Revision Exam 3 Paper 1

Form an equation:

Revision Exam 3 Paper 1

21. If A = {2, 3, 5, 7} and B = {4, 6, 8}, then

=

(A) { } (B) {2, 3, 4, 5, 6, 7, 8} (C) {2, 4, 6, 8}

(D) {3, 5, 7} ANSWER

Revision Exam 3 Paper 1

21.

= { } since there are no common elements.

Revision Exam 3 Paper 1

22. If P = {15, 17, 19, 21} and Q = {16, 18, 20}, then = (A) {19, 20, 21} (B) {15, 16, 17, 18}

(C) { } (D) {15, 16, 17, 18, 19, 20, 21} ANSWER

Revision Exam 3 Paper 1

22.

= {15, 16, 17, 18, 19, 20, 21}

Revision Exam 3 Paper 1

23.

In the figure above, PQRS is a rectangle with PQ = 7 cm and QR = 3 cm. If QSTU is a square, then its area, in cm2, is

(A) 9

(B) 49

(C) 58

(D) 100 ANSWER

Revision Exam 3 Paper 1

23.

QRS = 90 and RS = PQ = 7 cm Considering the right-angled QRS and using Pythagoras’ theorem:

Revision Exam 3 Paper 1



QS2 = QR2 + RS2 = 32 + 72 = 9 + 49 = 58 QS = cm

The area of the square QSTU, A = l2 = QS2 = = 58 cm2

Revision Exam 3 Paper 1

24.

1 cm : 25 km =

(A) 1 : 2 500 (B) 1 : 25 000 (C) 1 : 250 000 (D) 1 : 2 500 000 ANSWER

Revision Exam 3 Paper 1

24.

1 cm : 25 km = 1 cm : 25 000 m

= 1 cm : 2 500 000 cm = 1 : 2 500 000

Revision Exam 3 Paper 1

25. The width of a glass paper weight with rectangular cross-section is x cm. It height is

its width and its length is 7 times its height. What is its volume in cm3?

ANSWER

Revision Exam 3 Paper 1

25.

The width of the cuboid, b = x cm The height of the cuboid, h =

The length of the cuboid, l = 7

x cm

cm

Revision Exam 3 Paper 1

The volume of the cuboid, V = lbh

Revision Exam 3 Paper 1

26. A rectangular sheet of paper measures, to the nearest cm, 11 cm by 8 cm. What are the least possible values of its dimensions? (A) 11.0 cm by 8.0 cm (B) 11.5 cm by 8.5 cm

(C) 10.5 cm by 7.5 cm (D) 10 cm by 7 cm ANSWER

Revision Exam 3 Paper 1

26.

The least possible values of its dimensions = 10.5 cm by 7.5 cm.

Revision Exam 3 Paper 1

27.

The area of the trapezium above is 28 cm2. What is the value of x?

(A) 5 (B) 4 (C)

(D)

ANSWER

Revision Exam 3 Paper 1

27.

The area of the trapezium,

Revision Exam 3 Paper 1

Form an equation: Divide both sides by 2. 2(x + 9) = 28 Subtract 9 from both sides. So x + 9 = 14 i.e. x =14 – 9  x=5

Revision Exam 3 Paper 1

28. Which of the following shapes is a triangular pyramid

(A)

(C)

(B)

(D) ANSWER

Revision Exam 3 Paper 1

28. This shape is a triangular pyramid It has an apex and a triangular base.

Revision Exam 3 Paper 1

29.

The diagram above, shows a circle with centre O. A line OM is drawn perpendicular to the chord AB. If OA = 15 cm and AB = 24 cm, then the length, in cm, of OM is (A) 5

(B) 9

(C) 12

(D) 15

ANSWER

Revision Exam 3 Paper 1

29.

Revision Exam 3 Paper 1

Using Pythagoras’ theorem:

So

Revision Exam 3 Paper 1

30. A circular hole with diameter 5 cm is cut out from a circular piece of bristol board with a diameter of 9 cm. The area of the remaining bristol board in cm2 is (A) 4 (B) 5 (C) 9 (D) 14

ANSWER

Revision Exam 3 Paper 1

30.

A1 = r2 A2 = R2

Revision Exam 3 Paper 1

Revision Exam 3 Paper 1

Alternative Method:

Revision Exam 3 Paper 1

Revision Exam 3 Paper 1

Items 31–34 refer to the graph below. The graph shows the number of parcels for postage having a certain mass.

Revision Exam 3 Paper 1

31. How many parcels were awaiting postage?

(A) 91 (B) 67 (C) 55 (D) 9 ANSWER

Revision Exam 3 Paper 1

31. The total number of parcels, n = 5 + 8 + 10 + 15 + 23 + 17 + 7 + 4 + 2 = 91

Revision Exam 3 Paper 1

32. What is the frequency of the median mass?. (A) 1 (B) 5 (C) 15 (D) 23 ANSWER

Revision Exam 3 Paper 1

32.

f

5 8 10 15 23 17 7 4 2 n = 91

46th

Revision Exam 3 Paper 1

The number of parcels,

n = 91

The position of the median

= 46th

The frequency of the median mass = 23

Revision Exam 3 Paper 1

33. What is the median mass? (A) 6.0 kg (B) 6.1 kg (C) 6.2 kg (D) 6.3 kg ANSWER

Revision Exam 3 Paper 1

33. The position of the median mass = 46th  the median mass, Q2 = 6.1 kg

Q2 =

x (kg)

f

6.1

23

46th

Revision Exam 3 Paper 1

34. What is the modal mass? (A) 6.5 kg (B) 6.2 kg (C) 6.1 kg

(D) 5.7 kg ANSWER

Revision Exam 3 Paper 1

34.

The highest frequency = 23  the modal mass = 6.1 kg x (kg)

Modal mass = 6.1

f 23

Highest frequency

From the construction on the histogram, the median mass, Q2 = 6.1 kg

Revision Exam 3 Paper 1

35.

The weekly wages of 10 factory workers are $600, $700, $400, $500, $600, $800, $900, $700, $800 and $900. How many of these wages are below the mean wage for the group?

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

ANSWER

Revision Exam 3 Paper 1

35.

The sum of the wages = $(600 + 700 + 400 + 500 + 600 + 800 + 900 + 700 + 800 + 900) = $6 900 The mean wage = $690 The wages less than $690 are $600, $400, $500 and $600  the number of wages below the mean wage = 4

Revision Exam 3 Paper 1

36.

The pie chart shows how a student spent 20 hours each week studying Mathematics (M), Physics (P), Chemistry (C) and Biology (B). The number of hours spent studying Chemistry is approximately (A) 1.25

(B) 3.75

(C) 6.25

(D) 10

ANSWER

Revision Exam 3 Paper 1

36. So i.e. 

Revision Exam 3 Paper 1

37. Watermelons are sold at $p per kg. The total mass of 4 watermelons is 25 kg. What is the average cost of one watermelon? (A)

(B) (C) $4 p (D) $100 p

ANSWER

Revision Exam 3 Paper 1

37.

The cost of the 4 watermelons = $p  25  the average cost of one watermelon

= $25 p

=

Revision Exam 3 Paper 1

38. Given that –4x –5 = –7, then x = (A)

(B) (C) –3 (D) 3 ANSWER

Revision Exam 3 Paper 1

38. –4x –5 = –7 So –4x = –7 + 5 = –2 

Add 5 to both sides. Divide both sides by –4.

Revision Exam 3 Paper 1

39. Five times the product of two numbers p and q may be written as (A) 25pq

(B) 5p + 5q (C) 5p + q (D) 5pq

ANSWER

Revision Exam 3 Paper 1

39. The product of two numbers p and q = pq

Five times the product of two numbers p and q

= 5pq

Revision Exam 3 Paper 1

40. 7(x + y) – 2(x – y) = (A) 5xy (B) 5x + 5y

(C) 5x + 9y (D) 9x + 9y

ANSWER

Revision Exam 3 Paper 1

40.

7(x + y) – 2(x – y) Use the distributive law. = 7x + 7y – 2x + 2y Group like terms. = 7x – 2x + 7y + 2y Add like terms. = 5x + 9y

Revision Exam 3 Paper 1

41. If 3(x – 1) –4x = 2, then x = (A) 5 (B)

(C) –5 (D)

ANSWER

Revision Exam 3 Paper 1

41. 3(x – 1) – 4x = 2

3x – 3 – 4x = 2 So i.e. 

Use the distributive law.

Add 3 to both sides.

3x – 4x = 2 + 3 Add like terms. –x = 5 x = –5

Multiply by –1

Revision Exam 3 Paper 1

42. If m = – 3 and p = 2, then

ANSWER

Revision Exam 3 Paper 1

42.

Substitute –3 for m and 2 for p

Revision Exam 3 Paper 1

43. 5 k3l2  2 kl3 =

(A) 7 k4l5 (B) 10 k4l5 (C) 7 k3l6 (D) 10 k3l6 ANSWER

Revision Exam 3 Paper 1

43.

5 k3l2  2 kl3

Group like letters.

= 5  2  k3  k  l2  l3 Add powers with the same base. 3+1 2+3 = 10  k  l = 10  k4  l5

= 10k4l5

Revision Exam 3 Paper 1

44.

ANSWER

Revision Exam 3 Paper 1

44.

The LCM of the denominators 3 and 6 is 6.

Using the distributive law. Grouping like terms. Adding like terms.

Revision Exam 3 Paper 1

45.

ANSWER

Revision Exam 3 Paper 1

45.

The LCM of the denominators 2 and 5 is 10.

Invert the fractions.

Revision Exam 3 Paper 1

46. Which of the following relations represents a one-to-one mapping? (A) (B)

(C)

(D)

ANSWER

Revision Exam 3 Paper 1

46.

The diagram above represents a one-to-one mapping.

Revision Exam 3 Paper 1

47. If 2y = –4x + 3, then the y-intercept is (A) –2 (B) 2 (C) (D) ANSWER

Revision Exam 3 Paper 1

47. 2y = –4x + 3

The y-intercept,

Revision Exam 3 Paper 1

48. If f(x) = 3x2 + 2x – 4, then f(–3) = (A) –37 (B) 17 (C) 31

(D) 33 ANSWER

Revision Exam 3 Paper 1

48. If

f(x) = 3x2 + 2x – 4

then f(–3) = 3(–3)2 + 2(–3) –4 = 3(9) – 6 –4

= 27 – 10 = 17

Revision Exam 3 Paper 1

49.

Which of the following relations is represented by the graph shown above? (A) 2x + 3y + 12 = 0

(B) –2x + 3y + 12 = 0

(C) 2x – 3y + 12 = 0

(D) 2x + 3y – 12 = 0

ANSWER

Revision Exam 3 Paper 1

49.

The gradient of the line, The y-intercept, c = –4 The equation of a straight line: y = mx + c becomes 3: 3y = 2x – 12  –2x + 3y +12 = 0

Revision Exam 3 Paper 1

50. If 3x – 2  5x + 4, then (A) x  –3 (B) x  –3 (C) x < 3

(D) x > 3

ANSWER

Revision Exam 3 Paper 1

50.

3x – 2  5x + 4

Group like terms.

3x – 5x  4 + 2

Add like terms.

–2x  6

Multiply by –1.

So

2x  –6

i.e.



x  –3

Divide by 2.

Revision Exam 3 Paper 1

51. If the altitude, H, of a triangle is 5 cm more than twice its width, W, then the relation between L and W is (A) H = 2W + 5

(B) H > 2W + 5 (C) 2H + 5 = W (D) H + 3 > 2W ANSWER

Revision Exam 3 Paper 1

51. Twice the width of the triangle = 2W 5 cm more that twice the width of the rectangle = 2W + 5 The relation is:

H = 2W + 5

Revision Exam 3 Paper 1

52. The size of the exterior angle of a regular pentagon is

(A) 36 (B) 72 (C) 108 (D) 144 ANSWER

Revision Exam 3 Paper 1

52. Each exterior angle =

= 72

Revision Exam 3 Paper 1

53.

LMN is rotated anti-clockwise about L through 90 degrees. Which of the following diagrams is its likely image? (A)

(B)

(C)

(D)

ANSWER

Revision Exam 3 Paper 1

53.

The angle of rotation,

Revision Exam 3 Paper 1

54.

In the figure above, ABCD is a parallelogram. Which triangle is similar to QRC? (A) DCA

(B) BRA

(C) PDQ

(D) PAR

ANSWER

Revision Exam 3 Paper 1

54.

QRC is similar to BRA since: QRC = BRA vertically opposite angels QCR = BAR alternate s

CQR = ABR alternate s

Revision Exam 3 Paper 1

55.

A is due north of B, C is west of A, and AB = AC. What is the bearing of B from C? (A) 045

(B) 135

(C) 225

(D) 315

ANSWER

Revision Exam 3 Paper 1

55.

ABC is isosceles since AB = AC. So

NCB = 90 + 45 = 135

 the bearing of B from C is 135.

Revision Exam 3 Paper 1

56. In a regular polygon, the interior angle is twice the exterior angel. How many sides does the polygon have? (A) 8 (B) 7

(C) 6 (D) 5 ANSWER

Revision Exam 3 Paper 1

56.

2x + x = 180 So 3x = 180

Sum of s on a straight line. Divide by 3.

 = 60 The number of sides of the regular polygon =

Revision Exam 3 Paper 1

57.

In the diagram above, which of the following is true?

ANSWER

Revision Exam 3 Paper 1

57. Items 58–59 refer to the diagram below.

The diagram above, shows a circle with centre O. A line OM is drawn perpendicular to AB so that AO = 15 cm and AB = 18 cm.

Revision Exam 3 Paper 1

58. The length, in cm, of OM is (A) (B) 9 (C) 12 (D) 15 ANSWER

Revision Exam 3 Paper 1

58.

Using Pythagoras’ theorem:

AO2 = OM2 + AM2 152 = OM2 + 92 225 = OM2 + 81

Revision Exam 3 Paper 1

225 – 81 = OM2 144 = OM2 So

OM2 =144



OM2 =

cm

= 12 cm

Revision Exam 3 Paper 1

59. sin AÔM =

ANSWER

Revision Exam 3 Paper 1

59.

Revision Exam 3 Paper 1

60.

In the diagram, the image of triangle ABC under a transformation is triangle ABC. The transformation is

Revision Exam 3 Paper 1

60.

(A) an enlargement, centre O, with scale factor –1 (B) an anti-clockwise rotation of 270 about O (C) a translation with (D) a reflection in the line ANSWER

Revision Exam 3 Paper 1

60.

CN = CN. Object distance = Image distance The transformation is a reflection in the line and the broken line drawn is perpendicular to the line CC.

REVISION EXAMINATION 3 MATHEMATICS Paper 2 2 hours 40 minutes Answer ALL the questions

NEXT

Revision Exam 3 Paper 2

1. (a) Using a calculator, or otherwise, find the value of 7.38 × 1.042, writing your answer (i) exactly

(ii) correct to two significant figures

(3 marks)

ANSWER

Revision Exam 3 Paper 2

1. (a) (i) 7.38 × 1.042 = 7.982 208

(exactly)

(ii) 7.98 | 2 208 = 7.98 (correct to 2 s.f ) The digit 2, in the fourth significant figure, is less than five, so we do not add 1 to the third significant figure.

Revision Exam 3 Paper 2

1. (b) A sum of money was shared between Amy and Boris in the ratio 3:5. If Boris received $98 more than Amy, how much money was (3 marks) shared?

ANSWER

Revision Exam 3 Paper 2

1. (b) The total number of shares = 3 + 5 = 8 Boris’ share – Amy’s share = 5 – 3 = 2 So

2 shares = $98



8 shares

= $98 × 4 = $392 Hence, the sum of $392 was shared.

Revision Exam 3 Paper 2

1. (c) Each student in a class plays one sport, cricket, football or basketball. of the students play cricket and of them play basketball. Calculate (i) the fraction of the students in the class who play cricket and basketball (ii) the percentage that plays (4 marks) football. Total 10 marks

ANSWER

Revision Exam 3 Paper 2

1. (c) (i) The fraction of the students in the class who play cricket and basketball =

Revision Exam 3 Paper 2

1. (c) (ii) The fraction of the students who play football

Revision Exam 3 Paper 2

The percentage of the students who play football

Revision Exam 3 Paper 2

2. (a) Simplify completely: (i) (ii) (iii)

(4 marks)

ANSWER

Revision Exam 3 Paper 2

2. (a) (i)

(ii) (iii)

(Group like terms) (Add power with like base) (Subtract the power of the denominator from the power of the numerator). Use the distributive law to remove the brackets. Group and add like terms.

Revision Exam 3 Paper 2

2. (b) Given that of

calculate the value (3 marks)

ANSWER

Revision Exam 3 Paper 2

2. (b)

Revision Exam 3 Paper 2

2. (c) (i) Solve the inequality:

(ii) Given that x is a whole number, what is the largest possible value of x?

(3 marks) Total 10 marks

ANSWER

Revision Exam 3 Paper 2

2. (c) (i) 19 + 8x > 5 + 15x

Group like terms. Add like terms. Divide both sides by 7. Simplify.

 (ii) Since x < 2, and x is a whole number, then the largest possible value of x is 1.

Revision Exam 3 Paper 2

3. (a) The cash price of a cellphone is $1 240. It can be bought on hire purchase for a deposit of $120 plus 18 monthly payments of $75. Calculate the difference between the cash (3 marks) price and the hire purchase price.

ANSWER

Revision Exam 3 Paper 2

3. (a)

The deposit = $120 The monthly payments = $75 × 18 = $1 350 The hire purchase price = $(120 + 1 350) = $1 470

The difference between the cash price and the hire purchase price = $(1 470 – 1 240) = $230

Revision Exam 3 Paper 2

3. (b) The property tax for a property (house and land) are charged as follows: Land: Fixed charge of $250 House: 5% of the rateable value of the house Calculate the total property tax on a property if the house is valued at $150 000. (3 marks) ANSWER

Revision Exam 3 Paper 2

3. (b) The land tax The house tax

= $250 = 5% of $150 000

= $7 500 The total property tax = $(250 + 7 500) = $7 750

Revision Exam 3 Paper 2

3. (c) $8 000 was deposited in a savings account which offers 5% compound interest per annum. Calculate the total amount in the (4 marks) account after two years. Total 10 marks

ANSWER

Revision Exam 3 Paper 2

3. (c)

The amount in the account after one year = 105% of $8000

= $8 400 The amount in the account after two years = 105% of $8400

= $8 820

Revision Exam 3 Paper 2

4. (a) The diagram below is a map of a town showing, a school, Movietown and a Church.

1 cm on the map represents an actual distance of 2 km.

(3 marks)

Revision Exam 3 Paper 2

4. (a) The distance from the Church to Movietown, as shown on the map, is 4.9 cm. (i) Calculate the actual distance from the Church to Movietown. The actual distance from Movietown to the School is 14 km. (ii) What length, on the map, would be used to represent this distance? ANSWER

Revision Exam 3 Paper 2

4. (a) (i) The actual distance from the Church to Movietown = 4.9 × 2 km = 9.8 km (ii) The length on the map from Movietown to the School =

= 7 cm

Revision Exam 3 Paper 2

4. (b) The diagram below, not drawn to scale, shows a square PQRS joined to a triangle QRT. PS = 12 cm and PT = 21 cm.

(i) Write, in cm, the length of QT.

(7 marks) Total 10 marks

ANSWER

Revision Exam 3 Paper 2

4. (b) (i) The length of QT = PT − PQ = (21 − 12) cm

= 9 cm

Revision Exam 3 Paper 2

4. (b) The diagram below, not drawn to scale, shows a square PQRS joined to a triangle QRT. PS = 12 cm and PT = 21 cm. (ii) Calculate the area, in cm2, of a) QRT b) PQRS

ANSWER

Revision Exam 3 Paper 2

4. (b) (ii) a) The area of ∆ QRT, A =

Revision Exam 3 Paper 2

4. (b) (ii) b) The area of the square PQRS, A = l2 = (12 cm2) = 144 cm2 the area of the trapezium PTRS = (54 + 144) cm2 = 198 cm2

Revision Exam 3 Paper 2

Alternatively: b) The area of trapezium

Revision Exam 3 Paper 2

5. (a) Mr Arjoon works as a salesman for a basic wage of $540 for a 45-hour week.

(i) Calculate the basic hourly rate of pay. The overtime hourly rate of pay is twice the basic hourly rate. (ii) Calculate the overtime hourly rate of pay. During a certain week Mr Arjoon worked 60 hours. (iii) Calculate the total wage earned for the (5 marks) week.

ANSWER

Revision Exam 3 Paper 2

5. (a) (i) The basic hourly rate of pay

= = $12

(ii) The overtime hourly rate of pay

(iii) The overtime wage

= $12 × 2 = $24

= $24 × 15 = $360

The total wage earned for the week = $(540 + 360) = $900

Revision Exam 3 Paper 2

5. (b) During a sale, Dravid bought 5 books at $30 and 4 magazines priced at $20 each from a bookstore. (i) Calculate the total cost of the a) books b) magazines

The bookstore offered a discount of 9% on all books and magazines. (ii) Calculate the total amount Dravid actually paid for his purchases. (5 marks)

ANSWER

Revision Exam 3 Paper 2

5. (b) (i) a) The total cost of the books = $30 × 5 = $150 b) The total cost of the magazines = $20 × 4 = $80

Revision Exam 3 Paper 2

5. (b) (ii) The total cost of the books and magazines = $(150 + 80) = $230 The total amount Dravid actually paid for his purchases = (100 − 9)% of $230 = 91% of $230 = 0.91× $230 = $209.30

Revision Exam 3 Paper 2

6. (a) Solve the following simultaneous equations x + 3y = 2 2x + y = −6

(5 marks)

ANSWER

Revision Exam 3 Paper 2

6. (a)  × 3:  – 1: ÷ 5:

x + 3y 2x + y 6x + 3y 6x – x + 3y – 3y 5x

=2 — = –6 — = –18 — = –18 – 2 = –20

 Substitute

 Hence, x = –4 and y = 2.

x= –4 in :

Revision Exam 3 Paper 2

6. (b) A graph book costs g dollars and a marker costs 13 dollars more than the graph book. (i) Write an expression, in g, for a) the cost of a marker b) the total cost of a graph book and a marker.

The total cost of the graph book and the marker is 26 dollars. (ii) Calculate the value of g

(5 marks) Total 10 marks

ANSWER

Revision Exam 3 Paper 2

6. (b) (i) a) The cost of a marker = (g +13) dollars b) The total cost of a graph book and a marker = (g + g + 13) dollars = (2g + 13) dollars

(ii) The equation is: So  Hence, the value of g is 6.50.

Revision Exam 3 Paper 2

7. (a) Triangle ABC is an isosceles triangle, where AB = 5.6 cm and AC = BC = 7 cm. (i) Using only a pencil, a ruler and a pair of compasses, construct triangle ABC

(ii) Measure and write down the size of the angle BAC

(4 marks)

ANSWER

Revision Exam 3 Paper 2

7. (a) (i)

The constructed triangle ABC can be seen above. (ii) The size of angle BAC = 62.3°

Revision Exam 3 Paper 2

7. (b) The diagram below, not drawn to scale, shows two right-angled triangles, WXY and XYZ joined along the side XY. WY = 10 cm, WX = 24 cm and XYZ = 60°.

Calculate the length, in cm, of the side (6 marks) (ii) XZ (i) XY Total 10 marks

ANSWER

Revision Exam 3 Paper 2

7. (b) (i)

Using Pythagoras’ theorem:

Revision Exam 3 Paper 2

7. (b) (ii)

Hence, the length of XZ is 22.5 cm.

Revision Exam 3 Paper 2

8. (a) The diagram shows the distance-time graphs of the journeys of Maria and Albert between two towns, A and B.

Revision Exam 3 Paper 2

8. (a)

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions:

(i) What is the distance between the two towns? (1 mark)

ANSWER

Revision Exam 3 Paper 2

8.

(a)

(i) The distance between the two towns = 100 km

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions:

(ii) How long did Maria take to travel from Town B to (1 mark) Town A? ANSWER

Revision Exam 3 Paper 2

8.

(a)

(ii) The length of time Maria take to travel from Town B to Town A = 5 hours

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions:

(iii) How long did Albert take to travel from Town A to (1 mark) Town B?

ANSWER

Revision Exam 3 Paper 2

8.

(a)

(iii) The length of time Albert take to travel from Town A to Town B = 4 hours

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions: (iv) What is the meaning of the straight line PQ? (1 mark)

ANSWER

Revision Exam 3 Paper 2

8.

(a)

(iv) The straight line PQ means that Maria rested for 1 hour.

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions: (v) Calculate Albert’s average speed in kmh−1. (1 mark)

ANSWER

Revision Exam 3 Paper 2

8.

(a)

(v) Albert’s average speed

Revision Exam 3 Paper 2

8.

(a)

Use the distance-time graph to answer the following questions: (vi) Calculate Maria’s average speed in kmh−1. (1 mark)

ANSWER

Revision Exam 3 Paper 2

8.

(a)

(vi) Maria’s average speed

Revision Exam 3 Paper 2

8. (b) The line PQ passes through the point (3, 5) and is parallel to the line y = −2x + 1. (i) State the gradient of PQ.

(ii) Determine the equation of the line PQ.

(4 marks) Total 10 marks

ANSWER

Revision Exam 3 Paper 2

8. (b) (i) y = −2x + 1 is in the form y = mx + c  m = −2. Hence, the gradient of PQ is −2.

Revision Exam 3 Paper 2

8. (b) (ii) Using m = −2 and the point (3, 5) then y = mx + c becomes

The equation of the line PQ is: y = −2x + 11.

Revision Exam 3 Paper 2

9. The figure below shows triangle KLM and its image K′L′M′ after undergoing a certain transformation.

Revision Exam 3 Paper 2

9.

Revision Exam 3 Paper 2

9.

(a)

Write down the coordinates of the points K and K′.

(2 marks)

ANSWER

Revision Exam 3 Paper 2

9.

(a)

K is the point (1, 1) K′ is the point (4, −5)

Revision Exam 3 Paper 2

9.

(b)

A translation maps KLM to K′L′M′. State the column vector that represents this translation.

(2 marks)

ANSWER

Revision Exam 3 Paper 2

9.

(b)

The column vector that represents the translation

Revision Exam 3 Paper 2

9.

(c)

Draw and label K″L″M″, the image of KLM after an enlargement centre (0, 0), scale factor 2.

(4 marks)

ANSWER

Revision Exam 3 Paper 2

9.

(c)

K″L″M″ can be seen drawn and labelled in the diagram above.

Revision Exam 3 Paper 2

9.

(d)

The area of KLM is 3 square units. Find the area of triangle K″L″M″.

(2 marks) Total 10 marks

ANSWER

Revision Exam 3 Paper 2

9.

(d)

The area of triangle K″L″M″

= k2 × The area of triangle KLM

Revision Exam 3 Paper 2

10. (a) The pie chart below, not drawn to scale, shows the favourite music of a group of 80 students. The angle representing Soca is 90° and the angle representing Calypso is 144°.

(5 marks)

Revision Exam 3 Paper 2

10. (a)

Calculate (i) the number of students whose favourite music is Soca (ii) the percentage of students whose favourite music is Chutney.

ANSWER

Revision Exam 3 Paper 2

10. (a) (i) The number of students whose favourite music is Soca

Revision Exam 3 Paper 2

10. (a) (ii) The sector angle that represents Chutney = 360° – (90° + 144°) = 360° – 234° = 126° The number of students whose favourite music is Chutney = 28

Revision Exam 3 Paper 2

10. (a) (ii) The percentage of students whose favourite music is Chutney

= 35% Alternatively: (ii) The percentage of students whose favourite music is Chutney = 35%

Revision Exam 3 Paper 2

10.

(b)

The masses, in kg, of 20 fish are shown below. 1 5 3 4

2 1 4 3

3 5 1 2

2 1 4 3

4 5 4 6

(i) Copy and complete the frequency table below to represent the information given above. (5 marks) Total 10 marks

Mass

Frequency

1 2 3 4 5 6

4 3

Revision Exam 3 Paper 2

10.

(b)

The masses, in kg, of 20 fish are shown below. 1 5 3 4

2 1 4 3

3 5 1 2

2 1 4 3

4 5 4 6

(ii) Using the frequency table, determine

(a) the median mass of the fish (b) the probability that a fish chosen at random has a mass greater than 4 kg.

ANSWER

Revision Exam 3 Paper 2

10. (b) (i)

Mass

Frequency

1

4

2

3

3

5



 Highest frequency

5

3

6

1

The completed frequency table is shown above.

Revision Exam 3 Paper 2

10. (b) (ii) (a) The highest frequency is 5, so the median mass of the fish is 4 kg.

(b) The number of fish with a mass greater than 4 kg = 3 + 1 = 4  P (fish has a mass greater than 4 kg)

REVISION EXAMINATION 4 MATHEMATICS Paper 1 90 minutes Answer ALL the questions

NEXT

Revision Exam 4 Paper 1

1. –8 – (–3) = (A) –11 (B) –5 (C) 5 (D) 11 ANSWER

Revision Exam 4 Paper 1

1. –8 – (–3) = –8 + 3 = –5

Revision Exam 4 Paper 1

2. The decimal equivalent of

is

(A) 0.007 (B) 0.07 (C) 0.7 (D) 7.0

ANSWER

Revision Exam 4 Paper 1

2.

Revision Exam 4 Paper 1

3. 0.053 47 in standard form to 3 significant figures is (A) 5.34  10–1 (B) 5.34  10–2 (C) 5.35  10–1 (D) 5.35  10–2 ANSWER

Revision Exam 4 Paper 1

3.

in standard form in standard form to 3 significant figures.

Revision Exam 4 Paper 1

4. (A) 0.0117 (B) 0.117 (C) 1.17 (D) 11.7 ANSWER

Revision Exam 4 Paper 1

4.

Revision Exam 4 Paper 1

5. (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

5. Add the whole numbers and the fractions separately.

The LCM of 5 and 10 is 10.

Revision Exam 4 Paper 1

6. (A)

(B)

(C)

(D)

3 ANSWER

Revision Exam 4 Paper 1

6.

3

Revision Exam 4 Paper 1

7.

In ascending order:

Revision Exam 4 Paper 1

8. In the numeral 327nine, the value of the digit 3 is (A) 9

(B) 27

(C) 81

(D) 243

ANSWER

Revision Exam 4 Paper 1

8.

Place value

92

91

90

Face value

3

2

7

The value of the digit 3 = 92  3 = 81  3 = 243

Revision Exam 4 Paper 1

9. Which of the following is a prime number? (A) 39

(B) 40

(C) 41

(D) 42

ANSWER

Revision Exam 4 Paper 1

9. 41 is a prime number. It is only divisible by 1 and itself, that is, 41.

Revision Exam 4 Paper 1

10. Three disco lights flash at intervals of 5, 10 and 15 seconds respectively. The lights flash together. How long after will the lights next flash together again? (A) 5 seconds

(B) 10 seconds

(C) 15 seconds

(D) 30 seconds

ANSWER

Revision Exam 4 Paper 1

10.

2

5, 10, 15

3

5, 5, 15

5

5, 5, 5

1, 1, 1 The LCM of 5, 10 and 15 = 2  3 5 = 30 30 seconds after the lights will next flash together again.

Revision Exam 4 Paper 1

11. $x are divided among three students, Peter, James and John, in the ratio 3:5:7 respectively. If John received $80 more than Peter, what is the value of x? (A) 180

(B) 300

(C) 600

(D) 1 200

ANSWER

Revision Exam 4 Paper 1

11. The total number of proportional parts = 3 + 5 + 7 = 15 (7 – 3) proportional parts = $80 So

4 proportional parts = $80

i.e.

1 proportional parts

 

15 proportional parts = $20  15 = $300 x = 300

Revision Exam 4 Paper 1

12. A shopkeeper buys 50 cakes for a wholesale price of $1 250. At what price must he sell a cake to make a profit of 20% on his cost? (A) $30

(B) $35

(C) $40

(D) $45

ANSWER

Revision Exam 4 Paper 1

12. The wholesale price per cake The selling price per cake

= (100 + 20)% of $25 = 120% of $25

= $30

Revision Exam 4 Paper 1

13. After a 10% discount, an article is sold for $450. The price of the article before the discount was (A) $460

(B) $495

(C) $500

(D) $520

ANSWER

Revision Exam 4 Paper 1

13.

90% of the original selling price

= $ 450

 the original selling price of the article = = $50  10 = $500

Revision Exam 4 Paper 1

14. The simple interest on $240 for 6 months at 5% is (A) $6

(B) $36

(C) $48

(D) $72

ANSWER

Revision Exam 4 Paper 1

14. The simple interest,

Revision Exam 4 Paper 1

15. The marked price of a microwave was $1 250. A worker bought the microwave on hire-purchase by making a down payment of $225, and twelve monthly payments of $95 each. How much could the worker have saved if the microwave was bought at the marked price? (A) $110

(B) $115

(C) $320

(D) $335

ANSWER

Revision Exam 4 Paper 1

15.

The down payment

= $225

The sum of the monthly payments = $95  12 = $1140 The hire-purchase price = $(225 + 1 140) = $1 365

The amount saved

= $(1 365 – 1 250) = $115

Revision Exam 4 Paper 1

16. At a bank the interest rate of investments decreased from

per cent per annum to

per cent per

annum. The difference in annual interest on a fixed

deposit of $25 000 is (A) $3 125 (B) $1 875

(C) $1 250 (D) $250

ANSWER

Revision Exam 4 Paper 1

16. The difference in simple interest,

Revision Exam 4 Paper 1

17. In how many years would $400, invested at 6% per annum simple interest, amount to $568? (A) 5

(B) 6

(C) 7

(D) 8

ANSWER

Revision Exam 4 Paper 1

17.

The simple interest,

Revision Exam 4 Paper 1

18. In selling an article, a shopkeeper made a profit of 20% on his cost price of $150. What was the selling price? (A) $170

(B) $180

(C) $190

(D) $200

ANSWER

Revision Exam 4 Paper 1

18. The selling price = 120% of $150

Revision Exam 4 Paper 1

19. If U = {1, 2, 3, . . . , 15} and K = {4, 5, 6, 7, 8, 9}, then K1 = (A) {13, 14}

(B) {1, 2, 3}

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

(D) {1, 2, 3, 10, 11, 12, 13, 14, 15}

ANSWER

Revision Exam 4 Paper 1

19.

If

U = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

and

K = {4, 5, 6, 7, 8, 9}

then

K1 = {1, 2, 3, 10, 11, 12, 13, 14, 15}

Revision Exam 4 Paper 1

20. The set of numbers greater than −3 but less than or equal to 8 may be written as (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

20. The set of numbers greater than −3 but less than or equal to 8 =

Revision Exam 4 Paper 1

21.

In the figure above, the shaded portion represents

(A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

21. The shaded region =

Revision Exam 4 Paper 1

22. If P = {1, 2, 3, 4, 5}, then the number of subsets of P = (A) 16

(B) 32

(C) 64

(D) 128

ANSWER

Revision Exam 4 Paper 1

22. The number of subsets, N = 2n = 25 = 32

P = {1, 2, 3, 4, 5} n(P) = 5 = n

Revision Exam 4 Paper 1

23. How many grams are in 5 kilograms? (A) 50 g

(B) 500 g

(C) 5 000 g

(D) 50 000 g

ANSWER

Revision Exam 4 Paper 1

23. 1 kg = 1 000 g 5 kg = 1 000 g  5 = 5 000 g

Revision Exam 4 Paper 1

24.

The area of the parallelogram ABCD above is (A) 30 cm2

(B) 35 cm2

(C) 60 cm2

(D) 70 cm2

ANSWER

Revision Exam 4 Paper 1

24. The area of the parallelogram ABCD, A = bh = 10  6 cm2 = 60 cm2

Items 25 – 26 refer to the diagram below, which consists of a square and a triangle.

Revision Exam 4 Paper 1

25. The perimeter of the entire diagram is (A) 36 cm (B) 68 cm (C) 78 cm

(D) 84 cm

ANSWER

Revision Exam 4 Paper 1

25.

The perimeter, P = (10 + 10 + 16 + 16 + 16) cm = 68 cm

Revision Exam 4 Paper 1

26. The area of the entire diagram is (A) 276 cm2

(B) 281 cm2

(C) 304 cm2

(D) 306 cm2

ANSWER

Revision Exam 4 Paper 1

26. The area of the triangle,

Revision Exam 4 Paper 1

The area of the square, A2 = l2 = (16)2 cm2 = 256 cm2

The total area,

A = A 1 + A2 = (48 + 256) cm2 = 304 cm2

Revision Exam 4 Paper 1

27. A square has the same area as a rectangle with sides of length 27 centimetres and 3 centimetres. What is the length of a side of the square? (A) 9 cm

(B) 24 cm

(C) 30 cm

(D) 81 cm

ANSWER

Revision Exam 4 Paper 1

27. The area of the rectangle, A = lb = 27  3 cm2 = 81 cm2 The area of the square,

A = 81 cm2

 the length of the square, l = 81cm = 9 cm

Revision Exam 4 Paper 1

28. A rectangular aquarium is 100 cm long, 50 cm wide and 30 cm deep. The volume of water it will hold is (A) 1.5 litres

(B) 15 litres

(C) 150 litres

(D) 1 500 litres

ANSWER

Revision Exam 4 Paper 1

28. The volume of the aquarium, V = lbh = 100  50  30 cm3 = 150 000 cm3 The volume of water it will hold

Items 29−30 refer to the diagram below.

In the sector above, the radius is 8 cm and angle AOB = 22.5°.

Revision Exam 4 Paper 1

29. The length of the arc AB, in centimetres, is (A) 

(B) 2

(C) 4

(D) 16

ANSWER

Revision Exam 4 Paper 1

29.

Revision Exam 4 Paper 1

30. The area of the sector AOB, in square centimetres, is (A) 2

(B) 4

(C) 8

(D) 16

ANSWER

Revision Exam 4 Paper 1

30. The area of the sector,

Revision Exam 4 Paper 1

31.

The histogram above shows, the number of stars with a given number of points decorating a Christmas tree. The total number of stars is (A) 25

(B) 44

(C) 150

(D) 220

ANSWER

Revision Exam 4 Paper 1

31.

The total number of stars = 20 + 34 + 44 + 30 + 22 = 150 Items 32−34 refer to the information below. The following scores were obtained by eleven footballers in a school’s goal-shoot out competition: 4

1

7

9

4

2

8

4

10

3

7

Revision Exam 4 Paper 1

32. The modal score was (A) 1

(B) 4

(C) 7

(D) 10

ANSWER

Revision Exam 4 Paper 1

32. The modal score = 4, since it occurs the most frequent. It occurred three times.

Revision Exam 4 Paper 1

33. The median score was (A) 8

(B) 7

(C) 4

(D) 3

ANSWER

Revision Exam 4 Paper 1

33. The scores in ascending order:

The middle score which is in the 6th position is 4. The median score, Q2 = 4.

Revision Exam 4 Paper 1

34. The mean score was (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

34. The sum of the scores = 1 + 2 + 3 + 4 +4 + 4 + 7 + 7 + 8 + 9 + 10 = 59 The number of scores = 11  the mean score, Items 35–36 refer to the pie chart below.

The pie chart, not drawn to scale, shows the number of bikes stored in a warehouse. Angle POQ is a right angle. The warehouse has the same number of motorbikes as scooters and the total number of bikes were 96.

Revision Exam 4 Paper 1

35. How many bicycles were there? (A) 12

(B) 24

(C) 30

(D) 36

ANSWER

Revision Exam 4 Paper 1

35. The number of bicycles

Revision Exam 4 Paper 1

36. If a bike is chosen at random, what is the probability that it is a motorbike? (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

36. The sector angle that represents the number of motorbikes

Revision Exam 4 Paper 1

P(bike is a motorbike)

r

135 360 3 = 8 =

The number of motorbikes = 96 − 24 2 7 = 2 = 36 P(bike is a motorbike)

36 96 3 = 8

=

Revision Exam 4 Paper 1

37. 3x + 7y – 8x + 2y = (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

37.

3x + 7y – 8x + 2y = 3x – 8x + 7y + 2y = – 5x + 9y

Group like terms Add like terms

Revision Exam 4 Paper 1

38. (A) 3

(B) 10

(C) 12

(D) 18

ANSWER

Revision Exam 4 Paper 1

38.

Revision Exam 4 Paper 1

39. (A)

(B)

(C)

(D)

3

ANSWER

Revision Exam 4 Paper 1

39.

Revision Exam 4 Paper 1

40. (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

40.

Revision Exam 4 Paper 1

41. (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

41.

Revision Exam 4 Paper 1

42. If a and b are two integers, then 4(a – b)2 means (A) four times the difference of their squares

(B) four times the square of their difference (C) eight times their difference (D) the square of four times their difference

ANSWER

Revision Exam 4 Paper 1

42. Their difference =a–b The square of their difference = (a – b)2 Four times the square of their difference = 4(a – b)2

Revision Exam 4 Paper 1

43.

If x is an odd number, which of the following is not odd? (A) x + 2

(B) x – 2

(C) x + 1

(D) 2x + 3

ANSWER

Revision Exam 4 Paper 1

43. x + 1 is not odd, it is even.

Revision Exam 4 Paper 1

44. If 70 – y  20, then y could be (A) 30

(B) 40

(C) 50

(D) 60

ANSWER

Revision Exam 4 Paper 1

44.

Alternatively:

Revision Exam 4 Paper 1

45. Gary has $8 more than Andrew. Ken has twice as much as Gary and Andrew together. If Andrew has $x, then Ken has (A) $(x + 16)

(B) $(2x + 8)

(C) $(2x + 16)

(D) $(4x + 16)

ANSWER

Revision Exam 4 Paper 1

45.

Revision Exam 4 Paper 1

46. $x are divided among three girls, Alicia, Britney and Candida, in the ratio 3:5:8 respectively. If Candida gets $18 more than Britney, what is the value of x? (A) 90

(B) 96

(C) 108

(D) 288

ANSWER

Revision Exam 4 Paper 1

46.

Candida’s share – Britney’s share = $18

Revision Exam 4 Paper 1

47. Which of the relations represented by the arrow diagrams below are functions?

(A) I and II only

(B) I and III only

(C) II and III only (D) I, II and III ANSWER

Revision Exam 4 Paper 1

47.

1–1 relation which is a function. 1–1 relation which is a function.

many – many relation which is not a function.

Revision Exam 4 Paper 1

48. If f : x→ x3 + 5, then f (–2 ) is (A) –3

(B) –1

(C) 7

(D) 13

ANSWER

Revision Exam 4 Paper 1

48.

f : x→ x3 + 5 f : –2→ (–2)3 + 5 = –8 + 5 = –3

Revision Exam 4 Paper 1

49.

1 2 3 4

→ → → →

1 3 5 7

Which of the following functions could describe the mapping above? (A) f : x→ x + 1

(B) f : x→ 2x – 1

(C) f : x→ x2 – 1

(D) f : x→ 3x2 – 2 ANSWER

Revision Exam 4 Paper 1

49.

f : x → 2x –1 f : 1 → 2(1) – 1 = 2 – 1 = 1

f : 2 → 2(2) – 1 = 4 – 1 = 3 f : 3 → 2(3) – 1 = 6 – 1 = 5 f : 4 → 2(4) – 1 = 8 – 1 = 7

Items 50–52 refer to the statement given below. A straight line passes through the points A(–3, 8) and B(–5, 9).

Revision Exam 4 Paper 1

50.

The gradient, m, of the line is (A)

(B)

(C) 2

(D) –2

ANSWER

Revision Exam 4 Paper 1

50. Using the points A(−3, 8) and B(− 5, 9), the gradient of the line, m

Revision Exam 4 Paper 1

51. The line intersects the y-axis at c = (A) 13

(B) –13

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

51.

Revision Exam 4 Paper 1

52. The equation of the line is: (A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

52.

Revision Exam 4 Paper 1

53.

The triangle TAB shows the angle of elevation of the top, T, of a tower, TB, to be 29°. AB = 50 m. The height of the tower, in metres, is

(A) 50 sin 29

(B) 50 sin 61

(C) 50 tan 29

(D) 50 tan 61

ANSWER

Revision Exam 4 Paper 1

53.

Revision Exam 4 Paper 1

54.

A ship is travelling on a bearing of 135°. In what direction is it travelling? (A) north-east

(B) north-west

(C) south-east

(D) south-west

ANSWER

Revision Exam 4 Paper 1

54.

Revision Exam 4 Paper 1

55.

In the figure above, angle ABD = 150° and CD = BD. The angle BDE is (A) 30

(B) 60

(C) 120

(D) 150 ANSWER

Revision Exam 4 Paper 1

55.

Revision Exam 4 Paper 1

Alternatively:

Revision Exam 4 Paper 1

56.

In the rectangle above, if STR = 140°, then PQS =

(A) 20

(B) 40

(C) 45

(D) 50 ANSWER

Revision Exam 4 Paper 1

56.

Revision Exam 4 Paper 1

Alternatively:

Revision Exam 4 Paper 1

57.

The sum of four of the interior angles of a convex hexagon is 500°. If the remaining interior angles are equal, how much does each measure? (A) 90

(B) 100

(C) 110

(D) 120

ANSWER

Revision Exam 4 Paper 1

57. The sum of the interior angles of a convex polygon, Sn = (2n – 4) rt s The sum of the interior angles of a convex hexagon, S6 = (2  6 – 4) rt s = (12 – 4)  90 = 8  90 = 720 The sum of the two remaining equal interior angles = 720 – 500 = 220  the size of one of the remaining interior angles =

Revision Exam 4 Paper 1

58. The point P(4, – 7) is translated by a sector The coordinates of the image of P are

(A) (–2, –2)

(B) (–2, 2)

(C) (2, –2)

(D) (2, 2)

ANSWER

Revision Exam 4 Paper 1

58.

Revision Exam 4 Paper 1

59. When rotated through 90° about the origin in an anticlockwise direction, the image of the point (5, 2) is (A) (5, –2)

(B) (–5, 2)

(C) (2, –5)

(D) (–2, 5)

ANSWER

Revision Exam 4 Paper 1

59.

Revision Exam 4 Paper 1

60.

The diagram shown above represents the translation

(A)

(B)

(C)

(D)

ANSWER

Revision Exam 4 Paper 1

60.



T

REVISION EXAMINATION 4 MATHEMATICS Paper 2 2 hours 40 minutes Answer ALL the questions

NEXT

Revision Exam 4 Paper 2

1. (a) Calculate the exact value of

(3 marks)

ANSWER

Revision Exam 4 Paper 2

1. (a)

3  3 + 1 = 10 The LCM of 3 and 8 is 24.

Revision Exam 4 Paper 2

Invert the fraction that is the divisor and multiply instead of divide.

Revision Exam 4 Paper 2

Cancel common factors. Divide by 26 and write the result as a mixed number. Exact value.

Revision Exam 4 Paper 2

1. (b) A table shows the following rate of exchange.

US $1.00 = JM $70.50 Convert JM $9 000 to US $, giving your answer to two decimal places. (3 marks) ANSWER

Revision Exam 4 Paper 2

1. (b) Given JM $70.50 = US $1.00 then

JM $1.00 = US

So

JM $9 000 = US $ = US $127.66

Revision Exam 4 Paper 2

1. (c) The mean (average) mass of the 11 boys on a cricket team is 60 kg.

(i) Calculate the total mass, in kilograms, of the eleven – member team. The captain of the team has a mass of 80 kg. (ii) Calculate the mean (average) mass of the other 10 members of (4 marks) the team. Total 10 marks

ANSWER

Revision Exam 4 Paper 2

1. (c) (i) The total mass of the eleven-member team = 60 kg  11 = 660 kg (ii) The mass of the ten members = (660 – 80) kg = 580 kg

The mean mass of the ten members = 58 kg

Revision Exam 4 Paper 2

2. (a) Simplify (3 marks)

ANSWER

Revision Exam 4 Paper 2

2. (a)

= 3x9 – 4

= 3x5

3

Revision Exam 4 Paper 2

2. (b) If m = −2 and n = 5, calculate the value of mn − n2. (3 marks)

ANSWER

Revision Exam 4 Paper 2

2. (b)

mn – n2 = (–2) (5) – 52 = –10 – 25

= –35

Revision Exam 4 Paper 2

2. (c) Solve the equation (4 marks) Total 10 marks

ANSWER

Revision Exam 4 Paper 2

2. (c)

The LCM of 2 and 5 is 10.

So

6x – 10 = 5x

i.e.

6x – 5x = 10



x = 10

Revision Exam 4 Paper 2

3. (a) A company buys memory cards for $90.00 each and sells them for $120.00 each. (i) Calculate the profit made by selling one memory card. (1 mark)

ANSWER

Revision Exam 4 Paper 2

3. (a) (i) The profit made on one memory card = $(120.00 – 90.00) = $30.00

Revision Exam 4 Paper 2

3. (a) A company buys memory cards for $90.00 each and sells them for $120.00 each. (ii) Express your answer in (i) above as a percentage of the cost price.

(2 marks)

ANSWER

Revision Exam 4 Paper 2

3. (a) (ii) The profit as a percentage of the cost price

Revision Exam 4 Paper 2

3. (a)

A company buys memory cards for $90.00 each and sells them for $120.00 each. (iii) The company offers 15% discount on the memory cards to students, as shown in the advertisement below.

SPECIAL OFFER FOR STUDENTS 15% DISCOUNT ON MEMORY CARDS

A Student bought one memory card. Calculate a) the discount on the purchase b) the price paid for the memory card

(4 marks)

ANSWER

Revision Exam 4 Paper 2

3. (a) (iii) a) The discount on one memory card = 15% of $120.00

= 15  $1.20 = $18.00 b) The price paid for the memory card

= $(120.00 – 18.00) = $102.00

Revision Exam 4 Paper 2

3. (b) A customer can buy a stove on hire purchase by making a down payment of $600.00 and repaying the balance in 24 equal monthly instalments.

If the total cost of the stove is $4 380.00, calculate the amount of each monthly instalment. (3 marks) Total 10 marks ANSWER

Revision Exam 4 Paper 2

3. (b) The total amount paid in monthly instalments = $(4 380.00 – 600.00) = $3 780.00 The amount of each monthly instalments

= = $157.50

Revision Exam 4 Paper 2

4. (a) Solve for x and y: 4x – y = 14 3x + 5y = −1

(5 marks)

ANSWER

Revision Exam 4 Paper 2

4. (a)

Revision Exam 4 Paper 2

4. (a) Substitute x = 3 in : 4(3) – y = 14 So

12 – y = 14

i.e.

 Hence, x = 3 and y = –2.

– y = 14 – 12 = 2

y = –2

Revision Exam 4 Paper 2

4. (b) Nora was paid $x for a domestic job. Rita was paid $25.00 more than Nora. (i) Write an expression in x to represent the amount of money paid to Rita. (ii) The total amount of money paid to Nora and Rita was $281. a) Write an equation in x to represent this information.

b) Determine the value of x.

(4 marks) Total 10 marks

(1 mark) ANSWER

Revision Exam 4 Paper 2

4. (b) (i) The amount of money paid to Rita = $ (x + 25) (ii) a) The equation is: x + x + 25 = 281 (in dollars)

 b) Since then So

2x + 25 = 281 2x + 25 = 281 2x = 281 – 25 = 256 x =

Hence, x = 128.

= 128

Revision Exam 4 Paper 2

5. (a) Nyla invested $8 000 in a Savings Account at a Credit Union at 7% simple interest per annum. Calculate (i) the interest earned at the end of 18 months.

(2 marks)

ANSWER

Revision Exam 4 Paper 2

5. (a) (i) The principal,

P = $8 000

The rate % per annum,

R = 7%

The time,

T = 18 months

Revision Exam 4 Paper 2

The interest earned,

Hence, the interest earned at the end of 18 months is $840.

Revision Exam 4 Paper 2

5. (a) Nyla invested $8 000 in a Savings Account at a Credit Union at 7% simple interest per annum. Calculate

(ii) the amount of time that is required to earn $2 800 in interest. (2 marks)

ANSWER

Revision Exam 4 Paper 2

5. (a) (ii) The principal, The rate % per annum, The interest earned,

P = $8 000 R = 7% I = $2 800

The time,

Hence, a time of 5 years in required to earn $2 800 in interest.

Revision Exam 4 Paper 2

5. (b) The table below shows Mr George’s electricity bill for the month of February.

Calculate the missing values at (i), (ii), (iii), (iv), (v) and (vi). Previous Reading (i) _____ kwh Fixed Charge Energy Charge Fuel Charge Total Charge 5% Discount for Early Payment Net Amount Due

Present Reading 5 147 kwh

$1.08 per kwh $ 0.65 per kwh

kwh Used 457 $35.00 (ii) _____ (iii) _____ (iv) _____ (v)

(6 marks) Total 10 marks

(vi)

ANSWER

Revision Exam 4 Paper 2

5. (b) (i) The previous reading = (5 147 – 457) kwh = 4 690 kwh (ii) The energy charge

= $1.08  457 = $493.56

(iii) The fuel charge

= $0.65  457 = $297.05

(iv) The fuel charge

= $(35.00 + 493.56 + 297.05) = $825.61

Revision Exam 4 Paper 2

(v) 5% discount

= 5% of $825.61 = 0.05  $825.61 = $41.28 (correct to the nearest cent)

(vi) The net amount due = $(825.61 – 41.28) = $784.33

Revision Exam 4 Paper 2

6.

The diagram below, not drawn to scale, shows a quadrilateral KLMN.

(a) Using a pencil, a ruler and a protractor only, draw accurately the quadrilateral KLMN such that: KL = 9 cm, NK = 4 cm, NM = 6 cm and NKL = KNM = 90°. (5 marks)

ANSWER

Revision Exam 4 Paper 2

6. (a)

An accurate drawing of the quadrilateral KLMN is shown above.

Revision Exam 4 Paper 2

6.

The diagram below, not drawn to scale, shows a quadrilateral KLMN.

(b) Measure and state (i) the length of ML (ii) the size of angle KLM.

(2 marks) ANSWER

Revision Exam 4 Paper 2

6. (b) (i) The length of ML = 5 cm. (ii) The size of angle KLM = 53.1

Revision Exam 4 Paper 2

6.

The diagram below, not drawn to scale, shows a quadrilateral KLMN.

(c) State one reason why NM is parallel to KL.

(2 marks)

ANSWER

Revision Exam 4 Paper 2

6. (c) NM is parallel to KL since NKL = KNM = 90.

Revision Exam 4 Paper 2

6.

The diagram below, not drawn to scale, shows a quadrilateral KLMN.

(d) What type of quadrilateral in KLMN?

(1 mark) Total 10 marks ANSWER

Revision Exam 4 Paper 2

6. (d) Quadrilateral KLMN is a trapezium.

Revision Exam 4 Paper 2

7. The diagram below, not drawn to scale, shows a field, ABCO, whose boundaries are an arc ABC and two radii OA and OC of a circle.

(a) Calculate (i) the circumference of the complete circle ABCD (2 marks)

ANSWER

Revision Exam 4 Paper 2

7. (a) (i) The circumference of the complete circle ABCD, C = 2r

Revision Exam 4 Paper 2

7. The diagram below, not drawn to scale, shows a field, ABCO, whose boundaries are an arc ABC and two radii OA and OC of a circle.

(a) (ii) the length of the minor arc ADC. (2 marks) ANSWER

Revision Exam 4 Paper 2

7. (a) (ii) The length of the minor are ADC, l

Revision Exam 4 Paper 2

Alternatively:

Revision Exam 4 Paper 2

7. The diagram below, not drawn to scale, shows a field, ABCO, whose boundaries are an arc ABC and two radii OA and OC of a circle.

(a) (iii) the perimeter of the closed field ABCO. An athlete takes 1 minute and 20 seconds to run around the entire closed field, ABCO. (3 marks)

ANSWER

Revision Exam 4 Paper 2

7. (a) (iii) The length of the major are ADC,

Revision Exam 4 Paper 2

Alternatively 1:

Alternatively 2: l = (528 –132) m = 396 m

Revision Exam 4 Paper 2

7. The diagram below, not drawn to scale, shows a field, ABCO, whose boundaries are an arc ABC and two radii OA and OC of a circle.

(b) Calculate the speed of the athlete in ms–1.

(3 marks) Total 10 marks

ANSWER

Revision Exam 4 Paper 2

7. (b) The perimeter of the closed field ABCO, P = (396 + 84 + 84) m = 564 m The time taken, t

= 1 min 20 s = 80 s

The speed of the athlete, s =

Revision Exam 4 Paper 2

8. The graph below shows rectangle ABCD and its image A′B′C′D′.

Revision Exam 4 Paper 2

8.

(a)

Write down the coordinates of A and A′.

(2 marks)

ANSWER

Revision Exam 4 Paper 2

8.

(a)

The coordinates of A and A′ are (3, −2) and (−2, 5) respectively

Revision Exam 4 Paper 2

8. The graph below shows rectangle ABCD and its image A′B′C′D′.

Revision Exam 4 Paper 2

8.

(b)

Describe fully the transformation which maps ABCD onto A′B′C′D′. (3 marks)

ANSWER

Revision Exam 4 Paper 2

8.

(b)

A translation with onto A′B′C′D′.

maps ABCD

Revision Exam 4 Paper 2

8. The graph below shows rectangle ABCD and its image A′B′C′D′.

Revision Exam 4 Paper 2

8.

(c)

Draw and label A″B″C″D″, the image of A′B′C′D′, after a reflection in the line x = 2.

(5 marks) Total 10 marks

ANSWER

Revision Exam 4 Paper 2

8.

(c)

A″B″C″D″ is the image of A′B′C′D′ after a reflection in the line x = 2.

Revision Exam 4 Paper 2

9. (a) The diagram below shows the graph of the function y = 2x.

Revision Exam 4 Paper 2

9.

(a)

Using the graph, state the value of (i) x when

ANSWER (ii) y when x = 2

(2 marks)

Revision Exam 4 Paper 2

9.

(a)

(i) when

, then x = −1.

(ii) when x = 2, then y = 4.

Revision Exam 4 Paper 2

9.

(b)

A straight line passes through the points A (0, 5) and B (−3, 8). Determine (i) the gradient of the line segment AB

(2 marks)

ANSWER

Revision Exam 4 Paper 2

9.

(b)

(i) Let A (0, 5) = A (x1, y1) and B (−3, 8) = B (x2, y2), then the gradient of line segment AB is

Revision Exam 4 Paper 2

9.

(b)

(ii) the equation of the line segment AB

(2 marks)

ANSWER

Revision Exam 4 Paper 2

9.

(b)

(ii) Using m = −1 and A (0, 5) = A (x, y), then

y = mx + c becomes 5 = −1(0) + c  5=c So the equation of the line segment AB is:

i.e 

y = mx + c y = −1(x) + 5 y = −x + 5

where m = −1 and c = 5.

Revision Exam 4 Paper 2

9.

(b)

(iii) the coordinates of the point where AB cuts the x-axis

(2 marks)

ANSWER

Revision Exam 4 Paper 2

9.

(b)

(iii) On the x-axis y = 0 So i.e.

0 = −x + 5 x = 5.

The coordinates of the point where AB cuts the x-axis are (5, 0).

Revision Exam 4 Paper 2

9.

(b)

(iv) the equation of the line which is perpendicular to AB, and passes through the origin.

(2 marks) Total 10 marks

ANSWER

Revision Exam 4 Paper 2

9.

(b)

(iv) The gradient of a line which is perpendicular to AB is 1. Using m = 1 and 0 (0, 0), then y = mx + c becomes 0 = 1(0) + c 0=c The equation of the line which is perpendicular to AB, and passes through the origin is

i.e

y = 1(x) + 0 y=x

Revision Exam 4 Paper 2

10. The histogram below shows the ages of the students in an Internet Café. (a) Copy and complete the frequency table below to represent the data shown in the histogram.

Age (years) 12 13 14 15 16 17 Frequency

6

10 19

7 (2 marks)

ANSWER

Revision Exam 4 Paper 2

10. (a)

Age (years) Frequency

12

13

14

15

16

17

6

12

10

19

16

7

The completed frequency table is shown above.

Revision Exam 4 Paper 2

10. The histogram below shows the ages of the students in an Internet Café.

(2 marks)

(b) How many students are members of the Internet Café ?

ANSWER

Revision Exam 4 Paper 2

10. (b) The number of students who are members of the Internet Café = 6 + 12 + 10 + 19 + 16 + 7

= 70

Revision Exam 4 Paper 2

10. The histogram below shows the ages of the students in an Internet Café.

(2 marks)

(c) Calculate the mean age of the students in the Internet Café.

Anita is a member of the Internet Café. There are 23 students in the café who are older than Anita ANSWER

Revision Exam 4 Paper 2

10. (c) The mean age of the students in the Internet Café

Revision Exam 4 Paper 2

Alternatively: Age (x) 12 13 14 15 16 17

Frequency (f) fx 6 72 12 156 10 140 19 285 16 256 7 119 f = 70 fx = 1 028

Revision Exam 4 Paper 2

The mean age,

Revision Exam 4 Paper 2

10. The histogram below shows the ages of the students in an Internet Café.

(2 marks)

(d) How is old Anita? Anita is a member of the Internet Café. There are 23 students in the café who are older than Anita

ANSWER

Revision Exam 4 Paper 2

10. (d) If there are 23 students in the Internet Café who are older than Anita, then Anita is 15 years old.

Revision Exam 4 Paper 2

10. The histogram below shows the ages of the students in an Internet Café. (e) Calculate the probability that a student chosen at random is fourteen years old or younger. Anita is a member of the Internet Café. There are 23 students in the café who are older than Anita

(2 marks) Total 10 marks

ANSWER

Revision Exam 4 Paper 2

10. (e) The number of students who are fourteen years old or younger = 6 + 12 + 10 + 28  P (student’s age ≤ 14 years old)

REVISION EXAMINATION 5 MATHEMATICS Paper 1 90 minutes Answer ALL the questions

NEXT

Revision Exam 5 Paper 1

1. (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

1.

Invert the fraction that is the divisor and cancel common factors.

Revision Exam 5 Paper 1

2. 28 ÷ 22 = (A) 24

(B) 26

(C) 210

(D) 216

ANSWER

Revision Exam 5 Paper 1

2. 28 ÷ 22 = 28 − 2 = 26

Revision Exam 5 Paper 1

3. To the nearest hundred, 5 849 = (A) 5 800

(B) 5 850

(C) 5 900

(D) 6 000

ANSWER

Revision Exam 5 Paper 1

3. 5 8 4 9 = 5 800 to the nearest hundred The tens digit which is 4 is less than 5, so we do not add 1 to the hundreds digit which is 8.

Revision Exam 5 Paper 1

4. The decimal equivalent of (A) 0.009

(B) 0.09

is

(C) 0.9

(D) 9.0

ANSWER

Revision Exam 5 Paper 1

4.

Revision Exam 5 Paper 1

5. (7.8)2 − (2.2)2 = (A) 11.2

(B) 20.0

(C) 56.0

(D) 65.68

ANSWER

Revision Exam 5 Paper 1

5. (7.8)2 − (2.2)2

= (7.8 + 2.2) (7.8 – 2.2) = (10) (5.6) = 56.0

Revision Exam 5 Paper 1

6. If

then

(A) 4.35 × 104

(B) 2.95 × 104

(C) 4.35 × 102

(D) 2.95 × 102

ANSWER

Revision Exam 5 Paper 1

6.

Or

Revision Exam 5 Paper 1

7. The HCF of two different numbers less than 15 is 6. If one is 12, then the other number is (A) 3

(B) 6

(C) 9

(D) 15

ANSWER

Revision Exam 5 Paper 1

7.

The HCF of 6 and 12 is 6.

Revision Exam 5 Paper 1

8. The set of factors of 20 is (A) {1, 2, 4, 5, 10, 20}

(B) {2, 4, 5, 10}

(C) {10, 20}

(D) {20, 40, 60, . . .}

ANSWER

Revision Exam 5 Paper 1

8. 20 = 1 × 20

= 2 × 10 =4×5 The set of factors of 20 = {1, 2, 4, 5, 10, 20}

Revision Exam 5 Paper 1

9. Given that 1 500 × 19 = 28 500, then 1 499 × 19 = (A) 28 500 − 1

(B) 28 500 − 19

(C) 28 500 − 1 499

(D) 28 500 − 1 500

ANSWER

Revision Exam 5 Paper 1

9. 1 499 × 19 = (1 500 – 1) × 19

= 1 500 × 19 – 1 × 19 = 28 500 – 19

Revision Exam 5 Paper 1

10. The next two terms in the sequence 5, 1, –3, . . ., is (A) 1, 5

(B) −1, −5

(C) 7, 11

(D) −7, −11

ANSWER

Revision Exam 5 Paper 1

10. 5, 1, –3, . . . 1 – 5 = –4 and –3 – 1 = –4 –3 + (–4) = –3 – 4 = –7 –7 + (–4) = –7 – 4 = –11 The next two terms in the sequence are: –7 and –11.

Revision Exam 5 Paper 1

11. The hire-purchase price of a water heater is $2 075. It may be purchased by depositing $650 and making monthly payments of $95. How many months are required to complete payment? (A) 12

(B) 15

(C) 18

(D) 21

ANSWER

Revision Exam 5 Paper 1

11. The amount to be paid monthly = $(2 075 − 650) = $1 425 The number of monthly payments

Revision Exam 5 Paper 1

12. What price is paid for a grill marked $400 if a 15 per cent discount is given? (A) $460

(B) $385

(C) $340

(D) $300

ANSWER

Revision Exam 5 Paper 1

12. The price after discount

Revision Exam 5 Paper 1

13.

House Insurance $25 per $10 000 Contents Insurance $20 per $10 000 The table above shows the rates charged by an insurance company. How much insurance will a person pay if the house is valued at $150 000 and its contents at $75 000? (A) $52.50

(B) $525

(C) $5 250

(D) $52 500 ANSWER

Revision Exam 5 Paper 1

13. The amount paid for house insurance

The amount paid for contents insurance  the amount paid for insurance = $(375 + 150) = $525

Revision Exam 5 Paper 1

14. The selling price of an article costing $480 was $600. The gain per cent was (A) 20%

(B) 25%

(C) 75%

(D) 80%

ANSWER

Revision Exam 5 Paper 1

14.

The gain

The gain %

= $(600 − 480) = $120

Revision Exam 5 Paper 1

15. The simple interest on $650 after 4 years is $130. The rate per cent per annum is (A) 5

(B)

(C) 6

(D)

ANSWER

Revision Exam 5 Paper 1

15. The rate % per annum, R

Revision Exam 5 Paper 1

16. A calculator originally cost $60. It was sold to make a profit of 150%. For how much was it sold? (A) $210

(B) $200

(C) $150

(D) $90

ANSWER

Revision Exam 5 Paper 1

16.

Or

Revision Exam 5 Paper 1

Revision Exam 5 Paper 1

17. A merchant bought eggs at $10.00 per dozen and sold them at $12.50 per dozen. His percentage profit was (A) 20%

(B) 25%

(C) 75%

(D) 80%

ANSWER

Revision Exam 5 Paper 1

17.

The profit

The percentage profit

= $(12.50 − 10.00) = $2.50

Revision Exam 5 Paper 1

18. If

of a sum of money is $90, then the

sum of money is (A) $11.25

(B) $101.25

(C) $450

(D) $720

ANSWER

Revision Exam 5 Paper 1

18.

Revision Exam 5 Paper 1

19. Using set builder notation, the set {−5, −4, −3, −2, −1, 0, 1, 2} can be represented as (A)

(B) (C) (D)

ANSWER

Revision Exam 5 Paper 1

19.

Revision Exam 5 Paper 1

20. The set of numbers greater than −4 but less than 3 may be written as (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

20.

Revision Exam 5 Paper 1

21. If then

=

(A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

21.

Revision Exam 5 Paper 1

22.

In the Venn diagram above, the shaded region represents (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

22. The region representing

is shown shaded.

Revision Exam 5 Paper 1

The region representing shaded.

is shown

Revision Exam 5 Paper 1

23. A man leaves home at 23:45 h and reaches his destination at 05:30 h the following day. How many hours did the journey take? (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

23.

The number of minutes to mid-night The number of hours to mid-night The number of hours travelled after mid-night The total number of hours travelled

Revision Exam 5 Paper 1

24. How many litres of juice would a 500 cm3 bottle hold? (A) 0.05

(B) 0.5

(C) 5.0

(D) 50.0

ANSWER

Revision Exam 5 Paper 1

24.

Revision Exam 5 Paper 1

25.

The perimeter of the trapezium above, not drawn to scale, in cm, is (A) 6

(B) 20

(C) 31.7

(D) 35.7 ANSWER

Revision Exam 5 Paper 1

25.

Revision Exam 5 Paper 1

26. The perimeter of a square is 60 cm. What is its area in cm2? (A) 3 600

(B) 225

(C) 144

(D) 16

ANSWER

Revision Exam 5 Paper 1

26.

Revision Exam 5 Paper 1

Items 27–28 refer to the diagram below.

Revision Exam 5 Paper 1

27. The ratio of the area of the major sector to the area of the circle is (A) 4:1

(B) 4:3

(C) 1:4

(D) 3:4

ANSWER

Revision Exam 5 Paper 1

27.

Revision Exam 5 Paper 1

Alternatively:

Revision Exam 5 Paper 1

28. The ratio of the circumference of the circle to the length of the major arc is (A) 4:3

(B) 4:1

(C) 3:4

(D) 1:4

ANSWER

Revision Exam 5 Paper 1

28.

Revision Exam 5 Paper 1

Alternatively:

Revision Exam 5 Paper 1

29. The circumference of a circular lawn that has a radius of 3.5 metres is (A) 1.75π metres

(B) 3.5π metres

(C) 7π metres

(D) 14π metres

ANSWER

Revision Exam 5 Paper 1

29.

Revision Exam 5 Paper 1

30. If a circle has diameter d, and area A, then A = (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

30.

Revision Exam 5 Paper 1

31. In a box there are 5 red, 4 orange and 3 yellow marbles. If one of the marbles is chosen at random, what is the probability that it is not yellow?

(A) 1

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

31.

Revision Exam 5 Paper 1

32.

The histogram above shows the number of students who scored a given number of points in an archery competition. The total number of students is (A) 250

(B) 200

(C) 80

(D) 40

ANSWER

Revision Exam 5 Paper 1

32.

Revision Exam 5 Paper 1

Items 33–35 refer to the pie chart that follows, which shows the favourite games of 1 080 students.

Revision Exam 5 Paper 1

33. How many of the students preferred hockey?

(A) 120

(B) 210

(C) 240

(D) 360

ANSWER

Revision Exam 5 Paper 1

33. The number of students who preferred hockey

Revision Exam 5 Paper 1

34. If a student is chosen at random, the probability that he likes football is (A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

34. The number of students who like football

 P(student likes football)

Revision Exam 5 Paper 1

Alternatively: The sector angle that represents football 

P(student likes football)

Revision Exam 5 Paper 1

35. How many of the students preferred tennis? (A) 120

(B) 210

(C) 240

(D) 360

ANSWER

Revision Exam 5 Paper 1

35.

Revision Exam 5 Paper 1

36. Which of the following is not a statistical diagram? (A) Line graph

(B) Chronological bar chart (C) Proportionate bar chart (D) Class interval

ANSWER

Revision Exam 5 Paper 1

36. Class interval is not a statistical diagram.

Revision Exam 5 Paper 1

37. What is the value of x if 5x + 2 = 17? (A) −3

(B)

(C) 3

(D)

ANSWER

Revision Exam 5 Paper 1

37.

Revision Exam 5 Paper 1

38. If (A) 9

then (B) 7

is (C) 6

(D) 5

ANSWER

Revision Exam 5 Paper 1

38.

Revision Exam 5 Paper 1

39. 8x − 3y − 5x + 3y = (A) 3x

(B) 11x

(C) 3x − 6y

(D) 11x + 6y

ANSWER

Revision Exam 5 Paper 1

39.

Revision Exam 5 Paper 1

40. −3p × (−x) = (A) −3p − x

(B) 3p + x

(C) −3px

(D) 3px

ANSWER

Revision Exam 5 Paper 1

40.

Revision Exam 5 Paper 1

41.

(A)

(B)

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

41.

The common denominator is 7x.

Revision Exam 5 Paper 1

42. What is the value of 3x2 when x = −4? (A) −24

(B) −16

(C) 48

(D) 144

ANSWER

Revision Exam 5 Paper 1

42.

2

Revision Exam 5 Paper 1

43. If x = 52 × 23, then x4 = (A) 58 × 212

(B) 56 × 27

(C) 56 × 23

(D) 58 × 23

ANSWER

Revision Exam 5 Paper 1

43.

Revision Exam 5 Paper 1

44. 7x − 6(x + 3) = (A) x + 18

(B) x − 18

(C) 13x + 18

(D) 13x − 18

ANSWER

Revision Exam 5 Paper 1

44.

Revision Exam 5 Paper 1

45. If

which of the following must be true?

(A) xp = yq

(B) x − p = y − q

(C) x + p = y + q

(D) xq = yp

ANSWER

Revision Exam 5 Paper 1

45.

Revision Exam 5 Paper 1

46. Which of the following arrow diagrams represents a 1 − 1 relation? (A)

Revision Exam 5 Paper 1

46. (B)

Revision Exam 5 Paper 1

46. (C)

Revision Exam 5 Paper 1

46. (D)

ANSWER

Revision Exam 5 Paper 1

46.

This arrow diagram represents a 1 − 1 relation. Each element in the domain is mapped onto only one element in the range.

Revision Exam 5 Paper 1

47.

The diagram above shows the graphs of y + 2x = 4 and −2y − x = 1. The ordered pair (x, y) which satisfies both equations is (A) (−1, 0)

(B) (0, 4)

(C) (2, 0)

(D) (3, −2)

ANSWER

Revision Exam 5 Paper 1

47.

The point of intersection of the two lines is (3, −2). The ordered pair (3, −2) satisfies both equations.

Revision Exam 5 Paper 1

48.

In the figure above, P is the point with co-ordinates (A) (2, 0)

(B) (0, 2)

(C) (−3, 0)

(D) (0, −3) ANSWER

Revision Exam 5 Paper 1

48. P is the point with co-ordinates (−3, 0). That is x = −3 when y = 0.

Revision Exam 5 Paper 1

49.

In the diagram above, the gradient of the line PQ is (A)

(B)

(C) 2

(D)

ANSWER

Revision Exam 5 Paper 1

49.

Revision Exam 5 Paper 1

49. Alternatively:

Revision Exam 5 Paper 1

50. Which of the following sets is represented by the relation ? (A)

(B) (C) (D)

ANSWER

Revision Exam 5 Paper 1

50.

The set represented by the relation is

Revision Exam 5 Paper 1

51.

Which of the inequations in x describes the graph above? (A) x < −2

(B) x ≤ −2

(C) x > −2

(D) x ≥ −2 ANSWER

Revision Exam 5 Paper 1

51.

The inequation in x which describes the graph is x ≤ −2.

Revision Exam 5 Paper 1

52.

In the diagram above, the transformation which will map triangle PQR onto P′Q′R′ is

Revision Exam 5 Paper 1

52. (A) a reflection in the line y = −x (B) a reflection in the line y = x (C) a reflection in the origin O

(D) an enlargement of scale factor −1.

ANSWER

Revision Exam 5 Paper 1

52. The transformation is a reflection in the origin O.

Revision Exam 5 Paper 1

53.

In the right-angled triangle above, tan θ is

(A)

(B)

(C)

(D) ANSWER

Revision Exam 5 Paper 1

53.

Revision Exam 5 Paper 1

54.

The triangle ABC above is right-angled at B. CÂB = 25 and BC = 30 m. The length of AC, in m, is (A) 30 cos 25°

(B) 30 sin 25°

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

54.

Revision Exam 5 Paper 1

55.

In the diagram above, the letter ‘R’ and an image of it are shown. The image was obtained from a

Revision Exam 5 Paper 1

55. (A) reflection in the line x = 0 (B) reflection in the line y = 0 (C) rotation about 0 through 90° anticlockwise

(D) translation parallel to the x-axis

ANSWER

Revision Exam 5 Paper 1

55.

The image was obtained from a reflection in the y-axis, that is, the line x = 0.

Revision Exam 5 Paper 1

56.

In the diagram above, AB is parallel to CD, and EF is parallel to GH. The value of x, in degrees, is (A) 75

(B) 85

(C) 95

(D) 105 ANSWER

Revision Exam 5 Paper 1

56.

x = 95°

Revision Exam 5 Paper 1

57.

What type of transformation maps triangle ABC onto triangle A′B′C′ in the figure above? (A) Rotation

(B) Reflection

(C) Translation

(D) Enlargement.

ANSWER

Revision Exam 5 Paper 1

57. Triangle ABC is mapped onto triangle A′B′C′ by a translation.

Revision Exam 5 Paper 1

58. A plane is heading in a direction of 045° and changes course in a clockwise direction to 180°. The angle through which the plane turns is (A) 90°

(B) 135°

(C) 225°

(D) 315°

ANSWER

Revision Exam 5 Paper 1

58.

The angle through which the plane turns = 180° − 45° = 135°

Revision Exam 5 Paper 1

59. In a triangle, angle A = 2x° and angle B = 3x°. The size of angle C is (A) 30°

(B) 45°

(C)

(D)

ANSWER

Revision Exam 5 Paper 1

59.

Revision Exam 5 Paper 1

60.

In the diagram above CÂB = 38° and AC passes through the centre of the circle O. Angle ADB is (A) 38°

(B) 52°

(C) 76°

(D) 90° ANSWER

Revision Exam 5 Paper 1

60.

REVISION EXAMINATION 5 MATHEMATICS Paper 2 2 hours 40 minutes Answer ALL the questions

NEXT

Revision Exam 5 Paper 2

1. (a) Calculate the exact value of (4 marks)

ANSWER

Revision Exam 5 Paper 2

1. (a)

Invert the fraction that is the divisor and multiply Instead of divide.

1

Revision Exam 5 Paper 2

1. (b) The students and teachers of St Stephen’s College went on a field trip in three buses, A, B and C. The table below shows the number of teacher and students in each bus. Teachers Students Bus A

8

61

Bus B

5

58

Bus C

56

(i) The ratio of teachers to students on Bus C was 3:14. How many teachers were on Bus C?

ANSWER

(2 marks)

Revision Exam 5 Paper 2

1. (b) (i) Let the number of teachers on Bus C = x.

Hence, there were 12 teachers on Bus C.

Revision Exam 5 Paper 2

1. (b) The students and teachers of St Stephen’s College went on a field trip in three buses, A, B and C. The table below shows the number of teacher and students in each bus. Teachers Students Bus A

8

61

Bus B

5

58

Bus C

56

(ii) What is the total number of passengers who went on the field trip? (1 mark)

ANSWER

Revision Exam 5 Paper 2

1. (b) (ii)

Teachers Students

Bus A

8

61

Bus B

5

58

Bus C

12

56

Total

25

175

The total number of passengers who went on the field trip = 25 + 175 = 200

Revision Exam 5 Paper 2

1. (b) The students and teachers of St Stephen’s College went on a field trip in three buses, A, B and C. The table below shows the number of teacher and students in each bus. Teachers Students Bus A

8

61

Bus B

5

58

Bus C

56

(iii) What percentage of the passengers on the field trip were students? (3 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

1. (b) (iii) The percentage of the passengers on the field trip who were students

Revision Exam 5 Paper 2

2. (a) If a = 5, b = – 4 and c = 3, calculate the value of:

(i) 2a + b

(3 marks)

ANSWER

Revision Exam 5 Paper 2

2. (a) (i) 2a + b = 2(5) + (–4) = 10 – 4 =6

Revision Exam 5 Paper 2

2. (a) If a = 5, b = – 4 and c = 3, calculate the value of:

(ii)

(2 marks)

ANSWER

Revision Exam 5 Paper 2

2. (a) (ii)

Revision Exam 5 Paper 2

2. (b) Simplify the expression 5(4y + 3) – 7y

(2 marks)

ANSWER

Revision Exam 5 Paper 2

2. (b) 5 (4y + 3) – 7y = 20y + 15 – 7y = 20y – 7y + 15 = 13y + 15

Use the distributive law Group like terms. Add like terms.

Revision Exam 5 Paper 2

2. (c) Solve the inequality 2x + 10 ≥ 7x

(3 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

2. (c)

Revision Exam 5 Paper 2

3. In the diagram below, a digital camera is advertised for a cash price of $1 200. CASH PRICE $1200

(a) The digital camera can be bought on hire purchase by paying a deposit of $120 followed by 24 monthly payments of $54. (i) Calculate the total amount paid if the digital camera is bought on hire purchase.

(3 marks) ANSWER

Revision Exam 5 Paper 2

3. (a) (i) The deposit = $120 The total amount of the monthly payments = $54  24 = $1 296 The total amount paid if the digital camera is bought on hire purchase = $(120 + 1 296) = $1 416

Revision Exam 5 Paper 2

3. In the diagram below, a digital camera is advertised for a cash price of $1 200. CASH PRICE $1200

(ii) Calculate the extra amount paid when buying the digital camera on hire purchase compared to the cash price.

(1 mark)

ANSWER

Revision Exam 5 Paper 2

3. (a) (ii) The extra amount paid when buying the digital camera on hire purchase compared to the cash price = $(1 416 – 1 200) = $216

Revision Exam 5 Paper 2

3. (b) Daniel started a bank account with a deposit

of $150. He deposited $50 a month for 9 months. (i) Calculate the total amount deposited at the end of 9 months. The bank pays simple interest quarterly at a rate of 5% per annum. (2 marks) ANSWER

Revision Exam 5 Paper 2

3. (b) (i) The total amount deposited at the end of 9 months = $(150 + 50  9) = $(150 + 450) = $600

Revision Exam 5 Paper 2

3. (b) Daniel started a bank account with a deposit of $150. He deposited $50 a month for 9 months. (ii) Calculate the amount in the bank after the interest was added. (4 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

3. (b) (i) The total amount deposited at the end of 9 months

= $(150 + 50  9) = $(150 + 450) = $600

(ii) The simple interest earned,

The amount in the bank after the interest was added, A = P + I = $(600 + 22.50) = $622.50

Revision Exam 5 Paper 2

4. (a) Solve the simultaneous equations: x + 3y = 2 4x + y = 32

(5 marks)

ANSWER

Revision Exam 5 Paper 2

4. (a)

⎯⎯  ⎯⎯ 

  3:  – 1: ÷ 11:



⎯⎯ 

Revision Exam 5 Paper 2

4. (b) Jane is k years old. Jean-Paul is 5 years older than Jane. Paula is twice as old as Jean-Paul. Write an expression in terms of k for: (i) Jean-Paul’s age

(1 mark)

ANSWER

Revision Exam 5 Paper 2

4. (b) (i) Jean-Paul’s age = (k + 5) years

Revision Exam 5 Paper 2

4. (b) Jane is k years old. Jean-Paul is 5 years older than Jane. Paula is twice as old as Jean-Paul. Write an expression in terms of k for: (ii) Paula’s ages

(1 mark)

ANSWER

Revision Exam 5 Paper 2

4. (b) (ii) Paula’s age = 2 (k + 5) years

Revision Exam 5 Paper 2

4. (b) Jane is k years old. Jean-Paul is 5 years older than Jane. Paula is twice as old as Jean-Paul. Write an expression in terms of k for: (iii) The sum of the ages of Jane, Jean-Paul and Paula The sum of their ages is 75 years.

(1 mark)

ANSWER

Revision Exam 5 Paper 2

4. (b) (iii) The sum of the ages of Jane, Jean-Paul and Paula = [k + (k + 5) + 2(k + 5)] years = (k + k + 5 + 2k + 10) years = (2k + 2k + 5 + 10) years = (4k + 15) years

Revision Exam 5 Paper 2

4. (b) Jane is k years old. Jean-Paul is 5 years older than Jane. Paula is twice as old as Jean-Paul. Write an expression in terms of k for: (iv) Write an equation in terms of k to represent this statement and solve it to determine Jane’s age.

(2 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

4. (b) (iv) The equation is:

Revision Exam 5 Paper 2

5. (a) Mrs Fletcher is paid a basic weekly wage of $560. During a certain week she worked 6 hours overtime. Her total wages were $686. Calculate

(i) her overtime wage

(2 marks)

(ii) the overtime rate of pay.

(2 marks)

ANSWER

Revision Exam 5 Paper 2

5. (a) (i) Her overtime wage = $(686 – 560) = $126 (ii) the overtime rate of pay.

Revision Exam 5 Paper 2

5. (b) A tourist had CAN currency made up of 3 one hundred dollars 7 fifty dollars 6 twenty dollars 2 ten dollars 4 five dollars 3 one dollar (i) Calculate the total value of the CAN currency.

The tourist changed all her money into EC currency. The exchange rate of the bank is CAN $1.00 = EC $2.70. The bank also charges 2% on all exchange transactions. (2 marks) ANSWER

Revision Exam 5 Paper 2

5. (b) (i)

Number of Notes

Value of note ($)

Value of notes ($)

3

100

300

7

50

350

6

20

120

2

10

20

4

5

20

3

1

3 CAN $813

The total value of the CAN currency is $813.

Revision Exam 5 Paper 2

5. (b) A tourist had CAN currency made up of 3 one hundred dollars 7 fifty dollars 6 twenty dollars 2 ten dollars 4 five dollars 3 one dollar (ii) Calculate the value of EC currency the tourist received after the bank charges were deducted.

(4 marks) Total 10 marks ANSWER

Revision Exam 5 Paper 2

5. (b) (ii)

The value of EC currency the tourist received after the bank charges were deducted = EC $(2 195.10 – 43.90) = EC $2 151.20

Revision Exam 5 Paper 2

6. (a) (i) Using a pencil and a ruler only, construct triangle WXY in which WX = 8 cm, WY = 5 cm and XWY = 90° (ii) Measure and write down the size of the angle WXY.

(4 marks) (1 mark)

ANSWER

Revision Exam 5 Paper 2

6. (a) (i)

The constructed triangle WXY is shown above. (ii) The size of the angle WXY = 32°.

Revision Exam 5 Paper 2

6. (b) The figure below, not drawn to scale, shows a ladder leaning against a vertical wall. The ladder is 3.9 metres long and the foot of the ladder is 2.6 metres away from the wall.

Calculate, to 2 significant figures (i) the height, in metres, that the ladder reaches up the wall

ANSWER (2 marks)

Revision Exam 5 Paper 2

6. (b) (i)

Using Pythagoras’ theorem:

 Hence, the ladder reaches a height of 2.9 m up the wall.

Revision Exam 5 Paper 2

6. (b) The figure below, not drawn to scale, shows a ladder leaning against a vertical wall. The ladder is 3.9 metres long and the foot of the ladder is 2.6 metres away from the wall.

Calculate, to 2 significant figures (ii)

the measure of the angle, , that the ladder makes with the wall.

(3 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

6. (b) (ii)

Hence, the ladder makes an angle of 42° with the wall.

Revision Exam 5 Paper 2

7. (a) Xanadu and Sieve are two water parks which are 30 km apart. These parks are drawn on a map using a scale of 1:500 000. Calculate, in cm, the distance on the map between the two water parks.

(3 marks)

ANSWER

Revision Exam 5 Paper 2

7. (a) A scale of 1 : 500 000 means 1 cm : 500 000 cm So 1 cm : 5 km i.e. 1 cm  6 : 5 km  6  6 cm : 30 km Hence, the distance on the map between the two water parks is 6 cm.

Revision Exam 5 Paper 2

7. (b) The figure below, not drawn to scale, shows a quadrilateral JKLM. Angle MJK = 90°, JM = 7.0 cm, JK = 4.8 cm and KL = 3.6 cm.

(i) Calculate the area, in cm2, of triangle MJK. The area of the quadrilateral JKLM is 30.66 cm2.

(2 marks)

ANSWER

Revision Exam 5 Paper 2

7. (b) (i)

The area of triangle

Revision Exam 5 Paper 2

7. (b) The figure below, not drawn to scale, shows a quadrilateral JKLM. Angle MJK = 90°, JM = 7.0 cm, JK = 4.8 cm and KL = 3.6 cm.

(ii) Calculate the area, in cm2, of triangle KLM. The perimeter of the quadrilateral JKLM is 23.1 cm.

(2 marks)

ANSWER

Revision Exam 5 Paper 2

7. (b) (ii) The area of triangle KLM = (30.66 – 16.8) cm2

= 13.86 cm2

Revision Exam 5 Paper 2

7. (b) The figure below, not drawn to scale, shows a quadrilateral JKLM. Angle MJK = 90°, JM = 7.0 cm, JK = 4.8 cm and KL = 3.6 cm.

(iii) Calculate the Length, in cm, of LM.

(3 marks) Total 10 marks

ANSWER

Revision Exam 5 Paper 2

7. (b) (iii) The length of LM

= [23.1 – (3.6 + 4.8 + 7.0)] cm = (23.1 – 15.4) cm = 7.7 cm

Revision Exam 5 Paper 2

8. (a) The diagram below shows triangle PQR and its image triangle PQR after a transformation.

ANSWER (i) Write down the coordinates of P, Q and R.

(1 mark)

Revision Exam 5 Paper 2

8. (a) (i) The coordinates are P(6, –3), Q(10, –3) and R(6, –1).

Revision Exam 5 Paper 2

8. (a) The diagram below shows triangle PQR and its image triangle PQR after a transformation.

(ii) Describe fully the transformation which maps PQR onto PQR.

ANSWER (3 marks)

Revision Exam 5 Paper 2

8. (a) (ii) A translation with

Revision Exam 5 Paper 2

8. (b) Triangle PQR is an enlargement of triangle PQR With centre of enlargement (1, 3) and scale factor 2.

ANSWER (i) Label the centre of enlargement (1, 3)

(1 mark)

Revision Exam 5 Paper 2

8. (b) (i) The centre of enlargement was labelled C(1, 3) in the diagram above.

Revision Exam 5 Paper 2

8. (b) Triangle PQR is an enlargement of triangle PQR With centre of enlargement (1, 3) and scale factor 2.

(ii) Draw and label the triangle PQR the image of triangle PQR after the enlargement.

ANSWER (3 marks)

Revision Exam 5 Paper 2

8. (b) (ii) The triangle PQR can be seen drawn and labelled in the diagram above.

Revision Exam 5 Paper 2

8.

(2 marks) Total 10 marks (c) Copy and complete the statement below: Area of PQR = ___________ square centimetres Area of PQR = ___________ square centimetres

ANSWER

Revision Exam 5 Paper 2

8. (c)

Revision Exam 5 Paper 2

9. (a) The diagram below shows the line, l, which passes through the points A(–2, –3) and B(2, 3).

ANSWER (i) Determine the gradient of the line, l.

(2 marks)

Revision Exam 5 Paper 2

9. (a) (i) Using A(–2, –3) = A(x1, y1) and B(2, 3) B(x2, y2), then the gradient of AB,

Hence, the gradient of the line, l, is

Revision Exam 5 Paper 2

Alternatively: 9. (a) (i) The gradient of the line, l, is

Revision Exam 5 Paper 2

9. (a) The diagram below shows the line, l, which passes through the points A(–2, –3) and B(2, 3).

ANSWER (ii) Write down the equation of the line, l.

(1 mark)

Revision Exam 5 Paper 2

9. (a) (ii) The y-intercept, c = 0 Using

and c = 0, then the equation

of the line, l, is in the form

Revision Exam 5 Paper 2

9. (b) (i) Given that

, copy and

complete the table below for –4  x  3. x

–4

f(x)

1

–3

–2 –3

–1

0 –3

1

2

3

1 (2 marks) ANSWER

Revision Exam 5 Paper 2

9. (b) (i)

Revision Exam 5 Paper 2

Revision Exam 5 Paper 2

The completed table for –4  x  3 is shown below.

x

–4

f(x)

1

–3

–2 –3

–1

0 –3

1

2 1

3

Revision Exam 5 Paper 2

9. (b) (ii) In the graph of the line, l, and using the same scales and axes, draw the graph of (3 marks)

ANSWER

Revision Exam 5 Paper 2

9. (b) (ii)

The graph of for –4  x  3 can be seen draw on the graph of the line, l, in the diagram above.

Revision Exam 5 Paper 2

9. (b) (iii) Write down the coordinates of the points where the line, l, and the graph of (2 marks) intersect. Total 10 marks

ANSWER

Revision Exam 5 Paper 2

9. (b) (iii) The graphs of the line, l, and the function intersect at the points A(–2, –3) and C(3, 4.5).

Revision Exam 5 Paper 2

10. (a) Shiv’s mean score in five cricket matches was 62 runs. (i) How many runs did he score altogether? After six matches his mean score was 67 runs.

(1 mark)

ANSWER

Revision Exam 5 Paper 2

10. (a) (i) The total number of runs scored in five matches = 62  5 = 310

Revision Exam 5 Paper 2

10. (a) Shiv’s mean score in five cricket matches was 62 runs. (ii) How many runs did he score in the sixth match?

(2 marks)

ANSWER

Revision Exam 5 Paper 2

10. (a) (ii) The total number of runs scored in six matches = 67  5 = 402 The number of runs scored in the sixth match = 402 – 310 = 92

Revision Exam 5 Paper 2

10. (b) The pie chart, not drawn to scale, shows the Sunday morning activities of a group of 480 children.

(i) The sector for children who swim is represented by an angle of 120°. Determine the number of children who swim on Sunday mornings.

ANSWER (3 marks)

Revision Exam 5 Paper 2

10. (b) (i) The number of children who swim on Sunday mornings

Revision Exam 5 Paper 2

10. (b) The pie chart, not drawn to scale, shows the Sunday morning activities of a group of 480 children.

(ii) Given that 200 children sing on Sunday mornings, calculate the value of x.

(2 marks)

ANSWER

Revision Exam 5 Paper 2

10. (b) (ii) The value of

Revision Exam 5 Paper 2

10. (b) The pie chart, not drawn to scale, shows the Sunday morning activities of a group of 480 children.

(iii) Determine the probability that a child chosen (3 marks) at random, dances on Sunday mornings. Total 10 marks

ANSWER

Revision Exam 5 Paper 2

10. (b) (iii) The sector angle that represents the children who dance on Sunday mornings = 360 – (150 + 120)

= 360 – 270 = 90 P (child dances on Sunday mornings)

Revision Exam 5 Paper 2

MATHEMATICS: A COMPLETE COURSE WITH CXC QUESTIONS Text © Raymond Toolsie First Published in 1996 Reprinted in 1997, 1998, 1999, 2000, 2001, 2002, 2003 Second Edition November 2004 Third Edition 2009 ISBN: 976-8014-16-43

Reprinted in 2004, 2005, 2006, 2007, 2008 by Eniath’s Printing Company Limited 6 Gaston Street, Lange Park, Chaguanas, Trinidad, West Indies Designed by: diacriTech, www.diacritech.com All rights reserved. No part of this publication may be reproduced in any form by photostat, microfilm, xerography, or any other means, or incorporated into an information retrieval system, electronic or mechanical, without the written permission of the copyright owner. CARIBBEAN EDUCATIONAL PUBLISHERS LTD., Teddy’s Shopping Centre, Gulf View Link Road, Gulf View, La Romaine, Trinidad, West Indies email: [email protected], Telephone: 868 657 9613 Fax: 868 652 5620