( i )

2. 1. An advertising billboard is 5 m wider than it is high. The billboard is 204 m2 of advertising space. Letting x be the height of the billboard, write down the width of the billboard, in terms of x. (i) Width = height + 5 Width = x + 5

3. 1. An advertising billboard is 5 m wider than it is high. The billboard is 204 m2 of advertising space. Form an equation for the area of the billboard, in terms of x. (ii) Area = height  width 204 = x (x + 5) 204 = x2 + 5x

4. 1. An advertising billboard is 5 m wider than it is high. The billboard is 204 m2 of advertising space. Hence, find the dimensions of the billboard. (iii) Area = 204 = x2 + 5x 0 = x2 + 5x – 204 0 = (x + 17)(x – 12) x + 17 = 0 and x – 12 = 0 x = – 17 and x = 12 Lengths cannot be negative, so x = – 17 is rejected x = 12 m = height Width = x + 5 = 12 + 5 = 17 m

5. 2. A triangle has a height that is equal to its base. If the area of the triangle is 32 cm2, find the base and height of the triangle.

6. 3. The length of a rectangle is 7 cm, and the width is 4 cm. Find the area of the rectangle. (i) Area = Length  Width = 7  4 = 28 cm2

7. 3. If both the length and the width are increased by kcm, the area of the rectangle is increased by 102 cm2. Form an equation, in terms of k. (ii) Area = length  width 28 + 102 = (7 + k)  (4 + k) 130 = 7(4 + k) + k(4 + k) 130 = 28 + 7k + 4k + k2 130 = k2 + 11k + 28 0 = k2 + 11k – 102

8. 3. If both the length and the width are increased by kcm, the area of the rectangle is increased by 102 cm2. Solve this equation and hence find the length and width of the larger rectangle. (iii) 0 = k2 + 11k – 102 0 = (k + 17)(k – 6) k + 17 = 0 and k – 6 = 0 k = –17 and k = 6 Lengths cannot be negative, so k = – 17 is rejected. Length = 7 + k = 7 + 6 = 13 cm Width = 4 + k = 4 + 6 = 10 cm

9. 4. A garden has outside dimensions of 35 m in length and 27 m wide. A concrete path, of width wm, is installed all around the inside wall of the garden, leaving a rectangular lawn in the middle of the garden, as shown in the diagram. Find the length and width of the lawn, in terms of w. (i) Length = 35 – 2w Width = 27 – 2w

10. 4. A garden has outside dimensions of 35 m in length and 27 m wide. A concrete path, of width wm, is installed all around the inside wall of the garden, leaving a rectangular lawn in the middle of the garden, as shown in the diagram. Write an expression for the area of the lawn, in terms of w. (ii) Area = Length  Width = (35 – 2w)  (27 – 2w) = 35(27 – 2w) – 2w(27 – 2w) = 945 – 70w – 54w + 4w2 Area = 4w2 – 124w + 945

11. 4. A garden has outside dimensions of 35 m in length and 27 m wide. A concrete path, of width wm, is installed all around the inside wall of the garden, leaving a rectangular lawn in the middle of the garden, as shown in the diagram. If the area of the lawn is 660 m2, find the value of w. (iii) 660 = 4w2 – 124w + 945 0 = 4w2 – 124w + 285 0 = (2w – 5)(2w – 57) 2w – 5 = 0 and 2w – 57 = 0 2w = 5 and 2w = 57 w = 2·5 m and w = 28·5 m w = 28·5 m is rejected because the width of the lawn is only 27 m.Therefore, the path is 2·5 m wide.

12. 5. (i) If xis an even integer, write down the next consecutive even integer, in terms of x. To get to the next even integer, we add on 2: Next even integer = x + 2

13. 5. If the product of half the smaller number and three times the larger number is 72, find x.

14. 6. A rectangular piece of cardboard is 10 cm longer than it is wide. Squares, of side length 2 cm, are cut from each corner and then the sides are folded up to make an open box. Write down the height of the box. (i) Height = 2 cm

15. 6. A rectangular piece of cardboard is 10 cm longer than it is wide. Squares, of side length 2 cm, are cut from each corner and then the sides are folded up to make an open box. Write the length and width of the box, in terms of x. (ii) Length = x + 10 – 4 = x + 6 Width = x – 4

16. 6. A rectangular piece of cardboard is 10 cm longer than it is wide. Squares, of side length 2 cm, are cut from each corner and then the sides are folded up to make an open box. Find the volume of the box, in terms of x. (iii) Volume = Length  Width  Height = (x + 6)  (x – 4)  2 = [x(x – 4) + 6(x – 4)]  2 = [x2 – 4x + 6x – 24]  2 = (x2 + 2x – 24)  2 Volume = 2x2 + 4x – 48

17. 6. A rectangular piece of cardboard is 10 cm longer than it is wide. Squares, of side length 2 cm, are cut from each corner and then the sides are folded up to make an open box. If the volume of the box is 112 cm3, find the value of x. (iv) 112 = 2x2 + 4x – 48 0 = 2x2 + 4x – 160 (2) 0 = x2 + 2x – 80 0 = (x + 10)(x – 8) x + 10 = 0 and x – 8 = 0 x = – 10 and x = 8 Dimensions cannot be negative, so x = − 10 is rejected. Therefore, x = 8

18. 6. A rectangular piece of cardboard is 10 cm longer than it is wide. Squares, of side length 2 cm, are cut from each corner and then the sides are folded up to make an open box. Hence, write down the length and width of the original piece of cardboard. (v) Length = x + 10 = 8 + 10 = 18 cm Width = x = 8 cm

19. 7. Three hundred metres of fencing is available to enclose a rectangular field alongside a river. The river forms one side of the field, as shown in the diagram. Given the width of the field is w, find the length of the field in terms of w. (i) Perimeter = width + length + width = 300 m w + l + w = 300 2w + l = 300 Length of the field: l = 300 – 2w

20. 7. Three hundred metres of fencing is available to enclose a rectangular field alongside a river. The river forms one side of the field, as shown in the diagram. Find the area of the field, in terms of w. (ii) Area = Length  width = (300 – 2w) w Area = 300w – 2w2

21. 7. Three hundred metres of fencing is available to enclose a rectangular field alongside a river. The river forms one side of the field, as shown in the diagram. What dimensions will produce an area of 10,000 m2? (iii) 10,000 = 300w – 2w2 2w2 – 300w + 10,000 = 0 (÷ 2) w2 – 150w + 5000 = 0 (w – 100)(w – 50) = 0 w – 100 = 0 and w – 50 = 0 w = 100 and w = 50

22. 7. Three hundred metres of fencing is available to enclose a rectangular field alongside a river. The river forms one side of the field, as shown in the diagram. What dimensions will produce an area of 10,000 m2? (iii) Length: l = 300 – 2w w = 100 m l = 300 – 2(100) = 300 – 200 = 100 m w = 50 m l = 300 – 2(50) = 300 – 100 = 200 m The dimensions of the field are either: Length = 100 m and Width = 100 m or Length = 200 m and Width = 50 m

23. 8. The cost of producing backpacks with a school’s logo can be modelled by: where C(b) represents the cost, in euro, to produce bbackpacks. Find the cost of producing 50 backpacks. (i)

24. 8. The cost of producing backpacks with a school’s logo can be modelled by: where C(b) represents the cost, in euro, to produce bbackpacks. Find C(150) and explain in words what this value means. (ii) The cost of producing 150 backpacks is €7,700.

25. 8. The cost of producing backpacks with a school’s logo can be modelled by: where C(b) represents the cost, in euro, to produce bbackpacks. How many backpacks can a school get with a budget of €3,000? (iii)

26. 8. The cost of producing backpacks with a school’s logo can be modelled by: where C(b) represents the cost, in euro, to produce bbackpacks. How many backpacks can a school get with a budget of €3,000? (iii)

27. 8. The cost of producing backpacks with a school’s logo can be modelled by: where C(b) represents the cost, in euro, to produce bbackpacks. How many backpacks can a school get with a budget of €3,000? (iii) The number of backpacks cannot be negative, so b = – 65 is rejected. The school can produce 95 backpacks with a budget of €3,000.

28. 9. A cyclist travels 45 km at a speed of xkm/hr. A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist. Write an expression for the time it takes the cyclist to complete his journey, in terms of x. (i)

29. 9. A cyclist travels 45 km at a speed of xkm/hr. A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist. Write an expression for the time it takes the jogger to complete his journey, in terms of x. (ii) Speed of the jogger = speed of the cyclist – 6 Speed of the jogger = x – 6

30. 9. A cyclist travels 45 km at a speed of xkm/hr. A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist. The cyclist completes his journey 20 minutes ahead of the jogger. Use this information to form an equation, in terms of x. (iii)

31. 9. A cyclist travels 45 km at a speed of xkm/hr. A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist. Solve this equation to find the speed of the cyclist. (iv)

32. 9. A cyclist travels 45 km at a speed of xkm/hr. A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist. Solve this equation to find the speed of the cyclist. (iv) Reject x = 54 as too high a value since the speed of the jogger is at x – 6  54 – 6 = 48 km/hr, which is not possible. Therefore, the speed of the cyclist is 15 km/hr

33. 10. • The revenue from selling digital cameras can be modelled by • R(x) = −3x2 + 90x • Where R(x) represents the revenue, in thousands of euro, from sellingx thousand digital cameras. (i) Find the revenue from selling five thousand digital cameras. R(x) = – 3x2 + 90x R(5) = – 3(5)2 + 90(5) = – 3(25) + 450 = – 75 + 450 = 375 R(5) = €375,000

34. 10. • The revenue from selling digital cameras can be modelled by • R(x) = −3x2 + 90x • Where R(x) represents the revenue, in thousands of euro, from sellingx thousand digital cameras. (ii) Find R(4) and explain in words what this value means. R(4) = – 3(4)2 + 90(4) = – 3(16) + 360 = – 48 + 360 = 312 The revenue from selling four thousand digital cameras would be €312,000.

35. 10. • The revenue from selling digital cameras can be modelled by • R(x) = −3x2 + 90x • Where R(x) represents the revenue, in thousands of euro, from sellingx thousand digital cameras. (iii) How many cameras must the company sell to have a revenue of €600,000? Revenue = €600,000 R(x) = 600 600 = – 3x2 + 90x 3x2 – 90x + 600 = 0 x2 – 30x + 200 = 0 The company must sell either 10,000 or 20,000 to make a profit of €600,000. (x – 10)(x – 20) = 0 x – 10 = 0 and x – 20 = 0 x = 10 and x = 20

36. 11. • A rock is dropped from a 180-metre-high cliff, so that it falls into the sea below. • The height of the rock, ℎ, in metres, above the surface of the water, after tseconds, is given by the following function: • ℎ(t) = − 5t2 + 180 (i) Find the height of the rock after 2 seconds. h(t) = – 5t2 + 180 h(2) = – 5(2)2 + 180 = – 5(4) + 180 = – 20 + 180 h(2) = 160 m

37. 11. • A rock is dropped from a 180-metre-high cliff, so that it falls into the sea below. • The height of the rock, ℎ, in metres, above the surface of the water, after tseconds, is given by the following function: • ℎ(t) = − 5t2 + 180 (ii) Find the height of the rock after 3∙5 seconds. h(3·5) = – 5(3·5)2 + 180 = – 5(12·25) + 180 = – 61·25 + 180 h(3·5) = 118·75 m

38. 11. • A rock is dropped from a 180-metre-high cliff, so that it falls into the sea below. • The height of the rock, ℎ, in metres, above the surface of the water, after tseconds, is given by the following function: • ℎ(t) = − 5t2 + 180 (iii) After how many seconds will the rock be at a height of 65 m? Give your answer to one decimal place. 65 = – 5t2 + 180 Time = s 5t2 – 115 = 0 t = 4·8 s t2 – 23 = 0 Time cannot be negative so the is rejected.

39. 11. • A rock is dropped from a 180-metre-high cliff, so that it falls into the sea below. • The height of the rock, ℎ, in metres, above the surface of the water, after tseconds, is given by the following function: • ℎ(t) = − 5t2 + 180 (iv) After how many seconds will the rock hit the surface of the water? The rock will be at the surface of the water, when it has a height of 0 m. h(t) = 0 m t + 6 = 0 and t – 6 = 0 0 = – 5t2 + 180 t = – 6 and t = 6 0 = – t2 + 36 0 = t2 – 36 Time cannot be negative, t = – 6 is rejected. 0 = (t)2 – (6)2 Time = 6s to hit the water. 0 = (t + 6)(t – 6)

40. 12. • (a) The annual net income for a clothing retailer can be modelled by the function: • I(t) = − 1∙5t2 + 32t − 140 • whereI(t) represents the annual net income, in millions of euro, tyears since the company was founded in 2000. (i) Use this model to find the net income the company took in 2005. Explain your answer in words. I(t) = – 1·5t2 + 32t – 140 2005 means t = 5: I(5) = – 1·5(5)2 + 32(5) – 140 = – 1·5(25) + 160 – 140 = – 37·5 + 20 During 2005, the company had a net income of – €17,500,000 = – 17·5

41. 12. • (a) The annual net income for a clothing retailer can be modelled by the function: • I(t) = − 1∙5t2 + 32t − 140 • whereI(t) represents the annual net income, in millions of euro, tyears since the company was founded in 2000. (ii) Use this model to find the net income the company took in 2012. Explain your answer in words. 2012 means r = 12: I(12) = – 1·5(12)2 + 32(12) – 140 = – 1·5(144) + 384 – 140 = – 216 + 244 = 28 During 2012, the company had a net income of €28,000,000.

42. 12. • (a) The annual net income for a clothing retailer can be modelled by the function: • I(t) = − 1∙5t2 + 32t − 140 • whereI(t) represents the annual net income, in millions of euro, tyears since the company was founded in 2000. (iii) During which years did the company show a net income of €20 million? I (t) = 20 20 = – 1·5t2 + 32t – 140 1·5t2 – 32t + 160 = 0 (x 2) 3t2 – 64t + 320 = 0 (3t – 40)(t – 8) = 0 3t – 40 = 0 and t – 8 = 0 3t = 40 and t = 8  2008 (13 year 4 month)  2013 Therefore, the company showed a net income of €20 million, during 2008 and 2013

43. 12. (b) A second clothing retailer has an annual net income, which can be modelled by the function f (t) = 2t + 4 where f (t) represents the annual net income, in millions of euro, t years since the company was founded in 2000. (i) Use this model to find the net income this company took in 2005. Explain your answer in words. f(t) = 2t + 4 2005 means t = 5: f(5) = 2(5) + 4 = 10 + 4 = 14 During 2005, the company had a net income of €14,000,000

44. 12. (b) A second clothing retailer has an annual net income, which can be modelled by the function f (t) = 2t + 4 where f (t) represents the annual net income, in millions of euro, t years since the company was founded in 2000. (ii) In which year(s) did the two companies have the same net income? Justify your answers. I (t) = f(t) – 1·5t2 + 32t – 140 = 2t + 4 0 = 1·5t2 – 30t + 144 0 = 3t2 – 60t + 288 0 = t2 – 20t + 96 0 = (t – 12)(t – 8) t – 12 = 0 and t – 8 = 0 t = 12 and t = 8 Therefore, the companies had the same income in 2008 and 2012.

45. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. Based on the information given above, write a formula for finding the density of a substance. (i)

46. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. Rearrange this formula to make volume the subject of the formula. (ii)

47. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. Liquid A has a density of D kg/litre. Write an expression, in terms of D, for the volume of 3 kg of liquid A. (iii) Liquid B has a density 0∙2 kg/litre greater than liquid A. Write an expression, in terms of D, for the volume of 5∙25 kg of liquid B. (iv) Density of liquid B = D + 0·2

48. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. When 3 kg of liquid A is combined with 5∙25 kg of liquid B, the total volume of liquid present is 6∙25 litres. Use this information to write an equation, in terms of D. (v) VolA + VolB = 6·25 litres

49. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. Hence, find the value of D. (v)

50. 13. The density of a substance is defined as its ‘mass per unit volume’. The density is calculated by dividing the mass of a sample of the substance by its volume. Hence, find the value of D. (v)