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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

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  1. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Please decide which Unit you would like to revise: UNIT 1 UNIT 2 UNIT 3 UNIT 4 Calculations using % Volumes of Solids Linear Relationships Algebraic Operations Circles Trigonometry Simultaneous Linear Equations Graphs, Charts & Tables Statistics Algebraic Operations Quadratic Functions Further Trigonometry Calculations in a Social Context Logic Diagrams Formulae EXIT

  2. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK Calculations using Percentages Volumes of Solids UNIT 1 : Linear Relationships Algebraic Operations Circles EXIT

  3. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Calculations using Percentages UNIT 1 : Please choose a question to attempt from the following: 1 2 3 4 Back to Unit 1 Menu EXIT

  4. Calculations using Percentages : Question 1 • In 2001, John deposits £650 in the bank at an interest rate of 3.5%. • How much is his deposit worth after 1 year? • In 2002, the interest rate changed to 3.2%, and in 2003 it changed again to 4.1%. Calculate how much interest John will have earned at the end of 2003. Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using Percentages Menu Go to Main Menu EXIT

  5. £22.75 • £72.75 Calculations using Percentages : Question 1 • In 2001, John deposits £650 in the bank at an interest rate of 3.5%. • How much is his deposit worth after 1 year? • In 2002, the interest rate changed to 3.2%, and in 2003 it changed again to 4.1%. Calculate how much interest John will have earned at the end of 2003. Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using Percentages Menu Go to Main Menu EXIT

  6. Value after 1 year = 22.75 + 650 = £672.75 Total interest = 722.75 – 650 = £72.75 Back to Home Question 1 • 3.5% of 650 = 0.035 x 650 = 22.75 • In 2001, John deposits £650 in the bank at an interest rate of 3.5%. • How much is his deposit worth after 1 year? • In 2002, the interest rate changed to 3.2%, and in 2003 it changed again to 4.1%. Calculate how much interest John will have earned at the end of 2003. OR 650 x 1.035 = £672.75 (b) 3.2% of 672.75 = 0.032 x 672.75 = 21.53 Value at end of 2002 = 672.75 + 21.53 = £694.28 4.1% of 694.28 = 0.041 x 694.28 = 28.47 Value at end of 2003 = 694.28 + 28.47 = £722.75 Begin Solution Continue Solution OR 672.75 x 1.032 x 1.041 = 722.74 Markers Comments Total Interest = 722.74 – 650 = £72.74 Calculations using PercentagesMenu

  7. State calculation for percentage Clearly add interest and original amount Value after 1 year = 22.75 + 650 = £672.75 Interpret interest percentage Use the answer from part (a) Substitute correct value Clearly calculate total interest Total interest = 722.75 – 650 = £72.75 Back to Home Marker’s Comments • 3.5% of 650 = 0.035 x 650 = 22.75 OR 650 x 1.035 = £672.75 (b) 3.2% of 672.75 = 0.032 x 672.75 = 21.53 Value at end of 2002 = 672.75 + 21.53 = £694.28 4.1% of 694.28 = 0.041 x 694.28 = 28.47 Value at end of 2003 = 694.28 + 28.47 = £722.75 Next Comment OR 672.75 x 1.032 x 1.041 = 722.74 Calculations using PercentagesMenu Total Interest = 722.74 – 650 = £72.74

  8. Calculations using Percentages – Question 2 Joy buys a piano for £350. It depreciates by 15% in the first year and 20% in the second. How much is the piano worth after 2 years? Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using Percentages Menu Go to Main Menu EXIT

  9. Calculations using Percentages – Question 2 Joy buys a piano for £350. It depreciates by 15% in the first year and 20% in the second. How much is the piano worth after 2 years? £238.00 Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using PercentagesMenu Go to Main Menu EXIT

  10. Value after 2 years = 297.50 – 59.50 = £238.00 OR 350 x 0.85 x 0.8 = £238.00 Back to Home Question 2 15% of 350 = 0.15 x 350 = 52.50 Joy buys a piano for £350. It depreciates by 15% in the first year and 20% in the second. How much is the piano worth after 2 years? Value after 1 year = 350 - 52.50 = 297.50 20% of 297.50 = 59.50 Begin Solution Continue Solution Markers Comments Calculations using PercentagesMenu

  11. Know to subtract for depreciation Use 297.50 rather than 350 Value after 2 years = 297.50 – 59.50 = £238.00 Calculate depreciation term OR 350 x 0.85 x 0.8 = £238.00 Repeat decrease by a set percentage Back to Home Marker’s Comments 15% of 350 = 0.15 x 350 = 52.50 Value after 1 year = 350 - 52.50 = 297.50 20% of 297.50 = 59.50 Next Comment Calculations using PercentagesMenu

  12. Calculations using Percentages – Question 3 • Joe buys a house for £93,000. Three years later it is worth £120,000. • Calculate the percentage increase in the value of Joe’s house as a percentage of the original price. • Calculate the current value of the house as a percentage of the original price. Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using PercentagesMenu Go to Main Menu EXIT

  13. 29 % • 129 % Calculations using Percentages – Question 3 • Joe buys a house for £93,000. Three years later it is worth £120,000. • Calculate the percentage increase in the value of Joe’s house as a percentage of the original price. • Calculate the current value of the house as a percentage of the original price. Reveal answer only Go to full solution Go to Marker’s Comments Go to Straight Line Menu Go to Main Menu EXIT

  14. Percentage Increase = (b) Back to Home Question 3 • Joe buys a house for 93,000. Three years later it is worth £120,000. • Calculate the percentage increase in the value of Joe’s house as a percentage of the original price. • Calculate the current value of the house as a percentage of the original price. (a) Increase = 120 000 – 93 000 = 27 000 Begin Solution Continue Solution Markers Comments Calculations using PercentagesMenu

  15. Calculate actual increase Percentage Increase = Express answer as a percentage (b) Back to Home Marker’s Comments (a) Increase = 120 000 – 93 000 = 27 000 Next Comment Calculations using PercentagesMenu

  16. Calculations using Percentages – Question 4 Adam puts money into a bank. It increases by 5% and is now worth £596.40. How much money did Adam originally put in the bank? Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using PercentagesMenu Go to Main Menu EXIT

  17. £568.00 Calculations using Percentages – Question 4 Adam puts money into a bank. It increases by 5% and is now worth £596.40. How much money did Adam originally put in the bank? Reveal answer only Go to full solution Go to Marker’s Comments Go to Calculations using PercentagesMenu Go to Main Menu EXIT

  18. Back to Home Question 4 105% = 596.40 Adam puts money into a bank. It increases by 5% and is now worth £596.40. How much money did Adam originally put in the bank? 1% = 5.68 100% = 568.00 => £568 invested originally Begin Solution Continue Solution Markers Comments Calculations using PercentagesMenu

  19. Notice that £596.40 = 105% Calculate 1%, and similarly 100% Clearly state original investment Back to Home Marker’s Comments 105% = 596.40 1% = 5.68 100% = 568.00 => £568 invested originally Next Comment Straight Line Menu

  20. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Volumes of Solids UNIT 1 : Please choose a question to attempt from the following: 1 2 3 Back to Unit 1 Menu EXIT

  21. 15cm Volumes of Solids – Question 1 A spherical football has a radius of 15 cm. Find the volume of the football., to 3 significant figures. Reveal answer only Go to full solution Go to Marker’s Comments Go to Volume of Solids Menu Go to Main Menu EXIT

  22. 15cm Volumes of Solids – Question 1 A spherical football has a radius of 15 cm. Find the volume of the football, to 3 significant figures. Reveal answer only 14100 cm³ Go to full solution Go to Marker’s Comments Go to Volume of SolidsMenu Go to Main Menu EXIT

  23. Back to Home Question 1 V A spherical football has a radius of 15 cm. Find the volume of the football, to 3 significant figures. = 14137.17 = 14100 cm³ Begin Solution Continue Solution Markers Comments Volume of SolidsMenu

  24. Select appropriate formula Substitute values Clearly state answer Round answer to 3 s.f. as requested Back to Home Marker’s Comments V = 14137.17 = 14100 cm³ Next Comment Volume of SolidsMenu

  25. 15 cm 12cm 12cm B 18 cm A Volumes of Solids – Question 2 A and B are 2 different shaped candles. A is a cone and B is a cylinder. Both cost £4.00. Which candle is better value for money? (Justify your answer.) Reveal answer only Go to full solution Go to Marker’s Comments Go to Volume of SolidsMenu Go to Main Menu EXIT

  26. 15 cm 12cm 12cm B 18 cm A Volumes of Solids – Question 2 A and B are 2 different shaped candles. A is a cone and B is a cylinder. Both cost £4.00. Which candle is better value for money? (Justify your answer.) Reveal answer only Go to full solution Go to Marker’s Comments Go to Volume of SolidsMenu Go to Main Menu Candle B because it has a larger volume and therefore will burn for longer (or similar). EXIT

  27. Back to Home Question 2 A and B are 2 different shaped candles. A is a cone and B is a cylinder. Both cost £4.00. Which candle is better value for money? (Justify your answer.) Begin Solution Continue Solution Markers Comments Candle B is better value because it has a larger volume => it will burn longer. (Or similar) Volume of SolidsMenu

  28. State formula for volume of a cone Substitute values State volume in cm³ State formula for volume of a cylinder Substitute values State volume in cm³ Decide which is better value and give a relevant reason Back to Home Marker’s Comments Next Comment Candle B is better value because it has a larger volume => it will burn longer. (Or similar) Volume of SolidsMenu

  29. 94cm 52cm 120 cm Volumes of Solids – Question 3 A watering trough is shown in diagram A. Diagram B gives the dimensions of the cross-section. Calculate the volume of the trough, in litres, to 2 significant figures. Diagram A Reveal answer only 800 cm Go to full solution Go to Marker’s Comments Go to Volume of SolidsMenu Diagram A Go to Main Menu EXIT

  30. 94cm 52cm 120 cm Volumes of Solids – Question 3 A watering trough is shown in diagram A. Diagram B gives the dimensions of the cross-section. Calculate the volume of the trough, in litres, to 2 significant figures. Diagram A Reveal answer only 800 cm Go to full solution Go to Marker’s Comments Go to Volume of SolidsMenu Diagram A Go to Main Menu 4800 litres EXIT

  31. Back to Home Question 3 A watering trough is shown in diagram A. Diagram B gives the dimensions of the cross-section. Calculate the volume of the trough, in litres, to 2 significant figures. Begin Solution Continue Solution Markers Comments Candle B is better value because it has a larger volume => it will burn longer. (Or similar) Volume of Solids Menu

  32. State formula for volume of a cone Substitute values State volume in cm³ State formula for volume of a cylinder Substitute values State volume in cm³ Decide which is better value and give a relevant reason Back to Home Marker’s Comments Next Comment Candle B is better value because it has a larger volume => it will burn longer. (Or similar) Volume of SolidsMenu

  33. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Linear Relationships UNIT 1 : Please choose a question to attempt from the following: 1 2 3 Back to Unit 1 Menu EXIT

  34. Linear Relationships - Question 1 Draw the graph of Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  35. Linear Relationships - Question 1 Draw the graph of Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  36. X 0 2 4 6 Y 2 1 0 -1 Back to Home Question 1 Draw the graph of Begin Solution Continue Solution Markers Comments Linear Relationships Menu

  37. X 0 2 4 6 Y 2 1 0 -1 Back to Home Marker’s Comments Clearly mark at least 3 points on the grid Extend the line as far as the grid allows Next Comment Linear Relationships Menu

  38. Linear Relationships - Question 2 Find the equation of the straight line shown in the diagram Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  39. Linear Relationships - Question 2 Find the equation of the straight line shown in the diagram Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  40. Gradient Back to Home Question 2 Y-intercept = -2 Find the equation of the straight line shown in the diagram Begin Solution Continue Solution Markers Comments Linear Relationships Menu

  41. Clearly state y-intercept Clearly state gradient Finish with the equation of the straight line of the form y=mx +c Back to Home Marker’s Comments Y-intercept = -2 Gradient Next Comment Linear Relationships Menu

  42. Linear Relationships - Question 3 Does y = 3x -2 pass through the point (5,10)? Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  43. Linear Relationships - Question 3 Does y = 3x -2 pass through the point (5,10)? y = 3x – 2 does not pass through (5, 10) Reveal answer only Go to full solution Go to Marker’s Comments Go to Linear Relationships Menu Go to Main Menu EXIT

  44. Back to Home Question 3 Does y = 3x -2 pass through the point (5,10)? At (5, 10), x = 5 and y = 10 3 x 5 – 2 = 13 Therefore y = 3x – 2 does not pass through (5, 10) as 3 x 5 – 2 ≠ 10 Begin Solution Continue Solution Markers Comments Linear Relationships Menu

  45. Substitute values for x and y correctly State answer clearly Back to Home Marker’s Comments At (5, 10), x = 5 and y = 10 3 x 5 – 2 = 13 Therefore y = 3x – 2 does not pass through (5, 10) as 3 x 5 – 2 ≠ 10 Next Comment Linear Relationships Menu

  46. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK You have chosen to study: Algebraic Operations UNIT 1 : Please choose a question to attempt from the following: 1 2 3 Back to Unit 1 Menu EXIT

  47. Algebraic Operations - Question 1 Remove brackets and simplify this expression. 6 – 8(2x -7) Reveal answer only Go to full solution Go to Marker’s Comments Go to Algebraic Operations Menu Go to Main Menu EXIT

  48. Algebraic Operations - Question 1 Remove brackets and simplify this expression. 6 – 8(2x -7) 62 – 16X Reveal answer only Go to full solution Go to Marker’s Comments Go to Algebraic Operations Menu Go to Main Menu EXIT

  49. Back to Home Question 1 Remove brackets and simplify this expression. 6 – 8(2x -7) 6 – 8(2x -7) = 6 – 16x + 56 = 62 – 16x Begin Solution Continue Solution Markers Comments Algebraic Operations Menu

  50. Multiply bracket by -8 Simplify expression by collecting like terms Back to Home Marker’s Comments 6 – 8(2x -7) = 6 – 16x + 56 = 62 – 16x Next Comment Algebraic Operations Menu