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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

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## INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

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