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2003 Paper 4

2003 Paper 4. 1.(a) Use your calculator to work out 2 (2.3 + 1.8) x 1.07 Write down all the figures on your calculator display. Put brackets in the expression below so that its value is 45.024 1.6 + 3.8 x 2.4 x 4.2. Use your calculator to work out 2

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2003 Paper 4

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  1. 2003 Paper 4

  2. 1.(a) Use your calculator to work out • 2 • (2.3 + 1.8) x 1.07 • Write down all the figures on your calculator display. • Put brackets in the expression below so that its value is 45.024 • 1.6 + 3.8 x 2.4 x 4.2

  3. Use your calculator to work out 2 (2.3 + 1.8) x 1.07 2 (2.3+1.8)X x1.07= ANSWER 1798.67

  4. Put brackets in the expression below so that its value is 45.024 • 1.6 + 3.8 x 2.4 x 4.2 • (1.6 + 3.8) x 2.4 x 4.2= = 54.432 • 1.6 + (3.8 x 2.4) x 4.2 = 39.904 • 1.6 + 3.8 x (2.4 x 4.2) = 39.904 • (1.6 + 3.8 x 2.4) x 4.2 = 45.024 • ANSWER (1.6 + 3.8 x 2.4) x 4.2

  5. 2. Simon repairs computers. He charges £56.80 for the first hour he works on a computer and £42.50 for each extra hour’s work. (a) Yesterday Simon repaired a computer and charged a total of £269.30 Work out how many hours Simon worked yesterday on this computer. Simon reduces his charge by 5% when he is paid promptly. He was paid promptly for yesterday’s work on the computer. (b) Work out how much he was paid.

  6. £56.80 for the first hour he works on a computer and £42.50 for each extra hour’s work. (a) Yesterday Simon repaired a computer and charged a total of £269.30 Work out how many hours Simon worked yesterday on this computer. 1 hour at £5.60 £269.30 - £56.80 = £212.50 Number of hour worked at £42.50 per hour = £212.50÷£42.50 = 5hours Total hours worked = 5hours + 1 hour = 6 hours ANSWER 6 hours

  7. Simon reduces his charge by 5% when he is paid promptly. He was paid promptly for yesterday’s work on the computer. (b) Work out how much he was paid. £13.465 5% of £269.30 = 5 x £269.30 = 100 £13.47 Amount he was paid = £269.30 - £13.47 ANSWER £255.83

  8. 3. o X o y • This is part of a design of a pattern found at the theatre of Diana at Alexandria • It is made up of a regular hexagon, squares and equilateral triangles. • o • (a) Write down the size of the angle marked x • o • (b) Write down the size of the angle marked y • 2 • The area of each triangle is 2cm . • (c) Work out the area of the regular hexagon. • In the space below use a ruler and compass to construct an equilateral • triangle with sides of length 4 centimetres. • You must show all construction lines.

  9. o X o y This is part of a design of a pattern found at the theatre of Diana at Alexandria It is made up of a regular hexagon, squares and equilateral triangles. o Write down the size of the angle marked x The triangle is an equilateral triangle. All the sides and angles of the equilateraltriangle are equal triangle. o Angles in a triangle add up to 180 . o Angle x = 180 ÷ 3 . O ANSWER X = 60

  10. o (b) Write down the size of the angle marked y o y 4 3 1 2 0 Sum of angles in the hexagon = 4x180 0 Sum of angles in the hexagon = 720 0 Angle Y = 720 ÷ 6 O ANSWER Y = 120

  11. 2 The area of each triangle is 2cm . (c) Work out the area of the regular hexagon. Area Hexagon = 6 x Area Triangle 2 Area Hexagon = 6 x 2 cm O ANSWER Area hexagon = 12 cm

  12. (d) In the space below use a ruler and compass to construct an equilateral triangle with sides of length 4 centimetres. You must show all construction lines. 4 cm

  13. In 2002, Shorebridge Chess Club’s total income came from a council grant • and members’ fees. • Council grant £50 • Members fees 240 x £5 each • (a) (i) Work out the total income of the club for the year 2002. • (ii) Find the council grant as a fraction of the club’s total income • for the year 2002 • Give your answer in the simplest form. • In 2001, the club’s total income was £1000. • The club spent 60% of its total income on a hall. • It spent a further £250 on prizes. • (b) Work out the ratio • The amount spent on the hall : the amount spent on pizzas. • Give your answer in its simplest form.

  14. In 2002, Shorebridge Chess Club’s total income came from a council grant • and members’ fees. • Council grant £50 • Members fees 240 x £5 each • (a) (i) Work out the total income of the club for the year 2002. Total income = Council grant + Member’s fees Total income = £50 + 240x£5 Total income = £50 + £1200 ANSWER Total income = £1250

  15. (ii) Find the council grant as a fraction of the club’s total income • for the year 2002 • Give your answer in the simplest form. council grant = club’s total income £50 £1250 ANSWER 1 25

  16. In 2001, the club’s total income was £1000. The club spent 60% of its total income on a hall. It spent a further £250 on prizes. (b) Work out the ratio The amount spent on the hall : the amount spent on pizzas. Give your answer in its simplest form. amount spent on the hall = 60% of £1250 amount spent on the hall = 60 x £1250 100 amount spent on the hall = 6 x £125 amount spent on the hall = £750 The amount spent on the hall : the amount spent on prizes. £750:£250 = 3:1 ANSWER 3:1

  17. 5. 2 Flowerbed x x • The diagram represents a garden in the shape of a rectangle. • All measurements are given in metres. • The garden has a flowerbed in one corner. • The flowerbed is a square of side x. • Write down an expression, in terms of x, for the shortest side of the garden. • (b) Find an expression, in terms of x, for the perimeter of the garden. • Give your answer in its simplest form. • The perimeter of the garden is 20 metres. • (c) Find the value of x. 5

  18. 5. 2 Flowerbed x • Write down an expression, in terms of x, for the shortest side of the garden. x 5 ANSWER Shortest side = x + 2 (b) Find an expression, in terms of x, for the perimeter of the garden. Give your answer in its simplest form. Perimeter = (x + 2) + (x + 5) + (x + 2) + (x + 5) ANSWER Perimeter = 4X + 14

  19. 5. 2 Flowerbed x x 5 The perimeter of the garden is 20 metres. (c) Find the value of x. 4X + 14 = 20 4X = 20 - 14 4X = 6 X = 6 4 ANSWER X = 1.5

  20. (a) Simplify 5p + 2q – 3p – 3q • y = 5x – 3 • (b) Find the value of y when x = 4

  21. (a) Simplify 5p + 2q – 3p – 3q • 5p + 2q – 3p – 3q • 5p – 3p = • 2p • + 2q– 3q = • - 1q 5p + 2q – 3p – 3q • = 2p – 1q • = 2p – q • ANSWER 2p – q

  22. y = 5x – 3 (b) Find the value of y when x = 4 y = 5x4 – 3 y = 20 – 3 y = 17 ANSWER y = 17

  23. 7. The table shows some rows of a number pattern. (a) In the table, complete row 4 of the number pattern. (b) In the table, complete row 8 of the number pattern. (c) Work out the sum of the first 100 whole numbers. (d) Write down an expression, in terms of n, for the sum of the n whole numbers.

  24. 7. The table shows some rows of a number pattern. = 4x5 2 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2 (a) In the table, complete row 4 of the number pattern. (b) In the table, complete row 8 of the number pattern.

  25. 7. The table shows some rows of a number pattern. = 4x5 2 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2 (c) Work out the sum of the first 100 whole numbers. = 1 + 2 + 3 + 4 + ………………………. + 98 + 99 + 100 ANSWER 100x101 2

  26. 7. The table shows some rows of a number pattern. = 4x5 2 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 8x9 2 (d) Write down an expression, in terms of n, for the sum of the n whole numbers. = 1 + 2 + 3 + 4 + ………………………. + n ANSWER n(n+1) 2

  27. 8. 4.5 m 2.8 m 3.2 m The diagram represents a large tank in the shape of a cuboid. The tank has a base. It does not have a top. The width of the tank is 2.8 metres. The length of the tank is 3.2 metres. The height of the tank is 4.5 metres. The outside of the tank is going to be painted. 2 1 litre of paint will cover 2.5 m The cost of the paint is £2.99 per litre. Calculate the cost of the paint needed to paint the outside of the tank.

  28. 2 Area = 3.2x4.5 m 2 Area = 14.4 m 4.5 m 3.2 m 2 Area = 2.8x4.5 m 2 Area = 3.2x2.8 m 2 Area = 12.6 m 2.8 m 2 Area = 12.6 m 2 Area = 8.96 m 2 Area = 14.4 m

  29. 22 2 2 2 8.96 m + 12.6 m + 12.6 m 14.4 m + 14.4 m Total Area = 2 Total Area = 52.96 m Number of litres of paint = 52.96 ÷ 2.5 litres Number of litres of paint = 21.184 litres Cost of paint = 21.184 x £2.99 ANSWER Cost of paint = £63.34

  30. 2 2 9. Change 2.5 m to cm 1 m = 100 cm 1 m x 1m = 100 cm x 100 cm 22 1 m = 10 0000 cm 22 2.5 m = 2.5 x 10 0000 cm 2 2 ANSWER 2.5 m = 10 000 cm

  31. 10.Bhavana asked some people which region their favourite football team • came from. • The table shows her results. • Complete the accurate pie chart to show these results. • Use the circle given below. Midlands

  32. Angle 22 x 360 = 22 x 4 = 88 90 36 x 360 = 36 x 4 = 144 90 8 x 360 = 8 x 4 = 32 90 24 x 360 = 24 x 4 = 96 90 Total = 90

  33. Four teams, City, Rovers, Town and United play a competition to win a cup. Only one team can win the cup. The table below shows the probabilities of City or Rovers or Town winning the cup. 0.38 + 0.27 + 0.15 = 0.8 x = 1- 0.8 ANSWER x = 0.2

  34. Here are the times, in minutes, taken to change some tyres. • 5 10 15 12 8 7 20 35 24 15 • 20 33 15 25 10 8 10 20 16 10 • In the space below, draw a stem and leaf diagram to show these times. 0 5 7 8 1 0 0 0 0 2 5 5 5 6 0 0 0 4 5 2 3 3 5

  35. 14. 4 cm 10 cm The diagram shows a cylinder with a height of 10 cm and radius 4 cm. Calculate the volume of the cylinder. Give your answer correct to 3 significant figures. The length of a pencil is 13 cm. The pencil cannot be broken (b) Show that this pencil cannot fit inside the cylinder.

  36. 2 Volume = πr h = 2 3 Πx4x10 cm 3 Volume = 160π cm 3 ANSWER Volume = 502 cm

  37. 14. 4 cm 10 cm The length of a pencil is 13 cm. The pencil cannot be broken (b) Show that this pencil cannot fit inside the cylinder.

  38. Pencil will not fit into box 14. 10 cm 12.8 cm 10 cm 8 cm 8 cm Pythagoras Theorem 2 2 2 c = a + b 2 2 2 c = 8 + 10 2 c = 64 + 100 2 c = 164 c = 164 c = 233 c = 12.8 cm

  39. 13. (a) Express the following numbers as products of their prime factors. • 60 • 96 • (b) Find the Highest Common Factor of 60 and 96. • (c) Find the Lowest Common Multiple of 60 and 96.

  40. 13. (a) Express the following numbers as products of their prime factors. (i) 60 Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19 60 = 2 x 30 60 ÷ 2 = 30 60 = 2 x 2 x 15 30 ÷ 2 = 15 60 = 2 x 2 x 3 x 5 15 ÷ 3 = 5 2 ANSWER 60 = 2x3x5

  41. 13. (a) Express the following numbers as products of their prime factors. (ii) 96 Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19 96 = 2 x 48 96 ÷ 2 = 48 96 = 2 x 2 x 24 48 ÷ 2 = 24 96 = 2 x 2 x 2 x 12 24 ÷ 2 = 12 96 = 2 x 2 x 2 x 2 x 6 12 ÷ 2 = 6 96 = 2 x 2 x 2 x 2 x 2 x 3 6 ÷ 2 = 3 5 ANSWER 96 = 2x3

  42. (b) Find the Highest Common Factor of 60 and 96. 60 = 2 x 2 x 3 x 5 96 = 2 x 2 x 2 x 2 x 2 x 3 60 = 2 x 2 x 3 x 5 96 = 2 x 2 x 2 x 2 x 2 x 3 Highest Common Factor = 2 x 2 x 3 ANSWER Highest Common Factor = 12

  43. (c) Find the Lowest Common Multiple of 60 and 96. 1x60 = 60 1x96 = 96 2x60 = 120 2x96 = 192 3x96 = 288 3x60 = 180 4x60 = 240 4x96 = 384 5x96 = 480 5x60 = 300 5x96 = 480 6x60 = 360 7x60 = 420 8x60 = 480 8x60 = 480 ANSWER Lowest Common multiple = 480

  44. 14. A garage keeps records of the costs of repairs to its customers’ cars. The table gives information about the costs of all repairs which were less than £250 in one week. • (a) Find the class interval in which the median lies. • There was only one further repair that week, not included in the table. • That repair cost £1000. • Dave says “The class interval in which the median lies will change.” • (b) Is Dave correct? Explain your answer.

  45. Cumulative Frequency 4 4 + 8 =12 4 + 8 + 7 =19 4 + 8 + 7 + 10 =29 4 + 8 + 7 + 10 + 11 = 40 40 + 1 = 20.5 2 • (a) Find the class interval in which the median lies. ANSWER Median class interval 150<C≤200

  46. There was only one further repair that week, not included in the table. • That repair cost £1000. • Dave says “The class interval in which the median lies will change.” • (b) Is Dave correct? Explain your answer. Cumulative Frequency 4 4 + 8 =12 4 + 8 + 7 =19 4 + 8 + 7 + 10 =29 4 + 8 + 7 + 10 + 11 = 40 250<C 1 4 + 8 + 7 + 10 + 11 + 1 = 41 41 + 1 = 21 2 • (a) Find the class interval in which the median lies. ANSWER Median class interval same Dave is wrong

  47. The garage also sells cars. It offers discounts at 20% of the normal price for cash. Dave pays £5200 cash for a car. (c) Calculate the normal price of the car. Normal Price x Normal Percent = 100% Discount Percent = 100% - 20% = 80% Discount Price £5200 X = 100 £5200 80 X = 100 x £5200 80 ANSWER Normal Price £6500

  48. 15. x x A cuboid has a square of side x cm. The height of the cuboid is 1 cm more than the length cm. 3 The volume of the cuboid is 230 cm . 3 2 (a) Show that x + x = 230

  49. x + 1 x x Volume = length x width x height Volume = X x X x (X+1) = 230 2 Volume = X x (X+1) 3 2 Volume = X + X

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