3. 1.(a)The nth term of a sequence is n 2 + 5. List the first three terms of the sequence. When n = 1, n 2 + 5 = 6 When n = 2, n 2 + 5 = 9 When n = 3, n 2 + 5 = 14 6, 9, 14
6. 2.A recipe for making 10 fruit oat bars has the following ingredients. 10 Fruit Oat Bars 80 grams Butter 80 grams Brown Sugar 2 tablespoons Golden Syrup 130 grams Porridge Oats 140 grams Dried Fruit 2 tablespoons Sunflower Seeds (a) Gillian is making fruit oat bars for a charity stall. Complete the following table to show the quantity of each ingredient needed to make 150 fruit oat bars. 150 Fruit Oat Bars ................grams Butter ................grams Brown Sugar ................tablespoons Golden Syrup ................grams Porridge Oats ................grams Dried Fruit ................tablespoons Sunflower Seeds
7. 2.A recipe for making 10 fruit oat bars has the following ingredients. 10 Fruit Oat Bars 80 grams Butter 80 grams Brown Sugar 2 tablespoons Golden Syrup 130 grams Porridge Oats 140 grams Dried Fruit 2 tablespoons Sunflower Seeds 150 Fruit Oat Bars ...... 1200 ..........grams Butter ...... 1200 ........grams Brown Sugar ....... 30 .......tablespoons Golden Syrup ....... 1950 .........grams Porridge Oats ....... 2100 .........grams Dried Fruit ......... 30....... tablespoons Sunflower Seeds X 15
11. (c) A recipe book states that 1 ounce is equivalent to 25 grams. Using this information find whether 5 ounces of butter is sufficient to make 20 fruit oat bars. Show calculations to support your answer. 5oz = 5 X 25 = 125g We need 80g for 10 which is 160g for 20 NO 5oz is not enough!
42. When a drawing pin is thrown it either lands pin-up or pin-down. The relative frequency of the drawing pin landing pin-up was calculated after a total of 20 throws, 40 throws, 60 throws, 80 throws and 100 throws. The results are plotted on the graph below.
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44. When a drawing pin is thrown it either lands pin-up or pin-down. The relative frequency of the drawing pin landing pin-up was calculated after a total of 20 throws, 40 throws, 60 throws, 80 throws and 100 throws. The results are plotted on the graph below. (b) Using the graph, find how many times the drawing pin (i) landed pin-up in the first 40 throws, (ii) landed pin-down in the 100 throws. 0.75 means 75/100 75 300 100 = 400 = 30 out of 40 so 30 times 64 pin-up so 36 pin down
45. Solve the following simultaneous equations by an algebraic (not graphical) method. Show all your working. 3 x + 4 y = –7 2 x + 6 y = –3 Make the x co-efficients equal by multiplying EVERYTHING in the top equation by 2 And EVERYTHING in the bottom equation by 3
46. Solve the following simultaneous equations by an algebraic (not graphical) method. Show all your working. 3 x + 4 y = –7 2 x + 6 y = –3 6x + 8y = -14 6x + 18y = -9 Subtract because the signs are the same! -10y = -5 10y = 5 y = 0.5 If y = 0.5 3x + 4(0.5) = -7 3x + 2 = -7 3x = -9 x = -3
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48. (c) Write each of the following numbers in standard form. (i) 45 000 (ii) 0·0023
49. (c) Write each of the following numbers in standard form. (i) 45 000 (ii) 0·0023 45 000 = 4 . 5 X 10 4 0·0023 = 2 . 3 X 10 -3
50. (d) In a test Jayne was asked to write the answer to 20 multiplied by 490 in standard form. She wrote 9800 as her answer. Explain carefully why this was marked incorrect and give the expected answer. 9800 is the correct answer but it is not in standard form 9800 = 9 . 8 X 10 3
51. (e) Find, in standard form, the value of 4 . 6 X10 -6 2 X 10 -4 2.3 X 10 -2 Remember -6 - -4 = -6 + 4
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53. ¾ = 0.75 = 0.4 = 0.15 These fractions are not recurring so 7/11 is the recurring decimal.
56. (h) Two pieces of string have lengths 23 cm and 14 cm measured correct to the nearest cm. What is the maximum possible total length of the two pieces of string? 23cm ± 0.5cm 14cm ± 0.5cm
57. (h) Two pieces of string have lengths 23 cm and 14 cm measured correct to the nearest cm. What is the maximum possible total length of the two pieces of string? 23cm + 0.5cm = 23.5cm 14cm + 0.5cm = 14.5cm Total = 38cm
58. 12. When ten times a number x is added to three, the result is less than the result of subtracting x from seven. Write down and simplify an inequality which is satisfied by x.
59. 12. When ten times a number x is added to three, the result is less than the result of subtracting x from seven. Write down and simplify an inequality which is satisfied by x. 10x + 3 < 7- x 11x < 7 – 3 11x < 4 x <
60.
61.
62. 13. The histogram below represents the results of gathering and measuring the lengths of pieces of driftwood . (b) Use the histogram to find an estimate for the median. 160 100 120 Half of 520 is 260 so looking at 260 th – which is exactly 60
63. 14. The diagram shows a trapezium OABC with OA parallel to CB. The point Q is the mid-point of AB. Given that OA = x , OB = y and CB = 2 OA express, in their simplest form, each of the following in terms of x and y . (a) OC (b) CQ
64. 14. The diagram shows a trapezium OABC with OA parallel to CB. The point Q is the mid-point of AB. Given that OA = x , OB = y and CB = 2 OA express, in their simplest form, each of the following in terms of x and y . (a) OC (b) CQ x y 2x y-2x AB = -x + y CQ = CB + BQ = 2x – 0.5(-x+y) = 2x +0.5x – 0.5y = 2.5x - 0.5y OC = OB + BC = y – 2x
65. 15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag. (a) Calculate the probability that the two beads are both yellow. (b) Calculate the probability that at least one white bead is drawn.
66. 15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag. (a) Calculate the probability that the two beads are both yellow. (b) Calculate the probability that at least one white bead is drawn. y y y y w w w w b b b b 7 11 3 11 1 11 2 10 3 10 6 10 1 10 7 10 1 10 7 10 3 10 0
67. 15. A bag contains 7 yellow beads, 3 white beads and 1 black bead. Two beads are drawn at random without replacement from the bag. (a) Calculate the probability that the two beads are both yellow. (b) Calculate the probability that at least one white bead is drawn. y y y y w w w w b b b b 7 11 3 11 1 11 2 10 3 10 6 10 1 10 7 10 1 10 7 10 3 10 0 P(y and y) = 7/11 X 6/10 = 42/110 P(y and w) or (w and y) or (w and b) or (w and w) or (b and w) = (7/11 x 3/10) + (3/11 X 7/10) + (3/11 X 1/10) + (3/11 X 2/10) + (1/11 X 3/10) = 21/110 + 21/110 + 3/110 + 6/110 + 3/110 =54/110
68. 16. (a) The diagram shows a sketch of y = f ( x ). On the same diagram, sketch the curve y = f ( x ) + 5. Mark clearly the value of y at the point where this curve crosses the y -axis.
69. 16. (a) The diagram shows a sketch of y = f ( x ). On the same diagram, sketch the curve y = f ( x ) + 5. Mark clearly the value of y at the point where this curve crosses the y -axis.
70. 16. (a) The diagram shows a sketch of y = f ( x ). On the same diagram, sketch the curve y = f ( x ) + 5. Mark clearly the value of y at the point where this curve crosses the y -axis. 3
