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Question 1 percentages Question 2 solve equations Question 3 area of parallelogram

Question 1 percentages Question 2 solve equations Question 3 area of parallelogram Question 4 given the circumference, calculate the diameter Question 5 area of a photo frame Question 6 solve an equation using trial and improvement Question 7 find the length of a side in a triangle

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Question 1 percentages Question 2 solve equations Question 3 area of parallelogram

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  1. Question 1 percentages Question 2 solve equations Question 3 area of parallelogram Question 4 given the circumference, calculate the diameter Question 5 area of a photo frame Question 6 solve an equation using trial and improvement Question 7 find the length of a side in a triangle Question 8 solve an inequality Question 9 construct an angle bisector Question 10 multiply out and simplify Question 11 find an angle in a RAT; calculate a side in a RAT Question 12 find an angle in a circle; give reasons Question 13 solve a quadratic equation Question 14 use y = mx+c to identify parallel and perpendicular lines Question 15 solve a non-right angled triangle Question 16 from surface area of a sphere, calculate its volume Question 17 solve an equation involving algebraic fractions Grade boundaries

  2. 1. (a) Calculate 36% of £420 0.36 x 420 = 151.2 36% = 0.36 as a decimal Answer = £151.20 2 marks 1 (b)Mike says that 18% of £840 is equal to 36% of £420 Explain why Mike is correct. Do not calculate 18% of £840. Answer: the percentage has been halved; the amount has been doubled; the effect is to leave the answer the same 1 mark 1 (c) 9% of £x is equal to 36% of £420 Work out the value of x double 840 = 1680 the percentage has been halved again so the amount must be doubled again Answer x = 1680 2 marks

  3. 2 Solve the equations (a) 4(z + 3) = 8 (b) 3t + 4 = 19 – 2t divide by 4 z + 3 = 2 z = -1 3 marks OR multiply out the bracket by 4 4z + 12 = 8 4z = -4 z = -1 - 2t add 2t to both sides 5t + 4 = 19 subtract 4 5t = 15 divide by 5 t = 3 3 marks

  4. 3 XYZT is a parallelogram The base XT = 11.6cm and the perpendicular height is 7.7 cm Calculate the area of the parallelogram. Area of parallelogram = base x height Area of parallelogram = 11.6 x 7.7 = 89.32 = 89.3 cm2 to 3 sig figs 2 marks

  5. 4 The circumference of a circle measures 26.7 cm Work out the length of the diameter of the circle. Circumference = π x diameter so 26.7 = π x diameter so diameter = 26.7 ÷ π = 8.498873961 = 8.50 cm to 3 sig figs 2 marks

  6. 5 A rectangular photo is surrounded by a frame which is 4cm wide. The outer measurements of the frame are 21.5cm by 17.2cm. Calculate the area of the frame. This area is shaded in the diagram. Size of photo is 8cm less than size of frame Area of photo = 13.5 x 9.2 = 124.2 Area of frame and photo = 21.5 x 17.2 = 369.8 Area of frame = 369.8 – 124.2 = 245.6 = 246 cm2 (to 3 sig figs) 5 marks

  7. 6 Liam is using trial and improvement to find a solution to the equation x3 + 4x = 72 The table shows the first two trials Continue the table to find a solution to the equation. Give your answer to 1 decimal place. 3.5 3.53 + 4x3.5 = 56.875 too small 3.7 3.73 + 4x3.7 = 65.453 too small 3.8 3.83 + 4x3.8 = 70.072 too small 3.9 3.93 + 4x3.9 = 74.919 too large 3.85 3.853 + 4x3.85 = 72.466 too large x = 3.8 to 1 decimal place 3 marks

  8. 7 The diagram shows an equilateral triangle of side 18cm. Calculate the height of the triangle (marked h in the diagram) h2 = 182 - 92 h2 = 243 h = √243 h = 15.58845727 = 15.6 cm (to 3 sig figs) Find the RAT – use Pythagoras’ rule 4 marks

  9. 8 (a) Solve the inequality 5x + 3  18 Subtract 3 from both sides 5x  15 Divide by 5 x  3 2 marks (b) y is an integer Write down all the solutions of the inequality -6  2y  0 Divide by 2 -3  y  0 The list of solutions is -3, -2 and -1 3 marks

  10. 9 Using a ruler and compasses only, construct the bisector of angle DEF 2 marks

  11. 10 (a) Multiply out 2p2q3 x 3p5q 2p2q3 x 3p5q 2 x 3=6 p2 x p5 = p7 q3 x q = q4 Answer = 6p7q4 2 marks (b) Expand and simplify (p + 7)(p + 2) = p2 + 9p +14 p2 7p 2p 14 2 marks

  12. 10 (c) solve the equation x + 1 = 15 – 6x Multiply by 3 Add 6x 7x + 1 = 15 Subtract 1 7x = 14 x = 2 Divide by 7 3 marks

  13. RATs with angles means SOHCAHTOA 11 (a) In triangle ABC, angle A = 900, AB = 16cm and AC = 7cm. Calculate the value of x. opposite adjacent tan x = opposite ÷ adjacent tan x = 16 ÷ 7 x = tan-1(16 ÷ 7) x = 660 (to the nearest degree) 3 marks

  14. RATs with angles means SOHCAHTOA 11 (b) In triangle PQR, angle Q = 900, angle R = 370 and QR = 12.6cm. Calculate the length of PR. hypotenuse adjacent hypotenuse = adjacent ÷ cosine of the angle PR = 12.6 ÷ cos 37 PR = 15.77690929 PR = 15.8 cm (to 3 sig figs) 3 marks

  15. 12 In the diagram, A, P, Q and B are points on the circumference of the circle. Angle APB = 470. Find the value of x. Give a reason for your answer. angle APB = angle AQB These are equal angles in the same segment, standing on the same chord AB x = 470 2 marks

  16. 13 Solve the equation y2 – 3y - 14 = 0 Give your answer to 2 decimal places. You must show your working. … helpful formula on page 2 Compare y2– 3y - 14 = 0 with ax2 + bx + c = 0 a = 1, b = -3 and c = -14 x = -2.53 or x = 5.53 (to 2 dec pl) 3 marks

  17. 14 (a) The equations of four lines are given below Line P: y = 3x + 5 Line Q: y = 4 – 3x Line R: y + 3x = 8 Line S: y – 3x = 1 (i) Name the lines that are parallel to the line y = 3x. Rewrite in the form y = mx + c y = -3x + 4 y = -3x + 8 y = 3x + 1 1 mark lines P and S have gradient = 3 (ii) Which line goes through the point (2, 7) ? when x = 2, y = 7 only on line S 1 mark (b) Write down the gradient of the line y + 2x = 7 1 mark rewrite as y = -2x + 7 gradient = -2

  18. 15 D, E and F are three points on level ground. TF is the vertical side of a building. The angle of elevation of the top of the building from D is 160. The angle of elevation of the top of the building from E is 350. The distance DE is 30m. Calculate the height of the building marked h in the diagram. 5 marks

  19. 19 145 Use the sine rule to find x in this triangle DET x = 30 x sin 16 ÷ sin 19 x = 25.39904644 Use SOHCAHTOA to find h from x in the RAT, TEF h = 25.399 x sin 35 h = 14.6 cm (to 3 sig figs) 5 marks

  20. Helpful formulas from page 2 16 The surface area of a sphere is 1380 cm2. Calculate the volume of the sphere. State the units of your answer. 4πr2 = 1380  r2 = 1380 ÷ (4xπ) Use brackets on your calculator  r = √109.8169107 Volume = 4 π r3 ÷ 3  r = 10.47935641 = 4 π x 10.4793 ÷ 3 = 4820 cm3 (to 3 sig figs) 5 marks

  21. 17 Solve the equation Multiply by (y + 7) Multiply by (y + 6) Multiply by 2 6(y + 6) – 2(y + 7) = (y + 6)(y + 7) 6y + 36 – 2y - 14 = y2 +13y + 42 Multiply out 0 = y2 +9y + 20 Rearrange to quadratic equation Factorise to solve (y + 5)(y + 4) = 0 y = -5 or y = -4 5 marks

  22. Total: out of 70 a rough guide

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