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Question 1 solve equations Question 2 substitute numbers for letters

Question 1 solve equations Question 2 substitute numbers for letters Question 3 reflection and enlargement Question 4 n th term of a sequence Question 5 construct perpendicular bisector; locus Question 6 quadrilaterals – facts about sides, angles and diagonals

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Question 1 solve equations Question 2 substitute numbers for letters

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  1. Question 1 solve equations Question 2 substitute numbers for letters Question 3 reflection and enlargement Question 4 nth term of a sequence Question 5 construct perpendicular bisector; locus Question 6 quadrilaterals – facts about sides, angles and diagonals Question 7 table of values; draw graph; solve quadratic Question 8 solve RATs with Pythagoras and SOHCAHTOA Question 9 circle theorems Question 10 factorise quadratic expressions Question 11 calculate perimeter of a sector of a circle; volume of a prism Question 12 transformations of graph of y =sinx Question 13 volume of frustum of a pyramid Question 14 area of a triangle Question 15 complete the square; solve quadratic equation Grade boundaries

  2. 1. Solve these equations (a) 4x – 7 = 5 (Add 7) 4x = 12 ( c ) 7z + 2 = 9 – 3z 2 Marks x = 3 (Add 3z) 10z + 2 = 9 (b) 2( y + 5 ) = 28 ( - 2 ) 10z = 7 2y + 10 = 28 ( - 10 ) 2y = 18 3 Marks y = 9 3 marks

  3. 2 (a) Find the value of 5p + 2q when p = 4 and q = -7 5 x 4 + 2 x -7 20 +-14 2 Marks 6 (b) Find the value of u2 – v2 when u = 5 and v = 3 52 – 32 2 Marks 25 – 9 = 16

  4. x= 3 3. The diagram shows two shapes P and Q Describe fully the single transformation which takes shape P onto shape Q Reflection in x = 3 2 Marks

  5. 3 (b) the vertices of triangle T are ( 1, 1 ), ( 1,2 ) and ( 4,1 ) Enlarge triangle T by scale factor 2, with ( 0, 0 ) as the centre of enlargement. 3 Marks

  6. 4. A sequence of numbers is shown. • 5 8 11 14 • a) Find an expression for the nth term of the sequence. Goes up by 3 each time • 2 • 5 • 8 • 11 • 14 Answer 3n - 1 2 Marks n 3n - 1 b) Explain why 99 will not be a term in this sequence. The answer must be one less than a multiple of 3. 99 is a multiple of 3. 98 would be in the sequence so would 102. 3n – 1 = 99 does not give a whole number. 2 marks

  7. 5 (a) The line LM is drawn below. Use ruler and compass to construct the perpendicular bisector of LM. You must show clearly all your construction arcs. 2 Marks b) Complete the sentence. The perpendicular bisector of LM is the locus of points which are ……… Equidistant from the two fixed points L and M 1 Mark

  8. 6. Here is a list of quadrilaterals. Kite rectangle rhombus square trapezium For each of the following description, choose the correct name from the list. a) One pair of sides are parallel. The other two sides are not parallel. 1 Mark Trapezium 1 Mark b) All the angles are the same size. Only opposite sides are equal. Rectangle c) All the sides are the same length. The diagonals are not equal in length. 1 Mark Rhombus

  9. 7. a) Complete the table of values for y = 2x2 – 4 x – 1 5 -3 2 ( -1 )2 – 4 ( -1 ) - 1 2 ( 1 )2 – 4 (1) -1 2 Marks 2 + 4 – 1 = 5 2 – 4 – 1 = - 3 b) On the grid opposite, draw the graph of y = 2x2 – 4x – 1 for values of x from -2 to 3.

  10. b) On the grid, draw the graph of y = 2x2 – 4x – 1 for values of x from -2 to 3. 2 Marks An approximate solution of the equation 2x2 - 4x – 1 = 0 is x = 2.2 i) Explain how you can find this from the graph and ii) use your graph to find another solution to this equation Intersection with the x axis 1 Mark x = - 0.2 1 Mark

  11. 8 (a) The diagram shows a right angled triangle ABC. AB = 10cm and AC = 15cm. Calculate the length of BC. Leave your answer as a square root. C 15cm 102 + BC2 = 152 100 + BC2 = 225 BC2 = 225 – 100 BC2 = 125 BC = √ 125 or 5 √5 A B 10cm 3 Marks

  12. 8 (b) The diagram shows a right – angled triangle DEF. EF = 10cm. Angle F = 50º . Use the table to of data to work out the length of DE. D Not drawn accurately 50º F E 10cm Given Adjacent side Find Opposite side Use Tan Tan 50 = DE ÷10 10 X 1.192 = DE DE = 11 .92 cm 3 Marks

  13. 9 (a) In the diagram, O is the centre of the circle and P, q and R are points on the circumference. Angle P = 25º. Work out the size of angle R. Q Remember angle in a semi circle = 90º 25º R P O Angle R = 180 – ( 25 + 90) Angle R = 65º 2 Marks

  14. 9 (b) A, B, C and D are four points on the circumference of another circle. AC meets BD at X. Angle ABD = 56º and angle CXD = 80º Work out the value of d. You must show all your working. B Angles in the same segment Angle ACD = 56º C 56º 80º d = 180 – ( 80 + 56) d = 44º X d A D 3 Marks

  15. 10(a) Factorise x2 - 10x + 25 2 Marks Answer ( x – 5 ) ( x – 5 ) ( b ) Factorise 2x2 + 3x – 5 2 Marks Answer ( 2x + 5 ) ( x – 1 )

  16. Length of arc: x 2 x π x 9 = 4π 11 (a) The diagram shows a sector of a circle of radius 9 cm Find the perimeter of the sector. Give your answer in terms of π Perimeter of sector: 4π + 9 + 9 4π + 18 3 Marks

  17. Area of sector: x π x 92 = 18π 11(b). The cross section of a prism is a sector of a circle, of radius 9cm as shown in the diagram. The height of the prism is 10cm. Calculate the volume of the prism. Give your answer in terms of π. Volume of prism: = 18π x 10 = 180π cm3 1 mark for units 4 Marks

  18. 12The diagram shows the graph of y = sin xº for 0 x  360 (a) Sketch the graph of y = 2 sin xº for 0 x  360 1 mark

  19. 12(b) Sketch the graph of y= sin 2xº for 0 x  360 12(c) Sketch the graph of y = 2 + sin xº for 0 x  360 1 mark 1 mark

  20. 13. A square – based pyramid A is divided into 2 parts: a square-based pyramid B and a frustum C, as shown. Pyramid A is similar to pyramid B. The base of pyramid A is a square of side 10cm. The base of pyramid B is a square of side 5cm. The vertical height of pyramid A is 12cm.

  21. 13 You are given the formula Volume of a pyramid = ⅓ x area of base x vertical height Calculate the volume of the frustum C. Volume of A ⅓ x 10 x 10 x 12 = 400 Volume of B ⅓ x 5 x 5 x 6 = 50 Volume of C 400 – 50 = 350cm3 Dimensions of B are ½ the dimensions of A 4 Marks

  22. 14 The diagram shows a triangle ABC. AB = 6cm, BC = 5cm and angle B = 75º Calculate the area of the triangle. Give your answer to a suitable degree of accuracy. A You are given that sine 75º = 0.966 (3sf) Area of triangle = ½ ac sine B (from page 2) 6cm Area = ½ x 6 x 5 x sine75 = 15 x 0.966 =14.49 =14.5 cm2 ( 3sf) 75º B 5cm C 3 Marks

  23. 15 (a) Find the values of a and b such that x2 + 6x – 3 = ( x + a )2 + b Completing the square x2 + 6x – 3 = ( x + 3 )2 – 12 a= 3, b = –12 2 marks ( b) Hence or otherwise, solve the equation x2 + 6x – 3 = 0 giving your answer in surd form x2 + 6x – 3 = 0 ( x+ 3 )2 – 12 = 0 ( x + 3 )2 = 12 x + 3 = ±√ 12 x = –3 ±√12 3 Marks

  24. Total: out of 70 - a rough guide

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