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Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins

Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins. 1. In the diagram AB is parallel to CD. (a) State the value of x. Give a reason for your answer. Answer:- 130 degrees Reason:- Corresponding angle. 2 Marks.

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Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins

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  1. Module 5 Paper1 Higher Non Calculator Specimen Paper (non-coursework) Mock exam 2009 1hr 15mins

  2. 1. In the diagram AB is parallel to CD. (a) State the value of x. Give a reason for your answer. Answer:- 130 degrees Reason:- Corresponding angle 2 Marks (b) State the value of y. Give a reason for your answer. Answer:- 130 degrees Reason:- Alternate angle or vertically opposite 2 Marks ( c) Find the value of z Answer:- 50 degrees 1 Mark

  3. 2. Javed says that the triangle and parallelogram shown have the same area. Is he correct? You MUST show your working. Triangle area = ½ x 4 x 12 = 24cm2 Parallelogram area = 8 x 3 = 24cm2 He is correct 4 Marks

  4. 3 (a) The nth term of a sequence is 5n-2 Write down the first three terms of the sequence 5 (1) – 2 = 3 5 (2) – 2 = 8 5 (3) – 2 = 13 Answer:- 3, 8, 13 2 Marks (b) Matchsticks are used to make this pattern of triangles Find an expression for the number of matchsticks needed to make Pattern n Answer:- 2n + 1 2 Marks +2 How many complete triangles can be made with 40 matchsticks? 2n + 1 = 40 2n = 39 n = 19.5 19 complete triangles n 2n + 1 2 Marks

  5. 4. (a) Complete the table of values for y = x2 + x Needs to be a smooth curve 2 2 Marks 6 (b) Draw the graph of y = x2 + x for values of x from -4 to +3 x x (c ) Write down the values of x where the line y = 5 crosses the graph Y = 5 x Answer : 1.8 and -2.8 x 2 Marks (d) What happens top the value of y between x=-1 and x=0? x x 2 Marks x x Answer: It is negative 1 Mark

  6. Simplify (a) 2(q + 3) + 3(q – 4) 2q + 6 + 3q - 12 2 Marks = 5q - 6 (b) x2 X x5 Answer: x7 1 Mark (c) 1 Mark Answer: y3

  7. 6. (a) Describe fully the single transformation that takes shape A onto shape B Reflection in the line y = 2 2 Marks (b) Triangle C is rotated onto triangle D. (i) Write down the angle of rotation. Answer 180º (ii) Write down the coordinates of the centre of rotation Answer ( 3, 0 ) 2 Marks

  8. 6 (c) Draw the new position of shape L 2 Marks

  9. 7. (a) In triangle ABC, angle B = 90º, AB = 9cm and AC = 15cm Calculate the length of BC 152 = 92 + BC2 225 – 81 = BC2 BC2 = 144 BC =√ (144) 3 Marks BC = 12 cm

  10. 7(b) In triangle DEF , angle E = 90º, DF = 10cm and angle F = 70º hypotenuse opposite adjacent Use the table of data to work out the length of EF Given Hypotenuse Need to find Adjacent Use COSINE Cos 70º = Cos 70º= 10 x 0.342 = EF 3 Marks EF = 3.42 cm

  11. 8(a) Expand and simplify (x + 3)(x – 4) x2 + 3x - 4x - 12 x2 - x - 12 2 Marks (b) Factorise fully 6a2 – 9ab 3(2)aa - 3(3)ab 3a(2a – 3b) 2 Marks (c) Factorise x2 – 7x + 10 Hence solve x2 - 7x + 10 = 0 ( x - 2 ) ( x - 5 ) Either (x – 2) = 0 Or (x - 5) = 0 3 Marks x = 2 and x = 5

  12. The line AB has equation 2x – 5y = 10 Show that the gradient of AB is Need to write it in the form y=mx+c Rearrange: 2x – 10 = 5y ÷ each side by 5 Gradient m is 2 Marks 9(b) A second line PQ has equation 2y = 6 - 5x. Jade says that the line PQ is parallel to the line AB. Is she correct? You MUST show your working. Divide both sides by 2 Jade is wrong Gradient is y = 3 - 2 Marks

  13. 10. The diagram shows two containers. One container is a hemisphere of radius 6cm. The other container is a cone of radius 5cm and height 18cm. (a) Show that the volume of the hemisphere is 144π Volume of a sphere is π r3 Which container has the larger volume? Volume of a hemisphere is π r3 Volume of a cone is πr2h V = π x 5 x 5 x 18 V = π x 6 x 6 x 6 V = 144π V = 150π The cone has the larger volume 5 Marks

  14. 11. Prove that (n+3)2 – (n-2)2 = 5(2n+1) Multiply out (n + 3)(n + 3) - (n – 2)(n – 2) n2 + 3n + 3n + 9 - (n2 - 2n - 2n + 4) n2 + 6n + 9 – n2 + 4n - 4 10n + 5 Factorise 5( 2n + 1) Hence shown 3 Marks

  15. 12. OABC is a parallelogram. OA = p and OC = q T is a point outside the parallelogram such that AT = 2p + 3q (a) Find in terms of p and q, expressions for the following vectors. Give your answer in its simplest form. OB = OA + AB 2 Marks = p + q Explain what your answers to part (a) tell you BT = BA + AT = -q + 2p + 3q 2 Marks They are in a straight line and the length of BT is twice the length of OB = 2q + 2p 2 Marks

  16. 13. The diagram shows the graph of y = cos x for 0º ≤ x ≤ 360º One solution of the equation cos x = 0.809 is x = 36º Work out the other solution in this range. Other solution is 360 – 36 = 324º Write down the number of solutions in the same range for each equation. (i) 2cos x = -0.6 (ii)cos 2x = -0.6 cos x = -0.3 4 solutions 3 Marks 2 solutions

  17. Find the values of a and b such that x2 – 8x + 21 = (x – a) 2 + b x2 – 8x + 21 = ( x - 4 )2 + 5 a = 4 and b = 5 2 Marks Hence or otherwise write down (i) The minimum value of x2 - 8x + 21 Minimum value is the coordinate (4,5) (ii) The equation of the line of symmetry of x2 - 8x + 21 Answer : x = 4 2 Marks

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