Cambridge IGCSE Mathematics 0580 — 2009 May/June Paper 2 · Variant 1

1 For Examiner's Use (a) Write down the order of rotational symmetry of the diagram. Answer(a) [1] (b) Draw all the lines of symmetry on the diagram. [1]

Mark scheme: 1 (a) 2 1 Any length, can be freehand lines solid or dotted (b) 1 Mark lost if additional lines drawn or axes extended

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2 Write the following in order of size, smallest first. _1 9 5  4  72%   17 7  3  Answer < < < [2]

Mark scheme: 2 −1 2 M1 correct decimals 5 9  4  72%   0.727(6… ) 0.71(4… ) 0.72 0.75 7 17  3 

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3 At 05 06 Mr Ho bought 850 fish at a fish market for $2.62 each. 95 minutes later he sold them all to a supermarket for $2.86 each. (a) What was the time when he sold the fish? Answer(a) [1] (b) Calculate his total profit. Answer(b) $ [1]

Mark scheme: 3 (a) 06 41 1 Allow 6.41(am). 6:41 and 06:41 Not 6h41m or 641h or 6.41pm (b) $204 1

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4 Shade the region required in each Venn Diagram. For Examiner's Use A B A B C A ∩ B ∩ C A ∪ B′ [2]

Mark scheme: 4 1, 1

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 2 3  5 A =  _   4 5  Find A–1, the inverse of the matrix A.     Answer       [2]

Mark scheme: 5 1  5 − 3   5.2 − 5.1  2 M1 det A or |A| or 5×–2 – 4×–3 = 2 or  or     1  5 − 3  a b  2  4 − 2  2 − 1      or  seen  4 − 2  2  c d  Allow 5/2, –3/2, 4/2, –2/2 in matrix 7

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6 In 2005 there were 9 million bicycles in Beijing, correct to the nearest million. The average distance travelled by each bicycle in one day was 6.5 km correct to one decimal place. Work out the upper bound for the total distance travelled by all the bicycles in one day. Answer km [2]

Mark scheme: 6 62225000 or 6.2225 × 107 or 62.225 2 M1 9.5(million) and 6.55 seen million cao 3sf not appropriate for UB and not allowed for 2 marks

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7 Find the co-ordinates of the mid-point of the line joining the points A(2, –5) and B(6, 9). Answer ( , ) [2]

Mark scheme: 7 (4, 2) 2 2 + 6 − 5 + 9 M1 and oe 2 2 or a drawing used correctly IGCSE – May/June 2009 0580, 0581 21

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8 For Examiner's G F E D Use g O a A B C The diagram is made from three identical parallelograms. O is the origin. = a and = g. Write down in terms of a and g (a) , Answer(a) [1] (b) the position vector of the centre of the parallelogram BCDE. Answer(b) [1]

Mark scheme: 8 (a) 2a – g cao 1 –g + 2a (b) 2 12 a + 12 g oe cao 1 Allow 2.5 or 52 and 0.5 2

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9 Rearrange the formula to make y the subject. y x + = 1 9 Answer y = [3]

Mark scheme: 9 (9(1 – x))2 oe 3 M1 1 move completed correctly M1 1 more move completed correctly Mark 3rd move in answer space

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1 1 c _ d 10 Write + – as a single fraction in its simplest form. c d cd Answer [3]

Mark scheme: 10 2 3 M1 d + c – c + d or better M1 common denominator cd used c

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11 In January Sunanda changed £25 000 into dollars when the exchange rate was $1.96 = £1. For In June she changed the dollars back into pounds when the exchange rate was $1.75 = £1. Examiner's Calculate the profit she made, giving your answer in pounds (£). Use Answer £ [3]

Mark scheme: 11 £3000 3 M1 1.96 × 25000 M1 “49000” / 1.75

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12 Solve the simultaneous equations 2y + 3x = 6, x = 4y + 16. Answer x = y = [3]

Mark scheme: 12 x = 4 y = –3 3 M1 consistent multiplication and subtraction of their rearranged eqns. Any other answers must first score M1 to gain an A mark Substitution, matrix and equating methods also permitted 2

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13 A spray can is used to paint a wall. The thickness of the paint on the wall is t. The distance of the spray can from the wall is d. t is inversely proportional to the square of d. t = 0.2 when d = 8. Find t when d = 10. Answer t = [3]

Mark scheme: 13 0.128 3 M1 t = k/d 2 k is any letter except t, d or α A1 k = 12.8 or M1 0.2 × 82 = 12.8 11 9

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14 (a) There are 109 nanoseconds in 1 second. For Find the number of nanoseconds in 5 minutes, giving your answer in standard form. Examiner's Use Answer(a) [2] (b) Solve the equation 5 ( x + 3 × 106 ) = 4 × 107. Answer(b) x = [2]

Mark scheme: 14 (a) 3 × 1011 2 M1 60 × 5 × 109 or better (b) 5 000 000 or 5 × 106 or 5 million 2 M1 0.8 × 107 – 3 × 106 oe or M1 5x = 4 × 107 – 15 × 106 oe If m is used for a million it must be used consistently

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15 T NOT TO SCALE h 25° B A 80 m 18° C Mahmoud is working out the height, h metres, of a tower BT which stands on level ground. He measures the angle TAB as 25°. He cannot measure the distance AB and so he walks 80 m from A to C, where angle ACB = 18° and angle ABC = 90°. Calculate (a) the distance AB, Answer(a) m [2] (b) the height of the tower, BT. Answer(b) m [2]

Mark scheme: 15 (a) 24.7 2 M1 sin18 = AB/80 or cos72 = AB/80 Allow AB/sin18 = 80/sin90 (b) 11.5 2 M1 tan25 = h/(a) or h/sin25 = (a)/sin65

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16 Using a straight edge and compasses only, draw the locus of all points inside the quadrilateral For ABCD which are equidistant from the lines AC and BD. Examiner's Use Show clearly all your construction arcs. A B C D [4]

Mark scheme: 16 Angle bisector of angle in the middle 4 W1 correct bisector drawn W1 at least two arcs drawn on the arms and one Second angle bisector drawn pair of correct crossing arcs W1 as above W1 as above Accuracy ±1° but line must go from edge to edge. IGCSE – May/June 2009 0580, 0581 21

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17 For y Examiner's Use 8 7 6 5 A 4 3 2 B 1 x 0 1 2 3 4 5 6 7 8 (a) Describe fully the single transformation which maps triangle A onto triangle B. Answer(a) [2] (b) On the grid, draw the image of triangle A after rotation by 90° clockwise about the point (4, 4). [2]

Mark scheme: 17 (a) Reflection in y = x 2 M1 Reflection A1 correct description of the line (b) Triangle at (4,6), (4, 7), (7, 7) 2 M1 Rotation 90° clockwise A1 position 3

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18 Two similar vases have heights which are in the ratio 3 : 2. (a) The volume of the larger vase is 1080 cm3. Calculate the volume of the smaller vase. Answer(a) cm3 [2] (b) The surface area of the smaller vase is 252 cm2. Calculate the surface area of the larger vase. Answer(b) cm2 [2]

Mark scheme: 18 (a) 320 2 M1 1080 × 8/27 or (2/3)3 or 1080 ÷ 27/8 or (3/2)3 (b) 567 2 M1 252 × 9/4 or (3/2)2 or 252 ÷ 4/9 or (2/3)2 2

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19 For Examiner's Use A F NOT TO SCALE B E 12 cm 6 cm 40° O C H D G The diagram shows part of a fan. OFG and OAD are sectors, centre O, with radius 18 cm and sector angle 40°. B, C, H and E lie on a circle, centre O and radius 6 cm. Calculate the shaded area. Answer cm2 [4]

Mark scheme: 19 314 4 M1 π. 182 . 40/360 or OAD = 113 identified M1 π. 62 (or π. 62 . 40/360) or OBC …”……. M1 2 × (OAD – OBC) + circle oe OR M1 π. 182 . 40/360 M1 π. 62 . 140/360 M1 2 × OAD + 2 × BOE oe

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20 For Examiner's y Use 6 5 4 3 2 1 0 x 1 2 3 4 5 6 –1 –2 –3 –4 –5 (a) Draw the three lines y = 4, 2x – y = 4 and x + y = 6 on the grid above. [4] (b) Write the letter R in the region defined by the three inequalities below. y Y 4 2x – y [ 4 x + y [ 6 [1]

Mark scheme: 20 draw 2x – y = 4 2 W1 Line through (2,0) or (0,–4) draw x + y = 6 1 draw y = 4 1 R correct region identified by R 1 0 6

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21 For Examiner's Use  x 6   2 3  A =   B =    4 3   2 1  (a) Find AB.     Answer(a)       [2] (b) When AB = BA, find the value of x. Answer(b) x = [3] Question 22 is on the next page

Mark scheme: 21 (a)  2 x + 12 3 x + 6  2 M1 for any correct row or column    14 15  Allow 2(x + 6), 3(x + 2) (b) 5 3  2 x + 12 21  M1   one row (or column) correct  2 x + 4 15  M1 2x + 4 = 14 or 3x + 6 = 21

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