Cambridge IGCSE Mathematics 0580 — 2012 Oct/Nov Paper 2 · Variant 3

1 Samantha invests $600 at a rate of 2% per year simple interest. For Examiner's Use Calculate the interest Samantha earns in 8 years. Answer $ [2] 2 2 1 2

Mark scheme: 600 × 2 × 8 1 96 2 M1 for oe 100 If zero SC1 696 1 4

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2 Show that + = 0.17. 10 5 Write down all the steps in your working. Answer [2]

Mark scheme: 1 4 2 + or 0.12 + 0.42 oe M1 100 25 1 16 + = 0.17 or 0.01 + 0.16 = 0.17 M1 Independent 100 100 300× 12

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3 Jamie needs 300 g of flour to make 20 cakes. How much flour does he need to make 12 cakes? Answer g [2]

Mark scheme: 300× 12 3 180 2 M1 for oe 20 4 4

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4 Expand the brackets. y(3 O y3) Answer [2]

Mark scheme: 4 3y – y4 final answer 2 B1 for 3y or – y4 as part of two term expression

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5 Maria pays $84 rent. For The rent is increased by 5%. Examiner's Use Calculate Maria’s new rent. Answer $ [2]

Mark scheme: 5 88.2(0) 2 M1 for 84 × 1.05 oe

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6 R T Using a straight edge and compasses only, construct the locus of points which are equidistant from R and from T. [2]

Mark scheme: 6 Accurate perpendicular bisector of RT with 2 B1 for 2 pairs of correct arcs arcs. B1 for correct line

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7 Find the value of . 11.8 − 10.95 Give your answer correct to 4 significant figures. Answer [2]

Mark scheme: 7 8.471 cao 2 144 8 B1 for 8.47 or 8.4705 to 8.4706 or or 8 17 17

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8 A carton contains 250 ml of juice, correct to the nearest millilitre. Complete the statement about the amount of juice, j ml, in the carton. Answer Y j I [2]

Mark scheme: 8 249.5 [Y j<=] 250.5 cao 2 B1 for either, or both correct but reversed

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9 Shade the required region in each of the Venn diagrams. For Examiner's Use P Q A B R A' (P ∩ R) ∪ Q [2] _ 3

Mark scheme: 9 2 B1 for one correct

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 49  2 64 10 Without using a calculator, show that   = .  16  343 Write down all the steps in your working. Answer [2] 1

Mark scheme: 10 Correct working seen 2 M1 for correct step M1 for correct step 64 64

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4 11 Simplify (256w256) . Answer [2]

Mark scheme: 11 4w64 2 B1 for 4wn or kw64

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12 For Examiner's Mass of parcel Use 0 I m Y 0.5 0.5 I m Y 1.5 1.5 I m Y 3 (m kilograms) Frequency 20 18 9 The table above shows information about parcels in a delivery van. John wants to draw a histogram using this information. Complete the table below. Mass of parcel 0 I m Y 0.5 0.5 I m Y 1.5 1.5 I m Y 3 (m kilograms) Frequency density 18 [2]

Mark scheme: 12 40 6 2 B1 for one correct

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13 Write the following as a single fraction in its simplest form. x + 2 2 x − 1 _ + 1 3 4 Answer [3]

Mark scheme: 13 23 – 2x 3 M1 for two correct algebraic fractions with a 12 common denominator of 12 M1 for correctly collecting their terms M1 for dealing correctly with the 1 √

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14 y varies inversely as the square root of x. For When x = 9, y = 6. Examiner's Use Find y when x = 36. Answer y = [3]

Mark scheme: 14 3, –3 or ±3 3 M1 for y = k/√x oe A1 for 18 IGCSE – October/November 2012 0580 23 2 2

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15 A model of a ship is made to a scale of 1 : 200. The surface area of the model is 7500 cm2. Calculate the surface area of the ship, giving your answer in square metres. Answer m2 [3]

Mark scheme: 15 30 000 3 M2 for 7500 × 2002/1002oe or M1 for 2002 seen πx 2 − A M1 for one correct move

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16 Make y the subject of the formula. A = πx2 O πy2 Answer y = [3]

Mark scheme: πx A 16 oe 3 M1 for second correct move π M1 for third correct move 2 θ 4r

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17 For B Examiner's Use 5r NOT TO SCALE 4r O A The diagram shows a sector of a circle, centre O, radius 5r. The length of the arc AB is 4r. Find the area of the sector in terms of r, giving your answer in its simplest form. Answer [3]

Mark scheme: θ 4r 17 10r2 cao www 3 B1 for ( =) 360 2 × π × 5r 4 r M1 for × ( 5 r 2) π 2 × π × 5 r

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18 C 23° A NOT TO SCALE 13 cm 6 cm B In triangle ABC, AB = 6 cm, BC = 13 cm and angle ACB = 23°. Calculate angle BAC, which is obtuse. Answer Angle BAC = [4]

Mark scheme: 18 122.2 4 M2 for 13sin23/6 A1 57.8 or M1 for sin23 = sin A 6 13

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19 For Examiner's 5 Use 4 Speed 3 (metres per second) 2 1 0 2 4 6 8 10 12 14 16 18 Time (seconds) The diagram shows the speed-time graph for the last 18 seconds of Roman’s cycle journey. (a) Calculate the deceleration. Answer(a) m/s2 [1] (b) Calculate the total distance Roman travels during the 18 seconds. Answer(b) m [3]

Mark scheme: 19 (a) 0.625 or 5/8 1 (b) 62 3 M1 for area under graph implied M1 for correct, complete, area statement

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20 For D Examiner's Use E NOT TO d SCALE O c C In the diagram, O is the origin. = c and = d. E is on CD so that CE = 2ED. Find, in terms of c and d, in their simplest forms, (a) , Answer(a) = [2] (b) the position vector of E. Answer(b) [2]

Mark scheme: 20 (a) ⅓(c – d) oe 2 M1 for DC = c – d oe or correct route Their (a) + d simplified (b) ⅓c + ⅔d oe 2ft M1 for any correct route from O to E stated

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21 Simplify the following. For h 2 − h − 20 Examiner's Use h 2 − 25 Answer [4]  3 2 

Mark scheme: 21 h + 4 4 B2 for (h – 5)(h + 4) seen h + 5 B1 for (h – 5)(h + 5) If B2 not scored then SC1 for (h + a)(h + b) where a + b = –1 or ab = –20 1  1 −2  1  a b  1 −2 

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22 (a) M =   − 1 1  Find M–1, the inverse of M.     Answer(a)   [2]     (b) D, E and X are 2 × 2 matrices. I is the identity 2 × 2 matrix. (i) Simplify DI. Answer(b)(i) [1] (ii) DX = E Write X in terms of D and E. Answer(b)(ii) X = [1]

Mark scheme: 1  1 2  1  a b  1 2  22 (a)   2 B1 for   or k   seen 5  1 3  5  c d  1 3  (b)(i) D cao 1 (ii) D–1E cao 1

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23 f(x) = 3x + 5 g(x) = 4x O 1 For Examiner's Use (a) Find the value of gg(3). Answer(a) [2] (b) Find fg(x), giving your answer in its simplest form. Answer(b)fg(x) = [2] (c) Solve the equation. f –1(x) = 11 Answer(c) x = [1] Question 24 is printed on the next page.

Mark scheme: 23 (a) 43 2 M1 for g(11) or 4[4(3) – 1] –1 (b) 12x + 2 2 M1 for 3(4x – 1) + 5 (c) 38 1 2 2 2

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24 For Q P Examiner's Use 6 cm NOT TO SCALE C B 5 cm D 10 cm A The diagram shows a triangular prism. ABCD is a horizontal rectangle with DA = 10 cm and AB = 5 cm. BCQP is a vertical rectangle and BP = 6 cm. Calculate (a) the length of DP, Answer(a) DP = cm [3] (b) the angle between DP and the horizontal rectangle ABCD. Answer(b) [3]

Mark scheme: 24 (a) 12.7 3 M2 for 102 + 52 + 62 or M1 for one of 102 + 52 or 62 + 52 or 102 + 62 (b) 28.2 3 M2 for sin x = 6/(a) or M1 for identifying angle PDB 70

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