Cambridge IGCSE Mathematics 0580 — 2013 Oct/Nov Paper 4 · Variant 2

3 (a) Write as a single fraction in its simplest form. For Examiner′s Use - 2 x 1 3 x + 1 - 2 5 Answer(a) ............................................... [3] (b) Expand and simplify. (2x – 3)2 – 3x(x – 4) Answer(b) ............................................... [4] (c) (i) Factorise. 2x2 + 5x – 3 Answer(c)(i) ............................................... [2] (ii) Simplify. 2 + - 2x 5 x 3 2 - 2x 18 Answer(c)(ii) ............................................... [3] _____________________________________________________________________________________

Mark scheme: 4x 7 5( 2 x )1 2(3 x + )1 3 (a) final answer nfww 3 M2 for 10 2 × 5 5( 2 x − )1 2(3 x + )1 or – 5 × 2 5 × 2 or M1 for attempt to convert to common denominator of 10 or multiple of 10 with one error in numerator (b) x² + 9 final answer nfww 4 B3 for 4x² – 6x – 6x + 9 – 3x² +12x or correct answer given and then spoilt or B1 for 4x² – 6x – 6x + 9 seen and B1 for – 3x² +12x or – (3x² – 12x) seen (c) (i) (2x – 1)(x + 3) isw solving 2 M1 for (2x + a)(x + b) where ab = –3 or 2b + a = 5 with integers a and b 2 x − 1 2 x − 1 (ii) or 3 M2 for 2(x + 3)(x – 3) or (2x – 6)(x + 3) or 2( x − )3 2 x − 6 (2x + 6)(x – 3) seen final answer nfww or M1 for 2(x² – 9) seen 2 2

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4 For Examiner′s O 8 cm A Use 42° NOT TO 8 cm SCALE B h cm A wedge of cheese in the shape of a prism is cut from a cylinder of cheese of height h cm. The radius of the cylinder, OA, is 8 cm and the angle AOB = 42°. (a) (i) The volume of the wedge of cheese is 90 cm3. Show that the value of h is 3.84 cm correct to 2 decimal places. Answer(a)(i) [4] (ii) Calculate the total surface area of the wedge of cheese. Answer(a)(ii) ........................................ cm2 [5] (b) A mathematically similar wedge of cheese has a volume of 22.5 cm3. Calculate the height of this wedge. Answer(b) ......................................... cm [3] _____________________________________________________________________________________

Mark scheme: 4 (a) (i) 90 ÷ (42/360 × π × 82) o.e. M3 M2 for 42/360 × π × 82 × h = 90 or M1 for 42/360 × π × 82 3.836 to 3.837 A1 (ii) 131 or 130.75 to 130.9 nfww 5 M2 for 42/360 × π × 2 × 8 × 3.84 oe [22.48 to 22.53] or M1 for 42/360 × π × 2 × 8 oe soi [5.86 to 5.87] and M1 for 2 × (8 × 3.84) [61.37 to 61.44] and M1 for 2 × (42/360 × π × 82) [46.88 to 47] 22 5. .3 84 3 × 22 5. (b) 2.42 or 2.416 to 2.419 3 M2 for 3.84 × 3 oe or h = 3 90 90 22 5. 90 or M1 for 3 oe or 3 oe seen 90 22 5. .3 84 3 90 or 3 = oe h 22 5. IGCSE – October/November 2013 0580 42

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2 1 - - Examiner′s 3x .5 (a) Complete the table of values for y = Use x2 x x –3 –2 –1 –0.5 –0.3 0.3 0.5 1 2 3 y 9.6 6 26.5 18.0 –2 –6 –9.1 [3] 2 1 - - 3x for –3 Y x Y –0.3 and 0.3 Y x Y 3 . (b) Draw the graph of y = 2 x x y 30 25 20 15 10 5 x –3 –2 –1 0 1 2 3 –5 –10 [5] (c) Use your graph to solve these equations. For Examiner′s Use 2 1 - - (i) 2 3x = 0 x x Answer(c)(i) x = ............... [1] 2 1 - - - (ii) 2 3x 7. 5 = 0 x x Answer(c)(ii) x = ............... or x = ............... or x = ............... [3] 2 1 - - = - (d) (i) By drawing a suitable straight line on the graph, solve the equation 2 3x 10 3 x . x x Answer(d)(i) x = ............... or x = ............... [4] 2 1 - - = - (ii) The equation 2 3x 10 3 x can be written in the form ax2 + bx + c = 0 where x x a, b and c are integers. Find the values of a, b and c. Answer(d)(ii) a = ............... , b = ............... , c = ............... [3] _____________________________________________________________________________________

Mark scheme: 5 (a) 7, 11.5, 4.5 1,1,1 (b) Correct curve cao 5 B3FT for 10 correct plots, on correct vertical grid line and within correct 2 mm square vertically Or B2FT for 8 or 9 correct plots Or B1FT for 6 or 7 correct plots and B1 indep for two separate branches on either side of y-axis (c) (i) 0.69 < x < 0.81 1 (ii) –2.3 < x < –2.2 –0.8 < x < –0.6 0.35 < x < 0.5 3 B1 for each correct After 0 scored, allow SC1 for drawing line y = 7.5 long enough to cross curve at least once (d) (i) y = 10 – 3x ruled correctly B2 long enough to cross curve twice. B1 for ruled line gradient –3 or y intercept at 10 but not y = 10 Or B1 for ‘correct’ but freehand –0.55 < x < –0.45 B1dep Dependent on at least B1 scored for line 0.35 < x < 0.45 B1dep After 0 scored, SC2 for –0.5 and 0.4 [from solving equation] (ii) 10 1 –2 3 B2 for 2 – x – 10x2 [= 0] oe or –10 –1 2 2 1 Or B1 for x 2 −x − 10 = 0 oe Correctly eliminating – 3x Or B1 for 2 – x – 3x3 = 10x2 – 3x3 oe Correctly clearing fractions IGCSE – October/November 2013 0580 42 1 1 1

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6 For Examiner′s E N L A R G E M E N T Use Prettie picks a card at random from the 11 cards above and does not replace it. She then picks a second card at random and does not replace it. (a) Find the probability that she picks (i) the letter L and then the letter G, Answer(a)(i) ............................................... [2] (ii) the letter E twice, Answer(a)(ii) ............................................... [2] (iii) two letters that are the same. Answer(a)(iii) ............................................... [2] (b) Prettie now picks a third card at random. For Examiner′s Use Find the probability that the three letters (i) are all the same, Answer(b)(i) ............................................... [2] (ii) do not include a letter E, Answer(b)(ii) ............................................... [2] (iii) include exactly two letters that are the same. Answer(b)(iii) ............................................... [5] _____________________________________________________________________________________

Mark scheme: 1 1 1 6 (a) (i) oe 2 M1 for × 110 11 10 6  3  3 2 (ii) oe 2 M1 for × 110  55  11 10 8  4  2 1 (iii) oe 2FT FT their (a)(ii) + × correctly evaluated 110  55  11 10 2 1 or M1 their (a)(ii) + × 11 10 6  1  3 2 1 (b) (i) oe 2 M1 for × × 990  165  11 10 9 336  56  8 7 6 (ii) oe 2 M1 for × × 990  165  11 10 9 198  1   3 2 8   2 1 9   × × (iii) oe  990 11 10  9  oe  5  5 M4 for 3 11 × 10 × 9  + 3   3 2 8   2 1 9   × or M3 for 3 × ×  or 3 ×   11 10 9   11 10  9   oe Or 3 2 8 M1 for × × oe seen and M1 for 11 10 9 2 1 9  × × oe seen 11 10  9  IGCSE – October/November 2013 0580 42

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7 Noma fl ies from Johannesburg to Hong Kong. For Examiner′s Her plane leaves Johannesburg at 18 45 and arrives in Hong Kong 13 hours and 25 minutes later. Use The local time in Hong Kong is 6 hours ahead of the time in Johannesburg. (a) At what time does Noma arrive in Hong Kong? Answer(a) ............................................... [2] (b) Noma sleeps for part of the journey. The time that she spends sleeping is given by the ratio sleeping : awake = 3 : 4 . Calculate how long Noma sleeps during the journey. Give your answer in hours and minutes. Answer(b) ................... h ................... min [2] (c) (i) The distance from Hong Kong to Johannesburg is 10 712 km. For Examiner′s The time taken for the journey is 13 hours and 25 minutes. Use Calculate the average speed of the plane for this journey. Answer(c)(i) ...................................... km/h [2] (ii) The plane uses fuel at the rate of 1 litre for every 59 metres travelled. Calculate the number of litres of fuel used for the journey from Johannesburg to Hong Kong. Give your answer in standard form. Answer(c)(ii) ...................................... litres [4] (d) The cost of Noma’s journey is 10 148 South African Rand (R). This is an increase of 18% on the cost of the journey one year ago. Calculate the cost of the same journey one year ago. Answer(d) R ............................................... [3] _____________________________________________________________________________________

Mark scheme: 7 (a) 14 10 or 2 10 pm final answer 2 M1 for (0)8 10 oe or answer 14 hours and 10 minutes or answer 2 10 [am] (b) 5 hours 45 minutes cao 2 M1 for 345 [mins] seen or for 805 /7 × 3 oe or 5.75 seen 25 (c) (i) 798 or 798.2 to 798.4…. 2 M1 for 10712 / 13 or 10712 ÷ 13.4… 60 (ii) 1.82 × 105 4 B3 for 182000 or 181500 to 181600 seen or 1.815 × 105 to 1.816 × 105 or M2 for 10712000/59 oe or M1 for figs 10712/figs 59 soi by figs 182 or figs 1815 to 1816 and B1 FT for their number of litres correctly converted to standard form rounded to 3sf or better (d) 8600 3 M2 for 10148 ÷ 1.18 oe or M1 for 10148 associated with 118[%]

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7 (x ¸ –1) h(x) = x2 + 5x Examiner′s8 f(x) = 4x + 3 g(x) = x + 1 Use (a) Work out (i) h(–3), Answer(a)(i) ............................................... [1] (ii) hg(13). Answer(a)(ii) ............................................... [2] (b) Find f –1(x). Answer(b) f –1(x) = ............................................... [2] (c) (i) Solve the equation f(x) = 23. For Examiner′s Use Answer(c)(i) x = ............................................... [2] (ii) Solve the equation h(x) = 7. Show all your working and give your answers correct to 2 decimal places. Answer(c)(ii) x = ......................... or x = ......................... [5] _____________________________________________________________________________________

Mark scheme: 8 (a) (i) –6 1 (ii) 2.75 oe 2 M1 for [g(x) =] 0.5 or 7/14  7  2  7  Or   + 5  oe  x + 1   x + 1  x − 3 x 3 (b) or – Final answer 2 M1 for y – 3 = 4x or better or x = 4y + 3 or 4 4 4 better 3 y or = + x or flowchart with – 3 then ÷ 4 4 4 3 23 (c) (i) 5 2 M1 for 4x = 23 – 3 or x + = or better 4 4 (ii) x² + 5x – 7 = 0 B1 May be implied by correct values in formula − 5 ± 5 2 − 41()( −7 ) B1 oe B1 B1 for 52 − 41()(−7) or better [53] 2)1( + − p q p q If in form or , B1 for –5 and r r 2(1) or better No recovery of full line unless seen 1.14 and –6.14 final answers B1 Or SC1 for 1.1 or 1.140…. and –6.1 B1 or – 6.140 … Or answers –1.14 and 6.14 IGCSE – October/November 2013 0580 42

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9 For Examiner′s y Use 8 7 6 5 4 D C 3 2 B A 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 –4 –5 –6 –7 –8 (a) Describe fully the single transformation that maps triangle A onto (i) triangle B, Answer(a)(i) ...................................................................................................................... [2] (ii) triangle C, Answer(a)(ii) ..................................................................................................................... [2] (iii) triangle D. Answer(a)(iii) .................................................................................................................... [3] (b) On the grid, draw For Examiner′s Use (i) the rotation of triangle A about (6, 0) through 90° clockwise, [2] (ii) the enlargement of triangle A by scale factor –2 with centre (0, –1), [2] (iii) the shear of triangle A by shear factor –2 with the y-axis invariant. [2] (c) Find the matrix that represents the transformation in part (b)(iii). Answer(c) [2] f p _____________________________________________________________________________________ Question 10 is printed on the next page.

Mark scheme: 9 (a) (i) Reflection x = –2 oe 2 B1 for either (ii) Translation − 7    oe 2   2 B1 for either (iii) Stretch x-axis oe invariant [factor] 3 3 B1 for each (b) (i) Triangle with coords at (8, 2) 2 B1 for rotation about (6, 0) but 90° (7, 3) and (7, 5) anticlockwise Or for rotation 90° clockwise around any point (ii) Triangle with coords at 2 B1 for 2 correct points or for enlargement of (–2, –5) (–6, –5) and (–8, –7) SF –2 any centre (iii) Triangle with coords at (1, –1) 2 B1 for 2 correct points or coordinates of (4, –6) and (3, –5) 2 points shown  1 0  (c)   2 B1 for one row or one column correct but not  − 2 1  identity matrix.  1 − 2  Or SC1 for   0 1  

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