Cambridge IGCSE Mathematics 0580 — 2010 Oct/Nov Paper 4 · Variant 2

1 (a) Hansi and Megan go on holiday. For The costs of their holidays are in the ratio Hansi : Megan = 7 : 4. Examiner's Hansi’s holiday costs $756. Use Find the cost of Megan’s holiday. Answer(a) $ [2] (b) In 2008, Hansi earned $7800. (i) He earned 15% more in 2009. Calculate how much he earned in 2009. Answer(b)(i) $ [2] (ii) In 2010, he earns 10% more than in 2009. Calculate the percentage increase in his earnings from 2008 to 2010. Answer(b)(ii) % [3] (c) Megan earned $9720 in 2009. This was 20% more than she earned in 2008. How much did she earn in 2008? Answer(c) $ [3] (d) Hansi invested $500 at a rate of 4% per year compound interest. Calculate the final amount he had after three years. Answer(d) $ [3]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 432 2 M1 for 756 ÷ 7 × 4 oe (b) (i) 8970 2 M1 for 7800 × 1.15 oe After 0 scored, SC1 for 1170 as answer their 9867 ( −7800) (ii) (× 100) M2 Their 9867 is their (b)(i) × 1.1 7800 Implied by 1.265 or 0.265 or 126.5 or 1.15 × 1.10 or M1 for their (b)(i) × 1.10 (9867 seen or 2067 seen) 26.5 % cao A1 www3 (c) 8100 3 M2 for 9720 ÷ 1.2 oe or M1 for 120% = 9720 oe (d) 562.43 or 562 or 562.4(0) or 562.432 3 M2 for 500 × 1.04³ or alt complete method or M1 for 1.04² or 1.04³ oe soi e.g. $540.80 or 562.(43..) seen in working

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2 f(x) = 6 + x2 g(x) = 4x –1 For Examiner's (a) Find Use (i) g(3), Answer(a)(i) [1] (ii) f (–4 ). Answer(a)(ii) [1] (b) Find the inverse function g–1(x). Answer(b) g–1(x) = [2] (c) Find fg(x) in its simplest form. Answer(c) fg(x) = [3] (d) Solve the equation gg(x) = 3. Answer(d) x = [3]

Mark scheme: 2 (a) (i) 11 1 (ii) 22 1 x + 1 g ( x ) + 1 y + 1 (b) oe final answer 2 M1 for x + 1 = 4y or or 4 4 4 (c) 16x² – 8x + 7 final answer 3 M1 for 6 + (4x – 1)² and B1 for 16x² – 4x – 4x + 1 or better seen (d) 0.5 or ½ www 3 M2 for 16x – 4 – 1 = 3 or better or M1 for 4(4x – 1) – 1 (= 3) Alt method M2 allow g–1g–1(3) complete method or M1 for g(x) = g–1(3) IGCSE – October/November 2010 0580 42

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3 80 boys each had their mass, m kilograms, recorded. For The cumulative frequency diagram shows the results. Examiner's Use 80 60 Cumulative 40 frequency 20 0 m 30 40 50 60 70 80 90 Mass (kg) (a) Find (i) the median, Answer(a)(i) kg [1] (ii) the lower quartile, Answer(a)(ii) kg [1] (iii) the interquartile range. Answer(a)(iii) kg [1] (b) How many boys had a mass greater than 60kg? Answer(b) [2] (c) (i) Use the cumulative frequency graph to complete this frequency table. For Examiner's Use Mass, m Frequency 30 I m Y 40 8 40 I m Y 50 50 I m Y 60 14 60 I m Y 70 22 70 I m Y 80 80 I m Y 90 10 [2] (ii) Calculate an estimate of the mean mass. Answer(c)(ii) kg [4]

Mark scheme: 3 (a) (i) 63 to 63.5 1 (ii) 50 to 50.5 1 (iii) 21.5 to 22.5 1 (b) 46 2 B1 for 34 seen (could be on graph) (c) (i) 12, 14 1, 1 (ii) {35 × 8 + 45 × their 12 + 55 × 14 + 65 × 22 + 75 × their 14 + 85 × 10} M3 M1 for mid-values soi (allow 1 error/omit) with x in correct ÷ their 80 (or 80) and M1 for use of ∑fx boundary including both ends (at least 4 products) (4920 seen implies M2) and M1 depend on 2nd M for dividing by their 80 (or 80) (not 54 or less) 61.5 cao A1 www4

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4 (a) For Examiner's 4 cm Use NOT TO SCALE 13 cm The diagram shows a cone of radius 4 cm and height 13 cm. It is filled with soil to grow small plants. Each cubic centimetre of soil has a mass of 2.3g. (i) Calculate the volume of the soil inside the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = π r h .] 3 Answer(a)(i) cm3 [2] (ii) Calculate the mass of the soil. Answer(a)(ii) g [1] (iii) Calculate the greatest number of these cones which can be filled completely using 50 kg of soil. Answer(a)(iii) [2] (b) A similar cone of height 32.5 cm is used for growing larger plants. Calculate the volume of soil used to fill this cone. Answer(b) cm3 [3] (c) For Examiner's Use NOT TO SCALE 12 cm Some plants are put into a cylindrical container with height 12 cm and volume 550 cm3. Calculate the radius of the cylinder. Answer(c) cm [3]

Mark scheme: 4 (a) (i) 218 (217.7 to 218) 2 M1 for 1/3π × 42 × 13 (ii) 501 (500.7 to 501.4) 1ft ft their (a) × 2.3 (iii) 99 2ft ft 50 000 ÷ their (a)(ii) and truncated to whole number M1 for 50 000 ÷ their (a)(ii) oe or answers 99.8 or 100 3  325.  (b) their (a)(i) ×   oe M2 or 1/3π × 102 × 32.5  13  or M1 for (32.5 ÷ 13)³ (=15.625) seen or (13 ÷ 32.5)³ (= 0.064) seen 3400 or 3410 (3401 to 3407) A1 www3 (c) (r² =) 550 ÷ 12π M2 (14.58 to 14.6) or M1 for 12π r² = 550 or better 3.82 (3.818 to 3.821) A1 www3 IGCSE – October/November 2010 0580 42

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5 (a) For A Examiner's Use NOT TO SCALE 17 cm x cm B (x + 7) cm C In the right-angled triangle ABC, AB = x cm, BC = (x + 7) cm and AC = 17 cm. (i) Show that x2 + 7x – 120 = 0. Answer(a)(i) [3] (ii) Factorise x2 + 7x – 120. Answer(a)(ii) [2] (iii) Write down the solutions of x2 + 7x – 120 = 0. Answer(a)(iii) x = or x = [1] (iv) Write down the length of BC. Answer(a)(iv) BC = cm [1] (b) For Examiner's NOT TO Use SCALE 3x cm (2x + 3) cm (2x – 1) cm (2x + 3) cm The rectangle and the square shown in the diagram above have the same area. (i) Show that 2x2 – 15x – 9 = 0. Answer(b)(i) [3] (ii) Solve the equation 2x2 – 15x – 9 = 0. Show all your working and give your answers correct to 2 decimal places. Answer(b)(ii) x = or x = [4] (iii) Calculate the perimeter of the square. Answer(b)(iii) cm [1]

Mark scheme: 5 (a) (i) x² + (x + 7)² = 17² oe B1 Must be seen x² + x² + 7x + 7x + 49 = 17² B1 or better 2x² + 14x – 240 = 0 Must be shown – correct 3 terms x² + 7x – 120 = 0 E1 With no errors seen (ii) (x + 15)(x – 8) 2 M1 for (x + a)(x + b) where a and b are integers and a × b = –120 or a + b = 7 Ignore solutions after factors given (iii) –15 and 8 1ft Correct or ft dep on at least M1 in (ii) (iv) 15 1ft Correct or ft their positive root from (ii) + 7 dep on a positive and negative root given (b) (i) 3x(2x – 1) = (2x + 3)² oe M1 e.g. 6x² – 3x = 4x² + 12x + 9 must see equation before simplification 4x² + 6x + 6x + 9 or better seen B1 Indep 6x² – 3x = 4x² + 12x + 9 oe 2x² – 15x – 9 = 0 E1 With no errors seen and both sets of brackets expanded 2 1 In square root B1 for ((–)15)2 – 4(2)(–9) or ( − − )15 ± (( − )15) − 4( 2)( −9) (ii) oe 1 better (297) 2( 2) p + q p − q If in form or , r r B1 for –(–15) and 2(2) or better 8.06 and -0.56 cao 1, 1 SC1 for –0.6 or –0.558… and 8.1 or 8.058… (iii) 76.5 (76.46 to 76.48) 1ft ft 8 times a positive root to (b)(ii) add 12

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6 For L 5480 km Examiner's Use D NOT TO 165° 3300 km SCALE C The diagram shows the positions of London (L), Dubai (D) and Colombo (C). (a) (i) Show that LC is 8710 km correct to the nearest kilometre. Answer(a)(i) [4] (ii) Calculate the angle CLD. Answer(a)(ii) Angle CLD = [3] (b) A plane flies from London to Dubai and then to Colombo. For It leaves London at 01 50 and the total journey takes 13 hours and 45 minutes. Examiner's The local time in Colombo is 7 hours ahead of London. Use Find the arrival time in Colombo. Answer(b) [2] (c) Another plane flies the 8710 km directly from London to Colombo at an average speed of 800 km/h. How much longer did the plane in part (b) take to travel from London to Colombo? Give your answer in hours and minutes, correct to the nearest minute. Answer(c) h min [4]

Mark scheme: 6 (a) (i) 54802 + 33002 – 2 × 5480 × 3300 M2 (75 856 005) M1 for implicit version × cos165 8709.5.. E2 If E0, A1 for 75800000 to 75900000 sin 165 sin L sin 165 (ii) (sinL =) × 3300 M2 M1 for = oe (allow 8709.5.) 8710 3300 8710 (0.09806…) Could use cosine rule using 8710 or better – M2 for explicit form or M1 for implicit form (allow 5.6 to 5.63 for A mark) 5.6 (5.62 to 5.63) A1 www3 (b) 22 35 or 10 35 pm 2 Accept 22 35 pm B1 for 15 35 or 3 35 pm seen or answers 22h 35 mins or (0)8 35(am) or 10 35(am) (c) 8710 ÷ 800 M1 10.88 to 10.9 with no conversion to A1 Implied by correct final ans 2hrs 52 mins if not h/min shown or 10 (hrs) 52 (mins) to 10 (hrs) 54 (mins) oe 13 hrs 45 mins – their time in hrs and M1 Dep on first M1 mins oe e.g. 13 hrs 45mins – 11 hrs 29 mins or 13.75 – their decimal time and a or 13.75 – 10.9 then 2hrs 51 mins correct conversion to hrs and mins or minutes 2 hr 52 mins cao A1 www4 (2 hrs 51.75 mins) IGCSE – October/November 2010 0580 42

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2 2 For 7 (a) Complete the table for the function f(x) = − x . Examiner's x Use x –3 –2 –1 –0.5 –0.2 0.2 0.5 1 2 3 f(x) –9.7 –5 –10.0 10.0 3.75 1 –8.3 [3] (b) On the grid draw the graph of y = f(x) for –3 Y x Y –0.2 and 0.2 Y x Y 3. y 10 8 6 4 2 x –3 –2 –1 0 1 2 3 –2 –4 –6 –8 –10 [5] (c) Use your graph to For Examiner's (i) solve f(x) = 2, Use Answer(c)(i) x = [1] (ii) find a value for k so that f(x) = k has 3 solutions. Answer(c)(ii) k = [1] 2 2(d) Draw a suitable line on the grid and use your graphs to solve the equation − x = 5x. x Answer(d) x = or x = [3] (e) Draw the tangent to the graph of y = f(x) at the point where x = –2. Use it to calculate an estimate of the gradient of y = f(x) when x = –2. Answer(e) [3]

Mark scheme: 7 (a) –3, –4.25, –3 1, 1, 1 Allow – 4.2 or – 4.3 for – 4.25 (b) 10 correct points plotted P3ft P2ft for 8 or 9 correct P1ft for 6 or 7 correct Smooth curve through their 10 points C1 Correct shape not ruled, (curves could be joined) and correct shape Two separate branches B1ft Indep but needs two ‘curves’ on either side of y- axis (c) (i) 0.7 to 0.85 1 –1 each extra (ii) Any value of k such that k Y –3 1ft ft consistent with their graph and must be consistent with their (If curves are joined then k = –3 only) graph (d) y = 5x drawn L1 Ruled and long enough to meet curves – 0.6 to –0.75, 0.55 to 0.65 1, 1 Indep –1 each extra (e) Tangent drawn at x = –2 T1 Must be a reasonable tangent, not chord, no clear daylight y change / x change attempt M1 Depend on T and uses scales correctly. Mark intention – allow one slight slip e.g. sign error from coords but not scale misread If no working shown and answer is out of range – check their tangent for method 2.7 to 4.3 A1 Answer in range gets 2 marks after T1 earned 3   k 

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8 (a) For y Examiner's Use 8 6 4 A A 2 x –8 –6 –4 –2 0 2 4 6 8 –2 –4 –6 –8 Draw the images of the following transformations on the grid above.  3  (i) Translation of triangle A by the vector   . Label the image B. [2]  −7  (ii) Reflection of triangle A in the line x = 3. Label the image C. [2] (iii) Rotation of triangle A through 90° anticlockwise around the point (0, 0). Label the image D. [2] (iv) Enlargement of triangle A by scale factor –4, with centre (0, 1). Label the image E. [2] (b) The area of triangle E is k × area of triangle A. For Write down the value of k. Examiner's Use Answer(b) k = [1] (c) y 5 4 3 2 1 F x –5 –4 –3 –2 –1 0 1 2 3 4 5 –1 –2 –3 –4 –5 (i) Draw the image of triangle F under the transformation represented by the  1 3  matrix M =   . [3]  0 1  (ii) Describe fully this single transformation. Answer(c)(ii) [3] (iii) Find M–1, the inverse of the matrix M.     Answer(c)(iii)       [2]

Mark scheme: 8 (a) (i) Correct translation to (3, –5), 2 SC1 for translation of  3  or  k  or vertices  k   −7  (5, –6) and (4, –4) only (ii) Correct reflection to (4, 1), (5, 3) 2 SC1 for reflection in y = 3 or vertices only and (6, 2) (iii) Correct rotation to (–2, 0), (–1, 2) 2 SC1 for rotation 90 clockwise around (0, 0) and (–3, 1) or vertices only (iv) Correct enlargement to (0, –3), 2 SC1 for two correct points or vertices only (–8, 1) and (–4, –7) (b) 16 cao 1 (c) (i) Correct transformation to 3 B2 for 3 correct points shown in working but not (–4, 0), (5, 3) and (–2, 0) plotted or B1 for incorrect shear drawn with x-axis invariant or two correct points shown (ii) Shear only 1 If more than one transformation given – no marks available x-axis oe invariant 1 Accept fixed, constant oe for invariant (factor) 3 1  1 − 3  (iii) oe 2 B1 for determinant = 1 or k  1 − 3  oe  0 1  0 1 IGCSE – October/November 2010 0580 42 4 4

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9 A bag contains 7 red sweets and 4 green sweets. For Aimee takes out a sweet at random and eats it. Examiner's She then takes out a second sweet at random and eats it. Use (a) Complete the tree diagram. First sweet Second sweet 6 red 10 7 red 11 green .......... .......... red .......... green green .......... [3] (b) Calculate the probability that Aimee has taken (i) two red sweets, Answer(b)(i) [2] (ii) one sweet of each colour. Answer(b)(ii) [3] (c) Aimee takes a third sweet at random. For Calculate the probability that she has taken Examiner's Use (i) three red sweets, Answer(c)(i) [2] (ii) at least one red sweet. Answer(c)(ii) [3]

Mark scheme: 4 4 9 (a) , 1 Accept fraction, %, dec equivalents (3sf or 11 and10 better) throughout but not ratio or words 7 3 1, 1 i.s.w. incorrect cancelling/conversion to other 10 10 forms Pen –1 once for 2 sf answers 7 6 (b) (i) × M1 11 10 42  21  oe   A1 www2 0.382 (0.3818…) 110  55  7 4 4 7 (ii) × + × M2 ft their tree 11 10 11 10 M1 for either pair seen 56  28  oe   A1 www3 0.509(0..) 110  55  7 6 5 5 (c) (i) × × or their (b)(i) × M1 11 10 9 9 210  7  oe   A1 www2 0.212(1..) 990  33   4 3 2  (ii) 1 –  × ×  oe M2 Longer methods must be complete  11 10 9  M1 for 4/11, 3/10 and 2/9 seen 966  161  oe   A1 www3 0.976 (0.9757…) 990  165 

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10 In all the following sequences, after the first two terms, the rule is to add the previous two terms to For find the next term. Examiner's Use (a) Write down the next two terms in this sequence. 1 1 2 3 5 8 13 [1] (b) Write down the first two terms of this sequence. 3 11 14 [2] (c) (i) Find the value of d and the value of e. 2 d e 10 Answer(c)(i) d = e = [3] (ii) Find the value of x, the value of y and the value of z. O33 x y z 18 Answer(c)(ii) x = y = z = [5]

Mark scheme: 10 (a) 21 and 34 1 (b) –5 8 1 + 1 (c) (i) 4, 6 3 M1 for 2 + d = e oe or d + e = 10 oe seen and either M1 for a correct eqn in d or e seen e.g. 2e = 12 oe or 2d = 8 oe or B1 for either correct (ii) x = 28 5 B4 for any two correct y = –5 or M3 for any of 18 = 3x – 66 oe z = 23 or 3y + 33 = 18 oe or 33 – 3z = -36 oe or M1 for 2 of y = x – 33 oe or y + z = 18 oe or x + y = z oe and M1 for combining two of the previous equations correctly isw (does not have to be simplified) after 0 scored SC1 for –33 + their x = their y or their x + their y = their z or their y + their z = 18

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