Cambridge IGCSE Mathematics 0580 — 2008 Oct/Nov Paper 3 · Variant 1

1 Aida, Bernado and Cristiano need $30 000 to start a business. For Examiner's 2 Use (a) (i) They borrow of this amount. 5 Show that they still need $18 000. Answer (a)(i) [1] (ii) They provide the $18 000 themselves in the ratio Aida : Bernado : Christiano = 5 : 4 : 3. Calculate the amount each of them provides. Answer(a)(ii)Aida $ Bernado $ Cristiano $ [3] (b) (i) Office equipment costs 35 % of the $30 000. Calculate the cost of the equipment. Answer(b)(i)$ [2] (ii) Office expenses cost another $6500. Write this as a fraction of $30 000. Give your answer in its lowest terms. Answer(b)(ii) [2] (iii) How much remains of the $30 000 now? Answer(b)(iii)$ [1] (c) They invest $12 500. After one year this has increased to $15 500. Calculate this percentage increase. Answer(c) % [3]

Mark scheme: Qu Answers Mark Part Marks 1 (a) (i) 3 × 30 000 M1 Must see evidence of fractions 5 or 30 000 − 52 × 30 000 (ii) Aida $7500 5 or 4 or 3 W3 M1 for 5+ 4 + 3 × 18000 Bernado $6000 A1 for 1 correct answer Christiano $4500 (b) (i) 10 500 W2 M1 for 10035 × 30 000 or 0.35 × 30 000 (ii) 13 6500 60 W2 W1 for 30000 seen or other ‘correct’ fraction. (iii) ($)13 000 W1ft 15500 (c) 24 W3cao M1 for 15 500 − 12500 or 12500 × 100 M1 for 12500' 3000 ' × 100 or ‘124’− 100

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2 For F Examiner's Use NOT TO SCALE A B 12 m 55m 25° 18° E D C ABCD represents a building with a vertical flagpole, AF, on the roof. The points E, D and C are on level ground. EA = 55 metres. The angle of elevation of A from E is 18° and the angle of elevation of F from E is 25°. (a) Calculate (i) ED, Answer(a)(i) m [2] (ii) FD, Answer(a)(ii) m [2] (iii) DA. Answer(a)(iii) m [2] (b) Show that AF = 7.4 metres, correct to 1 decimal place. Answer(b) [1] (c) The width, AB, of the building is 12 metres. The top of the flagpole is attached to the point B by a rope. Calculate (i) the length of the rope, FB, Answer(c)(i) m [2] (ii) the angle of elevation of F from B. Answer(c)(ii) [2]

Mark scheme: 2 (a) (i) 52.3 art W2cao M1 for 55cos18° (ii) 24.4 art W2 ft M1 for ‘52.3’tan25°. Ft their ED (iii) 17.0 art W2cao M1 for 55sin18° or √(55 2 − ‘52.3’ 2 ) or ‘52.3’ tan18° Long methods, e.g. sine rule must be explicit and ‘correct’. (b) ‘24.4’ − ‘17.0’ (= 7.4) M1 Allow for clear attempt to find FD − AD. (c) (i) 14.1 art W2cao M1 for √( 12 2 + 7.4 2 ) or correct long methods 12 ÷ cos (tan −1 4.712 ) or 7.4 ÷ sin(tan −1 4.712 ) (ii) 31.7 art W2cao M1 for tan (FBA) = 4.712 oe or sin FBA = ' 4.7FB ' or cos FBA = ' FB12 '

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3 The table below shows the average daily sunshine, s, and the total monthly rainfall, r, for a city For during one year. Examiner's Use Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec s (hours) 6 7 7 9 10 12 12 12 9 8 6 5 r (mm) 70 52 72 41 20 6 1 4 16 52 65 67 (a) For s, find (i) the mode Answer(a)(i) hours [1] (ii) the range, Answer(a)(ii) hours [1] (iii) the median. Answer(a)(iii) hours [2] (b) On the grid below, plot the 10 points for March to December to complete the scatter diagram. r 70 60 50 Total 40 Monthly Rainfall (mm) 30 20 10 0 s 5 6 7 8 9 10 11 12 Average Daily Sunshine (hours) [3] (c) (i) Calculate the mean of s. For Examiner's Use Answer(c)(i) hours [2] (ii) The mean of r is 38.8 millimetres. On the grid, plot the point representing these means. Label this point M. [1] (d) (i) Draw a line of best fit on the grid. [1] (ii) What type of correlation does your scatter diagram show? Answer(d)(ii) [1]

Mark scheme: 3 (a) (i) 12 W1 (ii) 7 W1 (iii) 8.5 W2 M1 for Attempt at ordering the data. (b) 10 points correctly plotted W3 W2 for 8 or 9 points correctly plotted W1 for 6 or 7 points correctly plotted IGCSE – October/November 2008 0580 and 0581 03

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4 G D C NOT TO SCALE B 68° E A EG is a diameter of the circle through E,C and G. The tangent AEB is parallel to CD and angle AEC = 68°. Calculate the size of the following angles and give a reason for each answer. (a) Angle CEG = because [2] (b) Angle ECG = because [2] (c) Angle CGE = because [2] (d) Angle ECD = because [2]

Mark scheme: ( ) g 4 (a) 22° W1cao Degree symbol not essential throughout question. Tangent (and) radius/ W1 Allow perpendicular for 90° diameter (meet at) 90° (b) 90° W1cao (Angle in a) semi-circle W1 (c) 68° W1ft Ft is180 −( their (a) + their (b)) (Angles in a )triangle W1 or alternate segment (theorem) (=)180° (d) 68° W1cao Alternate or Z (angles) W1 Allow Z correctly placed on the diagram.

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5 Aminata and her brother live 18 kilometres from a shopping centre. For (a) Aminata leaves home at 09 00 and runs 3 kilometres to a bus stop. Examiner'sUse She arrives there at 09 30. Write down her average speed, in kilometres per hour. Answer(a) km / h [1] (b) She waits 15 minutes for the bus. The bus travels the remaining 15 kilometres to the shopping centre at an average speed of 20 km / h. (i) At what time does she arrive at the shopping centre? Answer(b)(i) [2] (ii) On the grid below, complete the travel graph showing her journey to the shopping centre. 20 Shopping Centre 18 16 14 12 Distance 10 from home (km) 8 6 4 2 Home 0 09 00 10 00 11 00 12 00 13 00 Time [2] (c) Her brother leaves home at 11 15. For He travels to the shopping centre by car at an average speed of 54 km / h. Examiner's Use (i) Work out how long, in minutes, he takes to travel to the shopping centre. Answer(c)(i) minutes [1] (ii) Show his journey on the grid. [1] (d) Aminata and her brother leave the shopping centre at 12 00. They travel home by car and arrive at 12 45. (i) Show their journey home on the grid. [1] (ii) Calculate the average speed of their journey home. Answer(d)(ii) km / h [2]

Mark scheme: 5 (a) 6 W1 (b) (i) 10 30 W2 M1 for 1520 SC1 for 10 15 (ii) Line from 09 30 to 0945 W1 accuracy ± 1mm Line to (‘10 30’, 18) W1ft (c) (i) 20 W1 (ii) Line (11 15, 0) to W1ft ft their time in (c)(i) provided in minutes and Y 45 ( their 11 35, 18) Line (11 15, 0) to (11 [15 + ‘20’], 18) (d) (i) Line (12 00,18) to (12 45,0) W1 (ii) 24 W2 M1 for 18 ÷ 0.75 Allow 18 ÷ 45 × 60 for method

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6 (a) 2y = 75 − 7x (i) Find y when x = 7. Answer(a)(i) y = [2] (ii) Find x when y = 6. Answer(a)(ii) x = [2] (b) Make x the subject of the equation 2y = 75 − 7x. Answer(b) x = [2] (c) Solve these simultaneous equations. 4x − y = 45 7x + 2y = 75 Answer(c) x = y = [3]

Mark scheme: 6 (a) (i) ( y =)13 W2 M1 for (2y =) 75 − 7 × 7 (ii) ( x =) 9 W2 M1 for 7x = 75 − 12 or −7x = 12 − 75 (b) 75 − 2 y or 2y−75−7 W2 M1 for 7x + 2y = 75. 7 7x = 75 − 2y or −7x = 2y − 75 or −7x − 2y = −75 IGCSE – October/November 2008 0580 and 0581 03

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7 (a) Complete the table of values for the equation y = x2 + x − 3. For Examiner's Use x −4 −3 −2 −1 0 1 2 3 y 9 −1 −3 −1 9 [3] (b) On the grid, draw the graph of y = x2 + x − 3. y 10 9 8 7 6 5 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 –1 –2 –3 –4 [4] (c) Write down the coordinates of the lowest point of the curve. Answer(c) ( , ) [2] (d) (i) Draw the line of symmetry of the graph. [1] (ii) Write down the equation of the line of symmetry. Answer(d)(ii) [1]

Mark scheme: y 7 (a) 3, −3, 3 W3 W1 for each correct value (b) 8 correctly plotted points W3ft W2 for 6 or 7 points, W1 for 4 or 5 points Smooth curve W1 Half square accuracy must go below line y = −3 (c) ( −0.5, −3.25) W2ft W1 for one coordinate correct Ft their graph but −1 < x < 0 and y < −3 Allow calculated if exact values (W2 or W1) (d) (i) Line x = −0.5 drawn W1cao Half square accuracy (ii) x = −0.5 oe W1ft Ft any vertical line only

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8 For y Examiner's Use 6 5 4 3 C 2 1 B x –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 –2 A –3 –4 –5 Triangle ABC is drawn on the grid. (a) (i) Write down the coordinates of A. Answer(a)(i) ( , ) [1] (ii) Write AB and BC as column vectors. Answer(a)(ii) AB = BC = [2]  4  (b) Translate triangle ABC by the vector  _  . Label the image T. [2]  3    (c) AP = 2AB and AQ = 2AC. (i) Plot the points P and Q on the grid. [2] (ii) Describe fully the single transformation which maps triangle ABC onto triangle APQ. Answer(c)(ii) [3] (d) Rotate triangle ABC through 180° about the midpoint of the side AB. Label the image R. [2]

Mark scheme: 8 (a) (i) (−3, −2) W1 (ii)  2   2   4  − 3  (AB =) , (BC =) SC1 for   and   W1,      4   −3  2  2   W1 (b) (1, −5), (5, −3), (2, −1) W2 W1 for 2 correct points plotted Must join points, with straight lines, for both marks. (c) (i) P( 5, 2), Q( −1, 6) W1, W1 (ii) Enlargement W1 (Scale factor) 2 W1 (Centre ) A or (−3, −2) W1ft Ft their (a)(i) Zero if not a single transformation (d) ( 0, −4) marked W1 Joined to A and B W1ft Their image of C joined to A and B.

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9 The quadrilateral ABCD is a scale drawing of a park. For Angle ABC = 90° and 1 centimetre represents 10 metres. Examiner's Use D A B C (a) Write down (i) the actual length, in metres, of the side CD, Answer(a)(i) m [1] (ii) the size of angle BAD. Answer(a)(ii) [1] (b) Two straight paths cross the park. One path is the same distance from AB as from BC. The other path is the same distance from A as from D. (i) Using a straight edge and compasses only, construct the lines which show each path. [4] (ii) Tennis courts in the park are situated in a region closer to AB than to BC and closer to A than to D. Label this region T. [1] (c) Keith cycles past the park, so that he is always 30 metres outside the boundary ABC. Construct the locus of points which shows this part of his route. [2]

Mark scheme: 9 (a) (i) 99 to 101 (metres) W1 (ii) 103° to 105° W1 (b) (i) Bisector of angle ABC W2 W1 correct bisector without arcs (45 ± 1 to BC) with arcs Bisector of AD with arcs W2 W1 correct bisector without arcs. Bisector about ±1mm from centre of AD 89° to 91° to AD by eye and centre within 2mm by and 89° to 91° to AD. eye. (ii) Closed region T indicated W1 Dependent on at least W1 for each bisector. Allow T omitted if region is clear. IGCSE – October/November 2008 0580 and 0581 03

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10 The first three diagrams in a sequence are shown below. For Each diagram has one more trapezium added on the right. Examiner's Use Diagram 1 Diagram 2 Diagram 3 (a) Complete the table which shows the number of lines and dots in each diagram. Diagram 1 2 3 4 Number of lines 4 7 Number of dots 4 6 [2] (b) Find the number of lines and dots in Diagram 10. Answer(b) lines and dots [2] (c) For Diagram n, write down in terms of n, the number of (i) lines, Answer(c)(i) [2] (ii) dots. Answer(c)(ii) [2] (d) Find the difference, in terms of n, between your answers to parts (c)(i) and (c)(ii). Simplify your answer. Answer(d) [2]

Mark scheme: 10 (a) (Lines) 10 and 13 W1 (Dots) 8 and 10 W1 (b) (Lines) 31, (Dots) 22 W1, W1 (c) (i) 3n + 1 oe SC1 for jn + 1 or 3n + k W2cao where j and k are integers. j ≠ 0 (ii) 2n + 2 oe SC1 for jn + 2 or 2n + k W2cao where j and k are integers. j ≠ 0 (d) n − 1 or 1 − n M1 for ‘(3n + 1)’ − ‘(2n + 2)’ or reversed W2ft Ft and M1 dependent on two linear algebraic expressions

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