Cambridge IGCSE Mathematics 0580 — 2010 May/June Paper 4 · Variant 1

1 A school has 220 boys and 280 girls. Examiner's Use (a) Find the ratio of boys to girls, in its simplest form. Answer(a) : [1] (b) The ratio of students to teachers is 10 : 1. Find the number of teachers. Answer(b) [2] (c) There are 21 students on the school’s committee. The ratio of boys to girls is 3 : 4. Find the number of girls on the committee. Answer(c) [2] (d) The committee organises a disco and sells tickets. 35% of the school’s students each buy a ticket. Each ticket costs $1.60. Calculate the total amount received from selling the tickets. Answer(d) $ [3] (e) The cost of running the disco is $264. This is an increase of 10% on the cost of running last year’s disco. Calculate the cost of running last year’s disco. Answer(e) $ [2]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 11:14 1 (b) 50 2 M1 for (220 + 280) ÷ 10 o.e. (c) 12 2 M1 for 21 ÷ (4 + 3) × 4 (or 3) o.e. (d) 280 3 M1 for 0.35 × their 500 (175) M1 dependent × 1.60 (e) 240 2 M1 for dividing 264 by 1.1 oe

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2 40 students are asked about the number of people in their families. Examiner's Use The table shows the results. Number of people in family 2 3 4 5 6 7 Frequency 1 1 17 12 6 3 (a) Find (i) the mode, Answer(a)(i) [1] (ii) the median, Answer(a)(ii) [1] (iii) the mean. Answer(a)(iii) [3] (b) Another n students are asked about the number of people in their families. The mean for these n students is 3. Find, in terms of n, an expression for the mean number for all (40 + n) students. Answer(b) [2]

Mark scheme: 2 (a) (i) 4 1 (ii) 5 1 (iii) 4.75 3 M1 for 1 × 2 + 1 × 3 + 17 × 4 + 12 × 5 + 6 × 6 + 3 × 7 condone one slip then M1 dependent result (190) ÷ 40 190 + 3n (b) 2 SC1 for their 190 + 3n 40 + n

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3 Examiner's y Use 8 7 T 6 5 4 3 2 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 P –4 –5 –6 Q –7 –8 (a) On the grid, draw the enlargement of the triangle T, centre (0, 0), scale factor 12 . [2] For − 1 0  Examiner's (b) The matrix   represents a transformation. Use  0 1  − 1 0  8 8 2  (i) Calculate the matrix product    .  0 1  4 8 8  Answer(b)(i) [2] (ii) On the grid, draw the image of the triangle T under this transformation. [2] (iii) Describe fully this single transformation. Answer(b)(iii) [2] (c) Describe fully the single transformation which maps (i) triangle T onto triangle P, Answer(c)(i) [2] (ii) triangle T onto triangle Q. Answer(c)(ii) [3] (d) Find the 2 by 2 matrix which represents the transformation in part (c)(ii).   Answer(d)   [2]    

Mark scheme: 3 (a) Triangle drawn with co-ords at (1, 4), 2 SC1 for 2 correct vertices or an enlargement sf (4, 2), (4, 4) 1 with wrong centre 2  − 8 − 8 − 2  (b) (i)  4 8 8  2 B1 each row (ii) Triangle drawn at (–8, 4), (–8, 8), (–2, 8) 2ft SC1 for 2 correct ft vertices. Can also be ft (i) correct regardless of (i) (iii) Reflection cao B1 Independent of (i) or (ii) Extra transformations lose all marks y – axis or x = 0 cao 2 B1 Independent of (i) or (ii) (c) (i) Translation B1 Extra transformations lose all marks  − 10   − 10  o.e. 2 B1 (ii) Rotation 3 B1 Extra transformations lose all marks (0, 0) B1 Allow word origin for (0, 0) 90° clockwise oe B1 Allow – 90° or 270° (anti-clockwise)  0 1  (d)  − 1 0  2 B1 each column IGCSE – May/June 2010 0580 41

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4 Examiner's Use A B Box A contains 3 black balls and 1 white ball. Box B contains 3 black balls and 2 white balls. (a) A ball can be chosen at random from either box. Complete the following statement. There is a greater probability of choosing a white ball from Box . Explain your answer. Answer(a) [1] (b) Abdul chooses a box and then chooses a ball from this box at random. The probability that he chooses box A is 23 . (i) Complete the tree diagram by writing the four probabilities in the empty spaces. BOX COLOUR 1 white 4 A 2 3 black white B black [4] For (ii) Find the probability that Abdul chooses box A and a black ball. Examiner's Use Answer(b)(ii) [2] (iii) Find the probability that Abdul chooses a black ball. Answer(b)(iii) [2] (c) Tatiana chooses a box and then chooses two balls from this box at random (without replacement). The probability that she chooses box A is 23 . Find the probability that Tatiana chooses two white balls. Answer(c) [2]

Mark scheme: 4 In (b) and (c) isw any cancelling or changing to other forms, after correct answer seen. Penalty of – 1 for 2 sf decimals or percentages. Do not accept ratio or worded forms. 2 1 (a) B and , oe 1 Allow any reasonable explanation, e.g. 2 out of 5 4 5 greater than 1 out of 4. 1 3 2 3 (b) (i) , , , 4 B1 B1 B1 B1 3 4 5 5 6 1 2 3 (ii) oe cao www 2 2 , 0.5 etc M1 for × their i.e. product of 12 2 3 4 correct branches on their tree 42 7 (iii) oe cao www2 2 , 0.7 etc 60 10 1 3 M1 for their (ii) + their × their from their 3 5 tree 2 1 (c) oe cao www2 2 , 0.0333(3……..) etc 60 30  2 1  1 2 1 M1 for  × × 0  + × ×  3 4  3 5 4

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5 Examiner's D Use 30° C NOT TO SCALE 24 cm 40° 40° A 26 cm B ABCD is a quadrilateral and BD is a diagonal. AB = 26 cm, BD = 24 cm, angle ABD = 40°, angle CBD = 40° and angle CDB = 30°. (a) Calculate the area of triangle ABD. Answer(a) cm2 [2] (b) Calculate the length of AD. Answer(b) cm [4] (c) Calculate the length of BC. Answer(c) cm [4] (d) Calculate the shortest distance from the point C to the line BD. Answer(d) cm [2]

Mark scheme: 5 (a) 200.5… to 201 www 2 2 M1 for 0.5 × 24 × 26 sin 40 oe A1 (b) 17.2 (0….) www 4 4 M2 for 262 + 242 – 2 × 26 × 24 cos 40 26 2 + 24 2 − BD 2 or M1 for cos 40 = 2 × 24 × 26 A2 or A1 for 295.976.. (c) 12.8 (12.77…) www 4 4 B1 for Angle C = 110 soi accept on diagram 24 sin 30 M2 for ( BC ) = oe or sin 110 sin 110 sin 30 M1 = oe i.e. a correct implicit 24 BC statement soi A1 (d) 8.208 to 8.230 www 2 2 M1 for their (c) × sin40 oe IGCSE – May/June 2010 0580 41

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6 The masses of 60 potatoes are measured. Examiner's The table shows the results. Use Mass (m grams) 10 I m Y 20 20 I m Y 40 40 I m Y 50 Frequency 10 30 20 (a) Calculate an estimate of the mean. Answer(a) g [4] (b) On the grid, draw an accurate histogram to show the information in the table. 2 Frequency density 1 m 0 10 20 30 40 50 Mass (grams) [3]

Mark scheme: 6 (a) 32.5 cao www4 4 M1 for mid-values seen M1 for use of Σfx with x’s anywhere in each interval (10 × 15 + 30 × 30 + 20 × 45) M1 ÷ 60 dependent on second M1 (b) Histogram drawn 3 B1 Bars correct positions and widths – no gaps B2 Heights of bars 1, 1.5 and 2 (B1 for any two correct or for heights in the ratio 2:3:4)

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7 (a) Calculate the volume of a cylinder of radius 31 centimetres and length 15 metres. Examiner's Give your answer in cubic metres. Use Answer(a) m3 [3] (b) A tree trunk has a circular cross-section of radius 31 cm and length 15 m. One cubic metre of the wood has a mass of 800 kg. Calculate the mass of the tree trunk, giving your answer in tonnes. Answer(b) tonnes [2] (c) NOT TO plastic SCALE sheet D C E The diagram shows a pile of 10 tree trunks. Each tree trunk has a circular cross-section of radius 31 cm and length 15 m. A plastic sheet is wrapped around the pile. C is the centre of one of the circles. CE and CD are perpendicular to the straight edges, as shown. For (i) Show that angle ECD = 120°. Examiner's Use Answer(c)(i) [2] (ii) Calculate the length of the arc DE, giving your answer in metres. Answer(c)(ii) m [2] (iii) The edge of the plastic sheet forms the perimeter of the cross-section of the pile. The perimeter consists of three straight lines and three arcs. Calculate this perimeter, giving your answer in metres. Answer(c)(iii) m [3] (iv) The plastic sheet does not cover the two ends of the pile. Calculate the area of the plastic sheet. Answer(c)(iv) m2 [1]

Mark scheme: 7 (a) 4.53 or 4.526 – 4.530…. 3 SC2 for figs 453 or 4526 – 4530 If SC0, M1 for π × (figs 31)2 × 15 (b) 3.62 to 3.624 ft 2ft M1 for their (a) × figs 8 oe (c) (i) 360 – 2 × 90 – 60 oe 2 E2 The 90’s and the 60 must be clearly justified. Accept in diagram. SC1 for 60 or two 90’s soi in correct positions oe e.g 360 ÷ 3 scores 0 (ii) 0.649 (0.6492 to 0.6493) 2 M1 for π × figs 62 ÷ 3 (iii) 7.53 (7.527 or 7.528….) 3 M1 for their (ii) × 3 M1 (indep) for 18 × figs 31 This M is spoiled by extra lengths. (iv) 112.9 to 113 ft 1ft ft their (iii) × 15

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8 (a) f(x) = 2x Examiner's Use Complete the table. x –2 –1 0 1 2 3 4 y = f(x) 0.5 1 2 4 [3] (b) g(x) = x(4 – x) Complete the table. x –1 0 1 2 3 4 y = g(x) 0 3 3 0 [2] For (c) On the grid, draw the graphs of Examiner's Use (i) y = f(x) for −2 Y x Y 4, [3] (ii) y = g(x) for −1 Y x Y 4. [3] y 16 14 12 10 8 6 4 2 x –2 –1 0 1 2 3 4 –2 –4 –6 (d) Use your graphs to solve the following equations. (i) f(x) = 10 Answer(d)(i) x = [1] (ii) f(x) = g(x) Answer(d)(ii) x = or x = [2] (iii) f -1(x) = 1.7 Answer(d)(iii) x = [1]

Mark scheme: 8 (a) 0.25, 8, 16 3 B1 B1 B1 (b) – 5, 4 2 B1 B1 (c) (i) 7 points plotted ft P2ft P1 for 5 or 6 points ft Curve through all 7 points exponential C1ft ft only if exponential shape shape (ii) 6 points plotted ft P2ft P1 for 5 points ft Curve through all 6 points parabola C1ft ft only if parabola shape shape (d) (i) 3.2 to 3.4 1 (ii) 0.3 to 0.4 and 2 2 B1 B1 (iii) 3.1 to 3.4 1

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9 (a) Solve the following equations. Examiner's Use 5 3 (i) = w w + 1 Answer(a)(i) w = [2] (ii) ( y + 1) 2 = 4 Answer(a)(ii) y = or y = [2] x + 1 x − 2 (iii) − = 2 3 5 Answer(a)(iii) x = [3] (b) (i) Factorise u2 −9u −10. Answer(b)(i) [2] (ii) Solve the equation u2 −9u −10 = 0. Answer(b)(ii) u = or u = [1] For (c) Examiner's Use NOT TO SCALE x + 1 x x x + 2 The area of the triangle is equal to the area of the square. All lengths are in centimetres. (i) Show that x2 −3x −2 = 0. Answer(c)(i) [3] (ii) Solve the equation x2 – 3x – 2 = 0, giving your answers correct to 2 decimal places. Show all your working. Answer(c)(ii) x = or x = [4] (iii) Calculate the area of one of the shapes. Answer(c)(iii) cm2 [1]

Mark scheme: 9 (a) (i) –2.5 oe 2 M1 for 5(w + 1) = 3w (ii) –3 or 1 2 B1 B1 (If 0, SC1 for y + 1 = ± 2) (iii) 9.5 oe B3 M2 for 5 x + 5 − 3 x + 6 = 2 × 15 Condone one slip (sign or numerical) on left hand side 5( x + )1 3( x − 2) or M1 for − or better, 15 15 condoning one sign or numerical slip. IGCSE – May/June 2010 0580 41

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