Cambridge IGCSE Mathematics 0580 — 2012 May/June Paper 3 · Variant 1

1 (a) Vince and Wendy share $2000 in the ratio Vince : Wendy = 19 : 21. For Examiner's Calculate the amount of money that Vince receives. Use Answer(a) $ [2] (b) Wendy has $265 to spend on some chairs. The chairs cost $37 each. Work out the largest number of chairs she can buy. Answer(b) [2] (c) Wendy shares $200 between her three children Jake, Karl and Lana. 2 She gives 27% of the money to Jake and of the money to Karl. 5 Work out the amount of money she gives to Lana. Answer(c) $ [3] (d) Wendy invests $500 at a rate of 4% per year compound interest. Calculate the total amount of interest she receives at the end of 2 years. Give your answer correct to the nearest dollar. Answer(d) $ [4]

Mark scheme: Qu. Answers Mark Part Mark 1 (a) 950 2 M1 for 2000 ÷ (19 + 21) 265 (b) 7 cao 2 M1 for seen oe e.g. adding up 37s 37 (c) 66 3 M1 for 54 seen M1 indep for 80 seen Or 33 67 M2 for × 200 or M1 for × 200 100 100 (d) 41 4 M1 for (500 × 1.04) × (1.04) oe A1 for 540.8 M1 dep for ‘their 540.8’ – 500 B1 ft for ‘their 40.8’ rounded to 41 Alt Method M1 for [500 + (500×0.04)] × 0.04 M1 dep ‘their 20’ + ‘their 20.8’ A1 for 40.8 B1 ft for ‘their 40.8’ rounded to 41

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2 For y Examiner's Use 8 6 4 A 2 x –6 –4 –2 0 2 4 6 8 –2 D –4 –6 –8 Two shapes A and D are shown on the grid. (a) (i) Reflect shape A in the line x = 0. Label this image B. [2] (ii) Rotate shape A through 180° about (2, 4). Label this image C. [2] (iii) Enlarge shape A with scale factor 2 and centre (3, 7). Label this image E. [2] (b) Describe fully the single transformation that maps shape D onto (i) shape B, Answer(b)(i) [2] (ii) shape C. Answer(b)(ii) [2]

Mark scheme: 2 (a) (i) Image at (–5,2), (–2,2), (–2,4), 2 B1 correct reflection in x = k, k ≠ 0 (–3,4), (–3,3), (–5,3) SC1 for totally correct reflection in x axis (ii) Image at (2,4), (2,6), (–1,6), (–1,5), 2 SC1 for 180° rotation not about (2,4) (1,5), (1,4) (iii) Image at (1,1), (3,1), (3, –1), 2 SC1 for correct size and orientation (7, –1), (7, –3), (1, –3) (b) (i) Reflection, y = 0 or x axis 1ft, 1ft Ft their (a)(i)  4  (ii) Translation,   1ft, 1ft Strict ft  8  Allow 4 right and 8 up IGCSE – May/June 2012 0580 31 1

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3 (a) Jon spins this 6-sided spinner. For 2 Examiner's 2 10 Use 4 8 6 The probability that the spinner lands on any of the six sides is equally likely. Write down the probability that the spinner lands on (i) the number 6, Answer(a)(i) [1] (ii) a prime number, Answer(a)(ii) [1] (iii) a number less than 11. Answer(a)(iii) [1] (b) Felix has a 12-sided spinner with the numbers 2, 4, 5, 7 and 9 written on it. It is equally likely to land on any side. The table shows the probability of the spinner landing on each number. Number on spinner 2 4 5 7 9 1 1 1 1 1 Probability 4 3 6 6 12 The diagram of the spinner has been completed for the number 2. Complete the diagram for the numbers 4, 5, 7 and 9. 2 2 2 [3] (c) Felix says that his spinner is more likely to land on a 2 than Jon’s spinner. Explain why he is wrong. Answer(c) [1] (d) Felix spins his 12-sided spinner 60 times and records the results. For Examiner's Use Number on spinner Frequency Pie chart sector angle 2 15 90° 4 20 120° 5 5 30° 7 12 9 8 (i) Complete the table by working out the sector angles for the numbers 7 and 9 . [3] (ii) Complete the pie chart. 2 4 [2] (iii) Write down the mode. Answer(d)(iii) [1] (iv) Calculate the mean. Answer(d)(iv) [3]

Mark scheme: 1 3 (a) (i) oe 1 Accept 0.167 or 16.7% or better 6 2 1 (ii) oe 1 Accept or 0.333 or 33.3% or better 6 3 (iii) 1 1 Accept “one” or 100% (b) (2,2,2), 4,4,4,4,5,5,7,7,9 seen on 3 B1 for 4,4,4,4 seen spinner B1 for 5,5 AND 7,7 seen B1 for ONE 9 seen. 3 (c) Felix’s probability is which is 1 Accept equivalent reasoning 12 2 less than Jon’s probability (of ) 6 4 which is oe 12 360 (d) (i) (90°, 120°, 30°), 72°, 48° 3 M1 for × f for one ‘Number’ correct 60 A1 for 1 correct answer If zero scored SC1 for their two answers totalling 120° (ii) 30° angle correct 1 72°, 48° 1ft (iii) 4 1 (iv) 4.85 3 M1 2 × 15 + 4 × 20 + 5 × 5 + 7 × 12 + 9 × 8 (allow 1 error) Σfx M1 dep for their 60

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4 In this question all the measurements are in centimetres. For Examiner's Use 11 – x NOT TO 2x + 3 SCALE 3x The diagram shows a triangle with sides of length 2x + 3, 11 – x and 3x. (a) Explain why x must be less than 11. Answer(a) [1] (b) Write down an expression, in terms of x, for the perimeter of the triangle. Give your answer in its simplest possible form. Answer(b) [2] (c) The perimeter of the triangle is 32 cm. (i) Write down an equation in terms of x and solve it. Answer(c)(i) x = [3] (ii) Work out the length of the shortest side of the triangle. Answer(c)(ii) cm [2]

Mark scheme: 4 (a) If x is more than 11 then 11 – x 1 would be negative oe (b) 14 + 4x cao 2 M1 for 2x + 3 + 11 – x + 3x accept 2(2x + 7) (c) (i) 4.5 cao 3 B1ft for “their (b)” = 32 M1ft for collecting their like terms correctly to give simplified expression of form ax = b b OR M1ft x = a (ii) 6.5 2ft M1ft for clear attempt at substituting their (c)(i) into 2 or more sides of triangle IGCSE – May/June 2012 0580 31 5 (a) Correct diagram: 4 rows & 6 1 columns

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5 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 The number of crosses in each Diagram forms a sequence. (a) On the grid draw Diagram 4. [1] (b) Write down the number of crosses needed to draw Diagram 5. Answer(b) [1] (c) Diagram 1 has 1 row of 3 crosses. Diagram 2 has 2 rows of 4 crosses. (i) Complete this statement for Diagram n. Diagram n has n rows of crosses. [1] (ii) Write down, in terms of n, how many crosses are needed to draw Diagram n. Answer(c)(ii) [1] (iii) Find the number of crosses needed to draw Diagram 20. Answer(c)(iii) [1]

Mark scheme: their (a)(i) (b) 5 1ft Ft is 34 (c) Said by 1.5 secs 3ft ' their (a)(ii)' M1ft (= 32.5) 4 ' their (a)(ii)' M1ft 34 – (34 – 32.5) 4 120 120 (d) (i) 67.4° 2 M1 ‘tan’= or ‘sin’= 50 their 130 50 or ‘cos’= their 130 (ii) 113° or 112.6° 1ft 180 – ‘their (d)(i)’

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6 For y Examiner's Use B 6 4 2 A E C x –4 –2 0 2 4 6 8 10 12 –2 –4 –6 Triangle ABC is drawn on a 1cm2 grid. E is the point (0, 0). (a) Write down the gradient of the line AB. Answer(a) [2] (b) The gradient of BC is – 0.5 . Write down the equation of the line BC in the form y = mx + c. Answer(b) y = [2] (c) Write down the ratio AE : EC. For Give your answer in its simplest form. Examiner's Use Answer(c) : [2] (d) Measure angle ABE. Answer(d) Angle ABE = [1] (e) Triangle ABE is similar to triangle BCE. Explain what the word similar tells you about the triangles ABE and BCE. Answer(e) [2] (f) Calculate the area of triangle ABC. Answer(f) cm2 [3] (g) ABCD is a rectangle. (i) Mark point D on the grid. [1] (ii) Write down the co-ordinates of D. Answer(g)(ii) ( , ) [1]

Mark scheme: (e) 6 × 10–3 4 M1 ‘50’ × ‘120’ figs seen in area calculation A1 for 6000 seen (implied by 0.006 later) M1 for dividing by 1000², 0.05 & 0.12 seen or ×10–6 oe somewhere B1 ft from ‘their 0.006’ provided SF power is –ve Or SC1 for 0.6 × 10–2 oe 9 (a) (i) 226 to 226.224 cm³ 3 M1 π × 3² × 8 B1 for units : cm³ (ii) 8 cao www 4 B1 1500 used M1ft 3 × their (a)(i) 4 their 1500 M1ft 3 × their (a)(i) 4 16 (b) 5.09 (5.092 to 5.10) 2 M1 π (c) 148 cm² 3 M2 for 2 × 4 × 5 + 2 × 4 × 6 + 2 × 5 × 6 SC1 for 2 × 4 × 5 oe or 4 × 5 + 4 × 6 + 5 × 6 implied by 40, 48, 60 or 74, or list of 20, 20, 24, 24, 30, 30 (d) (i) mv oe 1 (ii) msv oe 1ft Ft (d)(i) × s (iii) 1000 msv oe 1ft Ft (d)(ii) × 1000

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