Cambridge IGCSE Mathematics 0580 — 2012 Oct/Nov Paper 4 · Variant 3
0580/43/O/N/12 · 4 questions · 52 marks · ≈59 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper20 pages




















Mark scheme8 pages
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Questions as text
Q1 · The Martinez family travels by car to Seatown
1 (a) The Martinez family travels by car to Seatown. For The distance is 92 km and the journey takes 1 hour 25 minutes. Examiner's Use (i) The family leaves home at 07 50. Write down the time they arrive at Seatown. Answer(a)(i) [1] (ii) Calculate the average speed for the journey. Answer(a)(ii) km/h [2] (iii) During the journey, the family stops for 10 minutes. Calculate 10 minutes as a percentage of 1 hour 25 minutes. Answer(a)(iii) % [1] (iv) 92 km is 15% more than the distance from Seatown to Deecity. Calculate the distance from Seatown to Deecity. Answer(a)(iv) km [3] (b) The Martinez family spends $150 in the ratio For Examiner's fuel : meals : gifts = 11 : 16 : 3 . Use (i) Show that $15 is spent on gifts. Answer (b)(i) [2] (ii) The family buys two gifts. The first gift costs $8.25. Find the ratio cost of first gift : cost of second gift. Give your answer in its simplest form. Answer(b)(ii) : [2]
Mark scheme: Qu. Answers Mark Part Marks 1 (a) (i) [0]9 15 [am] 1 Any acceptable form of time (ii) 64.9 or 65.[0] or 64.92 to 2 M1 for 92 ÷ (1 and 25 mins) or 92/85 × 60 oe 64.98 or 92 ÷ (1.41 to 1.42) (iii) 11.76…or 11.8 1 (iv) 80 3 M2 for 92 ÷ 1.15 oe or M1 for 115% associated with 92 (b) (i) 150 ÷ (11 + 16 + 3) or M1 Correct first step 150 × 3 oe then × 3 or ÷ 30 E1 Correct conclusion (ii) 11 : 9 final answer 2 M1 for 8.25 : (15 – 8.25) oe For M1 e.g. allow 1 : 0.818 [0.8181 to 0.8182] or 1.22 : 1 [1.222…] After M0, SC1 for 9 : 11 as final answer −11 k
Q2 · For y Examiner's Use 8 7 X 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…
2 (a) For y Examiner's Use 8 7 X 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 Y –3 –4 –5 –6 –7 –8 –9 − 11 (i) Draw the translation of triangle X by the vector . [2] − 1 1 (ii) Draw the enlargement of triangle Y with centre (–6, – 4) and scale factor . [2] 2 (b) For y Examiner's Use 8 7 W X 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 Y –3 –4 Z –5 –6 –7 –8 –9 Describe fully the single transformation that maps (i) triangle X onto triangle Z, Answer(b)(i) [2] (ii) triangle X onto triangle Y, Answer(b)(ii) [3] (iii) triangle X onto triangle W. Answer(b)(iii) [3] (c) Find the matrix that represents the transformation in part (b)(iii). Answer(c) [2]
Mark scheme: − 11 k 2 (a) (i) Image at (– 3, 1), (– 7, 7), 2 SC1 for translation or k −1 (– 3, 7) (ii) Image at (– 4, – 1), (– 4, – 4), 2 SC1 for enlargement factor 0.5 and correct (– 2, – 4) orientation In each part of (b) must be one transformation only – if more then lose all marks for that part (b) (i) Reflection, y = 1 2 B1 B1 independent (ii) Rotation, (3, 2), 180 oe 3 B1 B1 B1 independent or enlargement, (3, 2), (factor) – 1 (iii) Stretch, (factor) 0.5, 3 B1 B1 B1 independent – must be clear on invariant Invariant line y-axis or x = 0 line IGCSE – October/November 2012 0580 43 5.0 0 (c) 2 ft ft their factor in (b)(iii) only if stretch not 0 or 1 0 1 k 0 SC1 for [k ≠ 0 or 1] or 0 1 1 0 ft their factor only if stretch in (b)(iii) 0 5.0
Q3 · A metal cuboid has a volume of 1080 cm3 and a mass of 8 kg
3 A metal cuboid has a volume of 1080 cm3 and a mass of 8 kg. For Examiner's (a) Calculate the mass of one cubic centimetre of the metal. Use Give your answer in grams. Answer(a) g [1] (b) The base of the cuboid measures 12 cm by 10 cm. Calculate the height of the cuboid. Answer(b) cm [2] (c) The cuboid is melted down and made into a sphere with radius r cm. (i) Calculate the value of r. 4 [The volume, V, of a sphere with radius r is V = πr 3.] 3 Answer(c)(i) r = [3] (ii) Calculate the surface area of the sphere. For Examiner's [The surface area, A, of a sphere with radius r is A = 4πr 2.] Use Answer(c)(ii) cm2 [2] (d) A larger sphere has a radius R cm. The surface area of this sphere is double the surface area of the sphere with radius r cm in part (c). R Find the value of . r Answer(d) [2]
Mark scheme: 3 (a) 7.407….. or 7.41 1 (b) 9 2 M1 for 1080 ÷ (12 × 10) oe 1080 (c) (i) 6.36 to 6.37 www 3 M2 for 3 4 oe π 3 1080 or M1 for 4 oe [ 257.7 to 258.7] π 3 Accept 4.18 to 4.19 for 4/3 π (ii) 508 to 510 2 M1 for 4 × π × (their (c)(i))2 (d) 2 or 1.41 [1.414…] www 2 Allow over 1 or 2 : 1 etc M1 for (R / r)2 = 2 oe or [R2 =] (2 × their (c)(ii))/4π or [R2 =] 2 × (their (c)(i))²
Q4 · For 4 f(x) = O 3x, x ≠ 0 Examiner's 2 x Use (a) Complete the table
2 For 4 f(x) = O 3x, x ≠ 0 Examiner's 2 x Use (a) Complete the table. x O3 O2.5 O2 O1.5 O1 O0.5 0.5 1 1.5 2 2.5 3 f(x) 9.2 7.8 6.5 5.4 9.5 6.5 O3.6 O5.5 O7.2 O8.8 [2] (b) On the grid, draw the graph of y = f(x), for O3 Y x Y O0.5 and 0.5 Y x Y 3 . y 10 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 [5] (c) Use your graph to solve the equations. For Examiner's (i) f(x) = 4 Use Answer(c)(i) x = [1] (ii) f(x) = 3x Answer(c)(ii) x = [2] (d) The equation f(x) = 3x can be written as x3 = k. Find the value of k. Answer(d) k = [2] (e) (i) Draw the straight line through the points (–1, 5) and (3, –9). [1] (ii) Find the equation of this line. Answer(e)(ii) [3] (iii) Complete the statement. The straight line in part (e)(ii) is a to the graph of y = f(x). [1]
Mark scheme: 4 (a) 5, – 1 2 B1 B1 (b) 12 points plotted ft P3ft P2ft for 10 or 11, P1ft for 8 or 9 Smooth curve through at least 12 C1 In absence of plot[s], allow curve to imply plot[s]. points No ruled sections Two separate branches B1 Not touching y-axis (c) (i) 0.55 to 0.65 1 (ii) 0.65 to 0.75 2 M1 for y = 3x drawn (ruled) to cross curve 1 0.3& (d) 2 Accept 0.333[3….] or 3 2 M1 for 2 − 3 x = 3 x or better x IGCSE – October/November 2012 0580 43 (e) (i) Ruled line through (– 1, 5) 1 and (3, – 9) (ii) y = −5.3 x + 5.1 oe final 3 B2 for y = kx + 5.1 [k ≠ 0] oe or y = −5.3 x + d oe answer B1 for gradient = – 3.5 oe accept integer/integer or y = kx + 4.1[ to 6.1] oe SC2 for answer − 5.3 x + 5.1 [no ‘y =’ ] (iii) Tangent 1 5 (a) 0.57 B4 Condone use of other variables M1 for 2 w + 3l = 6.3 oe and M1 for l = w + .0 25 oe A1 for correct aw = b or cl = d or M2 for 2 w + (3 w + .025) = 6.3 oe or 2(l − .025) + 3l = 6.3 oe or M1 for w + 0.25 or l – 0.25 seen A1 for 2 w + 3w = 6.3 − .075 or better or 2l + 3l = 6.3 + 5.0 or better l = 0.82 implies M2A1 trial & error scores B4 or zero accept answer 57 if written 57 cents after M0, SC3 if answer 57
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