Cambridge IGCSE Mathematics 0580 — 2009 May/June Paper 4 · Variant 1

1 Marcus receives $800 from his grandmother. For Examiner's (a) He decides to spend $150 and to divide the remaining $650 in the ratio Use savings : holiday = 9 : 4. Calculate the amount of his savings. Answer(a) $ [2] (b) (i) He uses 80% of the $150 to buy some clothes. Calculate the cost of the clothes. Answer(b)(i) $ [2] (ii) The money remaining from the $150 is 37 1 % of the cost of a day trip to Cairo. 2 Calculate the cost of the trip. Answer(b)(ii) $ [2] (c) (i) Marcus invests $400 of his savings for 2 years at 5 % per year compound interest. Calculate the amount he has at the end of the 2 years. Answer(c)(i) $ [2] (ii) Marcus’s sister also invests $400, at r % per year simple interest. At the end of 2 years she has exactly the same amount as Marcus. Calculate the value of r. Answer(c)(ii) r = [3]

Mark scheme: 1 (a) ($) 450 B2 M1 for 650 ÷ (9 + 4) × 9 (÷ 14 does not imply 9 + 4) (b) (i) ($) 120 B2 M1 for 0.8 × 150 o.e. (ii) ($) 80 ft B2 ft M1 for (150 – their(b)(i)) ÷ 0.375 o.e. only if +ve. After M0, SC1 for answer 320 (c) (i) ($) 441 B2 M1 for 400 × 1.052 o.e. or for answer 41 (ii) If use Simple Int in (i), M0, M0 in this part 1 their ( (i) − 400 ) ÷ 400 × 100 o.e. M2 i.e. a full explicit method for r 2 If M0, 400 × r × 2 5.125 or 5.13 or 5.12 c.a.o. www3 A1 M1 for = their (i) − 400 100 or their (i) ÷ 400 × 100 then – 100 their (i) - 400 or × 100 (s.o.i. by 10.25) 400 If still M0, SC1 for answers 55.125 or 55.12 or 55.13 or 55.1 or 0.05125 or 0.0512 or 0.0513 [11]

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2 A normal die, numbered 1 to 6, is rolled 50 times. For Examiner's 1 Use 2 4 The results are shown in the frequency table. Score 1 2 3 4 5 6 Frequency 15 10 7 5 6 7 (a) Write down the modal score. Answer(a) [1] (b) Find the median score. Answer(b) [1] (c) Calculate the mean score. Answer(c) [2] (d) The die is then rolled another 10 times. The mean score for the 60 rolls is 2.95. Calculate the mean score for the extra 10 rolls. Answer(d) [3]

Mark scheme: 2 (a) 1 B1 (b) 2.5 o.e. B1 (c) 2.96 c.a.o. B2 If B0, M1 for 15 × 1 + 10 × 2 + 7 × 3 + 5 × 4 + 6 × 5 + 7 × 6 (allow one slip) implied by 148 seen Ignore subsequent rounding (d) 60 × 2.95 (= 177) M1 their 177 – their 148 (or 50 × their 2.96) M1 Dependent on first M and only if positive or M1 for (Mean of new rolls =) 2.9 c.a.o. www3 A1 their 148( 50 × their .296 ) + x ( or 10 x ) = .295 60 then M1 for x ( or 10 x ) = 60 × .295 − their 148 (or 50 × their 2.96) and only if positive [7] IGCSE – May/June 2009 0580, 0581 04 48

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3 For Examiner's P Use NOT TO SCALE 10 cm 14 cm Q R In triangle PQR, angle QPR is acute, PQ = 10 cm and PR = 14 cm. (a) The area of triangle PQR is 48 cm2. Calculate angle QPR and show that it rounds to 43.3°, correct to 1 decimal place. You must show all your working. Answer (a) [3] (b) Calculate the length of the side QR. Answer(b) QR = cm [4]

Mark scheme: 48 3 (a) (sin P) = o.e. fraction M2 M1 for 0.5 × 10 × 14 sin P = 48 o.e. 5.0 × 10 × 14 Allow 0.5 × 10 × 14 sin 43.3 = 48 for M1 but no further credit P = 43.29…. cao A1 (b) 102 + 142 – 2 × 10 × 14cos 43.3 (= 92.2) M2 If M0, M1 for correct implicit statement Evaluating square root M1 M1 (dependent on M2) for square root of correct combination (not negative) i.e 16cos43.3 (11.64..) implies M2M0 (QR =) 9.6(0) (9.60 to 9.603…) c.a.o. ww2 A1 [7] 250 AB 250

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4 For Examiner's Use North NOT TO North SCALE A 126° B North 250 m 23° P The diagram shows three straight horizontal roads in a town, connecting points P, A and B. PB =250 m, angle APB = 23° and angle BAP = 126°. (a) Calculate the length of the road AB. Answer(a) AB = m [3] (b) The bearing of A from P is 303°. Find the bearing of (i) B from P, Answer(b)(i) [1] (ii) A from B. Answer(b)(ii) [2]

Mark scheme: 250 AB 250 4 (a) ( AB = ) × sin 23 (s.o.i by 120…) M2 M1 for = o.e. (implicit) sin 126 sin 23 sin 126 121 (120.7 to 121) (m) c.a.o. www3 A1 (b) (i) 280 B1 (ii) (0)69 c.a.o. B2 SC1 for answer 249 [6]

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x _ 25 (a) The table shows some values for the equation y = for – 4 Y x Y=–0.5 and 0.5 Y x Y 4. Examiner'sFor 2 x Use x –4 –3 –2 –1.5 –1 –0.5 0.5 1 1.5 2 3 4 y –1.5 –0.83 0 0.58 –3.75 –0.58 0 0.83 1.5 (i) Write the missing values of y in the empty spaces. [3] x _ 2 for – 4 Y x Y=–0.5 and 0.5 Y x Y 4. (ii) On the grid, draw the graph of y = 2 x y 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 4 –1 –2 –3 –4 [5] x _ 2 For (b) Use your graph to solve the equation = 1 . Examiner's 2 x Use Answer(b) x = or x = [2] (c) (i) By drawing a tangent, work out the gradient of the graph where x = 2. Answer(c)(i) [3] (ii) Write down the gradient of the graph where x = –2. Answer(c)(ii) [1] (d) (i) On the grid, draw the line y = – x for – 4 Y x Y4. [1] x _ 2 _ (ii) Use your graphs to solve the equation = x . 2 x Answer(d)(ii) x = or x = [2] (e) Write down the equation of a straight line which passes through the origin and does not x _ 2 intersect the graph of y = . 2 x Answer(e) [2]

Mark scheme: 5 (a) (i) 1.5, 3.75, –1.5 B1,B1,B1 (ii) 12 points plotted ft P3 ft P2 ft for 10 or 11 points, Curve through at least 10 points and correct P1 ft for 8 or 9 points shape over full domain C1 i.s.w. if two branches joined Two separate branches, one on each side of y-axis, neither in contact with y-axis B1 Independent (b) –1.4 ≤ x ≤ –1.1 and 3.1 ≤ x ≤ 3.4 B1,B1 i.s.w. 3rd answer if curve cuts y = 1 again (c) (i) Correct ruled tangent at x = 2 or x = –2 M1 Long enough to be able to find gradient Evidence of rise/run M1 Dependent – check their graph against gradient of 1 – must be correct side of 1 No tangent drawn M0M0 0.8 to 1.2 A1 (ii) 0.8 to 1.2 inc. or same answer as (i) ft B1 ft (d) (i) Correct ruled line to cut curve for all B1 Within ½ square of (–1, 1) and (1, –1) possible intersections (at least 2) (ii) –1.3 to –1.05, 1.05 to 1.3 inclusive B1, B1 i.s.w. any extra answers (e) y = kx with k ≥ 12 o.e. or x = 0 B2 If B0, allow SC1 for y = kx with k < 12 or for y-axis stated [19] IGCSE – May/June 2009 0580, 0581 04

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6 (a) For Examiner's B Use NOT TO SCALE (x + 1) cm A (x + 6) cm D (x + 2) cm C In triangle ABC, the line BD is perpendicular to AC. AD = (x + 6) cm, DC = (x + 2) cm and the height BD = (x + 1) cm. The area of triangle ABC is 40 cm2. (i) Show that x2 + 5x – 36 = 0. Answer (a)(i) [3] (ii) Solve the equation x2 + 5x – 36 = 0. Answer(a)(ii) x = or x = [2] (iii) Calculate the length of BC. Answer(a)(iii) BC = cm [2] (b) Amira takes 9 hours 25 minutes to complete a long walk. For 113 Examiner's (i) Show that the time of 9 hours 25 minutes can be written as hours. Use 12 Answer (b)(i) [1] (ii) She walks (3y + 2) kilometres at 3 km/h and then a further (y + 4) kilometres at 2 km/h. 9 y + 16 Show that the total time taken is hours. 6 Answer(b)(ii) [2] 9 y + 16 113 (iii) Solve the equation = . 6 12 Answer(b)(iii) y = [2] (iv) Calculate Amira’s average speed, in kilometres per hour, for the whole walk. Answer(b)(iv) km/h [3]

Mark scheme: 6 (a) (i) 0.5 [(x + 6) + (x + 2)] × (x +1) (= 40) or M1A1 M1 for any algebraic use of half base × better height (Brackets may be implied later) 0.5(2x + 8)(x + 1) (= 40) o.e. May be first line 0.5(2x2 +10x + 8) (= 40) o.e. If this first line, then M0 x2 + 5x + 4 = 40 o.e. E1 Dependent on M1A1. Fully established – x2 + 5x – 36 = 0 no errors throughout and at least 2 steps, one with 40 or 80, after first line (ii) –9, 4 B1,B1 If B0, SC1 for +9 and –4 (iii) (BC2 = ) (their x + 1)2 + (their x + 2)2 M1 Their x must be positive (BC = ) 7.81(0…)) c.a.o. www2 A1 Ignore any extra solutions 5 108 + 5 9 × 12 + 5 565 (b) (i) 9 or or or E1 Must be fractional form 12 12 12 60 Condone 113/12 × 60 = 565; 9 × 60 + 25 9 × 60 + 25 = 565 or seen Not for decimals 60 3 y + 2 y + 4 (ii) or o.e. B1 3 2 2 ( 3 y + 2 ) 3( y + 4 ) 6 y + 4 3 y + 12 + o.e. B1 or + o.e. 6 6 6 6 2 ( 9 y + 16 ) 113 (iii) = o.e. M1 o.e. means with common denominator or 12 12 better y = 4.5 c.a.o. www2 A1 (Trial and error scores 2 or 0.) (iv) (Total dist =) (3 × their y) + 2 + (their y) + 4 M1 (= 24) o.e. their 24 (Average speed = ) o.e. M1 (dependent) Must be km divided by hours 9 512 o.e. for full method 2.55 (km/h) (2.548 – 2.549) c.a.o. www 3 A1 Accept fractions in range [15] IGCSE – May/June 2009 0580, 0581 04 2 2

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7 For Examiner's NOT TO Use SCALE x cm x cm 250 cm A solid metal bar is in the shape of a cuboid of length of 250 cm. The cross-section is a square of side x cm. The volume of the cuboid is 4840 cm3. (a) Show that x = 4.4. Answer (a) [2] (b) The mass of 1 cm3 of the metal is 8.8 grams. Calculate the mass of the whole metal bar in kilograms. Answer(b) kg [2] (c) A box, in the shape of a cuboid measures 250 cm by 88 cm by h cm. 120 of the metal bars fit exactly in the box. Calculate the value of h. Answer(c) h = [2] (d) One metal bar, of volume 4840 cm3, is melted down to make 4200 identical small spheres. For Examiner's All the metal is used. Use (i) Calculate the radius of each sphere. Show that your answer rounds to 0.65 cm, correct to 2 decimal places. 4 3 [The volume, V, of a sphere, radius r, is given by V = πr .] 3 Answer(d)(i) [4] (ii) Calculate the surface area of each sphere, using 0.65 cm for the radius. [The surface area, A, of a sphere, radius r, is given by A = 4 πr 2 .] Answer(d)(ii) cm2 [1] (iii) Calculate the total surface area of all 4200 spheres as a percentage of the surface area of the metal bar. Answer(d)(iii) % [4]

Mark scheme: 7 (a) 250x2 = 4840 o.e. M1 Allow M1 for 250 × 4.42 = 4840 x² = 19.36 or (x =) 4840 ÷ 250 (= 4.4) E1 Then E1 for 250 × 19.36 = 4840 (b) 42.6 (kg) cao (42.592 or 42.59) B2 SC1 for figures 426 or 4259… (c) 26.4 (cm) c.a.o. B2 If B0, M1 for any of following 88 ÷ 4.4 = 20 and 120 ÷ 20 = 6 (accept 6 bars high o.e.) or 88h = 4.42 × 120 or 250 × 88 × h = 120 × 4840 (d) (i) 4840 ÷ 4200 (implied by 1.15(2)) M1 4200 × 4 3 π r3 = 4840 ÷ 4 3 π (implied by 0.274 to 0.276) M1 (r3 =) 4840 ÷ (4200 × 4 3 π ) 3 (seen or implied by correct answer to M1 3 Third M dependent on M1M1 dep more than 2 dp) 0.649 – 0.651 A1 Must be 3dp or better (ii) 5.31 (5.306 – 5.31) (cm2) B1 4200 × their (ii) (iii) × 100 M3 If M0, M1 for 4200 × their (ii) (22299) 2 × 4.4 2 + 4 × 4.4 × 250 and M1 (independent) for correct method for surface area of solid cuboid (4438.72) 501.9 – 503 (%) c.a.o. www4 A1 [15]

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8 For Examiner's First Second Third Use Calculator Calculator Calculator p F F F p q NF NF F F q NF NF NF F = faulty NF = not faulty The tree diagram shows a testing procedure on calculators, taken from a large batch. 1 Each time a calculator is chosen at random, the probability that it is faulty (F) is . 20 (a) Write down the values of p and q. Answer(a) p = and q = [1] (b) Two calculators are chosen at random. Calculate the probability that (i) both are faulty, Answer(b)(i) [2] (ii) exactly one is faulty. Answer(b)(ii) [2] (c) If exactly one out of two calculators tested is faulty, then a third calculator is chosen at random. For Examiner's Calculate the probability that exactly one of the first two calculators is faulty and the third one Use is faulty. Answer(c) [2] (d) The whole batch of calculators is rejected either if the first two chosen are both faulty or if a third one needs to be chosen and it is faulty. Calculate the probability that the whole batch is rejected. Answer(d) [2] (e) In one month, 1000 batches of calculators are tested in this way. How many batches are expected to be rejected? Answer(e) [1]

Mark scheme: 8 Throughout the question ratios score zero. If using decimals, 2 s.f. correct answers to parts (c) and (d) – penalty of 1 once Use of words e.g. 1 in 400 or 1 out of 400, Correct answers – penalty of one For method marks only accept probabilities p and q between 0 and 1 (a) p = 201 , q = 1920 o.e. B1 Could be on diagram (b) (i) 4001 o.e. c.a.o. B2 0.0025 allow M1 for (their p)2 o.e. (ii) 40038 o.e. c.a.o. B2 0.095 allow M1 for 2 (their p)( their q) o.e. (c) 800038 o.e. c.a.o. B2 0.00475 allow M1 for 2(their p)² (their q) o.e. including their (ii) × their p (d) their (b)(i) + their (c) M1 58 o.e. c.a.o. A1 0.00725 8000 (e) their (d) × 1000 = 7.25 o.e. ft B1 ft Accept 7 or 8 or an equivalent integer ft [10] IGCSE – May/June 2009 0580, 0581 04

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9 The heights of 100 students are measured. For Examiner's The results have been used to draw this cumulative frequency diagram. Use Cumulative frequency 100 90 80 70 60 50 40 30 20 10 0 0 150 155 160 165 170 175 180 185 190 Height (cm) (a) Find For Examiner's (i) the median height, Use Answer(a)(i) cm [1] (ii) the lower quartile, Answer(a)(ii) cm [1] (iii) the inter-quartile range, Answer(a)(iii) cm [1] (iv) the number of students with a height greater than 177 cm. Answer(a)(iv) [2] (b) The frequency table shows the information about the 100 students who were measured. Height (h cm) 150 < h Y=160 160 < h Y=170 170 < h Y=180 180 < h Y=190 Frequency 47 18 (i) Use the cumulative frequency diagram to complete the table above. [1] (ii) Calculate an estimate of the mean height of the 100 students. Answer(b)(ii) cm [4]

Mark scheme: 9 (a) (i) 174 to 174.25 (cm) c.a.o. B1 (ii) 167 (cm) c.a.o. B1 (iii) 12 (cm) c.a.o. B1 (iv) 37 c.a.o. B2 If B0, B1 for 63 seen in working space (b) (i) 10, 25 B1 (ii) 155, 165, 175, 185 M1 s.o.i. allow 1 slip (their 10 × 155 + their 25 × 165 + 47 × 175 M1 Use of Σ fx where the x’s are in/on their + 18 × 185) intervals (allow one more slip) (17 230) ÷ 100 M1 (dependent on second M) ÷ 100 172 or 172.3 (cm) c.a.o. www 4 A1 [10] 2

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10 f(x) = 2x – 1 g(x) = x2 + 1 h(x) = 2x For Examiner's Use (a) Find the value of _ 1 (i) , f( 2 ) Answer(a)(i) [1] (ii) g( _5 ) , Answer(a)(ii) [1] _ (iii) h ( 3) . Answer(a)(iii) [1] (b) Find the inverse function f –1(x). Answer(b) f –1(x) = [2] (c) g(x) = z. Find x in terms of z. Answer(c) x = [2] (d) Find gf(x), in its simplest form. Answer(d) gf(x) = [2] (e) h(x) = 512. For Find the value of x. Examiner's Use Answer(e) x = [1] (f) Solve the equation 2f(x) + g(x) = 0, giving your answers correct to 2 decimal places. Answer(f) x = or x = [5] (g) Sketch the graph of (i) y = f(x), (ii) y = g(x). y y x x O O (i) y = f(x) (ii) y = g(x) [3]

Mark scheme: 10 (a) (i) –2, B1 (ii) 26, B1 (iii) 18 o.e. B1 y + 1 (b) ( = x ) M1 If switch x and y first then 2 M1 for x = 2y – 1 or (f—1(x) = ) x + 1 o.e. www2 A1 If use a diagram/chart then 2 M1 for any evidence of +1 then result ÷ 2 (c) z = x 2 + 1 z − 1 = x 2 M1 Correct rearrangement at any stage for x or x2. (x = ) z − 1 www2 M1 Correct sq root at any stage Ignore +, – or ± in front of (d) ( 2 x − )1 2 + 1 M1 = 4 x 2 − 4 x + 2 or 2 ( 2 x 2 −x2 + )1 A1 Final answer but condone one minor www 2 factorising slip if first answer seen (e) 9 B1 (f) 2(2x – 1) + x2 + 1 (= 0) or better B1 (x2 + 4x – 1 = 0 ) 2 4 − 4 1()( −)1 or better seen 2 M1 − 4 ± 4 − 4 1()( −)1 ( x = ) ft p + or − q 2 × 1 M1 If in form for – 4 and 2 × 1 r or better Ft their 1, 4 and –1 from quadratic equation seen (x = ) – 4.24, 0.24 c.a.o. www 4 A1,A1 After A0A0, SC1 for – 4.2 or – 4.235 or (final answers) – 4.236… and 0.2 or 0.235 or 0.236….. The SC1’s www imply the M marks (g) (i) Straight line with positive gradient and L1 (ii) negative y-intercept U-shape Parabola C1 vertex on positive y-axis V1 Dependent [18] IGCSE – May/June 2009 0580, 0581 04

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11 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 The first four terms in a sequence are 1, 3, 6 and 10. They are shown by the number of dots in the four diagrams above. (a) Write down the next four terms in the sequence. Answer(a) , , , [2] (b) (i) The sum of the two consecutive terms 3 and 6 is 9. The sum of the two consecutive terms 6 and 10 is 16. Complete the following statements using different pairs of terms. The sum of the two consecutive terms and is . The sum of the two consecutive terms and is . [1] (ii) What special name is given to these sums? Answer(b)(ii) [1] n ( n +1) (c) (i) The formula for the nth term in the sequence 1, 3, 6, 10… is , k where k is an integer. Find the value of k. Answer(c)(i) k = [1] (ii) Test your formula when n = 4, showing your working. For Examiner's Answer (c)(ii) Use [1] (iii) Find the value of the 180th term in the sequence. Answer(c)(iii) [1] (d) (i) Show clearly that the sum of the nth and the (n + 1)th terms is (n + 1)2. Answer (d)(i) [3] (ii) Find the values of the two consecutive terms which have a sum of 3481. Answer(d)(ii) and [2]

Mark scheme: 11 (a) 15, 21, 28, 36 B2 B1 for 3 correct (b) (i) 10 + 15 = 25, 15 + 21 = 36 etc B1 Any two complete and correct statements (ii) Square B1 (c) (i) 2 B1 4 × 5 (ii) = 10 o.e. E1 2 (iii) 16 290 c.a.o. B1 ( n + 1)( n + 2 ) n 2 + 3n + 2 (d) (i) or seen M1 Denominator could be their k 2 2 May be implied by next line n(n +)1 (n +1)(n + )2 n 2 + n n 2 + 3n + 2 + or + M1 This line must be seen and at least one 2 2 2 2 more step, without any error, to gain the E 2n2 + 4n + 2 mark ( n + )1 ( n + n + 2) 2 2 ( n + 1)(2n + 2) n2 + 2n +1 2 2 (n +)1 2( n + 1)(n + )1 = ( n + 2)1 E1 Dependent on M1M1. Fully established – 2 no errors (ii) 1711 and 1770 final answers c.a.o. B2 SC1 for 59 or 58 or 1711 or 1770 seen [12] Graph for Question 5 4444 3333 2222 1111 0000 -8-8-8-8 -7-7-7-7 -6-6-6-6 -5-5-5-5 -4-4-4-4 -3-3-3-3 -2-2-2-2 -1-1-1-1 0000 1111 2222 3333 4444 5555 6666 7777 8888 -1-1-1-1 -2-2-2-2 -3-3-3-3 4444

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