Cambridge IGCSE Mathematics 0580 — 2013 Oct/Nov Paper 4 · Variant 3

1 (a) (i) In a camera magazine, 63 pages are used for adverts. For Examiner′s The ratio number of pages of adverts : number of pages of reviews = 7 : 5 . Use Calculate the number of pages used for reviews. Answer(a)(i) ............................................... [2] (ii) In another copy of the magazine, 56 pages are used for reviews and for photographs. The ratio number of pages of reviews : number of pages of photographs = 9 : 5 . Calculate the number of pages used for photographs. Answer(a)(ii) ............................................... [2] (iii) One copy of the magazine costs $4.90 . An annual subscription costs $48.80 for 13 copies. Calculate the percentage discount by having an annual subscription. Answer(a)(iii) ........................................... % [3] (b) In a car magazine, 25% of the pages are used for selling second-hand cars, For 1 Examiner′s 62 2 % of the remaining pages are used for features, Use and the other 36 pages are used for reviews. Work out the total number of pages in the magazine. Answer(b) ............................................... [4] _____________________________________________________________________________________

Mark scheme: Qu. Answers Mark Part Marks 1 (a) (i) 45 2 M1 for 5 × 63 ÷ 7 (ii) 20 2 M1 for 5 × 56 ÷ 14 13 × 9.4 – 48 8. (iii) 23.4 or 23.38 to 23.41 3 M2 for × 100 13 × 9.4 9.4 – 488. ÷ 13 or × 100 9.4 Or 13 × 9.4 – 48 8. 48 8. M1 for or × 100 or 76.6[…] 13 × 9.4 13 × 9.4 (b) 128 4 Using fractions (percentages / decimals): 3 3 9  75 M1 for × = or × 37 5. [= 28.125%] 4 8  32  100 9 A1 for or 28.125[%] 32 9 M1 for 36 ÷ oe 32 100 or 36 × oe 28 . 125 Partial percentages 100 × 36 M1 for (Remaining) [= 96] 375. A1 for 96 75 M1 for 96 ÷ oe 100 SC1 for 288 IGCSE – October/November 2013 0580 43 62 × sin 122

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2 A fi eld, ABCD, is in the shape of a quadrilateral. For Examiner′s A footpath crosses the fi eld from A to C. Use C 26° B NOT TO SCALE 55 m 65° 32° 122° A D 62 m (a) Use the sine rule to calculate the distance AC and show that it rounds to 119.9 m, correct to 1 decimal place. Answer(a) [3] (b) Calculate the length of BC. Answer(b) BC = ........................................... m [4] (c) Calculate the area of triangle ACD. For Examiner′s Use Answer(c) .......................................... m2 [2] (d) The fi eld is for sale at $4.50 per square metre. Calculate the cost of the fi eld. Answer(d) $ ............................................... [3] _____________________________________________________________________________________

Mark scheme: 62 × sin 122 2 (a) 119.94[…] nfww 3 M2 for sin 26 AC 62 or M1 for = oe sin 122 sin 26 SC2 for correct answer from alternative methods (b) 109 or 108.7 to 108.8 nfww 4 M2 for 119.9..2 + 552 – 2 × 119.9.. × 55cos65 A1 for 11827[·…] or 11834 to 11835[·…] or M1 for implicit version (c) 1970 or 1969 to 1970.4 2 M1 for ½ × 119.9.. × 62 × sin 32 (d) 22300 or 22310 to 22320 3 M2 for (their (c) + 0.5 × 55 × 119.9.. × sin65) × 4.5 or M1 for their (c) + 0.5 × 55 × 119.9.. × sin65

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3 A rectangular metal sheet measures 9 cm by 7 cm. For Examiner′s A square, of side x cm, is cut from each corner. Use The metal is then folded to make an open box of height x cm. 9 cm NOT TO x cm SCALE 7 cm x cm x cm (a) Write down, in terms of x, the length and width of the box. Answer(a) Length = ............................................... Width = ............................................... [2] (b) Show that the volume, V, of the box is 4x3 – 32x2 + 63x. Answer(b) [2] (c) Complete this table of values for V = 4x3 – 32x2 + 63x. x 0 0.5 1 1.5 2 2.5 3 3.5 V 0 35 36 30 9 0 [2] (d) On the grid opposite, draw the graph of V = 4x3 – 32x2 + 63x for 0 Y x Y 3.5 . Three of the points have been plotted for you. For V Examiner′s Use 40 35 30 25 20 15 10 5 x 0 0.5 1 1.5 2 2.5 3 3.5 [3] (e) The volume of the box is at least 30 cm3. Write down, as an inequality, the possible values of x. Answer(e) ............................................... [2] (f) (i) Write down the maximum volume of the box. Answer(f)(i) ........................................ cm3 [1] (ii) Write down the value of x which gives the maximum volume. Answer(f)(ii) ............................................... [1] _____________________________________________________________________________________

Mark scheme: 3 (a) 9 – 2x, 7 – 2x oe 2 B1 for each, accept in any order (b) x(9 – 2x)(7 – 2x) M1FT 4x3 – 32x2 + 63x A1 Correct expansion and simplification with no errors (c) 24 20 2 B1 for each correct value (d) Correct curve 3 B2FT for 5 correct plots or B1FT for 3 or 4 correct plots (e) 0.65 to 0.75 ≤ x ≤ 2 oe 2 B1 for 0.65 to 0.75 seen (f) (i) 36 to 37 1 (ii) 1.2 to 1.4 1

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4 (a) One angle of an isosceles triangle is 48°. For Examiner′s Use Write down the possible pairs of values for the remaining two angles. Answer(a) .................. and .................. .................. and .................. [2] (b) Calculate the sum of the interior angles of a pentagon. Answer(b) ............................................... [2] (c) Calculate the sum of the angles a, b, c, d, e, f and g shown in this diagram. a g b NOT TO SCALE f c e d Answer(c) ............................................... [2] (d) The trapezium, ABCD, has four angles as shown. For Examiner′s All the angles are in degrees. Use B C 3y – 20 4x – 5 NOT TO SCALE 2x + 5 x + y – 10 A D (i) Show that 7x + 4y = 390 . Answer(d)(i) [1] (ii) Show that 2x + 3y = 195 . Answer(d)(ii) [1] (iii) Solve these simultaneous equations. Answer(d)(iii) x = ............................................... y = ............................................... [4] (iv) Use your answer to part (d)(iii) to fi nd the sizes of all four angles of the trapezium. Answer(d)(iv) ................. , ................. , ................. , ................. [1] _____________________________________________________________________________________

Mark scheme: 4 (a) 48 and 84 2 B1 for each pair 66 and 66 (b) 540 2 M1 for 3 × 180 or (2 × 5 – 4) × 90 or 5 × (180 – 360 ÷ 5) oe (c) 1620 2 M1 for 7 × 360 – their 540 – 360 (d) (i) 2x + 5 + 3y – 20 + 4x – 5 + x + y – 1 Allow partial simplification but not 7x + 4y – 30 = 360 10 = 360 oe (ii) 2x + 5 + 3y – 20 = 180 1 (iii) [x =] 30, [y =] 45 nfww 4 M1 for correct multiplication M1 for correct elimination A1 x = 30 or y = 45 If 0 scored SC1 for correct substitution to find the other variable (iv) 65, 115, 115, 65 1 Accept in any order IGCSE – October/November 2013 0580 43

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5 (a) 80 students were asked how much time they spent on the internet in one day. For Examiner′s This table shows the results. Use Time (t hours) 0 < t Y 1 1 < t Y 2 2 < t Y 3 3 < t Y 5 5 < t Y 7 7 < t Y 10 Number of students 15 11 10 19 13 12 (i) Calculate an estimate of the mean time spent on the internet by the 80 students. Answer(a)(i) ..................................... hours [4] (ii) On the grid, complete the histogram to show this information. 16 14 12 10 Frequency 8 density 6 4 2 t 0 1 2 3 4 5 6 7 8 9 10 Time (hours) [4] 3 For (b) The probability that Chaminda uses the internet on any day is 5 . Examiner′s Use 3 The probability that Niluka uses the internet on any day is 4 . (i) Complete the tree diagram. Chaminda Niluka 3 Uses the 4 internet Uses the 3 internet 5 Does not ........ use the internet ........ Uses the internet ........ Does not use the internet Does not ........ use the internet [2] (ii) Calculate the probability, that on any day, at least one of the two students uses the internet. Answer(b)(ii) ............................................... [3] (iii) Calculate the probability that Chaminda uses the internet on three consecutive days. Answer(b)(iii) ............................................... [2] _____________________________________________________________________________________

Mark scheme: 5 (a) (i) 3.81 or 3.812 to 3.813 or 4 M1 for midpoints soi (condone 1 error or omission 3h 49min nfww and M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) and M1 (dep on 2nd M1) for ∑fx ÷ 80 (305 ÷ 80) (ii) Correct histogram 4 B1 for each correct block and B1 for correct widths 2 1 3 1 2 1 (b) (i) , , , oe 2 B1 for or both s in correct place 5 4 4 4 5 4 18  9  2 1 (ii) nfww 3 M2 FT for 1 – their × their 20  10  5 4 3 3 3 1 2 3 or × + × their + their × oe 5 4 5 4 5 4 or 2 1 M1 FT for their × their 5 4 3 1 2 3 or × their + their × oe 5 4 5 4 27 3 3 3 (iii) [0.216] 2 M1 for × × 125 5 5 5 2 2 2

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6 Sandra has designed this open container. For Examiner′s The height of the container is 35 cm. Use NOT TO SCALE 35 cm The cross section of the container is designed from three semi-circles with diameters 17.5 cm, 6.5 cm and 24 cm. 17.5 cm 6.5 cm NOT TO SCALE (a) Calculate the area of the cross section of the container. Answer(a) ........................................ cm2 [3] (b) Calculate the external surface area of the container, including the base. Answer(b) ........................................ cm2 [4] (c) The container has a height of 35 cm. For Examiner′s Use Calculate the capacity of the container. Give your answer in litres. Answer(c) ...................................... litres [3] (d) Sandra’s container is completely fi lled with water. All the water is then poured into another container in the shape of a cone. The cone has radius 20 cm and height 40 cm. 20 cm NOT TO r SCALE 40 cm h (i) The diagram shows the water in the cone. h Show that r = . 2 Answer(d)(i) [1] (ii) Find the height, h, of the water in the cone. 1 [The volume, V, of a cone with radius r and height h is V = 3 πr 2h.] Answer(d)(ii) h = ......................................... cm [3] _____________________________________________________________________________________

Mark scheme: 6 (a) 329.7 to 330 3 M2 for ½π(122 + 8.752 – 3.252) oe or M1 for ½π122 or ½π8.752 or ½π3.252 SC2 for answer 1318 to 1320 (b) 2970 or 2967 to 2969.[…] 4 M3 for ½π(24 + 17.5 + 6.5) × 35 + their (a) or M2 for ½π(24 + 17.5 + 6.5) × 35 or M1 for ½π × 24 or ½π × 17.5 or ½π × 6.5 SC3 for 3955 to 3960 dep on SC2 in (a) (c) 11.5 or 11.6 or 11.53 to 11.55 3FT M1 for their (a) × 35 A1 for 11500 or 11530 to 11550 IGCSE – October/November 2013 0580 43 r 20 r h Accept 20 : 40 = r : h leading to 40r = 20h [r = h/2] = = or (d) (i) 1 20 1 r 1 h 40 20 40 = and = 40 2 h 2 their 11545 × 12 (ii) 35.3 or 35.31 to 35.34 3 M2 for 3 oe or 2 × their r π or 1  h  2 M1 for their 11545 = × π ×   × h oe 3  2  1 2 or their 11545 = × π × r × 2 r oe 3 3 14 − ( − 4 ) 1 5

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7 (a) The co-ordinates of P are (–4, –4) and the co-ordinates of Q are (8, 14). For Examiner′s Use (i) Find the gradient of the line PQ. Answer(a)(i) ............................................... [2] (ii) Find the equation of the line PQ. Answer(a)(ii) ............................................... [2] (iii) Write as a column vector. Answer(a)(iii) = [1] f p (iv) Find the magnitude of . Answer(a)(iv) ............................................... [2] (b) For Examiner′s T Use A NOT TO SCALE R 4a O 3b B In the diagram, = 4a and = 3b. 1 R lies on AB such that = 5 (12a + 6b). 3 T is the point such that = 2 . (i) Find the following in terms of a and b, giving each answer in its simplest form. (a) Answer(b)(i)(a) = ............................................... [1] (b) Answer(b)(i)(b) = ............................................... [2] (c) Answer(b)(i)(c) = ............................................... [1] (ii) Complete the following statement. The points O, R and T are in a straight line because ................................................................ ........................................................................................................................................... [1] (iii) Triangle OAR and triangle TBR are similar. area of triangle TBR Find the value of . area of triangle OAR Answer(b)(iii) ............................................... [2] _____________________________________________________________________________________

Mark scheme: 3 14 ( 4 ) 7 (a) (i) or 1.5 2 M1 for oe 2 8 − ( −4 ) 3 3 (ii) y = x + 2 oe 2 B1 for y = their 2 x + c o.e. 2 or y = mx + 2, m ≠ 0 3 SC1 for x + 2 2 (iii)  12  1   18   (iv) 21.6 or 21.63[…] 2 M1 FT for their 122 + their 182 oe (b) (i) (a) 3b – 4a 1 1 1 (b) (6b – 8a) oe simplified 2 M1 for (12a + 6b) – 4a or AR = AO + OR 5 5 (c) 6a + 3b oe simplified 1 (ii) OR is parallel to OT 1 Dep on OT correct 9 2  3  (iii) or 2.25 2 M1 for   4  2  IGCSE – October/November 2013 0580 43 2 (s – ut )

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1 For 8 (a) Rearrange s = ut + 2 at2 to make a the subject. Examiner′s Use Answer(a) a = ............................................... [3] (b) The formula v = u + at can be used to calculate the speed, v, of a car. u = 15, a = 2 and t = 8, each correct to the nearest integer. Calculate the upper bound of the speed v. Answer(b) ............................................... [3] (c) The diagram shows the speed-time graph for a car travelling between two sets of traffi c lights. For Examiner′s Use 16 Speed (m/s) 0 10 20 25 Time (seconds) (i) Calculate the deceleration of the car for the last 5 seconds of the journey. Answer(c)(i) ....................................... m/s2 [1] (ii) Calculate the average speed of the car between the two sets of traffi c lights. Answer(c)(ii) ........................................ m/s [4] _____________________________________________________________________________________

Mark scheme: 2 (s ut ) 8 (a) 2 oe nfww 3 M1 for a correct rearrangement to isolate the a term t and M1 for a correct multiplication by 2 and M1 for a correct division by t2 (b) 36.75 cao 3 M2 for 15.5 + 2.5 × 8.5 B1 for two of 15.5, 2.5, 8.5 seen 16 (c) (i) or better [3.2] 1 5 (ii) 11.2 4 M2 for ½(25 + 10)16 (= 280) or M1 for appreciation of distance from area and M1 for their 280 ÷ 25 (dep on M1)

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9 The fi rst four diagrams in a sequence are shown below. For Examiner′s Use Diagram 1 Diagram 2 Diagram 3 Diagram 4 The diagrams are made from dots ( ) and squares ( ) joined by lines. (a) Complete the table. Diagram 1 2 3 4 5 n Number of dots 6 9 12 1 Number of squares 0 1 3 n(n – 1) 2 Number of triangles 4 9 16 3 Number of lines 9 18 30 45 63 (n + 1)(n + 2) 2 [9] (b) Which diagram has 360 lines? Answer(b) ............................................... [2] (c) The total number of lines in the fi rst n diagrams is For Examiner′s Use 1 2 n3 + pn2 + qn. 1 (i) When n = 1, show that p + q = 8 . 2 Answer(c)(i) [1] (ii) By choosing another value of n and using the equation in part (c)(i), fi nd the values of p and q. Answer(c)(ii) p = ............................................... q = ............................................... [5] _____________________________________________________________________________________ Question 10 is printed on the next page.

Mark scheme: 9 (a) 15 18 3n + 3 or 3(n + 1) 9 B2 for 15, 6, 25 6 10 or B1 for two correct values 25 36 (n + 1)2 B3 for 18, 10, 36 or B1 for each correct value B2 for 3n + 3 oe or M1 for 3n + k, for any k B2 for (n + 1)2 oe or M1 for a quadratic expression (b) 14 2 M1 for (n + 1)(n + 2) = 240 or better or 15 × 16 = 240 (c) (i) ½ + p + q = 9 1 (ii) [p = ] 3 5 B2 for 4p + 2q = 23 11 or B1 for ½ × 23 + p × 22 + q × 2 oe [q = ] 2 M1 for correct multiplication and subtraction of their equations 11 A1 for [p = ] 3 or [q = ] 2 If 0 scored then SC1 for either correct IGCSE – October/November 2013 0580 43 x

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10 (a) Simplify. For Examiner′s x2 - 3 x Use x2 - 9 Answer(a) ............................................... [3] (b) Solve. 15 20 – = 2 x x + 1 Answer(b) x = ............... or x = ............... [7]

Mark scheme: x 10 (a) cao 3 B1 for (x + 3)(x – 3) x + 3 B1 for x(x – 3) 3 (b) and –5 7 M2 for 15(x + 1) – 20x = 2x(x + 1) 2 or M1 for multiplication by one denominator only 15( x + )1 − 20 x or x ( x + )1 and B2 for 2x2 + 7x – 15 [= 0] or B1 for 15x + 15 – 20x or 2x2 + 2x and M2 for (2x – 3)(x + 5) or their correct factors or formula or M1 for (2x + a)(x + b) where ab = –15 or a + 2b = 7 3 A1 for x = and –5 2

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