Cambridge IGCSE Mathematics (9-1) 0980 — 2023 May/June Paper 4 · Variant 2

1 (a) 42° NOT TO SCALE x° The diagram shows an isosceles triangle with the base extended. Find the value of x. x = ................................................ [3] (b) The diagram shows three lines meeting at a point. The ratio a : b : c = 3 : 4 : 5. Find the value of c. a° NOT TO c° b° SCALE c = ................................................ [3] (c) A regular pentagon has an exterior angle, d. A regular hexagon has an interior angle, h. d Find the fraction . h Give your answer in its simplest form. ................................................. [4] (d) S R x° ( x + 20)° NOT TO SCALE ( 3x – 40)° Q ( 2x – 5)° P Show that PQRS is a cyclic quadrilateral. [5] (e) B A 50° 9 cm NOT TO O SCALE The diagram shows a circle of radius 9 cm, centre O. The minor sector AOB, with sector angle 50°, is removed from the circle. Calculate the length of the major arc AB. ............................................ cm [3]

Mark scheme: Question Answer Marks Partial Marks 1(a) 111 3  42 M2 for 180 –180 oe or 42 + 2 180  42 oe 2 180  42 or M1 for oe 2 1(b) 150 3 M1 for k ÷ (3 + 4 + 5) [×p] where p = 1, 3, 4 or 5 5 or oe 12 B1 for 360 used 1(c) 3 4 72 cao nfww B3 for 5 120 or B2 for [d = ] 72 or [h = ] 120 or M1 for 360 ÷ 5 oe isw or 180 – (360 ÷ 6) isw or for (6 – 2) × 180 [÷ 6] 1(d) x + 2x – 5 + x + 20 + 3x – 40 = 360 M1 Accept equivalent equation e.g. 7x – 25 = 360 7x = 360 + 5 – 20 + 40 or better M1 FT their equation, accept e.g. 7x = 385 x = 55 B1 55 and 125 B1dep Dep on M1M1B1 or 105 and 75 Accept 55 + 3 × 55 – 40 = 180 or 2 × 55 – 5 + 55 + 20 = 180 If B0 scored, SC1 for 55, 75, 105 and 125 Opposite angles sum to 180 oe A1 Dep on M1M1B1B1 [so PQRS is a cyclic quadrilateral ] 1(e) 48.7 or 48.69 to 48.70… 3 360  50 M2 for 2 π oe9 360 50 or M1 for  2  π oe9 360

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2 (a) Anil changes $830 into euros when the exchange rate is 1 euro = $1.16 . He spends 500 euros. He then changes the remaining money back into dollars at the same exchange rate. Work out how much, in dollars, Anil receives. $ ................................................ [3] (b) In 2021, Anil earns $37 000. (i) He spends $12 400 on bills in 2021. Calculate the percentage of his earnings he spends on bills. ............................................. % [2] (ii) His earnings of $37 000 increase by 3.2% in 2022. Calculate his earnings in 2022. $ ................................................ [2] (c) Anil invests $3500 in an account that pays a rate of 2.4% per year compound interest. (i) Calculate the total interest earned at the end of 5 years. $ ................................................ [3] (ii) Find the number of complete years before Anil has at least $5000 in this account. ........................................ years [3]

Mark scheme: 2(a) 249.98 to 250[.0…] 3 M2 for 830 – 500 × 1.16 or M1 for 500 × 1.16 OR M1 for 830 ÷ 1.16 M1 for (their 715.5… – 500 ) × 1.16 2(b)(i) 33.5 or 33.51… 2 12400 M1 for [ 100] oe 37000 If 0 scored, SC1 for answer 66.5 or 66.48 to 66.49 2(b)(ii) 38 184 cao 2  3.2  M1 for 37 000   1   oe  100  or B1 for 1184 2(c)(i) 441 or 440.6 3 B2 for answer 3941 or 3940.6 or 3940.64 or 440.64 to 440.65 to 3940.65  2.4  5 or M2 for 3500 ×  1   – 3500  100   2.4  5 or M1 for 3500 ×  1   oe isw  100  2(c)(ii) 16 3 B2 for 15[.0] nfww to 15.1  2.4 15 or M2 for 3500 ×  1   oe seen  100   2.4 16 or 3500 ×  1   oe seen  100  or M1 for  2.4  n (3500 or their 3941) ×  1    100  associated with 5000 oe

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3 C NOT TO SCALE ( x + 3) cm A ( 2x + 5) cm B The diagram shows a right-angled triangle ABC. (a) (i) The area of the triangle is 60 cm 2. Show that 2x 2 + 11 x - 105 = 0 . [3] (ii) Solve by factorisation. 2x 2 + 11x - 105 = 0 x = ....................... or x = ....................... [3] (iii) Calculate angle ACB. ................................................. [3] (b) Triangle ABC is similar to triangle DEF. Triangle DEF has an area of 93.75 cm 2. (i) Find the size of the smallest angle of triangle DEF. ................................................. [1] (ii) Find the length of the shortest side of triangle DEF. ............................................ cm [3]

Mark scheme: 3(a)(i)  x  3  (2 x  5) M1 Accept (x + 3)(2x + 5) = 2 × 60 or 120  60 Accept e.g. (x + 3) (x + 2.5) = 60 without 2 division by 2 shown for M1 (but not A1) 2x2 + 6x + 5x + 15 seen B1 Accept 2x2 + 11x + 15 seen 2x2 + 11x – 105 = 0 A1 Correct completion after M1B1 with the fraction seen removed with no errors or omissions seen 3(a)(ii) (2x + 21) (x – 5) [= 0] M2 M1 for partial factors 2x (x – 5) + 21(x – 5) [ = 0] or x (2x + 21) – 5 (2x + 21) [ = 0] OR (2x + a)(x + b) [ = 0] where ab = – 105 or 2b + a = 11 –10.5 and 5 B1 3(a)(iii) 61.9 or 61.92 to 61.93 3 2  their 5  5 M2 for tan = oe their 5  3 or B1FT for 2 × their 5 + 5 and their 5 + 3 3(b)(i) 28.1 or 28.07 to 28.08 1 FT their 90 – their (a)(iii) unless their (a)(iii) < 45, in which case FT their (a)(iii) 3(b)(ii) 10 3 93.75 M2 for (their 5  3)  oe 60 93.75 60 or M1 for or oe seen 60 93.75  their 5  3 2 60 oe or    x  93.75

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4 The table shows information about the heights of 80 children. Height 1.2 1 h G 1.4 1.4 1 h G 1.5 1.5 1 h G 1.65 1.65 1 h G 1. 8 1.8 1 h G 1.9 (h metres) Frequency 2 13 24 32 9 (a) (i) Write down the interval containing the median. .................... 1 h G .................... [1] (ii) Calculate an estimate of the mean height. ............................................. m [4] (b) (i) One of these children is chosen at random. Calculate the probability that they have a height of 1.4 m or less. ................................................. [1] (ii) Two of these children are chosen at random. Calculate the probability that both children are taller than 1.5 m but only one of them is taller than 1.8 m. ................................................. [3] (c) (i) Complete the cumulative frequency table for the heights. Height h G 1.4 h G 1.5 h G 1.65 h G 1.8 h G 1.9 (h metres) Cumulative 2 frequency [2] (ii) On the grid, draw the cumulative frequency diagram. 80 70 60 50 Cumulative 40frequency 30 20 10 0 h 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 Height (m) [3] (d) Use your diagram to find an estimate of (i) the interquartile range .............................................. m [2] (ii) the 60th percentile. ............................................ m [2]

Mark scheme: 4(a)(i) 1.65 < h ≤ 1.8 1 4(a)(ii) 1.63875 4 M1 for midpoints soi M1 for use of ∑fh with h in correct interval including both boundaries M1dep on 2nd M1 for ∑fh ÷ 80 4(b)(i) 1 1 oe 40 4(b)(ii) 63 3 56 9 oe M2 for  [ 2] oe 395 80 79 56 9 9 56 or B1 for or or or oe seen 80 79 80 79 If 0 or B1 scored, instead award SC2 for 117 answer oe 632 63 or SC1 for answer oe 400 4(c)(i) 15, 39, 71, 80 2 B1 for 3 correct or M1 for 1 error in addition with other values then consistent 4(c)(ii) Correct curve 3 B1 for correct horizontal placement for 5 plots B1FT for correct vertical placement for 5 plots B1FT dep on at least B1 for reasonable increasing curve or polygon through their 5 points If 0 scored SC1 FT for 4 out of 5 points correctly plotted 4(d)(i) Strict FT their UQ – their LQ 2dep B1dep for their UQ or their LQ seen Dep on increasing curve/polygon for 2 marks or B1 4(d)(ii) Strict FT their reading at 48 2dep B1 for 48 written

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5 (a) NOT TO 15 cm SCALE 8 cm A cone has base diameter 8 cm and perpendicular height 15 cm. (i) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 ......................................... cm3 [2] (ii) A label completely covers the curved surface area of the cone. Calculate the area of the label as a percentage of the total surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] ............................................. % [5] (b) NOT TO SCALE 0.45 m An empty cylindrical container has radius 0.45 m. 300 litres of water is poured into the container at a rate of 375 ml per second. (i) Find the time taken, in minutes and seconds, for all the water to be poured into the container. ............................. min ............................. s [3] (ii) Calculate the height of the water in the container. ............................................. m [3]

Mark scheme: 5(a)(i) 251 or 251.3 to 251.4 2 1 2 M1 for  π  4  15 oe 3 5(a)(ii) 79.5 or 79.51… 5 2 2 M3 for π 4 4  15 oe or M2 for 15 2  4 2 oe or M1 for [l2 = ] 42 + 152 oe or π×4× theirl M1 for their curved surfacearea [ 100] their curved surfacearea  π  4 2 oe 5(b)(i) 13 min 20 sec 3 40 B2 for 800 or oe seen 3 or M1 for figs 3 ÷ figs 375 or figs 3 ÷ 22 500 5(b)(ii) 0.472 or 0.4715 to 0.4716… 3 M2 for π  0.452  h  0.3 or π  45 2  h  300000 oe or M1 for π  figs45 2  h  figs3 oe

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6 (a) A sequence has nth term . 2n + 3 (i) Find the first three terms of this sequence. Give your answers as fractions. ......................... , ......................... , ......................... [2] 12 (ii) The kth term of this sequence is . 25 Find the value of k. k = ................................................ [2] (b) Find the nth term of each sequence. (i) 6, 13, 32, 69, 130, … ................................................. [2] (ii) 100, 50, 25, 12.5, 6.25, … ................................................. [2]

Mark scheme: 6(a)(i) 1 2 3 2 B1 for 2 correct terms isw , , final answer or for 0.2 and (0.286 or 0.2857…) and 5 7 9 0.333… 6(a)(ii) 36 2 12(2 k  3) M1 for k = or better 25 6(b)(i) n3 + 5 oe final answer 2 B1 for any cubic or common third differences of 6 (at least 2) or for correct answer seen and spoilt 6(b)(ii) 100 × 21–n oe final answer 2 n   k  [+k]  1  B1 for 2–n oe or   oe in answer  2  or for correct answer seen and spoilt

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7 North C NOT TO SCALE 60 km 87 km 38° B A The diagram shows the straight roads between town A, town B and town C. AC = 60 km , CB = 87 km and B is due east of A. The bearing of C from A is 038°. (a) Show that angle ACB = 95.1° , correct to 1 decimal place. [5] (b) Without stopping, a car travels from town A to town C then to town B, before returning directly to town A. The total time taken for the journey is 3 hours 20 minutes. Calculate the average speed of the car for this journey. Give your answer in kilometres per hour. ........................................ km/h [6]

Mark scheme: 7(a) Angle CAB = 52 B1 1 60sin their 52  M3 60sin their 52 180 – 52 – sin M2 for [sin[...]  ] oe    87  87 60 87 or M1 for  oe sin B sin their 52 95.08… A1 7(b) 77.1 or 77.08 to 77.11 6 B4 for dist travelled = 256.9 to 257[.0…] or B3 for [AB =] 109.9 to 110[.0…] or M3 for 60 + 87 + 60 2  87 2 – 2  60  87  cos 95.1 oe or M2 for 60 2  87 2 – 2  60  87  cos 95.1 oe or AB2 = 12093. … to 12097. … 87sin95.1 or oe sin their 52 or M1 for AB2 = 602 + 872 – 2 × 60 × 87 × cos 95.1 oe sin95.1 sin their 52 or  oe AB 87 20 M1 for their total distance ÷ 3 oe 60

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8 (a) (i) Show that the equation y = ( x - 4)( x + 1)( x - 2) can be written as y = x 3 - 5x 2 + 2x + 8 . [2] (ii) On the diagram, sketch the graph of y = x 3 - 5x 2 + 2x + 8 , indicating the values where the graph crosses the axes. y O x [4] (b) The graph of y = x 3 - 5x 2 + 2x + 8 has two tangents with a gradient of 10. Find the equations of these two tangents. You must show all your working and give your answers in the form y = mx + c . y = ................................................ y = ................................................ [7]

Mark scheme: 8(a)(i) Correct expansion of a pair of brackets M1 accept x2 – 4x + [1]x – 4 x2 – 3x – 4 or x2 – 4x – 2x + 8 or x2 – 6x + 8 or x2 + [1]x – 2x – 2 or x2 – [1]x – 2 x3 – 4x2 + x2 – 4x – 2x2 + 8x – 2x + 8 A1 Accept leading to and stating x3 – 3x2 – 4x – 2x2 + 6x + 8 [y = ] x3 – 5x2 + 2x + 8 or x3 – 6x2 +[1] x2 + 8x – 6x + 8 or x3 –[1] x2 – 2x – 4x2 + 4x + 8 leading to and stating [y = ] x3 – 5x2 + 2x + 8 8(a)(ii) Correct labelled sketch 4 positive cubic Crossing x-axis at –1, 2 and 4 only Crossing y – axis at 8 only B1 for positive cubic B2 for three intercepts only with x -axis labelled at – 1, 2 and 4 or B1 for 1 or 2 correctly labelled x – intercepts B1 for a single intercept on y-axis labelled at 8 but not if line y = 8 8(b) 3x2 – 10x – 8 [= 0] M3 B2 for derivative = 3x2 – 10x + 2 isw OR B1 for derivative with 3x2 or –10x given in expression isw M1dep on B1 for their first derivative = 10 2 B1 x = 4 and x =  3  2 112  B1 (4, 0) and   ,  oe  3 27  [y =] 10x – 40 B2 B1 for each and or for two different equations of the form 292 [y = ] 10x + c (c must be numeric) [y =] 10 x  27 292 or for c = –40 and 27

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9 (a) Simplify. (i) ( 3x 2 y 4 ) 3 ................................................. [2] 3 16 - 2 (ii) 16 8 e x y o ................................................. [3] (b) (i) Factorise. x 2 - 9 ................................................. [1] (ii) Simplify. x 2 - 9 2 xy - 6 y + 5 x - 15 ................................................. [3] (c) Solve the simultaneous equations. You must show all your working and give your answers correct to 2 decimal places. 2x + y = 7 y = 5x 2 + 2x - 13 x = .......................... , y = .......................... x = .......................... , y = .......................... [6]

Mark scheme: 9(a)(i) 27x6y12 final answer 2 B1 for two terms correct in answer e.g. 27x6yk or 27xky12 or kx6y12 or for correct answer seen then spoilt 9(a)(ii) x 24 y12 3 B2 for final answer with two correct final answer elements 64 64 641 or final answer or or x 24 y12 x 24 y  12 better or for correct answer seen or B1 for 64 or x24 or y12 seen in final answer k or final answer x 24 y 12 or M1 for first correct step seen 3  3  x16 y 8   2  4  eg   or  8 4  or  16   x y   1  2   4096     48 24   x y  9(b)(i) (x + 3)(x – 3) final answer 1 9(b)(ii) x  3 3 M2 for (x – 3)(2y + 5) final answer or M1 for 2y(x – 3) + 5(x – 3) 2 y  5 or x (2y + 5) – 3( 2y + 5) 9(c) 5x2 + 4x – 20 [= 0] oe M2 M1 for 7 – 2x = 5x2 + 2x – 13 oe seen or 2  7  y   7  y  5y2 – 78y + 221 [= 0] oe or y  5    2    13 oe seen  2   2  2 M2 FT their 3-term quadratic 4 4  4(5)( 20) oe 2(5) 2 or M1 for (4)  4(5)( 20) or better or 2 4 q 4 q 4  4  or for or   4   oe 2  5 2  5 10  10  2  4  or for x  oe    10  x = 1.64 y = 3.72 B2 B1 for one correct pair or both x-values and correct or both y – values correct x = – 2.44 y = 11.88

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10 (a) H G F E NOT TO SCALE D C x cm A x cm B ABCDEFGH is a cuboid with a square base of side x cm. CG = 20 cm and AG = 28 cm . Calculate the value of x. x = ................................................ [4] (b) R Q N P NOT TO SCALE M L J K The diagram shows a different cuboid JKLMNPQR. MR = 30 cm correct to the nearest centimetre. KR = 37 cm correct to the nearest centimetre. Calculate the lower bound of the angle between KR and the base JKLM of the cuboid. ................................................. [4]

Mark scheme: 10(a) 13.9 or 13.85 to 13.86 4 M3 for 2x2 = 282 – 202 or better or x  28 2  20 2 sin45 oe   or M2 for x2 + x2 + 202 = 282 oe x or sin45  28 2  20 2 ) or M1 for any correct Pythag in 2D or their AC × sin 45 oe dep on trig/Pythagoras attempt for AC 10(b) 51.9 or 51.87 to 51.88 4 29 to 30 30  0.5 M3 for sin = or oe 37  0.5 37 to 38 or M2 for correct trig statement for correct angle with values in range 29 to 31 and 36 to 38 or M1 for 30 + 0.5 or 30 – 0.5 or 37 + 0.5 or 37 – 0.5 seen or for identifying correct angle RKM

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