Cambridge IGCSE Mathematics (US) 0444 — 2020 Oct/Nov Paper 4 · Variant 3

Cambridge IGCSE™ This document has 20 pages. Blank pages are indicated. DC (PQ/CGW) 196713/2 © UCLES 2020 [Turn over * 4 3 3 6 1 2 5 5 1 8 * MATHEMATICS (US) 0444/43 Paper 4 (Extended) October/November 2020 2 hours 30 minutes You must answer on the question paper. You will need: Geometrical instruments INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, center number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You should use a calculator where appropriate. ● You may use tracing paper. ● You must show all necessary work clearly. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ● For r, use either your calculator value or 3.142. INFORMATION ● The total mark for this paper is 130. ● The number of marks for each question or part question is shown in parentheses [ ].

2 0444/43/O/N/20 © UCLES 2020 Formula List For the equation ax bx c 0 2 + + = x = b b a ac 2 4 2 ! - - Lateral surface area, A, of cylinder of radius r, height h. A rh 2r = Lateral surface area, A, of cone of radius r, sloping edge l. A rl r = Surface area, A, of sphere of radius r. A r 4 2 r = Volume, V, of pyramid, base area A, height h. V Ah 3 1 = Volume, V, of cone of radius r, height h. V r h 3 1 2 r = Volume, V, of sphere of radius r. V r 3 4 3 r = sin sin sin A a B b C c = = A c b a C B cos a b c bc A 2 2 2 2 = + - Area sin bc A 2 1 =

3 0444/43/O/N/20 © UCLES 2020 [Turn over 1 A B – 6 – 5 – 4 – 3 – 2 – 1 1 0 2 3 4 5 6 7 8 – 1 – 2 – 3 – 4 – 5 – 6 7 8 6 5 4 3 2 1 y x (a) On the grid, draw the image of (i) triangle A after a rotation of 90° counter-clockwise about (0, 0), [2] (ii) triangle A after a translation by the vector 3 5 - e o. [2] (b) Describe fully the single transformation that maps triangle A onto triangle B. … … [3]

4 0444/43/O/N/20 © UCLES 2020 2 (a) The Earth has a surface area of approximately 510 100 000 km2. (i) Write this surface area in scientific notation. … km2 [1] (ii) Water covers 70.8% of the Earth’s surface. Work out the area of the Earth’s surface covered by water. … km2 [2] (b) The table shows the surface area of some countries and their estimated population in 2017. Country Surface area (km2) Estimated population in 2017 Brunei 5.77 × 103 433 100 China 9.60 × 106 1 388 000 000 France 6.41 × 105 67 000 000 Maldives 3.00 × 102 374 600 (i) Find the total surface area of Brunei and the Maldives. … km2 [1] (ii) The ratio surface area of the Maldives : surface area of China can be written in the form 1 : n. Find the value of n. n = … [2] (iii) Find the surface area of France as a percentage of the surface area of China. … % [2]

5 0444/43/O/N/20 © UCLES 2020 [Turn over (iv) Find the population density of the Maldives. [Population density = population ÷ surface area] …people/km2 [2] (c) The population of the Earth in 2017 was estimated to be 7.53 × 109. The population of the Earth in 2000 was estimated to be 6.02 × 109. (i) Work out the percentage increase in the Earth’s estimated population from 2000 to 2017. … % [2] (ii) Assume that the population of the Earth increased exponentially by y% each year for these 17 years. Find the value of y. y = … [3]

6 0444/43/O/N/20 © UCLES 2020 3 P O S S I B I L I T Y Morgan picks two of these letters, at random, without replacement. (a) Find the probability that he picks (i) the letter Y first, … [1] (ii) the letter B then the letter Y, … [2] (iii) two letters that are the same. … [3] (b) Morgan now picks a third letter at random. Find the probability that (i) all three letters are the same, … [2]

7 0444/43/O/N/20 © UCLES 2020 [Turn over (ii) exactly two of the three letters are the same, … [5] (iii) all three letters are different. … [2]

8 0444/43/O/N/20 © UCLES 2020 4 (a) 162° O B A E F C D 42° x w y z NOT TO SCALE A, B, C and D are points on the circle, center O. EF is a tangent to the circle at D. Angle ADE = 42° and angle COD = 162°. Find the values of w, x, y, and z. w = … x = … y = … z = … [7]

9 0444/43/O/N/20 © UCLES 2020 [Turn over (b) NOT TO SCALE P T U Q R PQR is a triangle. T is a point on PR and U is a point on PQ. RQ is parallel to TU. (i) Explain why triangle PQR is similar to triangle PUT. Give a reason for each statement you make. … … … … [3] (ii) PT : TR = 4 : 3 (a) Find the ratio PU : PQ. … : … [1] (b) The area of triangle PUT is 20 cm2. Find the area of the quadrilateral QRTU. … cm2 [3]

10 0444/43/O/N/20 © UCLES 2020 5 (a) The average speeds, in km/h, of cars traveling along a road are recorded. The box plot shows this information. 40 45 50 55 60 65 Speed (km/h) 70 75 80 85 Find (i) the lowest speed recorded, … km/h [1] (ii) the median, … km/h [1] (iii) the interquartile range. … km/h [1] (b) Another car takes 18 seconds to travel 400 m along this road. Calculate the average speed of this car in km/h. … km/h [3]

11 0444/43/O/N/20 © UCLES 2020 [Turn over 6 (a) Find the integer values that satisfy the inequality x 2 2 10 1 G . … [2] (b) Factor. (i) y xy 6 15 2 - … [2] (ii) y x 9 2 2 - … [2] (c) Simplify. x x 1 3 2 1 2 - - + … [3]

12 0444/43/O/N/20 © UCLES 2020 7 G 11 cm NOT TO SCALE 5 cm 8 cm C B A E H D F ABCDEFGH is a closed hollow cuboid. AB = 8 cm, BC = 5 cm and CG = 11 cm. (a) (i) Work out the total surface area of the cuboid. … cm2 [2] (ii) The cuboid is made from thin metal and 1 cm2 of this metal has a mass of 0.73 grams. Work out the mass of the cuboid. … g [1] (b) Ivana has a rod of length 13 cm. (i) The total mass of this rod and the cuboid is 0.3 kg. Find the mass of the rod, giving your answer in grams. … g [2]

13 0444/43/O/N/20 © UCLES 2020 [Turn over (ii) Show that the rod fits completely inside the cuboid. [4] (c) Calculate angle CAB. Angle CAB = … [2]

14 0444/43/O/N/20 © UCLES 2020 8 (a) (i) Factor x x 24 5 2 + - . … [2] (ii) The diagram shows a sketch of y x x 24 5 2 = + - . c y a b x O NOT TO SCALE Work out the values of a, b, and c. a = … b = … c = … [3] (iii) The line y 18 = intersects the graph of y x x 24 5 2 = + - at P and Q. Find the length of PQ. PQ = … [4]

15 0444/43/O/N/20 © UCLES 2020 [Turn over (b) O x NOT TO SCALE f( )x The diagram shows a sketch of the graph of a cubic function ( )x f . The graph passes through the points (-5, 0), (-2, 0), (0, 0), and (1, 36). Find ( )x f in the form ax bx cx 3 2 + + . … [6]

16 0444/43/O/N/20 © UCLES 2020 9 (a) AB 6 1 = - e o BC 2 5 = - e o DC 2 3 = - e o Find (i) AC, AC = f p [2] (ii) BD, BD = f p [2] (iii) BC . … [2]

17 0444/43/O/N/20 © UCLES 2020 [Turn over (b) O E A C D B a b NOT TO SCALE In the diagram, OAB and OED are straight lines. O is the origin, A is the midpoint of OB and E is the midpoint of AC. a AC = and b CB = . Find, in terms of a and b, in its simplest form (i) AB, AB = … [1] (ii) OE, OE = … [2] (iii) the position vector of D. … [3]

18 0444/43/O/N/20 © UCLES 2020 10 ( ) f x x 4 3 = - ( ) g x x x2 = + ( ) h x 3x = (a) Find f(h(2)). … [2] (b) Find ( ) f x 1 - . ( ) f x 1 = - … [2] (c) Simplify. (i) ( ) f x 1 2 - … [2] (ii) ( ( )) ( ) g f g x x 9 - … [4]

19 0444/43/O/N/20 © UCLES 2020 (d) ( ) h x 1 9kx = Find the value of k. k = … [2] (e) ( ) ( )( ) x x x 1 2 j = + + The graph of ( )x g is mapped onto the graph of ( )x j by a translation. Find the column vector that represents this translation. f p [2] Question 11 is printed on the next page. [Turn over

20 0444/43/O/N/20 © UCLES 2020 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. 11 The table shows the first four terms in sequences A, B, C and D. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 4 9 14 19 B 3 10 29 66 C 1 4 16 64 D 17 3 26 4 37 5 50 6 Complete the table. [12]

This document consists of 9 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ MATHEMATICS (US) 0444/43 Paper 4 (Extended) October/November 2020 MARK SCHEME Maximum Mark: 130 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2020 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 2 of 9 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 3 of 9 Maths-Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. Abbreviations cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent SC Special Case nfww not from wrong working soi seen or implied

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 4 of 9 Question Answer Marks Partial Marks 1(a)(i) Triangle at (–3, 2) (–3, 3) (–5, 2) 2 B1 for correct rotation about incorrect point or for rotation 90 clockwise around (0, 0) 1(a)(ii) Triangle at (5, –2) (6, –2) (5, 0) 2 B1 for translation by 3 k    or 5 k     −   1(b) Enlargement [SF] 3 [Centre] (1, 4) 3 B1 for each 2(a)(i) 5.101[00…] × 108 final answer 1 2(a)(ii) 361 150 800 oe 2 FT their (a)(i) M1 for 70.8 100 × 510 100 000 or for 70.8 100 × their a(i) 2(b)(i) 6070 oe 1 2(b)(ii) 32 000 oe 2 B1 for figs 32 2(b)(iii) 6.68 or 6.677 … 2 M1 for 5 6 6.41 10 9.6[0] 10 × × [× 100] oe 2(b)(iv) 1250 or 1248 to 1249 oe 2 B1 for figs 125 or figs1248 to figs 1249 2(c)(i) 25.1 or 25.08… 2 M1 for 9 9 9 7.53 10 6.02 10 6.02 10     × − ×       ×   oe or 9 9 7.53 10 100 6.02 10   ×  ×   ×   2(c)(ii) 1.33 or 1.325… 3 M2 for 9 17 9 7.53[ 10 ] 6.02[ 10 ] × × or 17 ( )( ) 1 100 their c i + or M1 for 6.02[×109] × p17 = 7.53[×109] or p17 = ( )( ) 1 100 their c i + 3(a)(i) 1 11 oe 1

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 5 of 9 Question Answer Marks Partial Marks 3(a)(ii) 1 110 oe 2 M1 for 1 1 11 10 × oe 3(a)(iii) 4 55 oe 3 M2 for 2 1 11 10   ×     + 3 2 11 10   ×     oe or M1 for 2 1 11 10   ×     or 3 2 11 10   ×     seen oe 3(b)(i) 1 165 oe 2 M1 for 3 2 1 11 10 9 × × oe 3(b)(ii) 1 5 oe 5 M4 for 2 1 9 3 2 8 3 3 11 10 9 11 10 9      × × + × ×           oe or M3 for 3 2 8 3 11 10 9   × ×     or M2 for 2 1 9 3 11 10 9    × ×       or 3 2 8 11 10 9 × × oe or M1 for 2 1 11 10 9 k  × ×   where k is 3, 6 or 9 3(b)(iii) 131 165 oe 2 M1 for 1 – (their (b)(i) + their(b)(ii)) oe 4(a) 81 1 81 2 FT their w B1 for angle ODC = 9 123 2 FT their [y =] 42 + their x B1 for angle ADC = 48 and 9 oe 33 2 B1 for angle ACD = 42 or M1 for 180 – 81 – 48 – 9 – 9 oe 4(b)(i) Angle PTU = angle PRQ corresponding Angle PUT = angle PQR corresponding Angle RPQ is common oe M2 Accept in any order M1 for one correct pair with reason If 0 scored, SC1 for two correct pairs of equal angles identified with incorrect/no reasons Corresponding angles are equal oe A1

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 6 of 9 Question Answer Marks Partial Marks 4(b)(ii)(a) 4 : 7 oe 1 4(b)(ii)(b) 41.25 oe 3 M2 for 20 × 2 7 4       oe or 20 × 2 2 2 7 4 4 − oe or M1 for 2 7 4       or 2 4 7       or 2 2 2 7 4 4 − or 2 2 2 4 7 4 − 5(a)(i) 43 1 5(a)(ii) 65 1 5(a)(iii) 13 1 5(b) 80 3 M2 for 400 60 60 18 1000 × × oe Or M1 for 400 18 or for their speed in m/s 60 60 1000 × × or for 400 1000 and 18 60 60 × soi 6(a) 2, 3, 4, 5 2 B1 for 3 correct and no extra or 4 correct and one extra or M1 for 1 < x ⩽ 5 6(b)(i) 3y (2y – 5x) 2 B1 for 3(2y2 – 5xy) or y (6y – 15x) or for the correct answer seen and then spoiled 6(b)(ii) (y – 3x) (y + 3x) 2 B1 for (y + 3) (y – 3) 6(c) 4 5 ( 1)(2 1) x x x + − + or 2 4 5 2 1 x x x + − − final answer 3 M1 for 3(2x + 1) – 2(x – 1) oe isw M1 for (x – 1)(2x + 1) oe isw 7(a)(i) 366 2 M1 for (8 × 5 + 8 × 11 + 5 × 11) [ × 2] 7(a)(ii) 267.18 1 FT their (a)(i) 7(b)(i) 32.82 2 FT 300 – their (a)(ii) dep on (a)(ii) < 300 M1 for 300 – their (a)(ii) dep on (a)(ii) < 300

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 7 of 9 Question Answer Marks Partial Marks 7(b)(ii) 2 2 2 8 5 11 + + oe or 82 + 52 + 112 and 132 ALTERNATIVE 2 2 8 11 + or 82 + 112 and 132 M3 M2 for 82 + 52 + 112 or 82 + 112 oe or M1 for 82 + 52 or 52 + 112 oe Yes and 14.5 or 14.4 or 14.49… or Yes and 13.6[0…] A1 Accept equivalent conclusion 7(c) 32.0[…] 2 M1 for tan[..] = 5 8 oe 8(a)(i) (8 – x)(3 + x) 2 M1 for 8(3 + x) – x(3 + x) or 3(8 – x) + x(8 – x) or (a – x) (b + x) where ab = 24 or a – b = 5 8(a)(ii) [a = ] –3 [b = ] 8 [c = ] 24 3 FT their (a)(i) for a and b B1FT for each of a and b or both correct but reversed B1 for [c =] 24 8(a)(iii) 7 4 B3 for –1 and 6 or B2 for (x – 6)(x + 1) or fully correct formula or M1 for 24 + 5x – x2 = 18 8(b) 2x3 + 14x2 + 20x final answer 6 B5 for 2x(x2 + 7x + 10) or B4 for 2x(x + 5)(x + 2) or 3 2 [ ]( 7 10 ) k x x x + + or B3 for 2 [ ]( 7 10) kx x x + + or M2 for [ k ] x(x + 5)(x + 2) and M1 for k × 1 × 6 × 3 = 36 9(a)(i) 4 4    2 B1 for 4 k       or 4 k       9(a)(ii) 4 8 −       2 B1 for 4 k −       or 8 k       9(a)(iii) 5.39 or 5.385.. 2 M1 for (–2)2 + 52 oe 9(b)(i) a + b 1

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 8 of 9 Question Answer Marks Partial Marks 9(b)(ii) 3 2 a + b 2 M1 for a correct route, e.g.OA AE +   9(b)(iii) 2a + 4 3 b 3 M2 for unsimplified OD  or for 4 3 b or M1 for OD  attempted in terms of a and b or for CD  = 1 3 b or DB  = 2 3 b seen 10(a) –23 2 M1 for 4 – 3(3x) oe soi 10(b) 4 3 x − oe final answer 2 M1 for x = 4 – 3y or y + 3x = 4 or x + 3y = 4 or 4 3 3 y x = + − − oe or 4 3 3 x y = + − − oe 10(c)(i) 1 + 6x final answer 2 M1 for 4 – 3(1 – 2x) 10(c)(ii) 20 – 36x or 4(5 – 9x) final answer 4 B3 for 20 – 36x seen in working then spoiled OR M1 for (4 – 3x)2 + 4 – 3x – 9(x2 + x) or better B1 for [(4 – 3x)2=]16 – 12x – 12x + 9x2 or better B1 for answer 20 – kx or k – 36x oe or answer 20 – 36x + kx2 k ≠ 0 10(d) 1 2 − oe 2 M1 for (32)kx or 9kx = 1 2 9 x − oe 10(e) 1 0 −       2 B1 for 0 k    or fully correct description of translation

0444/43 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 9 of 9 Question Answer Marks Partial Marks 11 A 24 B1 5n – 1 oe B2 B1 for 5n – k or jn – 1 oe j ≠ 0 B 127 B1 n3 + 2 oe B2 B1 for n3 oe C 256 B1 4(n–1) oe B2 B1 for 4k oe D 7 65 B1 2 2 ( 3) 1 n n + + + oe B2 B1 for 2 any quadratic n + oe