Cambridge IGCSE Mathematics (9-1) 0980 — 2025 May/June Paper 2 · Variant 2

This document has 20 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™(9–1) MATHEMATICS 0980/22 Paper 2 Non-calculator (Extended) May/June 2025 2 hours 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, centre 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. ● Calculators must not be used in this paper. ● You may use tracing paper. ● You must show all necessary working clearly. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. * 1 0 1 4 9 7 4 3 3 1 * DC (SL) 353848 © UCLES 2025 ,










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2 0980/22/M/J/25 © UCLES 2025 List of formulas Area, A, of triangle, base b, height h. A bh 2 1 = Area, A, of circle of radius r. rr A 2 = Circumference, C, of circle of radius r. r C r 2 = Curved surface area, A, of cylinder of radius r, height h. rrh A 2 = Curved surface area, A, of cone of radius r, sloping edge l. r A rl = Surface area, A, of sphere of radius r. r A r 4 2 = Volume, V, of prism, cross-sectional area A, length l. V Al = Volume, V, of pyramid, base area A, height h. V Ah 3 1 = Volume, V, of cylinder of radius r, height h. r V r h 2 = Volume, V, of cone of radius r, height h. r V r h 3 1 2 = Volume, V, of sphere of radius r. r V r 3 4 3 = For the equation ax2 + bx + c = 0, where a ! 0, x a b b ac 2 4 2 ! = - - For the triangle shown, sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin ab C 2 1 Area = A c b a C B * 0000800000002 * ,










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3 0980/22/M/J/25 © UCLES 2025 [Turn over Calculators must not be used in this paper. 1 (a) Shade one more small square so that the diagram has one line of symmetry. [1] (b) Shade one more small square so that the diagram has rotational symmetry of order 2. [1] 2 The scale drawing shows the positions of two villages, P and Q. The scale is 1 cm represents 0.5 km. North P North Q (a) Find the actual distance between village P and village Q. … km [2] (b) Measure the bearing of village Q from village P. … [1] * 0000800000003 * ,










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4 0980/22/M/J/25 © UCLES 2025 3 NOT TO SCALE 45° 65° y° x° The diagram shows two straight lines intersecting two parallel lines. Find the value of x and the value of y. x = … y = … [3] 4 1 2 3 4 5 6 7 Samira picks one of these cards at random and replaces it. (a) Find the probability that she picks an odd number. … [1] (b) Samira repeats this 35 times. Calculate the number of times Samira is expected to pick an odd number. … [1] * 0000800000004 * ,










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5 0980/22/M/J/25 © UCLES 2025 [Turn over 5 y x 4 2 3 1 0 – 2 – 3 – 1 – 4 – 2 – 3 – 1 2 1 3 4 – 4 T U (a) Translate triangle T by the vector 0 2 - e o. [1] (b) Describe fully the single transformation that maps triangle T onto triangle U. … … [3] 6 Solve. (a) x 8 7 39 + = x = … [2] (b) ( ) y 2 5 1 24 - = y = … [3] * 0000800000005 * ,










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6 0980/22/M/J/25 © UCLES 2025 7 These are the first 4 terms of a sequence. 11 8 5 2 (a) Find the next term of this sequence. … [1] (b) Find the nth term of this sequence. … [2] 8 Find the highest common factor (HCF) of 36 and 54. … [2] * 0000800000006 * ,










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7 0980/22/M/J/25 © UCLES 2025 [Turn over 9 A is the point ( , ) 3 1 - . AB 2 4 = - e o (a) AC AB 2 = Find the coordinates of the point C. ( … , … ) [2] (b) The length of AB is k 5. Find the value of k. k = … [2] (c) P is a point on AB. AP PB 1 3 | | = Find the position vector of P. f p [2] * 0000800000007 * ,










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8 0980/22/M/J/25 © UCLES 2025 10 NOT TO SCALE 45° O 18 cm The diagram shows a sector of a circle, centre O. The length of the arc is r n cm . Find the value of n. n = … [2] 11 (a) Write 0.007 08 in standard form. … [1] (b) Work out . . ( ) ( ) 3 8 10 3 8 10 22 23 # # + . Give your answer in standard form. … [2] * 0000800000008 * ,










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9 0980/22/M/J/25 © UCLES 2025 [Turn over 12 NOT TO SCALE 74° P Q R P, Q and R lie on a circle. QR is a diameter. Find angle PRQ. Give geometrical reasons for your answer. Angle PRQ = … because … … [2] * 0000800000009 * ,










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10 0980/22/M/J/25 © UCLES 2025 13 (a) 100 students solve a puzzle. The table shows information about the time taken by each student to solve the puzzle. Time (t seconds) t 20 40 1 G t 40 60 1 G t 60 100 1 G Frequency 30 40 30 (i) Work out an estimate of the mean. …s [4] (ii) Complete the histogram to show the information in the table. t 3 4 2 1 0 40 30 50 70 90 20 60 Time (seconds) 80 100 Frequency density [2] * 0000800000010 * ,










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11 0980/22/M/J/25 © UCLES 2025 [Turn over (b) 80 adults solve the same puzzle as the students. The cumulative frequency table shows information about the time taken by each adult to solve the puzzle. Time (t seconds) t 20 G t 40 G t 60 G t 80 G t 100 G t 120 G Cumulative frequency 0 12 36 60 74 80 (i) On the grid, draw a cumulative frequency diagram. t 30 40 50 60 70 80 20 10 0 40 30 50 70 90 20 60 Time (seconds) 80 100 110 120 Cumulative frequency [3] (ii) Use your cumulative frequency diagram to find an estimate for (a) the median … s [1] (b) the lower quartile. … s [1] * 0000800000011 * ,









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12 0980/22/M/J/25 © UCLES 2025 14 Write .0 25o as a fraction. … [2] * 0000800000012 * ,









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13 0980/22/M/J/25 © UCLES 2025 [Turn over 15 y x 4 5 3 2 1 0 – 2 – 3 – 5 – 1 – 4 – 2 – 1 – 3 – 5 – 7 2 4 1 3 5 7 6 8 – 4 – 6 – 8 The diagram shows the graph of y x 2 1 = - . (a) Write down the coordinates of the point where the graph crosses the x-axis. ( … , … ) [1] (b) Write down the equation of each asymptote. … … [2] (c) By drawing a suitable straight line on the grid, solve x x 2 1 0 - - = . x = … or x = … [3] * 0000800000013 * ,









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14 0980/22/M/J/25 © UCLES 2025 16 6 cm 5 cm NOT TO SCALE The diagram shows a solid made by joining a hemisphere to a cylinder. The radius of both the hemisphere and the cylinder is 6 cm. The height of the cylinder is 5 cm. Find the total surface area of the solid. Give your answer in terms of r. … cm2 [4] 17 Find the value of (a) 125 3 2 … [2] (b) 4 2 5 - . … [2] * 0000800000014 * ,









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15 0980/22/M/J/25 © UCLES 2025 [Turn over 18 (a) 3 9 Rationalise the denominator. Give your answer in its simplest form. … [2] (b) ( )( ) c k 5 2 1 3 2 2 - + = + Find the value of c and the value of k. c = … k = … [2] 19 Write as a single fraction in its simplest form. (a) a a b 6 5 3 # … [2] (b) p t 2 4 3 + … [2] (c) x x 2 2 1 3 - - + … [3] * 0000800000015 * ,









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16 0980/22/M/J/25 © UCLES 2025 20 y x 1 \ (a) When , x y 9 2 = = . Find the value of y when x 36 = . y = … [3] (b) When x is increased by a factor of 4, the value of y changes by a factor of p. Find the value of p. p = … [1] * 0000800000016 * ,









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17 0980/22/M/J/25 © UCLES 2025 [Turn over 21 NOT TO SCALE y x A B O P Q The diagram shows the graph of y x x 3 3 = - . The graph crosses the x-axis at A, at O and at B. The turning points of the graph are at P and at Q. (a) Find the x-coordinate of A and the x-coordinate of B. Give your answers as exact values. x-coordinate of A … x-coordinate of B … [3] (b) (i) Differentiate x x 3 3 - . … [2] (ii) Find the coordinates of P and Q. P ( … , … ) Q ( … , … ) [4] * 0000800000017 * ,









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18 0980/22/M/J/25 © UCLES 2025 22 (a) Write down the exact value of tan 60°. … [1] (b) Solve sinx 2 1 0 - = for x 0 3 0 6 ° ° G G . x = … or x = … [3] 23 NOT TO SCALE A O B b a M C In the diagram, OA is parallel to BC. BC = 3OA M is the midpoint of AC. The position vector of A is a and the position vector of B is b. Find the position vector of M. Give your answer in terms of a and b, in its simplest form. … [3] * 0000800000018 * ,









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19 0980/22/M/J/25 © UCLES 2025 24 The line y x 7 3 = + intersects the curve y x x 5 12 2 = + - at the points A and B. Find the coordinates of A and B. A ( … , … ) B ( … , … ) [5] * 0000800000019 * ,









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20 0980/22/M/J/25 © UCLES 2025 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 Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE * 0000800000020 * ,










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This document consists of 11 printed pages. © Cambridge University Press & Assessment 2025 [Turn over Cambridge IGCSE™ (9–1) MATHEMATICS 0980/22 Paper 2 (Extended) May/June 2025 MARK SCHEME Maximum Mark: 100 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 May/June 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 2 of 11 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 descriptions 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.

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 3 of 11 Mathematics-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, non-integer 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 or sign 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 A or B 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.

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 4 of 11 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning More information required Accuracy mark awarded zero Accuracy mark awarded one Accuracy mark awarded two Accuracy mark awarded three Independent mark awarded zero Independent mark awarded one Independent mark awarded two Independent mark awarded three Benefit of the doubt Communication mark Incorrect Follow through Highlighter Highlight a key point in the working Ignore subsequent work Method mark awarded zero Method mark awarded one Method mark awarded two Method mark awarded three

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 5 of 11 Annotation Meaning Misread Omission Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Premature rounding/approximation Special case Indicates that work/page has been seen Transcription error Correct Correct answer from incorrect working MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 6 of 11 Question Answer Marks Partial Marks 1(a) or 1 1(b) 1 2(a) 4.4 to 4.6 2 B1 for 8.8 [cm] to 9.2 [cm] or M1 for 0.5 × their written measurement where their measurement is in the range 8 to 10 2(b) 108 to 112 1 3 [x =] 70 and [y =] 65 3 B2 for either correct or B1 for 45 and 65 correctly placed on diagram or M1 for 180 – 45 – 65 oe 4(a) 4 7 oe 1 4(b) 20 1 FT their (a) × 35 provided 0 < their (a) < 1 5(a) Triangle at (–1, –1), (–1, 1), (–2, –1) 1 5(b) Rotation 90 anticlockwise oe (2, 2) 3 B1 for each 6(a) 4 2 M1 for 8x = 39 – 7 or better 6(b) 2.6 or 13 5 oe 3 M1 for correct first step e.g. 5y – 1 = 24 2 or 10y – 2 = 24 or better M1 for correctly isolating terms in y FT their first step e.g. 5y = 12 + 1 or 10y = 24 + 2 7(a) –1 1

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 7 of 11 Question Answer Marks Partial Marks 7(b) 14 – 3n oe final answer 2 B1 for c – 3n or 14 – kn (k ≠ 0) or 14 – 3n seen then spoilt 8 18 2 B1 for answer 2, 3, 6 or 9 or M1 for answer 2 × 32 oe or for 2 × 2 × 3 × 3 and 2 × 3 × 3 × 3 or 2 correct factor trees or tables 9(a) (7, –9) 2 B1 for (7, k) or (k, –9) or 7 9     −   or for 4 8     −   seen or M1 for 3 2 2 1 4     +     − −     9(b) 2 2 M1 for 22 + ([–]4)2 oe or better 9(c) 3.5 2     −   oe 2 B1 for answer 3.5       k or 2     −   k or for 0.5 1     −   or 1 2 1         −   seen or M1 for 3 1     −   + 2 1 4 4     −   oe 10 9 2 oe 2 M1 for  45 2 π 18 360   oe 11(a) 7.08 × 10–3 cao 1 11(b) 4.18 × 1023 cao 2 B1 for figs 418 or M1 for 0.38 × 1023 or 38 × 1022 or 1022(3.8 + 3.8 × 10) or 1023(3.8 ÷ 10 + 3.8) oe 12 16 and angle in a semicircle = 90 and angle sum of a triangle = 180 2 B1 for 16 or angle in a semicircle = 90

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 8 of 11 Question Answer Marks Partial Marks 13(a)(i) 53 4 M1 for correct midpoints soi M1 for fx where x is in correct interval including boundaries M1 for fx ÷ 100 dep on second M1 13(a)(ii) Two correct bars with correct widths and heights 2 and 0.75 2 B1 for one correct bar or M1 for 40/20 oe and 30/40 oe soi 13(b)(i) Correct diagram 3 B1 for correct horizontal placement for 6 plots B1 for correct vertical placement for 6 plots B1FT dep on at least B1 for reasonable increasing curve or polygon through their 6 points If 0 scored, SC1 for 5 out of 6 points correctly plotted 13(b)(ii)(a) 62 to 64 1 FT their increasing curve or polygon reading at 40 13(b)(ii)(b) 46 to 48 1 FT their increasing curve or polygon reading at 20 14 23 90 oe fraction 2 M1 for 25.55… – 2.55… oe or for 90x = 23 oe or for 2 5 10 90 + oe 15(a) (2, 0) 1 15(b) x = 0, y = –1 2 B1 for each 15(c) y = x ruled B1 x = –2 and x = 1 B2 B1 for one correct or for two correct answers FT from their line

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 9 of 11 Question Answer Marks Partial Marks 16 168 π 4 B3 for answer 168 OR M3 for 2 π 6  + 2 1 4π 6 2   + 2π 6 5  oe OR To a maximum of 2 marks ignoring extra areas added or subtracted M1 for π × 62 M1 for 2 1 4 π 6 2  oe M1 for 2 × π × 6 × 5 17(a) 25 2 M1 for ( ) 2 3 125 or 3 2 125 or ( ) 2 3 3 5 or B1 for 3 125 = 5 17(b) [] 1 32 2 M1 for 5 1 2 or 32–1 or 1 1024 18(a) 3 3 cao 2 M1 for 9 3 × 3 3 oe 18(b) [c =] –1 [k =] 14 2 B1 for each or for 3 correct terms from 5 15 2 2 3 2 2 + − − oe 19(a) 5 2 b cao final answer 2 B1 for 15 6 ab a or better seen 19(b) 2 3 4 + p t cao final answer 2 M1 for adding two correct fractions with a common denominator e.g. 4 6 8 8 + p t 19(c) ( )( ) 8 2 1 − − + x x x or 2 8 2 − −− x x x cao final answer 3 B1 for 2(x + 1) – 3(x – 2) or better isw B1 for common denominator (x – 2)(x + 1) oe isw

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 10 of 11 Question Answer Marks Partial Marks 20(a) [] 1 3 M1 for 2 = 9 k oe M1 for 36 their k OR M2 for 2 9 36  =  y 20(b) 1 2 oe 1 21(a) [A =] 3 − oe [B =] 3 oe 3 B2 for – 3 oe or 3 oe or M1 for ( )  2 3 0 − = x x or better or 2 0 0 4 1 3 2 1  −− − oe 21(b)(i) 3 – 3x2 final answer 2 B1 for 3 or – 3x2 correct in an expression or for correct answer spoilt 21(b)(ii) [P =] (–1, –2) and [Q =] (1, 2) 4 B3 for (–1, –2) or (1, 2) or for two correct values of x or M2 for x2 = 1 or for [3](1 – x)(1 + x) [= 0] oe factorised or 2 0 0 4 3 3 2 3  −− − oe OR M1 for [3](1 – x2) [= 0] or their (b)(i) = 0 or for stating d d y x = 0 M1 for correct method to solve their quadratic 22(a) 3 1 22(b) 30, 150 3 B2 for 30 or 150 or M1 for sin x = 1 2 If 0 or M1 scored, SC1 for one acute angle and one obtuse angle with a sum of 180

0980/22 Cambridge IGCSE (9–1) – Mark Scheme PUBLISHED May/June 2025 © Cambridge University Press & Assessment 2025 Page 11 of 11 Question Answer Marks Partial Marks 23 2a + 1 2 b final answer 3 B2 for a correct route in terms of a and b not in its simplest form or for AM (or 1 2 AC ) = a + 1 2 b oe or B1 for AC = –a + b + 3a oe or M1 for correct route for OM using the lines of the diagram 24 (5, 38) and (–3, –18) 5 B4 for one correct coordinate or for x = 5 and x = –3 OR M2 for x2 – 2x – 15 [= 0] or y2 – 20y – 684 [= 0] or M1 for 7x + 3 = x2 + 5x – 12 oe or 2 3 3 5 12 7 7 − −     = + −         y y y M1 for correct method to solve their three- term quadratic (x – 5)(x + 3) ( ) ( ) 2 2 2 4 1 15 2 1 −−  − −−  oe If B0 scored and at least 2 method marks scored, SC1 for correct substitution of both of their x values or their y values into y = 7x + 3 or y = x2 + 5x – 12