Cambridge IGCSE Mathematics - International 0607 — 2019 Oct/Nov Paper 6 · Variant 1

This document consists of 12 printed pages. DC (SC/CGW) 170792/1 © UCLES 2019 [Turn over * 1 9 5 2 6 5 0 7 0 4 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) October/November 2019 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Graphics Calculator READ THESE INSTRUCTIONS FIRST Write your centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. Do not use staples, paper clips, glue or correction fluid. You may use an HB pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer both parts A (questions 1 to 3) and B (questions 4 to 5). You must show all relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 40. Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education

2 0607/61/O/N/19 © UCLES 2019 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 3) DECIMAL FORMS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation looks at the patterns when changing a fraction to its decimal form. Examples . . 3 2 0 666 0 6 f = = o This is a repeating decimal. . 4 3 0 75 = This is a terminating decimal. The fraction 8 5 has a numerator of 5 and a denominator of 8. 1 This question is about terminating decimals. (a) (i) Complete the table. Fraction 2 1 5 1 20 7 25 1 500 3 Equivalent fraction 10 10 100 100 1000 Decimal 0.5 0.2 (ii) What is always true about the denominators of equivalent fractions when the decimal form is a terminating decimal? … … (b) (i) Write each number as a product of its prime factors. The first two have been completed for you. 20 = 2 # 2 # 5 25 = 5 # 5 50 = 100 = 500 =

3 0607/61/O/N/19 © UCLES 2019 [Turn over (ii) Use your answers to part (i) to help you complete the table. Fraction Decimal Number of decimal places Denominator written as a product of primes using powers Larger power 20 1 0.05 2 2 5 2 # 2 25 7 0.28 2 52 2 50 9 0.18 2 100 19 0.19 2 200 13 0.065 3 2 5 3 2 # 3 500 11 0.022 5000 17 0.0034 2 5 3 4 # 4 (iii) A fraction with denominator 2 5 p q # , where q is greater than p, is changed to its decimal form. Write down the number of decimal places in the decimal form of this fraction. …

4 0607/61/O/N/19 © UCLES 2019 2 This question is about repeating decimals. The number of digits in the repeating pattern is called the repeat length. Example . . 13 1 0 076923 076923076923 0 076 923 f = = o o A C BBBB This is a repeating decimal with a repeat length of 6. (a) (i) Complete the table. Fraction 3 1 11 1 37 1 111 1 41 1 7 1 Equivalent fraction 9 3 99 9 999 99 999 999 999 Denominator of equivalent fraction 10 1 1 - 10 1 2 - 10 1 3 - Decimal .0 3o .0 09 o o .0 027 o o .0 142857 o o Repeat length 1 2 5 6 (ii) A repeating decimal has a repeat length of k. Write down an expression, in terms of k, for the denominator of this fraction. …

5 0607/61/O/N/19 © UCLES 2019 [Turn over (b) (i) 407 1 11 37 1 11 1 37 1 # # = = 407 1 is changed to its decimal form. Show that this has a repeat length that is equal to the lowest common multiple (LCM) of the repeat lengths of the decimal forms of 11 1 and 37 1 . (ii) Show how the lowest common multiple (LCM) of the repeat lengths of 7 1 and 37 1 gives the repeat length of 259 1 . (iii) m and n are different prime numbers. The decimal form of m 1 has a repeat length of 6. The decimal form of n 1 has a repeat length of 9. Find the repeat length of the decimal form of m n 1 # . …

6 0607/61/O/N/19 © UCLES 2019 3 Some decimals have non-repeating decimal parts followed by repeating decimal parts. Example . . 0 65 0 65555f = o In this decimal, the 6 does not repeat but the 5 does. (a) Show that adding the decimal forms of 5 1 and 3 1 gives a decimal of this type. (b) Complete the table. Fraction Decimal Number of non- repeating decimal places Repeat length Denominator written as a product of primes using powers 6 1 .0 16o 1 1 2 # 3 12 1 .0 083o 2 1 75 7 24 11 3 600 317 .0 5283o 23 # 52 # 3 1320 1 .0 000 75 o o 3 2 23 # 5 # 11 # 3 101750 50001 .0 491410 319 o o 3 6 2 # 53 # 11 # 37

7 0607/61/O/N/19 © UCLES 2019 [Turn over (c) A fraction is of the form c d 2 5 1 a b # # # . In the fraction a and b are positive integers and c and d are different prime numbers less than 90. The decimal form of this fraction has 5 non-repeating decimal places and a repeat length of 30. Using question 1(b) and question 2(a)(i), find a possible value for each of a, b, c and d. a = … b = … c = … d = … (d) m and n are different prime numbers. The decimal form of the fraction m 1 has a repeat length of q. The decimal form of the fraction n 1 has a repeat length of 3q. The decimal form of the fraction w 1 has • k non-repeating decimal places and • a repeat length of 3q. Find a possible expression for w, in terms of k, m and n. …

8 0607/61/O/N/19 © UCLES 2019 B MODELLING (QUESTIONS 4 to 5) FLOWERING TIMES (20 marks) You are advised to spend no more than 45 minutes on this part. This task is about when plants flower. The number of hours of darkness affects when plants flower. In this investigation, the number of hours of darkness + the number of hours of daylight = 24. 4 (a) The graph shows the approximate number of hours of daylight in Normandy, France for 2017. On the x-axis, 0 is 1st January 2017, 12 is 1st January 2018 and 24 is 1st January 2019. 1 1 3 5 7 9 11 13 15 17 19 21 23 4 2 0 6 8 10 12 14 16 18 20 F A J A O D F A J A O M M J J S N J M Month M J S N J D 22 24 0 3 2 5 4 7 6 8 10 9 12 11 Number of hours 14 13 15 17 y x 16 (i) The pattern for the number of hours of daylight remains the same each year. Complete the graph to show the approximate number of hours of daylight for 2018. (ii) On the same grid, draw the graph to show the number of hours of darkness for the two years. (iii) Describe fully the single transformation that maps the graph of the number of hours of daylight onto the graph of the number of hours of darkness. … (b) Pierre grows oats in Normandy, France. Oat plants flower when there are less than 12 hours of darkness. Find the earliest month when an oat plant flowers. …

9 0607/61/O/N/19 © UCLES 2019 [Turn over (c) Pierre models the number of hours of daylight, p, using ° . sin p x 12 3 9 30 2 31 20 = + - f e op where x has the following value at the start of each month. Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec x 0 1 2 3 4 5 6 7 8 9 10 11 (i) Using Pierre’s model, find the maximum number of hours of daylight and the month in which it occurs. Maximum hours of daylight … Month … (ii) Pierre thinks that his oat plants flower at their best when the number of hours of darkness is at its minimum. Find the minimum number of hours of darkness. … (iii) Using Pierre’s model, write down a model for the number of hours of darkness, q. … (d) On 20th March, the number of hours of darkness is the same as the number of hours of daylight. There are 31 days in March. Show that Pierre’s model finds this date accurately.

10 0607/61/O/N/19 © UCLES 2019 5 Alexa grows soybeans in Queensland, Australia. In Australia, there are more hours of darkness in June than there are in January. Alexa records the number of hours of darkness for 360 days, from 1st January to 26th December. She finds this information. Hours of darkness Maximum 13.8 Minimum 10.2 Alexa models the number of hours of darkness, y, by drawing the graph of ° cos y a b t = - where t is the day of the year. (a) (i) Write suitable calculations to show that a 12 = and . b 1 8 = . (ii) On the axes, sketch the graph of the model . ° cos y t 12 1 8 = - for t 0 360 G G . 0 360 y t

11 0607/61/O/N/19 © UCLES 2019 [Turn over (b) Soybeans flower when there are more than 12 hours of darkness. The flowers grow at the fastest rate when there are 13.6 or more hours of darkness. (i) Find the number of days when the flowers are growing at their fastest rate. … (ii) The table shows the value of t on the first day of each month. Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec t 1 32 60 91 121 152 182 213 244 274 305 335 On average, Alexa’s soybeans flower 53 days after she plants them. She wants to plant them so that, when they begin to flower, the flowers grow at the fastest rate. Use Alexa’s model to show that the latest date she should plant her soybeans is 3rd June. Question 5(c) is printed on the next page.

12 0607/61/O/N/19 © UCLES 2019 (c) (i) Alexa uses her model, . ° cos y t 12 1 8 = - , to find the first date when the number of hours of darkness is the same as the number of hours of daylight. Find this date. … (ii) The actual date when the number of hours of darkness was the same as the number of hours of daylight was 20th March. Alexa decides to change her model so that it finds this date accurately. She does this by a translation of the graph of her model. Find the new model. … 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.

This document consists of 7 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) October/November 2019 MARK SCHEME Maximum Mark: 40 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 2019 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 2 of 7 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 3 of 7 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

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION DECIMAL FRACTIONS 1(a)(i) 5 2 35 4 6 0.35 0.04 0.006 1 1(a)(ii) power of 10 1 1(b)(i) 2 × 5 × 5 2 × 2 × 5 × 5 2 × 2 × 5 × 5 × 5 1 1(b)(ii) 2 22 × 5 2 2 52 2 2 2[1] × 52 2 2 22 × 52 2 3 23 × 52 3 3 22 × 53 3 4 23 × 54 4 2 Strict FT their column 4 to get column 5 B1 for 4 correct entries 1(b)(iii) q Final answer 1 2(a)(i) 1 37 1 111 1 41 1 7 27 9 999 2439 142857 103 – 1 105 − 1 106 – 1 [ ] 0 .009   [0]. 024 39   3 3 2 B1 for 8, 9 or 10 correct entries 2(a)(ii) 10k – 1 Final answer 1 2(b)(i) 0.002457   oe and [LCM] (2, 3) = 6 oe 1 2(b)(ii) 259 = 7 × 37 oe and [LCM] (3, 6) = 6 oe and 0.003861   oe 2 B1 for one correct statement 2(b)(iii) 18 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 7 Question Answer Marks Partial Marks 3(a) 0.2 + 0.3 0.53 =   1 3(b) 2 1 22 × 3 0.093 2 1 3 × 52 0.4583 3 1 23 × 3 3 1 2 B1 for 6 or 7 correct entries 3(c) [a =] 5, [b =] 5, 4, 3, 2, 1, 0 or vice versa and [c =] 7, [d =] 41 or vice versa 2 B1 for a or b = 5 [and the other 5 or less] or for a correct c, d pair 3(d) Correct expression as final answer 1 Communication: Seen in two of the following questions 1 1(a)(i) method to find equivalent fractions e.g. 100 25 ÷ 2(a)(i) 999 ÷ 37 oe or 99999 ÷ 41 oe or 999999 ÷ 7 oe or method to find equivalent fractions 2(b)(iii) LCM [6 and 9] 18 3(d) LCM [q and 3q] 3q 3(b) method for finding factors e.g. factor trees

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING FLOWERING TIMES 4(a)(i) Correct continuation of graph 1 4(a)(ii) Correct reflection of their curve in the line y = 12 1 FT their (a)(i) 4(a)(iii) Reflection y = 12 2 B1 for each 4(b) March 1 4(c)(i) 15.9 1 June 1 4(c)(ii) 8.1 1 FT 24 – their (c)(i) 4(c)(iii) [q =]12 −3.9 sin 20 30 2 31 x     − °         oe 1 4(d) [x =] 2 20 31 or 2.64 to 2.65 seen 1 5(a)(i) leading to a = 12 and b = 1.8 2 B1 for both equations seen OR a = 13.8 10.2 2 + = 12 and b = 13.8 10.2 2 − = 1.8 or only if a calculated first b = 13.8 – 12 = 1.8 or b = 12 – 10.2 = 1.8 B1 for each 5(a)(ii) Correct sketch 1 5(b)(i) 54 to 56 2 B1 207 [. …] or 208 or 152[. …] or 153 13.8 10.2 a b a b = + = −

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 7 of 7 Question Answer Marks Partial Marks 5(b)(ii) 207 – 53 = 154 [leading to 3rd June] oe 1 5(c)(i) 31st March 1 If 0 scored, SC1 for 90 seen and answer March 30 5(c)(ii) [y =] 12 – 1.8 cos(t + k) where k = 11 or their 5(c)(i) date in March – 20 or their 5(c)(i) day number – 79 2 FT their 5(c)(i) date B1 for k = their 5(c)(i) date in March – 20 or their 5(c)(i) day number – 79 [y =] 12.34346 – 1.8 cos( )t B1 for 12.3[…] – 1.8 cos(t) Communication: Seen in two of the following questions 1 4(a)(iii) drawing the line y = 12, from (0, 12) to (24, 12) 4(c)(i) 5.6[…] seen or sketch 5(a)(ii) vertical scale to 13.8 or max point indicated at (180, 13.8) 5(b)(i) y = 13.6 drawn on their sketch of the model 5(c)(i) [t =] 90 or 90th day seen