Mathematics Higher Level

J.18/0 Pre-Junior Certificate Examination, 016 Mathematics Higher Level Marking Scheme Paper 1 Pg. Paper Pg. 36 Page 1 of 56

Name/version: Printed: Whom: exams Checked: Pre-Junior Certificate Examination, 016 To: Ret d: Updated: Whom: Mathematics Name/version: Complete (y/n): Whom: 005 Print Stamp.doc Higher Level Paper 1 Marking Scheme (300 marks) Structure of the Marking Scheme Students responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table: Scale label A B C D No. of categories 3 4 5 5 mark scale 0, 3, 5 0,, 4, 5 0,, 3, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response (no credit) correct response (full credit) B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit) 014 (Paper 1) Scale label A B C No of categories 3 4 5 mark scale 0, 5 0, 3, 5 0,, 10 mark scale 0, 10 0, 5, 10 0, 3, 7 15 mark scale 0, 15 0, 10, 15 0, 10, C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit) D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit) In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded. The * for units to be applied only if the student s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions. Unless otherwise specified, accept correct answer with or without work shown. Accept a students work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved. 016.1 J.18/0_MS /7 Page of 71 exams

Summary of Marks 016 JC Maths (Higher Level, Paper 1) Q.1 (a) 5C (0,, 4, 5) Q.8 (a) 5B (0, 3, 5) (b) (i) 5C (0,, 4, 5) (b) 5C* (0,, 4, 5) (ii) 5C (0,, 4, 5) 10 (c) 5D (0,, 3, 4, 5) 0 Q.9 (a) 5B (0, 3, 5) (b) 5D (0,, 3, 4, 5) Q. (a) (i) 5B (0, 3, 5) (c) 10D (0, 4, 6, 8, 10) (ii) 5B (0, 3, 5) 0 (b) (i) 5C (0,, 4, 5) (ii) 5C (0,, 4, 5) (iii) 5B (0, 3, 5) Q.10 (a) 10C (0, 4, 7, 10) (iv) 5C (0,, 4, 5) (b) (i) 5B (0, 3, 5) 30 (ii) 5B (0, 3, 5) (iii) 10D* (0, 4, 6, 8, 10) 30 Q.3 (a) 5C (0,, 4, 5) (b) (i) 5C (0,, 4, 5) (ii) 5C (0,, 4, 5) Q.11 (a) (i) 5C (0,, 4, 5) (iii) 5C (0,, 4, 5) (ii) 5C (0,, 4, 5) (iv) 5C (0,, 4, 5) (b) (i) 5C (0,, 4, 5) (v) 5C (0,, 4, 5) (ii) 5C (0,, 4, 5) 30 0 Q.4 (a) 5C* (0,, 4, 5) Q.1 (a) Values10D (0, 4, 6, 8, 10) (b) 5D* (0,, 3, 4, 5) Graph 5D (0,, 3, 4, 5 ) 10 (b) 5B* (0, 3, 5) (c) 5C* (0,, 4, 5) (d) 5B* (0, 3, 5) Q.5 (a) 5C (0,, 4, 5) 30 (b) 10C* (0, 4, 7, 10) 15 Q.13 (a) 10D (0, 4, 6, 8, 10) (b) (i) 5C (0,, 4, 5) Q.6 (a) 5C (0,, 4, 5) (ii) 10D* (0, 4, 6, 8, 10) (b) 5C (0,, 4, 5) 5 (c) 5C (0,, 4, 5) (d) 5C (0,, 4, 5) 0 Q.14 (a) 10D (0, 4, 6, 8, 10) (b) 5C (0,, 4, 5) (c) 5C (0,, 4, 5) Q.7 (a) 5B (0, 3, 5) 0 (b) 5C (0,, 4, 5) (c) 5C (0,, 4, 5) (d) 5C (0,, 4, 5) 0 Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice. 016.1 J.18/0_MS 3/7 Page 3 of 71 exams

exams Pre-Junior Certificate Examination, 016 Mathematics Higher Level Paper 1 Marking Scheme (300 marks) General Instructions 1. There are 14 questions on this examination paper. Answer all questions.. Questions do not necessarily carry equal marks. 3. Marks will be lost if all necessary work is not clearly shown. 4. Answers should include the appropriate units of measurement, where relevant. 5. Answers should be given in simplest form, where relevant. Q.1 (Suggested maximum time: 10 minutes) (0) 1(a) During a discussion in Maths class, Adam said: Prime numbers are a subset of Natural numbers (N). Is Adam correct? Explain your answer. Answer yes Explanation Any 1: prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and itself, therefore prime numbers are a subset of natural numbers // natural numbers are positive whole numbers; prime numbers are positive whole numbers greater than 1 that have no positive divisors other than 1 and itself, therefore prime numbers are a subset of natural numbers // etc. ** Accept other appropriate answers. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. lists example(s) of prime numbers. Correct answer but no explanation given. High partial credit: (4 marks) Correct answer, but incomplete or unsatisfactory explanation given. 016.1 J.18/0_MS 4/7 Page 4 of 71 exams

Q.1 (cont d.) 1(b) Coins in the euro currency are issued in eight different denominations. (i) Jennifer has one coin of each denomination in her pocket. How much money does she have in total? Total 1c + c + 5c + 10c + 0c + 50c + 1 + 1 + + 5 + 10 + 0 + 50 + 100 + 00 388c or 3 88 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. identifies at least two coin denominations of euro currency. High partial credit: (4 marks) Identifies all coin denominations correctly, but fails to add or adds incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. (ii) Jennifer would like the total value, in cent, of the coins she has in her pocket to be a prime number. What is the least amount of money she would have to add to her total to make it a prime number? Smallest prime number > 388 389 Least amount of money 389 388 1c or 0 01 ** Accept students answers from part (i) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. identifies any prime number greater than 388 or greater than answer from part (i). High partial credit: (4 marks) Identifies correct prime number, but fails to finish or finishes incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. 016.1 J.18/0_MS 5/7 Page 5 of 71 exams

Q.1 (cont d.) 1(c) Express 3 and 368 as a product of prime numbers. Hence, or otherwise, find the highest common factor of 3 and 368. 3 3 116 116 58 58 9 9 9 1 3 9 368 368 184 184 9 9 46 46 3 3 3 1 368 3 (5D) HCF of 3 and 368 8 Scale 5D (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes 3 and/or 368 as a product. Divides correctly 3 and/or 368 by any prime number. Middle partial credit: (3 marks) Finds correctly 3 and/or 368 as a product of prime numbers. High partial credit: (4 marks) Finds correctly 3 and 368 as a product of prime numbers, but fails to find or finds incorrectly the HCF of both numbers. 016.1 J.18/0_MS 6/7 Page 6 of 71 exams

Q. (Suggested maximum time: 10 minutes) (30) (a) (i) On the Venn diagram below, shade in the region that represents A \ (B C). Use set notation to give an alternative description of your shaded region. (5B) Shaded region A B C Alternative description Answer (A \ B) (A \ C) ** Accept set notation consistent with shaded region if not oversimplified. Scale 5B (0, 3, 5) Partial credit: (3 marks) Correct region on Venn diagram shaded, but fails to name or incorrectly names alternative set notation. Incorrect region on Venn diagram shaded, but correctly names alternative set notation for shaded region (if not oversimplified). (ii) On the Venn diagram below, shade in the region that represents (A \ C) (B \ C). Use set notation to give an alternative description of your shaded region. (5B) Shaded region A B C Alternative description Answer (A B) \ C ** Accept set notation consistent with shaded region if not oversimplified. Scale 5B (0, 3, 5) Partial credit: (3 marks) Correct region on Venn diagram shaded, but fails to name or incorrectly names alternative set notation. Incorrect region on Venn diagram shaded, but correctly names alternative set notation for shaded region (if not oversimplified). 016.1 J.18/0_MS 7/7 Page 7 of 71 exams

Q. (cont d.) (b) A group of 100 students were asked which of the following social networking websites they used in the previous week: Facebook (F), Instagram (I) or Snapchat (S). These are the results: 60 had used Facebook. 45 had used Instagram. 31 had used Snapchat. 1 had used all three websites. 5 had used Facebook and Instagram, but not Snapchat. 8 had used Facebook and Snapchat, but not Instagram. 16 had used exactly two of the websites listed. (i) Represent the above information on the Venn diagram. U [ 100 ] F [60] [60518] [35] [5] [8] [45513] [5] [1] [1658] [3] I [45] [31813] [8] S [31] Scale 5C (0,, 4, 5) Low partial credit: ( marks) Three elements correctly identified. High partial credit: (4 marks) Between four and six elements correctly identified. (ii) How many students did not use any of the social networking websites listed above? # students 100 (35 + 5 + 1 + 8 + 5 + 3 + 8) 100 96 4 ** Accept students answers from part (i) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. sum of three or more elements identified if not oversimplified. Identifies universal set equals 100. High partial credit: (4 marks) Finds sum of elements equals 96 or equivalent, but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 8/7 Page 8 of 71 exams

Q. (cont d.) (b) (cont d.) (iii) Find the probability that a student chosen at random from the group used Snapchat only. (5B) # students who used Snapchat only 8 P(student used Snapchat only) 8 100 5 or 0 08 ** Accept students answers from part (i) if not oversimplified. Scale 5B (0, 3, 5) Partial credit: (3 marks) Some work of merit, # Students who used Snapchat only e.g.. # Total students Identifies # students who used Snapchat only equal to 8 or equivalent. Finds any fraction with denominator equal to 100 or 5. (iv) Find the probability that a student chosen at random from those who used the social networking websites listed above used at least two of the websites. # students # students who used sites + # students who used 3 sites (5 + 8 + 3) + (1) 16 + 1 8 # students who used the social networking websites 100 4 96 or 35 + 5 + 1 + 8 + 5 + 3 + 8 96 P(student used at least two websites) 8 96 7 4 or 0 91666... ** Accept students answers from parts (i) and (ii) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. identifies all correct elements and stops or continues. Finds # students who used at least two websites equal to 8 or equivalent. Finds # students who used websites equal to 96 or equivalent. High partial credit: (4 marks) Finds both 8 and 96 or equivalent if not oversimplified, but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 9/7 Page 9 of 71 exams

Q.3 (Suggested maximum time: 15 minutes) (30) Last summer, Roisin and Tom attended a Young Entrepreneurs camp for transition year students. 3(a) The ratio of girls to boys who attended the camp was 5:3. If there were 60 more girls than boys, find the total number of students who attended the camp. Ratio of girls : boys 5 : 3 5 3 of students 5 + 3 difference between the number of girls and boys of students 8 60 1 60 of students 8 30 Total students 8 30 40 or Number of girls x Number of boys x 60 Ratio of girls : boys x : x 60 5: 3 x x 60 5 3 3x 5(x 60) 5x 300 3x 5x 300 x 300 x 300 x 150 Total students x + (x 60) 150 + (150 60) 150 + 90 40 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. 3 + 5 8. 5 Writes down number of girls (total) 8 and/or number of boys 8 3 (total). High partial credit: (4 marks) Finds ratio x : x 60 :: 5 : 3. 5 3 Finds difference (total) 60 8 or equivalent, but fails to finish or finishes incorrectly. x x 60 Finds, but fails to finish 5 3 or finishes incorrectly. Finds 30 (with work shown), but fails to multiply by 8. Finds 150 (with work shown), but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 10/7 Page 10 of 71 exams

Q.3 (cont d.) 3(b) As part of the camp, Roisin and Tom set up a small business enterprise selling cupcakes. They made 300 cupcakes and sold them at a local farmers market. (i) Each cupcake cost 35 cent to make and they planned to sell them for 1 40 each. Find the planned mark-up (profit as a percentage of cost price) on each cupcake. Profit 1 40 0 35 1 05 Planned mark-up 1 05 100 0 35 1 300% Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds correct profit on each cupcake (in cent or euro). Finds margin (profit as a percentage 1 05 100 of selling price), i.e. 75%. 1 40 1 High partial credit: (4 marks) Correct expression for mark-up, but fails to finish or finishes incorrectly. Incorrect profit, from substantially correct work, but finishes correctly. * No deduction applied for the omission of % symbol in final answer. (ii) A stall at the farmers market cost 40 to rent and the packaging for all the cupcakes cost 15. Find the total cost of making and selling all the cupcakes. Total cost (0 35 300) + 40 + 15 105 + 40 + 15 160 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds 0 35 300 105 and stops. High partial credit: (4 marks) Finds three elements correctly, i.e. 105, 40 and 15, but fails to finish or finishes incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. (iii) On the day of the farmers market, Roisin and Tom sold 40 cupcakes at the planned price. Towards the end of the day, they sold the remaining cupcakes at a reduced price to ensure that everything was sold. Given that the overall profit on the day s enterprise was 6, find the total sale price of the remaining cupcakes. Sales from 40 cupcakes 1 40 40 336 Interim profit Sales from 40 cupcakes Total cost 336 160 176 Overall profit 6 Sales from (300 40) cupcakes 6 176 Sales from 60 cupcakes 50 016.1 J.18/0_MS 11/7 Page 11 of 71 exams

Q.3 (cont d.) 3(b) (iii) (cont d.) ** Accept students answers from part (ii) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds sales from 40 cupcakes, 1 40 40 336 and stops. Subtracts 160 or equivalent from incorrect sales. High partial credit: (4 marks) Finds correct interim profit, i.e. 176, but fails to finish or finishes incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. (iv) The remaining cupcakes were packed together in equal amounts and sold in bundles. In how many different ways could the cupcakes be packed together? Different ways 1 60 // 60 1 // 30 // 30 // 3 0 // 0 3 // 4 15 // 15 4 // 5 1 // 1 5 // 6 10 // 10 6 # ways 1 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. lists 3 or more ways in which cupcakes could be packed. High partial credit: (4 marks) List all 1 ways in which cupcakes could be packed, but fails to state # ways. (v) Roisin and Tom sold the bundles of cupcakes for 5 each. Find the number of cupcakes in each bundle. Sales from 60 cupcakes 50 Price for each bundle 5 50 # bundles 5 10 # cupcakes per bundle 6 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. correct use in fraction form of either 50 or 5. High partial credit: (4 marks) Finds correct # bundles, i.e. 10, but fails to find or finds incorrect # cupcakes per bundle. 016.1 J.18/0_MS 1/7 Page 1 of 71 exams

Q.4 (Suggested maximum time: 5 minutes) (10) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. On 14 July 015, the New Horizons space probe became the first spacecraft to perform a flyby study of Pluto. During its mission, the space probe took detailed measurements and observations of the planet and its moons. It is estimated that the volume of Pluto is 6 970 000 000 km 3. 4(a) The volume of the Earth is 1 0831 10 1 km 3. How many times greater than the volume of Pluto is the volume of the Earth? Give your answer correct to the nearest whole number. Volume of Pluto 6,970,000,000 6 97 10 9 km 3 or # times 1 0831 10 9 6 97 10 1 0831 1 9 10 6 97 0 155410... 10 3 155 41039... 155 Volume of the Earth 1 0831 10 1 1,083,10,000,000 km 3 # times 1,083,10,000,000 6,970,000,000 155 41039... 155 1 (5C*) Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. indicates correct 1 division, 1 0831 10 or similar. 6,970,000,000 Equates 6,970,000,000 to 6 97 10 9 or 1 0831 10 1 to 1,083,10,000,000. High partial credit: (4 marks) Correct division with correct numerator 1 1 0831 10 and denominator, i.e. or 9 6 97 10 1,083,10,000,000, but fails to finish or 6,970,000,000 finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 13/7 Page 13 of 71 exams

Q.4 (cont d.) 4(b) Find the difference between the diameters of the two planets, in kilometres. Give your answer in the form a 10 n, where 1 a < 10 and n Z, correct to three significant figures. Volume of Pluto 4 πr 3 3 6,970,000,000 4 πr 3 6,970,000,000 3 r 3 6,970,000,000 3 π 4 1,663,964,930 05765... r 3 1,663,964,930 05765... 1,184 99010... 1,185 km Volume of the Earth 4 πr 3 3 1 0831 10 1 4 πr 3 1 0831 10 1 3 r 3 1 1 0831 10 3 π 4 0 58597... 10 1 r 3 1 0 58597... 10 0 637100... 10 4 6,371 006044... 6,371 km (5D*) Difference in diameters of the two planets (6,371 1,185) 1,74,370 10,37 km 1 037 10 4 1 04 10 4 km Scale 5D* (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. equates volume of either planet correctly, i.e. Pluto as 4 πr 3 4 6,970,000,000 or Earth as πr 3 3 3 1 0831 10 1. Middle partial credit: (3 marks) Finds correct radius or diameter of either Pluto or Earth. High partial credit: (4 marks) Finds correct radius or diameter of Pluto and Earth, but fails to find difference in diameters of the two planets or finds incorrect difference. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 14/7 Page 14 of 71 exams

Q.5 (Suggested maximum time: 5 minutes) (15) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. ** No deduction should be applied for the omission of or incorrect use of units involving currency. Jim invests a sum of money for one year in a deposit account which offers an annual compound interest rate of 1 5%. 5(a) At the end of the year, the investment is worth 5,58 50. Find the sum of money that Jim invested. F P(1 + i) t P(1 + 0 015) 1 5,58 50 P(1 015) 5,58 50 P 5,58 50 1 015 5,500 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. identifies correct relevant formula from Tables. High partial credit: (4 marks) Substitutes correctly into relevant formula, i.e. P(1 + 0 015) 1 5,58 50, but fails to finish or finishes incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. At the end of the year, Jim is offered a higher fixed rate of interest if he retains the money in the account for a further two years. Assuming he does not withdraw any money, Jim s investment will be worth 5865 11 at the end of this period. 5(b) Calculate the annual rate of interest that Jim is offered on his investment. Give your answer as a percentage, correct to one decimal place. F P(1 + i) t 5,865 11 5,58 50(1 + i) (1 + i) 5,865 11 5,58 50 1 05064... 1 + i 1 05064... 1 04999... i 1 04999... 1 0 04999... 499964... 5% (10C*) Scale 10C* (0, 4, 7, 10) Low partial credit: (4 marks) Some work of merit. Substitutes correctly into relevant formula, i.e. 5,58 50(1 + i) 5,865 11 and stops or continues. High partial credit: (7 marks) Finds (1 + i) 1 05064... or 1 + i 1 05064... / 1 04999..., but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of % symbol in final answer. 016.1 J.18/0_MS 15/7 Page 15 of 71 exams

Q.6 (Suggested maximum time: 10 minutes) (0) 6(a) Express 3 as a power of. 3 16 or 1 1 1 4 5 or 3 ( 3) 1 1 ( 5 ) 5 1 or Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. 3 16 or 3 ( 3) 1 and stops or continues. High partial credit: (4 marks) 1 Finds 3 1 or 3 ( 5 ), but fails to finish or finishes incorrectly. 6(b) Hence, solve the equation 5 3. x 5 x 3 5 x 3 5 5 x 5 (5 x)() 5 10 x 5 x 5 10 5 x 5 x 5 Scale 5C (0,, 4, 5) ** Accept students answers from part (a) if not oversimplified. Low partial credit: ( marks) High partial credit: (4 marks) 5 Some work of merit, e.g. 5 x. x Finds 5 x 5 3 or or student s own answer from (a) and stops or continues. Equates powers, i.e. 5 x 5 or power of student s own answer from part (a), but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 16/7 Page 16 of 71 exams

Q.6 (cont d.) 6(c) Verify your answer to part (b). 5 3 x 5 Let x 5 5 5 5 5 5 5 5 3 when x x 5 5 ** Accept students answers from parts (a) and (b) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down 5 x 5 5 3 or (one side x correctly evaluated). Substitutes student s answer from part (b) 5 5 correctly into equation, i.e. 5 and stops or continues. High partial credit: (4 marks) Correct substitution into equation and finishes, but values do not equate and no conclusion stated. 6(d) Use the properties of surds to show that 18 8 50 simplifies to a constant. 18 8 50 64 5 64 5 8 5 3 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down one or more of 18 64 or 8, or 50 5 or 5 or 8 4 or. Uses decimal versions to find constant. High partial credit: (4 marks) 64 5 8 5 Finds or, but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 17/7 Page 17 of 71 exams

Q.7 (Suggested maximum time: 10 minutes) (0) 7(a) Factorise fully x 6x. (5B) x 6x x(x 3) Scale 5B (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. any common factor identified. 7(b) Factorise fully 3a c 9a + 6c ac. 3a c 9a + 6c ac 3a c 9a ac + 6c 3a(ac 3) c(ac 3) (ac 3)(3a c) or 3a c 9a + 6c ac 3a c ac 9a + 6c ac(3a c) 3(3a c) (3a c)(ac 3) Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. one common factor or group indicated. High partial credit: (4 marks) Finds one correct factorisation - both inside and outside bracket, e.g. 3a(ac 3). Finds two correct factorisations, but with sign errors. 7(c) Use factors to simplify the following: x + 5x 1. 4x 9 x + 5x 1 4x 9 (x 3)( x + 4) (x + 3)(x 3) x + 4 x + 3 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. correct factors of 1 and/or 9 identified. High partial credit: (4 marks) Correct factorisation of numerator or denominator. Correct factorisation of both numerator and denominator, but with sign errors. 016.1 J.18/0_MS 18/7 Page 18 of 71 exams

Q.7 (cont d.) 7(d) Multiply out and simplify (x 3x + 5)(x 7). (x 3x + 5)(x 7) x (x 7) 3x(x 7) + 5(x 7) x 3 7x 3x + 1x + 5x 35 x 3 10x + 6x 35 or (x 3x + 5)(x 7) x(x 3x + 5) 7(x 3x + 5) x 3 3x + 5x 7x + 1x 35 x 3 10x + 6x 35 or x 3x +5x x x 3 3x 5x 7 7x 1x 35x (x 3x + 5)(x 7) x 3 7x 3x + 1x + 5x 35 x 3 10x + 6x 35 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. any correct multiplication, i.e. one correct term (sign, number and x coefficient). Sets up multiplication correctly, i.e. x (x 7) 3x(x 7) + 5(x 7) or x(x 3x + 5) 7(x 3x + 5). Sets up grid correctly. High partial credit: (4 marks) Finds four correct (sign, number and x coefficient). 016.1 J.18/0_MS 19/7 Page 19 of 71 exams

Q.8 (Suggested maximum time: 5 minutes) (10) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. ** No deduction should be applied for the omission of or incorrect use of units involving currency. 8(a) EuroMillions is a lottery game operated in several European countries. The winners of the largest lottery jackpot to date are Adrian and Gillian Bayford from Great Britain, who won 190 million on 10th August, 01. Given that the exchange rate was 1 0 784 sterling on that day, calculate the value of the jackpot in sterling. Jackpot 190 million 190,000,000 0 784 148,656,000 stg. (5B) Scale 5B (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. writes formula to calculate value using exchange rate. Incorrect use of exchange rate, i.e. 190,000,000 4,84,535 787... 0 784 Finds 190,000,000 0 784, but fails to finish or finishes incorrectly. * No deduction applied for omission of or incorrect use of units involving currency. 8(b) This was not the largest sterling amount won, as Chris and Colin Weir, also from Great Britain, scooped a 185 million jackpot on 1th July, 011 but received 1 997 million more than the Bayfords due to a more favourable exchange rate. Calculate the exchange rate on that day in the form 1. Exchange rate Value (in ) Value (in ) 148,656,000 + 1,997,000 185,000,000 161,653,000 185,000,000 0 8738 0 87 1 0 87 ** Accept students answers from part (a) if not oversimplified. Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. equates 185,000,000 161,653,000 stg. Finds total value of jackpot in sterling, i.e. 148,656,000 + 1,997,000 161,653,000 or answer from part (a) + 1,997,000. 016.1 J.18/0_MS 0/7 Page 0 of 71 exams (5C*) High partial credit: (4 marks) Finds inverse of exchange rate correctly, 185,000,000 i.e. 1 14446... 1 14. 161,653,000 Indicates division with correct numerator 161,653,000 and denominator, i.e., but 185,000,000 fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for omission of or incorrect use of units involving currency.

Q.9 (Suggested maximum time: 10 minutes) (0) The first three stages of a pattern are shown below. Each stage of the pattern is made up of bowling pins. 9(a) Draw the next two stages of the pattern. (5B) Scale 5B (0, 3, 5) Partial credit: (3 marks) One stage correctly drawn. 9(b) Show that the numbers of bowling pins in each stage form a quadratic sequence. (5D) Stage # of pins 1st Diff. nd Diff. 1 1 3 1 3 3 6 1 4 4 10 1 5 5 15 as the first differences are not all the same, but the second differences are constant numbers of bowling pins in each pattern form a quadratic sequence Scale 5D (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds one first difference. Middle partial credit: (3 marks) Finds one second difference. Finds at least three first differences and stops or continues. High partial credit: (4 marks) Finds at least two second differences correctly, but fails to explain or explains incorrectly significance of second difference. 9(c) Find a general formula for the number of bowling pins in stage n of the pattern, where n N. (10D) T n an + bn + c nd difference a 1 // 0 5 016.1 J.18/0_MS 1/7 Page 1 of 71 exams

Q.9 (cont d.) 9(c) (cont d.) 1 T 1 (1) + b(1) + c 1 1 (1) + b(1) + c 1 1 + b + c 1 1 b + c 1 0 5 T 1 () + b() + c 3 1 () + b() + c 3 + b + c 3 b + c 3 1 T 1 : b + c 0 5 T : b + c 1 b + c 0 5 ( 1) b + c 1 ( 1) b c 0 5 b + c 1. b 0 5 b + c 0 5 0 5 + c 0 5 c 0 5 0 5 0 or b + c 0 5 ( ) b + c 1 ( 1) b c 1 b + c 1. c 0 c 0 b + c 0 5 b + 0 0 5 b 0 5 T n an + bn + c 0 5n + 0 5n or b + c 1 (0 5) + c 1 1 + c 1 c 0 or b + c 1 b + 0 1 b 0 5 Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. writes down general formula for nth term, identifies coefficient of n. Middle partial credit: (6 marks) Finds one or both equations for T 1 and T, i.e. b + c 0 5 and/or b + c 1. Finds either variable (b or c) correctly by trial and error, but fails to verify in both equations or verifies incorrectly. High partial credit: (8 marks) Finds first variable (b or c), but fails to find second variable or finds incorrectly. Finds both variables (b and c) correctly with no work shown or by graphical means. Finds both variables (b and c) by trial and error, but does not verify in both equations or verifies incorrectly.. 016.1 J.18/0_MS /7 Page of 71 exams

Q.10 (Suggested maximum time: 15 minutes) (30) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. 10(a) Write the following as a single fraction in its simplest form. 3 4 x 1 3x + 3 4 x 1 3x + 3(3x + ) 4(x 1) (x 1)(3x + ) 9x + 6 8x + 4 (x 1)(3x + ) x + 10 or (x 1)(3x + ) 6x x + 10 + x (10C) Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) Some work of merit, e.g. identifies (x 1)(3x + ) as the denominator with some (incorrect) numerator. Correct numerator, i.e. multiplies 3(3x + ) 4(x 1), but no denominator. High partial credit: (7 marks) Finds 3(3x + ) 4(x 1) or similar, (x 1)(3x + ) but fails to finish or finishes incorrectly. 10(b) x is a real number. Number A is equal to 1 less than x. Number B is equal to greater than seven times x. (i) Write down the numbers A and B, in terms of x. (5B) A x 1 B 7x + Scale 5B (0, 3, 5) Partial credit: (3 marks) One correct answer. (ii) The product of these two numbers is. Use this information to write an equation in x. (5B) A B (x 1)(7x + ) Scale 5B (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. something or exhibits understanding that product means multiplication. 016.1 J.18/0_MS 3/7 Page 3 of 71 exams

Q.10 (cont d.) 10(b) (cont d.) (iii) Solve this equation to find the two possible values of x. Give each of your answers correct to two decimal places. (10D*) (x 1)(7x + ) 7x + x 7x 7x 5x 0 7x 5x 4 0 x x b ± b a ( 5) ± 4ac ( 5) 4(7)( 4) (7) 5 ± 5 + 11 14 5 ± 137 14 5 ± 11 704699... 14 x 5 + 11 704699... 5 11 704699..., x 14 14 16 704699... 6 704699... 14 14 1 19319... 0 478907... 1 19 0 48 Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. writes down b formula and identifies a, b and c. Finds 7x 5x 4 0 correctly, but not does not proceed to use b formula. Middle partial credit: (6 marks) Full correct substitution into b formula, ( 5) ± ( 5) 4(7)( 4) i.e. x. (7) Two of the following steps correct: correct formula and correctly identifies a, b and c, fully correct substitution, works out solution to surd form, finishes correctly to two decimal places. High partial credit: (8 marks) Three of the above steps correct. Finds only one value of x (1 19 or 0 48), but fails to find or finds incorrectly other value of x. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 4/7 Page 4 of 71 exams

Q.11 (Suggested maximum time: 10 minutes) (0) 11(a) Let g be the function g : x x 7, where x R. (i) Find the value of g( ). g(x) x 7 g( ) ( ) 7 (4) 7 8 7 1 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. some correct substitution into given function, i.e. ( ), (4) or 8. Correctly solves g(x). High partial credit: (4 marks) Full correct substitution into formula, but fails to finish or finishes incorrectly. (ii) Find the values of k for which g(k) 5. g(x) x 7 g(k) (k) 7 5 (k) 7 5 k 5 + 7 3 k 3 16 k ± 16 ±4 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. correctly equates g(k) 5 to find correct equation, i.e. (k) 7 5 or similar. High partial credit: (4 marks) Finds k 16, but fails to find values for k or only finds one value of k. 016.1 J.18/0_MS 5/7 Page 5 of 71 exams

Q.11 (cont d.) 4(4 C ) 11(b) Let F. C + 1 (i) Calculate the value of F when C 1. F 1 4 4 1 + 1 16 1 + 1 14 7 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. simplifies given equation, i.e. F 16 4C. C + 1 Some correct substitution into equation. High partial credit: (4 marks) Full correct substitution into formula, but fails to finish or finishes incorrectly. (ii) Write C in terms of F. F 4(4 C) C + 1 F(C + 1) 4(4 C) FC + F 16 4C FC + 4C 16 F C(F + 4) 16 F C 16 F or 16 F F + 4 ( F + ) or 16 F 4 + F or 16 ( + F F ) Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. one correct manipulation. High partial credit: (4 marks) Both sides fully expanded with or without C being isolated, i.e. FC + 4C 16 F or C(F + 4) 16 F, but fails to finish or finishes incorrectly. 016.1 J.18/0_MS 6/7 Page 6 of 71 exams

Q.1 (Suggested maximum time: 15 minutes) (30) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. An archer, standing on level ground, shoots an arrow into the air. The height, in metres, of the arrow above ground level, t seconds after the arrow is released, is given by h : t + 6t t. 1(a) On the grid below, draw the graph of y h(t) in the domain 0 t 6, where t R. Values Table (1st method) t 0 1 3 4 5 6 6t 0 6 1 18 4 30 36 t 0 1 4 9 16 5 36 y h(t) 7 10 11 10 7 (10D) or Table (nd method) h(t) 6t t y h(0) 0 0 h(1) 6 1 7 h() 1 4 10 h(3) 18 9 11 h(4) 4 16 10 h(5) 30 5 7 h(6) 36 36 or Substitution method h(t) + 6t t h(0) + 6(0) (0) h(1) + 6(1) (1) 7 h() + 6() () 10 h(3) + 6(3) (3) 11 h(4) + 6(4) (4) 10 h(5) + 6(5) (5) 7 h(6) + 6(6) (6) Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) One or two points, (x, y) or (y, x), correctly identified. Middle partial credit: (6 marks) Three or four points, (x, y) or (y, x), correctly identified. High partial credit: (8 marks) Five or six points, (x, y) or (y, x), correctly identified. 016.1 J.18/0_MS 7/7 Page 7 of 71 exams

Q.1 (cont d.) 1(a) (cont d.) Graph Points (0, ), (1, 7), (, 10), (3, 11), (4, 10), (5, 7), (6, ) 1 (5D) 11 10 ht () Height of Arrow in metres, h 8 6 4 076 1 3 4 55 4 6 Time in seconds, t 674 3 4 ** Accept values calculated from previous work (seven co-ordinates needed). ** If no points are worked out, but correctly graphed, award the marks for the graph in both parts. Scale 5D (0,, 3, 4, 5) Low partial credit: ( marks) One or two points, (x, y) or (y, x), correctly plotted, labelled or not. Points incorrectly calculated, but correctly plotted to form a line. Middle partial credit: (3 marks) Three or four points, (x, y) or (y, x), correctly plotted, labelled or not (joined or not). High partial credit: (4 marks) Five or six points, (x, y) or (y, x), correctly plotted, labelled or not (joined or not). All points, (x, y) or (y, x), correctly plotted, but joined together with inappropriate curve. 016.1 J.18/0_MS 8/7 Page 8 of 71 exams

Q.1 (cont d.) For parts (b), (c) and (d), you must show your working out on the diagram on the previous page. 1(b) Use your graph to estimate the height at which the speed of the arrow is zero after it is shot. Speed of arrow is zero @ maximum height From graph Maximum height 11 m ** Accept students answers based on fully plotted graph. Scale 5B* (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. draws vertical line from 3 s and intersects graph with or without horizontal line from point of intersection to the y-axis. Answer outside tolerance (1 grid box), but inside tolerance of ± grid boxes. Answer 3 seconds. (5B*) * Deduct 1 mark off correct answer only for omission of or incorrect use of units - apply only once per section (a), (b), (c), etc. of question. 1(c) Use your graph to estimate the length of time that the arrow is more than 6 m above ground level. From graph Time (> 6m) 5 4 0 76 4 48 seconds ** Accept students answers based on fully plotted graph. Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. draws horizontal line from 6 m and intersects graph with or without vertical line(s) from points of intersection to the x-axis. Answer outside tolerance (1 grid box), but inside tolerance of ± grid boxes. One correct x-intercept. High partial credit: (4 marks) Two correct x-intercepts, but fails to find or finds incorrect time interval. (5C*) * Deduct 1 mark off correct answer only for omission of or incorrect use of units - apply only once per section (a), (b), (c), etc. of question. 1(d) Imagine the archer shoots the same arrow standing on an elevated position 3 m above ground level. By extending your graph, estimate the time that it would take the arrow to hit the ground. From graph Time 6 74 seconds ** Accept students answers based on fully plotted graph. Scale 5B* (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. extends graph to y 3 or draws horizontal line at y 3 and intersects graph with or without vertical line from point of intersection to the x-axis. Answer outside tolerance (1 grid box), but inside tolerance of ± grid boxes. 016.1 J.18/0_MS 9/7 Page 9 of 71 exams (5B*) * Deduct 1 mark off correct answer only for omission of or incorrect use of units - apply only once per section (a), (b), (c), etc. of question.

Q.13 (Suggested maximum time: 10 minutes) (5) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. 13(a) Solve the following inequality and graph your solution on the number line. 6 < 4x + 14, x R. (10D) Solution 6 < 4x + 4x < + 6 4x < 8 x > < x 3 and 4x + 14 4x 14 1 1 x 4 3 or 6 < 4x + 14... subtract 6 < 4x + 14 8 < 4x 1 8 < 4x 1... divide by 4 8 4x 1 < 4 4 4 < x 3 < x 3 Number line 4 3 1 0 1 3 4 5 Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. substitutes in value for x. Middle partial credit: (6 marks) Finds x > or x 3 (accept with or without inequality sign). High partial credit: (8 marks) Solution to inequality fully correct or number line correctly graphed. 016.1 J.18/0_MS 30/7 Page 30 of 71 exams

Q.13 (cont d.) 13(b) The area of a swimming pool is given by x 3 + 6x 5x 5. (i) The width of the swimming pool is x + 5. Find the length of the pool, in terms of x. x 4x 5 x + 5 ) x 3 + 6x 5x 5 x 3 10x or 4x 5x 5 4x + 0x 5x 5 5x + 5 0 x 4x 5 x x 3 4x 5x 5 10x 0x 5 Answer x 4x 5 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. sets up division. Identifies one term correctly in answer. Inserts the x 3 or 5 term correctly in Method. High partial credit: (4 marks) Substantial work, but with one or more critical omissions, e.g. x identified on top with solution down to 4x 5x 5 stage correct. Finds correctly two terms in answer. (ii) Given that the length of the swimming pool is 5 m, find the possible widths of the pool. Which answer do you think is most appropriate? Give a reason for your answer. Possible widths x 4x 5 5 x 4x 5 5 0 x 4x 30 0 x x 15 0 (x 5)(x + 3) 0 x 5 0 x 5 x + 5 5 + 5 10 m Most appropriate answer Answer 10 m (10D*) or x + 3 0 x 3 x + 5 3 + 5 m Reason other answer, i.e. width m, would make the swimming pool very narrow, etc. ** Accept students answers from part (ii) if not oversimplified. Scale 10D* (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. correctly equates x 4x 5 5 to find quadratic equation. Middle partial credit: (6 marks) Correctly solves quadratic equation, but fails to find roots or only finds one root. High partial credit: (8 marks) Finds both roots, but fails to identify most appropriate answer or fails to give reason. * Deduct 1 mark off correct answer only for omission of or incorrect use of units - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 31/7 Page 31 of 71 exams

Q.14 (Suggested maximum time: 10 minutes) (0) The diagram below shows part of the graph of the function f : x x + bx + c, where x R and b, c Z. y 3 x The graph of f intersects the x-axis at the points where x and x 3. 14(a) Find the value of b and the value of c. (10D) y x + bx + c x (sum of roots)x + (product of roots) x ( + 3)x + ( )(3) x x 6 b 1 c 6 The graph intersects the x-axis at x and x 3 y 0 @ x and x 3 @ x 0 ( ) + b( ) + c 0 4 b + c b c 4 @ x 3 0 (3) + b(3) + c 0 9 + 3b + c 3b + c 9 b c 4 3b + c 9 5b 5 b 1 b c 4 ( 1) c 4 c 4 c 4 + 6 c 6 or 3b + c 9 3( 1) + c 9 3 + c 9 c 9 + 3 6 016.1 J.18/0_MS 3/7 Page 3 of 71 exams

Q.14 (cont d.) 14(a) (cont d.) Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. states correctly x (sum of roots)x + (product of roots) and stops or continues. Forms one correct simultaneous equation and stops or continues. Middle partial credit: (6 marks) Substitutes into sum and product rule correctly, but fails to find or finds incorrect values for b and c. Forms two correct simultaneous equations and stops or continues. High partial credit: (8 marks) Substitutes into sum and product rule correctly, but only finds correct value for either b or c. Solves simultaneous equations, but only finds correct value for either b or c. 14(b) Write down the co-ordinates of the point where the graph intersects the y-axis. Graph intersects the y-axis x 0 y x x 6 (0) (0) 6 6 ** Accept students answers from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. correctly identifies x 0. High partial credit: (4 marks) Correctly substitutes 0 into equation, but fails to finish or finishes incorrectly. Correctly identifies x c, but fails to find or finds incorrect value of c. 14(c) (t, 4t) is a point on the graph, where t Z. Find the two possible values of t. y x x 6 4t t t 6 t t 6 4t 0 t 5t 6 0 (t 6)(t + 1) 0 t 6 0 t 6 or t + 1 0 t 1 ** Accept students answers from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. correctly substitutes (t, 4t) into equation to find quadratic equation in terms of t. High partial credit: (4 marks) Correctly solves quadratic equation, but fails to find roots or only finds one root. 016.1 J.18/0_MS 33/7 Page 33 of 71 exams

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Name/version: Printed: Whom: exams Checked: To: Ret d: Pre-Junior Certificate Examination, 016 Updated: Name/version: Whom: Mathematics Complete (y/n): Whom: 005 Print Stamp.doc Higher Level Paper Marking Scheme (300 marks) Structure of the Marking Scheme Students responses are marked according to different scales, depending on the types of response anticipated. Scales labelled A divide students responses into two categories (correct and incorrect). Scales labelled B divide responses into three categories (correct, partially correct, and incorrect), and so on. These scales and the marks that they generate are summarised in the following table: Scale label A B C D No. of categories 3 4 5 5 mark scale 0, 5 0, 3, 5 0,, 4, 5 0,, 3, 4, 5 10 mark scale 0, 4, 7, 10 0, 4, 6, 8, 10 15 mark scale A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response (no credit) correct response (full credit) B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit) C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit) D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit) In certain cases, typically involving incorrect rounding, omission of units, a misreading that does not oversimplify the work or an arithmetical error that does not oversimplify the work, a mark that is one mark below the full-credit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, scale 10C* indicates that 9 marks may be awarded. The * for units to be applied only if the student s answer is fully correct. The * to be applied once only within each section (a), (b), (c), etc. of all questions. The * penalty is not applied to currency solutions. Unless otherwise specified, accept correct answer with or without work shown. Accept students work in one part of a question for use in subsequent parts of the question, unless this oversimplifies the work involved. 016.1 J.18/0_MS 36/7 Page 36 of 71 exams

Summary of Marks 016 JC Maths (Higher Level, Paper ) Q.1 (a) 5C* (0,, 4, 5) Q.7 (a) 5C (0,, 4, 5) (b) 5C* (0,, 4, 5) (b) 5B (0, 3, 5) (c) 5C (0,, 4, 5) (c) 5C (0,, 4, 5) (d) 5D* (0,, 3, 4, 5) (d) 5C (0,, 4, 5) (e) 5D* (0,, 3, 4, 5) (e) 5C (0,, 4, 5) 5 (f) 5C (0,, 4, 5) (g) 5C (0,, 4, 5) (h) 5C (0,, 4, 5) Q. (a) 5B (0, 3, 5) 40 (b) 5B (0, 3, 5) (c) 10D (0, 4, 6, 8, 10) (d) (i) 5B (0, 3, 5) Q.8 (a) 5C* (0,, 4, 5) (ii) (b) 5C* (0,, 4, 5) (e) (i) 5B (0, 3, 5) (c) 5C* (0,, 4, 5) (ii) 5C (0,, 4, 5) (d) 5C* (0,, 4, 5) (f) 5C (0,, 4, 5) 0 40 Q.9 (a) 5B (0, 3, 5) Q.3 (a) 5C (0,, 4, 5) (b) 5B (0, 3, 5) (b) 5B (0, 3, 5) (c) 5A (0, 5) (c) (i) 5C (0,, 4, 5) (d) 5C (0,, 4, 5) (ii) 5C (0,, 4, 5) 0 0 Q.10 (a) 5C (0,, 4, 5) Q.4 (a) (i) 5C* (0,, 4, 5) (b) 5C* (0,, 4, 5) (ii) 5C (0,, 4, 5) (c) 10C (0, 4, 7, 10) (b) (i) 5C (0,, 4, 5) 0 (ii) 5C* (0,, 4, 5) (iii) 5C (0,, 4, 5) 5 Q.11 (a) 5C (0,, 4, 5) (b) 10C (0, 4, 7, 10) (c) 10D (0, 4, 6, 8, 10) Q.5 5B (0, 3, 5) (d) 5C (0,, 4, 5) 5B (0, 3, 5) 30 10C (0, 4, 7, 10) 0 Q.1 (a) 5C (0,, 4, 5) (b) 5D (0,, 3, 4, 5) Q.6 (a) 10C (0, 4, 7, 10) (c) 10C (0, 4, 7, 10) (b) 5B (0, 3, 5) 0 (c) 5C* (0,, 4, 5) 0 Assumptions about these marking schemes on the basis of past SEC marking schemes should be avoided. While the underlying assessment principles remain the same, the exact details of the marking of a particular type of question may vary from a similar question asked by the SEC in previous years in accordance with the contribution of that question to the overall examination in the current year. In setting these marking schemes, we have strived to determine how best to ensure the fair and accurate assessment of students work and to ensure consistency in the standard of assessment from year to year. Therefore, aspects of the structure, detail and application of the marking schemes for these examinations are subject to change from past SEC marking schemes and from one year to the next without notice. 016.1 J.18/0_MS 37/7 Page 37 of 71 exams

exams Pre-Junior Certificate Examination, 016 Mathematics Higher Level Paper Marking Scheme (300 marks) General Instructions 1. There are 1 questions on this examination paper. Answer all questions.. Questions do not necessarily carry equal marks. 3. Marks will be lost if all necessary work is not clearly shown. 4. Answers should include the appropriate units of measurement, where relevant. 5. Answers should be given in simplest form, where relevant. Q.1 (Suggested maximum time: 15 minutes) (5) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. A standard DVD is in the shape of a circular disc of diameter 1 cm. It has a circular hole of diameter 1 8 cm in its centre, as shown. 1(a) Find the surface area of one side of a DVD. Give your answer in terms of π. Surface area πr 1 πr 1 π( ) 1 8 π( π(6) π(0 9) 36π 0 81π 35 19π cm ) (5C*) ** Accept students answers in mm (3519π mm ) or m (0 00003519π m ). Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct area formula from Tables. 1 1 8 Finds r 1 or 6 and/or r or 0 9 and stops. High partial credit: (4 marks) Finds π(6) π(0 9), but fails to finish or finishes incorrectly. Finds area using incorrect diameters correctly, i.e. π(1) π(1 8) 140 76π. Final answer not in terms of π. * Deduct 1 mark off correct answer only for the omission of or incorrect use of units (cm ) - apply only once per section (a), (b), (c), etc. of the question. 016.1 J.18/0_MS 38/7 Page 38 of 71 exams

Q.1 (cont d.) 1(b) The case for a single standard DVD is 19 cm long, 13 5 cm wide and 1 4 cm high. Find the volume of a single DVD case. Volume L W H 13 5 19 1 4 359 1 cm 3 (5C*) ** Accept students answers in mm (3519π mm ) or m (0 00003519π m ). Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct volume formula from Tables. Some correct substitution into relevant volume formula (not stated). High partial credit: (4 marks) Correct substitution into relevant volume formula, but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only for the omission of or incorrect use of units (cm 3 ) - apply only once per section (a), (b), (c), etc. of the question. 1(c) A DVD cake box holder is in the shape of a cylinder, as shown. It has an internal height of 6 1 cm. Given that the thickness of a DVD is 1 mm, find the maximum number of DVDs it can hold. 1 mm 1 cm 10 0 1 cm or # DVDs 6 1 0 1 50 83333... maximum # DVDs 50 6 1 cm 6 1 10 mm 61 mm # DVDs 61 1 50 83333... maximum # DVDs 50 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. converts 1 mm to 0 1 cm or 6 1 cm to 61 mm. High partial credit: (4 marks) 6 1 61 Finds or or 50 83333..., but 0 1 1 fails to round down correctly to 50. 016.1 J.18/0_MS 39/7 Page 39 of 71 exams

Q.1 (cont d.) 1(d) The external volume of the DVD cake box holder is 1,384 cm 3. It has an external height of 8 5 cm. Find the external diameter of the DVD cake box holder. Give your answer correct to one decimal place. External volume πr h 1,384 cm 3 πr (8 5) 1,384 r 1,384 8 5π 51 88339... r 51 88339... 7 199190... d r d 7 199190... 14 398380... 14 4 cm ** Accept students answers in mm (144 mm) or m (0 0144 m). Scale 5D* (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct volume formula from Tables. Some correct substitution into relevant volume formula (not stated). (5D*) Middle partial credit: (3 marks) Correct substitution into relevant volume formula, but fails to manipulate or manipulates incorrectly. High partial credit: (4 marks) Finds r 51 88339... or 7 199190..., but fails to find or finds incorrect diameter. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units ( cm ) - apply only once per section (a), (b), (c), etc. of question. 1(e) Find the volume saved by using a DVD cake box holder as a percentage of using single DVD cases to store the maximum number of DVDs that the DVD cake box holder can contain. Give your answer correct to two significant figures. Volume of 50 cases 50 359 1 17,955 cm 3 Volume saved 17,955 1,384 16,571 cm 3 % Saving Saving 100 Total Volume 1 16,571 100 17,955 1 0 9918... 100 9% ** Accept students answers from part (b) if not oversimplified. Scale 5D* (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds volume of 50 standard cases (ans. 17,555) and stops. Middle partial credit: (3 marks) Finds volume saved correctly without attempting to evaluate % Saving. (5D*) High partial credit: (4 marks) Finds 16,571 or 16,571 100, but fails 17,955 17,955 1 to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 40/7 Page 40 of 71 exams

Q. (Suggested maximum time: 0 minutes) (40) A simple random sample was taken from all the votes cast in last year s marriage referendum. The sample was taken from both urban and rural areas. The way in which votes were cast is shown in the two-way table below. Vote Urban Area Rural Area Total Yes 60 Y 1,014 No X 388 786 Total 1,000 800 1,800 (a) Find the missing values for X and Y in the table above. (5B) Value of X 60 + X 1,000 X 1,000 60 398 or X + 388 786 X 786 388 398 Value of Y 60 + Y 1,014 Y 1,014 60 41 or Y + 388 800 Y 800 388 41 Scale 5B (0, 3, 5) Partial credit: (3 marks) One value correct. (b) Explain what is meant by the term Simple Random Sample in the context of this question. (5B) Any 1: a simple random sample is where every person who voted in the referendum has an equal chance of being selected // a simple random sample is meant to be an unbiased representation of all voters in the referendum // etc. ** Accept other appropriate material. Scale 5B (0, 3, 5) Partial credit: (3 marks) Incomplete or unsatisfactory explanation, e.g. writes down every person who voted or equal chance of being selected, etc. 016.1 J.18/0_MS 41/7 Page 41 of 71 exams

Q. (cont d.) (c) Complete the pie chart to display the data from the sample above. Show all of your calculations clearly. (10D) Rural No Urban Yes Rural Yes Urban No Total number of votes 1,800 Angle per vote 360 1,800 0 Urban Yes vote 60 360 1,800 10 4 Urban No vote 398 360 1,800 79 6 Rural Yes vote 41 360 1,800 8 4 Rural No vote 388 360 1,800 77 6 or 60 0 10 4 or 398 0 79 6 or 41 0 8 4 or 388 0 77 6 ** Allow a tolerance of ± in pie chart. ** Accept students answers from part (a) if not oversimplified. Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. finds angle per vote (ans. 0 ), indicates that angles are fraction of 360. Finds one angle or fraction correct. Middle partial credit: (6 marks) Finds correct values of angles and stops or continues. High partial credit: (8 marks) Finds correct values of angles and two angles correct in pie chart (with labels). Finds correct values of angles and all angles correct in pie chart, but no labels or incorrect labels. Values of angles correct and correct pie chart (with labels), but no work shown. Finds correct values of angles and all angles correct in pie chart (with labels), but outside tolerance (± ). 016.1 J.18/0_MS 4/7 Page 4 of 71 exams

Q. (cont d.) (d) (i) Do you think that this is a suitable type of graph in which to illustrate this data? Give a reason for your answer. (5B) Answer no or Reason Any 1: 3 out of the 4 angles look very similar in the pie chart // figures (on vertical axis) of a bar chart / histogram may add weight to the point that the graph is being used to illustrate // etc. Answer yes Reason Any 1: easy to see that the Urban Yes vote is the largest sector in the pie chart // not cluttered by figures which may not be required for the point that the graph is being used to illustrate // not cluttered by figures which are only a sample of the overall vote // etc. ** Accept other appropriate answers. (ii) Identify another type of graph that could be more suitable to illustrate the data. Answer Any 1: bar chart // histogram ** Award no marks for line graph, line plot or stem-and-leaf plot in part (ii). Scale 5B (0, 3, 5) Partial credit: (3 marks) Correct reason, but no alternative suitable type of graph given. No reason or incorrect reason, but suitable type of graph given. (e) (i) A vote is chosen at random from the entire sample. Find the probability that it is a Yes vote. (5B) # Yes vote 1,014 #Total sample 1,800 P( Yes vote) 1,014 1,800 169 300 or 0 563333... Scale 5B (0, 3, 5) Partial credit: (3 marks) Correct numerator or denominator. 016.1 J.18/0_MS 43/7 Page 43 of 71 exams

Q. (cont d.) (e) (cont d.) (ii) If one vote is chosen at random from both the urban sample and the rural sample, find the probability that both are No votes. P(urban No vote) 398 1,000 199 500 P(rural No vote) P(both No vote) 388 800 97 00 199 97 500 00 19,303 or 0 19303 100,000 ** Accept students answers for 398 from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( mark) Finds 199 and/or 97, but does not 500 00 attempt to multiply them. Uses addition and finishes correctly [ans. 199 + 97 883 or equivalent]. 500 00 1, 000 High partial credit: (4 mark) Finds 199 97, but fails to finish 500 00 or finishes incorrectly. (f) Paula says, It is more likely that a person from an urban area voted Yes than a person from a rural area. Based on the data in the sample, do you agree with Paula? Give a reason for your answer. Answer yes Reason From the random sample: 60 P(urban Yes vote) 1,000 0 60 41 P(rural Yes vote) 800 0 515 as 0 60 > 0 4975, it is more likely that a person from an urban area voted Yes than a person from a rural area ** Accept students answers for 41 from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Correct answer, but no reason or incorrect reason given. High partial credit: (4 marks) Correct answer with incomplete or unsatisfactory explanation. Finds 60 and 41, or equivalent, but 1,000 800 no conclusion given. 016.1 J.18/0_MS 44/7 Page 44 of 71 exams

Q.3 (Suggested maximum time: 5 minutes) (0) 3(a) Explain what is meant by the term Trial in the context of probability and give an example. Explanation Any 1: a trial is the act of doing a random experiment in probability // a trial is any particular performance of a random experiment // a trial is any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as a sample space // etc. ** Accept other appropriate material. Example Any 1: tossing a coin // rolling a die // picking a card from a deck of 5 cards // etc. ** Accept other appropriate material. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Incomplete or unsatisfactory explanation with no example or incorrect example given. High partial credit: (4 marks) Correct explanation with no example or incorrect example given. Incomplete or unsatisfactory explanation with suitable example given. 3(b) A cinema offers three different combi-meal sizes on its menu, as shown below. A combi-meal consists of a soft drink and popcorn of the same size. Combi-Meal Menu Size Soft Drink Popcorn Small Orange Plain Medium Cola Salted Large Lemonade Buttered Ginger Ale Caramel Cream Soda Calculate the number of different combi-meal options that can be ordered. (5B) #options 3 5 4 60 Scale 5B (0, 3, 5) Partial credit: (3 marks) Some work of merit, e.g. mention of the Fundamental Principle of Counting. Writes a list of possible options. Finds 3 5 4, but fails to finish or finishes incorrectly. Uses addition instead of multiplication, i.e. # options 3 + 5 + 4 1. 016.1 J.18/0_MS 45/7 Page 45 of 71 exams

Q.3 (cont d.) 3(c) Below is part of a tree diagram showing the probabilities that a customer chooses a medium combi-meal with cola and popcorn. 9 Medium 1 3 Cola 1 1 6 1 1 1 4 Plain Popcorn Salted Popcorn Buttered Popcorn Caramel Popcorn (i) A person is chosen at random in the cinema. Find the probability that this person bought a medium combi-meal with cola and plain popcorn. P(medium, cola, plain popcorn) 1 1 9 3 1 or 54 7 or 0 037037... Scale 5C (0,, 4, 5) Low partial credit: ( mark) Some work of merit, e.g. identifies 1 1 and/or and/or, but does not 9 3 attempt to multiply them. Uses addition and finishes correctly 1 1 19 [ans. or equivalent]. 9 3 18 High partial credit: (4 mark) Finds 9 3 1 1, but fails to finish or finishes incorrectly. (ii) A customer is chosen at random from those who bought a medium combi-meal. Find the probability that this customer did not order a cola with the meal. 1 P(not Cola) 1 3 or 0 666666... 3 Scale 5C (0,, 4, 5) Low partial credit: ( marks) High partial credit: (4 marks) 1 Some work of merit, e.g. identifies 3 and stops or continues. 1 Finds 1, but fails to finish or finishes 3 incorrectly. 016.1 J.18/0_MS 46/7 Page 46 of 71 exams

Q.4 (Suggested maximum time: 10 minutes) (5) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. P 4(a) M, N and P are points on a circle with centre O. NPM 58, as shown. 58 O (i) Find OMN. NOM (58 ) 116 OMN 180 ( NOM + MNO ) but OMN MNO 1 OMN (180 NOM ) 1 OMN (180 116 ) 1 (64 ) 3 M N (5C*) Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down that the angle at the centre is equal to twice the angle at the circumference. Writes down NOM NPM or similar. Finds NOM 116, but does not attempt to find OMN. High partial credit: (4 marks) Finds NOM 64. * Deduct 1 mark off correct answer only for the omission of or incorrect use of units ( ) - apply only once per section (a), (b), (c), etc. of the question. (ii) Given that PMO 9, show that the triangle PMN is isosceles. OMN 3 PMN PMO + OMN 9 + 3 61 MNP 180 ( NPM + PMN ) 180 (58 + 61 ) 180 119 61 PMN ΔPMN is isosceles ** Accept students answers for OMN from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down that the sum of the three angles in a triangle is equal to 180. Finds PMN 61, but does not attempt to find MNP. High partial credit: (4 marks) Finds MNP 180 (58 + 61 ) or similar. Finds MNP 61, but no conclusion (i.e. MNP PMN, ΔPMN is isosceles ) or an incorrect conclusion given. 016.1 J.18/0_MS 47/7 Page 47 of 71 exams

Q.4 (cont d.) 4(b) In the diagram, [ AB ] is parallel to [ CD ]. [ AD ] and [ CB ] intersect at the point E. C D (i) Prove that triangles ABE and CED are similar. Give a reason for each of the statements that you make in your proof. In ΔABE and ΔCED, BEA CED... vertically opposite angles EAB EDC... alternate angles as [ AB ] is parallel to [ CD ] () ABE DCE... third angle in two triangles are equal if the other two angles are equal or... alternate angles as [ AB ] is parallel to [ CD ] ΔABE and ΔCED are similar A E B ** Must include a reason for all three steps for full credit. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. indicates that - vertically opposite angles are equal, - alternate angles are equal, - third angle in two triangles are equal if the other two angles are equal. One step correct with valid reason. High partial credit: (4 marks) Two steps correct with one valid reason. Three steps correct with no reasons or incorrect reasons given. (ii) Given that BEA 119 and DCE 3, find EAB. (5C*) DCE 3 ABE 3 EAB 180 ( ABE + BEA ) 180 (3 + 119 ) 180 151 9 Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down that the sum of the three angles in a triangle is equal to 180. States that EAB 180 ( ABE + BEA ) or similar. Finds ABE 3, but does not attempt to find EAB. High partial credit: (4 marks) Finds EAB 180 (119 + 3 ), but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only for the omission of or incorrect use of units ( ) - apply only once per section (a), (b), (c), etc. of question. 016.1 J.18/0_MS 48/7 Page 48 of 71 exams

Q.4 (cont d.) 4(b) (cont d.) (iii) AE 16 5, ED 55 and CD 90. Find AB. as ΔABE and ΔCED are similar AB AE CD ED AB 16 5 90 55 AB 16 5 90 55 1,485 55 7 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Any work of merit, e.g. correct relevant ratio, corresponding sides identified or indicates that corresponding sides are in proportion. Some correct substitution into correct relevant ratio with minor errors. Identifies one correct relevant ratio, AB 16 5 i.e. or. 90 55 High partial credit: (4 marks) AB 16 5 Finds or equivalent, but 90 55 fails to finish or finishes incorrectly. Finds AB 7, but no work shown. 016.1 J.18/0_MS 49/7 Page 49 of 71 exams

Q.5 (Suggested maximum time: 10 minutes) (0) Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. (5B, 5B, 10C) Diagram Given ΔABC where BAC 90. To prove BC AB + AC Construction Draw AD BC. Proof CAB BDA... both 90 ABC ABD... common angle triangles ABC and ABD are similar. BC AB 016.1 J.18/0_MS 50/7 Page 50 of 71 exams AB BD AB BC. BD Likewise, triangles ABC and ADC are similar. BC AC AC DC AC BC. DC AB + AC BC. BD + BC. DC BC.( BD + DC ) BC. BC BC or Some steps may be indicated on the diagram. Diagram Given Right-angled triangle with length of sides a, b, c, where c is the hypotenuse. To prove a + b c Construction Construct a square ABCD of side a + b Construct the point E on [AB] such that AE b (and hence EB a). Similarly construct points F, G and H on the other sides, as shown. Join E, F, G and H to divide the square ABCD into a quadrilateral and four triangles. Label the angles 1,, 3 and 4, as shown. Proof Each of the four inscribed triangles is congruent to the original triangle B D C... corresponding sides are proportional... corresponding sides are proportional... SAS Each side of the inner quadrilateral has length c 1 + 90... angle sum of triangle 1 3... corresponding parts in congruent triangles + 3 90 4 90... straight angle The inscribed quadrilateral is a square D b H a A A a c b c b G E 1 b a a 4 3 C a F b B

Q.5 (cont d.) Proof (cont d.) Area of large square 4(area of one triangle) + area of inscribed square (a + b) 4 (area of one triangle) + c (a + b) 4(½ ab) + c a + ab + b ab + c a + b c Some steps may be indicated on the diagram. Diagram/Construction Scale 5B (0, 3, 5) Partial credit: (3 marks) Right-angled triangle drawn only. Given/To prove Scale 5B (0, 3, 5) Partial credit: (3 marks) Given correct. To prove correct. Proof Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) Any correct step. High partial credit: (7 marks) One step missing or incomplete. 016.1 J.18/0_MS 51/7 Page 51 of 71 exams

Q.6 (Suggested maximum time: 10 minutes) (0) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. 6(a) Construct a right-angled triangle ABC, where: BAC 36 AB 8 cm CBA 90. (10C) C 99cm A 36 8cm B ** Allow a tolerance of ±0 cm or ±. Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) One side or one angle (36 or 90 ) correctly drawn. Sketch drawn with given measurements shown. High partial credit: (7 marks) Triangle correctly drawn, but unlabelled or incorrectly labelled. Triangle correctly drawn, but outside allowable tolerances. 6(b) On your diagram, measure the length of the side AC. Give your answer in cm, correct to one decimal place. From diagram: AC 9 9 cm (±0 cm) (5B) ** Accept students triangle from part (a) if not oversimplified. Scale 5B (0, 3, 5) Partial credit: (3 marks) Wrong side correctly measured. Triangle incorrect and unlabelled, but one side correctly measured. * No deduction applied for the omission of or incorrect use of units ( cm ). 016.1 J.18/0_MS 5/7 Page 5 of 71 exams

Q.6 (cont d.) 6(c) Using trigonometry, verify your answer to part (b) above. (5C*) cos BAC Adj Hyp AB AC cos 36 8 AC AC 8 cos36 8 0 809016... 9 888543... 9 9 cm or tan BAC Opp Adj BC AB tan 36 BC 8 BC 8 tan 36 8 0 7654... 5 81340... Hyp Opp + Adj... Pythagoras theorem AC AB + BC 8 + (5 81340...) 64 + 33 78398... 97 78398... AC 97 78398... 9 888543... 9 9 cm ** Accept students answers from parts (a) and (b) if not oversimplified. Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down trigonometric ratio (cos or tan) or states formula for Pythagoras theorem. Some correct substitution into correct trigonometric ratio (cos or tan). Some correct substitution into formula for Pythagoras theorem. Finds BC and stops (ans. 8 0 7654... or 5 81340...). High partial credit: (4 marks) Correct substitution into trigonometric ratio and correctly manipulated, e.g. AC 8. cos 36 Correct substitution into formula for Pythagoras, i.e. AC 8 + 5 81340.... 8 Finds AC, but fails to 0 809016... finish or finishes incorrectly. Finds AC 97 78398... or AC 97 78398..., but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of or incorrect use of units ( cm ). 016.1 J.18/0_MS 53/7 Page 53 of 71 exams

Q.7 (Suggested maximum time: 15 minutes) (40) 7(a) Plot the points A( 4, 1), B( 4, 3) and C( 1, 3) on the co-ordinate plane below. Join the points to form a triangle. y B 4 C 3 5 A 4 3 1 1 1 1 3 4 5 x 3 4 Scale 5C (0,, 4, 5) Low partial credit: ( marks) One or two points correctly plotted. All points reversed, i.e. (y, x), with or without triangle drawn. High partial credit: (4 marks) Three points correctly plotted, but not joined. 7(b) What type of triangle does ABC represent? Give a reason for your answer. (5B) Type of triangle right-angled triangle Reason Any 1: ABC is equal to 90 // [AB ] is perpendicular to [ BC ] // using Pythagoras theorem, AB + BC AC // etc. Type of triangle scalene triangle Reason Any 1: all three sides of triangle ABC have different lengths // triangle ABC has no two sides which are equal // etc. Scale 5B (0, 3, 5) Partial credit: (3 marks) Correct answer, but no reason or incorrect reason given. 016.1 J.18/0_MS 54/7 Page 54 of 71 exams

Q.7 (cont d.) 7(c) Find the area of triangle ABC. A( 4, 1), B( 4, 3) AB y y 1 3 1 or AB x1) + ( y 1) ( x y ( 0) + () 4 B( 4, 3), C( 1, 3) BC x x 1 4 1 4 1 3 or BC ( 4 ( 4)) + (3 1) ( 1 ( 4)) + (3 3) ( 3) + (0) 9 3 1 Area of the triangle Base Height 1 3 3 units Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct area formula from Tables. Finds correct length of one relevant side. Some correct substitution into correct area formula (not stated). High partial credit: (4 marks) Finds correct lengths of two relevant sides with some correct substitution into area formula. 1 1 Finds 3 or 4 9, but fails to finish or finishes incorrectly. * No deduction applied for the omission of or incorrect use of units ( units ). 016.1 J.18/0_MS 55/7 Page 55 of 71 exams

Q.7 (cont d.) 7(d) Draw in the image of triangle ABC under central symmetry in the origin on the co-ordinate plane above. y B C 4 3 5 A 4 3 1 1 1 1 3 4 5 A' x 3 4 C' B' ** Accept students triangle from part (a) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) One or two image points correctly plotted. All points reversed, i.e. (y, x), with or without triangle drawn. Correct image of ABC under axial symmetry in the x-axis or y-axis. High partial credit: (4 marks) Three image points correctly plotted, but not joined. Correct image of ABC under central symmetry not in the origin. 7(e) State whether triangle ABC and its image under the transformation above are similar or congruent. Give a reason for your answer. Answer congruent Reason SSS: AB A B, BC B C, AC A C // or SAS: AB A B, CBA C B A, BC B C // or RHS: CBA C B A, AC A C, AB A B or BC B C // etc. ** Accept other appropriate material. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Correct answer, but no reason or incorrect reason given. High partial credit: (4 marks) Correct answer, but reason given insufficient or incomplete. 016.1 J.18/0_MS 56/7 Page 56 of 71 exams

Q.7 (cont d.) 7(f) Is the area of the triangle ABC equal to the area of its image under the transformation above? Give a reason for your answer. Answer yes Reason Any 1: central symmetry preserves area // as ΔABC and its image are congruent, the areas of both triangle are the same // etc. ** Accept other appropriate material. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Correct answer, but no reason or incorrect reason given. High partial credit: (4 marks) Correct answer, but reason given insufficient or incomplete. 7(g) Find AC, giving your answer in surd form. A( 4, 1), C( 1, 3) AC or x1) + ( y 1) ( x y ( 1 ( 4)) + (3 1) ( 3) + () 9 + 4 13 units A( 4, 1), C( 1, 3) Using Pythagoras theorem AC AB + BC + 3 4 + 9 13 AC 13 units ** Accept students answers from part (c) if not oversimplified in method. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct distance formula from Tables. Some correct substitution into correct distance formula (not stated). High partial credit: (4 marks) Fully correct substitution into correct distance formula, but fails to finish or finishes incorrectly. Final answer given in decimal form (not surd form). * No deduction applied for the omission of or incorrect use of units ( units ). 016.1 J.18/0_MS 57/7 Page 57 of 71 exams

Q.7 (cont d.) 7(h) Sean says that cos ACB is equal to sin CAB. Is he correct? Give a reason for your answer. Answer yes Reason cos ACB Adj Hyp BC AC 3 13 sin CAB Opp Hyp BC AC 3 13 cos ACB sin CAB ** Accept students answers from parts (c) and (g) if not oversimplified. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Correct answer, but no reason or incorrect reason given. High partial credit: (4 marks) Correct answer, with one trigonometric ratio (cos ACB or sin CAB ) correctly evaluated. 016.1 J.18/0_MS 58/7 Page 58 of 71 exams

Q.8 (Suggested maximum time: 10 minutes) (0) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. The Monument of Light, alternatively known as the Spire, is a stainless steel, cone-shaped monument located in Dublin. The Spire is the world s tallest sculpture at a height of 11 m. It was officially unveiled in 003 at a total cost of 4,000,000 to Dublin City Council. 8(a) The circumference of the Spire at its base is 9 4 m. Find the radius of its base, correct to one decimal place. (5C*) Circumference πr 9 4 m πr 9 4 9 4 r π 1 496056... 1 5 m Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct circumference formula from Tables. High partial credit: (4 marks) Fully correct substitution into circumference formula, but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units ( m ) - apply only once per section (a), (b), (c), etc. of question. 8(b) Using the diagram, find l, the slant height of the Spire. Give your answer in metres, correct to two decimal places. (5C*) or Using Pythagoras theorem Hyp Opp + Adj l h + r (11 ) + (1 5) 14,689 44 + 5 14,691 69 l 14,691 69 11 0981... 11 1 m l tan Opp Adj 11 1 5 tan 1 (80 8) 89 90930... 016.1 J.18/0_MS 59/7 Page 59 of 71 exams

Q.8 (cont d.) 8(b) (cont d.) sin Opp Hyp sin 89 90930... 11 l l 11 sin89 90930... l 11 0 99993... 11 0981... 11 1 m ** Accept students answers from part (a) if not oversimplified. Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down Pythagoras theorem. Some correct substitution into Pythagoras theorem and stops or continues. Substitutes correctly into relevant trigonometric ratio, i.e. tan 11, 1 5 and stops or continues. High partial credit: (4 marks) Finds l 14,691 69 or l 14,691 69, but fails to finish or finishes incorrectly. Correct substitution into second relevant trigonometric ratio, i.e. sin 89 90930... 11 or cos 89 90930 1 5, but fails l l to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of or incorrect use of units ( m ). 8(c) Find the curved surface area of the Spire. Give your answer in m, correct to one decimal place. (5C*) Curved surface area πrl π(1 5)(11 1) 181 815π 571 188668... 571 m ** Accept students answers from parts (a) and (b) if not oversimplified. Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down correct formula for curved surface area of a cone from Tables. Some correct substitution into relevant formula (not stated) and stops or continues. High partial credit: (4 marks) Fully correct substitution into relevant formulae, but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of or incorrect use of units ( m ). 016.1 J.18/0_MS 60/7 Page 60 of 71 exams

Q.8 (cont d.) 8(d) The Spire cost the construction consortium thst were awarded the contract 3,675,41 to manufacture and erect. Find the percentage profit made by the consortium, correct to one decimal place. (5C*) Cost to DCC 4,000,000 Construction cost 3,675,41 Profit 4,000,000 3,675,41 34,579 % profit 34,579 3,675,41 100 1 0 088310... 100 8 883106... 8 9% Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds correct profit (i.e. 34,579), but does not attempt to evaluate percentage. High partial credit: (4 marks) 34,579 34,579 100 Finds or, 3,675,41 3,675,41 1 but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of or incorrect use of units ( % ). 016.1 J.18/0_MS 61/7 Page 61 of 71 exams

Q.9 (Suggested maximum time: 10 minutes) (0) The back-to-back stem-and-leaf plot below shows the systolic blood pressure of thirteen people before and two hours after taking a particular drug. Before After 10 9 11 1 6 8 6 1 5 7 4 13 1 7 9 7 0 14 5 6 8 6 4 3 15 9 7 5 16 0 8 17 0 18 1 Key: 15 9 means 159 mm Hg 9(a) Find the median blood pressure both before and after taking the drug. (5B) Median (before) 16, 18, 134, 140, 147, 149, 153, 154, 156, 165, 167, 178, 180 Median 153 mm Hg Median (after) 109, 111, 116, 15, 17, 131, 137, 145, 146, 148, 159, 160, 181 Median 137 mm Hg Scale 5B (0, 3, 5) Partial credit: (3 marks) One correct answer. * Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect use of units ( mm Hg ). 9(b) Find the range of blood pressures both before and after taking the drug. (5B) Range (before) 180 16 54 mm Hg Range (after) 181 109 7 mm Hg Scale 5B (0, 3, 5) Partial credit: (3 marks) One correct answer. * Accept correct answers with or without work shown. * No deduction applied for the omission of or incorrect use of units ( mm Hg ). 016.1 J.18/0_MS 6/7 Page 6 of 71 exams

Q.9 (cont d.) 9(c) What other measure of variability (spread) could have been used when examining this data? (5A) Answer Any 1: interquartile range // standard deviation Scale 5A (0, 5) Hit or miss. 9(d) Compare the blood pressure results both before and after taking the drug. Refer to at least one measure of central tendency and at least one measure of variability (spread) in your answer. Measure of central tendency systolic blood pressures are lower two hours after taking the drug Any 1: the median systolic blood pressure before taking the drug is 153 mm Hg, while median systolic blood pressure after taking the drug is 137 mm Hg // the mean systolic blood pressure before taking the drug is 15 mm Hg, while median systolic blood pressure after taking the drug is 138 mm Hg // etc. ** Accept other appropriate material. Measure of variability (spread) the spread of systolic blood pressure is greater two hours after taking the drug Any 1: the range of systolic blood pressure before taking the drug is 54 mm Hg, while range of systolic blood pressure after taking the drug is 7 mm Hg // the interquartile range of systolic blood pressure before taking the drug is 9 mm Hg, whereas the interquartile range of systolic blood pressure after taking the drug is 33 mm Hg // etc. ** Accept other appropriate material. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Mention of mean, mode, median, range or interquartile range in answer. High partial credit: (4 marks) Comparison using measure of central tendency or measure of variability only. * No deduction applied for the omission of or incorrect use of units ( mm Hg ). 016.1 J.18/0_MS 63/7 Page 63 of 71 exams

Q.10 (Suggested maximum time: 10 minutes) (0) ** Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded or for the omission of or incorrect use of units in part(s) of question asterisked - this deduction should be applied only once per section (a), (b), (c), etc. of the question. In 1975, an American daredevil known as Evel Knievel attempted to jump over thirteen single-deck buses on his motorcycle in front of 90,000 people at Wembley Stadium in London. To complete the jump, a launch ramp 3 5 m in height with a base 10 4 m in length was constructed, as shown below. 10 4 m 3 5 m 10(a) For Evel not to undershoot or overshoot the jump, he required the angle of elevation of the launch ramp to be between 15 and 0. Would you conclude that the ramp met this criteria? Use calculations to justify your answer. Answer yes Reason tan X Opp Adj 3 5 10 4 0 336538... X tan 1 (0 336538...) 18 600065... 18 6 ramp meets the criteria as 15 X 0 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Any work of merit, e.g. writes down correct trigonometric ratio (tan). Some correct substitution into correct trigonometric ratio (sin or cos). High partial credit: (4 marks) Correct substitution into trigonometric ratio with some manipulation, e.g. finds X tan 1 3 5 10 4 or tan 1 (0 336538...), but fails to finish or finishes incorrectly. Trigonometric ratio inverted, but finishes correctly [ans. X tan 1 10 4 71 4 ]. 3 5 Finds X 18 6, but no conclusion (i.e. meets criteria as 15 X 0 ) or an incorrect conclusion given. 016.1 J.18/0_MS 64/7 Page 64 of 71 exams

Q.10 (cont d.) 10(b) Find the length of the ramp. Give your answer in metres, correct to two decimal places. (5C*) Using Pythagoras theorem Hyp Opp + Adj l 3 5 + 10 4 1 5 + 108 16 10 41 l 10 41 10 973149... 10 98 m or sin Opp Hyp sin 18 6 3 5 l l 3 5 sin18 6 l 3 5 0 318959... 10 973186... 10 98 m or cos Adj Hyp cos 18 6 10 4 l l 10 4 cos18 6 l 10 4 0 947768... 10 973144... 10 98 m Scale 5C* (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down trigonometric ratio (sin or cos) or states formula for Pythagoras theorem. Some correct substitution into correct trigonometric ratio (sin or cos). Some correct substitution into formula for Pythagoras theorem. High partial credit: (4 marks) Correct substitution into formula for Pythagoras, i.e. l 3 5 + 10 4. Finds l 10 41 or l 10 41, but fails to finish or finishes incorrectly. Correct substitution into trigonometric ratio and correctly manipulated, e.g. l 3 5 or l 10 4. sin18 6 cos18 6 3 5 10 4 Finds l or l, 0 318959... 0 947768... but fails to finish or finishes incorrectly. * Deduct 1 mark off correct answer only if final answer is not rounded or is incorrectly rounded - apply only once per section (a), (b), (c), etc. of question. * No deduction applied for the omission of or incorrect use of units ( m ). 016.1 J.18/0_MS 65/7 Page 65 of 71 exams

Q.10 (cont d.) 10(c) The width of each bus was 5 m. Evel estimated that he needed to travel at an average speed of 90 km/h in mid-air to complete the jump. Find the minimum time he needed to travel in mid-air to complete the jump successfully. Give your answer in seconds. (10C) Length of mid-air jump 13 5 3 5 m Average Speed 90 km/h 90 1,000 m 60 60 s 90,000 3,600 5 m/s Distance Average speed Time Distance Time Average Speed 3 5 5 1 3 seconds Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) Some work of merit, e.g. writes down correct speed formula. Finds correct length of mid-air jump i.e. 13 5 or 3 5 m. Finds average speed required in mid-air i.e. 90 1,000 m or 90,000 or 5 m/s. 60 60 s 3,600 High partial credit: (7 marks) Substitutes correct length of mid-air jump and/or average mid-air speed into speed formula, but fails to finish or finishes incorrectly. * No deduction applied for the omission of or incorrect use of units ( seconds ). 016.1 J.18/0_MS 66/7 Page 66 of 71 exams

Q.11 (Suggested maximum time: 15 minutes) (30) The customers in a restaurant were surveyed to find out how long they each had to wait for a table on a particular night. The results are shown in the grouped frequency table below. Time (minutes) 0 5 5 10 10 15 15 0 0 5 Frequency 0 11 9 7 3 Note: 10 15 means at least 10 minutes but less than 15 minutes, etc. 11(a) What type of data was collected in the survey? Put a tick () in the correct box below. Explain your answer. Type of data Numerical Numerical Categorical Categorical Discrete Continuous Nominal Ordinal Explanation numerical: any data that can be represented by numbers // etc. continuous: can take any value between two points, e.g. height, weight, time, speed, temperature // etc. ** Accept other appropriate material. Scale 5C (0,, 4, 5) Low partial credit: ( marks) Correct answer, but no explanation or incorrect explanation given. High partial credit: (4 marks) Correct answer, but explanation given insufficient or incomplete. 11(b) Using mid-interval values, estimate the mean waiting time for a table on that particular night. ( 5 0) + (7 5 11) + (1 5 9) + (17 5 7) + ( 5 3) Mean 0 + 11 + 9 + 7 + 3 50 + 8 5 + 11 5 + 1 5 + 67 5 50 435 50 8 7 Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) Some work of merit, e.g. shows indication of division by 50. Finds one correct mid-interval value. High partial credit: (7 marks) Finds correct numerator, i.e. 435 or ( 5 0) + (7 5 11) + (1 5 9) + (17 5 7) + ( 5 3). Finds incorrect mean using consistent incorrect mid-interval values. 435 Finds, but fails to finish or finishes 50 incorrectly. (10C) 016.1 J.18/0_MS 67/7 Page 67 of 71 exams

Q.11 (cont d.) 11(c) Display the above data on a histogram. (10D) 0 15 Frequency 10 5 0 0 5 10 15 0 5 Time (minutes) Scale 10D (0, 4, 6, 8, 10) Low partial credit: (4 marks) Some work of merit, e.g. draws correctly labelled axes (with no bars inserted). One or two bars correct (each of consistent width). Middle partial credit: (6 marks) Three or four bars correct (each of consistent width). Bar chart correctly drawn, but both axes not numerated or labelled. High partial credit: (8 marks) Five bars correct (each of consistent width), but both axes not labelled. Bar chart correctly drawn. 11(d) What percentage of customers were waiting 15 minutes or longer for a table? # people waiting 15 minutes or longer 7 + 3 10 %people 10 100 50 1 0 100 0% Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. finds correct # people (i.e. 10), but does not attempt to evaluate percentage. High partial credit: (4 marks) 10 10 100 Finds or, but fails to finish 50 50 1 or finishes incorrectly. * No deduction applied for the omission of or incorrect use of percentage symbol ( % ). 016.1 J.18/0_MS 68/7 Page 68 of 71 exams

Q.1 (Suggested maximum time: 10 minutes) (0) 1(a) The equation of the line k is x + y 1 0. Find the value of t such that (t, 3t) is a point on the line k. k: x + y 1 0 (t, 3t) k (t) + ( 3t) 1 0 4t 3t 1 0 t 1 0 t 1 Scale 5C (0,, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down (t, 3t) k or similar. Some correct substitution into equation, i.e. (t) or ( 3t) or inverts x and y values in equation, i.e. ( 3t) +(t) 1 0 and stops or continues. High partial credit: (4 marks) Correct substitution into line equation, but fails to finish or finishes incorrectly. 1(b) The line l passes through the point (3, ) and is perpendicular to the line k. Find the equation of the line l in the form ax + by + c 0. (5D) k: x + y 1 0 y x +1 m k m l m k 1 m l 1, point (3, ) y y 1 m(x x 1 ) y 1 (x 3) y 4 x 3 x y 3 + 4 0 x y + 1 0 Scale 5D (0,, 3, 4, 5) Low partial credit: ( marks) Some work of merit, e.g. writes down k in the form y x +1 and stops. Finds correct slope for k, i.e.. Middle partial credit: (3 marks) Finds correct slope for l, i.e. 1, and stops. Some correct substitution into a line formula. High partial credit: (4 marks) Correct slope with one incorrect substitution into line formula, but continues correctly. Correct slope, but both x and y reversed in substitution, but continues correctly. Correct substitution into line formula, but fails to finish or finishes incorrectly. Final answer correct, but not in the form ax + by + c 0. 016.1 J.18/0_MS 69/7 Page 69 of 71 exams

Q.1 (cont d.) 1(c) Find the co-ordinates of the point of intersection of lines l and k. (10C) k: x + y 1 ( ) l: x y 1 ( 1) 4x + y x y 1 5x 1 1 x 5 x + y 1 or x y 1 y 1 x y 1 x 1 1 y 1 ( ) y (1 + x) 5 1 1 1 (1 + ) 5 5 3 3 5 5 1 3 l k (, ) 5 5 or x + y 1 ( 1) x y 1 ( ) x + y 1 x + 4y 5y 3 3 y 5 x + y 1 x 1 y x 1 (1 y) 1 3 (1 ) 5 1 5 l k 1 3 (, ) 5 5 or x y 1 x 1 + y 3 1 + ( ) 5 6 1 + 5 1 5 ** Accept students answers from part (b) if not oversimplified. Scale 10C (0, 4, 7, 10) Low partial credit: (4 marks) Any work of merit, e.g. multiplying equation by appropriate constant to facilitate cancellation of x or y term. Finds 5x 1 or 5y 3, but fails to finish or finishes incorrectly. Finds either variable (x or y) correctly by trial and error, but fails to verify in both equations or verifies incorrectly. High partial credit: (7 marks) Finds first variable (x or y) correctly, but fails to find second variable or finds incorrectly. Finds both variables (x and y) correctly, but no work shown. Finds both variables (x and y) by graphical means. Finds both variables (x and y) by trial and error, but does not verify into both equations. 016.1 J.18/0_MS 70/7 Page 70 of 71 exams

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