GCSE (9-1) MATHEMATICS J560/06 Paper 6 (Higher Tier) PRACTICE PAPER (SET 3) MARK SCHEME

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GCSE (9-1) MATHEMATICS J560/06 Paper 6 (Higher Tier) PRACTICE PAPER (SET 3) MARK SCHEME H Date – Morning/Afternoon GCSE (9-1) MATHEMATICS J560/06 Paper 6 (Higher Tier) PRACTICE PAPER (SET 3... ) MARK SCHEME Duration: 1 hour 30 minutes MAXIMUM MARK 100 Final This document consists of 15 pagesJ560/06 Mark Scheme Practice Paper (Set 3) 2 Subject-Specific Marking Instructions 1. M marks are for using a correct method and are not lost for purely numerical errors. A marks are for an accurate answer and depend on preceding M (method) marks. Therefore M0 A1 cannot be awarded. B marks are independent of M (method) marks and are for a correct final answer, a partially correct answer, or a correct intermediate stage. SC marks are for special cases that are worthy of some credit. 2. Unless the answer and marks columns of the mark scheme specify M and A marks etc, or the mark scheme is ‘banded’, then if the correct answer is clearly given and is not from wrong working full marks should be awarded. Do not award the marks if the answer was obtained from an incorrect method, i.e. incorrect working is seen and the correct answer clearly follows from it. 3. Where follow through (FT) is indicated in the mark scheme, marks can be awarded where the candidate’s work follows correctly from a previous answer whether or not it was correct. Figures or expressions that are being followed through are sometimes encompassed by single quotation marks after the word their for clarity, e.g. FT 180 × (their ‘37’ + 16), or FT 300 – √(their ‘52 + 72’). Answers to part questions which are being followed through are indicated by e.g. FT 3 × their (a). For questions with FT available you must ensure that you refer back to the relevant previous answer. You may find it easier to mark these questions candidate by candidate rather than question by question. 4. Where dependent (dep) marks are indicated in the mark scheme, you must check that the candidate has met all the criteria specified for the mark to be awarded. 5. The following abbreviations are commonly found in GCSE Mathematics mark schemes. - figs 237, for example, means any answer with only these digits. You should ignore leading or trailing zeros and any decimal point e.g. 237000, 2.37, 2.370, 0.00237 would be acceptable but 23070 or 2374 would not. - isw means ignore subsequent working after correct answer obtained and applies as a default. - nfww means not from wrong working. - oe means or equivalent. - rot means rounded or truncated. - seen means that you should award the mark if that number/expression is seen anywhere in the answer space, including the answer line, even if it is not in the method leading to the final answer. - soi means seen or implied.J560/06 Mark Scheme Practice Paper (Set 3) 3 6. In questions with no final answer line, make no deductions for wrong work after an acceptable answer (i.e. isw) unless the mark scheme says otherwise, indicated by the instruction ‘mark final answer’. 7. In questions with a final answer line following working space: (i) If the correct answer is seen in the body of working and the answer given on the answer line is a clear transcription error allow full marks unless the mark scheme says ‘mark final answer’. Place the annotation  next to the correct answer. (ii) If the correct answer is seen in the body of working but the answer line is blank, allow full marks. Place the annotation  next to the correct answer. (iii) If the correct answer is seen in the body of working but a completely different answer is seen on the answer line, then accuracy marks for the answer are lost. Method marks could still be awarded. Use the M0, M1, M2 annotations as appropriate and place the annotation  next to the wrong answer. 8. In questions with a final answer line: (i) If one answer is provided on the answer line, mark the method that leads to that answer. (ii) If more than one answer is provided on the answer line and there is a single method provided, award method marks only. (iii) If more than one answer is provided on the answer line and there is more than one method provided, award zero marks for the question unless the candidate has clearly indicated which method is to be marked. 9. In questions with no final answer line: (i) If a single response is provided, mark as usual. (ii) If more than one response is provided, award zero marks for the question unless the candidate has clearly indicated which response is to be marked. 10. When the data of a question is consistently misread in such a way as not to alter the nature or difficulty of the question, please follow the candidate’s work and allow follow through for A and B marks. Deduct 1 mark from any A or B marks earned and record this by using the MR annotation. M marks are not deducted for misreads.J560/06 Mark Scheme Practice Paper (Set 3) 4 11. Unless the question asks for an answer to a specific degree of accuracy, always mark at the greatest number of significant figures even if this is rounded or truncated on the answer line. For example, an answer in the mark scheme is 15.75, which is seen in the working. The candidate then rounds or truncates this to 15.8, 15 or 16 on the answer line. Allow full marks for the 15.75. 12. Ranges of answers given in the mark scheme are always inclusive. 13. For methods not provided for in the mark scheme give as far as possible equivalent marks for equivalent work. If in doubt, consult your Team Leader. 14. Anything in the mark scheme which is in square brackets […] is not required for the mark to be earned, but if present it must be correct.J560/06 Mark Scheme Practice Paper (Set 3) 5 Question Answer Marks Part marks and guidance 1 (a) 2250 1 1 AO2.1a Allow answers in range 2175 to 2325 (b) (i) 12 750 3 1 AO1.3b 1 AO2.1a 1 AO3.1a M2 for 17 × k oe where 725 ≤ k ≤ 775 Or M1 for 1 [tonne] = [£]750 soi or 17 × their ‘k’ oe where k is their weight per tonne Allow answers in range 12 325 to 13 175 May be in parts read from the graph, e.g. 4 [tonnes] = [£]3000 1 [tonne] = [£]750 17 [tonnes] = 4 × 3000 + [1 ×] 750 Attempt at unitary method Accept 725 – 775 (ii) The unit cost is the same for all amounts bought 1 1 AO3.4b See Appendix for further examples 2 (a) 0.45 oe 2 1 AO1.3a 1 AO2.1a M1 for 1 – (0.3 + 0.25) If 0 scored, SC1 for 0.72 (b) (i) 9 2 1 AO1.3a 1 AO2.1b M1 for 0.3 × 30 oe (ii) Correct explanation 1 1 AO3.4b e.g. May need to play more times or she may not be very good at this game 3 (a) Position correctly shown on map 3 1 AO1.3a 1 AO2.3a 1 AO2.3b M1 for bearing 130 to 134 drawn at mast Y M1 for bearing 250 to 254 drawn at mast Z Ignore other constructions for M marks For full marks, position of ‘mast X’ must be unambiguously markedJ560/06 Mark Scheme Practice Paper (Set 3) 6 Question Answer Marks Part marks and guidance (b) (i) 1 250 000 2 2 AO1.3a B1 for [2 :] 2 500 000 (ii) 122 to 125 2 1 AO1.3a 1 AO2.3a M1 for 12.5 × their measurement between 9.8 and 10 oe Method may be seen in stages e.g. Measurement = 9.9 9.9 ÷ 2 = 4.95 4.95 × 25 = 123.75 4 11 100 to 11 200 3 3 AO1.3a B2 for 124 400 000 to 124 450 000 seen Or M1 for 11 24 6 2 6.67 10 5.97 10 6.4 10 − × × × × × oe Or B1 for figs 124… seen May be in standard form: 1.244 × 108 to 1.2445 × 108 Exact answer: 11 155.14175 5 (a) Crompton, Harwood, Astley nfww 4 2 AO1.3b 1 AO3.1c 1 AO3.3 M3 for converting all three into comparable forms Or M2 for converting two into comparable forms Or M1 for correctly manipulating one into an equivalent alternative form Condone C, H, A as abbreviations for school names e.g. Working in decimals: H = 0.428… or 0.43 A = 0.44[4…] C = 0.42 or working in fractions: H = 1350 3150 , A= 3150 1400 , C = 1323 3150 See Appendix for further examples (b) If Beechfield has more pupils than Kenwood the number of girls may be more than at Kenwood 1 1 AO2.5b Comments must indicate that the pie charts show proportions, not quantityJ560/06 Mark Scheme Practice Paper (Set 3) 7 Question Answer Marks Part marks and guidance 6 (a) 1400 3 1 AO1.3a 1 AO2.3a 1 AO3.1c M2 for 24 3500 60 × oe Or B1 for 24 60 oe or 60 36 oe If 0 scored, SC1 for 2100 as final answer (b) Different age groups may not have the same opinion 1 1 AO3.5 7 ( ) ( 1) ( 2) ( 3) ( 4) 5 x x x x x + + + + + + + + = 5 10 5 x + = x + 2, which is the median 4 2 AO2.4a 1 AO3.1a 1 AO3.2 M1 for x, x + 1, x + 2, x + 3, x + 4 seen M1 for (x) + (x + 1) + (x + 2) + (x + 3) + (x + 4) M1 for their (5x + 10) ÷ 5 If 0 scored, allow SC2 for a numerical example of any 5 consecutive numbers with mean clearly evaluated, and median identified as the same value Or SC1 for a numerical example of any 5 consecutive numbers with mean clearly identified, or median clearly identified, or both identified with no conclusion Or equivalent algebraic representation of 5 consecutive numbers Alternative (non-algebraic) arguments also accepted for full marks e.g. M1 for “The numbers are the first number, 1 more, 2 more, 3 more and 4 more.” M1 for “So the mean of the differences is (1 + 2 + 3 + 4) ÷ 5 = 2.” M1 for “So the mean is the first number plus 2.” And the final mark for concluding this with “Which is the median”.J560/06 Mark Scheme Practice Paper (Set 3) 8 Question Answer Marks Part marks and guidance 8 (a) 3 1 AO2.1a 2 AO2.3b B1 for 18 correctly placed AND B2 for 13, 7 and 22 correctly placed Or M1 for 60 – (18 + their ‘13’ + their ‘22’) in overlap Their 7 must be ≥ 1 (b) 42 20 oe 2FT 1 AO2.1b 1 AO2.3a M1 for denominator 42 seen, FT from their ‘13’ + their ‘7’ + their ‘22’ FT from their Venn diagram Look for 10 21 or 0.476[190…] Condone 0.47 to 0.48 9 (a) k 5 1 1 AO1.3a (b) 9 12m2 2 2 AO1.3a B1 for 2 9 m or 1 4 m 2 or m4.5 Allow 1 4 12m 2 or 12m4.5 (c) 25p17 3 3 AO1.3b B2 for p17 seen Or B1 for 25 or p14 seen 10 (a) 1 12 30 12 6 18 24 ( ) 2 × × + − × = + = k k k k k 3 1 AO2.1a 1 AO2.2 1 AO2.3b M1 for 1 12 2 × × k AND M1 for (30 – 12) × k Or M2 for 1 2 × k × (18 + 30) Condone missing × signs 13 7 22 18J560/06 Mark Scheme Practice Paper (Set 3) 9 Question Answer Marks Part marks and guidance (b) 17.1 3 2 AO1.3a 1 AO2.1a M2 for [k =] 410 24 oe soi by 17.0[83…] Or M1 for 24k = 410 (c) (i) 0.272 3 1 AO1.3b 1 AO2.1a 1 AO3.1c M2 for 2 410 13 25 1 25 2 − × × oe Or M1 for 410 13 25 25 1 2 2 = × + × × a May be done in stages Substitutes numbers correctly into formula (or their attempt at a rearranged formula) (ii) 21 nfww 5 1 AO1.3b 3 AO3.1d 1 AO3.3 M1 for 410 15 0.4 1 2 2 = + × t t oe AND M2 for [t =] 15 15 4 0.2 410 2 2 0.2 − ± − × × − × Or M1 for [t =] 15 15 4 0.2 410 2 2 0.2 − ± − × × − × with at most 1 sign error AND A1 for [-96.2 to -96.3 or -96 and] 21.2 to 21.3 If no relevant working shown, SC3 for -96 and 21 as final answer Or SC2 for -96.2 to -96.3 and 21.2 to 21.3 as final answer oe includes 410 15 0.2 = + t t 2 or 2050 75 = + t t 2 M2 or M1 are FT from their 3 term quadratic Condone ‘short’ division line in working if seen correct at least once Would earn M1 only if only “+” or “-“ used instead of “±” Maximum 4 marks if unrounded and/or negative solution not rejectedJ560/06 Mark Scheme Practice Paper (Set 3) 10 Question Answer Marks Part marks and guidance 11 (a) 24x + 64 or 8(3x + 8) 2 2 AO1.3a M1 for 3x + 6 seen (b) -2 3 1 AO1.3a 2 AO3.1a M1 for 3‘x’ + 6 = 8(‘x’ + 2) oe M1 for 5‘x’ = -10 Form equation Begin to solve for ‘x’ 12 π 50 5 5 2.5 2.5 2 2 2.5 360 + + + + × × × × = 19.36[…] = 19.4 [m] 4 2 AO1.3b 1 AO2.2 1 AO2.3a B1 for 50 or 100 seen AND M2 for [2 ] 50 2 2.5 π 360 their × × × × Or M1 for 2 2.5 × × π Also allow 100 2 2.5 π 360 their × × × 13 26.3 to 26.4 6 2 AO1.3a 3 AO3.1b 1 AO3.2 M5 for [ ] π 2 1 2 sin124 2 100 r r r     × × × ×   × OR M1 for 180 – 56 oe AND M2 for 2 1 sin '124' 2     × × × × r r their   Or M1 for 1 sin '124' 2 × × × r r their AND M1 for their triangle area ÷ πr 2 [× 100] May see in stages May see evaluated for a particular radius. Award M marks for method seen with r = their consistent radiusJ560/06 Mark Scheme Practice Paper (Set 3) 11 Question Answer Marks Part marks and guidance 14 (a) “Sides =” 33, 56, 65 33² + 56² = 4225 4225 65 = 3 1 AO1.3a 1 AO2.4a 1 AO3.1a B1 for 56 and 65 seen M1 for 33² + 56² Alternative method M1 for (x x x + + = − 23 2 1 )2 2 2 ( ) M1 for x x 2 − − = 25 264 0 or equivalent quadratic = 0 A1 for x = 33 [and x ≠ -8] (b) 2x – 1 = x + 23 x = 24 2x – 1 = x x = 1 which does not give a triangle x = x + 23 which has no solution M1 B1 M1 A1 M1 A1 1 AO1.3a 3 AO3.1a 2 AO3.4b Could be in any order x = 1 must be rejected Needs to be explained 15 322.26 4 3 AO1.3b 1 AO3.3 B3 for 2722.26 to 2722.263 OR M2 for 2400 × 1.0324 OR M1 for 2476.8[0] or 1.032 soi M1 for their 2476.8 × 1.032 and their 2556.05 to 2556.06 × 1.032 and their 2637.85 to 2637.86 × 1.032 Steps towards first year calculated Further 3 years calculatedJ560/06 Mark Scheme Practice Paper (Set 3) 12 Question Answer Marks Part marks and guidance 16 (a) 5 815 = 13 8 = 3 8 = 2 2 2 AO2.2 M1 for 5 815 or 13 8 (b) 1 32 3 1 AO1.3b 1 AO2.2 1 AO3.1b B2 for 48 3 or equivalent fractional power Or M1 for 34 or ( ) 1 3 3 3 × 8 or 1 1 27 3 8 8 × 17 BP = PC (P midpoint of BC) Angle MBP = angle NPC (corresponding angles) Angle BPM = angle PCN (corresponding angles) Triangles [MBP and NPC] congruent by ASA. 4 1 AO1.1 3 AO2.4b B3 for three facts with conclusion Or B2 for three facts with missing or unclear conclusion or for two facts with conclusion Or B1 for one fact Accept any correct proof Each fact must be backed up with a reason for full marks 18 (a) 4 440 000 or 4 441 000 to 4 441 100 2 2 AO1.3a M1 for 30 000 × 2.36 Allow 4 400 000 provided correct working seenJ560/06 Mark Scheme Practice Paper (Set 3) 13 Question Answer Marks Part marks and guidance (b) 26 or 25.9[9…] 3 1 AO1.3a 2 AO3.1a M2 for ( 3 2 1 100 − × ) oe Or B1 for 3 2 or 1.2599… Or M1 for 3 100 2 100   + k   =   oe Alternative method using trial and improvement M2 for working towards calculating overall percentage increase over 3 years for both 25% and 26% e.g. As a minimum: 1.253 = 1.95[3…] and 1.263 = 2.000[…] Or M1 for working towards calculating a percentage increase over 3 years where k is between 20 and 30 e.g. 1.253 = 1.95[3…] Allow method marks for working in decimals, percentages or from their initial population k After 3 years k After 3 years 20 72.8 26 100.03… 21 77.15… 27 104.83… 22 81.58… 28 109.71… 23 86.08… 29 114.66… 24 90.66… 30 119.7 25 95.31…J560/06 Mark Scheme Practice Paper (Set 3) 14 APPENDIX Exemplar responses for Q1(b)(ii) Response Mark There are no discounts for buying more 1 There is enough aluminium 0 Exemplar responses for Q5(a) Working in fractions: 1 mark 1 mark Still only 1 mark 2 marks 2 marks 2 marks 3 marks H = 3 7 27 63 150 350 1350 3150 A 4 9 49 28 63 200 450 1400 3150 C 21 50 21 50 189 450 147 350 1323 3150 Working in ratios: 1 mark 1 mark Still only 1 mark 2 marks 2 marks 2 marks 2 marks 2 marks 2 marks 3 marks 3 marks H 3 : 4 3 : 4 12 : 16 21 : 28 15 : 20 87 : 116 84 : 112 435 : 580 A = 4 : 5 12 : 15 84 : 105 16 : 20 116 : 145 84 : 105 464 : 580 C 21 : 29 21 : 29 21 : 29 84 : 116 84 : 116 105 : 145 84 : 116 420 : 580J560/06 Mark Scheme 15 Assessment Objectives (AO) Grid Question AO1 AO2 AO3 Total 1(a) 0 1 0 1 1(b)(i) 1 1 1 3 1(b)(ii) 0 0 1 1 2(a) 1 1 0 2 2(b)(i) 1 1 0 2 2(b)(ii) 0 0 1 1 3(a) 1 2 0 3 3(b)(i) 2 0 0 2 3(b)(ii) 1 1 0 2 4 3 0 0 3 5(a) 2 0 2 4 5(b) 0 1 0 1 6(a) 1 1 1 3 6(b) 0 0 1 1 7 0 2 2 4 8(a) 0 3 0 3 8(b) 0 2 0 2 9(a) 1 0 0 1 9(b) 2 0 0 2 9(c) 3 0 0 3 10(a) 0 3 0 3 10(b) 2 1 0 3 10(c)(i) 1 1 1 3 10(c)(ii) 1 0 4 5 11(a) 2 0 0 2 11(b) 1 0 2 3 12 2 2 0 4 13 2 0 4 6 14(a) 1 1 1 3 14(b) 1 0 5 6 15 3 0 1 4 16(a) 0 2 0 2 16(b) 1 1 1 3 17 1 3 0 4 18(a) 2 0 0 2 18(b) 1 0 2 3 [Show More]

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