GCSE (9-1) Mathematics J560/05 Paper 5 (Higher Tier) PRACTICE PAPER (SET 3) MARK SCHEME

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GCSE (9-1) Mathematics J560/05 Paper 5 (Higher Tier) PRACTICE PAPER (SET 3) MARK SCHEME H Date – Morning/Afternoon GCSE (9-1) Mathematics J560/05 Paper 5 (Higher Tier) PRACTICE PAPER (SET 3... ) MARK SCHEME Duration: 1 hour 30 minutes MAXIMUM MARK 100 Final This document consists of 14 pagesJ560/05 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.J560/05 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.J560/05 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.J560/05 Mark Scheme Practice Paper (Set 3) 5 Question Answer Marks Part marks and guidance 1 No [correlation] Strong, negative [correlation] 1 2 1 AO2.1a 1 AO2.3a 1 AO2.3b Accept none, zero B1 for negative Accept good Do not accept ‘nothing’ 2 (a) 600 2 2 AO1.3b M1 for 2000 × 5 100 [× 6] oe soi (b) 300 3 3 AO1.3b M2 for 360 ÷ 100 20 100   +     oe Or M1 for 360 associated with (100 + 20)[%] seen 3 7 3 1 AO1.3b 2 AO3.1c 2 AO3.3 B2 for 20 3 oe isw Or M1 for 10 2 3 × Implied by answer 6 4 (a) 48 2 2 AO1.3b M1 for 160 ÷ (2 + 5 + 3) [× 3] oe (b) (i) She has calculated Rebecca’s share as a percentage of her share oe 1 1 AO3.4a Accept the fraction is upside down oe (ii) 5 3 or 23 66 to 67% 1 1 1 AO1.3a 1 AO2.5a 5 (a) Constructs angle bisector of angle ABC with two pairs of correct arcs 2 1 AO2.3a 1 AO2.3b B1 for correct bisector with no/incorrect arcs Use transparency for accuracy (± 2°)J560/05 Mark Scheme Practice Paper (Set 3) 6 Question Answer Marks Part marks and guidance (b) Arc or point on angle bisector 3 cm from A inside the triangle 1 1 AO2.3b Accept arc/point in range 2.8 cm to 3.2 cm 6 (a) 1.5 × 108 1 1 AO1.3a (b) 300 000 seen 150 000 000 ÷ their 300 000 oe 500 seconds Conclusion with comparison of 20 minutes to 500 seconds B1 M1 A1FT A1 2 AO1.3b 1 AO3.1d 1 AO3.4b Condone their (a) used here FT their (a) ÷ their 300 000 Must convert correctly or approximately to seconds or minutes to compare Accept correct full values used Condone their (a) rounded to 1 sf for M1 and A1FT e.g. 20 minutes = 1200 seconds, 500 seconds is between 8 and 9 minutes 7 2.5 3 2 AO1.3b 1 AO2.3a M2 for 7.5 ÷ (16.5 ÷ 5.5) oe Or M1 for 7.5 5.5 16.5 h = oe Or B1 for 3 or 1 3 oe seen Condone 2.5 cm 8 (a) 2 2 AO2.1a Allow rotations of multiples of 90° Allow vertical and horizontal reflections B1 for one error or addition Condone interior lines not shown (b) 50 2 1 AO2.1a 1 AO3.1a M1 for 4 × 4 × 4 oe soi 9 (a) x = 2 5y oe final answer 2 AO21.3a M1 for correct first step soiJ560/05 Mark Scheme Practice Paper (Set 3) 7 Question Answer Marks Part marks and guidance (b) 2 19 or 1 9 2 or 9.5 3 AO31.3b M Or2M1 for 5 for 5 x –x3x – = 13 + 6 3x = k oroe mx = 13 + 6 10 Pen £2.50 Notebook £4 5 1 AO1.2 1 AO2.3b 2 AO3.1d 1 AO3.3 M2 for both equations correct Or M1 for 5p + 8n = 44.50 or 10p + 3n = 37 AND M1 for scaling one/both equations M1 for correct method to eliminate 1 variable, allow 1 arithmetic error For method marks, condone use of 4450 and 3700 and use of any consistent variables Answers 250 and 400 imply M4 11 8 : 3 nfww 5 1 AO1.1 1 AO1.3b 2 AO3.1b 1 AO3.2 B2 for CD = 8 cm Or M1 for CD2 + 62 = 102 oe AND B2 for AC = 16 Or M1 for sin 30 = CD AC their oe Or B1 for sin 30 = 0.5 oe Could be on diagram 12 (a) 1 8 2 1 AO1.3a 1 AO2.3a M1 for 20 and 160 (b) He should be using 150 not 160 oe 1 1 AO3.4a Accept answer 37.5 as evidenceJ560/05 Mark Scheme Practice Paper (Set 3) 8 Question Answer Marks Part marks and guidance (c) Tangent at 11am drawn [-]50 to [-]36 Conclusion e.g. estimate is reasonable B1 B2dep B1dep 2 AO2.1b 1 AO3.1d 1 AO3.4b No daylight at 11am Dependent on tangent mark awarded Allow integer/integer if in range Or M1 for rise/run also dependent on tangent drawn or close attempt at tangent. Must see correct or implied calculation from a drawn tangent Dependent on at least B2 earned Look at the value first and check one unit horizontally for their tangent. Absolute value of gradient must be within 4 of your value. If no value then check working – must be correct Accept estimate is unreasonable depending on their gradient and dependent on B2 earned 13 (a) Sweets are replaced oe 1 1 AO3.5 (b) 5 4 3 3 2 1 10 9 8 10 9 8 × × + × × oe 66 720 = 11 120 or shows correct cancelling leading to 11 120 M4 A1 3 AO2.4a 1 AO3.1d 1 AO3.3 M3 for 5 4 3 10 9 8 × × or 3 2 1 10 9 8 × × Or M2 for 5 4 , 10 9 and 3 8 OR 3 2 , 10 9 and 18 seen Or M1 for RRR and BBB identified in tree diagram or elsewhere Dependent on M4 and no errors seen For M4 condone 2 1 0 10 9 8 × × in addition For M3 and M2 isw 14 (a) 0.63   2 2 AO1.3a M1 for 0.63... or 7 ÷ 11 shown in workingJ560/05 Mark Scheme Practice Paper (Set 3) 9 Question Answer Marks Part marks and guidance (b) 11 30 3 3 AO1.3b B2 for 33 90 Or M1 for 3.66... and 36.66... seen or answer 90 k Allow other correct values to equate decimals for M1 e.g. 0.366… and 3.66…. 15 (a) He could be correct with reference to not knowing the maximum or minimum values for the time so the range could lie between 20 and 50 oe 1 1 AO3.4b The maximum could be less than 50 minutes The exact data is not given for times on the histogram (b) 37 3 1 AO1.3b 1 AO2.1a 1 AO2.3a M2 for 10 × 2.1 + 5 × 3.2 oe Or M1 for correct interpretation of vertical scale e.g. 1 cm = 0.5 or area scale e.g. 1 cm² = 2.5 trains or 0.4 cm2 = 1 train e.g. 14.8 × 2.5 oe [1 cm2 = 2.5 trains] 16 (a) (i) y ≤ 9 and y > x 2 1 AO1.2 1 AO2.3a B1 for each Condone y ≥ x + 1 instead of y > xJ560/05 Mark Scheme Practice Paper (Set 3) 10 Question Answer Marks Part marks and guidance (a) (ii) x = 4 ruled y = 9 ruled y = x broken line Correct region left unshaded 11 1 1 2 AO1.3b 1 AO2.3a 1 AO2.3b Condone lines broken/solid Ignore any labels on lines All lines fit for purpose to enclose correct region Passes within 1 mm of (0, 0) and (9, 9), extended if necessary Condone y = x + 1 ruled only after y ≥ x + 1 in part (a)(i) Ignore additional incorrect lines drawn (as working possibly for part (b)) (b) 5 apples and 6 oranges 2 1 AO2.1b 1 AO3.1c M1 for a calculation shown of the form [0.]45x + [0.]3y where (x, y) is clearly in their region and both x and y are integers 17 (a) (i) 6 3 2 2 AO1.3b M1 for 3 12 seen (ii) 3 2 2 2 AO1.3b M1 for 6 2 2 2 × or better (b) [ ± ] 4 1 1 AO1.2J560/05 Mark Scheme Practice Paper (Set 3) 11 Question Answer Marks Part marks and guidance 18 (a) a = 1 b = 3 c = -9 113 1 AO1.1 1 AO1.3b 1 AO2.1a 2 AO3.1b M2 for b2 – 4ac = 45 Or M1 for b 2 − 4ac = 3 5 (b) There will be other values of a, b, c for a quadratic function that will give the same roots 1 1 AO3.4b e.g. there are many parabolas that can be drawn through (-1.5 – 1.5 5 , 0) and (-1.5 + 1.5 5 , 0) 19 (a) Incorrect as 6.25 > 5 oe 2 2 AO2.5a M1 for 22 + 1.52 (b) Gradient of tangent = 2 soi Equation of tangent: y = 2x + 5 oe [Area APO = (base × height) ÷ 2 =] (5 × 2) ÷ 2 = 5 M2 M2 A2 1 AO1.3b 1 AO2.2 2 AO3.1b 2 AO3.2 M1 for gradient of OP = - 1 2 After B0 allow M1 for gradient of tangent is negative reciprocal of their gradient of OP M1 for equation y = 2x + c or for substitution of (-2, 1) into their y = mx + c B1 for OA = 5 or A is (0, 5) If 0 scored, SC1 for recognition that method involves finding equation of tangentJ560/05 Mark Scheme Practice Paper (Set 3) 12 Question Answer Marks Part marks and guidance 20 OB = + b a  or BO = − − a b  2 2 OL + 3 3 = a b  or 2 2 LO 3 3 = − − a b  or 1 1 LB + 3 3 = a b  or 1 1 BL 3 3 = − − a b  OL 2LB =   oe B1 B3 A1 2 AO2.4b 2 AO3.1b 1 AO3.3 Implies previous B1 B1 for CM + 1 2 = − b a  or 1 MC 2 = − − b a  M1 for any correct route to find OL  , LO  , LB  or BL  Dependent on B1 and B3 Allow use of BO, LO and BL also in conclusion Condone poor vector notation e.g. arrows omitted throughoutJ560/05 Mark Scheme Practice Paper (Set 3) 13 APPENDIX Question 16(a)(ii) solutionJ560/05 Mark Scheme 14 Assessment Objectives (AO) Grid Question AO1 AO2 AO3 Total 1 0 3 0 3 2(a) 2 0 0 2 2(b) 3 0 0 3 3 1 0 2 3 4(a) 2 0 0 2 4(b)(i) 0 0 1 1 4(b)(ii) 1 1 0 2 5(a) 0 2 0 2 5(b) 0 1 0 1 6(a) 1 0 0 1 6(b) 2 0 2 4 7 2 1 0 3 8(a) 0 2 0 2 8(b) 0 1 1 2 9(a) 2 0 0 2 9(b) 3 0 0 3 10 1 1 3 5 11 2 0 3 5 12(a) 1 1 0 2 12(b) 0 0 1 1 12(c) 0 2 2 4 13(a) 0 0 1 1 13(b) 0 3 2 5 14(a) 2 0 0 2 14(b) 3 0 0 3 15(a) 0 0 1 1 15(b) 1 2 0 3 16(a)(i) 1 1 0 2 16(a)(ii) 2 2 0 4 16(b) 0 1 1 2 17(a)(i) 2 0 0 2 17(a)(ii) 2 0 0 2 17(b) 1 0 0 1 18(a) 2 1 2 5 18(b) 0 0 1 1 19(a) 0 2 0 2 19(b) 1 1 4 6 20 0 2 3 5 [Show More]

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