Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts: alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this Mark Scheme are available from aqa.org.uk Copyright © 2014 AQA and its licensors. All rights reserved. trial
2(a) 0.3 or or 30% B1 oe Q Additional Guidance Mark 2(c) (do not allow any other fractions) e.g. scores M0 No misreads allowed M1A0 3 0.15 × 65.5(0) or 9.825 or 9.82 or 9.83 or 6.55 + 3.27(5) or 6.55 + 3.28 or 6.5... + 3.2... M1 oe 0.85 or 85% seen or implied 55.675 or 65.5(0) – their 9.825 0.85 × 65.5 55.67 or 55.68 Q1 strand (i) correct money notation Q Additional Guidance Mark 3 Do not accept £55.67p for the Q mark If they use the wrong value with a fully correct method they can score M1M0Q0, e.g. 0.15 ×15.50 = 2.325 15.50 – 2.33 = 13.17
5(30 + 2 × 16) M1 310 A1 SC2 for 15 seen and 5(their 30 + 2 × their 16) correctly evaluated SC1 for 5(their 30 + 2 × their 16) correctly evaluated Q Additional Guidance Mark 12 Their 310 must be in correct money notation 13 Correctly evaluated M1 e.g. 2 4 – 10 = 6 Obtains 3 ≤ x ≤ 4 M1 Obtains 3.3 ≤ x ≤ 3.4 or Two correct trials [3.25, 3.35] which bracket 0 A1
Test 3.35 and concludes 3.3 or Two correct trials [3.25, 3.35] bracket 0 and 3.3 for final answer Q1 Strand (ii) Using 2 dp to ensure 1 dp Q Additional Guidance Mark 13 CORRECT ANSWER WITH NO WORKING SCORES ZERO → – 2 → – 1.4(…) → – 0.8(…) → – 0.15(…) or – 0.2 → 0.556(…) or 0.56 or 0.6 → 1.3(…) → 2.1(…) → 2.996(…) or 3 → 3.9(…) → 4.9(…) → 6 3.3 → – 0.15(…) or – 0.2 3.31 → – 0.082 3.32 → – 0.013 3.33 → 0.056 3.34 → 0.126 3.35 → 0.196 3.36 → 0.267 3.37 → 0.339 3.38 → 0.411 3.39 → 0.483 3.4 → 0.556(…) or 0.56 or 0.6 Also useful 3.25 → – 0.486 14 costs (£)1.2(0) B1 oe 1. M1 oe
9(3 + x) = 4 × 5 M1 3 + x = 4 × 27 + 9x = 20 M1dep 3 + x = 9x = 20 – 27 M1dep x = – 3 – or –0.77... or –0.78 A1 oe Alternative Method 2 4(3 + x) = 5 × 9 M1 3 + x = 9 × 12 + 4x = 45 3 + x = 4x = 45 – 12 M1dep x = – 3 or 8.25 A1 oe Alternative Method 3 4(3 + x) = 5 × 9 M1 3 + x = 5 × 12 + 4x = 45 M1dep 3 + x = 4x = 45 – 12 M1dep x = – 3 or 8.25 A1 oe
15(b) Cont. Alternative Method 4 5(3 + x) = 4 × 9 M1 3 + x = 4 × 15 + 5x = 36 M1dep 3 + x = 5x = 36 – 15 M1dep x = – 3 or 4.2 A1 oe Q Additional Guidance Mark 15(b) 3+ x =9 4 4 5 x =9 3− 4 5 4 x= If students use a different equation than that chosen in part (a) they can only score a maximum of 3 out of 4, unless they choose the correct equation M1 M1dep M1dep A1
16 Alternative Method 1 tan 50 = h 24 M1 tan 40 = 24 h 24 × tan 50 or [28.5, 29] M1dep 24 ÷ tan 40 or [28.5, 29] × 24 × 24 tan 50 M1dep × 24 × (24 ÷ tan 40) [342, 348] A1 340 or 350 B1ft Must ft from their 3sf answer if seen Alternative Method 2 cos 50 = 24 x M1 sin 40 = 24 x 24 ÷ cos 50 or [37.3, 37.5] M1dep 24 ÷ sin 40 or [37.3, 37.5] × 24 × 37.3 × sin 50 [342, 348] A1
340 or 350 B1ft Must ft from their 3sf answer if seen Alternative Method 3 x = 24 sin50 sin40 M1 x = 24 sin90 sin40 or [28.5, 29] M1dep or [37.3, 37.5] × 24 × 24 tan 50 M1dep × 24 × their 37.3 × sin 50 [342, 348] A1 340 or 350 B1ft Must ft from their 3sf answer if seen 17(a) Fully correct box plot with minimum = 65 LQ = 70 median = 80 UQ = 85 maximum = 95 B2 B1 for 3 correct Q Additional Guidance Mark 17(a) Minimum and maximum values can be marked with a cross or a plus 17(b) LQ = 75 B1 Need not be plotted
24 = k(4)2 or or 1.5 M1dep oe × 16 M1dep oe 900 A1 (R =) 30 or – 30 A1 Either value Alternative method 2 R² 4² M1 oe 4² R² 1350 = R² 24 4² M1dep oe 24 = 4² 1350 R² × 16 M1dep oe 900 A1 (R =) 30 or – 30 A1 Either value Q Additional Guidance Mark 20 y = kx2 y αkR2 900 scores 4 M1 M1
21 y = 2 + x B1 x = y – 2 2x2 + 5x + 1 = their (2 + x) M1 oe y = 2(y – 2)2 + 5(y – 2) + 1 2y2 – 8y + 8 + 5y – y – 10 + 1 = 0 2x2 + 4x – 1 = 0 M1dep 2y2 – 4y – 1 = 0 M1 −4 ± 42 − (4 × 2 ×−1) −−4 ± (−4)2 − (4 × 2 ×−1) 2 × 2 or 2 × 2 or 4 ± 4 24 x = –2.2(…) and x = 0.2(…) or x = –2.2(…) and y = –0.2(…) or x = 0.2(…) and y = 2.2(…) A1 y = 2.2(…) and y = –0.2(…) or y = 2.2(…) and x = 0.2(…) or y = –0.2(…) and x = –2.2(…) x = –2.2 and y = –0.2 and x = 0.2 and y = 2.2 A1 y = 2.2 and x = 0.2 and y = –0.2 and x = –2.2 Q Additional Guidance Mark 21 BEWARE, roots of 2x2 + 5x + 1 = 0 are – 0.22 and – 2.28 Correctly substituting their values from their quadratic scores M1, e.g. 2x2 + 5x + 1 = 0 −5± 52 −(4×2×1) scores M0M0M1A0A0 2×2 All four solutions are required to score full marks
22 (x – 4)(x + 4) B1 (2x + 3)(x – 4) or (2x + a)(x + b) M1 where ab = ±12 or 2b + a = –5 x + 4 2x + 3 A1 23 or or or or M1 oe × or × or × M1 oe 0.166... or 0.277... or 0.17 or 0.28 × + × + × M1 1 – × or A1 oe 0.72(2..) Q Additional Guidance Mark 23 Accept decimals 24 (3n + 5)(n – 1) or 3n × n M1 (3n + 5)(n – 1) – 3n × n M1dep 3n2 − 3n + 5n − 5 − 3n2 3n(n−1) or 3n2 + 2n− 5 − 3n2 3n(n−1) A1 Denominator used Q Additional Guidance Mark