PiXL AQA Style Paper 2H (November 2016) Mark Scheme

Transcription

1 PiXL AQA Style Paper 2H (November 2016) Mark Scheme Q Answer Mark Comments (a) 2x 2 (b) 2x +3 2 (c) (2x+3)/2 or x (a) Triangle with coordinates at ( 7, 1 ) ( 5, 1 ) and ( 5, 4 ) Sight of line y = x on diagram or correct reflection in any other diagonal line (if incorrect, the line used must be shown on the diagram) gains Correct answer only 3 (b) Triangle with coordinates at ( 1, 1 ) ( 3, 1 ) and ( 3, 4 ) Sight of line x = 4 on diagram or correct reflection in any other vertical or horizontal line (if incorrect, the line used must be shown on the diagram) or follow through from incorrect answer to 3 (a) gains Correct answer only Rotation 3 (c) Anti clockwise 90 or Clockwise 270 Accept about ( 4, 4 ) Centre ( 4, 4 ) 4 SAS

2 5 (a) =38 Sight of x = x2 = 82 points = 10, or similar, award 6 (a) The longer a student revises, the higher the mark (s)he would get or the higher the mark the pupil gets, the longer (sh)he is likely to have spent revising. 6 (b) 104 Evidence of use of line of best fit (line drawn from 120 minutes) [a,b] [30,40] 7 25% of 40% = 10% 32% of 60% = 19.2% = also / Yes it could x 50 = = 43 Form and solve simultaneous eqns 3x +4y = 43 x+y =12 X = 5 Alternative method 2.4 x 50 = = 43 Trial of pairs of values till 5 and 7 discovered and 5 given as answer 5 given with no working only given

3 10 Equation rearranged 3=2x -2 5 = 2x 2.5 =x Or 6x+5 = 8x 11 (a) 6/14 and 8/14 on top branches 8/15 for plain 7/14 for both of bottom pair of branches 11 (b) 8/15 x 7/14 Or their 8/15 x their 7/14 4/15 oe 12 Side lengths as 5cm Use of Pythagoras to get height of triangle Alternative method as height 4.33 x 0.3 =3.26 Their 4.33 x 0.3 Side lengths as 5cm Correct use of ½ ab sin(c) Alternative method 2 M x 0.3 = Trials of numbers with evidence of squaring and doubling. Trial of 7 with 63 as answer or Trial of 8 with 80 as answer 8 14 Lower bound of ,500 Lower bound of at least one interest rate Their 11,500 x their = % and 4.5% Or finding 3.5% and adding it Their x their Or finding 4.5% and adding it

4 15 (b) Frequency densities (frequency class width) calculated for each Appropriate scales and labels Alternative method 1 At least 4 correctly calculated for Horizontal scale must be continuous (B0 if class intervals, eg 0 < h 30 are used) All bars correctly drawn A2 At least 3 bars correctly drawn for. Area key used on histogram Appropriate scale and labels All bars correctly drawn Alternative method 2 A2 Must be labelled to indicate clearly the frequency represented by a given area. Horizontal scale must be continuous (B0 if class intervals, eg 0 < h 30 are used) At least 3 bars correctly drawn for. Maximum of possible if M0 earlier. 16(a) 2P+Q = 13 and P=6 or other credible method for finding dimensions of rectangle. Finds dimensions of rectangle to be 6 units horizontally by 1 unit vertically and attempts to find B. Calculation pertaining to x co-ordinates with 13 seen ( 11, 8 ) Correct answer only Calculation pertaining to y co-ordinate and 6 seen 17 (a) Attempts to use 1/2absinC. Area = [69] Additional guidance Allow up to if candidate attempts to find the height of triangle (including turning it to make base 2m or 1m and finding perpendicular height before using base height 2 base height 2 using 2m and 1m only (ie oe) M0 A0. Attempts to use the cosine rule 17 (b) Correctly substitutes into it For this allow omission of square root stage this must appear for award of following. Width =2.06[78...]

5 17 (c) ½ their width x height = their area Finds height 0.68[157...] Total height x+1 +x = 2x+1 (x 2 +2x+1) x 2 =2x+1 Correct simplification and evidence that conclusion clearly reached/understood by candidate. 19 Q is at 0, 10 Gradient of OP is 1 3 Gradient of AP is 1 their gradient of OP Equation of AP is y = 3x x + 10 = their y co-ordinate of Q Uses x = 0 to get to y 2 = 10 and then Q. May be seen later. Give for sight of 10 alone. May be implied by later working out. Allow recurring decimal equivalent. Accept alternative methods for finding the gradient of AP. Gradient of AP must now be correct. Allow follow through for value of c if method is OK from incorrect gradient. for attempt to solve for x co-ordinate of A. y = 3x + 10 must now be correct , 10 Both correct. Additional guidance Question asks for answers in surd form; decimal equivalents lose first mark and/or final mark, although award for, if surd is given in first (a) b a or a + b Correct answer only. States that MO = 1 2 OA Accept MO = 1 2 a. 20 (b) a 2b or 2b + 1 for 2 a A2 2 a, for 2b. For A2 there must be a correct sign (+ or ) between the two components. Additional guidance If A0 awarded allow instead for MC = MO + OC or MC = 1 2 OA 2OB after first.

6 21 x 2 + 4x -1 = 3x + 5 Factorises x 2 + x -6 =0 into 2 brackets completely correctly. (x+3)(x-2) =0 Either bracket right = Or correct substitution into the formula x=-3 and x=2 Both must be correct. Attempt is made to use y = 3x + 5 y = -4 and y = 11 ( -3, -4 ) and ( 2, 11 ) Matches correct x and y values in coordinate pairs (allow other notation if pairing is clear, although answer space signposts coordinates).