P1 A1 for 39(.0625) 3 construction B2 (B1) Correct construction showing all necessary arcs. (Pair of intersecting arcs centred on A and B) 4 (a) (b) –1.2 and 3.2 (1, –5) B2 (B1) B1 for both roots correct (for one correct root) cao 5 134 P1 P1 P1 P1 A1 Process to find the distance around one or both ends of the track, e.g. π × 54 (= 169.6460033) or ( π × 54) ÷ 2 (=84.82300165) (dep on P1) complete process to find the total length of the track, e.g. 40 × 2 + “169.6460033” (=249.6460033) Process to find the circumference of wheel, e.g. π × 590 (=1853.539666 mm) or π ×0.59 (= 1.85353966 m) Complete process to find the number of revolutions in consistent units, e.g. “249.64…” ÷ “1.85…” or unrounded answer of 134.6860863 cao 6 Elevation B2 (B1) Fully correct side elevation (a rectangle 4 high by 2 wide)
7(a) Shown M1 M1 A1 for distance ÷ speed to find time, e.g. (1.496×1011) ÷ (3×108) (=498.666) (dep) for conversion to hours, e.g. “498.666” ÷ (60 × 60) 0.1385185185… Explanation C1 Correct explanation, e.g. they have multiplied the indices rather than adding 8 y = 3x – 1 M1 A1 For y = 3x + c OR a line drawn with gradient 3 passing through A oe 9(a) Lauren £9537.20 £9545 P1 P1 P1 C1 Process to find the value of one car at the end of one year, e.g. 13995 × 0.88 or 14495 × 0.87 Process to find the value of one car at the end of 3 years, e.g. 13995 × (0.88)3 or 14495 × (0.87)3 Complete process to find the value of both cars at the end of 3 years,, e.g. 13995 × (0.88)3 and 14495 × (0.87)3 £9537.20(064) and £9545(.000985) and Lauren (b) Explanation C1 Appropriate explanation, e.g. explanation that her car will be worth less
Question Working Answer Mark Notes 15 Proof M1 A1 C1 Correct expansion or factorisation of a suitable expression for 2 consecutive integers,, e.g. (n + 1)2 – n2 = n2 + 2n + 1 – n2 or (n + 1)2 – n2 = (n + 1 + n)(n + 1 – n) Expansion or factorisation correctly simplified,, e.g. 2n + 1 or 2n + 3 Correct conclusion drawn from fully correct working 16 Enlargement, scale factor –2, centre (4, 6) B2 (B1) Enlargement, scale factor –2, centre (4, 6) (For 2 correct aspects) NB score B0 for more than one transformation 17 No with justification P1 P1 P1 C1 for one correct bound, e.g. 69.5, 70.5, 39.5, 40.5, 121.5, 122.5, 13.5, 14.5 for complete process to find the upper bound for the volume of the tank, e.g. 120.5 × 40.5 × 70.5 (= 344057.625) for complete process to find the upper bound for the number of buckets, (upper bound for volume of tank ÷ lower bound for volume of bucket) e.g. “344057.625” ÷ 13500. Must be in consistent units OR correct process to compare the lower bound for 25 buckets of water with the upper bound for the volume of the tank, e.g. 13.5 × 1000 × 25 (=337500) Correct conclusion based on correct calculations
18 2 M1 M1 A1 T = or 0.0096 = or T = Method to find u, e.g. cao 19 (-5, -7) M1 M1 A1 Method to start to complete the square, e.g. (x + 5)2 (x + 5)2 – 7 cao (dep on method seen) 20 M1 A1 C2 (C1) For ACD = 54°, or ADC= 66° (may be on diagram) For CAD = 60° from correct working C2 for all correct reasons stated (C1 for one appropriate reason linked to a circle theorem used) Alternate segment theorem. Opposite angles of a cyclic quadrilateral add up to 180. Angles in a triangle add up to 180 Angles on a straight line add up to 180 21 M1 A1 gf(x) = or f(4) = 48 oe