d 2 = Dep on M2 in (a) M1 = 625 Accept M1 dep d = 25 A1 [7] M10. (a) (0, 1) Generally marked B1 (b) Matching any (non zero for x) Values eg, a1 = 3, a2 = 9, etc Must show as power M1 a = 3 A1 [3] M11. Volume of one (or two) spheres Allow 10 for r for M1 M1 = 523.6 (1047.2) {524} {1048, 1050} oe 500π/3 or 1000π/3 (523.3 to 523.7) or (1046.6 to 1047.4) A1 Volume of cylinder Allow 10 for h or 10 for r for M1 (not both) M1
oe Due to different values of π an answer between 523.3 and 523.7 gets full marks ft If both Ms awarded and one value is correct. A1 ft [5] M12. Breaks down into areas of rectangles and areas of (quarter) circles Any combination of rectangles and circles or 12.56... or is enough evidence for area of circles NB 12.56 from 2 × π × 2, if seen is M0 NB 3.14 on its own is not evidence of the area of a quarter circle as it is π M1 Uses an ‘addition’ method (method 1) and finds Area of one (or 5) ‘external’ quadrants or or Uses a ‘subtraction’ method (methods 2 and 3) and finds 5 × area one quadrant or M1 dep = 0.8584..., {0.9, 0.86, 0.858} or = (× 5) 4.292... {4.3, 4.29} 15.71, 15.7 A1 52.3 or 52.29... 68 – 5π Allow 52 if 52.3 or 52.29... or a full method seen A1 [4]
Volume increases by 6 × so claim justified. Allow ‘Almost but not quite’ A1 [5] M36. Use of Must use 4 or 8 as radius. M1 (Volume hemisphere =) 133.9 to 134.1 (inclusive) 133.97 if π = 3.14 used. A1 (Volume paperweight =) 500+(their 134) (=634) If Ml awarded. A1 ft cm3 This mark is independent B1 [4] M37. Area semicircle – area circle Accept π × 102 ÷ 2 and/or π × 52 for M1 M1 π52 ÷ 2 – π(2.5)2 (= 12.5π – 6.25π = 39.27 – 19.63) Accept fractions, decimals or in terms of π A1 19.6(....)(= 6.25π) ft on one error only, e.g. Accept fractions, decimals or in terms of π. Use of π as 3.14 gives 19.625 Al Common errors e.g. π × 102 ÷ 2 – π × 52 = 157.1 – 78.5 = 78.6 = Ml,A0,A0ft.
A1 (c) Straight line because OD = a + b B1 3 times bigger because OF = 3a + 3b B1 [7] M47. 2.75 × 27 M1 for 33 M1 74.25..... Accept 74 or 74.3 Height first pyramid is 2.0625 Height second is 6.1875 Volume is × 62 × 6.1875 = 74.25 This line M1, A1. A1 [2] M48. Angle ATB = 13° B1 M1 for use of sine rule, A1 for correct substitution.
–5 and 6 => 35 and not prime –8 and 9 => 77 and not prime accept any indication of "not prime" B1 any correctly evaluated with no conclusion Examples – 3 and 4 => 17 –4 and 5 => 25 –5 and 6 => 35 –6 and 7 => 47 –7 and 8 => 61 –8 and 9 => 77 or incorrectly evaluated that gives a counter example with non-prime conclusion B2 [2] M53. Extra volume = 50 × 34 × 4.5 = 7650 M1 1912.5 A1
and 5cm (±2mm) Consistent use of scale 1 cm to 100m B3 max (-1 MR) B1 Construction arcs for perp. Bisector Two sets of intersecting arcs centres A and C B1 Perp. Bisector of AC (±2mm) B1 Line 2cm (±2mm) from path B1 Correct region shaded B1 [5] M86. (a) (i) 40 B1 (ii) 140 or 180 – (their x) Do not ft if answer = 140 in (a)(i) B1 ft (b) Logical and precise explanation (either written or as calculation) B1 for 1 angle labelled or stated correctly, no reason B2 (c) 24 ÷ 8 × 2 or OP = 6 M1 π(or 3.14) × (their 6)2 M1 36π or 36 × π or π × 36 allow π36
(b = ) oe A1 [4] M117. Evidence of searching for a pattern or r = 2 or d = 4 or 6 extra discs gives extra 12 cm eg, 4 → 10, 10 → 22 or 5 → 10, 11 → 22 or markings on diagram or diagram of 24 discs (2 rows) 6 extra discs gives extra 12 cm M1
Allow one error or [5 ± √59] ÷ 6 M1 (5 ± √ 109) ÷ 6 A1 2.57 and –0.91 SC1 2.3 and –0.45 A1 [3] M123. tan 38 = BC/7.21 or BC ⁄ sin 38 = 7.21 ⁄ sin 52 M1 7.21 × tan 38 or 7.21 × sin 38 ÷ sin 52 M1 dep 5.6 (3 …) A1 5.6 or 5.63 B1 ft [4] M124. (a) y = 2 sin x oe y = 2 f( x ) ‘y =’ or ‘x =’ missing penalize first time only B1 (b) y = 1 + sin x oe y = 1 + f( x ) B1 (c) y = – sin x
oe y = – f( x ) y = sin (x + 180) y = sin (x – 180) B1 [3] M125. π × 3 × 3 × 6 Cylinder M1 54π 169-170 A1 4 × π ×3 × 3 × 3 ÷ 3