AISS 4N Prelim 2023 P2 MS

1 2023 Sec 4NA Math Prelim Paper 2 Marking Scheme Qn Working Mark Awarded Sub- total Remarks 1 (a) 7.004869155 = 7.00 (to 3 s.f.) B1 1 (b)(i) 62.589 10 − B1 1 (ii) ( ) ( ) 625.97 10 7.34 10  = 38.133514986 10 = 38.13 10 (to 3 s.f.) M1 A1 2 2 Sum of interior angles in a pentagon = ( )5 2 180−   = 540o Angle FED = angle EDC = angle DCG = 540o ÷ 5 = 108o x = 180o – 108o (adj angles on a straight line) = 72o 540o – 90o – 108o – 108o – 108o = 126o (2y – 3) + (y + 15) = 126 3y + 12 = 126 3y = 114 y = 38o M1 M1 A1 M1 A1 5 M1 for equation formed. 3 (a)(i) The total number of students in each class may be different so it is not accurate to compare using the size of the sector or its angle. B1 1 (ii) Grey eyes in Class 1A → 30o 30o → 3 students 1o → 1 10 student 360o → 1 36010 = 36 students M1 A1 2 (iii) 45 100%360 = 12.5% M1 A1 2

2 Qn Working Mark Awarded Sub- total Remarks (b)(i) Median = 23 24 2 + = 23.5 B1 1 (b)(ii) Total = 08 + 09 + 12 + 15 + 15 + 16 + 18 + 18 + 21 + 23 + 24 + 24 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 30 = 427 Mean = 427 ÷ 20 = 21.35 [Alternative solution] Mean = (08 + 09 + 12 + 15 + 15 + 16 + 18 + 18 + 21 + 23 + 24 + 24 + 26 + 27 + 27 + 27 + 28 + 29 + 30 + 30) ÷ 20 = 21.35 M1 A1 [M1] [A1] 2 (b)(iii) P(scored more than 25 marks) 8 20 2 5 = = M1 A1 2 4 Let r be the radius of the water surface. 2 225r = r = 15 cm By Pythagoras’ Theorem, 2217 15− = 8 cm x = 17 – 8 = 9 M1 M1 M1 A1 4 5 (a) 60 000 cm = 60 000 / 100 000 km = 0.6 km B1 1

3 Qn Working Mark Awarded Sub- total Remarks (b) 13 0.5 = 7.8 km B1 1 (c) 1 cm to 0.6 km 1 cm2 to 0.62 km2 1 cm2 to 0.36 km2 4.8 1 130.36 3= cm2 M1 A1 2 6 (5x – 4) (x + 2) = 0 5x – 4 = 0 or x + 2 = 0 x = 4 5 or x = –2 M1 A2 3 No marks given for any other methods used. 7 (a) Volume of cone = 21 3 rh = 21 12 53 = 753.9822369 cm3 Volume of cylinder = 2rh = 212 8 = 3619.114737 cm3 Volume of composite solid = 753.9822369 + 3619.114737 = 4373.096974 = 4370 cm3 (to 3 significant figures) M1 M1 A1 3 (b) By Pythagoras’ Theorem, 225 12+ = 13 cm B1 1 (c) Curved surface area (cone) = rl = 12 13 = 490.088454 cm2 Curved surface area (cylinder) = 2 rh = 2 12 8 = 603.1857895 cm2 Surface area of circle (base) = 2r = 212 M1 M1 M1

4 Qn Working Mark Awarded Sub- total