2024 4NA Ahmad Ibrahim P1 answer

1 AISS 2024 Sec 4NA Prelim Math Marking Scheme Qn Working Mark Awarded Sub- total Remarks 1 9.962571767 = 9.963 (to 4 s.f.) B1 1 2 39%, 2 30.25 , 0.16 , 0.412, 21 50 B2 2 B1 if order is correct if one of the terms is covered. 3 x < –7.5 x = –8 M1 A1 2 4 (a) 5 units → 1.4 kg 1 unit →0.28 kg 7 units → 1.96 kg B1 1 (b) 2 : 3 = 8 : 12 4 : 9 = 12 : 27 8 : 12 : 27 B1 1 No marks if ratio is not simplified. 5 2 ky x= When x = 3, y = 15. 215 3 k= k = 15(32) k = 135 2 135y x= When x = 5, 2 135 5y= y = 5.4 M1 A1 2 M1 for finding k. 6 (a) 24x6 B1 1 (b) 9y4 B1 1

2 Qn Working Mark Awarded Sub- total Remarks 7 12 32 8 a− = 32 2 32a−−= 3522a−− = –a – 3 = 5 –3 – 5 = a a = –8 M1 A1 2 M1 for 2 –3 and 25. 8 7 2160% 10 50= 21 units → 21 marbles 1 unit → 1 marble 50 units → 50 marbles Or 760% 0.4210= 21 500.42 = M1 A1 [M1] [A1] 2 M1 for finding the fraction / percentage of marbles that are yellow. 9 Method 1: Substitution y = x – 5 --------- (1) 3x + 4y = 29 ---------- (2) Sub (1) into (2) 3x + 4(x – 5) = 29 3x + 4x – 20 = 29 7x = 49 x = 7 Sub x = 7 into (1) y = 7 – 5 y = 2 P(7, 2) Method 2: Elimination y = x – 5 x – y = 5 --------- (1) 3x + 4y = 29 ---------- (2) (1)  3: 3x – 3y = 15 ---------- (3) (2) – (3): (3x + 4y) – (3x – 3y) = 29 – 15 M1 A1 A1 [M1] M1 for correct substitution / elimination.

3 Qn Working Mark Awarded Sub- total Remarks 3x + 4y – 3x + 3y = 14 7y = 14 y = 2 Sub y = 2 into (1) x – 2 = 5 x = 7 P(7, 2) [A1] [A1] 3 10 5(x – 4) = 6 – (2 – x) 5x – 20 = 6 – 2 + x 4x = 24 x = 6 M1 A1 2 M1 for expanding correctly 11 2 2 41 6 13 5 y yy − −+ = (2 1)(2 1) (2 1)(3 5) yy yy +− −− = 21 35 y y + − M1 M1 A1 3 M1 for factorizing numerator M1 for factorizing denominator 12 (a) 19  0.5 = 9.5 km B1 1 (b) Map Actual 1 cm 0.5 km 12 cm2 = 1 cm2 0.52 km2 = 0.25 km2 4 cm2 1 km2 140 cm2 35 km2 Answer: 140 cm2 M1 A1 2 M1 for ratio of area (1 : 0.25) 13 Area of triangle = 1 sin2 ab C = 1 10 10 sin 602    = 43.30127019 cm2 Area of the 3 sectors M1 M1 for finding area of triangle.

4 Qn Working Mark Awarded Sub- total Remarks = area of semicircle = 2 2 r = 25 2  = 39.26990817 cm2 Area of shaded region = 43.