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 = 43.30127019 – 39.26990817 = 4.031362019 = 4.03 cm2 (to 3 s.f.) M1 A1 3 M1 for finding area of 3 sectors 14 1 100 n rAP =+  3 5000 800 5000 1 100 r+ = +  3 5800 5000 1 100 r=+  3 5800 15000 100 r=+ 3 5800 15000 100 r=+ 3 5800 15000 100 r−= 3 5800100 1 5000 r −=  3 5800100 1 5000r =−   r = 5.07175745 r = 5.07 (to 3 s.f.) M1 M1 A1 3 M1 for formula of compound interest, with correct components substituted. M1 for cube root. 15 (a) x2 – 18x + 19 = (x – 9)2 – 92 + 19 = (x – 9)2 – 62 n = –62 B1 1 (b) (x – 9)2 – 62 = 0 (x – 9)2 = 62 M1 M1 for  and square root.

5 Qn Working Mark Awarded Sub- total Remarks 9 62x− = 9 62x= x = 16.87400787 or x = 1.125992126 x = 16.9 (3 s.f.) x = 1.13 (3 s.f.) A1 2 16 (a) 13 1 2 bca bc += + 13(1.5)( 1.2) 1 2(1.5) ( 1.2)a −+= +− 412 9a=− B1 1 Accept 112 9− too. (b) 13 1 2 bca bc += + (2 ) 13 1a b c bc+ = + 2ab + ac = 13bc + 1 2ab – 13bc = 1 – ac b (2a – 13c) = 1 – ac 1 2 13 acb ac −= − M1 A1 2 M1 for correct expansion 17 (a) 21 57xx −=+ 2 1( 7) 5( 7) ( 7) xx x x x x +−=++ 2 1( 7) 5( 7) xx xx −+ =+ 2 1( 7) 5 ( 7)x x x x− + = + 22 7 5 35x x x x− − = + 20 5 34 7xx= + + 25 34 7 0xx+ + = 2 4 2 b b acx a −  −= 234 34 4(5)(7) 2(5)x −  −= 34 1016 10x −= x = –0.2125245099 or x = –6.58747549 x = –0.21 (to 2 d.p..) x = –6.59 (to 2 d.p) M1 M1 M1 A1 4 M1 for combining into a single fraction correctly. M1 for getting the equation. M1 for quadratic formula A1 for both answers, correct to 2 d.p.

6 Qn Working Mark Awarded Sub- total Remarks 18 (a) 1 + 3n B1 1 (b) 46 B1 1 (c) 1 + 3n = 547 3n = 546 n = 182, which is an integer. Hence, 547 is a term of the sequence. Yes, Chye Joo is correct. B1 1 B1 for finding n and stating that it is an integer / whole number. (d) Analyse the relationship between the two sequences. First sequence 4 7 10 13 16 + 1 + 4 + 9 + 16 + 25 Second sequence 5 11 19 29 41 Note that 1 = 1 2 , 4 = 2 2 , 9 = 3 2 , 16 = 4 2 and 25 = 52. Hence, using (b), 1 + 3n + n2. B1 1 19 (a) angle PSU = 180o – 65o (interior angles, PQ//SR) = 115o reflex angle PSU = 360o – 115o (angles at a point) = 245o M1 A1 2 Deduct marks for missing angle properties. (b) angle PTU = 121o (alternate angles, PQ//SR) B1 1 (c) angle PTS = 180o – 65o – 65o (base angle of an isosceles triangle) = 50o angle STU = angle PTU – angle PTS = 121o – 50o = 71o M1 A1 2 20 (a) Volume of hemisphere = 170 cm3 32 1703 r = 3 2170 3r = 3 2170 3r =  M1 M1 for formula involving volume of hemisphere.

7 Qn Working Mark Awarded Sub- total Remarks r = 4.329756132 r = 4.33 cm (to 3 s.f.) A1 2 (b) Volume of cylinder = 190 – 170 = 20 cm3 2 20rh = 21.5 20h  = 220 1.5h = h = 2.829421211 h = 2.83 cm (to 3 s.f.) M1 A1 2 M1 for 2 20rh = (c) Surface area of hemisphere and base of cylinder = 222 rr + = 23 r = 24.32973 561( 32) = 176.6843159 cm2 Surface area of curved surface of cylinder = 2 rh = 2.8294212112 1.5 = 26.6666666667 cm2 Total surface area = 176.6843159 + 26.6666666667 = 203 cm2 (to 3 s.f.) M1 M1 A1 3 21 (a) By Pythagoras’ Theorem, 202 = AB2 + 132 AB2 = 202 – 132 2220 13AB= − 2220 13AB=− (reject negative, length > 0) AB = 15.19868415 AB = 15.2 m (to 3 s.f.) M1 A1 2 Minus one mark if “By Pythagoras’ Theorem” is missing. (b) BC2 = 292 = 841 BE2 + CE2 = 202 + 212 = 841 M1 M1 for finding both BC2 and BE2 + CE2

8 Qn Working Mark Awarded Sub- total Remarks Hence, BC2 = BE2 + CE2 and by the Converse of Pythagoras’ Theorem, triangle BCE is a right-angled triangle (proven). A1 2 A1 for stating Converse of Pythagoras’ Theorem. (c) TOA CAH SOH angle CBE = 1 21sin 29 −   = 46.39718103 = 46.4o (to 1 d.p.) M1 A1 2 (d) cos50 21 CD=