2024 4NA ahmad ibrahim P2 answer

2024 4NA Prelim P2 Marking Scheme 1 Qn Solutions Marks SECTION A 1a 82.98 8.93 4.45 80 9 4 60  = = M1 A1 1b 6 1 1397074 100548 10 2.55 10 −  = M1 A1 2a 0(more than 1 error) M1 (1 error) M2(0 error) 2b 22 2 2 + = B1 2c 6 20 3 10= B1 3a ( ) ( ) 2 180 15 2 180 15 156o n n − −= = M1 A1 3b 180 156(base angle of an isosceles triang le) 12 360 156 156 20 12(sum of angle in a quadrila teral) 16 EFD DCF  = − =  = − − − − = M1 M1 A1 4a 2 240 2 120 2 60 2 30 3 15 5 5 1 42 3 5 B1

2024 4NA Prelim P2 Marking Scheme 2 Qn Solutions Marks 4b 4 1 1 4 1 1 1 1 240 2 3 5 240 2 3 5 (3 5 ) 35 15 aa a a =    =     = = B1 4c Green Black 2 240 210 3 120 105 5 40 35 8 7 LCM of 210 and 240 =2 3 5 7 8 1680 1680 7240 1680 8210 blue red     = == == M1 B1 B1 5a Price of the car at a 10% discount 22500 90% $20250 Percentage profit 28 888 20250= 100%20250 42.7% = = −  = M1 A1 5b Amount of deposit+ Instalment (Mary) =(20% 28888)+(58 510) $35 357.80 Amount of deposit+ Instalment (Jane) =(15% 28888)+(60 520) =$35 533.20  =  He should accept the offer from Jane as Jane will pay a higher amount of money ($35533.20) as compared to Mary(35357.80) M1 M1 A1 6a ( ) ( ) 22 2 5 5 9 5 units − + − = M1 A1

2024 4NA Prelim P2 Marking Scheme 3 Qn Solutions Marks 6b 95 52 4 3 4 3 49 (5)3 209 3 209 3 7 3 47 33 gradient y x c c c c c yx −= − = =+ =+ =+ =− = =+ M1 for either m or c A1 6c and has the same gradient. 5 3 8 9 4 13 (8 ,13) AB BC x y =+= = + = B1,B1 7a(i) 3 2 12 8 (3 2) 4(3 2) (3 2)( 4) uv v u v u u uv + − − = + − + = + − M1 A1 7a(ii) 22 11 5xx−+ 2x -1 -x x -5 -10x -11x (2 1)( 5)xx−− M1 A1 7b 32 3 2 3 2 93 48 98 43 72 12 6 xx x x x x x  = = = M1 A1 8a 20 2.5 20 2.5 8 t t ts = = = B1

2024 4NA Prelim P2 Marking Scheme 4 Qn Solutions Marks 8b 14 20 20 14 2020 14 m/s speed speed = = = M1 A1 8c Distance =Area under the graph ( )1 22 50 202 720 m  +  = M1 A1 9a 113 1 2.2 2 2.172(3) 6y or or= + − = B1 9b, B2 : all points correctly plotted B1: 4-7 points correctly plotted B1 points joined smoothly. 9c 0.17, 2.82 B1,B1 (FT) 9d Tangent drawn -4.56 Range -5 to - 4.33 B1 B1 10a 360 85 275 − = M1 A1 10b The total length of wire 2755.2 3.9 2 (3)360 23.5cm = + +   = M1 for 275 2 (3)360   A1 11a Angles and length plot correctly. 75, 122 or 6.2cm drawn correctly B2 B1 11b Perpendicular bisector with construction lines seen. B1 11c Accurate measurement of QX. (FT) B1

2024 4NA Prelim P2 Marking Scheme 5 Qn Solutions Marks 12a 50 510mean== B1 12b There is an extreme data of 23 and it is observed that 8 of the data are less than the mean of 5. Median will be a better measu