SCSS 2024 EM Prelim P1 MS

2024 4E5N EM Prelim P1 Mark Scheme 1 2 2 2 2 15 2 225 PQ PQ PQ += = 10.6PQ= cm M1 A1 2a 3 5 B1 2b New total = 35 = 15 Yellow balls left 15 8 5 2= − − = 7 2 5x=−= B1 3 2020 population = 100 5920000104 = 5690000 (nearest ten thousand) M1 A1 4 ( ) ( ) 322 34 8 9 2 12 8 9 18 12 xx xx xx −− −−= −+= 18 12 x−= M1 – Common denominator A1 5 4 34 3 6827 81 xx yy  = B1 - numerator B1 - denominator 6 3 2 24 2 3 20 2 5 15 3 5 = = = LCM 32 3 5=   Smallest length = 120 cm M1 A1 7a 9 B1 7b Angle = 3 360 6018= B1 8a 27 24 21.90 10 5.97 10 3.18 10   =  B1 8b Jupiter density = 27 241.90 10 1.43 10   =1328.6713 kg/m3 Earth density = 24 215.97 10 1.08 10   = 5527.7778 kg/m3 Earth has a higher density M1 A1

9 3 3 41 89989.12100 489989.12 1 100 P P += =  +  = $80000 M1 M1 A1 10a 196.5 g B1 10b 197.5 24.5 = 8.06 g M1 A1 11a 7 and 8 B1 11b Mean 4 5 6 3(7) 3(8) 10 10 + + + + += = 7 B1 11c Yes. The mean mark for the whole class, 7.5, was higher than the mean mark for the girls. Thus the mean mark for the boys would have to be higher than 7.5, which is better than the girls marks. Accept use of actual calculation of mean mark for boys to justify. B1 12a 42ab− ( )( ) 22a b a b= + − B1 12b ( ) 8 1 4 2 8 4 2 1 4 2 1 2 1 xy x y xy x y x y y − − + = − + − = − + − ( )( )2 1 4 1yx= − + M1 A1 13a B1 13bi {12, 42, 72} B1 13bii 0 B1 14a Exterior angle = 180 – 140 = 40 Number of sides 360 40= = 9 Or ( )2 180 140 180 360 140 40 360 nn nn n −  = −= = 9n= M1 A1 M1 A1 A B

14b Exterior angle of octagon 360 8 45=  = Exterior angle of equilateral triangle 180 60 120= − = 120 45 165x= + = M1 A1 15a Accept 3 700 000 to 3 900 000 B1 15b The total waste generated in the 2020s have decreased compared to the 2010s as the waste generated are lower on average. B1 15c Yes. The amount of waste disposed of remains generally constant over the last decade but the waste recycled was generally lesser in the 2020s, thus there is a decline in percentage of waste recycled in the 2020s. B1 B1 16a angle ACB = angle ADC (given) angle BAC = angle CAD (common angle) Since all corresponding angles of the triangle are equal, triangles ABC and ACD are similar M1 A1 16b 2 6.5 6.5 5.5 78 AB AC AC AD AC AC AC = = + = 8.83AC= cm M1 A1 17a ( )( ) 2 2 2 2 22 4 a b a b a a b a + − − = − − 24b=− M1 – correct expansion A1 17b ( )( ) ( )( ) ( ) 2 2 2 12 349 12 341 12 345 12 345 2 2 12 345 2 2 12 345 42 − +  − − =− 16=− M1 A1 18a No. of days 15 8 245=  = B1 18b T k l= % increase ( ) 1.5 100 1.5 1 100 k l k l kl kl kl −= − = =22.5% M1 A1 19a 4 cm : 1 km 4 cm : 100 000 cm 1 : 25 000 B1

19b 40 4 10= km B1 19c 4 cm : 1 km 16 cm2 : 1 km2 4.5 16 72= cm2 M1 A1 20a ( )( )2 2 262 56 73 2 56 73 cos BAC= + −  55.5845BAC= Bearing 180 55.5845=− = 124.4 M2 A1 A1 20b Area of triangle ABC ( )( )1 56 73 sin 55.58452= = 1686.2195 =1690 km2 M1 A1 21a 356 2 7= 2126 2 3 7=   56 126 4 2 22 3 7=   Since the powers of all prime factors are even, 56 126 is a perfect square. M1 M1 A1 21b 2 7 14= B1 21c 222 3 7k =   = 588 M1 A1 22a ( )( ) 29 6 8 0 3 2 3 4 0 24 or 33 xx xx xx + − = − + = = =− M1 A2 22b ( ) ( ) ( ) ( ) ( ) 2 2 2 9 1 6 14 0 9 1 6 1 6 14 0 9 1 6 1 8 0 yy yy yy − + − = − + − + − = − + − − = From (a) , 241 or 133yy− = − =− 211 or 33yy= =− M1 A1 23a B = 30 60 35 55   B1 23b C = 20 40 30 605230 30 35 55     +      

= 160 320 220 260   B1 23c Elements of C represent the total number people from each of the categories of adults and children, admitted to the theatre in the morning and afternoon respectively over one week. B1 23di D = 160 320 25 7840 220 260 12 8620      =         B1 23dii E =   7840131 1 2351862077  =  B1 23diii E represents the average amount of money collected per day from admissions over one week. B1 24a 3 4 1 12 x xx− − = − 3 3 2 02 x x− − = 3 6 4 0xx− − = B1 24b Draw tangent Gradient = 2 M1 A1 24ci 3 3 3 40 202 4 1 2 12 xx x x x xx −= −= − − =− − Equation of line: 21yx=− − B1

24cii 24cii Draw 21yx=− − x = – 2, 0, 2 M1 A1 25a 16 m/s B1 25b 2 m/s2 B1 25c Distance = area under graph = 100 + 200 + 200 = 500 m M1 A1 25d B1 for straight section correct B1 for curved section correct