NBSS EM Prelim P1 2024 v2 MS

1 Answer Scheme Sec 4E/5N Mathematics Preliminary Examination Paper 1, 2024 Qn Answer Marks Remarks 1a 3.58 B1 2a 2512xy B1 2b 68x− B1 3(2 4)x− not accepted 3 32 2 2 13 6 (2 13 6) (2 1)( 6) x x x x x x x x x −+ = − + = − − B1, B1 Fact. x or quadratic expression B1, B1 4 36.9 or 143.1 B1, B1 5a 7.75 B1

2 Qn Answer Marks Remarks 5b 3 6 7 8 9 9 10 10(12 8) ( 20 96 (6 ) )2 20 158 20 7.9 + + ++ += = + = + + + M1 A1 6a 12250 p q 3212250 2 5 7=   p = 7 q = 2 B1 B1, B1 6b 32 3 3 12250 2 5 7 2 5 3 HCF = 250 2 5 750 x x =   =   −−−−−−−−−−−−−−−− = = B1 7 2 4 2 2 3 81 9 27 (3 ) (3 ) 3 8 2 3 3 / 0.310 xx xx xx x = = += = or 4 1.59 9 9 5 1.5 0.3 3 / 0.310 xx x x x = = = = M1 A1 See base 3 or base 9

3 Qn Answer Marks Remarks 8 Top Perimeter 20 length + breadth 10 = = Paired No. Area of top Height 1, 9 Reject < 3 2, 8 Reject < 3 3, 7 Reject  3 4, 6 4 6 24= 120 524Height== 5, 5 Reject, cannot be same length = cube Or 120 2 2 2 3 5 =     By trial and error, Height = 5 cm M1 A1 M1 A1 Prime factors 9 ( ) 2 2 2 2 2 2 2 4 LT g TL g LT g Lg T     = = = = M1 M1 A1 Divide 2 Square both sides

4 Qn Answer Marks Remarks 10a 310 1 (309 – 311) B1 10b See behind (Construct 5 cm radius) 4/ 4.5 / 5 (Convert to km)km M1 A1 11ai 282 50 5.64 Mean= = B1 11aii - There is no exact data of each child’s time spent on playing online games - The average for time spent on playing games is used in the calculation - The time spent is given as a range of values - The mid value average is used when calculating the mean B1 11b 2 1906 282 50 50 2.51 SD =−  = M1 (ecf mean) A1 Show correct substitution of each value in formula

5 Qn Answer Marks Remarks 12 2 (1)42 3 19 4 (2) (1) 4 28 8 2 (3) sub (3) into (2) 3(8 2 ) 19 4 24 6 19 4 2.5 8 2( 2.5) 3 xy xy xy xy yy yy y x x − = −−−−−−−−− = + −−−−−−−−  −= = + −−−−−−−− + = + + = + =− = + − = M1 A1 A1 13a 5 ( 4)yx − B1 13bi AB  AB ()B A B B1

6 Qn Answer Marks Remarks 13bii B1 14a 521 9 15 14 45 −− = B1 A A B

7 Qn Answer Marks Remarks 14b 3 10 8 16 2 12 12 7 5 parts balls part balls parts balls n n balls −− + −− −− =− = OR 72 10 7 8 5 3 15 5 n n n n + =+ + + = = OR 3P(pick red and blue balls) 5 3 10 8 16 2 12 12 7 5 parts balls part balls parts balls n n balls = −−  + −−  −−  =− = M1 A1 15 3 100%53 37.5% + = M1 A1

8 Qn Answer Marks Remarks 16 Method 1 2 1 2 1 2 22 2 (5 ) 25 25 100% 1 0.04 100%1 96 % kx y kx y kx y kk yyPercentaage change k y = = = − = −= = M1 M1 A1

9 Qn Answer Marks Remarks Method 2 2 2 1 2 1 (5 ) 25 25 100% 96 % k xy xyx y xx xx Percentaage change x = = = − = = M1 M1 A1 17a 90 B1 17b ( 1) ( 1) 10 10 3 18 nL n n n n = − + − + + + =+ B1 (See 3n) B1 (See 18) 17c 3 18 3( 6) nLn n shown =+ =+ B1 (no ecf) 18a 8.4tan 5 59.2 ABC ABC =  =  OR AB = 9.77548 1 8.4sin ( )9.77548 59.2 ABC ABC −=  =  M1 A1

10 Qn Answer Marks Remarks 18b 8.4sin 48 11.3 AD AD = = M1 A1 19 2 22 2 2 5 16 0 ( 2.5) (2.5) 16 ( 2.5) 6.25 16 ( 2.5) 22.25 2.5 22.25 2.5 22.25 2.22 7.22 xx x x x x x x or + − = + − = + = + += + = =−  =− M1 M1 A1, A1 See completing the sq Sq rt 20a 3sin 5 4 AXC BX = = 3tan 4AXB= M1 A1 20b 3tan 42 1 2 ACB= + = M1 A1