NASS 2025 EMATH PRELIM P2 MS

1 Ngee Ann Secondary School Secondary 4&5 Elementary Mathematics-O 2025 Prelim paper 2 Marking Scheme Qn No. Qn Part Solutions Marks (Remarks) Total 1 (a)(i) 23( ) 1hybx b += − 23 (0 5)21 5 h += − 233 h= 2 1h = 1h=± M1: substitution A1: both answers [2] (a)(ii) 23( ) 1hybx b += − ( ) 23xb h y b b= +− 2233xb h y h b b=+− 2233xb b h b h y+− = 22( 13)3bx h hy+− = 2 2 3 13 hyb xh= +− M1: Multiply denominator throughout M1: Grouping and factorising b A1 [3] (b) 6 3 16 (1) 9 2 11 (2) xy xy − = −−− + = −−− (1) 2 12 6 32 (2) 3 27 6 33 xy xy × −= × += 39 65 213 x x = = 26 1 3 163 y −= 2y=− M1: Substitution/ Elimination A1 A1 [3] (c) 2 19 3 2 72 53 3 x xx x + +=−− −

2 19 3 2 7(2 1)( 3) 3 x xx x + +=+− − 219 3 2(2 1) 7(2 5 3)x x xx++ + = − − 219 3 4 2 14 35 21x x xx++ += − − 214 58 26 0xx− −= 258 ( 58) 4(14)( 26) 2(14)x ±− − −= 0.41x=− or 4.55x= M1 M1 A1 [3] 2 (a) 5.3 B1 [1] (b) Refer to graph B2: correct points plotted B1: any one error B1: Smooth curve [3] (c) a = 0.3 ; b = 3.7 B1, B1 [2] (d) – 1.13 to – 1.39 M1: drawing of correct tangent A1: Within the range [2] (e) 552 4 x x −+= 2085 xx+= − 28 20 5x xx+=−

3 2 3 20 0xx++= 1A= and 3B=− B1, B1 [2] 3 (a) 1500 x B1 [1] (b) 1500 50x− B1 [1] (c) 1500 1500 1 250 2xx −=− 251500 1500( 50) ( 50 ) 2x x xx− −= − 23000 3000( 50) 5( 50 )x x xx− −= − 23000 3000 150000 5 250xx xx−− = − 25 250 150000 0xx−+ = 2 50 30000 0xx−+ = (Shown) M1: Correct equation AG1 [2] (d) 2 50 30000 0xx−+ = ( ) 150 ( 200) 0xx+ −= x =200 or x = – 150 M1: factorisation or other method A1, A1 [3] (e) 200 63.7π = cm (3 sf) B1 [1]

4 4 (a)(i) 29 min B1 [1] (a)(ii) 34 – 23 =11 min M1 A1 [2] (a)(iii) 80 44 100%80 − × = 45% M1 A1 [2] (b) The new cumulative frequency curve will meet the given curve at the median. However, the gradient of the new cumulative frequency curve from the lower quartile to upper quartile will be less steep as compared to given curve. B1 [1] (c)(i) 26 25 65 80 79 632×= B1 [1] (c)(ii) 26 14 280 79×× = 91 790 M1 A1 [2] 5 (a)(i) DB   =16p – 5q B1 [1] (a)(ii) 1 4DP=   (16p – 5q) AP AD DP= +       = 5q + 1 4 (16p – 5q) = 15 4 q + 4p M1 A1 [2] (b) 4 3DC AB=    BC BA AP DC=++        = – 16p + 5q + 64 3 p = 16 3 p + 5q 4 3BC =   (15 4 q + 4p) M1: find BC  

5 = 4 3 AP   Since 4 3BC AP=     , they are parallel. A1