NASS 2025 EMATH PRELIM P1 MS

Ngee Ann Secondary School Secondary 4&5 Mathematics-O 2025 Prelim Examinations Paper 1 Marking Scheme 1 [B1] 2 8 96x− −= 8 15x−= 15 7or 1 or 1.87588x= −−− [B1; no mark for 1.88− ] 3 6 2410 1 2660100 r+= [M1] 6 26601 100 2410 r+= 1 626601 100 2410 r +=  1 62660 1100 2410 r = − 1 62660100 12410r = ×−  1 62660100 12410r = ×−  1.66= (3 s.f.) [A1] 4(a) 0.0050500 (5 s.f.) [B1] 4(b) 35.0500 10 −× [B1; accept 35.05 10 −× and allow for ecf] 5 32600 2 3 5= ×× For the base area to be a square, possible dimensions are 2 2 150×× , 5 5 24×× or ( ) ( )25 25 6×××× . [M1] Smallest possible height is 6 cm. [A1]

2 NAS/2025/Prelim/MA-O/P1 6(a) 72 160000360× [M1; for 72 1°± ° ] $32000= [A1] 6(b) The family’s annual income in 2024 might be less than that in 2023, such that even though they spent a great proportion of their income on food in 2024 (25% compared to 72 100% 20%360×= ), the total amount spent on food in 2024 might be less than that in 2023. [B1; for underlined response. Accept any reasonable response] 7(a) ( )( )52 34xy xy− − −+ =( )( )52 34xy xy−+ −+ or ( ) 22 15 20 6 8x xy xy y−− + + − [M1] = 2215 20 6 8x xy xy y− −+ = 2215 26 8x xy y−+ [A1] 7(b) 29 12 28 21ab ac bc b−+− = ( ) ( )33 4 73 4a bc b bc−− − or ( ) ( )33 4 74 3ab c bc b−+ − [M1] =( )( )3437bcab−− [A1] 8(a) 2 36yx= − 33 2yx= −+ Gradient is 3− . [B1] 8(b) The line in part (a) is parallel to the line in part (b) since they have the same gradient. [B1] Also, for the line in part (a), when 0x= , 3 2y= . The point (0, 1) does not lie on the line in part (a). [B1; must explain that (0, 1) does not lie on the line in part (a). Accept any reasonable response] Therefore, the two lines will not intersect. 9(a) 2243BC = − [M1] 7= 2.65= cm (3 s.f.) [A1] 9(b) ( ) 22 77BC = = ( ) ( ) 2222 2 37BD CD+=+ = [Must be two separate lines of working] Since 2 22BC BD CD= + , by the Converse of Pythagoras’ Theorem, triangle BCD is right-angled. [B1; alternative method: use Cosine Rule to find angle BDC exactly 90°]

3 NAS/2025/Prelim/MA-O/P1 10 2 3 4 19 30 32 50 xx xx +− − = ( )( ) ( )( ) 45 6 24 54 5 xx xx x −+ −+ or ( )( ) ( )( ) 45 6 4 5 8 10 xx xx x −+ −+ [B1: for numerator] [B1: for denominator] = ( ) 6 24 5 x xx + + [B1] 11 231 13 23 xx x −− −= − ( )( ) ( ) ( ) 2323 3 1 3 23x xx x− −−−=− [M1: removing denominator] 2223 6 93 3 6 9xx x x x− −+ −+ =− [M1: expand ( )( )323xx−− correctly] 11 9 6 9xx−=− 20 15x= 3 or 0.754x= [A1] 12 Each interior angle of pentagon ( )5 2 180 5 −×= [M1] 108= ° Each interior angle of octagon ( )8 2 180 8 −×= [M1] 135= ° Sum of angles of a, b and c ( ) ( )360 108 2 360 108 135= − + −− 486= ° [A1]

4 NAS/2025/Prelim/MA-O/P1 13 2d kt=