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= ( ) 2 320 8 k= [M1] 320 64k = 5k = 280 5 t= 2 80 165t = = 16 4 or 4 (N.A.)t = ±= − 4t = [A1] 14 Let the amount of money shared by $x. Amy’s original share: 2 9 x Amy’s final share: 32 1 49 6 xx = [M1] Amy’s transferred share: 12 1 4 9 18xx = Ben’s original share: 31 93xx= Ben’s final share: 21 1 7 3 3 18 27xx x += [M1] Ben’s transferred share: 11 1 7 3 3 18 54xx x += Cal’s original share: 4 9 x Cal’s final share: 4 7 31 9 54 54xx x+= [M1] Final ratio 1 7 31::6 27 54xxx= 9 : 14 : 31= [A1]

5 NAS/2025/Prelim/MA-O/P1 15(a) 15 5.8 5 14.4 10 x×= × + [M1] 10 15x= 1.5x= [A1] 15(b) Median would be more appropriate because the mean would be affected by the outlier of 50 points. The player who scored 50 points contributed about 69% of all points scored by starting players. [B1: must mention outlier of 50 points. Accept any reasonable response] 16(a) 2AC a= 2BD d= Area of rhombus ( )1 222 ad= × 2ad= [B1] 16(b) Area of triangle PQS ( )1 22 xy= × xy= [M1] 3MR x= Area of triangle RQS ( )1 322 xy= × 3xy= Area of kite 3xy xy= + 4xy= [A1]

6 NAS/2025/Prelim/MA-O/P1 17(a) 1 4 4 8 625x y −   1 2 1 5 x y − −= [M1: for either 1 5 , 1x− , 2y− ] 2 5 y x= [A1] 17(b) 888243 3 3 3a =++ ( ) 583 3111a = ++ ( ) 583 33a = 5933a = [M1: for getting 53 a or 93 ] By comparing indices, 59a= 94or 1 or 1.855a= [A1] 18(a) When xh≠ , ( ) 2 0xh−> and ( ) 2 y xh k k= − +> . When xh= , ( ) 2 0xh−= and ( ) 2 y xh k k= − += . (Or mention that coeff of x2 is positive) [B1: for highlighted part] So, y has its minimum value of k when xh= . 18(b) Since the graph of ( ) 2 y xh k= −+ is symmetrical about the vertical line passing through its minimum point, 24 2h −+= [B1] 1h= (shown) 18(c) Sub 4x= , 0y= , 1h= , ( ) 2 0 41 k=−+ 9k = − Min. value of y is 9− . [B1]

7 NAS/2025/Prelim/MA-O/P1 18(d) [C1: correct shape] [P1: correct coordinates for x-intercepts and min pt; must be coordinates] 19(a) When 1n= , ( )( )3 113 32 −= − So, 1 3T =− When 2n= , ( )( )3 223 32 −= − So, 12 3TT+= − 233 T−+ = − 2 0T = When 3n= , ( )( )3 333 02 −= So, 123 0TT T++= 330 0 T−++ = 3 3T = First 3 terms: 3, 0, 3− [B1: one correct term] [B1: all terms correct] 19(b) 1 3T =− Common difference = 3 63nTn= −+ [B1] 19(c) Since ( )63 3 2nT nn= −+ = − , [B1: must show factorization] each term in the sequence is a multiple of 3.

8 NAS/2025/Prelim/MA-O/P1 20 180 119CDA∠ = °− ° (∠s in opp. segments) 61= ° [B1: with reason] 90OAT∠= ° (tan rad⊥ ) 29OAD∠= ° (base sof isos. ∠∆ ) So, 90 29DAT∠ = °− ° 61= ° [B1: with reason for tan rad⊥ ] Since CDA DAT∠= ∠ , they are alternate angles and lines CD and ST are parallel [A1] 21(ai) { }' ' 0, 4, 6, 8, 10AB∩= [B1] 21(aii) [B1] 21(bi) 12, 16 [B1: do not accept {12, 16}] 21(bii) P = {16} Q = φ R = {12, 16} So, ( ) 'PQ R φ∪∩= ( )n '0PQ R∴ ∪∩ = [B1]

9 NAS/2025/Prelim/MA-O/P1 22 Method 1 Let distance QF be x m. tan 20.8 20 TF x°= + [M1] ( )20 tan 20.8x TF+ °= 20 tan 20.8 tan 20.8 TFx −°= ° tan 21.7 TF x°= [M1] tan 21.7 TFx= ° So, 20 tan 20.8 tan 20.8 tan 21.7 TF TF−° =°° [M1] tan 21.7 20 tan 20.8 tan 21.7 tan 20.8TF TF°− ° °= ° tan 21.7 tan 20.8 20 tan 20.8 tan 21.7TF TF °− °= ° ° ( )tan 21.7 tan 20.8 20 tan 20.8 tan 21.7TF °− ° = ° ° 20 tan 20.8 tan 21.7 tan 21.7 tan 20.8TF °°= °− ° = 167 m (3 s.f.) [A1] Method 2: 180 21.7 (adj s on a st. line)PQT∠ = °− ° ∠ 158.3= ° 180 158.3 20.8 ( sum of )PTQ∠ = °− °− ° ∠ ∆ 0.9= ° By Sine Rule, 20 sin158.3 sin 0.9 PT =°° [M1] 20sin158.3 sin 0.9PT °= ° 470.79555= [M1]

10 NAS/2025/Prelim/MA-O/P1 sin 20.8 470.79555 TF°= [M1] 470.79555sin 20.8TF = ° = 167 m (3 s.f.) [A1] 23(a) P = 546 879   [B1] 23(b) T = 120546 80879 100     = 1520 2420   [B2: B1 for each of the element] 23(c) The elements in T represent the total cost ($1520) and the total revenue ($2420; or selling price) respectively of the cakes sold by the shop in July. [B1: not necessary to state the values of the elements; do not accept the term “earnings”] 23(d) Q = ( )11− [B1] 24(a) Volume of one cup ( ) 2 1 6.9 9.132π=  113.4248= 113= cm3 (3 s.f.) [B1] 24(b) 18.9 l = 18 900 cm3 Max number of cups 18900 113.4248= [M1] = 166.63 = 166 cups that are completely filled [A1] 24(c) Let the curved surface area of smaller cone (i.e. cup) be A1, t and that of larger cone (i.e. cup + frustrum) be A2. By similar cones, 2 22 11 2dA dA  = =  [M1] 2 1 2d d =