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 [2] (c)  APB =  CBD (alternate  , AP//BC)  ABP =  CDB (alternate  , AB//DC) Triangle APB is similar to Triangle CBD by AA similarity. Triangle APB : Triangle BCD 3 2 : 42 9 : 16 OR Area of Triangle 3 Area of Triangle 4 APB BAD = Area of Triangle 3 Area of Triangle 4 BAD BCD = Area of Triangle Area of Triangle Area of Triangle Area of Triangle 33 44 9 16 APB APB BAD BAD× = × = B1: Any correct line B1: All correct A1 M1: Any of the ratio M1 A1 [3] (d) Since Triangle APD and Triangle APB have common height, Triangle APD : Triangle APB 1 : 3 Triangle APD : Trapezium ABCD 3 : 28 (Since 16 + 9 +3 = 28) OR M1 A1

6 Area of Triangle Area of Triangle Area of Triangle Area of Trapezium 13 47 3 28 APD BAD B AD ABC D× = × = M1 A1 [2] (6) (a)  POS =  QOR (vertically opposite  ) OS = OR (radii of circle)  OSP =  ORQ (tangent ⊥ radius) Triangle PSO is congruent to Triangle QRO (ASA). (tan )ORQ OSP rad= ⊥ OQ = OP (radii of circle) OR = OS (radii of circle) Triangle PSO is congruent to Triangle QRO (RHS). B1: For any line B1: All correct [2] (b)(i) 228 4.36QR= − = 6.707488353 Area of triangle: 1 4.36 6.7074883532×× = 14.6 cm2 (3 sf) M1 M1 A1 [3] (b)(ii) Shaded circular area: 22(8) (4.36)ππ − = 141.3415101 1 4.36cos 8ROQ − =   = 0.9944072121 Area of sector: 21 (4.36) (0.9944072121)2 =9.451641669 Total shaded area: 141.34151 – 2(14.62232461 – 9.451641669) + 2(9.451641669) = 149.9034 cm 2 = 150 cm2 (3 sf ) M1 M1 M1 A1 [4]

7 7 (a)(i) 75÷30 = 12 2 m/s2 B1 [1] (a)(ii) 75÷5 = 15 s B1 [1] (a)(iii) B1: correct shape for both sections B1: correct values on both axes [2] (b)(i) For 70 m/s, Time for acceleration: 70 ÷ 12 2 = 28 min Time for deceleration: 70 ÷ 5 = 14 min Distance covered by 70 m/s 11 28 70 70 1422××+ ×× = 1470 m The plane will be able to decelerate safely if the speed of the aircraft is 70m/s as the distance covered would be less than 1600 m. M1 M1 M1 A1: Must include compariso n with 1600 m [4] (b)(ii) V1 would be lower as there are less friction, and the runway is of a fixed length. B1 [1] 8 (a) Angle ABD: 180° – 76° – 50° = 54° Bearing of D from B: 360° – 54° – 76° = 230° M1 A1 [2] (b) Let AX be the shortest distance. 1687.5

8 55sin 54AX = ° = 44.49593469 10tan 44.49593469θ = 12.66618θ = ° = 12.7° M1 M1 A1 [3] (c) 55 sin 76 sin 50 BD =°° 55sin76 sin50BD °= ° = 69.664711 m Area of park: 11 55 69.664711 sin54 30 69.664711 sin13022× ×× ° + × ×× = 2350.392187 = 2350 m 2 M1 M1,M1: Area of each triangle A1 [4] (d) 2230 69.664711 2(30)(69.664711) cos130BC = +− ° = 91.86919365 m 55 sin54 sin50 AD =°° 55sin54 sin50AD °= ° = 58.08531749 m Perimeter: 55 +30 +91.86919365 +58.08531749 = 234.9545111 I disagree with Tony as the perimeter is less than 240m. M1: Find BC M1: Find AD A1: Need concluding statement with comparsio n with 240 m [3] 9 (a) 9880 – 1480 – 10 – 5 = $8385 B1 [1]

9 (b) Based on lowest rate, 54 000 × 1.280108 = SGD 69 125.83 OR Based on highest rate, 54 000 × 1.31154 = SGD 70 823.16 OR Based on average rate, 54 000 × ((1.280108+1.31154)/2) = SGD 69 974.50 M1 A1 [2] (c) Only Cash Components: Buyer Stamp Duty: 1800 + 180 000×0.02+ 640 000×0.03+ 300 000×0.04 = $36 600 Booking Fee: 0.05 × 1 300 000 = $65 000 Total: 3000 + 500 + 36 600 + 65 000 = $105 100 M1: BSD M1: Booking Fee A1 [3] (d) Assumption: Mr Sim used current account and matured fixed deposit to pay for the condominium. Outstanding amount: 0.95×1 300 000 = $1 235 000 Assuming that the contribution from his June pay is negligible Remaining amount to be covered by Bank loan 1 235 000 – 350 000 – (69 974.50 – 5100) = $820 125.50 Interest over 20 years: 820125.50 3.3 20 100 ×× = $541 282.83 M1: Find loan amount M1: Interest

10 Suggested condo instalment: 541282.83 820125.50 20