Northbrooks 4E5N E Math Prelim P2 2026 MS Final

1 Northbrooks Secondary School 4E5N Preliminary Examinations 2026 Mathematics Paper 2 Mark Scheme 1a 21 2 3 4 2(3 4) 1( 2) 6 8 2 5 10 2 xx xx xx x x =+− − = + − = + = = M1 – Cross multiply / common denominator A1 1b 22 22 3 48 3( 16 ) 3( 4 )( 4 ) xy xy x y x y − =− = + − M1 – Factor 3 seen A1 1c 85 15 15 15 15 100 16 100 16 84 33 333 3 3 3 3 −  ++ = = = M1 – 100 16 3 3 shown A1 1d Elimination / Substitution Method 3 2 2 (1) 7 3 43 (2) From (1) 3: 9 6 6 From (2) 2 : 14 6 86 9 14 86 6 23 92 4 5 xy xy xy xy xx x x y −= +=  − =  + = + = + = = = M1 For elimination – making coefficient the same and either + or − correctly For substitution – making x or y the subject correctly A1 A1

2 1e 2 2 2 2 2 2 2 ( )( 3 ) 3 ( )( 3 ) 3 ( )( ) ( )( 3 ) ( )( 3 ) ( )( 3 ) y x y x y x y y x y x y x y x y x y y x y x y x y x y y x y x y x y x x y x y −−+ − − −=+ + − − + − += +− +−= +− = +− M1 – Factorise -ve 3 ( 3 )y x x y− =− − M1 – attempt common denominator and correct expansion of numerator A1 2a 2 2 2 2( )( )cosQR PQ PR PQ PR RPQ= + −  222 3 2(2)(3)cos 60 = + −  = 7 (ratio) Hence 2 2 2: 3 : 7PR RQ = = 9 : 7 (shown) M1 – Apply correct Cosine rule M1 – Calculate to find ratio of QR2 A1 2bi 2 2 25 9 12cos 2 5 9 19 45 114.975 180 114.975 65.0 (1d.p) ADE ADE DAB DAB +−=  =− =  = − =   M1 – Apply correct Cosine rule M1 – Find angle ADE correctly A1 2bii 2 Area of 12 ( 5 13 sin 65.025 )2 58.9cm (3s.f ) ABCD =     =  M1 – Area ADB A1

3 2c 2 2 2 23 3 Vol of cylinder = (8) (14) 1Vol of cone = (8) (14)3 2Vol of remaining solid = (8) (14)3 24 (8) (14) 7 ( )33 64 4 r r r     = = = M1 Find volume of remaining solid M1 – Form equation A1 3ai 48 g B1 3aii 51.5 – 45 = 6.5 g M1 (for LQ or UQ) A1 3aiii 40 g B1 3b 260 – 200 = 60 260 – 195 = 65 60 59 59 300 299 1495= or 0.0395 65 64 16 300 299 345= M1 – Identify 60 cookies A1 3c Second batch Median = 44 g IQR = 50 – 40 = 10 g 1. Batch 1 cookies are heavier than Batch 2 cookies, as the median mass is higher (48 g > 44 g). 2. Batch 1 cookies are also more consistent than Batch 2 cookies, as the interquartile range is smaller (6 g < 10 g). B1 B1 4a 3174.20 1.18 $2690  = M1 – 1.18 or 118% A1 4b First Map Map : Actual (Comparing Length) 1: 50 000 1 cm : 0.5 km

4 Map : Actual (Comparing Area) ( ) ( ) 22 22 22 1 cm : 0.5 km 1 cm : 0.25 km 16 cm : 4 km Second Map Map : Actual (Comparing Area) 22 25 cm : 4 km Map : Actual (Comparing Length) 22 25 cm : 4 km 5 cm : 2 km 1 cm : 0.4 km 1 cm : 40 000 cm 1 : 40 000 M1 – Finding area ratio M1 – Finding actual area of building M1 – Converting area scale to linear scale A1 4c Dance : Music 4 : 7 Music : Art 3 : 5 Dance : Music : Art 12 : 21 : 35 Let the number of students in Dance, Music & Art be 12x, 21x and 35x. 1 (35 ) 12 955 19 95 5 68 340 xx x x x += = = = M1 – Find ratio of all three activities M1 – Equation formed to find one unit A1 4di 0.9 0 0 0 1 0 0 0 0.8    B1 Alternatively, Dance : Music : Art 12 + 7 : 21 : 35 – 7 19 : 21 : 28

5 4dii 0.9 0 0 50 0 1 0 40 0 0 0.8 75 45 40 60      =        = T T B1 4diii The discounted price of Apple Pie, Nutella Pie and Strawberry Pie respectively. B1 4div ( ) ( ) 4513 15 18 2265 4010 20 22 2570 60 22651 1 48352570      =          = Total amount = $4835 M1 – from their (ii) M1 A1 5ai Angle ABE = Angle ADC (given) Angle BAE = Angle DAC (common angle) By AA similarity, triangles ABE and ADC are similar. B1 – Need 2 correct statement A1 5aii 8 12 6 16 cm = 10 cm AC AE AD AB AC AC BC = = = M1 – Correct similarity ratio A1 5aiii Area ratio of triangles ABE and ADC = 1 : 4 Area ratio of quadrilateral BCDE and ADC = 3 : 4 B1 5b (5 2) 180Each int of 108 5A − = =  Each int of 360 108 90C = − −  = 162º M1 – Find int angle of pentagon M1 – Find int angle of C (FT)

6 Solve by int angle / ext angle ( 2) 180 162n n −   = n = 20 M1 – Equation formed (FT) A1 6a -0.3 B1 – Exact, do not award -0.25 6b B3 Or B2FT for 8 points plotted correctly Or B1FT for 6 or 7 points plotted correctly Tolerance  0.5mm (half a small grid) for plotting points and drawing curve through points 6c -2.25  0.5 B1 6d 3 3 34 3 2 24 x xk x xk −= − + = + 2yk=+ (From graph, y = 6 or – 2) k = 4 or – 4 B1, B1

7 6e 3 3 3 3 16 4 0 4 1 04 4 1 3 34 3 2 34 3 xx x x x x x x x xx yx − −  − −  − − + +  + − +  + =+ The intervals of x for which the curve is above the line: 3.9 0.25x−  − M1 – Correct equation M1 FT – Correct straight line drawn from their equation A1 FT - Correct inequality formed 7ai {6,12} B1 7aii 18 B1 7b PQ B1 7ci 16 B1 7cii 9 B1

8 7d full-time woman 2 287= = 8 2P(PT man, PT man) = 27 Let m represent the number of part-time men. 12 28 27 27 ( 1) 2 756 27 ( 1) 56 8 or 7 (rej) mm mm mm mm −= − = −= = =− FT man = 28 6 8 8 − − − = 6 Part-time employees Full-time employees Women 6 8 Men 8 6 B1 M1 – Equation form using probability given M1 – Find m A1 FT – from their value of m 8a ( ) ( )( ) 22 2 2 7 6 1 13 14 49 25 169 14 95 0 ( 19)( 5) 0 19 or 5 k kk kk kk k − + − − − = − + + = − − = − + = =− Since k < 0, k = –5 M1 – Equation formed using length A1 8b Area of triangle 2 1 (8 6) (7 5)2 84 units =  +  + = M1 A1

9 8c ( ) ( ) ( ) 81 3Gradient 0.75 5 7 4 385 4 17 ( 4.25)4 3 17 44 PQ or c c or yx −−= =− −−− =− − + = =− + M1 – gradient of PQ M1 – find y-intercept based on their m A1

10 PRWC 9a 5 34300 1 100 4984.878 $5000 (shown) + =  B1 9b 100 000 5(5000) $75 000 − = Accept more accurate value (to 2 d.p): 100 000 5(4984.878) $75 075.607 $75 075.61 − =  B1 9c To find the maximum loan amount, we consider the most expensive possible scenario, 1598cc petrol car Category A, Max COE Price  $120 000 Total cost price 120 000 120 000 $240 000 + = 208kW electric car Category B, Max COE Price  $130 000 – 140 000 Total cost price 130 000 130 000 30 000 $230 000 +− = For 208kW electric car, downpayment required for 60% bank loan 40% 230 000 $92 000  = Downpayment > $75 000 (unable to afford) For 1598cc petrol car, downpayment required for 70% bank loan 30% 240 000 $72 000  = Downpayment < $75 000 (able to afford)