Northbrooks 4E5N E Math Prelim P 1 MS final

2026 4E5N Math Prelim Paper 1 Marking Scheme 1 950 B1 2 4 1.8$55000 1 2 100  +  = $57006.89 (nearest cents) M1 A1 3 Possible answers: 7 cm by 9 cm by 16 cm 7 cm by 8 cm by 18 cm 7 cm by 12 cm by 12 cm B2 B1 for 2 correct answers 4 B1 B1 B1 Line bisector Angle bisector Shaded region 5a 1 day needs 12 workers 3 days need 12  3 = 4 workers B1 o.e. 5b 1 day needs 12 workers 5 days need 12  5 = 2.4 workers Number of structures = 36  2.4 = 15 M1 A1 o.e. 6 Let the mean COE price in 2022 be x. The mean COE price in 2023 = 1.254x. The mean COE price in 2024 = 1.254x × (1 – 0.0505) = 1.1907x The mean COE price in 2025 = 1.352x Price increase = 1.352x – 1.1907x = 0.1613x Percentage increase = 1.1907 100%their price increase x  M1 M1 A1 Accept FT

= 13.5% (3 s.fs.) OR 1.352 100%1.190673 = 113.549% Percentage increase = 113.549% – 100% = 13.5% M1 A1 7a The statement could be false because we do not know the total number of shirts sold in each store. If total shirts sold Outlet A are more than Outlet B, then Outlet A could have sold more blue shirts than Outlet B. B1 o.e. 7b Angle subtended by black shirts = (1351)o percentage of the black shirts sold = 135 100%360  = 37.5% M1 A1 37.2% (134o) 37.8%(136o) 8a Height = 2213 5− = 12 cm M1 A1 Use of P.T. 8b 21 10 123  = 400 cm2 M1 A1 Accept FT (a) 9a 4 3 5 2 21 28 5 3 20 p q px x q p q  = M1 A1 Or B2 9b 15ab – 3ac – 5b + c = 3a(5b – c) – (5b – c) = (3a – 1)(5b – c) M1 A1 10 3 12 p − = p = 5 y = – k(x + 3)(x – 5) , k > 0 At q, x = 0 q = – k(0 + 3)(0 – 5), k > 0 = 15k q is a multiple of 15 and q > 0. B1 B1 Accept any answer with k > 0 11a 20.08 25a or= − − B1

11b (5a – 1)(1 – 2c) = 3c 5a – 10ac – 1 + 2c = 3c 5a – 1 = c(1 + 10a) 51 1 10 ac a −= + M1 M1 A1 Remove denominator c as a common factor o.e. 12 Given PT = ST, SPT = x (base angle of isosceles ) QRS = 90o – x (angle sum of a ) Given QS = QR, QRS = QSR (base angle of isosceles ) = 90o – x RQS = 180o – 2(90o – x) (angle sum of a ) = 2x (shown) M1 M1 A1 13a PQ OQ OP=−  3 1 4 4 5 9 −     = − =     −−      B1 13b PR k PQ=  3 1 4 59 kh −     −=      −      4 = 4k 1k = h – 5 = –9k h – 5 = –9 h = – 4 B1 B1 Accept FT Accept FT 14a L.H.S R.H.S 2x – 7 < 25 25  3x – 2 x < 16 x  9  9  x < 16 M1 M1 A1 LHS RHS 14b 11 and 13 B1 Accept FT 15a 3 11 64 4 height of jar A height of jar C == Diameter of base of jar A : Diameter of base of jar B 1 : 4 B1

15b 2 3 9 64 900 Surface areaof jar B = Surface area of jar B = 243 cm2 (3 s.f.s) M1 A1 16 (2n – 1 )(2n + 3) = 4n2 + 6n – 2n – 3 = 4n2 + 4n – 3 = 4(n2 + n) – 3 Since 4(n2 + n) is an even number, 4(n2 + n) – 3 is odd. OR 2n is even. So (2n – 1 ) and (2n + 3) are odd numbers. The product of odd numbers is also odd. M1 A1 Express as 4k – 3 or 2k + 1 Justify the answer 17a 57 212 11 35 66   = M1 A1 17b 5 7 6 312 11 10 21 44     = M1 A1 18a y = (x – 5)2 + 4 B1 B1 Value of a Value of b 18b The minimum point is at (5, 4) which is above the x-axis. OR At y = 0, 45x =  − + , which is undefined. Therefor the graph will not pass through the x-axis. B1 19a 3 12 4 416 m n   9 38 m n= B1 B1 m6 8n3 19b 3343 3 21xx= 337 3 21xx= 321 21x = or 73x × 33x = 7 × 3 1 3x = M1 A1 20ai 360 = 23 × 32 × 5 B1

20aii 72, 90 B1B1 20b 2×32 × 7 =126 B1 21 2( 5) 5 (2 3) 1 (2 3)( 5) 2 x x x xx − + − =−− 2 2 2 10 10 15 1 2 10 3 15 2 x x x x x x − + − =− − + 2(10x2 – 13x – 10) = 2x2 – 13x + 15 18x2 – 13x – 35 = 0 213 ( 13) 4(18)( 35) 2(18)x  − − −= 13 2689 36x = x = 1.80, – 1.08 (2 d.p) M1 M1 M1 M1 A1 Make denominator same Simplify Reduce Correct substitution of values 22a AD = AB (length of square) 1 2DE DC= 1 2BF BC=  DE = BF ADE = ABF = 90o  ADE  ABF (SAS) M1 M1 A1 o.e. 22b 2DE = AD 1 2 DE AD = tan EAD = DE AD EAD = tan–2(0.5) = 26.565…o Since ADE  ABF , EAD = BAF, EAF = 90o – 2(26.565…o) = 36.9o (1 d.p.) M1 A1 23ai $82 B1 23aii $56 , $114 B1 23bia (2000×30+6000×38+10000×40+14000×35+1800×25+2500×22 +40000×10)  200 = 12 890 (shown) M1 B1 Mid-value for $30 000 < x ≤ $50 000 (max) is $40 000

23bib $9211 B1 23bii The group earning $x > $30,000 acts as "outliers." Because the mean is sensitive to these very high values, it gets pulled upward, making it higher than the median (the 100th/101st household's income). B1 23biii 4 25 B1 24ai AEB = 39o (s in the same segment) ABD = 180o – 86o – 39o (s in opposite segment) = 55o M1 A1 24aii CAB = 90o – 39o (s in semicircle) = 51o Since ABD  CAB, triangle AFB is not isosceles. M1 A1 24b Reflex KOL = 2 – 2.4 = 3.883185….. radians 440 = 0.5 × r2 × 3.883185….. r2 = (440 × 2)  3.883185….. r = 15.1 cm (3 s.f.s) M1 M1 A1 25a 220 5 m / s4 = B1 25b 3.6 4 20 x= x = 18 M1 A1 25c 18t = 10(t – 4 + t) t = 20 M1 A1 Accept FT from (b)