Dunman 2026 Emaths Paper 1 MS

4052/01/PRELIM/4E5NA/2026/MARKING SCHEME DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2026 SECONDARY 4 EXPRESS / 5 NORMAL ACADEMIC MATHEMATICS 4052/02 Paper 1 18 August 2026 2 hours 15 minutes Marking Scheme

Page 2 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME (Updated 1 Dec 2022 by HOD/Math aligned to Marking Scheme for Specimen Papers 4049)

4052/01/PRELIM/4E5NA/2026/MARKING SCHEME Question Answer Marks Partial Marks Guidance 1(a) 26 x− 1 B1 1(b) ( ) 346 12 12 22 ( ) 8 aa aa + =+ 2 M1 129a= A1 1(c) 3 24 5 1 4 2 31 514 9 21 6 4 9 xy y x a xa y a −− −  = 2 M1 Seen law 2 applied for 1 term. 2 424 9 xa y= A1 Accept 4 2 24 9 x a y − o.e. 2(a) HCF(252, 280) 2 M1 Can be implied = 28 A1 2(b) boys = 9, girls = 10 1 B1 3 21 6 2 3 2 ( 3 1) 3 1 k ab kb ka kb ka ka + − − =− − + − + 2 M1 (1 3 )(1 2 )ka kb= − − A1 4(a) ( ) 2 5 4.23 5( 204)a −= − 2 M1 correct substitution o.e. without rounding to 2 s.f. 0.013 (2 s.f.)= A1 4(b) 255ca b=− 3 M1 Simplify to one line equation Can be implied 2 55b ca=− M1 Make 2b the subject Can be implied 2 55b ca=− A1 Must have 

Page 4 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 5 22 77 4022x   − − + =       3 M1 7 33 24x −= M1 6.37 or 0.63xx== A1 6 8 2$ .26545.91 1 100P=+  3 M1 5500.00$P= M1 1045.91$I = A1 7 24 16 5 24 12 16 x x x x − + − += 3 M2 Or M1 for either common denominator, 16x or correct expansion, 24 16 5 24 12x x x− + − + 1 16= A1 8(a) 122000 4000 568020 x x + =+ 3 M1 Or M1 for attempt to use mean formula ( )122000 4000 5680 20xx+ = + M1 Remove denominator 5x= A1 8(b) $5000 1 B1 8(c) Median. $30 000 salary is an outlier/extreme value that affects the mean but not the median. 1 B1 9 2 2 2 2 2 S2 (2 ).A. of l (2o ) s id 20 )2 (4r r r r r r r       = + − = ++ 3 M1 22 ()20 2r r rl  =  + M1 their S.A. 9lr= A1

Page 5 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 10 1 24 6 4 b b = = 4 B1 11 14 715 x x−= M1 2 30 0 5 (rej.) or 6 x x xx − − = =− = No. of girls that did not win a prize = 6 A1 No. of girls that won a prize = 9 A1 Their x but do not accept if their x does not make sense (non-integer or negative values) 11(a) 2700k = 2 M1 S = 21.6 A1 11(b) m = 64 1 B1 12 The increase in area is not (directly) proportional to the height. 2 B1 This may mislead the readers that the total investment returns in 2025 is more than 3 times that in 2023. B1

Page 6 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 13 5 5 1radius 3 ..2 0=  Length of running track = ( )( )2 3.2 0.1 20 390 415 (3s.f.)   = − 14(a) Factors of 24 3 B1 14(b)  B1 14(c) 8 B1

Page 7 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 15 Disagree 2 B1 Not all rational numbers are integers (give an example). Or There are elements in set B that are not in A. For example, 1 2 B but 1 2 A . B1 16(a) ( ) 2 5 10 1 21 102T = − + = 1 B1 16(b) ( ) ( ) 2 2 1 4 1nT n n= − + + 3 M2 M1 for ( ) 2 21n− M1 for ( )41n+ 242n=+ A1 16(c) ( )1 2 4 1 2kTk+ = + + 1 M1 ( ) 22 1 4 8 6 4 2 84 kkT T k k k k + − = + + − + =+ A1 16(d) When 8 4 10k+= k = 0.75 Since k is not a positive integer, two consecutive terms of the sequence cannot have a difference of 10. 1 B1 17 Correct method to get rid of 1 variable using either substitution or elimination. 3 M1 M1 for seen attempt to use the correct method 2.5, 1xy= =− A2 A1 for each correct answer

Page 8 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 18 Statement True or False Example (if false) 21n+ is always odd. True The product of two distinct irrational numbers is always irrational False 2 2 2 4= 2 1 1 n n False ( ) 2 1 11 1 − −  3 B3 B1 for each correct row 19(a) Sub ( ),3M a a 2 M1 ( )2.4, 7.2M or 212 , 755M  A1 Accept improper fraction 19(b) ( )3, 0K − 3 M1 22 (2( .4 37.2 ) )0 ()− + − − M1 Correct formula for length with their M =9 units A1 20 5 180x=  3 M1 Exterior angle = 72 M1 360no. of sides 72 5 = = A1

Page 9 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 21 (corr , // ) (corr , // ) is an isos. EDC DAB s AB DC ECD CBA s AB DC EDC  =   =    3 M1 ( ) ( ) hence, is an isos. is an isos. EA EB EAB EA ED EB EC EDC =   − = −     M1 AD BC= A1 Alternatively, (given) 90 (height of trapezium) (AAS) DAE CBF AED CFB DE CF ADE BCF  =  = =  =   M2 M1 for any of the correct A, A, S AD BC= A1 Alternatively, sin sin hDAE AD hCBF BC = = M1 Since ,DAE CBF = hh AD BC= M1 AD BC = A1

Page 10 of 11 4052/01/PRELIM/4E5NA/2026/MARKING SCHEME 22(a) )70 alt / ( , /N N N WW Fs FWFF   = Or (alt , )84 //S p FFP FF PN N =  Or 20WWFF =  2 M1 M1 for any one of the following angles. Bearing of W from F 290= A1 22(b) 84 20 64 WPF = −  = 180 90 64 ( sum of ) 26 WFP s = − −    = 2 - sin 26 sin 64 100 WF = M1 205m (3 s.f.)WF= A1 Alternatively, 0t n 64 0a 2 FW= 2 M1 205 mFW = A1 22(c) 8 1.7 205.030tan −= 2 M1 1.8 = A1