Dunman 2026 Emaths Paper 2 MS

DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2025 SECONDARY 4 EXPRESS / 5 NORMAL (ACADEMIC) MATHEMATICS 4052/02 Paper 2 25 August 2026 2 hours 15 minutes Marking Scheme (Updated 1 Dec 2022 by HOD/Math aligned to Marking Scheme for Specimen Papers

Qn Answer Marks Partial Marks Guidance Markers’ Comments 1a 5 2 7 2 7 11 2 12 2 18 69 andxx xx xx  − −    3 M1 69 x M1 The integers are 6, 7 and 8 A1 1b 2280 45xy− 2 225(16 9 )xy=− M1 5(4 3 )(4 3 )x y x y= − + A1 1c 18 453 18 20 12 x x =− =− 2 M1 Rewrite equation to a non-fractional equation. 18 453 18 20 12 12 2 1 6 x x x x =− =− = = A1 DO NOT accept x = 0.167 or 0.1666 Can accept x= 0.16  (recurring decimal) 1d (i) 2( 3) 4x−− 2 2 6 9 4 65 xx xx = − + − = − + 2 M1 Show expansion and simplify to the general form of quadratic expression ( 1)( 5)xx= − − A1 If no working step shown, no marks

Page 3 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 1c(ii) 3 B3 B1 – positive quadratic curve sketched B1 – correct x and y intercepts marked B1 – min turning point marked No mark awarded if coordinates are not given 2a 120 (angle at circum = angle at centre)2ABD =   3 M1 Deduct 1 mark if any reason is not stated 37 20 17 CBD = − =   M1 180 2(17 )(angle sum of isos triangle ) =146 BCD BCD = −   A1 2b Let P be a point on the circle. 180 146 (angles in opposite segments) =34 BPD = −   2 M1 Deduct 1 mark if any reason is not stated 34 2(angle at centre = 2 angle at circum) =68 BOD =     A1 (3, −4)  (0, 5)  (1, 0)  (5, 0)  P

Page 4 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 2c 68 40 108 AOB = + =   4 M1 Area of sector OADCB 2 2 108 (7 )360 46.1814 cm = = M1 Area of triangle AOB 2 2 1 (7 )sin1082 23.300 cm = =  M1 Area of segment = 46.1819 − 23.300 = 22.9 cm2 (3 sig fig) A1 3a 23 18 39 17 31 44   1 B1 3b k=0.95 1 B1 3c ( ) ( ) 0.95 3700 3000 3515 2850 D = = 2 M1 ecf from 3b ( ) ( ) 23 18 393515 2850 17 31 44 129295 151620 262485 R =  = A1 3d The elements of R represents the total cost of the discounted packages sold at the travel fair on Friday, Saturday and Sunday respectively. 1 B1 3e Total cost of packages sold outside of travel fair = 132($3700) + 95 ($3000) = $773400 4 M1 Total cost of packages sold during travel fair = $129295 + $151620 + $262485 = $ 543400 M1 ecf from 3c Month’s total sales = $773400 + $543400 =$1316800 M1 ecf Percentage of month’s overall sales 543400 100%1316800= 41.3% (3 sig, fig.)= A1

Page 5 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 4a 2 3 2 AB OD n = =  2 M1 42 OB OA AB mn =+ =+  A1 4b 34AD n m=−  4 M1 ( )2 345AT n m=−  M1 ( ) ( ) 24 3 4 5 3 425 3 5 OT OA AT m n m mn OB =+ = + − =+ =   M1 Since 3 5OT OB=  and O is the common point, therefore O , T and B are collinear. A1 5a 1 1 1 24 2 2 1 ( 2) 4 xx x xx += − − =− 3 M1 2 4(2 2) ( 2) 8 8 2 x x x x x x − = − − = − M1 2 10 8 0xx− + = (shown) A1 5b 2( 10) ( 10) 4(1)(8) 2(1)x − −  − −= 3 M2 M1 for 2( 10) 4(1)(8)−− 9.12 or 0.88 (2 dec. places)x= A1 5c Time Taken =9.123 – 2 = 7.123 hours = 7 hours 7 minutes 1 B1

Page 6 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 6a p = 2.3 1 B1 6b 3 B3 P2 : all points plotted correctly Award P1 if 1 to 3 points are plotted wrongly C1 : Curve pass through all points smoothly 6c(i) 22 2 5 24 5 2 4 (2 ) (4 ) 5 0 ax b x x ax bx x x a x b x + = + − + = + − − − + + = 3 M1 2 1 and 4 9 15 ab ab − = + = == A2 6c(ii) 2 B2 1 mark for correct equation y = x + 5 1 mark for correct line drawn (ecf) 6c(iii) From the graph, x = 0.6 1 B1 Accept 0.55 or 0.65 7a 3 B3

Page 7 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 7b 7 6 7 8 11 1 1 18 42 ( 7)( 8) 11 ( 1) 18 nn n n n n nn nn −−   +=   −−    + − − =− 3 M1 Formulate the equation 218(42 15 56) 11 ( 1)n n n n+ − + = − M1 Simplify to a non-fractional equation 2 2 2 18(42 15 56) 11 ( 1) 7 259 1764 0 37 252 0 (shown) n n n n nn nn + − + = − − + = − + = A1 7c ( 9)( 28) 0nn− − = 2 M1 Or using general formula 9 or 28n= A1 7d Since there are more white beads than red beads, if n =9, then there is only 2 white beads which is less than the red beads. Therefore, n  9 1 B1 7e P(two beads chosen are of different colours) 111 18 7 18 =− = 1 B1 8a(i) 34 seconds 1 B1 8a(ii) ITQ = 60 – 22 = 38 1 B1 8b Total number of residents = 304 = 120 1 B1 8c(i) 15 < t  30 1 B1 8c(ii) Estimated mean waiting time 7.5(8) 22.5(27) 37.5(10) 52.5(13) 67.5(5) 82. 5(2) 8 27 10 13 5 2 + + + + += + + + + + 1 Working must be shown to be awarded B1 = 34.269 = 34.3 (3 sig. fig) B1 8c(iii) Standard Deviation 2100406.25 34.26965=− 1 Working must be shown to be awarded B1 = 19.3 (3 sig. fig.) B1 8c(iv) The lift at Blk 302 is more efficient because the median waiting time is in the interval of 15  t  30 is shorter than the median waiting time of Blk 301 which is 34s. 1 B1

Page 8 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments 9a (alt , AD//BC) (vert. opp s) ADX CBX AXD CXB  =   =  CBX is similar to ADX (AA Similarity test) 2 B2 B1 for each pair of corresponding angles indicated with reason. 9b(i) Area of DAX : Area of BCX = 4:25 1 B1 9b(ii) Note that ABD and ABC are triangles of the same height. Area of ABD : Area of ABC = AD : BC = 2 : 5 1 B1 9b(iii) Area of ABD : Area of ABC = Area of (ABX + DAX) : Area of (ABX + BCX) = 4 + p : 25 + p 42 25 5 20 5 50 2 3 30 10 p p pp p p + =+ + = + = = 2 M1 area of triangle ABX : area of trapezium ABCD = 10: (10+10+4+25) = 10 : 49 A1 9c 10 units rep 8 cm2 Area of shaded region 8 (4 25)10=  + 2 M1 =23.2 cm2 A1 9d AB = DC is the additional condition 2 B1 Alternatively the condition can be AX= DX or BX = CX Then AX= DX and BX = CX Therefore triangle BAX and triangle CDX will be congruent. (SSS) B1 State based on the condition The condition set will make trapezium ABCD an isosceles trapezium 10a(i) Total protein content = 1.2+3.5+18+2.5+1.0 =26.2g 1 B1 10a(ii) Amount of energy = 20 + 40 + 80 + 210 + 160 1 B1

Page 9 of 9 4052/02/4EXP5NA/PRELIM2026/MARKING SCHEME Qn Answer Marks Partial Marks Guidance Markers’ Comments = 510 kCal 10b Base : Brown Rice or Quinoa Toppings : Sunflower seeds and (Roasted Corn or Avocado Slices) Main : Tofu Medley Dressing : Balsamic Vinagrette 2 B2 B1 for Base & Toppings B1 for Main & Dressing 10c Ingredients of Signature Wellness Bowl Energy Protein Sodium Brown Rice 250 6.5 5 Smoked Duck 210 18 450 Hard-Boiled Egg 70 6 60 Roasted Corn 65 2.0 5 Creamy Caesar 160 1.0 310 Total 755 33.5 830 6 M3 B1 for each correct total calculated for energy, protein and sodium The Chef’s comment is correct as the total energy gained is 755kCAl and the sodium content 830mg have exceeded the standard requirements of between 500 kCal and 700kCal amount of energy and less than 600mg of sodium A1 Must indicate the 2 nutritional standards that are not met. Ingredients Energy Protein Sodium Brown Rice 250 6.5 5 Grilled Salmon Fillet 130 22 90 Ha