[FMS] [2023] 4E EM 4052 Prelims P1 MS

Pg 1 Fairfield Methodist School (Secondary) Secondary 4 Express / 5 Normal (Academic) Mathematics Paper 1 Marking Scheme Preliminary Examinations 2023 Qn No. Workings Description Mark Allocation AO 1(a) B1 AO1 1(b) m = 2 n = 5 B1 AO1 2(a) 901.16 B1 AO1 2(b) B1 AO1 3(a) 37 B1 AO1 3(b) 9 + 4n B1 AO1 3(c) Let 318 = 9 + 4n 4n = 309 n = 77 ¼ Since n is not a positive integer, 318 is not a term of the sequence. B1 AO3 4(a) 72 + 2x + 75 + 69 + x = 360 3x + 216 = 360 3x = 144 x = 48 B1 AO2 4(b) B1 AO1 5 y-intercept = 5( 2 5 )=2 Gradient of line graph 2 2 5 5 vert horizontal=− =− =− Equation of line is y = - 5x + 2 M1 M1 A1 AO2

Pg 2 Qn No. Workings Description Mark Allocation AO 6 Distance travelled by fuel in 24 mins = 2×24×60=2880m Capacity of fuel tank 2 3 822880 ( ) 100 14.476 14.5 (3 ) 145000 (3 ) m sf l sf = = = = M1 M1 A1 AO2 7a ( ) 3 4 4 3 3 4 3 3 4 4 33 3 81 1 () 81 11 (3 ) 11 3 1 27 x x x x x − −= = = = Apply () m n mnaa = M1 A1 AO1 7b 1 1 5 4 1 1 53 5 4 3 4 3 1 4 32 2 8 (2 ) 2 (2 ) 2 2 2 22 31 4 3 14 1 4 x x x x x x + = = = = += =− =− Apply () m n mnaa = M1 A1 AO1 8(a)(i) {3, 6, 9, 12, 15, 18, 21} No { } will result to no marks awarded B1 AO1 8(a)(ii) {4, 8, 10, 14, 16, 20, 22} B1 AO1 8(b)(i) B1 for each correct answer B1 B1 AO1 8(b)(ii) n[ ] = 5 B1 AO1

Pg 3 Qn No. Workings Description Mark Allocation AO 9 M1 for factorising out either common factor correctly M1 for factorising m2 – 5mn + 4 n2 correctly M1 M1 A1 AO1 10 M1 for converting to linear equation correctly M1 for expanding either side correctly M1 for simplifying either algebraic terms or constants correctly M1 M1 M1 A1 AO2 11a cos cos65 180 65 115 x x =−  = − =  B1 AO1 11b 1 1Area of ( )( )sin 2 115 (10)(6)sin2 15 30sin 15sin 30 15sin ( )30 30° or 150 PQR PQ PR QPR QPR QPR QPR QPR − = = = = = = M1 M1 A1 for both ans AO1 12a A2 = 4 6 4 6 0 2 0 2      −−   = 16 12 04   B1 AO1

Pg 4 Qn No. Workings Description Mark Allocation AO 12b A = 2B 46 02   − =2 2 01 k  − 46 02   − = 42 02 k  − 2k = 6 k = 3 B1 AO1 13a S = 34000 20100 14500 30000   B1 AO1 13b ( ) ( ) 34000 2010011 14500 30000 48500 50100   = B1 AO1 13c The total/combined sales in 2021 and 2022 (or for the 2 years) for IMic and Lenovo Laptops respectively • Keywords: Total/Combined , respectively, • 2021 and 2022 or past 2 years B1 AO2 14a Angle x = 180○ – 112○(int. angles, AB//CD) = 68○ Deduct 1 mark – wrong or missing reasons B1 AO1 14b Angle z = 85○ (corr. angles, AB//CD) Angle y = 180○ – 85○(adj. on a str. line) = 95○ M1 A1- AO1 15a y = x2 + 3 B1 AO1 15b y = x3 + 3 B1 AO1 15c y = 3x + 3 B1 AO1 15d y = 3x-2 B1 AO1 16a Refer to last page AO1 16b Refer to last page AO1 16c Refer to last page AO2

Pg 5 Qn No. Workings Description Mark Allocation AO 17(a) 3ax + 16by – 12ay – 4bx = 3ax – 12ay – 4bx + 16by = 3a(x – 4y) – 4b(x – 4y) = (3a – 4b)(x – 4y) or (4b – 3a)(4y – x) M1 for factorising any 2 terms correctly M1 A1 AO1 17(b) 3mn – 243mn5 = 3mn(1 – 81n4) = 3mn(1 + 9n2)(1 – 9n2) = 3mn(1 + 9n2)(1 + 3n)(1 – 3n) M1 for factorising out 3mn M1 for applying difference of squares to 1 – 81n4 M1 M1 A1 AO1 18a 2 2 2 2 2 16 30 16 1616 ( ) ( ) 3022 ( 8) 94 xx xx x +− = + + − − = + − B1 AO1 18b 2 2 2 16 30 0 ( 8) 94 0 ( 8) 94 8 94 94 8 or 94 8 1.70(2d.p.) or -17.70(2d.p.) xx x x x x x + − = + − = += + = = − − − = M1 A1 for both ans AO1 18c Turning Point = (-8, -94) When x = 0, y=(0 + 8)2 - 94 = –30 Correct Shape Correct y-intercept and turning point C1 P1 AO1 x y 0 -94 - 8 -30

Pg 6 Qn No. Workings Description Mark Allocation AO 19a Size of 1 int angle from octagon (8 2) 180 1358 −  = =  Size of 1 int angle from Hexagon (6 2) 180 1206 −  = =  360 135 120 (sum of s at a pt.) 105 a = − −   = M1 M1 A1 AO1 19b ( 2) 180 105 1052 180 105 2180 0.41667 2 4.8 n n nn nn n n −   = −=  −=  = = Since n is not a positive integer, that polygon does not exist. M1 M1 A1 AO3 20a % of student with revision > 12 hours = 10 12 100% 53.7%(3 . )41 sf+ = B1 AO1 20b Median position = 21st position Median = 12 h – 16hr or 12 hr  time  16 hr B1 AO1 20ci Est Mean 5(2) 6(6) 8(10) 10(14) 12(18) 41 11.756h 11.8h + + + += = = B1 AO1 20cii Std Deviation = 5.4495h = 5.45 h B1 AO1

Pg 7 Qn No. Workings Description Mark Allocation AO 21a Since AB=CD, OF = OE(equal chords) ---- S AF = FB = CG = GD ( ⊥ bisector of chord) 90AFO CGO = =  -------------- R EO is shared side of and OFE OGE ----H By RHS Congruency Test, OFE OGE As EF = CE, EA = EF – AF EC = EG – CG Since EF = CE and AF = CG, EA = EC (proven) M2 for all Evidences M1 for 2 evidences M1 for RHS Test AG1 AO3 21a Alternative Solution: (Intersecting Secant Theorem) ( ) ( ) ,given that = (proven) EA EB EC ED EA EA AB EC EC CD EA EA AB EC EC CD EA EA AB CD ABEC EC AB EA EC  =  + = + += + += + = State the theorem M1 M1 M1 AG1 21b 30 (OE is bisector of )2 15 40 202 15tan 15tan15 15 55.981tan15 20 55.981 20 36.0cm (3 s.f) BEO DEO BED AF cm OFOEF EF EF EF EF cm AE EF  = =   = ==  = = = ==  =− =− = M1 M1 A1 AO2 F G

Pg 8 Qn No. Workings Description Mark Allocation AO 21b Alternative solution: 2 2 2 1 180 30 2 75 (base s of an isos. Δ,EB=ED) 40 25 25cos 2(40)(25) 36.870 (5 ) 75 36.870 38.13 25 sin(180 (38.13 2))sin 38.13 39.331 (3 ) 39.331 sin 75sin 30 75.982 EBD ABO sf OBD BD cm sf EB cm − − = =    +−=  =  = − =  =  −  = =    = (3 ) 75.982 40 36.0 (3 ) sf AE cm sf =− = M1 M1 A1 22 Perimeter of major sector = 3 Perimeter of minor sector (2 ) 2 3( 2 ) 2 2 3 6 (2 2) (3 6) 4 2 6 2 4 2 4 2 0.571 (3 )2 x x x x x x x x x xx radians dp              − + = + − + = + − + = + = − + =− −== M1 for each correct perimeter M1 for making θ as the subject A1 AO2 23a 2 2 1Volume of cone (Base area)( height)3 1259.44 (5.6) ( height)3 259.44 3height (5.6) 7.9001cm   =⊥ =⊥ ⊥= = By Pythagoras’ Theorem, Slant height of cylinder 227.9001 5.6 9.68357 radius of sector OAB cm + = = Arc length AB = 2π(5.6) M1 M1 AO2

Pg 9 Distance (m) 0 16.875 48.375 79.875 Qn No. Workings Description Mark Allocation AO 9.68357( ) 2 (5.6) 2 (5.6) 9.68357 3.63 radian 208.2 (1 ) x x dp   = = = = A1 23b Volume of hemisphere 3 3 3 14 23 2 (5.6)3 367.809 Total volume of ornament = 367.809 + 259.44 = 627.249 = 627 cm (3 ) r sf   =  = = M1 M1 A1 AO1 24a 2 5 10= Hence v = 3 + 10 = 13 m/s B1 AO1 24b B1 for 2 parts drawn correctly. B2 for all 3 parts drawn correctly with correct distance AO2 5 8 14 Time(s) 40 79 118

Pg 10 Qn No. Workings Description Mark Allocation AO 16 Minus 1 mark for missing construction lines, incomplete perpendicular bisector. B1 for part (a), (b) and (c) respectively