Greendale Sec 4048 01 Answer Scheme

2022 4E5A Preliminary Examination Paper 1 Mark Scheme 1 22 33 3 3 5 ( 5) 5 25 25 5 55 55 25 125 55 150 55 0.475235 0.475 (3 )sf − −− × −−×= −× −× −−= −× −= −× = = B1 2 38 2( at centre = 2 times at circumferenc e) 76 180 76 (base angles of isosceles triangle)2 52 QOS OSQ ∠ =×∠ ∠ = ° −∠= = ° M1 A1 3 Diagram 4 B1 4a The size of the picture is different over the years. B1 (o.e) 4b The reader may be misled to think that a bigger picture represents more people. B1 (o.e) 5a 3 33 9 27 (3 ) 3 = = B1 5b 23 4 23 4 4 22 24 5 25 2 25 54 5 2 xy xz zy xy y z xz y xz ÷ = × = M1 (for any simplification of constants or variable) A1 6 Worker : Hours : Footbridge 18: 60 : 3 18: 20 : 1 1 : 360 : 1 1 : 2520 : 7 63 : 40 : 7 Additional workers = 63 18 45 workers−= M1 A1 7 6, 6, 9, 11, 18 B1 for any 3 B2 for all 5

8 41 125 20 5 2 110 10 18 5 10 5 6 xx xx x x + −= − + −= − += − =− M1 M1 A1 9a + 2 3 5 7 11 2 4 5 7 9 13 3 5 6 8 10 14 5 7 8 10 12 16 7 9 10 12 14 18 11 13 14 16 18 22 B1 for all correct 9bi 6 25 B1 (ecf) 9bii 5 25 B1 (ecf) 9c The spinner has to be fair so that the probabilities of obtaining each outcome is equal. B1 (o.e) 10a 2 22 2 2 2 75 7775 22 7 29 24 113724 xx xx x x −+  =−+ − +    = −− = −− M1 A1 10b 113,724  − B1 11a p = 7 q = –2 B1 B1 11b 7 2 100 7 102 14.571 n n n −< < < Largest n is14 B1 12a 4u -> 7.2 1u-> 1.8 3u-> $5.40 B1 12b 700 : 1750 2 : 5 B1

13a 2 2 17 1 7 ( 3) 1 116 16 r q r r r =+ =+− = = B1 13b 2 2 2 17 7 7 7 r q rq qr qr =+ = + = − = ±− M1 A1 14 Let the initial cost be $x. Cost price = 1.08x Selling price = 1.15 (1.08x) = 1.242x 1.242x -> $465.75 1.242 465.75 $375 x x → → M2 for 465.75 ()1.08 1.15 oe× B1 for 1.08 or 1.15 (oe seen) 15 3 3 511576.25 (1 ) 100 11576.25 (1.05) $10000 P P P = + = = B1 for 5% seen B1 for power 3 B1 16ai 3 , 4 , 5 , 7 , 10 , 11 B1 16aii 4 , 10 B1 16b ( )AB ′∪ or AB′′∩ B1 17 2228.6 5 5 2(5)(5)cos 118.63317 180 118.63317 61.36683 8cos 61.36683 16.69449 16.69449 5 11.69449 11.7 DCE DCE ACB AC AC AE m =+− ∠ ∠= ∠= − = = = = − = = M1 M1 M1 A1

18a 224.8 ( ) 7.5 ( ) 4.8 ( ) 7.5 ( ) 4.8 ( ) 7.5 1000 100 ( ) 7.5 1000 1001 () ( ) 4.8 1:125000 cm map km actual cm map km actual cm map cm actual cm map cm actual → → →×× ××→ M1 A1 18b 4.8 ( ) 7.5 ( ) 7.51 () ( ) 4.8 7.59 () 9 ( ) 4.8 11.25 cm map km actual cm map km actual cm map km actual km → → →× = B1 19a 12(2 ) 5( 4 )32 254 2032 3 58 23 xxy y xxy y xy −−− =−−+ = + M1 A1 19b 4 5 16 20 (4 5 ) 4 (4 5 ) (4 5 )( 4 ) am bm an bn m ab n ab a bm n −−+ = −− − = −− M1 A1 20ai 2495 3 5 11= ×× B1 20aii 2495 3 5 11= ×× N = ? HCF 15 3 5= × LCM 224950 2 3 5 11= ××× 112 02 3 5 11 150 N N =××× = M1 (HCF) M1 (LCM) A1 20b LCM of 50, 60 and 72 = 1800 No of can C = 1800 2572 = M1 A1

21ai Exterior angle 204 180 24 = − = ° Number of sides 360 24 15 = = M1 A1 21aii 180 156(interior angles) 24 RQT∠ = − = ° B1 21bi ( 2) 180 5 (2 2) 180 11 25 2 2 11 11 22 10 10 12 n n n n nn n −× =−× − =− −= − = M1 A1 21bii Exterior angle for Polygon Y 360 2(12) 15 = = ° B1 22a QT = TP (given) (S) TS = PR (opposite sides of parallelogram are equal) (S) (alt. s, base s of )STQ RPT PTQ∠ = ∠ ∠ ∠∆ (A) (SAS congruence test)TPR QTS∴∆ ≡∆ M1 A1 22bi 4 12 16 3 cm DE DE = = B1 22bii (vertically opposite ) (A) (alt. s, // ) (A) is similar to by AA similarity test DXE CXB s DEX CBX DE BC DXE CXB ∠= ∠ ∠ ∠= ∠ ∠ ∴∆ ∆ M1 A1 23a (10 10) (30 20) (50 40) (70 60) (90 20) 10 20 40 60 20 8700 58150 Mean Mean ×+×+×+×+×= ++++ = = B1 23bi From the diagram, 150 – 24 = 126 students scored more than 36 marks. B1 23biia 126 125 105 150 149 149×= B1 (ecf)

23biib Number of students who scored at most 64 marks = 82 students Number of students who scored more than 80 marks = 150 – 130 =20 students 82 20 328 2150 149 2235× ×= M1 for 82 or 20 seen A1 24a 14 34 3 1 BC = − −=− uuu r B1 24b 3 1 6 2 32 1 CB OA = = =  u u ur u u ur Since OA u u ur =2 CB u u ur , OA and CB are parallel and thus OABC is a trapezium. M1 A1 (A0 if OA u u ur =2 CB u u ur is not mentioned) 24ci 61 23 6 32 AX k k k −  = +  −   −= − uuur M1 A1 24cii 64 32 4 64 2 k k k k −−   =  −   −= − = B1 24ciii area of 4area of OAX CBX ∆ =∆ B1