2023 Springfield 3G3 Mathematics (4052) (P1-MS)

22 Marking Scheme 3Exp P1 1a 22484 2 11= Since the power of the factors are in multiple of 2, thus 484 is a perfect square. AO3 M1 A1 1b 2 66 2 3 11 150 2 3 5 HCF 6 =   =   = AO1 M1 A1 2a ( ) ( ) 2 22 2 2 62 66 222 3 9 2 3 11 11 xx x x x b −−    = − − −       = − − − = − −  =− AO1 M1 A1 2b The minimum point occur at x = 3, therefore the y value is a minimum value. AO3 B1 3 ( ) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 23 216 2 6 3 1 216 2 12 3 3 256 15 256 15 2 5 6 15 2 10 12 2 9 3 0 9 9 4(2)(3) 2(2) 0.363 or 4.14 xx xx xx xx xx x xx x x x x x x xx x xx −=+− − − + =+− − − − =−− −− =−− − − = − − − − = − − − + = − −  − −= == AO2 M1 M1 M1 A1

23 [Turn Over 4 3 5 15 5 53 5subst , 1, 3 3 53 (1) 3 4 3 54 33 yx yx m x y y mx c c c yx += =− + =− = =− =+ − =− + =− =− − AO2 M1 M1 A1 5a 3 3 3 2 3 3 2 27 8 a a a −   = = AO1 M1 A1 5b 2 32 53 125 10 5 2 xy xy xy −− = AO1 B2,1,0 6a (Given) 90 (vert. opp ) (AAS) SQ RW QST WRT QTS WTR STQ RTW =  = =   =     AO2 M2, 1, 0 A1 6b 180 90 62 28 180 90 28 62 PWS PWS RTW RTW  = − −   =   = − −   =  AO1 M1 A1

24 7a 1 26 1 3 1.0472 1.05rad    = = = AO1 M1 A1 7b 2 2 1area of sector 2 118 3 2 3 3.9894 3.99 r r r r   = =  = = AO1 M1 A1 7c ( ) Perimeter 2 Perimeter 3.9894 1.0472 2(3.9894) Perimeter 12.2 cm rr=+ =+ = AO2 M1 A1 8a 5 3(2 7) 9 5 6 21 9 30 xx xx x − − + = − + + =− + AO1 M1 A1 8b ( ) ( ) ( )( ) ( )( ) ( )( ) 37 2 1 1 3 1 7 2 1 2 1 1 3 3 14 7 2 1 1 10 11 2 1 1 xx xx xx xx xx x xx −−+ + − −= −+ + − += −+ −= −+ AO1 M1 M1 A1 9a 2 2 2 5 12 13 YW YW =+ = By converse of Pythagoras theorem, angle WZY is a right angle AO2 M1 A1 9b 2 2 2 9 12 15 WX WX =+ = AO1 M1 A1 9c cos cos 5cos 13 XYW WYZ XYW  =−   =− AO1 B1

25 [Turn Over 9d 2 1area of triangle 4 12 2 area of triangle 24 cm WXY WXY =   = AO1 B1 10a Correct curve – 1 mark x intercepts (0.414 and -2.414) and y intercepts (1) – 1 mark maximum point (-1 , 2) – 1 mark AO2 10b x = -1 AO1 B1 11a 3 – (-4) = 7 AO1 B1 11b 3 31=−− AO1 B1 11c ( ) ( ) 22 43 25 −+ = AO1 B1 12a 22 500 5 55000g  = AO1 B1

26 12b ( ) Supermarket A 55000no. of 3 packs set 2700 no. of 3 packs set 20.4 no. of 3 packs set 21 21 2700 56700 (more than the require amoun t) total cost 21 8.45 8.45 total cost $354.90 Supermarket B 55no. of 6 tins g = = = = =  + = = 000 1800 6 no. of 6 tins 5.09 5 6 1800 54000 total cost 5 91 16 total cost $471 g  =   = =  + = Supermarket A, not only will the owner get more milo powder and it cost lesser than Supermarket B AO3 M1 A1 M1 A1 A1 13a ( ) ( ) 22 length of 6 6 1 7 length of 14.422 length of 14.4 AB AB AB = − − + − − = = AO1 M1 A1 13b ( ) 1 2 1grad 6 6 4 126 4 1 2 11eqn : 42 BC y mx c c c yx −−==−− =+ =+ = =+ AO2 M1 M1 A1 14a ( ) 23 2 28 2 1 4 ab ab ab b + =+ AO1 B1 14b 22 5 3 (2 3)( 1) xx xx −+ = − − AO1 B1 14c 6 2 3 2 (3 ) (3 ) (3 )(2 ) ab ad bc cd a b d c b d b d a c − + − = − + − = − + AO1 M1,M1 A1

27 [Turn Over 15a 2 total amt 1 100 24768.80 1 100 $500 n rP P P =+  =+  = AO2 M1 A1 15b 22 total amt 1 100 24 2total amt 500 1 100 total amt $786.76 n rP  =+   =+    = AO2 M1,M1 (FT) A1 16a AO2 B1 B1 16b Plan A AO1 B1 16c 225 mins AO1 B1 17a Graph E AO1 B1 AO1 t (mins) C ($) Plan A Plan B

28 17b Graph C B1 17c Graph D AO1 B1 18a ( ) 325 270 55 52 sin 55 sin 90 18 44.788 44.8m ACB AB AB AB  = − =  = −  = = AO2 M1 M1 A1 18bi 180 55 125 1area of triangle sin 2 11256 (52)( )sin1252 58.973 59.0m ACD ACD ab C CD CD CD  = − =  = = = = AO2 M1 M1 A1 18bii 2 2 2 52 58.973 2(52)(58.973)cos125 98.487 98.5m AD AD AD = + −  = = AO2 M1 A1 19a 2 2 26 1.45 5.8222 5.82 m h h h =+ = = AO1 M1 A1 19b 1.45cos 6 76.02 76.0    = = = AO1 M1 A1 6 1.45

29 [Turn Over 19c 22 height 5.8222 0.6 height 5.2222 distance from the wall 6 5.2222 distance from the wall 2.9544 5.2222ratio 1.76762.9544 =− = =− = == Since the ratio of the height to the distance from the wall does not follow the 4 to 1 safety rule. Thus it is not safe to use the ladder. AO3 M1 M1 A1 6