GESS 2025 EM P1 MS

1(a) 1 3 3 33.17 55.2 ( 1) 647.9 13.08359745 10  − − + = = B1 1(b) 3 0.003 1 3600 m km s h = 10.8 /km h= B1 2(a) 22 2 10 (10 ) 10 2 11 4 11 m n m n− = = = B1 2(b) 1 2 3 2 7 ( ) 7 mn mn m −  36 2 7 7 m m n mn= 2 3 6 77m m n m n= 47 49 mn= M1 for either simplying or change of sign A1 3 1.25 New T k L T k L = = 1.25 Newk L k L= 1.25 NewLL= ( ) 2 1.25 NewLL= ( ) 2 % 100% 1.25 100% 0.5625 100% 56.25% New change LL L LL L −= −  = = M1 100% A1

4(a) 37 min B1 4(b) 25% of 28 = 25 28100 = 7 From the stem and leaf diagram, 7 workers took less than 19 minutes to assemble a toy. p = 19 min. A1 5 P = $12000, R = 7 3.5%2 = , n = 4 1 100 n RAP =+  4 3.512000 1 100A =+  13770.27601 $13770.28 = = Compound interest 13770.27601 $12000 $1770.27601 $1770.28(2 . )dp =− = = M2,1,0 A1 6 2 2 xyz += 22z x y=+ 22 22 4 4 z x y x z y =+ =− 24x z y= − M1 A1 (minus 1 m if no  ) 7(a)(i) 2 22 2 2 6 16 ( 3) 16 ( 3) ( 3) 16 9 ( 3) 25 xx x x x −− = − − − − = − − − = − − B2 7(a)(ii) ( )3, 25− B1

7(a)(iii ) 3x= B1 7(b) B1 – Shape + turning point (2,27) B1 – x-intercepts at 1x=− and 5x= , and y-Intercept at (0,15) 8 9(a) (3 2)sum n n=− 1st term  1 3(1) 2 1= − =

Sum of 2 terms  2 3(2) 2 8= − = 2nd term 8 1 7= − = Sum of 3 terms  3 3(3) 2 21= − = 3rd term 21 8 13= − = First 3 terms: 1, 7, 13 B2,1,0 9(b) 65n− B1 10(a) (5 2) 180 90 90 3 540 180 3 3 360 120 y y y y −  = + + = + = = 360 120 240x= − =  120 1 240 2 y x ==  M1 A1 10(b) (12 2) 180 1800 1800 15012 360 150 90 120 180 120 60 360 660 sides −  =   = − − =  − =   = M1 A1 11 1. The title is biased • It does not allow reader to make his/her own judgement OR 2. Horizontal axis does not start from zero • This exaggerates the differences between the data B1 – Misleading Feature AND – Effect of the misleading feature on interpretation of data

12(a) 32 15 5 6 CD OD OC=− −   =−   −    =− B1 12(b) 5 6 CE nCD n = = − 225 ( 6) 61 2 61( ) CE n CD n n given = =  + − =  = Since x-coordinate of C is greater than the x-coordinate of E, 61 2 61CE n=  =−  2n=− 10 12CE −= M1 61 2 61( )n seen= A1 13(a) ( 6, 5)− , and ( 3, 2)−− 5 ( 2) 6 ( 3) Gradient −−= − − − 7 3= − 7 3y x c=+− Sub ( 6, 5)− 75 ( 6)3 9 c c = − +− =− 7 93yx=−− M1 A1

13(b) 7 3y x c=+− Sub (9, 5) 75 (9) 3 c=+− 26c= 7 263yx=+− M1 A1 13(c) 9 ( 6) 15− − = 5 ( 2) 7− − = 1 15 7 52.52Area=   = M1 A1 13(d) (9, 5) ( 3, 2)−− 22(9 ( 3)) (5 ( 2))BC= − − + − − 144 49 193BC= + = 1 193 52.52Area d=   = 52.5 1 1932 7.558065 7.56(3 ) d sf =  = = M1 A1 13(e) 7 93yx=−− Sub 0y= 7 93 x=− 27 7x −= 27( ,0)7P − M1 A1 14(a) 2212 5 13AC = + = cos cos 5 13 ACD ACB  =−  =− M1 A1

14(b) Using angles in a semicircle, AC is diameter. Radius = 13 6.52== B1 15(a) 500 A1 15(b) 2500(4) 500 100%500 −  1500%= M1 A1 15(c) t=3.5 A1 16(a) 2 22 2 x xy x xy y + ++ 2 () () x x y xy += + () x xy= + M1 A1 16 (b) 210 5 2yz y xz xy− − + 5 (2 ) (2 )y z y x z y= − − − (5 )(2 )y x z y= − − M1 A1 17(a) 34,59 B1,0 17(b) (i) 3 4 2 2 5 9 5 3 8 15  +  = M1 A1 17(b) (ii) 351 59 2 3 − = M1 A1 18(a) Deposit 20 $1200 $240100=  = Remaining amount $1200 $240 $960= − = Monthly payment 1.5 $960 12100 $172.80 =   = Total amount paid $172.80 $1200 $1372.80 =+ = M1 A1

18(b)(i ) 3 B1 18b)(ii ) B1 19(a) 1 cm : 10 km 1 cm² : 100 km² Area (map) = 225 100 = 2.25 cm² M1 A1 19b) 36 cm² : 225 km² 6 cm : 15 km 1 cm : 2.5 km 1 cm : 250000 cm Map scale is 1:250000 M1 A1 20(a)(i ) 424624 2 17= B1 20(a)(i i) 424624 2 17pp qq =   17, 2pq== B1 20(b) 27, 29, 32a b c= = = B2 21(a) (vert. opp. s) 24 ( s in the same segment) 67 ( s in the same segment) AED BEC EAD EBC EDA ECB  =   = =    = =   ∆AED is similar to ∆BEC (AA similarity test) M1 (1 pair of correct corresponding angles with reasons) A1 21(b) 2 3.5 4 7 EB EB = = Alternative solution M1 A1 M1

4 sin 24 sin 67 4sin 67 9.05 (3 . )sin 24 EB EB cm s f = ==  A1 21(c) 180 67 67 46AOD = − − =  180 24 24 132BOC = − − =  46 2360 132 2360 46 132 23 66 r r     = = M1 A1 22(a) 2 Base area of Vase Base radius of Vase Base area of Vase Base radius of Vase BB AA = 2 Base area of Vase 3 50 1 B = 9Base area of Vase 50 4501B=  = 2cm M1 A1 22(b)(i ) Curved surface area of Vase Curved surface area of Vase C A Base area of Vase Base area of Vase C A= 800 16 50 1== The required ratio is 16: 1 A1 22(b)(i i) Base area of Vase 16 Base area of Vase 1 C A = height of Vase 16 4 height of Vase 1 1 C A == 3 Volume of Vase 4 Volume of Vase 1 C A = Volume of Vase 64 400 1 C = 64Volume of Vase 400 256001C =  = 3cm M1 A1

23 (a) () 60 0.1560 100 speedgradient acceleration time= −= =−− 0 0.1570 100 0.1530 v v − =−− =−− 4.5v= /ms M1 A1 23 (b)