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)−− 2