ANDSS 4E2024EM_Prelim_P1_MS

Sec 4 Express Mathematics 2024 Prelims Marking Scheme 1 ( )( )5 2 5 2a a b a b−+ ( ) ( )( ) 22 52a a b=− ( ) 2225 4a a b=− M1 Correct expansion 3225 4a ab=− A1 2 23 3 90xx ++= 23 3 3 90xx+  = M1 ( )3 1 9 90x += 39x = 233x = 2x= A1 3 2 22 2 4 3 6 26 x xy x y x xy y + − − +− ( ) ( ) 22 2 2 3 2 26 x x y x y x xy y + − += +− M1 Factorise by grouping for numerator ( )( ) 22 2 2 3 26 x y x x xy y +−= +− ( )( ) ( )( ) 2 2 3 2 3 2 x y x x y x y +−= −+ M1 Factorise denominator 23 23 x xy −= − A1 4 (a) LCM of A and B is 237mq + and at the same time the LCM of A and B is 33 5 7 . 1m= B1 5q= B1 (b) 337A= and 3 5 7B=   HCF of A and B 37= 21= B1

5 The size/height of the bar graph. B1 Stating that the graph does not start from zero is accepted too. It can be misleading because for example, the size/height of the graph in 2019 is twice that of the graph in 2010 but this does not represent their actual temperatures. B1 A relevant example has to be given. 6 (a) 2 9 3 627 y x − −   29 3 2236 333 y x −     − −  −       = 6 243 y x − −= M1 Multiply 2 3 − to each index 2 46 3 xy= A1 46 9 xy is also accepted (b) 2 25 5 125 y x− = ( ) 2 23 5 5 5 y x− = 2 63 5 5 5 y x− = 26355 xy−+ = M1 Any other equivalent form of 55mn= will be accepted too 3455xy− = Comparing index, 34xy−= 4 3 yx += A1

7 (a) 229 24 16x xy y++ ( ) 2 34xy=+ B1 (b) Let 4xa= , ( ) 8 8 4 2144 9 24 16a a a y y− + + ( ) 2 2 2144 9 24 16x x xy y= − + + ( ) ( ) 22 12 3 4x x y= − + M1 Attempt to factorise an expression in the form 22ab− , after making use of part (a) answer ( )( )12 3 4 12 3 4x x y x x y= + + − − ( )( )15 4 9 4x y x y= + − ( )( ) 4415 4 9 4a y a y= + − A1

8 (a) 2 kq r= , where k is the proportionality constant. q and r are the initial intensity and distance respectively. When the distance is reduced by 40%, the new distance is 3 5 r . New intensity 2 3 5 k r =  M1 29 25 k r= 2 25 9 k r= 25 9 q= A1 (b) Percentage difference New Intensity Initial Intensity 100%Initial Intensity −= 25 9 100% qq q − = 16 100%9= 178%= (3 s.f.) B1 7177 %9 will be accepted too.

9 B1 Both shape and y- intercept needs to be drawn and indicated correctly 10 SC RD= (given) No marks for this step SCD RDA = (alternate angles) M1 CD DA= (sides of a rhombus are equal) M1 By SAS congruency test, triangle SCD is congruent to triangle RDA. A1 11 2 10 2xx−− ( ) 2 5 25 2x= − − − ( ) 2 5 27x= − − M1 Minimum point is ( )5, 27− A1 y x ( )0,1

12 (a) ( )n  33= 9= B1 (b) P ( ) ( ) ( ) ( ) 1,0 , 0,