Beatty Sec EM_4E5N_P2_2020_MS

pg. 1 Beatty Secondary School Prelim Examination 2020 Sec 4E/5N Mathematics Paper 2 ( 4048/02) Marking Scheme 1(a) (i)   2 21 2A h a b    2 21 (6) 4 2.52A  A = 29.25 ------------------------- B1 (ii)   2 21 2A h a b  2 2 2Aa b h  ----------------------- M1 2 2Ab a h  or 2 2a h Aa h  --------------- A1 1(b) 2 7 2 4 4 1 2 1x x x    = 2 7 2 (2 1) 2 1x x   ---------------- B1 ( factorisation ) = 2 7 2(2 1) (2 1) x x    ------------------ M1 = 2 7 4 2 (2 1) x x    = 2 9 4 (2 1) x x   ------------------------- A1

pg. 2 1(c) (i) 5 4 4 7 2 x x   2(5 4 ) 7( 4)x x   --------------------------------- M1 10 8 7 28x x   15 18x  115x or 1.2x --------------------------- A1 (ii) Greatest integer 2x ---------------------------- B1 1(d) 7 2 2 5 x x x   5( 7) 4 10x x x   ------------------------------- M1 5 35 4 10x x x   35x --------------------------------- A1 [ 11 marks ] 2(a) 48 x hours --------------------------------------------- B1 2(b) Time taken = 18 60 hour ------------------------------ B1 = 3 10 hour 48 32 3 6 10x x  ----------------------------------- M1 348( 6) 32 ( 6) 10x x x x    210(48 288 32 ) 3 18x x x x    --------------- M1 2160 2880 3 18x x x   23 142 2880 0x x   ------------------------- A1

pg. 3 2(c) 2( 142) ( 142) 4(3)( 2880) 2(3)x       -------- M1 142 54724 6x  62.655x or 15.321 62.66x or 15.32 ---------------------- A2 2(d) Time taken = 48 32 62.655 62.655 6  = 1.2321 hours ------------------ B1 = 1 hour 14 minutes ---------------- B1 [ 10 marks ] 3(a) (i) Total mail in 2009 = 1 002 662 Total mail in 2012 = 764 780 Percentage decrease = 1002662 764780 100%1002662   ----- M1 = 23.7 % ------------------------------ A1 (ii) Mean number of mail per month for 2012 = 764780 12 ---------------------------- M1 = 63731.66 = 46.37 10 ( 3 sf) -------------------- A1 (iii) Year ratio/ fraction 2009 6.9223… 2010 5.6781… 2011 6.0349… --------------- M1 2012 5.9866… Years 2011 and 2012 --------------------- A1

pg. 4 3(b) Amount = 36 1.5 128500 1 100          ------------- M1 Interest = 36 1.5 128500 1 8500100           = $ 390.99 ------------------------ A1 3(c) THB 5400 = $ 5400 22.66 ------------------------- M1 = $ 238.305 Amount payable = 101.2 238.305100  --------- M1 = $ 241.16 = $241 ( nearest dollars ) ---- A1 [ 11 marks ] 4(a) p = 1.25