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 --------------------- B1 4(b) Drawing of graph - Correct plotting of points --------------------- G1 - Drawing of curve through all the points ---- G1 - Curve is smooth -------------------------------- G1 4(c) 3 3 3 1 2 04 3 1 24 x x x x        This equation can be solved by drawing the line y = 2 on the same grid. ----------------------------------------- B1 The line y = 2 meets the curve at 3 intersection points, hence there are 3 solutions for the equation 3 3 3 04 x x   . ---------------------------------- B1 4(d) Drawing of tangent at (3, 3.25) ------------------------- M1 Gradient = 3.75 ( Accept 3 to 4 ) ------------------------ A1

pg. 5 4(e) (i) 2 6y x  6 2 xy  Take any 2 or 3 points to draw the above line. x 4 0 4 y 5 3 1 Drawing of line on the grid. Ensure that line is drawn for 4 4 x   . ------------------------------ B1 (ii) 3 63 14 2 x x x    --------------------------------- M1 3 12 4 12 2x x x    3 10 16 0x x   ------------------------------------ A1 [ 11 marks ] 5(a) 11 8 44 19 n n   -------------------------------------- M1 19(11 ) 8(44 )n n   209 19 352 8n n   11 143n 13n ---------------------------------------- A1 Number of students in S2 = 31 ----------------- B1

pg. 6 5(b) (i) First card Second card First branch correct ------- B1 All correct ----------------- B1 (ii) (a) 6 5 18 17 = 5 51 ---------------------------------- B1 (b) 6 10 10 6 18 17 18 17   ------------------- M1 = 20 51 -------------------------------- A1 (c) P( no red cards) = 8 7 18 17 -------------- M1 = 28 153 P( at least one red card) = 281 153 = 125 153 -------------------- A1 Alternative method 6 5 6 2 2 6 2 11 18 17 18 17 18 17 18 17                  ----- M1 = 125 153 ------------------------------------------------ A1 [ 10 marks ] blue red blue blue blue yellow red yellow yellow red red yellow BB BR BY RB RR RY YB YR YY

pg. 7 6 (a) 6.8sin 40 AC 6.8 sin 40AC   --------------------- M1 10.5789AC  10.58AC  m ( 2 dp ) ------------- A1 (b) (i) 2 212.7 10.58 2(12.7)(10.58) cos 66AB    ---- M1 12.8032AB 12.8AB m ( 3sf) ------------------------------------ A1 Can use the given value of AC = 10.58 km (ii) Area = 1 1(12.7)(10.58)sin 66 (10.58)(6.8)sin 502 2   --- M2 = 88.930 =88.9 m2 ----------------------------------------------- A1 (c) tan 8 12.8032 h 12.8032 tan8h  --------------------------------------------- M1 1.7993h tan h AC  1 12.8032 tan 8tan 10.58        ---------------------------------- M1 9.652   Angle of depression is 9.7 -------------------------------- A1 [ 10 marks] 7 (a) F = 0.47 0.75 1.27         ---------------------------------------------------- B1 (b) T = 16 16 18 20 21 8 20 17 0         0.47 0.75 1.27         = 7.52 12 22.86 9.4 15.75 10.16 9.4 12.75 0            

pg. 8 = 42.38 35.31 22.15         ------------------------------------------------ B1 (c) 42.38 represents the total transport cost of all the employees in shift 1 when they come for work from the nearest MRT station. 35.31 represents the total transport cost of all the employees in shift 2 when they come for work from the nearest MRT station. 22.15 represents the total transport cost of all the employees in shift 3 when they come for work from the nearest MRT station. ------------------------------------------------------- B1 (d) N =  1 1 1 ----------------------------------------------- B1 NT =  1 1 1 42.38 35.31 22.15         =  99.84 Total fare = $99.84 ----------------------------------------- B1 [5 marks] 8 (a) (i) 2 25OAC    = 5  rad Angle at centre = 2  angle at circumference ----- B1 (ii)    2 2 2 36 6 2 6 6 cos 5AC        ----------------- M1 AC = 9.7082 = 9.71 cm (3sf) ------------------------------------ A1 (iii) M1 M1 Area of OCE =      21 1 39.7082 6 6 sin2 5 2 5            = 29.60927 – 17.119 = 12.4903 cm2 Therefore, Area of shaded region =   21 26 12.49032 5      -- M1 = 10.1 cm2 (3sf) ------------ A1

pg. 9 (b) (i) 48OCB   (radius perpendicular to tangent) --- M1 84COB   (triangl