BDKSS 4N Prelim P1 2023 MS

2023 4N Prelim P1 MS 1 33 0.02 0.25 0.2 11 • B2 Deduct 1 mark for each error 2a 2 : 5 B1 2b 1200 100% 150  = 800% M1 A1 3 28 4 x x − −  Smallest prime = 5 M1 A1 4 2 2 2 20 (2.5) 125 125 5 kP x k k P = = = = = 5 M1 A1 5a Acute x = 1sin (0.7) 44.4− = Obtuse x = 180-44.4=135.6° B1 5b y = 180-123 = 57° B1 6a 24( 5) 5m= − − = 95 B1 6b 2 2 2 45 54 5 4 5 4 ma ma m a m a =− += + = += 5 4 ma += M1 (isolate a2) A1

7a x = 180 – 140 = 40 They are interior angles between 2 parallel lines B1 (mention interior and parallel) 7b Angle a= = 62 (base angles of isos tri) Y = 180 – 62 (interior angles, // lines) = 118 M1 A1 Deduct 1 mark if missing/ no reason 8 The trend may not be decreasing There are missing data for the months of March to June and August. B2 9 2 7.5 1.5 4 7.5 3 PQ x x PQ = = PQ = 10 cm M1 (or find scale factor = 4/3) A1 10 x = 2(x + 3) x = 2x + 6 – 6 = x x = – 6 M1 (expand) A1 11 5 5 1 5 1 5 1592.19 1500(1 ) 100 1.06146 (1 ) 100 1.06146 1 100 1.06146 1 100 r r r r =+ =+ =+ −= R = 1.20% (3sf) M1 M1 (remove power 5) or attempt to solve A1 12ai 4405 3 5= B1 12aii HCF = 235 =45 B1 12b 3 3 6 2 3 6 12 (2 5 ) 25 n n Q Q = = 2425 nQ= M1 A1 13 Remainder = 11/20 Food and shopping = 60% of (11/20) * 100% = 33% M1 (find remainder) M1 (or find 40%) A1

14a See BUY rate as bank is buying CAD 1 CAD = 1.0028 SGD 250 CAD = 250 x 1.0028 = 250.70 SGD M1, A1 14b See SELL rate as bank is selling JYP 100 JPY = 0.9447 SGD 500 SGD = 500/0.9447 X 100 = 52926.855 = 52927 JYP M1 A1 15a B1 15bi 2n B1 15bii 5n+1 B1 15c 5 1 615 5 614 122.8 n n n += = = He is not correct as n is not a whole number when total has 615 arrows. M1 A1 16a ( 1) 33 1 33 1 9 3 48 2 k k k k kk k k −− =− + =− + = − = = P is (2, -1) M1 M1 or (attempt) to solve A1 16b 3 3(5) 3 15 18 y mx c c c c =+ − = + − = + −= 3 18yx=− M1 A1

17a B2 Deduct 1 for each error 17bi P (prime) = 6/36 = 1/6 B1 17bii P(at most 20) = 1-6/36 = 5/6 B1 17biii P(irrational) = 0 B1 18a 2 22 2 2 6 16 666 16 22 ( 3) 25 xx xx x −− −−   = − + − −       = − − a = -3 and b = -25 M1 A1 OR B2 18bi x = 3 B1 18bii When y = 0, 2 2 ( 3) 25 0 ( 3) 25 3 25 25 3 x x x x − − = −= − = = + X = 8 or -2 M1 (solve) A1 19a No he is not correct as the line AB has a positive gradient so it should be upward sloping. M1 (positive m) A1 (upward slope) 19b 42 12 ( 1 4)( 1 2) 9 x y −+= =− =− − + − − = Max pt (-1, 9) M1 (find x coord) or any other method A1 20a 225 18 349L= + = Curved area = (5)( 349)rL = =293.4489707 = 293 cm2 M1 M1 A1 20b 2293.4489707 ( 349)360  = 96.4 = (1dp) M1 A1

Or 21 ( 349) 2