SGSS 4N Prelim 2023 P1 MS

2023 E Math 4NA PRELIM Paper 1 Marking Scheme Solutions 1 4( 5) 3 2 4 20 3 2 22 xx xx x − = + − = + = 2 8 10 12 15 2 (4 5 ) 3 (4 5 ) (4 5 )(2 3 ) ab ay bx xy a b y x b y b y a x + − − = + − + = + − 3 34.5 6.3 30 6 8.76 9 24 3 8 −− = = = 4 5a 34 3 4 ( 2) 2 9 55 55 5 + − − −  = = 5b ( ) 1 3 1 2 3 93 33 2 3 n n n = = = 6a 10 3(7 4 ) 10 21 12 22 21 x x x x x − − = − + =− 6b 2 2 2 2 2 h k m m k h khm OR h k m hkm =− =− −= − =− −= − 7a p = 60 alternate angles

Solutions 7b ABC = 80 (adj s on a straight line)  60  ABC is not an equilateral triangle. 8a Modal age = 28 years 8b Median age = 32 34 2 + = 33 years 8c Probability = 22 50 = 11 25 9 4 13 (1) 7 33 (2) xy xy − = −−−−− + = −−−−− From (1), 4 13 (3)xy= + −−−−− Subst (3) into (2), 7(4 13) 33yy+ + = 28 91 33 29 58 2 yy y y + + = =− =− Subst y = −2 into (3), x = 4(−2) + 13 = 5 10a 2 2 , where is a constant 45 9 405 405 kyk x k k y k = = = = 10b 405 36 111.25 or 11 4 y= = 11a 43yx=− When x = 2, y = 4(2) – 3 = 5  (2, 5) lies on line l1. 11b 4y x c=+ Using (−1, 5), 5 = 4(−1) + c c = 9 y = 4x + 9 12 Area = 22160 1 π(15 ) (15 )sin160360 2  −  = 276 cm2 (3 s.f.)

Solutions 13 2 2 2 42 5 2 10 4 5 2 0 ( 5) ( 5) 4(1)( 2) 2(1) 5 33 2 5.37 or 0.372 x x xx xx x = − −= − − = − −  − − −= = =− (3 s.f.) (3 s.f.) 14a 2 2 2 2 2 10 9 10 25 25 9 or ( 5) 5 9 ( 5) 34 xx x x x x +− = + + − − + − − = + − 14b 2 2 2 10 9 0 ( 5) 34 0 ( 5) 34 5 34 34 5 0.83 or 10.83 xx x x x x + − = + − = += + = = − =− (2 d.p.) (2 d.p.) 15 Int.  for shape A = 60 Sum of int  for shape B= (n-2) × 360 = (10-2) × 360 = 1440 Int.  for shape B = 1440 10  = 144 Int.  for shape C = 360 − 60 − 144 = 156 1 ext. angle for C = 180- 156 = 24 Number of sides for C = 360 24   = 15 16a 1 cm : 50 000 cm 1 cm : 0.5 km n = 0.5 16b Map distance = 10 0.5 = 20 cm 16c 1 cm2 : 0.25 km2 Actual area = 0.25  20 = 5 km2

Solutions 17a 24, 29 17b 4 5( 1) 4 5 5 51 n n n +− = + − =− 17c 5(15) – 1 = 74 lines 17d 5 1 100 101 is not an integer5 n n −= = Not able to form a pattern with 100 lines 18 Surface area = 222π(8)(15) π(8 ) 2π(8 )++ = 240 + 64 + 128 = 432 = 1357.168 = 1360 cm2 (3 s.f.) 19a QR = BC = 3 cm Area of PQR = 1 (3)(3.8)sin 602  = 4.94 cm2 19b 4.5 3.8 3 4.5 3.83 5.7 cm XY YZ PQ QR XY XY = = = = 20a 20c A B C M 20b BAC = 110 ( 2) 20d BM = 4.9 cm ( 0.1 cm) 21a Particle is moving at a constant speed of 70 m/s.

Solutions 21b Retardation = 70 35 = 2 m/s2 21c Total distance = 1 (25 60)(70)2 + = 2975 m OR can add up the 2 areas Area Rect + area triangle 1750+1225 = 2975 Average speed = 2975 60 = 49.6 m/s or 74912 m/s