2023 ZHSS EMATH P1 PRELIM MS

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution Qn Solution Mark Remarks 1 –3.2001 2(a) 82 9 a 2(b) 2+10x 3(a) 27.95 Exact value only 3(c) 32.4 4(a) 37 4(b) 4n + 1 Accept equivalent form 4(c) Number of dots, n = 133 is a whole number/ positive integer/532 is a multiple of 4 5(a) P(yellow) = 0.25 5(b) Let the initial number of yellow balls be x. Total number of balls after 10 yellow balls are added = 36+x P(yellow) = 10 1 26 3 x x + =+ 3x + 30 = 36 + x x = 3 6 11 11 (1) 2.1 1.35 18.6 (2) (1) into (2) 2.1 1.35(11 ) 18.6 number of pens, 5 number of pencils, 6 yx yx xy sub xx x y += = − −−−−− + = −−−−− + − = = = Must use simultaneous eqns to solve. 7(a) (7 2)180 7 O FGA −= 900 7 O = 900 4502 77 OO OGF =  = Accept 64.3o

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution 7(b) Given 612 7 900 6 810127 7 7 450 810Then 180 77 By converse of interior angles, FH and GO are parallel. Given that GF//OH and FH//GO, this implies that Quad EGOH is a parallelogram. o ooo oo o EFH GFH OGF GFH =  = − =  + = + = Must work with exact values. 7(c) 360 8 Refle 10 360360 77 192 x .9 o oo o o AOH GFH GOA  = − − = − − = 8 HCF of 60 and 105 = 15 Size per group = 15 No. of groups = (60+105)  15 = 11 9 Let the length of square be x cm Since E is midpoint of BC, BE = 0.5x Area of triangle ABE = (0.5)(0.5x)(x) = 0.25x2 Given E is also midpoint of AF, Therefore area of triangle BEF = area of triangle ABE 0.25x2 = 9 x = 6 Accept any logical method 10(a) Accept Bearing of 164o to 166 o Length of AL = 10 to 10.1 cm 10(b) Bearing of Sandy’s path = 253 o to 257 o 11(a) 35(3r – s) 11(b) (2x – 3)(6x + 7) B1 for each correct factor 12(a) Chocolate : Mint : Toffees 4n ×13 : 7×13 13× 7: 6n× 7 52n : 91 : 42n Ratio of chocolates to toffees = 52n : 42n = 26 : 21

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution 12(b) Total number of sweets = 273 (52 91 42 )91 nn + + = 282n + 273 13(a) 10 3 v Accept any equivalent expression 13(b) 12( 7) 5 v+ Accept any equivalent expression 13(c) 12( 7) 5 v+ + 14 6 = 10 3 v 144( 7) 60 v+ + 250 6 = 200 60 v v = 22.46 = 22.5 (3 s.f.) Form equation Correct equation 14(a) (3 9) 2 553 p− + Do not accept any simplified form 14(b) (3 9) 2 553 p− + 3 53 56 p p −  p = 53 15 2 2 2 2 2 22 (2 3 )(3 2 ) (2 ) 9 4 (4 4 ) 5 4 5 x y y x y x y x y xy x y xy x + − − − = − − − + = + − For 2294yx− or 2244y xy x−+ 16 AB = AD (Given) BC = DC (Given) AC is a common length shared by triangle BAC and triangle DAC. Since there are 3 pairs of equal corresponding sides, triangle BAC is congruent to