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 triangle DAC. Alternatively, AB = AD (Given) Since triangle BAC and triangle DAC are isosceles, Angle ABC = angle DBC – angle DBA = angle BDC – angle BDA = angle ADC BC = DC (Given) Since there are 2 pairs of equal corresponding sides, and a pair of equal included angles, triangle BAC is congruent to triangle DAC. Must see the word included angles.

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution 17 (radius tangent)2 (radius tangent)2 OBD rad OCD rad    = ⊥  = ⊥ 2 (angle sum of quad) 22 = rad (angle at centre = 2 angle at circum ference) 2 Since ABDC is a kite, diagonal of kite bisects . Hence, (Base of BOC x x xBAC BAC ABO BAO s     = − − − − − =    =  isosceles triangle are equal) = 2 2 = rad4 x x   −  − Accept any logical proof with ALL reasons given. 18(a) Q = 330 1675 310 1710 3635 3380 xx++  B1 for each correct row 18(b) Each element of the 2nd row of matrix Q represents Dean’s expenditure if he buy from the shop and the online store respectively. 18(c) (310x + 1710) – (330x + 1675) = 25 x = 0.5 18(d) Amt saved = $(3635 – 3380) = $255 19 x = 17.5o or 162.5 o B1 for each answer 20 33 3 3 33 3 3 3 265 3 865 27 162 135 8 143 162 162 143 162 143 qpq qpq p q q qp pq pq −= −= −= = = = 21(a) 2 2 2 o1.4 1.3 2(1.4)(1.3)cos 72.6 1.60047 = 1.60 QS QS = + − =

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution 21(b) o o o o o o o sin sin 83.3 1.60047 2.0 52.633 180 52.633 83.3 = 44.06 = 44.1 QPS QPS PSQ  = =  = − − 22 Triangle ACJ is similar to triangle EFJ 1 o o 1 9 , 9 and 18 9tan 18 9tan 18 = 26.56 = 26.6 AC FJ FE CJ hence CJ cm AC cm AEH AEH − == == = = Must show how to find lengths 23 48 4 2 4 24 4 2 3 18 2 (3 ) 18 (2 3 ) 18 18 18 41 0.25 xx xx x x x x = = = = = = Make the base the same and do comparison of the indices 24(a) 2130.952130.95 1.08 = $157.85 GST =− 24(b) Dealer’s selling price without GST= $(2130.95-157.858) = $1973.101 Jen’s selling price to dealer = $(1973.101  1.15) = $1715.74 25(a) Men Distance(km) Days 6 10 5 1 30 90 6 30 5 No. of extra men hired = 18 – 6 = 12 I disagree with Dave as he needs to hire 12 more men instead of 9.

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution Alternatively, Men Distance(km) Days 6 10 5 1 10 30 15 30 6 I disagree with Dave. If he hires 9 more men, the job will need 6 days instead of 5 days to complete. Alternatively, Men Distance(km) Days 6 10 5 1 10 30 15 150 30 15 25 5 I disagree with Dave. If he hires 9 more men, the distance tarred will be 25 km instead of 30 km. 25(b) All men work at the same rate. 26(a) Shape Both coordinates of x-intercepts yintercept (b)(i) Let the height of triangle ABC = h 1 (5 ( 6)) 1212 22 hence, x-coordinates of is 24 or -20 h h C − −  = = Either 24 or -20

ZHONGHUA SECONDARY SCHOOL 2023 Emath PRELIM P1 solution b(ii) 22 ( 26 2) ( 16 ( 6)) = 29.732 = 29.7 Length of BD = − − + − − −