2024 HYSS Prelim Sec 4G2 Math MS P2

Name: Index Number: Class: HUA YI SECONDARY SCHOOL Preliminary Examination 2024 MATHEMATICS Paper 2 For Examiner’s Use This document consists of 7 printed pages including the cover page. © HYSS 2024 No part of this document may be reproduced in any form or transmitted in any form or by any means without the prior permission of Hua Yi Secondary School. [Turn Over 4G2 4G2 MARK SCHEME

2 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME Qn Suggested Solution Mark Allocation 1ai Not all the powers of the prime factors are even numbers/multiples of 2. --- B1 1aii 22 3 11 726 k =   = --- B1 1bi 2 2100 2 1050 3 525 5 175 5 35 7 7 1 222100 2 3 5 7=    --- M1 --- A1 1bii 223 12 HCF = = --- B1 (ecf) 2 (70 2.5) (2 ) 804.5 175 2 360 2 185 92.5 x x x x + = += = = --- M1 ---M1 --- A1 3 2 2 10506.25 10000 1 100 1.050625 1 100 1 1.025100 2.5 p p p p =+  =+ += = --- M1 --- M1 --- A1 4a *M1 – multiplication frame 22 9 5 (2 1)( 5)x x x x− − = + − --- M1 --- A1 4b 2 2 2 2 22 2 18 2( 9 ) 2 (3 ) 2( 3 )( 3 ) x y x y xy x y x y − = − =−  = + − ---M1 ---A1

3 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME 5a 5 :1 5 :100000 1: 20000 cm km cm cm --- B1 5b 22 22 5 :1 25 :1 200 :8 cm km cm km cm km Or 22 22 5 :1 1 : 0.2 1 : 0.04 200 :8 cm km cm km cm km cm km --- M1 --- A1 6a 2 2 2 2 22 20 99 10201 101 10201 XZ XY YZ + = + = == Since 2 2 2XZ XY YZ+= , XYZ is a right-angled triangle. --- B1 6b 99cos 101 11.4211 XYZ XYZ = = Or using area Let the shortest distance from X to YZ be h. sin11.4211 99 19.603 19.6 (3 ) h h h m sf = = = --- M1 --- M1 --- A1 7a --- A1 (for the 6 outcomes for the die) --- A1 (for the outcomes for the coin) *wont penalise students if they write the probabilities/write the probabilities wrongly 7bi 31 12 4= --- B1 7bii 0 --- B1 8a 3 --- B1

4 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME 8b 1m --- at least 8 points plotted correctly 1m --- smooth curve drawn 8c 2.1,4.6( 0.1)x=−  --- B2 8d The minimum/lowest point of the curve is at 7y=− , so there will not be any point below 7y=− . --- B1 8e Drawing of tangent correctly 1 ( 12) 4.2 0 3.0952 3.10(3 ) gradient sf −−= − = = --- B1 --- B1 9ai 300$ x   --- B1 9aii 300$ 1.5x  + --- B1 9b ( ) 2 2 300 1.5 30 360 9000300 1.5 45 360 9000 1.5 105 0 9000 1.5 105 0 70 6000 0 xx xx xx xx xx  + − = − + − = − + − = − + − = − − = --- M1 (form eqn) --- M1 (expansion) --- A1 9c 2 70 6000 0 ( 120)( 50) 0 120 50 xx xx x or x − − = − + = = =− --- M1 (or any other method) --- A1 9d x represents the number of T-shirts and it cannot be a negative number. --- B1 (or any logical explanation)

5 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME 10a 2 3 (1.75) (2.4) 23.0907 23.1 (3 ) Volume cm sf = = = --- B1 10b 3 1 3.5 2.4 3.5 29.4 Volumeof pocket cm =   = 3 6(29.4 23.0907) 37.8558 37.9 (3 ) Volumeof total air cm sf =− = = Alternative: can take the volume of whole box – volume of macarons --- M1 --- A1 10ci For 12 macarons, 30cos 65 1000 $1.95 t of almond flour = = 2.5cos 65 1000 $0.1625 t of powdered sugar = = 2.6cos 45 800 $0.14625 t of castor sugar = = 2.5cos 2 12 $0.4167 t of egg whites= = ( )1cos 1.95 0.1625 0.14625 0.41672 $1.3377 total t of ingredients = + + + = cos 1.3377 2.80 1.65 0.40 $6.1877 $6.19( ) total t price nearest cent = + + + = = --- M1 (any 2 correct) --- M1 (finding total cost of 6 macarons, ecf) --- M1 (add packing, ecf) --- A1 10cii 72 126 boxes= cos $6.1877 12 $10 $84.2524 total t price with delivery =  + = $84.2524 130% $109.528 $110( ) selling price nearest dollar = = = --- M1 (ecf from (cii)) --- M1 (ecf) --- A1

6 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME 11a 360 (180 105 ) 285 Bearing of A fromC = − −  = --- B1 11b 180 105 75 BAC = −  = 2 2 2 2 78 80 2(78)(80) cos 75 9253.938 96.197 BC BC BC = + −  = = sin sin 75 80 96.1973 sin 0.80328 53.445 53.4 (1 ) ACB ACB ACB dp  = = = = --- M1 (ecf) --- M1 (ecf) --- A1 11c 2 1 (78)(80)sin 752 3013.688 3010 (3 ) Area of ABC m sf  =  = = --- M1 --- A1 11d let the angle of elevation be x. 180tan 78 66.571 66.6 (1 ) x x dp = = = --- M1 --- A1 12ai *For Qn 12, deduct one mark overall if reasons are missing or wrong 44BDC =  Reason: angles in same segment --- B1 12aii 180 58 122ADC = − =  Reason: angles in opp segment --- B1 12bi 122 44 78 ADB = −  = 90BAD =  (angle in a semicircle) 180 90 78 12 ( ) ABD sum of  = − −  =    --- M1 (either step) --- A1 78 180 A C balloon x

7 4G2 Prelim 2024_Mathematics_P2_MARK SCHEME 12bii 180 44 58 ( ) 78 ACB sum of = − −    = Or 78 ( )ACB anglesin same segment =  78 2( 2 ) 156 AOB at centre at circumference =   =  = 360 156 90 90 (tan , ) 24 ATB gent radius sum of quad = − − −  ⊥  = Or any other method --- M1 --- M1 --- A1 12c ATO and BTO are congruent triangles. --- B1 13ai $55 or $54 --- B1 13aii 1 3 $38 $71 $71 $38 $33 Q Q IQR = = =− = --- M1 --- A1 13aiii 70% 160 112= 70th percentile $68= --- B1 13b Number of workers who earn more than $60 160 94 66= − = 66 65 143 160 159 848= --- M1 (either finding 66 or multiplying the probability (ecf) correctly) --- A1 13ci The inter-quartile range is lower in company A, so the wage is more consistent/has a smaller spread. --- B1 13cii The median wage in company B is higher so I will earn more. --- B1