Paper 2 MS

Name: Index Number: Class: HUA YI SECONDARY SCHOOL Preliminary Examination 2022 MATHEMATICS Paper 2 2022 2 h 30 min Candidates answer on the Answer Space provided. Mark Scheme This document consists of 15 printed pages including the cover page. © HYSS 2022 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 4E 4E 4048/2 Setter: Ms Lee Hui Ling

2 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 Mathematical Formulae Compound interest Total amount = n rP      + 1001 Mensuration Curved surface area of a cone = l rπ Surface area of a sphere = 24 rπ Volume of a cone = h r2 3 1π Volume of a sphere = 3 3 4 rπ Area of triangle ABC C absin2 1 Arc length = θr , where θ is in radians Sector area = θ2 2 1 r , where θ is in radians Trigonometry C c B b A a sin sin sin= = A bc c b acos22 2 2− + = Statistics Mean = f fx ∑ ∑ Standard deviation = 22       ∑ ∑−∑ ∑ f fx f fx

3 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 Answer all the questions. 1. (a) 42 x ------B1 (b) 42 0.5x+ -----B1 (c) (i) 2 2 42 42 1 10.5 4 84 0.5 1 2 168 0 1 Mxx xx M xx A − = −−−−−+ = + −−−−− + − = −−−−−− (ii) ( ) 21 1 4(2)( 168) 12(2) 8.919 8.92(3 )------A1 9.419 9.42 31 xM x sf or sf A −± − −= −−−−− = = − =− −−−−−− (iii) 42 4.459 4.46 (3 ) 18.919 0.5 = 4 h 28 min -------A1 h h sf M= = −−−−−+ 2. (a) (i) 1.25 ---------B1 (ii) Find area of triangle AOC: 0.5 (12)( 8 )( sin 1.25) = 45.55 ----M1 ecf Find area of sector AOB: 0.5 (122 )(1.25) = 90 ------M1 ecf Area of shaded region = 44.45 ----------A1 (iii) Find AC using cosine rule = 12.14 ---------M1 Perimeter = 31.1 ( 3sf) --------A1

4 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 (b) (i) 32+38 = 70 ( isosceles triangle AOD and AOB ) (ii) 140 ( angle at center = twice angle at circumference ) (iii) Find DCB using opposite angle of cyclic quad ( M1 ecf ) Angle ODC = 180 – angle DCB ( interior angles, // lines ) =180-110 =70-------A1 3. (a) . AD 2 = 42+ 122 --------M1 AD = 12.649 ( shown ) -----A1 (b) Area of ABCD = 96 -------M1 Lateral area = ( 10+12+6+12.649) x 16 =650.384 --------M1 Surface area = 842.384 =842 ( 3sf) -----A1 (c) Volume of prism = 96 ( 16) -----M1 = 1536 -----A1

5 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 4. (a) 3c - 3a -----B1 (b) 3a – 1.5c -----B1 (c) 11.5 (3 1.5 ) 13 21 AK M AK KP M = + − −−−− = = −−−−− c ac c+a     This implies that AK// KP and they have a common point K. Hence A, K and P lies on a straight line. -----A1 (d) 1/ 2 ------B1 (e) 1/6 -----B1 5. (a) Form equation : Area of trapezium 0.5(20 55) 450v+= ---------M1 Get v = 12 -----A1 (b) Acceleration = 1.2 ------M1 or use similar triangle Speed at t = 8 , 1.2(8)= 9.6 -----A1 (c) The motorcyclist is travelling at a constant speed . ------B1 [1]

6 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 (d) 0.48 ------B1 (e) A1: increasing curve (10,60) A1: straight line till (30,300) A1: decreasing curve till (55,450)

7 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 6. (a) p = 3.4 -----B1 (b) See graph (c) Draw any line that cuts y- axis at -2 and intersect the curve at 2 points for 44 x−≤≤ . One possible line is the line that passes through ( 2,-1.4) and ( 0, -2) m = 1.4 ( 2) 0.3 120 A− −− = −−−−−− 0.3 ≤ m ≤ 1.95 (d) Get y = 4 ----M1 The line y = 4 intersect the curve at 1 point, hence the equation only has one solution------A1 (e) B1 for graph (f) Form 3 21 252 xx x− += −+ ----M1 Balance equation -------M1 Get 32 15 10 0xx− −= ------A1 7. (a) Time taken = 11/3 h or 1 h 20 min -----M1 Time expected to arrive at B = 0005 or 12:05am ----A1 (b) Use cosine rule : PQ 2 = 3002+1202-2(300)(120)cos 116 -------M1 PQ = 368.73 =369 ( 3sf) (c) Form sine rule equation or cosine rule to find angle QPB-----M1 find angle QPB------M1 ( ecf )

8 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 bearing of Q from P = 064 - 017.00 =047.0----A1 8. (a) (i) 11.625---B1 (ii) It was because we do not have the actual timing for each customer. ---B1 (iii) 5.35 -----B1 (iv) I would go to Shop B although the the mean is slightly higher than shop A. But the smaller SD suggest that the more consistency in the waiting time. --- --B1 Or I would go to Shop A as the mean is smaller, meaning on average I will have a shorter waiting time. ----B1 Any reasoning that is logicial. (v) The mean will remain the same and the SD will decrease. B1 each (b) (i) 52 ---B1 (ii) 66----B1 (iii) 44 -----B1 (iv) 180-75 -------M1 ( Find 75) =105 ------B1 9. (a)

9 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 �84 90 56 92 60 61 � ---B1 (b) �1812 886 1673 853� -----B1 (c) 48.9% -----B1 (d) ( 1742.5 869.5 )--------A1 (e) The average sale for the 2 days is $1742.50 and the average profit is $ 869.50. ---B1 10. (a) (i) Find gradient or y intercept correct ---M1 𝑦𝑦 = 3 8 𝑥𝑥 + 3 1 4 − − − 𝐴𝐴1 She is not correct as the y intercept should be 3.25. (ii) Use Pythagoras’ thm or formula ------M1 AB = 5 units (iii) -4/5 -----B1 (iv) x = -1 or 5 (b) (i) 6 130000000 155 2.37 10 1 M A −−−− = × −−−−−−−−− (ii) (1.3 × 10 8) ÷ 70 ÷ 10 ÷ 150----M1 = 1238 ---- A1 11 (a)

10 Sec 4E Preliminary Examination 2022 Mathematics Paper 2 2 30interest = 40000 2000 1100 12 monthly instalment = 42000 30=1400---A1 M× × = −−− ÷ (b) 5100 1585 × ----M1 ( 85% is $5100, find 15%) = 900------A1 (c) Teaching staff : 12 (2)( 4 )( $70) + 14(2)(4)($90)= ($6720 + $10080) = $16 800 M1 Total staff cost : 16 800 + 1800 + 45(5)(4)=$19500 Rental and printing cost = 8400 Total operational cost per month = $19500 + 8000+400 =27900 M1 Total fee collected (assuming each class has the minimum number of students ) = 12(5)($200) + 14(5)($300) ---M1 =$33 000 ( $52800 if they find 8 students per class ) Can find for other number as well but they have to state. Min Profit per month = $5100 ---M1 Her target of a minimum of $5000 per month can be reached as her minimum profit per month is $5100. –A1 justification $5100 x 6 months = $30 600 < $45 100 + 2000 interest ----M1 She might miss her target of recovering her start up cost within 6 months as the total profit for 6 months assuming she get the minimal number of students per class is less than the start up cos