Kranji 4048 02 AS

1 2022 4E5N E Math Prelim Paper 2 (Marking Scheme) Qn Solution Remarks 1(a) Total= 5000 (1 + ( 4 12) 100) (3×12) = $5636 (nearest dollar) 1(b) 100 107 × 82500 = $77102.80 (nearest cent) 1(c) 7𝑚𝑏 = 3𝑏 + 2𝑚 𝑏(7𝑚 − 3) = 2𝑚 𝑏 = 2𝑚 7𝑚 − 3 all the terms with m on 1 side of the equation 1(d) 21𝑥2 = 2 − 11𝑥 21𝑥2 + 11𝑥 − 2 = 0 (7𝑥 − 1)(3𝑥 + 2) = 0 x = 1 7 or x = 2 3− 2(a) 500 n   jars of cookies 500 n   2(b) ( )500$ 2 3 nn  −+ or 1500$ 500 2 6 nn  + − − No need $ sign 2(c) ( ) 2 2 500 2 3 500 92 1500500 2 6 500 92 1500 2 98 0 1500 2 98 0 49 750 0 (Shown) nn nn nn nn nn  − + − = + − − − = − − = − − = +−= Simplify until RHS=0 then shown to given equation 2(d) ( ) ( )( ) ( ) 2 49 49 4 1 750 21n −  − −= n = 12.246 (3dp) or n = – 61.246 (3dp) Correct working with quadratic formula with correct values substituted in 2(e) n = – 61.246 needs to be rejected because n represents cost and cost cannot be negative 2(f) (Reject 61.246 as >0)nn=− Selling price 12.246 3=$15 (nearest dollar )=+

2 3(a) P 218 138 66 40   =   3(b) 0.75 0 0 0 218 163.5 0 0.75 0 0 138 103.5 0 0 0.85 0 66 56.1 0 0 0 0.85 40 34                =               3(c) T 35 120 75 16 60 93 112 27 = 3(d) S = TP 218 35 120 75 16 138 29780 60 93 112 27 66 34386 40      ==          3(e) 29780, 34386 represent the amount collected from the sales of tickets for the afternoon show and evening show respectively. OR 29780, 34386 represent the amount collected from thesales of tickets for each show. OR 29780 represents the amount collected from the sales of tickets for the afternoon show. 34386 represents the amount collected from the sales of tickets for the evening show. Accept any answer with equivalent meaning e.g. ticket sales, amount obtained, revenue, Do not accept: profit, price, cost, amount of tickets sold Must quote values from (c). 4(a)(i) angle BDC = 180° – 62° = 118° (s in opp. segments) 4(a)(ii) angle OCB = 118 592 = (OC bisects BCD) angle OCB = angle OBC = 59° (base ∠s of isosceles triangle) angle BOC = 180° – 59° – 59° = 62° ( sum of isosceles triangle) 4(b) Area of minor segment BC 2262 1 (6) 6 6 sin 62 3.58cm (3sf)360 2    =  −     =       Find area of minor sector. Find area of triangle.

3 5(a) Total surface area of frustum ( ) ( ) ( )( ) ( )( ) 22 22 2 40 25 48 30 12 40 48 2.5 272 12 25 30 27 42 6330cm =  +  + + + + + + = Area of either bases Any correct trapezium area or adding 6 faces 5(b)(i) Let height of original pyramid be h 27 40 48 48( 27) 40 48 1296 40 8 1296 1296 162 cm (Shown)8 h h hh hh h h − = −= −= = == 5(b)(ii) Volume 11(30 48)(135 27) (40 25)(135)33=  + −  332760 cm= 6(a) 48 1.5 ( 2) 1.5( 2) 12 6 x x x −− =−−− − + =− = OR Equation of line AB: y = – 1.5x + c 8 = – 1.5 (– 2) + c → c = 5 Hence, equation of line AB: y = – 1.5x + 5 – 4 = – 1.5x + 5 x = 6 6(b) Length of AB = √(−2 − 6)2 + (8 − (−4)) 2 = √208 = 14.4 units (3sf) 6(c) 6(−𝑏) + 9(4𝑏) = 5 → 𝑏 = 1 6 6(d) Line CD: 6𝑦 + 9𝑥 = 5 → 𝑦 = −1.5𝑥 + 5 6 Gradient of line CD = – 1.5 Since the gradients of line CD and line AB are equal, line CD is parallel to line AB. Thus, line CD does not intersect line AB. Reason: Gradients equal OR lines parallel

4 7(a) Trapezium 7(b)(i) ZX⃗⃗⃗⃗⃗ = ZW⃗⃗⃗⃗⃗⃗ + WX⃗⃗⃗⃗⃗⃗ = –b + a OR ZX⃗⃗⃗⃗⃗ = WX⃗⃗⃗⃗⃗⃗ – WZ⃗⃗⃗⃗⃗⃗ = a – b 7(b)(ii) WY⃗⃗⃗⃗⃗⃗ = WX⃗⃗⃗⃗⃗⃗ + XY⃗⃗⃗⃗⃗ = a + 3 2 b 7(b)(iii) 5TX⃗⃗⃗⃗⃗ = 3ZX⃗⃗⃗⃗⃗ TX⃗⃗⃗⃗⃗ = 3 5 ZX⃗⃗⃗⃗⃗ ZT⃗⃗⃗⃗ = 2 5 ZX⃗⃗⃗⃗⃗ WT⃗⃗⃗⃗⃗⃗ = WZ⃗⃗⃗⃗⃗⃗ + ZT⃗⃗⃗⃗ = WZ⃗⃗⃗⃗⃗⃗ + 2 5 ZX⃗⃗⃗⃗⃗ = b + 2 5 (a – b) WT⃗⃗⃗⃗⃗⃗ = 2 5 a + 3 5 b OR 1 5 (2a +3b) OR 2 5 (a + 3 2 b) OR WT⃗⃗⃗⃗⃗⃗ = WX⃗⃗⃗⃗⃗⃗ + XT⃗⃗⃗⃗⃗ WT⃗⃗⃗⃗⃗⃗ = WX⃗⃗⃗⃗⃗⃗ + 3 5 -( ZX⃗⃗⃗⃗⃗ ) WT⃗⃗⃗⃗⃗⃗ = a + 3 5 (b – a) WT⃗⃗⃗⃗⃗⃗ = 2 5 a + 3 5 b OR 1 5 (2a +3b) OR 2 5 (a + 3 2 b) 7(c) WY⃗⃗⃗⃗⃗⃗ = a + 3 2 b = 1 2 (2a +3b) WT⃗⃗⃗⃗⃗⃗ = 2 5 a + 3 5 b = 1 5 (2a +3b) ( ) ( ) 1 25 1 52 WT WY == WT⃗⃗⃗⃗⃗⃗ = 2 5 WY⃗⃗⃗⃗⃗⃗ WT⃗⃗⃗⃗⃗⃗ and WY⃗⃗⃗⃗⃗⃗ are parallel.W is a common point. Hence W, T and Y lie on a straight line. WT⃗⃗⃗⃗⃗⃗ = 2 5 WY⃗⃗⃗⃗⃗⃗ WT⃗⃗⃗⃗⃗⃗ and WY⃗⃗⃗⃗⃗⃗ are parallel AND W is a common point. 7(d)(i) Triangle WZT and Triangle YXT are similar 2 Area of triangle 2 4 Area of triangle 3 9 WZT YXT ==

5 7(d)(ii) 1 Area of triangle 2 2 1Area of triangle 3 2 ZT hYZT YXT XT h  ==  8(a)(i) angle SPR = 18° 1 1 ˆsin sin18 90 28 90sin18sin 28 90sin18or 180 sin 28 83.348 (3 ) or 96.652 (3 ) PSR PSR PSR PSR dp PSR dp − − = =   = −   =   =  (reject 83.348 (3 )PSR dp =  as PSR is obtuse) Hence, obtuse 96.652 (3 )PSR dp =  obtuse 96.7 (1 )PSR dp =  angle SPR = 18° seen or implied correct use of sine rule 8(a)(ii) bearing of P from R = 180° + 18° = 198° 8(b) ( )( )( ) ( )( ) 2 2 2 2 2 2 ˆ90 72 45 2 72 45 cos 72 45 90ˆcos 2 72 45 97.9032 97.903 (3 ) PQR PQR PQR dp = + − +−=  = =  Find value to at least 4 decimal places then show to given answer 8(c)(i) area of triangle PQR 21 72 45 sin 97.903 1604.613 1600m (3sf)2=    = = 8(c)(ii) Let shortest distance be BN. 1 90 1604.6132 35.658m 35.7m(3 sf) BN BN   = == 8(d) heighttan 8 35.658 height 35.658 tan 8 5.01m (3sf) = = =

6 9(a) h = 0.66 (2dp) c.o. 9(b) Deduct 1 mark for every incorrect and/or missing point. Allow ecf for answer in (a). Smooth curve passing through all plotted points. 9(c) ( ) ( ) ( ) 32 2 2 2 6 8 12 6 8 12 6 8 34 8 6 34 x x x x x x x xx x xx − + = − + = − + = − + = Hence, draw line 3y= . See graph in (b). x = 4.9 Manipulate given equation to x 4 8 - 6x + x2 ( ) = -3 Accept intersection of line y= 3 with their graph 9(d)(i) 1 2y x c=− + ( )113 2 c=− + → 5 2c= 15 22yx=− + Line of negative gradient. Correct line y = - 1 2 x + 5 2 drawn for -1£ x £ 5 : Passes through (–1, 3), (3, 1) and (5, 0) 9(d)(ii) x = 4.15 Read up to 0.05

7 9(d)(iii) ( ) ( ) 2 2 23 32 15864 2 2 8 6 2 10 8 6 2 10 6 10 10 0 x x x x x x x x x x x x x x x − + =− + − + =− + − + =− + − + − = 10(a)(i) 50 80 40 eggs100= → Median = 53g 10(a) (ii) 30 80 24 eggs100= → 30th percentile = 48g 10(a) (iii) 25 80 20 eggs100= → Lower quartile = 46g 75 80 60 eggs100= → Upper quartile = 58g Interquartile range = 58 – 46 = 12g 10(b) Number of eggs less than or equal 60 g = 69 Number of eggs more than 60g = 80 – 69 = 11 (accept 70) (accept 10) 10 (c)(i) Frequency for mass, x £ 45 = 18 Frequency for mass, x £ 35 = 6 m =18- 6 =12 (Shown) OR 6 + m+ 30+ 32 = 80 m = 80- 6 - 30-32 =12 (Shown) 10 (c)(ii) P(both eggs more than 45g 62 61 1891 0.5984(4 )80 79 3160 sf=  = = 10 (c)(iii) Mid value (x) 30 40 50 60 Mass (x g) 25 < x £ 35 35< x £ 45 45 < x £ 55 Frequency 6 12 30 32 Mean of masses of eggs at Farm A = 51g Standard deviation of masses of eggs at Farm A = 9.16515139=9.17g (3sf) 10(d) 1. The eggs at Farm B are heavier because the mean of the massess of eggs at Farm B, 53g, is more than that of Farm A, 51 g. 2. The masses of the eggs at Farm B are more consistent because the standard deviation of the massess of eggs at Farm B, 9g, is less than that of Farm A, 9.17g.

8 11(a) 420 14560  = 1015 calories 11(b) 2000 calorie diet need 2