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