2023 Greendale 4NA Prelim P2 MS

Suggested Marking Scheme 4NA Prelim P2 2023 1 Int. ∠ of B = (6 2) 180 6 −× ° = 120° Int. ∠ of A = 360 120 90°− °− ° = 150° Ext. ∠ of A = 180 150°− ° = 30° 360 30 12 n °= ° = 2 (a)(i) (a)(ii) (b) 190 – 156 = 34 cm Median 171 172 2 171.5cm += = Since the range is increased by 1, this means John is either 155 or 191 but since the median is decreased, this means that John’s height is lower than the median thus he is 155 cm. 3 Area of water in contact with the bowl ( )( ) ( )( ) 22 2 112 15 1522 1060.286625 1060 cm ππ= + = = 4 (a) 22( 6) 4( 7) 15.2315... 15.2 − +− = = (b) 22, 3.142, , 3.2 7 π−−  (ci) 410994 1.0994 10= × (cii) 3 10994 3750 7244 7.244 10 m − = = ×

5 (ai) 1: 200000 1 :2cm km Length of cycling path 3.5 2 1.75cm = ÷ = (aii) 22 1 :2 1 :4 cm km cm km Actual Area 2 84 32km = × = (b) 80 200000 160000100 1:160000 ×= 6 (a) 1 2 1 12 m yx = = − (b) Draw horizontal line y = 3 (c)(i) (8, 3) (c)(ii) 2210 5 125 11.2 units AX = + = = (d)(i) (0, −6) (d)(ii) Area = 12 (10)(4)2× = 40 units2 7 (a) 3 minutes (b) Average speed 3600m 30min (3600 1000)km (30 60)h 7.2km/h = ÷= ÷ =

(c) (d) Distance away from park = 3600 − 1760 = 1840 m 8 (a) (b) (c) (d) (e) 0.8, 3.5 0.75 (+/-0.05) - 2 (+/-0.2) x cannot be 0/ or is undefined.

9 (a) (b) (c) 2 2 2 2(0.5) 0.238 ( 1.1) 1.03 ah mn h h = + = − +− = 2 2 2 2 2 2 2 2 2 ah mn hm hn a hn a hm a hmn h a hmn h = + += = − −= −=± 2 (2 1) 5 2 50 1 1 4(2)( 5) 2(2) 1.85 or 1.35 xx xx x += +−= −± − −= =− 10 (ai) Total volume of one Bloobox 3 30 30 30 27000cm =×× = (aii) Total SA of Bloobox ( ) ( ) 2 30 30 5 30 25 5250cm = ××+ × = (b) Maximum volume of Type A 3 120 50 120 720 000cm = ×× = Maximum volume of Type B 3 95 120 120 200100 2 736 000cm =××× = Maximum volume of recyclables on the same day 3 27 000 6 16 2 592 000cm = ×× = Maximum volume of 2 Type A 3 120 50 120 2 1 440 000cm = ×× × =

John should request for 1 Type B Bloobin as it is enough to collect all the recyclables from the households in his block. 11 (ai) (aii) (aiii) (aiv) Median = $320 Online bookshop A because it has a greater median than online bookshop B. Lower quartile = 120 Upper quartile = 520 Inter-quartile range = 520 − 120 = $400 For online bookshop B, lower quartile = 120 upper quartile = 440 Inter-quartile range = 440 − 120 = $320 Online bookshop B has a more consistent amount of sales because it has a lower interquartile range. (bi) (bii) P (at least 1 green) = 1 – P (no green) = 1 – 65 10 9× = 3 2 Yellow Green Yellow Green Yellow Green