2023 Greendale 4NA Prelim P1 MS

Suggested Marking Scheme 4NA Prelim P1 2023 1 ( ) ( )3 215 36 55 81 1 8 xx xx x x += −− += − + =− =− 2 12units 1200cm 3units 300cm → → 3 1 10 11 11 y y y − <− − <− > Smallest prime = 13 4 Red box Blue box 1 A 1 B 1 C 1 D 2 A 2 B 2 C 2 D 3 A 3 B 3 C 3 D

5 6 5 seconds 15 characters 60 seconds 180 characters → → 7 ( )( ) 2 15 45 4, 5 yx x yx x bc = +− =−− = −= − 8 (a) ( ) ( ) 2 2 2 2 8 10 800 800 800 4 50 kh r k k h r h h = = = = = = (b) 9 (a) 4a= (b) 35 125= x y

10 (a) (b) ( ) 2 2 49 2 94 2, 13 xx x hk +− = + −− = =− -13 11 (a) 1 2 3 5 3 5 3 n n xxx xx n − − ×= = =− (b) 8 2 6 3 y y y y = = 12 48 (base angle of isosc. ) 48 (congruent) 48 48 (alt. s) 96 AED DEF x x ∠= ° ∆ ∠= ° °= °+ ° ∠ = 13 (a) (b) 6n-23 6 23 50 6 73 12.2 13 n n n k −> > > = 14 1 1 2.6tan 3 2.6tan 3 (alt. s) 2.6Obtuse 180 tan 3 139.1 DCA DCA CAB DCA EAB − − ∠= ∠=  ∠= ∠ ∠ ∠ = °−   = °

15 2215 9 12 22.5 5 15 7.5 22.5 12 15 18 AF DE DE AG AG = − = = = = = Area of triangle ADE = 1 (7.5)(18)2 = 67.5 cm2 16 (a) (b) ( ) 2 22 3 4 12 9x xx−=−+ ( ) ( ) 2 2 21 23 2 32 23 3 23 3(2 3) 23 69 23 6, 9 x x xx x x x x x x ab − − −−= − −= − −−= − −+= − = −= 17 22 25 47 7 13 31 7 2 5 Length 20 Breadth 6 20 6 20.9 x yx y xy xy yy y x BD += ++ += += −+ = = = = = = += 18 (a) Time (h hours) Mid- value x Frequency f fx

(b) 02 h<≤ 1 0 0 24 h<≤ 3 1 3 46 h<≤ 5 4 20 68 h<≤ 7 8 56 8 10h<≤ 9 9 81 10 12h<≤ 11 2 22 12 14h<≤ 13 1 13 Total 25 195 Estimated mean 195 25= = 7.8 The time is given in a range and we are not able to determine the exact amount of time spent studying for each student. 19 i ii 110 2 (80)360 153.5889... L L π= × =   154L cm= Area of big sector = 2110 (80)360L π= ×   Area of small sector = 2110 (45)360L π= ×   Area of display = 22110 110 (80) (45)360 360ππ× −×   = 4199.698513 = 4200 cm2

20 2 ( 5) 5 ( 5) 4 16 4 25 1 42 2(2 5) 1( 4) 4 10 4 36 2 AB ACmm m m m m mm mm m m = −− −−=−− −− + =−−− + = −−− +=+ =− =− 21 (a) (b) 2 π(10)(15) π(3)(4.5) 429 cm − = 2314ππ33 4 rh r hr = = 22 (a) (b) (c) Construct triangle: ∠CAB = 98° BC = 9.8 cm (with construction arc) Construct angle bisector for ∠CAB Construct arc 2 cm from A Shade correct region 23 (a) Total Cost ( ) ( )2 $34.20 2 $20.70 $109.80 = + = (b) Total cost (before Wildpass discount) ( ) ( )2 $43.20 $20.00 $106.40 = + = Total cost he pays 80 $106.40100 $85.12 = × = (c) Percentage increase 43.20 34.20 10034.20 26.3% −= × =