Greendale Sec 4048 02 Answer Scheme

1 2022 4E5A Preliminary Examination Paper 2 Mark Scheme 1a 3 12 1 34 12 4 6 3 12 6 3 4 116 xx xx xx x −+< −< + − <+ < M1 A1 1b ( ) ( ) ( ) ( ) ( ) 2 2 2 2 72 4343 7 24 3 43 7 86 43 13 8 43 x xx xx x xx x x x − −− −−= − −+= − −= − M1 A1 1c 1 10 2 6 1 63 2 10 5 49 49 7 m n nn mm −   = =  M1 A1 1d ( ) ( )( ) ( )( ) ( )( ) ( ) 2 2 2 50 32 5 11 12 2 25 16 54 3 25 4 5 4 54 3 25 4 3 x xx x xx xx xx x x − −− − = +− +−= +− −= − M1 M1 A1 1e 4 5 16xy−= ------ (1) 6 3 10xy+= ------ (2) (1) 3: 12 15 48 (3) (2) 5: 30 15 50 (4) (3) (4): 42 98 123 12 in(2),3 16 2 3 103 14 3 10 113 xy xy x x Substitute x y y y × −= → × +=→ += = =  += += =− M1 A1 A1

2 2a 8 15 10 15 6 8 1[ 20 30 20 35 13 14 ]3 6 5 12 6 9 5 24 38 1 53 793 27 16 2128 3 2117 2633 91 5 3 A A A        = ++           =     =   M1 A1 2b The number of bus trips per day made by each type of bus of the 2 companies. B1 2ci 3BA exists because B is a 1x3 matrix and A is a 3x1 matrix B1 2cii The elements in this 1x2 matrix represents the total number of bus trips made by Company A and B respectively. B1 2di ( ) 8 15 112 25 50 20 30 165     B1 for ( )12 25 50 B1 for 1 1   2dii [(8 12 ) (20 25 ) (6 50 )] $22 896 22 $19712 trips passengers trips passengers trips passengers ×+ ×+ ×× = × = M1 A1

3 3a 9.3 B1 3b P1-plot all points P1-join up points P1-smooth curve 3c 83 15 83 15 0 83 10 5 x x x x x x += +−= +− = Draw y = 5, From the graph, x = 0.6 or x = 4.4 B1, B1 3di Line 21yx= + drawn. x 0 3 6 y 0.5 2 3.5 P1 P1 3dii From the graph, x = 1 or x = 3.15 B1, B1 3diii 22 2 82 3 10 1 166 20 1 6 16 20 5 21 16 0 21, 16 xx x xx x x xx x xx BC  +− =+ +−= + +− =+ − += ∴= − = M1 M1 A1

4 4ai Bearing of from 360 (180 126) 306 AD =−− = ° B1 4aii 270 242 28 480 sin28 sin36 383.3821 383.4 (1 d.p) ABD AD AD m ∠= − = ° =°° = = M1 M1 A1 (b) Let be the angle of depression 4800tan 630 82.5226 82.5 x x x = = = ° M1 A1 (c) 180 36 36 108 383.3821 sin36 sin108 236.94317 237 (3 s.f) APD PD PD m ∠ = −− = ° =°° = = M1 M1 A1

5 5a 22 2 3 Let be the height of the cone that was r emoved. By similar triangles, 1 65 56 46 1.5 Volume of paint bottle 11= (5 )(20) ( 5 7.5) ( 1 1.5)33 1765.5755 1765.6 (shown) x x x xx x x cm ππ π =+ = + = = + ×× − ×× = = M1 A1 M1 (volume of cylinder), ecf M1 (volume of frustrum), ecf A1 (answer shown correctly) (b) 1765.6 1500 1001765.6 15.043 15.0% (3 s.f) − × = = B1 (c) Total mass 30 1500(3) 4530 4.53 g kg = + = = M1 A1 (d) 3 2 3 2 3 2 2 3 1 3 1 3 1 Base area of larger bottle 3= (5)1 163.3694... 163 (3 s.f) L S L S L S V V l l A A cm π = = =  × = = M1 A1

6 6ai 2 2 3tan 4 36.8699 180 36.8699 143.1301 Area of shaded region 143.1301 1= (8) (4)(8)sin143.1301360 2 70.3389 70.3 DOB DOB AOD cm π ∠= ∠= ∠= − = ×− = = [Accept finding area of smaller sector and using area of semicircle to subtract the area of sector and triangle] M1 M1 A1 (a)(ii) 222 143.1301 2 (8)360 19.98473 4 8 2(4)(8)cos143.1301 11.45426 Perimeter = 4 11.45426 19.98473 35.438987 35.4 AD ED ED cm π= × = =+− = ++ = = M1 M1 A1 (b)(i) 51 (alternate ) 51 2 ( at centre = 2 times at circumferenc e) 102 PRS s POS ∠= ∠ ∠ =×∠ ∠ = ° B1 B1 (b)(ii) ( s in the same segment) (alt. s) Since = , is an isosceles triangle QPR QSR PQS QSR QPR PQS PQT ∠= ∠ ∠ ∠= ∠ ∠ ∠ ∠∆ M1 A1 (b)(iii) in semicircle is 90 Opposite angles in quadrialteral are su pplementary OR OPUS ∠° Hence points O, P, U, S can lie on the circumference of a circle OR Yes it can. M1 (either one o.e) A1

7 7a Median position 91 52 += = Median for Group A = 51 minutes B1 7b IQR (Grp B) 61 62 31 42 22 61.5 36.5 25 min ++= − = − = M1 (for either correct Q1 or Q3) A1 7c Standard deviation 22 2 28324 476 99 18.70498... 18.7 (3 ) fx fx ff sf  = −  = −   = = ∑∑ ∑∑ B1 7d There is an outlier of 96 mins in Grp B, thus the interquartile range is more appropriate because it is less sensitive to outliers. B1

8 8i 68 22 2 4 PQ −− = −  −  = uuu r B1 (ii) 2 4 12 2 11 8 72 3 9 13 3 is not parallel to . does not lie on t he line . PQ PA PQ PA A PQ = =  −−  = −   −   −= −=  ∴ uuu r u u ur M1 M1 (or vector AQ) A1 (iii) 22 24 40 6 4 64 7.21 units PR PQ QR PR = + = + = = + = uuu r uuu r uuu r [Accept finding coordinates of R first] M1 A1 (iv) 46 02 2 2 ( 2, 2) 4 2 2 2 2( 2) 6 26 PQ RS OR R mm y mx c c c yx − = +   −= = − = = = = + =−+ = = + uuu r M1 (for finding R) M1 (either m or c correct) A1

9 9a) Cost 32 30 35 20 $1660=×+×= Revenue 40 (30 20) $2000 = ×+ = The shopkeeper made a gain. M1 (oe) A1 9bi 800 x B1 9bii 800( 2)( 2) xx −+ B1 (ecf) 9biii 2 2 800( 2)( 2) 800 99 1600800 2 4 800 99 1600 2 103 0 2 103 1600 0 2 103 1600 0 ( ) xx xx xx xx x x shown − +− = + − −− = −− = −− + = +−= M1 (ecf) M1 A1 9biv 2 2 2 103 1600 0 103 103 4(2)( 1600) 2(2) 12.5 64 xx x x or +−= −± − −= = − M1 (oe) A1/A1 9bv 800 2 6212.5−= B1

10 10ai (1107.8 1066.3 1123.6 1259 1249.5 1281.6) 6 7087.8 1181.36 kWh + + ++ + = = M1 A1 10aii) 1181.3 $0.2139 $252.68007 $252.68007 1.07 $270.3676749 $270.37 (3 ) Amount without GST Amount with GST sf = × = = × = = M1 (ecf) A1 10b) Total number solar panels 9 1.65 5.4545... 5 ÷= ≈ 414÷= 5 4 20×= 919÷= 4 1.65 2.4242... 2 ÷= ≈ 9 2 18×= The max number of solar panels is 20. B1 10c) Amount of electricity saved per month 19 20 380 kWh solar panels kWh = × = Average amount of electricity used after installation 1181.3 380 801.3kWh = − = Average cost of electricity per month (after installation) 801.3 0.2139 1.07 $183.3959349 $183.40 (2 )dp = ×× = = Average cost of solar panels per month 2 $6250 20 12 $52.083333 $52.08 (2 )dp ×= × = = Total average cost of electricity after installation of solar panels Ecf for earlier parts B1 B1 B1