PHSS 2022 EYE 3NA Paper 2 QP

3NA EYE 2022 Marking Scheme & Marker’s report Qns Suggested Workings Marks Remarks 1 3.14 • ,  , 3.14 , 3.142− , 22 7− B2 – all 4 B1 – 3 correct 2a 32792 2 3 11=   B1 2b 2 11 22 p= = B1 3a 125 3 6.45 x= = B1 3b 2 2 yx z yx z yz x = = = M1 A1 4 5 3.6interest 25000 1 25000100 $4835.88 or $4840 = + −  = M1, M1 A1 M1 – correct value in formula M1- subtract 25000 5a cos152 cos 180 152 28 z z =−  = − = B1 5b ( ) ( ) ( ) ( ) ( ) ( ) 11 1 12.8 8.2 sin 36.7362 2 36.736sin 12.8 8.2 2 36.736 2 36.736sin or 180 sin 12.8 8.2 12.8 8.2 44.427 or 180 44.427 44.4 or 135.6 XYZ XYZ XYZ −−     = =     = −         =− =   M1 A1, A1 Need to put units for angle, else deduct 1 mark 6a   let total no. of balls be 13 9 2 13 22 2 13 13( 22) 2 11 286 26 shown x x x x x xx x x −− = − = −= = = M1 A1 Need to show 11 286x=

or 11 units ---------- blue + red 13 91 units ---------- 11 13 units ---------- 2 13 26 +  = M1 A1 Need to show 11 units (blue+red) 6b ( ) 26 9not red 26 17 26 P −= = B1 7a 9 11 123 10 1.23 10  = B1 7b 8 11 3 10 1.23 10 0.0024390 m =2.44 mm  =  = M1 A1 8a 3 2 9 29 3 2 33 2gradient 3 yx xy yx −= += =+ = M1 A1 Do not accept 3 s.f 8b ( )0,3 B1 8c ( ) Evidence: : subs. 3 and 1 into 3 2 3( 1) 2(3) 9 concept: Conclusion: 2, 1 does not lies on the line . LHS x y y x LHS RHS l = =− − − − =−  − M1 A1 Need to explain LHS not equal to RHS 9ai, 9aii ai: B1 aii: B1 – angle bisector B1 – point P Need to show the construction lines No mark if never mark out P

9b ( )7.8 0.1 cm ecf. B1 10 interior angle of polygon 90 360180 5 108 360 90 108 162 ext. angle of polygon 180 162 18 360no. of sides in polygon 18 A B C C C = =− = = − − = =− = = 20= M1 M1 M1 A1 Find both A and B interior Find C interior Either using ext. or int. formula Use int. formula needs to show algebra manipulation 11 ( ) 1 Evidence: 115 1.65 m tan 115 1.65tan 125 42.2 Concept: 40 42.2 Conclusion: Since 40 42 , Xing Qin claims is not correc t. She is unable to see the top of the flag unless FD FDFED GX FED − =− = −=  =     she tilted her head to at least 42.2 . M1 M1 M1 A1 Need to explain using the angle. OR Evidence: tan 40 125 125 tan 40 height of flag from ground 125 tan 40 1.65 106.5 m Concept: 106.5 m 115 m Conclusion: Since 106.5 m 115 m, Xing Qin claims is no t correct. She could only see up to 106.5 m f FD FD = = =  + =   rom the ground if she tilted her head to 40 . M1 M1 M1 A1 Need to explain using the height of flag from ground.

OR ( ) Evidence: 115 1.65 m 113.35 m tiling @ 40 tan 40 125 125 tan 40 104.88 Concept: 104.88 m 113.35 m Conclusion: Since 104.88 m 113.35 m, Xing Qin claims i s not correct. The height of FD found by tiling a FD FD FD =− =  = = =   t 40 is not same as 113.35. M1 M1 M1 A1 Need to explain using citing the FD found are not same 12a 2 2 2 22 8 5 2(8)(5) cos 60 8 5 2(8)(5) cos 60 7 m AC AC = + −  = + −  = M1 A1 /B2 12b 1 sin sin 110 47 sin 110sin 4 7 32.5 ACD ACD −  =  =   = M1 A1 Need to put units for angle, else deduct 1 mark 13a 23 47difference 4 6 47 6 41 41 6 35 35 6 29 a b c −= =− =− = =− = =− = M1 for finding difference A1 13b 53 6 n− or 47 6( 1)n−− B1

13c Evidence: 53 6 123 29.3 Concept: 29.3 positive integer Conclusion: Since is not a positive integer, 123 is not in the sequence. n n n − =− =  − M1 A1 14a B2 – all 4 sides -1 for 1 side wrong 14b ( ) ( ) 52 42 5 2 2 4 3 8 10 6 10 cm nn nn n n PQ =+− − = + =+ = = M1 M1 A1 15a 2 5 : 2 : 3 AC AB AC CB = B1 15b 2 units ---------- 25 cm 251 unit ---------- 2 255 unit ---------- 52 62.5 cmAB  = 15bi 4 :100000 1: 25000 B1 15bii 2 2 2 2 2 2 4 cm : 1 km 4 cm : 1 km 11 cm -------------- 16 1250 cm ------------ 25016 15.625 km  = M1 A1 A

16a 800 700 100%800 % 12.5 x x −= = B1 Need to show % in working 16b 2red of 25%5 10% green 15% 15% 45 451% 15 45100% 100 15 total marbles 300 = = = −−−−−−− −−−−−−−− −−−−−−  = M1 M1 A1 17ai 31 XBZ BAC = = B1 No mark if never put degree 17aii 5.7 3.2 2.5 cm BX AB BC XZ XC BX BC = = =− =− = M1 A1 Either identify BX or BC 17b ( ) Evidence: (congruent property) Concept: converse of alt. angle Conclusion: Since , by converse of alternate angle. is parallel to . BXZ ABC BXZ ABC BXZ ABC XZ AB  =  =  = B1 B1 State congruent State alternate angle OR ( ) Evidence: (congruent triangle) 31 81 68 =180 Concept: 180 converse of sum of interior angles of parallel lines Conclusion: Since 180 , by convers XBZ BAC XBZ BXZ XZB XBZ BXZ XZB XBZ BXZ XZB  =  + + = + +    + + =   + + =  e sum of interior angles of parallel lines, is parallel to .XZ AB B1 B1 State congruent State sum of interior angles of parallel lines

18a 4 23 4 1 4 ( 1) 5 3 3 3 3 3 3 − − −− = = = B1 18b 2 3 6 4 8 4 a a   = M1 A1 18c ( ) 1 31 5 8 32 22 3( 1) 5 3 3 5 2 3 m m m m m + + = = += += = M1 M1 A1 For convert to common base 19a ( ) ( ) 2 2 2 2 6 20 3 20 3 3 29 3 29 xx x x a b +− = + − − = + − = =− M1 A1 19b ( ) ( ) 2 2 3 29 0 3 29 3 29 2.39 or 8.39 x x x x + − = += + = =− M1 A1 19b ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 5 10 3 6 5 2 3 2 2 5 3 dimension = 2 cm by 5 3 cm or dimension = 5 3 cm by 2 cm ax a x a x x xa xa ax − − + = − − − = − − −− −− M1 M1 √A1 20a ( ) ( )2 2 1 3 3 2 4 2 9 6 58 yy yy y + − − = + − + =− + M1 A1

20bi ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 23 22 322 22 2 3 2 2 83 2 xx x xx x x x x − −− −=− −− −−= − −= − M1 A1 20bii ( ) ( ) ( ) ( ) 2 2 2 2 2 83 1 2 2 8 3 4 4 8 3 40 1 1 4(1)( 4) 2(1) 1 17 2 2.56 or 1.56 x x xx x x x xx x − = − − = − − + = − − − = − −  − − −= = =− √M1 √M1 A1 Correct expansion Need to show 1 17 2  21a 826 2 0 866 6 1.3 p q = + − = = + − = B1 B1

21b B1 for all correct points B1 for smooth curve 21ci 0.2 to 0.4 y=− − or √B1 21cii 2 or 4y= B1 21d ( ) ( ) ( ) 0, 1.8 and 11,3 3 1.8gradient 11 0.436 0.3 0.7m − += =   M1 A1 Need to show correct tangent line drawn 22a 16.8 5.6diameter or 62 2.8 cm = = B1 22b ( ) 3 3 vol. of empty space 4 2.816.8 5.6 3 12 32 144.31 144 cm   =   −   = = M1 A1

END OF PAPER 22c ( ) 10 12No. of repack boxes 3 40 Amt. spent on buying the chocolate and boxes 10 12.5 0.95 40 $163 100 = = =  +  = % $163 1631% 100 163125% 125 100 Amt. for 25% profit $203.75 $203.75Selling price of each box 40 5 .093 She needs to sell each box at at least −−−−−−−−− −−−−−−−−−− −−−−−−−−−  = = = $6. M1 M1 M1 A1 no. of repack boxes