solutions_2023 Beatty Sec 4NA_Math_P2

1 2023 4N(A) Mathematics Prelim Exam Paper 2 Mark Scheme S/n Solution Mark Allocation 1a 2 3540 2 3 5   B1 1b Greatest Integer 2 22 3 36    B1 (B0 if left in indices form) 2 12 mths ------ 3.2% 6 mths ------ 3.2 612 1.6%   Thus r = 1.6%, and n = 3x2 = 6. 6 1 100 1.610000 10000 1 100 10000 10999.229 $999.229 $999 (3 ) n rA P I I I I sf                 B1 (either r or n correct) M1 (correct formula for compound interest) A1 3a STR PQR  48  180 60 48RST     72  B1 3b 7.3 8 10 10 7.3 8 10 7.38 PQ QR PR ST TR SR ST ST ST       9.125 cm M1 (correct values in ratio) A1 (exact) (A0 if in 3sf) 4 V olume of 4 balls 3 3 44 3 16 3 r r       M1 (either formula of sphere or cylinder correct with values substituted) V olume of cylinder   2 2 3 8 8 r h r r r       Accept alternative working 10 sin 60 sin 72 10sin 60 sin 72 9.1059 9.11cm (3sf) ST ST ST ST      

2 Percentage of empty space 3 3 3 3 3 168 3 100%8 8 3 100%8 133 %3 r r r r r            Thus is it 1/3 of the volume of the tube. M1 (percentage formed) A1 (A0 if no conclusion of 1/3) 5a        2 2 4 2 4 4 4 2 7 2 2 4 72 4 3.121 1.1213 3.12 1.12 (3 ) b b acx a x x x or x x or x sf             M1 (correct substitution in formula) A1, A1 (minus 1m if one or both answers are not in 3sf) 5b From eqn 2: 2 3x y  ---2a Sub eqn 2a into eqn 1:  4 2 3 6 8 12 6 9 18 2 y y y y y y         Sub y = 2 into eqn 2a x = 1 Thus x = 1, y = 2. M1 (substitution into one eqn) M1 (manipulation to solve for one variable) A1 (A0 if 1 value is incorrect) 6a Angle in a semi-circle property, since CE is the diameter of the circle. B1 6b 180 68ABC    112  B1 (cao) 6c 180 90 68CAD     22  Since OAB is isosceles, 30 22OBA BAO     52  M1 (finding CAD or other suitable angle leading to OBA ) A1 (cao) 7a In ABD , 8sin 12ADB  1 8sin 12ADB        0.72972ADB  = 0.730 radians (3sf) M1 (forming sine ratio) A1

3 7b In ABC , 8tan1 BC 8 tan1BC  5.13674 5.14 (3 ) BC BC cm sf   M1 (forming sine ratio) A1 7c In ABD , 2 2 212 8 BD  2 2 2 12 8 80 8.94427 BD BD BD     (reject negative value) Area ABC  1 82 AD   2 1 80 5.13674 82 15.2301 15.2 (3 )cm sf     B1 (BD value correct) M1 (correct substitution of values, no penalty if AD does not have extra accuracy) A1 (cao) 8a     2 3 2a b b a b     2 2 2 2 9 6 2 9 4 9 4 a ab b ab b a ab or a a b         M1, M1 (correct expansion for each term) A1 8bi 412 3 x        2 22 2 2 2 3 4 3 2 3 2 2 x x x x 