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             M1 (factorise common term) A1 8bii 4 10 6 15y yz z          2 2 5 3 2 5 2 5 2 3 y z y y z        M1 (factorise common term for each term) A1

4 9a y = 1.2 B1 (B0 if not left in 1dp) 9b B1 (at least 6 plots correct) B1 (smooth curve) B1 (shape – composite graph) 9c Minimum Point =  3.5, 1.2 B1 (accept 3.5 4 x  , 1.4 1.1y    , must be correct to 0.1 decimal only) 9d Gradient 3.6 0 9 6.1 1.24 (3 )sf    M1 (draw tangent at x = 1.5) A1 (accept 1.1 1.5m  ) 9e 2 2 12 79 12 6 7 69 1 x x x x y        From the graph, 1.8x or 6.9x M1 (obtain 1y graphically or algebraically) A1 (any 1 answer correct) (accept 1.6 1.9x  and 6.7 6.9x  , must be correct to 0.1 decimal only) 10a Choosing age less than 30yrs old, walking speed 4.82 1000 1 60 60 1.3388 1.34 / (3 )m s sf      B1 (choose 4.82 km/h) B1 (must be in m/s) 10b By measuring the arc length AB, from 4T1 to 5N1, I estimated it to be about 16.5 cm. 1cm ----- 20m 16.5cm ----- 16.5 20 330 m   M1 (accept 14 18arc AB  ) M1 (proportion is calculated correctly) 10c Time to walk from classroom to field 400 1.3388 298.775 4 min 59sec    M1 (distance / 10a speed)

5 Time to form up BT70 logo 330 1.3388 246.489 4 min 6sec    Assume time to walk back from field to classroom is the same as time to walk from classroom to field = 4min 59sec Total time taken after flag-raising to end of activity = 2(4min 59sec) + 4min 6sec = 14min 4sec Time permitted for photo-taking factoring time wastage = 55min – 14min 4sec = 40min 56sec It is reasonable to assume that photo taking factoring time wastage will not exceeed 40min 56sec. Thus, we conclude that the students should be able to return by 8.30am. M1 (10b distance / 10a speed) M1 (reasonable attempt to calculate total time taken) A1 (reasonable explanation and conclusion) 11a i Mathematics test Lowest mark 40 Highest mark 89 Range 49 Lower Quartile 58 Upper Quartile 81 Interquartile Range 23 Median mark 71 B1 (range cao) B1 (UQ – LQ)√ B1 (Median cao) 11a ii No, I do not agree, because the spread of the lowest quartile (bottom 25%) is much wider than the spread of highest quartile (top 25%). B1 (any reasonable explanation) 11a iii Students score higher marks in the Mathematics test than English test, because the Medianmath > Medianenglish. or The spread of marks is wider for the Mathematics test than English test, because IQRmath > IQRenglish and Rangemath > Rangeenglish B1 (any 1 comparison) √ (no need to state Range) 11b i Probability of both blue balls 4 3 7 6  2 7 or 0.286 (3sf) B1 11b ii Probability of both balls of different colours = 1 – Probability of both blue balls – Probability of both red balls 2 11 7 7   M1 (or equivalent) A1

6 4 7 or 0.571 (3sf) 12a 144 68BAC    76  Using cosine rule,    2 2370 510 2 370 510 cos 76 552.900 553 (3 ) BC BC BC m sf       B1 M1 (cosine rule applied correctly) A1 12b Using sine rule, 1 sin sin 76 510 552.900 510sin 76sin 552.900 510sin 76sin 552.900 63.509 ABC ABC ABC ABC                Thus, Bearing 360 63.509 112    = 184.5° (1d.p) M1 (sine rule applied correctly) A1 B1√ 12c In ACD , sin 76 510 CD 510sin 76 494.850 495 (3 ) CD CD CD m sf     Thus the shortest distance is 495m M1 A1 (need to state shortest distance)