NBSS EM 2024 Prelim P2 MS

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 1 of 18 Qns Solution Marking Scheme Remarks 1(a) 23 654 4 5(2 3) 620 4 10 15 120 14 120 15 14 105 7.5 xx xx xx x x x −+ =− +− =− + − =− =− + =− =− M1 A1 1(b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 3 3 10 23 10 3 23 10 3 1 1 1 or or 5 3 5 3 15 5 p q q p rr pq r r qp pq r r pq p q q p q p −−  −= − −= − −−= − − − M1 A1

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 2 of 18 1(c) Method 1: 7 2 5 191 45 7 2 7 2 5 191 and 4 4 5 4( 1) 7 2 5(7 2) 4(5 19) 4 4 7 2 35 10 20 76 3 6 35 20 76 10 2 15 86 11515 11 862 5 or 2 15 15 xxx x x xx x x x x x x x x x x x xx x xx −++   − − ++   +  − −  + +  − −  + − − −  +        OR M1 --- 2x M1 --- 11515x A1 --- 1125 15x Deduct 1m for 2 5.73x No marks given for 115215 x

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 3 of 18 Method 2: 7 2 5 191 45 20( 1) 5(7 2) 4(5 19) 20 20 35 10 20 76 20 30 35 20 86 30 15 86 11 862 5 or 2 15 15 xxx x x x x x x x x x x xx −++   +  −  + +  −  + +   +      M1 M1 A1 1(d) 2 2 2 7 9 4 81 yy y +− − ( )( ) ( )( ) 2 9 1 2 9 2 9 yy yy +−= +− 1 29 y y −= − M1 --- correct numerator M1 --- correct denominator A1 [10] Deduct 1m for changing y to x: ( )( ) ( )( ) 2 9 1 1 2 9 2 9 2 9 xx x x x x +− −=+ − − 2(a)(i) First $10 000: Interest = 10000 4 0.05 $20100  = Next $20 000: Interest = 20000 4 0.95 $760100  = Total amount = $30 000 + $20 + $760 = $30 780 M1 M1 A1

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 4 of 18 2(a)(ii) Bank L: Total amount = 4 0.830000 1 100 + = $30 971.58 Since the total amount in Bank L > total amount of Bank H after 4 years, I disagree with Cheryl’s claim. M1 A1 A1 If 2a(i) or 2a(ii) wrong, no marks given for correct conclusion.

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 5 of 18 2(b)(i) Hong Kong Hotel: HK$1 = S$0.17 HK$825 = S$(825 x 0.17) = S$140.25 Guangzhou Hotel: S$1 = CNY$5.33 CNY$825 = 825$ 5.33S =S$154.78 The Hong Kong hotel charges a cheaper rate per night. OR HK$1 = S$0.17 S$1 = 1$ 0.17HK =HK$5.8823 Since S$1 can get HK$5.8823, by comparison with S$1 = CNY$5.33, CNY$ is stronger than HK$. Thus the Hong Kong hotel charges a cheaper rate per night. A1 M1 A1 2(b)(ii) Total cost = 1(4 825 0.17) (2 825 ) 5.33  +   =561 + 309.5684 = S$870.5684 870.5684 870.56 890 100 2.23%(3 8 ) 4 . k k s f −= = M2 – Total cost in SGD M1 A1 [12] 3(a)(i) 13 B1 M1 (either $140.25 or $154.78)

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 6 of 18 3(a)(ii) Interquartile range = 15 – 10 = 5 M1 A1 3(a) (iii) Percentage 500 450 100500 10% −= = M1 A1 3(b) Median in February 2021 = 13 is less than median in February 2022 = 16. Hence, the patients generally stayed longer in February 2022. Interquartile range in Feb 2021 = 5 is less than interquartile range of Feb 2022 = 12. Hence, the number of days stayed in February 2022 is generally less consistent/more wide spread. B1 B1 B1 – state the figures correctly If IQR in 3a(ii) wrong, answer will be wrong. Need to state the figures on the answer line. 3(c) 450 9 500 10= B1 3(d) Probability 125 500 125 500 125 125 500 499 500 499 375 998 −−   =  +        = M1 A1 [11] 1m given for either of the fractions is correct. 4(a) x – 2 – 1 0 1 2 3 4 5 y – 3.6 0.8 2 1.2 – 0.4 – 1.6 – 1.2 2 B2 (B1 for each correct value)

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 7 of 18 4(b) Refer to the graph. P2 - Plotting of correct points (P1 for at most 2 points plotted incorrectly). C1 - Smooth curve. 4(c) Refer to graph for tangent drawn. Using (3, – 2.8) and (5, 0.4), Gradient 0.4 ( 2.8) 53 −−= − 1.6= (Accept 1.1 to 2.2) B1 – tangent drawn B1 4(d) 3 2 215 x x− + = Refer to graph for 1y= drawn. Using the graph, x = – 0.9 (accept – 0.8 to –1.0) or x = 1.1 (accept 1.0 to 1.2) or x = 4.75 (accept 4.6 to 4.9) B1 - For drawing of 1y= . B1 4(e) True: 3 intersections - 1.7 0 k−   False: 1.7k− Accept all negative values B1 [11] 5(a)(i) 32 32 23( )3 2 2 (shown) r r h r r h rh   = = = M1 A1 B1

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 8 of 18 5(a) (ii) 2 22 2 2 (8 ) 2 16 2 16 r r r r r r r      +− = + − = M1 for either 22 or 2 (8 ) seenr r r − A1 5(b)(i) 25 64 5 16 8 5 168 10 B C B B B h h h h h = = = = M1 – Square root A1 5(b) (ii) 3 8 16 1 450 8 1 4508 56.25 A C A A A V V V V Vg = = = = M1 - 3 8 16   OR 3 1 2   seen A1 [8] 6(a) AB = AD (tangents drawn from ext. point) angle BAC = angle DAC (tangents drawn from ext. point) AC is a common side. (SAS)ABC ADC  M1 (for only 1 condition) M2 (for all 3 conditions) A1

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 9 of 18 6(b)(i) 90 8cos 42.5 5.8982186 oABO AB AB cm = = = Area of triangle 1 8 5.8982186sin 42.52=   = 15.93911 = 15.9 cm2 (3s.f) OR 8sin 42.5 5.40472 OB OB = = Area of triangle 2 1 2 1 5.8982186 5.404722 15.9 AB OB cm =   =   = M1 (for either AB or OB) M1 A1 M1 (for either AB or OB) M1 A1 6(b) (ii) 180 90 42.5 47.5 oAOB = − − = 90 8sin 42.5 5.404721 oABO OB OB cm = = = Area of sector OBE M1 (allow e.c.f) OR 228 5.8982186 5.404721 OB OB cm =− = [M1]

NAVAL BASE SECONDARY SCHOOL PRELIMINARY EXAMINATION, 2024 Paper 2 Marking Scheme Page 10 of 18 2 2 47.5 5.404721360 12.10842 12.1cm =   = = Area of shaded region = 15.93911 – 12.10842 = 3.83 cm2 (3 s.f.) M1 (allow e.c.f) A1 [9] 7(a) DV = VB 221 682 5 cm =+ = EV 2215 5 250 15.8113883008 cm =15.8 cm (3s.f) (shown) =+ = = M1 A1 7(b) EC 2215 8 17 cm =+ = If EV = 15.8 cm cos ACE = cosVCE M1 M1