2024 GSS 4E5N Prelim Paper 2 MS

2024 Preliminary Examination Mathematics (Syllabus 4052/2) Setter: Mrs Li Geok Eng 1(a) 3 2 32 4 52 64 54 62 5 3 a ab b a b a a b b ab a b    B1 (b) 2 50 8 400 20 x x x x    B2 (c) 241 2 10 5 (2 1)(2 1) ( 5)(2 1) 21 5 v pv p v vv pv v p         M1 M1 A1 (2 1)(2 1)vv ( 5)(2 1)pv (d) 2 2 2 2 2 3(3 1) 2 9 3 2 9 3 0 ( 9) ( 9) 4(2)(3) 2(2) 4.14 or 0.36 xx xx xx x x            M1 M2 A1 Quadratic eqn formed M1 for b2 – 4ac correct

2(a) 3 3.5% 10.5% 10.5% $10374 10374100% 10010.5 $98 800     M1 A1 OR I = P x r% x T (b) Total petrol consumption 16992 6.7100 1138.464l   Total amount paid 1138.464 $2.72 $3096.62   M1 A1 (c)(i) The decreased of 5% is compounded. The value of the car (base) for each year is lower than the previous year. B1 (c)(ii) Year 0 120500 Year 1 0.95 120500 $114 475 Year 2 0.95 x 114 475 = $108 751.25 Year 3 0.95 x 108 751.25 = $103 313.69 Ans $103 314 (nearest dollars) M1 M1 A1 Year 1 Year 2

3(a)(i) {2, 3, 5, 7,11,13}A {3, 6, 9,12}B  ' {4,8,10,14}AB B1 (a)(ii)   {3 }AB B1 (a)(iii) Any subset with 3 elements from {2,5,7,11,13} B1 3(b)(i) 3 1 1 24 2 19.0 d d    M1 A1 (b)(ii) Let r be the radius of cone P. V olume of cone P, v 2 2 31 (24) = 8 cm3 rr V olume of cone T = 2 2 3 11 (2 ) ( 24)33 4 83 4 cm3 r r v        M1 A1 (c) CD is common. DA = DB (tangent from an external point) Since, angle CAB = angle CBA  triangle CAB is isosceles, Hence, AC = BC. Therefore, triangle ACD and triangle BCD are congruent (SSS). B1 B1 B1

4(a) 2 2 2 22 (18 ) 12 324 36 144 36 468 13 rr r r r r r cm          M1 A1 (b) Angle AOB 1 o 122 tan 5 134.76     M1 A1 Or ½ absinC (c) Reflex angle AOB = o o o360 134.76 225.24 Area of major sector = 2225.24 (13)360  Area of triangle AOB = 21 13 sin134.762 Area of segment = 2225.24 (13)360  + 21 13 sin134.762 = 392.18 = 392 cm2 M1 A1 M1 A1 R (d) Volume of water = 392.18 40 15687.2 cm3 2 15687.2 (13) 29.547cm 29.5cm h h h    M1 A1

5(a)(i) 8 3 8(2) 19 8 19 Gradient c c yx       M1 A1 (a)(ii) 22(2 3) (3 5) 8.06 units     OR 22 3 2 1 5 3 8 1 ( 8) 8.06 AB AB                       M1 A1 M1 A1 (a)(iii) 3 4 1 5 3 2 ( 1, 2) OC C                    M1 A1 5(b)(i) OB 2 3ac B1 5(b)(ii) 2 (3 2 )3OT  ca B1 (b)(ii)   22 (3 2 )3 22 3 2 33 3 AT a AT AT AM       ca ca ca ca 2 3AT AM AT is parallel to AM and A is common. Therefore A, T and M lies on a straight line. M1 M1 A1 or orAT AM TM AT k AM

6(a) 2 2 2 o60 37.5 2(60)(37.5) cos 63 54.436 54.4m LB LB LB      M2 A1 (b) o sin sin 63 37.5 54.436 37.865 37.9 LBA LBA LBA     M1 A1 (c) otan1037.5 6.6123m = 6.61m LT LT LT   M1 A1 (d) o3 sin 60 3 sin 60 3.46m d d d    M1 A1

7(a) 0.8 B1 (b) P2 P1 C1 Plot all points correctly Plot at least 6 points correctly Smooth curve (c) Tangent drawn correctly 2 0.2 M1 A1 (d)(i) Straight line passing through (0,6) and (4,0) B1 (d)(ii) 0.6 0.05 , 4.2 0.05 B1 B1 (d)(iii) 22 2 92(2 11) 12 3 184 22 12 3 4 18 22 12 3 7 34 18 0 34 18 xx x xx x x x x x xx A B                  M1 M1 A1

8(a) Year Lower quartile Median Upper quartile Interquartile range 2022 150.5 157 6.5 2023 151 M1 A1 B1 LQ or UQ IQR Median (b) No, I do not agree. The mean time for the runners in 2023 is lower than the mean time in 2022. B1 (c) Time (minutes) Frequency 135 140t 4 140 145t 26 145 150t 32 150 155t 28 155 160t 21 160 165t 9 B1 B1 8(b)(i)(a) 1 2 2 3 3 9 B1 (b)(i)(b) 1 1 2 2 3 3 3 3 5 9              M1 A1 (b)(ii) 2 5 2 9 9 9 28 81   M1 A1

10(a) Amount paid before GST (737 692 749) 0.29893 726 0.2989 $217    M1 A1 (b) 9 1.65 5 4 1 4 5 4 20    M1 A1 (c) Average amount of electricity produced = 20 x 19 = 380 kWh P1 Average cost per month after solar energy savings = (726-380) x $0.2989 =$103.42 C1 Average cost of installing & maintenance of solar panel per month (2 $5950 20 $500) (20 12) $91.25        I1 Total average amount paid per month after installation =$103.42 + $91.25 =$194.67 (< $217) T1 Since the average amount paid by Mr Faizal after installing the solar panels is less than what he is currently paying, he should proceed with the installation. A1 Conclusion Assumption: The average electricity consumption remains the same, The price of tariff did not increase. A1 Either one Method 2 (Total cost based on 20 years) (c) Average amount of electricity produced = 20 x 19 = 380 kWh P1 Cost for 20 years before installation = $217x20x12= $52080 C1 Cost of installation & maintenance (2 $5950 20 $500)    =$21900 I1 Total cost for 20 years after installation =(726-380) x $0.2989x12x20 +21900 = 46720.66 < $52080 T1 Since the total amount paid by Mr Faizal after installing the solar panels is less than what he is currently paying, he should proceed with the installation. A1