Dunman Sec 2025 Prelims 4EXP Math Paper 2 Solutions

DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2025 SECONDARY 4 EXPRESS/ 5 NORMAL ACADEMIC MATHEMATICS 4052/02 Paper 2 2 hours 15 minutes Solutions

Question Answer 1(a) 5921 2 xx +− 4 2 5 9xx−  + 75 x− 7 5x− or 1.4x− 1(b)(i) ( ) 2 2 25 212 3 np b np b − = − = 1(b)(ii) ( ) ( ) ( ) ( ) 2 2 22 22 22 2 2 25 25 2 10 10 2 10 2 2 10 2 10 np b np b n p n p bn bp n p p bp n bn p b n bn n bnp b n bnp b  − = − − = − − = − − = − − = − −= − −= − 1(c) ( ) ( ) ( )( ) ( ) 22 22 2 32 72 1 5 3 5 2 2 1 72 1 5 15 3 4 2 7 10 2 5 3 11 2 77 14 35 0 11 66 37 0 x xx x x x xx x x x x x x x x x x xx −=−− − − − =−− − − + = − − + − + + − + + = − + = ( ) ( ) ( )( ) ( ) 2 66 66 4 11 37 2 11 66 2728 22 5.37 (3 s.f.) or 0.626 (3 s.f.) x − −  − −= = =

Page 3 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS Question Answer 2(a)(i) Difference = (3.553 – 3.3064) million tonnes 6 5 0.2466 10 2.47 10 tonnes = = 2(a)(ii) 3.553 3.040percentage increase 100%3.040 16.875% 16.9% (3 s.f.) −= = = 2(a)(iii) 6 6 amount of waste in 2022 3.553 10 84.8 100 4.19 10 (3 s.f.) =    = 2(b)(i) deposit = 20% 120000 24000= total monthly payment = 84 1400 117600= total amount = 24000 117600 141600+= 2(b)(ii) 5value of car 5 years later = 120000 0.85 5 3244.6375= 120000 53244.6375percentage decrease 100% 120000 55.6% (3 s.f.) −= = Question Answer 3(a) h = 0.15 3(b) 3(c) 3.5 hours 3(d)(i) Gradient = -0.5 (± 0.1) 3(d)(ii) Since gradient is negative, the daily growth rate decreases with additional sunlight. 3(e)(i) Straight line passing through (0, 0) and (4, 2) 3(e)(ii) ( ) 2 23 32 1 462 43 4 3 0 ttt t t t t t t −= −= − + = Question Answer 4(a)(i) ( ) 23 1 234 13 24 AB AX =− + = − + =− + ab ab ab

Page 4 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS 4(a)(ii) 132 24 33 24 OX = − + =+ a a b ab 4(b) ( ) ( ) 3 2 6 3 2 3 64 99 24 BY XY XB BY == =+ = − + + =+ aa a b a ab 4(c) 1 3OX XY= Since OX is parallel to XY, and X is a common point, O, X and Y lie on a straight line. 4(d) Area of triangle 3 Area of triangle 4 Area of triangle 2 4 Area of triangle 3 6 Area of triangle 3 3 Area of triangle 4 6 10 OBX BX OAB AB OAB OA ABC BC OBX OABC == = = = == + 4(e) 33OC =+ab ( )2 335 66 55 OW =+ =+ ab ab

Page 5 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS Question Answer 5(a) ( )( ) ( )( ) 2 2 2 2 2 2 3 8 7 2 8 7 cos 873cos 2 8 7 EFI EFI = + −  +−= 13 14= 1 13cos 21.7867... 21.814EFI −  = = =   5(b) 3 1vol of prism = 8 3 7 8 sin 21.7867 122 412.70717.. 412.71m   +      = = 5(c) height = 3 7sin 21.8 5.5995... 5.60 m (3 s.f.) + = = 5(d) 22 1 Let be the point vertically below . 7 cos 21.8 6.4994... 12 6.4994 13.647... 5.5995tan 13.647 5.5995tan 13.647 = 22.3 (1 d.p.) XJ CX BX JBX JBX − = = = + = = =  

Page 6 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS Question Answer 6(a) 90 (tangent radius)OCT =  ⊥ 360 90 90 32 ( sum of quadrilateral) =148 AOC = − − −    148 2 ( at centre = 2 at circumference) = 74 ADC =     6(b) 180 74 ( in opposite segment) =106 CBA s = −    180 148 (base of isosceles triangle)2 =16 OCA s −  =   16 38 (base of isosceles triangle) = 54 OCB s = +    106 54 = 160 180 OCB CBA + = +    Therefore, by converse of interior angles, OC is not parallel to AB. 6(c) 38 2 ( at centre = 2 at circumference) = 76 AOB =     Area of sector ( ) 276 190π 10 = π360 9= Area of triangle ( )( )1 10 10 sin 76 48.5147...2= = Area of shaded region = 2190 π 48.5147 17.8 cm (3 s.f.)9 −=

Page 7 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS Question Answer 7(a)(i) n + 20 7(a)(ii) Product of top left and bottom right = ( ) 220 20n n n n+ = + Product of top right and bottom left = ( )( ) 22 18 20 36n n n n+ + = + + Difference = ( ) ( ) 22 20 36 20 36n n n n+ + − + = 7(a)(iii) Sum = ( ) ( ) ( ) ( )2 10 18 20 5 50n n n n n n+ + + + + + + + = + Let 5 50 1715 333nn+ =  = If n = 333, the cross will be 333 … 335 … 343 … 351 … 353 However, 333 = 9 × 37, the first number will be in the last column of the number grid, therefore, the sum cannot be 1715. 7(b)(i) ( ) 2 10 2 10 10 3 193T = − + = 7(b)(ii) ( ) ( ) ( ) 2 1 2 2 2 1 1 3 2 2 1 1 3 2 3 4 nT n n n n n nn + = + − + + = + + − − + = + + ( ) 222 3 4 2 3 41 D n n n n n = + + − − + =+ 7(b)(iii) Since D is a linear expression in n and the coefficient of n is 4, it means that the difference increases by 4 each time n increases by 1.

Page 8 of 8 4052/02/PRELIM/4E/2025/SOLUTIONS Question Answer 8(a)(i) 7.5 hours (±0.1) 8(a)(ii) Q1 = 6, Q3 = 9.4 IQR = 9.4 – 6 = 3.4 (±0.1) 8(b) 32 students spent 5 hours or less. 168 students spent at least 5 hours. Probability = 168 21 0.84200 25== 8(c) In general, teenagers in Country X spent less daily screen time with median of 6.5 hours compared to Singapore’s median of 7.5 hours. The amount of daily screen time spent by teenagers in Country X has a larger spread with an IQR of 5.2 hours as compared to that of Singapore with IQR of 3.4 hours. Question Answer 9(a) Arc length = ( ) 142π 2 π 2.513... 2.5 m (1 d.p.)55 = = = 9(b) Volume of air = ( ) 2 π 2.5 0.5 50 60 8.1812... 8.18 (3 s.f.)   = = 9(c) For air circulation Minimum airflow requirement 330 20 8 5 960 m / min=    = VFR = ( ) 2 π 3 0.6 5.4π 16.9646... = = Total airflow per minute of fans = VFR 70 2 756 π 2375.044...  = = Therefore, 2375.044... 960, minimum airflow requirement is met. For comfortable cooling environment VPS = 5.4π 45 60 4.05π  = Air velocity = ( ) 2 4.05π π 3 0.45 m/s > 0.3 m/s= Therefore, comfortable cooling environment is met. Hence, the recommendation is suitable.