Dunman Sec 2025 Prelims 4EXP Math Paper 1 Solutions

DUNMAN SECONDARY SCHOOL CANDIDATE NAME CLASS INDEX NUMBER PRELIMINARY EXAMINATION 2025 SECONDARY 4 EXPRESS/ 5 NORMAL ACADEMIC MATHEMATICS 4052/01 Paper 1 2 hours 15 minutes Solutions

Qn Answer 1(a) Inconsistent y-axis scale 1(b) The revenue growth appears to be more than doubled from 2020 compared to 2024 when in actual fact only increased from about $500 000 to $800 000. Qn Answer 2 In Italy, 830 1.4893 $1236.119= In Singapore, 830 0.6747 $1230.176375= Difference = $5.94 Qn Answer 3(a) 1sin 3x= 19.47122... or 160.5287... =19.5 or 160.5 (1 d.p.) x=  3(b) 26 0.45378... 0.454 (3 s.f.)= = Qn Answer 4 Amount of liquid in bottles B and C = 300 15 13 260  = ml Amount of liquid in bottle A = 40 ml Amount of liquid left in bottle A = 260 65 7 28  = ml n = 40 – 28 = 12 OR 2 : 8 : 5 is equivalent to 10 : 40 : 25 75 units rep. 300 ml 1 unit rep. 4 ml n = 4 × (10 – 7) = 12

Page 3 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Qn Answer 5(a) 2 12 3 3 2 3 3 12 2 8 27 27 9 a b b a b a −   = = 5(b) ( ) 24 44 4 42 13 9 4 3 3 13 3 4 3 3 3 13 4 3 3 3 3 6 p p p p p  −  =  −  = −= = = Qn Answer 6 ( ) 2 1 2 22 11 9 9 93 kF r k k kFF rrr = = = = =  He is not correct since the force should be 1 9 instead. Qn Answer 7(a) 300 km 2hour 150 km/h= 7(b) Since distance is rounded down and time is rounded up, this speed will be lower than the actual speed. 7(c) 227 km/h 227 1000 3600 63.1 m/s =   = Qn Answer 8(a) ( ) 2216 8 82 a b ab ab a b − =− 8(b) ( ) ( ) ( )( ) 3 4 3 4 3 1 4 1 3 4 1 c d cd c d d cd − + − = + − + = − +

Page 4 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Qn Answer 9 22 5 29 cm : 728.6 km 29 cm : 728.6 km 5.3852 cm : 26.993 km (5 s.f.) 1 cm : 5.0124 km (5 s.f.) 1 : 501244 1: 5 10 55 10n= Qn Answer 10 ( ) ( ) 215 5 12 18 54 27 15 5 12 4 1 54 10 12 4 4 54 68 54 34 27 xxx x x x xx x x −+−− − − − −= − − += −= −= Qn Answer 11 (radius of circle) (radius of circle) (equal chords) OD OB OC OA CD AB = = = Therefore, triangles OAB and OCD are congruent by SSS congruency test.

Page 5 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Qn Answer 12 ( )( ) ( ) ( )( ) ( )( ) ( )( ) ( )( )( ) ( )( ) 2 4 222 22 2 2 2 7 15 16 81 2 3 5 49 2 3 5 4 9 4 9 2 3 5 2 3 2 3 4 9 5 2 3 4 9 xx x xx x xx xx xx x x x x xx +− − −+= − −+= −+ −+= − + + += ++ Question Answer 13(a) ( ) ( ) 2 2 2 2 2 7 10 10 7 5 5 7 5 18 xx xx x x −+ = − + = − − + = − − 13(b) x = 5 Question Answer 14(a)(i) 1, 4, 6, 8, 9, 10, 12, 14 14(a)(ii) Since perfect squares have more than 2 factors, they will never be prime numbers. Hence, XY = 14(b) A B

Page 6 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Question Answer 15 Question Answer 16 ( ) 2 3 32 4 1 2π 2 2 π 103 3 3 32π 80π 3 27 32 27 3 80 864 240 864 240 : 18 : 5 r r h r r h rh rh h r hr =    =  =  = = = Question Answer 17 angle 108 angle 360 108 100 ( at a point) 152 ABC FBC s = = − −  = exterior angle 180 152 28 360for regular polygon 28 12.85... not whole number = − =  =  = For a regular polygon, they have equal angles. Hence, 360 is not divisible by 28 , the second polygon is not regular. y x O -2 7 14

Page 7 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Question Answer 18(a) ( ) ( ) 22 4 3 5− + − = 18(b) 33gradient = 44 − =− Equation: ( ) 3 4 352 4 13 2 y x c c c =+ − = + =− 3 13 1 or 6 or 6.5 4 2 2yx= − − − 18(c) x-coordinates = ( )32 4 54+= y-coordinates = ( )335 3 244− + =− R = (5, -2.75) Question Answer 19 016 16ka k=  = 2 2 64 16 1 4 1 (rej. -ve)2 a a a −= = =

Page 8 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Question Answer 20 2 11 22 2 1 1 1 4 5 30 4 5 30 430 26.832 cm (5 s.f.)5 length = 30 26.832 3.17 (3 s.f.) Al Al l l l =  = = =  = −= Question Answer 21(a) Probability = 5 4 4 3 4 9 8 9 8 9 +  = 21(b) Probability = 5 4 4 5 5 9 8 9 8 9 +  = Question Answer 22(a) mean = 15 4.25 29 4.75 12 5.25 34 5.75 10 6.25 100 5.225  +  +  +  +  = mean = $5225 22(b) Standard deviation = $637.87 or $638 22(c) Mean remains unchanged, standard deviation becomes smaller Question Answer 23(a) Angle NAB = 52° Angle N1BA = 180° − 52° = 128° (interior angles, NA // N1B) Bearing of A from B = 360° − 128° = 232° 23(b) BP = 2.1 cm

Page 9 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Question Answer 24(a) 5 6 2yx=− × 2: 10 12 4yx=− -- (1) 2 5 6yx=− × 5: 10 25 30yx=− -- (2) (2) – (1): 13 26 0x−= 2x= , y = 2 ⸫ (2, 2) 24(b) For 31py x=− not to intersect with 5 6 2yx=− , they must be parallel. 31yx pp=− 3 6 5 52 pp =  = 3 62: 55l y x =− Since line 1l and 3l have the same equation, they will intersect. Therefore, it is not possible to find a value of p that the lines will not intersect. Question Answer 25(a) 1.20 1.50 1.80  = Q 25(b) 1.2060 80 50 1.5050 90 70 1.80 72 120 90 60 135 126 282 321  =   ++= ++ = T = PQ 25(c) 282 and 321 represents the total revenue by Class A and class B respectively. 25(d) ( )0.8 0.9=D

Page 10 of 11 4052/01/PRELIM/4E/2025/SOLUTIONS Question Answer 26(a) 32 3 54= and 322 3 72= 26(b)(i) All prime factors have even indices/powers Or ( ) 24 6 2 32 3 2 3 =  and therefore, a perfect square 26(b)(ii) 46 46 35 611664 623 23 6 23 m m m m  =   =   =   m = 3