[Wlands Ring] [2023] 4E EM 4052 Prelims P1 MS

1 Woodlands Ring Secondary School Secondary 4E Mathematics Prelim P1 2023 – Worked Solutions Qns. No. Working 1(a) 34 7.23 8.9 706.5031.305 707 (3 s.f.) += = 1(b) 7.07  102 2(a) 2 6 84 7 28a a a= 2(b) 8 5(1 2 ) 8 5 10 3 10 bb b − − = − + =+ 3(a) 2 4 12 ( 2)( 6)x x x x− − = + −  2a= , 6b= 3(b) Line of symmetry is 2x= 4(a) Range = 31 – 13 = 18 4(b) Median = 19 22 2 + = 20.5 5 The line graph for Star-One looks steeper than that for Galaxy as the scales for the vertical axes are different. 6(a) Diagram 2 6(b) Height of cylindrical part of the container 7(a) Sequence: − 2, 5, 12, 19, … Multiples of 7: 7, 14, 21, 28, … , 7n  nth term = 7n − 9 7(b)(i) 1, 7, 17 7(b)(ii) For any integer n, 2n2 is always a positive even number. Since 1 is an odd number, when it is subtracted from 2n2, 2n2 – 1 will be odd.

2 Qns. No. Working 8(a) 2546 3 12 3 2 5 18 xx xx −−   −  −  12 3 2 5xx−  − and 2 5 18x− 5 17x− − 2 23x 23 5x 1112x [or 3.4x and 11.5x ]  213 1152 x [or 3.4 11.5x or 17 23 52 x ] 8(b) Smallest integer = 4 9(a) P(cash) = 111 45−− = 11 20 9(b) P(at least one of them uses mobile wallet (M) = 1 – P(no M) = 331 44− = 7 16 9(c) P(only the third customer uses cash) = P(no cash, no cash, cash) = 9 9 11 20 20 20 = 891 8000 10(a) Map scale = 5 cm : 2.5 km = 1 cm : 0.5 km = 1 : 50 000 10(b) On the map, PQ = 8 cm (accept 7.9 – 8.1 cm) Actual length of PQ = 8  0.5 = 4 km (accept 3.95 – 4.05 km) Other accepted working: Map scale = 1 cm : 0.5 km = 8 cm : 4 km

3 Qns. No. Working 10(c) Area scale = 12 cm2 : 0.52 km2 = 1 cm2 : 0.25 km2 Area on the map = 3.5 0.25 = 14 cm2 11(a) Gradient of line l 15 5 10 2 −= − 5 4= 11(b) Sub (2, 5) into 5 4y x c=+ : 55 (2)4 c=+ 55 2 5 2 c=− = Equation of line l. is 55 42yx=+ OR 4 5 10yx=+ … (1) 11(c) Length of AB 22(10 2) (15 5)= − + − 164 12.8 units (3 s.f.) = = 11(d) Since the vertical line passes through the point (18, 2) and point N, the x-coordinate of N is 18. Sub 18x= into (1): 55(18)42y=+ = 25  N is (18, 25) 12 2 kR x= New x = 0.4x New 2(0.4 ) kR x= 2 2 0.16 1 0.16 6.25 k x k x R =   =     =

4 Qns. No. Working Percentage increase 6.25 100%RR R −= 5.25 100%R R= 525%= 13 2126 2 3 7=   290 2 3 5=   354 2 3= HCF of 126, 90 and 54 = 223 = 18  number of cubes required = 126 90 54 18 18 18 = 105 14(a) 14 35 7 (2 5)gh g g h− = − 14(b) 2 2 2 2 2 22 ( 3 ) 8 6 9 8 6 r s s r rs s s r rs s − − = − + − = − + 15 4 3 220ca+= … (1) 3 5 275ca+= … (2) (1)  3: 12 9 660ca+= … (3) (2)  4: 12 20 1100ca+= … (4) (4) – (3): 11 440a= 40a= Sub 40a= into (1): 4 3(40) 220c+= 4 100c= 25c=  cost of a ticket for a child = $25 cost of a ticket for an adult = $40 16 1 interior angle of the polygon B at point O = 360 – 90 – 115 = 155° (s at a point) 1 exterior angle of the polygon B at point O = 180 – 155 = 25° (adj. s on a str. line)

5 Qns. No. Working If polygon B is a regular polygon, number of sides = 360 25 = 14.4 Since n is not an integer, polygon B cannot be a regular polygon. 17 3sin 2 0 2sin 3 41.8 ,138.2 (1 d.p.) x x x −= = =   18(a) 1, 4, 6, 8, 9, 10 or 'A = {1, 4, 6, 8, 9, 10} 18(b) AC = {2, 3, 4, 5, 6, 7, 8, 10, 11} ( ) 'AC = {1, 9} n ( ) 'AC = 2 18(c)  A B 3 5 7 1 11 2 4 6 9 8 10 C 19 21 2 2 3 1 4 3 3 16 64 (4 ) (4 ) 44 ww ww ww − − − = = = Comparing the indices: 4 3 3 73 3 7 ww w w =− = =

6 Qns. No. Working 20 1. (given) 2. 180 (adj. on a str. line) 180 (base s of isos ) = (adj. s on a str. line) 3. (given) BQ CB ABQ ABC ACB ABC PCB AB AC PC AQB =  = −  = −    = =    (SAS) PBC 21 2 2 2 2 215 3 3 45 2 2 45 3 45 3 2 45 3 2 pq qp pq qp qp =− =− =− −= −= 22 22 2 2 2 2 33 (2 5) 5 2 (2 5) 2 5 3(2 5) (2 5) 6 15 (2 5) 15 5 (2 5) 5(3 ) (2 5) yy y y y y yy y yy y y y y y + = −− − − − −−= − −+= − −= − −= − 23(a) In TVU, 2 2 2100 125 167cos 2(100)(125) 283 2264 or 3125 25000 TVU +−= =− −  TVU = 95.1958° = 95.2° (1 d.p.) (shown)

7 Qns. No. Working 23(b)  a = 58° (alt s, // lines)  b = 95.1958 – 58 = 37.1958°  the bearing of V from U = 180 + 37.1958 = 217.2° (1 d.p.) 23(c) tan 24.7 100 100 tan 24.7 45.99486 m 46.0 m (3 s.f.) DT DT = = = = 24(a) Q = 20 4 5 34    24(b) R = PQ = 205.2 6.6 7.3 4 55.5 6.4 7.4 34     5.2 2 6.6 4 7.3 3 5.2 0 6.6 5 7.3 4 5.5 2 6.4 4 7.4 3 5.5 0 6.4 5 7.4 4  +  +   +  + =  +  +   +  +  58.7 62.2 58.8 61.6 = N NU a b b N D T V 100 m 24.7° (angle of depression) 24.7°

8 Qns. No. Working 24(c) The elements in R represent the total amount that Derrick and Elaine need to pay if they purchase the drinks at Café X and Café Y respectively. 24(d)(i) S = 1.1 0 0 0 1.15 0 0 0 0.95    24(d)(ii) 1.1 0 0 2 05.2 6.6 7.3 0 1.15 0 4 55.5 6.4 7.4 0 0 0.95 3 4 205.72 7.59 6.935 4 5 6.05 7.36 7.03 34 62.605 65.69 62.63 64.92                     =   = Derrick spent $62.61 at Café X. Elaine spent $64.92 at Café Y. 25(a) Arc 8.4(1.5) 12.6 cm PQ= =  perimeter of the shaded region = 11.5 + 12.6 = 24.1 cm (3 s.f.) 25(b) Area of the shaded region = Area of sector OPQ – Area of OPQ 2 2 11(8.4) (1.5) (8.4)(8.4)sin1.522 52.92 35.1916 17.728 17.7 cm (3 s.f.) =− =− = = 25(c) In OPR: cos1.5 8.4 8.4cos1.5 0.59419 cm (5 s.f.) OR OR = = =  RQ = 8.4 – 0.59419 = 7.80581 (6 s.f.) = 7.81 cm (3 s.f.)

9 Qns. No. Working 26(a) Since speed is constant from the 20th second to the 45th second, acceleration of the object at the 30th second = 0 m/s2 26(b) Let v m/s be the speed at the 7th second. 35 7 20 12.25 v v = = The required speed is 12.25 m/s. 26(c) Total distance = 1.925 km  1 (45 20) (35) 19252 35( 25) 3850 25 110 85 k k k k − + = += += = 27 (c) (b ) (a) v 7 A B