[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=