North Vista 4E EM Prelim P1 2024 Solutions

4E5N Mathematics Prelim Paper 1/2024 Qn Answer AO Marks 1 12.57 10 ….B1 AO1 1 2 14.75 20.2054............... 1 0.73 21.99 20.2054 $1.78................. 1(with ) 21.99 0.73 16.0527.............. 1 16.0527 14.75 1.30.............. 1(with )£ M A units or M A units = − = −= AO1 2 3(a) 3 4 2 68 69 5(3 5 ) 5(3 5 ) 3 5 ................ 1 B  = = AO1 1 3(b) 100 97 100 2 97 100 99 99 2 4 2 2 2 2 2 2 . 2 2 2 ............ 1 2 (2 1) 2 99...... 1 k k k k M kA −  = −  = −= − =  = AO2 2 4(a) Factors of 60 :1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 60........... 1pB= AO1 1 4(b)(i ) 2525 3 5 7=   ……..B1 AO1 1 4(b)(i i) 15 3 5 1=   35 1 5 7=   221 5 1 or 3 5 1xx=   =   2525 3 5 7=   25,x= 75 ……B1, B1 AO2 2 5 10 10 0.1 4540+1328.54 = 4540(1 ) ....... 1100 (1 ) 1.2926299100 (1 ) 1.2926299 ...... 1100 2.6000 2.60(3 )........ 1 r M r r M r sf A + += += = AO1 3 6 Size of each exterior angle = 180 4 809 o= …..B1 AO3 2

Since the number of sides = 360 4.5 80 o o = is not a positive integer, therefore it is not possible to form a regular polygon…..B1 OR Let n be number of sides. ( 2) 180 :360n− ( 2) : 2n− 2 4 : 5n− 2 4 5 4.5 n n −= = OR Exterior + Interior angle = 180 degrees ( 2) 180 5 (180)9 n n − = n = 4.5 7(a) 22 22 17 ( 6) 17 12 36 Comparing 12...... 1 17 36 19....... 1 x ax x b x ax x x b aB b b B + + = − + + + = − + + =− = +  =− AO2 2 7(b) 22 17 ( 6)x ax x b+ + = − + Since the coefficient of x2 > 0, 2 17x ax++ is minimum when 2( 6) 0x−= therefore x = 6. AO3 1 8 New Selling price 136= 140...... 176 $250.53...... 1 M A  = AO2 2 9 Let the number of yellow balls be Number of blue balls is 4 51 ................... 14 10 6 6( 5) 4 10 6 30 4 10............ 1 2 40 20 Number of yellow balls = 20..........A1 x x x Mx xx x x M x x − =+ − = + − = + = = AO2 3

OR 1 : 4 20 : 80 15 : 90 1 : 6 Ans: 20 OR 1 : 6 15 : 90 20 : 80 1 : 4 Ans: 20 OR Before After 1 : 4 1 : 6 4 : 16 3 : 18 1 unit = 5 4 units = 20 OR Let initial yellow balls be x Let initial blue balls be y 4x = y …. (1) 6(x – 5) = y + 10 … (2) x = 20 OR Let initial yellow balls be x Let new yellow balls be y x – 5 = y … (1) 4x + 10 = 6y … (2) x = 20 10(a) 4x=− ………B1 AO1 1 10(b) 5 10 25 5 .............. 152 12.5.............. 1 y mx myx m M mA += =− = = AO2 2 11(a) 78, a, b, c, 42,….. AO2 2

Common difference = 78 42 94 − = 78 9 69 69 9 60 60 9 51....... 2 / 1 for any 2 correct a b c B B = − = = − = = − = 11(b) General term = 87 9 n− ……B1 AO2 1 11(c) 87 9 0............ 1 . 9 87 29 103 negative term = 87 90 3......... 1 n M o e n nn First A − − −   = − =− AO2 2 12(a) 'PQ AO1 1 12(b) (i) ε = {integer x : 1 15x } A = {1,4,9} B = {2,3,5,7,11,13} '()AB = {6, 8, 10, 12, 14} '( ) 5n A B= …..B1 AO1 1 12(b) (ii) B’ = {1,4,6,8,9,10,12,14} ' { 1,4,9}AB= ……..B1 (no mark award for m