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 missing curly bracket) AO1 1 12(b) (iii) C = {1}, {4}, {9}, {1,4}, {1,9} or {4,9} any other possible answers………….B1 AO1 1 13 Shape with correct y – intercept (0,8) + x intercepts (-2,0) & (4,0) ……………. B2 Coordinates of turning point (1,9) ………….. B1 AO1 3 14(a) 63………..B1 AO1 1

14(b) 72 – 56 = 16………….M1, A1 AO1 2 14(c) The spread of marks for the group of foreign students is wider since the interquartile range is higher. The cumulative frequency curve will be less steep than the original curve and passes through (63, 80) since both groups have the same median. AO3 1 15 2 2 2 43 9 81 ( 9) (4 3) ( 9) 4 3 or 1( 9)( 9) ( 9)( 9) ( 9)( 9) 9 4 3 ( 9)( 9) 13 3 ................ 1( 9)( 9) xx xx x x x x x x Mx x x x x x x x x xx xx Axx +−+− − − + − +=− + − + − + − − − −= +− −−= +− AO1 2 16(a) 22 22 (2 3 )(7 5 ) 14 10 21 15 .......... 1 14 11 15 ................ 1 x y x y x xy xy y M x xy y A +− = − + − = + − AO1 2 16(b) (i) 3 3 3 32 3 ( 1)......... 1 ( 1)( 1).............. 1 x y xy xy x B xy x x B − =− = + − AO1 2 16(b) (ii) 5 3 10 6 (5 3 ) 2 (5 3 )........ 1 (5 3 )( 2 )............... 1 ax ay cx cy a x y c x y M x y a c A − − + = − − − = − − AO1 2 17 (Given) 90 (int angle of a square) ( , sides of square) ................... 1(for all statements and reasons) AW BX WAZ XBW AD DZ AB AW AW DZ given AD AB AZ BW B =  = =  − = − == = cB oy gSAS triangle is n ru t, ent to riangle AWZ BXW …..B1 AO3 2 18(a) 3 3 3 1 1 7 1.......... 1 0.5 0.5 124 1 2.5.......... 1 y k x kM k yA =+ =+ = = + = AO1 2

18(b) 2 2 2 F ....... 1 (0.5 ) 44 kF d kNew B d k F d = = ==  New value of y becomes 4 times of the original value……..B1 Other good answers : F increases to 400% of the original value F increases by 300% F increases by 3 times F is increased by a factor of 4 Acceptable answers: F is multiplied by 4 AO2 2 19(a) 22 1 6 1/ .1 0 62 / 51 or ms sm AO1 1 19(b) Speed 0.5(30 13)16 13 8 ........ 1( correct distance)24 448 24 218 / ............. 13 Ave M m s A + + = = = AO2 2 20(a) 1250 250 100 600 1400 200 90 690 1 AB     ==         ………..B2 AO1 2 20(b) The elements in AB represent the total amount of flour, butter and sugar (or ingredients) used in making a pandan cake and a marble cake respectively…..B1 AO3 1 20(c) 0.2 0.3 Dx  = ……..B1 AO1 1 20(d) 250 2500.2 0.3 4.25100 100 $1.38............ 1 x xB  +  + = = AO1 1

21 Perpendicular Bisector – B1 Angle bisector - B1 With correct position of E – B1 AO1 3 22 22 22 2 2 2 2 2 2 2 2 height = (12 ) (17 ) 433 ...... 1 (12 ) (12 )( 433 ) 4 ( ) ..... 1 144 12 433 4 4 (144 12 433) ..... 1 (144 12 433) 98.425954 9.92(3 ) ............. 1 ( 9.92rejected) slant x x x M x x x rx M x x x r x r x M r r sf cm A        += += += =+ +== =− 7 AO2 4 23(a) 2 3 tan 5 ...... 12.1 Length of water level = 0.9 2(2.1tan 5 )... ..... 1 1Area of trapezium = (0.9 0.9 2(2.1tan 5 )) 2. 1 2.2758 m .... 12 Volume of water = 3.9464 100 227.5825 228m . ........... 1 o o o x M M M A = + + +  =  =  AO2 4 23(b) 227.5825 60 12.64340.3 12 hours 38.60 mins 12 hours 39 mins.......B1 h= = = AO1 1 24(a) (i) ( ) 1 3 1 ........ 13 PL PT PL m p B = =− AO1 1

24(a) (ii) 2 1 1 3 3 3 21 ................. 133 KL KO OP PL KL m p m p KL p m B = + + =− + + − =− AO2 1 24(b) 2 2 4 3 3 3 4 ....... 13 4 ............... 13 OM OK KM OM m m p OM p B OM OP B =+ = − + = = Since 4 3OM OP= and O is a common point, therefore M lies on OP extended.B1 AO3 2 24(c) 2 1 2 ............... 2( . ) 3 3 9 Area KTL Area KTL Area KPT Area OTP Area KPT Area OTP Area KTL B o eArea OTP = =  = AO2 1 25(a) (8 2) 180 of each int angle = 135 ........ 18 o oSize M − = Bearing of from = 360 90 135 135 ......... .. 1o o o oH A A − − = AO2 2 25(b) 2 2 2 20.65 0.65 2(0.65) cos135 ........... 1 1.44250........... 1 1.2010 1.20 ......... 1 oBH M BH M BH km A = + − = = AO1 3 25(c) 2 1 of 0.65 1.2010 sin112.5........... 12 of 0.36061 0.361 ......... 1 Area BHG M Area BHG km A =    = AO2 2