North Vista 4E 2024 Prelim Paper 2 Solutions

1 North Vista Secondary Secondary 4 Express Mathematics (4052) 2024 Paper 2 Marking Scheme Qn Solutions Mark AO Total 1(a) 5 3 2 13 3 3 15 15 x x x −  −  −   −   3 3 and 3 15 1 and 5 15 xx xx x −   −   −   M1 A1 AO1 2 1(b) 1 5 --- (1)2 2 3 13 --- (2) xy xy += −= (1) 4 2 4 20 --- (3)xy  += (2) – (3) 77 1 y y − =− = When 1y= , ( )2 3 1 13 8 x x −= = 8, 1xy = = M1 A1 A1 AO1 3 1(c)(i) When 7r = and 15q=− , AO1 1

2 Qn Solutions Mark AO Total 3 3 41 7 15 4(7) 1 2 3 rqp r += − −= − =− B1 1(c)(ii) ( ) ( ) 3 3 3 33 33 33 3 3 41 41 41 4 4 41 41 rqp r rqp r p r r q p r p r q p r r p q r p p q pqr p += − += − − = + − = + − = + − = + += − M1 M1 A1 AO1 3 1(d) ( ) 2 15 321 2 1 ( 3) 15 2 5 18 0 (2 9)( 2) 0 1422 xx xx xx xx x or x =+− − + = + − = + − = =− = M1 M1 A1 AO1 3 2(a)(i) Total Food Waste Output AO1 1

3 Qn Solutions Mark AO Total 6 565 000 605 000  640 000 1.81 10 tonnes = + + = B1 2(a)(ii) Percentage increase (2009-2010) 99000 75000 10075000 32% −= = Food Waste Recycled in 2011 132 99000100 130680 tonnes = = M1 A1 AO1 2 2(a)(iii) Per capital food waste 565000 1000 4.84 1000000 365 0.319823 0.320 kg/day =  = = Per capital food waste 565000 1000 4.84 1000000 0.318949 kg y 366 0 d.3 /a19 =  = = M1 A1 AO1 2 2(a)(iv) Food Waste Output 509000 100100 8.6 556892.779 557000 tonnes = − = = M1 A1 AO1 2 2(b)(i) Perimeter of the lake on map A 1700 9612 25 cm = = M1 A1 AO1 2 2(b)(ii) Actual area of lake ( ) 2 2 36 68 166464 m = = M1 AO2 2

4 Qn Solutions Mark AO Total Area of lake on Map B 2 2 166464 51 64 cm = = A1 3(a) 2 = 120 4 120 4 2 2(60 2 ) 2 60 2 (shown) rx xr x x     − −= −= −= M1 A1 AO2 2

5 Qn Solutions Mark AO Total 3(b) ( ) 2 2 2 2 22 2 60 2 60 2 3600 240 4 (4 ) 240 3600 0 (shown) xx xx x x x xx      −=  −= =−+ − − + = M1 A1 AO2 2 3(c) ( ) ( ) ( )( ) 2 240 240 4 4 3600 2(4 ) 240 45238.93421 2(4 ) 263.68 or 15.90 x    − −  − − −= − = − = M1 M1 A1, A1 AO1 4 3(d) Reject 263.68x= because - the perimeter of the square (4x) must be less than the length of the wire - the radius of the circle cannot be negative B1 AO3 1 4(a)(i) Mean = 166 cm B1 AO1 1 4(a)(ii) Standard deviation ( ) 2331532 16612 8.47 cm =− = B1 AO1 1

6 Qn Solutions Mark AO Total 4(a)(iii) 1. Since the mean height of Group B (168 cm) is greater mean height of Group A (166 cm), the students in Group B are generally taller on average. 2. Since standard deviation of heights of Group B (9.5 cm) is greater standard deviation of heights of Group A (8.47 cm), the heights of students in Group B have a wider spread / less consistent. B1 B1 AO3 2 4(b)(i) 12 4 21 7= B1 AO1 1 4(b)(ii) 98 17 16 9 34  = M1 A1 AO1 2 4(b)(iii) 23 17 16 340 39 38 391 1235     = M1 A1 AO1 2 5(a) 4− B1 AO1 1

7 Qn Solutions Mark AO Total 5(b) P2 (9 points plotted correctly) P1 (7-8 points plotted correctly) C1 (Smooth curve through at least 7 points) Tolerate 1mm for plotting and drawing curve through points AO1 3 5(c) 5 or 5kk=− = B2 AO2 2 5(d) 32 32 32 6 2 16 0 31 404 2 2 31 44 2 2 x x x xx x xx x + − − = + − − = + − = Draw the line 1 2yx= . (Must label equation of line on graph) 5.85, 1.7,1.6 ( 0.1)x=− −  M1 M1 A1 AO2 3 5(e) 3.75 ( 0.5) M1 – tangent line A1 AO2 2 6(a)(i) 2 2 n nT = (or 4n nT = ) B1 AO2 1 3213 442y xx= +− 1 2y x=

8 Qn Solutions Mark AO Total 6(a)(ii) 31 4 n nR += (or 26 2 n nR += ) B1 AO2 1 6(a)(iii)(a) 31 2 2(3 1) 2 6 2 2 42 4 2 2 2 2 2 (shown) n n n n n nn n Q + + +− + = = = = B1 AO2 1 6(a)(iii)(b) 42 4 2 7 2 128 22 4 2 7 1.25 n n n n + + = = += = 128 is not a term of sequence nQ as n = 1.25 is not a positive integer. (or positive whole number) M1 A1 AO3 2 6(b)(i) 22 10 4 4 12 6 8 6 ( 8) 10 units AB AB    =−       =− = + − = or 22 22 (4 10) (12 4) 10 units (10 4) (4 12) 10 units AB OR AB = − + − = = − + − = M1 A1 AO1 2

9 Qn Solutions Mark AO Total 6(b)(ii) 1 2 46 1 12 8 2 34 4 12 1 16 BAAC OC OC = −   −=       −   =+       = (1,16)C M1 A1 AO1 2 7(a) 1 sin sin 95 30 38 30sin 95sin 38 51.856 180 95 51.856 33.144 33.1 ACB BAC ABC −  = =  =  = − − = = M1 M1 A1 AO2 3 7(b) Let the shortest distance from A to BC be h and greatest angle of depression be  . sin 33.144 30 30 sin 33.144 16.402 m h h = = = M1[ECF on ABC ] AO2 3

10 Qn Solutions Mark AO Total 2.5tan 16.402 8.6663 8.7 (1 d.p.)   = = = Note: Students who found the length of AC in 7(b) will be awarded M2 in 7c only if AC is used to calculate the speed of Ali, otherwise, M2 is not awarded for working to find AC in 7b M1 A1 7(c) 38 sin 33.144 sin 95 38sin 33.144 sin 59 20.855 m AC AC = = = 38Time taken by Ken = 4 9.5 s= 20.855Speed of Ali = 9.5 3 3.2085 m/s =3.21 m/s (3 s.f.) − = M1 [ECF on ABC ] M1 M1 A1 AO2 4 8(a) o Reflex 360 92( s at a point) 268 COA = −  = o 268 (angle at centre = 2 angle at circumferene)2 134 CDA= = M1 2