WGSS 2024 Prelim EMath 4052(2) MS

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 1 Q/N Solution Marks Remarks 1 (a) (i) 2 1 32 3 1 4 2 =+ − =+ − = ab ac b b B1 (ii) 2 1 2( 1) ( 1) ( 1) ( 1) 2 2 22 22 ( 2) 2 22 or 22 =+ − −=+ −− − = + − − = + − − − = − − − = − −−= − − − + ab ac ac ab a c c a bc a ac a abc bc ac a abc ac a bc a bc c bc bc bca bc c bc c M1 M1 A1 Combine fractions, common denominator Grouping of ‘a’ terms and factorising (b) 4 11− =−xy -------- (1) 5 3 1+ =−xy -------- (2) (1) 3 12 3 33− =−xy ------- (3) (2) + (3) 17 34 2 =− =− x x Sub 2=−x into (1) 4( 2) 11 11 8 3 − − =− − =− + = y y y M1 A1 A1 Elimination method Alternate solution : solving by substitution 4 11− =−xy -------- (1) 5 3 1+ =−xy -------- (2) From (1) 4 11 4 11 − =− =+ xy yx -------- (3) M1 Substitution method

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 2 Sub (3) into (2) 5 3(4 11) 1 5 12 33 1 7 34 2 + + =− + + =− =− =− xx xx x x Sub 2=−x into (3) 4( 2) 11 3 = − + = y y A1 A1 (c) 2 22 2 2 2 5 13 1 2 3 (2 3) 5(3 1) 1(3 1)(2 3) (3 1)(2 3) 2 3 15 5 1(3 1)(2 3) 2 3 15 5 6 9 2 3 4 19 8 0 4 2 19 (19) 4(4)( 8) 2(4) 19 489 8 0.38916 or 5.1391 0.389 or −=−+ +− −=− + − + + − + =−+ + − + = + − − + − = −  −= −  − −= −= =− = x xx x x x x x x x x x x xx x x x x x x xx b b acx a x x x x 5.139 (to 3 dp)− M1 M1 M1 A1 Combine fraction, common denominator 24 19 8 0+ − =xx 19 489 8 −=x Both answers 2 (a) Plan B. Plan B pays a higher interest amount as it is compounded yearly. Or the principal sum increases every year. B1 (b) Total simple interest 3.58000 12100 $3360 =   = Total amount = 8000 + 3360 = $11360 M1 A1 Interest = $3360

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 3 (c) 14 50000 1 65320100 1 1.01927100 1.92744 1.93% (to 3 sf) r r r r += += = = M1 M1 A1 M1: forming equations using compound interest M1: 1.01927 (d) Discounted price in THB 56000 0.85 47600 THB= Total cost in THB 47600 1.02 48552 THB= Cost in SGD 48552 27.16 1787.6288 $1787.63 (to nearest cent) = = = M1 M1 A1 47600 THB (after discount) and Multiply by 1.02 Divide by exchange rate 3 (a) 5 7 55 7 15 1 = + − = + − =− =− yx x y k B1 (b) Plotting of graph G1 G1 G1 G1: 0 – 4 points plotted correctly G1: all points plotted correctly G1: smooth curve

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 4 (c) 5 5 5 7 5 7 2 += + − = − =− x x x x y Intersection points 1.4 0.1 3.6 0.1 = = x x M1 A1 A1 M1: draw 2=−y (d) Drawing of suitable tangent line at ( )4, 1.8− Estimated gradient = 0.688 0.1 (3sf as its an estimate of the gradient) M1 A1 (e) 22 2 5 73 5 7 3 2 10 5 0 2 and 10 + − =− + + − =− + − + = = =− xx x x x x x xx PQ M1 A1 A1 Equating both equations

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 5 4 (a) (i) - Angle ACB equal to angle OCD (Common angle) (A) - Angle ABC = 90  (angle in semi circle) and Angle ODC = 90  (tangent to radius) (A) Hence using AA test, triangles ABC and ODC are similar.(AA) B1 B1 * minus 1 mark if student did not state the test used. (ii) 2 2 Area triangle 2 Area triangle OD 1 Area triangle 4 15 1 Area triangle 60 Trapezium = 60 15 45cm = = = −= ABC C ABC ABC ABDO M1 A1 M1: 60 (b) (i) angle QRS 156 782= =  (angle at centre is twice angle at circumference) B1 *minus maximum of 1 mark if no reasons are give for whole of Q4(b). But to circle and highlight to student importance of following question and give reasons to support answer. (ii) angle SRO 78 50 28 = −  = (Isosceles triangle) angle RSO 28= (Isosceles triangle) angle PSO 73 28 45 = −  = M1 A1 *also accept other correct methods of finding answer e.g. angles in opp segment (longer method)

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 6 (iii) angle PQR 180 73 107 = −  = (Angles in opposite segment) angle PQO 107 50 57 = −  = M1 A1 (iv) angle PQR + angle SRQ 107 78 185 = +  = angle PQR + angle SRQ is not equal to 180  , using the rule of interior angles in parallel lines, PQ is not parallel to SR. M1 A1 Add up both angles to get 185  5 (a) (i) Median = 74kg B1 (ii) Q3 = 78kg Q1= 70kg Interquartile range 78 70 8kg =− = M1 A1 M1: Q3 – Q1 (b) Yes I agree because the median of factory B is larger than the median of factory A. B1 Larger median for factory B (c) Interquartile range for factory B 92 78 14kg =− = Since the IQR for A is smaller than B, factory A is more consistent. M1 A1 IQR for factory B (d) Factory A, more than 80kg 400 340 60 =− = or 400 330 70 =− = Factory B, more than 80kg 400 140 260 =− = or 400 130 270 =− = M1 M1: demo understanding of finding number of steel bars more than 80kg

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 7 P(both more than 80kg) 60 260 400 400 39 400 = = **Also accept 81 91 189,,800 800 1600 M1 A1 M1: multiplication of probability 6 (a) (i) 53 48 3 5 8 8 4 4 64 16 80 8.94 (to 3sf) AB OB OA OA OA OA =− −   =−       +   ==   −    = + = = M1 A1 M1: 3 5 8 8 4 4OA +   ==   −    (ii) Gradient AP = Gradient PB 10 4 10 8 83 6 18 2 16 42 1 2 kk kk k k −− =−− − = − = = M1 A1 M1: equating the gradients Also accept AP PB= where  is a constant (b) (i) 4 7 4 78 OQ OR RQ OQ OQ =+ = + + =+ r p r pr B1 (ii)

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 8 7 8 3 48 PQ OQ OP PQ PQ =− = + − =+ p r p pr B1 (iii) 43 34 PR OR OP PR PR =− =− =− + rp pr 3 5 3 ( 3 4 )5 9 12 55 9 12 55 9 12 355 6 12 55 PB PR PB PB OB OP OB OB = =  − + =− + − =− + =− + + =+ pr pr pr p r p pr M1 M1 A1 M1: 34PR=− +pr M1: 9 12 55PB=− +pr (d) 6 12 6 ( 2 )5 5 5OB= + = +p r p r 78OQ=+ pr Points O, B and Q are not colinear because OB cannot be expressed as a scalar multiple of OQ. B1 7 (a) 2 2 2 2 2 2 2 2( )( ) cos 700 550 2(700)(550) cos115 1117916.062 1057.32m (shown) AC AB BC AB BC ABC AC AC AC =+− = + − = = M1 A1 Cosine rule

Woodgrove Secondary School - Mathematics Department 2024 4E5N O Level Prelim EM Paper 2 Marking Scheme Setter: Phillip Tan 9 (b) sin sin115 700 1057.315 sin 0.6000250 36.871 BCA BCA BCA = = = Bearing