GMSS 2024 4N Prelim EM P2 Marking Scheme

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 1 of 9 SECONDARY 4 NORMAL (ACADEMIC) MARKING SCHEME PRELIMINARY EXAMINATION 2024 ELEMENTARY MATHEMATICS PAPER 2 SECTION A Question Answer Marks Guidance 1(a)(i) 42.1 million = 4 21 000 00 = 42 100 thousand B1 1(a)(ii) 0.005 187 62 ≈ 0.0052 (4 decimal places) B1 1(b) 2 2 6.582 0.891 435.18 7 1 400 0.35    M1 A1 Total 4 Question Answer Marks Guidance 2(a) 5 2 5 7 49 7 7 7 7 2 5 3 a a a a         M1 A1 2(b) 3 4 4 3 4 4 3 4 4 4 3 16 16 2 8 x x x x                M1 A1 Total 4

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 2 of 9 Question Answer Marks Guidance 3(a) p = 40 – (3 + 7 + 12 + 8 + 5) p = 5 B1 3(b) 12 8 5p(more than 2) = 40 25 = 40 0.625    M1 A1 3(c) (0 3) (1 7) (2 5) (3 12) (4 8) (5 5)mean = 40 110 = 40 = 2.75            M1 A1 Total 5 Question Answer Marks Guidance 4(a) 15 units --- 180 1801 unit --- 15 = 12 2 units (exterior angle) ---- 12 2 = 24 360 24 = 15 n         M1 A1

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 3 of 9 4(b) 180 50angle   = 2 = 65 (isosceles trian gle) angle    = 360 230 = 130 (angles at a po int) angle = 180 130 20 = 30 (angles in sum of t CDE BAC ABC               riangle) angle 35 30 = 65 Conclusion: Since angle = 65 and angle C = 65 , by corresponding angles, line is parall el to line . ABD ABD DE AB CD         M1 M1 M1 M1 A1 Total 6 Question Answer Marks Guidance 5(a) For the first part of the journey, remaining distance = 300 - 180 = 120 km Time taken for remaining journey = 120 = 75 D s =1.6 h T otal DistanceAverage speed for entire journey = Total Time 300 = 1.8 1.6  88. 2 3529412 88.2 km/h (3 sig fig)   M1 M1 A1 Accept answer in fraction: 488 km/h17

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 4 of 9 5(b) 2 2 Total surface area = 2 (7.6)(2.5) 2 (7.6) = 482.2973042 482 cm (3 si g. fig)     M1 A1 Total 5 Question Answer Marks Guidance 6(a) 70% × $149 = $104.30 60% × $95 = $57 Total = $104.3 + $57 = $ 161.30 M1 M1 A1 6(b) Let x be the price of the 1st handbag paid after discount. x + (x − $25.80) = $184.20 2x = $210 x = $105 70% --- $105 $105100% 10070 = $150   M1 M1 A1 Total 6 Question Answer Marks Guidance 7(a)(i) 16.2 1 8.52 3.811764706 3.81 (3 sig.fig) h h     M1 A1 7(a)(ii) 2 2 If angle is a right-angle, then by PT, 9.2 8.5 3.519943181 3.52 (3 sig.fig) Since length of is not equal to 3.81 (a nswer in a(i)), is not the perpendicular height from t o . Thus, triang Z XZ XZ XZ X YZ     le is not a right-angled triangle. XYZ M1 A1

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 5 of 9 7(b) heighttan 35 80 height = tan 35 80 56.01660306 56.0 (3 sig.fig)        M1 A1 Total 6 Question Answer Marks Guidance 8(a) 2 2 2 5 3 7 5 3 7 0 ( 3) ( 3) 4(5)( 7) 2(5) 3 149 10 1.520655562 or 0.920655616 1.52 or 0.92 x x x x x x x x x x                 M1 M1 A1 8(b) 2 6 31 2(8 12 ) 6 31 16 24 6 31 30 15 0.5 when 0.5, 8 12( 0.5) 14 x y y y y y y y y y x x                 M1 A1 A1 8(c) Amy : x years old Mr Pang: 3x years old 15 years time Amy: x + 15 Mr Pang: 3x + 15 3 15 2( 15) 3 15 2 30 15 x x x x x        M1 M1 A1 Total 9

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 6 of 9 Question Answer Marks Guidance 9(a) h = −3 k = 5 B1 B1 9(b) Plotting all the points correctly. Smooth curve. B2 B1 Award B1 for only at most 1 incorrect plotted point. 9(c) Drawing the correct tangent at x = −0.5. Gradient = 4.5 ( 1)  M1 A1 Award A1 based on finding the gradient correctly with their own tangent 9(d) −3 < m < 5 (see the shaded region above, where the horizontal line will cut exactly at 3 points) B1, B1 B1 for correct min value B1 for correct max value Total 9 y = 2x 3− 6x +1 9(c) tangent line

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 7 of 9 Question Answer Marks Guidance 10(a)(i) Comfort Strides Flag down fare (Premium) $4.10 (take the average between $3.9-$4.3) $4.20 (take the average between $4.1 to $4.3) Distance Rate fare 12 km = 12 000m 12 000 /400 = 30 $0.35 ×30 = $10.50 $0.34 ×30 = $10.20 Metered Fare $4.10 + $10.5 = $14.60 $4.20 + $10.20 = $14.40 Hence, Strides is a cheaper option as compared to ComfortDelGro by $0.20. M1 M1 A1 Award M1 for taking the maximum value/minimum value as the flag-down fare (instead of taking average) Award M1 for metered fare based on the chosen flag down fare A1 for writing down the conclusion statement 10(a)(i) Booking Fee $3.30 Timed Based Surcharge 25% × $14.40 = $3.60 Total metered fare (MF + BF + TBS) $14.40 + $3.30 + $3.60 = $21.30 M1 A1 10(b) 10% × $25.40 = $2.54 9% GST = 9% × ($25.40 + $2.54) = $2.5146 Payment surcharges = $2.54 + $2.5146 = $5.0546 Total payment = $25.40 + $5.0546 = $30. 4546 (≈ $30.45) M1 M1 A1 Total 8

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 8 of 9 SECTION B Question Answer Marks Guidance 11(a)(i) 27 min B1 11(a)(ii) LQ (200) = 14 min UQ (600) ≈ 25.5 min IQR = 25.5 – 14 = 11.5 min B1 No marks for UQ to be exact at 25. Award B1 for their estimated UQ in the following range: 25 UQ 25.5  . 11(a)(iii) 800 – 240 = 560 B1 11(b)(i) 15 15 x B1 11(b)(ii) 2 15 15 15 1 (15 )( ) 210 15 210 x x x x x x      M1 A1 11(b)(iii) 15P(first ribbon gold) = 15 15 1P(second ribbon gold) = 15 1 14 = 14 14 4 14 7 14 8 6 x x x x x x             Total 8

Marking Scheme GMS(S)/4N/EMath/Prelim P2/2024 Page 9 of 9 SECTION B Question Answer Marks Guidance 12(a)(i) 70 angle 2 35 (angle at centre = 2 angle at circumference) BEC      B1 12(a)(ii) angle 31 (angle in same segment)ACE  B1 12(a)(iii) angle 180 70 = 110 (adjacent angles on a straight line) angle = 360 110 90 90 = 70 (2 external t angents from a point) BOA AGB               M1 A1 12(b)i) 2 2 2 1area of semi-circle = (6)2 = 18 cm 18area sector = 3 = 6 cm     M1 A1 12(b)ii) 21 (12) 62 72 = 6 0.2617993878 0.262 (3 sig.fig)         M1 A1 Total 8