2024 GSS 4E Math Prelim P1 Marking Scheme FINAL

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 1 2024 GSS Sec 4E5N Mathematics Preliminary Examination Paper 1 Item Worked Solutions Marks Awarded Remarks 1(a) 9.52 (3 s.f.) B1 Accept more exact answers. 1(b) $24.49 B1 2(a) 912 = 27x (32)12 = (33)x 324 = 33x 24 = 3x x = 8 M1 A1 Express all in powers of 3. 2(b) 4 4 6500 5000 1 100 65001 100 5000 0.06778997100 6.78% (3 s.f.) p p p p      M1 A1 3(a) Brand C’s and Brand D’s sectors add up to 60% but is shown as half of the pie chart (which should be 50%). OR The total add up to 110% instead of 100%. B1 3(b) Recalculate/Check the percentages for Brand C and Brand D so that the sectors of the pie chart should be proportional to the actual percentage. B1 Accept “Recalculate all values to get the correct percentages.” 4 (2x + 1)(3x – 2) B2 M1 for multiplication frame or B1 for each correct factor 5 23 2 2 1 2(2 1) 3( 2) ( 2)(2 1) 4 2 3 6 ( 2)(2 1) 8 ( 2)(2 1) xx xx xx xx xx x xx            M1 A1 M1 for combining fractions. 6 12,13,14, 16, 16, 19 B2 B1 for 14, 16, 16 in correct places. B1 for 12, 13, 19 in

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 2 correct places. 7 2 3 5 45 5(2 3) 4(5 ) 10 15 20 4 14 35 35 14 2.5 xx xx xx x x x            M1 A1 M1 for multiplying 20 on both sides 8(a) 240180 9   rad B1 B1 for correct answer 8(b) 23 km/h = 23 1000 60 60   = 115 7 6 6.3918 18 m/s B1 B1 for correct answer 9 Total surface area 22 2 1 4 (6) (6)2 72 36 108 339 cm (3 s.f.)          M1 A1 M1 for hemisphere + circle 10 2 2 2 ( 1) (2 ) (2 ) 1 2 1 bccb a ac ab b c ab b ac c b a ac c ac cb a ac cb a               M1 M1 A1 M1 for cross- multiplication. M1 for isolating b. Accept answers with – sign in numerator. 11 162 + 632 = 4225 = 652 By the converse of Pythagoras’ Theorem, triangle ABC is a right-angled triangle. A, B and C are also points on a circle by angle in a semicircle property. Yes, A, B and C lie on the circumference of a circle. M1 A1 A1 M1 for showing P.T. A1 for P.T. A1 for circle property 12(a) ab2(ab2 – 1) B1 12(b)     2 22 2 4 3 42 + 42 9 12 8 4 52 – 0 xxxx xx xx x        M2 A1 M1 for 29 12 4xx M1 for 284xx 13(a) Area of hexagon = 16 7 7 sin(60 )2     M1

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 3 = 127.3057344 cm2 = 127 cm2 (3 s.f.) A1 13(b) (15 2) 180 15615   B1 14(a) 3 33 33 3 original new 8 2 2Percentage change 100 100% y k x y k x k x k x k x kx     M1 A1 14(b) 6 men takes 50 hours to paint a mural. 4 men takes 6 50 75 hours4 to paint the same mural. M1 A1 15(a) P(yellow) = 1 0.2 0.1 0.352   (shown) B1 15(b) Total number of counters 1 140.35 40   M1 A1 15(c) P(yellow) = 14 14 40 3 37 B1 16 Reflex Angle AOC = 100 2 200   (Angle at centre is twice the angle at circumference) Obtuse angle AOC = 360 200 160   (Angles at a point) Angle OAC = 180 160 102    (Angles of an isosceles triangle OAC) M1 M1 M1 A1 M1 awarded with correct reasoning 17(a) 2+4(n – 1) = 4n – 2 B1 17(b) 4n – 2 = 82 4n = 84 n = 21 B1 17(c) If 4n – 2 = 360, n = 90.5 which is not an integer, so 360 is not a term in the sequence. B1 17(d) 8th term = 22[ 4(8) 54(8)] [ 4(7) 54(7)]     = 176 – 182 = –6 M1 A1 M1 for subtraction

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 4 18(a) x2 – 4x + 5 = (x – 2)2 + 1 a = 2 b = 1 B1 B1 18(b) x = 2 B1 18(c) B1 B1 B1 Correct shape Correct turning point Correct y-intercept 19 5 2 111 (2) 30 (1) (2) 2 (1) : 3 51 17 Sub into (1) : 13 xy xy x x y           Amount of money Siti has = 17 5 $85 B1 B1 M1 A1 Forming correct equations. Solving. 20(a) 7.75 B1 20(b) 4.18 (3 s.f.) B1 20(c) The mean would be increased by 3. The standard deviation will remain the same. B1 B1 21(a) 232 5 11 B1 21(b) p = 2 q = 11 B1 B1 21(c) LCM of 50, 60 and 75 = 300 min = 5 hours They will meet again at 11 am. M1 A1 M1 for LCM 22(a) Bearing of B from A = 180 + 040 = 220o M1 A1 22(b) Bearing of C from A = 180 – (65 – 040) = 155o M1 A1 y = x2 – 4x + 5

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 5 23(a) 1100 1000 1200 1400 1200 1300   B1 23(b) 110 80 x    B1 23(c) 1100 1000 1200 1400 1200 1300   110 80 x    201000 1200 250000 1300 x x   M1 A1 23(d) The elements represent the amount of ticket sales for each day (Saturday, Sunday). B1 23(e) 201000 1200 250000 1300 688500 2500 323000 2500 237500 95 xx x x x        M1 A1 24(a) Area of triangle ABC = 0.5 × 4 × 7 = 14 unit2 B1 24(b) (6, 6) B1 24(c) Area of parallelogram = 14 × 2 = 28 unit2 B1 24(d) 1 4Angle tan 8 26.6 (1 d.p.) BAC     B1 24(e) undefined B1 24(f) 1 p B1 Using cosine rule to get 2 31 14 p p  also accepted.

Marking Scheme For 2024 GSS Sec 4E5N Mathematics Preliminary Examinations Paper 1 6 25(a) B1 Bisector constructed accurately with construction arcs 25(b) B1 Perpendicular bisector constructed accurately with construction arcs on both sides of AB 25(c) B1 Correct region shaded 25(d) B1 Correct position of T