2023 GSS 3E EOY Math Express P1 Marking Scheme

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 Item Worked Solutions Marks Awarded Remarks 1 53 3 158 14%6 −7.8 B2 1m – at least 2 consecutive correct order 2 100 PRTI     99800 2.58 5 100 $12874.20   Total value 99800 12874.20 $112674.20   M1 A1 Correct simple interest working Total value (overall minus 1m if $112674.2 given) 3 46 0.07035 22.34 3.1 4.45 50 0.07 20 3 4 3.5 14 0.25       M1 A1 for at least 3 number corrected to 1 sig. fig. 4a 330 180 5.76 rad   B1 Accepts 51 6 or 11 6  4b 5 180 8 112.5    B1 5   22 22 2 (8 3) ( 6) 34 (8 3) ( 6) 34 6 34 25 36 39 n n n n or               M1 A2 Apply length of line formula correctly Both values of n found

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 6 38.5 141.5or    (1 d.p.) A2 Both values found 7 1 2 2 112 177 49 7 7 7 77 11 2 p p p p       M1 A1 Award M1 if either 1 2 277 or seen Accepts 3 2 8 Total Amount 7 3.136850 1 100 $8499.349649   Interest $8499.349649 $6850 $1649.35   M1 M1 A1 A1 awarded only if corrected to nearest cent 9a 3yx B1 9b 5xy B1 9c 2 2y x B1 10 2 2 2( 5) 9( 5) 0 5 2 9 5 0 ( 5)(2 1) 0 5 (not applicable) 1 2 a a xx xx x or x               M1 M1 A1 Sub x = −5 M1 awarded only if a value = 5 found. Factorised form or use of general formula Award A1 only if 1 2x seen No A1 awarded if 5x is not rejected OR Both answers written on answer line 11a 2 15 19x 2 34x 17x B1 No B1 if only 17 or other form of inequalities seen

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 11b No, because there are 7 prime numbers less than or equal to 17 (2, 3, 5, 7, 11, 13 and 17). M1 A1 students must list all the 7 prime numbers. 12a 4 5 B1 12b 16sin 20 53.1 ABC ABC     M1 A1 13a 2 0 3 6 66 12 6( ) 6 6 a b a aa a      M1 A1 Award M1 if 66a is seen. No M1 if 606ab or any other factors multiplied to 66a 13b 1 15 3 1 3 15 5 125 125 5 a a a      M1 A1 Award M1 if 5 1 3125 a  seen Not accepted 55a 5 0.2 a 14 (A) (common angle) (S) 12 3 9 (given) (A) 90 (given) By ASA congruence test, triangle is congruent to triangle CAB SAR AB AR ABC ARS ABC ARS           M2 A1 M1 for any 2 statements Conclusion with correct test chosen. Award if ‘  ’ sign used 15   2 22 22 22 33 9 9 9 9 9 18 9 9( ) 18( ) 9(23) 18(17) 99 xy x xy xy y x xy y x y xy              M1 A1 16a ( 5)(2 ) Subs 0, 10 10 (0 5)(2(0) ) 2 y x x b xy b b         B1 16b (5, 0) B1

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 16c ( 5)(2 ) 0 51 x x b x or x       Line of symmetry is 5 ( 1) 22x  (2 5)(2(2) 2) 18y    M1 A1 Line of symmetry found 17a 80s to 90s B1 17b Speed = Gradient Speed Distance Time50 30 80 35 4 m/s or9 0.444 m/s      M1 A1 17c Average Speed Total Distance Total Time 50 50 120 5 m/s or 0.833 m/s6    M1 A1 18a G1 G1 Correct shape and y- intercept Correct turning point 18b   2 21x     2 2 3 2x   Draw the line y = 2. M1 A1 Show attempt to obtain y = 2 Conclusion Do not accept y x 0 (2, 3) x = 2 7

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 Since the line y = 2 does not intersect the graph of   2 12yx   , the equation has no real solution. “intercept the graph” or “points of interception” 19a 2 22 2 2 2 13 9 13 13 922 13 169 924 13 133 24 13 1 3324 xx x x x or x                          M1 A1 Accept if without negative in the square 2 13 2   Accept either answer in improper or mixed number 19b 2 2 2 13 9 0 13 133 024 13 133 24 13 133 24 12.3 0.734 (3 s.f.) xx x x x or            M1 M1 A1 Bringing of the term 133 4 to the right-hand side Taking square root on both sides Both answers found 20a 77 7 9 9 141 10 3.12 10 137.88 10 1.3788 10 or 1.38 10 (3 s.f)       M1 A1 Accept answers in 5 s.f. or 3 s.f 20b 91.41 10 3287263 428.93 429    M1 A1 21a 2211 12 265 B1 Accept if “=265” not included 21b n2 + (n + 1)2 or 2n2 + 2n + 1 B1

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 21c n = 64 or 2 2 2 2 2 1 8321 2 2 8320 0 4160 0 ( 65)( 64) 0 65( ) 64 nn nn nn nn n rej n               B1 21d For any two consecutive numbers, one number would be even and the other one odd. Since the square of an even number is even and the square of an odd number is odd, one of the squares of the two consecutive numbers will be even, and the other will be odd. Hence, their sum will be odd. B1 Accept if students explains: From Tn, the sum of an odd number and an even number is always odd. 22a G1 G1 Shape of quadrilateral ABCD drawn with construction marks at C and D. All lengths and angle ABC correctly measured 22bi G1 Perpendicular bisector constructed accurately

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 with construction arcs on both sides of AB 22bii G1 Bisector of angle constructed accurately with construction arcs 22c G1 P region indicated 23a When y = 8, 5(8) 4 20 0 5 (shown) p p     B1 p shown 23b Gradient of l2, 8 ( 2) 50 2    M1 A1 23c Length of AB 22(0 5) ( 2 8) 125      Area of triangle= 1 652 11 125 6 522 2.68 units d d       M1 M1 A1 Length of AB found Equate area of triangle 24a tan 32 12 7.4984 7.50 TP TP m    B1 24b 229 12 15SQ PR    Since angle of depression of R from T = angle of elevation of T from R Let  be the angle of elevation of T from R M1 M1 A1 Apply Pythagoras’ Theorem Apply Trigo ratio

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 tan 15 7.4984tan 15 26.560 26.6 (1 . ) TP dp         25a 2.2 11 5 cm OP    B1 25b Perimeter of the shaded region PQRS = 12+12+11+17(2.2) = 72.4 cm M1 A1 25c Area of the shaded region PQRS = Area of sector OQR – Area of sector OPS 22 2 11(17) 2.2 (5) 2.222 317.9 27.5 290.4 cm      M1 A1 Area of at least one sector found 26a Acceleration 22 14 0 24 0 7 m/s or 0.583 m/s12    B1 26b Distance travelled by the van in 30 seconds = Area under graph 1 [30 (30 24)](14)2 252 m     M1 A1 26c Let h be the speed of the bike at the instant when it overtakes the van 1 (30 12) 2522 28 m/s h h      Disagree, since the required speed at the instant to overtake is 28 m/s, 26 m/s is lower. M1 A1 Conclusion

Marking Scheme For 2023 Sec 3E Mathematics EOY Paper 1 Alternative: Disagree. If the speed at this instant is 26 m/s, distance travelled by bike = 1 (30 12) 262   = 234 m. However, this is lesser than the distance travelled by the van (252 m), so the bike would not have overtaken the van at this speed.