Beatty Sec EM 4E5N P1 2020 MS

1 Mark Scheme for Paper 1 1 2 3 4x y   4 6 8x y  4 2x y  Subtracting, 5y = 10, so y = 2 Then x = –1 M1 accept substitution A1, A1 2  3 2 1 8x x x   2 3 2 1 0x x     3 1 1 0x x   Hence x = 1/3 or x = –1 M1 A1 3 6u = 180° u = 30° (exterior angle) So n = 360 30  12 M1 A1 4(a) Either 1. Yes does not indicate response is to choose petrol or milk. 2. No scale is provided for the bars to compared against the given percentages. B1 4(b) The values on the y-axis have been inverted, causing the graph to be inverted as well. Hence the graph now shows the opposite trend. B1 5 10 1000 1 2000100 r     10 1 2100 r     1 101 2100 r  1 10100 2 1r        7.18% (3 sf) M1 A1 6(a) 3 10 5 6ap by apy b   = 3 5 10 6ap apy by b   =    3 5 2 5 3ap y b y   =    3 5 2 3 5ap y b y   =   2 3 5ap b y  M1 A1 6(b) 2 2 9 2 2 x y =   2 21 92 x y =   1 3 32 x y x y  M1 A1

2 7 Map 1: 1 : 25000 = 1 cm : 0.25 km Area ratio = 1 cm2 : 0.0625 km2 30 cm2 : 1.875 km2 Map 2: 2 cm : 3 km = 1 cm : 1.5 km Area ratio = 1 cm2 : 2.25 km2 5/6 cm2 : 1.875 km2 Accept 0.833 cm2 (3 sf) M1 M1 A1 8(a) 3 ky x , where k is any non-zero real number. B1 8(b)(i) Sub p = 27 and q = 9 into p k q , 27 9 k  27 9 9 k   Hence 9p q When x = 2, 3 125 2y   515 8 or 15.625 M1 A1 8(b)(ii) Sub p = 81 into 9p q , 81 9 q  81 99q   29q  81 B1 9 2 15 5 15 3 15 115u u u      10 45 115u  10 70u 7u Hence longest part = 5(7) + 15 = 50 m 50 100%115  1143 %23 or 43.5% (3 sf) M1 A1 M1, A1 10(a) 2 21 3x x  = 2 2 21 21 32 2x            = 2 1 110 1072 4x     = 2 1 110 1072 4 x      M1 A1 10(b) 2 21 3x x  = 0  2 1 110 1072 4 x      = 0 1 110 1072 4 x   1 110 1072 4x  = 0.14 or 20.86 (2 dp) M1 (/) A1 (both correct)

3 11(a) There are 72 data, so the median terms are the 36th and 37th terms. Median is in the interval 20 30x  . B1 11(b) Mean = 2010 72  27.9 years old SD = 2 69400 2010 72 72      13.6 years old (3 sf) No working is necessary. B1 B1 12(a) B1 12(b) ' 'X Y or   'X Y B1 12(c) A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, 41, 43, 47} B = {1, 2, 5, 10, 25, 50} 'A B  {3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47} Hence  ' 13n A B  and    n B n A B1, B1 13. X : (X + Y + Z) = 13 : 33 = 1 : 27 X : (X + Y) = 13 : 23 = 1 : 8 Hence Y = 7 units Then k = 7 27 M1 (either ratio) A1 14(a) 3 1 3 2 6 3q p p q            = 3 2 3 6 3 q p q p      = 3 3 2 6 3q p p q      = 2 11 648q p M1 (either) A1 14(b) 2 3 10 1na a a a a     6 10na a  6 10n  16n B1 (working should be seen) 14(c) 38 64x       1 3 6 32 2 x  3 22 2x  3 2x 2 3x B1 (working should be seen)

4 15(a)  35 13 124c    13 12 25p   13 12 1r    1 12 11s   11 12 23t    B2 (all correct) B1 – (two correct) B0 – (all wrong, or only one correct) 15(b)   25 1 12nT n    37 12 n  B1 15(c) 8 terms They are 11, 23, 35, 47, 59, 71, 83, 95 B1 16(a) x y G1 – correct shape and TP G1 – y –intercept correct 16(b) Sub (0, 600):   0600 a b a  Sub 274, 18932      :   427189 60032 b 4 81 256b  1 481 256b      3 4 B1 B1 17(a) 2 10 = 1 5 B1 17(b) Choose MALAYSIA COVID VIRUS + ITALY Consonants = 9, vowels = 6, P(vowel) = 6 2 4 15 5 9  COVID VIRUS + MALAYSIA Consonants = 10 vowels = 8, P(vowel) = 8 4 18 9 COVID VIRUS + SINGAPORE Consonants = 11, vowels = 8, P(vowel) = 8 4 19 9 M1, A1 (–1, 3) 2 0.732 –2.73

5 18   2 2 50 50100r h h      2 2 50 50 1250100r   1250r  35.4 cm (3 sf) M1 A1 19 To find A, set x = 0: y = 4 Hence A(0 , 4) To find B, set y = 0: 40 4 3 x  3x Hence B(3, 0) AB = 2 23 4 5  Hence cos = 3 5 B1 B1 B1 20(a) OQ = OP (radii of circle) (S) Angle QOR=angle POS (vertically opposite angles are equal)(A) OR = OS (radii of circle) (S) Hence by SAS, triangle OQR is congruent to triangle OPS. (also accept ASA with correct reasons) M1 (any one) A1 (with correct conclusion) 20(b) 2 1018r  1018r   18.001 m (5 sf)   2180 100 18.001360    2226 9 m2 or 226 m2 (3 sf) Alternatively, 360° – 1018 80° – 226 m2 (3 sf) M1 A1 21(a)      22 3 4 2 10k       2 3 36 100k   ,   2 3 64k  3 64 8k   11k  (reject) or –5 M1 A1 21(b) Gradient PQ =   7 1 3 6 4 5    Gradient QR = 1 5 3 3 5      Let R(x, 0) and set gradient QR = 0 7 5 6 3x    21 5 6 5 30x x     5 51x 110 5x Hence R 110 , 05      M1 M1 A1

6 22(a) 8 6.41.25 k = 6 + 6.4 = 12.4 s B1 22(b) Total dist =      1 14 8 2 8 6.4 82 2    = 16 + 16 + 25.6 = 57.6 m Average speed = 57.6 12.4  204 31 m/s or 4.65 m/s (3 sf) M1 A1 22(c) G1 – points are correct G1 – all shapes are correct 23(a) Correct A – B1 Correct B – B1 Arcs at A and D must be seen, else minus 1 mark. Lines AB, AD and DC must be drawn, else minus 1 mark. 23(b) 244° (accept 243° and 245°) B1 23(c) B1 – correct construction 23(d) B1 – correct construction 23(e) B1 – correct region shaded (k, 57.6) (6, 32) (4, 16)