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