KSS 2020 P1 (MS)

1 2020 EYE 4NA MA4045 P1 Marking Scheme 1a 3.45 Accept also 3.451612903, 14331 B1 1b 6.55 B1 2a 25 B1 2b 23 B1 3 Small size 150 $7.50 10 $0.50 ml ml   Large size 200 $10.20 10 $0.51 ml ml   The small size provides better value as it costs $0.50 per 10 ml. A1(w orkin g to justif y answ er) B1 4a 5 2( 1) 7 2 n n T n T n      B1 4b 7 2 27 2 34 17 p p p     M1 A1 5 104% $4160 100% $4000   M1 A1 6a 12 12 4 2 16 2 2 12 4 3 a a a a     B1 6b 5 3 2 5 2 3 5 7 14 7 14 2 p q q p q q p q     B1 7 Time = 34 min Distance

2 3424 60 13.6 km    M1 for findin g distan ce A1 8 ky x When x = 4, y = 2 2 4 8 8 k k y x    When x = 1/2, 8 12 16 y y   M1 A1 9a 7 300000 1 210 0000 21 1 M cm km A    or 1 300000 3 1 7 7 3 21 1 cm cm km M cm km km A      M1 A1 9b 2 2 2 2 1 9 2 18 cm km cm km   Area of lake is 2 cm2. M1 A1 10a 2 2 2 64 49 (8 ) (7) (8 7)(8 7) a a a a       B1 10b 5 10 2 5 ( 2) ( 2) ( 2)(5 ) my m ny n m y n y y m n           M1 A1 11a P(ball is red)

3 10 25 2 5   B1 11b P(ball is not green) 18 25 B1 11c P(ball is black) = 0 B1 12a Surface area 2(5)(12) (5) 85       M1 A1 12b 2 2 25 12 10.9 h h    B1 13 Sum of interior angles (6 2) 180 720       160 2 120 10 40 100 720 5 550 110 x x x x x x            M1 M1 A1 14a 1 50 1002 4 v v     M1 A1 14b Deceleration 4 0 50 20 2 15    M1 A1 14c Speed at 12 s is 4 1220 2.4    Distance 1 2.4 122 14.4     M1 A1

4 15 2 2 2 2 2 Area of sector 150 (6)360 15 Area of triangle 1 6 sin1502 9 Area of shaded area = 15 9 cm cm cm             M1 for area of sector M1 for area of triang le A1 16a The total number of students in Nora’s class and Eunice’s class are different so it is not accurate to compare using the size of each sector. B1 16b 135 9 91 135 9360 360 24135         M1 A1 17a 180 128 (interior angles, parallel lines) = 52 x  B1 17b 44 128 (exterior angle of triangle) 84 y y    B1 18a 6 2 8 3 4 x x       Possible values of x are -2, -1, 0, 1, 2, 3 and 4 M1 A1 18b 7 26 26 7 537 x x x     B1

5 19 21 1E y x B  11A y Bx 221C y Bx 2 1xD y B B1 B1 B1 B1 for each graph 20a 16.25%16 B1 20b median 162 1632162.5 B1 21a 22 2228 218 88 212 2( 4) 54, 5x xx xxp q                    M1, A1 or B1, B1

6 21b 2 2 2 8 21 9 ( 4) 5 9 ( 4) 4 4 4 2 4