7(a) Shown M1 for distance ÷ speed to find time, e.g. (1.496×1011) ÷ (3×108) (=498.666) M1 (dep) for conversion to hours, e.g. ÷ (60 × 60) “498.666” A1 0.1385185185… Explanation C1 Correct explanation, e.g. they have multiplied the indices rather than adding 8 = 3x 1 M1 For = 3x + OR a line drawn with gradient 3 passing through y y c A – A1 oe 9(a) Lauren P1 Process to find the value of one car at the end of one year, e.g. 13995 × 0.88 or 14495 × 0.87 £9537.20 Process to find the value of one car at the end of 3 years, e.g. 13995 × £9545 P1 (0.88)3 or 14495 × (0.87)3 Complete process to find the value of both cars at the end of 3 years,, e.g. P1 13995 × (0.88)3 and 14495 × (0.87)3 £9537.20(064) and £9545(.000985) and Lauren C1 (b) Explanation C1 Appropriate explanation, e.g. explanation that her car will be worth less
7(a) Shown M1 M1 A1 for distance ÷ speed to find time, e.g. (1.496×1011) ÷ (3×108) (=498.666) (dep) for conversion to hours, e.g. “498.666” ÷ (60 × 60) 0.1385185185… Explanation C1 Correct explanation, e.g. they have multiplied the indices rather than adding 8 y = 3x – 1 M1 A1 For y = 3x + c OR a line drawn with gradient 3 passing through A oe 9(a) Lauren £9537.20 £9545 P1 P1 P1 C1 Process to find the value of one car at the end of one year, e.g. 13995 × 0.88 or 14495 × 0.87 Process to find the value of one car at the end of 3 years, e.g. 13995 × (0.88)3 or 14495 × (0.87)3 Complete process to find the value of both cars at the end of 3 years,, e.g. 13995 × (0.88)3 and 14495 × (0.87)3 £9537.20(064) and £9545(.000985) and Lauren (b) Explanation C1 Appropriate explanation, e.g. explanation that her car will be worth less
GCSE%201MA1%20Mark%20Scheme%202H%20Mock%20Set%202.docx