39 a M1 2 M1 for appropriate iterations to find 2A H × 1.04n = 2A A Divide both sides by A. 1.04n 2 ≥ 1.0410 = 1.48 (2 dp) 1.0415 = 1.80 (2 dp) 1.0420 = 2.19 (2 dp) 1.0417 = 1.95 (2 dp) 18 years A1 A1 cao 1.0418 = 2.03 (2 dp) b M1 M1 for appropriate iterations to find number of bounces 0.6n = 0.1 0.62 = 0.36 0.65 = 0.077 76 0.64 = 0.1296 4 bounces A1 A1 cao c Own problem C1 C1 for clarity, relevance and accuracy of own question 5 40 a 26 = 64 M1 2 M1 for correct iterations H f(x) = a(b)x The population doubles each day. A1 A1 cao Day Number of bacteria 1 2 4 8 16 32 64 = 26 b C3 and are constants. C3: one mark for each explanation of and a b a, b x is the starting size of the population a and so change. doesn’t is the multiplier (by how much the b population grows each day) and the value of this change. doesn’t is a variable as it represents the x changing number of days. 5
39 a b c A × 1.04n = 2A Divide both sides by A. 1.04n ≥ 2 1.0410 = 1.48 (2 dp) 1.0415 = 1.80 (2 dp) 1.0420 = 2.19 (2 dp) 1.0417 = 1.95 (2 dp) 1.0418 = 2.03 (2 dp) 0.6n = 0.1 0.62 = 0.36 0.65 = 0.077 76 0.64 = 0.1296 18 years 4 bounces Own problem M1 A1 M1 A1 C1 2 M1 for appropriate iterations to find 2A A1 cao M1 for appropriate iterations to find number of bounces A1 cao C1 for clarity, relevance and accuracy of own question H 5 40 a b f(x) = a(b)x Day Number of bacteria 1 2 4 8 16 32 64 = 26 26 = 64 The population doubles each day. a and b are constants. a is the starting size of the population and so doesn’t change. b is the multiplier (by how much the population grows each day) and the value of this doesn’t change. x is a variable as it represents the changing number of days. M1 A1 C3 2 M1 for correct iterations A1 cao C3: one mark for each explanation of a, b and x H 5
Higher_Skills_Ratio_MarkScheme_Edexcel.docx