14 a b c d With a reduction of 15%, the sale price (B) is A × 0.85. A = Yes, the new value will always be the original value multiplied by a percentage, calculated from the percentage change. For a reduction, the multiplier is (100 – the percentage reduction)%, for an increase it is (100 + the percentage increase)%. Percentage change problem, for example: The cost of a new car was £A. In the new financial year, it increased by 5% to £B. Write a formula to describe the proportional change. B = A × 1.05 and A = . M1 P1 C1 C1 2 M1 for correct formula P1 for correct rearrangement of ÷ by 0.85 C1 for clear explanation C1 for clarity of communication of question M 4 15 a b c A × 1.5 × 1.5 = A × 1.52 = A × 2.25 80% discount gives a price of A × 0.20. 60% followed by 20% gives a price of A × 0.4 × 0.8 = A × 0.32. A × 0.75 × 1.20 = 0.9A A × 1.20 × 0.75 =0.9A No, an increase to A of 50% followed by another increase of 50% gives 2.25A. Doubling would give 2A and 2A ≠ 2.25A. An 80% discount off the price of A gives a new price of 0.2A. A 60% discount off the price of A, followed by a further 20% discount, gives a new price of 0.32A so the 80% discount is better value. If the original cost is A, the cost after a discount of 25% is 0.75A and paying VAT at 20% gives a new price of 0.9A. If VAT is added first, the price is 1.2A. A 25% reduction gives a new price of 0.9A. Because multiplication is commutative, the final prices are the same. It makes no difference. C1 P1 C1 2 C1 for clear explanation with calculated justification oe P1 for clear explanation with calculated justification oe C1 for clear explanation with calculated justification oe M 3
29 B × 0.8n < Divide both sides by B. 0.8n < 0.83 = 0.512 0.84 = 0.4096 OR £100 × 0.8 =£80 £80 × 0.8 = £64 £64 × 0.8 = £51.20 £51.20 × 0.8 = £40.96 4 weeks P1 M1 P1 A1 2 P1 for choosing a starting a position, either a variable such as B or a specific amount such as £100 M1 for working through the weeks in some way P1 for the process of finding amounts for weeks 3 and 4 to show the point at which the bank account first dips below 50% of the original balance cao H 4 30 graph d graph e graph b graph c graph f graph a B6 2 B1 for each correctly identified graph with reference to why, for example: f(x) x2 is graph d as points are (–2, 4), (–1, 1), (0, 0), (1, 1), (2, 4) and it is a parabola f(x) = 2x , x >0 f(x) = –2x , x < 0 is graph e as it is linear and has no negative f(x) values; the gradient is 2 and –2 H 6 31 a b c y = xy = k where k is the constant of proportionality Inverse proportion describes the relationship between two variables such that as one increases the other decreases. xy = k or y = Own problem, for example: It takes 5 men 10 days to dig a hole. The number of men, y, is inversely proportional to the number of days, x. How long would it take for ten men to dig the same hole? (5 days) C1 M1 C1 2 C1 for clear explanation of inverse proportion M1 for correct equation C1 for clear question H 3
32 a b c r = 6 × 103 m F1 = F2 = = 0.996 Fg = 0.996 to 3 dp The difference is too small (reference part b). M1 B1 A1 C1 2 M1 for correct function B1 for calculation of F1 and F2 A1 cao C1 for correct interpretation of a scale factor close to 1 H 4 33 The speed of the faster car is 40 mph. T = = So they meet after 30 minutes. Speeds are in the ratio 1 : 2 = 20 : 40 = 10 : 20 So the cars meet when the slower car has travelled 10 miles and the faster car has travelled 20 miles. It will take half an hour for a car travelling at 20 mph to go a distance of 10 miles. M1 C1 A1 2 M1 for recognising and using the ratio of the speeds C1 for clarity of reasoning and explanation, diagram oe A1 cao H 3
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
41 i ii iii F(x) = a(b)x b < 1 the population decreases. b = 1 the population stays the same. b > 1 the population increases. C3 2 C1 for each correct explanation of the impact on the population as b varies H 3 42 a b Epidemic started by a single carrier so x0 = 1. Considering infection after 10 days so t = 10. xn + 1 = Rtx0 x10 = R10 Newspaper headline to engage readers with the story of this epidemic e.g. how long before x people are infected. M1 A1 C1 2 M1 for correct interpretation of , and use of, the iterative formula A1 for cao C1 for relevant , informative headline H 3 43 0.43 bar M1 A1 2 M1 for correct use of formula A1 cao H 2 44 a b x = 1 + x(x – 3) = x – 3 + 11 x2 – 3x = x + 8 x2 – 4x – 8 = 0 if x1 = 5 x2 = 1 + = 1 + = 6.5 x3 = 1 + = 4.14286 … Show that… as workings. x = –1.46 to 2 dp M1 C1 M1 B1 A1 3 M1 for algebraic manipulation C1 for clarity of justification M1 for correct use of iteration B1 for substitutions A1 for one root found (no credit if both roots are found) Using x = 5 as the first iteration, after 19 iterations you arrive at x = –1.46 to 2 dp Likewise, if the first trial is –1, 11 iterations lead to the solution x = –1.46 to 2 dp. H 5