Note: The sample answers provided are illustrative of one possible approach to answering the particular question. Students may adopt different but equally valid approaches and should be encouraged to compare their approaches when reviewing the mock exams. In all questions the focus should be on the skills and knowledge required to answer the question and the marking should focus on the extent to which these have been demonstrated in the student s answer. LCHL Paper 1 Q2 (25 marks) A tyre has pressure P = 300 kpa, but it has a slow leak and loses 5% of its pressure every hour. (a) What will its pressure be after ten hours? (b) Find a formula for P after t hours. (c) After how long will P = 150 kpa? Give your answer in hours and minutes, to the nearest minute. Skills/Knowledge being assessed: Recognises this as a geometric sequence and can find P using the generalised formula or by completing a table. Can generalise the sequence. Solving problems using the rules of logarithms. (a) 300(0.95) 10 = 179.62 kpa [0, 3, 7, 10] (b) P = 300(1.05) t = 300(0.95) t [0, 2, 5] (c) 150 kpa = 300(0.95) t 0.5 = Log0.5 = t log 0.95 13 hours 31 minutes. [0, 3, 7, 10] 1
LCHL Q5 (50 marks) Question 5 Students were investigating how a spring behaved when it was compressed by placing different masses on top of it. They measured the length of a spring with no mass on top of it; then compressed the spring by placing different masses on top of it. They recorded the length each time. Here are the values they obtained. Mass (g) Length of Spring (cm) 0 21.5 50 20.2 100 19.4 200 16.8 400 13.2 600 8.8 800 5 (a) Analyse this data and arrive at a mathematical model that explains the relationship between the mass placed on the spring and the length of the spring. Conclude with a formula that expresses Length of spring (cm) as a function of Mass (g). (b) Use your formula to predict the length of the spring when it is compressed by placing a mass of 550g on top of it. Skills/Knowledge being assessed: Ability to apply their knowledge of functions/linear relationships to solve problems in unfamiliar contexts. Identifies patterns. Uses equation of line to arrive at a generalised formula. Makes predictions based on the line of best fit. Student does not have to draw a graph, but it is useful to establish a linear relationship Student must establish that relationship is linear. Line of best fit drawn by eye. [0, 5, 10, 15, 20] 2
Selects any two data points say (800, 5) and (0, 21.5) and calculates slope: = Uses y y 1 = m(x x 1 ) To get y = - 0.02x +21 or uses y = mx + c to get y = -0.02x +21.5, where y = length and x = mass. [0, 3, 7, 10] (b) Using formula L = 10.5 cm [0, 2, 5] LCOL Q3 (25 marks) The graph below relates to a bike journey taken by Adrian. (a) When was Adrian cycling fastest? Explain your answer. (b) Adrian fell off his bike. How long into the journey do you think this happened? Explain your answer. (c) Describe the whole journey in words. (d) What was Adrian s average speed for the whole journey? (e) Sketch how the graph might look if his average speed was 6km/hour. 3
Skills/Knowledge being assessed: Understanding of the concept of slope as related to the concept of distance time in this context. Describing in words the story of the graph. Ability to read from the graph to get the relevant values. Knowledge of relationship between distance speed and time. (a) Adrian was cycling fastest in the last 45 minutes of his journey. He covered a distance the most distance in the shortest time in this section of the journey. [0,2,5] Or The last section of the graph is the steepest. (b) This happened between half an hour and an hour and a half He covered no distance at this time, the line has no slope; or student may choose after 4 hours, for fifteen minutes. [0,2,5] (c) Adrian covered a distance of 6 km in the first half hour. Then he fell off his bike and had to stop for an hour. He then began cycling again and covered 2 km in 1.5 hours. After 3.5hrs his speed increased and he covered 6 km in half an hour. He took a break for 15mins and continued home. [0,2,5] (d) Total distance: 6 + 2 + 6 + 11 = 25 km Total time: 5 hours Average speed = 25 5 = 5 km / hr. [0,2,5] (e) There are many possibilities here, here are three possibilities: [0,2,5] 4
LCOL Q5 (50 marks) The manager of a group called Girlzone has presented them with a choice between the following two contracts for their first CD recording. Contract A: The recording company will pay them 1.50 per CD, but they must pay production costs of 50,000 from their earnings. Contract B: The recording company will pay them 0.50 per CD but they do not have to pay any production costs. (a) How many CDs will the group need to sell in order to make the same profit from each contract? (b) The group was optimistic about sales of their CD so they chose Contract A. Two weeks after it had been on sale they received this note: Congratulations! Sales of your CD have covered the production costs and have left you with a net profit of 1000. Your account will be credited with 1000 today. How many CDs were sold in the first two weeks? 5
(c) After the success of their first CD, the group is offered the choice of two new options for a better contract. One option includes a signing bonus (a bonus given before any CDs are sold). The graph below represents net earnings under both options (i) Which option plan is better if 50,000 is the projected figure for CD sales? (ii) How much is the signing bonus? Explain how you arrived at your answer. (iii) What is the band s profit per CD under each option, not counting the signing bonus? Explain how you arrived at your answer. Skills/knowledge being assessed: Ability to choose a suitable strategy to solve the problem: graphical, numeric, algebraic or mental. Understands the significance of start value on the graph. Understands the meaning of slope, step value in this context. (a) 50,000 CDs (b) Student could use an equation: 1.5x 50,000 = 1,000 1.5x = 51,000 X = 34,000 CD s (c) (i) Lots of acceptable strategies could be adopted here Algebraic: Payment C: y = 2x y = 2(50,000) = 100,000 Payment D: y = 1.5x y = 1.5(50,000) y = 75,000 Payment plan C is the better option. Taking data points on graph: In payment plan C the band get 2 for every CD sold. If they sell 50,000 then they will get 100,000. In payment plan D they get one a half times the sales (ignoring signing fee) so they would get 75,000. Therefore payment plan C is the better option. Analysing graph: Payment plan C is the better option, as it overtakes payment plan D after the intersection, therefore will always earn more than D. 6
(c)(ii) 50, because 50 is the start value for Option D. Option C has no signing bonus. 7