Free practice · VCE Units 3 & 4
These are ten original VCE General Mathematics practice questions written to the Units 3 and 4 study design — data analysis, recursion and financial modelling, matrices, and networks. They are VCAA-style for practice, not real VCAA questions, and ATARMAxxing is not affiliated with the VCAA.
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A data set of monthly rainfall (mm) has a mean of 48 and a standard deviation of 6. Assuming an approximately normal distribution, between what two values do about 95% of the months lie?
1 mark for applying the 95% rule (mean plus or minus 2 standard deviations): 48 - 2(6) and 48 + 2(6). 1 mark for the correct interval: between 36 mm and 60 mm.
The number of hours studied (x) and the test score (y) for a group of students gives a least-squares line y = 12 + 4.5x. Interpret the slope of this line in context.
1 mark for identifying the slope as 4.5. 1 mark for a correct contextual interpretation: for each additional hour studied, the test score is predicted to increase, on average, by 4.5 marks. 'On average / predicted' is required for the mark.
A standardised test score has a mean of 70 and a standard deviation of 8. Calculate the standardised (z) score for a student who scored 86, and state what it tells you.
2 marks for the calculation: z = (86 - 70) / 8 = 16 / 8 = 2 (1 mark for correct substitution, 1 mark for the answer). 1 mark for interpretation: the student's score is 2 standard deviations above the mean.
A savings account earns interest under the recurrence relation V(n+1) = 1.004 x V(n), with V(0) = 5000 dollars, where n is the number of months. State the monthly interest rate and the annual nominal interest rate.
1 mark for recognising the multiplier 1.004 corresponds to a monthly rate of 0.4%. 2 marks for the annual nominal rate: 0.4% x 12 = 4.8% per annum. Award 1 of the 2 if the method is shown but arithmetic is wrong.
A loan is modelled by V(n+1) = 1.005 x V(n) - 600, with V(0) = 20000 dollars. Calculate the balance after the first two repayments.
1 mark for V(1): 1.005 x 20000 - 600 = 20100 - 600 = 19500. 1 mark for V(2): 1.005 x 19500 - 600 = 19597.5 - 600 = 18997.5. 1 mark for both correct, i.e. 19500 dollars after the first repayment and 18997.50 dollars after the second.
Two stores order stock described by the matrix product. Store orders are O = [3 5] (drinks, snacks) and unit prices are P as a column [2.50; 1.20]. Calculate the total cost O x P and state what it represents.
2 marks for the matrix multiplication: (3 x 2.50) + (5 x 1.20) = 7.50 + 6.00 = 13.50 (1 mark for correct method, 1 mark for the answer). 1 mark for interpretation: 13.50 dollars is the total cost of that store's order.
A 2 x 2 transition matrix describes customers switching between Brand A and Brand B each week. Explain what the column of the transition matrix headed 'A' must sum to, and why.
1 mark: each column must sum to 1 (or 100%). 1 mark for the reason: the column lists the proportions of current Brand A customers who move to each state next week, and every customer must go somewhere, so the proportions account for the whole group.
A connected graph has 6 vertices and 9 edges. Using Euler's formula for a connected planar graph, determine the number of faces.
1 mark for stating Euler's formula: v - e + f = 2. 1 mark for substitution: 6 - 9 + f = 2. 1 mark for solving: f = 5 faces. (Includes the outer/infinite face.)
Explain the difference between an Eulerian trail and a Hamiltonian path in a graph.
1 mark for Eulerian trail: a route that uses every edge of the graph exactly once. 1 mark for Hamiltonian path: a route that visits every vertex exactly once. The edge-versus-vertex distinction must be explicit.
A project has activities with durations on a network. The forward scan gives the project an earliest finish time of 18 days, and one activity has an earliest start time of 7 and a latest start time of 10. Calculate the float of that activity and state whether it is on the critical path.
2 marks for the float: float = latest start - earliest start = 10 - 7 = 3 days (1 mark method, 1 mark answer). 1 mark for the conclusion: because the float is greater than 0, the activity is not on the critical path.
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