General Mathematics Exam 1: Fri 30 Oct, 2:00pm — 20 days away

ATARMAxxing · General Mathematics

VCE General Mathematics Practice Questions

64 exam-style questions · full worked solutions

The 64 practice questions inside the VCE General Mathematics Mastery Pack, grouped by area of study. Every question comes with a full worked solution.

  1. AOS132 questions · 73 marks
    • Multiple choice × 20
    • Calculate × 4
    • Determine × 7
    • Solve × 1
  2. AOS232 questions · 73 marks
    • Multiple choice × 20
    • Determine × 8
    • Calculate × 2
    • Construct × 1
    • Explain × 1
Sample question
The Brunswick Striders is a running club with 60 members. The club records fitness and training data for every member. (a) The resting heart rates of club members are approximately normally distributed, with a mean of 66 beats per minute (bpm) and a standard deviation of 8 bpm. (i) Using the 68-95-99.7% rule, determine the percentage of members expected to have a resting heart rate between 58 bpm and 82 bpm. (1 mark) (ii) Calculate the standardised score (z-score) for a member with a resting heart rate of 54 bpm. (1 mark) (b) The weekly training distances, in kilometres, of the 60 members have the following five-number summary: minimum = 12, Q1 = 24, median = 32, Q3 = 40, maximum = 68. Use an appropriate calculation to show that the maximum value of 68 km would be shown as an outlier on a boxplot of this data. (2 marks) (c) Eight members completed a 10-km race. Their average weekly training distance, x (km), and race time, y (minutes), are shown below. weekly distance (km): 20 24 28 32 36 40 44 48 race time (minutes): 57.8 56.1 54.6 52.4 51.9 49.2 47.6 47.1 Use your CAS to determine the equation of the least squares regression line that enables race time to be predicted from weekly training distance. Round the values of the intercept and the slope to three significant figures. (2 marks) (d) Interpret the slope of this regression line in terms of the variables race time and weekly training distance. (1 mark) (e) (i) Determine the value of the correlation coefficient, r, rounded to three decimal places. (1 mark) (ii) Determine the value of the coefficient of determination, as a percentage rounded to one decimal place, and interpret it in terms of the variables. (1 mark) (f) (i) Use the least squares regression line to predict the 10-km race time of a member whose average weekly training distance is 38 km. Round your answer to one decimal place. (1 mark) (ii) The member who trained 36 km per week ran the race in 51.9 minutes. Determine the residual value for this member, rounded to two decimal places. (1 mark) (g) The club also runs group fitness classes, and quarterly attendance is seasonal. The seasonal indices are: Q1 (Jan-Mar) 1.15 | Q2 (Apr-Jun) 0.95 | Q3 (Jul-Sep) 0.78 | Q4 (Oct-Dec) unknown (i) Determine the seasonal index for Quarter 4. (1 mark) (ii) The actual attendance in Q1 2026 was 2415. Determine the deseasonalised attendance for this quarter. (1 mark) (iii) A least squares trend line fitted to the deseasonalised attendance data is: deseasonalised attendance = 1980 + 28 x quarter number, where quarter number 1 is Q1 2025. Use this trend line to forecast the ACTUAL attendance in Q3 2027. Round to the nearest whole number. (2 marks)
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Answer: Worked solution

(a)(i) 1. 58 = 66 - 8 (1 SD below mean); 82 = 66 + 2x8 (2 SD above mean). 2. Percentage = 34 + 34 + 13.5 = 81.5%. (a)(ii) z = (54 - 66)/8 = -1.5. (b) 1. IQR = 40 - 24 = 16. 2. Upper fence = Q3 + 1.5 x IQR = 40 + 1.5 x 16 = 64. 3. Since 68 > 64, the maximum of 68 km is an outlier. (c) CAS linear regression gives intercept 65.657, slope -0.39911. To 3 significant figures: race time = 65.7 - 0.399 x weekly distance. (d) On average, 10-km race time decreases by 0.399 minutes (about 24 seconds) for each additional kilometre of weekly training distance. (e)(i) r = -0.993 (CAS: -0.99312; negative, matching the negative slope). (ii) r^2 = (-0.99312)^2 = 0.98629, so 98.6% of the variation in race time is explained by the variation in weekly training distance. (f)(i) time = 65.657 - 0.39911 x 38 = 50.49 -> 50.5 minutes (interpolation, since 38 is within 20-48). (ii) 1. Predicted = 65.657 - 0.39911 x 36 = 51.29. 2. Residual = actual - predicted = 51.9 - 51.29 = +0.61 minutes (the member was slower than predicted). [Accept 0.56 if the 3-sig-fig equation is used.] (g)(i) Seasonal indices sum to 4: SI(Q4) = 4 - (1.15 + 0.95 + 0.78) = 1.12. (ii) Deseasonalised = actual / SI = 2415 / 1.15 = 2100. (iii) 1. Q3 2027 is quarter number 11 (2025 = 1-4, 2026 = 5-8, 2027 = 9-12). Trend value = 1980 + 28 x 11 = 2288. 2. Actual forecast = 2288 x 0.78 = 1784.64, so approximately 1785 attendances.
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General Mathematics · 64 practice questions