This set has eleven original questions, ordered from easier to harder, covering all five lessons in International Mathematics modelling. Questions 1 to 3 are about setting up, 4 to 7 about fitting and interpreting, and 8 to 11 about comparing and extrapolating.
Attempt each question on paper and write full sentences where the question asks you to explain. Only then open the answer. Use your mistake log and retest queue to record the ones you get wrong.
Questions
1. A courier charges a fixed fee of RM6 plus RM2.50 for each kg of parcel weight. Define suitable variables and write a model for the cost.
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Let C be the cost in RM and w the weight in kg, with w ≥ 0. The fixed fee is paid once and the rate stays RM2.50 per kg.
Model: C = 6 + 2.5w.
2. Use the model in question 1 to find the cost of an 8 kg parcel, and the weight of a parcel that costs RM41.
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For w = 8: 2.5 × 8 = 20, and C = 6 + 20 = RM26.
For C = 41: 6 + 2.5w = 41, so 2.5w = 35 and w = 35 ÷ 2.5 = 14 kg. Check: 2.5 × 14 = 35 and 35 + 6 = 41.
3. State two assumptions behind the model in question 1, and say what would happen to the model if one of them failed.
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Two assumptions: the rate stays RM2.50 per kg for every weight, and the fixed fee is the same for every parcel. If the courier gave a discount for heavy parcels, the rate would change, and a single straight-line equation would no longer describe the cost.
4. A plant’s height h (cm) was measured at the end of weeks 1 to 4: 8.2, 10.1, 11.9 and 14.2. Fit a straight-line model using the first and last readings, and find the gaps at weeks 2 and 3.
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Gradient = (14.2 − 8.2) ÷ (4 − 1) = 6 ÷ 3 = 2. Using (1, 8.2): 8.2 = 2 + c, so c = 6.2. Model: h = 2w + 6.2.
Week 2: model 10.2, observed 10.1, gap −0.1. Week 3: model 12.2, observed 11.9, gap −0.3. The gaps are small, so the line is a fair fit.
5. A bottle holds 500 mL of sanitiser. Readings show 440 mL after 10 days and 380 mL after 20 days. Find a linear model V = 500 − kt, where V is in mL and t in days.
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From 500 to 440 is a fall of 60 mL in 10 days, so k = 6. Check with day 20: 500 − 6 × 20 = 500 − 120 = 380, which matches.
Model: V = 500 − 6t.
6. Interpret the numbers 500 and 6 in question 5, with units, and find when the bottle is empty.
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500 is the volume at t = 0, so the bottle holds 500 mL at the start. The 6 means the volume decreases by 6 mL each day.
Set V = 0: 6t = 500, so t = 500 ÷ 6 = 83.33…, about 83.3 days.
7. A car park charges P = 4.5 + 0.8h, where P is in RM and h is the parking time in hours. Interpret the 4.5 and the 0.8, and find the charge for 3.5 hours.
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4.5 is the charge at h = 0, so it is a fixed entry charge of RM4.50. 0.8 means the charge rises by RM0.80 for each extra hour.
For h = 3.5: 0.8 × 3.5 = 2.8, so P = 4.5 + 2.8 = RM7.30.
8. A bus journey model is T = 10 + 2.4d, where T is in minutes and d is the distance in km. For 15 km the real time was 52 minutes. Find the residual and the percentage error, and say whether the model over- or underestimates.
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Prediction: 10 + 2.4 × 15 = 10 + 36 = 46 minutes. Residual = 52 − 46 = +6 minutes.
Percentage error = 6 ÷ 52 × 100 = 11.53…, about 11.5%. The residual is positive, so the model underestimates.
9. For a different model, the residuals at x = 1, 2 and 3 are +2, +5 and +9. What does this pattern suggest?
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All residuals are positive and increasing, so the model underestimates by more and more as x grows. The gradient is probably too small, or the real relationship is curved. A good model has residuals on both sides of zero with no pattern.
10. The model in question 5 is used to predict the volume at t = 100 days. Calculate it and comment.
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V = 500 − 6 × 100 = 500 − 600 = −100 mL. A volume cannot be negative, so the prediction is impossible.
The data ran only to day 20, so this is extrapolation. The bottle is empty at about 83.3 days, and the model applies only for 0 ≤ t ≤ 83.3.
11. Two readings of a quantity are (2, 11) and (5, 20). Find the linear model, predict y at x = 10, and comment on the reliability.
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Gradient = (20 − 11) ÷ (5 − 2) = 9 ÷ 3 = 3. Using (2, 11): 11 = 6 + c, so c = 5. Model: y = 3x + 5.
At x = 10: 3 × 10 + 5 = 35. The data cover x = 2 to 5 only, so this is extrapolation. With only two points, there is no test of whether a straight line suits the data, so the prediction is weak.
If you got these wrong
Match the type of error to the lesson that repairs it.
- Questions 1 to 3, setting up: go to define variables and assumptions for a model.
- Questions 4 and 5, fitting a line: go to fit a simple model to supplied data.
- Questions 6 and 7, units and meaning: go to interpret parameters with units.
- Questions 8 and 9, residuals: go to compare a prediction with an observed value.
- Questions 10 and 11, predictions outside the data: go to explain why extrapolation may fail.
The percentage-base explorer can help with question 8, and the non-calculator working trainer with the arithmetic in questions 2 and 6. If you want a teacher to review your written explanations, see online one-to-one Mathematics tuition.