NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Accuracy and measurement errors

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In a hydrological study in the Murray River region, researchers measure the depth of a river at a point using a calibrated sonar device. The device records a depth of 4.25 metres with a maximum error of ±0.03 metres.

The researchers use this depth to calculate the cross-sectional area of the river using the formula A = w × d, where w is the width of the river and d is the depth. The width is measured as 18.6 metres with a maximum error of ±0.05 metres.

What is the exact maximum possible value of the cross-sectional area, in square metres, when both depth and width are at their maximum measured values? Express your answer as a simplified fraction.

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Worked solution

  1. The depth measurement is 4.25 m with a maximum error of ±0.03 m, so the maximum possible depth is 4.25 + 0.03 = 4.28 m.
  2. The width measurement is 18.6 m with a maximum error of ±0.05 m, so the maximum possible width is 18.6 + 0.05 = 18.65 m.
  3. The cross-sectional area is calculated using A = w × d, where w is width and d is depth.
  4. Substitute the maximum values: A = 18.65 × 4.28.
  5. Convert both numbers to fractions to maintain exactness: 18.65 = 1865/100 = 373/20, and 4.28 = 428/100 = 107/25.
  6. Multiply the fractions: (373/20) × (107/25) = (373 × 107) / (20 × 25).
  7. Calculate numerator: 373 × 107. Use distributive property: 373 × (100 + 7) = 373×100 + 373×7 = 37300 + 2611 = 39911.
  8. Calculate denominator: 20 × 25 = 500.
  9. The area is 39911 / 500 square metres.
  10. Check if the fraction can be simplified. Find GCD of 39911 and 500. 500 = 2^2 × 5^3. 39911 is odd, so not divisible by 2. 39911 ÷ 5 = 7982.2, not integer. GCD is 1.
  11. The fraction 39911/500 is already in simplest form.
  12. The maximum possible cross-sectional area is 39911/500 m².

Reference answer: 39911/500

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