Error Bounds and Approximations
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Error Bounds and Approximations — Free MYP4 Mathematics (Extended) Practice Questions

1QuestionCalculations Using Rounded ValuesAssessment Practice
2 marks~3 minCriterion D
A homeowner measures a bathroom floor as 2.02.0 m long and 1.51.5 m wide, rounded to the nearest 0.50.5 m. Square tiles measure 0.20.2 m by 0.20.2 m, rounded to the nearest 0.050.05 m. The homeowner estimates the number of tiles needed using floor areatile area\dfrac{\text{floor area}}{\text{tile area}}.

Advise the homeowner whether this calculation gives a reliable tile count for purchasing purposes, with reference to the direction of both rounding errors. [2]
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Solutions

2QuestionUnderstanding Absolute and Percentage ErrorAssessment Practice
8 marks~12 minCriterion A
A manufacturing engineer measures two components of a mechanical assembly. Component a=5.0±0.1a = 5.0 \pm 0.1 cm and component b=3.0±0.1b = 3.0 \pm 0.1 cm.

Use: absolute error of a product =a0 Δb+b0 Δa+Δa Δb= a_0\,\Delta b + b_0\,\Delta a + \Delta a\,\Delta b, where a0a_0, b0b_0 are measured values and Δa\Delta a, Δb\Delta b are absolute errors.
a
Calculate the absolute error and percentage error for the sum a+ba + b and the product a×ba \times b. [2]
b
A second engineer measures c=6.0±0.2c = 6.0 \pm 0.2 cm and d=4.0±0.2d = 4.0 \pm 0.2 cm. Deduce whether the percentage error in the product c×dc \times d is greater than, equal to, or less than the percentage error in a×ba \times b, and explain why. [3]
c
Prove algebraically that when two measured quantities are added, the maximum absolute error of the sum equals the sum of the individual absolute errors. Then advise the quality control team whether the assembly, with total length a+ba + b, should be approved for a specification requiring the total to lie within ±0.15\pm 0.15 cm of 8.0 cm. [3]
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Solutions

3QuestionCalculations Using Rounded ValuesAssessment Practice
2 marks~3 minCriterion B
When two numbers are each rounded to nn decimal places, the sum is also expressed to nn decimal places.

The four sums below illustrate this:

3+4=73 + 4 = 7 (rounded to 0 decimal places)
3.2+4.3=7.53.2 + 4.3 = 7.5 (rounded to 1 decimal place)
3.25+4.31=7.563.25 + 4.31 = 7.56 (rounded to 2 decimal places)
3.253+4.314=7.5673.253 + 4.314 = 7.567 (rounded to 3 decimal places)

Analyse the relationship between the number of decimal places in the addends and in their sum. Hence, formulate a conjecture for the maximum possible error in the sum when both addends are rounded to nn decimal places. [2]

Solutions

4QuestionUnderstanding Absolute and Percentage ErrorAssessment Practice
4 marks~6 minCriterion C
A student measures the time for a ball to fall from a fixed height, repeating the experiment four times. The accepted value, determined by a precision instrument, is 3.03.0 seconds.

Trial 1: 3.03.0 s
Trial 2: 3.23.2 s
Trial 3: 2.82.8 s
Trial 4: 3.13.1 s

Percentage error=∣Approximate value−Exact value∣∣Exact value∣×100%\text{Percentage error} = \frac{|\text{Approximate value} - \text{Exact value}|}{|\text{Exact value}|} \times 100\%
a
Calculate the percentage error for Trials 2, 3, and 4. [2]
b
The student claims their method is reliable because the mean of all four trials equals the accepted value. Calculate the mean of the four trials, then justify whether this claim is valid. [2]

Solutions

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