CIE IGCSE Mathematics (0580) Limits of accuracy Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Types of number Study Notes
LEARNING OBJECTIVE
- Give upper and lower bounds for data rounded to a specified accuracy
Key Concepts:
- Limits of accuracy
Limits of Accuracy
Limits of Accuracy
What Are Limits of Accuracy?
When a measurement is rounded, the exact value could be slightly more or less than the rounded value. The upper and lower bounds describe the range of values that the original (exact) value could have been.
Why Use Bounds?
To identify the maximum and minimum possible values of a quantity when it has been rounded, especially in measurement and error calculation contexts.
Key Terms
- Upper bound: The highest possible actual value before it would round up to the next unit.
- Lower bound: The lowest possible actual value before it would round down to the previous unit.
How to Find Bounds
- If a number is rounded to the nearest whole number, the bounds are found by adding and subtracting 0.5.
- If rounded to the nearest tenth (1 decimal place), add and subtract 0.05.
- If rounded to the nearest hundred, add and subtract 50.
General Rule:
If a number is rounded to the nearest \( n \), then:
- Lower bound = rounded value \( – \frac{n}{2} \)
- Upper bound = rounded value \( + \frac{n}{2} \)
Important:
- Bounds are usually written as inequalities or intervals.
- For calculations involving measurements, use bounds to find the maximum and minimum possible result.
Example:
Write down the upper bound of a length measured as 57 m, correct to the nearest metre.
▶️ Answer/Explanation
Step 1: Since the length is rounded to the nearest metre, the accuracy is \( \pm 0.5 \, \text{m} \).
Step 2: Add 0.5 to get the upper bound:
\( 57 + 0.5 = 57.5 \, \text{m} \)
Final Answer: Upper bound = \( 57.5 \, \text{m} \)
Example:
A length is measured as 84 cm, correct to the nearest centimetre. Find the upper and lower bounds of the actual length.
▶️ Answer/Explanation
Step 1: Since the value is rounded to the nearest cm, the accuracy is \( \pm 0.5 \, \text{cm} \).
Step 2: Use bounds formula:
- Lower bound: \( 84 – 0.5 = 83.5 \, \text{cm} \)
- Upper bound: \( 84 + 0.5 = 84.5 \, \text{cm} \)
Final Answer: The actual length lies between \( 83.5 \, \text{cm} \leq \text{length} < 84.5 \, \text{cm} \)
Example:
A rectangular field is said to be 30 m long and 12 m wide, both correct to the nearest metre. Find the maximum and minimum possible area of the field.
▶️ Answer/Explanation
Step 1: Since both are to the nearest metre:
- Length bounds: \( 29.5 \leq L < 30.5 \)
- Width bounds: \( 11.5 \leq W < 12.5 \)
Step 2: Calculate minimum and maximum area:
- Minimum area = \( 29.5 \times 11.5 = 339.25 \, \text{m}^2 \)
- Maximum area = \( 30.5 \times 12.5 = 381.25 \, \text{m}^2 \)
Final Answer: The area lies between \( 339.25 \, \text{m}^2 \leq \text{Area} < 381.25 \, \text{m}^2 \)