Approximation and Estimation Revision Notes
Subject: Mathematics | Level: GCSE | Exam Board: OCR
Master the art of Approximation and Estimation, a foundational topic that underpins almost every paper in GCSE Mathematics. From avoiding the common traps of significant figures to calculating error bounds for complex measurements, this guide provides everything you need to secure method marks and tackle multi-step problems with confidence.
Revision Notes & Key Concepts
Key Terms & Definitions
- Significant Figure
- The digits in a number that contribute to its precision, starting from the first non-zero digit.
- Decimal Place
- The position of a digit to the right of a decimal point.
- Estimation
- Finding a value that is close enough to the right answer, usually by rounding to 1 significant figure.
- Error Interval
- The range of possible true values that a rounded number could represent, expressed as an inequality.
- Upper Bound
- The largest possible value a measurement could be before it was rounded down.
- Lower Bound
- The smallest possible value a measurement could be before it was rounded up.
Worked Examples
Worked Example
Question: Calculate the value of $\sqrt{4.5^2 - 2.1^3}$. Give your answer to 3 significant figures. (2 marks)
Solution: Step 1: Calculate the exact value using a calculator: $\sqrt{20.25 - 9.261} = \sqrt{10.989} = 3.314966...$ Step 2: Identify the first 3 significant figures: $3.31$. Step 3: Look at the 4th significant figure ($4$). Since $4 < 5$, round down. Final answer: $3.31$
Worked Example
Question: Work out an estimate for the value of $\frac{41.3 \times 29.6}{0.198}$. (3 marks)
Solution: Step 1: Round each number to 1 significant figure. $41.3 \approx 40$ $29.6 \approx 30$ $0.198 \approx 0.2$ Step 2: Substitute the rounded values into the expression: $\frac{40 \times 30}{0.2}$ Step 3: Calculate the numerator: $40 \times 30 = 1200$ Step 4: Divide by the denominator: $1200 \div 0.2 = 6000$ Final answer: $6000$
Worked Example
Question: A rectangular field has a length of $85$ m and a width of $42$ m, both measured to the nearest metre. Calculate the upper bound for the area of the field. (3 marks)
Solution: Step 1: Find the upper bound for the length. Nearest metre means degree of accuracy is $1$. Half is $0.5$. Upper bound = $85 + 0.5 = 85.5$ m. Step 2: Find the upper bound for the width. Upper bound = $42 + 0.5 = 42.5$ m. Step 3: To find the maximum area, multiply the upper bounds together: $\text{Area} = 85.5 \times 42.5$ Step 4: Calculate: $85.5 \times 42.5 = 3633.75$ Final answer: $3633.75$ m$^2$
Practice Questions
Question: Round $0.05681$ to 2 significant figures. (1 mark)
Answer:
Question: Estimate the value of $\frac{6.8 \times 191}{0.051}$. (3 marks)
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Question: A car travels a distance of $240$ km, correct to the nearest $10$ km. It takes $3.5$ hours, correct to the nearest $0.1$ hours. Calculate the upper bound for the average speed of the car. (4 marks)
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Question: Write down the error interval for $y$ if $y = 15.3$ truncated to 1 decimal place. (2 marks)
Answer:
Question: Use estimation to show that the exact answer to $3.1^2 \times 8.9$ is less than $100$. (2 marks)
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