Study Guide

Degree of Accuracy

Edexcel International GCSE Mathematics AΒ· 1.8Β· 18 min read

1. Rounding to Decimal Places, Powers of 10 and Significant Figuresβ˜…β˜…β˜†β˜†β˜†β± 5 min

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Rounding reduces the precision of a value to match a required degree of accuracy. You will be asked to round to a given power of 10, number of decimal places (dp), or significant figures (sf).

πŸ“˜ Definition

Significant Figures

Digits that carry meaning contributing to a value's precision. Leading zeros are not significant; trailing zeros after a decimal point are significant.

Example:

4.02 has 3 sf; 0.0051 has 2 sf; 1200 has 2 sf if rounded to the nearest 100.

πŸ“ Worked Example

Round 2746 to the nearest 100, 3.14159 to 2 dp, and 0.00487 to 2 sf.

  1. 1

    Round 2746 to nearest 100: Look at the tens digit (4), which is less than 5, so round down. Result = 2700.

  2. 2

    Round 3.14159 to 2 dp: Look at the 3rd decimal digit (1), less than 5, so keep the first 2 decimals. Result = 3.14.

  3. 3

    Round 0.00487 to 2 sf: First non-zero digit is 4 (1st sf), next digit 8 (2nd sf), next digit 7 β‰₯5 so round the 8 up to 9. Result = 0.0049.

Exam tip:

If you are asked to round to a power of 10 (e.g. nearest 10, 100, 1000), circle the digit in that place value first before checking the next digit to decide if you round up or down.

2. Estimation using 1 Significant Figureβ˜…β˜…β˜†β˜†β˜†β± 4 min

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Estimation is used to quickly check if a calculation result is reasonable. For Edexcel IGCSE, you will always round every value in the calculation to 1 significant figure first, then compute the result.

πŸ“ Worked Example

Estimate the value of (392 Γ— 0.71) / 19.8

  1. 1

    Round each value to 1 sf: 392 β†’ 400, 0.71 β†’ 0.7, 19.8 β†’ 20

  2. 2
    Estimate=400Γ—0.720Estimate = \frac{400 Γ— 0.7}{20}
  3. 3

    Calculate numerator first: 400 Γ— 0.7 = 280

  4. 4

    Divide by 20: 280 / 20 = 14. Final estimated value is 14.

βœ“ Quick check

  1. Estimate 512 Γ— 2.9 / 10.3

    • 1.5

    • 15

    • 150

    Reveal answer
    150 β€”

    Round 512β†’500, 2.9β†’3, 10.3β†’10. (500Γ—3)/10 = 1500/10 = 150.

Exam tip:

Estimation questions are usually worth 2 marks: 1 mark for rounding all values to 1 sf correctly, 1 mark for the final approximate result.

3. Identifying Upper and Lower Boundsβ˜…β˜…β˜…β˜†β˜†β± 4 min

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Any value given to a specific degree of accuracy has a range of possible original values. The minimum possible value is the lower bound (LB), the maximum possible is the upper bound (UB). For a value rounded to the nearest unit u, bounds are value Β± u/2.

πŸ“˜ Definition

Bounds Interval

The range of values a rounded measurement can take, written as LB ≀ x < UB. Note the upper bound is not included, as values equal to the UB would round up.

Example:

A length given as 6 cm to the nearest cm has bounds 5.5 cm ≀ x < 6.5 cm.

πŸ“ Worked Example

Find the upper and lower bounds of a mass given as 2.4 kg, rounded to 2 significant figures.

  1. 1

    First identify the precision of the rounding: 2 sf for 2.4 kg means the value is rounded to the nearest 0.1 kg, so u = 0.1 kg.

  2. 2
    LB=2.4βˆ’0.05=2.35kgLB = 2.4 - 0.05 = 2.35 kg
  3. 3
    UB=2.4+0.05=2.45kgUB = 2.4 + 0.05 = 2.45 kg
  4. 4

    Final bounds: 2.35 kg ≀ m < 2.45 kg

Exam tip:

Always state the precision unit u first when calculating bounds, as this will help you avoid mistakes with significant figure bound questions.

4. Higher Tier: Solving Multi-Step Bound Problemsβ˜…β˜…β˜…β˜…β˜†Higher only⏱ 5 min

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For Higher tier questions, you will combine bounds for calculations. To find the maximum result of a product, multiply upper bounds of both values. For minimum product, multiply lower bounds. For quotients, maximum result = UB of numerator / LB of denominator, minimum result = LB of numerator / UB of denominator.

πŸ“ Worked Example

A rectangle has length 6.2 cm and width 4.8 cm, both rounded to 2 significant figures. Calculate the maximum possible area of the rectangle, giving your answer to 3 significant figures.

  1. 1

    First find bounds for length and width: both rounded to 2 sf, so precision 0.1 cm.

  2. 2

    Length bounds: 6.15 cm ≀ l < 6.25 cm, UB of l = 6.25 cm

  3. 3

    Width bounds: 4.75 cm ≀ w < 4.85 cm, UB of w = 4.85 cm

  4. 4
    Maximumarea=UB(l)Γ—UB(w)=6.25Γ—4.85=30.3125cm2Maximum area = UB(l) Γ— UB(w) = 6.25 Γ— 4.85 = 30.3125 cmΒ²
  5. 5

    Round final answer to 3 sf: 30.3 cmΒ²

Exam tip:

Always state the bounds you are using first in your working, as examiners award marks for correct bound identification even if your final calculation is wrong.

5. Common Pitfalls

Wrong move:

Rounding intermediate steps of a calculation before getting the final answer

Why:

This introduces rounding errors that can make your final answer inaccurate, leading to lost marks

Correct move:

Keep full precision for all intermediate working, only round the final answer to the requested degree of accuracy.

Wrong move:

Including the upper bound in the possible value range (e.g. writing 5.5 ≀ x ≀ 6.5 for a value rounded to nearest 1 cm)

Why:

A value of exactly 6.5 would round up to 7 cm, so it is not part of the range for values rounded to 6 cm

Correct move:

Always use strict inequality for the upper bound: 5.5 ≀ x < 6.5

Wrong move:

Using upper bounds for both numerator and denominator when calculating the minimum value of a quotient

Why:

Dividing by a larger number gives a smaller result, so you need the upper bound of the denominator to get the minimum quotient

Correct move:

Minimum quotient = LB(numerator) / UB(denominator); Maximum quotient = UB(numerator) / LB(denominator)

Wrong move:

Counting leading zeros as significant figures, e.g. saying 0.0023 has 4 significant figures

Why:

Leading zeros only show the place value of the number, they do not contribute to its precision

Correct move:

Start counting significant figures from the first non-zero digit from the left: 0.0023 has 2 significant figures.

Wrong move:

Calculating maximum area as UB(length) Γ— LB(width) for a rectangle

Why:

Area is a product, so to maximise the result you need to multiply the largest possible values of both sides

Correct move:

Maximum product = UB of both factors; Minimum product = LB of both factors.

6. Quick Reference Cheatsheet

Concept

Rule

Example

Rounding to nearest u

Value < u/2 above midpoint: round down; β‰₯ u/2: round up

Round 346 to nearest 10: 350

1 sf Estimation

Round all values to 1 sf first, then calculate

(39Γ—0.21) β†’ (40Γ—0.2) = 8

Bounds of rounded value

LB = value - u/2, UB = value + u/2, u = precision unit

2.7 cm (1 dp): 2.65 ≀ x < 2.75

Max product

Multiply UB of all factors

Max area = UB(length) Γ— UB(width)

Min quotient

Divide LB of numerator by UB of denominator

Min speed = LB(distance) / UB(time)

Going deeper

What's Next

Now that you have mastered degree of accuracy, you can apply these skills to a wide range of Edexcel IGCSE Math A topics, including measurement problems, geometry calculations, and statistical data analysis. Rounding and bound reasoning are frequently tested across both Foundation and Higher papers, so practice these skills regularly to avoid easy marks lost to small errors. For Higher tier students, you will use bound reasoning in more complex topics such as trigonometry and kinematics problems where measurement precision is important.

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