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Error intervals

Free Printable Error intervals PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 3)

Download free error intervals worksheets (PDF), GCSE maths past paper questions and worked solutions for Error intervals (3A). This is a Grade 3 number topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from MathsLinks, Maths Genie, Corbett Maths, Maths Made Easy, 1st Class Maths, Dr Austin, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our error intervals revision notes and worked examples with diagrams, try real GCSE past paper error intervals questions with step-by-step solutions, or download a PDF of all error intervals worksheet links.

Free Error intervals PDF Worksheets & Answers

Grade 3 • Topic 3A
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Error intervals questions & answers
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Grade 3 topics
3A - Error intervals3B - Operations with fractions
3C - Estimation3D - Writing and simplifying ratio
3E - Ratio3F - Proportion recipes
3G - Percentages3H - Percentage change
3J - Exchange rates3K - Conversions and units
3L - Scale drawings3M - Best buy questions
3N - Substitution3P - Solving linear equations
3Q - Straight line graphs3R - Area and circumference
3S - Rotations3T - Reflections
3U - Enlargements3V - Translations
3W - Mixed transformations3X - Area of compound shapes
3Y - Frequency trees3Z - Two way tables
GCSE Maths Grade 3 Revision Notes & Method

Guide to Error intervals: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

An error interval shows the range of possible actual values of a number before it was rounded or truncated. For a rounded measurement, find half of the degree of accuracy and subtract it for the lower bound, then add it for the upper bound.

For a value x, the error interval is always written using the inequalities LB ≤ x < UB. If a number is truncated (chopped off without rounding), the lower bound is the truncated number itself and the upper bound is the next unit up.

Example 1: A length, L cm, is 8.4 cm rounded to 1 d.p. Write down the error interval for L.

8.35 ≤ L < 8.45 8.3 8.35 8.4 8.45 8.5
Half of 0.1 cm = 0.05 cm → LB = 8.4 − 0.05 = 8.35, UB = 8.4 + 0.05 = 8.45
Answer: 8.35 ≤ L < 8.45

Example 2: A number, y, is truncated to 2 decimal places to give 4.72. Write down the error interval for y.

Solution:

Truncated at 2 d.p. means digits after 0.01s were cut off.
Smallest possible value = 4.72
Upper bound (next hundredth up) = 4.72 + 0.01 = 4.73
Answer: 4.72 ≤ y < 4.73
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Error intervals Questions & Answers

Past Paper Example 1 GCSE Jun 2025 3F Q26 / 3H Q5 2 marks
Full Paper (PDF) • Mark Scheme (PDF)
Karim rounds a number, n, to 1 decimal place.
The result is 23.8
Complete the error interval for n.
Worked Solution & Mark Scheme (2 marks)
Half the unit of rounding: 0.1 ÷ 2 = 0.05 M1
Lower bound = 23.8 – 0.05 = 23.75
Upper bound = 23.8 + 0.05 = 23.85. A1
Answer: 23.75 ≤ n < 23.85
Past Paper Example 2 GCSE Jun 2023 2F Q26 / 2H Q7 2 marks
Full Paper (PDF) • Mark Scheme (PDF)
Work out
A number, d, is rounded to 1 decimal place.
The result is 12.7
Complete the error interval for d.
Worked Solution & Mark Scheme (2 marks)
12.65 ⩽ d < 12.75 M1 A1
Answer: 12.65, 12.75
Past Paper Example 3 AQA Jun 2025 2F Q23 / 2H Q5 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
(a) Error interval for 160 cm to nearest 10 cm. [2 marks]
5 (b) Panel is 2 m to nearest 10 cm. Show total of 3 panels < 6.2 m. [2 marks]
(a) The length of a fence panel is 160 cm to the nearest 10 cm
Complete the error interval. [2 marks]
(b) A different fence panel measures 2 metres to the nearest 10 cm
Kim says that the total length of **three** of these fence panels must be less than 6.2 m
Show that Kim is correct. [2 marks]
Worked Solution & Mark Scheme (4 marks)
Part (a) Solution:
155 ≤ length < 165 M1 A1
Answer (a): Answer 155, 165 cm
Part (b) Solution:
Maximum single length = 2.05 m. 3 × 2.05 = 6.15 m M1
Since 6.15 < 6.2, Kim is correct C1
Answer (b): Upper bound = 3 × 2.05 m = 6.15 m < 6.2 m

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