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Bounds

Free Printable Bounds PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free bounds worksheets (PDF), GCSE maths past paper questions and worked solutions for Bounds (7B). This is a Grade 7 number topic from the GCSE questions by topic collection for 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, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, 1st Class Maths, Metatutor, Maths Takeaway, Dr Austin, GCSE Maths Rocket, E Classroom and PMT, along with past paper question booklets.

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

Free Bounds PDF Worksheets & Answers

Grade 7 • Topic 7B
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Grade 7 topics
7A - Surds7B - Bounds
7C - Proportion (harder)7D - Algebraic fractions
7E - The quadratic formula7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs7H - Rearranging (harder)
7J - Functions (harder)7K - Iteration
7L - Area of any triangle7M - The sine rule
7N - The cosine rule7P - Congruent triangles
7Q - 3D pythagoras / trig.7R - Histograms
7S - Conditional probability7T - Exact trig. values (harder)
7U - Area of a segment
GCSE Maths Grade 7 Revision Notes & Method

Guide to Bounds: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

When a measurement is rounded to a given unit, the Lower Bound (LB) is half a unit below the rounded value and the Upper Bound (UB) is half a unit above, giving the error interval LB ≤ x < UB. If a value is truncated rather than rounded, the LB is the stated value itself and the UB is one full unit above.

In calculations, choose the bounds of each variable carefully: UB(A + B) = UB(A) + UB(B), UB(A × B) = UB(A) × UB(B), UB(A − B) = UB(A) − LB(B), and UB(A ÷ B) = UB(A) ÷ LB(B). To justify a final answer to a suitable degree of accuracy, round both UB and LB until they agree.

Example 1: d = 140 m (2 s.f.) and t = 8.4 s (1 d.p.). Find the upper bound of speed s = dt (2 d.p.).

LB = 135 135 ≤ d < 145 UB = 145 130 135 140 145 150

Solution:

d (nearest 10 m): LB = 135, UB = 145
t (nearest 0.1 s): LB = 8.35, UB = 8.45
UB(s) = UB(d)LB(t) = 1458.35 = 17.3652...
Answer: 17.37 m/s (2 d.p.)

Example 2: Using Example 1, find s to a suitable degree of accuracy and give a reason.

Solution:

LB(s) = LB(d)UB(t) = 1358.45 = 15.9763...
UB(s) = 17.3652... and LB(s) = 15.9763...
To 1 s.f., both UB and LB round to 20 m/s
(To 2 s.f., UB = 17 and LB = 16, which differ)
Answer: 20 m/s (both bounds round to 20 to 1 s.f.)
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Bounds Questions & Answers

Past Paper Example 1 IGCSE Jun 2025 2H Q22 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
The diagram shows a square inside rectangle ABCD
ABCDDiagram NOTaccurately drawn
The total area of the region shown shaded in the diagram is X cm2
AB = 11.5 cm correct to the nearest 0.5 cm
BC = 9.2 cm correct to 2 significant figures
side of square = 4.1 cm correct to 2 significant figures
By considering bounds, work out the value of X to a suitable degree of accuracy.
Show your working clearly.
Worked Solution & Mark Scheme (4 marks)
(The area of a rectangle is length × width)
UB of X = 11.75 × 9.25 – 4.05² = 108.6875 – 16.4025 = 92.285 M2
LB of X = 11.25 × 9.15 – 4.15² = 102.9375 – 17.2225 = 85.715 M1
Both round to 90 (1 s.f.) A1
Answer: X = 90
Past Paper Example 2 IGCSE Nov 2021 1H Q13 3 marks
Full Paper (PDF) • Mark Scheme (PDF)
Here is a triangle XYZ.
YXZ25°130°16 cmDiagram NOTaccurately drawn
The length XZ and the angles YXZ and XYZ are each given correct to 2 significant figures.
Calculate the upper bound for the length YZ.
Give your answer correct to one decimal place.
Show your working clearly.
Worked Solution & Mark Scheme (3 marks)
UB(XZ) = 16.5 cm, LB(∠YXZ) = 125° (which maximises sin ∠YXZ on [125°, 135°]), LB(∠XYZ) = 24.5° M2
By Sine Rule: UB(YZ) = (16.5 × sin 125°) / sin 24.5° = 13.516... / 0.41469... = 32.593... = 32.6 cm A1
Answer: 32.6 cm
Past Paper Example 3 GCSE Jun 2023 2H Q19 3 marks
Full Paper (PDF) • Mark Scheme (PDF)
ABC is a right-angled triangle.
x9.3 cm12.6 cmABC
AB = 9.3 cm correct to the nearest mm.
AC = 12.6 cm correct to the nearest mm.
Calculate the lower bound for the size of the angle marked x.
You must show all your working.
Worked Solution & Mark Scheme (3 marks)
LB of AB = 9.25 cm, UB of AC = 12.65 cm M2
LB of x = sin⁻¹(9.25 / 12.65) = 46.9899...° (or 46.98 to 47) A1
Answer: 46.99

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