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).
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.).
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.)
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
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
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