Bounds Quiz (Grade 7)

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Score: 0 / 10

Bounds — Notes & Worked Examples Show / Hide

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.).

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.)

One Page Guide (PDF) Bounds Worksheets

Question 1

[1 mark]

Work out
A rectangle has length 19 cm and width 11 cm, both measured correct to the nearest centimetre.
Calculate the upper bound for the area of the rectangle.

cm²

Worked Solution & Answer

(The area of a rectangle is length × width)
UB = 19.5 × 11.5 = 224.25 cm² B1

Answer: 224.25 cm² cm²

Question 2

[1 mark]

Work out
A rectangle has length 18 cm and width 11 cm, both measured correct to the nearest centimetre.
Calculate the upper bound for the area of the rectangle.

cm²

Worked Solution & Answer

(The area of a rectangle is length × width)
UB = 18.5 × 11.5 = 212.75 cm² B1

Answer: 212.75 cm² cm²

Question 3 IGCSE Jan 2022 2F Q19 / 2H Q4

[2 marks]

Each side of a regular octagon has a length of 18 mm, correct to the nearest 0.5 mm

(a) Write down the lower bound of the length of each side of the octagon. [1 mark]

(b) Write down the upper bound of the length of each side of the octagon. [1 mark]

Worked Solution & Answer

(a)

18 – 0.25 = 17.75 mm B1

Answer (a): 17.75 mm

(b)

18 + 0.25 = 18.25 mm B1

Answer (b): 18.25 mm

Question Paper (PDF) • Mark Scheme (PDF)

Question 4 IGCSE Nov 2021 1H Q13

[3 marks]

Here is a triangle XYZ.

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 & Answer

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

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[3 marks]

ABC is a right-angled triangle.

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 & Answer

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

Question Paper (PDF) • Mark Scheme (PDF)

Question 6 IGCSE Jun 2025 2H Q22

[4 marks]

The diagram shows a square inside rectangle ABCD

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.

X =

Worked Solution & Answer

(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

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 Specimen Set 2 2H Q17

[4 marks]

Work out
a is 8.3 cm correct to the nearest mm
b is 6.1 cm correct to the nearest mm
Calculate the upper bound for c.
You must show your working.

a is 8.3 cm correct to the nearest mm
b is 6.1 cm correct to the nearest mm
Calculate the upper bound for c.
You must show your working.

Worked Solution & Answer

Upper bound of a = 8.35 cm M2
Lower bound of b = 6.05 cm M1
Upper bound of c = √(8.35² – 6.05²) = √(69.7225 – 36.6025) = √33.12 = 5.7549978... cm. A1

Answer: 5.755 cm

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 IGCSE Jan 2022 1HR Q22

[4 marks]

The diagram shows triangle ABC

c = 11.5 correct to one decimal place
x = 80 correct to the nearest whole number
y = 75 correct to the nearest whole number
Calculate the upper bound for the value of b
Show your working clearly.
Give your answer correct to 3 significant figures.

Worked Solution & Answer

To maximise b = AC: by Sine Rule, b / sin B = c / sin y ⇒ b = c × sin(180 – x – y) / sin y M2
Use UB(c) = 11.55, LB(x) = 79.5°, LB(y) = 74.5° so UB(B) = 180 – (79.5 + 74.5) = 26° M1
UB(b) = 11.55 × sin 26° / sin 74.5° = 5.2543... = 5.25. A1

Answer: 5.25

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 AQA Nov 2017 2H Q25

[5 marks]

The dimensions of a rectangular floor are to the nearest 0.1 metres.
A force of 345 Newtons is applied to the floor.
The force is to the nearest 5 Newtons.

Work out the upper bound of the pressure.
Give your answer to 4 significant figures.
You must show your working.

pressure = forcearea

Answer N/m²

Worked Solution & Answer

(Pressure = force ÷ area)
To maximise pressure, find upper bound of force and lower bound of area: M1
Force = 345 ± 2.5 ⇒ UB(force) = 347.5 N M1
Length = 6.4 ± 0.05 ⇒ LB(length) = 6.35 m M1
Width = 2.6 ± 0.05 ⇒ LB(width) = 2.55 m M1
LB(area) = 6.35 × 2.55 = 16.1925 m²
UB(pressure) = 347.516.1925 ≈ 21.46055… ≈ 21.46 N/m² (to 4 s.f.) A1

Answer: Answer 21.46 N/m²

Question Paper (PDF) • Mark Scheme (PDF)

Question 10

[5 marks]

D = u²2a
u = 26.2 correct to 3 significant figures
a = 4.3 correct to 2 significant figures

(a) Calculate the upper bound for the value of D.
Give your answer correct to 6 significant figures.
You must show all your working. [3 marks]

(b) By considering bounds, write down the value of D to a suitable degree of accuracy.
You must give a reason for your answer. [2 marks]

Worked Solution & Answer

(a)

UB of u = 26.25, LB of a = 4.25 M2
UB of D = 26.25² / (2 × 4.25) = 689.0625 / 8.5 = 81.066176... = 81.0662 A1

Answer (a): 81.0662

(b)

80, because both the upper bound (81.0662) and the lower bound (78.6003) round to 80 to 1 significant figure (nearest 10). M1 C1

Answer (b): 80, because both the upper bound (81.0662) and the lower bound (78.6003) round to 80 to 1 significant figure (nearest 10).

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 32 marks)

Bounds Worksheets Next: Proportion (harder) →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment