Learn | Grasp Maths

Lesson overview

Number — measurement and accuracy

When a measurement is rounded, the true value lies within a bounded interval. For a value rounded to the nearest $n$ , the lower bound is $x - \frac{n}{2}$ and the upper bound is $x + \frac{n}{2}$ . Error intervals describe the range of possible true values: if a length is given as $5$ cm (rounded to 1 cm), the true length lies in the interval $[4.5, 5.5)$ cm. In compound problems (e.g., area, speed, density), maximum and minimum values occur at specific combinations of bounds. Understanding which bound to use in which context is crucial for exam performance.

Lower and upper bounds from rounding to the nearest 1

A mass is recorded as $75$ kg (rounded to the nearest 1 kg). Write the error interval.

Rounding to the nearest 1 means ±0.5. The upper bound is not included (≥ would exceed the rounding).

Bounds when rounding to 1 decimal place

A length is $8.3$ m (rounded to 1 d.p.). State the bounds.

Rounding to $n$ decimal places uses ± 0.5 × 10^(-n).

Maximum area from length bounds

A rectangle has length $12$ cm and width $8$ cm, both rounded to the nearest cm. Find the maximum area.

For products, maximum area uses the upper bounds of both dimensions.

Minimum speed from distance and time bounds

A car travels $150$ km (to nearest 10 km) in $2.5$ hours (to nearest 0.1 h). Find the minimum speed.

For a quotient (speed = distance ÷ time), minimum uses min numerator and max denominator.

Worked example — combined bounds problem

The length of a rectangle is $18.4$ cm (to 1 d.p.) and the width is $9.6$ cm (to 1 d.p.). Calculate the range of possible areas. Give your answer to 2 d.p.

Identify the rounding precision: both measurements are to 1 d.p., so use ±0.05.
Calculate lower bounds: length ≥ 18.35 cm, width ≥ 9.55 cm.
Calculate upper bounds: length < 18.45 cm, width < 9.65 cm.
Minimum area = 18.35 × 9.55 = 175.2425 ≈ 175.24 ${cm}^{2}$ (2 d.p.).
Maximum area = 18.45 × 9.65 = 178.0425 ≈ 178.04 ${cm}^{2}$ (2 d.p.).
State the range: the area lies in the interval [175.24, 178.04] ${cm}^{2}$ .

Try it

Use the examples carefully, then choose the answer.

Question 1

A mass is 52 kg (rounded to the nearest kg). What is the lower bound?

💡 Rounding to the nearest 1 kg means ±0.5 kg. Lower bound = 52 − 0.5.

Question 2

A length is 7.8 m (to 1 d.p.). What is the error interval?

💡 To 1 d.p. means ±0.05. Interval is [7.8 − 0.05, 7.8 + 0.05).

Question 3

A square has side 5.2 cm (to 1 d.p.). What is the maximum area?

💡 Maximum side = 5.25 cm. Max area = 5.25 × 5.25 = 27.5625 ${cm}^{2}$ .

Question 4

A speed is calculated as distance ÷ time. Distance = 100 km (to nearest 10 km), time = 2 hours (to nearest 0.1 h). Find the minimum speed (to 2 d.p.).

💡 Min speed uses min distance and max time: 95 ÷ 2.05.

Question 5

A rectangle has length 12.5 m and width 8.3 m (both to 1 d.p.). Which calculation gives the maximum area?

💡 Maximum area uses upper bounds of both length and width.

Common mistakes

Watch for these when working through the lesson.

Using ±1 instead of ±0.5 for rounding to the nearest 1.
Getting the direction wrong: thinking max product uses min bounds.
Forgetting that the upper bound is not included (using [ ] instead of [ )).
For quotients (speed, density), using min bounds for both numerator and denominator instead of min top / max bottom.

Bounds and error intervals in exam contexts

Lesson overview

Worked example — combined bounds problem

Try it

Common mistakes

Related topics

Independent practice will plug in here