Grasp Maths

Year 10

Upper and lower bounds

Use error intervals to express the range of possible values for measurements, and apply bounds when performing calculations involving measured quantities.

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Lesson overview

Number — bounds and error intervals

When a length is measured as 5 cm 'to the nearest cm', the true length lies in the range: 4.5 cm≤length<5.5 cm4.5 \text{ cm} \leq \text{length} < 5.5 \text{ cm}. The lower bound is 4.5 cm (inclusive) and the upper bound is 5.5 cm (exclusive). When calculations involve measurements with bounds, we find the maximum and minimum possible results by using the worst-case values: maximum results use upper bounds for values being added or multiplied (and lower bounds for values being divided); minimum results use the opposite.

Finding bounds for a rounded measurement

A distance is given as 12 m, rounded to the nearest metre. Find the lower and upper bounds.

For rounding to the nearest unit, subtract and add 0.5. The lower bound is inclusive (≥), the upper bound is exclusive (<).

Bounds for a measurement to 1 decimal place

A mass is recorded as 3.2 kg to 1 decimal place. Find the bounds.

For 1 decimal place, the bounds are ±0.05. Divide the place value by 2.

Maximum and minimum area

A rectangle has length 5 cm and width 3 cm, both measured to the nearest cm. Find the maximum and minimum possible areas.

For maximum, use the upper bounds of both dimensions. For minimum, use the lower bounds.

Worked example — bounds in calculation

A rectangle has length 8.5 cm (to 1 d.p.) and width 6 cm (to the nearest cm). Calculate the maximum possible area.

  1. Identify the precision of each measurement.
  2. Length is to 1 d.p.: subtract and add 0.05 to get bounds 8.45 ≤ length < 8.55.
  3. Width is to the nearest cm: subtract and add 0.5 to get bounds 5.5 ≤ width < 6.5.
  4. For maximum area, use the upper bounds: 8.55 × 6.5.
  5. 8.55×6.5=55.575cm2.8.55 × 6.5 = 55.575 {cm}^{2}.
  6. The maximum possible area is 55.575 cm2{cm}^{2} (or 55.58 to 2 d.p.).

Try it

For each measurement, determine the precision and calculate the correct bounds. Then apply them to find maximum and minimum results.

Question 1

A height is recorded as 160 cm, to the nearest cm. What is the lower bound?

💡 Subtract 0.5 from the recorded value.

Question 2

A speed is given as 25 m/s to the nearest m/s. What is the upper bound?

💡 Add 0.5 to the recorded value.

Question 3

A time is recorded as 12.3 seconds to 1 decimal place. Find the lower bound.

💡 Subtract 0.05 (half of 0.1) from the recorded value.

Question 4

A rectangle has length 10 cm and width 6 cm, both to the nearest cm. What is the maximum possible area?

💡 Maximum area uses upper bounds: 10.5 × 6.5.

Question 5

A number is given as 7.8 to 1 d.p. Which inequality represents the bounds?

💡 For 1 d.p., add and subtract 0.05. Use ≤ for lower, < for upper.

Common mistakes

Watch for these when working through the lesson.

  • Confusing which bounds to use: for maximum area, use upper bounds for both dimensions; for minimum, use lower bounds.
  • Forgetting that the upper bound is exclusive (<), not inclusive (≤).
  • Using the wrong adjustment: for the nearest cm, add/subtract 0.5; for 1 d.p., add/subtract 0.05.
  • Only finding bounds for simple measurements, not applying them through calculations.

Related topics

These ideas fit closely with this lesson.

  • Rounding and place value
  • Standard form and significant figures
  • Estimating and checking answers

Practice next

Independent practice will plug in here

This lesson builds the understanding first. Deeper adaptive practice can sit here later.