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If a number or measurement has been rounded, it can be important to consider what possible values the exact value could have been. For example, if a cooker is 600 mm wide to the nearest 10 mm, it could actually be any width from 595 mm up to 605 mm – so it might not fit into a gap that is exactly 600 mm wide.

To describe all the possible values that a rounded number could be, we use limits of accuracy.

Limits of accuracy or bounds are used to describe all the possible values that a rounded number could be.

The lower limit or lower bound is the smallest value that would round up to the estimated value.
The upper limit or upper bound is the smallest value that would round up to the next estimated value.

To calculate the upper and lower bounds,

• halve the degree of accuracy,
• add this value to the rounded value for the upper limit,
• subtract this value from the rounded value for the lower limit.

Example

Find the upper and lower limits for a mass of 62 kg, measured to the nearest kg.

Solution

• halve the degree of accuracy 1 kg ÷2 = 0.5 kg
• add this value the rounded value for the upper limit
  62 + 0.5 = 62.5 kg
• subtract this value from the rounded value for the lower limit 62 – 0.5 = 61.5 kg

There is a difference of 1 kg between the upper and lower limits because the accuracy of the measurement was to the nearest kg.

Example

Find the upper and lower bounds for a shelf of length 70 cm, measured to the nearest 10 cm.

Solution

• 10 cm ÷2 = 5 cm
• Upper bound is 70 + 5 = 75 cm
• Lower bound is 70 – 5 = 65 cm

There is a difference of 10cm between the upper and lower limits because the accuracy of the measurement was to the nearest 10cm.

Example

It was reported that there were 1800 fans at a match. This has been estimated to the nearest hundred. Calculate the minimum number of fans at the match.

Solution

• 100 ÷2 = 50
• 1800 – 50 = 1750
• The least number of fans at the match is 1750.

Measurements are often combined, for example, when finding the perimeter or the area of a shape.
If the calculation involves adding or multiplying then:

• to find the maximum use the upper limits of each measurement,
• to find the minimum use the lower limits of each measurement.

This is not the same for subtraction and division, which is covered in the 'apply and interpret limits of accuracy' guide.

Example

Calculate the maximum perimeter of this triangle. All measurements have been rounded to 1 decimal place.

To find the maximum use the upper limits of each measurement

0.1 ÷ 2 = 0.05

Upper limits are 6.45 cm, 3.85 cm and 5.25 cm. Maximum perimeter = 6.45 + 3.85 + 5.25 = 15.55 cm