Rounding, Estimation and Bounds

Number

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What the bounds are

A rounded value stands for a range of true values, and the bounds are the ends of that range
The lower bound is the smallest value that would still round to the figure given
The upper bound is the value the true figure must stay below
Both lie half a unit of accuracy away from the rounded value, so they are the midpoints between it and its neighbours
A true value may sit exactly on the lower bound, but reaching the upper bound would round it the other way, so only one end of the range is inclusive

Number line showing a value rounded to the nearest hundred as 24 800, with 24 700 and 24 900 marked either side, the lower bound 24 750 and upper bound 24 850 marked at the midpoints, and the range written as 24 750 is less than or equal to x, which is less than 24 850 — Source: Bounds & error intervals by Save My Exams

Finding them

Halve the degree of accuracy the value was rounded to
Subtract that half-unit for the lower bound and add it for the upper bound
- A mass of 6.4 kg given to one decimal place has a half-unit of 0.05
- The bounds are 6.35 ⩽ m < 6.45

Exam tip

Half a unit either way, using the accuracy the question states and not the number of digits you happen to see. A length given as 24 cm to the nearest cm runs from 23.5 to 24.5. The lower bound is reachable and the upper bound is not, which is why it is written with a strict inequality even though 24.5 is the number you use.

Worked example

Writing down a pair of bounds

A length is 8.6 cm, correct to 1 decimal place. Write down its lower and upper bounds.

Solution:

1 decimal place means the value was rounded to the nearest 0.1
Half of 0.1 is 0.05, so the true length is within 0.05 cm of 8.6
Lower bound = 8.6 − 0.05 = 8.55 cm
Upper bound = 8.6 + 0.05 = 8.65 cm
The lower bound is included and the upper bound is not, so 8.55 ⩽ length < 8.65
8.65 itself would round up to 8.7, which is why it is excluded even though it is written as the upper bound