When a number is rounded or truncated, the original value could have been anything within a range either side of it. An error interval states that range as an inequality:

where LB is the lower bound (smallest possible original value) and UB is the upper bound (smallest value that would not round/truncate to the given figure). Notice the upper bound uses a strict inequality $<$, never $\le$ — the upper bound itself rounds up to the next value, so it can't be included.

1. Identify the degree of accuracy (the place value rounded to).
2. Halve it.
3. Subtract this half-interval from the rounded value for the lower bound; add it for the upper bound.

Example: A letter has a mass of 58 g to the nearest gram. Degree of accuracy $= 1$ g, half $= 0.5$ g.

This works identically for decimal places and significant figures — only the "degree of accuracy" changes. For $14.8$ (1 d.p.), the accuracy is $0.1$, giving $14.75 \le y < 14.85$. For $3.69$ (3 s.f.), the accuracy is in the hundredths place ($0.01$), giving $3.685 \le x < 3.695$.

Truncating just chops off the digits — it never rounds up. So the given value is the lower bound, and you add one full unit of the degree of accuracy to get the upper bound.

Example: $x$ is truncated to 1 d.p., giving 6.2. Then $\text{LB} = 6.2$ and $\text{UB} = 6.2 + 0.1 = 6.3$, so $6.2 \le x < 6.3$.

Check which word the question uses — "rounded"/"nearest" means halve the accuracy; "truncated" means the given value is already the lower bound.

When a result is calculated from two rounded measurements, you can't just use both upper bounds or both lower bounds — you must pick whichever combination of bounds gives the actual maximum or minimum of the result:

For subtraction and division, the bounds "cross over" — to get the smallest possible difference or quotient, you subtract/divide by the largest possible second value.

A rectangle has length $20$ cm and width $6$ cm, both correct to the nearest cm. Find the error interval for the perimeter.

1. Bounds: $19.5 \le L < 20.5$, and $5.5 \le W < 6.5$.
2. Minimum perimeter: $2 \times (19.5 + 5.5) = 50$ cm.
3. Maximum perimeter: $2 \times (20.5 + 6.5) = 54$ cm.
4. Error interval: $50 \le P < 54$ cm.

For a motorbike travelling $110$ m (nearest 10 m) in $5$ s (nearest second), minimum speed uses the smallest distance over the largest time: $105 \div 5.5 \approx 19.1$ m/s; maximum speed uses $115 \div 4.5 \approx 25.6$ m/s.