How to Calculate the Upper and Lower Bounds
Upper and lower bounds tell you the range a rounded measurement could really lie in. In Edexcel IGCSE Maths these questions test whether you can spot the degree of accuracy, halve it to find the tolerance, and write a precise error interval using inequalities. This page shows the method step by step, explains why the lower bound is included but the upper bound is not, and works through exam-style examples involving length, area and speed. Scroll down for auto-marked practice questions at five levels.
What Are Upper and Lower Bounds?
When a value is rounded, the true value lies somewhere in a range around the recorded number. The lower bound is the smallest value that still rounds to the recorded number; the upper bound is the smallest value that would round to the next recorded value.
The method is always the same: identify the degree of accuracy (nearest 10, 1 d.p., 2 s.f., nearest 5 mm, and so on), halve it to get the tolerance, then subtract and add to get the bounds. Finish with the error interval \( \text{lower} \le x \lt \text{upper} \).
Why is the lower bound included but the upper bound excluded?
- Lower is inclusive. A value exactly at the lower bound still rounds to the recorded value. If a number is recorded as \(3.2\) to 1 d.p., the interval begins at \(3.15\). Since \(3.15\) rounds to \(3.2\), it belongs in the interval: \(3.15 \le x \ldots\)
- Upper is exclusive. Including the upper would let in values that round to the next recorded value. With \(3.2\) to 1 d.p., the upper boundary is \(3.25\). If we wrote \(x \le 3.25\), then \(3.25\) would round to \(3.3\), not \(3.2\). Writing \(x \le 3.24\) instead would wrongly shut out values like \(3.2499\ldots\), which do round to \(3.2\). So the interval keeps everything just below the boundary while excluding the first value that rounds up, \(3.25\).
So the rule is: include the lower, exclude the upper: \(\text{lower} \le x \lt \text{upper}\).
The Fast Method
Upper \(= \text{recorded} + \text{tol}\)
Finding Upper and Lower Bounds
Example 1: Number to 1 d.p.
A value is recorded as \(x = 6.4\) to 1 decimal place. Find the error interval.
\[ \begin{array}{rcl} \text{accuracy} &=& 0.1 \\ \text{tolerance} &=& 0.05 \\ \text{lower} &=& 6.4 - 0.05 \\ &=& 6.35 \\ \text{upper} &=& 6.4 + 0.05 \\ &=& 6.45 \end{array} \]
\[ 6.35 \le x \lt 6.45 \]
What's happening?
The value is given to the nearest \(0.1\), so the tolerance is half of that: \(0.05\).
Subtract \(0.05\) for the lower bound; add \(0.05\) for the upper bound.
The upper is excluded because \(6.45\) rounds to \(6.5\), not \(6.4\).
Example 2: 2 significant figures
A value is recorded as \(x = 3600\) to 2 significant figures. Find the error interval.
\[ \begin{array}{rcl} \text{last s.f. place} &=& 100 \\ \text{tolerance} &=& 50 \\ \text{lower} &=& 3600 - 50 \\ &=& 3550 \\ \text{upper} &=& 3600 + 50 \\ &=& 3650 \end{array} \]
\[ 3550 \le x \lt 3650 \]
What's happening?
The two significant figures are the \(3\) and the \(6\), so the last one sits in the hundreds column: the accuracy is \(100\).
Half of \(100\) is \(50\), the tolerance either side.
\(3650\) rounds to \(3700\) (to 2 s.f.), so it is excluded.
Bounds in Calculations
Example 3: Area of a rectangle
A rectangle has length \(L = 85\) mm and width \(W = 60\) mm, each measured to the nearest 5 mm. Find the error interval for the area.
\[ \text{tolerance} = \dfrac{5}{2} = 2.5 \text{ mm} \]
\[ 82.5 \le L \lt 87.5 \]
\[ 57.5 \le W \lt 62.5 \]
\[ \begin{array}{rcl} A_{\min} &=& 82.5 \times 57.5 \\ &=& 4743.75 \\ A_{\max} &=& 87.5 \times 62.5 \\ &=& 5468.75 \end{array} \]
\[ 4740 \le A \lt 5470 \text{ mm}^2 \]
What's happening?
Find the bounds of each measurement first.
To make a product smallest, use the smallest value of each factor; to make it largest, use the largest of each.
The final interval is stated to 3 significant figures with units.
Example 4: Speed from rounded distance and time
Distance \(d = 140\) m (nearest 10 m) and time \(t = 8\) s (nearest 1 s). Find the error interval for the speed \(s = \dfrac{d}{t}\).
\[ 135 \le d \lt 145 \]
\[ 7.5 \le t \lt 8.5 \]
\[ \begin{array}{rcl} s_{\min} &=& 135 \div 8.5 \\ &=& 15.9 \\ s_{\max} &=& 145 \div 7.5 \\ &=& 19.3 \end{array} \]
\[ 15.9 \le s \lt 19.3 \text{ m/s} \]
What's happening?
For a quotient \(\dfrac{a}{b}\): the smallest result uses the smallest \(a\) and the largest \(b\).
The largest result uses the largest \(a\) and the smallest \(b\), the opposite of the product rule.
\(135 \div 8.5 = 15.882\ldots\), which rounds to \(15.9\) (3 s.f.).
Key Ideas
- Tolerance is half the accuracy. Subtract for the lower bound, add for the upper.
- Always write \(\text{lower} \le x \lt \text{upper}\): include the lower, exclude the upper.
- For a product: the minimum uses lower times lower; the maximum uses upper times upper.
- For a quotient \(\tfrac{a}{b}\): the minimum uses \(\tfrac{a_{\min}}{b_{\max}}\); the maximum uses \(\tfrac{a_{\max}}{b_{\min}}\).
- State units, and round a calculation answer to 3 significant figures.
Common Pitfalls
- Using the full accuracy as the tolerance instead of halving it.
- Writing \(\le\) at the upper end instead of \(\lt\).
- Mixing up the min and max rules for division: the smallest quotient uses the largest denominator, not the smallest.
- Rounding a bound early, before the final calculation, which compounds the error.
- Forgetting the units in the final error interval.