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Cambridge IGCSE Mathematics 0580 · Number

Limits of accuracy

What a rounded measurement really tells you: the smallest and largest values it could have come from, the lower and upper bounds. Extended students go on to find the bounds of a calculation, such as the greatest possible area of a rectangle or the least possible speed.

C1.10E1.10Core and ExtendedFree lesson

What the syllabus says

C1.10 (Core)

Give upper and lower bounds for data rounded to a specified accuracy.

Notes and examples

e.g. write down the upper bound of a length measured correct to the nearest metre.
Candidates are not expected to find the bounds of the results of calculations which have used data rounded to a specified accuracy.

E1.10 (Extended)

1 Give upper and lower bounds for data rounded to a specified accuracy.
2 Find upper and lower bounds of the results of calculations which have used data rounded to a specified accuracy.

Notes and examples

e.g. write down the upper bound of a length measured correct to the nearest metre.
Example calculations include:
calculate the upper bound of the perimeter or the area of a rectangle given dimensions measured to the nearest centimetre
find the lower bound of the speed given rounded values of distance and time.

Quoted from the Cambridge IGCSE Mathematics 0580 syllabus for 2025 to 2027.

Why a rounded value has bounds

Every measurement is rounded. A tape measure that says a wall is 23 m long to the nearest metre does not mean exactly 23 m: it means the length is closer to 23 m than to 22 m or 24 m.

The values that round to 23 run from 22.5 up to, but not including, 23.5.

22.5 is the lower bound: the smallest value that rounds to 23 (by the usual rule that 5 rounds up).
23.5 is the upper bound: the value where rounding would switch to 24.

A length of 23.5 m itself would round to 24, so the true length is always less than the upper bound. Even so, the upper bound is written as 23.5, not 23.49 or $23.4 \dot{9}$ : there is no "largest number below 23.5", so the halfway value is the agreed boundary. On the exam, 23.49 for the upper bound scores no marks.

The bounds on a number line

Every value on the shaded line rounds to 23. The filled circle at 22.5 means 22.5 is included; the open circle at 23.5 means 23.5 is not, because it would round up to 24.

Finding the bounds

The rule: the bounds are half a unit of accuracy either side of the rounded value.

Write down the unit the value was rounded to: 1 m, 0.1 kg, 10 people, 0.01 s.
Halve it.
Subtract the half to get the lower bound; add it to get the upper bound.

Some examples:

23 m to the nearest metre: half of 1 is 0.5, so 22.5 m and 23.5 m.
2.6 kg to 1 decimal place: the unit is 0.1, half is 0.05, so 2.55 kg and 2.65 kg.
4800 to 2 significant figures: the second significant figure is the hundreds digit, so the unit is 100 and half is 50. The bounds are 4750 and 4850.
40 g to the nearest 5 g: half of 5 is 2.5, so 37.5 g and 42.5 g.

Significant figures. Find which place the last significant figure is in; that is the unit. $0.36$ to 2 significant figures has its last figure in the hundredths, so the bounds are $0.355$ and $0.365$ .

A check that always works: the bounds should be the same distance from the rounded value, and the gap between them should be exactly one unit.

Worked example 1

(a) A length is 23 m, correct to the nearest metre. Write down its upper bound. (b) A mass is 2.6 kg, correct to 1 decimal place. Write down its lower and upper bounds. (c) The height of a hill is 4800 m, correct to 2 significant figures. Write down its lower bound.

(a) The unit is 1 m, so half a unit is 0.5 m. Upper bound: $23 + 0.5 = 23.5$ m.
(b) The unit is 0.1 kg; half is 0.05 kg. Lower bound $2.6 - 0.05 = 2.55$ kg; upper bound $2.6 + 0.05 = 2.65$ kg.
(c) The second significant figure, 8, is in the hundreds place, so the unit is 100 m and half is 50 m. Lower bound: $4800 - 50 = 4750$ m.

Answer: (a) 23.5 m (b) 2.55 kg and 2.65 kg (c) 4750 m

Writing bounds as an inequality

The range of possible values is often written as an inequality, sometimes called the error interval. If the length $L$ m is 23 m to the nearest metre,
$22.5 ⩽ L < 23.5$

Note the two different signs. The lower bound is included ( $⩽$ ), because 22.5 rounds up to 23. The upper bound is not ( $<$ ), because 23.5 rounds up to 24.

Counted data. Some values can only be whole numbers, such as a number of people. If a crowd is 3400 to the nearest hundred, the smallest possible crowd is 3350 and the largest is 3449. Exam questions about counted data usually ask for "the least possible number" or "the greatest possible number", and expect a whole number.

Check yourself

The length of a key is 3.7 cm, correct to 1 decimal place. Write down the lower bound and the upper bound of the length.

Answer it in the portal, where your answer is checked as you type →

Worked example 2

(a) The mass, $m$ grams, of a parcel is 40 g correct to the nearest 5 g. Write the possible values of $m$ as an inequality. (b) The number of people at a concert is 3400, correct to the nearest hundred. Write down the least possible number of people.

(a) The unit is 5 g, so half a unit is 2.5 g.
Lower bound $40 - 2.5 = 37.5$ ; upper bound $40 + 2.5 = 42.5$ .
So $37.5 ⩽ m < 42.5$ .
(b) Half of 100 is 50, so the lower bound is $3400 - 50 = 3350$ .
3350 rounds to 3400 (5 rounds up), so the least possible number is 3350.

Answer: (a) $37.5 ⩽ m < 42.5$ (b) 3350

Check yourself

The mass, $m$ grams, of a bag of rice is 250 g, correct to the nearest 10 g. Complete the inequality for $m$ , typed like 1.5 <= m < 2.5.

Answer it in the portal, where your answer is checked as you type →

Bounds of a calculation Extended

Extended candidates also find the bounds of a value worked out from rounded measurements. The question is always the same: which combination of bounds makes the answer as big, or as small, as it can be?

Adding. To make $a + b$ as big as possible, use the biggest $a$ and the biggest $b$ . Upper bound of $a + b$ = upper $a$ + upper $b$ . The lower bound uses both lower bounds.

Multiplying (positive numbers). The same idea: upper times upper gives the upper bound, lower times lower gives the lower bound. This is how the bounds of a perimeter or an area are found.

Subtracting. To make $a - b$ as big as possible, start with the biggest $a$ and take away as little as possible. Upper bound of $a - b$ = upper $a$ $-$ lower $b$ . The lower bound is lower $a$ $-$ upper $b$ .

Dividing. To make $\frac{a}{b}$ as big as possible, use the biggest top and the smallest bottom, because dividing by a smaller number gives a bigger answer. Upper bound of $\frac{a}{b}$ = $\frac{upper a}{lower b}$ , and the lower bound is $\frac{lower a}{upper b}$ .

Subtraction and division are where marks are lost: it feels natural to use "upper with upper", and for these two operations that is wrong.

This is why speed causes trouble. Speed is $\frac{distance}{time}$ , so the lower bound of the speed uses the lower bound of the distance and the upper bound of the time.

Which bounds to use Extended

Combining bounds (for positive values)
Calculation	Upper bound uses	Lower bound uses
$a + b$	upper $a$ + upper $b$	lower $a$ + lower $b$
$a \times b$	upper $a$ $\times$ upper $b$	lower $a$ $\times$ lower $b$
$a - b$	upper $a$ $-$ lower $b$	lower $a$ $-$ upper $b$
$a \div b$	upper $a$ $\div$ lower $b$	lower $a$ $\div$ upper $b$

Do not learn the table by heart: for each calculation ask which way each value should go to make the answer as large, or as small, as possible.

Worked example 3 Extended

A rectangle is 12 cm long and 7 cm wide, each measured correct to the nearest centimetre. (a) Calculate the upper bound of the perimeter. (b) Calculate the lower bound of the area.

Bounds of the length: 11.5 cm and 12.5 cm. Bounds of the width: 6.5 cm and 7.5 cm.
(a) The perimeter is $2 \times (length + width)$ , a sum, so use both upper bounds: $2 \times (12.5 + 7.5) = 2 \times 20 = 40$ cm.
(b) The area is length times width, so the lower bound uses both lower bounds: $11.5 \times 6.5 = 74.75 {cm}^{2}$ .
The answer is the exact bound. Do not round it unless the question asks you to.

Answer: (a) 40 cm (b) $74.75 {cm}^{2}$

Worked example 4 (exam style) Extended

A runner covers a distance of 200 m, correct to the nearest 10 m, in a time of 24.6 seconds, correct to 1 decimal place. Find the lower bound of her average speed. Give your answer correct to 3 significant figures.

Bounds of the distance: 195 m and 205 m. Bounds of the time: 24.55 s and 24.65 s.
Speed $= \frac{distance}{time}$ . For the smallest speed, use the smallest distance and the largest time.
Lower bound $= \frac{195}{24.65} = 7.9107 \dots$
Correct to 3 significant figures: 7.91 m/s.
Using $\frac{195}{24.55}$ instead gives 7.94, the most common wrong answer: dividing by the smaller time makes the speed bigger, not smaller.

Answer: $7.91$ m/s

Check yourself Extended

$a = 8.6$ and $b = 3.2$ , both correct to 1 decimal place. Calculate the upper bound of $a - b$ .

Answer it in the portal, where your answer is checked as you type →

Giving an answer to a suitable accuracy Extended

Bounds also tell you how accurately a calculated answer can honestly be given. Work out the lower and upper bounds of the answer, then round both: the answer can be given to the greatest accuracy at which the two bounds agree.

For example, if a density lies between 7.62 and 8.32 ${g/cm}^{3}$ , then to 2 significant figures the bounds are 7.6 and 8.3, which disagree. To 1 significant figure both are 8. So the density is 8 ${g/cm}^{3}$ to 1 significant figure, and claiming any more accuracy is not justified by the measurements.

When a question asks you to give an answer "to an appropriate degree of accuracy" and explain, show both bounds rounded and say that they agree to that accuracy and no further.

Key points

A value rounded to a unit of accuracy has bounds half a unit either side.
The upper bound is written as the halfway value (23.5), never 23.49, even though the true value is less than it.
As an inequality: lower bound $⩽ x <$ upper bound.
For significant figures, find the place value of the last significant figure: that is the unit.
Extended: for $+$ and $\times$ , combine upper with upper and lower with lower.
Extended: for $-$ and $\div$ , combine the upper bound of one with the lower bound of the other.
Extended: an answer can be given to the accuracy at which its upper and lower bounds agree.

Where marks are lost

Adding and subtracting a whole unit instead of half a unit: 2.6 to 1 d.p. is not between 2.5 and 2.7.
Writing the upper bound as 23.49 or 23.4999. It is 23.5.
Getting the unit wrong for significant figures: 4800 to 2 s.f. has unit 100, so the bounds are 4750 and 4850.
Using $⩽$ at both ends of an error interval.
Extended: dividing upper by upper for the upper bound of a speed or density.
Extended: subtracting upper from upper for the upper bound of a difference.
Extended: rounding the bounds of the measurements before using them, or rounding an exact bound such as 74.75 when the question did not ask.

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An original lesson written to the Cambridge IGCSE Mathematics 0580 syllabus (2025 to 2027). The worked examples and questions are our own, not Cambridge questions. Back to every IGCSE Mathematics 0580 lesson.

Limits of accuracy

Why a rounded value has bounds

The bounds on a number line

Finding the bounds

Worked example 1

Writing bounds as an inequality

Check yourself

Worked example 2

Check yourself

Bounds of a calculation Extended

Which bounds to use Extended

Worked example 3 Extended

Worked example 4 (exam style) Extended

Check yourself Extended

Giving an answer to a suitable accuracy Extended

Key points

Where marks are lost

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Ratio and proportion

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Standard form