Estimation and Limits of Accuracy | IGCSE Maths Revision

Syllabus coverage

Round values to a specified degree of accuracy, including decimal places and significant figures
Make estimates for calculations involving numbers, quantities and measurements
Round answers to a reasonable degree of accuracy in the context of a given problem
Give upper and lower bounds for data rounded to a specified accuracy
Write down the upper bound of a length measured correct to the nearest metre
Core candidates are not expected to find the bounds of the results of calculations which have used data rounded to a specified accuracy
Extended only: find upper and lower bounds of the results of calculations, for example the upper bound of the perimeter or the area of a rectangle given dimensions measured to the nearest centimetre, or the lower bound of a speed given rounded values of distance and time

The band of values a length given as 7 m correct to the nearest metre could have: 6.5 m is included, 7.5 m is not.

Rounding to a given degree of accuracy

Rounding replaces a number with a nearby, simpler one. The degree of accuracy says how simple: to the nearest thousand, to 1 decimal place, to 3 significant figures. Cambridge always states which one it wants, and the mark is for the value rounded to exactly that accuracy - not close to it.

Decimal places count digits after the decimal point. $27.638$ to 1 decimal place is $27.6$ , because the second digit after the point is 3 and rounds down. Significant figures count digits from the first non-zero digit. The same number to 4 significant figures is $27.64$ : the figures are 2, 7, 6 and 3, and the next digit 8 rounds the 3 up to 4.

The two are not interchangeable and questions swap between them. $0.0463$ to 2 significant figures is $0.046$ , not $0.05$ : the leading zeros are place holders, so the first two significant figures are 4 and 6.

Value	Nearest whole number	To 2 sf	To 3 sf
5764	5764	5800	5760
27.638	28	28	27.6
0.0463	0	0.046	0.0463
0.5995	1	0.60	0.600

Count significant figures from the first non-zero digit and decimal places from the point. A question asking for 3 significant figures is not asking for 3 decimal places, and the two answers are usually different.

Estimating a calculation: round every number to 1 significant figure

An estimate is a value you can work out in your head. The method Cambridge expects is to write every number in the calculation correct to 1 significant figure, then do the arithmetic with those rounded values.

Take the estimate of $\frac{38.7 \times 0.51}{19.6}$ . To 1 significant figure $38.7$ is $40$ , $0.51$ is $0.5$ and $19.6$ is $20$ . Then $40 \times 0.5 = 20$ and $20 \div 20 = 1$ , so the estimate is $1$ . The exact value is about $1.007$ , which is why the estimate is worth having: it tells you the answer is about 1 and not about 10.

Estimates catch calculator slips. For $4.87 \times 612$ , rounding gives $5 \times 600 = 3000$ , so an answer of $298044$ or $29.8$ is wrong before you check anything else.

Round every number to 1 significant figure before you multiply or divide
Keep the structure of the calculation: if the question divides, you divide
The estimate is not the exact answer, so do not write down every digit from the display

The method mark in an estimation question is for the rounded values and the arithmetic with them. An answer written with no working cannot show that 1 significant figure was used.

Choosing a sensible degree of accuracy for an answer

The syllabus also asks you to round answers to a reasonable degree of accuracy in the context of a given problem. When the question does not state an accuracy, the context decides it: money goes to 2 decimal places, a count of people or objects to a whole number, and a measurement no more accurately than the measurements you were given.

So a calculation that produces $199.2$ people is written as $199$ - half a person is not an answer - and a cost of $6.2719$ dollars is written as $6.27$ dollars, because a price has two decimal places.

Never copy the whole display. Writing $6.2719345$ claims a precision the data does not have, and in a question that asks for a reasonable accuracy an over-precise answer is not what was asked for.

Money: 2 decimal places, to the nearest cent
People, chairs, cars: a whole number
If the question states an accuracy, that instruction wins over the context

Upper and lower bounds: the band a rounded value hides

A measurement is never exact. If a length is given as $7$ m correct to the nearest metre, the true length lies in a band either side of 7 m. The lower bound is the smallest value that would still round to 7 m, and the upper bound is the largest.

The width of the band comes from the accuracy. For the nearest metre the rounding unit is $1$ m, and the band reaches half a unit - $0.5$ m - either side. That gives $6.5 \leq x < 7.5$ .

Look at the signs. The lower bound is included, because a length of exactly $6.5$ m rounds up to $7$ m. The upper bound is not, because exactly $7.5$ m rounds up to $8$ m. Cambridge expects $6.5 \leq x < 7.5$ , not $6.5 \leq x \leq 7.5$ .

Stated accuracy	Half a unit is ...	Example
Nearest whole number	0.5	$7$ gives $6.5 \leq x < 7.5$
Nearest 10	5	$340$ gives $335 \leq x < 345$
Nearest 0.1 (1 decimal place)	0.05	$12.6$ gives $12.55 \leq x < 12.65$
Nearest 0.01 (2 decimal places)	0.005	$0.72$ gives $0.715 \leq x < 0.725$
Nearest 0.001 (3 decimal places)	0.0005	$5.406$ gives $5.4055 \leq x < 5.4065$

Half the unit, never the whole unit. The upper bound of a length measured as 7 m to the nearest metre is 7.5 m - not 8 m, and not 7.1 m.

Significant figures and the half-unit

The half-unit rule still works when the accuracy is given in significant figures, but first you have to see which column the last significant figure sits in. For $4.62$ given to 3 significant figures, the last figure is in the hundredths column, so the rounding unit is $0.01$ and half a unit is $0.005$ : $4.615 \leq x < 4.625$ .

The trap is a number whose significant figures are not in the columns you expect. In $360$ given to 2 significant figures the 6 is in the tens column, so the rounding unit is $10$ , half of that is $5$ , and the bounds are $355 \leq x < 365$ - not $359.5$ and $360.5$ .

Work out the column, then halve. Writing the rounding unit down before you halve it is the whole difference between the two answers.

$4.62$ to 3 sf: unit $0.01$ , half a unit $0.005$ , so $4.615 \leq x < 4.625$
$360$ to 2 sf: unit $10$ , half a unit $5$ , so $355 \leq x < 365$
$0.0428$ to 3 sf: unit $0.0001$ , half a unit $0.00005$ , so $0.04275 \leq x < 0.04285$
$1.5$ to 2 sf: unit $0.1$ , half a unit $0.05$ , so $1.45 \leq x < 1.55$

Bounds of a calculation (Extended only)

Core candidates are not expected to find the bounds of the results of calculations that used rounded data. Extended candidates are, and the two examples Cambridge gives are the perimeter or the area of a rectangle measured to the nearest centimetre, and the speed from rounded values of distance and time.

The method is to choose the values that push the result the way you want. For the upper bound of a sum or a product, use the upper bound of every measurement; for the lower bound, use the lower bound of every measurement. For a quotient, the largest answer comes from the largest numerator over the smallest denominator, and the smallest answer from the smallest numerator over the largest denominator.

A rectangle is measured as $8$ cm by $5$ cm, each to the nearest centimetre, so $7.5 \leq l < 8.5$ and $4.5 \leq w < 5.5$ . The upper bound of the area is $8.5 \times 5.5 = 46.75$ square centimetres and the lower bound is $7.5 \times 4.5 = 33.75$ square centimetres. The upper bound of the perimeter is $2 (8.5 + 5.5) = 28$ cm and the lower bound is $2 (7.5 + 4.5) = 24$ cm.

For a speed, a distance of $200$ m to the nearest 10 m gives $195 \leq d < 205$ and a time of $25$ s to the nearest second gives $24.5 \leq t < 25.5$ . The lower bound of the speed is $195 \div 25.5 = 7.65$ metres per second to 3 significant figures, and the upper bound is $205 \div 24.5 = 8.37$ metres per second.

The upper bound of a quotient is the biggest top over the smallest bottom. Pairing those the wrong way round is the standard error on this question.

A bounds question, start to finish

A square tile is measured as $12$ cm along one side, correct to the nearest centimetre. Step 1: the rounding unit is $1$ cm, so half a unit is $0.5$ cm. Step 2: the side length lies between $11.5$ and $12.5$ , written $11.5 \leq s < 12.5$ . Step 3: the perimeter is $4 s$ , so the lower bound of the perimeter is $4 \times 11.5 = 46$ cm and the upper bound is $4 \times 12.5 = 50$ cm.

The area spreads even further, because a product of two bounds moves further than either of them. The measured area is $144$ square centimetres, but the lower bound is $11.5 \times 11.5 = 132.25$ square centimetres and the upper bound is $12.5 \times 12.5 = 156.25$ square centimetres.

Write the bounds as an inequality with the smaller value first, and keep the unit on each value. Both parts of the answer come from the same first step, so an error in halving the unit costs every mark in the question.

$12$ cm to the nearest centimetre: $11.5 \leq s < 12.5$
Lower bound of the perimeter: $4 \times 11.5 = 46$ cm
Upper bound of the perimeter: $4 \times 12.5 = 50$ cm
Lower bound of the area: ${11.5}^{2} = 132.25$ square centimetres
Upper bound of the area: ${12.5}^{2} = 156.25$ square centimetres

Half the rounding unit is the whole topic. Get that one step right and the rest of the question is ordinary arithmetic.

Key terms you must be able to define

Round: Replace a number with a nearby value that has fewer figures: 5764 rounded to the nearest hundred is 5800.

Degree of accuracy: How precise a rounded value is - to the nearest thousand, to 1 decimal place, or to 3 significant figures.

Decimal place (dp): A digit position after the decimal point. 27.638 to 1 decimal place is 27.6.

Significant figure (sf): A digit counted from the first non-zero digit. 0.0463 has three significant figures: 4, 6 and 3.

Estimate: A value found by writing every number in a calculation correct to 1 significant figure, so the arithmetic can be done in your head.

Upper bound: The largest value that would still round to the given value. For 7 m measured to the nearest metre it is 7.5 m.

Lower bound: The smallest value that would still round to the given value. For 7 m measured to the nearest metre it is 6.5 m.

Limits of accuracy: The two values between which the true measurement must lie, written as an inequality such as $6.5 \leq x < 7.5$ .

Rounding unit: The place value of the stated accuracy: 1 for the nearest whole number, 0.01 for 2 decimal places, 10 for the nearest 10.

Half-unit: Half of the rounding unit, and the distance from the rounded value to each bound: 0.05 for 1 decimal place, 5 for the nearest 10.

Inequality notation for bounds: Cambridge writes the lower bound included and the upper bound excluded, as in $1.5 \leq x < 2.5$ .

Reasonable degree of accuracy: An accuracy that suits the context: money to 2 decimal places, a count of people to a whole number.

Bounds of a calculation: The range of possible results when the measurements used in the calculation have themselves been rounded (Extended).

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

Three accuracies, one number

Take $0.0463$ . To 2 significant figures it is $0.046$ , to 3 significant figures it is $0.0463$ , and to the nearest whole number it is $0$ . The leading zeros are place holders, so the first significant figure is the 4 and not the first digit you can see.

Estimating a division

Estimate $\frac{38.7 \times 0.51}{19.6}$ . Each number to 1 significant figure: $40$ , $0.5$ and $20$ . Then $40 \times 0.5 \div 20 = 1$ . The exact value is $1.007$ , so the estimate is close enough to confirm a calculator answer and far enough out to expose a decimal point in the wrong place.

A bound from the nearest metre

A path is measured as $7$ m correct to the nearest metre. The rounding unit is $1$ m and half a unit is $0.5$ m, so the length lies between $6.5$ m and $7.5$ m. Write it as $6.5 \leq x < 7.5$ : the lower bound is included because $6.5$ m rounds up to $7$ m.

A bound from significant figures

A bag of flour has a mass of $1.5$ kg correct to 2 significant figures. The second significant figure is the tenths digit, so the rounding unit is $0.1$ kg and half a unit is $0.05$ kg. The mass lies between $1.45$ kg and $1.55$ kg, written $1.45 \leq m < 1.55$ .

Extended: the bounds of an area

A rectangle is measured as $8$ cm by $5$ cm, each correct to the nearest centimetre. The upper bound of the area uses both upper bounds: $8.5 \times 5.5 = 46.75$ square centimetres. The lower bound uses both lower bounds: $7.5 \times 4.5 = 33.75$ square centimetres. Both are further from the measured $40$ square centimetres than the bounds of either length.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1Round

5764

to the nearest thousand.

6000

. The digit in the hundreds column is 7, so the 5 rounds up.

Q2Round

5764

to 3 significant figures.

5760

. The first three significant figures are 5, 7 and 6, and the next digit 4 rounds down.

Q3Round

0.0463

to 2 significant figures.

0.046

. The leading zeros are not significant, so the first two significant figures are 4 and 6.

Q4Round

6.2784

to 2 decimal places.

6.28

. The third digit after the point is 8, so the 7 rounds up to 8.

Q5How many significant figures are there in

0.0305

Three: 3, 0 and 5. The two leading zeros are place holders.

Q6What is the upper bound of a length measured as

7

m correct to the nearest metre?

7.5

m. Half of the

1

m rounding unit each side of 7 m gives

6.5 \leq x < 7.5

Q7Write the limits of accuracy of

8.4

correct to 1 decimal place.

8.35 \leq x < 8.45

. The rounding unit is

0.1

, so half a unit is

0.05

Q8Write the limits of accuracy of

360

correct to 2 significant figures.

355 \leq x < 365

. The second significant figure is in the tens column, so the rounding unit is

10

Q9Write the limits of accuracy of

0.72

correct to 2 decimal places.

0.715 \leq x < 0.725

. The rounding unit is

0.01

and half a unit is

0.005

Q10Estimate

4.87 \times 612

5 \times 600 = 3000

. Each number is written correct to 1 significant figure first.

Q11Estimate

\frac{38.7 \times 0.51}{19.6}

40 \times 0.5 \div 20 = 1

. The exact value is

1.007

Q12Which bound is included when you write limits of accuracy?

The lower bound is included and the upper bound is not, because a value exactly at the upper bound rounds up to the next value:

1.5 \leq x < 2.5

Q13A rectangle is measured to the nearest centimetre. Which values give the upper bound of its area?

Both upper bounds. For

8

cm by

5

cm:

8.5 \times 5.5 = 46.75

square centimetres (Extended).

Q14What is the upper bound of a value of

0.0428

given to 3 significant figures?

0.04285

. The last significant figure is in the fourth decimal column, so the rounding unit is

0.0001

and half a unit is

0.00005

0 of 14 marked as known

Exam-style questions

8 questions, 17 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

Write downAO11 mark

Write down

27.638

correct to 4 significant figures.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

B1 $27.64$ : the first four significant figures are 2, 7, 6 and 3, and the next digit 8 rounds the 3 upnot checked yet

Examiner tip: Write down means no working is needed. Count significant figures from the first non-zero digit - 2, 7, 6, 3 - then look at the digit after them to decide whether to round up.

Work outAO12 marks

By writing each number correct to 1 significant figure, work out an estimate for

\frac{38.7 \times 0.51}{19.6}

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1writes $38.7$ , $0.51$ and $19.6$ correct to 1 significant figure as $40$ , $0.5$ and $20$
A1 $40 \times 0.5 \div 20 = 1$ not checked yet

Examiner tip: The method mark is for the rounded values, so write them down before you multiply.

0.51

to 1 significant figure is

0.5

and not

1

, because the leading zero is not a significant figure.

StateAO11 mark

The length of a corridor is given as

7

m correct to the nearest metre. State the upper bound of the length.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

B1 $7.5$ m, because the rounding unit is $1$ m and the band reaches $0.5$ m above 7 mnot checked yet

Examiner tip: Half the unit, not the whole unit. The upper bound of a value rounded to the nearest metre is never the next whole metre.

CalculateAO22 marks

A bag of flour has a mass of

1.5

kg correct to 2 significant figures. Calculate the lower bound and the upper bound of the mass.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1the second significant figure is in the tenths column, so the rounding unit is $0.1$ kg and half a unit is $0.05$ kg
A1lower bound $1.45$ kg and upper bound $1.55$ kg, so $1.45 \leq m < 1.55$ not checked yet

Examiner tip: The accuracy is given in significant figures, so find the column of the second significant figure first - here it is the tenths. The bounds are

0.05

kg either side of

1.5

kg, not

0.5

kg either side.

ExplainAO12 marks

A length is given as

7

m correct to the nearest metre. Explain why

7.5

m is not included in the possible values of the length.

Your answer

Write a short answer in words.

Mark scheme — tick each point you actually made

B1a length of exactly $7.5$ m rounds up to $8$ m, not down to $7$ m
B1so the values are $6.5 \leq x < 7.5$ : the lower bound is included and the upper bound is notnot checked yet

Examiner tip: The reason is about the rounding, not about the number being large.

7.49

m is fine because it rounds down to

7

7.5

m is not, because it rounds up to

8

CalculateAO23 marks

A car travels

200

m, correct to the nearest 10 m, in a time of

25

seconds, correct to the nearest second. Calculate the lower bound of the average speed, in metres per second, correct to 3 significant figures.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1lower bound of the distance is $195$ m and upper bound of the time is $25.5$ s
M1smallest distance divided by largest time: $195 \div 25.5$
A1 $7.65$ metres per second, to 3 significant figuresnot checked yet

Examiner tip: This part is Extended content. The smallest speed comes from the smallest distance and the largest time, so the lower bound of a quotient takes the lower bound on top and the upper bound underneath.

CalculateAO24 marks

A rectangular field is measured as

120

m long and

80

m wide, each correct to the nearest 10 m. Calculate the upper bound of the area of the field, in square metres.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

M1upper bound of the length is $125$ m
M1upper bound of the width is $85$ m
M1multiplies the two upper bounds: $125 \times 85$
A1 $10625$ square metresnot checked yet

Examiner tip: This part is Extended content. The largest possible area uses the largest possible length and the largest possible width, so both measurements take their upper bound.

ExplainAO22 marks

A calculator gives the number of people attending a concert as

199.2

. Explain why this is not a sensible answer, and state the value that should be written down instead.

Your answer

Write a number, a fraction such as 3/4, or an expression such as 3x+2.

Mark scheme — tick each point you actually made

B1a count of people must be a whole number, so an answer with a decimal part cannot be right in this context
B1 $199$ , to the nearest whole numbernot checked yet

Examiner tip: This is the syllabus requirement to round answers to a reasonable degree of accuracy in the context of the problem. The context decides the accuracy when the question does not state one.

Exam technique for this topic

Half the unit is the whole rule for bounds. Write the rounding unit down first - 1 m, 0.01, 10 - then halve it, and remember that the upper bound of a quotient needs the largest numerator over the smallest denominator.
Use the inequality form Cambridge expects: lower bound included, upper bound excluded, so $1.5 \leq x < 2.5$ . Writing $\leq$ at both ends changes the answer, because a value of exactly $2.5$ rounds up to $3$ and cannot be in the band.
In an estimation question, show the rounded values you used. The method mark is for writing each number correct to 1 significant figure, and it is still earned if your arithmetic afterwards slips.
Do not write every digit from the display. Give the accuracy the question asks for, or, if it does not say, an accuracy that fits the context - 2 decimal places for money, a whole number for people.
Check which column the last significant figure is in before you halve. $360$ correct to 2 significant figures is accurate to the nearest $10$ , so its bounds are $355$ and $365$ , not $359.5$ and $360.5$ .

Why students lose marks here

A student wroteThe upper bound of a length measured as

7

m correct to the nearest metre is

8

Why it loses marksThe band reaches half a unit, not a whole one. A length of

7.4

m still rounds down to

7

m, so

8

m is too far out.

The answer that scoresThe upper bound is

7.5

m and the lower bound is

6.5

m, so

6.5 \leq x < 7.5

A student wrote

0.0463

correct to 2 significant figures is

0.05

Why it loses marksThe two leading zeros are place holders, not significant figures, so the first significant figure is the 4. Counting from the decimal point gives the wrong two digits.

The answer that scores

0.0463

to 2 significant figures is

0.046

A student wroteI estimated

4.87 \times 612

2980.44

, which is what my calculator gave.

Why it loses marksThat is the exact value, not an estimate. The syllabus method is to write each number correct to 1 significant figure first, and the mark is for that step.

The answer that scores

4.87

rounds to

5

and

612

rounds to

600

, so the estimate is

5 \times 600 = 3000

A student wrote

360

correct to 2 significant figures has bounds

359.5

and

360.5

Why it loses marksHalving

1

gives

0.5

, but the rounding unit here is not

1

. The second significant figure is in the tens column, so the unit is

10

The answer that scoresHalf of

10

5

, so

355 \leq x < 365

A student wroteThe lower bound of

12.6

cm measured correct to 1 decimal place is

12.5

cm.

Why it loses marksThe rounding unit for 1 decimal place is

0.1

, so half a unit is

0.05

and not

0.1

. The error is the half-unit step again.

The answer that scoresThe bounds are

12.55 \leq x < 12.65

Where this fits in IGCSE Maths

This page is one topic of Cambridge IGCSE Maths 0580. IGCSE Maths is the course home: every area and every topic in Cambridge's own order, the papers you sit and what each one tests, and the exam-practice tools. The full topic list, with each topic linked to its revision page, is on the IGCSE Maths syllabus page.