Estimation, Bounds & Accuracy · Notes

1.Lesson overview

Syllabus focus

Cambridge IGCSE syllabus reference

1.9 Estimation
1.10 Limits of accuracy

Edexcel IGCSE syllabus reference

1.8 Degree of accuracy

AQA IGCSE syllabus reference

1.1 Structure and calculation

An accuracy statement defines an interval, not a single exact value: its rounding unit fixes the half-width and its endpoints determine valid bounds. Use estimation to test whether an exact calculation is plausible before accepting it.

By the end of this lesson

1 Round values to a specified degree of accuracy.
2 Make estimates for calculations involving numbers, quantities and measurements.
3 Round answers to a reasonable degree of accuracy in the context of a given problem.
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.

2.Learning map

Follow the precision chain rounding interval → chosen endpoint → calculated bound → final rounding. Estimation checks size; bounds prove the greatest or least possible value.

Accuracy and bounds decision route

Accuracy describes an interval

Review standard form and calculator rounding before tackling bounds. Big idea. A stated measurement is usually a range of possible values, not an exact value.

Significant Figures

Significant figures (s.f.) tell us how many meaningful digits a number has. Understanding which digits are "significant" is the first step to rounding correctly.

Rounding to Significant Figures

To round a number to a given number of significant figures:

Rounding to Decimal Places

Decimal places (d.p.) count the number of digits after the decimal point .

3.Core idea

Review standard form and calculator rounding before tackling bounds.

Big idea. A stated measurement is usually a range of possible values, not an exact value.

Key relationship

lower bound \leq x < upper bound

Worked example

Example. A length is $8.4 cm$ , correct to the nearest $0.1 cm$ .

The true length satisfies $8.35 cm \leq l < 8.45 cm$ . The upper endpoint is excluded because 8.45 would round to 8.5.

4.Significant Figures

Significant figures (s.f.) tell us how many meaningful digits a number has. Understanding which digits are "significant" is the first step to rounding correctly.

Rules for Counting Significant Figures

1
All non-zero digits are significant: 4,732 has 4 s.f.
2
Zeros between non-zero digits are significant: 3{,}051 has 4 s.f.
3
Leading zeros (before the first non-zero digit) are not significant: 0.0042 has 2 s.f.
4
Trailing zeros after a decimal point are significant: 2.50 has 3 s.f.
5
Trailing zeros in a whole number may or may not be significant (ambiguous without context): 4500 could be 2, 3 or 4 s.f.

Counting Significant Figures

Number	Significant Figures	Explanation
45.6	3 s.f.	All digits non-zero
3{,}072	4 s.f.	Zero between 3 and 7 counts
0.00803	3 s.f.	Leading zeros don't count; 8, 0, 3 are significant
0.0040	2 s.f.	Leading zeros don't count; trailing zero after decimal does: 4, 0
70.10	4 s.f.	All digits significant (sandwiched zero + trailing zero after decimal)
600	1, 2 or 3 s.f.	Ambiguous — context needed

The Zero Trap

The zero in 0.0042 just shows how small the number is — it is a placeholder, not a significant figure. But the zero in 3{,}051 is sandwiched between significant digits and does count.

Counting and Rounding Significant FiguresOpen full screen

5.Rounding to Significant Figures

To round a number to a given number of significant figures:

Example 1: Round 47 832 to 3 s.f.

1
First 3 significant digits: 4, 7, 8
2
Decision digit: 3 (less than 5 → round down)
3
Answer: 47 800

Example 2: Round 47 862 to 3 s.f.

1
First 3 significant digits: 4, 7, 8
2
Decision digit: 6 (5 or more → round up)
3
Answer: 47 900

Example 3: Round 0.005 473 to 2 s.f.

1
First 2 significant digits: 5, 4 (leading zeros don't count)
2
Decision digit: 7 (round up)
3
Answer: 0.0055

Example 4: Round 0.039 61 to 1 s.f.

1
First significant digit: 3
2
Decision digit: 9 (round up, 3 becomes 4)
3
Answer: 0.04

Example 5: Round 996 to 2 s.f.

1
First 2 significant digits: 9, 9
2
Decision digit: 6 (round up, 99 becomes 100)
3
Answer: 1000

Key Point

After rounding a whole number, you must keep placeholder zeros to maintain the value. 47 832 rounded to 3 s.f. is 47 800, not 478.

Rounding to a Given Power of Ten

Whole numbers are also rounded to a stated power of 10 — the nearest ten, hundred, thousand, and so on — using the same "look at the next digit" rule, rather than a count of significant figures.

Example: write 5764 correct to the nearest thousand.

1
The thousands digit is 5; look at the next digit (hundreds) to decide: it is 7.
2
7 is 5 or more, so round up: 5764 rounds to 6000.

This overlaps with significant figures — 5764 to the nearest thousand and 5764 to 1 s.f. both give 6000 here, because the thousands digit is also the first significant digit. They are not always the same: 5764 to the nearest hundred is 5800, which is 5764 to 2 s.f.

6.Rounding to Decimal Places

Decimal places (d.p.) count the number of digits after the decimal point.

To round to

n

decimal places:

Example 1: Round 3.4572 to 2 d.p.

1
Keep 2 digits after the point: 3.45
2
Decision digit: 7 (round up)
3
Answer: 3.46

Example 2: Round 12.983 to 1 d.p.

1
Keep 1 digit after the point: 12.9
2
Decision digit: 8 (round up)
3
Answer: 13.0

Example 3: Round 0.06847 to 3 d.p.

1
Keep 3 digits after the point: 0.068
2
Decision digit: 4 (round down)
3
Answer: 0.068

Significant Figures vs Decimal Places

These are different:

0.06847 to 3 d.p. = 0.068 (count from the decimal point)
0.06847 to 3 s.f. = 0.0685 (count from the first non-zero digit)

Decimal places count position; significant figures count meaningful digits.

7.Estimating Calculations

To estimate the answer to a calculation, round every number to 1 significant figure first, then perform the simplified calculation.

The Estimation Method

1
Round each number in the calculation to 1 s.f.
2
Perform the calculation with these simpler numbers
3
This gives an approximate answer to check against

Example 1 (Syllabus): Estimate

\frac{6.24 \times 37.2}{0.389}

Round to 1 s.f.:

1
$6.24 \approx 6$
2
$37.2 \approx 40$
3
$0.389 \approx 0.4$

Estimate:

$\frac{6 \times 40}{0.4} = \frac{240}{0.4} = 600$

Estimated answer: 600

(Actual answer: 596.6... — very close!)

Example 2: Estimate

\frac{48.7 \times 0.523}{9.84}

1
Round to 1 s.f.: $48.7 \approx 50$ , $0.523 \approx 0.5$ , $9.84 \approx 10$
2
$\frac{50 \times 0.5}{10} = \frac{25}{10} = 2.5$
3
Estimated answer: 2.5
4
(Actual: 2.59...)

Example 3: Estimate

3.87 \times 24.6

1
Round to 1 s.f.: $3.87 \approx 4$ , $24.6 \approx 20$
2
$4 \times 20 = 80 \approx 9$
3
Estimated answer: 9
4
(Actual: 9.76...)

Example 4: Estimate

\frac{78. 3 ^{2}}{4.12 + 15.9}

1
Round to 1 s.f.: $78.3 \approx 80$ , $4.12 \approx 4$ , $15.9 \approx 20$
2
$\frac{8 0 ^{2}}{4 + 20} = \frac{6400}{24} \approx 267$
3
Estimated answer: about 270
4
(Actual: 306.0...)

Show Your Rounding!

In exams, you must show each value rounded to 1 s.f. to get full marks. Just writing the final estimate is not enough.

Reasonable Accuracy

When a question says "give your answer to a reasonable degree of accuracy", round to 3 significant figures as a default — unless the context suggests otherwise (e.g. money to 2 d.p., people as whole numbers).

8.Bounds Overview

When we measure something, the measurement is never perfectly exact. Every measurement has limits — a range within which the true value must lie. Understanding these limits of accuracy (also called bounds) is essential for working with real-world data.

Why This Matters

Engineers, scientists and surveyors must know how accurate their measurements are. A bridge designed with lengths "to the nearest metre" has very different safety margins from one measured "to the nearest millimetre". Bounds tell us the worst-case scenarios.

9.Upper & Lower Bounds

When a value is rounded (or truncated) to a given degree of accuracy, the true value could be anywhere within a range. The endpoints of this range are called the lower bound and upper bound.

A number line from 23 to 25 with the interval from 23.5 to 24.5 marked, a filled circle at 23.5 and an open circle at 24.5, for a length recorded as 24 cm to the nearest cm. — Figure 1: A length of 24 cm to the nearest cm can be as small as 23.5 cm, but it can never quite reach 24.5 cm.

Finding Bounds

If a measurement is given to a certain accuracy:

1
Find half of the degree of accuracy
2
Lower bound = given value - half the accuracy
3
Upper bound = given value + half the accuracy

Bounds Formula

Lower bound = value - \frac{1}{2} \times accuracy Upper bound = value + \frac{1}{2} \times accuracy

Example 1: Length

4.8 c m t o t h e n e a r es t 0.1 c m

1
Degree of accuracy 0.1 cm, so $ha l f = 0.05$ cm
2
Lower bound = $4.8 - 0.05$ 4.75 cm
3
Upper bound = $4.8 + 0.05$ 4.85 cm
4
The true length $l$ satisfies: $4.75 ⩽ l < 4.85$

Example 2: Mass

72 k g t o t h e n e a r es t k g

1
Degree of accuracy 1 kg, so $ha l f = 0.5$ kg
2
Lower bound = $72 - 0.5$ 71.5 kg
3
Upper bound = $72 + 0.5$ 72.5 kg
4
The true mass m satisfies: $71.5 ⩽ m < 72.5$

Example 3: Distance

200 m t o t h e n e a r es t 10 m

1
Degree of accuracy 10 m, so $ha l f = 5$ m
2
Lower bound = $200 - 5$ 195 m
3
Upper bound = $200 + 5$ 205 m
4
The true distance $d$ satisfies: $195 ⩽ d < 205$

Example 4: Time

3.60 s t o t h e n e a r es t 0.01 s

1
Degree of accuracy 0.01 s, so $ha l f = 0.005$ s
2
Lower bound = $3.60 - 0.005$ 3.595 s
3
Upper bound = $3.60 + 0.005$ 3.605 s
4
The true time $t$ satisfies: $3.595 ⩽ t < 3.605$

Example 5: Population 45 000 to the nearest 1000

1
Degree of accuracy 1000, so $ha l f = 500$
2
Lower bound = $45000 - 500 = 44500$
3
Upper bound = $45000 + 500 = 45500$

Common Mistake

Don't confuse "to the nearest 10" with "to 1 significant figure". For example, 400 to the nearest 10 gives bounds 395 and 405, but 400 to 1 s.f. gives bounds 350 and 450 (accuracy = 100).

10.Bounds Notation

Bounds are expressed using inequality notation. There is an important reason why we use a strict inequality (<) for the upper bound.

Standard Notation

lower bound ⩽ x < upper bound

Why is the upper bound strict (<)?

The lower bound is included $(⩽)$ because a value exactly on the lower bound would round up to the given value
The upper bound is excluded (<) because a value exactly on the upper bound would round up to the next value

For example, if a length is 4.8 cm to the nearest 0.1 cm:

4.75 rounds to 4.8 Correct: — so 4.75 is included
4.85 rounds to 4.9 ✗ — so 4.85 is NOT included

Example: Write the bounds using inequality notation

A length $l$ is 15 cm to the nearest cm.

1
Lower bound 14.5, upper bound = $15.5$
2
$14.5 ⩽ l < 15.5$

Summary of Common Bounds

Given Value	Accuracy	Lower Bound	Upper Bound	Inequality
7.3	1 d.p.	7.25	7.35	$7.25 ⩽ x < 7.35$
150	nearest 10	145	155	$145 ⩽ x < 155$
2.40	2 d.p.	2.395	2.405	$2.395 ⩽ x < 2.405$
8000	nearest 1000	7500	8500	$7500 ⩽ x < 8500$
0.6	1 s.f.	0.55	0.65	$0.55 ⩽ x < 0.65$

Exam Tip

When the question says "correct to 2 significant figures" or "correct to 1 decimal place", first work out the degree of accuracy, then find half of it. This is the step many students skip.

11.Calculations with Bounds

When we use rounded values in calculations, the answer also has bounds. The key is knowing which combination of upper and lower bounds gives the maximum or minimum result.

Rules for Bounds in Calculations

Operation	Maximum Result	Minimum Result
Addition $a + b$	upper(a) + upper(b)	lower(a) + lower(b)
Subtraction $a - b$	upper(a) - lower(b)	lower(a) - upper(b)
Multiplication $a \times b$	$u pp er (a) \times u pp er (b)$	$l o w er (a) \times l o w er (b)$
Division $a \div b$	$u pp er (a) \div l o w er (b)$	$l o w er (a) \div u pp er (b)$

Key Insight

For subtraction and division, opposite bounds give the extreme values. To get the biggest difference, use the biggest top and the smallest bottom. To get the biggest quotient, divide the biggest numerator by the smallest denominator.

Example 1: Area (Multiplication)

A rectangle has length $l = 4.8$ cm and width $w = 0.4$ cm, both measured to 1 decimal place.

1
Bounds for length: $4.75 ⩽ l < 4.85$
2
Bounds for width: $0.35 ⩽ w < 0.45$
3
Upper bound of area = $4.85 \times 0.45$ = 2.1825 $c m^{2}$
4
Lower bound of area = $4.75 \times 0.35$ = 1.6625 $c m^{2}$
5
So the area A satisfies: $1.6625 ⩽ A < 2.1825 c m^{2}$

Example 2: Speed (Division)

A car travels $d = 120$ km (nearest 10 km) in $t = 2.0$ hours (nearest 0.1 hour).

1
Bounds for distance: $115 ⩽ d < 125$
2
Bounds for time: $1.95 ⩽ t < 2.05$
3
Maximum speed = $\frac{upper ( d )}{lower ( t )} = \frac{125}{1.95} = 64.1$ ... $km/h$
4
Minimum speed = $\frac{lower ( d )}{upper ( t )} = \frac{115}{2.05} = 56.1$ ... $km/h$

Example 3: Perimeter (Addition)

A rectangle has length $l = 12$ cm and width $w = 8$ cm, both to the nearest cm.

1
Perimeter $P = 2 (l + w) = 2 (l + w)$
2
Bounds for $l$ : $11.5 ⩽ l < 12.5$
3
Bounds for $w$ : $7.5 ⩽ w < 8.5$
4
Upper bound of P = $2 (12.5 + 8.5) = 2 \times 21 = 42 cm$
5
Lower bound of P = $2 (11.5 + 7.5) = 2 \times 19 = 38 cm$

Example 4: Subtraction

Two weights are $A = 5.4$ kg and $B = 3.2$ kg, both to 1 d.p. Find the bounds of $A - B$ .

1
Bounds for A: $5.35 ⩽ A < 5.45$
2
Bounds for B: $3.15 ⩽ B < 3.25$
3
Maximum of $A - B$ = $5.45 - 3.15$ 2.30
4
Minimum of $A - B$ = $5.35 - 3.25$ 2.10

Memory Aid

For maximum: make the answer as big as possible. For multiplication, use both upper bounds. For division, divide something big by something small. For subtraction, subtract something small from something big.

Planned interactivecore

Standard Form, Rounding and Bounds

Build a rounding interval and propagate its endpoints through a calculation.

Connect rounded values, inequality notation and worst-case result bounds.

Inputs & controls

value
decimal or significant-figure precision
operation
second interval
endpoint selector

Learner actions

drag a true value
choose endpoints
test maximum and minimum

Visible outputs

number-line interval
bound inequality
lower and upper result
relative error

Implementation note

Default to positive IGCSE cases; show open versus closed endpoints and keep units visible.

Propagate Bounds Through a CalculationOpen full screen

12.Challenge Questions

Three stepping-stone questions that combine bounds with addition, division and subtraction — the operations where the max/min pairing rule actually matters.

Challenge 1 — bounds of a sum (perimeter)

Question. A rectangle has length 8.4 cm and width 5.2 cm, both measured to 1 decimal place. Find the upper and lower bounds of the perimeter.

1
Bounds of each side: $8.35 ⩽ l < 8.45$ and $5.15 ⩽ w < 5.25$ .
2
$P = 2 (l + w)$ . For addition, the maximum total uses the maximum of every part: $P_{ma x} = 2 (8.45 + 5.25) = 2 \times 13.70 = 27.40$ .
3
The minimum total uses the minimum of every part: $P_{min} = 2 (8.35 + 5.15) = 2 \times 13.50 = 27.00$ .
4
$27.00 cm ⩽ P < 27.40 cm$ .

Challenge 2 — bounds of a division (average speed)

Question. A car travels a distance of 120 km, correct to 2 significant figures, in a time of 2.5 hours, correct to 1 decimal place. Find the upper and lower bounds of the average speed, in km/h to 3 significant figures.

1
Bounds: $115 ⩽ d < 125$ and $2.45 ⩽ t < 2.55$ .
2
For division, the largest answer comes from the largest numerator paired with the smallest denominator — not the largest of both.
3
$speed_{ma x} = \frac{125}{2.45} = 51.0$ (3 s.f.).
4
The smallest answer pairs the smallest numerator with the largest denominator: $speed_{min} = \frac{115}{2.55} = 45.1$ (3 s.f.).
5
$45.1 km/h ⩽ speed < 51.0 km/h$ .

This is the step most students get backwards: for division, the bounds of the two quantities are crossed, not matched.

Challenge 3 — bounds of a subtraction (difference)

Question. Two masses are measured as 350 g and 180 g, each correct to the nearest 10 g. Find the upper and lower bounds of the difference between the masses.

1
Bounds: $345 ⩽ m_{1} < 355$ and $175 ⩽ m_{2} < 185$ .
2
For subtraction, like division, the bounds cross: the maximum difference is the largest $m_{1}$ minus the smallest $m_{2}$ : $355 - 175 = 180$ .
3
The minimum difference is the smallest $m_{1}$ minus the largest $m_{2}$ : $345 - 185 = 160$ .
4
$160 g ⩽ difference < 180 g$ .

The pairing rule for all four operations

Addition and multiplication — match like with like: max with max gives the maximum, min with min gives the minimum.

Subtraction and division — cross them: the maximum comes from (largest) $-$ or $\div$ (smallest), and the minimum comes from (smallest) $-$ or $\div$ (largest).

13.Exam-Style Worked Examples

Finding bounds for a calculation

1
Find each bound
Add and subtract half the degree of accuracy to each measurement.
2
Decide the operation
Are you adding, multiplying or dividing the quantities?
3
Adding or multiplying
Upper with upper for the maximum; lower with lower for the minimum.
4
Subtracting or dividing
Mix them — largest numerator with smallest denominator for the maximum.
5
Check the sense
The upper bound must be larger than the lower — if not, the combination is wrong.

Worked example 1 — estimation (3 marks)

Question. Estimate the value of $\frac{38.7 \times 5.14}{0.196}$ by rounding each number to 1 significant figure.

1
Round each: $38.7 \approx 40$ , $5.14 \approx 5$ , $0.196 \approx 0.2$ .
2
$\frac{40 \times 5}{0.2} = \frac{200}{0.2}$ .
3
$= 1000$ .

Marking. 1 mark per point. Show the rounded values before calculating — that is where the method mark is. Dividing by $0.2$ is the same as multiplying by $5$ .

Worked example 2 — bounds (4 marks)

Question. A rectangle has length $12.4 cm$ and width $8.6 cm$ , each measured to the nearest $0.1 cm$ . Find the upper and lower bounds of its area.

1
Each measurement has a half-interval of $0.05 cm$ .
2
Length: $12.35 ⩽ l < 12.45$ . Width: $8.55 ⩽ w < 8.65$ .
3
Upper bound of area $= 12.45 \times 8.65 = 107.6925 cm^{2}$ .
4
Lower bound of area $= 12.35 \times 8.55 = 105.5925 cm^{2}$ .

Marking. 1 mark per point. For an area both bounds go the same way — upper $\times$ upper. For a division they go opposite ways.

Worked example 3 — bounds in a division (4 marks)

Question. A car travels $250 m$ (to the nearest $10 m$ ) in $18 s$ (to the nearest second). Find the upper bound of its average speed.

1
Distance: $245 ⩽ d < 255$ . Time: $17.5 ⩽ t < 18.5$ .
2
Speed $= \frac{distance}{time}$ , so the largest speed needs the largest distance and the smallest time.
3
Upper bound $= \frac{255}{17.5}$ .
4
$= 14.571 \dots \approx 14.6 m/s$ .

Marking. 1 mark per point. Think about which combination makes the answer largest rather than applying a rule — for division the bounds are mixed.

14.Exam Tips & Common Misconceptions

Exam Tips

Estimate = round every value to 1 significant figure, then calculate. Don't round to 2 s.f. "to be safe".
Half the smallest unit gives bounds: 4.8 cm to 1 d.p. means 4.75 ≤ l < 4.85.
For maximum of $a - b$ : use max(a) - min(b). For minimum: min(a) - max(b).
For maximum of $a \div b$ : use max(a $) \div min ($ b). Opposite for minimum.
Quote answers to at least 3 s.f. unless the question specifies otherwise — and never round mid-calculation.

Common Misconceptions

"Bounds = ±0.5" — only when measuring to the nearest whole; halve the smallest unit of the actual accuracy.
Upper bound uses strict <: the upper bound is excluded (a value rounded to 4.8 cm cannot have been 4.85 cm exactly).
Using $ma x (a) \times min (b)$ for $ma x (a \times b)$ — for multiplication of positive values, $ma x \times ma x$ gives the maximum.
Estimating then rounding the answer — keep the estimate rough; don't try to "polish" it to 3 s.f.

15.Worked method — 4.2 correct to 1 decimal place defines a half-open interval

The lower midpoint rounds up to 4.2; the upper midpoint rounds to 4.3, so the upper endpoint is open.

Transfer the method

Reasoning prompt. A speed is distance divided by time. Which bounds create its greatest possible value?

1
Use the upper bound for the distance.
2
Use the lower bound for the time.
3
Keep the upper endpoint excluded when stating a bound interval.

Self-check

An estimate is deliberately approximate; a bound is a guaranteed limiting value.

16.From a rounding interval to a bound on a result

A reliable limits-of-accuracy workflow

1
Build intervals
A distance of 120 km to the nearest kilometre gives $119.5 \leq d < 120.5$ .
2
Choose endpoints
For positive speed $d / t$ , the lower bound uses the smallest distance and largest time.
3
Calculate unrounded
If $t = 2.4$ h to the nearest 0.1 h, then $t_{U} = 2.45$ and $v_{L} = 119.5/2.45$ .
4
State appropriately
$v_{L} = 48.7755 \dots$ km/h; do not round this upward and claim a value that may not be guaranteed.

Endpoint logic is operation-specific

The shortcut “lower with lower, upper with upper” fails for division. With positive quantities, making the denominator larger makes the quotient smaller. If an interval crosses zero or contains negative values, reason from the operation rather than applying a memorised endpoint rule.

Input intervals pass through an operation to produce result bounds

1
First interval
$2.5 \leq a < 3.5$ from 3 to nearest integer
2
Second interval
$4.5 \leq b < 5.5$ from 5 to nearest integer
3
Choose endpoints
positive product: · lower×lower, upper×upper
4
Result interval
$11.25 \leq ab < 19.25$ · state assumptions

For positive multiplication the smallest endpoints give the lower bound and the largest endpoints approach the upper bound.

17.Summary

Key Points

Central principle. A stated measurement is usually a range of possible values, not an exact value
Significant figures (s.f.) tell us how many meaningful digits a number has. Understanding which digits are "significant" is the first step to rounding correctly.
All non-zero digits are significant: 4,732 has 4 s.f.
Zeros between non-zero digits are significant: 3{,}051 has 4 s.f.
Leading zeros (before the first non-zero digit) are not significant: 0.0042 has 2 s.f.
Key relationship: $lower bound \leq x < upper bound$
Bounds Formula: $Lower bound = value - \frac{1}{2} \times accuracy Upper bound = value + \frac{1}{2} \times accuracy$
Standard Notation: $lower bound ⩽ x < upper bound$