C1.10 Understanding Bounds (Limits of Accuracy)

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C1.10 Understanding Bounds (Limits of Accuracy) – Enhanced IGCSE Mathematics Study Notes

C1.10 Understanding Bounds (Limits of Accuracy)
Enhanced IGCSE Mathematics Study Notes #

What You Need to Know: Sometimes, numbers are rounded to make them easier to work with. But the real number could have been a little smaller or bigger. In this topic, we learn how to find the smallest and biggest values it could have been — we call these the lower and upper bounds.

What Does Accuracy Mean? #

Before we start with limits of accuracy, let’s understand what accuracy means in mathematics:

Understanding Accuracy #

When we say a number is “accurate to the nearest 10”, it means the number has been rounded to the nearest multiple of 10.

For example:

  • $47$ rounded to the nearest 10 becomes $50$
  • $132$ rounded to the nearest 10 becomes $130$
  • $285$ rounded to the nearest 10 becomes $290$

What Are Understanding Bounds (Limits of Accuracy)? #

When a number has been rounded, we want to know what the original number could have been before it was rounded. This gives us two limits:

Upper Bound: The largest value the original number could have been
Lower Bound: The smallest value the original number could have been

Understanding Bounds with Simple Examples #

Example 1: Length measured to the nearest metre #

A length is measured as $5 \text{ m}$ to the nearest metre.

What could the actual length have been?

Step 1: First, let’s understand what “to the nearest metre” means.

This means the accuracy is $1$ metre. Any number that is closer to $5$ than to $4$ or $6$ will round to $5$.

Step 2: Work out half of the accuracy.

Half of the accuracy = $\frac{1}{2} \times 1 = 0.5$ metres

Step 3: Find the lower bound by subtracting the half-accuracy from the given number.

Lower bound = $5 – 0.5 = 4.5$ metres

This means any number from $4.5$ metres and above would round up to $5$ metres.

Step 4: Find the upper bound by adding the half-accuracy to the given number.

Upper bound = $5 + 0.5 = 5.5$ metres

This means any number less than $5.5$ metres would round down to $5$ metres.

Step 5: Put it all together.

The actual length could be anywhere from $4.5$ metres up to (but not including) $5.5$ metres.

Lower bound = $4.5 \text{ m}$
Upper bound = $5.5 \text{ m}$ (but not including $5.5$)
This is written as: $4.5 \leq x < 5.5$

How to Find Upper and Lower Bounds #

Step-by-Step Method for Finding Bounds #

Step 1: Work out half of the accuracy

Step 2: For the upper bound, add this half-accuracy to the given number

Step 3: For the lower bound, subtract this half-accuracy from the given number


Formula:

Half-accuracy = $\frac{1}{2} \times \text{accuracy}$

Upper bound = Given number + Half-accuracy

Lower bound = Given number – Half-accuracy

Important: The upper bound is not included in the possible values. We write it as “less than” (<) the upper bound, not "less than or equal to" (≤).

Working Through Different Types of Accuracy #

Example 2: Rounded to the nearest 10 #

A number is $470$ to the nearest 10. Find the upper and lower bounds.

Step 1: Identify the accuracy.

The number is rounded “to the nearest 10”, so the accuracy is $10$.

Step 2: Work out half of the accuracy.

Half of the accuracy = $\frac{1}{2} \times 10 = 5$

Step 3: Find the lower bound by subtracting the half-accuracy from the given number.

Lower bound = $470 – 5 = 465$

Step 4: Find the upper bound by adding the half-accuracy to the given number.

Upper bound = $470 + 5 = 475$

Lower bound = $465$
Upper bound = $475$ (but not including $475$)
We can write this as: $465 \leq x < 475$

🚀 NEW SECTION: Calculating with Upper and Lower Bounds #

When numbers are written to a specific degree of accuracy, calculations involving those numbers also give a range of possible answers. This is a crucial skill for IGCSE Extended candidates.

Key Rules for Calculations with Bounds:
When combining bounds in calculations, we need to think about which combination gives us the maximum and minimum possible results.

Simple Multiplications #

Worked Example: Multiplication with Bounds #

Calculate the upper and lower bounds for $34 \times 65$, given that each number is given to the nearest whole number.

Step 1: Find the bounds for each number.

$34$ lies in the range $33.5 \leq x < 34.5$

$65$ lies in the range $64.5 \leq x < 65.5$

Step 2: Understand what we’re calculating.

We want to find bounds for $34 \times 65$

This means: first number × second number, where the actual values lie within our bounds

Step 3: For multiplication, think about what makes the result smallest and largest.

• To get the smallest result: multiply the two smallest possible values

• To get the largest result: multiply the two largest possible values

Step 4: Calculate the lower bound.

Lower bound = smallest × smallest = $33.5 \times 64.5 = 2160.75$

Step 5: Calculate the upper bound.

Upper bound = largest × largest = $34.5 \times 65.5 = 2259.75$

Lower bound = $2160.75$
Upper bound = $2259.75$
Range: $2160.75 \leq \text{result} < 2259.75$
Rule for Multiplication: #

Lower bound: Multiply the two lower bounds together

Upper bound: Multiply the two upper bounds together

Working with Fractions (Division) #

Worked Example: Division with Bounds #

Calculate the upper and lower bounds for $\frac{33.5}{22.0}$ given that each number is accurate to 1 d.p.

Step 1: Find the bounds for each number.

$33.5$ lies in the range $33.45 \leq x < 33.55$

$22.0$ lies in the range $21.95 \leq x < 22.05$

Step 2: Calculate the lower bound.

Lower bound = $\frac{\text{smallest numerator}}{\text{largest denominator}} = \frac{33.45}{22.05} = 1.52$ (to 2 d.p.)

Step 3: Calculate the upper bound.

Upper bound = $\frac{\text{largest numerator}}{\text{smallest denominator}} = \frac{33.55}{21.95} = 1.53$ (to 2 d.p.)

Lower bound = $1.52$ (to 2 d.p.)
Upper bound = $1.53$ (to 2 d.p.)
Range: $1.52 \leq \text{result} < 1.53$
REMEMBER: #

To get the LOWER BOUNT: $\frac{\text{Smallest numerator}}{\text{Largest denominator}}$

To get the UPPER BOUND: $\frac{\text{Largest numerator}}{\text{Smallest denominator}}$

Perimeter Calculations #

Example: Rectangle Perimeter (From Exercise 2.7, Question 2) #

Calculate upper and lower bounds for the perimeter of the rectangle shown below, if its dimensions are correct to 1 d.p.

6.8 cm
4.2 cm
→

Rectangle with length 6.8 cm and width 4.2 cm (both to 1 d.p.)

Step 1: Find the bounds for each dimension.

Length $6.8$ cm lies in the range $6.75 \leq l < 6.85$

Width $4.2$ cm lies in the range $4.15 \leq w < 4.25$

Step 2: Write the perimeter formula.

Perimeter = $2 \times (\text{length} + \text{width})$

Perimeter = $2 \times (6.8 + 4.2)$

Perimeter = $2 \times (l + w)$ where $l$ and $w$ are the actual values

Step 3: Calculate the lower bound of the perimeter.

Use the smallest possible values: $l = 6.75$, $w = 4.15$

Lower bound = $2 \times (6.75 + 4.15) = 2 \times 10.9 = 21.8$ cm

Step 4: Calculate the upper bound of the perimeter.

Use the largest possible values: $l = 6.85$, $w = 4.25$

Upper bound = $2 \times (6.85 + 4.25) = 2 \times 11.1 = 22.2$ cm

Lower bound = $21.8$ cm
Upper bound = $22.2$ cm
Range: $21.8 \leq \text{perimeter} < 22.2$ cm
RMEMBER: #

Lower bound: Add all the lower bounds together

Upper bound: Add all the upper bounds together

Area Calculations #

Example: Rectangle Area (From Exercise 2.7, Question 4) #

Calculate upper and lower bounds for the area of the rectangle shown below, if its dimensions are accurate to 1 d.p.

10.0 cm
7.5 cm
Area = ?
→

Rectangle with length 10.0 cm and width 7.5 cm (both to 1 d.p.)

Step 1: Find the bounds for each dimension.

Length $10.0$ cm lies in the range $9.95 \leq l < 10.05$

Width $7.5$ cm lies in the range $7.45 \leq w < 7.55$

Step 2: Write the area formula.

Area = Length × Width

Area = $10.0 \times 7.5$

Area = $l \times w$ where $l$ and $w$ are the actual values

Step 3: Calculate the lower bound of the area.

Use the smallest possible values: $l = 9.95$, $w = 7.45$

Lower bound = $9.95 \times 7.45 = 74.1275$ cm²

Step 4: Calculate the upper bound of the area.

Use the largest possible values: $l = 10.05$, $w = 7.55$

Upper bound = $10.05 \times 7.55 = 75.8775$ cm²

Lower bound = $74.1$ cm² (to 1 d.p.)
Upper bound = $75.9$ cm² (to 1 d.p.)
Range: $74.1 \leq \text{area} < 75.9$ cm²

Finding Unknown Lengths (From Exercise 2.7, Question 6) #

Complex Example: Finding x when Area is Given #

Calculate upper and lower bounds for the length marked $x$ cm in the rectangle below. The area and length are both given to 1 d.p.

x cm
4.2 cm
Area = 55.8 cm²
→

Rectangle with unknown length x cm, width 4.2 cm, and area 55.8 cm²

Step 1: Find the bounds for the given measurements.

Width $4.2$ cm lies in the range $4.15 \leq w < 4.25$

Area $55.8$ cm² lies in the range $55.75 \leq A < 55.85$

Step 2: Start with the area formula and rearrange to find $x$.

Area = Length × Width

$55.8 = x \times 4.2$

$\frac{55.8}{4.2} = x$ (dividing both sides by 4.2)

So: $x = \frac{55.8}{4.2}$

Step 3: Now apply bounds to this formula $x = \frac{\text{Area}}{\text{Width}}$

Step 4: Calculate the lower bound of $x$.

For the smallest value of $x$: use smallest area ÷ largest width

Lower bound = $\frac{55.75}{4.25} = 13.12$ cm (to 2 d.p.)

Step 5: Calculate the upper bound of $x$.

For the largest value of $x$: use largest area ÷ smallest width

Upper bound = $\frac{55.85}{4.15} = 13.46$ cm (to 2 d.p.)

Lower bound = $13.12$ cm
Upper bound = $13.46$ cm
Range: $13.12 \leq x < 13.46$ cm

Practice Questions from Exercise 2.6 and 2.7 #

Exercise 2.6, Question 1(a): Simple Multiplication #

Calculate lower and upper bounds for $14 \times 20$, if each of the numbers is given to the nearest whole number.

Solution:

$14$ lies in range $13.5 \leq x < 14.5$

$20$ lies in range $19.5 \leq x < 20.5$

Lower bound = $13.5 \times 19.5 = 263.25$

Upper bound = $14.5 \times 20.5 = 297.25$

Answer: $263.25 \leq \text{result} < 297.25$

Exercise 2.6, Question 1(d): Simple Division #

Calculate lower and upper bounds for $\frac{40}{10}$, if each of the numbers is given to the nearest whole number.

Solution:

$40$ lies in range $39.5 \leq x < 40.5$

$10$ lies in range $9.5 \leq x < 10.5$

Lower bound = $\frac{39.5}{10.5} = 3.762$ (to 3 d.p.)

Upper bound = $\frac{40.5}{9.5} = 4.263$ (to 3 d.p.)

Answer: $3.76 \leq \text{result} < 4.27$ (to 2 d.p.)

Exercise 2.6, Question 2(b): Decimal Multiplication #

Calculate lower and upper bounds for $6.3 \times 4.8$, if each of the numbers is given to 1 d.p.

Solution:

$6.3$ lies in range $6.25 \leq x < 6.35$

$4.8$ lies in range $4.75 \leq x < 4.85$

Lower bound = $6.25 \times 4.75 = 29.6875$

Upper bound = $6.35 \times 4.85 = 30.7975$

Answer: $29.69 \leq \text{result} < 30.80$ (to 2 d.p.)

Exercise 2.7, Question 1: Combined Mass #

The masses to the nearest 0.5kg of two parcels are 1.5kg and 2.5kg. Calculate the lower and upper bounds of their combined mass.

Solution:

First parcel: $1.5$ kg lies in range $1.25 \leq m_1 < 1.75$

Second parcel: $2.5$ kg lies in range $2.25 \leq m_2 < 2.75$

Lower bound = $1.25 + 2.25 = 3.5$ kg

Upper bound = $1.75 + 2.75 = 4.5$ kg

Answer: $3.5 \leq \text{combined mass} < 4.5$ kg

Exercise 2.7, Question 5: Area with Significant Figures #

Calculate upper and lower bounds for the area of the rectangle whose dimensions are correct to 2 s.f.: 600 m by 120 m.

Solution:

$600$ m (2 s.f.) lies in range $595 \leq l < 605$

$120$ m (2 s.f.) lies in range $115 \leq w < 125$

Lower bound = $595 \times 115 = 68,425$ m²

Upper bound = $605 \times 125 = 75,625$ m²

Answer: $68,425 \leq \text{area} < 75,625$ m²

Summary Table – Enhanced #

Operation Lower Bound Formula Upper Bound Formula Example
Addition ($a + b$) Lower $a$ + Lower $b$ Upper $a$ + Upper $b$ Perimeter calculations
Subtraction ($a – b$) Lower $a$ – Upper $b$ Upper $a$ – Lower $b$ $17.6 – 4.2$ (both to 1 d.p.)
Multiplication ($a \times b$) Lower $a$ × Lower $b$ Upper $a$ × Upper $b$ Area calculations
Division ($\frac{a}{b}$) $\frac{\text{Lower } a}{\text{Upper } b}$ $\frac{\text{Upper } a}{\text{Lower } b}$ Finding unknown lengths

More Practice Questions from Exercises 2.6 and 2.7 #

Exercise 2.6, Question 2(f): Complex Fraction #

Calculate lower and upper bounds for $\frac{7.7 – 6.2}{3.5}$, if each number is given to 1 d.p.

Solution:

Step 1: Find bounds for each number.

$7.7$ lies in range $7.65 \leq x < 7.75$

$6.2$ lies in range $6.15 \leq x < 6.25$

$3.5$ lies in range $3.45 \leq x < 3.55$

Step 2: Work out the expression step by step.

We have: $\frac{7.7 – 6.2}{3.5}$

Let $N = 7.7 – 6.2$ (the numerator)

So our expression becomes: $\frac{N}{3.5}$

Step 3: Find bounds for the numerator $N = 7.7 – 6.2$.

For subtraction: Lower bound = Lower $a$ – Upper $b$

Lower bound of $N$ = $7.65 – 6.25 = 1.4$

Upper bound of $N$ = $7.75 – 6.15 = 1.6$

So: $1.4 \leq N < 1.6$

Step 4: Now find bounds for $\frac{N}{3.5}$.

Lower bound = $\frac{1.4}{3.55} = 0.394$ (to 3 d.p.)

Upper bound = $\frac{1.6}{3.45} = 0.464$ (to 3 d.p.)

Answer: $0.394 \leq \text{result} < 0.464$

Exercise 2.6, Question 3(a): Significant Figures #

Calculate lower and upper bounds for $64 \times 320$, if each number is given to 2 s.f.

Solution:

$64$ (2 s.f.) lies in range $63.5 \leq x < 64.5$

$320$ (2 s.f.) lies in range $315 \leq x < 325$

Lower bound = $63.5 \times 315 = 20,002.5$

Upper bound = $64.5 \times 325 = 20,962.5$

Answer: $20,003 \leq \text{result} < 20,963$ (to nearest whole number)

Exercise 2.7, Question 3: Perimeter with 2 d.p. #

Calculate upper and lower bounds for the perimeter of the rectangle whose dimensions are accurate to 2 d.p.: 4.86 m by 2.00 m.

Solution:

$4.86$ m lies in range $4.855 \leq l < 4.865$

$2.00$ m lies in range $1.995 \leq w < 2.005$

Perimeter = $2(l + w)$

Lower bound = $2(4.855 + 1.995) = 2 \times 6.85 = 13.70$ m

Upper bound = $2(4.865 + 2.005) = 2 \times 6.87 = 13.74$ m

Answer: $13.70 \leq \text{perimeter} < 13.74$ m

Key Rules Summary #

Essential Rules for Calculations with Bounds:
  • Addition: Add lower bounds for minimum, upper bounds for maximum
  • Subtraction: For $a – b$: Min = Lower $a$ – Upper $b$, Max = Upper $a$ – Lower $b$
  • Multiplication: Multiply lower bounds for minimum, upper bounds for maximum
  • Division: For $\frac{a}{b}$: Min = $\frac{\text{Lower } a}{\text{Upper } b}$, Max = $\frac{\text{Upper } a}{\text{Lower } b}$

Exam Tips #

Remember for your IGCSE exam:
  • Always identify what type of accuracy is being used (nearest 10, 1 decimal place, 2 significant figures, etc.)
  • Use the formula: bounds = given number ± (half the accuracy)
  • The upper bound is never included in the possible values
  • For calculations, think about which combination gives maximum and minimum results
  • Write your answer clearly showing both bounds with inequality notation
  • Include units if the question has units
  • Round your final answer appropriately for the context
⚠️ Watch Out! These Mistakes Are Common:
  • Don’t include the upper bound as a possible value
  • Don’t forget to halve the accuracy when finding bounds
  • Don’t mix up significant figures with decimal places
  • Don’t forget which combinations give maximum/minimum in division
  • Don’t forget to include units in your answer
  • Don’t round intermediate calculations too early

Quick Reference #

Step-by-Step Method – Always Follow These Steps #

Step 1: Identify the accuracy (what the number was rounded to)

Step 2: Work out half of the accuracy

Step 3: Find the lower bound = Given number – Half-accuracy

Step 4: Find the upper bound = Given number + Half-accuracy

Step 5: For calculations, apply the appropriate rule

Step 6: Write your final answer clearly with inequality notation


Remember: Half-accuracy = $\frac{1}{2} \times \text{accuracy}$

Notation: $\text{lower bound} \leq x < \text{upper bound}$

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