Bounds, estimation and error intervals | OCR GCSE Maths Notes

Visual model

Bounds show the interval a rounded value could be in

7.45

7.55

7.45 \le \times < 7.55

Gold-standard guide

20 mins

What you will learn

Estimate answers and use limits of accuracy.

Use a clear step-by-step method for bounds, estimation and error intervals.

Check your answer and avoid the most common exam mistake.

Useful before you start

Core number skillsEarlier number skillsShowing clear working

Core knowledge

Know the rule, then use it

These are the short notes. Read each one, then check you can use it in the worked example below.

Method

Upper bound of a product: use the upper bound of BOTH measurements

Step 1

Find the upper bound of each measurement

Measured to nearest 0.1 cm, so accuracy = 0.1 and half-accuracy = 0.05

Step 2

Calculate the upper bound of the area

Upper bound of area = 8.45 × 5.15

Step 3

Evaluate

8.45 × 5.15 = 43.5175 cm²

Watch out

Students use the upper bound for one measurement and the lower bound for the other

Error interval

$rounded value - half unit \le \times < rounded value + half unit.$

Product bounds

maximum product uses upper x upper; minimum product uses lower x lower.

Worked example

A rectangle has length 8.4 cm and width 5.1 cm, each measured to the nearest 0.1 cm. Find the upper bound of the area.

05. 45 cm. 15 cm. We add half the accuracy because any value up to (but not including) this rounds to the given value.

15. We use the upper bounds of BOTH because making both as large as possible gives the largest possible area.

Evaluate: 8.45 × 5.15 = 43.5175 cm².

Final answer

43.5175 cm²

Question ladder

Build up to the hardest questions

Do them in order. If you miss a step, read the solution, then redo the question without looking.

Workedreasoning

A rectangle has length 8.4 cm and width 5.1 cm, each measured to the nearest 0.1 cm. Find the upper bound of the area.

3 marks4 minsbounds-and-estimation-worked

Show solution

Worked solution

1.05. 45 cm. 15 cm. We add half the accuracy because any value up to (but not including) this rounds to the given value.
2.15. We use the upper bounds of BOTH because making both as large as possible gives the largest possible area.
3.Evaluate: 8.45 × 5.15 = 43.5175 cm².

Final answer

43.5175 cm²

Mark points

M1: find the upper bound of each measurement
M1: calculate the upper bound of the area
M1: evaluate
A1: 43.5175 cm²

Watch out

Students use the upper bound for one measurement and the lower bound for the other.This does not give the maximum or minimum — it gives something in between.For a maximum product, use BOTH upper bounds. For a minimum product, use BOTH lower bounds. g.speed = d/t), use upper bound of numerator and lower bound of denominator.

Diagnosticrecall

Write the error interval for a length of 12 cm measured to the nearest cm

1 mark2 minsbounds-and-estimation-q1

Show solution

Worked solution

1.Spot the skill: Upper bound of a product: use the upper bound of BOTH measurements.
2.Use the find the upper bound of each measurement stage first, then calculate the upper bound of the area.
3.Keep the final answer visible: 11.5 ≤ l < 12.5.

Final answer

11.5 ≤ l < 12.5

Mark points

M1: use the correct upper bound of a product: use the upper bound of both measurements.lower bound of a product: use the lower bound of both measurements.error interval for each measurement: value ± half the degree of accuracy.
A1: 11.5 ≤ l < 12.5

Watch out

Easyprocedure

Estimate 19.7 × 4.1 ÷ 2.03 by rounding each value to 1 significant figure

2 marks3 minsbounds-and-estimation-q2

Show solution

Worked solution

1.Spot the skill: Upper bound of a product: use the upper bound of BOTH measurements.
2.Use the calculate the upper bound of the area stage first, then evaluate.
3.Keep the final answer visible: 40.

Final answer

Mark points

M1: use the correct upper bound of a product: use the upper bound of both measurements.lower bound of a product: use the lower bound of both measurements.error interval for each measurement: value ± half the degree of accuracy.
A1: 40

Watch out

Mediumreasoning

A length is 7.3 m to the nearest 0.1 m. Find the upper and lower bounds.

3 marks4 minsbounds-and-estimation-q3

Show solution

Worked solution

1.Spot the skill: Upper bound of a product: use the upper bound of BOTH measurements.
2.Use the evaluate stage first, then find the upper bound of each measurement.
3.Keep the final answer visible: Upper: 7.35 m, Lower: 7.25 m.

Final answer

Upper: 7.35 m, Lower: 7.25 m

Mark points

M1: use the correct upper bound of a product: use the upper bound of both measurements.lower bound of a product: use the lower bound of both measurements.error interval for each measurement: value ± half the degree of accuracy.
A1: Upper: 7.35 m, Lower: 7.25 m

Watch out

Hardproblem solving

A journey takes 2.4 hours to the nearest 0.1 hour at a speed of 60 km/h to the nearest 10 km/h. Find the upper bound of the distance.

3 marks5 minsbounds-and-estimation-q4

Show solution

Worked solution

1.Spot the skill: Upper bound of a product: use the upper bound of BOTH measurements.
2.Use the find the upper bound of each measurement stage first, then calculate the upper bound of the area.
3.Keep the final answer visible: 159.75 km.

Final answer

159.75 km

Mark points

M1: use the correct upper bound of a product: use the upper bound of both measurements.lower bound of a product: use the lower bound of both measurements.error interval for each measurement: value ± half the degree of accuracy.
A1: 159.75 km

Watch out

Exam-stylemulti-step

A density is calculated from mass = 48 g (nearest gram) and volume = 6 cm³ (nearest cm³). Find the upper bound of the density.

4 marks6 minsbounds-and-estimation-q5

Show solution

Worked solution

1.Spot the skill: Upper bound of a product: use the upper bound of BOTH measurements.
2.Use the calculate the upper bound of the area stage first, then evaluate.
3.Keep the final answer visible: 48. $\frac{5}{5}$ .5 ≈ 8.818... g/cm³.

Final answer

48. $\frac{5}{5}$ .5 ≈ 8.818... g/cm³

Mark points

M1: use the correct upper bound of a product: use the upper bound of both measurements.lower bound of a product: use the lower bound of both measurements.error interval for each measurement: value ± half the degree of accuracy.
A1: 48. $\frac{5}{5}$ .5 ≈ 8.818... g/cm³

Watch out

Grade 9 stretchproblem solving

A rectangle has length 8.4 cm and width 5.2 cm, each correct to the nearest 0.1 cm. Find the upper bound for its area.

4 marks7 minsbounds-g9

Show solution

Worked solution

1.Find the upper bound of each measurement.
2.Multiply the upper bounds.

Final answer

44.3625 cm²

Mark points

M1: use 8.45 and 5.25
A1: 44.3625 cm²

Watch out

Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.

Hard exam-stylemulti-step problem

A rectangle has length 12.4 cm and width 5.8 cm, each correct to the nearest 0.1 cm. Find the lower bound for its area.

3 marks6 minsbounds-paper

Show solution

Worked solution

1.Find the lower bound for each measurement.
2.Multiply the two lower bounds.
3.Do not round the bound unless asked.

Final answer

71.0125 cm²

Mark points

M1: use 12.35 and 5.75
A1: obtain 71.0125 cm²

Watch out

Read the full question before calculating. Keep each stage of your working visible.

Timed checkpoint

12 mins - 9 marks

Switch between skills

Set a timer and attempt all four questions before opening any answers. This is closer to the way skills appear in a real paper.

1Bounds, estimation and error intervals - 2 marksWrite the error interval for a length of 12 cm measured to the nearest cmMark answer

Answer

11.5 ≤ l < 12.5

2Calculations and order of operations - 2 marksWork out (6 + 2) × 3 − 5Mark answer

Answer

3Integers, decimals and place value - 2 marksOrder these from smallest to largest: 0.3, 0.03, 0.303, 0.033Mark answer

Answer

0.03, 0.033, 0.3, 0.303

4Fractions - 3 marksA recipe needs ¾ cup of sugar. How much sugar is needed for 2½ batches?Mark answer

Answer

1⅞ cups

Mastery check

I can explain the method for bounds, estimation and error intervals.
I can show clear working without skipping key steps.
I can avoid this mistake: Students use the upper bound for one measurement and the lower bound for the other.This does not give the maximum or minimum — it gives something in between.For a maximum product, use BOTH upper bounds. For a minimum product, use BOTH lower bounds. g.speed = d/t), use upper bound of numerator and lower bound of denominator.

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