Bounds show the interval a rounded value could be in
What you will learn
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
Find the upper bound of each measurement
Measured to nearest 0.1 cm, so accuracy = 0.1 and half-accuracy = 0.05
Calculate the upper bound of the area
Upper bound of area = 8.45 × 5.15
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
maximum product uses upper x upper; minimum product uses lower x lower.
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.
43.5175 cm²
Build up to the hardest questions
Do them in order. If you miss a step, read the solution, then redo the question without looking.
WorkedreasoningA 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-workedShow solution
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.
- 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.
43.5175 cm²
- M1: find the upper bound of each measurement
- M1: calculate the upper bound of the area
- M1: evaluate
- A1: 43.5175 cm²
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.
DiagnosticrecallWrite the error interval for a length of 12 cm measured to the nearest cm
1 mark2 minsbounds-and-estimation-q1Show solution
Write the error interval for a length of 12 cm measured to the nearest cm
- 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.
11.5 ≤ l < 12.5
- 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
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.
EasyprocedureEstimate 19.7 × 4.1 ÷ 2.03 by rounding each value to 1 significant figure
2 marks3 minsbounds-and-estimation-q2Show solution
Estimate 19.7 × 4.1 ÷ 2.03 by rounding each value to 1 significant figure
- 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.
40
- 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
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.
MediumreasoningA length is 7.3 m to the nearest 0.1 m. Find the upper and lower bounds.
3 marks4 minsbounds-and-estimation-q3Show solution
A length is 7.3 m to the nearest 0.1 m. Find the upper and lower bounds.
- 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.
Upper: 7.35 m, Lower: 7.25 m
- 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
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.
Hardproblem solvingA 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-q4Show solution
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.
- 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.
159.75 km
- 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
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.
Exam-stylemulti-stepA 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-q5Show solution
A density is calculated from mass = 48 g (nearest gram) and volume = 6 cm³ (nearest cm³). Find the upper bound of the density.
- 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..5 ≈ 8.818... g/cm³.
48..5 ≈ 8.818... g/cm³
- 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..5 ≈ 8.818... g/cm³
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.
Grade 9 stretchproblem solvingA 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-g9Show solution
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.
- 1.Find the upper bound of each measurement.
- 2.Multiply the upper bounds.
44.3625 cm2
- M1: use 8.45 and 5.25
- A1: 44.3625 cm2
Do not rush straight into arithmetic. Select the relevant method and show a complete chain of working.
Hard exam-stylemulti-step problemA 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-paperShow solution
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.
- 1.Find the lower bound for each measurement.
- 2.Multiply the two lower bounds.
- 3.Do not round the bound unless asked.
71.0125 cm2
- M1: use 12.35 and 5.75
- A1: obtain 71.0125 cm2
Read the full question before calculating. Keep each stage of your working visible.
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
11.5 ≤ l < 12.5
2Calculations and order of operations - 2 marksWork out (6 + 2) × 3 − 5Mark answer
19
3Integers, decimals and place value - 2 marksOrder these from smallest to largest: 0.3, 0.03, 0.303, 0.033Mark 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
1⅞ cups
- 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.
This guide follows the Pearson Edexcel GCSE Mathematics 1MA1 specification. Practice questions are original Learnova questions shaped around official content and exam skills.