The big idea: When a number is rounded, the real value could be slightly smaller or slightly bigger.
So the true value lies in a range called the bounds.
Worked example
A length is given as 6.4 cm correct to 1 decimal place.
Find the bounds.
Step by step
- Correct to 1 decimal place means rounded to the nearest 0.1.
- The number can move halfway before it rounds to a different value. Half of 0.1 is 0.05.
- Subtract 0.05 to get the smallest possible value.
- Add 0.05 to get the largest possible value.
- But 6.45 would round to 6.5, not 6.4, so the value must be less than 6.45.
Final answer
6.35 ≤ L < 6.45
Easy memory trick: Find the rounding unit → take HALF → subtract for the lower bound → add for the upper bound.
Why the symbols are different: Use ≤ for the lower bound because 6.35 still rounds to 6.4.
Use < for the upper bound because 6.45 rounds to 6.5 instead.
IB-style worked example
Scenario: A park fence post has a square top of side 20 cm.
The slanted edge from the apex to a base corner is measured as 14.6 cm, correct to the nearest 0.1 cm.
Write bounds clearly
Given AC = 14.6 cm correct to the nearest 0.1 cm, find lower and upper bounds. [2 marks]
Step by step
- Nearest 0.1 means the rounding step is 0.1.
- Half of 0.1 is 0.05, so subtract and add 0.05.
- Lower bound = 14.6 - 0.05 = 14.55 cm.
- Upper bound = 14.6 + 0.05 = 14.65 cm.
- 14.65 is not included because it would round to 14.7 instead.
Final answer
14.55 ≤ AC < 14.65
Exam tips for bounds questions:
- Always write the bounds interval before substituting values into formulas.
- Find the rounding unit first, then take HALF of it.
- Use ≤ for the lower bound and < for the upper bound.
- Check whether the upper bound would round to the next number.
Worked example
A mass is 370 g correct to 2 significant figures.
Find the bounds.
Step by step
- 2 significant figures here means rounded to the nearest 10 g.
- Half a step is 5 g.
Final answer
365 ≤ m < 375
Look at place value first: With significant figures, first decide what place value the last kept digit is in.
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| Rounded value | Accuracy | Bounds |
|---|---|---|
| 8 cm | nearest cm | 7.5 ≤ x < 8.5 |
| 2.7 s | 1 d.p. | 2.65 ≤ t < 2.75 |
| 540 | 2 s.f. | 535 ≤ n < 545 |
Always include units when appropriate: If the quantity is in cm, kg, or seconds, keep the units with your interval statement.
Why bounds matter: In real problems, bounds help you work out best-case and worst-case possibilities from rounded measurements.
Worked example
A time is 12.6 s correct to 1 decimal place.
What is the greatest possible true time?
Step by step
- Bounds are 12.55 ≤ t < 12.65.
- So the greatest possible true time is just less than 12.65 s.
Final answer
Upper bound 12.65 s