N15: Rounding & Error Intervals
Round to decimal places and significant figures; use error intervals from truncation and rounding
📋 Key Concepts
Decimal Places (d.p.): Count the number of digits after the decimal point.
Significant Figures (s.f.): Count all non-zero digits, and zeros between non-zero digits. Leading zeros don't count.
Rounding Rule:
• Look at the digit AFTER the required place
• If 5 or more, round UP
• If less than 5, round DOWN
• Look at the digit AFTER the required place
• If 5 or more, round UP
• If less than 5, round DOWN
📝 Rounding to Decimal Places
Example 1
Round 3.472 to 1 decimal place
Solution:
Look at the 2nd decimal place: 7
7 ≥ 5, so round up
Answer: 3.5
Example 2
Round 12.8439 to 2 decimal places
Solution:
Look at the 3rd decimal place: 3
3 < 5, so round down
Answer: 12.84
Example 3
Round 7.996 to 2 decimal places
Solution:
Look at the 3rd decimal place: 6
6 ≥ 5, so round up: 7.99 → 8.00
Answer: 8.00
📝 Rounding to Significant Figures
| Number | 1 s.f. | 2 s.f. | 3 s.f. |
|---|---|---|---|
| 3847 | 4000 | 3800 | 3850 |
| 0.0472 | 0.05 | 0.047 | 0.0472 |
| 56.38 | 60 | 56 | 56.4 |
Example 4
Round 3847 to 2 significant figures
Solution:
First 2 digits: 3 and 8
Look at the 3rd digit: 4
4 < 5, so round down
Answer: 3800
Example 5
Round 0.0472 to 1 significant figure
Solution:
First significant figure is 4
Look at the next digit: 7
7 ≥ 5, so round up
Answer: 0.05
📝 Error Intervals
Error Interval: The range of values that would round to a given value. Shows the possible actual values.
Finding Error Intervals:
For a number rounded to x (nearest whole, tenth, etc.):
• Lower bound: x - half the rounding unit
• Upper bound: x + half the rounding unit
Error interval: lower ≤ value < upper
For a number rounded to x (nearest whole, tenth, etc.):
• Lower bound: x - half the rounding unit
• Upper bound: x + half the rounding unit
Error interval: lower ≤ value < upper
Example 6
Find the error interval for a length of 8 cm rounded to the nearest cm
Solution:
Half the rounding unit = 0.5 cm
Lower bound: 8 - 0.5 = 7.5 cm
Upper bound: 8 + 0.5 = 8.5 cm
Error interval: 7.5 cm ≤ length < 8.5 cm
Example 7
A number is 3.4 rounded to 1 decimal place. Find the error interval.
Solution:
Half the rounding unit = 0.05
Lower bound: 3.4 - 0.05 = 3.35
Upper bound: 3.4 + 0.05 = 3.45
Error interval: 3.35 ≤ x < 3.45
📝 Truncation
Truncation: Cutting off digits without rounding. Always makes the number smaller (in absolute value).
Example 8
Truncate 7.892 to 1 decimal place
Solution:
Simply cut off after 1 decimal place
Answer: 7.8
Compare: rounding would give 7.9
Example 9
Find the error interval for a truncated value of 5.2 (1 d.p.)
Solution:
Truncation error is between 0 and the next digit's place
Lower bound: 5.2 (nothing added)
Upper bound: 5.3 (anything less rounds up)
Error interval: 5.2 ≤ x < 5.3
📝 Upper and Lower Bounds in Context
Example 10
A rectangle has length 12 cm and width 8 cm, both measured to the nearest cm. Find the minimum and maximum possible area.
Solution:
Length bounds: 11.5 ≤ l < 12.5
Width bounds: 7.5 ≤ w < 8.5
Minimum area: 11.5 × 7.5 = 86.25 cm²
Maximum area: just below 12.5 × 8.5 = 106.25 cm²
🧮 Worked Examples — Exam Style
Error intervals answer one question: 'what values would round to this number?' The rule: lower bound included (≤), upper bound NOT included (<).
A length is measured as 8.4 cm to 1 decimal place. Write the error interval.
Half of the last placed unit: 0.05
Lower: 8.4 − 0.05 = 8.35 (included — it rounds UP to 8.4)
Upper: 8.4 + 0.05 = 8.45 (NOT included — it would round to 8.5)
8.35 ≤ L < 8.45
M1 for 8.35 and 8.45, A1 for the interval with correct symbols. The < on the upper bound is the examined detail.
A number is truncated to 2.7. Write the error interval.
Truncation chops digits off — no rounding up.
2.7 ≤ T < 2.8
M1 for 2.7 and 2.8, A1 for correct symbols. Truncated values can grow to the next digit but never shrink.
Exam wording to expect: "Write the error interval for …", "rounded to …", or "truncated to …".
❓ Practice Questions
Q1: Round 4.728 to 2 decimal places
Q2: Round 0.0847 to 2 significant figures
Q3: Round 4729 to 2 significant figures
Q4: Find the error interval for 15 rounded to the nearest whole number
Q5: A number is 0.72 rounded to 2 decimal places. Write the error interval.
Q6: Write the error interval for 260 rounded to 2 significant figures
Q7: A mass is 3.05 kg to 2 d.p. Write the error interval.
✅ Answers
- 4.73
- 0.085
- 4700
- 14.5 ≤ n < 15.5
- 0.715 ≤ x < 0.725
🧮 Worked Examples — Harder Exam Style
A number x satisfies 40 ≤ x < 50. State the value it could have been rounded to, and the rounding unit.
Any value in [40, 50) rounds to 40 to 1 significant figure — the rounding unit is 10, giving 40 − 5 ≤ x < 40 + 5, i.e. 35 ≤ x < 45.
Wait — check: 40 to 1 s.f. gives 35 ≤ x < 45, NOT [40, 50). So [40, 50) cannot come from 1 s.f. rounding; it could come from truncation to 1 s.f. (40 ≤ x < 50).
This distinction — rounding vs truncation — is a genuine AO2 discriminator. Rounding spreads ±½ unit around the value; truncation only reaches upward.
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. Third decimal is 8 (≥5), round up: 4.73
Q2. First two significant figures are 8 and 4; next digit 7 rounds: 0.085
Q3. First two sig figs 4, 7; next digit 2 stays: 4700 — keep the zeros as placeholders
Q4. Nearest whole → ±0.5 → 14.5 ≤ n < 15.5
Q5. 2 d.p. → ±0.005 → 0.715 ≤ x < 0.725
Q6. 2 s.f. → nearest 10 → 255 ≤ n < 265
Q7. 2 d.p. → ±0.005 → 3.045 ≤ m < 3.055
📌 Key Facts to Memorise
- Rounded to the nearest unit U → interval (U − ½unit) ≤ x < (U + ½unit)
- Rounded to n decimal places → add/subtract 5 in the (n+1)th place
- Lower bound INCLUDED (≤), upper bound EXCLUDED (<)
- Truncation: value can only be bigger — U ≤ x < U + 1 unit
- Rounding 5s: 4.725 → 4.73 (round half up); significant figures count from the first non-zero digit
🗺️ In the Exam
Rounding: 1 mark almost every Paper 1. Error intervals: 2 marks, increasingly common on Foundation Papers 2/3. The excluded upper bound is the discriminator between grades 4 and 5.
Links to: estimation (N14), bounds in calculations (N16), units (N13).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always check what you're rounding to (d.p. or s.f.)
- For significant figures, remember leading zeros don't count
- Error intervals use ≤ for lower bound and < for upper bound
- Truncation is NOT the same as rounding - it's cutting off
- When finding bounds, think: what values would round to this?
🧠 Problem-Solving Strategies
Problem-Solving
For error intervals, first identify what the number was rounded to (nearest whole, tenth, hundredth, etc.). The error interval is always lower ≤ value < upper. For significant figures, work out what place value the rounding was to, then halve it. For truncation, the lower bound is the truncated value itself and the upper bound adds one place value unit.
Multi-Step Problem
A number x is rounded to 2 significant figures and gives 4700. Find the error interval for x.
Solution: 4700 to 2 s.f. means rounded to the nearest 100. Half = 50. Lower bound = 4700 − 50 = 4650. Upper bound = 4700 + 50 = 4750. Error interval: 4650 ≤ x < 4750.
⚠️ Common Errors
Watch Out!
1. Wrong: Error interval uses ≤ on both sides, e.g. 7.5 ≤ x ≤ 8.5 Correct: Upper bound is NOT included: 7.5 ≤ x < 8.5
2. Wrong: 0.0500 has 2 significant figures (counting the leading zeros) Correct: Leading zeros don't count — 0.0500 has 3 significant figures (5, 0, 0)
3. Wrong: Truncating 6.789 to 2 d.p. gives 6.79 (rounding instead) Correct: Truncation means cutting off: 6.789 truncated to 2 d.p. = 6.78
✍️ 6-Mark Exam Question
Extended Answer
6 marks: (a) Round 0.08472 to 2 significant figures. (b) Round 0.08472 to 3 decimal places. (c) A number is rounded to 1 decimal place and gives 3.0. Find the error interval. (d) A different number is truncated to 1 decimal place and also gives 3.0. Write its error interval and explain how it differs from your answer to (c).
(a) 0.085 (first 2 sig figs are 8 and 4; next digit is 7 ≥ 5, so round up).
(b) 0.085 (4th decimal place is 7 ≥ 5, so round up the 3rd place from 4 to 5).
(c) Rounded to 1 d.p.: half unit = 0.05. Error interval: 2.95 ≤ x < 3.05.
(d) Truncated to 1 d.p.: lower bound = 3.0, upper bound = 3.1. Error interval: 3.0 ≤ x < 3.1. The truncated interval is much wider (0.1 wide vs 0.1 wide but shifted). The key difference is that truncation always rounds down, so the lower bound is 3.0 itself, whereas rounding's lower bound is 2.95.
Mark scheme: 1 mark each for (a) and (b), 2 marks for (c), 2 marks for (d) including comparison
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A thermometer reads 37.2°C to 1 decimal place. A digital thermometer truncates to 1 decimal place and also reads 37.2°C.
(a) Write the error interval for each thermometer.
(b) A patient's actual temperature is 37.24°C. What would each thermometer display?
(c) A doctor says "The truncated thermometer is less useful because it could miss a fever." A fever is 37.5°C or above. Is the doctor's concern valid? Explain with reference to the error intervals.
Answers: (a) Rounding: 37.15 ≤ t < 37.25. Truncation: 37.2 ≤ t < 37.3. (b) Actual = 37.24°C. Rounded: 37.2°C. Truncated: 37.2°C. Both display the same. (c) For the rounded thermometer, 37.2 could mean up to 37.25 — well below 37.5. For the truncated thermometer, 37.2 could mean up to 37.3 — still below 37.5. However, a truncated reading of 37.4 could mean up to 37.5, which is at the fever threshold. So yes, truncation could mask a value just at the boundary — the rounded version is more precise about what the true value could be.
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