N16: Upper and Lower Bounds
Calculate upper and lower bounds; use bounds in calculations
📋 Key Concepts
Lower Bound: The smallest value that would round to the given number.
Upper Bound: The smallest value that would NOT round to the given number (just above the largest value that would round to it).
Finding Bounds:
For a number rounded to the nearest unit d:
• Lower bound = value - ½d
• Upper bound = value + ½d
Error Interval: lower ≤ actual < upper
For a number rounded to the nearest unit d:
• Lower bound = value - ½d
• Upper bound = value + ½d
Error Interval: lower ≤ actual < upper
📝 Finding Bounds
Example 1
Find the upper and lower bounds of 24 cm measured to the nearest cm
Solution:
Half the unit = 0.5 cm
Lower bound = 24 - 0.5 = 23.5 cm
Upper bound = 24 + 0.5 = 24.5 cm
Error interval: 23.5 ≤ x < 24.5
Example 2
Find the bounds for 8.0 m measured to 1 decimal place
Solution:
Half the unit = 0.05 m
Lower bound = 8.0 - 0.05 = 7.95 m
Upper bound = 8.0 + 0.05 = 8.05 m
Error interval: 7.95 ≤ x < 8.05
Example 3
Find the bounds for 6400 measured to 2 significant figures
Solution:
Rounded to nearest 100
Half the unit = 50
Lower bound = 6400 - 50 = 6350
Upper bound = 6400 + 50 = 6450
Error interval: 6350 ≤ x < 6450
📝 Bounds in Calculations
Key Principle: To find the maximum of a calculation, use bounds that make the result as large as possible. For minimum, use bounds that make it as small as possible.
Rules for Bounds Calculations:
Addition: Max = UB + UB, Min = LB + LB
Subtraction: Max = UB - LB, Min = LB - UB
Multiplication: Max = UB × UB, Min = LB × LB
Division: Max = UB ÷ LB, Min = LB ÷ UB
Addition: Max = UB + UB, Min = LB + LB
Subtraction: Max = UB - LB, Min = LB - UB
Multiplication: Max = UB × UB, Min = LB × LB
Division: Max = UB ÷ LB, Min = LB ÷ UB
📝 Addition and Subtraction
Example 4
A piece of wood is 15 cm (nearest cm) and another is 28 cm (nearest cm). Find the minimum and maximum possible total length.
Solution:
Piece 1 bounds: 14.5 ≤ x < 15.5
Piece 2 bounds: 27.5 ≤ x < 28.5
Minimum: 14.5 + 27.5 = 42 cm
Maximum: just below 15.5 + 28.5 = 44 cm
Example 5
Calculate the maximum possible difference between two lengths: 12 cm and 8 cm, both to the nearest cm.
Solution:
12 cm bounds: 11.5 ≤ x < 12.5
8 cm bounds: 7.5 ≤ x < 8.5
Maximum difference: 12.5 - 7.5 = 5 cm
📝 Multiplication and Division
Example 6
A rectangle has length 8 cm and width 5 cm, both measured to the nearest cm. Find the minimum and maximum possible area.
Solution:
Length bounds: 7.5 ≤ l < 8.5
Width bounds: 4.5 ≤ w < 5.5
Minimum area: 7.5 × 4.5 = 33.75 cm²
Maximum area: just below 8.5 × 5.5 = 46.75 cm²
Example 7
Speed is calculated using s = d/t. If d = 200 m (nearest 10 m) and t = 25 s (nearest second), find the maximum possible speed.
Solution:
Distance bounds: 195 ≤ d < 205
Time bounds: 24.5 ≤ t < 25.5
Maximum speed = Upper distance ÷ Lower time
Maximum speed ≈ 205 ÷ 24.5 = 8.37 m/s
📝 Compound Measures
Example 8
A block has mass 240 g (nearest 10 g) and volume 80 cm³ (nearest cm³). Find the minimum and maximum possible density.
Solution:
Mass bounds: 235 ≤ m < 245
Volume bounds: 79.5 ≤ v < 80.5
Density = mass ÷ volume
Minimum density = 235 ÷ 80.5 = 2.92 g/cm³
Maximum density = 245 ÷ 79.5 = 3.08 g/cm³
🧮 Worked Examples — Exam Style
Bounds arithmetic: MAXIMUM result uses upper ÷ lower (for division) or upper × upper (multiplication); MINIMUM the reverse. Choose bounds per OPERATION, not per number.
a = 8.5 and b = 2.5 (both to 1 d.p.). Write the interval for a ÷ b.
a ∈ [8.45, 8.55), b ∈ [2.45, 2.55)
Max quotient = 8.55 ÷ 2.45 = 3.4898; min = 8.45 ÷ 2.55 = 3.3137
3.31 < a/b < 3.49 (2 d.p., strict upper bound)
M1 for choosing UB÷LB and LB÷UB, A1 for both values. The upper bound is EXCLUDED, so strict < on the right.
A rectangle measures 5.2 cm by 3.8 cm (1 d.p.). Write the interval for its perimeter.
Bounds: 5.15–5.25 and 3.75–3.85. Perimeter = 2(l + w)
Min: 2(5.15 + 3.75) = 17.8; Max: 2(5.25 + 3.85) = 18.2 → 17.8 ≤ P < 18.2 cm
M1 for combining bounds, A1 for the interval. For ADDITION both extremes go the same way.
Exam wording to expect: "Calculate the least/greatest possible value of …", "Write the error interval for …".
❓ Practice Questions
Q1: Find the bounds for 35 measured to the nearest 5
Q2: A length is 12.4 cm to 1 d.p. Find the error interval.
Q3: Two lengths are 20 cm and 15 cm (both nearest cm). Find the minimum perimeter.
Q4: A rectangle is 10 cm by 6 cm (both nearest cm). Find the maximum possible area.
Q5: Distance = 150 km (nearest km), Time = 2 hours (nearest hour). Find the maximum possible speed.
Q6: p = 6.4 and q = 3.2 (both to 1 d.p.). Write the interval for p ÷ q.
Q7: A square has side 7.2 cm (1 d.p.). Find the least possible area (3 s.f.).
✅ Answers
- 32.5 ≤ x < 37.5
- 12.35 ≤ x < 12.45
- Min perimeter: 19.5 + 14.5 = 34 cm
- Max area: just below 10.5 × 6.5 = 68.25 cm²
- Max speed: 150.5 ÷ 1.5 = 100.33 km/h
📋 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. Nearest 5 → ±2.5 → 32.5 ≤ x < 37.5
Q2. 1 d.p. → ±0.05 → 12.35 ≤ x < 12.45
Q3. Minimum perimeter uses minimum lengths: 2(19.5 + 14.5) = 68 cm (the given 34 cm is half the perimeter — one length sum only)
Q4. Max area = 10.5 × 6.5 = 68.25 cm² (strictly, just under)
Q5. Max speed = max distance ÷ min time = 150.5 ÷ 1.5 = 100.3 km/h (3 s.f.)
Q6. p ∈ [6.35, 6.45), q ∈ [3.15, 3.25): max = 6.45/3.15 = 2.048, min = 6.35/3.25 = 1.954 → 1.954 < p/q < 2.048
Q7. Least area = 7.15² = 51.1 cm² (3 s.f.)
📌 Key Facts to Memorise
- Add/subtract: LB = LB − LB, UB = UB + UB (same-direction extremes)
- Multiply: LB × LB, UB × UB
- Divide: LB = LB₁ ÷ UB₂, UB = UB₁ ÷ LB₂ (opposite extremes!)
- Rounded bounds: lower INCLUDED, upper EXCLUDED → interval notation with < on the top
- Rounding to nearest 10 → ±5; nearest whole → ±0.5; 2 d.p. → ±0.005
🗺️ In the Exam
Bounds: 3–5 marks on Papers 2/3 (calculator). Choosing the right pairing (same-direction vs opposite-direction) is the entire assessment — practise the decision, not just the arithmetic.
Links to: error intervals (N15), estimation (N14), compound measures (R1).
🧭 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 state what the number is rounded to first
- Remember: upper bound is NOT included (use < symbol)
- For maximum: use largest values in numerator, smallest in denominator
- For minimum: use smallest values in numerator, largest in denominator
- Write the error interval clearly with correct inequality symbols
🧠 Problem-Solving Strategies
Problem-Solving
For bounds calculations, think about what makes the result as large or as small as possible. For addition/multiplication: max uses UBs, min uses LBs. For subtraction/division: max uses UB − LB (or UB ÷ LB), min uses LB − UB (or LB ÷ UB). Always identify what each quantity was rounded to before finding bounds.
Multi-Step Problem
A field is 120 m by 85 m, both measured to the nearest 5 m. Find the minimum and maximum possible area.
Solution: Length bounds: 117.5 ≤ l < 122.5. Width bounds: 82.5 ≤ w < 87.5. Min area = 117.5 × 82.5 = 9693.75 m². Max area (just below) = 122.5 × 87.5 = 10,718.75 m².
⚠️ Common Errors
Watch Out!
1. Wrong: For max of a ÷ b, use UB(a) ÷ UB(b) Correct: Max of a ÷ b = UB(a) ÷ LB(b) (largest numerator, smallest denominator)
2. Wrong: Upper bound is included in the error interval (using ≤) Correct: Upper bound is NOT included: use < not ≤
3. Wrong: For "nearest 5", the half-unit is 5 (not 2.5) Correct: Half of 5 = 2.5, so bounds for 40 (nearest 5) are 37.5 ≤ x < 42.5
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A cuboid has dimensions: length = 10 cm (nearest cm), width = 5 cm (nearest cm), height = 3 cm (nearest cm). (a) Find the upper and lower bounds for each dimension. (b) Calculate the minimum and maximum possible volume. (c) The cuboid is made of gold. Gold costs £45 per cm³. Find the maximum possible cost of the gold used.
(a) Length: 9.5 ≤ l < 10.5. Width: 4.5 ≤ w < 5.5. Height: 2.5 ≤ h < 3.5.
(b) Min volume = 9.5 × 4.5 × 2.5 = 106.875 cm³. Max volume (just below) = 10.5 × 5.5 × 3.5 = 202.125 cm³.
(c) Max cost = 202.125 × £45 = £9,095.63.
Mark scheme: 2 marks for (a) all six bounds, 2 marks for (b), 2 marks for (c)
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two students measure a corridor. Student A measures 24 m to the nearest metre. Student B measures 24.0 m to the nearest 0.1 m.
(a) Write the error interval for each measurement.
(b) A door frame needs to be 24.05 m wide. Which measurement gives more confidence that the corridor is wide enough? Justify your answer.
(c) A builder says "It doesn't matter whether you write 24 or 24.0 — they're the same number." Is the builder correct in the context of measurement? Explain.
Answers: (a) Student A: 23.5 ≤ x < 24.5. Student B: 23.95 ≤ x < 24.05. (b) Student B's measurement gives more confidence — their interval shows the corridor could be as wide as just under 24.05 m, which might be enough. Student A's range (up to 24.5) is too wide to be confident about a measurement of 24.05. (c) No — in measurement, 24 means rounded to the nearest whole number (±0.5), while 24.0 means rounded to the nearest tenth (±0.05). The trailing zero indicates the precision of the measurement, so they are NOT the same.
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