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c059741
NotesMath AI HLTopic 1.6
Unit 1 · Number & Algebra · Topic 1.6

IB Math AI HL — Approximating and estimating

IB Mathematics AI SL topic covering core concepts and exam-style applications.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Exam technique guidePractice questions

Key concepts in Approximating and estimating

Key Idea: Every real-world measurement is an approximation. This topic teaches you to round correctly (to decimal places or significant figures), measure how wrong an approximation is (percentage error), and find the range of values a rounded number could represent (upper and lower bounds).

Three things IB tests on this topic:

Round correctly

  • To d.p.: count digits after decimal
  • To s.f.: count from first non-zero digit
  • e.g. 0.04863 to 3 s.f. = 0.0486

Percentage error

  • ε = |vA − vE| ÷ vE × 100%
  • Always divide by the EXACT value
  • e.g. approx 240, exact 256 → 6.25%

Upper and lower bounds

  • Rounded to nearest unit: ±0.5 unit
  • Lower bound = value − 0.5 unit
  • Upper bound = value + 0.5 unit

🎯 Rounding: d.p. vs s.f.


📐 Percentage error

ε=∣vA−vEvE∣×100%\varepsilon = \left|\frac{v_A - v_E}{v_E}\right| \times 100\%ε=​vE​vA​−vE​​​×100%
vA = approximate value, vE = exact (true) value. The absolute value makes error always positive.
Upper bound
The largest value the true measurement could be. If a length is 15 cm to the nearest cm, upper bound = 15.5 cm.
Lower bound
The smallest value the true measurement could be. Same measurement: lower bound = 14.5 cm. The true value satisfies: 14.5 ≤ length < 15.5
Combining bounds
Max sum = upper + upper. Min sum = lower + lower. Max product = upper × upper. Min quotient = lower ÷ upper. For differences: max difference = upper − lower.
Tip: Always divide by the exact (true) value vE — not the approximate one. Approx = 240, Exact = 256: ✓ ε = |240 − 256| ÷ 256 × 100 = 6.25% ✗ ε = |240 − 256| ÷ 240 × 100 = 6.67% ← wrong denominator

✏️ Worked examples

Round to significant figures

Round 0.048632 to 3 significant figures.

Step by step:

  1. First significant figure: 4 (first non-zero digit)

  2. 2nd s.f.: 8, 3rd s.f.: 6

  3. 4th digit is 3 → do not round up

  4. Answer: 0.0486

Final answer:

0.0486

Calculate percentage error

Estimated cost $240, actual cost $256. Find the percentage error.

Step by step:

  1. vA = 240 (approximate), vE = 256 (exact)

  2. ε = |240 − 256| ÷ 256 × 100

  3. = 16 ÷ 256 × 100

  4. = 6.25%

Final answer:

6.25%

Find bounds and max area

A box: length 15 cm, width 8 cm, each to the nearest cm. Find the maximum possible area.

Step by step:

  1. Upper bound of length = 15.5 cm, upper bound of width = 8.5 cm

  2. Maximum area = upper × upper = 15.5 × 8.5

  3. = 131.75 cm²

Final answer:

Maximum area = 131.75 cm²

IB default: Give answers to 3 significant figures unless told otherwise. This applies after every GDC calculation. Percentage error: Identify the 'exact' or 'true' value — it goes in the denominator. The approximate value is what was estimated or measured. Bounds precision: The half-unit depends on the rounding used. Nearest cm → ±0.5 cm. Nearest 0.1 → ±0.05. Nearest 10 → ±5. Paper 1: Rounding questions are often 1–2 marks. Write clearly and show the digit you're rounding at.

IB-style question — model estimate and percentage error [5 marks]

A falling object's distance (in metres) after t seconds is modelled by d = 5t². (a) Use the model to estimate the distance fallen after 4 seconds. (b) The distance is actually measured as 76 m. Find the percentage error of the model's estimate, giving your answer to 3 significant figures.

Step by step:

  1. (a) Substitute t = 4 into the model.

    d=5×42=5×16=80 md = 5 \times 4^2 = 5 \times 16 = 80\ \text{m}d=5×42=5×16=80 m
  2. (b) Use the percentage error formula with the model value as the approximation and the measured value as the exact value.

    ε=∣vA−vEvE∣×100%\varepsilon = \left|\frac{v_A - v_E}{v_E}\right| \times 100\%ε=​vE​vA​−vE​​​×100%
  3. Substitute the model estimate (80) and the measured value (76).

    ε=∣80−7676∣×100%\varepsilon = \left|\frac{80 - 76}{76}\right| \times 100\%ε=​7680−76​​×100%
  4. Evaluate and round to 3 significant figures.

    ε=476×100%=5.263…%≈5.26%\varepsilon = \frac{4}{76} \times 100\% = 5.263\ldots\% \approx 5.26\%ε=764​×100%=5.263…%≈5.26%
Final answer:

(a) 80 m. (b) Percentage error ≈ 5.26%.

What you'll learn in Topic 1.6

  • 1.6.1 Rounding and Approximation
  • 1.6.2 Absolute and Relative Error
  • 1.6.3 Upper and Lower Bounds
  • 1.6.4 Percentage Error in Context
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 1.6 Approximating and estimating

1.6.1

Rounding and Approximation

Notes
1.6.2

Absolute and Relative Error

Notes
1.6.3

Upper and Lower Bounds

Notes
1.6.4

Percentage Error in Context

Notes

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Topic 1.6 Approximating and estimating forms a core part of Unit 1: Number & Algebra in IB Math AI HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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