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IB DP Maths AI · Topic 1 Number and algebra · SL 1.1, 1.6 · ~6 min read

Approximation, standard form and errors

Standard form and calculator E-notation, rounding to decimal places and significant figures, upper and lower bounds of measurements and calculations, estimation, and percentage error, all set in the real contexts IB Maths AI uses.

Standard form a × 10ᵏGDC E-notationDecimal places & significant figuresUpper and lower boundsBounds of calculationsEstimationPercentage error
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Standard form: a × 10k with 1 ≤ a < 10 and k an integer. Your GDC writes 1.28E4 for 1.28 × 104, but you must write 1.28 × 104 in your answer.
  • Rounding: decimal places count after the point; significant figures count from the first non-zero digit. IB's default is 3 s.f. unless the question says otherwise.
  • Bounds: a measurement rounded to the nearest unit lies within half a unit either way. 12.4 cm (1 d.p.) means 12.35 ≤ L < 12.45.
  • Percentage error ε = ǀvA − vEǀǀvEǀ × 100%, where vA is the approximate value and vE the exact value.
  • This note is the same at SL and HL, and unchanged for 2029.

By the end you can

  • Write and calculate with numbers in standard form, and read GDC E-notation.
  • Round to a given number of decimal places or significant figures.
  • Find upper and lower bounds of rounded measurements and of calculations that use them.
  • Estimate answers and find percentage errors.

1Standard form

  • Big numbers: 384 000 km = 3.84 × 105 km. Small numbers: 0.000 052 m = 5.2 × 10−5 m.
Worked example

Calculate (3.2 × 105) × (4 × 10−2) and give the answer in standard form.

  1. 3.2 × 4 = 12.8 and 105 × 10−2 = 103 M1
  2. 12.8 × 103 = 1.28 × 104 A1
Worked example

Light travels at 3 × 105 km s−1. The Sun is 1.496 × 108 km from Earth. How many minutes does sunlight take to reach Earth?

  1. time = 1.496 × 1083 × 105 = 498.67 s M1A1
  2. ≈ 8.31 minutes A1
Common mistakeWriting a GDC answer such as 1.28E4 as your final answer. Examiners don't accept calculator notation: write 1.28 × 104.
GDC (calculator) methodEnter powers of ten with the EE (or ×10ⁿ) key, not "× 10 ^". Using brackets avoids order-of-operations slips with negative powers.

2Rounding and significant figures

Number3 s.f.2 d.p.
4567.89145704567.89
0.003 04560.003 050.00
12.004912.012.00
  • Zeros: zeros between non-zero digits count (2005 has 4 s.f.); leading zeros never count (0.004 has 1 s.f.).
  • Exam rule: give exact answers or answers correct to 3 s.f. unless told otherwise. A wrong level of accuracy can lose a mark, so keep full values on the GDC and round only at the end.
Common mistakeRounding in the middle of a calculation. Rounding 498.67 s to 499 s and then dividing by 60 still gives 8.32, which is wrong at 3 s.f. Store values in memory instead.

3Upper and lower bounds

12.212.312.412.5LB 12.35UB 12.4512.4 cm (1 d.p.)Every length from 12.35 up to (but not including) 12.45 rounds to 12.4: 12.35 ≤ L < 12.45
The lower bound is included; the upper bound is the edge of the interval and is not.
  • Rule: if a value is rounded to the nearest u, the bounds are value ± ½u.
Worked example

A rectangle is 12.4 cm by 5.0 cm, both correct to 1 d.p. Find the upper and lower bounds of its area.

  1. length 12.35 to 12.45, width 4.95 to 5.05 A1
  2. upper bound = 12.45 × 5.05 = 62.9 cm2 (62.8725) M1A1
  3. lower bound = 12.35 × 4.95 = 61.1 cm2 (61.1325) A1
  • Dividing: the upper bound of a ÷ b is (upper a) ÷ (lower b); the lower bound is (lower a) ÷ (upper b).
Worked example

A runner covers 100 m (to the nearest metre) in 12.3 s (to the nearest 0.1 s). Find the bounds of her average speed.

  1. upper = 100.512.25 = 8.20 m s−1 M1A1
  2. lower = 99.512.35 = 8.06 m s−1 A1
Exam tipSubtraction works the same way: the largest a − b is (upper a) − (lower b). Write out which bound you use for each value before you calculate.

4Estimation and percentage error

ε = ǀvA − vEvEǀ × 100% FORMULA BOOKLET
percentage error; divide by the exact value
Worked example

Estimate √48.7 × 19.60.52 by rounding each number, then find the percentage error of your estimate.

  1. estimate ≈ 7 × 200.5 = 280 M1A1
  2. exact value = 263.04 (GDC) A1
  3. ε = ǀ280 − 263.04ǀ263.04 × 100% = 6.45% M1A1
Worked example

A student uses 227 for π. Find the percentage error.

  1. ε = ǀ22/7 − πǀπ × 100% M1
  2. = 0.0402% A1
Worked example

A crowd is estimated as 50 000. The true number is 47 300. Find the percentage error.

  1. ε = ǀ50 000 − 47 300ǀ47 300 × 100% M1
  2. = 5.71% A1
Common mistakeDividing by the approximate value. The formula always divides by the exact value vE.
  • Why estimate? Estimating first catches GDC slips: if your estimate is 280 and the GDC says 2630, you have mistyped something.
Linking questionBounds and errors return in Statistics, correlation and regression (how accurate is a prediction?) and in Integration, areas and kinematics (how good is the trapezoidal rule?). Standard form appears in Exponents, logarithms and equations, for example in pH and decibel scales.
Idea for your IA / explorationMeasure something many times (the period of a pendulum, the height of classmates) and analyse how rounding and measurement error affect a calculated result, such as g from a pendulum.

Test yourself — cover the answers first

Quick check

  1. Write 0.000 708 in standard form.
  2. Round 0.049 96 to 3 s.f.
  3. A mass is 64 kg to the nearest kg. Give its bounds.
  4. What does the GDC display 3.5E−3 mean?
  5. Find the percentage error when 9.8 is used for 9.81.

Exam-style

  1. A square has side 8 cm, correct to the nearest cm. Find the upper and lower bounds of its perimeter and of its area.
  2. The population of a city is 1.84 × 107 and its area is 4.35 × 103 km2. Find the population density, and the percentage error if the density is estimated as 4000 people per km2.
Show answers
  1. 7.08 × 10⁻⁴
  2. 0.0500
  3. 63.5 ≤ m < 64.5 kg
  4. 3.5 × 10⁻³
  5. 0.102%
  6. side 7.5 to 8.5 cm
    perimeter 30 ≤ P < 34 cm
    area 56.25 ≤ A < 72.25 cm²
  7. density = 1.84 × 1074.35 × 103 = 4229.9 ≈ 4230 people per km²
    ε = ǀ4000 − 4229.9ǀ4229.9 × 100% = 5.43%
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.