IB Maths AI HL Number Toolkit Paper 1 & 2 s.f. & d.p. ~6 min read

Approximation

Approximation means rounding a number to a stated accuracy — significant figures or decimal places. The rule is always the same: keep the digits you want, then look at the next digit to decide whether to round up. Some real-world contexts also force you to round a particular way regardless of that digit.

📘 What you need to know

Significant figures

The significant figures of a number are its meaningful digits, starting from the first non-zero digit. Leading zeros are not significant — they only fix the size of the number — but zeros between significant digits are.

Which digits are significant? 0.0 4 0 8 7 leading zeros — not significant 1st s.f. 2nd s.f. 3rd s.f. deciding digit7 ≥ 5, so 8 rounds up to 9 ≈ 0.0409 (3 s.f.)
Counting starts at the first non-zero digit. The deciding digit (7) sits just past the last figure kept — 7 ≥ 5, so the 8 rounds up.

Decimal places

The number of decimal places is simply how many digits sit after the decimal point. Rounding to d decimal places means keeping d digits there, then using the next digit to decide whether the last kept digit rounds up.

The rounding rule next digit ≥ 5  ⇒  round up next digit ≤ 4  ⇒  round down (leave it) the same rule applies to significant figures and to decimal places

Significant figures and decimal places are different counts: 0.0408 has 3 s.f. but 4 d.p. Always check which one the question asks for.

Rounding up in context

Sometimes the situation — not the deciding digit — tells you which way to round. If you need enough seats, containers or vehicles for everyone, you must round up, even when the decimal part is small.

Working accuracy: to be sure a final answer is correct to 3 s.f., carry 4 s.f. or more (or exact values) through your working — round only at the very end.

🧭 Recipe — rounding a number

  1. Read the accuracy asked for — s.f. or d.p.; if none is given, use 3 s.f.
  2. Identify the last digit to keep — count significant figures from the first non-zero digit, or decimal places from the point.
  3. Look at the next digit: 5 or more rounds up, 4 or less rounds down.
  4. Keep the place value — replace dropped whole-number digits with zeros so the number stays the right size.
  5. Check the context — if the situation demands it, round up (or down) regardless of the digit.

Worked examples

WE 1

Rounding to significant figures

A grain of sand has a mass of 0.0046382 g. Write this mass correct to 3 significant figures.

count 3 s.f. from the first non-zero digit 0.004 6 382 → keep 4, 6, 3 deciding digit is 8 8 ≥ 5, so 3 rounds up to 4 ≈ 0.00464 g (3 s.f.) leading zeros are not significant — counting starts at the 4.
WE 2

Rounding to decimal places

A wooden plank measures 7.38561 m. Write this length correct to 2 decimal places.

keep 2 digits after the decimal point 7.38561 → keep 3, 8 deciding digit is 5 5 ≥ 5, so 8 rounds up to 9 ≈ 7.39 m (2 d.p.) a deciding digit of exactly 5 still rounds up.
WE 3

Significant figures in a large number

A concert was attended by 48 627 people. Write this attendance correct to 2 significant figures.

count 2 s.f. from the first digit 4 8627 → keep 4, 8 deciding digit is 6 6 ≥ 5, so 8 rounds up to 9 replace the dropped digits with zeros ≈ 49 000 (2 s.f.) keep the place value — the zeros hold the number at the right size.
WE 4

Rounding a calculated result

A circle has radius 6.4 cm. Find its area, giving your answer correct to 3 significant figures.

area = πr2 A = π × 6.42 = π × 40.96 A = 128.679… cm2 round to 3 s.f. — deciding digit 6 A ≈ 129 cm2 (3 s.f.) keep the full value through the working, then round only the final answer.
WE 5

Rounding up in context

310 students are going on a trip. Each coach seats 48 students. How many coaches are needed?

divide the students by the seats 31048 = 6.458… 6 coaches seat only 288 — not enough round up so every student has a seat 7 coaches the context forces a round up — 6.458 does not round to 6 here.
WE 6

Full question: exact value then rounding

A square photo frame has sides of 8 cm. (a) Find the exact length of the diagonal. (b) Give the diagonal correct to 2 decimal places. (c) Give the diagonal correct to 3 significant figures.

(a) diagonal by Pythagoras d = √(82 + 82) = √128 = 8√2 cm (b) 8√2 = 11.31370… — deciding digit 3 round down ⇒ 11.31 cm (c) 3 s.f. of 11.31370… — deciding digit 1 round down ⇒ 11.3 cm (a) 8√2 cm · (b) 11.31 cm · (c) 11.3 cm leave (a) as a surd — that is the exact value; round only when asked.

💡 Top tips

âš  Common mistakes

Next up: Upper & Lower Bounds — the smallest and largest values a rounded number could really have. Approximation works the other way round there: instead of rounding a number, you ask what range an already-rounded number came from.

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