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Year 9 Maths Advanced Indices and Scientific Notation

Significant Figures

20 practice questions 1 video lesson Theory + worked examples

Significant figures show how precise a number is: \(0.0056050\) rounded to two significant figures is \(0.0056\). This NSW Year 9 Advanced maths subtopic rounds numbers of any size to a set number of significant figures (NESA outcome MA5‑MAG‑C‑01).

You will round large numbers, small decimals and tricky cases such as \(0.04995\), and use a calculator to give answers in scientific notation to three significant figures. You will estimate the grains of sand in a truck load and the distance to the Sun, see how rounding too early changes an area, and find the limits of a length or a crowd size that was rounded. It is part of Year 9 Advanced Indices and Scientific Notation.

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Theory

Significant figures are the digits of a number that carry information about its size and accuracy, counted from the first non-zero digit. This Year 9 Advanced maths page covers counting and rounding to significant figures, calculator answers in scientific notation, rounding errors and upper and lower bounds.

The significant figures (s.f.) of a number start at the first non-zero digit. They show how accurate the number is.

Rules for zeros: leading zeros (as in \(0.0052\)) are never significant; they only place the decimal point. Zeros between non-zero digits (as in \(4.07\)) are always significant. Trailing zeros after the decimal point (as in \(3.50\)) are significant, because they show the number was measured or rounded to that place.

Trailing zeros in a whole number are unclear: \(7000\) could have \(1\), \(2\), \(3\) or \(4\) significant figures. The context or the question tells you which. Scientific notation removes the doubt: \(7.0 \times 10^3\) has \(2\).

To round to \(n\) significant figures, keep \(n\) digits from the first non-zero digit. If the next digit is \(5\) or more, round up. Fill any places before the decimal point with zeros so the size of the number does not change.

A rounded value stands for a range of possible values. The lower bound and upper bound are half a unit of the last significant place below and above it.

Which digits of 0.004070 are significantThe number 0.004070. The leading zeros 0.00 are grey and not significant. The digits 4, 0, 7 and the final 0 are highlighted: the zero between 4 and 7 counts and the final zero after the decimal point counts, so the number has 4 significant figures. 0 . 0 0 4 0 7 0 0.004070 has 4 significant figures leading zeros: not significant 4, 0, 7, 0: significant zeros between digits count; a final zero after the decimal point counts (it shows accuracy)
Count from the first non-zero digit. Zeros in front only place the point; zeros after it can carry information.
Bounds of a measurement given to 2 significant figuresA number line from 6.3 to 6.5. A measurement of 6.4 m to 2 significant figures could be anywhere from 6.35, shown with a filled dot, up to but not including 6.45, shown with an open dot. 6.3 6.35 6.4 6.45 6.5 6.4 m, correct to 2 significant figures lower bound upper bound 6.35 ≀ length < 6.45
A rounded value stands for an interval: half a unit of the last significant place either side.
NumberSignificant figuresWhy
\(8.13\)\(3\)all non-zero digits count
\(5006\)\(4\)zeros between digits count
\(0.029\)\(2\)leading zeros do not count
\(1.80\)\(3\)a final zero after the point counts
\(4.30 \times 10^{5}\)\(3\)only the front number is counted
Bounds. If a value is rounded to a place worth \(u\) (such as \(u = 0.1\) or \(u = 100\)), then the true value \(v\) satisfies \(\text{rounded} - \dfrac{u}{2}\) \(\le v\) \(\lt \text{rounded} + \dfrac{u}{2}\).
Counting whole items. If a count of people is \(5200\) to \(2\) s.f., it was rounded to the nearest \(100\): the smallest count is \(5150\) and the largest is \(5249\), because a count is a whole number.
Greatest and least results. For a sum or product of positive measurements, the greatest result uses the upper bounds and the least uses the lower bounds.

How to round to \(n\) significant figures

  1. Find the first non-zero digit and count \(n\) digits from it, including zeros.
  2. Look at the next digit: \(5\) or more, round up; \(4\) or less, leave it. A round-up can carry: \(0.09996 \to 0.100\) (3 s.f.).
  3. Keep the size: in a whole number, fill places up to the point with zeros. In a decimal, drop the digits after the last one kept.
  4. Write the final zeros that are significant: \(3.10\), not \(3.1\), for \(3\) s.f.

Working with measurements

  1. Calculator answers: write the full value in scientific notation, then round the front number (Example 2).
  2. Round once, at the end: keep full calculator values in every step (Example 3).
  3. Rounded to the nearest...: rounding $4 682 519 to the nearest million gives $5 000 000, which has \(1\) significant figure.
  4. Bounds: find the place the last significant figure is in, take half of it, and go below and above (Example 4).
  5. Bounds of a calculation: work out the result with the bounds of each measurement.
Example 1 β€” Rounding to significant figures
Round:
(i) \(0.003\,072\,9\) to \(3\) s.f.
(ii) \(59\,962\) to \(2\) s.f.
(iii) \(2.6996\) to \(4\) s.f.
Solution

Count from the first non-zero digit, then look at the next digit to decide whether to round up.

\(0.003\,07|29\)\(\approx\)\(\mathbf{0.003\,07}\)
\(59|962\)\(\approx\)\(\mathbf{60\,000}\)
\(2.699|6\)\(\approx\)\(\mathbf{2.700}\)
2.6996≈2.700

(i) The next digit is \(2\), so round down. (ii) The next digit is \(9\): \(59 \to 60\), and zeros fill the places up to the point. (iii) Rounding up carries all the way: \(2.699 \to 2.700\). Both final zeros must be written to show \(4\) significant figures.

Example 2 β€” Calculator answers in scientific notation
Use a calculator. Give each answer in scientific notation, correct to \(3\) significant figures:
(i) \(\dfrac{7.8 \times 10^{9}}{3.3 \times 10^{-2}}\)
(ii) \(0.0418^3\).
Solution

Write the full calculator value in scientific notation first, then round the front number.

(i)

\(\dfrac{7.8 \times 10^{9}}{3.3 \times 10^{-2}}\)\(=\)\(2.3636\ldots \times 10^{11}\)
\(\)\(\approx\)\(\mathbf{2.36 \times 10^{11}}\)

(ii)

\(0.0418^3\)\(=\)\(0.000\,073\,034\ldots\)
\(\)\(=\)\(7.3034\ldots \times 10^{-5}\)
\(\)\(\approx\)\(\mathbf{7.30 \times 10^{-5}}\)
0.04183≈7.30×10−5

The final \(0\) in \(7.30\) is the third significant figure, so it stays.

Example 3 β€” Rounding too early
A cube-shaped tank has side \(2.36\) m.
(i) Find its volume correct to \(2\) s.f.
(ii) Ava rounds the side to \(2\) s.f. first, then cubes it and rounds. How far out is she?
Solution

(i) Use the exact side and round once, at the end.

\(2.36^3\)\(=\)\(13.144\ldots\)
\(\)\(\approx\)\(\mathbf{13}\text{ m}^3\)

(ii) Ava's side is \(2.4\) m.

\(2.4^3\)\(=\)\(13.824\)
\(\)\(\approx\)\(14\text{ m}^3\)
\(14 - 13\)\(=\)\(\mathbf{1}\text{ m}^3\)
2.363≈13

Cubing roughly triples the percentage error: about \(1.7\%\) in the side becomes about \(5\%\) in the volume. Round only the final answer.

Example 4 β€” Upper and lower bounds
A square tile has side \(30\) cm, correct to \(2\) significant figures. Find the range of possible values of
(i) the side
(ii) the perimeter.
Solution

To \(2\) s.f., \(30\) means the side was rounded to the nearest \(1\) cm. Go half a unit either side.

(i)

\(30 - 0.5\)\(=\)\(29.5\)
\(30 + 0.5\)\(=\)\(30.5\)
\(\)\(\)\(\mathbf{29.5} \le s < \mathbf{30.5}\text{ cm}\)

(ii) The perimeter is \(4s\), so multiply both bounds by \(4\).

\(\text{lower}\)\(=\)\(4 \times 29.5 = 118\)
\(\text{upper}\)\(=\)\(4 \times 30.5 = 122\)
\(\)\(\)\(\mathbf{118} \le P < \mathbf{122}\text{ cm}\)
118≤P<122

A side of exactly \(30.5\) cm would round to \(31\), so the upper bound is not included.

Common pitfalls

Leading zeros are not significant. \(0.0083\) has \(2\) significant figures, not \(5\).
Keep the size of a whole number. \(83\,617\) to \(2\) s.f. is \(84\,000\), not \(84\).
Do not drop significant final zeros. \(5.997\) to \(3\) s.f. is \(6.00\). Writing \(6\) shows only \(1\) s.f.
Round once, at the end. Rounding inputs first can change the answer, especially after squaring or cubing.
The upper bound is not reached. \(4.6\) to \(2\) s.f. means \(4.55 \le v \lt 4.65\): exactly \(4.65\) would round to \(4.7\).

Frequently asked questions

What are significant figures?

They are the digits of a number that show its size and accuracy, counted from the first non-zero digit. Leading zeros do not count; zeros between other digits do.

Are zeros significant figures?

Zeros between non-zero digits are significant. Zeros at the start of a decimal are not. Zeros at the end of a decimal are. Zeros at the end of a whole number are unclear unless the question says how it was rounded.

How do you round to 3 significant figures?

Start at the first non-zero digit and keep three digits. If the next digit is \(5\) or more, round the third digit up. Fill with zeros up to the decimal point if needed.

What is the difference between decimal places and significant figures?

Decimal places count digits after the point. Significant figures count from the first non-zero digit. For example \(0.0472\) has \(4\) decimal places but \(3\) significant figures.

What are upper and lower bounds?

They are the smallest and largest values that round to a given number. For \(6.4\) to \(2\) s.f., the lower bound is \(6.35\) and the upper bound is \(6.45\), which itself is not included.