What is the difference between rounding to 3 decimal places and 3 significant figures?

Decimal places count digits after the decimal point; significant figures count from the first non-zero digit. They often agree for numbers near $1$ — $2.3456$ is $2.346$ to both 3 d.p. and 3 s.f. — but they diverge elsewhere. For $12\,345$, three significant figures give $12\,300$, while '3 d.p.' is meaningless (there are no decimals). For $0.0023456$, three significant figures give $0.00235$, but 3 d.p. gives $0.002$. Always read which one the question asks for.

Why do we divide by the exact value $v_E$ and not the approximate value $v_A$?

Percentage error measures how far the approximation strays from the truth, expressed as a fraction of that truth. The exact value is the fixed reference point, so it belongs in the denominator. Dividing by $v_A$ would change the yardstick every time the approximation changed, and it is the most common way students lose the accuracy mark here.

Can percentage error be negative?

No. The formula contains absolute-value bars, so the result is always positive. Percentage error reports the size of the discrepancy, not whether the estimate was too high or too low. If your calculator shows a negative number, you have forgotten the modulus — take its size.

How do I enter a number like $3.2\times10^{-5}$ on my GDC?

Use the dedicated scientific-notation key, labelled 'EE' or 'EXP' or '$\times10^{x}$'. Type $3.2$, then the key, then $(-)5$. Do not type $3.2\times10\wedge-5$: without brackets the calculator can misread the exponent, and even with them the extra keystrokes invite mistakes. The scientific-notation key enters the whole power as one object.

What does it mean to give the upper and lower bounds of a rounded value?

A rounded measurement stands for a whole interval of possible true values. If a mass is $50$ g 'to the nearest gram', the true mass is anything from $49.5$ g up to (but not including) $50.5$ g, so the lower bound is $49.5$ g and the upper bound is $50.5$ g. In general the bounds sit half a rounding-unit either side: to the nearest $10$ they would be $\pm 5$, to 1 decimal place $\pm 0.05$.

Standard form, approximation and percentage…

Revise Standard form, approximation and percentage error for IB Mathematics: Applications and Interpretation (1.1)

Looking for free IB Mathematics: Applications and Interpretation notes on Standard form, approximation and percentage error? This Standard Level lesson covers official syllabus point 1.1 on Mathematics: Applications and Interpretation SL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Standard form, approximation and percentage error. Built for students searching "IB Maths AI 1.1", "Standard form, approximation and percentage error IB revision", or a free alternative to paid IB note sites.

Standard form, approximation and percentage error

Every quantitative subject — physics, finance, biology, engineering — has to handle numbers that are awkwardly large, awkwardly small, or known only approximately. Topic 1.1 gives you three tools for exactly that: standard form to tame the size of a number, rounding and significant figures to shorten it honestly, and percentage error to say how close an approximation is to the truth. This lesson covers the full syllabus item, from writing and operating in standard form, through significant figures, decimal places, estimation and bounds, to percentage error. Because Applications and Interpretation allows the GDC on every paper, the calculator does the arithmetic — so the skill being marked is choosing the right form, quoting it to the right accuracy, and writing down the method. As in every calculation topic, the worked examples below are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1Write numbers in standard form $a\times10^{k}$ , where $1\le a<10$ and $k\in\mathbb{Z}$ , and convert back to ordinary form.
2Operate with numbers in standard form (multiply, divide, add and subtract), giving the answer in standard form to a stated accuracy.
3Round a number to a specified number of significant figures or decimal places, and use estimation to check that an answer is reasonable.
4State the upper and lower bounds of a value that has been rounded to a given accuracy.
5Calculate the percentage error between an approximate value and an exact value, showing the method that protects the marks.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Big, small, and 'close enough'

This topic gives you a tidy way to write enormous or tiny numbers (standard form), a disciplined way to shorten numbers (rounding and significant figures), and an honest way to say how far an approximation sits from the truth (percentage error).

THINK OF IT LIKE…

Picture guessing the number of sweets in a giant jar. You say 500; the real count is 520. Your guess is an approximation, and percentage error puts a number on how good it was — here about $3.85\%$ off. Standard form is the shorthand for a number too long to say aloud, like the atoms in the jar: instead of $602\,200\,000\,000\,000\,000\,000\,000$ you write $6.022\times10^{23}$ .

1
For standard form, rewrite the number so it starts with a single non-zero digit, then a decimal point. Count how many places the point moved to find the power of $10$ ; moving left gives a positive power, moving right gives a negative power.
2
To round, find the digit you are keeping last, then look at its right-hand neighbour: $5$ or more rounds up, $4$ or less leaves it unchanged.
3
For percentage error, subtract the exact value $v_E$ from the approximate value $v_A$ , then divide the difference by the exact value $v_E$ .
4
Take the absolute value, multiply by $100$ and add a ' $\%$ ' sign: $\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%$ .

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$a\times10^{k}\quad\text{where}\quad 1\le a<10\quad\text{and}\quad k\in\mathbb{Z}$

$\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$

Full topic notes

Formal explanation with the rigour you need for the exam.

Standard form (scientific notation)

Standard form writes any number as a value between $1$ and $10$ multiplied by a power of $10$ . It removes long strings of zeros and makes the size of a number instantly readable from its exponent. The definition rests on a single, strict condition on the coefficient.

$a\times10^{k}\quad\text{where}\quad 1\le a<10\quad\text{and}\quad k\in\mathbb{Z}$

To operate in standard form on a calculator you simply enter each number with the scientific-notation key and combine as usual; the GDC returns the result, which you then present in standard form to the required accuracy. It is still worth understanding the underlying index laws: to multiply, multiply the coefficients and add the powers; to divide, divide the coefficients and subtract the powers; then adjust so that the coefficient satisfies $1\le a<10$ .

The condition $1\le a<10$ is what makes the form correct. $6.3\times10^{5}$ is standard form; $63\times10^{4}$ and $0.63\times10^{6}$ are the same number but NOT in standard form because $a$ is out of range.
A positive power means a large number. $k>0$ : e.g. the mass of Earth $\approx 5.97\times10^{24}$ kg. To convert to ordinary form, move the decimal point $k$ places right.
A negative power means a small number. $k<0$ : e.g. the mass of an electron $\approx 9.11\times10^{-31}$ kg. Move the decimal point $|k|$ places left.
Enter standard form with the GDC's 'EE' or ' $\times10^{x}$ ' key, not by typing ' $\times$ ', ' $1$ ', ' $0$ ', ' $\wedge$ ', which risks order-of-operations errors.

Approximation: significant figures and decimal places

Approximation replaces a number with a nearby, shorter one that is accurate enough for the purpose. The two instructions you must distinguish are 'to a number of significant figures' and 'to a number of decimal places'. They are different rules, and answering with the wrong one is a needless lost mark.

Significant figures (s.f.): count from the first non-zero digit on the left. Leading zeros are placeholders and never count; trailing zeros after a decimal point do count. So $0.0459$ has 3 s.f. and $4.500$ has 4 s.f.
Decimal places (d.p.): count digits to the right of the decimal point, regardless of whether they are zero.
Rounding rule: look at the digit just after the last one you keep — $5$ or more rounds up, $4$ or less leaves it unchanged. Round once, from the original.
Worked instances (3 s.f.): $12\,851\approx 12\,900$ ; $0.045923\approx 0.0459$ ; $23.97\approx 24.0$ (the trailing zero is significant and must be written).

EXAM TIP

In multi-step calculations do not round intermediate results. Keep the full value in the calculator's memory (or use the 'Ans' key) and round only the final answer, to 3 significant figures unless the question says otherwise. Premature rounding is one of the most common reasons a correct method produces a final digit the mark scheme will not accept.

Upper and lower bounds

A rounded value is not a single number but a shorthand for an interval. When a measurement is given to a stated accuracy, the true value could be anything within half a rounding-unit either side. The lower bound is the smallest value that would round to the stated figure; the upper bound is the smallest value that would round up to the next one.

If a value $v$ is rounded to the nearest $u$ , then $v-\tfrac{u}{2}\le v_{\text{true}}<v+\tfrac{u}{2}$ . \n For 'nearest cm' $u=1$ ; for '1 decimal place' $u=0.1$ ; for 'nearest $10$ ' $u=10$ .

Percentage error

Percentage error measures the discrepancy between an approximate (or measured) value and an exact (or known) value, expressed as a percentage of the exact value. Dividing by the exact value makes the error relative — a $2$ cm error matters far more on a $10$ cm object than on a $10$ m one, and percentage error captures that.

$\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ \n where $v_A$ is the approximate (measured) value and $v_E$ is the exact (true) value.

The exact value $v_E$ goes on the bottom — always. Dividing by $v_A$ is the single most penalised error in this topic.
The bars make the answer positive. Percentage error reports the size of the error, not its direction, so it is never negative.
$v_A$ vs $v_E$ : $v_A$ is what was measured, estimated or read off; $v_E$ is the true or accepted value. If you swap them the answer is wrong.
Round only at the end, to 3 s.f. unless told otherwise — so a percentage error of $0.8\%$ is quoted as $0.800\%$ to 3 s.f.

Common mistakes examiners penalise

Leaving the coefficient outside $1\le a<10$ — $63\times10^{4}$ and $0.63\times10^{6}$ are not standard form. Adjust the coefficient and the power together until $1\le a<10$ .
A coefficient of exactly $10$ after multiplying — e.g. writing $10\times10^{4}$ ; bump the power to get $1.0\times10^{5}$ .
Confusing significant figures with decimal places — they are different instructions. $0.0023456$ is $0.00235$ to 3 s.f. but $0.002$ to 3 d.p.
Miscounting significant figures around zeros — leading zeros never count ( $0.0459$ has 3 s.f.); trailing zeros after a decimal point do ( $4.500$ has 4 s.f.), so $23.97\approx24.0$ must keep its zero.
Rounding in stages — round once, from the original number. Rounding $2.4491$ to $2.45$ and then to $2.5$ is wrong; to 1 d.p. it is $2.4$ .
Dividing by the approximate value in percentage error — the exact value $v_E$ must be the denominator, not $v_A$ .
Dropping the modulus — a negative percentage error means the absolute-value bars were forgotten.
Premature rounding in multi-step work — carry full GDC figures and round only the final answer, to 3 s.f. unless told otherwise.
Using the wrong bound — the upper bound is 'the stated value plus half a unit'; using the lower bound where the largest possible value is needed gives the wrong extreme.

Model answer — marked the way our engine marks it

In IB Mathematics the marks are analytic: each is tied to a specific line of working — a method mark (M) or an accuracy mark (A) — and an accuracy mark depends on the method mark it follows. Follow-through (FT) means a value that is wrong earlier need not cost you the marks that depend on it, provided the later step is done correctly on your own figure. The engine accepts any equivalent form and any correctly-rounded value, and ignores subsequent working once a correct answer appears. Because Applications and Interpretation allows the GDC on every paper, you may do the arithmetic on the calculator — but that protection only exists if the method is written on the page. Study how each mark below is earned by a specific line.

Where this leads

Standard form, rounding and percentage error are the number sense the rest of the course leans on. Standard form reappears whenever you handle scientific or financial magnitudes and when you interpret a GDC display that switches to scientific notation. The discipline of rounding only at the end, and to 3 significant figures, governs every calculation topic from trigonometry to finance. Percentage error returns whenever a model is compared with data — the gap between a predicted and an observed value is exactly a percentage error, and the same $\left|\frac{v_A-v_E}{v_E}\right|$ idea underlies how you judge whether a model is good enough. Master these three habits — pick the right form, quote it to the right accuracy, and show the method — and the calculation-heavy topics ahead become variations on skills you already own.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

The distance from the Earth to the Sun is approximately $1.496\times10^{8}$ km. Light travels at $3.00\times10^{5}$ km s $^{-1}$ . Calculate the time, in minutes, for light to travel from the Sun to the Earth, giving your answer to 3 significant figures. [4]

1
Use $\text{time}=\dfrac{\text{distance}}{\text{speed}}$ , entering both numbers with the GDC's scientific-notation key.\n\nSet up. $\text{time}=\dfrac{1.496\times10^{8}}{3.00\times10^{5}}$ seconds. [M1: correct quotient of distance over speed]\n\nEvaluate (in seconds). The GDC gives $\text{time}=498.66\ldots$ s. Dividing the coefficients gives $\dfrac{1.496}{3.00}=0.4986\ldots$ and subtracting the powers gives $10^{8-5}=10^{3}$ , so $0.4986\ldots\times10^{3}=498.6\ldots$ s — a useful check on the calculator. [A1]\n\nConvert to minutes. Divide by $60$ : $\dfrac{498.66\ldots}{60}=8.3111\ldots$ minutes. [M1: divide seconds by 60]\n\nRound to 3 s.f. $\text{time}=8.31$ minutes. [A1]\n\nCarry the full figure $498.66\ldots$ into the division by $60$ ; rounding to $499$ first would still round to $8.31$ here, but in general premature rounding shifts the final digit and loses the accuracy mark.

✓ FULLY WORKED

EXAMPLE 2

Evaluate $(4.0\times10^{7})\times(2.5\times10^{-3})$ , giving your answer in standard form. [3]

1
Multiply the coefficients and add the powers.\n\nCoefficients. $4.0\times2.5=10$ . [M1: multiply coefficients and add powers]\n\nPowers. $10^{7}\times10^{-3}=10^{7+(-3)}=10^{4}$ , so the product is $10\times10^{4}=10\times10^{4}$ .\n\nAdjust to standard form. The coefficient $10$ breaks the rule $1\le a<10$ , so write $10=1\times10^{1}$ and combine: $1\times10^{1}\times10^{4}=1.0\times10^{5}$ . [A1: correct value]\n\nAnswer: $1.0\times10^{5}$ . [A1: presented in valid standard form]\n\nThe last step is where marks are lost: a coefficient of exactly $10$ (or more) is not yet standard form — bump the power up by one.

✓ FULLY WORKED

EXAMPLE 3

A rectangular field measured $86.4$ m by $53.7$ m. \n (a) Estimate its area by rounding each length to 1 significant figure. \n (b) Calculate the exact area, then round it to 3 significant figures. [4]

1
(a) Estimate. Round each length to 1 s.f.: $86.4\approx 90$ and $53.7\approx 50$ .\n $\text{estimated area}=90\times50=4500\text{ m}^{2}$ . [M1: both lengths to 1 s.f. and multiplied] [A1]\n\n**(b) Calculated area.** $86.4\times53.7=4639.68\text{ m}^{2}$ .\nRounded to 3 s.f.: $4640\text{ m}^{2}$ . [A1]\n\nThe estimate $4500$ is the right order of magnitude, confirming the calculated $4640$ is reasonable — a quick guard against a keying error. [A1: estimate used to check the calculation]\n\nEstimation is not busywork: a 1-s.f. estimate catches a slipped decimal point or a wrong power of ten before it costs you the answer.

✓ FULLY WORKED

EXAMPLE 4

The length of a metal rod is measured as $24$ cm, correct to the nearest centimetre. \n (a) Write down the lower and upper bounds of the true length. \n (b) The rod is cut into $5$ equal pieces. Find the upper bound for the length of one piece. [4]

1
(a) Bounds of the rod. The rounding unit is $1$ cm, so the true length lies within $0.5$ cm of $24$ cm.\nLower bound $=24-0.5=23.5$ cm. [A1]\nUpper bound $=24+0.5=24.5$ cm, so $23.5\le \ell<24.5$ . [A1]\n\n**(b) One piece.** The longest a piece can be comes from the longest possible rod, so divide the upper bound by $5$ :\n $\dfrac{24.5}{5}=4.9$ cm. [M1: divide the upper bound by 5] [A1]\n\nAnswer: the upper bound for one piece is $4.9$ cm. Note that the upper bound is written as $24.5$ even though the true length is strictly less than $24.5$ ; using $23.5$ here — the lower bound — would give the wrong extreme.

✓ FULLY WORKED

EXAMPLE 5

A student estimates the value of $\pi$ as $\frac{22}{7}$ and uses $\pi=3.14159265\ldots$ as the exact value. Calculate the percentage error in the estimate, giving your answer to 3 significant figures. [3]

1
Formula. $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ with $v_A=\dfrac{22}{7}=3.142857\ldots$ and $v_E=3.141592\ldots$ . [M1: formula with correct $v_A$ and $v_E$ ]\n\nSubstitute. $\varepsilon=\left|\dfrac{3.142857\ldots-3.141592\ldots}{3.141592\ldots}\right|\times100\%$ . [M1: substitution, exact value in the denominator]\n\nEvaluate. $=\left|\dfrac{0.001264\ldots}{3.141592\ldots}\right|\times100\%=0.04024\ldots\%$ .\n\nRound to 3 s.f. $\varepsilon=0.0402\%$ . [A1]\n\nKeep the full decimal for $v_E$ throughout; rounding $\pi$ to $3.14$ before subtracting would give a wrong difference and a wrong final digit.

✓ FULLY WORKED

EXAMPLE 6

A student measures a length as $24.8$ cm; the true length is $25.0$ cm. Calculate the percentage error, giving your answer to 3 significant figures. [3]

1
Model answer — full working.\n\nHere the approximate value is the measurement, $v_A=24.8$ cm, and the exact value is the true length, $v_E=25.0$ cm.\n\nFormula. $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ .\n\nSubstitute. $\varepsilon=\left|\dfrac{24.8-25.0}{25.0}\right|\times100\%$ .\n\nEvaluate. $=\left|\dfrac{-0.2}{25.0}\right|\times100\%=0.008\times100\%=0.8\%$ .\n\nRound to 3 s.f. $\varepsilon=0.800\%$ .\n\n---\nHow our marking engine awards the 3 marks:\n\n- M1 — percentage-error formula. A method mark for writing $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ with the correct roles: the measured $24.8$ as $v_A$ and the true $25.0$ as $v_E$ in the denominator. Putting $24.8$ on the bottom does NOT earn this mark.\n- M1 — substitution. A second method mark for the correct substitution $\left|\dfrac{24.8-25.0}{25.0}\right|\times100$ , exact value on the bottom. The engine checks the denominator is $25.0$ , not $24.8$ .\n- A1 — accuracy. Awarded for the final value $0.800\%$ (equivalently $0.8\%$ seen to the required accuracy). This A-mark depends on the two M-marks above.\n\nFollow-through (FT). A candidate who reached the correct difference but slipped in the final division still earns the A-mark on their own figure, provided the method lines are present and the answer is correctly rounded from their working.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $0.8\%$ , $0.800\%$ , or $0.80\%$ as correctly-rounded equivalents to 3 s.f., and accepts the fraction written as $\dfrac{0.2}{25}$ or $\dfrac{25.0-24.8}{25.0}$ (the modulus makes the order of subtraction irrelevant). Once the correct answer appears, ISW means a later restatement will not lose the mark.\n\nBottom line: of the 3 marks, two are method marks that survive an arithmetic slip, and the accuracy mark is shielded by follow-through — but only if the formula and substitution are on the page. A bare ' $0.8\%$ ' with no working risks losing both method marks.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAStandard form, approximation and percentage error

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

Standard form

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ANSWER

A number written as $a\times10^{k}$ with $1\le a<10$ and $k\in\mathbb{Z}$ . The single condition $1\le a<10$ is what makes the form unique: $a$ has exactly one non-zero digit before the decimal point.

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The condition $1\le a<10$ is what makes the form correct. $6.3\times10^{5}$ is standard form; $63\times10^{4}$ and $0.63\times10^{6}$ are the same number but NOT in standard form because $a$ is out of range.
✓
A positive power means a large number. $k>0$ : e.g. the mass of Earth $\approx 5.97\times10^{24}$ kg. To convert to ordinary form, move the decimal point $k$ places right.
✓
A negative power means a small number. $k<0$ : e.g. the mass of an electron $\approx 9.11\times10^{-31}$ kg. Move the decimal point $|k|$ places left.
✓
Enter standard form with the GDC's 'EE' or ' $\times10^{x}$ ' key, not by typing ' $\times$ ', ' $1$ ', ' $0$ ', ' $\wedge$ ', which risks order-of-operations errors.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Get a calculation marked: work a percentage-error or standard-form question with full method marks and follow-through

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

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✓

Checkpoint

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Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Get a calculation marked: work a percentage-error or standard-form question with full method marks and follow-through on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

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Standard form, approximation and percentage error

BY THE END, YOU CAN…

1Write numbers in standard form $a\times10^{k}$ , where $1\le a<10$ and $k\in\mathbb{Z}$ , and convert back to ordinary form.
2Operate with numbers in standard form (multiply, divide, add and subtract), giving the answer in standard form to a stated accuracy.
3Round a number to a specified number of significant figures or decimal places, and use estimation to check that an answer is reasonable.
4State the upper and lower bounds of a value that has been rounded to a given accuracy.
5Calculate the percentage error between an approximate value and an exact value, showing the method that protects the marks.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Big, small, and 'close enough'

THINK OF IT LIKE…

1
For standard form, rewrite the number so it starts with a single non-zero digit, then a decimal point. Count how many places the point moved to find the power of $10$ ; moving left gives a positive power, moving right gives a negative power.
2
To round, find the digit you are keeping last, then look at its right-hand neighbour: $5$ or more rounds up, $4$ or less leaves it unchanged.
3
For percentage error, subtract the exact value $v_E$ from the approximate value $v_A$ , then divide the difference by the exact value $v_E$ .
4
Take the absolute value, multiply by $100$ and add a ' $\%$ ' sign: $\varepsilon=\left|\frac{v_A-v_E}{v_E}\right|\times100\%$ .

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

$a\times10^{k}\quad\text{where}\quad 1\le a<10\quad\text{and}\quad k\in\mathbb{Z}$

$\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$

Full topic notes

Formal explanation with the rigour you need for the exam.

Standard form (scientific notation)

$a\times10^{k}\quad\text{where}\quad 1\le a<10\quad\text{and}\quad k\in\mathbb{Z}$

The condition $1\le a<10$ is what makes the form correct. $6.3\times10^{5}$ is standard form; $63\times10^{4}$ and $0.63\times10^{6}$ are the same number but NOT in standard form because $a$ is out of range.
A positive power means a large number. $k>0$ : e.g. the mass of Earth $\approx 5.97\times10^{24}$ kg. To convert to ordinary form, move the decimal point $k$ places right.
A negative power means a small number. $k<0$ : e.g. the mass of an electron $\approx 9.11\times10^{-31}$ kg. Move the decimal point $|k|$ places left.
Enter standard form with the GDC's 'EE' or ' $\times10^{x}$ ' key, not by typing ' $\times$ ', ' $1$ ', ' $0$ ', ' $\wedge$ ', which risks order-of-operations errors.

Approximation: significant figures and decimal places

Significant figures (s.f.): count from the first non-zero digit on the left. Leading zeros are placeholders and never count; trailing zeros after a decimal point do count. So $0.0459$ has 3 s.f. and $4.500$ has 4 s.f.
Decimal places (d.p.): count digits to the right of the decimal point, regardless of whether they are zero.
Rounding rule: look at the digit just after the last one you keep — $5$ or more rounds up, $4$ or less leaves it unchanged. Round once, from the original.
Worked instances (3 s.f.): $12\,851\approx 12\,900$ ; $0.045923\approx 0.0459$ ; $23.97\approx 24.0$ (the trailing zero is significant and must be written).

EXAM TIP

Upper and lower bounds

If a value $v$ is rounded to the nearest $u$ , then $v-\tfrac{u}{2}\le v_{\text{true}}<v+\tfrac{u}{2}$ . \n For 'nearest cm' $u=1$ ; for '1 decimal place' $u=0.1$ ; for 'nearest $10$ ' $u=10$ .

Percentage error

$\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ \n where $v_A$ is the approximate (measured) value and $v_E$ is the exact (true) value.

The exact value $v_E$ goes on the bottom — always. Dividing by $v_A$ is the single most penalised error in this topic.
The bars make the answer positive. Percentage error reports the size of the error, not its direction, so it is never negative.
$v_A$ vs $v_E$ : $v_A$ is what was measured, estimated or read off; $v_E$ is the true or accepted value. If you swap them the answer is wrong.
Round only at the end, to 3 s.f. unless told otherwise — so a percentage error of $0.8\%$ is quoted as $0.800\%$ to 3 s.f.

Common mistakes examiners penalise

Leaving the coefficient outside $1\le a<10$ — $63\times10^{4}$ and $0.63\times10^{6}$ are not standard form. Adjust the coefficient and the power together until $1\le a<10$ .
A coefficient of exactly $10$ after multiplying — e.g. writing $10\times10^{4}$ ; bump the power to get $1.0\times10^{5}$ .
Confusing significant figures with decimal places — they are different instructions. $0.0023456$ is $0.00235$ to 3 s.f. but $0.002$ to 3 d.p.
Miscounting significant figures around zeros — leading zeros never count ( $0.0459$ has 3 s.f.); trailing zeros after a decimal point do ( $4.500$ has 4 s.f.), so $23.97\approx24.0$ must keep its zero.
Rounding in stages — round once, from the original number. Rounding $2.4491$ to $2.45$ and then to $2.5$ is wrong; to 1 d.p. it is $2.4$ .
Dividing by the approximate value in percentage error — the exact value $v_E$ must be the denominator, not $v_A$ .
Dropping the modulus — a negative percentage error means the absolute-value bars were forgotten.
Premature rounding in multi-step work — carry full GDC figures and round only the final answer, to 3 s.f. unless told otherwise.
Using the wrong bound — the upper bound is 'the stated value plus half a unit'; using the lower bound where the largest possible value is needed gives the wrong extreme.

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Worked examples

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EXAMPLE 1

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Use $\text{time}=\dfrac{\text{distance}}{\text{speed}}$ , entering both numbers with the GDC's scientific-notation key.\n\nSet up. $\text{time}=\dfrac{1.496\times10^{8}}{3.00\times10^{5}}$ seconds. [M1: correct quotient of distance over speed]\n\nEvaluate (in seconds). The GDC gives $\text{time}=498.66\ldots$ s. Dividing the coefficients gives $\dfrac{1.496}{3.00}=0.4986\ldots$ and subtracting the powers gives $10^{8-5}=10^{3}$ , so $0.4986\ldots\times10^{3}=498.6\ldots$ s — a useful check on the calculator. [A1]\n\nConvert to minutes. Divide by $60$ : $\dfrac{498.66\ldots}{60}=8.3111\ldots$ minutes. [M1: divide seconds by 60]\n\nRound to 3 s.f. $\text{time}=8.31$ minutes. [A1]\n\nCarry the full figure $498.66\ldots$ into the division by $60$ ; rounding to $499$ first would still round to $8.31$ here, but in general premature rounding shifts the final digit and loses the accuracy mark.

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EXAMPLE 2

Evaluate $(4.0\times10^{7})\times(2.5\times10^{-3})$ , giving your answer in standard form. [3]

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Multiply the coefficients and add the powers.\n\nCoefficients. $4.0\times2.5=10$ . [M1: multiply coefficients and add powers]\n\nPowers. $10^{7}\times10^{-3}=10^{7+(-3)}=10^{4}$ , so the product is $10\times10^{4}=10\times10^{4}$ .\n\nAdjust to standard form. The coefficient $10$ breaks the rule $1\le a<10$ , so write $10=1\times10^{1}$ and combine: $1\times10^{1}\times10^{4}=1.0\times10^{5}$ . [A1: correct value]\n\nAnswer: $1.0\times10^{5}$ . [A1: presented in valid standard form]\n\nThe last step is where marks are lost: a coefficient of exactly $10$ (or more) is not yet standard form — bump the power up by one.

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EXAMPLE 3

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(a) Estimate. Round each length to 1 s.f.: $86.4\approx 90$ and $53.7\approx 50$ .\n $\text{estimated area}=90\times50=4500\text{ m}^{2}$ . [M1: both lengths to 1 s.f. and multiplied] [A1]\n\n**(b) Calculated area.** $86.4\times53.7=4639.68\text{ m}^{2}$ .\nRounded to 3 s.f.: $4640\text{ m}^{2}$ . [A1]\n\nThe estimate $4500$ is the right order of magnitude, confirming the calculated $4640$ is reasonable — a quick guard against a keying error. [A1: estimate used to check the calculation]\n\nEstimation is not busywork: a 1-s.f. estimate catches a slipped decimal point or a wrong power of ten before it costs you the answer.

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EXAMPLE 4

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(a) Bounds of the rod. The rounding unit is $1$ cm, so the true length lies within $0.5$ cm of $24$ cm.\nLower bound $=24-0.5=23.5$ cm. [A1]\nUpper bound $=24+0.5=24.5$ cm, so $23.5\le \ell<24.5$ . [A1]\n\n**(b) One piece.** The longest a piece can be comes from the longest possible rod, so divide the upper bound by $5$ :\n $\dfrac{24.5}{5}=4.9$ cm. [M1: divide the upper bound by 5] [A1]\n\nAnswer: the upper bound for one piece is $4.9$ cm. Note that the upper bound is written as $24.5$ even though the true length is strictly less than $24.5$ ; using $23.5$ here — the lower bound — would give the wrong extreme.

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EXAMPLE 5

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Formula. $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ with $v_A=\dfrac{22}{7}=3.142857\ldots$ and $v_E=3.141592\ldots$ . [M1: formula with correct $v_A$ and $v_E$ ]\n\nSubstitute. $\varepsilon=\left|\dfrac{3.142857\ldots-3.141592\ldots}{3.141592\ldots}\right|\times100\%$ . [M1: substitution, exact value in the denominator]\n\nEvaluate. $=\left|\dfrac{0.001264\ldots}{3.141592\ldots}\right|\times100\%=0.04024\ldots\%$ .\n\nRound to 3 s.f. $\varepsilon=0.0402\%$ . [A1]\n\nKeep the full decimal for $v_E$ throughout; rounding $\pi$ to $3.14$ before subtracting would give a wrong difference and a wrong final digit.

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EXAMPLE 6

A student measures a length as $24.8$ cm; the true length is $25.0$ cm. Calculate the percentage error, giving your answer to 3 significant figures. [3]

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Model answer — full working.\n\nHere the approximate value is the measurement, $v_A=24.8$ cm, and the exact value is the true length, $v_E=25.0$ cm.\n\nFormula. $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ .\n\nSubstitute. $\varepsilon=\left|\dfrac{24.8-25.0}{25.0}\right|\times100\%$ .\n\nEvaluate. $=\left|\dfrac{-0.2}{25.0}\right|\times100\%=0.008\times100\%=0.8\%$ .\n\nRound to 3 s.f. $\varepsilon=0.800\%$ .\n\n---\nHow our marking engine awards the 3 marks:\n\n- M1 — percentage-error formula. A method mark for writing $\varepsilon=\left|\dfrac{v_A-v_E}{v_E}\right|\times100\%$ with the correct roles: the measured $24.8$ as $v_A$ and the true $25.0$ as $v_E$ in the denominator. Putting $24.8$ on the bottom does NOT earn this mark.\n- M1 — substitution. A second method mark for the correct substitution $\left|\dfrac{24.8-25.0}{25.0}\right|\times100$ , exact value on the bottom. The engine checks the denominator is $25.0$ , not $24.8$ .\n- A1 — accuracy. Awarded for the final value $0.800\%$ (equivalently $0.8\%$ seen to the required accuracy). This A-mark depends on the two M-marks above.\n\nFollow-through (FT). A candidate who reached the correct difference but slipped in the final division still earns the A-mark on their own figure, provided the method lines are present and the answer is correctly rounded from their working.\n\n**'Accept equivalent forms and correct rounding.'** The engine accepts $0.8\%$ , $0.800\%$ , or $0.80\%$ as correctly-rounded equivalents to 3 s.f., and accepts the fraction written as $\dfrac{0.2}{25}$ or $\dfrac{25.0-24.8}{25.0}$ (the modulus makes the order of subtraction irrelevant). Once the correct answer appears, ISW means a later restatement will not lose the mark.\n\nBottom line: of the 3 marks, two are method marks that survive an arithmetic slip, and the accuracy mark is shielded by follow-through — but only if the formula and substitution are on the page. A bare ' $0.8\%$ ' with no working risks losing both method marks.

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MAIN IDEAStandard form, approximation and percentage error

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The condition $1\le a<10$ is what makes the form correct. $6.3\times10^{5}$ is standard form; $63\times10^{4}$ and $0.63\times10^{6}$ are the same number but NOT in standard form because $a$ is out of range.
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A positive power means a large number. $k>0$ : e.g. the mass of Earth $\approx 5.97\times10^{24}$ kg. To convert to ordinary form, move the decimal point $k$ places right.
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A negative power means a small number. $k<0$ : e.g. the mass of an electron $\approx 9.11\times10^{-31}$ kg. Move the decimal point $|k|$ places left.
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Enter standard form with the GDC's 'EE' or ' $\times10^{x}$ ' key, not by typing ' $\times$ ', ' $1$ ', ' $0$ ', ' $\wedge$ ', which risks order-of-operations errors.

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Revise Standard form, approximation and percentage error for IB Mathematics: Applications and Interpretation (1.1)

In simple terms

Explore the concept

Key formulas

Full topic notes

Standard form (scientific notation)

Approximation: significant figures and decimal places

Upper and lower bounds

Percentage error

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

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Key takeaways

Practice — then mark it

Get a calculation marked: work a percentage-error or standard-form question with full method marks and follow-through

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Standard form (scientific notation)

Approximation: significant figures and decimal places

Upper and lower bounds

Percentage error

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a calculation marked: work a percentage-error or standard-form question with full method marks and follow-through

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Standard form, approximation and percentage error