Decimals and Rounding
- Round numbers to given decimal places and significant figures
- Distinguish truncation from rounding and compare their errors
- Convert recurring decimals to fractions using algebraic methods
- Estimate calculations and identify whether results are over- or underestimates
- Order decimals and fractions on a number line
🔑 Core Concepts
Decimal Place Values
Every digit in a decimal number occupies a place value position. Understanding these positions is the foundation for all rounding and ordering work.
| Thousands | Hundreds | Tens | Units | . | Tenths | Hundredths | Thousandths |
|---|---|---|---|---|---|---|---|
| 1000 | 100 | 10 | 1 | . | $\frac{1}{10}$ | $\frac{1}{100}$ | $\frac{1}{1000}$ |
| $10^3$ | $10^2$ | $10^1$ | $10^0$ | . | $10^{-1}$ | $10^{-2}$ | $10^{-3}$ |
Rounding to Decimal Places
Rounding to $n$ decimal places means keeping exactly $n$ digits after the decimal point. The key decision is based on the digit immediately after the last kept digit.
Example: Round $3.7465$ to 2 decimal places.
The 2nd decimal place is 4. The next digit is 6 ($\geq 5$), so round up: $3.75$.
Rounding to Significant Figures
Significant figures count from the first non-zero digit, regardless of the position of the decimal point. This makes sig figs more useful for very large or very small numbers.
| Number | 1 s.f. | 2 s.f. | 3 s.f. |
|---|---|---|---|
| $36{,}472$ | $40{,}000$ | $36{,}000$ | $36{,}500$ |
| $0.004076$ | $0.004$ | $0.0041$ | $0.00408$ |
| $1.0050$ | $1$ | $1.0$ | $1.01$ |
Truncation vs Rounding
Truncation and rounding both reduce precision, but they differ crucially: truncation always discards without adjusting, while rounding adjusts based on the next digit. This distinction is important for error analysis at Grade 9.
| Number | Rounded to 2 d.p. | Truncated to 2 d.p. |
|---|---|---|
| $3.476$ | $3.48$ | $3.47$ |
| $3.471$ | $3.47$ | $3.47$ |
| $9.999$ | $10.00$ | $9.99$ |
| $2.345$ | $2.35$ (or $2.34$ by some conventions) | $2.34$ |
When a value $x$ is truncated to $n$ decimal places, the truncation error is always in the range $[0, 10^{-n})$. The truncated value is always an underestimate of the true value (for positive numbers).
For rounding, the error lies in $(-\tfrac{1}{2} \cdot 10^{-n},\; +\tfrac{1}{2} \cdot 10^{-n}]$, and the estimate can be either above or below.
Recurring Decimals
A recurring decimal has one or more digits that repeat infinitely. The repeating block is indicated with a dot above the first and last repeating digit.
$0.\dot{1}\dot{4} = 0.141414\ldots$ (block 14 repeats)
$0.1\dot{6} = 0.16666\ldots$ (only the 6 recurs, not the 1)
$0.\dot{1}4\dot{2} = 0.142142142\ldots$ (block 142 repeats)
Converting Recurring Decimals to Fractions (Algebraic Method)
This is a Grade 9 technique. The strategy is to multiply the recurring decimal by a power of 10 that aligns the repeating blocks, then subtract to eliminate the recurring part.
- Let $x$ equal the recurring decimal.
- Multiply by $10^n$ where $n$ is the length of the repeating block, to shift one full cycle.
- Subtract the original equation to eliminate the recurring part.
- Solve for $x$ and simplify the fraction.
If the recurring block has length $k$ and starts at the $m$th decimal place, multiply by $10^m$ and $10^{m+k}$, then subtract. This always eliminates the recurring part. Always simplify your final fraction using the HCF of numerator and denominator.
Estimating by Rounding
Estimation involves rounding each value in a calculation (usually to 1 s.f.) and then computing the simplified result. This gives a quick check on whether an answer is sensible.
Whether an estimate is an overestimate or underestimate depends on what was rounded and where it appears in the calculation:
- Rounding a number up in the numerator → overestimate.
- Rounding a number down in the numerator → underestimate.
- Rounding a number up in the denominator → underestimate.
- Rounding a number down in the denominator → overestimate.
For products, round one up and one down: the net effect depends on the relative magnitudes.
🗺 Visual Notes
Rounding
- Look at digit $n+1$
- $\geq 5$: round up
- $< 5$: round down
- Carry propagates through 9s
- Count from first non-zero
- Leading zeros: not sig
- Trailing zeros after dp: sig
- Zeros between digits: sig
- Dot notation for repeating blocks
- Multiply by $10^{\text{block length}}$
- Subtract to eliminate recurrence
- Simplify the fraction
- Always drops without adjusting
- Always $\leq$ original (positive)
- Error: $[0, 10^{-n})$
- Used in floor/integer division
- Round to 1 s.f. typically
- Simplify then calculate
- Identify over/underestimate
- Check reasonableness
- Compare digit by digit
- Pad with trailing zeros
- Convert to same form
- Use number line for negatives
Rounding vs Truncation Comparison
| Property | Rounding | Truncation |
|---|---|---|
| Method | Look at $(n+1)$th digit; adjust if $\geq 5$ | Simply drop all digits after $n$th place |
| Direction of error | Can be positive or negative | Always $\geq 0$ (for positive numbers) |
| Maximum error | $\tfrac{1}{2} \times 10^{-n}$ | Just under $10^{-n}$ |
| Example: $3.47$ to 1 d.p. | $3.5$ (rounded up) | $3.4$ (dropped) |
| Example: $3.41$ to 1 d.p. | $3.4$ (same) | $3.4$ (same) |
| Bias | Symmetric about true value | Always below true value |
Decision Tree: How to Convert a Recurring Decimal
No: first multiply by $10^m$ (shift non-recurring part), then by $10^{m+k}$
Over- and Underestimate Summary
| Position in calculation | Rounded Up | Rounded Down |
|---|---|---|
| Numerator (top) | Overestimate | Underestimate |
| Denominator (bottom) | Underestimate | Overestimate |
| Being added | Overestimate | Underestimate |
| Being subtracted | Underestimate | Overestimate |
✏ Worked Examples
(b) Hence determine whether $0.3\dot{1}\dot{5}$ is greater than, less than, or equal to $\dfrac{7}{22}$.
(c) An estimate for $\dfrac{\sqrt{26} \times 4.8}{0.49}$ is computed by rounding each value to 1 s.f. State the estimate and explain whether it is an overestimate or underestimate.
❓ Exam Questions
Write $0.003706$ correct to 2 significant figures.
Mark scheme: B1 for $0.0037$ (first sig fig is 3, second is 7; the next digit is 0 so round down).
A number $n$ is truncated to 1 decimal place to give $4.7$. Write down the error interval for $n$.
Mark scheme: M1 for recognising truncation gives a lower bound equal to the truncated value. A1 for the complete interval $4.7 \leq n < 4.8$. Note: rounding would give $4.65 \leq n < 4.75$.
Convert $0.\dot{4}3\dot{2}$ to a fraction. Give your answer in its simplest form.
Mark scheme:
M1: Let $x = 0.\dot{4}3\dot{2} = 0.432432\ldots$; recognise block length 3, multiply by $10^3 = 1000$.
M1: $1000x = 432.432432\ldots$; subtract $x$: $999x = 432$, so $x = \dfrac{432}{999}$.
A1: HCF(432, 999) = 27; $x = \dfrac{16}{37}$.
Note: $\dfrac{16}{37}$ is the correct simplified answer.
Work out an estimate for $\dfrac{(19.7)^2}{0.048 \times 312}$. You must show all your rounding. State whether your estimate is an overestimate or underestimate, giving a reason.
Mark scheme:
M1: Round $19.7 \approx 20$, $0.048 \approx 0.05$, $312 \approx 300$.
M1: Numerator: $20^2 = 400$; denominator: $0.05 \times 300 = 15$.
A1: $\dfrac{400}{15} = 26.\overline{6}$ (accept $26.7$ or $\frac{80}{3}$).
A1: Overestimate — $19.7$ was rounded up (larger numerator), $0.048$ was rounded up (larger denominator, reducing the fraction) and $312$ was rounded down (smaller denominator, increasing the fraction). Overall the numerator effect dominates: $19.7^2 = 388.09 < 400$, but $0.048 \times 312 = 14.976 < 15$, giving true value $\approx 25.9$, so estimate $26.7$ is an overestimate.
(a) Convert $0.1\dot{8}$ to a fraction in its simplest form. [3]
(b) Prove that $0.\dot{9} = 1$. [2]
(c) Without a calculator, show that $0.\dot{1}\dot{8}$ lies between $\dfrac{1}{6}$ and $\dfrac{1}{5}$. [1]
(a) $x = 0.1888\ldots$; $10x = 1.888\ldots$; $100x = 18.888\ldots$; $100x - 10x = 17$; $90x = 17$; $x = \dfrac{17}{90}$. (3 marks: M1 for setup, M1 for correct subtraction, A1 for $\frac{17}{90}$)
(b) Let $x = 0.\dot{9} = 0.999\ldots$; $10x = 9.999\ldots$; $10x - x = 9$; $9x = 9$; $x = 1$. Therefore $0.\dot{9} = 1$. (2 marks: M1 for algebraic method, A1 for conclusion)
(c) $0.\dot{1}\dot{8} = \dfrac{18}{99} = \dfrac{2}{11}$. Check: $\dfrac{1}{6} = \dfrac{11}{66}$, $\dfrac{2}{11} = \dfrac{12}{66}$, $\dfrac{1}{5} = \dfrac{13.2}{66}$. Since $11 < 12 < 13.2$, we have $\dfrac{1}{6} < \dfrac{2}{11} < \dfrac{1}{5}$. (1 mark: B1)
⭐ Grade 9 Model Answers
Full annotated answer for Q5 — the hardest question on this topic.
Setting up correctly (M1): Identifying that $0.1\dot{8}$ has a non-recurring digit "1" followed by the recurring digit "8" is the critical first step. This means you need two multiplications, not one. Writing $x = 0.1888\ldots$ clearly earns the first method mark.
Correct subtraction (M1): The key insight is to multiply by 10 (to shift past the non-recurring "1") and by 100 (to shift one full recurring cycle). Subtracting $10x$ from $100x$ gives $90x = 17$. This earns the second method mark.
Final answer (A1): $x = \frac{17}{90}$. You must verify this is in lowest terms: HCF(17, 90) = 1 (17 is prime, and 90 = 2 × 3² × 5 shares no factor with 17). No further simplification is needed.
This is a Grade 9 proof question. The key is to use the exact algebraic method, not a hand-wavy argument:
Let $x = 0.999\ldots$ Then $10x = 9.999\ldots$ Subtracting: $10x - x = 9.999\ldots - 0.999\ldots = 9$. So $9x = 9$, giving $x = 1$.
This is an equality, not an approximation. The recurring decimal $0.\dot{9}$ is another way of writing the number $1$. Students who write "it gets infinitely close to 1 but never equals it" are incorrect and score no marks on part (b).
"Hence" in a question means you should use earlier results. Here, $0.\dot{1}\dot{8} = \frac{2}{11}$ (derived from the same algebraic technique used in part (a)).
Converting all three fractions to a common denominator (66 works: LCM of 6, 11, 5 is 330, but 66 also works for the first two): $\frac{1}{6} = \frac{11}{66}$, $\frac{2}{11} = \frac{12}{66}$. For comparison with $\frac{1}{5}$: use 55 as denominator for $\frac{1}{5} = \frac{11}{55}$ vs $\frac{2}{11} = \frac{10}{55}$... actually compare using decimals: $\frac{1}{6} \approx 0.1\overline{6}$, $\frac{2}{11} = 0.\overline{18}$, $\frac{1}{5} = 0.2$. Since $0.166\ldots < 0.1818\ldots < 0.2$, the inequality holds.
Full marks require a clear comparison step — just writing the three fractions without showing the comparison is insufficient.
📋 Revision Sheet
| Term | Meaning |
|---|---|
| Decimal place | Position of a digit after the decimal point |
| Significant figure | Meaningful digit; count from first non-zero |
| Rounding | Reducing precision by adjusting last kept digit |
| Truncation | Dropping digits without adjustment |
| Recurring decimal | Decimal with an infinitely repeating block |
| Overestimate | Approximate value > true value |
| Underestimate | Approximate value < true value |
Truncation error: $0 \leq x - T < 10^{-n}$
Rounding error: $|x - R| \leq \tfrac{1}{2} \times 10^{-n}$
Single-digit recurring: $x = 0.\dot{d} \Rightarrow 9x = d \Rightarrow x = \frac{d}{9}$
Two-digit recurring: $x = 0.\dot{d_1d_2}\dot{} \Rightarrow 99x = d_1d_2 \Rightarrow x = \frac{d_1d_2}{99}$
Mixed: $x = 0.p\dot{q_1}\dot{q_2}$: multiply by $10^m$ and $10^{m+k}$, subtract
Ordering: pad with zeros, compare digit by digit from left
- "FNZ = F1" — First Non-Zero digit = Figure 1 (for sig figs)
- "Drop, don't bump" — truncation just drops digits
- "Block length = power" — multiply by $10^{\text{block length}}$ for recurring
- "Up top, up result" — rounding numerator up = overestimate
- "Up below, down result" — rounding denominator up = underestimate
- "Dots mark the block" — dots sit over first and last digit of recurring block
- For significant figures: always count from the first non-zero digit — zeros before it don't count
- When rounding a 9 up, carry the 1 all the way through (e.g. $9.995 \to 10.0$ to 1 d.p.)
- Write dot notation correctly — dots over first and last of repeating block only
- Always simplify recurring decimal fractions using HCF
- When asked "over or underestimate", state which values you rounded and in which direction
- Truncation error interval: lower bound = truncated value (not $-0.5 \times$ last digit)
- Check: does your converted fraction give the original decimal when divided? Quick sanity test
🔄 Flashcards
Click a card to reveal the answer.
✗ Common Mistakes
What students do: Write $0.0047$ to 2 s.f. as $0.0$ or $0.00$, treating the zeros after the decimal point as the first two significant figures.
Why marks are lost: The answer is wrong; no marks for the final answer, and possibly no method mark either.
How to avoid: Remember: leading zeros are place holders, not significant. Scan from the left until you find the first non-zero digit — that is Figure 1. For $0.0047$: first sig fig is 4, second is 7. Answer: $0.0047$ to 2 s.f. is $0.0047$ (unchanged, as there are only 2 sig figs to begin with).
What students do: When asked to truncate $7.89$ to 1 d.p., write $7.9$ (which is the rounded value), not $7.8$.
Why marks are lost: A1 answer mark lost (method mark may still be awarded if working is shown).
How to avoid: Truncation = "cut off" — like scissors, it removes the extra digits without any decision. The result is always $\leq$ the original positive value. Only rounding looks at the next digit.
What students do: For $0.\dot{1}4\dot{2}$, multiply by 10 instead of 1000 (block "142" has length 3, not 1).
Why marks are lost: The subtraction step does not eliminate the recurring part, so the method fails entirely — zero marks.
How to avoid: Always count the digits in the repeating block. Use dot notation to identify the full block. For $0.\dot{1}4\dot{2}$: the dots are over 1 and 2, so the block is "142" — length 3 — multiply by $10^3 = 1000$.
What students do: Leave $\dfrac{27}{99}$ as the final answer instead of simplifying to $\dfrac{3}{11}$.
Why marks are lost: Questions almost always ask for the fraction "in its simplest form" — the final A mark is for simplification. An unsimplified correct fraction typically earns one mark fewer.
How to avoid: Always check HCF(numerator, denominator) before writing the final answer. Use prime factorisation if necessary: $27 = 3^3$, $99 = 3^2 \times 11$, HCF $= 9$.
What students do: Given that $n$ is truncated to 2 d.p. giving $3.14$, write the error interval as $3.135 \leq n < 3.145$ (which is the rounding error interval).
Why marks are lost: Both bounds are wrong. Truncation always uses the truncated value as the lower bound: $3.14 \leq n < 3.15$.
How to avoid: Memorise: truncation lower bound = the truncated value itself (no subtraction of half a unit). Upper bound = truncated value + one unit in the last place ($10^{-n}$).
What students do: Write "it is an overestimate" without explaining which values were rounded up or down, or without analysing the position (numerator vs denominator).
Why marks are lost: The justification mark (typically the second of 2 marks on such questions) requires explicit reference to which value was rounded in which direction and the effect on the result.
How to avoid: Always write: "I rounded [value] up/down in the [numerator/denominator], which [increases/decreases] the estimate." Analyse each rounded value and state the net effect on the calculation.
✅ Final Checklist
Tick each item as you master it. Progress is saved automatically.
- I can identify the place value of any digit in a decimal number
- I can round a decimal to any given number of decimal places
- I can handle the carry when rounding a 9 (e.g. $9.996 \to 10.0$)
- I can identify the first significant figure of any number (including small decimals)
- I can round to any given number of significant figures
- I understand what truncation means and can truncate to any decimal place
- I can write the error interval for a number that has been rounded
- I can write the error interval for a number that has been truncated
- I can use dot notation correctly to represent recurring decimals
- I can convert a recurring decimal with a single repeating digit to a fraction
- I can convert a recurring decimal with a repeating block (e.g. $0.\dot{1}4\dot{2}$) to a fraction algebraically
- I can convert a mixed recurring decimal (non-recurring prefix) to a fraction
- I can estimate a calculation by rounding values to 1 significant figure
- I can determine whether an estimate is an overestimate or underestimate with full justification
- I can order a mixed set of decimals and fractions on a number line