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This lesson covers the foundational number skills required for the Edexcel GCSE Mathematics (1MA1) specification. Understanding place value, ordering numbers, rounding to various degrees of accuracy, truncation, bounds and estimation underpins almost every other topic across all three papers. These skills are tested on both Paper 1 (non-calculator) and Papers 2 and 3 (calculator).
Every digit in a whole number has a place value determined by its position. Our number system is a base-10 (decimal) system where each column is worth ten times the column to its right.
| Millions | Hundred Thousands | Ten Thousands | Thousands | Hundreds | Tens | Units |
|---|---|---|---|---|---|---|
| 1,000,000 | 100,000 | 10,000 | 1,000 | 100 | 10 | 1 |
For example, in the number 5,274,809:
Edexcel Exam Tip: When a question asks "what is the value of the digit 8 in 5,274,809?", the answer is 800 — give the full place value, not just the column name.
The place value system extends to the right of the decimal point. Each column is one-tenth of the column to its left.
| Units | . | Tenths | Hundredths | Thousandths | Ten-thousandths |
|---|---|---|---|---|---|
| 1 | . | 0.1 | 0.01 | 0.001 | 0.0001 |
In the number 3.4087:
To order numbers — including decimals and negative numbers — compare digit by digit from the highest place value.
Put these numbers in ascending order: 0.72, 0.7, 0.072, 0.702
Step 1: Write each number with the same number of decimal places by adding trailing zeros:
| Number | Rewritten |
|---|---|
| 0.72 | 0.720 |
| 0.7 | 0.700 |
| 0.072 | 0.072 |
| 0.702 | 0.702 |
Step 2: Compare as whole numbers: 72, 700, 720, 702→ rearranging: 72, 700, 702, 720
Answer: 0.072, 0.7, 0.702, 0.72
Remember that on a number line, numbers increase from left to right. A negative number further from zero is smaller.
Order from smallest to largest: -4, 2, -9, 7, -1
On a number line: -9 is furthest left, then -4, then -1, then 2, then 7.
Answer: -9, -4, -1, 2, 7
To round to a given number of decimal places (d.p.):
Round 3.4762 to 2 decimal places.
Answer: 3.48
Significant figures (s.f.) count from the first non-zero digit.
| Rule | Example |
|---|---|
| All non-zero digits are significant | 345 has 3 s.f. |
| Zeros between non-zero digits are significant | 3045 has 4 s.f. |
| Leading zeros are NOT significant | 0.0052 has 2 s.f. |
| Trailing zeros after a decimal point ARE significant | 2.50 has 3 s.f. |
| Trailing zeros in a whole number may or may not be significant | 3400 could be 2, 3 or 4 s.f. |
Round 0.004567 to 2 significant figures.
Answer: 0.0046
Round 27,849 to 3 significant figures.
Answer: 27,800
Truncation means cutting off digits without rounding — simply removing them.
Truncate 4.6789 to 2 decimal places.
Answer: 4.67
Key Difference: Rounding 4.6789 to 2 d.p. gives 4.68 (rounded up), but truncating gives 4.67.
Estimation means rounding each number to one significant figure (unless told otherwise) to make a calculation simpler.
Estimate the value of (4.87×21.3)/0.053
Step 1: Round each number to 1 significant figure:
Step 2: Calculate:
Answer: Approximately 2000
Edexcel Exam Tip: On Paper 1 (non-calculator), estimation questions always appear. Show each rounded value clearly — you earn method marks for showing your 1 s.f. approximations. The command word "estimate" tells you to round first.
When a number has been rounded, the lower bound is the smallest value that would round to the given number, and the upper bound is the smallest value that would round to the next number up.
A length is 4.7 cm, correct to 1 decimal place. Find the lower and upper bounds.
We write: 4.65≤ length < 4.75
Note: the lower bound is included (≤) but the upper bound is excluded (<), because a value of exactly 4.75 would round up to 4.8.
| Mistake | Correction |
|---|---|
| Confusing "value of a digit" with "position name" | Always give the full value (e.g. 5000, not "thousands") |
| Adding trailing zeros changes a number | 0.5 = 0.50 = 0.500 — the value is the same |
| Rounding 4.95 to 1 d.p. gives 4.9 | The decider is 5, so round UP to 5.0 |
| Truncation and rounding are the same | They are not — truncation simply removes digits |
| Leading zeros are significant | They are NOT — 0.003 has only 1 s.f. |
| Upper bound is included in an error interval | The upper bound uses < (strictly less than) |
Write down the value of each underlined digit in 46,308.5792.
Working from the decimal point rightwards, the positions are tenths, hundredths, thousandths, ten-thousandths. The digit 7 sits in the hundredths column, so its value is 1007=0.07.
For completeness, the 4 in 46,308 is worth 40,000 (ten thousands), the 6 is worth 6,000, the 3 is worth 300, the 0 contributes 0 tens, the 8 is worth 8 units, the 5 tenths gives 0.5, the 7 hundredths gives 0.07, the 9 thousandths gives 0.009, and the 2 ten-thousandths gives 0.0002. The total value of the number is 40,000+6,000+300+8+0.5+0.07+0.009+0.0002=46,308.5792, which acts as a check on the place-value decomposition.
Arrange in descending order: −0.08, 0.8, −0.088, 0, −0.808, 0.08.
Step 1 — rewrite with a common number of decimal places: −0.080, 0.800, −0.088, 0.000, −0.808, 0.080.
Step 2 — sort positives first, then zero, then negatives (closer to zero is larger):
| Rank | Value |
|---|---|
| 1 (largest) | 0.800 |
| 2 | 0.080 |
| 3 | 0.000 |
| 4 | −0.080 |
| 5 | −0.088 |
| 6 (smallest) | −0.808 |
Answer: 0.8,0.08,0,−0.08,−0.088,−0.808.
Round 6.9962 to 2 decimal places.
The first two decimals are 9 and 9, giving 6.99. The decider is the third decimal, 6, so we round up. Rounding 6.99 up moves the 9 in the hundredths column to 10, which carries into the tenths column: 6.99+0.01=7.00.
Answer: 7.00 (the trailing zeros must be kept to show the required degree of accuracy).
A piece of timber is measured as 2.347 m. Compare (i) truncating to 2 d.p. and (ii) rounding to 2 d.p.
(i) Truncation removes any digits after the second decimal without adjusting — result 2.34 m.
(ii) Rounding looks at the decider (7) which is ≥5, so we round up — result 2.35 m.
A carpenter using the truncated value systematically under-measures, so truncation is only used when an under-estimate is safer (e.g. "how many full 2.34 m lengths can I cut?"). For reporting a measurement, rounding is conventional.
A stadium attendance is reported as 27,500, correct to the nearest 50. Write the error interval.
Half the degree of accuracy =25.
Error interval: 27,475≤attendance<27,525. Because attendance is a discrete quantity (a whole number of people), the upper bound can be tightened to 27,524 — but Edexcel mark schemes accept the continuous-style inequality, so the standard answer is the one above.
Estimate 0.19798.7×40.2 to 1 significant figure.
0.210×40=0.2400=2000.
Answer: Approximately 2000. The actual calculator value is ≈2023, so the estimate is within 2% — good enough for an exam sanity-check.
Exam-style question: A length is given as 3.6 m, correct to 1 decimal place. Work out the error interval for the length. Then explain, using the word "integer" and the phrase "upper/lower bound", why the upper bound uses a strict inequality.
Grades 3–4 response: Lower bound =3.6−0.05=3.55 m. Upper bound =3.6+0.05=3.65 m. The error interval is 3.55≤l<3.65. The upper bound has "<" because 3.65 would round up to 3.7.
Grades 5–6 response: The precision of the measurement is 0.1 m, so half the precision is 0.05 m. The lower bound is 3.55 m (included, because 3.55 rounds to 3.6). The upper bound is 3.65 m (excluded, because 3.65 rounds up to 3.7). The error interval is 3.55≤l<3.65. Writing the interval this way shows the length can be any real number in that range.
Grades 7–9 response: Since length is a continuous (non-integer) measurement, we use a half-open interval [3.55,3.65). The lower bound 3.55 satisfies the rounding rule "rounds to 3.6 (1 d.p.)" and is therefore included. The upper bound 3.65 is the infimum of the next rounding class (values that round to 3.7), so any value strictly less than 3.65 rounds to 3.6, while 3.65 itself rounds to 3.7 under "round half up" — hence the strict "<". Had we been truncating rather than rounding, the interval would instead be 3.6≤l<3.7, a unit-wide interval anchored on the given value. This distinction between rounding-based and truncation-based upper/lower bounds is a common 3-mark discriminator on Paper 1.
Edexcel alignment: This content is aligned with Edexcel GCSE Mathematics (1MA1) specification — specifically Topic N1 Place value and ordering of integers and decimals, N2 Four operations applied to decimals, N14 Rounding to decimal places and significant figures, N15 Estimation and checking, and N16 Bounds and error intervals [H]. Assessed on Paper 1 (non-calc), Paper 2 and Paper 3 (calc).