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Accuracy rounding bounds and standard form - Mathematics Year 10

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Introduction

Two different rounding instructions can sometimes produce the same number. Explain why that coincidence does not make decimal places and significant figures the same rule.

Round ten values, write four error intervals and solve three standard-form calculations. Keep original data, explicit tie conventions, units and exact intermediate values so that each result can be checked independently.

Complete ten rounding decisions and explain matching outputs

Significant figures start at the first nonzero digit. Decimal places count fractional positions after the point, including leading zeros in those positions. The value.04786 has its first nonzero digit 4 in the hundredths place. To two significant figures retain 4 and 7, then use 8 to round 7 upward, giving.048.

To three decimal places for the same value, retain the tenths 0, hundredths 4 and thousandths 7. The next digit 8 again rounds the thousandths upward, giving.048. Both instructions select the same original rounding place here. They are different rules whose selected place happens to agree for this value.

For comparison, .004876 to two significant figures gives.0049, while to three decimal places it gives.005. This counterexample shows why the agreement in the source example is not a general identity. Locate the retained place before deciding which digit controls the rounding.

Ten complete rounding tasks. Throughout these positive-number examples, use nearest rounding with halfway ties upward. Other tie conventions exist; do not silently transfer our endpoint conclusions to an unspecified convention.

Task	Original	Required accuracy	Original rounding unit	Reported result
1	0.04786	2 significant figures	0.001	0.048
2	0.04786	3 decimal places	0.001	0.048
3	6745	3 significant figures	10	6750
4	6745	Nearest 100	100	6700
5	8.764	2 decimal places	0.01	8.76
6	0.009876	2 significant figures	0.0001	0.0099
7	125.55	Nearest 0.1	0.1	125.6
8	0.0008321	3 significant figures	0.000001	0.000832
9	9746	2 significant figures	100	9700
10	0.999	Nearest 0.01	0.01	1.00

In row 3, 6745 to three significant figures retains the tens position. The ones digit 5 is a halfway tie, so 6750 is the result. To show the stated three-digit precision unambiguously, write 6.75×10³. Row 4 instead asks for nearest 100; the tens digit 4 rounds downward to 6700. The original rounding unit therefore changes from 10 to 100.

The leading zeros in.0008321 locate its size but do not start the significant-figure count. Retain 8, 3, 2 and inspect 1, giving.000832. In row 10, the fixed hundredths request crosses 1: a carry gives 1.00, still explicitly to nearest.01. Do not relabel a fixed-place result as a different significant-figure task after carrying.

The original Corbettmaths tutorial uses examples such as 8.32→8.3 and.89→.9 at one decimal place. Identify the retained place, then compare our.04786 task where 2 significant figures and 3 decimal places happen to agree. The tutorial numbers are separate examples. Original: corbettmaths, Rounding to 1 and 2 Decimal Places - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Keep the original value until the final required rounding. Repeatedly rounding an intermediate report can change the final outcome: 1.449 to two decimal places is 1.45, and rounding that report to one decimal place gives 1.5, whereas rounding the original 1.449 directly to one decimal place gives 1.4. The reported precision belongs to a decision process, not just a neat-looking string of digits.

Write four error intervals with justified endpoints

A rounded value represents an interval of possible original values under its stated rule. For a positive original x, reported value r and fixed rounding increment u, round-half-up gives $r - \frac{u}{2} \leq x < r + \frac{u}{2}$ . The half-width comes from u, not from half the reported value r.

Four complete error intervals. Each unit applies to both endpoints of its row. The left endpoint is included and the right endpoint excluded under the convention just stated.

Task	Reported value	Nearest increment	Included lower endpoint	Excluded upper endpoint
1	4.7 cm	0.1 cm	4.65 cm	4.75 cm
2	350 s	10 s	345 s	355 s
3	0.048 kg	0.001 kg	0.0475 kg	0.0485 kg
4	2.40 m	0.01 m	2.395 m	2.405 m

For source 4.7 cm to nearest.1 cm, half-width.05 cm gives 4.65 cm≤x<4.75 cm. The lower tie 4.65 rounds upward to 4.7, so it is included. The upper tie 4.75 rounds upward to 4.8, so it is excluded. A value such as 4.749 lies inside and rounds to 4.7, but the endpoint 4.75 itself does not.

For 350 s to nearest 10 s, the half-width is 5 s, so 345 s≤x<355 s. A whole-number report ending in zero does not by itself identify precision; nearest 10 s is explicitly supplied. Using 350/2=175 as the half-width would confuse the reported size with its rounding increment.

For.048 kg to nearest.001 kg, the interval is.0475 kg≤x<.0485 kg. Multiplying every value by 1000 gives 47.5 g≤x<48.5 g, with a 1 g rounding increment. Convert both endpoints and the original increment together rather than changing only the central value.

For 2.40 m to nearest.01 m, the half-width.005 m gives 2.395 m≤x<2.405 m. The last zero records the stated hundredths accuracy in this report. Subtracting an entire.01 m on each side would create an interval twice as wide as intended.

An independent check reconstructs the midpoint and width of the interval. The first pair averages 4.7 and differs by.1; the fourth averages 2.40 and differs by.01. The upper endpoint is an upper bound, but it is not an attained maximum for numbers rounding to the reported value. Values can approach it arbitrarily closely from below.

These intervals describe rounding uncertainty only. They do not automatically include instrument calibration errors, biased observations or uncertain sampling. Treat the supplied reports as mathematical data rather than evidence that an actual measurement process was performed or certified.

Calculate in standard form and check exact values

Standard form separates a nonzero magnitude into a coefficient and a power of ten. For positive values, use A×10ⁿ with 1≤A<10 and integer n. The coefficient interval is a normalisation convention, not permission to change the represented value. Negative powers describe reciprocal scales, not negative numbers.

For multiplication and division of like-base powers, use $1 0^{m} 1 0^{n} = 1 0^{m + n}, \frac{1 0^{m}}{1 0^{n}} = 1 0^{m - n}$ . A product or quotient also needs coefficient arithmetic. Powers of an entire product apply to both the coefficient and scale factor. In the following table, ^ introduces an exponent and brackets state what is raised to a power.

Three complete calculations. Cover normalised and ordinary forms, show an intermediate step and choose a different representation for the check.

Task	Expression	Normalised exact result	Ordinary decimal
1	(4.5×10^6)×(2×10^-4)	9×10^2	900
2	(7.2×10^-5)÷(3×10^-2)	2.4×10^-3	0.0024
3	(6×10^3)^2	3.6×10^7	36000000

In row 1, coefficients 4.5×2=9 and exponents 6+(−4)=2 give 9×10²=900. The ordinary route 4500000×.0002=900 agrees. The small multiplier makes the large input smaller, so an answer 90000000000 would fail an immediate magnitude check.

For row 2, coefficient 7.2/3=2.4 and exponent −5−(−2)=−3, giving 2.4×10⁻³=.0024. Multiply the result by the original divisor 3×10⁻² to reconstruct 7.2×10⁻⁵. Subtracting the negative exponent is adding its opposite, not simply combining the written digits 5 and 2.

For row 3, square the whole 6×10³: 6²×10⁶=36×10⁶. The value is correct but coefficient 36 is not normalised. Rewrite 36 as 3.6×10 to get 3.6×10⁷=36000000. The ordinary square 6000² provides an independent check. Leaving coefficient 6 unchanged would square only part of the original product.

Addition uses a different structure. For 2.4×10⁴+3×10³, align scales to(2.4+.3)×10⁴=2.7×10⁴. Adding exponents would incorrectly apply a multiplication rule to a sum. Name the operation before choosing a law.

For unfamiliar transfer, (8×10⁻³)/(2×10²)=4×10⁻⁵=.00004. Reconstruct.008 by multiplying.00004×200. A rough scale estimate catches factor-of-ten errors, but retain exact forms or enough guard digits until the requested final rounding. Your record should contain ten rounding decisions, four fully signed intervals, three standard-form calculations, independent checks and a repaired first error.

Interactive Tasks

Quiz: Test Your Knowledge

0.04786 to 2 significant figures? (0.048) (!0.04) (!0.047) (!0.050)

Same value to 3 decimal places? (0.048) (!0.04786) (!0.047) (!0.05)

6745 to 3 significant figures? (6750) (!6700) (!6740) (!6800)

0.009876 to 2 significant figures? (0.0099) (!0.01) (!0.0098) (!0.00)

0.999 to nearest 0.01? (1.00) (!0.99) (!0.999) (!1.01)

4.7 cm nearest 0.1 cm: included lower endpoint? (4.65 cm) (!4.6 cm) (!4.75 cm) (!4.7 cm)

Same report: upper 4.75 cm? (Excluded because it rounds to 4.8 cm) (!Included because it is an upper bound) (!The rounding increment) (!The only original value)

350 s nearest 10 s: interval width? (10 s) (!5 s) (!20 s) (!350 s)

4.5×10^6 times 2×10^-4? (900) (!90000000000) (!90) (!0.0009)

Square the whole 6×10^3 product. (3.6×10^7) (!6×10^6) (!3.6×10^6) (!1.2×10^4)

Memory Game

Significant figures	Count beginning with the first nonzero digit
Decimal places	Fractional positions after the decimal point
Rounding increment	Fixed place unit used in the stated rule
Error interval	Set of values compatible with a rounded report
Normalised	Positive standard-form coefficient at least 1 and below 10
Guard digits	Extra retained precision before final rounding

Drag and Drop

0.048	Source rounded result under both stated requests
6750	6745 to three significant figures
4.65 cm	Source included lower bound
9×10^2	First standard-form product
2.4×10^-3	Second standard-form quotient

Crossword Puzzle

ROUNDING	Reporting at selected accuracy
INCREMENT	Step defining a rounding place
INTERVAL	Set with stated endpoints
COEFFICIENT	Leading standard-form multiplier
EXPONENT	Power index
NORMALISATION	Value-preserving adjustment to coefficient convention

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Compare the two source rounding requests: Explain why.04786 happens to give.048 under both.
Locate the retained place: Use leading zeros and the specified accuracy in three table rows.
Read the first interval: Check 4.65 and 4.75 separately under the tie rule.
Read a power-of-ten scale: Explain why 10^-3 is positive and small.

Standard

Round all ten values: Retain requested accuracies, original increments and final displays.
Write four complete intervals: Use endpoints, units and correct inclusion signs.
Solve three standard-form expressions: Show coefficient arithmetic, index operation and normalisation.
Use independent checks: Rebuild interval midpoints, widths and a quotient numerator.

Advanced

Challenge double rounding: Compare 1.449 rounded directly with two-stage rounding.
Transfer a new quotient: Find 4×10^-5 and reconstruct.008.
Distinguish mathematical from physical uncertainty: State what rounding intervals do and do not include.
Retain a complete evidence set: Keep ten/four/three tasks, a first-error repair and an actual delayed response.

Learning Check

Ten rounded results: State the retained place and final display.
Four intervals: Use positive round-half-up endpoints and units.
Endpoint and conversion check: Explain exclusion and convert the small mass interval.
Three standard-form results: Show operation and an independent check.
Transfer and first errors: Explain new data and invalid rules.
Scope and portfolio: Retain full tasks and actual evidence.

Worked checks and assessment guidance

Results.048, .048, 6750, 6700, 8.76, .0099, 125.6, .000832, 9700, 1.00. The first two requests coincide for this value but differ in general.1.00 is explicitly nearest.01 after carrying.
4.65≤x<4.75 cm; 345≤t<355 s; .0475≤m<.0485 kg; 2.395≤L<2.405 m. Each unit applies to both endpoints. Midpoints recover reports and widths recover increments.
4.65 rounds to 4.7 while 4.75 rounds to 4.8, so only the lower tie is included. The mass interval becomes 47.5≤m<48.5 g, with 1 g increment. An excluded upper bound is not an attained maximum.
Product 9×10²=900; quotient 2.4×10^-3=.0024; square 3.6×10^7=36000000. Use ordinary multiplication, reverse multiplication and 6000² respectively.
New.008/200=.00004=4×10^-5; reverse product restores.008. Align scales for 2.4×10⁴+3×10³=2.7×10⁴. Double rounding 1.449→1.45→1.5 differs from direct 1.4 at one decimal place.
Keep ten rounding tasks, four signed intervals, three standard-form calculations, checks and first-error repair. State tie conventions, units and original values. Rounding intervals alone do not certify instrument accuracy or include every physical error; record later work only when it occurs.

Learning evidence

Retain ten rounding decisions, four signed unit-aware error intervals, three complete standard-form calculations, independent midpoint/inverse checks, double-rounding error analysis and unfamiliar transfer.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Number - Measures and accuracy; Structure and calculation. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

Decimal Place Value Chart (emphasizing the One's Column).jpg, GregWelch8, CC0. Unchanged original; caption and learning activities are original.

Dienes Material (Dyskalkulie).jpg, Rhetos, CC0. Unchanged original; caption and learning activities are original.