Place Value, Ordering and Rounding

Place value is the bedrock of the whole Number strand of OCR GCSE Mathematics (J560). Before you can confidently add, multiply, round or estimate, you need to know exactly what each digit in a number is worth. This lesson covers place value in integers and decimals, ordering numbers (including negatives), rounding to a number of decimal places and to significant figures, truncation, and estimation by rounding each value to one significant figure. These skills surface on every paper, both the non-calculator Paper 1 and the calculator Papers 2 and 3, and they underpin the bounds work you will meet later in this course.

This lesson mainly builds AO1 fluency (using and applying standard techniques accurately), with a strand of AO2 reasoning when you justify why a digit has a particular value, and AO3 problem-solving in the estimation worked examples where you decide how to simplify a messy calculation.

Key Vocabulary

Term	Meaning
Place value	The value a digit has because of its position in a number
Integer	A whole number — positive, negative, or zero
Decimal	A number written using a decimal point and place value to the right of it
Significant figure (s.f.)	A digit that contributes to the precision of a number, counted from the first non-zero digit
Decimal place (d.p.)	A digit's position after the decimal point
Truncate	To cut a number short by removing digits, without rounding
Estimate	An approximate answer, usually found by rounding each value to 1 s.f.
Ascending / descending	Smallest-to-largest / largest-to-smallest order

Place Value in Integers and Decimals

Our number system is base-10: each column is worth ten times the column to its right, and one tenth of the column to its left. That single fact governs everything in this lesson.

Worked Example 1

Write down the value of the digit 6 in the number $4{,}628{,}503$ .

Reading the columns from the right, the 6 sits in the hundred-thousands column.

So the 6 is worth $600{,}000$ (six hundred thousand).

Answer: $600{,}000$ .

Common error: Writing "hundred thousands" as the answer. OCR wants the value — the number $600{,}000$ — not the name of the column.

The same idea runs to the right of the decimal point, where the columns are tenths, hundredths, thousandths and so on.

Worked Example 2

In the number $7.3094$ , write down the value of the digit 9.

After the decimal point the columns are tenths ( $3$ ), hundredths ( $0$ ), thousandths ( $9$ ), ten-thousandths ( $4$ ).

The 9 is in the thousandths column, so it is worth $\dfrac{9}{1000} = 0.009$ .

Answer: $0.009$ .

A neat check is that the digits add back to the whole number: $7 + 0.3 + 0.00 + 0.009 + 0.0004 = 7.3094$ .

Ordering Numbers, Including Negatives

To order numbers, compare them column by column starting from the highest place value. With decimals it helps to give every number the same number of decimal places first by writing in trailing zeros — this does not change a number's value, since $0.5 = 0.50 = 0.500$ .

Worked Example 3

Write these numbers in ascending order: $0.61$ , $0.6$ , $0.069$ , $0.601$ .

Pad to three decimal places: $0.610$ , $0.600$ , $0.069$ , $0.601$ .

Now compare as if they were the whole numbers $610$ , $600$ , $69$ , $601$ . In order: $69, 600, 601, 610$ .

Answer: $0.069, \; 0.6, \; 0.601, \; 0.61$ .

Worked Example 4

Arrange these temperatures from coldest to warmest: $-3$ , $5$ , $-11$ , $0$ , $-1$ (in °C).

On a number line, values increase from left to right, and a negative further from zero is smaller. The coldest is the most negative.

Answer: $-11, \; -3, \; -1, \; 0, \; 5$ .

Common error: Treating $-11$ as larger than $-3$ "because 11 is bigger than 3". With negatives, the larger the digit, the colder (smaller) the value.

OCR Exam Tip: When a question mixes positive and negative decimals, write them all to the same number of decimal places, sketch a quick number line, and place the negatives to the left of zero. The command word here is usually "Write down" or "Order".

Rounding to Decimal Places and Significant Figures

To round to a given number of decimal places:

Count that many digits after the decimal point.
Look at the very next digit — the decider.
If the decider is $5$ or more, round up; if it is $4$ or less, leave the last kept digit unchanged.

Worked Example 5

Work out $5.7468$ rounded to 2 decimal places.

Two decimal places gives $5.74\ldots$ The decider (third decimal) is $6$ , which is $\ge 5$ , so round up.

Answer: $5.75$ .

Significant figures are counted from the first non-zero digit. Leading zeros are never significant; zeros sandwiched between non-zero digits always are.

Number	Significant figures
$508$	3 (the middle zero counts)
$0.00420$	3 (the leading zeros do not count; the trailing zero does)
$63{,}000$	at least 2 (trailing zeros in an integer are ambiguous)

Worked Example 6

Round $0.0073812$ to 2 significant figures.

The first significant figure is $7$ , the second is $3$ . The decider is the next digit, $8$ , which is $\ge 5$ , so round up the $3$ to a $4$ .

Answer: $0.0074$ .

Worked Example 7

Round $48{,}615$ to 3 significant figures.

The first three significant figures are $4$ , $8$ , $6$ . The decider is the next digit, $1$ ( $<5$ ), so leave the $6$ unchanged. Replace the remaining digits with zeros to hold place value.

Answer: $48{,}600$ .

Common error: Writing $486$ instead of $48{,}600$ . You must keep the place-holding zeros, or the number changes from forty-eight thousand to four hundred and eighty-six.

Truncation

To truncate a number, you simply chop off the unwanted digits — you do not round.

Worked Example 8

Truncate $9.2867$ to 2 decimal places.

Keep the first two decimals and discard the rest: $9.28$ . We do not look at the next digit at all.

Answer: $9.28$ .

Contrast this with rounding $9.2867$ to 2 d.p., which gives $9.29$ (because the decider $6 \ge 5$ ). Truncation always rounds towards the lower value, so it can only ever stay the same or go down.

Estimation by Rounding to 1 Significant Figure

To estimate the answer to a calculation, round every number to 1 significant figure, then work out the simpler sum. This is a guaranteed Paper 1 skill.

Worked Example 9

Work out an estimate for $\dfrac{59.3 \times 4.8}{0.21}$ .

Round each value to 1 s.f.: $59.3 \approx 60$ , $4.8 \approx 5$ , $0.21 \approx 0.2$ .

$\dfrac{60 \times 5}{0.2} = \dfrac{300}{0.2} = 1500.$

Answer: approximately $1500$ . (A calculator gives $\approx 1355$ , so the estimate is sensible.)

OCR Exam Tip: Show each " $\approx$ " rounding explicitly — those lines earn the method marks even if your final arithmetic slips. Dividing by a number smaller than 1, such as $0.2$ , increases the result, which catches many candidates out.

Extended Worked Examples

Worked Example 10: A "9-boundary" rounding case

Round $4.7961$ to 2 decimal places.

The first two decimals are $7$ and $9$ , giving $4.79\ldots$ The decider is $6$ ( $\ge 5$ ), so round up. Rounding the $9$ in the hundredths column up makes it $10$ , which carries: $4.79 + 0.01 = 4.80$ .

Answer: $4.80$ — and the trailing zero must stay to show the answer is correct to 2 d.p.

Worked Example 11: Estimation with a square root

Work out an estimate for $\dfrac{\sqrt{96} \times 31.4}{0.49}$ .

$\sqrt{96} \approx \sqrt{100} = 10$ , $31.4 \approx 30$ , $0.49 \approx 0.5$ .

$\dfrac{10 \times 30}{0.5} = \dfrac{300}{0.5} = 600.$

Answer: approximately $600$ .

Worked Example 12: Rounding money to a sensible degree of accuracy

A bill of £47.3826 is to be rounded to the nearest penny. Work out the rounded amount.

The nearest penny means 2 decimal places. The first two decimals are $3$ and $8$ ; the decider is $2$ ( $<5$ ), so leave the $8$ unchanged.

Answer: £47.38.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: Work out an estimate for the value of $\dfrac{812 \times 0.038}{0.0096}$ . Give a reason why your answer is only an estimate.

Grades 3–4 response: $812 \approx 800$ , $0.038 \approx 0.04$ , $0.0096 \approx 0.01$ . So $\dfrac{800 \times 0.04}{0.01} = \dfrac{32}{0.01} = 3200$ . It is an estimate because I rounded the numbers.

Grades 5–6 response: Rounding each value to 1 s.f.: $812 \approx 800$ , $0.038 \approx 0.04$ , $0.0096 \approx 0.01$ . Then $\dfrac{800 \times 0.04}{0.01} = \dfrac{32}{0.01} = 3200$ . The answer is an estimate because each number was rounded to 1 significant figure before calculating, so the result is approximate rather than exact.

Grades 7–9 response: Round to 1 s.f.: $\dfrac{800 \times 0.04}{0.01}$ . The numerator is $800 \times 0.04 = 32$ ; dividing by $0.01$ multiplies by $100$ , giving $3200$ . It is an estimate because rounding each factor introduces error; rounding $0.0096$ up to $0.01$ enlarges the denominator, which pulls the estimate below the true value, so I would expect the exact answer to be a little above $3200$ . (The calculator value is $\approx 3214$ , confirming the direction of the rounding error.)

Examiner-style commentary: the Grades 3–4 answer scores the method and a bare reason; the Grades 5–6 answer states the "1 s.f." rule explicitly; the Grades 7–9 answer reasons about the direction of the rounding error, which is the AO2 discriminator.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: Write the numbers $-0.4$ , $\dfrac{1}{3}$ , $-\dfrac{1}{2}$ , $0.34$ in order, starting with the smallest. Give a reason for the position of $\dfrac{1}{3}$ .

Grades 3–4 response: $\dfrac{1}{3} = 0.33$ and $-\dfrac{1}{2} = -0.5$ . In order: $-0.5, -0.4, 0.33, 0.34$ , so $-\dfrac{1}{2}, -0.4, \dfrac{1}{3}, 0.34$ .

Grades 5–6 response: Convert all to decimals: $-0.4$ , $\dfrac{1}{3} = 0.3\dot{3}$ , $-\dfrac{1}{2} = -0.5$ , $0.34$ . Smallest first: $-0.5, -0.4, 0.333\ldots, 0.34$ . So the order is $-\dfrac{1}{2}, \; -0.4, \; \dfrac{1}{3}, \; 0.34$ . $\dfrac{1}{3}$ comes before $0.34$ because $0.333\ldots < 0.34$ .

Grades 7–9 response: Writing each as a decimal: $-\dfrac{1}{2} = -0.5$ , $-0.4$ , $\dfrac{1}{3} = 0.3\dot{3}$ , $0.34$ . Ordering ascending: $-0.5, -0.4, 0.3\dot{3}, 0.34$ . $\dfrac{1}{3}$ sits just below $0.34$ : comparing to three decimal places, $0.333\ldots < 0.340$ , and the gap is only about $0.0067$ . The negatives are ordered "in reverse" of their digits — $-0.5$ is smaller than $-0.4$ because it is further from zero.

Examiner-style commentary: converting everything to a common form (decimals) is the key move; the top band quantifies why $\dfrac{1}{3}$ and $0.34$ are close and handles the negative ordering with confidence.

Worked Example 13: Rounding to the nearest ten, hundred and thousand

Round $26{,}748$ to (a) the nearest $10$ , (b) the nearest $100$ , (c) the nearest $1000$ .

For each, find the column you are rounding to and use the next digit as the decider.

(a) Nearest $10$ : the tens digit is $4$ , the decider (units) is $8 \ge 5$ , so round the $4$ up to $5$ and fill the units with a zero: $26{,}750$ .

(b) Nearest $100$ : the hundreds digit is $7$ , the decider (tens) is $4 < 5$ , so leave the $7$ and replace the lower digits with zeros: $26{,}700$ .

(c) Nearest $1000$ : the thousands digit is $6$ , the decider (hundreds) is $7 \ge 5$ , so round up to $7$ : $27{,}000$ .

Answers: (a) $26{,}750$ , (b) $26{,}700$ , (c) $27{,}000$ .

Common error: Rounding the same number "in stages" — first to the nearest $10$ , then using that answer to round to the nearest $100$ . Always round the original number directly, or you can drift to the wrong value.

Worked Example 14: An ordering problem in context

Five rivers have lengths $346$ km, $34.6$ km, $364$ km, $3460$ m and $0.346$ km. Write them in order, shortest first.

The trap is the mixed units: $3460$ m must be converted to kilometres first, $3460 \text{ m} = 3.46$ km. Now compare like with like (all in km): $0.346$ , $3.46$ , $34.6$ , $346$ , $364$ .

Answer: $0.346$ km, $3460$ m, $34.6$ km, $346$ km, $364$ km.

This shows why "compare from the highest place value" only works once every quantity is in the same unit — converting first is the essential first step whenever units are mixed.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: The number of people at a festival is reported as $48{,}000$ , correct to 2 significant figures. Show that the number rounded to the nearest $1000$ could be different from $48{,}000$ , and explain your reasoning.

Grades 3–4 response: $48{,}000$ to 2 s.f. means the first two figures $4$ and $8$ are fixed. The real number could be something like $47{,}600$ , which rounds to $48{,}000$ to 2 s.f. but to $48{,}000$ to the nearest thousand too, or $48{,}400$ . So it could be different.

Grades 5–6 response: To 2 s.f., any value from $47{,}500$ up to just under $48{,}500$ rounds to $48{,}000$ . For example $47{,}500$ rounds to $48{,}000$ (2 s.f.) but to $48{,}000$ to the nearest $1000$ as well; however $47{,}501$ rounds to $48{,}000$ to 2 s.f. yet to $48{,}000$ to the nearest thousand. A value like $48{,}499$ rounds to $48{,}000$ to 2 s.f. but to $48{,}000$ to the nearest thousand — so the two accuracies usually agree, but the interval of possible values is wider for 2 s.f.

Grades 7–9 response: " $48{,}000$ to 2 s.f." fixes only the first two significant figures, so the true value lies in $47{,}500 \le n < 48{,}500$ . Within that interval, values such as $47{,}600$ round to $48{,}000$ to the nearest $1000$ , but a value like $47{,}500$ rounds to $48{,}000$ to 2 s.f. while $47{,}499$ would round to $47{,}000$ . The key point is that "2 s.f." and "nearest $1000$ " define different rounding intervals: 2 s.f. allows anything in a $1000$ -wide band, so the underlying number is not pinned down to a single nearest-thousand value. Stating the degree of accuracy therefore matters — it tells the reader how much the figure can be trusted.

Examiner-style commentary: the discriminator is recognising that a stated accuracy defines an interval of possible true values, not a single number; the top band states that interval precisely and links it to why degree of accuracy must always be reported.

Going Further

Place value, rounding and estimation are the silent workhorses of science and finance. In Physics and Chemistry, every measured quantity carries a degree of accuracy, and results are quoted to a sensible number of significant figures — you would never report a mass as $2.7384910$ g from a balance reading $2.7$ g. Estimation to 1 s.f. is exactly how scientists do a "back-of-the-envelope" sanity check before trusting a calculator. In personal finance, rounding to the nearest penny governs every bill, while rounding to 2 or 3 s.f. is how budgets, populations and national statistics are reported in the news. The same skills also feed directly into the Bounds and Error Intervals lesson later in this course, where "correct to 1 d.p." becomes a precise statement about the range a true value can take.

Common Mistakes and Misconceptions

Giving the name of a column ("thousands") when asked for the value of a digit ( $3000$ ).
Believing trailing zeros change a decimal — they do not: $0.7 = 0.70 = 0.700$ .
Thinking $-11 > -3$ : with negatives, the larger the digit, the smaller the number.
Dropping the place-holding zeros when rounding integers to significant figures, turning $48{,}600$ into $486$ .
Confusing truncation with rounding — truncation never looks at the next digit.
Forgetting that dividing by a value below $1$ makes an estimate larger, not smaller.
Rounding too early in a multi-step calculation, so errors accumulate; keep full accuracy until the final step.

Exam Tips

AO1 (use and apply standard techniques): practise rounding and ordering until they are automatic. State the degree of accuracy you are working to ("to 2 s.f.") so the examiner can follow your method.
AO2 (reason, interpret and communicate): when a question says "Give a reason for your answer", explain why — for example, "it is an estimate because each value was rounded to 1 s.f." A bare numerical answer will not earn the communication mark.
AO3 (solve problems): in estimation problems you choose how to simplify. The mark scheme rewards showing each " $\approx$ " rounding line; even if the final arithmetic slips, the rounded values earn method marks.
Watch the OCR command words: "Write down" expects a quick fact with no working; "Work out" and "Calculate" expect method; "Show that" expects every step to a given answer.

Summary

Each digit's place value depends on its position; give the value, not the column name.
To order numbers, compare from the highest place value; pad decimals with trailing zeros and treat negatives carefully.
Rounding to d.p. or s.f. uses the decider digit: $\ge 5$ rounds up, otherwise leave unchanged.
Keep place-holding zeros when rounding integers to significant figures.
Truncation chops digits off without rounding, so it never increases a value.
Estimate by rounding each number to 1 s.f.; dividing by a number below $1$ increases the result.
State your degree of accuracy and show " $\approx$ " lines to secure the method marks.

This content is aligned with the OCR GCSE Mathematics (J560) specification.

Place Value, Ordering and Rounding

Place Value, Ordering and Rounding

Key Vocabulary

Place Value in Integers and Decimals

Worked Example 1

Worked Example 2

Ordering Numbers, Including Negatives

Worked Example 3

Worked Example 4

Rounding to Decimal Places and Significant Figures

Worked Example 5

Worked Example 6

Worked Example 7

Truncation

Worked Example 8

Estimation by Rounding to 1 Significant Figure

Worked Example 9

Extended Worked Examples

Worked Example 10: A "9-boundary" rounding case

Worked Example 11: Estimation with a square root

Worked Example 12: Rounding money to a sensible degree of accuracy

Answering at different grade levels

Answering at different grade levels

Worked Example 13: Rounding to the nearest ten, hundred and thousand

Worked Example 14: An ordering problem in context

Answering at different grade levels

Going Further

Common Mistakes and Misconceptions

Exam Tips

Summary

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