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N15

Round numbers and measures to an appropriate degree of accuracy (decimal places or significant figures); use inequality notation to specify simple error intervals due to truncation or rounding

Rounding and error intervals

Worked answers, methods and verified real exam appearances for N15 on Edexcel GCSE Maths 1MA1.

Explanation

  • Decimal places count digits after the decimal point; significant figures begin at the first non-zero digit. Identify the deciding digit immediately after the required place: 55 or more rounds up, while 44 or less leaves the retained digit unchanged.
  • Choose a degree of accuracy appropriate to the context, such as money to the nearest penny.
  • A rounded value also represents an error interval.
  • For rounding to a unit uu, subtract and add u2\dfrac u2; include the lower boundary but exclude the upper boundary because that endpoint rounds to the next value.
  • Truncation instead keeps values from the stated number up to the next truncation step.
The error interval for 12.612.6 correct to one decimal place.

Worked example

A positive number yy is truncated to 4.374.37 at two decimal places. Write its error interval and find the greatest integer value of 100y100y.

  1. 1.Truncation gives 4.37y<4.384.37\leq y<4.38.
  2. 2.Multiply the whole interval by 100100: 437100y<438437\leq100y<438.
  3. 3.The greatest integer in this interval is 437437.

Answer: 4.37y<4.384.37\leq y<4.38 and the greatest integer value is 437437.

Common mistakes

  • Don't count leading zeros as significant figures in 0.0078460.007846.
  • Don't include the upper boundary of a rounding interval.
  • Don't use half a rounding unit for a truncation interval.

Exam tip

State the rounding or truncation unit first; it determines both interval endpoints and which endpoint is excluded.

Worked practice

Q1
Tier 1 · Easy

1

Write 0.0078460.007846 correct to 22 significant figures.

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • 0.00780.0078
1The first two significant digits are 77 and 88. The next digit is 44, so the 88 stays unchanged and the rounded value is 0.00780.0078.
Q2
Tier 2 · Standard

2

A number xx is 12.612.6 correct to 11 decimal place. Write the error interval for xx.

(2)

(Total for Question 2 is 2 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • 12.55x<12.6512.55\leq x<12.65
2The rounding unit is 0.10.1, so half a unit is 0.050.05. Subtract and add 0.050.05 to get the boundaries 12.5512.55 and 12.6512.65; include the lower boundary only.
Q3
Tier 3 · Hard

3

A positive number yy is truncated to 4.374.37 at 22 decimal places. Write its error interval and find the greatest possible integer value of 100y100y.

(3)

(Total for Question 3 is 3 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • 4.37y<4.384.37\leq y<4.38
  • Greatest possible integer value of 100y100y is 437437.
3Truncation to 22 decimal places keeps every value from 4.374.37 up to but not including 4.384.38, so 4.37y<4.384.37\leq y<4.38. Multiplying by 100100 gives 437100y<438437\leq100y<438, whose greatest possible integer value is 437437.
Q4
Tier 1 · Easy

4

Write 5374953\,749 correct to 22 significant figures.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • 5400054\,000
1The first two significant digits are 55 and 33. The next digit is 77, so round the 33 up to 44 to get 5400054\,000.
Q5
Tier 2 · Standard

5

A number nn is 730730 correct to the nearest 1010. Write the error interval for nn.

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • 725n<735725\leq n<735
2Half of the rounding unit 1010 is 55. The lower boundary is 7305=725730-5=725 and is included; the upper boundary is 730+5=735730+5=735 and is excluded.
Q6
Tier 3 · Hard

6

A length is 155155 mm correct to the nearest 55 mm. Salma writes 152.5<x157.5152.5<x\leq157.5. Write the correct error interval and explain both changes to Salma's inequality signs.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • 152.5x<157.5152.5\leq x<157.5; 152.5152.5 is included because it rounds to 155155 mm, but 157.5157.5 is excluded because it rounds to 160160 mm.
3Half of the rounding unit 55 mm is 2.52.5 mm. The lower bound 152.5152.5 mm rounds to 155155 mm, so it is included. The upper bound 157.5157.5 mm rounds to 160160 mm, so it is excluded. Hence 152.5x<157.5152.5\leq x<157.5.
Q7
Tier 2 · Standard

7

Write 12.486712.4867 correct to 33 significant figures and correct to 22 decimal places.

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • 12.512.5 to 33 significant figures and 12.4912.49 to 22 decimal places
2For 33 significant figures, keep 1,2,41,2,4 and use the next digit 88, giving 12.512.5. For 22 decimal places, keep 12.4812.48 and use the next digit 66, giving 12.4912.49.
Q8
Tier 3 · Hard

8

A positive number xx is 2.42.4 correct to 11 decimal place. Work out all the possible integer values of 20x20x.

(3)

(Total for Question 8 is 3 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • 4747 and 4848
32.35x<2.452.35\leq x<2.45. Multiplying the interval by 2020 gives 4720x<4947\leq20x<49. The integer values in this interval are 4747 and 4848.
Q9
Tier 3 · Hard

9

A positive number xx is 0.0080.008 correct to 11 significant figure and 0.00760.0076 correct to 22 significant figures. Write the error interval for xx that satisfies both statements.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • 0.00755x<0.007650.00755\leq x<0.00765
4The first statement gives 0.0075x<0.00850.0075\leq x<0.0085. The second gives the narrower interval 0.00755x<0.007650.00755\leq x<0.00765, which lies wholly inside the first interval. Their intersection is therefore 0.00755x<0.007650.00755\leq x<0.00765.
Q10
Tier 3 · Hard

10

A positive integer is 6800068\,000 correct to 22 significant figures. The integer is a multiple of 400400. Work out the least and the greatest possible values of the integer.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • Least =67600=67\,600; greatest =68400=68\,400.
4The rounding interval is 67500n<6850067\,500\leq n<68\,500. The first multiple of 400400 in this interval is 6760067\,600. The last is 6840068\,400, because the next multiple, 6880068\,800, is outside the interval.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2019-063HQ72AllowedHigherQPMS
2019-063FQ11AllowedFoundationQPMS
2022-062FQ11AllowedFoundationQPMS
2023-113FQ21AllowedFoundationQPMS
2022-062HQ82AllowedHigherQPMS
2019-062HQ62AllowedHigherQPMS
2022-062FQ232AllowedFoundationQPMS
2023-062FQ11AllowedFoundationQPMS
2021-111HQ104Non-calculatorHigherQPMS
2019-111FQ21Non-calculatorFoundationQPMS
2022-063HQ163AllowedHigherQPMS
2019-112FQ21AllowedFoundationQPMS
2021-113HQ112AllowedHigherQPMS
2019-112HQ22AllowedHigherQPMS
2024-061FQ11Non-calculatorFoundationQPMS
2024-113HQ72AllowedHigherQPMS
2022-113HQ122AllowedHigherQPMS
2019-062FQ252AllowedFoundationQPMS
2021-111FQ272Non-calculatorFoundationQPMS
2019-113HQ23AllowedHigherQPMS
2023-112FQ183AllowedFoundationQPMS
2019-113FQ233AllowedFoundationQPMS
2021-112FQ123AllowedFoundationQPMS
2022-112FQ253AllowedFoundationQPMS
2022-113FQ41AllowedFoundationQPMS
2022-062HQ32AllowedHigherQPMS
2019-063HQ195AllowedHigherQPMS
2022-113FQ173AllowedFoundationQPMS
2024-062HQ92AllowedHigherQPMS
2023-062HQ72AllowedHigherQPMS
2023-062FQ262AllowedFoundationQPMS
2023-112HQ82AllowedHigherQPMS
2022-063FQ182AllowedFoundationQPMS
2019-112FQ222AllowedFoundationQPMS
2024-062FQ193AllowedFoundationQPMS
2021-111FQ41Non-calculatorFoundationQPMS

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