MathematicsFoundation18 min read

Estimation and Limits of Accuracy

Rounding on purpose, and knowing how wrong an answer might be

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01

Rounding, and the difference between the two kinds

Decimal places counts digits after the point. Significant figures counts digits from the first non-zero one, wherever the point happens to be. They give different answers for the same number, and questions specify which is wanted.

0.004 638 to 2 decimal places is 0.00 — everything has vanished. To 2 significant figures it is 0.0046, which actually says something. For small numbers, significant figures are almost always the sensible measure.

Number2 d.p.2 s.f.3 s.f.
3.141593.143.13.14
0.0046380.000.00460.00464
58 62158 621.0059 00058 600
0.99711.001.00.997

Zeros before the first significant figure do not count

In 0.00408 the first significant figure is the 4. The zeros before it are only holding the decimal point in place. But the zero between the 4 and the 8 does count, because it is between two significant digits — so 0.00408 to 2 s.f. is 0.0041.

02

Estimating before calculating

An estimate is not a rough answer offered instead of the real one. It is a check on the real one, and the syllabus asks for it explicitly.

The method: round every number to 1 significant figure, then do the arithmetic mentally. If the calculator says something far from the estimate, a key was pressed wrongly — and finding that out takes ten seconds rather than losing every mark that followed.

Worked example

Estimate (48.7 × 6.13) ÷ 0.198, then say what it tells you if a calculator returns 15.1.

  1. Round each to 1 significant figure: 50, 6 and 0.2.One significant figure is enough — the point is a check, not a second answer.
  2. Numerator: 50 × 6 = 300.Easy mental arithmetic is the whole reason for rounding this far.
  3. Divide: 300 ÷ 0.2 = 1500.Dividing by 0.2 is the same as multiplying by 5.
  4. A calculator answer of 15.1 is a hundred times too small, so something was mistyped — most likely 0.198 entered as 19.8.The estimate does not give the right answer; it tells you the one you have is wrong.

About 1500. An answer of 15.1 is out by a factor of 100, so re-enter the calculation.

03

Limits of accuracy

A length recorded as 8 cm to the nearest centimetre was not exactly 8 cm. It was anything that rounds to 8 — from 7.5 cm up to (but not including) 8.5 cm.

Those two values are the lower and upper bounds, and the rule is always the same: half of the rounding unit either side. Rounded to the nearest 10, the bounds are ±5. Rounded to 1 decimal place, ±0.05.

value rounded to unit u:lower = x − u/2upper = x + u/28 cm to the nearest cm7.5 ≤ length < 8.52.4 kg to 1 d.p.2.35 ≤ mass < 2.45350 to the nearest 10345 ≤ value < 355the lower bound is included and the upper bound is not, which is why one sign is ≤ and the other <

A recorded value stands for a whole interval, not a point. The finer the instrument, the narrower that interval — which is what "more precise" actually means.

04

Bounds in calculations

When bounded values are combined, the bounds of the answer come from choosing the right combination — and it is not always the obvious one.

For a maximum: add two upper bounds, or multiply two upper bounds. But to make a subtraction as large as possible, take the largest first value and the smallest second. To make a division as large as possible, take the largest numerator and the smallest denominator.

To find theUse
maximum of a + bupper a + upper b
minimum of a + blower a + lower b
maximum of a − bupper a − lower b
minimum of a − blower a − upper b
maximum of a × bupper a × upper b
maximum of a ÷ bupper a ÷ lower b
minimum of a ÷ blower a ÷ upper b

Why subtraction and division reverse

Taking away less leaves more, so the biggest difference comes from subtracting the smallest possible amount. Dividing by less gives more, so the biggest quotient comes from the smallest possible denominator. Both follow from thinking about what makes the answer large, rather than from memorising the table.

Before you leave this chapter

  1. Decimal places count after the point; significant figures count from the first non-zero digit.
  2. Leading zeros are not significant; zeros between significant digits are.
  3. To estimate, round everything to 1 s.f. and check the calculator against it.
  4. Bounds are half the rounding unit either side: 8 cm to the nearest cm means 7.5 ≤ x < 8.5.
  5. For a maximum difference or quotient, use the SMALLEST second value.
05

How many figures should an answer have?

An answer cannot be more accurate than the data it came from. If a measurement was given to 3 significant figures, quoting a result to eight is claiming a precision that does not exist — and examiners penalise it.

The convention: round the final answer to the same number of significant figures as the least accurate value used, or to 3 s.f. if the question does not indicate otherwise. Angles are usually given to 1 decimal place, and money to 2.

Given dataSensible answer
12.4 and 3.7 (3 s.f. and 2 s.f.)2 significant figures
no accuracy stated3 significant figures
money2 decimal places
an angle1 decimal place
a number of people or objectsa whole number

Round once, at the end

Carry the full calculator value through every intermediate step and round only when you write the final answer. Rounding at each stage compounds the error, and a chain of three or four steps can shift the last figure. If a question asks for working, write the intermediate values to more figures than the final answer needs.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Write 0.040 782 to 3 significant figures.
Model answer

The first significant figure is the 4, so the three are 4, 0 and 7. The next digit is 8, which rounds up: 0.0408.

Examiner tip. The leading zeros do not count, but the zero between 4 and 7 does. Counting from the first non-zero digit is the rule.

SQ2[2 marks]
A length is 24 cm to the nearest centimetre. Write down its lower and upper bounds.
Model answer

Half a centimetre either side: 23.5 ≤ length < 24.5 cm.

Examiner tip. Use ≤ for the lower bound and < for the upper — 24.5 would round up to 25, so it is excluded.

SQ3[2 marks]
Estimate the value of (0.51 × 397) ÷ 19.8.
Model answer

Rounding to 1 s.f.: (0.5 × 400) ÷ 20 = 200 ÷ 20 = 10.

Examiner tip. Round every number, including the divisor. Rounding only some of them defeats the purpose and often makes the arithmetic harder.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A rectangle measures 12 cm by 8 cm, each to the nearest centimetre. Find the upper and lower bounds of its area.
Full working
  1. Bounds of the length: 11.5 ≤ l < 12.5[1]
  2. Bounds of the width: 7.5 ≤ w < 8.5[1]
  3. Maximum area = 12.5 × 8.5 = 106.25 cm²both upper bounds, since multiplying[1]
  4. Minimum area = 11.5 × 7.5 = 86.25 cm²[1]

86.25 cm² ≤ area < 106.25 cm²

Examiner tip. The nominal area is 96 cm², and the true value could be ten square centimetres either side of that. Rounding two measurements is enough to make the answer surprisingly uncertain.

N2[4 marks]
A car travels 150 m, measured to the nearest 10 m, in 12 s, measured to the nearest second. Find the maximum possible average speed.
Full working
  1. Distance bounds: 145 ≤ d < 155 mhalf of 10 either side[1]
  2. Time bounds: 11.5 ≤ t < 12.5 shalf of 1 either side[1]
  3. Maximum speed uses the largest distance and the smallest timethis reversal is the point of the question[1]
  4. 155 ÷ 11.5 = 13.5 m/s (3 s.f.)[1]

13.5 m/s

Examiner tip. Dividing by a smaller number gives a larger answer. Using 12.5 for the time would give the minimum speed instead, and it is the standard error here.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
A square tile has side 15 cm measured to the nearest centimetre. A floor 3 m by 2 m is to be covered.
  1. Write down the bounds of the tile's side length.
  2. Calculate the bounds of its area.
  3. Explain why the number of tiles needed cannot be calculated exactly, and state which bound a builder should use when ordering.
Mark scheme
  1. 14.5 ≤ s < 15.5 cm[1]
  2. Minimum area = 14.5² = 210.25 cm²[1]
  3. Maximum area = 15.5² = 240.25 cm²[1]
  4. The floor area is 60 000 cm², so the number of tiles lies between 60 000 ÷ 240.25 ≈ 250 and 60 000 ÷ 210.25 ≈ 285note the reversal — larger tiles mean fewer needed[1]
  5. The exact number cannot be found because the true tile size is unknown within that range[1]
  6. A builder should order for the smallest tile size, that is 285, since ordering too few would leave the floor unfinishedthe practical judgement is the mark[1]

(a) 14.5 ≤ s < 15.5 (b) 210.25 to 240.25 cm² (c) between 250 and 285 tiles; order 285

Examiner tip. The final part asks for judgement, not arithmetic. In any practical bounds question, ask which error would actually cost something — here, running out of tiles.