1MA1 · N14–N16NumberFoundation & HigherGrades 3–7

Rounding, Estimation & Bounds

Q: Why is the upper bound not included?

A value exactly at the upper bound would round up to the next value, so it cannot have produced the rounded number.

Q: What does "to a suitable degree of accuracy" mean?

Round the upper and lower bounds; the most accurate rounding on which they agree is the suitable answer.

Q: Should estimates always use 1 significant figure?

Usually, unless a different rounding gives much easier arithmetic, such as 63.8 64 = 8.

A shelf "2.4 m long" is not exactly 2.4 m — it is somewhere between 2.35 m and 2.45 m, rounded. Most measurements you use are rounded, and this lesson covers the whole story: rounding accurately, estimating answers quickly, and working out the range of values a rounded number could really be.

55 min Number Fundamentals

The big picture

Rounding throws information away on purpose, to make numbers easier to use. Estimation uses rounding to check whether an answer is sensible. Bounds put the lost information back, by stating the smallest and largest values the original could have been.

Edexcel tests this at both tiers: error intervals at Foundation, and at Higher, upper and lower bounds of calculations — where you have to decide which bound of each input produces the largest or smallest possible result.

What you'll learn

Round to a given number of decimal places or significant figures
Estimate answers by rounding to 1 significant figure
Write error intervals for rounded and truncated values
Calculate upper and lower bounds of calculations (Higher)

Decimal places and significant figures

To round to a number of $\textbf{decimal places}$ , look at the next digit to the right: 5 or more rounds up, otherwise the digit stays. $3.14159$ to 2 d.p. is $3.14$ .

$\textbf{Significant figures}$ count from the first non-zero digit. $0.004\,719$ to 2 s.f. is $0.0047$ ; $83\,650$ to 2 s.f. is $84\,000$ .

Keep the number the same size: rounding $83\,650$ to 2 s.f. gives $84\,000$ , not $84$ .

Example 1Round

0.060\,95

to (a) 3 decimal places and (b) 3 significant figures.

1(a) Third decimal place is 0, next digit 9: round up to

0.061

2(b) First significant figure is the 6:

6, 0, 9

, next digit 5, round up to

0.0610

Answer(a)

0.061

(b)

0.0610

Your turnSame method, new numbers — try it before moving on.

Round $0.004\,752$ to 2 significant figures.

Your answer

Show worked solution

Solution

1The first significant figure is 4: $0.0047|52$ .
2The next digit is 5, so round up: $0.0048$ .

Answer: $0.0048$

Tip — The trailing zero in $0.0610$ matters: it shows the value is accurate to 3 significant figures. Leaving it off changes the stated accuracy.

Estimating

To estimate a calculation, round every number to 1 significant figure, then calculate. $\dfrac{612 \times 4.87}{0.189} \approx \dfrac{600 \times 5}{0.2} = 15\,000$ .

Say whether an estimate is an overestimate or underestimate by checking how each rounding affected the result — rounding a divisor up makes the answer smaller, for example.

Example 1Estimate

\dfrac{38.2 \times 9.71}{0.52}

and state whether your estimate is too large or too small.

1Round:

\dfrac{40 \times 10}{0.5} = 800

2Both numbers on top were rounded up, and the denominator was rounded down.

Answer

800

— an overestimate.

Your turnSame method, new numbers — try it before moving on.

By rounding each number to 1 significant figure, estimate $\dfrac{61.4 \times 0.49}{2.97}$ .

Your answer

Show worked solution

Solution

1 $\dfrac{60 \times 0.5}{3}$ .
2 $= \dfrac{30}{3} = 10$ .

Answer: $10$

Error intervals

If $x = 7.3$ correct to 1 decimal place, $x$ could be anything that rounds to $7.3$ : from $7.25$ up to, but not including, $7.35$ . The error interval is $7.25 \le x < 7.35$ .

The $\textbf{lower bound}$ is half a unit below the rounded value; the $\textbf{upper bound}$ is half a unit above. The upper bound itself is excluded, because it would round up.

For $\textbf{truncated}$ values (digits simply chopped off), the interval runs from the value up to one unit more: $x = 7.3$ truncated to 1 d.p. gives $7.3 \le x < 7.4$ .

Example 1A length is 450 cm correct to the nearest 10 cm. Write the error interval.

1Half of 10 cm is 5 cm.

2Lower bound

445

, upper bound

455

Answer

445 \le l < 455

Your turnSame method, new numbers — try it before moving on.

A mass is 3.6 kg, correct to 1 decimal place. Write down its upper bound.

Your answer

Show worked solution

Solution

1Half of 0.1 is 0.05.
2 $3.6 + 0.05 = 3.65$ kg.

Answer: $3.65$ kg

Why this works

The strict inequality at the top is the detail examiners look for. $455$ would round to $460$ , so it cannot be a possible value — but anything a fraction below $455$ can.

Bounds of calculations

To find the $\textbf{largest}$ possible result, choose each input’s bound to push the answer up; for the $\textbf{smallest}$ , push it down.

For addition and multiplication, use upper bounds for the maximum. For subtraction $a - b$ , the maximum uses upper $a$ and lower $b$ . For division $\dfrac{a}{b}$ , the maximum uses upper $a$ and $\textit{lower}$ $b$ .

To give a final answer "to a suitable degree of accuracy", round both bounds; the answer is the value to which they agree.

Example 1A car travels 125 km, correct to the nearest km, in 1.6 hours, correct to 1 decimal place. Find the upper bound of its average speed.

1Distance bounds:

124.5 \le d < 125.5

. Time bounds:

1.55 \le t < 1.65

2Maximum speed: largest distance over smallest time.

\dfrac{125.5}{1.55} = 80.97\dots

AnswerUpper bound

80.97

km/h (2 d.p.)

Your turnSame method, new numbers — try it before moving on.

A square has side 7.2 cm, correct to 1 decimal place. Work out the lower bound of its area.

Your answer

Show worked solution

Solution

1Lower bound of the side: 7.15 cm.
2 $7.15^2 = 51.1225$ cm².

Answer: $51.1225$ cm²

Tip — Write the bounds of every input in a small table before calculating. It makes choosing the right combination almost automatic.

Think like an examiner

Write error intervals with $\le$ on the left and $<$ on the right.
Show which bounds are used in each part of a calculation.
Round estimates to 1 significant figure, and show the rounded calculation.

Bounds

x \pm \tfrac12 \text{ unit}

Bounds of a rounded value.

\max\left(\tfrac{a}{b}\right) = \tfrac{a_{\text{upper}}}{b_{\text{lower}}}

Upper bound of a quotient.

\max(a - b) = a_{\text{upper}} - b_{\text{lower}}

Upper bound of a difference.

Watch out for these

Rounding

0.004\,719

to 2 s.f. as

0.00

Count from the first non-zero digit:

0.0047

Writing the error interval as

7.25 \le x \le 7.35

The upper bound is excluded:

7.25 \le x < 7.35

Using both upper bounds to maximise a division.

Divide the upper bound by the lower bound.

Treating a truncated value like a rounded one.

Truncation gives

x \le \text{value} < x + 1 \text{ unit}

Stretch yourself

A rectangle has length $12$ cm and width $7$ cm, both correct to the nearest cm. Find, to a suitable degree of accuracy, the length of its diagonal, justifying your choice of accuracy.

Hint — Use Pythagoras with the lower bounds and with the upper bounds, then round both results and see how far they agree.

Reveal solution

SolutionBounds:

11.5 \le l < 12.5

and

6.5 \le w < 7.5

. Lower bound of the diagonal:

\sqrt{11.5^{2} + 6.5^{2}} = \sqrt{174.5} = 13.21

cm. Upper bound:

\sqrt{12.5^{2} + 7.5^{2}} = \sqrt{212.5} = 14.58

cm. To 1 s.f. both round to

10

cm, but to 2 s.f. they give

13

and

15

, which differ. So the only accuracy both bounds agree on is 1 significant figure: the diagonal is

10

cm to 1 s.f. The large uncertainty comes from rounding both sides to the nearest cm.

\blacksquare

Questions students ask

Why is the upper bound not included?

A value exactly at the upper bound would round up to the next value, so it cannot have produced the rounded number.

What does "to a suitable degree of accuracy" mean?

Round the upper and lower bounds; the most accurate rounding on which they agree is the suitable answer.

Should estimates always use 1 significant figure?

Usually, unless a different rounding gives much easier arithmetic, such as

\sqrt{63.8} \approx \sqrt{64} = 8

Key takeaways

Decimal places count from the point; significant figures from the first non-zero digit.
Estimate by rounding to 1 s.f. and say whether it is an over- or underestimate.
Error interval: lower bound $\le x <$ upper bound, half a unit either side.
Truncation intervals run up by a whole unit.
Maximise or minimise calculations by choosing bounds deliberately.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

A number $x$ is truncated to an integer. The result is 18. Write down the error interval for $x$ .[2]
Show mark scheme
- B1 $18$ and $19$ seen
- B1 $18 \le x < 19$
Answer: $18 \le x < 19$

Question 2Grade 32 marks

Estimate the value of $\sqrt{48.7} \times 3.12^2$ .[2]
Show mark scheme
- M1Rounds to $\sqrt{49} \times 3^2$
- A1 $63$
Answer: $63$

Question 3Grade 73 marks

A metal block has mass 540 g, correct to the nearest 10 g, and volume 60 cm³, correct to the nearest cm³.

Work out the upper bound for the density of the metal. Give your answer to 3 significant figures.[3]
Show mark scheme
- B1Bounds 545 g and 59.5 cm³ seen
- M1 $545 \div 59.5$
- A1 $9.16$ g/cm³
Answer: $9.16$ g/cm³

Unlimited practice

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