GCSE · Number

Bounds and Error Intervals

A ruler says a pencil is 7 cm long, to the nearest centimetre. It isn't exactly 7 cm: it could be 6.8 cm, or 7.4 cm. Every rounded measurement hides a range of possible true values. This page shows how to write that range down, and then (Higher) how to work out the biggest and smallest possible answers when you calculate with rounded numbers.

DfE GCSE subject content (Number): round numbers and measures to an appropriate degree of accuracy; use inequality notation to specify simple error intervals due to truncation or rounding; apply and interpret limits of accuracy; Higher tier: including upper and lower bounds

What you need from KS3

Rounding to the nearest 10, 100, whole number, to a number of decimal places (d.p.) and to a number of significant figures (s.f.). A 5 or more in the deciding digit rounds up. Recap on rounding and estimation.
Inequality symbols: < means "less than", ≤ means "less than or equal to". 3 ≤ x < 5 means x is at least 3 but less than 5.
Number lines: a filled circle includes its end point, a hollow circle doesn't.

1 · Rounding hides a range

Slide the true length of the pencil and watch what it rounds to. Notice that a whole band of true values all round to the same answer.

true length cm

Every length from 6.5 cm up to (but not including) 7.5 cm rounds to 7 cm. The band goes half a unit either side of the rounded value, because "nearest centimetre" means "closer to 7 than to 6 or 8", and the halfway points are 6.5 and 7.5.

The smallest possible true value is the lower bound: 6.5 cm.
The value at the top end is the upper bound: 7.5 cm. Strictly, 7.5 itself rounds up to 8, so the length is less than 7.5; but 7.5 is still called the upper bound.

2 · Error intervals

An error interval uses inequalities to say exactly which true values are possible. For the pencil:

6.5 ≤ l < 7.5

Use ≤ on the left, because 6.5 rounds up to 7, so it's included; and < on the right, because 7.5 would round to 8. The recipe for any accuracy:

Work out the unit you rounded to (nearest 10 → 10; 1 d.p. → 0.1; 2 s.f. on 4700 → 100).
Halve it.
Lower bound = value − half a unit; upper bound = value + half a unit.

value rounded to

Significant figures: find the unit from the place value of the last significant figure. 4700 to 2 s.f.: the 7 is in the hundreds column, so the unit is 100 and the interval is 4650 ≤ x < 4750. 0.36 to 2 s.f.: the 6 is in the hundredths column, so 0.355 ≤ x < 0.365.

Try it

≤ x <

Common mistakes with error intervals

Going a whole unit either side. 7 to the nearest whole number is 6.5 ≤ x < 7.5, not 6 ≤ x < 8.
Writing 7.49 or 7.4999 for the upper bound. The upper bound is 7.5 exactly; the < sign does the job of leaving it out.
Using ≤ at both ends, or < at the lower end. It is always ≤ … < for rounding.

3 · Truncation intervals

Truncating means chopping off digits without rounding: 7.86 truncated to a whole number is 7, and truncated to 1 d.p. is 7.8. Ages work like this (you're 15 right up until your 16th birthday), and so do many calculator and computer displays.

So if a number has been truncated to 7, the true value could be anything from 7 up to (not including) 8: 7 ≤ x < 8. The interval starts at the value and goes a whole unit up, instead of half a unit each way. Compare the two:

accuracy true value

4 · Bounds in calculations Higher

A rectangle measures 12 cm by 8 cm, both to the nearest cm. Its area is "about 96 cm²", but how big could it really be? Each length has its own interval: 11.5 ≤ length < 12.5 and 7.5 ≤ width < 8.5. The largest area comes from the two largest lengths: 12.5 × 8.5 = 106.25 cm². The smallest from the two smallest: 11.5 × 7.5 = 86.25 cm².

Adding and multiplying are straightforward like that. Subtracting and dividing are the traps: to make a − b as big as possible you want a big a but a small b. Explore all four combinations of bounds below; the table always shows the true maximum and minimum.

situation operation

To find	Maximum	Minimum
a + b	UB + UB	LB + LB
a − b	UB − LB	LB − UB
a × b	UB × UB	LB × LB
a ÷ b	UB ÷ LB	LB ÷ UB

Exam traps: using UB ÷ UB for the maximum of a division (dividing by a bigger number makes the answer smaller), and using the rounded values themselves. In context, think about what makes the answer big: the fastest speed = longest distance ÷ shortest time.

5 · Choosing a suitable degree of accuracy Higher

Once you have a lower and upper bound for an answer, how accurately can you honestly state it? Round both bounds to the same accuracy. If they agree, you can trust the answer to that accuracy. Try more and more significant figures until they stop agreeing; the last accuracy where they agree is the one to use.

calculation s.f.

On the number line, the dashed lines are where rounding changes. If a dashed line falls between the two bounds, they round to different values, so that accuracy claims more than you know.

6 · Practice

New questions each time. Bounds-in-calculation questions are Higher tier. Give answers exactly, or to 3 s.f. if they don't terminate.

question type

≤ x <

answer

Worked examples

A pencil is 7 cm long to the nearest cm. Write the error interval for its length l.

Unit: nearest cm → 1 cm. Half a unit: 0.5 cm.

Bounds: 7 − 0.5 = 6.5 and 7 + 0.5 = 7.5.

Error interval: 6.5 ≤ l < 7.5.

The population of a village is 4700, correct to 2 s.f. Write the error interval.

Unit: the second significant figure (7) is in the hundreds column, so the unit is 100. Half a unit is 50.

Bounds: 4700 − 50 = 4650 and 4700 + 50 = 4750.

Error interval: 4650 ≤ p < 4750. (A population must be a whole number, so in practice 4650 ≤ p ≤ 4749, but the GCSE form is fine.)

x = 5.2, truncated to 1 d.p. Write the error interval for x.

Truncation chops, so any number from 5.2 up to just below 5.3 truncates to 5.2: 5.29 → 5.2, but 5.3 → 5.3.

Error interval: 5.2 ≤ x < 5.3. (Rounding would have given 5.15 ≤ x < 5.25.)

A rectangle is 12 cm by 8 cm, each to the nearest cm. Find the upper and lower bounds of its area. Higher

Bounds of each side: 11.5 ≤ length < 12.5 and 7.5 ≤ width < 8.5.

Upper bound of area = UB × UB = 12.5 × 8.5 = 106.25 cm².

Lower bound of area = LB × LB = 11.5 × 7.5 = 86.25 cm².

Two masses are 54 kg and 31 kg, each to the nearest kg. Find the greatest possible difference. Higher

Bounds: 53.5 ≤ a < 54.5 and 30.5 ≤ b < 31.5.

Biggest difference: biggest a minus smallest b = 54.5 − 30.5 = 24 kg.

Trap: UB − UB = 54.5 − 31.5 = 23 kg is not the maximum.

A runner covers 100 m (to the nearest m) in 12.3 s (to 1 d.p.). Find her speed to a suitable degree of accuracy. Higher

Bounds: 99.5 ≤ d < 100.5 and 12.25 ≤ t < 12.35.

Upper bound of speed = UB(d) ÷ LB(t) = 100.5 ÷ 12.25 = 8.2040… m/s.

Lower bound of speed = LB(d) ÷ UB(t) = 99.5 ÷ 12.35 = 8.0566… m/s.

Compare roundings: to 2 s.f. they give 8.2 and 8.1 (different); to 1 s.f. both give 8.

Answer: speed = 8 m/s (to 1 s.f.), because both bounds agree to 1 s.f. but not to 2 s.f.

Quick check

What you've learned

A rounded value stands for a whole interval of possible true values.
For rounding, go half a unit either side: value − ½ unit ≤ x < value + ½ unit. Use ≤ at the lower bound and < at the upper bound.
Find the unit from the accuracy: nearest 10 → 10, 1 d.p. → 0.1, and for s.f. the place value of the last significant figure.
For truncation, the interval starts at the value and goes a whole unit up: value ≤ x < value + 1 unit.
(Higher) For the maximum of a calculation: + and × use UB with UB; − and ÷ use UB with LB (big first number, small second). Swap everything for the minimum.
(Higher) A suitable degree of accuracy is the most accurate rounding on which the upper and lower bounds agree.