Bounds and Error Intervals (Higher)

This is a Higher-tier [H] lesson throughout. When a measurement is given "correct to" a certain accuracy, the true value lies in a range around it. This lesson covers writing error intervals from rounding (and from truncation), finding upper and lower bounds, and carrying out calculations with bounds to find the largest and smallest possible results. These ideas form part of the Number content of OCR GCSE Mathematics (J560) and are a frequent source of higher-mark questions on all three papers.

This lesson mainly builds AO1 fluency in finding bounds, with substantial AO2 reasoning about strict vs non-strict inequalities and AO3 problem-solving when you decide which bounds to combine for a maximum or minimum.

Key Vocabulary

Term	Meaning
Lower bound (LB)	The smallest value that would round to the given figure
Upper bound (UB)	The smallest value that would round up to the next figure
Error interval	The range of possible true values, written as an inequality
Degree of accuracy	The precision the value was rounded to (e.g. nearest $10$10, $1$1 d.p.)
Truncation	Cutting digits off without rounding, which gives a different interval

Why Bounds Matter

Every real measurement is rounded. When a ruler reads "$7$7 cm to the nearest centimetre", the true length is not exactly $7$7 cm — it is somewhere in a band around $7$7. Bounds make that band precise. This matters in the real world: an engineer machining a part to "$50$50 mm, correct to the nearest mm" needs to know the component could be anywhere from $49.5$49.5 mm to just under $50.5$50.5 mm, and must check the largest and smallest possible sizes still fit. The mathematics below is exactly that reasoning, made systematic.

Bounds from Rounding

When a value is rounded to a given degree of accuracy, the true value can be up to half that accuracy either side. So:

$\text{LB} = \text{value} - \dfrac{\text{accuracy}}{2}, \qquad \text{UB} = \text{value} + \dfrac{\text{accuracy}}{2}.$LB=value−2accuracy​,UB=value+2accuracy​.

The lower bound is included and the upper bound is excluded, because a value exactly at the upper bound would round up to the next figure.

Worked Example 1

A length is $24$24 cm, correct to the nearest centimetre. Write the error interval.

The accuracy is $1$1 cm, so half is $0.5$0.5 cm. LB $= 24 - 0.5 = 23.5$=24−0.5=23.5; UB $= 24 + 0.5 = 24.5$=24+0.5=24.5.

$23.5 \le \text{length} < 24.5.$23.5≤length<24.5.

Answer: $23.5 \le l < 24.5$23.5≤l<24.5 (cm).

Common error: Using $\le$≤ at both ends. The upper bound is strict ($<$<): $24.5$24.5 would round up to $25$25.

Worked Example 2

A mass is $4.7$4.7 kg, correct to $1$1 decimal place. Write the error interval.

Accuracy $= 0.1$=0.1 kg, so half is $0.05$0.05. LB $= 4.7 - 0.05 = 4.65$=4.7−0.05=4.65; UB $= 4.7 + 0.05 = 4.75$=4.7+0.05=4.75.

$4.65 \le m < 4.75.$4.65≤m<4.75.

Answer: $4.65 \le m < 4.75$4.65≤m<4.75 (kg).

Worked Example 3

A crowd is $35{,}000$35,000, correct to the nearest $1000$1000. Write the error interval.

Accuracy $= 1000$=1000, half is $500$500. LB $= 35{,}000 - 500 = 34{,}500$=35,000−500=34,500; UB $= 35{,}000 + 500 = 35{,}500$=35,000+500=35,500.

$34{,}500 \le \text{crowd} < 35{,}500.$34,500≤crowd<35,500.

Answer: $34{,}500 \le c < 35{,}500$34,500≤c<35,500.

Error Intervals from Truncation

If a value has been truncated (digits chopped off) rather than rounded, the interval is different: the given value is the lower bound, and the upper bound is one whole unit of accuracy higher.

Worked Example 4

A reading of $8.3$8.3 has been truncated to $1$1 decimal place. Write the error interval.

Truncation keeps everything from $8.3$8.3 up to (but not including) $8.4$8.4, because, say, $8.3999$8.3999 truncates to $8.3$8.3.

$8.3 \le x < 8.4.$8.3≤x<8.4.

Answer: $8.3 \le x < 8.4$8.3≤x<8.4.

Common error: Treating a truncated value like a rounded one ($8.25 \le x < 8.35$8.25≤x<8.35). For truncation, the value itself is the lower bound.

Calculating with Bounds

To find the bounds of a calculated quantity, combine the bounds of the inputs cleverly. The rules depend on the operation:

Operation	Maximum result	Minimum result
Addition $a + b$a+b	$\text{UB}_a + \text{UB}_b$UBa​+UBb​	$\text{LB}_a + \text{LB}_b$LBa​+LBb​
Multiplication $a \times b$a×b	$\text{UB}_a \times \text{UB}_b$UBa​×UBb​	$\text{LB}_a \times \text{LB}_b$LBa​×LBb​
Subtraction $a - b$a−b	$\text{UB}_a - \text{LB}_b$UBa​−LBb​	$\text{LB}_a - \text{UB}_b$LBa​−UBb​
Division $a \div b$a÷b	$\text{UB}_a \div \text{LB}_b$UBa​÷LBb​	$\text{LB}_a \div \text{UB}_b$LBa​÷UBb​

The key insight for subtraction and division is that to make the result as big as possible, you take a big numerator and a small denominator (and the reverse for the smallest).

Worked Example 5

$a = 12$a=12 and $b = 5$b=5, both to the nearest integer. Work out the upper bound of $a + b$a+b.

Bounds: $11.5 \le a < 12.5$11.5≤a<12.5 and $4.5 \le b < 5.5$4.5≤b<5.5. Maximum sum uses both upper bounds: $12.5 + 5.5 = 18$12.5+5.5=18.

Answer: upper bound of $a + b$a+b is $18$18.

Worked Example 6

Using the same $a$a and $b$b, work out the lower bound of $a - b$a−b.

Minimum of $a - b$a−b uses the smallest $a$a and the largest $b$b: $\text{LB}_a - \text{UB}_b = 11.5 - 5.5 = 6$LBa​−UBb​=11.5−5.5=6.

Answer: lower bound of $a - b$a−b is $6$6.

Common error: Using $\text{LB}_a - \text{LB}_b$LBa​−LBb​. For subtraction, the smallest result comes from subtracting the biggest possible $b$b.

Extended Worked Examples

Worked Example 7: Bounds of a speed (division)

A car travels $150$150 m (to the nearest $10$10 m) in $8.0$8.0 s (to the nearest $0.1$0.1 s). Work out the upper bound of its average speed.

Distance bounds: $145 \le d < 155$145≤d<155. Time bounds: $7.95 \le t < 8.05$7.95≤t<8.05.

Speed $= \dfrac{d}{t}$=td​. The maximum speed uses the largest distance and the smallest time:

$\text{UB of speed} = \dfrac{\text{UB}_d}{\text{LB}_t} = \dfrac{155}{7.95} = 19.4968\ldots \approx 19.50 \text{ m/s}.$UB of speed=LBt​UBd​​=7.95155​=19.4968…≈19.50 m/s.

Answer: upper bound of the speed is $19.50$19.50 m/s (to 2 d.p.).

Worked Example 8: Bounds of an area (multiplication)

A rectangle measures $6.4$6.4 cm by $3.5$3.5 cm, each to $1$1 decimal place. Work out the lower bound of its area.

Bounds: $6.35 \le L < 6.45$6.35≤L<6.45 and $3.45 \le W < 3.55$3.45≤W<3.55. Minimum area uses both lower bounds:

$\text{LB of area} = 6.35 \times 3.45 = 21.9075 \text{ cm}^{2}.$LB of area=6.35×3.45=21.9075 cm2.

Answer: lower bound of the area is $21.9075$21.9075 cm².

Worked Example 8b: Upper bound of a perimeter (addition)

A triangle has sides $5.2$5.2 cm, $6.8$6.8 cm and $9.1$9.1 cm, each correct to 1 decimal place. Work out the upper bound of the perimeter.

For a sum, the maximum uses every upper bound. Each side's upper bound is the value $+ 0.05$+0.05: $5.25$5.25, $6.85$6.85 and $9.15$9.15.

$\text{UB of perimeter} = 5.25 + 6.85 + 9.15 = 21.25 \text{ cm}.$UB of perimeter=5.25+6.85+9.15=21.25 cm.

Answer: $21.25$21.25 cm. (The lower bound would use all three lower bounds: $5.15 + 6.75 + 9.05 = 20.95$5.15+6.75+9.05=20.95 cm.)

Worked Example 9: Deciding a value to a suitable accuracy

A field is $48$48 m by $35$35 m, each to the nearest metre. By finding the upper and lower bounds of the area, give the area to a suitable degree of accuracy.

Bounds: $47.5 \le L < 48.5$47.5≤L<48.5, $34.5 \le W < 35.5$34.5≤W<35.5.

• UB of area $= 48.5 \times 35.5 = 1721.75$=48.5×35.5=1721.75 m².
• LB of area $= 47.5 \times 34.5 = 1638.75$=47.5×34.5=1638.75 m².

The bounds agree to the leading "$1{,}7$1,7..." but differ in the hundreds, so the only safe rounded statement is that the area is about $1700$1700 m² (to 2 s.f.) — we cannot justify more precision.

Answer: area $\approx 1700$≈1700 m² (to 2 s.f.), because the upper and lower bounds round to the same value only at that accuracy.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: A number $n$n is given as $6.2$6.2, correct to $1$1 decimal place. Write down the error interval for $n$n, and explain why the upper bound uses a strict inequality.

Grades 3–4 response: LB $= 6.15$=6.15, UB $= 6.25$=6.25. So $6.15 \le n < 6.25$6.15≤n<6.25. The upper bound has "$<$<" because $6.25$6.25 rounds up to $6.3$6.3.