Bounds and Error Intervals
In Rounding and Estimation, you learnt how to replace a number with a nearby value. Now we will work backwards: which actual measurements could have produced a recorded rounded value?
Foundation & Higher: start with simple measurements, number lines and error intervals. Later Higher work uses bounds in calculations with totals, differences, areas and speeds.
A rounded measurement can hide detail
A length is recorded as 8 cm to the nearest centimetre. The unit cm means centimetres. The actual length need not be exactly 8 cm.
For example, 7.6 cm, 8 cm and 8.4 cm all round to 8 cm. The halfway points are 7.5 cm and 8.5 cm.
Using the positive-number rounding rule from Rounding and Estimation, 7.5 cm rounds to 8 cm, but 8.5 cm rounds to 9 cm.
So the actual length can be from 7.5 cm up to, but not including, 8.5 cm.
The lower bound is 7.5 cm. The upper bound is 8.5 cm.
The upper bound marks the top edge of the range. It is not a length that rounds to 8 cm.
The rounded values are 1 cm apart. The halfway point on each side is 0.5 cm away. This is why we use half the rounding unit.
Example Question:
A length is recorded as 12.4 cm to the nearest 0.1 cm. Find its lower and upper bounds.
First identify half of 0.1 cm. Example 3 shows how to use it.
Exam Tip
Underline the stated accuracy. A measurement recorded as 12.4 cm to the nearest 0.1 cm has a different accuracy from 12 cm to the nearest cm.
Keep full values during bounds calculations. Do not round a bound early and accidentally cut off possible values.
An upper endpoint marked with < is excluded. Do not describe it as an actual value the quantity can reach.
Method — Find half a rounding unit
1. Identify the rounding unit. This is the gap between neighbouring rounded values.
2. Halve that unit.
3. Subtract the half-unit from the recorded value. This gives the lower bound.
4. Add the half-unit to the recorded value. This gives the upper bound.
| Rounded to nearest… | Half-unit |
|---|---|
| 1 | 0.5 |
| 10 | 5 |
| 100 | 50 |
| 0.1 (1 decimal place) | 0.05 |
| 0.01 (2 decimal places) | 0.005 |
Writing an error interval
An error interval states the whole range of possible values. “Error” here means the detail lost through rounding, not a mistake someone made.
Let x be the actual numerical length in centimetres for the recorded 8 cm measurement.
7.5 ≤ x < 8.5
The symbol ≤ means “less than or equal to”. So 7.5 ≤ x says x is at least 7.5.
The symbol < means “less than”. So x < 8.5 says x is below 8.5.
The lower endpoint is included; the upper endpoint is excluded. On the number line, a filled circle means included and an open circle means excluded.
This lesson uses positive measurements rounded to the nearest stated unit, with halfway values rounding up.
Worked Examples
Example 1 — Nearest whole number
A length x cm is recorded as 8 cm to the nearest cm. Find its error interval.
Rounding unit = 1 cm
Half-unit = 0.5 cm
8 − 0.5 = 7.5
8 + 0.5 = 8.5
7.5 ≤ x < 8.5, with x measured in cm.
Example 2 — Nearest 10
A mass x grams is recorded as 120 g to the nearest 10 g. Find its bounds.
The symbol g means grams. The rounding unit is 10 g, so the half-unit is 5 g.
120 − 5 = 115
120 + 5 = 125
Lower bound = 115 g; upper bound = 125 g.
115 ≤ x < 125, with x measured in g.
125 g rounds to 130 g, so it is not included.
Example 3 — One decimal place
A length x cm is recorded as 12.4 cm to the nearest 0.1 cm.
One decimal place means the rounding unit is 0.1, not 1.
0.1 ÷ 2 = 0.05
12.4 − 0.05 = 12.35
12.4 + 0.05 = 12.45
12.35 ≤ x < 12.45, with x measured in cm.
Example 4 — Two decimal places
A time x seconds is recorded as 3.46 s to 2 decimal places. Write its error interval.
The symbol s means seconds. The rounding unit is 0.01 s.
0.01 ÷ 2 = 0.005
3.46 − 0.005 = 3.455
3.46 + 0.005 = 3.465
3.455 ≤ x < 3.465, with x measured in s.
Three decimal places are needed to write these endpoints exactly.
Example 5 — Significant figures
A distance x metres is recorded as 4300 m to 2 significant figures. Find its error interval.
The symbol m means metres. The first significant digit is 4 and the second is 3. The 3 is in the hundreds place.
So this has been rounded to the nearest 100 m. Find the place first, then use the usual method.
100 ÷ 2 = 50
4300 − 50 = 4250
4300 + 50 = 4350
4250 ≤ x < 4350, with x measured in m.
The stated “2 significant figures” tells us the accuracy. The written zeros alone would not be enough.
Truncation and practical accuracy — both tiers
Truncating means cutting off digits after the stated place without rounding. For example, 6.278 truncated to one decimal place is 6.2, although it rounds to 6.3.
Worked example — a truncated measurement
A positive length x cm is 6.2 cm when truncated to one decimal place. It could be 6.20, 6.21 or 6.299 cm, but not 6.3 cm.
6.2 ≤ x < 6.3, with x in cm. The recorded value is the lower endpoint; do not subtract half a unit. By contrast, rounding to 6.2 cm would give 6.15 ≤ x < 6.25.
Quick practice — a positive mass x kg is 4.37 kg when truncated to two decimal places. Write its error interval.
Cutting off after hundredths leaves 4.37 for values from 4.370… up to, but not including, 4.38.
4.37 ≤ x < 4.38, with x in kg. The interval width is 0.01 kg.
Worked example — will an object definitely fit?
A box is 12 cm wide and a gap is 13 cm wide, each rounded to the nearest cm. The box is less than 12.5 cm wide. The gap is at least 12.5 cm wide. The box is therefore narrower than the gap, although the clearance could be very small.
Quick practice — a box and a gap are both recorded as 12 cm wide to the nearest cm. Is the box certain to fit?
Both widths lie from 11.5 cm up to, but not including, 12.5 cm. The box could be 12.4 cm while the gap is 11.6 cm.
No. The rounded values do not guarantee that the box is narrower. More accurate measurements are needed.
When the quantity is a whole-number count — both tiers
A car count n is 200 to the nearest hundred. The rounding interval is 150 ≤ n < 250. Because n must be an integer, the greatest possible count is 249. The same count range can be written 150 ≤ n ≤ 249. A continuous measurement would not have this last whole-number restriction.
Quick practice — a count of tickets is 80 to the nearest ten. Find the smallest and largest possible counts.
The rounding interval is 75 ≤ n < 85. Only whole tickets are counted.
The smallest count is 75 and the largest is 84.
Higher calculations — choose the right endpoints
Now combine two rounded measurements. In these examples, both quantities are positive and their measurements can vary independently.
First write each error interval. Then ask how changing each quantity makes the result larger or smaller.
A calculated lower bound is not always an attainable minimum. A calculated upper bound is not an attainable maximum when it uses an excluded endpoint. We will state the inequality signs carefully.
Example 6 — Add two lengths
Two rods are recorded as 8 cm and 5 cm, each to the nearest cm. Let a and b be their actual lengths in cm. Find bounds for the total length T in cm.
7.5 ≤ a < 8.5
4.5 ≤ b < 5.5
The smallest total uses both lower bounds. These can both be reached:
7.5 + 4.5 = 12
The upper bound uses both upper bounds:
8.5 + 5.5 = 14
12 ≤ T < 14.
The minimum total is 12 cm. The upper bound is 14 cm, but the actual total is below 14 cm.
Example 7 — Subtract two lengths
Use the same rods: 7.5 ≤ a < 8.5 and 4.5 ≤ b < 5.5, in cm. Let D = a − b be the difference in cm.
To make the difference small, start with a small a and subtract a large b:
7.5 − 5.5 = 2
To make the difference large, start with a large a and subtract a small b:
8.5 − 4.5 = 4
2 < D < 4.
Both endpoints are excluded. Reaching 2 would require b = 5.5, which is not allowed. Reaching 4 would require a = 8.5, which is not allowed.
So 2 cm and 4 cm are limiting bounds, not attainable minimum and maximum differences.
Example 8 — Multiply to bound an area
A rectangle has recorded length 8 cm and width 5 cm, both to the nearest cm. Find bounds for its area A in cm2.
The length runs from 7.5 cm up to, but not including, 8.5 cm. The width runs from 4.5 cm up to, but not including, 5.5 cm.
Both dimensions are positive. Increasing either one increases the area.
Lower bound: multiply the two lower dimensions.
7.5 × 4.5 = 33.75
Upper bound: multiply the two upper dimensions.
8.5 × 5.5 = 46.75
33.75 ≤ A < 46.75.
The minimum area is 33.75 cm2. The upper bound is 46.75 cm2, excluded. The unit cm2 means square centimetres.
Example 9 — Divide to bound a speed
A distance is recorded as 100 m to the nearest 10 m. A time is recorded as 20 s to the nearest 10 s. Find bounds for the average speed.
Let d be the actual distance in m, and t the actual time in s.
95 ≤ d < 105
15 ≤ t < 25
Average speed = distance ÷ time. Let v be its numerical value in m/s, meaning metres per second.
For the lower speed bound, use a small distance and a large time:
95 ÷ 25 = 3.8
For the upper speed bound, use a large distance and a small time:
105 ÷ 15 = 7
3.8 < v < 7.
Both endpoints are excluded. The lower calculation used the excluded time 25 s; the upper calculation used the excluded distance 105 m.
The bounds are 3.8 m/s and 7 m/s. Do not use lower distance ÷ lower time automatically for a lower speed bound.
Example 10 — Higher: choose bounds for a perimeter
A rectangle is recorded as 12.4 cm long and 8.2 cm wide, each to 1 decimal place. Find lower and upper bounds for its perimeter.
Let l and w be its actual length and width in cm. Each half-unit is 0.05 cm.
12.35 ≤ l < 12.45
8.15 ≤ w < 8.25
Perimeter is the distance around the rectangle. Its numerical value P in cm is 2 × (l + w). The 2 is an exact count of two pairs of sides; it is not a rounded measurement.
Adding positive lengths gives a larger total when either length increases. Use both lower bounds for the lower perimeter:
12.35 + 8.15 = 20.5
2 × 20.5 = 41
Use both upper bounds for the upper perimeter:
12.45 + 8.25 = 20.7
2 × 20.7 = 41.4
41 ≤ P < 41.4.
The minimum perimeter is 41 cm. The upper bound is 41.4 cm, and is excluded.
Choose bounds by the operation
For two positive quantities A and B with positive lower bounds, these endpoint choices give the required bounds. “Lower A” means the lower bound of A, and similarly for the other labels.
| Calculation | Choose these endpoints |
|---|---|
| A + B | Lower: lower A + lower B Upper: upper A + upper B |
| A − B | Lower: lower A − upper B Upper: upper A − lower B |
| A × B | Lower: lower A × lower B Upper: upper A × upper B |
| A ÷ B | Lower: lower A ÷ upper B Upper: upper A ÷ lower B |
The subtraction and division choices differ because making the second quantity larger makes the result smaller.
Always check whether an endpoint can be reached before using ≤. Do not transfer the original interval’s signs automatically to a calculation.
Check the interval and the units
For a positive measurement rounded to the nearest stated unit, the recorded value sits halfway between the interval endpoints. A truncated positive value instead equals the included lower endpoint.
For a directly rounded measurement, the interval width is one complete rounding unit. For 12.4 to the nearest 0.1:
12.45 − 12.35 = 0.1
For a positive measurement rounded to the nearest stated unit, the lower bound is below the recorded value and the upper bound is above it. This check does not apply unchanged to truncation or calculated bounds.
For calculated bounds, the lower result must not exceed the upper result. Their gap need not equal an original rounding unit.
Keep units consistent: add lengths in the same unit; length × width gives area; distance ÷ time gives speed.
Try rounding an allowed value back to the stated accuracy. At an excluded upper endpoint, you should get the next rounded value.
Think Like a Mathematician
Why include 7.5 but exclude 8.5 when a length rounds to 8 cm?
7.5 is halfway between 7 and 8. Our rounding rule sends it to 8.
8.5 is halfway between 8 and 9. The same rule sends it to 9.
Each boundary belongs to just one rounding interval. If both neighbouring intervals included the shared endpoint, the same length would be assigned two different rounded values.
That is why a filled circle belongs at 7.5 and an open circle at 8.5.
Common mistakes to avoid
Using a whole rounding unit. Nearest 10 needs 5 on either side, not 10.
Halving the wrong place. Two decimal places means a unit of 0.01 and a half-unit of 0.005.
Including the upper endpoint. With the rounding convention here, the upper endpoint rounds to the next recorded value.
Using lower − lower for a lower difference. Subtract a large second quantity from a small first quantity instead.
Using lower ÷ lower for every lower bound. A small positive quotient needs a small numerator and a large denominator. Numerator means the amount being divided; denominator means the amount you divide by.
Treating a recorded measurement as exact. Use its stated accuracy before calculating.
Losing units or rounding early. Keep full endpoint values and state the correct units in the answer.
Try it yourself
Start by identifying each rounding unit. Show working and state units. Questions 9–12 are Higher calculations with bounds; Questions 1–8 build the both-tier interval skills.
1. Find the half-unit for rounding to the nearest 10, nearest 0.1 and nearest 0.01.
10 ÷ 2 = 5
0.1 ÷ 2 = 0.05
0.01 ÷ 2 = 0.005
The half-units are 5, 0.05 and 0.005.
2. A length x cm is recorded as 6 cm to the nearest cm. Find its bounds and write the error interval.
The rounding unit is 1 cm, so the half-unit is 0.5 cm.
6 − 0.5 = 5.5
6 + 0.5 = 6.5
Lower bound = 5.5 cm; upper bound = 6.5 cm.
5.5 ≤ x < 6.5, with x in cm.
3. A mass x grams is recorded as 240 g to the nearest 10 g. Write its error interval.
Half of 10 g is 5 g.
240 − 5 = 235
240 + 5 = 245
235 ≤ x < 245, with x in g.
4. A distance x metres is recorded as 3700 m to the nearest 100 m. Write its error interval.
Half of 100 m is 50 m.
3700 − 50 = 3650
3700 + 50 = 3750
3650 ≤ x < 3750, with x in m.
5. A length x cm is recorded as 4.7 cm to 1 decimal place. Write its error interval.
The rounding unit is 0.1 cm, so the half-unit is 0.05 cm.
4.7 − 0.05 = 4.65
4.7 + 0.05 = 4.75
4.65 ≤ x < 4.75, with x in cm.
6. A mass x kilograms is recorded as 2.36 kg to 2 decimal places. Write its error interval.
The unit kg means kilograms. The rounding unit is 0.01 kg, so its half is 0.005 kg.
2.36 − 0.005 = 2.355
2.36 + 0.005 = 2.365
2.355 ≤ x < 2.365, with x in kg.
7. A distance x metres is recorded as 5600 m to 2 significant figures. Find its error interval.
The second significant digit, 6, is in the hundreds place. The rounding unit is 100 m.
Half-unit = 50 m
5600 − 50 = 5550
5600 + 50 = 5650
5550 ≤ x < 5650, with x in m.
8. A length x cm satisfies 7.5 ≤ x < 8.5. Which of 7.5 cm, 8.49 cm and 8.5 cm are allowed?
7.5 cm is allowed because the lower endpoint has ≤.
8.49 cm is allowed because it is above 7.5 and below 8.5.
8.5 cm is not allowed because the upper endpoint has <.
At the nearest cm, 7.5 and 8.49 round to 8, but 8.5 rounds to 9.
9. Two rods are recorded as 10 cm and 4 cm, each to the nearest cm. Find bounds for their total T in cm.
The actual lengths a and b, in cm, satisfy:
9.5 ≤ a < 10.5
3.5 ≤ b < 4.5
Lower total: 9.5 + 3.5 = 13
Upper total: 10.5 + 4.5 = 15
13 ≤ T < 15. The lower total can be reached, but the upper endpoint cannot.
10. Use the rods from Question 9. Find bounds for their difference D = a − b in cm.
For the lower bound, subtract the upper b from the lower a:
9.5 − 4.5 = 5
For the upper bound, subtract the lower b from the upper a:
10.5 − 3.5 = 7
5 < D < 7.
Both endpoints are excluded, because each calculation uses an excluded upper length.
11. Higher: a rectangle is recorded as 6 cm by 4 cm, both to the nearest cm. Find lower and upper bounds for its area A in cm².
Length: from 5.5 cm up to, but not including, 6.5 cm.
Width: from 3.5 cm up to, but not including, 4.5 cm.
For positive dimensions, use lower × lower for the lower area:
5.5 × 3.5 = 19.25
Use upper × upper for the upper area:
6.5 × 4.5 = 29.25
19.25 ≤ A < 29.25, with A in cm2.
The minimum area is 19.25 cm2. The upper bound 29.25 cm2 is excluded.
12. Higher: a distance is recorded as 80 m to the nearest 10 m and a time as 30 s to the nearest 10 s. Find the upper bound for the average speed.
The distance is from 75 m up to, but not including, 85 m.
The time is from 25 s up to, but not including, 35 s.
To make distance ÷ time large, choose the upper distance and the lower time:
85 ÷ 25 = 3.4
The upper speed bound is 3.4 m/s.
The actual speed is below 3.4 m/s because the distance cannot equal 85 m. This is an upper bound, not an attainable maximum.
Frequently Asked Questions
Does the word error mean my measurement is wrong?
No. Rounding leaves out detail, so several actual values can share the same recorded measurement.
Why halve the rounding unit?
The rounding changes at halfway points between neighbouring recorded values. Each halfway point is half a unit away.
Can the actual value equal the upper bound?
Not for the directly rounded intervals used here. The upper endpoint rounds to the next value, so we use <.
Why do decimal bounds sometimes need an extra decimal place?
Half of 0.1 is 0.05, and half of 0.01 is 0.005. Writing those halfway points exactly needs an extra place.
Is a calculated lower bound always a possible minimum?
No. For example, the lower difference in Example 7 needs the second rod’s excluded upper endpoint. The actual difference stays above that lower bound.
Do the calculation rules work for negative quantities too?
Not automatically. This lesson uses positive quantities with positive lower bounds. Negative values need additional care beyond the methods introduced here.
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