Limits of Accuracy
When a measurement is rounded to a certain degree of accuracy (such as to the nearest cm, nearest 0.1 cm, or to 2 significant figures), the true value lies within a range. The smallest and largest possible true values are called the lower bound and upper bound.
Finding Bounds for Rounded Measurements
When a measurement is rounded, the actual value could be slightly less or slightly more. The lower bound is the smallest value that would round up to the given measurement. The upper bound is the smallest value that would round up to the next measurement.
Rule: To find the bounds, halve the degree of accuracy, then subtract from and add to the rounded value.
Lower Bound = Rounded value − ½ × degree of accuracy
Upper Bound = Rounded value + ½ × degree of accuracy
Example 1: Cement Bag Weight
A bag of cement is marked as 50 kg, rounded to the nearest 10 kg. Degree of accuracy = 10 kg. Half of this = 5 kg.
Lower bound = 50 − 5 = 45 kg
Upper bound = 50 + 5 = 55 kg
True weight: 45 ≤ weight < 55
The upper bound uses a < sign because if the value equals the upper bound, it would be rounded up to the next unit (60 kg in this case).
Different Degrees of Accuracy
To find bounds for any rounded measurement:
- Identify the degree of accuracy (e.g., nearest cm = 1, nearest 0.1 = 0.1, nearest 5 = 5)
- Halve the degree of accuracy to find how much to add/subtract
- Subtract from the rounded value to get the lower bound
- Add to the rounded value to get the upper bound
- Write the error interval using: LB ≤ value < UB
| Rounded Value | Degree of Accuracy | Half of Degree | Lower Bound | Upper Bound |
|---|---|---|---|---|
| 15 cm | nearest cm (1) | 0.5 cm | 14.5 cm | 15.5 cm |
| 6.8 m | nearest 0.1 m (0.1) | 0.05 m | 6.75 m | 6.85 m |
| 3.2 kg | nearest 0.1 kg (0.1) | 0.05 kg | 3.15 kg | 3.25 kg |
| 120 g | nearest 5 g (5) | 2.5 g | 117.5 g | 122.5 g |
| 4000 m | nearest 100 m (100) | 50 m | 3950 m | 4050 m |
Bounds with Significant Figures
For measurements given to a certain number of significant figures, the same principle applies: find the smallest and largest possible values that would round to the given number.
Rule: Identify the place value of the last significant figure. This becomes your degree of accuracy.
Example 2: Distance to 2 Significant Figures
A distance is given as 4500 m to 2 significant figures.
4500 to 2 s.f. means the first two digits (4 and 5) are significant. The 5 is in the hundreds place (4500). So the degree of accuracy = 100. Half of this = 50.
Lower bound = 4500 − 50 = 4450 m
Upper bound = 4500 + 50 = 4550 m
True distance: 4450 ≤ distance < 4550
Example 3: Decimal to 3 Significant Figures
A length is given as 0.0000639 m to 3 significant figures.
0.0000639 to 3 s.f. means the digits 6, 3, and 9 are significant. The last significant figure (9) is in the hundred-millionths place (0.00000009). So the degree of accuracy = 0.0000001. Half of this = 0.00000005.
Lower bound = 0.0000639 − 0.00000005 = 0.00006385 m
Upper bound = 0.0000639 + 0.00000005 = 0.00006395 m
True length: 0.00006385 ≤ length < 0.00006395
NB: For decimals, the same rule applies: halve the degree of accuracy (the place value of the last significant figure) and add/subtract from the rounded value.
Calculations with Bounds
When we use rounded numbers in calculations, the answer will also have a range of possible values. To find the maximum or minimum possible result, we use the upper and lower bounds of the measurements.
Key Rules for Bounds in Calculations:
Addition: Max = UB + UB, Min = LB + LB
Subtraction: Max = UB − LB, Min = LB − UB
Multiplication: Max = UB × UB, Min = LB × LB
Division: Max = UB ÷ LB, Min = LB ÷ UB
Example 4: Maximum and Minimum Area of a Rectangle
A rectangle has length 12 cm (to nearest cm) and width 5 cm (to nearest cm).
Step 1: Find bounds for each measurement
Degree of accuracy = 1 cm. Half = 0.5 cm.
Length: 11.5 ≤ L < 12.5
Width: 4.5 ≤ W < 5.5
Step 2: Maximum area (UB × UB)
Max area = 12.5 × 5.5 = 68.75 cm²
Step 3: Minimum area (LB × LB)
Min area = 11.5 × 4.5 = 51.75 cm²
Example 5: Multiple Items (Multiplication)
A bag of cement is marked 50 kg (to nearest 10 kg). Find the total mass of 5 bags.
Step 1: Find bounds for one bag
Degree of accuracy = 10 kg. Half = 5 kg.
Lower bound = 45 kg, Upper bound = 55 kg
Step 2: Total mass for 5 bags
Minimum total = 5 × 45 = 225 kg
Maximum total = 5 × 55 = 275 kg
Answer: 225 ≤ total < 275 kg
NB: For multiplication, maximum uses the largest possible values (UB × UB), while minimum uses the smallest possible values (LB × LB).
Speed, Distance, Time Calculations with Bounds
For speed problems, remember the formula: Speed = Distance ÷ Time
To find maximum speed, use the largest possible distance (UB) and smallest possible time (LB). To find minimum speed, use the smallest possible distance (LB) and largest possible time (UB).
Example 6: Maximum and Minimum Speed
A car travels 150 km (to nearest km) in 2 hours (to nearest hour).
Step 1: Find bounds for each measurement
Distance: nearest km → half of 1 = 0.5
149.5 ≤ D < 150.5
Time: nearest hour → half of 1 = 0.5
1.5 ≤ T < 2.5
Step 2: Maximum speed (UB ÷ LB)
Max speed = 150.5 ÷ 1.5 = 100.33 km/h
Step 3: Minimum speed (LB ÷ UB)
Min speed = 149.5 ÷ 2.5 = 59.8 km/h
Error interval: 59.8 ≤ speed < 100.33 km/h
NB: For division, maximum uses UB ÷ LB, while minimum uses LB ÷ UB.