Upper and Lower Bounds
Section: Number and Calculation | Syllabus: Cambridge Lower Secondary Checkpoint Mathematics (0862)
What are Upper and Lower Bounds?
Every rounded or truncated measurement hides a small range of possible actual values. The upper and lower bounds mark the edges of that range.
- Lower bound: the smallest possible value the actual measurement could be.
- Upper bound: the largest possible value the actual measurement could be.
Why Are Bounds Important?
In real life, all measurements are approximations:
- A ruler might measure to the nearest mm.
- Digital scales round to the nearest gram.
- Speedometers round to the nearest km/h.
- Understanding bounds tells us the accuracy limits of any calculation built on these measurements.
Finding Upper and Lower Bounds
Lower Bound = x − (half the degree of accuracy)
Upper Bound = x + (half the degree of accuracy)
Lower Bound ≤ Actual Value < Upper Bound
The lower bound is included in the range; the upper bound is not
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Question: A length is measured as 15 cm to the nearest cm. Find the bounds.
- Step 1: Degree of accuracy = 1 cm, so half the degree = 0.5 cm
- Step 2: Lower Bound = 15 − 0.5 = 14.5 cm; Upper Bound = 15 + 0.5 = 15.5 cm
- Answer: 14.5 ≤ actual length < 15.5
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Question: A mass is 7.3 kg to 1 decimal place. Find the bounds.
- Step 1: Degree of accuracy = 0.1 kg, so half the degree = 0.05 kg
- Step 2: Lower Bound = 7.3 − 0.05 = 7.25 kg; Upper Bound = 7.3 + 0.05 = 7.35 kg
- Answer: 7.25 ≤ actual mass < 7.35
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Question: A time is 12.45 seconds to 2 decimal places. Find the bounds.
- Step 1: Degree of accuracy = 0.01 s, so half the degree = 0.005 s
- Step 2: Lower Bound = 12.45 − 0.005 = 12.445 s; Upper Bound = 12.45 + 0.005 = 12.455 s
- Answer: 12.445 ≤ actual time < 12.455
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Question: A distance is 2400 m to 2 significant figures. Find the bounds.
- Step 1: The 2nd significant figure is the 4 (hundreds place), so the degree of accuracy = 100 m
- Step 2: Half the degree = 50 m, so Lower Bound = 2400 − 50 = 2350 m; Upper Bound = 2400 + 50 = 2450 m
- Answer: 2350 ≤ actual distance < 2450
Quick Reference Table
| Rounded To | Degree of Accuracy | Half the Degree | Example |
|---|---|---|---|
| Nearest whole number | 1 | 0.5 | 15 → [14.5, 15.5) |
| Nearest 10 | 10 | 5 | 350 → [345, 355) |
| Nearest 100 | 100 | 50 | 1200 → [1150, 1250) |
| 1 decimal place | 0.1 | 0.05 | 8.4 → [8.35, 8.45) |
| 2 decimal places | 0.01 | 0.005 | 3.67 → [3.665, 3.675) |
| 3 decimal places | 0.001 | 0.0005 | 5.234 → [5.2335, 5.2345) |
Bounds in Calculations
When two rounded measurements are combined, the maximum and minimum possible results depend on the operation - not just on plugging in both upper bounds or both lower bounds.
Addition - Maximum and Minimum Results
- Maximum sum = Upper bound of A + Upper bound of B
- Minimum sum = Lower bound of A + Lower bound of B
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Question: Two lengths are measured as 12 cm and 8 cm (both to the nearest cm). Find the bounds for their total length.
- Step 1: 12 cm: Lower = 11.5 cm, Upper = 12.5 cm. 8 cm: Lower = 7.5 cm, Upper = 8.5 cm
- Step 2: Maximum total = 12.5 + 8.5 = 21 cm; Minimum total = 11.5 + 7.5 = 19 cm
- Answer: 19 ≤ total length < 21
Subtraction - Maximum and Minimum Results
- Maximum difference = Upper bound of A − Lower bound of B
- Minimum difference = Lower bound of A − Upper bound of B
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Question: A rectangle has length 15 cm and width 9 cm (both to the nearest cm). Find the bounds for the difference.
- Step 1: 15 cm: Lower = 14.5 cm, Upper = 15.5 cm. 9 cm: Lower = 8.5 cm, Upper = 9.5 cm
- Step 2: Maximum difference = 15.5 − 8.5 = 7 cm; Minimum difference = 14.5 − 9.5 = 5 cm
- Answer: 5 ≤ difference < 7
Multiplication - Maximum and Minimum Results
- Maximum product = Upper bound of A × Upper bound of B
- Minimum product = Lower bound of A × Lower bound of B
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Question: A rectangle has length 8 cm and width 5 cm (both to the nearest cm). Find the bounds for the area.
- Step 1: 8 cm: Lower = 7.5 cm, Upper = 8.5 cm. 5 cm: Lower = 4.5 cm, Upper = 5.5 cm
- Step 2: Maximum area = 8.5 × 5.5 = 46.75 cm²; Minimum area = 7.5 × 4.5 = 33.75 cm²
- Answer: 33.75 ≤ area < 46.75 cm²
Division - Maximum and Minimum Results
- Maximum result = Upper bound of A ÷ Lower bound of B
- Minimum result = Lower bound of A ÷ Upper bound of B
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Question: Distance = 120 km (to the nearest 10 km), Time = 2 hours (to the nearest hour). Find the bounds for average speed.
- Step 1: 120 km: Lower = 115 km, Upper = 125 km. 2 hours: Lower = 1.5 h, Upper = 2.5 h
- Step 2: Maximum speed = 125 ÷ 1.5 = 83.33... km/h; Minimum speed = 115 ÷ 2.5 = 46 km/h
- Answer: 46 ≤ speed < 83.33 km/h
Summary Table of Operations
| Operation | Maximum Result | Minimum Result |
|---|---|---|
| A + B | Upper(A) + Upper(B) | Lower(A) + Lower(B) |
| A - B | Upper(A) - Lower(B) | Lower(A) - Upper(B) |
| A × B | Upper(A) × Upper(B) | Lower(A) × Lower(B) |
| A ÷ B | Upper(A) ÷ Lower(B) | Lower(A) ÷ Upper(B) |
Truncation vs Rounding
- Rounding: the actual value can be above or below the rounded number.
- Truncation (cutting off): the actual value is ONLY equal to or above the truncated number.
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Question: A number is truncated to 4.7. Find the bounds.
- Step 1: Truncation only cuts digits off, it never rounds up, so the lower bound is exactly 4.7.
- Step 2: The upper bound is 4.8, but not including 4.8.
- Answer: 4.7 ≤ actual value < 4.8
Worked Example - Complete Problem
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