Upper and Lower Bound ~ You’ve Got This!
Quick and absolute solution for the bounds! Mastering bounds comes down to one golden rule: find the unit of precision, cut it in half, and apply it. Once you lock in that pattern, you can calculate upper and lower bounds instantly.
How to follow the lesson:
- Step 1. Study and copy the notes. Submit it to your tutor and get remark if possible. Send the feedback.
- Step 2. Work on practice quizzes. Score will be added.
- Step 3. Work on exam style questions. Add your score.
- Step 4. Total score will be sent out to UT and to yourself.
Let’s get started!!!
Stage 1: Study & Copy Notes 📝💡
Please read through these key concepts carefully and copy them into your exercise book before moving forward. 📓
Core Level 🔵
1. Core Concept: Rounding & Degree of Accuracy
In real life, measurements are never 100% exact. When a quantity is rounded (e.g., to the nearest 1 cm, 10 g, or 50 m), the true value lies inside a range bounded by a Lower Bound (LB) and an Upper Bound (UB). 🔍
- Lower Bound (LB) ⬇️: The smallest possible value that rounds up to your number.
- Upper Bound (UB) ⬆️: The maximum cutoff value that rounds down to your number.
⚡ The Golden Rule for Bounds (Works for ANY rounding rule!)
- Step 1: Take the degree of accuracy and divide it by 2. ➗
- Step 2:
- Lower Bound = Given Value − Half-Unit ➖
- Upper Bound = Given Value + Half-Unit ➕
📊 Visual Diagram 1: Number Line Representation for 162 cm (nearest 1 cm)
Note: The solid circle (●) means included (≤), while the hollow circle (◯) means excluded (<).
💡 Inequality Notation Rule:
Always write the range as 161.5 ≤ height < 162.5.
• Lower bound uses ≤ because 161.5 rounds UP to 162.
• Upper bound uses < because 162.5 would round up to 163! 🛑
Always write the range as 161.5 ≤ height < 162.5.
• Lower bound uses ≤ because 161.5 rounds UP to 162.
• Upper bound uses < because 162.5 would round up to 163! 🛑
Extended Level 🟠
2. Extended Concept: Combining Bounds in Calculations 🧮
When performing arithmetic operations with rounded values, choose extreme combinations to find maximum or minimum outcomes:
📊 Visual Diagram 2: Extended Arithmetic Combination Matrix
🎯 Extended Worked Example (Speed = Distance ÷ Time):
Distance D = 150 km (nearest 10 km) ➔ LB = 145 km, UB = 155 km
Time T = 2 hours (nearest 0.1 hr) ➔ LB = 1.95 hrs, UB = 2.05 hrs
To calculate Maximum Speed, take the Largest Distance divided by the Smallest Time:
Max Speed = UB of Distance ÷ LB of Time = 155 ÷ 1.95 = 79.49 km/h (2 d.p.) 🚗💨
Distance D = 150 km (nearest 10 km) ➔ LB = 145 km, UB = 155 km
Time T = 2 hours (nearest 0.1 hr) ➔ LB = 1.95 hrs, UB = 2.05 hrs
To calculate Maximum Speed, take the Largest Distance divided by the Smallest Time:
Max Speed = UB of Distance ÷ LB of Time = 155 ÷ 1.95 = 79.49 km/h (2 d.p.) 🚗💨
Tutor Verification 👨🏫
Have you completed copying these notes into your exercise notebook?
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