IB MYP 4-5 Maths- Upper and lower bounds- Study Notes - New Syllabus
IB MYP 4-5 Maths- Upper and lower bounds – Study Notes
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- Upper and lower bounds
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Upper and Lower Bounds
Upper and Lower Bounds
In measurements and rounding, we often use approximations. The upper bound and lower bound define the range within which the true value lies. When a number is rounded to a certain degree of accuracy (like to the nearest 10, 0.1, or 2 significant figures), its true value could be slightly more or slightly less. Bounds define that range.
- Lower Bound: The smallest possible value that could round to the given number.
- Upper Bound: The largest possible value that could round to the given number.
These bounds are based on half of the degree of accuracy used in rounding.
For example: If a number is rounded to the nearest 10, the possible values could vary by ±5.
2. How to Find Bounds
To find bounds of a rounded value:
- Determine the level of accuracy (nearest unit, 0.1, whole number, etc.)
- Calculate \( \frac{\text{accuracy}}{2} \)
- $\text{Lower Bound = Rounded Value − Half of Accuracy}$
- $\text{Upper Bound = Rounded Value + Half of Accuracy}$
3. Decimal Place and Significant Figure Accuracy
- Nearest whole number: ±0.5
- 1 decimal place: ±0.05
- 2 decimal places: ±0.005
- 1 significant figure like 300: ±50
Example :
A number is given as 24, correct to the nearest whole number. Find the upper and lower bounds.
▶️ Answer/Explanation
Accuracy: Nearest 1 → Half of 1 = 0.5
Lower Bound = 24 − 0.5 = 23.5
Upper Bound = 24 + 0.5 = 24.5
Answer: 23.5 ≤ x < 24.5
Example :
A measurement is recorded as 3.47 to 2 decimal places. Find its bounds.
▶️ Answer/Explanation
2 decimal places → accuracy = 0.01 → half = 0.005
Lower Bound = 3.47 − 0.005 = 3.465
Upper Bound = 3.47 + 0.005 = 3.475
Answer: 3.465 ≤ x < 3.475
4. Applying Bounds in Calculations
When using measurements in formulas, use the maximum and minimum values (bounds) to estimate the maximum or minimum result.
- For multiplication: $\text{Max value = upper × upper, Min = lower × lower}$
- For division: $\text{Max = upper ÷ lower, Min = lower ÷ upper}$
Example :
A rectangle has length 12.4 cm (to 1 d.p.) and width 6.8 cm (to 1 d.p.). Find upper and lower bounds for the area.
▶️ Answer/Explanation
Example :
A rectangle has a length of 16 cm (measured to the nearest cm) and a width of 7.5 cm (measured to the nearest 0.5 cm). What are the upper and lower bounds for the area and perimeter of the shape?
▶️ Answer/Explanation
Step 1: Find the bounds
- For 16 cm (nearest cm):
Lower Bound $= 16 − 0.5 = 15.5 cm$
Upper Bound $= 16 + 0.5 = 16.5 cm$
- For 7.5 cm (nearest 0.5 cm):
Lower Bound $= 7.5 − 0.25 = 7.25 cm$
Upper Bound $= 7.5 + 0.25 = 7.75 cm$
Step 2: Calculate bounds for area
Maximum area $= 16.5 × 7.75 = 127.875$ cm²
Minimum area $= 15.5 × 7.25 = 112.375$ cm²
Step 3: Calculate bounds for perimeter
Maximum perimeter $= 2 × (16.5 + 7.75) = 2 × 24.25 =$ 48.5 cm
Minimum perimeter $= 2 × (15.5 + 7.25) = 2 × 22.75 =$ 45.5 cm