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IB MYP 4-5 Maths- Upper and lower bounds – Study Notes

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

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:

  1. Determine the level of accuracy (nearest unit, 0.1, whole number, etc.)
  2. Calculate \( \frac{\text{accuracy}}{2} \)
  3. $\text{Lower Bound = Rounded Value − Half of Accuracy}$
  4. $\text{Upper Bound = Rounded Value + Half of Accuracy}$

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

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