Limits of accuracy
Section: Number 2 | Syllabus: Cambridge IGCSE Mathematics (0580)
Upper and Lower Bounds
Once a value has been rounded, the true value could be anywhere within half a unit of the rounded figure. These two edges are called the upper bound and the lower bound.
- A rounded value could be anywhere from half a unit below to half a unit above the value given
- The lower bound is the smallest value that would round to the given value
- The upper bound is the smallest value that would round to the next value up - so the upper bound itself is not actually included
- This is written with a strict inequality: lower bound true value < upper bound
Worked Example: Finding Bounds
- Question: A length is measured as 8 cm, correct to the nearest cm. Give the lower and upper bounds.
- Step 1: Half of the rounding unit (1 cm) is 0.5 cm
- Step 2: Lower bound: 8 - 0.5 = 7.5. Upper bound: 8 + 0.5 = 8.5
- Answer: 7.5 length < 8.5
- Question: A mass is measured as 6.4 kg, correct to the nearest 0.1 kg. Give the lower and upper bounds.
- Step 1: Half of the rounding unit (0.1 kg) is 0.05 kg
- Answer: 6.35 mass < 6.45
Common Mistakes
MistakeUsing ⩽ for both bounds, e.g. writing 7.5 ⩽ length ⩽ 8.5
Fixthe upper bound uses a strict <, since a value equal to the upper bound would actually round up to the next figure: 7.5 ⩽ length < 8.5
MistakeUsing the wrong size of "half a unit" when the rounding accuracy is a decimal, e.g. using ± 0.5 for a value rounded to the nearest 0.1
Fixalways halve the actual rounding unit stated in the question - for the nearest 0.1, half a unit is 0.05, not 0.5
Bounds of a Calculated Quantity
Extended Only
When rounded values are combined in a calculation, the result also has a range of possible values. Choosing the right combination of upper and lower bounds for each part is the key skill.
- To find the maximum possible result of a sum or product, use the upper bound of every value involved
- To find the minimum possible result of a sum or product, use the lower bound of every value involved
- For a division, the numerator and denominator need opposite treatment:
- To maximise a quotient: (upper bound of numerator) ÷ (lower bound of denominator)
- To minimise a quotient: (lower bound of numerator) ÷ (upper bound of denominator)
Worked Example: Maximising an Area
- Question: A rectangle has length 12 cm and width 5 cm, both correct to the nearest cm. Find the upper bound of the area.
- Step 1: Find the bounds of each side: 11.5 l < 12.5, 4.5 w < 5.5
- Step 2: To maximise the area (a product), use the upper bound of both sides: 12.5 × 5.5 = 68.75
- Answer: 68.75 cm²
Worked Example: Minimising a Quotient
- Question: A cuboid has length 3 cm and width 6 cm, both correct to the nearest integer. Its volume is 68.3 cm³, correct to 1 decimal place. Find the lower bound of the height.
- Step 1: Find the bounds of each given value: 2.5 l < 3.5, 5.5 w < 6.5, 68.25 V < 68.35
- Step 2: Height is volume divided by area, h = (V/l × w) - to minimise this quotient, use the lower bound of V and the upper bound of l and w
- Step 3: lower bound of h = (68.25/3.5 × 6.5) = (68.25/22.75) = 3
- Answer: 3 cm
Common Mistakes
MistakeUsing the upper bound of the denominator to maximise a quotient, e.g. using the upper bound of both l and w when minimising the height above
Fixa larger denominator makes a fraction smaller - to minimise a quotient, the denominator needs its upper bound, not its lower bound, while the numerator needs its lower bound
MistakeUsing the same type of bound (all upper, or all lower) for every value in a calculation, regardless of whether that value is being added, multiplied or divided
Fixcheck each value's role in the calculation separately - values being divided by need the opposite bound to values being multiplied or added
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