大象传媒

Apply and interpret limits of accuracy - Higher

When more than one value in a calculation has been rounded, the upper or lower bounds for each value should be used in the calculation to find the greatest or least possible values that the calculation can take.

Example

A bag of flour has a mass of 1500 g (to the nearest 10 g). Find the maximum mass of 10 bags of flour.

The upper bound of 1500 g rounded to the nearest 10 g is 1505 g. The upper bound for 10 bags of flour is therefore \(10 \times 1505~\text{kg} = 15050~\text{g}\) ( = 15.05 kg).

Example

Jack is 1.8 m tall (rounded to the nearest 10 cm). Ella is 1.63 m tall (rounded to the nearest cm). What is the smallest possible difference in their heights?

To find the smallest difference, we need to use the lower bound for Jack鈥檚 height and the upper bound for Ella鈥檚 height. The lower bound of 1.8 m (rounded to the nearest 10 cm) is 1.75 m. The upper bound of 1.63 m (rounded to the nearest cm) is 1.635 m.

The smallest possible difference in Jack and Ella鈥檚 height is \(1.75~\text{m} - 1.635~\text{m} = 0.115~\text{m} = 11.5~\text{cm}\)

Example

A piece of A4 paper is 21.1 cm by 29.7 cm measured to the nearest 0.1cm. What are the lower and upper bounds of its area, in square centimetres, correct to 1 dp?

This problem requires the biggest and smallest areas possible. What are the upper and lower bounds of the measurements and how should they be combined to achieve the correct outcomes?

In this example, the maximum area is found by multiplying the two upper bounds. The minimum area is found by multiplying the two lower bounds.

Maximum area

\(21.15~\text{cm} \times 29.75~\text{cm} = 629.2125~\text{cm}^2\)

\(629.2125~\text{cm}^2 \approx 629.2~\text{cm}^2 \:\text{(1 dp)}\)

Minimum area

\(21.05~\text{cm} \times 29.65~\text{cm} = 624.1325~\text{cm}^2\)

\(624.1325~\text{cm}^2 \approx 624.1~\text{cm}^2 \:\text{(1 dp)}\)

The following rules help to decide which bounds to use when doing combinations and calculations.

OperationRule
Adding\(\text{Upper bound} + \text{upper bound} = \text{upper bound}\)\(\text{Lower bound} + \text{lower bound} = \text{lower bound}\)
Subtracting\(\text{Upper bound} - \text{lower bound} = \text{upper bound}\)\(\text{Lower bound} - \text{upper bound} = \text{lower bound}\)
Multiplying\(\text{Upper bound} \times \text{upper bound} = \text{upper bound}\)\(\text{Lower bound} \times \text{lower bound} = \text{lower bound}\)
Dividing\(\text{Upper bound} \div \text{lower bound} = \text{upper bound}\)\(\text{Lower bound} \div \text{upper bound} = \text{lower bound}\)
Adding
Rule\(\text{Upper bound} + \text{upper bound} = \text{upper bound}\)\(\text{Lower bound} + \text{lower bound} = \text{lower bound}\)
Subtracting
Rule\(\text{Upper bound} - \text{lower bound} = \text{upper bound}\)\(\text{Lower bound} - \text{upper bound} = \text{lower bound}\)
Multiplying
Rule\(\text{Upper bound} \times \text{upper bound} = \text{upper bound}\)\(\text{Lower bound} \times \text{lower bound} = \text{lower bound}\)
Dividing
Rule\(\text{Upper bound} \div \text{lower bound} = \text{upper bound}\)\(\text{Lower bound} \div \text{upper bound} = \text{lower bound}\)

Question

A = 34 cm to the nearest cm.

B = 11.2 cm to 1 decimal place.

C = 200 cm to 1 significant figure.

Calculate:

  1. the maximum value for \(A + B\)
  2. the minimum value for \(C - B\)
  3. the minimum value for \(A \times C\)
  4. the maximum value for \(C \div B\)