Approximate measurements - Intermediate & Higher tier - WJECAddition and subtraction

Problems involving addition and subtraction

When solving problems involving upper and lower bounds we have to:

• identify what the question is asking us to calculate
• find either the upper or lower bounds of the measurements
• work out your problem using the correct bounds
• ask yourself if your answer make sense?

Addition

Example

For the rectangle below, calculate the greatest possible value for the perimeter if all the measurements are to the nearest cm.

If we are finding the greatest value for the perimeter we will need the greatest lengths and widths.

We are rounding to the nearest cm so, to find the greatest lengths, we need to go 0.5 cm above.

Maximum length = 5.5 cm

Maximum width = 3.5 cm

Maximum perimeter = 5.5 + 3.5 + 5.5 + 3.5 = 18 cm or

Maximum perimeter = 2 × (5.5 + 3.5) = 18 cm

Subtraction

Example

The Liberty Stadium in Swansea has a capacity of 20,500 people to the nearest 100. On the first home game of the season there was an estimated 18,000 people at the match to the nearest 1,000. What is the maximum number of seats that could possibly have been left empty?

The number of seats was correct to the nearest 100.

Error in measurement = ½ of 100 = 50

The maximum number of seats = 20,500 + 50 = 20,550

The most seats would be left empty if the smallest number of people turned up.

The number of people was correct to the nearest 1,000.

Error in measurement = ½ of 1,000 = 500

The smallest number of people = 18,000 – 500 = 17,500 people.

To find the number of seats left empty we need to subtract the two values.

Maximum number of empty seats = 20,550 – 17,500 = 3,050