When dealing with data, we work with bounds of uncertainty. For example, you could estimate to the nearest 10 students to how many seats are in a classroom if we know the class can hold around 30 students. In GCSE Maths, bounds are a very important concept. I will help you find the easiest and best way for working with bounded intervals.
With GCSE Maths, you will be working with two types of data:
1. Discrete data
2. Continuous data
1. Discrete GCSE Maths Bounds
When you are working with discrete data, it means that the data has integer (whole number) values. Examples of discrete data are: counts of students/cars, money or football goals.
For example, if you calculate the number of football goals it will always be a whole number and can never be to a decimal place. It is the same with counting the number of people/students or cars.
Example of Discrete Bounds
If you know that a car park has a capacity of around 150 cars, how many cars could there be to the nearest 20 cars?
The key thing to the question is realising that the data is discrete, so the answer will be in whole numbers.
When a question says “to the nearest” it means that there is some inaccuracy with the exact number it could be. You can workout this range by using the number it gives you, heher 20 and dividing it by 2 as that is how many bounds we have.
Since we have to tell what the range of what the car park capacity is, we do:
20 ÷ 2 (2 because we have 2 bounds – upper and lower) = 10
Therefore, both bounds will be either ± 10
So, the answer is 150 -10 ≤ car park capacity ≤ 150 +10 which is:
140 ≤ car park capacity ≤ 160
2. Continuous GCSE Maths Bounds
When you are working with continuous data, it means that the data can take any range of values within a bound.
Examples of continuous data include: time, length, mass or speed. For example, you could have a measurement of length to any number of decimal places or in fact measurements of any continuous variable such as the 4 mentioned above.
Example of Discrete Bounds
What are the limits of the length 180cm to the nearest cm?
Since we are dealing with length, the numbers can take any value.
This time, we have to “the nearest cm” which means that the limit is 1cm in total.
So we do: 1 ÷ 2 (2 because we have 2 bounds – upper and lower) = 0.5cm
Therefore, both bounds will be either ± 0.5cm
So, the answer is 179.5cm ≤ car park capacity ≤ 180.5cm
Addition of GCSE Maths Bounds
You may have multiple variables and then have to find the maximum/minimum values they take. Let’s say we are interested in finding the maximum/minimum difference between two football players:
Football player one (goalkeeper) has a bound of: 190.5cm ≤ player 1 ≤ 191.5cm
Football player two (midfielder) has a bound of: 165.5cm ≤ player 2 ≤ 166.5cm
The maximum difference is equal to: largest value – lowest value = 191.5cm – 165.5cm = 26cm
The minimum difference is: lowest value – largest value = 190.5cm – 166.5cm = 24cm
Multiplcation of GCSE Maths Bounds
You may have multiple variables and then have to find the maximum/minimum values they take. Let’s say we are interested in finding the maximum/minimum distance travelled:
With distance you have a formula which involves time and speed. This is distance = speed × time
Hence, we have to multiple the speed and time values together.
The speed has the following bound: 47.5mph ≤ speed ≤ 52.5mph
Time has a bound of: 15 minutes ≤ time ≤ 45 minutes
When calculating distances, we need to convert time from minutes to hours into decimal format. We can do this by x minutes ÷ 60.
The maximum distance is equal to: maximum speed × maximum time = 52.5 × (45 ÷ 60) = 39.375 miles
The minimum distance is equal to: minimum speed × minimum time = 47.5 × (15÷ 60) = 11.875 miles
Division of GCSE Maths Bounds
This time we will look at speed instead of distance. This is since speed is calculated using division, while distance is calculated using multiplication.
Speed = distance ÷ time
Distance has a bound of: 30 miles ≤ distance ≤ 40 miles
Time has a bound of: 45 minutes ≤ time ≤ 1 hour 15 minutes
The maximum speed is equal to: maximum distance ÷ minimum time = 40 ÷ (45 ÷ 60) = 53.33 mph
The minimum speed is equal to: minimum distance ÷ maximum time = 30 ÷ ((60+15)÷ 60) = 24 mph
The reason why we do 60+15 is we rewrite 1 hour 15 minutes into minutes and then convert into a decimal.
Summary of Maths Bounds For GCSE
It is important to realise that discrete data is bounded by integer (whole number) values while continuous data has no integer bounds. Then we can work out different calculations given the bounds. Just remember what operation you are carrying out and whether you want a minimum or maximum value. You can adjust the values accordingly to make sure the number is either a minimum or maximum.