Welcome to Compound Measures!
Have you ever checked a speedometer in a car, looked at the label on a heavy parcel, or wondered why walking on snow is easier with snowshoes? All of these everyday situations involve compound measures.
A compound measure is simply a measurement made by combining two or more single measures (such as distance, time, mass, or area). Don't worry if this sounds complicated at first — if you can multiply and divide, you already have all the math tools you need to master this topic!
1. Understanding the Big Three Compound Measures
In your GCSE Mathematics exam, there are three key compound measures you need to know inside and out:
• Speed: Combines distance and time.
• Density: Combines mass and volume.
• Pressure: Combines force and area.
Helpful Tip: The unit itself always tells you the formula! For example, speed is measured in miles per hour (\(\text{mph}\) or \(\text{miles/hour}\)). In mathematics, the word per means divide by. So, speed is distance divided by time!
2. Speed, Distance, and Time
Speed is a measure of how fast an object moves over a given amount of time.
The Formula Triangle
You can remember the relationship between Distance (\(D\)), Speed (\(S\)), and Time (\(T\)) using a formula triangle where \(D\) is at the top, and \(S\) and \(T\) are at the bottom:
• \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)
• \(\text{Distance} = \text{Speed} \times \text{Time}\)
• \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
Memory Trick: Remember the phrase "Donkeys Slow Traffic" (\(D\) at the top, \(S\) and \(T\) at the bottom).
Crucial Skill: Converting Time to Decimals
Warning — The Most Common Exam Mistake!
Time is base 60, not base 10. That means \(1\text{ hour and } 30\text{ minutes}\) is not \(1.3\text{ hours}\)!
To convert minutes into decimal hours, divide the minutes by \(60\):
• \(15\text{ minutes} = \frac{15}{60} = 0.25\text{ hours}\)
• \(30\text{ minutes} = \frac{30}{60} = 0.5\text{ hours}\)
• \(45\text{ minutes} = \frac{45}{60} = 0.75\text{ hours}\)
• \(24\text{ minutes} = \frac{24}{60} = 0.4\text{ hours}\)
To turn a decimal back into minutes, multiply the decimal part by \(60\):
• \(0.6\text{ hours} = 0.6 \times 60 = 36\text{ minutes}\)
Worked Example: Calculating Time
Question: A train travels a distance of \(140\text{ km}\) at an average speed of \(56\text{ km/h}\). How long does the journey take in hours and minutes?
Step 1: Write down what you know.
\(\text{Distance} = 140\text{ km}\)
\(\text{Speed} = 56\text{ km/h}\)
Step 2: Choose the correct formula.
\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
Step 3: Calculate.
\(\text{Time} = \frac{140}{56} = 2.5\text{ hours}\)
Step 4: Convert the decimal into minutes.
\(0.5\text{ hours} = 0.5 \times 60 = 30\text{ minutes}\)
Answer: \(2\text{ hours and } 30\text{ minutes}\)
Important Concept: Average Speed
If a journey has multiple parts, you cannot just add the speeds together and divide by two. Instead, always use the total values:
\(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}\)
Key Takeaway: Always check your time units before calculating! Convert minutes to fractions of an hour by dividing by \(60\).
3. Density, Mass, and Volume
Density tells us how tightly packed the matter inside an object is. It is the mass of an object per unit of volume.
Analogy: Imagine a cardboard box filled with feathers versus the exact same box filled with solid lead. The volume is identical, but the lead box has far more mass because lead has a much higher density than feathers!
The Formula Triangle
With Mass (\(M\)) at the top, and Density (\(D\)) and Volume (\(V\)) at the bottom:
• \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\)
• \(\text{Mass} = \text{Density} \times \text{Volume}\)
• \(\text{Volume} = \frac{\text{Mass}}{\text{Density}}\)
Memory Trick: Think of "Massive Dancing Vampires" (\(M\) on top, \(D\) and \(V\) underneath).
Standard Units for Density
• Grams per cubic centimetre: \(\text{g/cm}^3\)
• Kilograms per cubic metre: \(\text{kg/m}^3\)
Worked Example: Finding the Mass of a Metal Block
Question: A solid copper cylinder has a volume of \(45\text{ cm}^3\). The density of copper is \(8.96\text{ g/cm}^3\). Calculate the mass of the cylinder.
Step 1: Identify the known values.
\(\text{Volume} = 45\text{ cm}^3\)
\(\text{Density} = 8.96\text{ g/cm}^3\)
Step 2: Choose the formula for Mass.
\(\text{Mass} = \text{Density} \times \text{Volume}\)
Step 3: Substitute and solve.
\(\text{Mass} = 8.96 \times 45 = 403.2\text{ g}\)
Answer: \(403.2\text{ g}\)
Key Takeaway: Ensure that your units match! If density is in \(\text{g/cm}^3\), your mass must be in grams (\(\text{g}\)) and your volume in \(\text{cm}^3\).
4. Pressure, Force, and Area
Pressure measures how much force is applied over a particular surface area.
Analogy: If someone steps on your foot wearing a flat trainer, it hurts a little. If they step on your foot wearing a sharp stiletto heel, it hurts a lot! The force (the person's weight) is identical, but the stiletto concentrates that force into a tiny area, creating very high pressure.
The Formula Triangle
With Force (\(F\)) at the top, and Pressure (\(P\)) and Area (\(A\)) at the bottom:
• \(\text{Pressure} = \frac{\text{Force}}{\text{Area}}\)
• \(\text{Force} = \text{Pressure} \times \text{Area}\)
• \(\text{Area} = \frac{\text{Force}}{\text{Pressure}}\)
Memory Trick: "Forces Push Anywhere" (\(F\) at the top, \(P\) and \(A\) below).
Units of Pressure
• Newtons per square metre: \(\text{N/m}^2\) (also called Pascals, \(\text{Pa}\))
• Newtons per square centimetre: \(\text{N/cm}^2\)
Worked Example: Calculating Pressure
Question: A heavy crate exerts a downward force of \(900\text{ N}\) on the floor. The rectangular base of the crate measures \(1.5\text{ m}\) by \(2\text{ m}\). Find the pressure exerted on the floor.
Step 1: Calculate the area of the base.
\(\text{Area} = 1.5 \times 2 = 3\text{ m}^2\)
Step 2: State known variables.
\(\text{Force} = 900\text{ N}\)
\(\text{Area} = 3\text{ m}^2\)
Step 3: Apply the pressure formula.
\(\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{900}{3} = 300\text{ N/m}^2\)
Answer: \(300\text{ N/m}^2\) (or \(300\text{ Pa}\))
Key Takeaway: Larger area means lower pressure; smaller area means higher pressure.
5. Other Rates of Change (Flow Rates)
Sometimes GCSE questions include other compound measures, such as rate of flow.
• \(\text{Rate of Flow} = \frac{\text{Volume}}{\text{Time}}\)
• Units can be litres per second (\(\text{litres/s}\)), \(\text{cm}^3\text{/s}\), or gallons per minute.
Example: If a hose fills a \(240\text{ litre}\) paddling pool in \(8\text{ minutes}\), the rate of flow is \(\frac{240}{8} = 30\text{ litres per minute}\).
6. Summary & Exam Checklist
Before tackling exam questions on compound measures, run through this quick checklist:
• Check the units first: Do the distance/speed/time or mass/density/volume units correspond with each other?
• Watch out for time: Never write \(45\text{ mins}\) as \(0.45\text{ hours}\). Divide by \(60\) to get \(0.75\text{ hours}\).
• Use your triangles: Cover up the letter you need to find to see the exact calculation required.
• Include units in your final answer: Even if not explicitly asked, writing the correct unit (e.g., \(\text{km/h}\), \(\text{g/cm}^3\), \(\text{N/m}^2\)) ensures full credit!