CIE IGCSE Mathematics (0580) Rates Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Rates Study Notes
LEARNING OBJECTIVE
- Rates
Key Concepts:
- Measures of Rate
- Average speed
Common Measures of Rate
What Is a Rate?
A rate is a ratio comparing two quantities with different units. Common rates compare amounts over time, distance, or quantity.
1. Hourly Rate of Pay
Hourly rate is the amount earned per hour. To find total earnings: \( \text{Earnings} = \text{Hourly Rate} \times \text{Hours Worked} \) To find hourly rate: \( \text{Hourly Rate} = \frac{\text{Total Pay}}{\text{Hours Worked}} \)
2. Exchange Rates Between Currencies
Exchange rates tell you how much one currency is worth in another. To convert: \( \text{Amount in New Currency} = \text{Original Amount} \times \text{Exchange Rate} \) For example, if \$1 = €0.85, then \$100 = €85.
3. Flow Rates
Flow rate measures the volume of fluid that moves per unit of time. Typical units: litres per minute (L/min), millilitres per second (mL/s), etc. Formula: \( \text{Flow Rate} = \frac{\text{Volume}}{\text{Time}} \) or \( \text{Volume} = \text{Flow Rate} \times \text{Time} \)
4. Fuel Consumption
Fuel consumption tells how much fuel is used per distance travelled. Common units: – Litres per 100 kilometres (L/100 km) – Miles per gallon (mpg) Formulas: \( \text{Fuel Used} = \frac{\text{Distance Travelled} \times \text{Rate}}{100} \) or \( \text{Fuel Efficiency} = \frac{\text{Distance}}{\text{Fuel Used}} \)
Important:
- Always check the units — convert them if necessary before using formulas.
- Rates are used in real-world financial, travel, engineering, and science contexts.
- If rates are given per 1 unit (e.g. per hour), simply multiply or divide accordingly.
Example:
A car uses fuel at a rate of 6 L/100 km. How much fuel is needed to travel 350 km?
▶️ Answer/Explanation
Step 1: Use the formula:
\( \text{Fuel Used} = \frac{\text{Distance} \times \text{Rate}}{100} \)
\( = \frac{350 \times 6}{100} = 21 \) litres
Answer: 21 litres
Example:
Jane earns \$234 for working 18 hours. What is her hourly rate of pay?
▶️ Answer/Explanation
Step 1: Use the formula:
\( \text{Hourly Rate} = \frac{\text{Total Pay}}{\text{Hours Worked}} \)
\( = \frac{234}{18} = \$13 \)
Answer: \$13 per hour
Example:
An amount of \$600 is converted to euros at an exchange rate of \$1 = €0.84. How many euros will you receive?
▶️ Answer/Explanation
Step 1: Multiply by the exchange rate:
\( \text{Euros} = 600 \times 0.84 =\)€\(504\)
Answer: €504
Other Measures of Rate
Other Measures of Rate
1. Pressure
Pressure is defined as the force applied per unit area. The formula is: \( \text{Pressure} = \frac{\text{Force}}{\text{Area}} \)
where:
- Force is measured in newtons (N)
- Area is measured in square metres (m²)
- Pressure is measured in pascals (Pa), where \( 1 \, \text{Pa} = 1 \, \text{N/m}^2 \)
2. Density
Density describes how much mass is packed into a given volume. The formula is: \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
where:
- Mass is measured in kilograms (kg) or grams (g)
- Volume is measured in cubic metres (m³) or cubic centimetres (cm³)
- Density is commonly measured in \( \text{kg/m}^3 \) or \( \text{g/cm}^3 \)
3. Population Density
Population density measures how many people live in a unit area of land. The formula is: \( \text{Population Density} = \frac{\text{Population}}{\text{Area}} \)
Typical units: people per square kilometre (people/km²).
Important Tips:
- Always keep units consistent (e.g., convert cm² to m² if needed).
- Use the triangle method for rearranging formulas if one variable is missing.
- These measures are widely used in physics, geography, and real-world contexts.
Example:
A small town has a population of 12,000 people and covers an area of \( 15 \, \text{km}^2 \). Calculate the population density.
▶️ Answer/Explanation
Step 1: Use the formula:
\( \text{Population Density} = \frac{12000}{15} = 800 \, \text{people/km}^2 \)
Answer: \( 800 \, \text{people/km}^2 \)
Example:
A cylindrical metal rod has a volume of \( 500 \, \text{cm}^3 \) and is made of a material with a density of \( 7.85 \, \text{g/cm}^3 \). Calculate the mass of the rod in kilograms.
▶️ Answer/Explanation
Step 1: Use the formula:
\( \text{Mass} = \text{Density} \times \text{Volume} = 7.85 \times 500 = 3925 \, \text{g} \)
Step 2: Convert grams to kilograms:
\( 3925 \, \text{g} = \frac{3925}{1000} = 3.925 \, \text{kg} \)
Answer: \( 3.925 \, \text{kg} \)
Example:
A block of metal has a mass of 540 g and a volume of \( 90 \, \text{cm}^3 \). Find its density in \( \text{g/cm}^3 \).
▶️ Answer/Explanation
Step 1: Use the formula:
\( \text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{540}{90} = 6 \, \text{g/cm}^3 \)
Answer: \( 6 \, \text{g/cm}^3 \)
Average Speed
Average Speed
Average speed is a measure of how far something travels over a given period of time. It is calculated using the formula:
\(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \)
Important Notes:
- Distance and time must be in compatible units (e.g. km/h, m/s).
- Convert mixed units like minutes into decimals (e.g. 45 minutes = 0.75 hours).
- Speed is often given in:
- km/h (kilometres per hour)
- m/s (metres per second)
Related Formulas (triangle method):
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}}, \quad \text{Distance} = \text{Speed} \times \text{Time}, \quad \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
Example:
A cyclist travels 45 km in 3 hours 45 minutes. What is their average speed in km/h?
▶️ Answer/Explanation
Step 1: Convert time into decimal hours:
\( 3 \, \text{hours} \, 45 \, \text{minutes} = 3 + \frac{45}{60} = 3.75 \, \text{hours} \)
Step 2: Use the formula for average speed:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{45}{3.75} = 12 \, \text{km/h} \)
Answer: \( 12 \, \text{km/h} \)
Example:
A runner completes a race of 1500 metres in 5 minutes. What is their average speed in metres per second (m/s)?
▶️ Answer/Explanation
Step 1: Convert time to seconds:
\( 5 \, \text{minutes} = 5 \times 60 = 300 \, \text{seconds} \)
Step 2: Use the formula for average speed:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1500}{300} = 5 \, \text{m/s} \)
Answer: \( 5 \, \text{m/s} \)