GCSE Maths

Numbers
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Integers
Integers Quiz
Prime Numbers and Composite Numbers
Positive and Negative Numbers
Order of Operations
Square Numbers and Cube Numbers
The Laws of Indices
Square Roots and Cube Roots
Triangular Numbers
The Fibonacci Sequence
Factors
Multiples
The Highest Common Factor
Fractions
Adding and Subtracting Fractions
Multiplication of Fractions
Division of Fractions
Fractional Indices
Decimal Places
Fractions and Decimals
Expressing Fractions as Decimal Numbers
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Algebra
29 Topics | 1 Quiz
Algebraic Notation
Algebraic Vocabulary
Collecting Like Terms
Substituting Values into Formulae
Rearranging Equations
Algebraic Fractions
Expanding Brackets and Simplifying
Expanding Double Brackets
Factorisation
Factorising Quadratic Expressions
Factorising Quadratics with a Coefficient of x Squared Greater than 1
The Quadratic Formula
The Difference of Two Squares
Subject of Formula
Solving Linear Equations
Quadratic Equations
Completing the Square
Sequences and Nth Term
Arithmetic and Geometric Sequences
Quadratic Sequences
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Ratio, Proportion and Rates of Change
7 Topics
Ratios
Scaling Ratios
Direct Proportion
Inverse Proportion
Reverse Percentages
Units and Conversions
Scale Factors
Geometry and Measures
16 Topics | 3 Quizzes
2D Shapes
2D Shapes Quiz
3D Shapes
3D Shapes Quiz
Types of angles
Angle Properties
Angle in Parallel Lines
Triangles
Triangles Quiz
Angles in Triangles
Polygons
Lines of Symmetry
Bearings
Loci and constructions
Constructing Triangles
Transformations
Enlargement
Plans and elevations
Areas and Volumes in Map Scale Problems
Probability
10 Topics
Introduction to Probability
Relative Frequency
Sample Space
Probability of Single Events
Probability of Combined Events
Probability Trees
Conditional probability
Set Notation
Venn diagrams
The Complement of a Set
Statistics
2 Topics
Population and samples
Representing data
Vectors
Mensuration and Calculation
7 Topics
Area
Volume
Congruence and similarity
Pythagoras’ theorem
Sine and cosine rule
Area of a triangle
Standard units of measure
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Inverse Proportion

GCSE Maths Ratio, Proportion and Rates of Change Inverse Proportion

An inverse proportion is a relationship between two quantities in which one quantity increases as the other decreases. In other words, the two values change in opposite directions, with the product of their values remaining constant.

When two variables are inversely proportional, we can use the proportionality symbol to represent their relationship. If the variable x is inversely proportional to the variable y, we write x \propto \frac{1}{y}. To convert this relationship into an equation, we introduce a constant of proportionality, denoted by k. The resulting equation for an inverse proportion is xy = k.

To identify an inverse proportion from the information you are given, look for a relationship in which the product of the two variables remains constant as one variable increases and the other decreases. In other words, if the product of the two variables is always the same value, then they are inversely proportional.

Solving Inverse Proportion Problems

Finding the constant of proportionality

To find the constant of proportionality, use the provided information about the relationship between the two variables. Plug these values into the inverse proportion equation, xy = k, and solve for the constant, k.

For example, if x = 4 when y = 3, then the constant of proportionality, k, equals 4 \times 3 = 12.

Using the inverse proportion equation

Once you have determined the constant of proportionality, you can use the inverse proportion equation to address problems involving the two variables. Insert the value of k and the known variable value into the equation, and solve for the unknown variable.

For example, consider a situation where x = 6 and the constant of proportionality, k, is 12. To find the value of y, follow these steps:

xy = k

6y = 12

y = \frac{12}{6} y = 2

In this case, when x = 6, the corresponding value of y is 2.

Examples

Example 1:

Suppose the time taken to complete a task is inversely proportional to the number of people working on it. If 3 people can complete the task in 8 hours, how long will it take for 6 people to complete the same task?

Given that the time taken (t) is inversely proportional to the number of people (p) working on the task, we can write the relationship as:

pt = k

We are given that 3 people can complete the task in 8 hours, so we can find the constant k as follows:

p = 3, t = 8

3 \times 8 = k

k = 24

Now we have the constant of proportionality, k = 24. Let’s find out how long it would take for 6 people to complete the task:

p = 6

6t = 24

Now, solve for t:

t = \frac{24}{6}

t = 4

It would take 6 people 4 hours to complete the same task.

Example 2:

The brightness of a light source is inversely proportional to the square of the distance from the source. If the brightness is 100 units at a distance of 2 metres, what will be the brightness at a distance of 5 metres?

Write the inverse proportion equation: y = \frac{k}{x^2}

Using the given information, x = 2 and y = 100:

100 = \frac{k}{2^2}

k = 100 \times 4 = 400

Now, find the brightness at a distance of 5 metres, with x = 5:

y = \frac{400}{5^2} = \frac{400}{25} = 16

The brightness at a distance of 5 metres will be 16 units.

Example 3:

The speed of a car is inversely proportional to the time taken to travel a fixed distance. If a car takes 6 hours to travel the distance at a speed of 30 km/h, how long will it take to travel the same distance at a speed of 60 km/h?

Given that the speed (s) of a car is inversely proportional to the time taken (t) to travel a fixed distance, we can write the relationship as:

st = k

We are given that the car takes 6 hours to travel the distance at a speed of 30 km/h, so we can find the constant k as follows:

s = 30 km/h, t = 6 hours

30 \times 6 = k

k = 180

Now we have the constant of proportionality, k = 180. Let’s find out how long it will take to travel the same distance at a speed of 60 km/h:

s = 60 km/h

60 \times t = 180

Now, solve for t:

t = \frac{180}{60}

t = 3

It would take 3 hours to travel the same distance at a speed of 60 km/h.

Example 4:

Suppose y is inversely proportional to \sqrt{x} and y=10, when x=25. Find the equation connecting x and y, then use it to find the value of y when x=100.

y \propto \dfrac{1}{\sqrt{x}}

To convert this into an equation, we introduce a constant of proportionality, k:

y = \dfrac{k}{\sqrt{x}}

We are given that y = 10 when x = 25. Let’s use this information to find the constant k:

10 = \dfrac{k}{\sqrt{25}}

10 = \dfrac{k}{5}

k = 10 \times 5

k = 50

Now we have the constant of proportionality, k = 50. The equation connecting x and y is:

y = \dfrac{50}{\sqrt{x}}

To find the value of y when x = 100, substitute x into the equation:

y = \dfrac{50}{\sqrt{100}}

y = \dfrac{50}{10}

y = 5

So, when x = 100, y = 5.

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