Proportion

1. Overview

Proportion describes how two or more quantities relate to each other. In IGCSE Mathematics, you'll learn to express these relationships algebraically, using a "constant of proportionality" ($k$) to form equations. This allows you to calculate unknown values when one of the related quantities changes. The key is to identify whether the relationship is direct or inverse proportion before applying the correct formula.

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Key Definitions

• Proportional ($\propto$): A symbol indicating that two quantities maintain a constant ratio or product.
• Direct Proportion: As one variable increases, the other increases at a constant rate. The graph is always a straight line through the origin $(0,0)$.
• Inverse Proportion: As one variable increases, the other decreases. Their product remains constant.
• Constant of Proportionality ($k$): The fixed value that relates the two variables in a proportion equation.

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Core Content

There are no specific Core-only objectives for this topic; the algebraic representation of proportion is an Extended curriculum requirement.

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Extended Content (Extended Curriculum Only)

A. Direct Proportion

When $y$ is directly proportional to $x$, we write $y \propto x$. This is expressed as the equation:

$\qquad \boxed{y = kx}$

Method to solve:

1. Set up the equation using $k$.
2. Substitute the given values of $x$ and $y$ to calculate $k$.
3. Rewrite the equation with the numerical value of $k$.
4. Use the new equation to find the unknown value requested in the question.

Worked example 1 — Direct Proportion with Squares

$y$ is directly proportional to the square of $x$. When $x = 3$, $y = 45$. Find $y$ when $x = 5$.

Step 1: Write the algebraic form $y = kx^2$

Step 2: Substitute values to find $k$ $45 = k(3)^2$ $45 = 9k$ $k = \frac{45}{9}$ $k = 5$

Divide both sides by 9

Step 3: State the full formula $y = 5x^2$

Step 4: Solve for the new value $y = 5(5)^2$ $y = 5 \times 25$ $y = 125$

$y = 125$

Worked example 2 — Direct Proportion with Square Roots

$A$ is directly proportional to the square root of $B$. When $B = 16$, $A = 12$. Find $A$ when $B = 81$.

Step 1: Write the algebraic form $A = k\sqrt{B}$

Step 2: Substitute values to find $k$ $12 = k\sqrt{16}$ $12 = 4k$ $k = \frac{12}{4}$ $k = 3$

Divide both sides by 4

Step 3: State the full formula $A = 3\sqrt{B}$

Step 4: Solve for the new value $A = 3\sqrt{81}$ $A = 3 \times 9$ $A = 27$

$A = 27$

B. Inverse Proportion

When $y$ is inversely proportional to $x$, we write $y \propto \frac{1}{x}$. This is expressed as the equation:

$\qquad \boxed{y = \frac{k}{x}}$

Worked example 3 — Inverse Proportion

$P$ is inversely proportional to $Q$. When $P = 10$, $Q = 2$. Find $Q$ when $P = 5$.

Step 1: Write the algebraic form $P = \frac{k}{Q}$

Step 2: Substitute values to find $k$ $10 = \frac{k}{2}$ $k = 10 \times 2$ $k = 20$

Multiply both sides by 2

Step 3: State the full formula $P = \frac{20}{Q}$

Step 4: Solve for the new value $5 = \frac{20}{Q}$ $5Q = 20$ $Q = \frac{20}{5}$ $Q = 4$

$Q = 4$

Worked example 4 — Inverse Proportion with Squares

$R$ is inversely proportional to the square of $S$. When $S = 2$, $R = 9$. Find $R$ when $S = 3$.

Step 1: Write the algebraic form $R = \frac{k}{S^2}$

Step 2: Substitute values to find $k$ $9 = \frac{k}{2^2}$ $9 = \frac{k}{4}$ $k = 9 \times 4$ $k = 36$

Multiply both sides by 4

Step 3: State the full formula $R = \frac{36}{S^2}$

Step 4: Solve for the new value $R = \frac{36}{3^2}$ $R = \frac{36}{9}$ $R = 4$

$R = 4$

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Key Equations

These formulas are not provided on the IGCSE formula sheet and must be memorized.

Relationship	Proportionality Statement	Equation
$y$ is proportional to $x$	$y \propto x$	$y = kx$
$y$ is proportional to $x^n$	$y \propto x^n$	$y = kx^n$
$y$ is proportional to $\sqrt{x}$	$y \propto \sqrt{x}$	$y = k\sqrt{x}$
$y$ is inversely proportional to $x$	$y \propto \frac{1}{x}$	$y = \frac{k}{x}$
$y$ is inversely proportional to $x^n$	$y \propto \frac{1}{x^n}$	$y = \frac{k}{x^n}$

• $k$: Constant of proportionality (can be any real number).
• Units: Ensure all units are consistent before calculating $k$.

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Common Mistakes to Avoid

• ❌ Wrong: Using the direct proportion formula when the relationship is inverse, or vice versa. For example, using $y = kx$ when the question states "$y$ is inversely proportional to $x$". ✓ Right: Carefully read the question to identify whether the relationship is "directly proportional" or "inversely proportional" before writing any equations. Use $y = kx$ for direct proportion and $y = \frac{k}{x}$ for inverse proportion.
• ❌ Wrong: Forgetting to apply the power (square, cube, square root, etc.) to the variable as stated in the question. For example, if $y$ is directly proportional to the square of $x$, using $y = kx$ instead of $y = kx^2$. ✓ Right: Pay close attention to phrases like "square of", "cube of", or "square root of" and ensure you include the correct power in your equation.
• ❌ Wrong: Solving for $k$ correctly but then failing to use that value to answer the question. For example, finding $k=5$ when $y=kx^2$, but not calculating the value of $y$ for a new value of $x$. ✓ Right: Always reread the question after finding $k$ to make sure you've answered the specific question asked (e.g., find $y$ when $x=...$, or find a formula for $y$ in terms of $x$).
• ❌ Wrong: Not squaring the entire variable when it is inside brackets. For example, if $y$ is proportional to $(x+1)^2$, writing $y = kx^2 + 1$ instead of $y = k(x+1)^2$. ✓ Right: If a variable expression is squared, ensure the entire expression is squared, not just the variable itself.

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Exam Tips

• Command Words: Look for "Find a formula for... in terms of...". This means your final answer for that part should look like $y = 5x^2$, not just the value of $k$.
• Calculator Tip: When calculating $k$ for inverse proportion, you will usually multiply the two known values. For direct proportion, you will usually divide them.
• Contexts: Real-world contexts often include:
  • Light intensity ($I \propto \frac{1}{d^2}$)
  • Time taken for a journey vs speed (Inverse)
  • Cost of fabric vs length (Direct)
• Show Your Working: markers often award 1 mark for correctly identifying the relationship (e.g., $y = kx^2$) and 1 mark for finding the correct value of $k$. Even if your final answer is wrong, you can gain these marks.