Ratio, Proportion and Variation Problems Solved

Ratio, proportion and variation questions are some of the most practical problems in algebra. They appear in Matric and O Level papers, entry tests, and everyday life, from sharing money to working out speeds and costs. This guide covers each idea clearly and solves the typical exam problems step by step.

Ratio

A ratio compares two quantities of the same kind. The ratio of \(a\) to \(b\) is written \(a : b\) or \(\frac{a}{b}\). A ratio has no units, so both quantities must be in the same unit before comparing.

For example, the ratio of 50 paisa to Rs 2 is \(50 : 200 = 1 : 4\).

Ratios are simplified like fractions: \(18 : 24 = 3 : 4\) (dividing both by 6).

Dividing a Quantity in a Given Ratio

To divide an amount in the ratio \(a : b : c\):

1. Add the parts: \(a + b + c\).
2. Find the value of one part: total ÷ sum of parts.
3. Multiply each ratio term by the value of one part.

Worked Example 1: Sharing Money

Divide Rs 4500 among Ahmed, Bilal and Hina in the ratio \(2 : 3 : 4\).

Step 1: Total parts \(= 2 + 3 + 4 = 9\).

Step 2: One part \(= \frac{4500}{9} = 500\).

Step 3: Each share:

• Ahmed: \(2 \times 500 = \text{Rs } 1000\)
• Bilal: \(3 \times 500 = \text{Rs } 1500\)
• Hina: \(4 \times 500 = \text{Rs } 2000\)

Check: \(1000 + 1500 + 2000 = 4500\). Correct.

Proportion

A proportion is a statement that two ratios are equal:

$$ a : b :: c : d \quad \text{means} \quad \frac{a}{b} = \frac{c}{d} $$

Here \(a\) and \(d\) are the extremes and \(b\) and \(c\) are the means. The key property is:

$$ \text{product of extremes} = \text{product of means} \qquad ad = bc $$

Important Terms

• Fourth proportional: \(d\) in \(a : b :: c : d\), so \(d = \frac{bc}{a}\).
• Third proportional: \(c\) in \(a : b :: b : c\), so \(c = \frac{b^2}{a}\).
• Mean proportional: \(m\) in \(a : m :: m : b\), so \(m = \pm\sqrt{ab}\) (usually the positive value is taken).

Worked Example 2: Proportionals

(a) Find the fourth proportional to 4, 6 and 10.

Set up \(4 : 6 :: 10 : x\):

$$ 4x = 6 \times 10 \quad \Rightarrow \quad x = \frac{60}{4} = 15 $$

(b) Find the third proportional to 3 and 12.

Set up \(3 : 12 :: 12 : x\):

$$ 3x = 144 \quad \Rightarrow \quad x = 48 $$

(c) Find the mean proportional between 4 and 9.

$$ m^2 = 4 \times 9 = 36 \quad \Rightarrow \quad m = 6 $$

Check (c): \(4 : 6 = 2 : 3\) and \(6 : 9 = 2 : 3\). The ratios match.

Variation

Variation describes how one quantity changes when another changes. Every variation problem uses a constant of variation \(k\).

Direct Variation

\(y\) varies directly as \(x\) (written \(y \propto x\)) if

$$ y = kx $$

When \(x\) doubles, \(y\) doubles. Example: cost varies directly with the number of items bought.

Inverse Variation

\(y\) varies inversely as \(x\) (written \(y \propto \frac{1}{x}\)) if

$$ y = \frac{k}{x} \quad \text{or} \quad xy = k $$

When \(x\) doubles, \(y\) halves. Example: time taken varies inversely with speed.

Joint and Combined Variation

• Joint variation: \(z = kxy\), meaning \(z\) varies directly as both \(x\) and \(y\).
• Combined variation: for example, \(z = \frac{kx}{y^2}\), direct with \(x\) and inverse with \(y^2\).

The Four-Step Method

1. Write the variation as an equation with \(k\).
2. Substitute the given values to find \(k\).
3. Write the equation with the value of \(k\).
4. Substitute the new values to find the unknown.

Worked Example 3: Direct Variation With a Square

\(y\) varies directly as \(x^2\), and \(y = 12\) when \(x = 2\). Find \(y\) when \(x = 5\).

Step 1: \(y = kx^2\).

Step 2: \(12 = k(2)^2 = 4k \Rightarrow k = 3\).

Step 3: \(y = 3x^2\).

Step 4: When \(x = 5\): \(y = 3(25) = 75\).

Worked Example 4: Inverse Variation (Workers and Days)

6 workers can build a wall in 15 days. How many days will 10 workers take, working at the same rate?

Step 1: More workers means fewer days, so days \(d\) vary inversely with workers \(w\): \(d = \frac{k}{w}\).

Step 2: \(15 = \frac{k}{6} \Rightarrow k = 90\).

Step 3: \(d = \frac{90}{w}\).

Step 4: When \(w = 10\): \(d = \frac{90}{10} = 9\) days.

Worked Example 5: Combined Variation

\(z\) varies directly as \(x\) and inversely as \(y^2\). If \(z = 6\) when \(x = 3\) and \(y = 2\), find \(z\) when \(x = 5\) and \(y = 4\).

Step 1: \(z = \dfrac{kx}{y^2}\).

Step 2: \(6 = \dfrac{3k}{4} \Rightarrow 3k = 24 \Rightarrow k = 8\).

Step 3: \(z = \dfrac{8x}{y^2}\).

Step 4: \(z = \dfrac{8(5)}{16} = \dfrac{40}{16} = 2.5\).

Common Mistakes

• Comparing quantities in different units. Convert first (e.g. minutes to hours, paisa to rupees).
• Confusing direct and inverse variation. Ask: if one increases, does the other increase (direct) or decrease (inverse)?
• Forgetting to find \(k\) before substituting the new values.
• Missing the power. "Varies as the square of \(x\)" means \(x^2\), not \(x\).

Key Takeaways

• To divide in a ratio, find the value of one part and multiply.
• In a proportion, the product of extremes equals the product of means.
• Direct variation: \(y = kx\); inverse variation: \(y = \frac{k}{x}\).
• Solve every variation problem by finding \(k\) first, then substituting new values.

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