Ratio, Proportion & Average

Syllabus Unit: Quantitative Aptitude

Why it matters

Chapter overview

Ratio and proportion is the algebraic backbone behind partnership, mixture/alligation, average and age problems on the SSC CGL Tier-I paper — nearly all of them reduce to setting up one ratio equation and cross-multiplying. This topic builds fluency in combining ratios, the alligation cross-rule for mixtures, weighted averages, and translating age word-problems into a single solvable equation.

Tier-I Examination Syllabus Quantitative Aptitude 25% weightage

Live Animation1 / 4

Ratio & Proportion

1Simplify RatioReduce a:b to lowest terms

2Combine RatiosScale shared term to LCM

3Mean Proportionalb = √(ac)

4Componendo-Dividendo(a+b)/(a−b)=(c+d)/(c−d)

1Ratio & Proportion

A ratio compares magnitude; a proportion equates two ratios — cross-multiplication is the one tool that unlocks partnership, mixture and speed-ratio problems alike.

Learning Objectives

Combine two ratios sharing a common term (a:b and b:c → a:b:c) by scaling to the LCM of the shared term.
Solve direct and inverse proportion word problems by setting up a/b = c/d and cross-multiplying.
Distinguish continued proportion (a:b = b:c) and compute the mean proportional b = √(ac).
Apply the componendo-dividendo shortcut to simplify equations of the form a/b = c/d quickly.

Key Terms

RatioA comparison of two quantities of the same kind, written a:b, equal to the fraction a/b.

ProportionAn equality of two ratios, a:b :: c:d, valid when the product of extremes equals the product of means (ad = bc).

Mean ProportionalFor a continued proportion a:b = b:c, the value b = √(ac) is the mean proportional between a and c.

Componendo-DividendoIf a/b = c/d, then (a+b)/(a−b) = (c+d)/(c−d) — a shortcut for simplifying certain proportion equations.

A ratio is just a fraction in disguise; a proportion says two such fractions are equal. Nearly every SSC 'ratio' word problem reduces to writing one proportion and cross-multiplying.

Combining two ratios

a:b = 2:3 = 8:12, b:c = 4:5 = 12:15 (LCM of 3, 4 is 12) ⟹ a:b:c = 8:12:15

Continued proportion and mean proportional

If a, b, c are in continued proportion (a:b = b:c), then b² = ac, so b = √(ac) — the mean proportional. This appears directly in geometric mean and similar-triangle problems.

a/b = c/d ⟹ (a+b)/(a−b) = (c+d)/(c−d) [componendo-dividendo]

Common MistakeTrap: a:b:c is NOT found by simply writing the two ratios side by side unless the shared term (b) is already equal in both — always scale first.