Ratio and Proportion: Formulas, Tricks, and Worked Examples

Ratio and proportion appear across every major placement aptitude section: profit and loss, speed-distance-time, mixtures, and age problems all reduce to ratio manipulation once you see the structure.

Ratio: Core Rules and Types

A ratio compares two quantities of the same kind. In the ratio a:b, the first term (a) is the antecedent and the second (b) is the consequent. Ratios are dimensionless. Both quantities must share the same unit before forming a ratio.

Ratio type	Definition	Example
Duplicate	a²:b²	3:4 becomes 9:16
Triplicate	a³:b³	2:3 becomes 8:27
Sub-duplicate	√a:√b	9:16 becomes 3:4
Sub-triplicate	∛a:∛b	8:27 becomes 2:3
Compound	(a×c):(b×d)	(2:3)×(5:4) = 10:12 = 5:6

Adding or subtracting the same positive constant x to both terms shifts the ratio toward 1:

• If a > b, then (a+x):(b+x) is less than a:b.
• If a < b, then (a+x):(b+x) is greater than a:b.
• If a = b, the ratio stays 1:1 regardless of x.

This rule appears in questions that ask whether a ratio increases or decreases after a fixed addition to both terms.

Proportion and Its Four Properties

When a:b = c:d, the four quantities are in proportion. The fundamental rule is the product of means equals the product of extremes: a × d = b × c. This cross-product identity is the primary tool for finding an unknown fourth term.

Four derived operations apply to any valid proportion a/b = c/d:

Operation	What it gives	When to use it
Invertendo	b/a = d/c	Numerator and denominator swapped on both sides
Alternendo	a/c = b/d	Middle two terms exchanged
Componendo	(a+b)/b = (c+d)/d	Denominator added to numerator on each side
Dividendo	(a-b)/b = (c-d)/d	Denominator subtracted from numerator
Componendo-dividendo	(a+b)/(a-b) = (c+d)/(c-d)	Sum divided by difference on each side

Componendo-dividendo is the one that most students skip memorising and then need at the worst moment. Worked Example 7 below shows exactly where it saves two steps.

Two other proportion forms that appear in placement tests:

• Direct proportion: x increases as y increases at the same rate. Formula: x/y = k (constant).
• Inverse proportion: x increases as y decreases. Formula: x times y = k.

Continued Proportion and Mean Proportional

Three numbers a, b, c are in continued proportion when a:b = b:c, which means b² = a×c. Four derived quantities follow from this:

• Mean proportional between a and c: b = √(a×c)
• Third proportional to a and b (i.e., c in a:b = b:c): c = b²/a
• Fourth proportional to a, b, c (i.e., d in a:b = c:d): d = (b×c)/a
• Continued proportion chain: a:b = b:c = c:d means each term is the geometric mean of its neighbours.

A typical problem reads: “Find a number such that the ratio of 4 to that number equals the ratio of that number to 25.” Set up b = √(4 × 25) = √100 = 10.

Speed Tricks for Placement Tests

Four setups that cut time per problem in a timed aptitude round.

Ratio chain linking. To combine a:b and b:c into a:b:c, scale until the shared term matches. If a:b = 2:3 and b:c = 4:5, scale a:b to 8:12 and b:c to 12:15, giving a:b:c = 8:12:15.

Division shortcut. To divide total T in ratio m:n, the first share is T × m/(m+n). Sum the parts once, then multiply. For a three-way split m:n:p, the same logic extends: first share = T × m/(m+n+p).

Cross-multiplication comparison. Is a/b larger than c/d? Compute a×d and b×c. If a×d > b×c, then a/b > c/d. Two multiplications, no fraction arithmetic.

Proportional liquid formula. When a container holds x litres and y litres are replaced with water in n rounds, the pure liquid remaining is x × (1 - y/x)^n litres. This appears in mixtures problems where dilution happens in repeated stages.

For a timed drill, IndiaBix’s Ratio and Proportion practice set provides over 100 questions sorted by difficulty, which is a practical follow-up to working through the formulas above.

Worked Examples

These seven problems cover the question types that appear most in placement screens. All answers are re-derived from first principles. For test context, TCS NQT’s Numerical Ability section draws from this exact topic cluster alongside percentages and time-work.

Example 1: Dividing a Sum in a Ratio

• Q: Divide ₹840 among A, B, and C in the ratio 3:4:5.
• Total parts: 3 + 4 + 5 = 12.
• Unit value: 840 / 12 = 70.
• Answer: A gets 3 × 70 = ₹210; B gets 4 × 70 = ₹280; C gets 5 × 70 = ₹350.
• Check: 210 + 280 + 350 = 840. Correct.

Example 2: Income Split with Equal Savings

• Q: A and B each save ₹6,000 per month. Their incomes are in ratio 5:4 and their expenses are in ratio 3:2. Find each person’s income.
• Let incomes = 5k and 4k; expenses = 3m and 2m.
• Equation (i): 5k - 3m = 6,000.
• Equation (ii): 4k - 2m = 6,000, which simplifies to 2k - m = 3,000, so m = 2k - 3,000.
• Substitute into (i): 5k - 3(2k - 3,000) = 6,000, giving 5k - 6k + 9,000 = 6,000, so k = 3,000.
• Answer: A’s income = 5 × 3,000 = ₹15,000. B’s income = 4 × 3,000 = ₹12,000.
• Verify: A’s expenses = 3 × 3,000 = 9,000; saves 15,000 - 9,000 = 6,000. Correct.

Example 3: Mixture with Added Water

• Q: A container has milk and water in ratio 5:2. After 14 litres of water are added, the ratio becomes 5:4. Find the original total volume.
• Let milk = 5k litres and water = 2k litres.
• After adding: 5k / (2k + 14) = 5/4.
• Cross-multiply: 20k = 10k + 70, so 10k = 70, giving k = 7.
• Answer: Original total = 7k = 7 × 7 = 49 litres.
• Verify: 35 milk, 14 water; add 14 water; 35:28 = 5:4. Correct.

Example 4: Age Ratio Problem

• Q: Father’s and son’s ages are in ratio 7:2 today. After 10 years the ratio will be 9:4. Find their current ages.
• Let current ages = 7x (father) and 2x (son).
• Equation: (7x + 10) / (2x + 10) = 9/4.
• Cross-multiply: 28x + 40 = 18x + 90, so 10x = 50, giving x = 5.
• Answer: Father = 35 years. Son = 10 years.
• Verify: (35 + 10):(10 + 10) = 45:20 = 9:4. Correct.

Example 5: Mean Proportional

• Q: Find the mean proportional between 9 and 25.
• Mean proportional = √(9 × 25) = √225 = 15.
• Answer: 15.
• Verify: 9:15 = 15:25 (both equal 3:5). Correct.

Example 6: Third Proportional

• Q: Find the third proportional to 4 and 6.
• Third proportional c satisfies 4:6 = 6:c, so c = (6 × 6)/4 = 36/4 = 9.
• Answer: 9.
• Verify: 4:6 = 2:3 and 6:9 = 2:3. Correct.

Example 7: Componendo-Dividendo Application

• Q: If (5a + 3b)/(5a - 3b) = 7/3, find a:b.
• Apply componendo-dividendo: [(5a + 3b) + (5a - 3b)] / [(5a + 3b) - (5a - 3b)] = (7 + 3)/(7 - 3).
• Left side: 10a / 6b. Right side: 10/4.
• So a/b = (10/4) × (6/10) = 60/40 = 3/2.
• Answer: a:b = 3:2.
• Verify: (5×3 + 3×2)/(5×3 - 3×2) = 21/9 = 7/3. Correct.

Getting the Reps In

Seven examples covers the standard problem types. The next step is exposure to how these types appear in actual screening contexts. The campus placement evaluation test guide breaks down which question categories appear in first-round screens at service and product companies. Time and Work problems follow the same two-step method: assign variables, build the rate equation. For analytics-firm screens that run heavier quant rounds, the Mu Sigma MuApt breakdown shows how multi-step ratio problems appear in that format.

The same define-variables, build-equation habit that resolves a componendo-dividendo problem in two steps is the foundation for structured reasoning in general. TinkerLLM at ₹299 is a live LLM API sandbox where that same constraint-and-test discipline applies to prompt design, which is a natural next step once the aptitude section is solid.