CA Foundation Quantitative Aptitude: Ratio, Proportion, Indices and Logarithms

CA Foundation Quantitative Aptitude’s Ratio, Proportion, Indices, and Logarithms form one of the most important and high-scoring areas. This topic builds the mathematical foundation required to solve numerical problems efficiently in the exam, covering concepts such as ratio comparison, proportion laws, alligation and mixture, laws of indices, fractional exponents, and logarithmic operations. A clear understanding of these concepts helps students apply formulas correctly, simplify complex expressions, and improve speed and accuracy in objective-type questions.

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Ratio, Proportion, Indices, and Logarithms in CA Foundation Exam

Ratio, Proportion, Indices, and Logarithms play a crucial role in the CA Foundation Quantitative Aptitude paper, contributing a significant number of objective questions every attempt. These topics test not only conceptual clarity but also calculation speed, logical thinking, and the ability to apply formulas under time pressure. Questions are often interlinked, where ratio concepts merge with indices or logarithms, making conceptual understanding more important than rote learning.

In the CA Foundation exam, problems from this unit are usually moderate in difficulty but time-sensitive, meaning students who understand shortcuts, properties, and transformation techniques can solve them much faster. Common question formats include:

• Finding unknown terms using ratio and proportion laws

• Simplifying expressions involving indices and fractional powers

• Solving logarithmic equations using base change and identities

• Application-based questions from alligation and mixture

Mastering this unit ensures accuracy in direct formula-based questions and confidence in handling multi-step numerical problems.

How to Prepare Ratio, Proportion, Indices, and Logarithms for CA Foundation?

Preparing Ratio, Proportion, Indices, and Logarithms effectively requires a balance between conceptual clarity and consistent practice. Since these topics are interconnected, students should follow a structured preparation strategy rather than studying them in isolation.

Start with ratio fundamentals, ensuring clarity on simplification, compound ratios, continued ratios, and substitution techniques. Once ratios are clear, move to proportion, focusing on laws like componendo, dividendo, and mean proportional, as these are frequently used in problem-solving.

For indices, emphasis should be placed on laws of exponents, fractional indices, and simplification of complex expressions. Students should practice identifying patterns and avoiding common traps, especially in cyclic exponent questions. In logarithms, understanding the conversion between exponential and logarithmic forms, base change formula, and application-based problems is essential.

To score well:

• Practice questions with smart value substitution

• Memorize core identities, not lengthy formulas

• Solve previous year and mock test questions under time limits

• Focus on accuracy first, then speed

A strong grip on this unit can significantly improve overall Quantitative Aptitude performance in CA Foundation.

Introduction to Ratio

A ratio compares two or more quantities, provided they are of the same kind and in the same units. Without these conditions, a formal ratio cannot be established. Quantities are defined as anything measurable. The ratio of A to B is expressed as A:B or A/B, where A is the antecedent and B is the consequent. Ratios are always unitless due to unit cancellation during division.

Definition of Ratio

A ratio is a comparison of two or more quantities. For a comparison to be a ratio, two conditions are essential:

• Quantities must be of the same kind.

• Quantities must be in the same units.

A quantity is anything that can be measured. If two quantities, A and B, meet these conditions, their ratio is A:B or A/B. A is the antecedent (numerator), and B is the consequent (denominator). Ratios are unitless.

Example: Ratio of 2 km to 500 m. Convert 2 km to 2000 m. The ratio is 2000 m / 500 m = 4/1 or 4:1.

Important Properties and Calculations

Simplest Form: A ratio must always be expressed in its simplest form, meaning no common factors exist between the antecedent and consequent.

Example: The ratio 92:56 simplifies to 23:14 (dividing both by 4).

Dividing a Quantity in a Given Ratio: To divide a total quantity into parts according to a ratio (e.g., A:B), sum the ratio parts (Total Parts = A+B).

Amount of A = (Part of A / Total Parts) × Total Quantity

Example: A class of 150 students has boys to girls in a 3:2 ratio. Total parts = 3+2=5. Boys = (3/5) × 150 = 90. Girls = (2/5) × 150 = 60.

Types of Ratios

Given a base ratio A : B:

Commensurable vs. Incommensurable Ratios

Ratios are classified by the nature of the numbers.

Effect of Multiplication and Division on a Ratio

Multiplying or dividing both the antecedent and the consequent of a ratio by the same non-zero number keeps the ratio unchanged.

A / B = (A × k) / (B × k) and A / B = (A ÷ k) / (B ÷ k)

Compound Ratio

The compound ratio is found by multiplying all antecedents and all consequents of two or more ratios.

For A:B, C:D, and E:F, the compound ratio is (A × C × E) : (B × D × F).

Example: Compound ratio of 2:3, 5:6, and 7:8 is (2×5×7) : (3×6×8) = 70:144, which simplifies to 35:72.

Continued Ratio

A continued ratio expresses the relationship between three or more quantities in a single expression (e.g., A:B:C).

Example: Given A:B = 2:3 and B:C = 4:5, find A:B:C.

• A: 2 × 4 = 8

• B: 3 × 4 = 12

• C: 3 × 5 = 15
  The continued ratio A:B:C is 8:12:15.
  (Memory Tip: Use the "Down-Up-Down" method for A:B and B:C: (A*B₂):(B₁*B₂):(B₁*C₂))

Simplifying Ratios with Fractions

To convert a ratio with fractions (e.g., 1/2 : 1/3 : 1/4) to a simple integer ratio:

1. Multiply all denominators: 2 × 3 × 4 = 24.

2. Multiply each term by this product: (1/2)×24 = 12, (1/3)×24 = 8, (1/4)×24 = 6.

3. The ratio is 12:8:6, which simplifies to 6:4:3.

Finding a Combined Ratio for Four Terms (A:B:C:D)

For A:B, B:C, and C:D, sequentially multiply:

• A: First terms of all three ratios (e.g., 2×4×6 = 48)

• B: Second term of first, first of second, first of third (e.g., 3×4×6 = 72)

• C: Second term of first, second of second, first of third (e.g., 3×5×6 = 90)

• D: Second terms of all three ratios (e.g., 3×5×7 = 105)
  (Memory Tip: The method involves a sequence of "downward" multiplications)
  The ratio 48:72:90:105 simplifies to 16:24:30:35.

Substitution in Ratio Expressions

Condition for Direct Substitution

When given a ratio (e.g., a/b = 3/4) and an expression (e.g., (2a + 3b) / (3a + 4b)), you can directly substitute values (a=3, b=4) only if every single term in the expression has the same total power (degree).

• Example 1: For (2a + 3b) / (3a + 4b), all terms (2a, 3b, 3a, 4b) have power 1. Direct substitution (a=3, b=4) gives 18/25.

• Example 2: For (x²y + xy²) / (x³ + y³), all terms (x²y, xy², x³, y³) have total power 3. Direct substitution (x=3, y=4) gives 12/13.

When Direct Substitution is Not Allowed

If terms have different degrees (e.g., (a² + b) / …), direct substitution is invalid. Use a constant of proportionality: a = 3k, b = 4k. The k will not cancel out.

Problem-Solving Strategies for Ratio Questions

Using Options to Verify: For word problems, testing given options can be faster than algebraic solution.

Example: Numbers in ratio 2:3. If 4 subtracted, new ratio 3:5. Check option (16, 24). 16/24 = 2/3 (passes). (16-4)/(24-4) = 12/20 = 3/5 (passes).

Using Smart Value Substitution: For complex algebraic problems, choose simple numerical values that satisfy conditions.

Example: Given 2x/3y is duplicate ratio of (2x-k)/(3y-k). Choose x=2, y=3. ((4-k)/(9-k))² = 4/9. Solving for k yields k=-6, so k²=36.

Introduction to Proportion

Definition of Proportion

When two ratios are equal to each other, they are in proportion. If A, B, C, D are in proportion, written as A:B = C:D or A/B = C/D.

• A: First Proportional, B: Second, C: Third, D: Fourth.

Fundamental Property of Proportions

If A/B = C/D, then A × D = B × C. This is the Product of Extremes = Product of Means.

• Extremes: A and D; Means: B and C.

Continuous Proportion and Mean Proportional

If only three terms A, B, C are in proportion, it's a continuous proportion: A/B = B/C.

This implies B² = A × C. B is the Mean Proportional between A and C.

• Example: Mean proportional to 4 and 9. Let B be the mean. B² = 4 × 9 = 36. So B = 6.

Properties of Proportions (Laws of Proportion)

If A/B = C/D, then:

Applications and Examples of Proportions

• Finding a Missing Term: If 2, 3, x, 9 are in proportion: 2/3 = x/9 implies x = 6.

• Mean Proportional: Find mean proportional between 25 and 81. b² = 25 × 81 implies b = 45.

• Fourth Proportional: Find fourth proportional to 2a, a², c. 2a/a² = c/D implies D = ac/2.

Solving Proportions with Multiple Equal Ratios

The 'k' Method

If a/1 = b/2 = c/5, let them all equal k. Then a=k, b=2k, c=5k. The ratio a:b:c is 1:2:5.

Direct Substitution Shortcut

If a/x = b/y = c/z, then a:b:c = x:y:z. This allows direct substitution (e.g., a=x, b=y, c=z) if all terms in the expression being evaluated have the same degree.

• Example: Given a/3 = b/4 = c/7, find (a+b+c)/c. Substitute a=3, b=4, c=7. (3+4+7)/7 = 14/7 = 2.

Alligation and Mixture

Concept

Alligation finds the ratio in which two or more ingredients of different values should be mixed to achieve a desired mean value.

Core Formula & Terminology

Let C = Cheaper value, D = Dearer value, M = Mean value.

Quantity of Cheaper / Quantity of Dearer = (D - M) / (M - C)

Example: Mix item at ₹80 (C) with item at ₹140 (D) to get a ₹100 (M) mixture.

Ratio (Cheaper:Dearer) = (140 - 100) : (100 - 80) = 40 : 20 = 2:1.

• Losses are treated as negative values in calculations. For 10% profit (D), 5% loss (C), 7% profit (M), the ratio is (10-7):(7-(-5)) = 3:12 = 1:4.

Introduction to Indices

Fundamental Terminology

• Indices / Exponent / Power: In aⁿ, a is the base, n is the power.

• aⁿ means a multiplied by itself n times.

• a⁻ⁿ = 1/aⁿ. (Memory Tip: To change the sign of an exponent, take the reciprocal of the base.)

• Negative base: (-a)ᴱᵛᵉⁿ = Positive, (-a)ᴼᵈᵈ = Negative.

Laws of Indices

• Zero Power Rule: a⁰ = 1 (where a ≠ 0).

• Product Rule (Same Base): aᵐ × aⁿ = aᵐ⁺ⁿ.

• Quotient Rule (Same Base): aᵐ / aⁿ = aᵐ⁻ⁿ.

• Power of a Power Rule: (aᵐ)ⁿ = aᵐⁿ.

• Equality Rule (Same Base): If aᵐ = aⁿ, then m = n (where a ≠ 1).

• Equality Rule (Same Power): If aᵐ = bᵐ, then a = b (where m ≠ 0).

Principle of Equal Exponents

If xⁿ = aⁿ, then x = a. The powers being equal implies the bases are equal.

• Example: x³ = 64. Since 64 = 4³, then x³ = 4³, so x = 4.

Fractional Indices

Definition of Fractional Indices (a¹/ⁿ)

a¹/ⁿ is the n-th root of a, written as ⁿ√a.

• If n is ODD: a can be any real number.

• If n is EVEN: a must be a positive real number.

• a¹/² = √a, a¹/³ = ³√a.

Conceptual Meaning and The Denominator Rule

a¹/ⁿ means finding a number that, when multiplied n times, equals a. The denominator n of the exponent is key. You must rewrite the base as a number raised to the power n.

• (Memory Tip: The fractional exponent ¹/ⁿ tells the base number: "Until you are broken down into a power of 'n', I will not leave you.")

• Example: Simplify 81¹/⁴. Rewrite 81 as 3⁴. Then (3⁴)¹/⁴ = 3¹ = 3.

General Fractional Index (aᵐ/ⁿ)

aᵐ/ⁿ can be ⁿ√(aᵐ) or (ⁿ√a)ᵐ. The denominator n is the root, and the numerator m is the power.

• Example: Simplify (27/8)²/³. Denominator is 3. 27=3³, 8=2³.
  ((3/2)³)²/³ = (3/2)² = 9/4.

Worked Examples and Applications

1. Basic Simplification: (x² * y³ / z⁴)⁴ = x⁸ * y¹² / z¹⁶.
   (x * y⁶ / z¹²)¹/³ = x¹/³ * y² / z⁴.

2. Cyclic Exponents (Caution): xᵃ⁻ᵇ * xᵇ⁻ᶜ * xᶜ⁻ᵃ = x⁰ = 1. Be wary of "trap" questions where the pattern is broken.

3. Advanced Simplification: Nested Roots and Powers: Work from innermost to outermost.
   [ ᵐ√(ⁿ√a) ]ᵐⁿ = a.
   [1 - {1 - (1 - x²)⁻¹}⁻¹]⁻¹/² = x. (Requires step-by-step algebraic simplification.)

4. Cubic Expressions: Based on (p ± 1/p)³ identities.

• If x = p + 1/p, then p³ + 1/p³ = x³ - 3x.

• If x = p - 1/p, then p³ - 1/p³ = x³ + 3x.
  (Memory Tip: Notice the sign inversion in the final expression)

• Example: If x = 5¹/³ - 5⁻¹/³, find 4x³ + 12x. Here p=5¹/³, so x = p - 1/p. Thus x³ + 3x = p³ - 1/p³ = 5 - 1/5 = 24/5.
  4x³ + 12x = 4(x³ + 3x) = 4(24/5) = 96/5.

Solving Exponential Equations of the form aˣ = bʸ = cᶻ

The Fundamental Method

If aˣ = bʸ = cᶻ = k, then a = k¹/ˣ, b = k¹/ʸ, c = k¹/ᶻ.

Use a given condition relating a, b, c (e.g., abc=1) to form an equation of exponents.

• Example: If abc=1, then k^(1/x + 1/y + 1/z) = k⁰ implies 1/x + 1/y + 1/z = 0.

The Shortcut Method

If the relationship between bases is aᵐ * bⁿ = cᵖ, then the relationship between exponents (x, y, z) is m/x + n/y = p/z.

• Example: 2ˣ = 3ʸ = 6⁻ᶻ. Base relation: 2¹ * 3¹ = 6¹. So 1/x + 1/y = 1/(-z). This means 1/x + 1/y + 1/z = 0.

• Example: 4ˣ = 5ʸ = 20ᶻ. Base relation: 4¹ * 5¹ = 20¹. So 1/x + 1/y = 1/z, which means z = xy / (x+y).

Simplifying Exponential Expressions

To simplify (x/y)^(a² + 4) = [ (y/x) ]⁻⁵ᵃ:

1. Harmonize bases: (y/x) = (x/y)⁻¹. So (x/y)^(a² + 4) = (x/y)⁵ᵃ.

2. Equate exponents: a² + 4 = 5a -> a² - 5a + 4 = 0.

3. Solve: (a-1)(a-4) = 0. Solutions are a=1 or a=4.

Introduction to Logarithms

From Exponential to Logarithmic Form

Logarithms are the inverse of exponentiation.

• Exponential Form: aˣ = y (a = base, x = exponent)

• Logarithmic Form: x = logₐ y (read as "x equals log base 'a' of y").
  logₐ y asks: "To what power must base 'a' be raised to get 'y'?"

Restrictions on Logarithms

For logₐ y to be defined:

1. Base a > 0.

2. Base a ≠ 1.

3. Value y > 0.

Examples

• log₅ 25 = 2 (because 5² = 25).

• log₂ 64 = 6 (because 2⁶ = 64).

• logₓ 32 = 5 implies x⁵ = 32, so x = 2.

• log₂ x = 7 implies x = 2⁷ = 128.

Laws of Logarithms and Applications

(Base a assumed same for all terms)

1. Log of the Base: logₐ a = 1.

2. Log of One: logₐ 1 = 0.

3. Value Power Rule: logₐ (xⁿ) = n ⋅ logₐ x.

4. Base Power Rule: logₐⁿ (x) = (1/n) ⋅ logₐ x.

5. Product Rule: logₐ x + logₐ y = logₐ (xy).

6. Quotient Rule: logₐ x - logₐ y = logₐ (x/y).

7. Inverse Property: a^(logₐ x) = x.

8. Common Logarithm: log x implies base 10.

Common Mistakes: log(x+y) is NOT log x + log y.

The Base Change Formula

Two cases:

Applications of the Base Change Formula

Example 1: log_a(b) * log_b(c) = log_a(c).

Example 2: [log_b(a) * log_c(b) * log_a(c)]³ = 1.

Example 3 (Chain Multiplication): log_3(4) * log_4(5) * ... * log_8(9). This simplifies to log_3(9) = 2.

Example 4: log_2(x) + log_4(x) + log_16(x) = 21/4.
Convert bases to 2: log_2(x) + (1/2)log_2(x) + (1/4)log_2(x) = 21/4.
log_2(x) * (1 + 1/2 + 1/4) = 21/4 -> log_2(x) * (7/4) = 21/4.
log_2(x) = 3 -> x = 2³ = 8.

Example 5 (Advanced): 1/(1 + log_a(bc)) + 1/(1 + log_b(ac)) + 1/(1 + log_c(ab)).
Rewrite 1 as log_a(a), etc. The expression becomes log_abc(a) + log_abc(b) + log_abc(c) = log_abc(abc) = 1.

Practice Problems

Problem: log₄(x² + x) - log₄(x + 1) = 2

log₄[ x(x+1)/(x+1) ] = 2 -> log₄(x) = 2 -> x = 4² = 16.

Problem: If log₁₀(2) = x, log₁₀(3) = y, express log₁₀(60).

log₁₀(60) = log₁₀(2 * 3 * 10) = log₁₀(2) + log₁₀(3) + log₁₀(10) = x + y + 1.

Problem: Simplify a^[ log_a(b) * log_b(c) * log_c(d) * log_d(t) ].

The exponent simplifies to log_a(t). The expression becomes a^[log_a(t)] = t.

Problem: Simplify log₅(1 + 1/5) + log₅(1 + 1/6) + ... + log₅(1 + 1/624).

log₅(6/5) + log₅(7/6) + ... + log₅(625/624) = log₅[ (6/5)*(7/6)*...*(625/624) ].

Cancellations leave log₅(625/5) = log₅(125) = 3.

Problem: If log(x) + log(x²) + log(x³) = 14, find log(x⁶). (Assuming log x + 2log x + 3log x = 14).

6log(x) = 14. Thus log(x⁶) = 14. (If coefficients are 1, 4, 9, then 14 log x = 14, so log x = 1. Then log(x⁶) = 6 log x = 6).

Problem: Given log((a+b)/4) = (1/2)log(ab), find a/b + b/a.

log((a+b)/4) = log(√(ab)). So (a+b)/4 = √(ab).

Square both sides: (a+b)²/16 = ab -> a² + b² + 2ab = 16ab.

a² + b² = 14ab.

a/b + b/a = (a² + b²) / ab = 14ab / ab = 14.