9.2 Ratio, Proportion & Variation Techniques

9.2 Ratio, Proportion & Variation Techniques

Ratio and proportion represent the absolute cornerstone of quantitative reasoning in banking examinations. Nearly every domain—from partnership profit-sharing and mixture-alligation problems to time-work dynamics, age calculations, and Data Interpretation (DI) bar and pie charts—relies on ratio transformations. Mastering structural ratio manipulations, algebraic proportional theorems, and variation models empowers candidates to dismantle complex word problems within the stringent 20-minute sectional timeframe of the SBI Clerk examination.

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1. Fundamentals and Classifications of Ratios

A ratio is an abstract mathematical comparison of the relative magnitudes of two or more quantities of the exact same kind and expressed in identical units. In the notation $a : b$ (or $\frac{a}{b}$):

• The first term $a$ is designated the antecedent.
• The second term $b$ is designated the consequent ($b \neq 0$).

Scale Invariance Property

Multiplying or dividing both antecedent and consequent by the same non-zero real number $k$ leaves the ratio unchanged: $\frac{a}{b} = \frac{k \times a}{k \times b} = \frac{a / k}{b / k} \qquad (k \neq 0)$

Key Ratio Nomenclature and Types

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2. Proportion Principles & Proportional Terms

When two ratios are equivalent, the four participating quantities are said to be in proportion. This relationship is denoted as: $a : b :: c : d \quad \iff \quad \frac{a}{b} = \frac{c}{d}$ Here, $a$ and $d$ are the extremes, while $b$ and $c$ are the means.

The Fundamental Law of Proportions

$\text{Product of Extremes} = \text{Product of Means} \implies a \times d = b \times c$

Proportional Terms Reference

1. Fourth Proportional: Given three quantities $a, b, c$, their fourth proportional $x$ satisfies $a : b = c : x$: $x = \frac{b \times c}{a}$

2. Third Proportional: Given two quantities $a$ and $b$, their third proportional $x$ satisfies continued proportion $a : b = b : x$: $x = \frac{b^2}{a}$

3. Mean Proportional (Geometric Mean): The mean proportional $x$ between two quantities $a$ and $b$ satisfies $a : x = x : b$: $x^2 = a \times b \implies x = \sqrt{a \times b}$ SBI Clerk Example: The mean proportional between 0.32 and 0.02 is $\sqrt{0.32 \times 0.02} = \sqrt{0.0064} = 0.08$.

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3. Algebraic Transformation Rules: Componendo & Dividendo

In competitive exams, algebraic fraction equations can be solved instantaneously using classical proportion theorems.

Transformation Catalog

Given $\frac{a}{b} = \frac{c}{d}$:

• Invertendo: $\frac{b}{a} = \frac{d}{c}$
• Alternando: $\frac{a}{c} = \frac{b}{d}$
• Componendo: $\frac{a + b}{b} = \frac{c + d}{d}$
• Dividendo: $\frac{a - b}{b} = \frac{c - d}{d}$
• Componendo and Dividendo (C&D): $\frac{a + b}{a - b} = \frac{c + d}{c - d}$

[!NOTE] The Componendo-Dividendo Shortcut in Action:
When an equation exhibits symmetric linear sums and differences such as $\frac{\sqrt{x + 5} + \sqrt{x - 5}}{\sqrt{x + 5} - \sqrt{x - 5}} = \frac{3}{2}$, apply Componendo & Dividendo directly: $\frac{(\sqrt{x+5} + \sqrt{x-5}) + (\sqrt{x+5} - \sqrt{x-5})}{(\sqrt{x+5} + \sqrt{x-5}) - (\sqrt{x+5} - \sqrt{x-5})} = \frac{3 + 2}{3 - 2} \implies \frac{2\sqrt{x+5}}{2\sqrt{x-5}} = \frac{5}{1}$ Squaring both sides yields $\frac{x + 5}{x - 5} = 25$. Applying C&D once more: $\frac{x}{5} = \frac{25 + 1}{25 - 1} = \frac{26}{24} = \frac{13}{12} \implies x = \frac{65}{12}$ This method bypasses radical rationalization and quadratic expansion.

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4. Techniques for Combining Ratios

A frequent challenge in SBI Clerk word problems is unifying distinct pair ratios ($A : B$ and $B : C$) into a single unified continuum ($A : B : C$ or $A : B : C : D$).

Method 1: The Inverted-N Multiplier

To combine $A : B = a : b$ and $B : C = c : d$:

1. Multiply vertically downwards on the left: $A = a \times c$
2. Multiply diagonally upwards: $B = b \times c$
3. Multiply vertically downwards on the right: $C = b \times d$

$A : B : C = (a \times c) : (b \times c) : (b \times d)$

Worked Illustration: If $A : B = 3 : 4$ and $B : C = 8 : 9$:

• $A = 3 \times 8 = 24$
• $B = 4 \times 8 = 32$
• $C = 4 \times 9 = 36$
• Simplifying by dividing by 4: $A : B : C = 6 : 8 : 9$.

Method 2: The Grid Alignment Method (For 4-Variable Ratios $A:B:C:D$)

When combining $A:B = 2:3$, $B:C = 4:5$, and $C:D = 6:7$, write each ratio on a new row and fill empty lateral slots with the nearest adjacent value:

Dividing through by 3 gives the reduced ratio: $A : B : C : D = 16 : 24 : 30 : 35$.

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5. Direct, Inverse, and Joint Variation

Variation models describe how mathematical variables react to mutual adjustments.

1. Direct Variation ($y \propto x$):
   $y = kx \implies \frac{y}{x} = k$ (constant). If $x$ increases, $y$ increases proportionally. Exemplified by simple purchase cost: total cost varies directly with quantity bought.

2. Inverse Variation ($y \propto \frac{1}{x}$):
   $y = \frac{k}{x} \implies x \times y = k$ (constant). If $x$ increases, $y$ decreases such that their product remains fixed. This governs speed-time problems (distance fixed) and men-days work schedules (work output fixed): $M_1 \times D_1 = M_2 \times D_2$

3. Joint Variation:
   When variable $z$ varies directly with $x$ and inversely with $y$: $z \propto \frac{x}{y} \implies z = k \frac{x}{y} \implies \frac{z_1 y_1}{x_1} = \frac{z_2 y_2}{x_2}$

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6. Coin and Currency Denomination Framework

A staple of the SBI Clerk Quantitative section is the coin-denomination problem. Candidates must avoid confusing the ratio of the number of coins with the ratio of their monetary values.

The Governing Vector Equation

$\text{Total Monetary Value (Rs.)} = \sum_{i=1}^{n} \Big[\text{Denomination}_i \text{ (in Rs.)} \times \text{Number of Coins}_i\Big]$

Denomination Conversions

• 50 Paise coin $= \text{Rs. } 0.50 = \text{Rs. } \frac{1}{2}$
• 25 Paise coin $= \text{Rs. } 0.25 = \text{Rs. } \frac{1}{4}$
• 10 Paise coin $= \text{Rs. } 0.10 = \text{Rs. } \frac{1}{10}$

Fully Worked Banking Cash Box Problem

Problem: A cash collection bag at an SBI branch contains Rs. 400 in coins consisting of Rs. 1, 50 paise, and 25 paise coins in the numerical ratio of $4 : 5 : 6$. Determine the exact number of 50 paise coins contained in the bag.

1. Assign Variable Multipliers to Coin Counts:
   Let the number of Rs. 1 coins $= 4x$, 50 paise coins $= 5x$, and 25 paise coins $= 6x$.
2. Express Monetary Values in Rupee Terms:
   • Value from Rs. 1 coins $= 4x \times 1 = 4x$
   • Value from 50 paise coins $= 5x \times 0.50 = 2.5x$
   • Value from 25 paise coins $= 6x \times 0.25 = 1.5x$
3. Sum Total Monetary Worth:
   $\text{Total Worth} = 4x + 2.5x + 1.5x = 8x$
4. Solve for Multiplier $x$:
   $8x = 400 \implies x = \frac{400}{8} = 50$
5. Calculate Target Coin Count:
   Number of 50 paise coins $= 5x = 5 \times 50 = \mathbf{250 \text{ coins}}$.
   (Verification: $4(50) \times 1 + 5(50) \times 0.50 + 6(50) \times 0.25 = 200 + 125 + 75 = \text{Rs. } 400$.)