4.1 Ratios, Duplicate/Sub-Duplicate Ratios, Direct and Inverse Variations & Proportional Sharing

Fundamental Definitions of Ratio and Proportion

The Mathematical Concept of Ratio

A ratio is an expression of the relative magnitude of two homogeneous quantities of the same kind, expressed in identical units. For two quantities $a$ and $b$ (where $b \neq 0$), the ratio is denoted as:

$$a : b = \frac{a}{b}$$

• The first term $a$ is the antecedent.
• The second term $b$ is the consequent.
• Scale Invariance: Multiplying or dividing both terms of a ratio by any non-zero constant $k$ leaves the ratio unchanged:

$$\frac{a}{b} = \frac{k \times a}{k \times b} = \frac{a / k}{b / k} \quad (k \neq 0)$$

The Mathematical Concept of Proportion

A proportion is an equation stating that two ratios are equal. If $\frac{a}{b} = \frac{c}{d}$, the four quantities $a, b, c, d$ are in proportion, written as:

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

• Terms $a$ and $d$ are called the extremes.
• Terms $b$ and $c$ are called the means.

$$\mathbf{ad = bc \iff \text{Product of Extremes} = \text{Product of Means}}$$

The displayed root-based proportional formulas use positive quantities. All ratio transformations require their denominators to be nonzero, including newly formed differences in componendo/dividendo.

Proportional Types and Formulas

When quantities are in continuous proportion ($a : b :: b : c$), term $b$ serves as the geometric bridge between $a$ and $c$:

Continuous Proportion Flow: a : b = b : c
   Extremes: a, c  ---> Product = a * c
   Means:    b, b  ---> Product = b^2
   Equating: b^2 = ac  ===>  b = sqrt(ac)

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Derived and Compounded Ratios

Railway quantitative questions frequently invoke classical power relationships that transform basic ratios:

Compounded Ratio

When two or more ratios are multiplied term-wise, antecedent by antecedent and consequent by consequent, the resulting ratio is the compounded ratio:

$$\text{For } (a : b), (c : d), \text{ and } (e : f), \quad \mathbf{\text{Compounded Ratio} = (a \times c \times e) : (b \times d \times f)}$$

Power and Root Transformations

1. Duplicate Ratio: The ratio of the squares of the original terms:

$$\text{Duplicate of } (a : b) = \mathbf{a^2 : b^2}$$

1. Sub-Duplicate Ratio: The ratio of the square roots of the terms:

$$\text{Sub-duplicate of } (a : b) = \mathbf{\sqrt{a} : \sqrt{b}}$$

1. Triplicate Ratio: The ratio of the cubes of the terms:

$$\text{Triplicate of } (a : b) = \mathbf{a^3 : b^3}$$

1. Sub-Triplicate Ratio: The ratio of the cube roots of the terms:

$$\text{Sub-triplicate of } (a : b) = \mathbf{\sqrt[3]{a} : \sqrt[3]{b}}$$

1. Inverse (Reciprocal) Ratio: For a two-term ratio $a : b$, the inverse is $\frac{1}{a} : \frac{1}{b} = b : a$. For three terms $a : b : c$, the reciprocal ratio is:

$$\frac{1}{a} : \frac{1}{b} : \frac{1}{c} = \mathbf{bc : ca : ab}$$

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Transformation Theorems: Invertendo, Alternando, Componendo & Dividendo

Given the proportion $\frac{a}{b} = \frac{c}{d}$, algebraic manipulations yield five standard classical transformation identities:

1. Invertendo: Inverting antecedents and consequents:

$$\mathbf{\frac{b}{a} = \frac{d}{c}}$$

1. Alternando: Swapping the inner terms (means):

$$\mathbf{\frac{a}{c} = \frac{b}{d}}$$

1. Componendo: Adding consequents to antecedents:

$$\frac{a + b}{b} = \frac{c + d}{d}$$

1. Dividendo: Subtracting consequents from antecedents:

$$\frac{a - b}{b} = \frac{c - d}{d}$$

1. Componendo and Dividendo (C&D Rule): Dividing the Componendo identity by the Dividendo identity:

$$\mathbf{\frac{a + b}{a - b} = \frac{c + d}{c - d}}$$ $$\frac{2\sqrt{x + p}}{2\sqrt{x - p}} = \frac{k + 1}{k - 1} \implies \frac{\sqrt{x + p}}{\sqrt{x - p}} = \frac{k + 1}{k - 1}$$

Squaring both sides and applying C&D a second time isolates $x$ in seconds, eliminating lengthy polynomial cross-multiplication.

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Multi-Ratio Continuous Bridging ($A:B:C:D$)

In railway allocation logistics, inputs are often quoted in overlapping pairs: for instance, locomotive allocations between Shed A and Shed B ($A : B$), and between Shed B and Shed C ($B : C$). Candidates must bridge these into a single continuous chain $A : B : C$.

The Bridging Mechanics

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

1. Identify the common linking variable (here, $B$).
2. Equate the value of $B$ by scaling both ratios to the Lowest Common Multiple (LCM) of $b$ and $c$.
3. Multiply ratio $(A : B)$ by $c$, and ratio $(B : C)$ by $b$:

$$A : B = (a \times c) : (b \times c)$$ $$B : C = (b \times c) : (b \times d)$$ $$\mathbf{A : B : C = ac : bc : bd}$$

The "Empty Cell" Grid Shortcut for $A : B : C : D$

For multi-tier chains such as $A : B = 2 : 3$, $B : C = 4 : 5$, and $C : D = 6 : 7$, populate a 4-column grid and fill vacant cells by extending adjacent boundary digits:

  A      B      C      D
[ 2  :   3 ]   (3)    (3)     <--- extend 3 to the right
 (4)   [ 4  :   5 ]   (5)     <--- extend 4 left, 5 right
 (6)    (6)   [ 6  :   7 ]    <--- extend 6 to the left
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Multiply columns downward:
 A = 2 * 4 * 6 =  48
 B = 3 * 4 * 6 =  72
 C = 3 * 5 * 6 =  90
 D = 3 * 5 * 7 = 105
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Divide by common factor 3:
 A : B : C : D = 16 : 24 : 30 : 35

Equality of Linear Multiples ($pA = qB = rC$)

When problems specify that multiples of allocations are equal, do NOT equate them to $p : q : r$. Instead, equate to a constant $k$:

$$pA = qB = rC = k \implies A = \frac{k}{p}, \quad B = \frac{k}{q}, \quad C = \frac{k}{r}$$ $$\mathbf{A : B : C = \frac{1}{p} : \frac{1}{q} : \frac{1}{r} = qr : pr : pq}$$

If $2A = 3B = 4C$, then $A : B : C = \frac{1}{2} : \frac{1}{3} : \frac{1}{4}$. Multiplying by $\text{LCM}(2, 3, 4) = 12$ yields:

$$A : B : C = 6 : 4 : 3$$

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Direct, Inverse, and Joint Variations

The following are idealized variation models. Fuel examples assume constant fuel burn per hour, and travel-time examples assume fixed distance and constant speed:

1. Direct Variation ($y \propto x$)

When quantity $y$ increases or decreases in direct proportion to $x$:

$$y = kx \quad (k = \text{constant of variation})$$ $$\mathbf{\frac{y_1}{x_1} = \frac{y_2}{x_2}}$$

Operational Example: The fuel consumed by an idling diesel locomotive varies directly with idling hours: $F \propto t$.

2. Inverse Variation ($y \propto \frac{1}{x}$)

When an increase in $x$ causes a proportional reduction in $y$:

$$y = \frac{k}{x} \implies xy = k$$ $$\mathbf{x_1 y_1 = x_2 y_2}$$

Operational Example: The time taken to clear a yard fouling section varies inversely with the shunting speed: $T \propto \frac{1}{S}$.

3. Joint Variation ($z \propto \frac{x}{y}$ or $z \propto xy$)

When a dependent variable relies on multiple independent variables:

$$z = k \times \frac{x}{y} \implies \mathbf{\frac{z_1 y_1}{x_1} = \frac{z_2 y_2}{x_2}}$$

Stipulated model: In a fictional production exercise, output $z$ varies directly with machines $x$ and inversely with time per item $y$: $z=kx/y$. The variation must be supplied or justified; do not assume this model describes real train braking.

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Proportional Partitioning and Coin Denomination Systems

Dividing Quantities Proportionally

To divide a total quantity $Q$ among three entities $A, B, C$ in the ratio $a : b : c$:

1. Compute the sum of ratio units: $S = a + b + c$.
2. Value of one ratio unit $= \frac{Q}{S}$.
3. Allocate shares:

$$A = \frac{a}{S} \times Q, \quad B = \frac{b}{S} \times Q, \quad C = \frac{c}{S} \times Q$$

Coin Denomination Mathematics

Coin or token exercises may stipulate mixed face values (for example 5, 2, 1, 0.50 and 0.25 units). These are numerical denominations, not a claim about current legal tender. Questions may provide either:

• The ratio of the number of coins, and the total monetary value in rupees; or
• The ratio of the monetary values, and the total number of coins.

Coin Problem Dual-Vector Structure:

[Coin Count Vector]      c_1    :    c_2    :    c_3       (Ratio of pieces)
        x                 x           x           x
[Face Value Vector]      v_1         v_2         v_3       (Rupees per piece)
        =                 =           =           = 
[Cash Value Vector]    (c_1*v_1) : (c_2*v_2) : (c_3*v_3)   (Rupees contribution)

Sum of Cash Values = Total Rupee Amount

The Solution Protocol for Coin Problems

1. Express the number of coins of each denomination as $c_1 x, c_2 x, c_3 x$.
2. Convert all face values to uniform units (preferably rupees):
   • ₹5 coin $= \text{INR } 5.00$
   • ₹2 coin $= \text{INR } 2.00$
   • ₹1 coin $= \text{INR } 1.00$
   • 50 paise $= \text{INR } 0.50 = \text{INR } \frac{1}{2}$
   • 25 paise $= \text{INR } 0.25 = \text{INR } \frac{1}{4}$
3. Formulate the total cash equation:

$$\sum (\text{Count}_i \times \text{Face Value}_i) = \text{Total Cash Value}$$ $$x \left(c_1 v_1 + c_2 v_2 + c_3 v_3\right) = \text{Total Cash Value}$$

1. Solve for multiplier $x$, then calculate individual coin counts or monetary values.

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Comprehensive Worked Operational Problems

Worked Problem 1: Freight Wagon Allocation Across Three Marshalling Sidings

Scenario: A divisional railway operating control office receives 1,770 empty BOXN wagons to be distributed among three loading sidings: Alpha, Beta, and Gamma. The ratio of wagons allocated to Siding Alpha and Siding Beta is $3 : 4$, while the ratio between Siding Beta and Siding Gamma is $5 : 6$. Calculate the exact number of wagons allocated to each siding.

Step-by-Step Mathematical Resolution:

1. Identify the Given Ratios:

$$A : B = 3 : 4$$ $$B : C = 5 : 6$$

1. Bridge into a Continuous 3-Way Ratio ($A : B : C$): The common variable is $B$. The values are $4$ and $5$. Their LCM is $20$.
   • Scale $A : B$ by multiplying by $5$:

$$A : B = (3 \times 5) : (4 \times 5) = 15 : 20$$

• Scale $B : C$ by multiplying by $4$:

$$B : C = (5 \times 4) : (6 \times 4) = 20 : 24$$

• Combined ratio:

$$A : B : C = 15 : 20 : 24$$

1. Compute the Total Ratio Parts:

$$\text{Sum of parts} = 15 + 20 + 24 = 59$$

1. Determine the Value of One Ratio Part:

$$\text{Unit Part} = \frac{1770}{59} = 30\text{ wagons}$$

1. Calculate Individual Siding Allocations:

   • Siding Alpha ($A$): $15 \times 30 = 450\text{ wagons}$
   • Siding Beta ($B$): $20 \times 30 = 600\text{ wagons}$
   • Siding Gamma ($C$): $24 \times 30 = 720\text{ wagons}$

2. Verification:

$$450 + 600 + 720 = 1,770\text{ wagons (matches total perfectly)}$$

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Worked Problem 2: Signal Cable Tendering via Componendo-Dividendo

Scenario: In an electronic interlocking procurement tender, the cost analysis parameter $x$ satisfies the following radical equation:

$$\frac{\sqrt{x + 5} + \sqrt{x - 5}}{\sqrt{x + 5} - \sqrt{x - 5}} = 3$$

Determine the exact value of $x$.

Step-by-Step Mathematical Resolution:

1. Express the Right-Hand Side as a Fraction:

$$\frac{\sqrt{x + 5} + \sqrt{x - 5}}{\sqrt{x + 5} - \sqrt{x - 5}} = \frac{3}{1}$$

1. Apply the Componendo and Dividendo Rule:

$$\frac{(\sqrt{x + 5} + \sqrt{x - 5}) + (\sqrt{x + 5} - \sqrt{x - 5})}{(\sqrt{x + 5} + \sqrt{x - 5}) - (\sqrt{x + 5} - \sqrt{x - 5})} = \frac{3 + 1}{3 - 1}$$

1. Simplify the Sums and Differences:

$$\frac{2\sqrt{x + 5}}{2\sqrt{x - 5}} = \frac{4}{2} = 2$$ $$\frac{\sqrt{x + 5}}{\sqrt{x - 5}} = 2$$

1. Square Both Sides to Eliminate Radicals:

$$\frac{x + 5}{x - 5} = 4 = \frac{4}{1}$$

1. Apply Componendo and Dividendo a Second Time (or Cross-Multiply):

$$\frac{(x + 5) + (x - 5)}{(x + 5) - (x - 5)} = \frac{4 + 1}{4 - 1}$$ $$\frac{2x}{10} = \frac{5}{3} \implies \frac{x}{5} = \frac{5}{3} \implies x = \frac{25}{3} = 8\frac{1}{3}$$

1. Verification:

$$\sqrt{25/3 + 15/3} = \sqrt{40/3} = \frac{2\sqrt{10}}{\sqrt{3}}$$ $$\sqrt{25/3 - 15/3} = \sqrt{10/3} = \frac{\sqrt{10}}{\sqrt{3}}$$ $$\frac{\frac{2\sqrt{10}}{\sqrt{3}} + \frac{\sqrt{10}}{\sqrt{3}}}{\frac{2\sqrt{10}}{\sqrt{3}} - \frac{\sqrt{10}}{\sqrt{3}}} = \frac{3\sqrt{10}}{\sqrt{10}} = 3 \quad \text{(verified)}$$

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Worked Problem 3: Counts and values have different weights

A fictional box contains tokens worth 100, 50 and 25 units in count ratio 3:4:5. Their total value is 9000 units. Write their counts as 3k, 4k and 5k. The total is 300k + 200k + 125k = 625k. This would require k = 14.4, which is impossible for the stipulated integer ratio of token counts. The data are inconsistent, even though a numerical equation can be solved.

For a consistent version, change the total to 12500 units. Then k = 20 and the counts are 60, 80 and 100. Their values are 6000, 4000 and 2500, totaling 12500. Always test that a count result is a nonnegative integer; never present a fractional number of coins or people as a physical answer.

Common Exam Traps & Operational Pitfalls