4.3 Rule of Alligation, Removal-Replacement Mixtures & Partnership Capital-Time Sharing

The Rule of Alligation: Theory, Diagram & Mathematical Derivation

Conceptual Etymology and Scope

The term alligation originates from the Latin alligare, meaning "to tie together" or "to bind". It represents a graphical and algebraic technique for determining the proportion in which two ingredients of differing unit prices (or concentrations) must be blended to produce a mixture of an intermediate target price (or concentration).

Mathematical Derivation of Alligation

Let:

• $c$ = unit price (or concentration) of the cheaper ingredient
• $d$ = unit price (or concentration) of the dearer ingredient
• $m$ = mean unit price (or concentration) of the resulting mixture ($c < m < d$)
• $q_c$ = quantity of the cheaper ingredient
• $q_d$ = quantity of the dearer ingredient

The total expenditure of the blend is the sum of its individual components:

$$c \cdot q_c + d \cdot q_d = m (q_c + q_d)$$ $$c \cdot q_c + d \cdot q_d = m \cdot q_c + m \cdot q_d$$

Rearranging terms to group like quantities:

$$d \cdot q_d - m \cdot q_d = m \cdot q_c - c \cdot q_c$$ $$q_d (d - m) = q_c (m - c)$$ $$\mathbf{\frac{q_c}{q_d} = \frac{d - m}{m - c} = \frac{\text{Dearer Price} - \text{Mean Price}}{\text{Mean Price} - \text{Cheaper Price}}}$$

The Alligation Cross-Subtraction Diagram

   Cheaper Price (c)                Dearer Price (d)
           \                               /
            \                             /
             --->    Mean Price (m)   <---
            /                             \
           /                               \
Quantity of Cheaper (d - m)   :    Quantity of Dearer (m - c)

Cardinal Rules for Valid Alligation

1. Identical Dimensional Baselines: All three rates ($c, d, m$) must share identical reference denominators. You can alligate cost per kilogram with cost per kilogram, or speed in $\text{km/h}$ with speed in $\text{km/h}$. You cannot alligate cost per liter with cost per kilogram.
2. Bounded Mean Value: The mean value $m$ must lie strictly between the cheaper and dearer inputs: $c < m < d$.
3. Cost Price vs. Selling Price Discipline: If a problem states that the mixture is sold at a specific selling price ($SP$) yielding a profit of $P\%$, you MUST convert $SP$ to its true Cost Price ($m$) before alligating:

$$\mathbf{m = \frac{SP}{1 + \frac{P}{100}} = \frac{SP \times 100}{100 + P}}$$

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For these liquid models, assume complete uniform mixing before each withdrawal, additive volumes, no chemical reaction and $0<b<a$.

Successive Replacement and Progressive Dilution

A classic railway quantitative pattern involves progressive drainage and dilution: a container initially holds pure fluid (such as transformer cooling oil, locomotive lubricant, or chemical weed killer), and a fixed volume is repeatedly drawn off and replaced with an adulterant (typically water or solvent).

Progressive Dilution Mechanics:
Stage 0: Pure Liquid = a
Draw 'b' units of mixture, replace with 'b' units of water.
Each extraction removes a fraction (b/a) of the remaining pure liquid.
After n operations, Pure Liquid remaining = a * (1 - b/a)^n

The Governing Repeated-Dilution Formula

Let a vessel initially contain $a$ units of pure liquid. If $b$ units of the mixture are drawn off and replaced with pure water, and this operation is repeated $n$ times in total, the quantity of pure liquid remaining in the vessel is:

$$\mathbf{Q_{\text{pure}} = a \left(1 - \frac{b}{a}\right)^n}$$ $$\mathbf{\frac{Q_{\text{pure}}}{\text{Total Vessel Volume } (a)} = \left(1 - \frac{b}{a}\right)^n}$$

Evaluating the Adulterant (Water) Quantity

Because the total volume of liquid in the vessel remains constant at $a$ units throughout all $n$ replacement cycles:

$$Q_{\text{water}} = a - Q_{\text{pure}} = a \left[1 - \left(1 - \frac{b}{a}\right)^n\right]$$ $$\mathbf{\frac{Q_{\text{pure}}}{Q_{\text{water}}} = \frac{\left(1 - \frac{b}{a}\right)^n}{1 - \left(1 - \frac{b}{a}\right)^n}}$$

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Solution Blending and Alloy Composition Merging

When combining two distinct pre-mixed solutions or metal alloys (e.g., bronze or brass bearings in railway wagon wheelsets), candidates must determine the final composition ratio of the combined bulk.

Method 1: Fractional Concentration Aggregation

Let Vessel 1 contain liquid $A$ and liquid $B$ in ratio $x_1 : y_1$, with volume $V_1$. Let Vessel 2 contain liquids $A$ and $B$ in ratio $x_2 : y_2$, with volume $V_2$:

• Amount of $A$ from Vessel 1: $A_1 = V_1 \left(\frac{x_1}{x_1 + y_1}\right)$
• Amount of $A$ from Vessel 2: $A_2 = V_2 \left(\frac{x_2}{x_2 + y_2}\right)$
• Combined volume of $A$: $A_{\text{total}} = A_1 + A_2$
• Combined volume of $B$: $B_{\text{total}} = (V_1 + V_2) - A_{\text{total}}$
• Final ratio: $A_{\text{total}} : B_{\text{total}}$

Method 2: LCM Equalization Shortcut (For Equal Mixing)

When equal volumes of two mixtures are blended:

1. Note total parts in Vessel 1 ($S_1 = x_1 + y_1$) and Vessel 2 ($S_2 = x_2 + y_2$).
2. Find $L = \text{LCM}(S_1, S_2)$.
3. Multiply ratio terms of Vessel 1 by $\frac{L}{S_1}$, and Vessel 2 by $\frac{L}{S_2}$.
4. Sum the scaled antecedents and consequents directly.

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Commercial Partnership Mathematics

In railway commercial ventures, freight siding leases, container depot consortiums, and station food court concessions, multiple partners pool financial capital over varying tenures. Partnership arithmetic allocates net commercial profits or losses among stakeholders in proportion to their effective investment.

The Stipulated Capital-Time Sharing Rule

Use capital-time profit sharing only when the question states or adopts that convention. Actual partnership agreements can specify other shares; this is an aptitude model, not a general legal rule. Profit distribution is directly proportional to both the magnitude of financial capital invested ($C$) and the time duration ($T$) for which that capital remains active in the enterprise:

$$\mathbf{\text{Profit Ratio} = (C_A \times T_A) : (C_B \times T_B) : (C_C \times T_C)}$$

Types of Commercial Partnerships

1. Simple Partnership: All partners invest capital for identical durations ($T_A = T_B = T_C$). Profit distributes strictly according to capital:

$$P_A : P_B : P_C = C_A : C_B : C_C$$

1. Compound (Complex) Partnership: Partners invest differing capital amounts for differing time durations. Profit distributes according to capital-time products:

$$P_A : P_B : P_C = (C_A T_A) : (C_B T_B) : (C_C T_C)$$

1. Variable / Staggered Capital Investments: When partners inject or withdraw funds during the financial year, compute the effective capital-months:

$$\mathbf{C_{\text{effective}} = \sum (C_i \times t_i)}$$

Working (Active) vs. Sleeping (Silent) Partners

• Sleeping Partner: Provides capital only, performing no operational duties.
• Working Partner: Contributes capital (or expertise) and actively manages day-to-day operations. As compensation, the working partner receives an agreed salary, management allowance, or commission percentage deducted from gross profit before remaining net profits are divided in the capital-time ratio.

Profit Distribution Architecture with a Working Partner:

[Gross Annual Profit: P_gross]
             |
             v
[Deduct: Working Partner Management Allowance (M = alpha * P_gross)]
             |
             v
[Remaining Net Distributable Profit: P_net = P_gross - M]
             |
             v
[Divide P_net among Partners in (C * T) Ratio]
             |
             +---> Working Partner Final Income = M + Share of P_net
             +---> Sleeping Partner Final Income = Share of P_net only

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Comprehensive Worked Operational Problems

Worked Problem 1: Alligation in Locomotive Diesel Fuel Blending

Scenario: A railway fueling depot blends bio-diesel costing ₹84 per liter with premium high-speed diesel costing ₹96 per liter. The resulting fuel blend is supplied to a private siding freight operator at ₹99 per liter, realizing a profit of $10\%$. In what ratio must the fueling depot mix the bio-diesel and high-speed diesel?

Step-by-Step Mathematical Resolution:

1. Determine the True Cost Price ($m$) of the Mixture: The selling price of ₹99 includes a $10\%$ profit margin.

$$m = \frac{SP}{1 + \frac{P}{100}} = \frac{99}{1 + 0.10} = \frac{99}{1.10} = \text{INR } 90\text{ per liter}$$

1. Identify Alligation Parameters:

   • Cheaper component ($c$) [Bio-diesel]: ₹$84$/L
   • Dearer component ($d$) [High-speed diesel]: ₹$96$/L
   • Mean cost price ($m$): ₹$90$/L

2. Apply the Rule of Alligation:

$$\frac{q_c}{q_d} = \frac{d - m}{m - c}$$ $$\frac{q_{\text{bio}}}{q_{\text{high-speed}}} = \frac{96 - 90}{90 - 84} = \frac{6}{6} = \frac{1}{1}$$

1. Conclusion: The depot must mix bio-diesel and high-speed diesel in an exact ratio of $1 : 1$.

2. Verification: Cost of 1 liter bio-diesel ($\text{INR } 84$) + 1 liter high-speed diesel ($\text{INR } 96$) = ₹$180$ for 2 liters, or ₹$90$/L. Selling 2 liters at ₹$99$/L yields ₹$198$. Profit $= 198 - 180 = \text{INR } 18$, which is $\frac{18}{180} \times 100 = 10\%$ (verified).

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Worked Problem 2: Successive Replacement in Track Lubricant Drum

Scenario: A maintenance depot drum contains 60 liters of pure synthetic wheel-flange lubricant. An apprentice technician draws off 12 liters of the lubricant and replaces it with standard thinning solvent. This operation is repeated two more times (for a total of 3 cycles). Calculate:

1. The quantity of pure synthetic lubricant remaining in the drum.
2. The final ratio of pure lubricant to thinning solvent.

Step-by-Step Mathematical Resolution:

1. Identify the Given Parameters:

   • Initial volume of pure liquid ($a$) $= 60\text{ liters}$
   • Volume extracted and replaced in each cycle ($b$) $= 12\text{ liters}$
   • Total operational cycles ($n$) $= 3$

2. Compute the Fractional Extraction Factor:

$$\frac{b}{a} = \frac{12}{60} = \frac{1}{5} = 0.20$$ $$1 - \frac{b}{a} = 1 - \frac{1}{5} = \frac{4}{5} = 0.80$$

1. Calculate Pure Lubricant Remaining ($Q_{\text{pure}}$):

$$Q_{\text{pure}} = a \left(1 - \frac{b}{a}\right)^n = 60 \times \left(\frac{4}{5}\right)^3$$ $$Q_{\text{pure}} = 60 \times \frac{64}{125} = 12 \times \frac{64}{25} = \frac{768}{25} = 30.72\text{ liters}$$

1. Calculate Volume of Thinning Solvent ($Q_{\text{solvent}}$):

$$Q_{\text{solvent}} = 60 - 30.72 = 29.28\text{ liters}$$

1. Determine the Final Ratio ($Q_{\text{pure}} : Q_{\text{solvent}}$):

$$\frac{Q_{\text{pure}}}{Q_{\text{solvent}}} = \frac{30.72}{29.28} = \frac{768 / 25}{732 / 25} = \frac{768}{732}$$

Dividing both numerator and denominator by their greatest common divisor ($12$):

$$\frac{768 / 12}{732 / 12} = \frac{64}{61}$$

1. Verification via Direct Proportions: After 3 cycles, pure liquid fraction is $(4/5)^3 = 64/125$. The solvent fraction is $1 - 64/125 = 61/125$. The ratio is strictly $64 : 61$.

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Worked Problem 3: Multi-Partner Container Siding Venture with Working Partner

Scenario: Two logistics entrepreneurs, Partner A and Partner B, invest in a joint railway container leasing venture:

• Partner A invests ₹$40,000$ for the entire 12-month annual period.
• Partner B invests ₹$60,000$, but withdraws from the venture after 8 months.
• Partner A actively manages operational dispatching and receives an agreed management allowance of $15\%$ of the gross annual profit.
• The remaining profit is divided between A and B in proportion to their capital-time investments.
• If the gross profit at year-end is ₹$72,000$, determine the total income received by Partner A and Partner B respectively.

Step-by-Step Mathematical Resolution:

1. Evaluate Working Partner Management Allowance:

$$M = 15\% \times 72,000 = \frac{15}{100} \times 72,000 = \text{INR } 10,800$$

1. Compute Remaining Net Distributable Profit:

$$P_{\text{net}} = 72,000 - 10,800 = \text{INR } 61,200$$

1. Calculate Capital-Time Investment Ratios:
   • Partner A's Capital-Time: $C_A \times T_A = 40,000 \times 12 = 480,000\text{ rupee-months}$
   • Partner B's Capital-Time: $C_B \times T_B = 60,000 \times 8 = 480,000\text{ rupee-months}$
   • Profit Sharing Ratio:

$$P_A : P_B = 480,000 : 480,000 = 1 : 1$$

1. Divide Net Distributable Profit: Since the ratio is $1 : 1$, each partner receives an equal share of $P_{\text{net}}$:

$$\text{A's share of net profit} = \frac{61,200}{2} = \text{INR } 30,600$$ $$\text{B's share of net profit} = \frac{61,200}{2} = \text{INR } 30,600$$

1. Calculate Total Income for Each Partner:
   • Partner A Total Income:

$$\text{Management Allowance} + \text{Net Profit Share} = 10,800 + 30,600 = \text{INR } 41,400$$

• Partner B Total Income:

$$\text{Net Profit Share only} = \text{INR } 30,600$$

1. Verification:

$$41,400 + 30,600 = \text{INR } 72,000\text{ (matches gross profit exactly)}$$

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Common Exam Traps & Operational Pitfalls