10.2 Alligation Method & Solution Replacement Problems
Key Takeaways
- The Rule of Alligation is an arithmetic shortcut derived from weighted averages: the quantity ratio of cheaper to dearer ingredients is inversely proportional to their deviations from the mixture mean (q_c / q_d = (d - m) / (m - c)).
- The reference denominator of the rate determines the alligation output units: alligating price/kg yields weight ratios, profit % on cost yields cost price ratios, and speed in km/h yields travel time ratios.
- Under repeated removal and replacement with pure solvent, the residual original substance follows the geometric decay formula: Q_n = Q_0 * (1 - x / V)^n.
- When removing x liters from an existing solution of ratio a : b and replacing with liquid B, focus tracking exclusively on the undisturbed component A to avoid compound fractional errors.
- The mean price or concentration in any valid alligation must strictly satisfy c < m < d; a calculated mean outside this boundary indicates an arithmetic contradiction.
10.2 Alligation Method & Solution Replacement Problems
The Rule of Alligation is among the most versatile calculation shortcuts in competitive aptitude examinations. Derived directly from the principles of weighted averages, alligation enables banking aspirants to solve complex mixture problems, blended interest rate allocations, weighted average speeds, and two-tier profit/loss scenarios without setting up cumbersome multi-variable algebraic equations. In the SBI Clerk Examination, alligation questions appear frequently in both Quantitative Aptitude word problems and Data Interpretation caselets.
1. Theoretical Foundations & Derivation of Alligation
Consider two varieties of a commodity: a cheaper variety with unit price (or concentration) $c$ and quantity $q_c$, and a dearer variety with unit price (or concentration) $d$ and quantity $q_d$. When blended together, they form a mixture of quantity $(q_c + q_d)$ with a mean price $m$.
By equating the total cost of the mixture to the sum of individual constituent costs:
Expanding and grouping like terms:
The Alligation Cross-Diagram
Cheaper Value (c) Dearer Value (d)
\ /
\ /
Mean Value (m)
/ \
/ \
(d - m) (m - c)
| |
Ratio Part (Cheaper) Ratio Part (Dearer)
[!IMPORTANT] The Bounded Mean Condition: The mean value $m$ must strictly satisfy the inequality $c < m < d$. It can never be smaller than the cheaper component or larger than the dearer component. If a question prompts a mean outside this range, the mixture is physically impossible.
2. The Denominator Reference Principle & Commercial Applications
A critical exam pitfall is misunderstanding what the resulting ratio $(d - m) : (m - c)$ actually represents.
The Denominator Rule
| Variable Being Alligated | Rate Expression | Denominator Entity | Resulting Alligation Ratio |
|---|---|---|---|
| Price per kg | $\frac{\text{Rupees}}{\text{kg}}$ | Weight in kg | Ratio of Weights ($kg_1 : kg_2$) |
| Profit / Loss Percentage | $\frac{\text{Profit}}{\text{Cost Price}} \times 100$ | Cost Price (Rs.) | Ratio of Cost Prices ($CP_1 : CP_2$) |
| Speed | $\frac{\text{Distance}}{\text{Time}} = \frac{\text{km}}{\text{hour}}$ | Time in hours | Ratio of Travel Times ($t_1 : t_2$) |
| Simple Interest Rate | $\frac{\text{Annual Interest}}{\text{Principal}} \times 100$ | Principal (Rs.) | Ratio of Principals ($P_1 : P_2$) |
| Percentage Marks | $\frac{\text{Marks Scored}}{\text{Total Marks}} \times 100$ | Maximum Marks | Ratio of Maximum Marks |
Step-by-Step Worked Banking Portfolio Problem
Problem: An SBI branch manager disburses an aggregate priority-sector agricultural credit portfolio of Rs. 80,000 between two farmer groups. Group 1 receives credit under a subsidized crop-development scheme at an annual simple interest rate of 6% per annum, while Group 2 receives funds under an allied agribusiness scheme at 10% per annum. At the end of one year, the total combined interest received by the bank is Rs. 6,800. Determine the principal amount disbursed at 10% per annum.
- Calculate the Overall Effective Mean Interest Rate ($m$):
- Construct the Alligation Cross-Diagram:
- Subsidized Scheme ($c$) = $6%$
- Commercial Scheme ($d$) = $10%$
- Mean Rate ($m$) = $8.5%$
- Left deviation: $d - m = 10 - 8.5 = 1.5$
- Right deviation: $m - c = 8.5 - 6 = 2.5$
- Derive Principal Ratio:
- Apportion the Total Principal (Rs. 80,000):
- Total parts = $3 + 5 = 8$ parts
- Value of 1 part = $\frac{80,000}{8} = \text{Rs. } 10,000$
- Commercial Scheme Principal ($P_{\text{commercial}}$) = $5 \times 10,000 = \mathbf{\text{Rs. } 50,000}$ (Verification: $30,000 \times 0.06 + 50,000 \times 0.10 = 1,800 + 5,000 = \text{Rs. } 6,800$.)
3. Successive Dilution & Repeated Replacement Mechanics
A hallmark question type in the SBI Clerk Mains exam involves a vessel containing an initial volume of pure substance from which a fixed quantity is repeatedly withdrawn and replenished with pure water (or an alternative diluent).
The Governing Formula
Let a container initially hold $V$ units of a pure liquid (or initial quantity $Q_0$). If $x$ units are withdrawn and replaced with pure water, and this entire process is executed $n$ successive times:
Step-by-Step Decay Schedule ($V = 60\text{ L}, x = 12\text{ L}$)
| Cycle ($n$) | Operation | Remaining Pure Liquid ($Q_n$) | Liquid Removed in Step | Replacement Diluent Added | Ratio (Pure : Diluent) |
|---|---|---|---|---|---|
| Initial ($n=0$) | Baseline | $60.0\text{ L}$ | — | $0.0\text{ L}$ | Pure ($1 : 0$) |
| Cycle 1 ($n=1$) | Withdraw 12 L, add 12 L water | $60(1 - 0.2)^1 = 48.0\text{ L}$ | $12.0\text{ L}$ | $12.0\text{ L}$ | $48 : 12 = 4 : 1$ |
| Cycle 2 ($n=2$) | Withdraw 12 L mix, add 12 L water | $60(0.8)^2 = 38.4\text{ L}$ | $9.6\text{ L}$ | $21.6\text{ L}$ | $38.4 : 21.6 = 16 : 9$ |
| Cycle 3 ($n=3$) | Withdraw 12 L mix, add 12 L water | $60(0.8)^3 = 30.72\text{ L}$ | $7.68\text{ L}$ | $29.28\text{ L}$ | $30.72 : 29.28 = 64 : 61$ |
4. Multi-Component Solution Replacement (Removal & Replacement with Another Liquid)
In more intricate SBI Clerk word problems, the container starts not with a pure liquid, but with an existing mixture of liquids A and B in ratio $a : b$, from which $x$ liters are drawn off and replaced with pure liquid B.
The Invariant Component Rule
Because liquid B is being added, the entry of pure B complicates tracking substance B. Substance A, however, only experiences exit; no new A is ever introduced!
- Initial volume of mixture = $V$.
- Initial fraction of substance A = $\frac{a}{a+b}$.
- When $x$ liters of mixture are drawn off, substance A is removed in strict proportion to its concentration:
- Replenishing with $x$ liters of liquid B restores the total volume back to $V$.
- If the new ratio of A to B is $a' : b'$, the new fraction of A is $\frac{a'}{a'+b'}$.
- Equating the absolute quantity of A:
Solving this single linear equation yields the total volume $V$ directly without multi-variable substitution.
A lender splits Rs. 1,20,000 between two loans, one at 8% per annum simple interest and the other at 13% per annum. If the total interest after one year is Rs. 11,100, how much was lent at 13%?
A tank holds 80 litres of pure milk. 20 litres are drawn off and replaced with water, and this operation is carried out three times in all. How much pure milk is left in the tank?
A container contains a mixture of liquid A and liquid B in the ratio 7 : 5. When 9 liters of the mixture are drawn off and the container is completely refilled with liquid B, the ratio of liquid A to liquid B becomes 7 : 9. What was the initial total volume of liquid contained in the container?