3.4 Averages, Mixtures & Alligations

Averages, Mixtures & Alligations

Averages and mixture calculations represent heavily tested arithmetic concepts in SBI PO exams. The Rule of Alligation is not merely a formula but a visual cross-subtraction technique that simplifies weighted average equations across price blending, solution concentrations, and interest rate problems.

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1. Averages Fundamentals & Deviation Method

Basic Definitions & Weighted Averages

• Arithmetic Mean (Average): $\text{Average} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} = \frac{\sum x_i}{N}$
• Weighted Average: When different groups of sizes $n_1, n_2, \dots, n_k$ have individual averages $A_1, A_2, \dots, A_k$, the combined weighted average $A_{w}$ is: $A_{w} = \frac{n_1 A_1 + n_2 A_2 + \dots + n_k A_k}{n_1 + n_2 + \dots + n_k}$

Assumed Mean (Deviation) Shortcut Method

Instead of summing large numbers, assume a convenient benchmark mean $A_0$ and compute net deviations: $\text{Actual Average} = A_0 + \frac{\sum (x_i - A_0)}{N}$

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2. The Rule of Alligation

Alligation is a visual rule for solving weighted average problems involving two ingredients of different rates or concentrations.

The Alligation Cross Diagram

Let Cheap Ingredient Rate $= c$, Dear (Costly) Ingredient Rate $= d$, and Mean (Blended) Rate $= m$.

Cheap Rate (c)           Dear Rate (d)
         \               /
          \             /
           Mean Rate (m)
          /             \
         /               \
(d - m)                     (m - c)

Mathematical Ratio Formula

$\frac{\text{Quantity of Cheap}}{\text{Quantity of Dear}} = \frac{d - m}{m - c}$

Critical Requirement: Alligation can only be applied to intensive variables (e.g., price per kg, percentage concentration, speed in km/h, interest rate %). The resulting cross-subtraction ratio gives the ratio of the corresponding extensive quantities (e.g., weight in kg, volume in liters, distance, principal amount).

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3. Successive Dilution & Replacement Formula

When a container has $V$ units of pure liquid, and $x$ units are drawn off and replaced with water $n$ times successively, the remaining quantity of pure liquid is given by: $\text{Final Quantity of Pure Liquid} = V \left(1 - \frac{x}{V}\right)^n$

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4. Step-by-Step Worked Math Examples

Worked Example 1: Correcting Misread Observations in Averages

Problem: The average weight of a class of 40 students was calculated as 55 kg. Later it was discovered that two weights were misread as 48 kg and 52 kg instead of their correct values of 64 kg and 76 kg. Find the correct average weight.

Step-by-Step Solution:

1. Initial incorrect sum of weights $= 40 \times 55 = 2,200 \text{ kg}$.
2. Incorrect values included $= 48 + 52 = 100 \text{ kg}$.
3. Correct values to be included $= 64 + 76 = 140 \text{ kg}$.
4. Net difference to be added $= 140 - 100 = +40 \text{ kg}$.
5. Corrected sum of weights $= 2,200 + 40 = 2,240 \text{ kg}$.
6. Correct average weight $= \frac{2,240}{40} = 56 \text{ kg}$.

Worked Example 2: Alligation for Price Blending

Problem: In what ratio must a grocer mix tea at ₹60 per kg with tea at ₹75 per kg so that the mixture is worth ₹65 per kg?

Step-by-Step Solution:

1. Identify values: Cheap rate $c = 60$, Dear rate $d = 75$, Mean rate $m = 65$.
2. Apply cross subtraction:
   • Cheap part $= d - m = 75 - 65 = 10$
   • Dear part $= m - c = 65 - 60 = 5$
3. Ratio of Cheap to Dear quantity $= \frac{d - m}{m - c} = \frac{10}{5} = \frac{2}{1}$.
4. The grocer must mix the teas in the ratio 2 : 1.

Worked Example 3: Successive Dilution of Wine with Water

Problem: A cask contains 80 liters of pure wine. 8 liters of wine are drawn out and replaced with water. This operation is performed two more times. Find the amount of pure wine remaining in the cask.

Step-by-Step Solution:

1. Total volume $V = 80 \text{ liters}$.
2. Volume replaced per step $x = 8 \text{ liters}$.
3. Total number of operations $n = 1 + 2 = 3$.
4. Apply replacement formula: $\text{Remaining Wine} = V \left(1 - \frac{x}{V}\right)^n$.
5. Substitute values: $\text{Remaining Wine} = 80 \left(1 - \frac{8}{80}\right)^3 = 80 \left(1 - \frac{1}{10}\right)^3 = 80 \left(\frac{9}{10}\right)^3$.
6. Calculate exponent: $\left(\frac{9}{10}\right)^3 = \frac{729}{1,000} = 0.729$.
7. Remaining Wine $= 80 \times 0.729 = 58.32 \text{ liters}$.

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5. Common Traps & Exam Strategies

• Wrong Base in Alligation: Never apply alligation using absolute profit or cost values. Always convert to percentage profit/loss or per-unit cost rates before cross-subtraction.
• Multiple Component Blending: When mixing three items A, B, C with prices $p_a < p_b < p_c$ to get mean $p_m$, split into two pairs: $(A, C)$ and $(B, C)$, apply alligation individually, and sum common component ratios.