🧪 Allegations and Mixtures

Day 2 | ⏱️ Part of 2-Hour Commercial Math Block | TCS Interview Prep

📌 Subtopics

• Alligation rule (mean price concept)
• Mixture of two quantities at different prices
• Replacement of part of a mixture
• Successive dilution/replacement

📖 Comprehensive Notes

1. The Alligation Rule (Mean Price Concept)

Alligation is a rule that enables us to find the ratio in which two or more ingredients at their respective prices must be mixed to produce a mixture of a specified mean price.

• Let Cost Price of cheaper item $= c$

• Let Cost Price of dearer item $= d$

• Let Mean Price of the mixture $= m$ $\text{Ratio of Cheaper Quantity } (Q_c) \text{ to Dearer Quantity } (Q_d) = \frac{Q_c}{Q_d} = \frac{d - m}{m - c}$

• Visual representation:

  Cheaper CP (c)      Dearer CP (d)
              \        /
               Mean CP (m)
              /        \
          (d - m)     (m - c)

• Note: The values $c, d,$ and $m$ must always be of the same unit and represent the same quantity type (e.g., all Cost Prices, or all profit percentages, or all concentrations).

Example: In what ratio must wheat at $₹32/\text{kg}$ be mixed with wheat at $₹40/\text{kg}$ so that the mixture is worth $₹35/\text{kg}$? $c = 32$, $d = 40$, $m = 35$. Ratio $= (40 - 35) : (35 - 32) = 5 : 3$.

2. Mixture of Two Quantities

Alligation can be used for things other than prices, such as percentage concentration (milk and water mixtures), average scores of two classes, or percentage profit/loss.

Example: A jar contains a mixture of milk and water in the ratio $4:1$. If $5$ liters of water is added, the ratio becomes $4:3$. Find the initial quantity of milk. Let milk $= 4x$, water $= x$. Adding water: $\frac{4x}{x + 5} = \frac{4}{3} \implies 12x = 4x + 20 \implies 8x = 20 \implies x = 2.5$. Initial milk $= 4(2.5) = 10\text{ liters}$.

3. Replacement of Part of a Mixture

If a part of a mixture is replaced by water or another ingredient:

• The concentration of the active ingredient decreases.
• Set up equations based on the volume of the ingredient that remains unchanged or trace the change in ratio.

Example: A vessel contains $80$ liters of milk. $10$ liters of milk is taken out and replaced by water. Find the amount of milk left. Milk left $= \text{Total Volume} \times \left(1 - \frac{\text{Replaced Volume}}{\text{Total Volume}}\right) = 80 \times \left(1 - \frac{10}{80}\right) = 80 \times \frac{7}{8} = 70\text{ liters}$.

4. Successive Dilution/Replacement

If a vessel contains $V$ liters of pure liquid (e.g., milk), and $x$ liters are taken out and replaced by water, and this operation is repeated $n$ times: $\text{Remaining Pure Liquid} = V \times \left(1 - \frac{x}{V}\right)^n$ $\text{Ratio of Pure Liquid to Total Liquid} = \left(1 - \frac{x}{V}\right)^n$

Example: From a cask containing $100$ liters of wine, $10$ liters are drawn and replaced with water. This process is repeated $3$ times. Find the volume of wine remaining. $V = 100$, $x = 10$, $n = 3$. $\text{Wine left} = 100 \times \left(1 - \frac{10}{100}\right)^3 = 100 \times (0.9)^3 = 100 \times 0.729 = 72.9\text{ liters}$.

🌐 Learning Resources

✅ Practice Links

📝 Key Formulas

Alligation Rule:
  Cheaper        Dearer
     c              d
         \      /
          Mean (m)
         /      \
   (d-m)         (m-c)

Ratio = (d - m) : (m - c)

Successive Replacement:
After n replacements of 'x' litres from 'V' litres:
  Quantity left = V × (1 - x/V)^n

🔗 Related Topics

• Percentages — Mixture concentrations are often expressed as percentages
• Ratios and Proportion — Finding ratio of ingredients to mix
• Averages — Mean price concept is a weighted average
• Profit and Loss — Cost price and selling price details in mixture problems
• Data Interpretation — Charts showing mixture compositions