Mixtures & Alligation

Overview

Mixtures problems involve combining two or more quantities with different attributes (price, concentration, average) to find the resultant attribute of the combined mixture. The core technique is weighted averaging — accounting for how much of each component contributes to the whole.

On CAT, Mixtures questions appear in Quant as 1–2 standalone problems per exam. They often combine with ratios, percentages, and alligation. On GMAT, they appear as mid-difficulty Problem Solving questions.

Core Concept: Weighted Average

When quantities Q₁ and Q₂ with attributes A₁ and A₂ are mixed:

Resultant attribute A = (Q₁ × A₁ + Q₂ × A₂) / (Q₁ + Q₂)

This is the weighted mean — each component contributes proportionally to its quantity.

Example: 24 kg rice at Rs.42/kg mixed with 12 kg at Rs.27/kg: A = (24×42 + 12×27) / (24+12) = (1008+324)/36 = 1332/36 = Rs.37/kg

Alligation (Cross Method)

The alligation shortcut finds the ratio in which two components must be mixed to achieve a target attribute value.

Cheaper (C)           Dearer (D)
      \               /
       \             /
        Target (T)
       /             \
      /               \
  (D - T)           (T - C)

Ratio of cheaper to dearer = (D − T) : (T − C)

Example: Mix wheat at Rs.6/kg with wheat at Rs.8/kg to get Rs.7.2/kg.

• (8 − 7.2) : (7.2 − 6) = 0.8 : 1.2 = 2 : 3

Three or More Components

Use weighted average directly:

A = (Q₁A₁ + Q₂A₂ + Q₃A₃) / (Q₁ + Q₂ + Q₃)

Or apply alligation in two stages (combine any two first, then the result with the third).

Concentration Problems

A mixture has a component (e.g., milk) at concentration c (fraction of total). When volumes V₁ and V₂ of two solutions with concentrations c₁ and c₂ are mixed:

c_resultant = (V₁c₁ + V₂c₂) / (V₁ + V₂)

For equal-volume mixtures (V₁ = V₂): c_resultant = (c₁ + c₂) / 2

Example: Two equal containers — container 1 has milk:water = 3:1 (milk = 3/4), container 2 has milk:water = 5:2 (milk = 5/7). Mixed equally:

• Milk fraction = (3/4 + 5/7)/2 = (21/28 + 20/28)/2 = 41/56
• Water fraction = (1/4 + 2/7)/2 = (7/28 + 8/28)/2 = 15/56
• Ratio milk:water = 41:15

Adding to an Existing Mixture

If a mixture has milk:water = m:w (total = m+w parts), and x liters of water are added, the new ratio changes only the water part.

New total water = (original water) + x, milk stays constant.

Set up equation from the new ratio and solve for unknown.

Repeated Replacement Formula

From a container of volume V containing pure liquid, each cycle removes d liters and replaces with another liquid. After n cycles:

Remaining pure liquid = V × (1 − d/V)ⁿ

Example: 40L wine, 4L removed and replaced with water, 3 times: Remaining wine = 40 × (36/40)³ = 40 × (9/10)³ = 40 × 0.729 = 29.16 L

Profit on Mixing (Free Component)

When water (cost = 0) is mixed with milk (cost = C per unit) and the mixture is sold at milk's price:

Gain% = (water added / milk quantity) × 100

So if gain% = g%, water:milk = g:100.

Common Mistakes

• Using simple average of attributes when quantities are unequal — always weight by quantity
• In concentration problems, confusing "ratio milk:water" with "fraction of milk" (3:1 means milk is 3/4 not 3)
• In repeated replacement, applying the reduction only once instead of repeatedly
• Alligation gives ratio of components, not quantities — multiply by a common factor if quantity is needed

Exam Tips

• For finding the ratio of mixing: draw the alligation cross — fastest method
• For 3-component mixtures, use weighted average directly (faster than double alligation)
• Repeated replacement: use (1 − d/V)ⁿ, don't compute step by step
• If a mixture's concentration is given as a ratio (a:b), convert to fraction a/(a+b) before computing
• For CAT, concentration problems often have "replaced k times" — memorize the formula