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Mixtures & Alligations - Aptitude Mastery Guide

Section titled “Mixtures & Alligations - Aptitude Mastery Guide”

Category: Quantitative Aptitude
Generated on: 2025-07-15 09:17:10
Source: Aptitude Mastery Guide Generator

Mixtures & Alligations: The Ultimate Guide

Section titled “Mixtures & Alligations: The Ultimate Guide”

This comprehensive guide will equip you with the knowledge and techniques needed to master the topic of Mixtures & Alligations, a crucial area within Quantitative Aptitude for various competitive exams. We’ll cover everything from foundational concepts to advanced problem-solving strategies, ensuring you’re well-prepared to tackle any challenge.

1. Foundational Concepts

Section titled “1. Foundational Concepts”

What are Mixtures and Alligations?

  • Mixture: A mixture is formed when two or more different ingredients (solids, liquids, or gases) are combined in some proportion. The individual ingredients retain their distinct properties, though they appear together as a single entity.

  • Alligation: Alligation is a rule that helps determine the ratio in which two or more ingredients at given prices must be mixed to produce a mixture of a desired price. Essentially, it’s a simplified and faster method of solving problems related to mixtures, especially when dealing with finding the proportion of ingredients.

Understanding the “Why”: The Principle of Weighted Averages

At its core, the concept of mixtures and alligations revolves around the idea of a weighted average. Think of it this way: The final price (or concentration, etc.) of the mixture is somewhere between the individual prices (or concentrations) of the ingredients. The closer the final price is to one ingredient’s price, the more of that ingredient is present in the mixture. Alligation leverages this relationship to quickly find the proportions.

Key Terms:

  • Mean Price/Value: The cost price (or value) of a unit quantity of the mixture.
  • Cheaper Ingredient: The ingredient with a lower price or value.
  • Dearer Ingredient: The ingredient with a higher price or value.

2. Key Tricks & Shortcuts (The Core of the Guide)

Section titled “2. Key Tricks & Shortcuts (The Core of the Guide)”

This section is the heart of this guide. Mastering these shortcuts will significantly improve your speed and accuracy.

  • Trick 1: The Alligation Rule (The Cross Method)

    • How it works: This is the most fundamental and widely used trick. It provides a visual and efficient way to determine the ratio of ingredients.

    • When to use: Whenever you have two ingredients with known values (price, concentration, etc.) and a desired mean value of the mixture.

    • Steps:

      1. Write the price/value of the dearer ingredient (D) above and to the left.
      2. Write the price/value of the cheaper ingredient (C) above and to the right.
      3. Write the mean price/value (M) in the center.
      4. Subtract diagonally: (D - M) and (M - C). Always subtract the smaller value from the larger.
      5. The ratio of the dearer to cheaper ingredient is (M - C) : (D - M).
                D                 C
                     \             /
                      \           /
                         M
                      /           \
                     /             \
              (M - C)            (D - M)

  • Trick 2: Component Multiplication and Ratio Adjustment

    • How it works: When you know the ratio of two ingredients in a mixture and need to adjust the ratio by adding more of one ingredient (or both), this trick helps maintain accurate proportions.

    • When to use: When the problem asks how much of ingredient X needs to be added to a mixture to change the existing ratio to a new ratio.

    • Steps:

      1. Represent the initial ratio of ingredient A to ingredient B as A:B.
      2. Represent the new ratio as A':B'.
      3. If only one ingredient is changed (e.g., ingredient B is added), then the quantity of the unchanged ingredient (A) must remain the same in both ratios. Adjust the ratios by multiplying them by appropriate constants to make the quantities of the unchanged ingredient equal.
      4. Compare the adjusted quantities of the changing ingredient to determine how much was added/removed.

    • Example: A mixture contains milk and water in the ratio 5:3. How much water should be added to the mixture so that the ratio of milk and water becomes 5:4? Here, only water is added. The amount of milk is constant.

  • Trick 3: Successive Replacement (Repeated Dilution)

    • How it works: This applies when a quantity of mixture is repeatedly removed and replaced with another liquid (usually pure solvent).

    • When to use: Problems involving repeated removal and replacement of a mixture.

    • Formula: If a container contains x units of a liquid from which y units are taken out and replaced by water. After n operations, the quantity of the original liquid left in the container is given by:

      Quantity of original liquid left = x * (1 - y/x)^n

  • Trick 4: Vedic Maths - Base Method (for close values)

    • How it works: If the values of the ingredients and the mean value are close to each other, the base method simplifies the calculations.

    • When to use: When the difference between the ingredient values and the mean value is small.

    • Steps:

      1. Choose a base value close to all the values (e.g., the mean value itself).
      2. Calculate the deviations of each value from the base value (deviation = value - base).
      3. Apply the alligation rule using the deviations instead of the actual values. This will result in smaller numbers, making the calculation easier.

  • Trick 5: Assuming Extreme Scenarios

    • How it works: Assume the entire mixture is composed of only one of the ingredients. Calculate the result. Then assume the entire mixture is composed of only the other ingredient. The actual result will lie between these two extremes. This can help eliminate answer choices or provide a ballpark estimate.

    • When to use: When you’re stuck and need to make an educated guess, or when the problem involves percentages and you need a quick estimate.

  • Trick 6: Percentage-to-Fraction Conversion (for Quick Calculations)

    • How it works: Convert percentages to their equivalent fractions. This often simplifies calculations, especially when dealing with ratios and proportions.

    • When to use: Always, especially when you see common percentages like 25%, 50%, 75%, 33.33%, 66.66%, etc.

    • Example: 25% = 1/4, 50% = 1/2, 33.33% = 1/3, 66.66% = 2/3.

  • Trick 7: Component Concentration Method

    • How it works: Focus on the concentration of a specific component (e.g., alcohol, milk) in each mixture. Use alligation based on these concentrations.
    • When to use: When the problem gives the percentage or proportion of a particular component in different mixtures.
    • Steps:
      1. Identify the component you want to focus on.
      2. Calculate the percentage or proportion of that component in each mixture.
      3. Apply the alligation rule using these percentages/proportions and the desired percentage/proportion in the final mixture.

3. Essential Formulas & Rules

Section titled “3. Essential Formulas & Rules”
Formula/RuleDescription
Alligation Rule: (Quantity of Dearer) / (Quantity of Cheaper) = (Mean Price - Cheaper Price) / (Dearer Price - Mean Price)Fundamental rule for finding the ratio of ingredients.
Concentration = (Amount of Solute / Total Amount of Solution) x 100%Defines the concentration of a substance in a solution.
New Ratio (after adding ‘x’ of ingredient A to a mixture with ratio A:B): (A+x):BCalculates the new ratio when one ingredient is added.
Quantity of original liquid left after n replacements: x * (1 - y/x)^nx = initial quantity, y = quantity replaced, n = number of replacements
If A:B is a ratio and we multiply both by ‘k’, the ratio remains the same: kA : kB = A : BMultiplying both sides of a ratio by a constant doesn’t change the ratio.
Weighted Average = (Weight1 * Value1 + Weight2 * Value2) / (Weight1 + Weight2)General formula, also applies to mixtures.

4. Detailed Solved Examples (Variety is Key)

Section titled “4. Detailed Solved Examples (Variety is Key)”

Example 1: Basic Alligation Application [Easy]

  • Problem: A shopkeeper mixes two varieties of rice, one costing ₹50/kg and the other costing ₹60/kg, in such a way that the mixture costs ₹54/kg. In what ratio does he mix the two varieties?

  • Solution:

    1. Identify: We have two ingredients (rice varieties) and a mean price. The Alligation Rule is perfect here.

    2. Apply Alligation Rule:

                60                 50
                     \             /
                      \           /
                         54
                      /           \
                     /             \
                  (54-50)           (60-54)
                     4                 6

    3. Ratio: The ratio of the dearer rice (₹60/kg) to the cheaper rice (₹50/kg) is 4:6, which simplifies to 2:3.

    4. Answer: The shopkeeper mixes the two varieties in the ratio 2:3.

Example 2: Reverse Question & Component Multiplication [Medium]

  • Problem: Two vessels contain milk and water in the ratios 7:3 and 5:1 respectively. In what ratio should quantities be taken from the two vessels so as to form a mixture that contains milk and water in the ratio 3:1?

  • Solution:

    1. Identify: We need to find the ratio in which to mix two mixtures to get a desired mixture. We can use the Component Concentration Method.

    2. Focus on Milk: Choose milk as the component to focus on.

      • Vessel 1: Milk concentration = 7/(7+3) = 7/10
      • Vessel 2: Milk concentration = 5/(5+1) = 5/6
      • Desired Mixture: Milk concentration = 3/(3+1) = 3/4

    3. Apply Alligation Rule:

               7/10               5/6
                     \           /
                      \         /
                        3/4
                      /         \
                     /           \
          (3/4 - 7/10)     (5/6 - 3/4)
           (15-14)/20        (10-9)/12
                1/20               1/12

    4. Ratio: The ratio is (1/20) : (1/12) = 12:20 = 3:5.

    5. Answer: The quantities should be taken from the two vessels in the ratio 3:5.

Example 3: Successive Replacement [Medium]

  • Problem: A container has 80 liters of milk. From this container, 8 liters of milk are taken out and replaced by water. This process is repeated two more times. How much milk is now in the container?

  • Solution:

    1. Identify: This is a successive replacement problem.

    2. Apply Formula: x = 80 liters, y = 8 liters, n = 3 (since the process is repeated two more times, totaling 3 times).

      Quantity of milk left = x * (1 - y/x)^n
                             = 80 * (1 - 8/80)^3
                             = 80 * (1 - 1/10)^3
                             = 80 * (9/10)^3
                             = 80 * (729/1000)
                             = 58.32 liters

    3. Answer: There are now 58.32 liters of milk in the container.

Example 4: Combining Multiple Concepts [Hard]

  • Problem: A container contains a mixture of two liquids A and B in the ratio 5:3. If 24 liters of the mixture are taken out and the same quantity of liquid B is added, the ratio becomes 3:5. What was the initial quantity of liquid A in the container?

  • Solution:

    1. Identify: This combines ratio concepts with mixture removal and addition.

    2. Initial Quantities: Let the initial quantity of A be 5x and the initial quantity of B be 3x. The total initial quantity is 8x.

    3. Quantities Removed: When 24 liters of the mixture are removed, the ratio of A to B in the removed mixture is still 5:3. Therefore:

      • Quantity of A removed = (5/8) * 24 = 15 liters
      • Quantity of B removed = (3/8) * 24 = 9 liters

    4. Quantities After Removal:

      • Quantity of A remaining = 5x - 15
      • Quantity of B remaining = 3x - 9

    5. Quantity After Adding B: After adding 24 liters of B, the quantity of B becomes (3x - 9 + 24) = (3x + 15).

    6. New Ratio: The new ratio of A to B is (5x - 15) : (3x + 15) = 3:5

    7. Solve for x: Cross-multiply: 5(5x - 15) = 3(3x + 15)

      • 25x - 75 = 9x + 45
      • 16x = 120
      • x = 7.5

    8. Initial Quantity of A: Initial quantity of A = 5x = 5 * 7.5 = 37.5 liters.

    9. Answer: The initial quantity of liquid A in the container was 37.5 liters.

5. Practice Problems (Graded Difficulty)

Section titled “5. Practice Problems (Graded Difficulty)”

Solve these to reinforce your understanding.

  • [Easy] Two types of sugar are mixed in the ratio of 2:3. The cost of the first type of sugar is ₹12 per kg and the cost of the second type of sugar is ₹15 per kg. What is the cost per kg of the mixed sugar?

  • [Medium] How many kilograms of wheat costing ₹8 per kg must be mixed with 36 kg of wheat costing ₹5.40 per kg, so that the mixture may be worth ₹6.20 per kg?

  • [Medium] A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

  • [Hard] A can contains two liquids A and B in the ratio 7:5. When 9 liters of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many liters of liquid A was contained by the can initially?

  • [Medium] A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. Find the quantity sold at 18% profit.

  • [Hard] A jar contains a mixture of 36 liters of milk and 9 liters of water. 25% of the mixture is taken out of the jar and some amount of pure milk is added to the jar. If the resultant mixture contains 50% water, what was the quantity of pure milk added to the jar?

  • [Easy] In what ratio must water be mixed with milk to gain 20% by selling the mixture at cost price?

6. Advanced/Case-Based Question

Section titled “6. Advanced/Case-Based Question”

Problem:

A dairy owner has three types of milk: A, B, and C. Milk A contains 80% milk and 20% water, milk B contains 60% milk and 40% water, and milk C contains 90% milk and 10% water. The dairy owner wants to create a new type of milk, X, that contains 75% milk and 25% water.

  • Case 1: If the owner only wants to mix milk A and milk B, what ratio should they be mixed in to create milk X?
  • Case 2: If the owner only wants to mix milk B and milk C, what ratio should they be mixed in to create milk X?
  • Case 3: The owner decides to mix all three types of milk (A, B, and C) to create milk X. If the owner uses 1 part of milk A and 2 parts of milk B, how many parts of milk C should be added to create milk X?