Practice Problems: Alligation Medial With Mixed Types And Solutions

Understanding the concept of alligation medial is essential for solving problems related to mixture and concentration. This article provides a variety of practice problems with different types and detailed solutions to help students grasp the concept effectively.

Introduction to Alligation Medial

Alligation medial is a method used to find the average value or concentration when mixing two or more quantities with different values. It is particularly useful in problems involving mixtures of liquids, solutions, or solids with varying concentrations.

Basic Concept and Formula

The alligation medial rule involves comparing the quantities and their concentrations to determine the average concentration of the mixture. The key formula is:

Average concentration = (Sum of individual quantities’ contributions) / Total quantity

Practice Problems with Solutions

Problem 1: Mixing of Two Solutions

A chemist mixes 10 liters of a 30% acid solution with 20 liters of a 50% acid solution. Find the concentration of the resulting mixture.

Solution:

  • Quantity of first solution (Q₁) = 10 liters, concentration (C₁) = 30%
  • Quantity of second solution (Q₂) = 20 liters, concentration (C₂) = 50%

Total quantity = Q₁ + Q₂ = 10 + 20 = 30 liters

Contribution of each solution to the acid content:

  • First solution: 10 × 30% = 3 liters of pure acid
  • Second solution: 20 × 50% = 10 liters of pure acid

Total pure acid = 3 + 10 = 13 liters

Concentration of the mixture:

= (Total pure acid / Total quantity) × 100 = (13 / 30) × 100 ≈ 43.33%

Problem 2: Alligation of Three Solutions

A farmer has three types of water: one with 10% salt, another with 20% salt, and the third with 30% salt. He wants to prepare 100 liters of a mixture with 20% salt. How much of each type should he use?

Solution:

  • Let x liters of 10% salt solution
  • Let y liters of 20% salt solution
  • Let z liters of 30% salt solution

Total volume: x + y + z = 100 liters

Desired salt content: 20% of 100 liters = 20 liters

Equation for salt contribution:

  • 0.10x + 0.20y + 0.30z = 20
  • x + y + z = 100

Express z in terms of x and y:

z = 100 – x – y

Substitute into the salt equation:

0.10x + 0.20y + 0.30(100 – x – y) = 20

0.10x + 0.20y + 30 – 0.30x – 0.30y = 20

Combine like terms:

(0.10x – 0.30x) + (0.20y – 0.30y) = 20 – 30

-0.20x – 0.10y = -10

Multiply through by -1:

0.20x + 0.10y = 10

Multiply entire equation by 10 to clear decimals:

2x + y = 100

Express y in terms of x:

y = 100 – 2x

Calculate z:

z = 100 – x – y = 100 – x – (100 – 2x) = 100 – x – 100 + 2x = x

Choose x such that y and z are positive:

For y ≥ 0: 100 – 2x ≥ 0 ⇒ x ≤ 50

For z = x ≥ 0: x ≥ 0

Thus, x can vary between 0 and 50 liters. For example, if x = 20 liters:

y = 100 – 2(20) = 60 liters

z = x = 20 liters

Conclusion

Alligation medial is a powerful tool for solving mixture problems efficiently. Practice with various types of problems to strengthen your understanding and improve problem-solving speed.