How to Solve Mixture & Alligation Problems

How to Solve Mixture & Alligation Problems

(For SSC, Bank, Railway Exams – Ace Your Quant Section!)

? Introduction

"Master mixture problems, and you’ll unlock 3-5 easy marks in every SSC/Bank/Railway exam—marks that most students lose because they panic at ‘mix’ or ‘alligation’! (These questions appear in Tier-1, Prelims, and Mains. One wrong step = zero marks. One right step = full marks.)

? WHAT YOU NEED TO KNOW FIRST

Before you start, you must understand: 1. Percentage basics – How to convert % to fractions/decimals (e.g., 20% = 0.2). 2. Ratio & proportion – How to simplify ratios (e.g., 3:2 = 5 parts). 3. Basic algebra – Solving for one variable (e.g., if 2x + 3 = 7, then x = 2).

(If you’re shaky on these, pause here and review them first!)

? KEY VOCABULARY

⚡ FORMULAS TO KNOW

1. Alligation Rule (For Two Ingredients)

Formula:

(Quantity of Cheaper) / (Quantity of Dearer) = (Dearer Price – Mean Price) / (Mean Price – Cheaper Price)

Variables: - Cheaper Price (C) = Price of the cheaper ingredient. - Dearer Price (D) = Price of the expensive ingredient. - Mean Price (M) = Average price of the final mixture.

MEMORISE THIS – Exams won’t give this formula!

2. Replacement Formula (For Repeated Replacement)

Formula:

Final Quantity = Initial Quantity × (1 – Replacement Fraction)^n

Variables: - Replacement Fraction = Amount replaced / Total quantity. - n = Number of times replacement is done.

MEMORISE THIS – Critical for "milk-water" replacement problems.

3. Mixture Equation (For Two Ingredients)

Formula:

(C₁ × Q₁) + (C₂ × Q₂) = Cₘ × (Q₁ + Q₂)

Variables: - C₁, C₂ = Concentration of ingredients 1 and 2. - Q₁, Q₂ = Quantities of ingredients 1 and 2. - Cₘ = Concentration of final mixture.

Given on exam sheet? Sometimes, but memorise it to save time.

? STEP-BY-STEP METHOD

? For Alligation Problems (Two Ingredients)

Step 1: Identify the cheaper price (C), dearer price (D), and mean price (M). Step 2: Write the alligation rule:

(Cheaper Quantity) / (Dearer Quantity) = (D – M) / (M – C)

Step 3: Plug in the numbers and solve for the ratio. Step 4: If quantities are given, use the ratio to find exact amounts.

? For Replacement Problems (Milk & Water)

Step 1: Find the initial quantity and replacement fraction. Step 2: Use the formula:

Final Quantity = Initial Quantity × (1 – Replacement Fraction)^n

Step 3: Calculate the remaining pure substance. Step 4: Subtract from total to find the added substance.

? For Mixture Equations (Three+ Ingredients)

Step 1: Write the equation:

(C₁ × Q₁) + (C₂ × Q₂) + (C₃ × Q₃) = Cₘ × (Q₁ + Q₂ + Q₃)

Step 2: Plug in known values. Step 3: Solve for the unknown.

✏️ WORKED EXAMPLES

? Example 1 – Basic Alligation

Question: A shopkeeper mixes two types of rice costing ₹20/kg and ₹30/kg. What is the ratio of the two in a mixture worth ₹25/kg?

Solution: Step 1: Identify prices: - Cheaper (C) = ₹20 - Dearer (D) = ₹30 - Mean (M) = ₹25

Step 2: Apply alligation rule:

(Cheaper Quantity) / (Dearer Quantity) = (D – M) / (M – C)
= (30 – 25) / (25 – 20)
= 5 / 5
= 1 / 1

Step 3: Ratio = 1:1

What we did and why: We used the alligation rule to find the ratio of cheap to expensive rice. The difference from mean price gives the ratio directly.

? Example 2 – Medium (Replacement Problem)

Question: A 40L milk solution is 10% water. If 10L is removed and replaced with water, what is the new % of water?

Solution: Step 1: Initial water = 10% of 40L = 4L Step 2: After removing 10L, water left = (4L / 40L) × 10L = 1L Step 3: New water = 1L (remaining) + 10L (added) = 11L Step 4: New % = (11L / 40L) × 100 = 27.5%

What we did and why: We calculated the remaining water after removal, then added the new water. The key was finding how much water was removed proportionally.

? Example 3 – Exam-Style (Disguised Problem)

Question: A vessel has 60L of a 40% alcohol solution. If 20L is taken out and replaced with 20L of 20% alcohol, what is the new alcohol %?

Solution: Step 1: Initial alcohol = 40% of 60L = 24L Step 2: Alcohol removed = 40% of 20L = 8L Step 3: Alcohol left = 24L – 8L = 16L Step 4: Alcohol added = 20% of 20L = 4L Step 5: New alcohol = 16L + 4L = 20L Step 6: New % = (20L / 60L) × 100 = 33.33%

What we did and why: We treated the removal as a proportional reduction, then added the new alcohol. The trap was assuming the removed liquid had the same % as the final mix.

❌ Common Mistakes

? EXAM TRAPS