Alligation & Mixture - Theory & Concepts
⚗️ Alligation & Mixture - Complete Theory
Master mixing problems using the powerful Alligation Rule!
🎯 What is Alligation?
Alligation is a method to find the ratio in which two or more ingredients at different prices/strengths should be mixed to get a mixture at a desired price/strength.
Key Concept: It’s a shortcut alternative to weighted average!
📐 Basic Concepts
1. Mixture
Mixture = Combination of two or more substances.
Types:
- Simple Mixture: Two ingredients only
- Compound Mixture: More than two ingredients
2. Alligation
Alligation = Faster method to solve mixture problems using a visual rule.
Used When:
- Finding ratio of mixing
- Finding mean/average price
- Finding replacement calculations
⚖️ Rule of Alligation
Visual Method (Butterfly/Cross Method)
Cheaper (C) Dearer (D)
\ /
\ /
\ /
\ /
Mean Price (M)
/ \
/ \
/ \
/ \
(D - M) (M - C)
Ratio of C : D = (D - M) : (M - C)
Formula:
Cheaper quantity : Dearer quantity = (Dearer Price - Mean Price) : (Mean Price - Cheaper Price)
C : D = (D - M) : (M - C)
💡 Solved Examples
Example 1: Basic Alligation
Q: In what ratio must tea at ₹60/kg be mixed with tea at ₹80/kg to get a mixture worth ₹70/kg?
Solution:
Using Alligation Rule:
Cheaper (C) = ₹60 Dearer (D) = ₹80
\ /
\ /
\ /
Mean (M) = ₹70
/ \
/ \
/ \
(80-70) (70-60)
10 10
Ratio = 10 : 10 = 1 : 1
Answer: 1:1 ratio
Example 2: Finding Mean Price
Q: Two varieties of rice at ₹40/kg and ₹60/kg are mixed in ratio 2:3. Find the price of mixture.
Solution:
Using Weighted Average:
Mean Price = (2×40 + 3×60) / (2 + 3)
= (80 + 180) / 5
= 260 / 5
= ₹52/kg
Answer: ₹52/kg
Example 3: Three Ingredients
Q: Milk at ₹50/L, ₹60/L, and ₹70/L are mixed in ratio 2:3:5. Find price per liter of mixture.
Solution:
Mean Price = (2×50 + 3×60 + 5×70) / (2 + 3 + 5)
= (100 + 180 + 350) / 10
= 630 / 10
= ₹63/L
Answer: ₹63/L
🔄 Mixture Replacement
Formula for Replacement
When quantity is replaced:
If 'Q' liters of mixture is replaced by water/other liquid:
After 1st operation:
Remaining quantity of original = Q × (1 - q/Q)
After 2nd operation:
Remaining quantity = Q × (1 - q/Q)²
After n operations:
Remaining quantity = Q × (1 - q/Q)ⁿ
Where:
Q = Total mixture quantity
q = Quantity replaced each time
Example 4: Single Replacement
Q: A vessel contains 80 liters of milk. 8 liters are removed and replaced with water. What is the quantity of milk now?
Solution:
Original quantity = 80 liters
Removed = 8 liters
Fraction removed = 8/80 = 1/10
Remaining milk = 80 × (1 - 1/10)
= 80 × 9/10
= 72 liters
Answer: 72 liters
Example 5: Multiple Replacements
Q: 100 liters of milk. 10 liters removed and replaced with water. This is done twice. Find final quantity of milk.
Solution:
After 1st replacement:
Milk = 100 × (1 - 10/100) = 100 × 0.9 = 90 liters
After 2nd replacement:
Milk = 100 × (1 - 10/100)² = 100 × (0.9)² = 81 liters
Or directly:
Milk = 100 × (9/10)² = 100 × 81/100 = 81 liters
Answer: 81 liters
🧪 Concentration/Strength Problems
Finding Final Concentration
Formula:
If two solutions of concentrations C₁% and C₂% are mixed in ratio a:b:
Final Concentration = (a×C₁ + b×C₂) / (a + b)
Example 6: Mixing Solutions
Q: 40% alcohol solution is mixed with 70% alcohol solution in ratio 2:3. Find concentration of final solution.
Solution:
C₁ = 40%, C₂ = 70%
Ratio = 2:3
Final Concentration = (2×40 + 3×70) / (2 + 3)
= (80 + 210) / 5
= 290 / 5
= 58%
Answer: 58% alcohol
Example 7: Water Addition
Q: 20 liters of 60% acid solution. How much water must be added to make it 40% solution?
Solution:
Pure acid in solution = 20 × 60% = 12 liters
(This remains constant)
Let x liters of water be added.
Total solution = 20 + x
Final concentration = 40%
12 / (20 + x) = 40/100
12 = 0.4(20 + x)
12 = 8 + 0.4x
4 = 0.4x
x = 10 liters
Answer: 10 liters
Shortcut using Alligation:
Water (0%) Original (60%)
\ /
\ /
\ /
Mean (40%)
/ \
/ \
/ \
(60-40) (40-0)
20 40
Ratio = 20:40 = 1:2
Original : Water = 1:2
For 20 liters original, water = 20 × 2/1 = 40 liters total
But we already have 20, so add = 40 - 20 = 20 liters
Wait, let me recalculate:
Original 20L : Water x = 1:2
This means for every 1 part original, 2 parts total mixture
Actually: 20L : Water x = 2:1 (inverted)
20/x = 2/1
x = 10 liters ✓
📊 Important Patterns
Pattern 1: Equal Quantity Mixing
If equal quantities of two mixtures with concentrations C₁% and C₂% are mixed:
Final Concentration = (C₁ + C₂) / 2
Example: Equal quantities of 30% and 50% solutions mixed.
Final = (30 + 50) / 2 = 40%
Pattern 2: Pure Substance Addition
If pure substance (100% concentration) is added to a mixture:
Use Alligation with one ingredient at 100%
Pattern 3: Water Addition
Adding water = Adding 0% concentration
Use Alligation with water at 0%
⚡ Quick Shortcuts
Shortcut 1: 50-50 Mixing
If two items mixed in equal quantities (1:1):
Mean Price = (Price₁ + Price₂) / 2
Shortcut 2: Doubling Check
After adding equal quantity of water to a solution:
Concentration becomes half
Example: 100L of 60% solution + 100L water = 200L of 30%
Shortcut 3: Replacement Formula
For repeated equal replacements:
Final quantity = Original × (1 - fraction)ⁿ
Where n = number of times
🔢 Advanced Problems
Example 8: Milk-Water Mixture
Q: A vessel contains milk and water in ratio 3:2. If 10 liters is removed and replaced with water, ratio becomes 2:3. Find original quantity.
Solution:
Original ratio = 3:2 (Total = 5 parts)
Let original quantity = 5x
Milk = 3x, Water = 2x
After removing 10L:
Milk removed = (3/5) × 10 = 6L
Water removed = (2/5) × 10 = 4L
After replacement with 10L water:
Milk = 3x - 6
Water = 2x - 4 + 10 = 2x + 6
New ratio = 2:3
(3x - 6) : (2x + 6) = 2:3
3(3x - 6) = 2(2x + 6)
9x - 18 = 4x + 12
5x = 30
x = 6
Original quantity = 5x = 30 liters
Answer: 30 liters
Example 9: Profit Problem with Alligation
Q: A shopkeeper mixes two types of rice. If he sells at cost price of dearer rice, he gains 20%. If he sells at cost price of cheaper rice, he loses 20%. Find ratio of mixing.
Solution:
Let Cheaper = C, Dearer = D, Mean = M
Gain 20% when selling at D:
D = M × 1.2
M = D/1.2 = 5D/6
Loss 20% when selling at C:
C = M × 0.8
M = C/0.8 = 5C/4
From both:
5D/6 = 5C/4
D/6 = C/4
4D = 6C
D = 1.5C
Using Alligation:
C : D = (D - M) : (M - C)
We know M = 5D/6
D - M = D - 5D/6 = D/6
M - C = 5D/6 - C
Since D = 1.5C:
M = 5(1.5C)/6 = 7.5C/6 = 1.25C
D - M = 1.5C - 1.25C = 0.25C
M - C = 1.25C - C = 0.25C
Ratio = 0.25C : 0.25C = 1:1
Answer: 1:1
⚠️ Common Mistakes
❌ Mistake 1: Wrong Alligation Direction
Wrong: Using (C - M) and (M - D) ✗
Right: Always (D - M) and (M - C) ✓
Remember: Cross-multiply diagonally!
❌ Mistake 2: Replacement Confusion
Wrong: After removing x liters, x liters milk remains ✗
Right: Milk removed = (Milk fraction) × x ✓
❌ Mistake 3: Adding Concentrations
Wrong: 30% + 40% = 70% solution ✗
Right: Must use weighted average based on quantities ✓
❌ Mistake 4: Multiple Operations
Wrong: Treating multiple replacements as single replacement ✗
Right: Use (1 - q/Q)ⁿ formula ✓
📝 Practice Problems
Level 1:
- Mix ₹40/kg and ₹60/kg tea in what ratio to get ₹50/kg tea?
- Equal quantities of 20% and 40% solutions mixed. Find final concentration.
- 100L milk. 10L removed and replaced with water. Find milk remaining.
Level 2:
- Three liquids at ₹30, ₹40, ₹50 per liter mixed in ratio 2:3:5. Find mean price.
- 80L of 75% alcohol. How much water to add to make it 60%?
- 100L mixture. 20L removed twice and replaced with water. Find final milk if original had 80L milk.
Level 3:
- Milk and water in 4:1 ratio. 15L removed and replaced with water makes ratio 3:2. Find original quantity.
- Two alloys: 40% gold and 60% gold. In what ratio to mix for 50% gold alloy?
- A vessel has 60L mixture (milk:water = 2:1). x liters removed and replaced with water makes ratio 1:2. Find x.
🔗 Related Topics
Prerequisites:
- Ratio & Proportion - Foundation for alligation
- Percentage - For concentration calculations
- Average - Weighted average concept
Related:
- Profit & Loss - Mean price problems
- Partnership - Uses ratio concepts
Practice:
🎯 Continue Your Learning Journey
Master Alligation - The visual method makes complex mixing problems simple! ⚗️
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