Alligation or Mixture
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Solved Examples
Study MaterialSolved Examples – Alligation or Mixture
Solved examples help students understand the practical application of alligation, ratios, percentages, replacement concepts, and concentration problems in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, Insurance, Defence, CAT, CDS, NDA, and various aptitude examinations.
Topics Covered in Solved Examples
- Alligation Rule Problems
- Milk and Water Mixtures
- Mean Price Problems
- Replacement of Liquids
- Concentration and Dilution
- Percentage-Based Mixtures
- Profit and Cost Mixtures
- Advanced Mixture Applications
Example 1: Basic Alligation Problem
Question: In what ratio should rice costing ₹20 per kg be mixed with rice costing ₹30 per kg so that the mixture costs ₹24 per kg?
Solution:
Using alligation:
Cheaper price = ₹20
Dearer price = ₹30
Mean price = ₹24
Required ratio:
= (30 − 24) : (24 − 20)
= 6 : 4
= 3 : 2
Answer: 3 : 2
Example 2: Mean Price Problem
Question: A shopkeeper mixes 20 kg sugar costing ₹40/kg with 30 kg sugar costing ₹50/kg. Find the mean price of the mixture.
Solution:
Total cost:
= (20 × 40) + (30 × 50)
= 800 + 1500
= ₹2300
Total quantity:
= 20 + 30
= 50 kg
Mean price:
= 2300/50
= ₹46/kg
Answer: ₹46 per kg
Example 3: Milk and Water Problem
Question: In a mixture of 60 litres, the ratio of milk and water is 2 : 1. Find the quantity of milk.
Solution:
Total ratio:
= 2 + 1
= 3
Milk quantity:
= (2/3) × 60
= 40 litres
Answer: 40 litres
Example 4: Pure Alcohol Problem
Question: How many litres of pure alcohol should be added to 100 litres of 20% alcohol solution to make it 25% alcohol?
Solution:
Alcohol in original solution:
= 20% of 100
= 20 litres
Let x litres of pure alcohol be added.
Then:
0.25(100 + x) = 20 + x
25 + 0.25x = 20 + x
5 = 0.75x
x = 20/3 litres
Answer: 20/3 litres
Example 5: Replacement Problem
Question: A vessel contains 40 litres of milk. From it, 10 litres are removed and replaced with water. Find the quantity of milk left after one operation.
Solution:
Milk left:
= 40 × [(40 − 10)/40]
= 40 × 30/40
= 30 litres
Answer: 30 litres
Example 6: Repeated Replacement Problem
Question: A vessel contains 64 litres of pure milk. 16 litres are removed and replaced with water. This process is repeated twice more. Find the quantity of milk left.
Solution:
Using replacement formula:
Milk left:
= 64 × [(64 − 16)/64]3
= 64 × (48/64)3
= 64 × (3/4)3
= 64 × 27/64
= 27 litres
Answer: 27 litres
Example 7: Mean Price by Alligation
Question: Tea worth ₹45/kg is mixed with tea worth ₹60/kg in the ratio 2 : 3. Find the price of the mixture.
Solution:
Mean price:
= [(2 × 45) + (3 × 60)] / (2 + 3)
= (90 + 180)/5
= 270/5
= ₹54/kg
Answer: ₹54 per kg
Example 8: Water Addition Problem
Question: A container has 50 litres of milk. How much water should be added so that milk becomes 80% of the mixture?
Solution:
Let x litres water be added.
Then:
50/(50 + x) = 80/100
50/(50 + x) = 4/5
250 = 200 + 4x
4x = 50
x = 12.5 litres
Answer: 12.5 litres
Example 9: Ratio Problem
Question: In what ratio should two solutions containing 25% and 40% alcohol be mixed to obtain a 30% alcohol solution?
Solution:
Using alligation:
Required ratio:
= (40 − 30) : (30 − 25)
= 10 : 5
= 2 : 1
Answer: 2 : 1
Example 10: Percentage Mixture Problem
Question: A 40 litre mixture contains 25% water. Find the quantity of water.
Solution:
Water quantity:
= 25% of 40
= (25/100) × 40
= 10 litres
Answer: 10 litres
Example 11: Cost-Based Mixture Problem
Question: How many kg of rice costing ₹35/kg should be mixed with 15 kg rice costing ₹50/kg so that the mixture costs ₹45/kg?
Solution:
Using alligation:
Ratio:
= (50 − 45) : (45 − 35)
= 5 : 10
= 1 : 2
Cheaper : Dearer = 1 : 2
Dearer quantity = 15 kg
Cheaper quantity:
= 15 × 1/2
= 7.5 kg
Answer: 7.5 kg
Example 12: Wine and Water Problem
Question: A vessel contains 80 litres of wine. 20 litres are removed and replaced with water. Find the quantity of wine left after one replacement.
Solution:
Wine left:
= 80 × [(80 − 20)/80]
= 80 × 60/80
= 60 litres
Answer: 60 litres
Example 13: Advanced Replacement Problem
Question: A cask contains 100 litres of milk. 20 litres are removed and replaced with water. The operation is repeated twice more. Find the quantity of milk left.
Solution:
Milk left:
= 100 × [(100 − 20)/100]3
= 100 × (80/100)3
= 100 × (4/5)3
= 100 × 64/125
= 51.2 litres
Answer: 51.2 litres
Example 14: Equal Quantity Mixture Problem
Question: Equal quantities of sugar costing ₹40/kg and ₹60/kg are mixed. Find the mean price.
Solution:
Equal quantities mean average price:
= (40 + 60)/2
= ₹50/kg
Answer: ₹50 per kg
Example 15: Pure Substance Problem
Question: A solution contains 30% alcohol. Find the quantity of alcohol in 90 litres solution.
Solution:
Alcohol quantity:
= (30/100) × 90
= 27 litres
Answer: 27 litres
Important Exam Tips
- Use cross-difference method directly.
- Remember mean lies between cheaper and dearer values.
- Focus on pure substance quantity.
- Practice replacement formula regularly.
- Improve percentage calculation speed.
- Simplify ratios immediately.
- Practice previous year aptitude questions.
Common Mistakes to Avoid
- Using incorrect alligation ratios.
- Confusing cheaper and dearer quantities.
- Ignoring concentration changes.
- Calculation mistakes in percentages.
- Using wrong replacement powers.
Practicing solved examples regularly improves conceptual clarity, calculation speed, and logical analysis in solving Alligation or Mixture aptitude questions in competitive examinations.