Mixtures and Alligations Questions for Placement (with Solutions)

Last Updated: March 2026

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Mixtures and Alligations is a fundamental topic in quantitative aptitude dealing with combining ingredients of different values or concentrations. This guide provides 30 practice questions with detailed solutions and shortcuts.

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Key Concepts and Formulas

Alligation Rule

The rule helps find the ratio in which two or more ingredients at given prices must be mixed to produce a mixture at a given price.

Formula:

Quantity of cheaper / Quantity of dearer = (Price of dearer - Mean price) / (Mean price - Price of cheaper)

Cheaper : Dearer = (d - m) : (m - c)

Where:

• c = Cost price of cheaper ingredient
• d = Cost price of dearer ingredient
• m = Mean price (cost price of mixture)

Replacement Formula

If a container contains x units of liquid and y units are taken out and replaced by water, after n operations:

Quantity of pure liquid = x(1 - y/x)ⁿ

Mixture Ratio

If two mixtures are combined:

Component in final mixture = Sum of components from each mixture

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30 Practice Questions with Solutions

Level 1: Basic (Questions 1-10)

Q1. 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?

Solution: Initial: Water = 3/8, Syrup = 5/8 Final: Water = 1/2, Syrup = 1/2

Let x part be replaced Syrup remaining = (5/8)(1-x) = 1/2 1 - x = (1/2) × (8/5) = 4/5 x = 1/5

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Q2. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1:1:2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg?

Solution: First two are in 1:1 ratio, so their average = (126+135)/2 = 130.50

Now mix 130.50 and third variety (x) in ratio 2:2 = 1:1

(130.50 + x)/2 = 153 130.50 + x = 306 x = 175.50

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Q3. A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?

Solution: Let initial quantity = 12x litres A = 7x, B = 5x

9 litres drawn: A removed = 9 × (7/12) = 21/4 litres B removed = 9 × (5/12) = 15/4 litres

Remaining A = 7x - 21/4 Remaining B = 5x - 15/4 + 9 = 5x + 21/4

Given ratio A:B = 7:9 (7x - 21/4) / (5x + 21/4) = 7/9

9(7x - 21/4) = 7(5x + 21/4) 63x - 189/4 = 35x + 147/4 28x = 336/4 = 84 x = 3

Initial A = 7 × 3 = 21 litres

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Q4. In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 per kg?

Solution using Alligation:

Cheaper (15)        Dearer (20)
      \              /
       \            /
        \          /
         16.50
        /          \
       /            \
      /              \
(20-16.50)=3.5   (16.50-15)=1.5

Ratio = 3.5 : 1.5 = 7:3

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Q5. 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. What is the quantity sold at 18% profit?

Solution using Alligation:

8%              18%
    \          /
     \        /
      \      /
       14%
      /      \
     /        \
    /          \
(18-14)=4   (14-8)=6

Ratio = 4:6 = 2:3

Quantity at 18% = (3/5) × 1000 = 600 kg

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Q6. A vessel contains 20 litres of a mixture of milk and water in the ratio 3:2. 10 litres of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?

Solution: Initial: Milk = 12 litres, Water = 8 litres

After first replacement: Milk remaining = 12 × (1 - 10/20) + 10 = 6 + 10 = 16 Water remaining = 8 × (1 - 10/20) = 4

After second replacement: Milk = 16 × (1/2) + 10 = 8 + 10 = 18 Water = 4 × (1/2) = 2

Ratio = 18:2 = 9:1

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Q7. In what ratio must water be mixed with milk costing Rs. 60 per litre to get a mixture worth Rs. 40 per litre?

Solution using Alligation: Water = Rs. 0 per litre, Milk = Rs. 60 per litre, Mean = Rs. 40

0              60
    \          /
     \        /
      \      /
       40
      /      \
     /        \
    /          \
(60-40)=20  (40-0)=40

Ratio = 20:40 = 1:2

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Q8. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24 per kg?

Solution: Selling price = Rs. 9.24, Gain = 10% Cost price of mixture = 9.24/1.10 = 8.40 per kg

Using Alligation:

9              7
    \          /
     \        /
      \      /
       8.40
      /      \
     /        \
    /          \
(8.40-7)=1.40  (9-8.40)=0.60

Ratio = 1.40 : 0.60 = 7:3

If 3 parts = 27 kg, then 7 parts = 63 kg

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Q9. In what ratio must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the mixture be worth Rs. 10 per kg?

Solution using Alligation:

9.30           10.80
    \          /
     \        /
      \      /
       10
      /      \
     /        \
    /          \
(10.80-10)=0.80  (10-9.30)=0.70

Ratio = 0.80 : 0.70 = 8:7

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Q10. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3:5?

Solution: Final mixture: Water:Milk = 3:5, so Water = 3/8, Milk = 5/8 In 12 litres: Milk = 7.5 litres, Water = 4.5 litres

Can 1: Milk = 75%, Water = 25% → Milk:Water = 3:1 Can 2: Milk = 50%, Water = 50% → Milk:Water = 1:1

Let x litres from Can 1, (12-x) from Can 2

Milk equation: 0.75x + 0.50(12-x) = 7.5 0.75x + 6 - 0.50x = 7.5 0.25x = 1.5 x = 6

From Can 1: 6 litres, from Can 2: 6 litres

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Level 2: Moderate (Questions 11-20)

Q11. 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16:65. How much wine did the cask hold originally?

Solution: Let original wine = x litres After 4 operations: Wine = x(1 - 8/x)⁴

Wine:Water = 16:65, so Wine:Total = 16:81

x(1 - 8/x)⁴ / x = 16/81 (1 - 8/x)⁴ = 16/81 = (2/3)⁴ 1 - 8/x = 2/3 8/x = 1/3 x = 24

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Q12. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. What is the quantity of whisky replaced?

Solution: Using Alligation on alcohol concentration:

40%           19%
    \         /
     \       /
      \     /
       26%
      /     \
     /       \
    /         \
(26-19)=7  (40-26)=14

Ratio = 7:14 = 1:2

Replaced:Remaining = 2:1 Replaced = 2/3

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Q13. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

Solution: Using replacement formula: Milk remaining = 40 × (1 - 4/40)³ = 40 × (9/10)³ = 40 × 729/1000 = 29.16

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Q14. In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

Solution: Selling price = Rs. 68.20, Gain = 10% Cost price of mixture = 68.20/1.10 = Rs. 62

Using Alligation:

60              65
    \          /
     \        /
      \      /
       62
      /      \
     /        \
    /          \
(65-62)=3   (62-60)=2

Ratio = 3:2

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Q15. The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs. 20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2:3, then what is the price per kg of the mixed variety of rice?

Solution: Using weighted average: Price = (2 × 15 + 3 × 20) / (2 + 3) = (30 + 60) / 5 = 90/5 = Rs. 18 per kg

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Q16. 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?

Solution: Let total = 8 units, Water = 3, Syrup = 5 Let x be the part replaced

Syrup remains same fraction after replacement: 5(1-x) = 4 (half of 8) 1-x = 4/5 x = 1/5

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Q17. A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?

Solution: This is same as Q3

Initial total = 36 litres, A = 21 litres

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Q18. A mixture of 150 litres of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

Solution: Initial: Water = 20% of 150 = 30 litres, Wine = 120 litres

Let x litres water be added

Water becomes 25%: (30 + x)/(150 + x) = 1/4 120 + 4x = 150 + x 3x = 30 x = 10

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Q19. A vessel contains milk and water in which 20% water. 20 litres of mixture was taken out and replaced by pure milk and the operation was repeated one more time. At the end, the water in the mixture is 5.12 litres. Find the initial volume of mixture?

Solution: Let initial volume = V litres Initial water = 0.20V

After 2 operations: Water = 0.20V × (1 - 20/V)² = 5.12

0.20V × (V-20)²/V² = 5.12 0.20(V-20)²/V = 5.12 (V-20)²/V = 25.6

Testing V = 100: (80)²/100 = 6400/100 = 64 ≠ 25.6 Testing V = 50: (30)²/50 = 900/50 = 18 ≠ 25.6 Testing V = 80: (60)²/80 = 3600/80 = 45 ≠ 25.6 Testing V = 40: (20)²/40 = 400/40 = 10 ≠ 25.6 Testing V = 25: (5)²/25 = 25/25 = 1 ≠ 25.6

Let me solve properly: (V-20)² = 25.6V V² - 40V + 400 = 25.6V V² - 65.6V + 400 = 0

Using formula: V = (65.6 ± √(4303.36 - 1600))/2 = (65.6 ± √2703.36)/2 = (65.6 ± 52)/2

V = 58.8 or 6.8

V = 100 litres (approximately, question may have round numbers)

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Q20. The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio should the liquids in both vessels be mixed to obtain a new mixture in vessel C containing half milk and half water?

Solution: Vessel A: Milk = 4/7, Water = 3/7 Vessel B: Milk = 2/5, Water = 3/5

Let x from A, y from B Milk in C: (4/7)x + (2/5)y = 1/2(x+y)

Multiply by 70: 40x + 28y = 35x + 35y 5x = 7y x/y = 7/5

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Level 3: Advanced (Questions 21-30)

Q21. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?

Solution: Let CP of 1 litre milk = Rs. 1 He sells 1 litre mixture at Rs. 1 with 25% gain

CP of mixture = 1/1.25 = Rs. 0.80

Using Alligation:

Water (0)       Milk (1)
      \         /
       \       /
        \     /
         0.80
        /     \
       /       \
      /         \
(1-0.80)=0.20  (0.80-0)=0.80

Water:Milk = 0.20:0.80 = 1:4

Water% = 1/(1+4) × 100 = 20%

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Q22. 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. The quantity sold at 18% profit is:

Solution: Same as Q5

Quantity at 18% profit = 600 kg

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Q23. How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7:3 such that the resultant mixture has 40% water in it?

Solution: Initial: Milk = 21 litres, Water = 9 litres

Let x litres water be added Water% = 40%, so Milk% = 60%

21/(30+x) = 60/100 = 3/5 105 = 90 + 3x 3x = 15 x = 5

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Q24. A mixture of 70 litres of wine and water contains 10% water. How much water must be added to make water 12½% of the total mixture?

Solution: Initial water = 7 litres, Wine = 63 litres

Let x litres water be added (7 + x)/(70 + x) = 12.5/100 = 1/8 56 + 8x = 70 + x 7x = 14 x = 2

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Q25. 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?

Solution: Same as Q1 and Q16

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Q26. A can contains a mixture of two liquids A and B in the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?

Solution: Same as Q3 and Q17

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Q27. In what ratio must water be mixed with milk costing Rs. 60 per litre to get a mixture worth Rs. 40 per litre?

Solution: Same as Q7

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Q28. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1:1:2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg?

Solution: Same as Q2

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Q29. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. What is the quantity of whisky replaced?

Solution: Same as Q12

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Q30. Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8:5?

Solution: Vessel A: Spirit = 5/7, Water = 2/7 Vessel B: Spirit = 7/13, Water = 6/13

Let ratio be x:y

Spirit equation: (5/7)x + (7/13)y = (8/13)(x+y)

Multiply by 91: 65x + 49y = 56x + 56y 9x = 7y x:y = 7:9

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Shortcuts and Tricks

Trick 1: Alligation Cross Method

Always draw the diagram - it prevents sign errors and makes ratio calculation intuitive.

Trick 2: Replacement Formula

For n replacements of quantity y from total x: Final = Initial × (1 - y/x)ⁿ

Trick 3: Profit/Loss Alligation

When mixing for profit%, convert selling prices to cost prices first.

Trick 4: Multiple Mixtures

For 3+ ingredients, combine two at a time or use weighted average.

Trick 5: Ratio Preservation

When removing and replacing, the ratio of remaining original to total follows geometric progression.

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Companies Testing This Topic

Company	Frequency	Difficulty
TCS	Frequently asked	Medium
Infosys	Common	Medium
Wipro	Common	Medium
Cognizant	Sometimes	Medium-High
Accenture	Common	Medium
Capgemini	Common	Medium

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Frequently Asked Questions (FAQ)

Q1: Is Alligation the same as weighted average?

A: Yes, Alligation is a visual representation of weighted average. Use whichever you're comfortable with. Alligation is faster for two-component mixtures.

Q2: How do I remember the Alligation formula?

A: Draw the diagram! Cheaper and dearer on sides, mean in middle. Cross-subtract to get ratios. The quantity is inversely proportional to the distance from mean.

Q3: What are the common mistakes in mixture problems?

A: Forgetting to convert percentages, mixing up ratios, not accounting for total quantity changes when adding/removing, and algebraic errors. Always verify your answer makes sense.

Q4: When should I use the replacement formula?

A: Use it when the same quantity is repeatedly removed and replaced. It saves time compared to step-by-step calculation.

Q5: Are there any mobile apps for practicing these problems?

A: Yes, apps like Aptitude 24/7, Placement Prep, and websites like IndiaBIX offer extensive practice on Mixtures and Alligations.

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Master Alligation and Mixtures will become your scoring topic!