Mixtures AND Alligations FOR Placement

Meta Description: Master Mixtures & Alligations for 2026 placements. Learn alligation rule, successive dilution & profit formulas. Solve 18+ TCS, Infosys, Wipro exam PYQs.

Introduction

Mixtures and Alligations is a high-scoring, frequently tested arithmetic topic in campus recruitment drives. Every year, top IT and consulting companies like TCS, Infosys, Wipro, Capgemini, Accenture, and Cognizant include 2 to 4 direct questions from this chapter in their quantitative aptitude sections. The topic carries a typical weightage of 8–12% in placement papers, making it essential for candidates aiming for top percentiles.

Unlike complex algebra, Mixtures and Alligations relies on logical ratio-building and quick mental math. It bridges basic percentage, profit-loss, and ratio concepts, appearing in real-world scenarios like milk-water replacement, chemical blending, and commodity trading. With systematic practice, students can solve these questions in under 45 seconds. This guide covers two-ingredient mixtures, the alligation cross method, wine-water replacement patterns, successive dilution, and profit/loss applications, complete with 18+ solved problems and company-specific patterns for 2026.

Key Formulas & Concepts

Understanding the core formulas is crucial before attempting problems. Here are the essential concepts formatted for quick recall:

1. Alligation Cross Rule (Mean Price Concept)

Used to find the ratio in which two or more ingredients at different prices/quantities/concentrations must be mixed to achieve a desired mean value.

Cheaper Ingredient (C)     Higher Ingredient (H)
          \                       /
           \                     /
            \                   /
             Mean Price (M)

Ratio Formula: [ \frac{\text{Quantity of Cheaper}}{\text{Quantity of Higher}} = \frac{H - M}{M - C} ] (Note: Always subtract diagonally and place the result opposite to the ingredient.)

2. Successive Dilution / Repeated Replacement Formula

When x quantity is removed and replaced with a diluent (like water) n times from an initial quantity V: [ \text{Final Quantity of Original Liquid} = V \times \left(1 - \frac{x}{V}\right)^n ] Concentration after n replacements: [ \text{Final Concentration %} = \text{Initial %} \times \left(1 - \frac{x}{V}\right)^n ]

3. Mixture Cost & Profit/Loss

• Cost Price (CP) of Mixture = (\frac{(C_1 \times Q_1) + (C_2 \times Q_2) + \dots}{Q_1 + Q_2 + \dots})
• Selling Price (SP) = (CP \times \left(1 + \frac{\text{Profit %}}{100}\right)) or (CP \times \left(1 - \frac{\text{Loss %}}{100}\right))

4. Wine-Water Type Problems

When two mixtures are combined:

• Total Quantity = Sum of individual quantities
• Total Solute (Milk/Wine/Chemical) = Sum of individual solutes
• Final Concentration = (\frac{\text{Total Solute}}{\text{Total Quantity}} \times 100)

Solved Examples (Basic Level)

Q1. Two types of rice cost ₹40/kg and ₹60/kg. In what ratio should they be mixed to get a mixture costing ₹48/kg? Solution: Using Alligation Cross: Cheaper (₹40) | Higher (₹60) | Mean (₹48) Ratio = (60 - 48) : (48 - 40) = 12 : 8 = 3 : 2

Q2. A 40L mixture contains 25% milk. How much water must be added to make it 20% milk? Solution: Milk remains constant. Initial milk = 25% of 40 = 10L. Let final mixture = x L. 20% of x = 10 → x = 50L. Water added = 50 - 40 = 10L

Q3. From a 50L container of pure milk, 10L is removed and replaced with water. This is done twice. Find the final milk quantity. Solution: Using successive replacement: V = 50, x = 10, n = 2 Final Milk = (50 \times (1 - 10/50)^2 = 50 \times (4/5)^2 = 50 \times 16/25 = 32L)

Q4. A shopkeeper mixes tea at ₹120/kg and ₹150/kg in ratio 2:3. Find CP of mixture. Solution: CP = (\frac{(120 \times 2) + (150 \times 3)}{2 + 3} = \frac{240 + 450}{5} = \frac{690}{5} = ₹138/kg)

Q5. A 60L solution contains 30% acid. How much acid must be added to make it 50% acid? Solution: Initial acid = 18L, water = 42L. Water remains constant. In final mixture, 50% acid → 50% water = 42L → Total = 84L Acid added = 84 - 60 = 24L

Practice Questions (Medium Level)

Q6. In what ratio must a grocer mix two types of sugar costing ₹35/kg and ₹45/kg to sell at ₹42/kg and gain 5%? Solution: Target CP = (42 / 1.05 = ₹40/kg) Alligation: (45 - 40) : (40 - 35) = 5 : 5 = 1 : 1

Q7. A vessel contains 72L of milk. 8L is drawn and replaced with water. The process is repeated 3 times. Find water in final mixture. Solution: Final Milk = (72 \times (1 - 8/72)^3 = 72 \times (8/9)^3 = 72 \times 512/729 ≈ 50.57L) Water = 72 - 50.57 = 21.43L

Q8. Alloy A has Cu:Zn = 3:2. Alloy B has Cu:Zn = 4:3. Equal quantities are mixed. Find Cu:Zn in new alloy. Solution: Assume 10 units each (LCM of 5 and 7 = 35, but 10 works for ratio). A: Cu=6, Zn=4 | B: Cu=40/7≈5.71, Zn=30/7≈4.29 Better: Take 35g each. A: Cu=21, Zn=14 | B: Cu=20, Zn=15 Total Cu=41, Zn=29 → 41 : 29

Q9. A milkman buys milk at ₹50/L, adds 20% water, and sells at ₹60/L. Find profit %. Solution: 100L milk → CP = ₹5000. Adds 20L water → Total 120L. SP = 120 × 60 = ₹7200. Profit = 2200. Profit % = (2200/5000)×100 = 44%

Q10. Two solutions have 40% and 60% alcohol. Mixed to get 48% alcohol. Find ratio. Solution: Alligation: (60-48) : (48-40) = 12 : 8 = 3 : 2

Q11. A 100L tank has 10% salt. 20L is removed and replaced with pure water twice. Final salt %? Solution: Final salt = (10 \times (1 - 20/100)^2 = 10 \times 0.64 = 6.4L) % = (6.4/100)×100 = 6.4%

Q12. Wheat A (₹28/kg) and B (₹36/kg) mixed. 100kg sold at ₹40 with 25% profit. Find A:B. Solution: CP = 40/1.25 = ₹32/kg Alligation: (36-32):(32-28) = 4:4 = 1:1

Q13. A container has 30L juice. 5L removed, replaced with water. Repeated 4 times. Juice left? Solution: (30 \times (1 - 5/30)^4 = 30 \times (5/6)^4 = 30 \times 625/1296 ≈ 14.47L)

Advanced Questions

Q14. Three varieties of sugar cost ₹30, ₹40, and ₹50/kg. How should they be mixed to get ₹42/kg mixture? Solution: Pairwise alligation or weighted average. Let quantities be x, y, z. (30x + 40y + 50z = 42(x+y+z)) Simplifies to (12x + 2y - 8z = 0) → (6x + y = 4z). Infinite solutions, but common ratio: x:y:z = 2:4:4 (or 1:2:2)

Q15. From 80L pure wine, 20L is replaced with water. Then 16L of mixture replaced with water. Find final wine:water ratio. Solution: Step 1: Wine = 60L, Water = 20L (Ratio 3:1) Step 2: Removing 16L removes (16 \times 3/4 = 12L) wine, (4L) water. Wine left = 60-12 = 48L. Water = 20-4+16 = 32L. Ratio = 48:32 = 3:2

Q16. A trader mixes 3 types of pulses at ₹45, ₹55, ₹70/kg in ratio 3:2:1. Sells at 20% profit. Find SP/kg. Solution: CP = ((45×3 + 55×2 + 70×1)/6 = (135+110+70)/6 = 315/6 = ₹52.50) SP = 52.5 × 1.20 = ₹63/kg

Q17. A 50L milk-water mix has ratio 3:2. How much milk must be added to make it 4:3? Solution: Initial: Milk=30L, Water=20L. Water constant. Final ratio 4:3 → Water (3 parts) = 20L → 1 part = 20/3 Milk needed (4 parts) = 80/3 ≈ 26.67L Added = 26.67 - 30? Wait, ratio changes. Actually, 4:3 means water is 3/7. 20/(20+M) = 3/7 → 140 = 60+3M → M=80/3. Added = 80/3 - 30 = -10/3? Impossible. Recheck: 3:2 → 30M, 20W. Target 4:3 means M/W=4/3 → M= (4/3)×20 = 80/3. But initial M=30=90/3. Already more! Question implies adding water, not milk. If question meant adding milk to change ratio from 3:2 to 7:5, it works. Let's fix logic: Add milk → ratio becomes higher. 3:2 to 4:3 is actually decreasing milk proportion. So question is flawed as stated. Corrected version: Change to 7:5. Then M/W=7/5 → M=28. Wait, initial 30. Still wrong. Let's use standard: How much water to add to change 3:2 to 1:1? Water=30. Added=10L. I'll adjust Q17 to a valid one: Q17 (Revised). A 60L mixture has milk:water = 5:1. How much water must be added to make it 2:1? Solution: Milk = 50L. Target ratio 2:1 → Water = 25L. Initial water = 10L. Added = 15L

Q18. Two tanks: A has 40% alcohol, B has 70%. 10L from A and 15L from B mixed, then 5L water added. Final alcohol %? Solution: Alcohol from A = 4L, from B = 10.5L. Total = 14.5L. Total mix = 10+15+5=30L. % = (14.5/30)×100 = 48.33%

Q19. Successive replacement: 36L milk. x L replaced twice. Final milk = 20.25L. Find x. Solution: (20.25 = 36 \times (1 - x/36)^2) → ((1 - x/36)^2 = 20.25/36 = 0.5625 = (0.75)^2) (1 - x/36 = 0.75) → (x/36 = 0.25) → x = 9L

Common Mistakes to Avoid

• Inverting Alligation Ratios: Placing (M-C):(H-M) instead of (H-M):(M-C). Always remember: Higher - Mean goes to Cheaper quantity, Mean - Cheaper goes to Higher quantity.
• Forgetting Constant Component: In wine-water or milk-water problems, always identify what remains constant (usually water or original liquid) before replacement.
• Misapplying Dilution Formula: Using (x/V)^n instead of (1 - x/V)^n for remaining original liquid.
• Mixing CP and SP in Alligation: Alligation works only on Cost Prices or pure concentrations, never on Selling Prices unless profit/loss is removed first.
• Ignoring Unit Consistency: Mixing liters with gallons, or percentages with absolute quantities without conversion.
• Assuming Equal Ratios in Multi-Ingredient Mixtures: For 3+ ingredients, pairwise alligation requires careful balancing; weighted average is safer.

Shortcut Tricks

1. Cross Diagram Memory Trick: Draw an "X". Top-left = Cheaper, Top-right = Higher, Center = Mean. Bottom-left gets (Higher - Mean), Bottom-right gets (Mean - Cheaper). Instant ratio.
2. 100L Assumption Method: For % concentration problems, assume 100L total. Converts percentages directly to liters, making addition/removal arithmetic trivial.
3. Replacement Shortcut (2 Steps): If same quantity x is replaced twice, remaining fraction = (\left(\frac{V-x}{V}\right)^2). Memorize squares of common fractions (e.g., 4/5 → 16/25, 9/10 → 81/100) for 2-second calculations.
4. Profit-Alligation Combo Trick: When SP and profit % are given, mentally divide SP by (1 + P/100) to get CP before applying alligation. Saves writing steps.
5. Constant Solute Rule: If only solvent is added/removed, (\text{Initial Conc} \times \text{Initial Vol} = \text{Final Conc} \times \text{Final Vol}). Rearrange instantly: (\text{Final Vol} = \frac{\text{Initial Conc} \times \text{Initial Vol}}{\text{Final Conc}}).

Previous Year Questions

Q1. (TCS NQT 2024 Pattern) A container has pure milk. 10L is drawn out and replaced with water. This is done 3 times. Ratio of milk to water becomes 64:61. Find initial volume. Solution: Milk ratio = 64/125. ((1 - 10/V)^3 = 64/125 = (4/5)^3) → (1 - 10/V = 4/5) → (10/V = 1/5) → V = 50L

**Q2. (Infosys 2025 Pattern) Two alloys A (Cu: