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

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.
100L Assumption Method: For % concentration problems, assume 100L total. Converts percentages directly to liters, making addition/removal arithmetic trivial.
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.
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.
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:

Frequently Asked Questions

What is the salary range for placements that test Mixtures & Alligations in 2026?

Mixtures & Alligations is typically part of the aptitude/quant section used for shortlisting, so it appears across many salary bands rather than being tied to one specific CTC. In 2026 drives, candidates who clear the overall aptitude and reasoning cutoffs (not just this topic) commonly land in the standard mass-recruitment ranges offered by companies like TCS, Infosys, and Wipro.

Who is eligible to attempt Mixtures & Alligations questions for placement exams in 2026?

Eligibility is generally based on meeting the company’s overall criteria (education, CGPA/percentage, and backlogs policy), not on prior mastery of this topic. For aptitude preparation, any student comfortable with basic arithmetic (ratio, percentage, LCM/HCF) can start Mixtures & Alligations and improve quickly with practice.

How difficult are Mixtures & Alligations questions in placement exams?

Most placement questions are moderate in difficulty and focus on correct application of the alligation rule, successive dilution, and profit/loss logic. The difficulty usually increases when problems combine multiple steps (e.g., two mixtures plus a change in concentration), so speed and accuracy depend on practice and formula recall.

What are the best preparation tips for Mixtures & Alligations for 2026 placements?

Start by mastering the core rules: alligation (weighted average using differences), successive dilution, and profit/concentration conversions. Then practice a structured set of problems daily, begin with single-step mixtures, move to successive dilution, and finally attempt mixed sets with time limits to build exam speed.

How many interview rounds typically include aptitude topics like Mixtures & Alligations?

In most campus drives, Mixtures & Alligations is tested in the written/online assessment (aptitude/quant) rather than in technical interviews. Some companies may also include a short quantitative section in subsequent rounds, but the main scoring opportunity is usually in the initial screening test.

What are the most common topics asked under Mixtures & Alligations in placement exams?

Common patterns include alligation (mixing two liquids with different concentrations), successive dilution (mixing multiple times), and profit/loss using mixture cost price and selling price logic. You’ll also frequently see ratio-based mixture questions where the final concentration or quantity must be derived using difference methods.

How can I apply and practice effectively using Mixtures & Alligations resources for 2026 placements?

Use a placement-focused practice plan: collect PYQs and topic-wise sets, solve them in timed mode, and maintain an error log for recurring mistakes (sign errors, wrong difference pairing, or unit confusion). If you’re using a course or question bank, prioritize “formula + solved examples + mixed practice” sequences and revise weekly.

What is the selection rate impact of performing well in Mixtures & Alligations?

There isn’t a single fixed “selection rate” tied only to this topic, but strong performance in quant aptitude improves your chances of clearing sectional and overall cutoffs. Since Mixtures & Alligations is high-scoring when solved accurately, consistent practice can directly raise your total score and improve shortlisting odds.