Quantitative Aptitude · Lecture Series

Mixture Problems in Mathematics

A complete lecture sheet covering theory, formulas, alligation method, and 15 graded practice problems — aligned with GMAT & Bank PO / SSC exam patterns.

GMAT Bank PO / SSC 15 Solved Problems 3 Difficulty Levels

Contents

Part I — Theory & Core Concepts

Part II — Practice Problems

Level 1: Easy (Q1–Q5) | Level 2: Medium (Q6–Q10) | Level 3: Hard (Q11–Q15)

Part I

Theory & Core Concepts

§1 What are Mixture Problems?

A mixture problem involves combining two or more substances (or groups of items with different properties) to form a new combined entity — and then reasoning about the properties of that combination. In competitive exams, these substances are typically:

Liquids — milk & water, acid & water, alcohol solutions
Dry commodities — grains, spices, tea, coffee at different prices
Metals & alloys — gold, silver, copper in different proportions

The core question is always: What is the resulting concentration, ratio, or cost after mixing? Or inversely: In what proportion must we mix to achieve a desired result?

On the GMAT, mixture problems appear in Problem Solving and Data Sufficiency formats and test algebraic reasoning. In Bank PO / SSC exams, they heavily test the Alligation shortcut and weighted averages.

§2 The Alligation Method

Alligation is a graphical/arithmetic technique used to find the ratio in which two ingredients at different prices (or concentrations) must be mixed to produce a mixture at a desired intermediate price (or concentration).

The Alligation Rule:

C₁

Cheaper Price

C₂

Dearer Price

↙ M ↘

Mean / Target

C₂ − M

Quantity of Cheaper

M − C₁

Quantity of Dearer

Formula: Quantity of Cheaper : Quantity of Dearer = (C₂ − M) : (M − C₁)

Worked Example:

Rice at ₹40/kg is mixed with rice at ₹65/kg to get a mixture at ₹50/kg. Find the mixing ratio.

Cheaper : Dearer = (65 − 50) : (50 − 40) = 15 : 10 = 3 : 2

So for every 3 kg of the cheaper variety, mix 2 kg of the dearer variety.

§3 Weighted Average

When two or more groups with different averages are combined, the combined average is the weighted average, not the simple arithmetic mean. The weight of each group is its size (quantity).

General Formula:

Weighted Average = (n₁ × A₁ + n₂ × A₂ + ... + nₖ × Aₖ) / (n₁ + n₂ + ... + nₖ)

where n_i = quantity/weight of group i, and A_i = attribute value (price, concentration, etc.) of group i

Key Insight:

The alligation rule is simply the inverse of the weighted average — instead of finding the average given quantities, it finds the quantities given the average. They are two sides of the same coin.

Rule of thumb: The weighted average always lies between the two individual values. The mixture leans toward the ingredient present in larger quantity.

§4 Concentration & Solution Problems

These problems involve liquid mixtures where a solute (e.g., acid, alcohol, salt) is dissolved in a solvent (usually water). The concentration is the fraction or percentage of the solute in the total solution.

Core Equations:

Concentration (%) = (Amount of solute / Total solution) × 100

Mixing: Amount of solute remains conserved when two solutions are mixed.

Adding pure solute: Solute increases; total volume increases.

Adding pure solvent (water): Solute stays same; total volume increases; concentration drops.

Standard Setup (two solutions mixed):

V₁ litres of solution at c₁% + V₂ litres at c₂% = (V₁+V₂) litres at c%

⟹ V₁ × c₁ + V₂ × c₂ = (V₁ + V₂) × c

§5 Replacement Problems

These problems involve a container with a mixture from which a fixed quantity is repeatedly removed and replaced with a pure substance (usually water). This is one of the most tested GMAT and Bank PO problem types.

⚡ Master Formula (Replacement after n iterations):

Amount of original substance remaining = A × (1 − r/A)ⁿ

where A = initial quantity of mixture, r = quantity removed each time, n = number of replacements

Example:

A 100-litre pure milk container: 20 litres removed and replaced with water, 3 times.

Milk remaining = 100 × (1 − 20/100)³ = 100 × (0.8)³ = 100 × 0.512 = 51.2 litres

§6 Quick Formula Reference Sheet

Formula Cheatsheet

Alligation Ratio	Cheaper : Dearer = (C₂ − M) : (M − C₁)
Weighted Average	(n₁A₁ + n₂A₂) / (n₁ + n₂)
Solution Mixing (concentration)	V₁c₁ + V₂c₂ = (V₁ + V₂) × c
Replacement (n times)	Remaining = A × (1 − r/A)ⁿ
Adding Pure Solvent (water)	New conc. = (Old solute amount) / (Old volume + Water added)
Adding Pure Solute (acid/alcohol)	New conc. = (Old solute + Added solute) / (Old volume + Added solute)

Part II

Practice Problems

Each problem includes a complete step-by-step solution and a highlighted shortcut tip.

Level 1 · Easy

Question 01 · Alligation / Basic

Easy

A shopkeeper mixes two varieties of tea: one costing ₹40 per kg and the other costing ₹65 per kg. In what ratio should he mix them so that the mixture costs ₹50 per kg?

Solution

Step 1: Identify values. C₁ = ₹40 (cheaper), C₂ = ₹65 (dearer), M = ₹50 (mean/target).

Step 2: Apply Alligation Rule.

Quantity of Cheaper : Quantity of Dearer = (C₂ − M) : (M − C₁)

= (65 − 50) : (50 − 40) = 15 : 10 = 3 : 2

Answer: 3 : 2 (3 kg of cheaper tea for every 2 kg of dearer tea)

⚡ Shortcut: Draw the alligation cross. The difference between the higher price and the mean goes with the cheaper ingredient, and vice versa. Always cross-subtract diagonally.

Question 02 · Ratio in Mixture

Easy

A jar contains a mixture of milk and water in the ratio 5 : 3. If the total volume of the mixture is 64 litres, how many litres of milk does the jar contain?

Solution

Step 1: Total ratio parts = 5 + 3 = 8.

Step 2: Milk = (5/8) × 64 = 40 litres

Verification: Water = (3/8) × 64 = 24 litres. 40 + 24 = 64 ✓

⚡ Shortcut: Fraction of any component = (its ratio number) ÷ (sum of all ratio numbers). Multiply by total volume. No need to set up algebra.

Question 03 · Adding Pure Solute

Easy

A solution contains 20% alcohol. How many litres of pure alcohol must be added to 10 litres of this solution to make the resulting solution 50% alcohol?

Solution

Step 1: Alcohol in original 10 L = 20% × 10 = 2 litres.

Step 2: Let x litres of pure alcohol be added.

Total alcohol = 2 + x; Total solution = 10 + x

Step 3: Set up equation: (2 + x) / (10 + x) = 0.50

2 + x = 5 + 0.5x → 0.5x = 3 → x = 6 litres

⚡ Shortcut (Alligation): Treat 20% solution and 100% pure alcohol as two sources; target = 50%. Ratio = (100−50):(50−20) = 50:30 = 5:3. So alcohol to add = 10 × (3/5) = 6 litres. ✓

Question 04 · Two Alloys Mixed

Easy

Alloy X contains gold and silver in the ratio 2 : 3. Alloy Y contains gold and silver in the ratio 3 : 2. If 10 kg of each alloy are melted together, what is the ratio of gold to silver in the resulting alloy?

Solution

Step 1 — Alloy X (10 kg): Gold = (2/5) × 10 = 4 kg; Silver = (3/5) × 10 = 6 kg

Step 2 — Alloy Y (10 kg): Gold = (3/5) × 10 = 6 kg; Silver = (2/5) × 10 = 4 kg

Step 3 — Combined: Total Gold = 4 + 6 = 10 kg; Total Silver = 6 + 4 = 10 kg

Ratio of Gold : Silver = 10 : 10 = 1 : 1

⚡ Shortcut: When two alloys have "mirror" ratios (2:3 and 3:2) and are mixed in equal weight, the result is always 1:1. The symmetry collapses the ratios. Spot this pattern in 5 seconds on exam day.

Question 05 · Mixing with Water (free)

Easy

In what ratio must water (costing nothing) be mixed with a syrup that costs ₹18 per litre, so that the resulting mixture can be sold at ₹12 per litre and still yield a profit of 20%?

Solution

Step 1: SP = ₹12/litre. Profit = 20%. So CP of mixture = 12 / 1.20 = ₹10 per litre

Step 2: Apply Alligation. Water (₹0) and Syrup (₹18), mean = ₹10.

Water : Syrup = (18 − 10) : (10 − 0) = 8 : 10 = 4 : 5

Answer: Water : Syrup = 4 : 5

⚡ Shortcut: First find the true CP using SP and profit %, then apply alligation. The profit % step is where many students err — always compute CP before drawing the cross.

Level 2 · Medium

Question 06 · Two Vessels

Medium

Vessel A contains milk and water in the ratio 4 : 1. Vessel B contains milk and water in the ratio 3 : 2. In what ratio must the contents of A and B be mixed to obtain a mixture in which milk and water are in the ratio 7 : 3?

Solution

Step 1: Express milk fractions.

Milk fraction in A = 4/5 = 0.80; Milk fraction in B = 3/5 = 0.60; Target = 7/10 = 0.70

Step 2: Apply Alligation on milk fractions.

A : B = (Target − Fraction_B) : (Fraction_A − Target)

= (0.70 − 0.60) : (0.80 − 0.70) = 0.10 : 0.10 = 1 : 1

Answer: A and B should be mixed in the ratio 1 : 1

⚡ Shortcut: Convert all ratios to fractions of the same component (milk or water) first, then apply alligation directly. Choosing milk here gives 4/5, 3/5, 7/10 — easy fractions to cross-subtract.

Question 07 · Replacement Problem

Medium

A container holds 40 litres of pure milk. 4 litres of milk are removed and replaced with water. This process is repeated 3 times in total. What percentage of the final mixture is milk?

Solution

Step 1: Use the replacement formula: Remaining milk = A × (1 − r/A)ⁿ

A = 40 litres, r = 4 litres, n = 3

Step 2: Milk remaining = 40 × (1 − 4/40)³ = 40 × (0.9)³

= 40 × 0.729 = 29.16 litres

Step 3: % milk = (29.16 / 40) × 100 = 72.9%

⚡ Shortcut: The fraction of milk after n replacements = (1 − r/A)ⁿ. Here that's (9/10)³ = 729/1000 = 72.9%. You never need to track water separately. Total is always 40.

Question 08 · Three-component Weighted Average

Medium

Three types of wheat costing ₹30, ₹40, and ₹50 per kg are mixed in the ratio 1 : 2 : 3. What is the average cost of the resulting mixture per kg?

Solution

Step 1: Total parts = 1 + 2 + 3 = 6

Step 2: Apply weighted average formula.

Avg cost = (1×30 + 2×40 + 3×50) / (1+2+3)

= (30 + 80 + 150) / 6 = 260 / 6 = ₹43.33 per kg

⚡ Shortcut: For 3+ components, weighted average is your only tool. Note the answer (₹43.33) is closer to ₹50 than ₹30 because the ₹50 type has the highest weight (3 parts). Sanity-check your answer this way.

Question 09 · Two Acid Solutions

Medium

A chemist has two solutions of hydrochloric acid: Solution A is 25% pure acid and Solution B is 50% pure acid. How many litres of each should be mixed to prepare exactly 10 litres of a 40% acid solution?

Solution

Step 1 (Alligation): C₁ = 25%, C₂ = 50%, M = 40%

A : B = (50 − 40) : (40 − 25) = 10 : 15 = 2 : 3

Step 2: Total = 10 litres. Total parts = 2 + 3 = 5

Solution A (25%) = (2/5) × 10 = 4 litres

Solution B (50%) = (3/5) × 10 = 6 litres

Verification: (4 × 0.25) + (6 × 0.50) = 1 + 3 = 4 litres acid in 10 L = 40% ✓

⚡ Shortcut: Target (40%) is closer to 50% than to 25%, so you need more of the 50% solution. The alligation ratio 2:3 confirms this — more B than A. Use this direction check to instantly eliminate wrong-ratio answer choices.

Question 10 · Adulteration & Profit

Medium

A milkman purchases pure milk at ₹40 per litre. He secretly mixes water (free of cost) in the ratio of milk : water = 4 : 1, then sells the adulterated mixture at the original cost price of ₹40 per litre. What is his effective profit percentage?

Solution

Step 1: Assume he prepares 5 litres of mixture (4 litres milk + 1 litre water).

Step 2: Cost incurred = 4 litres × ₹40 = ₹160 (water is free)

Step 3: Revenue from selling 5 litres at ₹40 = ₹200

Step 4: Profit = ₹200 − ₹160 = ₹40

Profit % = (40/160) × 100 = 25%

⚡ Shortcut: Profit % from adulteration = (Water fraction / Milk fraction) × 100 = (1/4) × 100 = 25%. This formula works whenever the adulterant is free and the mixture is sold at pure price.

Level 3 · Hard

Question 11 · Double Replacement

Hard

A 81-litre container is filled with pure milk. 27 litres are drawn out and replaced with water. From the resulting mixture, 27 litres are again drawn out and replaced with water. Find the ratio of milk to water in the final mixture.

Solution

Step 1: After 1st replacement:

Milk = 81 × (1 − 27/81)¹ = 81 × (54/81) = 54 litres; Water = 27 litres

Step 2: After 2nd replacement (draw 27 litres of mixture):

Milk fraction in mixture = 54/81 = 2/3. Milk removed = 27 × (2/3) = 18 litres

Milk remaining = 54 − 18 = 36 litres; Water = 81 − 36 = 45 litres

Ratio of Milk : Water = 36 : 45 = 4 : 5

⚡ Shortcut: Milk after n replacements = 81 × (1 − 27/81)² = 81 × (2/3)² = 81 × 4/9 = 36. Formula gives the answer in one line — critical for GMAT time management. Note: 81 = 3⁴ and 27/81 = 1/3. Perfect powers are a strong signal that the replacement formula is intended.

Question 12 · Three Vessels Combined

Hard

Three vessels A, B, and C of equal capacity contain mixtures of milk and water in the ratios 1 : 2, 2 : 3, and 3 : 4 respectively. All three vessels are poured into a single large container. What is the ratio of milk to water in the final mixture?

Solution

Step 1: Let capacity of each vessel = 105 litres (LCM of 3, 5, 7 — the total ratio parts).

Vessel A (1:2): Milk = (1/3)×105 = 35 L; Water = 70 L

Vessel B (2:3): Milk = (2/5)×105 = 42 L; Water = 63 L

Vessel C (3:4): Milk = (3/7)×105 = 45 L; Water = 60 L

Step 2: Total Milk = 35 + 42 + 45 = 122 L; Total Water = 70 + 63 + 60 = 193 L

Ratio of Milk : Water = 122 : 193

⚡ Shortcut: Use LCM of (sum of ratio terms) as the reference volume per vessel to avoid fractions. Here denominators are 3, 5, 7 → LCM = 105. This converts all fractions to integers and prevents arithmetic errors under exam pressure.

Question 13 · Alligation + Profit

Hard

A shopkeeper has two varieties of sugar costing ₹36/kg and ₹54/kg. He plans to mix them and sell the blend at ₹50/kg, earning a profit of 25% on cost. In what ratio should he mix the two varieties?

Solution

Step 1: Find target cost price (CP) of the mixture.

SP = ₹50, Profit = 25% → CP = SP / 1.25 = 50 / 1.25 = ₹40 per kg

Step 2: Apply Alligation. C₁ = ₹36, C₂ = ₹54, M = ₹40

Cheaper (₹36) : Dearer (₹54) = (54 − 40) : (40 − 36) = 14 : 4 = 7 : 2

Answer: Mix ₹36 and ₹54 sugar in the ratio 7 : 2

⚡ Shortcut: This is a two-step problem — profit % first, alligation second. Many students accidentally alligation on SP directly (₹50). Always compute CP before drawing the cross. On GMAT DS, the presence of a profit % is an extra given that determines CP; confirm you've used it.

Question 14 · Adding Water to reach Target Ratio

Hard

In a 729-litre mixture of milk and water, the ratio of milk to water is 7 : 2. How many litres of water must be added to the mixture to change this ratio to 7 : 3?

Solution

Step 1: Current quantities. Total = 729 litres, Ratio = 7:2, Total parts = 9

Milk = (7/9) × 729 = 567 litres; Water = (2/9) × 729 = 162 litres

Step 2: Key observation: adding water does NOT change the amount of milk.

Step 3: In new ratio 7:3, milk is still 567 litres.

Water in new mixture = 567 × (3/7) = 243 litres

Water to add = 243 − 162 = 81 litres

⚡ Shortcut: Notice that the numerator (milk part) stays at 7 in both ratios. This is the anchor. When the milk ratio term doesn't change, milk quantity is constant across both states — use it as your reference. Find new water from new ratio × fixed milk quantity.

Question 15 · Three-Component Replacement

Hard

A 100-litre mixture of milk, water, and juice contains these three in the ratio 5 : 3 : 2. A certain quantity of this mixture is removed and replaced entirely with pure milk, until the percentage of milk in the container reaches exactly 70%. How many litres were removed and replaced?

Solution

Step 1: Initial composition. Total = 100 L, ratio = 5:3:2 (total parts = 10)

Milk = 50 L; Water = 30 L; Juice = 20 L

Step 2: Let x litres of the mixture be removed. Since the removed portion is a proportional sample of all three components:

Milk removed = (50/100) × x = x/2 litres

Step 3: Pure milk is added to bring total back to 100 L. New milk content:

Milk = (50 − x/2) + x = 50 + x/2

Step 4: Set new milk percentage = 70%

(50 + x/2) / 100 = 0.70 → 50 + x/2 = 70 → x/2 = 20 → x = 40 litres

Verification: Remove 40 L (milk=20, water=12, juice=8). Remaining: milk=30, water=18, juice=12. Add 40 L milk → Milk = 70, Water = 18, Juice = 12. Milk % = 70% ✓

⚡ Shortcut: When a mixture is replaced by a pure form of one of its own components, track only that component (milk here). The "net gain" in milk per x litres removed = x − x/2 = x/2. You need a net gain of 70 − 50 = 20 litres. So x/2 = 20 → x = 40. No need to track water or juice at all.

Problem Set Summary

Q	Topic	Level	Answer
1	Alligation — two teas	Easy	3 : 2
2	Ratio extraction	Easy	40 litres
3	Adding pure solute	Easy	6 litres
4	Two alloys mixed equally	Easy	1 : 1
5	Water + syrup with profit	Easy	4 : 5
6	Two vessel mixing ratio	Medium	1 : 1
7	Replacement (3 times)	Medium	72.9%
8	Weighted average (3 types)	Medium	₹43.33/kg
9	Two acid solutions	Medium	4 L (25%) + 6 L (50%)
10	Adulteration profit %	Medium	25%
11	Double replacement	Hard	4 : 5
12	Three vessels combined	Hard	122 : 193
13	Alligation + profit %	Hard	7 : 2
14	Add water to reach ratio	Hard	81 litres
15	3-component replacement	Hard	40 litres

Mixture Problems in Mathematics

Quantitative Aptitude · Lecture Series · GMAT & Bank PO / SSC

15 Problems · 3 Levels · All Solved

Alligation · Weighted Average · Replacement · Concentration

NextPrepBD

Mixture Problems in Mathematics

PDF Version Restricted

Mixture Problems in Mathematics

Theory & Core Concepts

§1 What are Mixture Problems?

§2 The Alligation Method

§3 Weighted Average

§4 Concentration & Solution Problems

§5 Replacement Problems

§6 Quick Formula Reference Sheet

Practice Problems

💬Community Discussion

0 Comments