Alligation and mixtures aptitude questions with solutions Wisdom Helps

Mixture & Alligation

Complete Concepts, Important Formulas, Shortcuts, Solved Examples & MCQs

Wisdom Helps | Maths Classes by Mahesh Sir

What will you learn?

Understand how mixtures are formed, how their average value changes, how to apply the alligation rule, and how to solve replacement, milk-water, profit and concentration problems quickly.

01. ConceptsMixtures, ratios and weighted averages

02. FormulasAlligation, replacement and concentration

03. ShortcutsFast methods for competitive exams

04. PracticeType-wise examples and interactive MCQs

📚 Table of Contents

Basic ConceptsImportant Formulas Rule of AlligationType-wise Solved Examples Replacement FormulaImportant Formulas Common MistakesPractice MCQs with Answers

1. Basic Concepts of Mixture & Alligation

What is a mixture?

A mixture is formed when two or more substances are combined. The substances may have different prices, concentrations or strengths.

Examples:

• Rice costing ₹40/kg mixed with rice costing ₹60/kg.
• Milk mixed with water.
• A 20% acid solution mixed with a 50% acid solution.
• Tea costing ₹200/kg mixed with tea costing ₹300/kg.

What is alligation?

Alligation is a shortcut method used to find the ratio in which two ingredients with different values must be mixed to obtain a required mean value.

Important condition: for a mixture of two ingredients, the required mean value must lie between their two values when both quantities are positive.

Weighted Average Concept

Mean value = Total value of all ingredientsTotal quantity

C = Q₁C₁ + Q₂C₂Q₁ + Q₂

Here:

• C₁, C₂ = values per unit of the two ingredients.
• Q₁, Q₂ = quantities of the ingredients.
• C = mean value of the final mixture.

Example: 2 kg of rice costing ₹40/kg is mixed with 3 kg of rice costing ₹60/kg.

1. Total cost = (2 × 40) + (3 × 60) = ₹260.
2. Total quantity = 2 + 3 = 5 kg.
3. Mean price = 2605 = ₹52/kg.

Shortcut: Mean price = 2 × 40 + 3 × 605 = ₹52/kg.

2. Important Mixture & Alligation Formulas

Formula 1: Mean Price

Mean price = Total costTotal quantity

Use when quantities and prices of the ingredients are given.

Formula 2: Alligation Ratio

Cheaper quantity : Dearer quantity
= (Dearer price − Mean price) : (Mean price − Cheaper price)

In symbols, if C₁ < C < C₂:

Q₁ : Q₂ = (C₂ − C) : (C − C₁)

Formula 3: Ingredient Quantity

Ingredient quantity = Total mixture × Its ratio partTotal ratio parts

Formula 4: Pure Substance in a Solution

Pure substance = Total solution × Concentration %100

Formula 5: Water Added to Pure Milk

Final mixture = Milk quantityRequired milk fraction
Water added = Final mixture − Milk quantity

If milk is pure initially and only water is added, the quantity of milk remains unchanged.

Formula 6: Replacement

Original liquid remaining = V (1 − xV)ⁿ

Here V = initial volume, x = volume removed and replaced each time, and n = number of replacement operations. The container is assumed to be well mixed before each removal and refilled to its original volume.

Formula 7: Final Concentration

Concentration % = Quantity of pure substanceTotal solution × 100

Formula 8: Profit and Selling Price

Profit % = Selling price − Cost priceCost price × 100

For mixtures, calculate the weighted average cost first, then compare the selling price with that cost.

3. Rule of Alligation — The Main Shortcut

Suppose two ingredients have values C₁ and C₂, where C₁ is cheaper and C₂ is dearer. The required mean value is C.

ALLIGATION CROSS METHOD

Ingredient	Value	Cross difference
Cheaper	C1	C2 − C
Mean	C	—
Dearer	C2	C − C1

Cheaper : Dearer = (C₂ − C) : (C − C₁)

How to Read the Cross Diagram

1. Write the cheaper value (C₁) at the top-left and the dearer value (C₂) at the bottom-left.
2. Write the required mean value (C) in the centre, between them.
3. Subtract diagonally, always taking the larger number minus the smaller number: C₂ − C and C − C₁.
4. Each difference is written on the right side and belongs to the ingredient at the opposite end of the line. So C₂ − C is the quantity of the cheaper ingredient, and C − C₁ is the quantity of the dearer ingredient.
5. Read the ratio as Cheaper : Dearer = (C₂ − C) : (C − C₁) and reduce it to its simplest form.

Why “opposite”? The mean is closer to the ingredient that is present in the greater quantity, so the smaller gap must be paired with the larger quantity. That is why the cross-differences are swapped.

Why Does Alligation Work? (Proof)

Alligation is simply the weighted-average formula rearranged. Start with the mean value of the mixture:

C = Q₁C₁ + Q₂C₂Q₁ + Q₂

1. Multiply both sides by (Q₁ + Q₂):   C(Q₁ + Q₂) = Q₁C₁ + Q₂C₂
2. Move the like terms together:   Q₁C − Q₁C₁ = Q₂C₂ − Q₂C
3. Take out the common factors:   Q₁(C − C₁) = Q₂(C₂ − C)
4. Divide both sides by Q₂(C − C₁):

Q₁Q₂ = C₂ − CC − C₁  ⇒  Q₁ : Q₂ = (C₂ − C) : (C − C₁)

Key idea: the quantities are in the inverse ratio of the distances of the mean from the two values. The ingredient whose value is nearer to the mean is used in the larger quantity.

What Can the “Values” Be?

The word “value” is flexible. Alligation works whenever a quantity is averaged using weights. Match the correct “value” with the correct “quantity” before applying the cross.

Situation	Value (C)	Quantity (Q)
Mixing items of different prices	Price per kg or per litre	Weight or volume
Mixing solutions	Concentration (%) or milk fraction	Volume of each solution
Profit or discount on two parts of a stock	Profit % or discount %	Quantity or marked price of each part
Loans at two interest rates	Rate of interest (% p.a.)	Amount borrowed at each rate
Average of two groups	Average of each group (weight, marks, age)	Number of members in each group

When Can You Use It? Conditions & Checks

• The values must be in the same unit (all ₹/kg, or all %).
• The mean must lie strictly between C₁ and C₂. If it does not, no positive mixture is possible.
• The rule applies directly to two ingredients. For three or more, use the weighted-average formula, or combine two of them first.
• If the mean equals one of the values, only that ingredient is used.
• When the mean is exactly midway between C₁ and C₂, the quantities are equal, ratio 1 : 1.

Always verify your answer: put the ratio back into the weighted-average formula. For Example 1 below the ratio is 3 : 1, so

Mean = 3 × 40 + 1 × 603 + 1 = 1804 = ₹45/kg ✓

The result matches the required mean, so the ratio is correct.

Example 1: Two Types of Rice

Rice costing ₹40/kg and ₹60/kg is mixed to obtain a mixture costing ₹45/kg. Find the ratio.

1. Cheaper price = ₹40; dearer price = ₹60; mean = ₹45.
2. Difference on the cheaper side = 60 − 45 = 15.
3. Difference on the dearer side = 45 − 40 = 5.
4. Ratio = 15 : 5 = 3 : 1.

₹40 rice : 3₹60 : 1

The cheaper ingredient is used in the greater quantity because the mean price is closer to ₹40.

Example 2: Find the Mean Price

Rice costing ₹30/kg and ₹50/kg is mixed in the ratio 2 : 3. Find the mean price.

Mean = 30 × 2 + 50 × 35 = 2105 = ₹42/kg

Example 3: Find Ingredient Quantities

Two varieties of rice costing ₹20/kg and ₹35/kg are mixed to obtain rice costing ₹29/kg. If the total mixture is 60 kg, find each quantity.

1. Ratio = (35 − 29) : (29 − 20) = 6 : 9 = 2 : 3.
2. Total parts = 2 + 3 = 5.
3. Cheaper rice = 60 × 25 = 24 kg.
4. Dearer rice = 60 × 35 = 36 kg.

4. Type-wise Solved Examples

TYPE 1

Finding the Mixing Ratio

Question: Tea costing ₹120/kg and ₹180/kg is mixed to obtain a mixture costing ₹150/kg. Find the ratio.

Solution:

(180 − 150) : (150 − 120) = 30 : 30 = 1 : 1

Answer: 1 : 1

TYPE 2

Finding the Mean Price

Question: 4 kg of sugar costing ₹30/kg is mixed with 6 kg costing ₹40/kg. Find the mean price.

Solution:

4 × 30 + 6 × 4010 = 36010 = ₹36/kg

Answer: ₹36/kg

Always multiply each price by its quantity before adding. Do not simply average the prices unless the quantities are equal.

TYPE 3

Finding the Quantity of Each Ingredient

Question: Rice costing ₹25/kg and ₹45/kg is mixed to obtain a mixture costing ₹35/kg. Find the quantities required for 80 kg.

1. Ratio = (45 − 35) : (35 − 25) = 10 : 10 = 1 : 1.
2. Each ingredient = 802 = 40 kg.

Answer: 40 kg and 40 kg.

TYPE 4

Milk and Water

Question: A 40-litre mixture contains milk and water in the ratio 3 : 1. How much water is present?

1. Total parts = 3 + 1 = 4.
2. Water = 40 × 14 = 10 litres.
3. Milk = 40 × 34 = 30 litres.

Answer: 10 litres of water.

TYPE 5

Adding Water to Milk

Question: A milkman has 30 litres of pure milk. How much water should be added to make the mixture 75% milk?

1. Milk remains 30 litres.
2. 30 litres represents 75% of the final mixture.
3. Total mixture = 30 × 10075 = 40 litres.
4. Water added = 40 − 30 = 10 litres.

Answer: 10 litres.

If pure milk is M and the required milk fraction is p, total mixture = Mp. Water added = Mp − M.

TYPE 6

Replacing Some Quantity of a Mixture

Question: A vessel contains 40 litres of milk. Eight litres are removed and replaced with water. Find the quantity of milk left.

1. Fraction of milk remaining after one operation = 1 − 840 = 45.
2. Milk left = 40 × 45 = 32 litres.

Answer: 32 litres of milk.

TYPE 7

Repeated Replacement

Question: A vessel contains 100 litres of milk. Ten litres are removed and replaced with water twice. Find the milk remaining.

Solution:

Milk remaining = 100 × (1 − 10100)²
= 100 × (910)² = 100 × 81100 = 81 litres

Answer: 81 litres.

This formula assumes the mixture is thoroughly mixed before each removal.

TYPE 8

Concentration of a Solution

Question: 20 litres of a 30% acid solution are mixed with 10 litres of a 60% acid solution. Find the final acid concentration.

1. Acid in first solution = 20 × 30100 = 6 litres.
2. Acid in second solution = 10 × 60100 = 6 litres.
3. Total acid = 12 litres; total solution = 30 litres.
4. Concentration = 1230 × 100 = 40%.

Answer: 40% acid.

TYPE 9

Profit Using a Mixture

Question: A trader mixes 10 kg of rice costing ₹40/kg with 10 kg costing ₹60/kg. He sells the mixture at ₹55/kg. Find the profit percentage.

1. Total cost = (10 × 40) + (10 × 60) = ₹1,000.
2. Total quantity = 20 kg.
3. Cost price per kg = 1,00020 = ₹50.
4. Selling price per kg = ₹55.
5. Profit percentage = 550 × 100 = 10%.

Answer: 10% profit.

TYPE 10

Adding Water to Change a Ratio

Question: A 50-litre mixture contains milk and water in the ratio 4 : 1. How much water must be added to make the ratio 2 : 1?

1. Milk = 50 × 45 = 40 litres.
2. Water initially = 50 × 15 = 10 litres.
3. For a 2 : 1 ratio, water must be 402 = 20 litres.
4. Water to add = 20 − 10 = 10 litres.

Answer: 10 litres.

5. Replacement Formula — Detailed Explanation

Replacement questions involve removing a quantity from a vessel and filling it again with another liquid. A part of the original liquid is lost during every operation.

Original liquid remaining after n replacements = V (1 − xV)ⁿ

• V: initial volume in the vessel.
• x: volume removed at each operation.
• n: number of replacement operations.

Example: Three Replacement Operations

A vessel contains 80 litres of milk. Eight litres are removed and replaced with water three times.

Milk remaining = 80 × (1 − 880)³
= 80 × (910)³ = 80 × 7291000 = 58.32 litres

Answer: 58.32 litres of milk.

Remember: the formula applies when the vessel is restored to its original volume after each operation, and the mixture is uniform before the next removal.

6. Important Formulas

Important Formula Revision Table

Question type	Formula / Method
Mean price	Total costTotal quantity
Mixing ratio	(Dearer − Mean) : (Mean − Cheaper)
Ingredient quantity	Total quantity × Ratio partTotal parts
Concentration	Pure substanceTotal solution × 100
Water added	Final volume − Initial volume, if only water is added
Repeated replacement	V (1 − xV)n
Profit percentage	ProfitCost price × 100

7. Common Mistakes to Avoid

1. Reversing the alligation ratio: the cross-difference opposite the cheaper ingredient gives the cheaper quantity.
2. Using a simple average for unequal quantities: use a weighted average instead.
3. Mixing up price and quantity: ₹/kg is a unit price; kg is a quantity.
4. Confusing concentration with volume: 30% of 20 litres is 6 litres of the pure substance.
5. Ignoring the number of replacements: repeated replacement requires a power, not just one subtraction.
6. Using an impossible mean: a positive mixture of two values cannot have a mean outside the range of those values.
7. Calculating profit on selling price: standard profit percentage is calculated on cost price unless the question states otherwise.

8. Practice Questions — Type-wise with Answers, Shortcuts & Explanations

Practice every question type below. Tap an option to check it instantly, then use the Answer, Shortcut and Explanation tabs under each question.

Type 1Type 2Type 3Type 4Type 5Type 6Type 7Type 8Type 9

Attempted: 0 / 39  |  Correct: 0

TYPE 1

Price Alligation — Finding the Mixing Ratio

Cheaper : Dearer = (Dearer − Mean) : (Mean − Cheaper)

1.1

In what ratio must a shopkeeper mix two types of rice worth ₹60 per kg and ₹80 per kg so that the average cost of the mixture is ₹75 per kg?

3 : 22 : 31 : 31 : 2

Correct answer: C) 1 : 3

Shortcut: Mean 75 is much closer to 80, so the dearer rice is used more. Differences 5 and 15 reduce to 1 : 3.

Explanation:

Cheaper : Dearer = (80 − 75) : (75 − 60)
= 5 : 15 = 1 : 3

1.2

In what ratio must a supplier mix two varieties of pulses costing ₹10 per kg and ₹30 per kg respectively, so as to get a mixture worth ₹15 per kg?

3 : 11 : 24 : 52 : 3

Correct answer: A) 3 : 1

Shortcut: Mean 15 is close to 10, so the cheaper pulse is more. Gaps: 15 and 5, so 3 : 1.

Explanation:

₹10 pulse : ₹30 pulse = (30 − 15) : (15 − 10)
= 15 : 5 = 3 : 1

1.3

In what ratio should rice costing ₹67.5 per kg be mixed with rice costing ₹60 per kg to get an average price of ₹64 per kg?

8 : 79 : 86 : 55 : 6

Correct answer: A) 8 : 7

Shortcut: Remove the decimal by doubling the gaps: 4 : 3.5 = 8 : 7.

Explanation:

₹67.5 rice : ₹60 rice = (64 − 60) : (67.5 − 64)
= 4 : 3.5 = 8 : 7

1.4

In what ratio must a grocer mix two varieties of rice costing ₹128 and ₹143 per kg respectively so as to get a mixture priced at ₹137.75 per kg?

7 : 1313 : 73 : 55 : 3

Correct answer: A) 7 : 13

Shortcut: Gaps are 5.25 and 9.75. Multiply by 4 to get 21 : 39, then divide by 3.

Explanation:

₹128 rice : ₹143 rice = (143 − 137.75) : (137.75 − 128)
= 5.25 : 9.75 = 21 : 39 = 7 : 13

TYPE 2

Alligation with Profit or Gain

Cost price of mixture = Selling price × 100100 + Profit % — find this first, then apply alligation.

2.1

In what ratio does a grocer mix tea at ₹26 per kg and ₹32 per kg so that by selling the mixture at ₹30 per kg he gains 10%?

26 : 77 : 167 : 2616 : 7

Correct answer: A) 26 : 7

Shortcut: Cost of mixture ≈ ₹27.27 lies close to 26, so the ₹26 tea is used more — only 26 : 7 fits.

Explanation:

Cost price of mixture = 30 × 100110 = 30011
₹26 tea : ₹32 tea = (32 − 30011) : (30011 − 26) = 5211 : 1411
= 52 : 14 = 26 : 7

2.2

In what ratio should tea costing ₹192 per kg be mixed with tea costing ₹150 per kg so that the mixture, when sold for ₹194.40 per kg, gives a profit of 20%?

2 : 523 : 55 : 35 : 2

Correct answer: A) 2 : 5

Shortcut: Cost of mixture = 194.40 ÷ 1.2 = 162. Take gaps from 162: 12 and 30, which reduce to 2 : 5.

Explanation:

Cost price of mixture = 194.40 × 100120 = ₹162
₹192 tea : ₹150 tea = (162 − 150) : (192 − 162)
= 12 : 30 = 2 : 5

2.3

How many kg of rice costing ₹60 per kg must be mixed with 24 kg of rice costing ₹42 per kg so that there may be a gain of 12% by selling the mixture at ₹56 per kg?

21.2 kg19.2 kg22 kg20 kg

Correct answer: B) 19.2 kg

Shortcut: Mixture cost = 56 ÷ 1.12 = 50. Cheaper : dearer = 10 : 8 = 5 : 4, so dearer = 24 × 4/5.

Explanation:

Cost price of mixture = 56 × 100112 = ₹50
₹42 rice : ₹60 rice = (60 − 50) : (50 − 42) = 10 : 8 = 5 : 4
₹60 rice = 24 × 45 = 19.2 kg

TYPE 3

Weighted Average & Unknown Value

Total value = Mean × Total parts. Or use alligation: Q₁ : Q₂ = (C₂ − C) : (C − C₁).

3.1

Tea worth ₹126 per kg and ₹135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth ₹153 per kg, the price of the third variety per kg will be:

₹133.5₹195.5₹175.5₹182.5

Correct answer: C) ₹175.5

Shortcut: Total value = 153 × 4 = 612. Subtract 126 + 135 = 261 → 351 for 2 parts → 175.5.

Explanation:

Total parts = 1 + 1 + 2 = 4
126 + 135 + 2x = 153 × 4 = 612
2x = 351, so x = ₹175.5

3.2

The average weight of the whole class is 50 kg. The average weight of boys is 60 kg and of girls is 44 kg. What could be the ratio of boys and girls in the class respectively?

3 : 55 : 32 : 33 : 2

Correct answer: A) 3 : 5

Shortcut: Boys : Girls = (Mean − Girls) : (Boys − Mean) = 6 : 10 = 3 : 5.

Explanation:

Boys : Girls = (50 − 44) : (60 − 50)
= 6 : 10 = 3 : 5

TYPE 4

Part-Whole Ratios & Alloys

Quantity of a part = Total × Its ratio partTotal ratio parts. For alloys, find the fraction of each metal first.

4.1

A mixture contains spirit and water in the ratio 3 : 2. If it contains 3 litres more spirit than water, the quantity of spirit in the mixture is:

10 litres12 litres8 litres9 litres

Correct answer: D) 9 litres

Shortcut: Difference of parts = 3 − 2 = 1 part = 3 L, so spirit = 3 parts = 9 L.

Explanation:

1 part = 3 litres
Spirit = 3 parts = 3 × 3 = 9 litres

4.2

60 kg of alloy A is mixed with 100 kg of alloy B. Alloy A has lead and tin in the ratio 3 : 2 and alloy B has tin and copper in the ratio 1 : 4. The amount of tin in the new alloy is:

44 kg50 kg80 kg27 kg

Correct answer: A) 44 kg

Shortcut: Tin fraction: A = 2/5, B = 1/5. Tin = 24 + 20 = 44 kg.

Explanation:

Tin in A = 60 × 25 = 24 kg
Tin in B = 100 × 15 = 20 kg
Total tin = 24 + 20 = 44 kg

4.3

X and Y are two alloys of copper and aluminium prepared by mixing the metals in the ratio 6 : 5 and 5 : 9 respectively. If equal quantities of the alloys are melted to form a third alloy Z, what is the ratio of copper and aluminium in Z?

154 : 163139 : 154165 : 139139 : 169

Correct answer: D) 139 : 169

Shortcut: For equal quantities, add the fractions of each metal. Copper = 6/11 + 5/14, aluminium = 5/11 + 9/14. Use denominator 154.

Explanation:

Copper = 611 + 514 = 84 + 55154 = 139154
Aluminium = 511 + 914 = 70 + 99154 = 169154
Cu : Al = 139 : 169

TYPE 5

Adding Water or Milk to Change the Ratio

The quantity that is NOT added stays constant. Fix it, find its new ratio parts, then subtract.

5.1

20 litres of a mixture contains milk and water in the ratio 3 : 1. The amount of milk to be added so as to have milk and water in the ratio 4 : 1 is:

6 L5 L7 L4 L

Correct answer: B) 5 L

Shortcut: Water stays 5 L. For 4 : 1, milk must be 20 L. Add 20 − 15 = 5 L.

Explanation:

Milk = 20 × 34 = 15 L, water = 5 L
New ratio 4 : 1 → milk = 4 × 5 = 20 L
Milk to add = 20 − 15 = 5 L

5.2

A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water the ratio of milk and water becomes 5 : 2. The quantity of milk in the mixture is:

22.75 L32.5 L16 L25 L

Correct answer: D) 25 L

Shortcut: Milk is constant. Water goes from 1 part to 2 parts, so 1 part = 5 L. Milk = 5 parts = 25 L.

Explanation:

Let milk = 5k and water = k
5kk + 5 = 52 → 10k = 5k + 25 → k = 5
Milk = 5k = 25 litres

5.3

The milk and water in a mixture are in the ratio 7 : 5. When 15 litres of water are added, the ratio of milk to water becomes 7 : 8. The total quantity of water in the new mixture is:

25 litres35 litres40 litres45 litres

Correct answer: C) 40 litres

Shortcut: Milk stays 7 parts. Water changes 5 → 8 parts, so 3 parts = 15 L, 1 part = 5 L. New water = 8 × 5 = 40 L.

Explanation:

Let milk = 7k, water = 5k
7k5k + 15 = 78 → 8k = 5k + 15 → k = 5
New water = 5k + 15 = 40 litres

5.4

In a mixture of 55 litres, fruit juice and water are in the ratio 4 : 1. How much water (in litres) must be added to make the ratio 2 : 1?

9101112

Correct answer: C) 11

Shortcut: Juice (44 L) stays. For 2 : 1 water must be 22 L. Add 22 − 11 = 11 L.

Explanation:

Juice = 55 × 45 = 44 L, water = 11 L
Required water = 442 = 22 L
Water to add = 22 − 11 = 11 litres

5.5

The ratio of water and milk in a mixture of 178 litres is 5 : 7. How much water should be added to make the ratio of water and milk 3 : 4?

2 2324 L3 1724 L2 1124 L3 1324 L

Correct answer: B) 3 1724 L

Shortcut: Milk is constant. New water = ¾ × milk. Subtract the old water.

Explanation:

Water = 178 × 512 = 4456, Milk = 178 × 712 = 6236
New water = 34 × 6236 = 186924
Water to add = 186924 − 178024 = 8924 = 3 1724 litres

5.6

In a 40-litre mixture of milk and water the ratio of milk to water is 7 : 1. To make the ratio of milk and water 3 : 1, the quantity of water (in litres) to be added will be:

20313299274

Correct answer: A) 203

Shortcut: Milk = 35 L stays. For 3 : 1 water = 35/3. Add 35/3 − 5 = 20/3.

Explanation:

Milk = 40 × 78 = 35 L, water = 5 L
Required water = 353
Water to add = 353 − 5 = 203 litres

5.7

A vessel contains 56 litres of milk and water in the ratio 4 : 3. If Y litres of water are added, the ratio of milk to water becomes 8 : 7. Find Y.

24536

Correct answer: B) 4

Shortcut: Milk = 32 L stays. 8 parts = 32 → 1 part = 4, so water = 28 L. Y = 28 − 24 = 4.

Explanation:

Milk = 56 × 47 = 32 L, water = 24 L
3224 + Y = 87 → 24 + Y = 28
Y = 4

5.8

Zinc and copper are in the ratio 5 : 3 in 200 g of an alloy. How many grams of copper must be added to make the ratio 3 : 5?

100 12133 13144 1425 34

Correct answer: B) 133 13

Shortcut: Zinc (125 g) stays. For Zn : Cu = 3 : 5, copper = 125 × 5/3 = 208⅓. Add 208⅓ − 75.

Explanation:

Zinc = 200 × 58 = 125 g, copper = 75 g
Required copper = 125 × 53 = 6253
Copper to add = 6253 − 75 = 4003 = 133 13 g

TYPE 6

Percentage & Concentration Problems

Concentration % = Pure substanceTotal solution × 100. The pure substance stays constant when only water is added.

6.1

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

1 L2 L3 L4 L

Correct answer: B) 2 L

Shortcut: Wine = 63 L stays. 12½% = 1/8 water → water : wine = 1 : 7 → water = 9 L. Add 9 − 7 = 2 L.

Explanation:

Water = 7 L, wine = 63 L
Water = 18 of total → water : wine = 1 : 7
Required water = 637 = 9 L → add 9 − 7 = 2 litres

6.2

A 40-litre mixture contains 25% alcohol and 75% water. If 10 litres of water are added, the percentage of alcohol in the new mixture is:

20%18%25%27%

Correct answer: A) 20%

Shortcut: Alcohol stays 10 L. New total = 50 L. 10 out of 50 is 20%.

Explanation:

Alcohol = 40 × 25100 = 10 L
New total = 40 + 10 = 50 L
Alcohol % = 1050 × 100 = 20%

6.3

25 litres of a mixture contain 30% spirit and the rest water. If 5 litres of water are mixed in, the percentage of spirit in the new mixture is:

1212%3313%25%45%

Correct answer: C) 25%

Shortcut: New % = old % × old totalnew total = 30 × 25/30 = 25%.

Explanation:

Spirit = 25 × 30100 = 7.5 L
New total = 25 + 5 = 30 L
Spirit % = 7.530 × 100 = 25%

6.4

140 litres of a solution is formed by mixing two solutions A and B. The concentrations of milk in A and B are 40% and 75% respectively. If the resultant solution has 52% milk, the quantity of A in the resultant solution is:

48 litres65 litres92 litres57 litres

Correct answer: C) 92 litres

Shortcut: A : B = (75 − 52) : (52 − 40) = 23 : 12. Total parts 35, so A = 140 × 23/35 = 92 L.

Explanation:

A : B = (75 − 52) : (52 − 40) = 23 : 12
Total parts = 35
A = 140 × 2335 = 92 litres

TYPE 7

Mixing Two Vessels or Containers

Write the milk (or acid/alcohol) fraction of each vessel. For equal capacities add the fractions; otherwise use alligation on the fractions.

7.1

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?

7 : 55 : 72 : 33 : 2

Correct answer: A) 7 : 5

Shortcut: Milk fractions 4/7 and 2/5, mean 1/2. A : B = (½ − ⅖) : (4/7 − ½) = 1/10 : 1/14 = 7 : 5.

Explanation:

Milk in A = 47, milk in B = 25, required = 12
A : B = (12 − 25) : (47 − 12) = 110 : 114
= 14 : 10 = 7 : 5

7.2

Two vessels of equal capacity, both completely filled, contain mixtures of milk and water in the ratio 5 : 1 and 6 : 5 respectively. If both are mixed thoroughly, the ratio of milk to water in the new mixture is:

105 : 6791 : 4128 : 4357 : 41

Correct answer: B) 91 : 41

Shortcut: Add milk fractions 5/6 + 6/11 and water fractions 1/6 + 5/11 over the common denominator 66.

Explanation:

Milk = 56 + 611 = 55 + 3666 = 9166
Water = 16 + 511 = 11 + 3066 = 4166
Milk : Water = 91 : 41

7.3

Two containers have acid and water mixed in the ratio 3 : 1 and 5 : 3 respectively. In what ratio must the two be mixed to get a new mixture with acid to water as 2 : 1?

1 : 22 : 13 : 22 : 3

Correct answer: A) 1 : 2

Shortcut: Acid fractions 3/4 and 5/8, mean 2/3. First : Second = (⅔ − ⅝) : (¾ − ⅔) = 1/24 : 1/12 = 1 : 2.

Explanation:

Acid in 1st = 34, acid in 2nd = 58, required = 23
1st : 2nd = (23 − 58) : (34 − 23) = 124 : 112
= 1 : 2

7.4

There are two containers of equal capacity. The ratio of milk to water in the first is 3 : 1 and in the second 5 : 2. If they are mixed, the ratio of milk to water in the mixture will be:

28 : 4141 : 2815 : 4141 : 15

Correct answer: D) 41 : 15

Shortcut: Add fractions: milk = 3/4 + 5/7, water = 1/4 + 2/7, denominator 28.

Explanation:

Milk = 34 + 57 = 21 + 2028 = 4128
Water = 14 + 27 = 7 + 828 = 1528
Milk : Water = 41 : 15

7.5

Two equal glasses filled with alcohol and water in the proportions 2 : 1 and 3 : 2 are emptied into a third glass. The proportion of alcohol and water in the third glass will be:

13 : 1719 : 1713 : 1119 : 11

Correct answer: D) 19 : 11

Shortcut: Add fractions: alcohol = 2/3 + 3/5, water = 1/3 + 2/5, denominator 15.

Explanation:

Alcohol = 23 + 35 = 10 + 915 = 1915
Water = 13 + 25 = 5 + 615 = 1115
Alcohol : Water = 19 : 11

7.6

Three utensils contain equal quantities of mixtures of milk and water in the ratio 6 : 1, 5 : 2 and 3 : 1 respectively. If all the solutions are mixed together, the ratio of milk and water in the final mixture is:

65 : 2865 : 1919 : 6519 : 28

Correct answer: B) 65 : 19

Shortcut: Add the three milk fractions (6/7, 5/7, 3/4) and water fractions (1/7, 2/7, 1/4). Denominator 28.

Explanation:

Milk = 67 + 57 + 34 = 24 + 20 + 2128 = 6528
Water = 17 + 27 + 14 = 4 + 8 + 728 = 1928
Milk : Water = 65 : 19

TYPE 8

Replacement of Mixtures

Liquid left after n operations = V (1 − xV)ⁿ, where x is removed and replaced each time.

8.1

A container contains 40 litres of milk. From it, 4 litres were taken out and replaced by water. This process was repeated two more times. How much milk is now in the container?

30 litres34.23 litres29.16 litres32 litres

Correct answer: C) 29.16 litres

Shortcut: Three operations in total: 40 × (0.9)³ = 40 × 0.729 = 29.16.

Explanation:

Milk left = 40 (1 − 440)³ = 40 (910)³
= 40 × 7291000 = 29.16 litres

8.2

A vessel full of pure acid holds 10 litres. 2 litres are withdrawn and the vessel is filled with water. Again 2 litres of the mixture are withdrawn and the vessel is filled with water. The ratio of the acid left to the original quantity is:

1 : 54 : 54 : 2516 : 25

Correct answer: D) 16 : 25

Shortcut: Each operation keeps 4/5 of the acid. Two operations: (4/5)² = 16/25.

Explanation:

Acid left = 10 (1 − 210)² = 10 (45)² = 10 × 1625
Left : Original = 1625 = 16 : 25

8.3

A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of the 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 did the can contain initially?

18212425

Correct answer: B) 21

Shortcut: Only B is added back, so A's fraction falls from 7/12 to 7/16: 7/12 × (1 − 9/x) = 7/16 → 1 − 9/x = 3/4 → x = 36. A = 36 × 7/12 = 21.

Explanation:

Let the total be x litres. After the change A : total = 7 : 16
712 (1 − 9x) = 716 → 1 − 9x = 34 → x = 36
A initially = 36 × 712 = 21 litres

TYPE 9

Profit, Discount & Interest Alligation

Treat profit %, discount % or interest rate as the 'price'. Then quantity ratio = (higher − mean) : (mean − lower).

9.1

A merchant borrowed ₹2500 from two money lenders. For one loan he paid 12% p.a. and for the other 14% p.a. The total interest paid for one year was ₹326. How much did he borrow at each rate?

₹1200 at 12% and ₹1300 at 14%₹1300 at 12% and ₹1200 at 14%₹1000 at 12% and ₹1500 at 14%₹1250 at 12% and ₹1250 at 14%

Correct answer: A) ₹1200 at 12% and ₹1300 at 14%

Shortcut: Assume all at 12% → interest 300. Extra 26 ÷ 2% = ₹1300 at 14%. Balance ₹1200 at 12%.

Explanation:

Mean rate = 3262500 × 100 = 13.04%
12% : 14% = (14 − 13.04) : (13.04 − 12) = 0.96 : 1.04 = 12 : 13
At 12% = 2500 × 1225 = ₹1200; at 14% = 2500 × 1325 = ₹1300

9.2

A seller sells two watches at 8% and 15% discount. The total marked price of both items is ₹2800. Find the difference between the marked prices of both items if the total discount is 12%.

₹300₹400₹450₹500

Correct answer: B) ₹400

Shortcut: MP ratio = (15 − 12) : (12 − 8) = 3 : 4. Total 7 parts = 2800, so 1 part = 400 = the difference.

Explanation:

MP (8% item) : MP (15% item) = (15 − 12) : (12 − 8) = 3 : 4
1 part = 28007 = ₹400
Difference = 4 − 3 = 1 part = ₹400

9.3

Kiran buys a pen at 4% discount and a book at 12% discount. Overall she gets a discount of 10%. If the total marked price of both is ₹180, find the marked price of the book.

₹120₹135₹150₹140

Correct answer: B) ₹135

Shortcut: Pen : Book (marked) = (12 − 10) : (10 − 4) = 1 : 3. Book = 180 × 3/4 = ₹135.

Explanation:

MP pen : MP book = (12 − 10) : (10 − 4) = 2 : 6 = 1 : 3
Book = 180 × 34 = ₹135

9.4

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:

400 kg560 kg600 kg640 kg

Correct answer: C) 600 kg

Shortcut: 8% : 18% = (18 − 14) : (14 − 8) = 4 : 6 = 2 : 3. The 18% part = 3/5 of 1000.

Explanation:

Qty at 8% : Qty at 18% = (18 − 14) : (14 − 8) = 2 : 3
Qty at 18% = 1000 × 35 = 600 kg

9.5

A merchant has 840 kg of rice, part of which he sells at 9% profit and the rest at 16% profit. He gains 12% on the whole. What is the quantity sold at 16% profit?

280 kg630 kg360 kg420 kg

Correct answer: C) 360 kg

Shortcut: 9% : 16% = (16 − 12) : (12 − 9) = 4 : 3. The 16% part = 3/7 of 840.

Explanation:

Qty at 9% : Qty at 16% = (16 − 12) : (12 − 9) = 4 : 3
Qty at 16% = 840 × 37 = 360 kg

9.6

A trader has 1600 kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole, find the quantity sold at 12%.

1200 kg1400 kg1600 kg800 kg

Correct answer: A) 1200 kg

Shortcut: 8% : 12% = (12 − 11) : (11 − 8) = 1 : 3. The 12% part = 3/4 of 1600.

Explanation:

Qty at 8% : Qty at 12% = (12 − 11) : (11 − 8) = 1 : 3
Qty at 12% = 1600 × 34 = 1200 kg

9. Final Revision — Remember These Rules

1. Mean value is a weighted average.
2. Alligation ratio = (Dearer − Mean) : (Mean − Cheaper).
3. Use the total ratio parts to calculate actual quantities.
4. For concentration questions, calculate the pure substance first.
5. For repeated replacement, use V (1 − xV)ⁿ.
6. Check whether the answer is reasonable before selecting an MCQ option.

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