Mixtures, Alligation and Replacement Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Mixtures, Alligation and Replacement short notes for CAT: 43 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Weighted Average Formula

The master formula

Weighted average = \frac{q _{1} v _{1} + q _{2} v _{2} + \dots + q _{n} v _{n}}{q _{1} + q _{2} + \dots + q _{n}}

Symbol	Meaning
$q_{i}$	quantity of the item
$v_{i}$	value per unit: price, percentage, concentration, strength

Exam memory: First multiply, then add, then divide. Never average the rates directly unless quantities are equal.

Why Alligation Gives a Ratio

Alligation is distance from the average

For two values $L$ and $H$ mixed to get mean $M$ :

Quantity of low : Quantity of high = (H - M) : (M - L)

The distances are crossed. That is why alligation is often shown as a cross method.

The 4-Step Alligation Method

When two values make one average

Place the lower and higher values.
Example: ₹240/kg and ₹320/kg.

Put the final average between them.
If it is not between them, something else is happening.

Subtract diagonally.
Quantity ratio comes from opposite distance.

Convert ratio into actual quantity.
Use total quantity, cost, or percentage as required.

L : M : H \Rightarrow q_{L} : q_{H} = (H - M) : (M - L)

Alligation vs Weighted Average

Choose the right weapon

Situation	Best method	Why
Two prices or two percentages	Alligation	Fast ratio from differences
Three or more items	Weighted average	Direct total value method
Profit, marked price, discount	Cost table	Need total cost and total revenue

Rule: Alligation is a shortcut. Weighted average is the parent idea.

Mixtures, Alligation and Replacement: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Ability NAT

A trader buys $100$ kg of sugar at Rs $40$ /kg and $60$ kg of jaggery at Rs $60$ /kg. He sells $40$ kg of the sugar separately at a $10%$ profit. He then mixes the remaining sugar with all the jaggery. To achieve an overall profit of $25%$ on the entire original purchase, at what price per kg must he sell the mixture?

Correct Answer:

64.5

Step-by-Step Solution

Key idea: This is a partial sale + mixture profit problem. The trap is averaging selling prices. We must track Total Cost Price (CP) and Required Total Selling Price (SP).

Step 1: Calculate Total CP.

Sugar CP = $100 \times 40 = 4000$ .

Jaggery CP = $60 \times 60 = 3600$ .

Total CP = $4000 + 3600 = 7600$ .

Step 2: Calculate Required Total SP.

Overall Profit = $25%$ .

Required Total SP = $7600 \times 1.25 = 9500$ .

Step 3: Account for Revenue Already Earned.

Sold Sugar = $40$ kg. SP/kg = $40 \times 1.10 = 44$ .

Revenue from Sugar = $40 \times 44 = 1760$ .

Step 4: Determine Required Revenue from Mixture.

Remaining Revenue Needed = $9500 - 1760 = 7740$ .

Step 5: Find Mixture Selling Price.

Mixture Composition:

Remaining Sugar = $100 - 40 = 60$ kg.

Jaggery = $60$ kg.

Total Mixture Weight = $60 + 60 = 120$ kg.

Required SP/kg = $\frac{7740}{120} = 64.5$ .

Answer: 64.5

Question 2 · Quantitative Ability NAT

A trader buys $80$ kg of dry fruit A and $120$ kg of dry fruit B.

Per kg, A costs $25%$ more than B.

Before selling, $10%$ of A and $20%$ of B are found damaged and cannot be sold.

The remaining undamaged stock is mixed and sold at Rs $132$ per kg.

If the trader makes an overall profit of $20%$ on the entire purchase, what is the cost price of B, in rupees per kg?

Correct Answer:

Step-by-Step Solution

Key idea: this is an overall-profit mixture question with spoilage, recognisable because the selling revenue comes only from undamaged stock, but the profit percentage is calculated on the whole purchase cost.

Step 1: Let the cost price of B be $b$ rupees per kg.

Since A costs $25%$ more than B,

$cost of A = 1.25 b .$

Step 2: Compute the total cost price of the entire purchase.

Cost of A:

$80 \times 1.25 b = 100 b .$

Cost of B:

$120 \times b = 120 b .$

Total cost:

$100 b + 120 b = 220 b .$

Step 3: Find the undamaged quantities.

Damaged A:

$10% of 80 = 8 kg .$

Undamaged A:

$80 - 8 = 72 kg .$

Damaged B:

$20% of 120 = 24 kg .$

Undamaged B:

$120 - 24 = 96 kg .$

Total sellable mixture:

$72 + 96 = 168 kg .$

Step 4: Compute total selling revenue.

The mixture is sold at Rs $132$ per kg:

$S P = 168 \times 132.$

Calculate:

$168 \times 132 = 168 (100 + 30 + 2) = 16800 + 5040 + 336 = 22176.$

Step 5: Use the overall profit condition.

Overall profit is $20%$ on the entire purchase, so

$S P = 1.2 \times C P .$

Therefore

$22176 = 1.2 \times 220 b .$

Compute:

$1.2 \times 220 b = 264 b .$

$264 b = 22176.$

Step 6: Solve for $b$ .

$b = \frac{22176}{264} = 84.$

Answer: 84.

More short notes in this unit

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Mixtures, Alligation and Replacement Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Mixtures, Alligation and Replacement short notes for CAT: 43 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions

A question from this chapter

Question 1

A trader buys 100 kg of sugar at Rs 40 /kg and 60 kg of jaggery at Rs 60 /kg. He sells 40 kg of the sugar separately at a 10% profit. He then mixes the remaining sugar with all the jaggery. To achieve an overall profit of 25% on the entire original purchase, at what price per kg must he sell the mixture?

Question 2

A trader buys 80 kg of dry fruit A and 120 kg of dry fruit B. Per kg, A costs 25% more than B. Before selling, 10% of A and 20% of B are found damaged and cannot be sold. The remaining undamaged stock is mixed and sold at Rs 132 per kg. If the trader makes an overall profit of 20% on the entire purchase, what is the cost price of B, in rupees per kg?

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