Mixtures, Alligation and Replacement Practice Questions for CAT: 168+ Solved Questions with Step-by-Step Solutions

Solve 168+ Mixtures, Alligation and Replacement practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Mixtures, Alligation and Replacement

CAT Quant • Arithmetic

Mixtures, Alligation and Replacement

A chapter about average value, concentration changes, and transfer logic.

chapter PYQs

⚖️

t1 — Alligation, Blending and Weighted Cost

Master weighted average, two-price mixing, ingredient percentage, and profit-linked blends.

Why first? This is the foundation of the whole chapter.

7 PYQs

High value

🔁

t2 — Successive Replacement and Dilution

Handle remove-and-refill cases using repeated fraction left.

8 PYQs

Very likely

🧪

t3 — Two-Vessel Transfer Problems

Track what moves between two containers without losing total quantity logic.

3 PYQs

Selective

End goal: When CAT gives prices, percentages, ratios, or ingredients, you should immediately ask: what is the weighted average?

Topic Hero: Alligation, Blending and Weighted Cost

Selected Topic • t1

Alligation, Blending and Weighted Cost

The art of mixing values without doing long equations every time.

⚖️

Core idea

Final value is a weighted average.

₹

CAT use

Cost price, selling price, profit, and marked price blends.

Ingredient use

Coffee, cocoa, sugar, silver, copper, milk, syrup.

One-line hook

If the mixture average is closer to one item, that item must be used more.

Mixtures, Alligation and Replacement: Solved Questions with Step-by-Step Explanations (5 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.

Question 3 · Quantitative Ability NAT

Vessels $A$ and $B$ contain $80$ litres of pure alcohol and $80$ litres of pure water, respectively.

A certain volume $x$ is taken from $A$ and poured into $B$ . After stirring, the same volume $x$ is taken from $B$ and poured back into $A$ .

If the final ratio of alcohol to water in $A$ is $4$ to $1$ , find $x$ in litres.

Correct Answer:

Step-by-Step Solution

Key idea: this is an unknown-transfer-volume question, recognisable because the final ratio is given but the transferred volume is not.

Step 1: Let the transferred volume be $x$ .

Initially:

$A$ : $80$ alcohol, $0$ water.
$B$ : $0$ alcohol, $80$ water.

Step 2: First transfer.

$x$ litres of pure alcohol move from $A$ to $B$ .

$B$ now has $x$ alcohol and $80$ water.

Total volume in $B$ is $80 + x$ .

Step 3: Water fraction in $B$ before the return.

$water fraction = \frac{80}{80 + x} .$

Step 4: Return transfer.

The $x$ litres returned from $B$ contains water equal to

$x \times \frac{80}{80 + x} = \frac{80 x}{80 + x} .$

This is the final water in $A$ .

Step 5: Use the final ratio in $A$ .

The final ratio alcohol to water in $A$ is $4 : 1$ .

The final total in $A$ is again $80$ litres.

So water in $A$ is

$\frac{1}{5} \times 80 = 16 litres .$

Step 6: Solve for $x$ .

$\frac{80 x}{80 + x} = 16.$

Multiply by $80 + x$ :

$80 x = 16 (80 + x) .$

$80 x = 1280 + 16 x .$

$64 x = 1280.$

$x = 20.$

Answer: 20.

Question 4 · Quantitative Ability NAT

Blend A contains $20%$ tea and costs Rs $220$ per kg.

Blend B contains $40%$ tea and costs Rs $290$ per kg.

Both blends contain only tea and chicory.

A manufacturer takes $12$ kg of blend B and adds pure chicory to it to obtain a blend costing Rs $230$ per kg.

How many kg of pure chicory must be added?

Correct Answer:

Step-by-Step Solution

Key idea: this is a hidden-pure-price mixture question, recognisable because two blends give both percentage and price, and later a pure ingredient is added.

Step 1: Let the price of pure tea be $t$ and the price of pure chicory be $c$ .

Step 2: Use blend A.

Blend A is $20%$ tea and $80%$ chicory, so

$0.2 t + 0.8 c = 220.$

Multiply by $5$ :

$t + 4 c = 1100. (1)$

Step 3: Use blend B.

Blend B is $40%$ tea and $60%$ chicory, so

$0.4 t + 0.6 c = 290.$

Multiply by $5$ :

$2 t + 3 c = 1450. (2)$

Step 4: Solve for $c$ .

From (1),

$t = 1100 - 4 c .$

Substitute into (2):

$2 (1100 - 4 c) + 3 c = 1450.$

$2200 - 8 c + 3 c = 1450.$

$2200 - 5 c = 1450.$

$5 c = 750.$

$c = 150.$

So pure chicory costs Rs $150$ per kg.

Step 5: Let $x$ kg of pure chicory be added to $12$ kg of blend B.

Cost of $12$ kg of blend B:

$12 \times 290 = 3480.$

Cost of added chicory:

$150 x .$

Final total weight:

$12 + x .$

Final cost is Rs $230$ per kg, so

$\frac{3480 + 150 x}{12 + x} = 230.$

Step 6: Solve for $x$ .

$3480 + 150 x = 230 (12 + x) .$

$3480 + 150 x = 2760 + 230 x .$

$720 = 80 x .$

$x = 9.$

Answer: 9.

Question 5 · Quantitative Ability NAT

A container holds $80$ litres of a $25%$ acid solution. In each step, $20$ litres of the solution are removed and replaced with $20$ litres of a $10%$ acid solution. What is the minimum number of such operations required for the acid concentration in the container to drop below $15%$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a successive replacement with non-zero inflow problem asking for a threshold count. We model concentration as a sequence and solve the inequality.

Step 1: Set up the recurrence for concentration $C_{n}$ .

Volume $V = 80$ , Replacement $R = 20$ , Inflow $C_{in} = 10%$ .

Retention factor $k = 1 - 20/80 = 0.75$ .

$C_{n + 1} = k C_{n} + (1 - k) C_{in}$

$C_{n + 1} = 0.75 C_{n} + 0.25 (10) = 0.75 C_{n} + 2.5$

Step 2: Solve for general term or iterate.

Fixed point (equilibrium) $C^{*}$ : $C^{*} = 0.75 C^{*} + 2.5 ⟹ 0.25 C^{*} = 2.5 ⟹ C^{*} = 10%$ .

General form: $C_{n} - C^{*} = (C_{0} - C^{*}) k^{n}$ .

$C_{n} - 10 = (25 - 10) (0.75)^{n}$ .

$C_{n} = 10 + 15 (0.75)^{n}$ .

Step 3: Solve inequality $C_{n} < 15$ .

$10 + 15 (0.75)^{n} < 15$

$15 (0.75)^{n} < 5$

$(0.75)^{n} < \frac{1}{3}$

Step 4: Test integer values.

$n = 1 : 0.75 > 0.333$

$n = 2 : 0.5625 > 0.333$

$n = 3 : 0.421875 > 0.333$

$n = 4 : 0.31640625 < 0.333$

Minimum $n = 4$ .

Answer: 4

Mixtures, Alligation and Replacement Practice Questions for CAT: 168+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Mixtures, Alligation and Replacement

Mixtures, Alligation and Replacement

Topic Hero: Alligation, Blending and Weighted Cost

Alligation, Blending and Weighted Cost

Mixtures, Alligation and Replacement: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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