Mixtures, Alligation and Replacement Previous Year Questions (PYQs) for CAT: 15+ Solved Questions with Step-by-Step Solutions

Solve 15+ Mixtures, Alligation and Replacement previous year 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 MCQ

A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

A.
653
B.
688
C.
692
D.
675

Correct Answer:

Step-by-Step Solution

Key idea: This is a mixture costing question with a partial sale and an overall profit target. The safe method is to track total cost price and total selling price, not to average selling prices directly.

Why this method applies: The tea qualities are bought in a quantity ratio, mixed, and then only part of the mixture is sold first. The remaining selling price must be chosen so that the total revenue gives 50% profit on the total cost.

Step 1: Let the quantities bought be $2 x$ , $3 x$ , and $5 x$ kg.

Total quantity:

$2 x + 3 x + 5 x = 10 x kg$

Step 2: Compute total cost price.

Cost of first quality:

$2 x \times 800 = 1600 x$

Cost of second quality:

$3 x \times 500 = 1500 x$

Cost of third quality:

$5 x \times 300 = 1500 x$

Total CP:

$1600 x + 1500 x + 1500 x = 4600 x$

Step 3: Overall profit is 50%, so total selling price must be:

$1.5 \times 4600 x = 6900 x$

Step 4: One-sixth of the mixture is sold at ₹700 per kg.

One-sixth quantity:

$\frac{1}{6} \times 10 x = \frac{5 x}{3}$

Revenue from this part:

$700 \times \frac{5 x}{3} = \frac{3500 x}{3}$

Step 5: Find the revenue needed from the remaining tea.

Remaining revenue:

$6900 x - \frac{3500 x}{3}$

$= \frac{20700 x - 3500 x}{3} = \frac{17200 x}{3}$

Step 6: Find the remaining quantity.

$10 x - \frac{5 x}{3} = \frac{30 x - 5 x}{3} = \frac{25 x}{3}$

Step 7: Required selling price per kg of remaining tea:

$\frac{\frac{17200 x}{3}}{\frac{25 x}{3}} = \frac{17200}{25} = 688$

Answer: B. The remaining tea should be sold at ₹688 per kg.

Common trap: Do not simply set the remaining price equal to the 50% profit average price. Since one-sixth is already sold at ₹700, the remaining price must be found from total revenue balance.

Question 2 · Quantitative Ability NAT

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Correct Answer:

Step-by-Step Solution

Key idea: this is a successive replacement question. When a well-mixed liquid is drawn out and replaced, the original liquid left after each step is multiplied by a survival fraction.

Step 1: Let the capacity of the container be $C$ litres.

Initially, the container is full of milk, so initial milk $= C$ .

Step 2: Understand one replacement step.

If 9 litres are removed from $C$ litres, the fraction removed is $9/ C$ .

So the fraction of milk remaining after that step is:

$1 - \frac{9}{C} .$

Step 3: Apply two successive replacements.

After the first draw and refill, milk left is:

$C (1 - \frac{9}{C}) .$

After the second draw and refill, the milk left is multiplied again by the same fraction:

$C (1 - \frac{9}{C})^{2} .$

Step 4: Use the final milk-water ratio.

The final ratio of milk to water is $16 : 9$ .

Therefore, milk is $\frac{16}{16 + 9} = \frac{16}{25}$ of the total container.

So:

$\frac{milk left}{C} = (1 - \frac{9}{C})^{2} = \frac{16}{25} .$

Step 5: Solve for $C$ .

Take the positive square root because $C > 9$ and the survival fraction is positive:

$1 - \frac{9}{C} = \frac{4}{5} .$

Then:

$\frac{9}{C} = 1 - \frac{4}{5} = \frac{1}{5} .$

Hence:

$C = 45.$

Answer: 45.

Common trap: subtracting $9 + 9 = 18$ litres of milk directly. The second removal is not pure milk; it removes a mixture of milk and water.

Question 3 · Quantitative Ability NAT

The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Correct Answer:

200

Step-by-Step Solution

Key idea: this is a single-replacement mixture question, recognisable because a fixed-volume bottle loses some solution and receives an equal volume of another solution.

Step 1: Let $x$ cc be the volume removed from the first bottle and replaced by solution from the second bottle. The first bottle still contains $800$ cc after replacement.

Step 2: Strength is being used like a concentration, so amount of indigo is proportional to strength $\times$ volume. We can keep the percentage numbers as they are because every term has the same hidden factor $\frac{1}{100}$ .

Step 3: Initially, the first bottle has strength $33%$ . After removing $x$ cc, the amount of original solution left is $800 - x$ cc, carrying strength $33%$ .

Step 4: The added volume is $x$ cc from the second bottle, with strength $17%$ . The final strength of the first bottle is $21%$ .

Step 5: Set up the amount balance:

$33 (800 - x) + 17 x = 21 (800) .$

Step 6: Solve:

$26400 - 33 x + 17 x = 16800$

$26400 - 16 x = 16800$

$16 x = 9600$

$x = 600.$

Step 7: So $600$ cc is taken from the second bottle. The second bottle initially had $800$ cc, so volume left in it is:

$800 - 600 = 200.$

Answer: $200$ cc.

Question 4 · Quantitative Ability MCQ

A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of this solution with water, then replaces 10% of the resulting solution with acid, and finally replaces 15% of the solution thus obtained, with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to

A.
23
B.
25
C.
29
D.
27

Correct Answer:

Step-by-Step Solution

This is a successive replacement question, but with a twist: the three replacements use different liquids (water, then acid, then water), not the same liquid each time. Recognise it because the question performs three separate remove-and-refill steps in a row, each with a different percentage and a different added liquid.

Key idea: whenever you remove a fraction $f$ of a well-mixed solution, you remove the same fraction $f$ of every ingredient in it. So track the acid amount (not the percentage) step by step.

Start: total volume $= 200$ L, acid $= 30% \times 200 = 60$ L, water $= 140$ L.

Step 1 — replace 20% with water:

Removing 20% of the solution removes 20% of the acid present.

$Acid left = 60 \times (1 - 0.20) = 48 L$

Water is added back, so total volume stays $200$ L. Acid $= 48$ L.

Step 2 — replace 10% of this solution with acid:

First, removing 10% removes 10% of the current acid:

$48 \times (1 - 0.10) = 43.2 L kept$

Then $10%$ of $200$ L $= 20$ L of pure acid is added back (not water this time):

$Acid now = 43.2 + 20 = 63.2 L$

Total volume is still $200$ L.

Step 3 — replace 15% of this solution with water:

Removing 15% removes 15% of the current acid:

$63.2 \times (1 - 0.15) = 53.72 L$

Water is added back, total volume $= 200$ L, acid $= 53.72$ L.

Final percentage of acid:

$\frac{53.72}{200} \times 100 = 26.86% \approx 27%$

Common trap: students often use the same "multiply by surviving fraction" shortcut for Step 2 as well, forgetting that Step 2 adds acid (not water), so it is not a pure multiplication — you must add the fresh acid back in litres after removing the fraction.

The final answer is 27%, option D.

Question 5 · Quantitative Ability MCQ

Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is

A.
1680
B.
1176
C.
2520
D.
1440

Correct Answer:

Step-by-Step Solution

Key idea: this is a three-item mixture question with a common cost relation and a partial discounted sale, recognisable because costs are linked by equivalent weights and profit is given in rupees.

Step 1: Let cost per kg of cashews be $C$ , peanuts be $P$ , and almonds be $A$ .

Step 2: The relation says:

$7 C = 30 P = 9 A .$

Let this common value be $K$ .

Step 3: Then:

$C = \frac{K}{7}, P = \frac{K}{30}, A = \frac{K}{9} .$

Step 4: Total cost price of what she bought:

$C P = 4 C + 14 P + 6 A .$

Substitute:

$C P = 4 \cdot \frac{K}{7} + 14 \cdot \frac{K}{30} + 6 \cdot \frac{K}{9} .$

Step 5: Simplify:

$C P = \frac{4 K}{7} + \frac{7 K}{15} + \frac{2 K}{3} .$

With denominator $105$ :

$C P = \frac{60 K + 49 K + 70 K}{105} = \frac{179 K}{105} .$

Step 6: Let marked price per kg of the mixture be $M$ . Total mixture weight is:

$4 + 14 + 6 = 24 kg .$

Step 7: If the whole mixture were sold at marked price, profit would be ₹1752:

$24 M = C P + 1752.$

Step 8: Actually, she sells 4 kg at marked price and the remaining 20 kg at a 20% discount. So actual revenue is:

$4 M + 20 (0.8 M) = 4 M + 16 M = 20 M .$

Actual profit is ₹744:

$20 M = C P + 744.$

Step 9: Subtract the second equation from the first:

$24 M - 20 M = (C P + 1752) - (C P + 744) .$

$4 M = 1008.$

$M = 252.$

Step 10: Use $20 M = C P + 744$ :

$20 \times 252 = C P + 744.$

$5040 = C P + 744.$

$C P = 4296.$

Step 11: Now use $C P = \frac{179 K}{105}$ :

$\frac{179 K}{105} = 4296.$

Since $4296 = 179 \times 24$ :

$K = 24 \times 105 = 2520.$

Step 12: Amount spent on almonds:

$6 A = 6 \cdot \frac{K}{9} = \frac{2 K}{3} .$

Substitute $K = 2520$ :

$\frac{2 \times 2520}{3} = 1680.$

Answer: option A.

Mixtures, Alligation and Replacement Previous Year Questions (PYQs) for CAT: 15+ 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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