Question 1

A grocer mixes two types of rice costing ₹40 per kg and ₹60 per kg. In what ratio should they be mixed so that the cost of the mixture is ₹50 per kg?

Accepted Answer

Step 1: Use the alligation method. Cost of cheaper rice = ₹40/kg, cost of dearer rice = ₹60/kg, cost of mixture = ₹50/kg. Step 2: Difference between mixture cost and cheaper cost = 50 − 40 = 10. Step 3: Difference between dearer cost and mixture cost = 60 − 50 = 10. Step 4: Ratio of cheaper to dearer rice = 10:10 = 1:1.

Question 2

Two containers have milk and water in ratios 3:1 and 5:3 respectively. If equal quantities from each container are mixed, what is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Container A has milk:water = 3:1, so in 4 parts, milk = 3 parts, water = 1 part. Step 2: Container B has milk:water = 5:3, so in 8 parts, milk = 5 parts, water = 3 parts. Step 3: Let equal quantity = LCM(4,8) = 8 units from each. Step 4: From A (8 units): milk = 6 units, water = 2 units. Step 5: From B (8 units): milk = 5 units, water = 3 units. Step 6: Final mixture: milk = 6+5 = 11 units, water = 2+3 = 5 units. Step 7: Ratio = 11:5.

Question 3

Two containers have milk and water in ratios 3:1 and 5:3 respectively. If equal quantities from each container are mixed together, what is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Container 1 has milk:water = 3:1, so in 4 units: milk = 3, water = 1. Step 2: Container 2 has milk:water = 5:3, so in 8 units: milk = 5, water = 3. Step 3: Take equal quantity (say 8 units) from each. From Container 1 (8 units): milk = 6, water = 2. From Container 2 (8 units): milk = 5, water = 3. Step 4: Total milk = 6 + 5 = 11, total water = 2 + 3 = 5. Step 5: Ratio = 11:5.

Question 4

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. What is the cost price of the mixture per kg?

Accepted Answer

Step 1: Rice A costs ₹40/kg, Rice B costs ₹60/kg, mixed in ratio 3:2. Step 2: Let quantity of A = 3 kg, quantity of B = 2 kg. Step 3: Cost of A = 3 × 40 = ₹120. Step 4: Cost of B = 2 × 60 = ₹120. Step 5: Total cost = 120 + 120 = ₹240. Step 6: Total quantity = 3 + 2 = 5 kg. Step 7: Cost per kg = 240/5 = ₹48/kg.

Question 5

A vessel contains 60 litres of a solution with alcohol and water in ratio 7:5. How much pure alcohol must be added to make the ratio 3:1?

Accepted Answer

Step 1: Original ratio alcohol:water = 7:5, total parts = 12. Step 2: In 60 litres: alcohol = 60 × (7/12) = 35 litres, water = 60 × (5/12) = 25 litres. Step 3: Let x litres of pure alcohol be added. Step 4: New alcohol = 35 + x, water remains 25. Step 5: New ratio is 3:1, so (35+x)/25 = 3/1. Step 6: 35 + x = 75 → x = 40 litres.

Question 6

A mixture of petrol and diesel is in the ratio 4:1. If 10 litres of diesel is added, the ratio becomes 4:3. Find the original quantity of petrol.

Accepted Answer

Step 1: Original ratio petrol:diesel = 4:1. Let petrol = 4x, diesel = x. Step 2: After adding 10 litres of diesel: petrol = 4x, diesel = x + 10. Step 3: New ratio is 4:3, so 4x/(x+10) = 4/3. Step 4: Cross-multiply: 3(4x) = 4(x+10) → 12x = 4x + 40 → 8x = 40 → x = 5. Step 5: Original petrol = 4x = 4(5) = 20 litres.

Question 7

A solution contains alcohol and water in the ratio 2:3. If 10 litres of water is added to 25 litres of this solution, what is the new ratio of alcohol to water?

Accepted Answer

Step 1: Original solution is 25 litres with alcohol:water = 2:3. Total parts = 5. Step 2: Alcohol in original = 25 × (2/5) = 10 litres. Water in original = 25 × (3/5) = 15 litres. Step 3: After adding 10 litres of water: alcohol remains = 10 litres, water becomes = 15 + 10 = 25 litres. Step 4: New ratio = 10:25 = 2:5.

Question 8

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. At what price per kg should he sell the mixture to gain 25% profit?

Accepted Answer

Step 1: Cost price of mixture per kg using alligation: Let ratio = 3:2 (total parts = 5). Step 2: Cost from first rice = (3/5) × 40 = ₹24 per kg of mixture. Cost from second rice = (2/5) × 60 = ₹24 per kg of mixture. Step 3: Total cost price = 24 + 24 = ₹48 per kg. Step 4: For 25% profit, selling price = 48 × 1.25 = ₹60 per kg.

Question 9

A merchant has two types of sugar: type A at ₹50 per kg and type B at ₹80 per kg. He wants to sell a mixture at ₹65 per kg. In what ratio should he mix them?

Accepted Answer

Step 1: Use alligation method. Cost of A = ₹50, Cost of B = ₹80, Selling price (mean) = ₹65. Step 2: Difference between B and mean = 80 - 65 = 15. Difference between mean and A = 65 - 50 = 15. Step 3: Ratio of A to B = 15:15 = 1:1.

Question 10

A container has 120 litres of a mixture of petrol and diesel in the ratio 5:3. If 30 litres of the mixture is removed and replaced with pure petrol, what is the new ratio of petrol to diesel?

Accepted Answer

Step 1: Original ratio 5:3 (total 8 parts). Petrol = (5/8)×120 = 75 litres, Diesel = (3/8)×120 = 45 litres. Step 2: Remove 30 litres: petrol removed = (5/8)×30 = 18.75 litres, diesel removed = (3/8)×30 = 11.25 litres. Step 3: After removal: petrol = 75 - 18.75 = 56.25 litres, diesel = 45 - 11.25 = 33.75 litres. Step 4: Add 30 litres pure petrol: petrol = 56.25 + 30 = 86.25 litres, diesel = 33.75 litres. Step 5: New ratio = 86.25:33.75 = 345:135 = 23:9.

Question 11

A chemist has three solutions: Solution X (30% acid), Solution Y (50% acid), and Solution Z (80% acid). He mixes X and Y in ratio 3:2 to get Mixture M. He then mixes Mixture M with Solution Z in ratio 4:1 to get the final mixture. What is the acid percentage in the final mixture?

Accepted Answer

Step 1: Mixture M from X and Y in ratio 3:2. Acid% in M = (3×30 + 2×50)/(3+2) = (90+100)/5 = 190/5 = 38%. Step 2: Final mixture from M and Z in ratio 4:1. Acid% in final = (4×38 + 1×80)/(4+1) = (152+80)/5 = 232/5 = 46.4%.

Question 12

A jeweller has gold of two different purities: 18-carat (75% pure) and 22-carat (91.67% pure). To make 100 grams of 20-carat gold (83.33% pure), how many grams of 18-carat gold should be used?

Accepted Answer

Step 1: Let 18-carat gold = x grams and 22-carat gold = y grams. Step 2: x + y = 100 (total weight). Step 3: Purity equation: 0.75x + 0.9167y = 0.8333(100) = 83.33. Step 4: From equation 1: y = 100 − x. Step 5: Substitute: 0.75x + 0.9167(100 − x) = 83.33 → 0.75x + 91.67 − 0.9167x = 83.33 → −0.1667x = −8.34 → x = 50. Step 6: Verify: 0.75(50) + 0.9167(50) = 37.5 + 45.835 = 83.335 ≈ 83.33. ✓

Question 13

Two containers A and B have milk and water in ratios 3:1 and 5:3 respectively. If equal volumes are mixed from both containers, what is the ratio of milk to water in the resulting mixture?

Accepted Answer

Step 1: Container A has milk:water = 3:1, so in any volume V, milk = 3V/4, water = V/4. Step 2: Container B has milk:water = 5:3, so in volume V, milk = 5V/8, water = 3V/8. Step 3: When equal volumes V are mixed, total milk = 3V/4 + 5V/8 = 6V/8 + 5V/8 = 11V/8. Step 4: Total water = V/4 + 3V/8 = 2V/8 + 3V/8 = 5V/8. Step 5: Ratio of milk to water = (11V/8):(5V/8) = 11:5.

Question 14

A chemist has two solutions of acid with concentrations 20% and 50%. In what ratio must they be mixed to obtain 30 litres of 35% acid solution?

Accepted Answer

Step 1: Let volume of 20% solution = x litres and 50% solution = y litres. Step 2: x + y = 30 (total volume). Step 3: Acid content: 0.20x + 0.50y = 0.35(30) = 10.5. Step 4: From equation 1: y = 30 − x. Step 5: Substitute: 0.20x + 0.50(30 − x) = 10.5 → 0.20x + 15 − 0.50x = 10.5 → −0.30x = −4.5 → x = 15. Step 6: y = 30 − 15 = 15. Step 7: Ratio = 15:15 = 1:1.

Question 15

A milkman mixes two types of milk: Type A costing ₹40 per litre and Type B costing ₹60 per litre. He sells the mixture at ₹55 per litre and makes a profit of 10% on the cost price of the mixture. In what ratio must he mix Type A and Type B milk?

Accepted Answer

Step 1: Selling price = ₹55/litre with 10% profit means Cost Price of mixture = 55 ÷ 1.10 = ₹50/litre. Step 2: Using alligation method, place cost prices on a line: Type A (₹40) and Type B (₹60) with mean (₹50). Step 3: Difference from Type A = 50 - 40 = 10. Difference from Type B = 60 - 50 = 10. Step 4: Ratio = 10:10 = 1:1.

SSC CHSL Mixture & Alligation

Mixture & Alligation— Rules & Concept

Key Points to Remember

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Mixture & Alligation — Practice Questions

60-Second Revision — Mixture & Alligation