Question 1

Two alloys have gold and silver in the ratio 7:5 and 5:3 respectively. Equal quantities of both alloys are melted together. What is the ratio of gold to silver in the resulting alloy?

Accepted Answer

Step 1: Let equal quantity = 12 kg (LCM of 12 and 8 for easy calculation). Step 2: Alloy 1 (ratio 7:5): gold = (7/12) × 12 = 7 kg, silver = (5/12) × 12 = 5 kg. Step 3: Alloy 2 (ratio 5:3): gold = (5/8) × 12 = 7.5 kg, silver = (3/8) × 12 = 4.5 kg. Step 4: Total gold = 7 + 7.5 = 14.5 kg, total silver = 5 + 4.5 = 9.5 kg. Step 5: Ratio = 14.5 : 9.5 = 29 : 19.

Question 2

A mixture contains sugar and salt in the ratio 4:3. If 21 kg of salt is present, how much sugar is in the mixture?

Accepted Answer

Step 1: Ratio of sugar to salt is 4:3. Step 2: If salt is 3 parts and equals 21 kg, then 1 part = 21/3 = 7 kg. Step 3: Sugar is 4 parts, so sugar = 4 × 7 = 28 kg.

Question 3

Two alloys contain copper and tin in the ratios 3:2 and 4:1 respectively. Equal weights of both alloys are melted together. What is the percentage of copper in the final alloy?

Accepted Answer

Step 1: Alloy 1 has copper:tin = 3:2, so copper fraction = 3/5. Step 2: Alloy 2 has copper:tin = 4:1, so copper fraction = 4/5. Step 3: When equal weights are mixed, average copper fraction = (3/5 + 4/5)/2 = 7/10. Step 4: Percentage of copper = (7/10) × 100 = 70%.

Question 4

A chemist mixes solution A (30% acid) and solution B (50% acid) in the ratio 2:3. What is the percentage of acid in the final mixture?

Accepted Answer

Step 1: Let quantities be 2x and 3x. Step 2: Acid from solution A = 2x × 30% = 0.6x. Step 3: Acid from solution B = 3x × 50% = 1.5x. Step 4: Total acid = 0.6x + 1.5x = 2.1x. Step 5: Total quantity = 2x + 3x = 5x. Step 6: Percentage of acid = (2.1x / 5x) × 100 = (2.1/5) × 100 = 0.42 × 100 = 42%.

Question 5

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 per kg of the mixture?

Accepted Answer

Step 1: Let the quantities be 3x kg and 2x kg respectively. Step 2: Total cost = (3x × 40) + (2x × 60) = 120x + 120x = 240x. Step 3: Total quantity = 3x + 2x = 5x kg. Step 4: Cost per kg = 240x / 5x = ₹48 per kg.

Question 6

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. If he wants to create a mixture costing ₹50 per kg, in what ratio should he mix the two types?

Accepted Answer

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

Question 7

In what ratio must sugar costing ₹15 per kg be mixed with sugar costing ₹24 per kg to get a mixture costing ₹18 per kg?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper sugar = ₹15, cost of dearer sugar = ₹24, mean cost = ₹18. Step 2: Difference between mean and cheaper = 18 - 15 = 3. Step 3: Difference between dearer and mean = 24 - 18 = 6. Step 4: Ratio = 6:3 = 2:1 (cheaper to dearer).

Question 8

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: Let the quantities be 3x kg and 2x kg respectively. Step 2: Cost of first type = 3x × 40 = 120x. Step 3: Cost of second type = 2x × 60 = 120x. Step 4: Total cost = 120x + 120x = 240x. Step 5: Total quantity = 3x + 2x = 5x. Step 6: Cost per kg = 240x / 5x = 48.

Question 9

Two containers have milk and water in the ratio 3:2 and 5:3 respectively. If 20 litres from the first container and 30 litres from the second container are mixed together, what is the ratio of milk to water in the resulting mixture?

Accepted Answer

Step 1: In first container (ratio 3:2), milk fraction = 3/5. In 20 litres: milk = 20 × 3/5 = 12 litres, water = 20 × 2/5 = 8 litres. Step 2: In second container (ratio 5:3), milk fraction = 5/8. In 30 litres: milk = 30 × 5/8 = 18.75 litres, water = 30 × 3/8 = 11.25 litres. Step 3: Total milk = 12 + 18.75 = 30.75 litres, total water = 8 + 11.25 = 19.25 litres. Step 4: Ratio = 30.75 : 19.25 = 123 : 77.

Question 10

A mixture contains alcohol and water in the ratio 4:1. If 10 litres of water is added to the mixture, the ratio becomes 4:3. What was the original quantity of alcohol in the mixture?

Accepted Answer

Step 1: Let original alcohol = 4x litres and water = x litres. Step 2: After adding 10 litres of water: alcohol = 4x, water = 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 alcohol = 4x = 4(5) = 20 litres.

Question 11

A container has 60 litres of a mixture of milk and water in the ratio 7:5. How much milk should be added to make the ratio 3:2?

Accepted Answer

Step 1: Original ratio 7:5 means milk = (7/12) × 60 = 35 litres, water = (5/12) × 60 = 25 litres. Step 2: Let x litres of milk be added. New milk = 35 + x, water remains 25. Step 3: New ratio is 3:2, so (35+x)/25 = 3/2. Step 4: Cross-multiply: 2(35+x) = 3(25) → 70 + 2x = 75 → 2x = 5 → x = 2.5 litres.

Question 12

A merchant has two types of sugar: Type A at ₹30 per kg and Type B at ₹50 per kg. He mixes them to get 80 kg of mixture at ₹40 per kg. How many kilograms of Type A sugar did he use?

Accepted Answer

Step 1: Let Type A sugar = x kg, then Type B sugar = (80 - x) kg. Step 2: Total cost equation: 30x + 50(80 - x) = 40 × 80. Step 3: 30x + 4000 - 50x = 3200. Step 4: -20x = -800. Step 5: x = 40 kg.

Question 13

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. In what ratio should they be mixed to get a mixture costing ₹50 per kg?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper rice = ₹40, cost of dearer rice = ₹60, cost of mixture = ₹50. 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 = 10:10 = 1:1.

Question 14

A container has 120 litres of a mixture 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 mixture has alcohol:water = 7:5, total parts = 12. Step 2: Alcohol = 120 × (7/12) = 70 litres, water = 120 × (5/12) = 50 litres. Step 3: Let x litres of pure alcohol be added. Step 4: New alcohol = 70 + x, water remains 50. Step 5: New ratio = (70 + x):50 = 3:1. Step 6: Cross multiply: 70 + x = 3 × 50 → 70 + x = 150 → x = 80 litres.

Question 15

A chemist has two solutions of acid with concentrations 30% and 50%. In what ratio should they be mixed to obtain 100 litres of 42% acid solution?

Accepted Answer

Step 1: Use alligation method. Concentration of weaker solution = 30%, stronger = 50%, target = 42%. Step 2: Difference between target and weaker = 42 − 30 = 12. Step 3: Difference between stronger and target = 50 − 42 = 8. Step 4: Ratio of weaker to stronger = 8:12 = 2:3. Step 5: Total parts = 2 + 3 = 5. Step 6: Weaker solution needed = 100 × (2/5) = 40 litres, stronger = 100 × (3/5) = 60 litres.

Question 16

A container has 120 litres of a mixture of petrol and diesel in the ratio 7:5. How many litres of diesel should be added to make the ratio 1:1?

Accepted Answer

Step 1: Original ratio = 7:5, total parts = 12. Step 2: Petrol = (7/12) × 120 = 70 litres. Diesel = (5/12) × 120 = 50 litres. Step 3: For ratio 1:1, petrol and diesel must be equal. Step 4: Diesel needed = 70 - 50 = 20 litres.

Question 17

A chemist has three solutions with acid concentrations of 20%, 30%, and 50% respectively. He mixes the first two in equal quantities and then adds the third solution in such a way that the final mixture has 40% acid concentration. In what ratio should the third solution be added to the mixture of the first two?

Accepted Answer

Step 1: Mix first two solutions in equal quantities (1:1). Resulting concentration = (20 + 30)/2 = 25%. Step 2: Now we have a 25% solution that needs to be mixed with 50% solution to get 40%. Step 3: Using alligation: difference between 50% and 40% = 10. Difference between 40% and 25% = 15. Step 4: Ratio of 25% mixture to 50% solution = 10:15 = 2:3.

Question 18

A merchant blends two varieties of tea costing ₹180 per kg and ₹240 per kg in the ratio 5:3. What is the cost per kg of the blended tea?

Accepted Answer

Step 1: Tea variety 1 costs ₹180/kg, variety 2 costs ₹240/kg, mixed in ratio 5:3. Step 2: Total parts = 5 + 3 = 8. Step 3: Cost contribution from variety 1 = 180 × (5/8) = 900/8 = 112.5. Step 4: Cost contribution from variety 2 = 240 × (3/8) = 720/8 = 90. Step 5: Total cost per kg = 112.5 + 90 = 202.5 per kg.

Question 19

A goldsmith has two gold alloys: one with 80% purity and another with 60% purity. He mixes them in the ratio 2:3 to create a new alloy. What is the purity percentage of the resulting alloy?

Accepted Answer

Step 1: Ratio of alloys = 2:3, total parts = 5. Step 2: Gold from first alloy = (2/5) × 80% = 32%. Step 3: Gold from second alloy = (3/5) × 60% = 36%. Step 4: Total purity = 32% + 36% = 68%.

Question 20

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 make a 25% profit?

Accepted Answer

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

RRB NTPC Mixture & Alligation

Mixture & Alligation— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Mixture & Alligation — Practice Questions

60-Second Revision — Mixture & Alligation