Question 1

Two containers have alcohol solutions of 30% and 50% respectively. In what ratio should they be mixed to get a 40% alcohol solution?

Accepted Answer

Step 1: Use alligation method. Desired concentration = 40%. Step 2: Difference between 50% and 40% = 10. Difference between 40% and 30% = 10. Step 3: Ratio = (Difference from higher):(Difference from lower) = 10:10 = 1:1. Step 4: Mix equal quantities of both solutions to get 40% concentration.

Question 2

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 weighted average. Ratio = 3:2, total parts = 5. Step 2: Cost = (3 × 40 + 2 × 60)/5 = (120 + 120)/5 = 240/5 = ₹48 per kg. Step 3: For 25% profit, Selling Price = Cost Price × (1 + 25/100) = 48 × 1.25 = ₹60 per kg.

Question 3

A mixture of 60 litres contains milk and water in the ratio 2:1. How much water should be added to make the ratio 1:1?

Accepted Answer

Step 1: Current ratio = 2:1, total parts = 3. Milk = (2/3) × 60 = 40 litres. Water = (1/3) × 60 = 20 litres. Step 2: Let x litres of water be added. New water = 20 + x. Step 3: For ratio 1:1, Milk = Water. Step 4: 40 = 20 + x → x = 20 litres.

Question 4

A chemist has two solutions of acid with concentrations 25% and 75%. In what ratio should they be mixed to obtain a 60% acid solution?

Accepted Answer

Step 1: Use alligation method. Desired concentration = 60%. Step 2: Difference between 75% and 60% = 15. Difference between 60% and 25% = 35. Step 3: Ratio = (Difference from higher):(Difference from lower) = 15:35 = 3:7. Step 4: Mix the 75% solution and 25% solution in ratio 3:7.

Question 5

Two containers have alcohol solutions of 30% and 50% concentration respectively. In what ratio must they be mixed to obtain 40% concentration?

Accepted Answer

Step 1: Use alligation method. Desired concentration = 40%. Step 2: Difference between 50% and 40% = 10 parts. Step 3: Difference between 40% and 30% = 10 parts. Step 4: Ratio of 30% solution to 50% solution = 10:10 = 1:1.

Question 6

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 using alligation. Ratio = 3:2, so total parts = 5. Step 2: CP of mixture = (3 × 40 + 2 × 60)/5 = (120 + 120)/5 = 240/5 = ₹48 per kg. Step 3: For 25% profit, SP = CP × 1.25 = 48 × 1.25 = ₹60 per kg.

Question 7

A vessel contains 120 litres of a mixture of petrol and diesel in the ratio 7:5. How much of the mixture should be removed and replaced with pure petrol so that the ratio becomes 3:1?

Accepted Answer

Step 1: Original mixture: petrol = 70L, diesel = 50L (ratio 7:5, total 12 parts = 120L). Step 2: Let x litres be removed and replaced with petrol. Step 3: Petrol removed = 7x/12, diesel removed = 5x/12. Step 4: New petrol = 70 - 7x/12 + x = 70 + 5x/12. New diesel = 50 - 5x/12. Step 5: Ratio equation: (70 + 5x/12)/(50 - 5x/12) = 3/1. Step 6: 70 + 5x/12 = 3(50 - 5x/12) → 70 + 5x/12 = 150 - 15x/12 → 20x/12 = 80 → x = 48 litres.

Question 8

Three containers have milk solutions with milk concentrations of 20%, 30%, and 50% respectively. They are mixed in the ratio 2:3:5. What is the milk concentration in the final mixture?

Accepted Answer

Step 1: Ratio of containers = 2:3:5, total parts = 10. Step 2: Assume 2L, 3L, and 5L from each container respectively. Step 3: Milk from first = 2 × 0.20 = 0.4L. Milk from second = 3 × 0.30 = 0.9L. Milk from third = 5 × 0.50 = 2.5L. Step 4: Total milk = 0.4 + 0.9 + 2.5 = 3.8L. Total mixture = 2 + 3 + 5 = 10L. Step 5: Concentration = 3.8/10 = 0.38 = 38%.

Question 9

A goldsmith has two alloys of gold. The first alloy contains 60% gold and the second contains 80% gold. If he mixes them in the ratio 5:3, what is the percentage of gold in the resulting alloy?

Accepted Answer

Step 1: Ratio of alloys = 5:3, total parts = 8. Step 2: Assume 5 units of first alloy and 3 units of second alloy. Step 3: Gold from first alloy = 5 × 60% = 3 units. Gold from second alloy = 3 × 80% = 2.4 units. Step 4: Total gold = 3 + 2.4 = 5.4 units. Total alloy = 5 + 3 = 8 units. Step 5: Percentage of gold = (5.4/8) × 100 = 67.5%.

Question 10

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%. If he prepares 100 litres of the mixture, how many litres of Type A milk did he use?

Accepted Answer

Step 1: Selling price = ₹55/litre, profit = 10%, so cost price = 55 ÷ 1.10 = ₹50/litre. Step 2: Using alligation method: Type A (₹40) and Type B (₹60) mixed to get ₹50. Step 3: Difference from A = 50 - 40 = 10; Difference from B = 60 - 50 = 10. Step 4: Ratio A:B = 10:10 = 1:1. Step 5: In 100 litres, Type A = (1/2) × 100 = 50 litres.

Question 11

A jeweller has gold of two different purities: 18-carat (75% pure) and 22-carat (91.67% pure). He wants to create 240 grams of 20-carat gold (83.33% pure). How many grams of 18-carat gold should he use?

Accepted Answer

Step 1: Using alligation method with purities: 18-carat = 75%, 22-carat = 91.67%, Target = 83.33%. Step 2: Difference from 18-carat = 83.33 - 75 = 8.33; Difference from 22-carat = 91.67 - 83.33 = 8.34 (approximately 8.33). Step 3: Ratio of 18-carat to 22-carat = 8.33 : 8.33 = 1 : 1. Step 4: In 240 grams, 18-carat gold = (1/2) × 240 = 120 grams.

Question 12

A chemist mixes three solutions with alcohol concentrations of 20%, 40%, and 60% in the ratio 2:3:5 respectively. He then removes 30% of the mixture and replaces it with a 50% alcohol solution. What is the final alcohol concentration in the mixture?

Accepted Answer

Step 1: Initial mixture ratio = 2:3:5 (total parts = 10). Step 2: Weighted average concentration = (2×20 + 3×40 + 5×60)/10 = (40 + 120 + 300)/10 = 460/10 = 46%. Step 3: When 30% of mixture is removed, alcohol removed = 0.30 × 46% = 13.8%. Step 4: Remaining alcohol concentration = 46% - 13.8% = 32.2% of the remaining 70% volume. Step 5: Actual alcohol remaining = 0.70 × 46% = 32.2% (in terms of original volume). Step 6: After adding 50% solution (30% of original volume): Final concentration = (0.70 × 46 + 0.30 × 50)/100 = (32.2 + 15)/100 = 47.2%.

Question 13

A merchant has two types of tea: Type P at ₹120 per kg and Type Q at ₹180 per kg. He mixes them and sells at ₹150 per kg. Due to wastage during packing, 5% of the mixture is lost. If he wants a profit of 20% on the cost price, what should be the ratio of Type P to Type Q in the mixture before packing?

Accepted Answer

Step 1: Selling price = ₹150/kg, profit = 20%, so cost price = 150 ÷ 1.20 = ₹125/kg. Step 2: Due to 5% wastage, if 100 kg is mixed, only 95 kg is sold. Step 3: To sell 95 kg at ₹150/kg and make 20% profit on cost, the effective cost per kg of mixture = 125/kg. Step 4: Using alligation: Type P (₹120) and Type Q (₹180) should mix to give ₹125/kg. Step 5: Difference from P = 125 - 120 = 5; Difference from Q = 180 - 125 = 55. Step 6: Ratio P:Q = 55:5 = 11:1.

IBPS RRB PO Mixture & Alligation

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