Question 1

Two containers have sugar solutions of concentrations 20% and 50% respectively. In what ratio should they be mixed to get a solution of 35% concentration?

Accepted Answer

Step 1: Use alligation method. Concentration of solution 1 = 20%, solution 2 = 50%, desired = 35%. Step 2: Difference from desired: 35 - 20 = 15 (for 20% solution), 50 - 35 = 15 (for 50% solution). Step 3: Ratio = 15:15 = 1:1.

Question 2

A solution contains acid and water in the ratio 2:3. If 10 litres of water is added to 50 litres of this solution, what is the new ratio of acid to water?

Accepted Answer

Step 1: Original solution = 50 litres, ratio acid:water = 2:3, total parts = 5. Acid = (2/5) × 50 = 20 litres, Water = (3/5) × 50 = 30 litres. Step 2: After adding 10 litres of water: Acid = 20 litres (unchanged), Water = 30 + 10 = 40 litres. Step 3: New ratio = 20:40 = 1:2.

Question 3

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: Rice type 1 costs ₹40/kg, quantity ratio = 3 parts. Rice type 2 costs ₹60/kg, quantity ratio = 2 parts. Step 2: Total cost = (40 × 3) + (60 × 2) = 120 + 120 = 240. Step 3: Total quantity = 3 + 2 = 5 parts. Step 4: Cost per kg = 240/5 = ₹48/kg.

Question 4

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. If the mixture costs ₹50 per kg and the quantity of the first type is 30 kg, find the quantity of the second type.

Accepted Answer

Step 1: Use alligation method. Cost of type 1 = ₹40, cost of type 2 = ₹60, mixture cost = ₹50. Step 2: Difference from type 1: 50 - 40 = 10. Difference from type 2: 60 - 50 = 10. Step 3: Ratio of quantities = 10:10 = 1:1. Step 4: If type 1 = 30 kg, then type 2 = 30 kg.

Question 5

Two containers have alcohol and water in ratios 3:2 and 4:1 respectively. If 20 litres from the first container and 30 litres from the second are mixed together, what is the ratio of alcohol to water in the final mixture?

Accepted Answer

Step 1: Container 1 has ratio 3:2 (total parts = 5). In 20L: alcohol = 20 × (3/5) = 12L, water = 20 × (2/5) = 8L. Step 2: Container 2 has ratio 4:1 (total parts = 5). In 30L: alcohol = 30 × (4/5) = 24L, water = 30 × (1/5) = 6L. Step 3: Final mixture: alcohol = 12 + 24 = 36L, water = 8 + 6 = 14L. Step 4: Ratio = 36:14 = 18:7.

Question 6

Three containers have sugar solutions with concentrations 20%, 30%, and 50% respectively. If 10 litres from the first, 15 litres from the second, and 25 litres from the third are mixed, what is the concentration of sugar in the final mixture?

Accepted Answer

Step 1: Sugar from container 1: 10 × (20/100) = 2 litres. Step 2: Sugar from container 2: 15 × (30/100) = 4.5 litres. Step 3: Sugar from container 3: 25 × (50/100) = 12.5 litres. Step 4: Total sugar = 2 + 4.5 + 12.5 = 19 litres. Step 5: Total mixture = 10 + 15 + 25 = 50 litres. Step 6: Concentration = (19/50) × 100 = 38%.

Question 7

A merchant has two types of tea costing ₹80 per kg and ₹120 per kg. He mixes them to get a blend costing ₹100 per kg. If he uses 15 kg of the cheaper tea, how much of the expensive tea should he use?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper tea = ₹80, cost of expensive tea = ₹120, blend cost = ₹100. Step 2: Difference from cheaper: 100 - 80 = 20. Difference from expensive: 120 - 100 = 20. Step 3: Ratio of quantities = 20:20 = 1:1. Step 4: If cheaper tea = 15 kg, then expensive tea = 15 kg.

Question 8

A chemist mixes two solutions of acid with concentrations 40% and 60%. After mixing, she realizes the resulting mixture has 50% concentration. She then adds 10 litres of pure acid to this mixture, making the concentration 55%. What was the original quantity of the 40% solution?

Accepted Answer

Step 1: Let the 40% solution = a litres, 60% solution = b litres. Step 2: After mixing: 0.4a + 0.6b = 0.5(a + b). Step 3: Simplify: 0.4a + 0.6b = 0.5a + 0.5b → 0.1b = 0.1a → a = b. Step 4: So equal quantities were mixed initially. Let a = b = x. Step 5: Mixture before adding pure acid = 2x litres with 50% concentration = x litres of acid. Step 6: After adding 10 litres of pure acid: total acid = x + 10, total volume = 2x + 10. Step 7: New concentration = (x + 10)/(2x + 10) = 0.55. Step 8: x + 10 = 0.55(2x + 10) → x + 10 = 1.1x + 5.5 → 4.5 = 0.1x → x = 45. Step 9: Original 40% solution = 45 litres.

Question 9

A container has 120 litres of a mixture of alcohol and water in the ratio 7:5. Some of this mixture is replaced with pure alcohol such that the new ratio becomes 3:2. How many litres of the mixture was replaced?

Accepted Answer

Step 1: Original mixture = 120 litres with ratio 7:5. Alcohol = 120 × (7/12) = 70 litres, Water = 120 × (5/12) = 50 litres. Step 2: Let x litres of mixture be removed and replaced with pure alcohol. Step 3: After removal: Alcohol = 70 - 7x/12, Water = 50 - 5x/12. Step 4: After adding x litres of pure alcohol: Alcohol = 70 - 7x/12 + x = 70 + 5x/12, Water = 50 - 5x/12. Step 5: New ratio = 3:2, so (70 + 5x/12)/(50 - 5x/12) = 3/2. Step 6: Cross multiply: 2(70 + 5x/12) = 3(50 - 5x/12). Step 7: 140 + 10x/12 = 150 - 15x/12. Step 8: 25x/12 = 10. Step 9: x = 120/25 = 4.8 litres.

Question 10

A merchant has two types of tea costing ₹80 per kg and ₹120 per kg. He mixes them and sells the mixture at ₹117 per kg, making a profit of 30%. In what ratio should he mix the two types to achieve this profit?

Accepted Answer

Step 1: Selling price = ₹117 per kg with 30% profit. Step 2: Cost price = 117/1.30 = ₹90 per kg. Step 3: Using alligation method for cost price: Tea A costs ₹80, Tea B costs ₹120, mixture cost = ₹90. Step 4: Difference for Tea A = 90 - 80 = 10. Step 5: Difference for Tea B = 120 - 90 = 30. Step 6: Ratio = 30:10 = 3:1 (inverse of differences). Step 7: Therefore, mix Tea A (₹80) and Tea B (₹120) in ratio 3:1.

IBPS Clerk Mixture & Alligation

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