Question 1

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

Accepted Answer

Step 1: Use alligation method. Concentration 1 = 20%, Concentration 2 = 50%, Target = 35%. Step 2: Difference from target: |20 - 35| = 15, |50 - 35| = 15. Step 3: Ratio = 15:15 = 1:1.

Question 2

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

Question 3

A container has 60 litres of a solution with acid and water in ratio 7:5. How many litres of water must be added to make the ratio 1:1?

Accepted Answer

Step 1: Original ratio acid:water = 7:5. Total parts = 12. Step 2: Acid = 60 × (7/12) = 35 litres, Water = 60 × (5/12) = 25 litres. Step 3: For ratio 1:1, acid = water. So water needed = 35 litres. Step 4: Water to add = 35 - 25 = 10 litres.

Question 4

A goldsmith mixes gold and copper in the ratio 4:1 by weight. If the total weight of the alloy is 50 kg, how much copper is present in the alloy?

Accepted Answer

Step 1: Gold:Copper = 4:1. Total parts = 4 + 1 = 5. Step 2: Total weight = 50 kg. Step 3: Copper weight = 50 × (1/5) = 10 kg.

Question 5

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. In what ratio should they be mixed?

Accepted Answer

Step 1: Use alligation method. Cost 1 = ₹80, Cost 2 = ₹120, Target = ₹100. Step 2: Difference from target: |80 - 100| = 20, |120 - 100| = 20. Step 3: Ratio = 20:20 = 1:1.

Question 6

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. If the mixture is sold at ₹55 per kg and no profit or loss is made, in what ratio were the two types mixed?

Accepted Answer

Step 1: Use alligation method. Cost of rice A = ₹40/kg, Cost of rice B = ₹60/kg, Selling price = ₹55/kg (which equals cost price for no profit/loss). Step 2: Difference between B's cost and mean price = 60 - 55 = 5. Difference between mean price and A's cost = 55 - 40 = 15. Step 3: By alligation, ratio of A to B = 5:15 = 1:3.

Question 7

A container has 120 litres of a mixture of alcohol and water in the ratio 7:5. How much pure alcohol must be added so that the ratio becomes 3:2?

Accepted Answer

Step 1: Original mixture has alcohol:water = 7:5, total parts = 12. Alcohol = (7/12) × 120 = 70 litres, Water = (5/12) × 120 = 50 litres. Step 2: Let x litres of pure alcohol be added. New alcohol = 70 + x, water remains 50. Step 3: New ratio is 3:2, so (70 + x)/50 = 3/2. Step 4: 2(70 + x) = 3 × 50 → 140 + 2x = 150 → 2x = 10 → x = 5 litres.

Question 8

Two containers have milk and water in ratios 3:2 and 5:4 respectively. If equal volumes are mixed from both containers, the resulting mixture has milk and water in what ratio?

Accepted Answer

Step 1: Container A has milk:water = 3:2, so in any volume V, milk = 3V/5, water = 2V/5. Step 2: Container B has milk:water = 5:4, so in volume V, milk = 5V/9, water = 4V/9. Step 3: When equal volumes V are mixed: total milk = 3V/5 + 5V/9 = (27V + 25V)/45 = 52V/45. Step 4: Total water = 2V/5 + 4V/9 = (18V + 20V)/45 = 38V/45. Step 5: Ratio = 52V/45 : 38V/45 = 52:38 = 26:19.

Question 9

A merchant mixes two types of rice costing ₹40/kg and ₹60/kg in a certain ratio. The mixture is sold at ₹55/kg, earning 10% profit. In what ratio were the two types mixed?

Accepted Answer

Step 1: Selling price = ₹55/kg with 10% profit means cost price = 55/1.1 = ₹50/kg. Step 2: Use alligation method. Cost of rice A = ₹40/kg, cost of rice B = ₹60/kg, mean cost = ₹50/kg. Step 3: Difference for A = 50 - 40 = 10. Difference for B = 60 - 50 = 10. Step 4: Ratio = 10:10 = 1:1.

Question 10

A container has 120 litres of a mixture with alcohol and water in ratio 7:5. How much pure alcohol must be added so that the ratio becomes 3:2?

Accepted Answer

Step 1: Original mixture: alcohol = 120 × (7/12) = 70 litres, water = 120 × (5/12) = 50 litres. Step 2: Let x litres of pure alcohol be added. New alcohol = 70 + x, water remains 50 litres. Step 3: New ratio is 3:2, so (70 + x)/50 = 3/2. Step 4: 2(70 + x) = 3(50) → 140 + 2x = 150 → 2x = 10 → x = 5 litres.

Question 11

Three containers have milk and water in ratios 1:1, 2:1, and 3:2 respectively. Equal volumes from each are mixed. What is the percentage of milk in the final mixture?

Accepted Answer

Step 1: Container 1 (1:1): milk fraction = 1/2. Container 2 (2:1): milk fraction = 2/3. Container 3 (3:2): milk fraction = 3/5. Step 2: Let V litres be taken from each. Total milk = V(1/2 + 2/3 + 3/5). Step 3: Find LCD of 2, 3, 5 = 30. Milk = V(15/30 + 20/30 + 18/30) = V(53/30). Step 4: Total volume = 3V. Milk percentage = (53V/30)/(3V) × 100 = (53/90) × 100 = 58.89% ≈ 58.9% or 589/10 %.

SSC MTS Mixture & Alligation

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60-Second Revision — Mixture & Alligation