Alligation or Mixture Aptitude

Alligation or Mixture problems involve combining two or more ingredients with different properties (like price, quality, concentration) to get a mixture with a desired property. The principle of alligation helps in finding the ratio in which the ingredients are to be mixed to achieve a specific quality or price in the resulting mixture. These problems are common in real-world scenarios such as cooking, chemistry, and economics.

1. A solution of 40 liters contains milk and water in the ratio 7:3. How much water must be added to make the ratio 3:7?

- Initial quantity of milk = (7/10) * 40 = 28 liters.
- Initial quantity of water = 40 - 28 = 12 liters.
- Let x liters of water be added. The new ratio = (28)/(12+x) = 3/7.
- Solving the equation, x = 50 liters.

2. Two types of wheat costing $100 and $80 per kg are mixed together in the ratio 2:3. What is the cost price of the mixed variety per kg?

- Cost of 1 kg of mixed variety = (2*$100 + 3*$80) / (2+3) = $90.

3. A 20-liter mixture of milk and water contains milk and water in the ratio 3:2. How much more milk should be added to make the ratio 2:1?

- Quantity of milk = (3/5) * 20 = 12 liters.
- Quantity of water = 20 - 12 = 8 liters.
- Let x liters of milk be added. The new ratio = (12+x)/8 = 2/1.
- Solving the equation, x = 10 liters.

4. A trader has 100 kg of rice, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. What is the quantity sold at 18% profit?

- Let x kg be sold at 18% profit.
- Using the alligation rule, (18 - 14) : (14 - 8) = (4):(6) = 2:3.
- Therefore, the quantity at 18% profit = (2/5) * 100 kg = 40 kg.

5. A container contains a mixture of two liquids A and B in the ratio 7:5. When 9 liters of the mixture is drawn off and the container is filled with B, the ratio of A and B becomes 7:9. How many liters of liquid A was contained by the container initially?

- Let the initial total quantity be x liters.
- Quantity of A initially = (7/12) * x.
- Quantity of A left = (7/12) * x - (7/16) * 9.
- New ratio of A:B = 7:9 implies (7/12) * x - (7/16) * 9 / (5/12) * x + 9 = 7/9.
- Solving, x = 48 liters. Thus, A initially = (7/12) * 48 = 28 liters.

6. A milk vendor has 2 types of milk: one costing $20 per liter and another $25 per liter. How much of each type should be mixed to get 20 liters of milk worth $22 per liter?

- Using alligation, the ratio of the mix is calculated as (25 - 22) : (22 - 20) = 3 : 2.
- Based on the alligation calculation, the quantities of each type of milk required are:
   - Quantity of $20 milk = (3/5) * 20 = 12 liters
   - Quantity of $25 milk = (2/5) * 20 = 8 liters

7. In what ratio must a grocer mix two types of pulses costing $15/kg and $20/kg, so that the mixture costs $16.50/kg?

- Using alligation, (20 - 16.50) : (16.50 - 15) = 3.50 : 1.50 = 7:3.

8. How many liters of water should be added to a 60-liter mixture of milk and water containing 45% water to make the water content 50% in the new mixture?

- Water in the original mixture = 45% of 60 = 27 liters.
- Let x liters of water be added. New quantity of water = 27 + x liters.
- New ratio (water) = 50%, so (27 + x) / (60 + x) = 1/2. Solving for x, x = 6 liters.

9. A 40-liter mixture contains milk and water in the ratio 2:3. How much milk should be added to make it a 3:2 ratio?

- Milk = (2/5) * 40 = 16 liters, Water = 24 liters.
- Let x liters of milk be added. New ratio (milk:water) = 3:2.
- So, (16 + x)/24 = 3/2. Solving, x = 10 liters.

10. A mixture of 150 kg of wine and water contains 20% water. How much more water should be added so that water may be 25% of the new mixture?

- Water in the mixture = 20% of 150 = 30 kg.
- Let x kg of water be added. New % of water = 25%.
- (30 + x) / (150 + x) = 25/100. Solving, x = 10 kg.

11. A merchant has coffee of two types, one costing $50 per kg and another $80 per kg. In what ratio should he mix them so the mixture costs $65 per kg?

- Using alligation, (80 - 65) : (65 - 50) = 15 : 15 = 3:2.

12. A container contains 40 liters of alcohol. From this container, 4 liters of alcohol is taken out and replaced with water. This process is repeated two more times. How much alcohol is now contained in the container?

- Alcohol left after each operation = (1 - 4/40) * previous quantity.
- After 3 operations, alcohol = 40 * (1 - 4/40)^3 ≈ 29.16 liters.

13. A solution contains 30% alcohol and the rest water. If 3 liters of water is added to the solution, the percentage of alcohol in the new solution becomes 25%. Find the initial volume of the solution.

- Let the initial volume be x liters.
- Alcohol in the initial solution = 0.30 * x.
- New volume = x + 3, Alcohol percentage = 25%.
- 0.30 * x = 0.25 * (x + 3). Solving, x = 15 liters.

14. A milkman has two types of milk: one with 15% water and another with 30% water. He wants to sell milk with 25% water content. In what ratio should he mix the two?

- Using alligation, (30 - 25) : (25 - 15) = 5 : 10 = 1:2.

15. A mixture contains wine and water in the ratio 4:1. On adding 5 liters of water, the ratio of wine to water becomes 4:3. Find the initial quantity of the mixture.

- Let the initial quantity of the mixture be x liters.
- Wine = (4/5) * x, Water = (1/5) * x.
- New ratio (wine:water) = 4:3, New water = (1/5) * x + 5.
- (4/5) * x / [(1/5) * x + 5] = 4/3. Solving, x = 20 liters.

16. How many liters of a 60% alcohol solution must be mixed with 80 liters of a 15% alcohol solution to get a 45% alcohol solution?

- Using alligation, (60 - 45) : (45 - 15) = 15 : 30 = 1:2.
- Ratio of the mix = 1:2.
- Quantity of 60% solution = (1/3) * (80 liters) = 40 liters.

17. In a 150-liter mixture of milk and water, the ratio of milk to water is 4:1. How much water must be added to make the ratio of milk to water 2:1?

- Milk = (4/5) * 150 = 120 liters, Water = 30 liters.
- Let x liters of water be added. New ratio (milk:water) = 2:1.
- 120 / (30 + x) = 2/1. Solving, x = 30 liters.

18. A 24-liter mixture contains lemonade and soda in the ratio 5:3. How much soda must be added to make the ratio 5:5?

- Lemonade = (5/8) * 24 = 15 liters, Soda = 9 liters.
- Let x liters of soda be added. New ratio (lemonade:soda) = 5:5 = 1:1.
- 15 / (9 + x) = 1/1. Solving, x = 6 liters.

19. A chemist has one solution that is 50% alcohol and another that is 75% alcohol. How many liters of each should be mixed to get 10 liters of a 60% alcohol solution?

- Using alligation, (75 - 60) : (60 - 50) = 15 : 10 = 3:2.
- Ratio of the mix = 3:2.
- Quantity of 50% solution = (2/5) * 10 liters = 4 liters, 75% solution = 6 liters.

20. In a 40-liter mixture of milk and water, the ratio is 7:1. How much water must be added to make the water content 25% of the mixture?

- Milk = (7/8) * 40 = 35 liters, Water = 5 liters.
- Let x liters of water be added. New ratio = 35 / (5 + x).
- For water to be 25%, (5 + x) / (40 + x) = 25/100.
- Solving, x = 11 liters.

21. A mixture of 90 kg of fruit and nuts contains 70% fruit. How many kilograms of nuts should be added to make the percentage of fruit 60%?

- Fruit = 70% of 90 = 63 kg, Nuts = 27 kg.
- Let x kg of nuts be added. New percentage of fruit = 63 / (90 + x).
- 63 / (90 + x) = 60/100. Solving, x = 15 kg.

22. A shopkeeper mixes two types of rice, one costing $18 per kg and the other $25 per kg, in the ratio 2:3. What is the average cost of the mixed rice per kg?

- Cost of 1 kg of mixed rice = (2*$18 + 3*$25) / (2+3) = $21.40.