Mixtures and Alligation

In 55 litres with ratio 7:4, milk = 7/11 x 55 = 35L, water = 4/11 x 55 = 20L. Let x be water added. New ratio 7:6 means 35/(20+x) = 7/6, solving gives 210 = 140 + 7x, so 7x = 70 and x = 10L. Option B is correct despite the typo 'litere'.

Initial mixture: 60L with milk:water = 3:7. Milk = (3/10) × 60 = 18L, Water = (7/10) × 60 = 42L. Let x liters of water be added. New ratio: 18/(42+x) = 2/5. Cross-multiplying: 90 = 84 + 2x, so 2x = 6 and x = 3L. Verification: 18/(42+3) = 18/45 = 2/5.

Let the mixing ratio be a:b. Vessel A has acid ratio 3/7, Vessel B has acid ratio 5/9. For the final mixture to have 50% acid (ratio 1:1), use allegation: (3/7)(a) + (5/9)(b) = 1/2(a+b). Solving: 3a/7 + 5b/9 = (a+b)/2. Multiply by 126: 54a + 70b = 63a + 63b. So 70b - 63b = 63a - 54a, or 7b = 9a. Thus a/b = 7/9. Option D (7:9) is correct. Options A, B, and C don't satisfy the equation.

Using allegation method: 25% and 50% alcohols mix to give 40%. Differences: 50 - 40 = 10, 40 - 25 = 15. Ratio of 25% : 50% = 10 : 15 = 2 : 3. Option C matches.

Each operation removes 8/80 = 1/10 of the milk and replaces it with water. After 3 operations: milk remaining = 80 × (0.9)^3 = 80 × 0.729 = 58.32 litres. This is a standard dilution problem using the concept of successive percentage decrease.

Using allegation: Cost difference between cheaper (6-5.40=0.60) and dearer (8-6=2) gives ratio 2:0.60 = 20:6 = 10:3. Mix rice at Rs 5.40 and Rs 8 in ratio 10:3 to get Rs 6/kg.

Initial solution has 3 liters × 10% = 0.3 liters of salt. Let x liters of water be added. Total volume becomes (3 + x) liters with 5% concentration: 0.3/(3 + x) = 0.05. Solving: 0.3 = 0.15 + 0.05x, so 0.15 = 0.05x, giving x = 3 liters. Adding 3 liters of water doubles the volume, halving the concentration.

Using alligation: Initial syrup (40% honey) mixed with replacement syrup (19% honey) gives final (26% honey). The ratio of initial to replacement is |26-19| : |40-26| = 7:14 = 1:2. So 2/3 of the original was replaced.

Initial mixture 80L with milk:water = 3:2. Milk = (3/5)×80 = 48L, Water = 32L. Let added water = x. New ratio 48:(32+x) = 2:3. Cross-multiply: 144 = 64 + 2x, so 2x = 80, x = 40L. Adding water only changes the water quantity, milk stays constant.

Initial mixture: 30L total, 90% milk = 27L milk and 3L water. After adding x liters of water, total volume = (30+x) liters. Milk percentage = 27/(30+x) = 80% = 0.8. Solving: 27 = 0.8(30+x) → 27 = 24 + 0.8x → 0.8x = 3 → x = 3.75 liters. Option C is correct.

To achieve 50% profit when selling at Rs. 40/litre, the cost price must be 40/1.5 = Rs. 26.67/litre. Using alligation, water (cost 0) and chemical (cost Rs. 50) must mix in ratio 8:7 to get cost price of Rs. 26.67, since (50-26.67):26.67 = 23.33:26.67 = 7:8, giving water:chemical = 8:7.

Selling price Rs. 68.2 at 10% profit means cost price = 68.2÷1.1 = Rs. 62. Using alligation: Rs. 60 and Rs. 65 must mix to get Rs. 62 average. The differences are (65-62)=3 and (62-60)=2, giving ratio 3:2. This means 3 parts of Rs. 60 tea and 2 parts of Rs. 65 tea - so ratio 3:2 is correct.

Initial: 100L mixture with milk:water = 7:3, so milk = 70L, water = 30L. Let x liters of milk be added. New ratio = (70+x)/(30) = 3/1. Solving: 70 + x = 90, so x = 20L. Adding 20L milk gives 90L milk and 30L water, which is 3:1.

After each of 6 customers, the petrol concentration reduces by 90%. After 6 rounds, concentration = (0.9)⁶ ≈ 0.5314. The 7th customer receives 40 litres, of which pure petrol = 40 × 0.5314 ≈ 21.25 litres.

Take capacities in ratio 5:4:5. Milk from A = 1/6 × 5 = 5/6, from B = 3/8 × 4 = 12/8, from C = 5/12 × 5 = 25/12. Total milk = 53/12. Total volume = 5 + 4 + 5 = 14. Water = 14 - 53/12 = 115/12. Milk:Water = 53:115.