Mixtures and Alligation

Solution A has milk:water = 5:1, so 6 parts total with 5 parts milk. Solution B has milk:water = 3:1, so 4 parts total with 3 parts milk. When mixed in 4:3 ratio, take 4 units of A (20 parts milk, 4 parts water) and 3 units of B (9 parts milk, 3 parts water). Total milk = 20+9 = 29 parts. Total water = 4+3 = 7 parts. Wait, let me recalculate: For A (5:1), in 4 units: milk = 5/6 × 4 = 20/6 = 10/3, water = 1/6 × 4 = 4/6 = 2/3. For B (3:1), in 3 units: milk = 3/4 × 3 = 9/4, water = 1/4 × 3 = 3/4. Total milk = 10/3 + 9/4 = 40/12 + 27/12 = 67/12. Total water = 2/3 + 3/4 = 8/12 + 9/12 = 17/12. Ratio = 67:17. This matches option B.

Let initial mixture be M litres. Milk = 0.28M. After adding 16L water, total = M + 16. Milk percentage becomes: 0.28M/(M + 16) = 0.20. Solving: 0.28M = 0.20M + 3.2, 0.08M = 3.2, M = 40. Initial mixture was 40 litres.

Original mixture: 28L milk + 8L water = 36L total. After adding x liters each: (28+x)+(8+x)=36+2x total. 40% of new mixture = 20L, so new mixture = 20/0.4 = 50L. Therefore 36+2x=50, 2x=14, x=7.

Total cost = (20 × 6.50) + (30 × 7.00) = 130 + 210 = Rs 350. Total quantity = 20 + 30 = 50 kg. Average cost = 350/50 = Rs 7 per kg. Profit = Rs 60 on entire quantity, so profit per kg = 60/50 = Rs 1.20. Selling price = Rs 7 + Rs 1.20 = Rs 8.20. But checking options, Rs 8.00 gives Rs 50 total profit (8 × 50 = 400, 400-350 = 50), Rs 8.80 gives Rs 90. Rs 8.00 is closest to match Rs 60 profit. Actually, if SP = Rs 8/kg, total SP = 400, profit = 400-350 = 50. If we need exactly Rs 60 profit, SP should be (350+60)/50 = Rs 8.20, which isn't an option. Rs 8.00 is the best available option.

Calculate total cost price: (12 × 14.50) + (15 × 13) = 174 + 195 = Rs 369. Total quantity: 12 + 15 = 27 kg. Selling price: 27 × 16 = Rs 432. Gain = SP - CP = 432 - 369 = Rs 63. This is a straightforward weighted average mixture problem.

Let the original quantity be x kg. Sugar initially = 0.16x. After evaporating 64 kg, remaining solution = x - 64 kg. New sugar percentage is 24%, so 0.16x = 0.24(x - 64). Solving: 0.16x = 0.24x - 15.36, which gives 0.08x = 15.36, so x = 15.36/0.08 = 192 kg. Since 1 kg of water ≈ 1 liter, the answer is approximately 192 liters. The key principle: the amount of sugar remains constant; only water evaporates.

Let the initial mixture quantity be x liters. Water = 25% of x = 0.25x, Spirit = 75% of x = 0.75x. After adding 15 liters of water: New water = 0.25x + 15, Total mixture = x + 15. Given that water becomes 40%: (0.25x + 15)/(x + 15) = 0.4. Solving: 0.25x + 15 = 0.4x + 6, so 15 - 6 = 0.4x - 0.25x, giving 9 = 0.15x, therefore x = 60. Spirit = 0.75 × 60 = 45 liters. Option B is correct.

Initial: 10L milk + 40L water = 50L mixture (20% milk). Replace x liters with pure milk to get 40% milk. Remaining milk = 10 - 0.2x, new milk added = x. Final milk = 10 - 0.2x + x = 10 + 0.8x. Final volume = 50L. For 40% concentration: (10 + 0.8x)/50 = 0.4, giving 10 + 0.8x = 20, so x = 12.5L.

This is a dilution problem. When 8 litres are drawn and replaced with water 4 times, the remaining milk is initial × (1 - 8/V)⁴, where V is initial volume. Given milk remaining/total = 81/625 after 4 operations. Since each operation removes the same fraction, (1 - 8/V)⁴ = 81/625. Taking fourth root: 1 - 8/V = 3/5 (since 3⁴ = 81 and 5⁴ = 625). So 8/V = 2/5, giving V = 20 litres.

Let the ratio be a:b:c. Using weighted average: (50a + 60b + 100c)/(a + b + c) = 80. This simplifies to 50a + 60b + 100c = 80a + 80b + 80c, giving 30a + 20b = 20c, or 3a + 2b = 2c. Checking option A (2:2:5): 3(2) + 2(2) = 10 = 2(5). This satisfies the equation. Verification: (2×50 + 2×60 + 5×100)/(2+2+5) = (100+120+500)/9 = 720/9 = Rs. 80/kg.

Let total = 5k. X = 2k, Y = 3k. After operation: X = 2k+6, Y = 3k-6. (2k+6)/(3k-6) = 4/3, so k = 7. Initial X = 14. Option B is correct.

माना initial mixture में water = 7x, milk = 12x, total = 19x। 38 L निकालने पर water में से 38×7/19 = 14 L और milk में से 38×12/19 = 24 L निकलता है। Remaining: water = 7x-14, milk = 12x-24। 19 L water add करने पर water = 7x+5, milk = 12x-24। New ratio: (7x+5)/(12x-24) = 25/32। Cross multiply: 32(7x+5) = 25(12x-24), 224x+160 = 300x-600, 76x = 760, x = 10। Initial = 19×10 = 190 L।

प्रत्येक चरण में 9/90 = 1/10 दूध हटाया जाता है, इसलिए दूध का 9/10 भाग बना रहता है। तीन चरणों के बाद दूध = 90 × (9/10)^3 = 90 × 729/1000 = 65.61 लीटर। पानी = 90 - 65.61 = 24.39 लीटर। अनुपात = 65.61 : 24.39 = 729 : 271। यह डायलूशन प्रॉब्लम का एक मानक प्रश्न है जहां प्रत्येक चरण में एक समान अनुपात में द्रव हटाया और बदला जाता है।

Each replacement keeps 75% milk. After 5 iterations, milk = initial × (0.75)^5 = initial × 243/1024. Given final milk is 972 ml, initial = 972 × 1024/243 = 4096 ml = 4.096 L.

Let total weight = 100 units. Container alone = 25 units, so fluid = 75 units. After removal, total weight = 50 units, meaning remaining fluid = 50-25 = 25 units. Fluid removed = 75-25 = 50 units. Fraction removed = 50/75 = 2/3.