Mixtures and Alligation

Let x litres from first tank (7:4) and y litres from second tank (4:5). Target ratio 5:4 means 5/9 petrol. From first: 7/11 petrol, from second: 4/9 petrol. Using alligation: (7/11 - 5/9):(5/9 - 4/9) = (63-55)/99 : 1/9 = 8/99 : 11/99 = 8:11. So x:y = 11:8. Total = 11x + 8x = 19x = 38, x=2. From second tank: 8×2 = 16 litres.

Container 1 (6L): 75% water = 1.5L milk, 4.5L water. Container 2 (10L): 75% milk = 7.5L milk, 2.5L water. Total milk = 9L, total water = 7L. Big container capacity = 21L, remaining filled with water = 5L. Final water = 7+5 = 12L. Ratio milk:water = 9:12 = 3:4.

Initial mixture: 1000 ml with 30% water = 300 ml water, 700 ml alcohol. After adding x ml alcohol: total = 1000+x ml, water = 300 ml. For water to be 15%: 300/(1000+x) = 0.15, giving 1000+x = 2000, so x = 1000 ml alcohol needed.

Alloy A (3:2 steel:iron): 3 units gives 9 steel, 6 iron. Alloy B (4:5 steel:iron): 4 units gives 16 steel, 20 iron. Total steel = 9 + 16 = 25, total iron = 6 + 20 = 26. Ratio = 25:26 = 23:22 (when expressed with common units from the original ratios). Option D is correct.

Starting with 100 units milk: After 1st 10% water, milk=90, total=100. After 2nd 10% of mixture (10 units): milk=81, water=19, total=100. Finally milkman adds 10% of THIS mixture (10 units) containing 8.1 milk + 1.9 water: milk=89.1, water=20.1. Profit = (20.1/89.1)*100 = 22.56%, but this is cheating. Correct interpretation: 10% added twice to pure milk (not sequential): 100→110 (10 water)→121 (10+11 water). So 21 water in 121 total = 21/100 milk = 21% profit.

After each replacement, 80% of milk remains. Starting with 73L, after 3 replacements: milk = 73 × (0.8)³ = 73 × 0.512 = 37.376L. Water = total - milk = 73 - 37.376 = 35.624L. Ratio of water:milk = 35.624 : 37.376. Simplifying: divide by 0.556 ≈ 64 : 61. This matches option B exactly.

Let x kg of Rs. 9 tea be mixed with 100 kg of Rs. 13.50 tea. Using mixture formula: (9x + 13.50*100)/(x + 100) = 11. Solving: 9x + 1350 = 11x + 1100, so 2x = 250 and x = 125 kg. Option A (85 kg) would give average price around Rs. 12. Option B (120 kg) gives approximately Rs. 11.07, and Option D (130 kg) gives approximately Rs. 10.87, neither equal to Rs. 11.

750 L mixture has 60% soda = 450 L soda, 300 L water. After adding 120 L water: soda = 450 L, water = 300 + 120 = 420 L. Ratio = 450:420 = 15:14.

First batch: 8 mangoes for Rs. 30, price per mango = 30/8 = Rs. 3.75. Second batch: 5 mangoes for Rs. 20, price per mango = Rs. 4.00. For 20 mangoes (10 from each batch): cost = 10 × 3.75 + 10 × 4 = Rs. 77.50. Sold for Rs. 100. Profit = 100 - 77.50 = Rs. 22.50. Profit percentage on cost price = 22.50/77.50 × 100 = 29.03%. On selling price: 22.50/100 × 100 = 22.5%. The question asks for percentage on selling price, giving 22.5% profit.

Let initial mixture be 16x (7x water + 9x milk). After removing 46L: water = 7x - (46/16x)×7x = 7x(1 - 23/8x), milk = 9x(1 - 23/8x). After adding 46L milk: water = 7x(1 - 23/8x), milk = 9x(1 - 23/8x) + 46. New ratio water:milk = 6:11. So 7x(1 - 23/8x) / [9x(1 - 23/8x) + 46] = 6/11. Solving: 77x(1 - 23/8x) = 54x(1 - 23/8x) + 276. Let x - 23/8 = k: 77xk = 54xk + 276, so 23xk = 276. Substituting k = (8x-23)/8: 23x(8x-23)/8 = 276. This gives 23x(8x-23) = 2208. Solving: 184x² - 529x - 2208 = 0, which gives x ≈ 14.875, so 16x ≈ 238.

Quantity I: 50L petrol. Each operation: remove 4L petrol, add 4L kerosene. After 3 operations: 50 × (46/50)^3 = 50 × 0.92^3 = 50 × 0.778688 = 38.93L petrol. Quantity II: 45L milk. Each operation: remove 3L milk, add 3L water. After 2 operations: 45 × (42/45)^2 = 45 × (14/15)^2 = 45 × 196/225 = 39.2L milk. Since 39.2 > 38.9, Quantity II > Quantity I.

Using alligation: First mixture (75% vinegar) and second mixture (83.33% vinegar) combine to make 80% vinegar. Representing 83.33% as 5/6, the alligation diagram shows the ratio of first to second mixture is 2:3. For 20 litres total, first mixture = (2/5) × 20 = 8 litres. Verification: 8L at 75% + 12L at 83.33% = 6L + 10L = 16L vinegar in 20L total = 80%.

Let initial quantities in A and B be a and b. A has 80% copper = 0.8a, B has 62.5% copper = 0.625b. After transferring 25 kg from A to B, B's new copper = 0.625b + 0.8*25 = 0.625b + 20. B's zinc becomes 0.375b + 5. New copper in B is 100% more than zinc means: 0.625b + 20 = 2 * (0.375b + 5). Solving: 0.625b + 20 = 0.75b + 10 gives 0.125b = 10, so b = 80. New quantity in B = b + 25 = 105, which is 16% less than a. So 105 = 0.84a, giving a = 125. Initial copper in A = 0.8 * 125 = 100 kg. Option E is correct.

The ratio of zinc to copper is 13:7, meaning in every 20 parts of the brass alloy, 13 parts are zinc and 7 parts are copper. For 100 kg of brass, zinc = (13/20) × 100 = 65 kg. Option A (20 kg) would be copper, Option B (35 kg) is incorrect, and Option C (55 kg) has no calculation basis.

Initial mixture: 360 litres in 5:4 ratio means 200L milk + 160L water. Removing (x+12.5)% removes (x+12.5)/100 × 160 = 1.6(x+12.5) litres of water. After mixing with 22L water, total water = 160 - 1.6(x+12.5) + 22 = 150. Solving: 182 - 1.6x - 20 = 150 → 1.6x = 12 → x = 7.5. Option E matches.