Mixtures and Alligation

Let x be gallons of water added. The salt amount stays constant: 0.15 gallons (15% of 1 gallon). Final concentration: 0.15/(1+x) = 0.10. Solving: 0.15 = 0.10 + 0.10x, so 0.05 = 0.10x, giving x = 0.5. The dilution formula is: C1V1 = C2V2 where C2 is the target concentration and V2 is the final total volume.

Use alligation: Brand A at Rs.9 and Brand B at Rs.4 to get mixture at Rs.7. The alligation chart gives ratios 3:2 (A:B). For 40 kg total, Brand A = 3/5 × 40 = 24 kg. Verification: (24×9 + 16×4) ÷ 40 = (216 + 64) ÷ 40 = 280 ÷ 40 = Rs.7/kg.

Initial sugar = 4% of 6L = 0.24L. After 1L evaporates, the total volume is 5L. The sugar amount remains 0.24L. The new percentage is (0.24 / 5) * 100 = 4.8%. 4.8% is equivalent to the fraction 24/5%.

Starting with 10 gallons at 10% water: 1 gallon water, 9 gallons paint. After adding x gallons water: total water = 1 + x, total volume = 10 + x. Need (1+x)/(10+x) = 0.20. Solve: 1+x = 2 + 0.2x → 0.8x = 1 → x = 1.25. Adding 1.25 gallons creates the 80/20 paint-to-water ratio.

Using allegation method: Gold is 19 parts, copper is 9 parts, alloy is 15 parts. Gold contributes 4 parts above 15 (19-15), copper contributes 6 parts below 15 (15-9). The ratio is the inverse: gold:copper = 6:4 = 3:2. Alternatively: 19g + 9c = 15(g+c), so 19g + 9c = 15g + 15c, giving 4g = 6c, so g:c = 6:4 = 3:2.

In 729 ml with milk:water = 7:2, milk = (7/9)*729 = 567 ml, water = 162 ml. Let x ml water be added to make ratio 1:1. Then 567/(162+x) = 1/1, so 567 = 162+x, giving x = 405 ml.

Let x be the kg of expensive coffee (Rs.60) and (20-x) be the cheap coffee (Rs.30). Current cost: 60x + 30(20-x). If only expensive: 60 * 20 = 1200. Difference: 1200 - (30x + 600) = 450. Solving gives 600 - 30x = 450, so 30x = 150, x = 5. Ratio of cheap to expensive is 15:5, which is 3:1.

Let the water:liquid ratio be k:1. For every k units of water (free), we have 1 unit of liquid costing Rs. 18. Total mixture = k + 1 units, so cost per unit = 18/(k + 1). Selling at Rs. 19.2 with 20% profit means cost price should be 19.2/1.2 = Rs. 16. So 18/(k + 1) = 16, giving k + 1 = 18/16 = 9/8, so k = 1/8. Ratio water:liquid = 1:8. Option D is correct.

In 40L mixture with 3:1 ratio: milk = 30L, water = 10L. Let water added = x. New ratio = 30/(10+x) = 2/1. Cross-multiply: 30 = 2(10+x) → 30 = 20+2x → 2x = 10 → x = 5L.

Initially, milk = 40L and water = 20L (2:1). To make the ratio 1:2, the milk (40L) must represent 1 part, meaning the total water must become 80L (2 parts). Since there is already 20L of water, the amount to be added is 80 - 20 = 60 litres.

Using allegation method: difference from 47% gives (50-47):(47-40) = 3:7. So Solution I (40%) and Solution II (50%) mix in ratio 3:7. Alternatively: (40x + 50y)/(x+y) = 47 gives x/y = 3/7.

C.P. of 80 kg of wheat = Rs. 540. C.P. of 120 kg of wheat = Rs. 960. Since C.P. of 200 kg of mixture = Rs. 540 + 960 = Rs. 1500, C.P. of 1 kg of mixture = Rs. 7.50. Gain% = 20%. On calculating, we get S.P. = Rs. 9.

In the mixture of 5 grams, A makes up 7% of 3 grams and 14.5% of 2 grams, or  (21/100 + 29/100) grams or 50/100 grams, or 1/2 gram. So, its percentage in the mixture of 5 grams is (1/2)/5 × 100 =  1/10 × 100 = 10%

Initial mixture has 4L water (10% of 40L) and 36L milk. After adding x liters of water, total becomes (40+x) liters with (4+x) liters water. For water to be 20%: (4+x)/(40+x) = 0.20. Solving: 4+x = 8+0.2x, so 0.8x = 4, giving x = 5 litres.

Let a kg from A and b kg from B. From A: coffee = (5/8)a, chicory = (3/8)a. From B: coffee = (7/10)b, chicory = (3/10)b. Total coffee: (5/8)a + (7/10)b = 6. Total chicory: (3/8)a + (3/10)b = 3. Multiply coffee equation by 40: 25a + 28b = 240. Multiply chicory equation by 40: 15a + 12b = 120. Solving: From chicory eq, 5a + 4b = 40, so a = (40-4b)/5. Substituting: 25(40-4b)/5 + 28b = 240 gives 200 - 20b + 28b = 240, so 8b = 40, b = 5. Then a = 4.