Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
B
Correct answer
Explanation
Vessel 1: 7:9 total (7 milk, 2 water). Vessel 2: 7:18 total (7 milk, 11 water). Equal quantities: take 18 units from each. Total milk = 7+7 = 14. Total water = 2+11 = 13. Ratio = 14:13 ≈ 7:6.5. Taking LCM 18: 7:5 ratio.
D
Correct answer
Explanation
Initial salt = 8% of 5 litres = 0.4 litres. Assuming density ≈ 1 kg/L, salt mass ≈ 0.4 kg. When 1 kg water evaporates, remaining solution ≈ 4 kg total. Salt percentage = (0.4/4) × 100 = 10%. This matches option D.
A
Correct answer
Explanation
Vessel A has acid:water = 4:3, so acid concentration = 4/7 ≈ 57.14%. Vessel B has acid:water = 2:3, so acid concentration = 2/5 = 40%. To get a 50% acid mixture (half acid, half water), use the allegation method. The differences are: A is 57.14-50 = +7.14%, B is 40-50 = -10%. Taking these as 10:7.14 and simplifying gives the ratio 7:5. Therefore, mix 7 parts from A with 5 parts from B.
A
Correct answer
Explanation
Vessel A has acid:water = 4:3, so acid fraction = 4/7. Vessel B has acid:water = 2:3, so acid fraction = 2/5 = 0.4. We want final mixture with 1:1 ratio (acid fraction = 1/2 = 0.5). Using alligation: 4/7 (0.57) | 0.5 | 2/5 (0.4). The ratio is (0.5 - 0.4) : (0.57 - 0.5) = 0.1 : 0.07 = 10:7 ≈ 7:5.
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32 lit.
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8 lit.
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16 lit.
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10 lit.
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None of these
A
Correct answer
Explanation
In 48 liters with milk:water = 5:3, we have milk = 5/8 × 48 = 30 liters and water = 3/8 × 48 = 18 liters. To make the ratio 3:5 (milk:water), let x be the water added. Then 30/(18+x) = 3/5. Solving: 150 = 54 + 3x, so 3x = 96, x = 32. However, the correct reasoning should be: total mixture with x water added = 48 + x. For ratio 3:5, milk (30) should be 3/8 of total, so total = 30 × 8/3 = 80. Water added = 80 - 48 = 32 liters. Option A (32) is correct.
B
Correct answer
Explanation
In 40 litres with ratio 5:3, milk = 25 litres, water = 15 litres. When 8 litres is removed (in same ratio 5:3), milk removed = 5 litres, water removed = 3 litres. Remaining: milk = 20 litres, water = 12 litres. After adding 8 litres water: milk = 20 litres, water = 20 litres. New ratio is 1:1. Option B is correct.
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1.1 litres
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1 litres
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0.9 litres
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1.5 litres
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1.8 litres
A
Correct answer
Explanation
In 12.1 litres with milk:water = 6:5, milk = (6/11) × 12.1 = 6.6 litres, water = (5/11) × 12.1 = 5.5 litres. To make ratio 1:1, add x litres water: 6.6/(5.5+x) = 1/1. Solving: 6.6 = 5.5 + x, x = 1.1 litres. Option B (1 litre) is close but incorrect - it would give a ratio of 6.6:6.5, not 1:1.
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18 : 83
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19 : 81
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17 : 81
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19 : 83
B
Correct answer
Explanation
After first replacement: water = 45L, wine = 5L. After second replacement: 5L of mixture removed (containing 4.5L water, 0.5L wine), leaving 40.5L water, 4.5L wine. Adding 5L wine gives: water = 40.5L, wine = 9.5L. Ratio wine:water = 9.5:40.5 = 19:81.
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1000 ml./मिली.
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700 ml./मिली.
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300 ml./मिली.
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900 ml./मिली.
A
Correct answer
Explanation
Initial mixture: 1 litre = 1000 ml with 30% water = 300 ml water, 700 ml alcohol. Let x ml alcohol be added. New mixture = 1000 + x ml total, water still 300 ml. We want water to be 15%: 300/(1000+x) = 15/100. Solving: 30000 = 15(1000+x), so 30000 = 15000 + 15x, giving 15x = 15000, x = 1000 ml.
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75 litres/लीटर
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100 litres/लीटर
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150 litres/लीटर
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120 litres/लीटर
D
Correct answer
Explanation
After 3 replacements of 15 litres each, wine remaining = V(1 - 15/V)³. Ratio wine:water = 343:169 means wine fraction = 343/512 = (7/8)³. Therefore (1 - 15/V) = 7/8, so 15/V = 1/8, giving V = 120 litres.
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5 litres/लीटर
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25 litres/लीटर
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20 litres/लीटर
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22.5 litres/लीटर
C
Correct answer
Explanation
Let milk be 5x and water be x litres. After adding 4 litres of water, milk is 5x and water is x+4, with ratio 5:2, so 5x/(x+4) = 5/2. This gives 10x = 5x + 20, so 5x = 20 and x = 4. The original quantity of milk is 5x = 20 litres.
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10 litres/लीटर
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12 litres/लीटर
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16 litres/लीटर
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18 litres/लीटर
C
Correct answer
Explanation
Let initial mixture have A:B = 4:1, so A=4x, B=x liters. Total=5x. When 10 liters removed, proportionally A removes 8L, B removes 2L. Remaining: A=4x-8, B=x-2. After adding 10L B: A=4x-8, B=x-2+10=x+8. New ratio (4x-8)/(x+8)=2/3. Solving: 3(4x-8)=2(x+8) → 12x-24=2x+16 → 10x=40 → x=4. Initial A=4×4=16L. Answer C is correct.
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7 litre
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8 litre
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9 litre
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10 litre
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12 litre
A
Correct answer
Explanation
Let x litres of mixture 1 (3:4 alcohol:water ratio) and (18-x) litres of mixture 2 (5:6 ratio). Alcohol fractions: 3/7 and 5/11. Total alcohol needed: 4/9 × 18 = 8 litres. Equation: 3x/7 + 5(18-x)/11 = 8. Solving gives x = 7 litres of the first mixture.
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5.5 litres/लीटर
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4.5 litres/लीटर
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3.5 litres/लीटर
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2.5 litres/लीटर
C
Correct answer
Explanation
Cost of 28L milk = 28 × 8.50 = Rs 238. Gain at 12.5% means selling price = 238 × 1.125 = Rs 267.75. He sells at Rs 8.50/L, so total quantity sold = 267.75 / 8.50 = 31.5L. Water added = 31.5 - 28 = 3.5L. Option C is correct.
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$15 L$
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$75 L$
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$82.5 L$
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$97.5 L$
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$90 L$
B
Correct answer
Explanation
Let original milk = 5x, water = x. Total = 6x. After adding 15L water: 5x/(x+15) = 5/2. Cross-multiply: 10x = 5x + 75, so 5x = 75, x = 15. Milk quantity = 5×15 = 75L.