Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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350 liters
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500 liters
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420 liters
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240 liters
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560 liters
C
Correct answer
Explanation
Let initial quantity be 7x. Petrol = 4x, Kerosene = 3x. Remove 105L: Petrol = 4x - 60, Kerosene = 3x - 45. Add 105L Kerosene: Petrol = 4x - 60, Kerosene = 3x - 45 + 105 = 3x + 60. Ratio reversed: (4x - 60) / (3x + 60) = 3 / 4. 16x - 240 = 9x + 180. 7x = 420.
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1.5
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2.0
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1.75
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2.5
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None of these
B
Correct answer
Explanation
Initial solution = 10L, salt = 150g. 25% of solution evaporates, so 7.5L remains. The amount of salt remains 150g. Percentage of salt = (150g / 7500g) * 100 = 2%.
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10 litres
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12 litres
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15 litres
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16 litres
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17 litres
D
Correct answer
Explanation
Let the initial amount of A be 4x and B be x. Total = 5x. When 10L is removed, 8L of A and 2L of B are removed. Remaining: A = 4x - 8, B = x - 2 + 10 = x + 8. Ratio (4x-8)/(x+8) = 2/3. 12x - 24 = 2x + 16, so 10x = 40, x = 4. Initial A = 4*4 = 16 litres.
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50% (50 Percent)
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40% (40 Percent)
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20% (20 Percent)
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10% (10 Percent)
C
Correct answer
Explanation
Concentration is calculated as (mass of solute / mass of solution) * 100. Here, (110 / 550) * 100 = (1/5) * 100 = 20%.
D
Correct answer
Explanation
Mixture A has milk/water ratio 3:Y. 75L of A contains 75 * 3/(3+Y) milk and 75 * Y/(3+Y) water. Added to 75L of pure water, total milk = 225/(3+Y), total water = 75 + 75Y/(3+Y). Ratio 3:7 implies 225/(3+Y) / (75 + 75Y/(3+Y)) = 3/7. Solving gives Y=2.
B
Correct answer
Explanation
Initial alcohol = 80%. Removing 20% of the solution leaves 80% of the original alcohol. This is done twice. Final alcohol = 80% * 0.8 * 0.8 = 80% * 0.64 = 51.2%.
B
Correct answer
Explanation
Let milk be 4x and water be x. Adding 6 liters of water makes the ratio 4x / (x+6) = 3/2. 8x = 3x + 18, so 5x = 18, x = 3.6. Total initial solution = 5x = 5 * 3.6 = 18 liters.
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700
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810
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729
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None of these
C
Correct answer
Explanation
Remaining milk = Initial * (1 - removed/total)^n = 1000 * (1 - 100/1000)^3 = 1000 * (0.9)^3 = 1000 * 0.729 = 729.
B
Correct answer
Explanation
The concentration of milk after n replacements is C = Initial * (1 - r/100)^n. Here, n=3 and r=10. C = 100 * (0.9)^3 = 100 * 0.729 = 72.9%.
D
Correct answer
Explanation
Starting with 100 units of 80% milk (80 units). After 25% removal, 75% remains: 80 * 0.75 = 60. After 20% removal, 80% remains: 60 * 0.8 = 48. After 10% removal, 90% remains: 48 * 0.9 = 43.2.
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1 : 2
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2 : 1
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1 : 4
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None of these
D
Correct answer
Explanation
This is a classic mixture problem. Since the volumes are equal and the process is repeated 3 times, the amount of juice remaining follows a specific ratio. Given the options and the nature of the problem, 'None of these' is the correct choice as the ratio is not a simple integer ratio.
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7 : 2
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5 : 3
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8 : 5
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None of these
C
Correct answer
Explanation
Initial: Lime = 21, Soda = 9. Let x be the amount of soda added. New ratio: 21 / (9 + x) = 3 / 7. Cross-multiplying: 147 = 27 + 3x, so 3x = 120, x = 40.
C
Correct answer
Explanation
Mixture 1: 3L has 1.8L milk, 1.2L water. Mixture 2: ratio 4:5, let x be volume, milk = 4x/9, water = 5x/9. Total milk = 1.8 + 4x/9, total water = 1.2 + 5x/9. Equating: 1.8 + 4x/9 = 1.2 + 5x/9. 0.6 = x/9, x = 5.4 L.