Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
C
Correct answer
Explanation
Container: 1000 oil. Vessel: 1000 vinegar. Step 1: 300 oil to vessel. Vessel now has 1000 vinegar + 300 oil (total 1300). Step 2: 300 mixture back to container. Mixture is 300/1300 oil and 1000/1300 vinegar. Amount of vinegar returned = 300 * (1000/1300) = 3000/13. Amount of oil remaining in vessel = 300 - 300 * (300/1300) = 300 - 900/13 = 3000/13. Ratio is 1:1.
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36 liters
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38.4 liters
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40.2 liters
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42 liters
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43.2 liters
B
Correct answer
Explanation
The formula for remaining amount after n replacements is Initial * (1 - replaced/total)^n. Here, 60 * (1 - 12/60)^2 = 60 * (4/5)^2 = 60 * 16/25 = 60 * 0.64 = 38.4 liters.
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7 : 5
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5 : 7
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4 : 5
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5 : 4
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3 : 2
A
Correct answer
Explanation
Jar A contains milk and water in fractions 4/7 and 3/7, while jar B contains 2/5 and 3/5. Equating the excess milk from A with the water deficit from B gives x/14 = y/10, so x:y = 7:5.
B
Correct answer
Explanation
Initial oil = 0.6 * 200 = 120 liters. New oil = 120 + 10 = 130 liters. New total = 200 + 10 = 210 liters. New percentage = (130 / 210) * 100 = 61.9%. Rounded, this is 62%.
C
Correct answer
Explanation
Total alcohol = (0.12 * 10) + (0.18 * 20) = 1.2 + 3.6 = 4.8 liters. Total volume = 10 + 20 = 30 liters. Percentage = (4.8 / 30) * 100 = 16%.
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20 liters
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25 liters
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30 liters
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35 liters
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40 liters
C
Correct answer
Explanation
Using the alligation method: Wine A (8%) and Wine B (16%) mixed to get 12%. The ratio of volumes is (16-12) : (12-8) = 4 : 4 = 1 : 1. Since we have 30 liters of Wine A, we need 30 liters of Wine B.
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3 kilograms
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4 kilograms
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5 kilograms
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6 kilograms
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7 kilograms
B
Correct answer
Explanation
Initial sugar = 0.3 * 10 = 3 kg. Let x be the sugar added. New sugar = 3 + x. New total weight = 10 + x. We want (3 + x) / (10 + x) = 0.5. 3 + x = 5 + 0.5x. 0.5x = 2, so x = 4.
C
Correct answer
Explanation
Initial: 6 liters, 10% alcohol = 0.6L alcohol, 5.4L water. Replace 3 liters of solution: 3 liters removed contains 0.3L alcohol and 2.7L water. Remaining: 0.3L alcohol, 2.7L water. Add 3 liters pure alcohol: Total alcohol = 0.3 + 3 = 3.3L. Total volume = 6L. Percentage = (3.3 / 6) * 100 = 55%.
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− q + 2 r + 5 t 4
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− q + 2 r − 5 t 4
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q − 2 r − 5 t 6
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q − 2 r + 5 t 6
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− q − 2 r + 5 t 6
A
Correct answer
Explanation
0.02p + 0.05q + 0.08r + 0.11t = 0.06(p + q + r + t). Multiply by 100: 2p + 5q + 8r + 11t = 6p + 6q + 6r + 6t. Rearrange: 4p = -q + 2r + 5t. p = (-q + 2r + 5t) / 4.
C
Correct answer
Explanation
This is a mixture problem where the amount of syrup remaining after n replacements is 80 * (1 - 8/80)^n = 80 * (0.9)^n. We want 80 * (0.9)^n < 40, or (0.9)^n < 0.5. For n=6, 0.9^6 = 0.531. For n=7, 0.9^7 = 0.478. Thus 7 repetitions are needed.
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30 litres
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42 litres
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49 litres
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56 litres
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63 litres
B
Correct answer
Explanation
Initial: P=5x, Q=7x. (5x + 12) / (7x - 14) = 3/2. 10x + 24 = 21x - 42. 11x = 66, x = 6. Initial Q = 7 * 6 = 42 litres.
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1.47%
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4.17%
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4.71%
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14.67%
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41.67%
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28 liters
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50 liters
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42 liters
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70 liters
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75 liters
D
Correct answer
Explanation
Initial ratio 5:2 (5x, 2x). Adding 22L water: 5x / (2x + 22) = 7/5. 25x = 14x + 154, 11x = 154, x = 14. Alcohol = 5x = 5 * 14 = 70.
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6 litres
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8 litres
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12 litres
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Can't be determined
D
Correct answer
Explanation
Insufficient information is provided to determine the individual volumes of milk in containers B and C. The average milk and mixture volumes don't uniquely identify the distribution across containers.