Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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9 : 1
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10 : 1
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11 : 1
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None of these
B
Correct answer
Explanation
Let the cost price of spirit be 60 and water be 0. Selling price is 75 with 37.5% profit, so cost price of mixture is 75 / 1.375 = 54.54. Using the allegation rule, the ratio of spirit to water is (54.54 - 0) / (60 - 54.54) = 54.54 / 5.46, which is approximately 10:1.
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15 litres
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18 litres
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20 litres
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22 litres
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24 litres
C
Correct answer
Explanation
Initial alcohol is 24L and water is 16L. Let x be the water added: 24 / (16 + x) = 2 / 3. Solving gives 72 = 32 + 2x, so 2x = 40, x = 20.
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7 : 4
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5 : 3
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4 : 5
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3 : 5
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None of these
D
Correct answer
Explanation
Batch 1: V:J = 1:2 (V=1/3). Batch 2: V:J = 2:3 (V=2/5). Target: J:V = 5:3 (V=3/8). Use alligation on V: 1/3 and 2/5 to get 3/8. |2/5 - 3/8| = 1/40. |1/3 - 3/8| = 1/24. Ratio = (1/40) : (1/24) = 24:40 = 3:5.
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46.45 litres
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45.89 litres
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50.56 litres
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47.06 litres
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48.2 litres
D
Correct answer
Explanation
Using the formula for repeated dilution: Final amount = Initial * (1 - removed/total)^n. Here, 50 * (1 - 1/50)^3 = 50 * (49/50)^3 = 50 * 117649 / 125000 = 47.0596, which rounds to 47.06.
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43.25%
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46.25%
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47.15%
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48.25%
B
Correct answer
Explanation
Total spirit = (100 * 0.50) + (60 * 0.40) = 50 + 24 = 74 litres. Total mixture = 100 + 60 = 160 litres. Concentration = (74 / 160) * 100 = 46.25%.
A
Correct answer
Explanation
Initial acid = 10 * 0.3 = 3 ml. Let x be the amount of pure acid added. (3 + x) / (10 + x) = 0.5. 3 + x = 5 + 0.5x. 0.5x = 2. x = 4.
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6, 10
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4, 12
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2, 14
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8, 8
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5, 11
C
Correct answer
Explanation
Let x be kg from A and y be kg from B. Wheat: (2/5)x + (3/10)y = 5. Rice: (3/5)x + (7/10)y = 11. Multiply first by 2: 0.8x + 0.6y = 10. Multiply second by 2: 1.2x + 1.4y = 22. Solving the system: 4x + 3y = 50 and 6x + 7y = 110. Multiply first by 7 and second by 3: 28x + 21y = 350 and 18x + 21y = 330. Subtracting gives 10x = 20, so x = 2. Then 8 + 3y = 50, 3y = 42, y = 14.
A barrel contains a mixture of 80 litres petrol, 50 litres kerosene and 20 litres diesel. ______ litres of the given mixture is drawn and _______ litres of kerosene and ______ litres of diesel are added. The final mixture has 64 litres petrol, 65 litres kerosene and 66 litres diesel. Which of the following options satisfies the three blanks in the question? A. 30, 25, 50 B. 20, 50, 40 C. 22, 44, 50 D. 30, 40, 50
D
Correct answer
Explanation
Initial: P=80, K=50, D=20. Total=150. Let x be amount drawn. Remaining: P=80-80x/150, K=50-50x/150, D=20-20x/150. Add k_add and d_add. Final: 80 - 8x/15 = 64 => 8x/15 = 16 => x = 30. K: 50 - 50(30)/150 + k_add = 65 => 50 - 10 + k_add = 65 => k_add = 25. D: 20 - 20(30)/150 + d_add = 66 => 20 - 4 + d_add = 66 => d_add = 50. Matches option A.
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32.58%
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33.59%
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34.15%
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None of these
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45 litres
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49 litres
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51 litres
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64 litres
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70 litres
A
Correct answer
Explanation
Initial: A = 15, B = 30 (Total 45). To change ratio to 2:1, let x be the amount of A added. (15+x)/30 = 2/1 => 15+x = 60 => x = 45.
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220 : 149
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229 : 141
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239 : 161
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251 : 163
C
Correct answer
Explanation
Let drum 1 ratio be 18:7 (total 25) and drum 2 be x:y (total x+y). Mixture ratio 3:4. (3/7 * 18/25 + 4/7 * x/(x+y)) / (3/7 * 7/25 + 4/7 * y/(x+y)) = 13/7. Solving this leads to the ratio 239:161.
A
Correct answer
Explanation
Initial ratio 5:3 means 250g zinc and 150g copper. To make the ratio 5:4, the zinc remains 250g. For 5:4, 250/x = 5/4 => x = 200g of copper. Since we started with 150g, we must add 200 - 150 = 50g.
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20 litres
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18 litres
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22 litres
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16 litres
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25 litres
B
Correct answer
Explanation
Let the capacity be V. After two replacements of 4 litres, the remaining milk is V * (1 - 4/V)^2. Given this is 49/81 V, we have (1 - 4/V)^2 = 49/81. Taking the square root, 1 - 4/V = 7/9. Thus, 4/V = 2/9, so V = 18.
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32.1876 litres
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34.2 litres
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35.45 litres
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36.8125 litres
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38 litres