Question 1

A shopkeeper mixes two types of tea costing ₹80 per kg and ₹120 per kg in the ratio 3:2. At what price per kg should he sell the mixture to gain 25% profit?

Accepted Answer

Step 1: Find the cost price of the mixture using alligation. In ratio 3:2, the mixture contains 3 parts of ₹80 tea and 2 parts of ₹120 tea. Step 2: Cost price of mixture = (3 × 80 + 2 × 120) / (3 + 2) = (240 + 240) / 5 = 480 / 5 = ₹96 per kg. Step 3: For 25% profit, selling price = 96 × 1.25 = ₹120 per kg. Therefore, the answer is ₹120 per kg.

Question 2

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. At what price per kg should he sell the mixture to gain 25% profit?

Accepted Answer

Step 1: Let quantities be 3x kg and 2x kg respectively. Step 2: Cost of first type = 3x × 40 = ₹120x. Cost of second type = 2x × 60 = ₹120x. Step 3: Total cost = 120x + 120x = ₹240x. Total quantity = 3x + 2x = 5x kg. Step 4: Cost price per kg = 240x / 5x = ₹48 per kg. Step 5: For 25% profit, selling price = 48 × (1 + 25/100) = 48 × 1.25 = ₹60 per kg.

Question 3

A container has milk and water in the ratio 3:1. How much water must be added to 40 litres of this mixture to make the ratio 2:1?

Accepted Answer

Step 1: In 40 litres, milk = 40 × 3/4 = 30 litres, water = 40 × 1/4 = 10 litres. Step 2: Let x litres of water be added. New ratio: 30/(10 + x) = 2/1. Step 3: 30 = 2(10 + x) → 30 = 20 + 2x → 2x = 10 → x = 5. Therefore, 5 litres of water must be added.

Question 4

A goldsmith mixes gold and silver in the ratio 2:3 to make an alloy. If he wants to make 500 grams of alloy, how much gold is needed?

Accepted Answer

Step 1: Gold:Silver = 2:3, so total parts = 2 + 3 = 5. Step 2: Gold fraction = 2/5 of total alloy. Step 3: Gold needed = (2/5) × 500 = 200 grams.

Question 5

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. What is the cost price of the mixture per kg?

Accepted Answer

Step 1: Rice type 1 costs ₹40/kg in ratio 3 parts. Rice type 2 costs ₹60/kg in ratio 2 parts. Step 2: Total parts = 3 + 2 = 5. Step 3: Cost of 3 parts of type 1 = 3 × 40 = ₹120. Step 4: Cost of 2 parts of type 2 = 2 × 60 = ₹120. Step 5: Total cost = 120 + 120 = ₹240 for 5 parts. Step 6: Cost per kg = 240 ÷ 5 = ₹48/kg.

Question 6

Two alloys contain copper in the percentages 30% and 50% respectively. In what ratio should they be mixed to get an alloy with 40% copper?

Accepted Answer

Step 1: Use alligation method. Alloy 1 has 30% copper, Alloy 2 has 50% copper. Target is 40% copper. Step 2: Difference between Alloy 2 and target = 50 - 40 = 10. Step 3: Difference between target and Alloy 1 = 40 - 30 = 10. Step 4: Ratio = (Alloy 1):(Alloy 2) = 10:10 = 1:1.

Question 7

Two containers have alcohol and water in the ratios 3:1 and 5:3 respectively. If equal quantities from both containers are mixed, what is the ratio of alcohol to water in the final mixture?

Accepted Answer

Step 1: Container 1 has alcohol:water = 3:1, so in 4 units: alcohol = 3, water = 1. Step 2: Container 2 has alcohol:water = 5:3, so in 8 units: alcohol = 5, water = 3. Step 3: Take equal quantities (LCM of 4 and 8 = 8 units from each). Step 4: From Container 1 (8 units): alcohol = 6, water = 2. Step 5: From Container 2 (8 units): alcohol = 5, water = 3. Step 6: Final mixture: alcohol = 6 + 5 = 11, water = 2 + 3 = 5. Step 7: Ratio = 11:5.

Question 8

Two vessels contain alcohol and water in the ratios 3:1 and 5:3 respectively. If equal quantities from each vessel are mixed together, what is the ratio of alcohol to water in the final mixture?

Accepted Answer

Step 1: Let equal quantity from each vessel = L litres. Step 2: Vessel 1 (ratio 3:1, total 4 parts): alcohol = (3/4)L, water = (1/4)L. Step 3: Vessel 2 (ratio 5:3, total 8 parts): alcohol = (5/8)L, water = (3/8)L. Step 4: Total alcohol in final mixture = (3/4)L + (5/8)L = (6/8)L + (5/8)L = (11/8)L. Step 5: Total water in final mixture = (1/4)L + (3/8)L = (2/8)L + (3/8)L = (5/8)L. Step 6: Ratio of alcohol to water = (11/8)L : (5/8)L = 11:5.

Question 9

In what ratio must sugar costing ₹15 per kg be mixed with sugar costing ₹24 per kg so that the mixture costs ₹18 per kg?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper sugar = ₹15, cost of dearer sugar = ₹24, cost of mixture = ₹18. Step 2: Difference between mixture price and cheaper price = 18 - 15 = ₹3. Step 3: Difference between dearer price and mixture price = 24 - 18 = ₹6. Step 4: By alligation rule, ratio of cheaper to dearer = (24 - 18) : (18 - 15) = 6:3 = 2:1.

Question 10

Two containers have alcohol solutions of 20% and 50% respectively. In what ratio should they be mixed to get a 35% solution?

Accepted Answer

Step 1: Use alligation method. Mark 20% and 50% on a number line with target 35% in between. Step 2: Distance from 20% to 35% = 15 parts. Distance from 35% to 50% = 15 parts. Step 3: Ratio of 20% solution to 50% solution = (50 − 35):(35 − 20) = 15:15 = 1:1.

Question 11

A chemist has two solutions with alcohol concentrations of 20% and 50%. In what ratio should these be mixed to get a solution with 35% alcohol concentration?

Accepted Answer

Step 1: Use alligation method. Mark the concentrations: 20%, 35%, 50%. Step 2: Difference between 35% and 20% = 15. Difference between 50% and 35% = 15. Step 3: Ratio = (50 - 35):(35 - 20) = 15:15 = 1:1.

Question 12

Two containers have milk and water in ratios 3:1 and 5:3 respectively. If equal volumes are mixed from both containers, what is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Container A has milk:water = 3:1, so milk fraction = 3/4, water fraction = 1/4. Step 2: Container B has milk:water = 5:3, so milk fraction = 5/8, water fraction = 3/8. Step 3: Let V litres be taken from each. Step 4: Total milk = V(3/4) + V(5/8) = V(6/8 + 5/8) = 11V/8. Step 5: Total water = V(1/4) + V(3/8) = V(2/8 + 3/8) = 5V/8. Step 6: Ratio = (11V/8):(5V/8) = 11:5.

Question 13

A chemist has two solutions of acid: Solution A contains 30% acid and Solution B contains 70% acid. In what ratio must these two solutions be mixed to obtain 50 litres of a 54% acid solution?

Accepted Answer

Step 1: Let the quantity of Solution A (30% acid) be x litres and Solution B (70% acid) be y litres. Step 2: Set up equations: x + y = 50 (total volume) and 0.30x + 0.70y = 0.54 × 50 = 27 (acid content). Step 3: From equation 1, x = 50 - y. Substitute into equation 2: 0.30(50 - y) + 0.70y = 27 → 15 - 0.30y + 0.70y = 27 → 0.40y = 12 → y = 30 litres. Step 4: Therefore x = 50 - 30 = 20 litres. Step 5: Ratio of Solution A to Solution B = 20:30 = 2:3.

Question 14

A shopkeeper has two types of tea: Type A costing ₹80 per kg and Type B costing ₹120 per kg. He wants to create a mixture that costs ₹100 per kg. If he uses 15 kg of Type A tea, how many kg of Type B tea should he mix to achieve the desired cost per kg?

Accepted Answer

Step 1: Use the alligation method. The cost difference between Type A and the mixture is 100 - 80 = 20. The cost difference between Type B and the mixture is 120 - 100 = 20. Step 2: The ratio of Type A to Type B is 20:20 = 1:1. Step 3: If Type A is 15 kg and the ratio is 1:1, then Type B must also be 15 kg. Verification: (15 × 80 + 15 × 120) ÷ 30 = (1200 + 1800) ÷ 30 = 3000 ÷ 30 = ₹100 per kg. Therefore, the answer is 15 kg.

Question 15

A vessel contains 80 litres of a mixture of alcohol and water in the ratio 3:5. How much pure alcohol should be added to make the ratio 1:1?

Accepted Answer

Step 1: Original mixture = 80 litres in ratio 3:5. Alcohol = 80 × (3/8) = 30 litres, water = 80 × (5/8) = 50 litres. Step 2: Let x litres of pure alcohol be added. Step 3: New alcohol = 30 + x, water = 50 (unchanged). Step 4: For ratio 1:1: (30 + x)/50 = 1/1. Step 5: 30 + x = 50 → x = 20 litres.

Question 16

A vessel contains a mixture of milk and water in the ratio 7:3. 20 litres of the mixture is taken out and replaced with pure milk. If the resulting ratio of milk to water becomes 9:1, what is the original volume of the mixture in the vessel?

Accepted Answer

Step 1: Let the original volume = V litres. Water in original mixture = 3V/10. Step 2: When 20 litres of mixture is removed, water removed = 20 × (3/10) = 6 litres. Step 3: Water remaining after removal = 3V/10 − 6. Since only milk is added back, water content stays the same. Step 4: In the new mixture, water/total = 1/10. So (3V/10 − 6)/V = 1/10. Step 5: 3V/10 − 6 = V/10 → 2V/10 = 6 → V/5 = 6 → V = 30 litres. Therefore, the original volume is 30 litres.

Question 17

A chemist has three containers of alcohol solutions: Container A has 60 litres of 40% alcohol, Container B has 40 litres of 70% alcohol, and Container C has 50 litres of 20% alcohol. The chemist mixes all three containers together, then removes 30 litres of the resulting mixture and replaces it with pure water. What is the percentage concentration of alcohol in the final mixture?

Accepted Answer

Step 1: Calculate total pure alcohol from all three containers. Container A: 60 × 0.40 = 24 litres. Container B: 40 × 0.70 = 28 litres. Container C: 50 × 0.20 = 10 litres. Total pure alcohol = 24 + 28 + 10 = 62 litres. Step 2: Calculate total volume = 60 + 40 + 50 = 150 litres. Initial concentration = 62/150. Step 3: When 30 litres of mixture is removed, alcohol removed = 30 × (62/150) = 12.4 litres. Remaining alcohol = 62 − 12.4 = 49.6 litres. Step 4: After adding 30 litres of pure water, total volume = 150 litres (unchanged). Final concentration = (49.6/150) × 100 = 33.067% ≈ 33.07%. Rounding to nearest practical value: approximately 33.07% or 33⅑%.

Question 18

A merchant mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. At what price per kg should the mixture be sold to gain 25% profit?

Accepted Answer

Step 1: Cost price of mixture per kg using alligation: CP = (3 × 40 + 2 × 60)/(3 + 2) = (120 + 120)/5 = 240/5 = ₹48 per kg. Step 2: For 25% profit, selling price = CP × (1 + 25/100) = 48 × 1.25 = ₹60 per kg.

Question 19

A container has 60 litres of a solution with alcohol and water in ratio 7:5. How much pure alcohol must be added to make the ratio 3:1?

Accepted Answer

Step 1: Original solution has ratio 7:5, total parts = 12. Alcohol = (7/12)×60 = 35 litres, Water = (5/12)×60 = 25 litres. Step 2: Let x litres of pure alcohol be added. New alcohol = 35 + x, water remains 25 litres. Step 3: New ratio is 3:1, so (35+x)/25 = 3/1. Step 4: 35 + x = 75 → x = 40 litres. Step 5: Verification: (35+40):25 = 75:25 = 3:1 ✓

Question 20

A container has 40 litres of milk. 8 litres of milk is taken out and replaced with water. What is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Milk removed = 8 litres from 40 litres, so milk remaining = 40 - 8 = 32 litres. Step 2: Water added = 8 litres. Step 3: Ratio of milk to water = 32 : 8 = 4 : 1. Therefore, the answer is 4 : 1.

Question 21

In what ratio should sugar costing ₹20 per kg be mixed with sugar costing ₹30 per kg to get a mixture costing ₹24 per kg?

Accepted Answer

Step 1: Use alligation method. Cost of cheaper sugar = ₹20, cost of dearer sugar = ₹30, cost of mixture = ₹24. Step 2: Difference between mixture cost and cheaper cost = 24 - 20 = 4. Step 3: Difference between dearer cost and mixture cost = 30 - 24 = 6. Step 4: Ratio of cheaper to dearer = 6:4 = 3:2. Step 5: Verification: (20×3 + 30×2)/(3+2) = (60+60)/5 = 120/5 = 24 ✓

Question 22

Two containers have milk and water in ratios 3:1 and 5:3 respectively. If equal quantities from each container are mixed, what is the ratio of milk to water in the final mixture?

Accepted Answer

Step 1: Container 1 has milk:water = 3:1, so in 4 parts, milk = 3 parts, water = 1 part. Step 2: Container 2 has milk:water = 5:3, so in 8 parts, milk = 5 parts, water = 3 parts. Step 3: Let equal quantity = L litres from each. From Container 1: milk = 3L/4, water = L/4. From Container 2: milk = 5L/8, water = 3L/8. Step 4: Total milk = 3L/4 + 5L/8 = 6L/8 + 5L/8 = 11L/8. Total water = L/4 + 3L/8 = 2L/8 + 3L/8 = 5L/8. Step 5: Ratio = (11L/8):(5L/8) = 11:5.

Question 23

A container has 120 litres of a solution with alcohol and water in the ratio 7:5. How much pure alcohol should be added to make the ratio 3:1?

Accepted Answer

Step 1: Original mixture: alcohol = 120 × (7/12) = 70 litres, water = 120 × (5/12) = 50 litres. Step 2: Let x litres of pure alcohol be added. New alcohol = 70 + x, water = 50 (unchanged). Step 3: New ratio is 3:1, so (70 + x)/50 = 3/1. Step 4: 70 + x = 150 → x = 80 litres.

Question 24

A goldsmith mixes gold of purity 80% with gold of purity 60% in the ratio 3:2. What is the purity of the resulting mixture?

Accepted Answer

Step 1: Gold purity 80% means 80 parts pure gold per 100 parts. Gold purity 60% means 60 parts pure gold per 100 parts. Step 2: Ratio of mixing is 3:2. Let quantities be 3 kg and 2 kg respectively. Step 3: Pure gold from first = 3 × (80/100) = 2.4 kg. Pure gold from second = 2 × (60/100) = 1.2 kg. Step 4: Total pure gold = 2.4 + 1.2 = 3.6 kg. Total mixture = 3 + 2 = 5 kg. Step 5: Purity of mixture = (3.6/5) × 100 = 72%.

Question 25

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg. If the mixture costs ₹50 per kg and he uses 15 kg of the cheaper rice, how much of the costlier rice should he use?

Accepted Answer

Step 1: Use alligation method. Cheaper rice = ₹40/kg, costlier rice = ₹60/kg, mixture = ₹50/kg. Step 2: Alligation distances: (60 − 50) = 10 and (50 − 40) = 10. Step 3: Ratio of cheaper to costlier = 10:10 = 1:1. Step 4: If cheaper rice = 15 kg, then costlier rice = 15 kg (since ratio is 1:1).

Question 26

A mixture contains milk and water in the ratio 5:3. If 16 litres of water is added to the mixture, the ratio becomes 5:7. What is the quantity of milk in the original mixture?

Accepted Answer

Step 1: Let the original mixture have milk = 5x and water = 3x litres. Step 2: After adding 16 litres of water, milk remains 5x and water becomes 3x + 16. Step 3: New ratio is 5:7, so 5x/(3x + 16) = 5/7. Step 4: Cross-multiply: 7(5x) = 5(3x + 16) → 35x = 15x + 80 → 20x = 80 → x = 4. Step 5: Quantity of milk = 5x = 5(4) = 20 litres.

Question 27

A container has 60 litres of a solution with alcohol and water in the ratio 7:5. How much of the solution should be removed and replaced with pure alcohol so that the alcohol content becomes 80% of the total?

Accepted Answer

Step 1: Original solution has alcohol = (7/12) × 60 = 35L and water = (5/12) × 60 = 25L. Step 2: Let x litres of solution be removed and replaced with pure alcohol. Step 3: After removal, alcohol = 35 - (7/12)x and water = 25 - (5/12)x. Step 4: After adding x litres of pure alcohol: alcohol = 35 - (7/12)x + x = 35 + (5/12)x. Step 5: Total volume remains 60L. For 80% alcohol: [35 + (5/12)x]/60 = 0.8 → 35 + (5/12)x = 48 → (5/12)x = 13 → x = 13 × (12/5) = 31.2L.

Question 28

A vessel contains 60 litres of a mixture of alcohol and water in the ratio 7:5. How much of the mixture should be removed and replaced with pure alcohol so that the ratio becomes 3:1?

Accepted Answer

Step 1: Original mixture: alcohol = 60 × (7/12) = 35L, water = 60 × (5/12) = 25L. Step 2: Let x litres be removed and replaced with pure alcohol. Step 3: After removal: alcohol = 35 − 7x/12, water = 25 − 5x/12. Step 4: After adding x litres of pure alcohol: alcohol = 35 − 7x/12 + x = 35 + 5x/12, water = 25 − 5x/12. Step 5: New ratio is 3:1, so (35 + 5x/12)/(25 − 5x/12) = 3/1. Step 6: 35 + 5x/12 = 3(25 − 5x/12) → 35 + 5x/12 = 75 − 15x/12 → 20x/12 = 40 → x = 24.

Question 29

Two containers A and B contain alcohol solutions of 40% and 60% respectively. If 20 litres from A and 30 litres from B are mixed together, what is the percentage of alcohol in the resulting mixture?

Accepted Answer

Step 1: Alcohol from container A = 20 × 0.40 = 8 litres. Step 2: Alcohol from container B = 30 × 0.60 = 18 litres. Step 3: Total alcohol = 8 + 18 = 26 litres. Step 4: Total mixture = 20 + 30 = 50 litres. Step 5: Percentage = (26/50) × 100 = 52%.

Question 30

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. He then sells the mixture at ₹55 per kg. What is his profit or loss percentage?

Accepted Answer

Step 1: Let the mixture contain 3 kg of ₹40 rice and 2 kg of ₹60 rice. Step 2: Cost price of mixture = (3 × 40) + (2 × 60) = 120 + 120 = ₹240 for 5 kg. Step 3: Cost price per kg = 240/5 = ₹48 per kg. Step 4: Selling price per kg = ₹55. Step 5: Profit per kg = 55 - 48 = ₹7. Step 6: Profit% = (7/48) × 100 = 14.583...% ≈ 14.58% or 175/12 % ≈ 14.6%.

Question 31

Two alloys A and B contain copper in the ratio 2:3 and 3:4 respectively. Equal weights of both alloys are melted together. What is the ratio of copper to non-copper in the final alloy?

Accepted Answer

Step 1: In alloy A, copper:non-copper = 2:3, so copper fraction = 2/5. Step 2: In alloy B, copper:non-copper = 3:4, so copper fraction = 3/7. Step 3: Let equal weight = w each. Step 4: Copper from A = 2w/5, copper from B = 3w/7. Step 5: Total copper = 2w/5 + 3w/7 = (14w + 15w)/35 = 29w/35. Step 6: Total weight = 2w, so non-copper = 2w - 29w/35 = (70w - 29w)/35 = 41w/35. Step 7: Ratio of copper to non-copper = (29w/35):(41w/35) = 29:41.

Question 32

A vessel contains 80 litres of a mixture of alcohol and water in the ratio 3:5. How much pure alcohol must be added to make the ratio 1:1?

Accepted Answer

Step 1: Original mixture has alcohol:water = 3:5, total parts = 8. Step 2: Alcohol = (3/8) × 80 = 30 litres, water = (5/8) × 80 = 50 litres. Step 3: Let x litres of pure alcohol be added. Step 4: New alcohol = 30 + x, water remains 50. Step 5: For ratio 1:1: 30 + x = 50 → x = 20 litres.

Question 33

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. At what price per kg should he sell the mixture to gain 25% profit?

Accepted Answer

Step 1: Let quantities be 3x and 2x kg respectively. Step 2: Cost of first type = 3x × 40 = 120x. Cost of second type = 2x × 60 = 120x. Step 3: Total cost = 120x + 120x = 240x. Step 4: Total quantity = 3x + 2x = 5x. Step 5: Cost price per kg = 240x / 5x = 48 per kg. Step 6: Selling price with 25% profit = 48 × 1.25 = 60 per kg.

Question 34

A merchant has two types of rice costing ₹40 per kg and ₹60 per kg. He mixes them and sells the mixture at ₹55 per kg, making a profit of 10%. If he uses 15 kg of the cheaper rice, how much of the expensive rice should he mix?

Accepted Answer

Step 1: Selling price = ₹55 per kg with 10% profit. Cost price = 55/1.10 = ₹50 per kg (average). Step 2: Using alligation: cheaper rice at ₹40, expensive rice at ₹60, average cost ₹50. Step 3: Difference from average: expensive rice = 60 - 50 = 10, cheaper rice = 50 - 40 = 10. Step 4: Ratio of cheaper to expensive = 10:10 = 1:1. Step 5: If cheaper rice = 15 kg, then expensive rice = 15 kg.

Question 35

Two alloys A and B contain copper in the ratio 3:1 and 1:3 respectively. If 20 kg of alloy A is mixed with 30 kg of alloy B, what is the percentage of copper in the resulting mixture?

Accepted Answer

Step 1: In alloy A (ratio 3:1), copper fraction = 3/4. In alloy B (ratio 1:3), copper fraction = 1/4. Step 2: Copper from A = 20 × (3/4) = 15 kg. Step 3: Copper from B = 30 × (1/4) = 7.5 kg. Step 4: Total copper = 15 + 7.5 = 22.5 kg. Step 5: Total mixture = 20 + 30 = 50 kg. Step 6: Percentage = (22.5/50) × 100 = 45%.

Question 36

A merchant blends two types of tea: Premium (₹500/kg) and Standard (₹300/kg). He sells the blend at ₹420/kg and makes a 20% profit. If he uses 15 kg of Premium tea, how many kg of Standard tea must he use?

Accepted Answer

Step 1: Selling price = ₹420/kg with 20% profit means cost price = 420/1.20 = ₹350/kg. Step 2: Let Standard tea used = x kg. Total weight = 15 + x kg. Step 3: Total cost = 15(500) + x(300) = 7500 + 300x. Step 4: Average cost price = (7500 + 300x)/(15 + x) = 350. Step 5: 7500 + 300x = 350(15 + x). Step 6: 7500 + 300x = 5250 + 350x. Step 7: 2250 = 50x. Step 8: x = 45 kg.

Question 37

A milkman mixes milk from two sources. Source A has 85% purity and Source B has 60% purity. He wants to create 50 litres of milk with 72% purity. How many litres from Source A must he use?

Accepted Answer

Step 1: Use alligation method. Set up the ratio of deviations from the mean (72%). Step 2: Source A deviation = 85 - 72 = 13. Source B deviation = 72 - 60 = 12. Step 3: Ratio of A:B = 12:13 (inverse of deviations). Step 4: Total parts = 12 + 13 = 25. Step 5: Quantity from A = (12/25) × 50 = 24 litres.

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