Mixing quantities at different prices or concentrations.
45 solved questions · 15 easy · 20 medium · 10 hard
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A 40-litre mixture contains milk and water in the ratio 3 : 1. How much water does it contain?
Answer: A. 10 litres
Solution
Total parts = 3 + 1 = 4 Water = 1/4 of 40 = 10 litres Answer: 10 litres
4 kg of sugar at ₹30 per kg is mixed with 6 kg of sugar at ₹40 per kg. What is the price per kg of the mixture?
Answer: D. ₹36
Solution
Total cost = 4 × 30 + 6 × 40 = 120 + 240 = ₹360 Total weight = 10 kg Mean price = 360 ÷ 10 = ₹36 per kg Answer: ₹36
In what ratio must wheat at ₹20 per kg be mixed with wheat at ₹30 per kg to get a mixture worth ₹24 per kg?
Answer: C. 3:2
Solution
Alligation: cheaper : dearer = (30 − 24) : (24 − 20) = 6 : 4 = 3 : 2 Answer: 3:2 (cheaper : dearer)
Shortcut: Cross-subtract along the diagonals: 30 − 24 = 6 and 24 − 20 = 4.
20 litres of milk is mixed with 5 litres of water. What percentage of the mixture is milk?
Answer: B. 80%
Solution
Total mixture = 20 + 5 = 25 litres Milk % = 20/25 × 100 = 80% Answer: 80%
How much water must be added to 20 litres of pure milk so that the mixture contains 80% milk?
Answer: B. 5 litres
Solution
Milk stays 20 litres and should be 80% of the new total. New total = 20 ÷ 0.8 = 25 litres Water to add = 25 − 20 = 5 litres Answer: 5 litres
3 litres of a 20% acid solution is mixed with 2 litres of a 45% acid solution. What is the percentage of acid in the mixture?
Answer: A. 30%
Solution
Acid in first = 20% of 3 = 0.6 litre Acid in second = 45% of 2 = 0.9 litre Total acid = 1.5 litres in 5 litres Percentage = 1.5/5 × 100 = 30% Answer: 30%
A 35-litre mixture of milk and water is in the ratio 4 : 3. What is the quantity of milk?
Answer: D. 20 litres
Solution
Total parts = 4 + 3 = 7 One part = 35 ÷ 7 = 5 litres Milk = 4 × 5 = 20 litres Answer: 20 litres
Tea costing ₹200 per kg and tea costing ₹300 per kg are mixed in equal quantities. What is the price of the mixture per kg?
Answer: C. ₹250
Solution
For equal quantities the mean price is the simple average. (200 + 300)/2 = ₹250 per kg Answer: ₹250
25 kg of sugar at ₹40 per kg is mixed with 15 kg of sugar at ₹48 per kg. What is the cost per kg of the mixture?
Answer: C. ₹43
Solution
Total cost = 25 × 40 + 15 × 48 = 1,000 + 720 = ₹1,720 Total weight = 40 kg Mean price = 1,720 ÷ 40 = ₹43 per kg Answer: ₹43
In what ratio should rice at ₹30 per kg be mixed with rice at ₹40 per kg to obtain a mixture worth ₹36 per kg?
Answer: B. 2:3
Solution
Alligation: cheaper : dearer = (40 − 36) : (36 − 30) = 4 : 6 = 2 : 3 Answer: 2:3 (cheaper : dearer)
A mixture has 40 litres of milk and 10 litres of water. What is the percentage of milk?
Answer: A. 80%
Solution
Total = 40 + 10 = 50 litres Milk % = 40/50 × 100 = 80% Answer: 80%
A 48-litre mixture of milk and water is in the ratio 5 : 3. How many litres of water does it contain?
Answer: D. 18 litres
Solution
Total parts = 5 + 3 = 8 One part = 48 ÷ 8 = 6 litres Water = 3 × 6 = 18 litres Answer: 18 litres
A 12-litre salt solution contains 25% salt. How much salt does it contain?
Answer: D. 3 litres
Solution
Salt = 25% of 12 = 12/4 = 3 litres Answer: 3 litres
Liquid A costs ₹60 per litre and liquid B costs ₹80 per litre. They are mixed in the ratio 1 : 3. What is the cost per litre of the mixture?
Answer: C. ₹75
Solution
Cost of 1 L of A + 3 L of B = 60 + 3 × 80 = ₹300 Total quantity = 4 litres Mean price = 300 ÷ 4 = ₹75 per litre Answer: ₹75
Rice at ₹50 per kg and rice at ₹100 per kg are mixed in the ratio 2 : 3. What is the price of the mixture per kg?
Answer: B. ₹80
Solution
Cost = 2 × 50 + 3 × 100 = 100 + 300 = ₹400 Total = 5 kg Mean price = 400 ÷ 5 = ₹80 per kg Answer: ₹80
In what ratio should a tea costing ₹42 per kg be mixed with a tea costing ₹52 per kg so that the mixture is worth ₹48 per kg?
Answer: A. 2:3
Solution
Alligation: cheaper : dearer = (52 − 48) : (48 − 42) = 4 : 6 = 2 : 3 Answer: 2:3 (cheaper : dearer)
How many kg of sugar at ₹40 per kg must be mixed with 20 kg of sugar at ₹25 per kg so that the mixture is worth ₹30 per kg?
Answer: A. 10 kg
Solution
Alligation: dearer : cheaper = (30 − 25) : (40 − 30) = 5 : 10 = 1 : 2 Cheaper quantity = 20 kg corresponds to 2 parts, so 1 part = 10 kg Dearer sugar needed = 10 kg Answer: 10 kg
A vessel contains 80 litres of pure milk. 20 litres is taken out and replaced by water. This is done once more. How much milk is now left in the vessel?
Answer: D. 45 litres
Solution
Fraction removed each time = 20/80 = 1/4, so 3/4 remains each time. Milk left = 80 × (3/4) × (3/4) = 80 × 9/16 = 45 litres Answer: 45 litres
A 60-litre mixture contains milk and water in the ratio 2 : 1. How much water must be added to make the ratio 1 : 2?
Answer: C. 60 litres
Solution
Milk = 2/3 × 60 = 40 litres, water = 20 litres. For ratio 1 : 2 the water must be 2 × 40 = 80 litres. Water to add = 80 − 20 = 60 litres Answer: 60 litres
A dishonest milkman mixes 20 litres of water with 100 litres of pure milk and sells the mixture at the cost price of pure milk. What is his profit percentage?
Answer: B. 20%
Solution
Cost of 100 litres of milk = 100 units (say ₹1 per litre = ₹100). He sells 120 litres at ₹1 per litre = ₹120. Profit = 120 − 100 = ₹20 Profit % = 20/100 × 100 = 20% Answer: 20%
How much water must be added to 40 litres of a 25% acid solution to make it a 20% solution?
Answer: B. 10 litres
Solution
Acid in the solution = 25% of 40 = 10 litres (stays constant). New total = 10 ÷ 0.20 = 50 litres Water to add = 50 − 40 = 10 litres Answer: 10 litres
6 litres of a 30% alcohol solution is mixed with 4 litres of an x% alcohol solution to get a 42% solution. Find x.
Answer: A. 60%
Solution
Total alcohol needed = 42% of 10 = 4.2 litres Alcohol in first = 30% of 6 = 1.8 litres Alcohol from second = 4.2 − 1.8 = 2.4 litres in 4 litres x = 2.4/4 × 100 = 60% Answer: 60%
A trader mixes tea at ₹100 per kg and tea at ₹150 per kg in the ratio 3 : 2. At what price per kg should he sell the mixture to earn a profit of 25%?
Answer: D. ₹150
Solution
Cost of 5 kg = 3 × 100 + 2 × 150 = ₹600 Mean cost = 600 ÷ 5 = ₹120 per kg Selling price = 120 × 125/100 = ₹150 per kg Answer: ₹150
A milk seller adds 5 litres of water to 20 litres of milk costing ₹50 per litre and sells the mixture at ₹50 per litre. What is his gain percentage?
Answer: C. 25%
Solution
Cost of 20 litres of milk = 20 × 50 = ₹1,000 Sale of 25 litres at ₹50 = ₹1,250 Gain = 1,250 − 1,000 = ₹250 Gain % = 250/1,000 × 100 = 25% Answer: 25%
Vessel A has 20 litres of milk and water in the ratio 3 : 2 and vessel B has 30 litres in the ratio 1 : 4. If both are poured into a larger vessel, what is the ratio of milk to water in the new mixture?
Answer: C. 9:16
Solution
Vessel A: milk = 3/5 × 20 = 12, water = 8 Vessel B: milk = 1/5 × 30 = 6, water = 24 Total milk = 18, total water = 32 Ratio = 18 : 32 = 9 : 16 Answer: 9:16
In what ratio must a 30% acid solution be mixed with a 50% acid solution to obtain a 42% solution?
Answer: B. 2:3
Solution
Alligation: (30% solution) : (50% solution) = (50 − 42) : (42 − 30) = 8 : 12 = 2 : 3 Answer: 2:3
From a 50-litre vessel of pure milk, 10 litres is drawn out and replaced by water. Then 10 litres of the mixture is drawn out and again replaced by water. What is the final ratio of milk to water?
Answer: A. 16:9
Solution
Milk after two operations = 50 × (4/5)² = 50 × 16/25 = 32 litres Water = 50 − 32 = 18 litres Ratio = 32 : 18 = 16 : 9 Answer: 16:9
An alloy of 20 kg contains gold and copper in the ratio 3 : 2. How much copper must be added to make the ratio 1 : 1?
Answer: D. 4 kg
Solution
Gold = 3/5 × 20 = 12 kg, copper = 8 kg. For ratio 1 : 1, copper must also be 12 kg. Copper to add = 12 − 8 = 4 kg Answer: 4 kg
A merchant mixes two types of rice costing ₹30 and ₹36 per kg in the ratio 2 : 1 and sells the mixture at ₹38.40 per kg. What is his profit percentage?
Answer: D. 20%
Solution
Cost of 3 kg = 2 × 30 + 36 = ₹96 Mean cost = 96 ÷ 3 = ₹32 per kg Profit = 38.40 − 32 = ₹6.40 Profit % = 6.40/32 × 100 = 20% Answer: 20%
25 litres of a mixture has milk and water in the ratio 3 : 2. If 5 litres of water is added, what is the new ratio of milk to water?
Answer: C. 1:1
Solution
Milk = 3/5 × 25 = 15 litres, water = 10 litres. After adding 5 litres of water: water = 15 litres. Ratio = 15 : 15 = 1 : 1 Answer: 1:1
12 litres of a 40% solution is mixed with 18 litres of a 60% solution. What is the percentage strength of the resulting mixture?
Answer: B. 52%
Solution
Solute in first = 40% of 12 = 4.8 litres Solute in second = 60% of 18 = 10.8 litres Total = 15.6 litres in 30 litres Strength = 15.6/30 × 100 = 52% Answer: 52%
Two kinds of oil, one at ₹60 per litre and the other costing more, are mixed in the ratio 3 : 2, giving a mixture worth ₹70 per litre. What is the price per litre of the costlier oil?
Answer: A. ₹85
Solution
Total cost of 5 litres = 5 × 70 = ₹350 Cost of 3 litres of cheaper = 3 × 60 = ₹180 Cost of 2 litres of costlier = 350 − 180 = ₹170 Price per litre = 170 ÷ 2 = ₹85 Answer: ₹85
Milk costs ₹60 per litre. A seller mixes water with it and sells the mixture at ₹60 per litre, making a profit of 20%. What is the ratio of water to milk in the mixture?
Answer: A. 1:5
Solution
Selling price of the mixture = ₹60 per litre with 20% profit. Cost price of the mixture = 60 × 100/120 = ₹50 per litre Alligation with water (₹0) and milk (₹60) for a mean of ₹50: water : milk = (60 − 50) : (50 − 0) = 10 : 50 = 1 : 5 Answer: 1:5
In what ratio should tea at ₹80 per kg be mixed with tea at ₹110 per kg so that selling the mixture at ₹110.40 per kg gives a profit of 15%? (Give the ratio cheaper : dearer.)
Answer: D. 7:8
Solution
Cost price of the mixture = 110.40 × 100/115 = ₹96 per kg Alligation: cheaper : dearer = (110 − 96) : (96 − 80) = 14 : 16 = 7 : 8 Answer: 7:8
A 100-litre mixture contains 40% milk. How many litres of pure milk must be added to make it 50% milk?
Answer: C. 20 litres
Solution
Milk now = 40 litres. Let x litres of pure milk be added. (40 + x)/(100 + x) = 1/2 80 + 2x = 100 + x x = 20 Answer: 20 litres
A container holds 64 litres of pure milk. One-fourth of the liquid is drawn out and replaced by water. This is repeated two more times (three operations in all). How much milk remains?
Answer: B. 27 litres
Solution
Each operation leaves 3/4 of the milk. Milk left = 64 × (3/4)³ = 64 × 27/64 = 27 litres Answer: 27 litres
Shortcut: If the capacity is a perfect cube multiple, (3/4)^3 = 27/64 gives the answer directly.
A vessel is full of pure milk. 10 litres is drawn out and replaced by water, and then 10 litres of the mixture is drawn out and replaced by water. The final ratio of milk to water is 9 : 7. What is the capacity of the vessel?
Answer: B. 40 litres
Solution
Milk fraction after two operations = (1 − 10/x)² Milk : total = 9 : 16, so (1 − 10/x)² = 9/16 1 − 10/x = 3/4 10/x = 1/4, x = 40 Answer: 40 litres
10 litres of a 20% acid solution, 20 litres of a 35% acid solution and x litres of a 50% acid solution are mixed to get a 40% acid solution. Find x.
Answer: A. 30 litres
Solution
Acid: 0.20 × 10 + 0.35 × 20 + 0.50x = 2 + 7 + 0.5x = 9 + 0.5x Total volume = 30 + x 9 + 0.5x = 0.4(30 + x) = 12 + 0.4x 0.1x = 3, x = 30 Answer: 30 litres
A dishonest dealer buys milk at ₹48 per litre and mixes it with water in the ratio 5 : 1 (milk : water). He sells the mixture at ₹54 per litre. What is his profit percentage?
Answer: D. 35%
Solution
Take 6 litres of mixture = 5 litres milk + 1 litre water. Cost = 5 × 48 = ₹240 Selling price = 6 × 54 = ₹324 Profit = 324 − 240 = ₹84 Profit % = 84/240 × 100 = 35% Answer: 35%
A can contains milk and water in the ratio 7 : 5. When 9 litres of the mixture is drawn out and replaced by water, the ratio becomes 7 : 9. What is the capacity of the can?
Answer: C. 36 litres
Solution
Let the capacity be T litres. Milk = 7T/12 initially. Milk removed = 9 × 7/12, so milk left = 7(T − 9)/12. New ratio 7 : 9 means milk is 7/16 of T (T is unchanged). 7(T − 9)/12 = 7T/16 16(T − 9) = 12T, 4T = 144, T = 36 Check: milk 21 → 15.75, water 15 → 11.25 + 9 = 20.25, ratio 15.75 : 20.25 = 7 : 9. Answer: 36 litres
Vessel A has milk and water in the ratio 5 : 3 and vessel B has them in the ratio 3 : 1. If equal quantities from A and B are mixed, what is the ratio of milk to water in the mixture?
Answer: C. 11:5
Solution
Take 8 litres from each. From A: milk = 5, water = 3. From B (3:1 means 6 and 2): milk = 6, water = 2. Total milk = 11, total water = 5 Ratio = 11 : 5 Answer: 11:5
A shopkeeper mixes two grades of rice costing ₹24 and ₹30 per kg. He sells 72 kg of the mixture at ₹30.80 per kg and earns a profit of 10%. How many kg of the cheaper grade did he use?
Answer: B. 24 kg
Solution
Cost price of the mixture = 30.80 × 100/110 = ₹28 per kg Alligation: cheaper : dearer = (30 − 28) : (28 − 24) = 2 : 4 = 1 : 2 Cheaper grade = 1/3 of 72 = 24 kg Answer: 24 kg
A 20-litre solution contains 40% alcohol. 4 litres of it is removed and replaced with 4 litres of pure alcohol. What is the new percentage of alcohol?
Answer: A. 52%
Solution
Alcohol in 20 litres = 40% of 20 = 8 litres After removing 4 litres, alcohol left = 8 − 4 × 0.40 = 6.4 litres Add 4 litres of pure alcohol: 6.4 + 4 = 10.4 litres in 20 litres Percentage = 10.4/20 × 100 = 52% Answer: 52%
20 kg of sugar solution contains 15% sugar. How many kg of water must evaporate to make it a 25% solution?
Answer: D. 8 kg
Solution
Sugar = 15% of 20 = 3 kg (does not change on evaporation). New total = 3 ÷ 0.25 = 12 kg Water evaporated = 20 − 12 = 8 kg Answer: 8 kg
Three varieties of pulses costing ₹20, ₹40 and ₹50 per kg are mixed so that the quantities of the ₹20 and ₹40 varieties are equal and the mixture is worth ₹36 per kg. What is the ratio of the quantities of the three varieties (₹20 : ₹40 : ₹50)?
Answer: D. 7:7:6
Solution
Equal amounts of ₹20 and ₹40 behave like one variety with mean price (20 + 40)/2 = ₹30 per kg. Alligation of ₹30 and ₹50 for mean ₹36: (₹30 group) : (₹50) = (50 − 36) : (36 − 30) = 14 : 6 = 7 : 3 Split the 7 equally between ₹20 and ₹40: 3.5 : 3.5 : 3 = 7 : 7 : 6 Check: (7 × 20 + 7 × 40 + 6 × 50)/20 = 720/20 = ₹36. Answer: 7:7:6
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