Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
All Direct removal and addition 448 Alligation cross rule 270 Ratios, quantities, and percentages 233 Weighted mean and concentration 195 Equal quantity replacement 189 Three or more components 174 Repeated equal-volume replacement 154 Profit-based alligation 153

Mixtures and Alligation Questions

Multiple choice

In one alloy there is 60% steel in its total mass, while in another alloy it is 35%. 15 kg of the first alloy was melted together with 30 kg of the second one to form a third alloy. Find the ratio of steel and other metals in the new alloy?
  1. $13:17$
  2. $11:25$
  3. $11:17$
  4. $13:19$

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Steel from first alloy: 60% of 15 kg = 9 kg. Steel from second alloy: 35% of 30 kg = 10.5 kg. Total steel = 19.5 kg in 45 kg alloy. Other metals = 45 - 19.5 = 25.5 kg. Ratio steel:other = 19.5:25.5 = 13:17 after simplifying.

Multiple choice

A tank which contains a mixture of syrup and water in ratio 15:6 . 42 liters of mixture is taken out from the tank and 8 liters of pure water and 7 liters of syrup is added to the mixture. If resultant mixture contains 33.33% water, what was the initial quantity of mixture in the tank before the replacement (in liters) ?

  1. 105

  2. 110

  3. 205

  4. 125

  5. None of These
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial ratio syrup:water = 15:6, so water is 6/21 = 28.57%. 42L removed removes same ratio. Adding 8L water + 7L syrup. Final water % = 33.33% = 1/3. Let initial total = x. Water initially = 6x/21 = 2x/7. After removal: water = (2x/7)(x-42)/x = 2(x-42)/7. After adding: water = 2(x-42)/7 + 8. Total = x - 42 + 15 = x - 27. So [2(x-42)/7 + 8]/(x-27) = 1/3. Solving: 6(x-42) + 168 = 7(x-27), 6x-252+168 = 7x-189, -84 = x-189, x = 105. Option A is correct.

Multiple choice

A container contains 80 liter milk and water and percentage of milk is 31.25%. How much more milk should be added so that milk and water becomes equal?

  1. 24 liter

  2. 30 liter

  3. 36 liter

  4. 25 liter
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initial mixture: 80 liters with 31.25% milk = 25 liters milk, 55 liters water. Let x be additional milk. New milk = 25 + x, water = 55. For equal quantities: 25 + x = 55, so x = 30. Total becomes 110 liters with 55 liters each of milk and water (50%).

Multiple choice

From container A containing 54 liter of mixture of milk and water in ratio of 8 : 1 , 18 liter of the mixture is taken out and poured into container B in which ratio of milk to water is 3 : 1. If difference between total milk and total water in container B is 30 liter then find the quantity of initial mixture in container B.

  1. 30

  2. 28

  3. 32

  4. 36

  5. 40
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Container A: 54L, milk:water = 8:1, so milk = 48L, water = 6L. 18L taken out (same ratio: 16L milk, 2L water). Added to container B (milk:water = 3:1). Let B initial have 3x milk, x water = 4x total. After adding: milk = 3x+16, water = x+2. Difference = (3x+16) - (x+2) = 2x+14 = 30. So 2x = 16, x = 8. Initial B = 4x = 32L. Answer C is correct.

Multiple choice

Two containers contain a mixture of milk and water. In the first container, the concentration of milk is 33.33\%, while in the second container, the concentration of milk is 25\%. 5 liters of container 1 is mixed with 7 liters of container 2. Find the ratio of milk and water in the resultant mixture.
  1. $41:105$
  2. $41:103$
  3. $53:105$
  4. $43:98$
  5. $None of these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Container 1: 5L, milk = 33.33% = 1/3. Milk = 5/3 = 1.6667L, water = 10/3 = 3.3333L. Container 2: 7L, milk = 25% = 1/4. Milk = 7/4 = 1.75L, water = 21/4 = 5.25L. Total milk = 5/3 + 7/4 = (20 + 21)/12 = 41/12 L. Total water = 10/3 + 21/4 = (40 + 63)/12 = 103/12 L. Ratio milk:water = 41:103. This matches option B.

Multiple choice

A contains 55 litres of milk, from this container 11 litres of milk was taken out and replaced by water. This process is repeated further two times. How much milk is now contained by the container ?

  1. 26.16 Liters

  2. 28.84 Liters

  3. 26.84 Liters

  4. 28.16 Liters

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Starting with 55 litres of milk, each operation removes 11 litres (which is 11/55 = 1/5 of the mixture) and replaces it with water. After each operation, only 4/5 of the milk remains. After 3 operations: 55 × (4/5)^3 = 55 × 64/125 = 28.16 litres. This uses the formula for dilution after repeated replacement.

Multiple choice

In a 80 litre of solution containing milk and water in the ratio 2: 3. How much milk should be added to the solution so that the new ratio becomes 4:1?

  1. 160

  2. 192

  3. 132

  4. 32

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Current solution: 80 liters with milk:water = 2:3. If ratio is 2:3, then milk = (2/5) × 80 = 32 liters, water = (3/5) × 80 = 48 liters. Let x liters of milk be added. New ratio = (32+x):48 = 4:1. So (32+x)/48 = 4/1, giving 32+x = 192, x = 160 liters. After adding 160 liters milk: total milk = 192 liters, water = 48 liters, ratio = 192:48 = 4:1. Therefore A (160) is correct.

Multiple choice

4 litre of a solution having 10% acid was mixed with 5 litre of a solution having 16% acid. Six litre of pure acid was then added to the resulting solution. The concentration of acid in the final solution is:

  1. 48%

  2. 50%

  3. 54%

  4. 60%

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First solution: 4L × 10% = 0.4L acid. Second solution: 5L × 16% = 0.8L acid. Total after mixing: 9L with 1.2L acid. Adding 6L pure acid gives total acid = 7.2L in 15L solution. Concentration = 7.2/15 × 100 = 48%. The key is tracking the absolute amount of acid, not just percentages.

Multiple choice

In what ratio must a shopkeeper have to mix the two varieties of rice costing Rs 28 and Rs 40 per kg respectively so as to get a mixture worth Rs 36 per kg ?

  1. 4 : 7

  2. 2 : 3

  3. 1 : 2

  4. 5 : 8

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Use allegation method: The difference between mean price (36) and each price gives the ratio. Cost of cheaper rice: Rs 28, Cost of costlier rice: Rs 40, Mean price: Rs 36. Using allegation: (40 - 36) : (36 - 28) = 4 : 8 = 1 : 2. So the ratio is 1 : 2.

Multiple choice

In a mixture percentage of water is 65%. When 18 liter mixture containing 30% water is mixed in this the percentage of water becomes 50%. Find the initial quantity of mixture. (in liters)

  1. 18

  2. 36

  3. 27

  4. 30

  5. None of these
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let initial quantity be x liters with 65% water. Water in initial mixture = 0.65x. Added mixture has 30% water in 18L = 5.4L water. Final mixture: (x + 18) liters with 50% water = 0.5(x + 18) water. Equation: 0.65x + 5.4 = 0.5(x + 18). Solving: 0.65x + 5.4 = 0.5x + 9, giving 0.15x = 3.6, so x = 24L. Since 24 is not in options A-D, answer is E (None of these).

Multiple choice

A bottle contains a mixture of two liquids A and B in the ratio 5: 3. If 6 litres of the mixture is replaced by 6 litres of liquid B, then the ratio of the two liquids becomes 5: 9. How much of the liquid A was there in the bottle? (Litres)

  1. 8

  2. 8.25

  3. 8.75

  4. 9.25

  5. 9.75
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial quantities be 5x and 3x. When 6 litres is removed, the ratio remains 5:3, so A removed = 5/8 × 6 = 3.75 L, B removed = 2.25 L. After adding 6 L of B, new A = 5x - 3.75, new B = 3x - 2.25 + 6 = 3x + 3.75. New ratio (5x - 3.75)/(3x + 3.75) = 5/9. Solving: 45x - 33.75 = 15x + 18.75, so 30x = 52.5, x = 1.75. Initial A = 5x = 8.75 litres.

Multiple choice

A vessel contains a 32 litre solution of acid and water in which the ratio of acid and water is 5 : 3. If 12 litres of the solution is taken out and 7.5 litres of water is added to it. Then what is the ratio of acid and water in the resulting solution?

  1. 4 : 7

  2. 5 : 6

  3. 4 : 9

  4. 8 : 11
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Initially, the 32L solution has 20L acid and 12L water (5:3 ratio). When 12L is removed, the ratio stays the same, so 7.5L acid and 4.5L water are removed, leaving 12.5L acid and 7.5L water. After adding 7.5L water, the final amounts are 12.5L acid and 15L water, which simplifies to 5:6. The key insight is that removal maintains the ratio, while addition only affects the component being added.

Multiple choice

When 45 liter water is mixed in a mixture containing 60\% milk, concentration of milk in mixture becomes 45\%. Find the initial quantity of mixture.

  1. 100

  2. 90

  3. 135

  4. 145
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial mixture be x liters containing 60% milk, so milk = 0.6x. After adding 45 L water, total = x + 45 and milk concentration becomes 45%, so 0.6x / (x + 45) = 0.45. Solving: 0.6x = 0.45(x + 45) = 0.45x + 20.25, so 0.15x = 20.25, x = 135. Verification: Initial milk = 0.6 × 135 = 81 L. After adding water: 81/180 = 45%.

Multiple choice

Ratio of water and milk in a container is 7 : 12, 38 litre of mixture taken out and replaced by 19 litre of water then ratio become 25: 32 . Then find the initial quantity of mixture .

  1. 156

  2. 133

  3. 152

  4. 190
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let initial quantity be 19k liters (7:12 ratio). After removing 38L (which is 2×19), water becomes 7k-14, milk becomes 12k-24. After adding 19L water: new water = 7k-14+19 = 7k+5. New ratio (7k+5)/(12k-24) = 25/32. Solving: 32(7k+5) = 25(12k-24) → 224k+160 = 300k-600 → 76k = 760 → k=10. Initial mixture = 19×10 = 190L.