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

Three containers of equal volume contains milk and water. The ratios of milk and water are 2 : 3, 3 : 4 and 4 : 5 respectively. Contents of these containers are poured in a large container. The ratio of milk and water in the large container is

  1. 411 : 540

  2. 401 : 544

  3. 417 : 564

  4. 407 : 560
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Milk in containers: 2/5, 3/7, 4/9. Water in containers: 3/5, 4/7, 5/9. Total milk = (2/5 + 3/7 + 4/9) = 126/315 + 135/315 + 140/315 = 401/315. Total water = (3/5 + 4/7 + 5/9) = 189/315 + 180/315 + 175/315 = 544/315. Ratio milk:water = 401:544.

Multiple choice

A milkman has three containers A, B and C. The ratio of milk and water in containers A, B and C respectively are 3:1, 4:1 and 5:1. Each container contains 60 liter mixture of milk and water. The Milkman boiled the milk one by one container, so 20% of water evaporated from each container. If the milkman mixed all mixtures in a container D, then what is the ratio of milk and water in container D?
  1. $615: 148$
  2. $715: 147$
  3. $715: 148$
  4. $715: 142$
  5. $515: 148$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Container A has 45L milk and 15L water. After 20% evaporation, water becomes 12L. Container B has 48L milk and 12L water. After 20% evaporation, water becomes 9.6L. Container C has 50L milk and 10L water. After 20% evaporation, water becomes 8L. Total milk = 143L, total water = 29.6L. Ratio = 143:29.6 = 715:148.

Multiple choice

Three containers A, B and C contain wine and water the ratio respectively 5:7, 7:9 and 4:5. Each container has equal volume of the mixture. If he took two-third of the quantity from container A, four – fifth quantity from container B and half quantity from container C and mixed in another container D. What is the ratio of wine to water in container D?

  1. 51: 67

  2. 53: 68

  3. 52:71

  4. 53:73

  5. 51:47
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let each container have V liters. Container A (wine:water = 5:7): taking 2/3 gives wine = 5/12 × 2V/3 = 10V/36, water = 7/12 × 2V/3 = 14V/36. Container B (7:9): taking 4/5 gives wine = 7/16 × 4V/5 = 28V/80, water = 9/16 × 4V/5 = 36V/80. Container C (4:5): taking 1/2 gives wine = 4/9 × V/2 = 4V/18, water = 5/9 × V/2 = 5V/18. Total wine = 612V/720, total water = 804V/720. Ratio = 612:804 = 51:67.

Multiple choice

$Find ratio of milk and water in the resultant$ I. two mixtures contain milk and water in the ratio 3:2 and 4:3 II. 20 litres from 1st container is taken out and x litres from 2nd is taken out and poured in third container III. value of x is 40 litres IV. the mixture from two containers are mixed in the ratio 3:2

  1. I,II and III

  2. I and IV

  3. I,II and III or I and IV

  4. All required

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

Using statements I, II, III: 20L from first container (3:2) has 12L milk and 8L water. 40L from second (4:3) has 160/7 ≈ 22.86L milk and 120/7 ≈ 17.14L water. Total: 12 + 160/7 = 244/7 milk, 8 + 120/7 = 176/7 water. Ratio = 244:176 = 61:44. Using statements I and IV: mixing in 3:2 ratio gives (3/5)(3/2) : (3/5)(2/5) + (2/5)(4/7) : (2/5)(3/7) which also yields a valid ratio. Either combination is sufficient.

Multiple choice

Six containers contained 567.975 kg, 369.005 kg, 509.945 kg, 129.915 kg, 39.995 kg and 800.150 kg of rice, costing Rs. 40 per kg. If all the rice in the six containers is mixed and equally placed into four containers, then what is the approximate cost of rice in each container?

  1. Rs.28120

  2. Rs.29090

  3. Rs. 24170

  4. Rs. 20170
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total rice = 567.975 + 369.005 + 509.945 + 129.915 + 39.995 + 800.150 = 2416.985 kg. Total cost at Rs. 40/kg = 2416.985 × 40 = Rs. 96,679.40. When divided equally among 4 containers, each contains rice worth 96,679.40 / 4 = Rs. 24,169.85 ≈ Rs. 24,170.

Multiple choice

To gain 10\% on selling sample milk at the cost price of pure milk, the quantity of water to be mixed with 50 kg of pure milk is

  1. 2.5 kg

  2. 5 kg

  3. 7.5 kg

  4. 10 kg
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Selling 50 kg milk mixed with x kg water at pure milk's CP means we effectively sell (50+x) kg at price of 50 kg. Water costs nothing, so profit comes from selling extra quantity. 10% profit means (50+x)/50 = 1.10, so x = 5 kg. Option A (2.5 kg) gives only 5% profit, while C (7.5 kg) and D (10 kg) would give 15% and 20% respectively.

Multiple choice

In a 45 liters mixture of milk and water, the ratio of the milk to water is 2 : 1. When some quantity of water is added to the mixture, this ratio becomes 1 : 2. The quantity of water added is

  1. 10 litres

  2. 21 litres

  3. 35 litres

  4. 45 litres
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Initial: 45L with milk:water = 2:1, so milk = 30L, water = 15L. Final ratio 1:2 means water = 2 × milk = 60L. Water added = 60 - 15 = 45L. The claimed answer D (45L) is correct.

Multiple choice

A person bought some articles at the rate of 5 per rupee and the same number at the rate of 4 per rupee. He mixed both the types and sold at the rate of 9 for 2 rupees. In this business he suffered a loss of Rs 3. The total number of articles bought by him was

  1. 1090

  2. 1080

  3. 540

  4. 545
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let n articles of each type. Cost = n/5 + n/4 = 9n/20. SP = 2n/9. Loss = 9n/20 - 2n/9 = 81n/180 - 40n/180 = 41n/180 = 3, so n = 540. Total articles = 2n = 1080. The claimed answer B (1080) is correct.

Multiple choice

Two mixtures A and B are prepared by mixing Alcohol and Soda in the ratio 5 : 8 and 3 : 7. Now equal quantities of these mixtures are mixed and 13 litres of water is also added in it. If the final mixture has 20% water concentration, then find the ratio of Alcohol, Soda and Water in the final mixture.
  1. $98 : 171 : 65$
  2. $89 : 171 : 65$
  3. $98 : 71 : 165$
  4. $89 : 71 : 165$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mixture A has Alcohol:Soda = 5:8, so 5/13 alcohol and 8/13 soda. Mixture B has Alcohol:Soda = 3:7, so 3/10 alcohol and 7/10 soda. Equal quantities (let's say x units each) are mixed. Total alcohol = x(5/13 + 3/10) = x(98/130) = 49x/65. Total soda = x(8/13 + 7/10) = x(171/130) = 171x/130. Total volume = 2x. Water added = 13 liters. Final water concentration = 20%, so 13/(2x + 13) = 0.2. Solving: 13 = 0.4x + 2.6, so x = 26. Then alcohol = 49×26/65 = 19.6, soda = 171×26/130 = 34.2, water = 13. Ratio = 19.6 : 34.2 : 13 = 196 : 342 : 130 = 98 : 171 : 65 after simplification.

Multiple choice

The ratio of oil and kerosene in the container is 3 : 2 when 10 L of the mixture is taken out and replaced by kerosene, the ratio becomes 2 : 3. Find the quantity of the mixture in the container initially.

  1. 20

  2. 30

  3. 35

  4. 40
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let initial oil = 3x, kerosene = 2x. After removing 10L (6L oil, 4L kerosene) and adding 10L kerosene: oil = 3x - 6, kerosene = 2x - 4 + 10 = 2x + 6. New ratio (3x-6)/(2x+6) = 2/3. Solving: 9x - 18 = 4x + 12; 5x = 30; x = 6. Initial mixture = 5x = 30L.

Multiple choice

A and B are solutions of acid and water. The ratios of water and acid in A and B are (4 : 5) and (1 : 2) respectively. If (x) liters of A is mixed with (y) liters of B, then the ratio of water and acid in the mixture becomes (8 : 13). What is (x : y?)

  1. 3 : 4

  2. 5 : 6

  3. 2 : 3

  4. 2 : 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution A has water:acid = 4:5, meaning water is 4/9 and acid is 5/9 of x liters. Solution B has water:acid = 1:2, meaning water is 1/3 and acid is 2/3 of y liters. Mixture ratio is 8:13, meaning water is 8/21 and acid is 13/21 of (x+y) liters. Setting up equations: (4x/9) + (y/3) = 8(x+y)/21 and (5x/9) + (2y/3) = 13(x+y)/21. Solving yields x:y = 3:4. This is a standard mixture problem requiring ratio algebra.

Multiple choice

A container contains 20 L mixture in which there is 10\% sulphuric acid. Find the quantity of sulphuric acid to be added in it to make the solution to contain 25\% sulphuric acid.
  1. $3$
  2. $4$
  3. $2$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In 20 L mixture with 10% acid, acid = 20 × 0.10 = 2 L. Let x liters of pure acid be added. New volume = 20 + x, new acid = 2 + x. Required: (2 + x)/(20 + x) = 0.25. Solving: 2 + x = 0.25(20 + x) = 5 + 0.25x, so 0.75x = 3 and x = 4 L.

Multiple choice

The price of a variety of a commodity is Rs. 7/kg and that of another is Rs. 12/kg. Find the ratio in which two varieties should be mixed so that the price of the mixture is Rs. 10/kg.

  1. 2 : 5

  2. 2 : 3

  3. 3: 4

  4. 4: 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the alligation method: First variety at Rs. 7/kg, target price Rs. 10/kg, second variety at Rs. 12/kg. The ratio is |12-10| : |10-7| = 2 : 3. Verification: (2×7 + 3×12)/(2+3) = (14+36)/5 = 50/5 = 10.

Multiple choice

A solution contains acid and water in the ratio of 4 : 5. If 20% of the solution is replaced by water, then what will be the ratio of acid and water in the new solution?

  1. 8 : 15

  2. 5 : 17

  3. 10 : 7

  4. 16 : 29
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Initially, the solution has acid and water in 4:5 ratio, so 40 units acid and 50 units water (total 90 units). Removing 20% means removing 18 units of solution, which contains 8 acid and 10 water (maintaining the 4:5 ratio). After removal: 32 acid, 40 water remain. Adding 18 water gives 32 acid and 58 water. The new ratio is 32:58 = 16:29. The critical step is that the removed portion maintains the original ratio.