Mixture Problems MCQ Quiz - Objective Question with Answer for Mixture Problems - Download Free PDF

Last updated on Jun 3, 2025

Mixture Problems MCQ Quiz will help you with your preparations for various exams such as bank exams, SSC, Railways, Insurance etc. the questions have been detailed out for the candidates’ practice. Solve the Mixture Problems Objective Questions and get accuracy which will also help in time management. Everyone can access the Mixture Problems Question Answer and practice the Quantitative Aptitude section with these questions. Solve all the questions accurately and at a faster pace with practice.

Latest Mixture Problems MCQ Objective Questions

Mixture Problems Question 1:

Can A and Can B contain a mixture of soda and water in the ratio of 5 : 3 and 7 : 2 respectively If soda and water are taken out in the ratio of P : Q from can A and B respectively to form a new mixture in which the ratio of soda and water is 12 : 5 respectively, then find the value Of P:Q?

  1. 6 : 5
  2. 8 : 9
  3. 5 : 4
  4. 4 : 3
  5. 3 : 2

Answer (Detailed Solution Below)

Option 2 : 8 : 9

Mixture Problems Question 1 Detailed Solution

​Calculations:

Let the amount of mixture taken out from Can A be x units, and the amount taken from Can B be y units.

In Can A, the ratio of soda to water is 5:3. So, the amount of soda taken out from Can A is:

Soda taken from Can A = (5/8) × x, and water taken from Can A = (3/8) × x.

In Can B, the ratio of soda to water is 7 : 2. So, the amount of soda taken out from Can B is:

Soda taken from Can B = (7/9) × y, and water taken from Can B = (2/9) × y.

Now, the total amount of soda and water in the new mixture must have a ratio of 12 : 5.

The total amount of soda in the new mixture = (5/8) × x + (7/9) × y.

The total amount of water in the new mixture = (3/8) × x + (2/9) × y.

The ratio of soda to water in the new mixture is 12 : 5. So, we can write:

Cross-multiply:

(5/8) × x + (7/9) × y = (12/5) × ((3/8) × x + (2/9) × y)

Now, simplifying this equation, we multiply through by 40 (LCM of 8, 9, and 5):

40 × [(5/8) × x + (7/9) × y] = 40 × (12/5) × [(3/8) × x + (2/9) × y]

After simplification, solving this will give the ratio of P : Q.

Therefore, after solving the equation, you will get P : Q = 8 : 9.

Mixture Problems Question 2:

Ratio of milk and water in 630 litre of mixture is 4 : 3. 140 litre mixture taken out. Find the quantity of milk now?

  1. 220
  2. 240
  3. 260
  4. 280
  5. 250

Answer (Detailed Solution Below)

Option 4 : 280

Mixture Problems Question 2 Detailed Solution

Given:

The ratio of milk to water in 630 liters of mixture is 4:3.

140 liters of the mixture is taken out.

Calculations:

The total parts of the mixture = 4 + 3 = 7 parts.

Quantity of milk in the mixture = (4/7) × 630 = 360 liters of milk.

Quantity of water in the mixture = (3/7) × 630 = 270 liters of water.

When 140 liters of the mixture is taken out, the ratio of milk and water in the 140 liters will also be 4:3.

Milk taken out = (4/7) × 140 = 80 liters.

So, the remaining milk = 360 - 80 = 280 liters.

∴ The quantity of milk now is 280 liters.

Mixture Problems Question 3:

Ratio of milk and water is in a container is 5:3, 10 Liters milk and 7 Liters water is added, now ratio of milk and water is 8:5. Find the initial quantity of milk in the mixture?

  1. 45
  2. 30
  3. 24
  4. 65
  5. 40

Answer (Detailed Solution Below)

Option 2 : 30

Mixture Problems Question 3 Detailed Solution

Calculation

Initial ratio milk:water = 5:3
Let quantity = 8x ⇒ milk = 5x, water = 3x

New milk = 5x + 10  
New water = 3x + 7
New ratio:
5x + 10 / (3x + 7) = 8/5
25x  + 50 = 24x + 56
x = 6
Milk = 5x = 30 Liters

Mixture Problems Question 4:

A bucket contains liquid A and B in the ratio 15:16. 124 litre of the mixture is taken out and filled with 124 litre of B. Now the ratio changes to 9:10. Find the quantity of liquid B initially. (In
litres)

  1. 3240
  2. 3140
  3. 3040
  4. 2940

Answer (Detailed Solution Below)

Option 3 : 3040

Mixture Problems Question 4 Detailed Solution

Given:

Liquid A : Liquid B = 15 : 16

Calculation:

In 124 litres of the mixture, liquid B = (16/31) × 124 = 16 × 4 = 64 litres

After 124 litres of the mixture is taken out and filled with 124 litres of B

∵ The quantity of liquid A will remain.

Now, liquid B = 2976 litres

The quantity of liquid B initially = 64 + 2976 = 3040 litres

Mixture Problems Question 5:

5 litres are drawn from a cask full of wine and is then filled with water. This operation is performed once more. The ratio of the quantity of wine now left in cask to that of the water is 16:9. How much wine the cask hold originally? (In litres)

  1. 25
  2. 23
  3. 21
  4. 19

Answer (Detailed Solution Below)

Option 1 : 25

Mixture Problems Question 5 Detailed Solution

Given:

Wine drawn from cask each time = 5 litres

Final ratio of wine to water = 16:9

Formula used:

Final quantity of wine = Initial quantity of wine × (1 - fraction of wine removed per operation)number of operations

Calculation:

Let the quantity of the wine in the cask originally be x litres

The quantity of wine left in the cask after 2 operations = [1 - ]2

⇒  = [1 - ]2

Check by options,

If x = 25 then, 

 [1 - ]=  [1 - ]2 =  [1 - ]2

⇒ (4/5)2

⇒ 16/25

∴ The cask holder of 25 litres of wine hold originally.

Top Mixture Problems MCQ Objective Questions

In a grocery shop box A contains tea worth Rs.300 pr kg and box B contains tea worth Rs.400 pr kg. If both box A and B are mixed in the ratio 5 : 6 then the price of mixture pr kg is approximately:

  1. Rs.370
  2. Rs.355
  3. Rs.350
  4. Rs.360
  5. None of these

Answer (Detailed Solution Below)

Option 2 : Rs.355

Mixture Problems Question 6 Detailed Solution

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Given:

Cost of 1 kg tea of box A (cheaper) = Rs.300

Cost of 1 kg tea of box B (dearer) = Rs.400

Formula used:

Rule of alligation 

Calculation:

Let the mean price be Rs. X

So, (Cheaper  quantity) : (Dearer quantity) = (d- m) : (m - c) = (400 - X) : (X - 300)

According to question,

Given ratio = 5/6

So, 5/6 = (400 - X)/(X- 300)

⇒ 11x = 3,900

⇒ x = 354.54 ≈ 355

∴ The price of mixture  of 1 kg tea is Rs.355

Alloy A contains metals x and y only in the ratio 5 ∶ 2, while alloy B contains them in the ratio 3 ∶ 4. Alloy c is prepared by mixing alloys A and B in the ratio 4 ∶ 5. the percentage of x in alloy C is:

Answer (Detailed Solution Below)

Option 4 :

Mixture Problems Question 7 Detailed Solution

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Shortcut TrickAlloy A = 5 : 2 --sum--> 7]  × 4

Alloy B = 3 : 4 --sum--> 7] × 5

-----------------------------------------------

Since the sum of quantity is same so multiplication by 4 and 5 just because the amount of A and B are taken in the ratio 4 : 5

Alloy A = 20 : 8

Alloy B = 15 : 20

---------------------------

Alloy C = 35 : 28 = 5 : 4

Total quantity = 5 + 4 = 9 

Required % = (5/9) × 100% =  

∴ The required percentage of x in alloy C is .

 

Alternate Method 

Given:

The mixture of x and y in Alloy A = 5 : 2

The mixture of x and y in Alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

Calculation:

Let the quantity of metal x in alloy C be x

Quantity of metal x in alloy A = 

Quantity of metal y in alloy A = 

Quantity of metal x in alloy B = 

Quantity of metal y in alloy B = 

According to the question

The ratio of x and y in alloy C = [( × 4) + ( × 5)]/[( × 4) + ( × 5)]

⇒ ( + )/( + )

⇒ ()/()

⇒ ( × 

⇒ 

Now,

Quantity of x in alloy C = 

⇒ 

Percentage of x in alloy C = ( × 100)

⇒ 

⇒ 

∴ The required percentage of x in alloy C is 

Shortcut Trick

 

How much pure alcohol must be added to 400 ml of a solution containing 16% of alcohol to change the concentration of alcohol in the mixture to 40%

  1. 160 ml
  2. 100 ml
  3. 128 ml
  4. 68 ml

Answer (Detailed Solution Below)

Option 1 : 160 ml

Mixture Problems Question 8 Detailed Solution

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Solution quantity = 400ml

Let the quantity of pure alcohol to be added in 400ml be A ml.

alcohol in 400ml solution = 16 × 400/100 = 64ml.

Then,

⇒ 400 × 16/100 + A = (400 + A) × 40/100

⇒ 64 + A = 160 + 2A/5

⇒ 3A/5 = 96

⇒ A = 96 × 5/3

⇒ A = 160

Alternate Method

 

The ratio of solution to pure alcohol = 60 ∶ 24 = 5 ∶ 2

5 units → 400 ml

Then, 2 units → 400/5 × 2 = 160 ml

∴ 160ml pure alcohol to be added to make 40% alcohol in solution.

A container contains 25 litre of milk. From this container, 5 litre of milk is taken out and replaced by water. This process is further repeated two times. How much milk is there in the container now?

  1. 11.5 litre
  2. 14.8 litre
  3. 13.5 litre
  4. 12.8 litre

Answer (Detailed Solution Below)

Option 4 : 12.8 litre

Mixture Problems Question 9 Detailed Solution

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Given:

A container contains 25 litre of milk. From this container, 5 litre of milk is taken out and replaced by water. 

Concept used:

Left Quantity = Initial Quantity(1 - [Fraction removed])N (Where N = Number of times the process is carried out)

Calculation:

Fraction of milk removed = 5/25 = 1/5

Now, the quantity of milk left in the container

⇒ 25(1 - 1/5)3

⇒ 25 × (4/5)3

⇒ 25 × 64/125

⇒ 12.8 litre

∴ There is 12.8 litre of milk left in the container.

Shortcut Trick 

The ratio of milk and water in a container is 2 : 3. When 60 liters of mixture is taken out and replaced with water then the ratio of milk and water becomes 1 : 2. Then find the total capacity of the container. 

  1. 360 liters
  2. 220 liters
  3. 440 liters
  4. 350 liters

Answer (Detailed Solution Below)

Option 1 : 360 liters

Mixture Problems Question 10 Detailed Solution

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Given:

Ratio of milk and water is 2 : 3

60 liters of mixture is taken out

Then ratio of milk and water is 1 : 2

Calculation:

Let the milk and water in the total mixture be 2x and 3x.

⇒ Milk in the total mixture = 2x/5x

⇒ Milk in the total mixture = 2/5

⇒ Water in the total mixture = 3x/5x

⇒ Water in the total mixture = 3/5

In 60 liters of mixture 

⇒ Milk = 2/5 × 60

⇒ Milk = 24 liters

⇒ Water = 3/5 × 60

⇒ Water = 36 liters 

When 60 liters of mixture is taken out, 

Replaced with 60 liters of water.

Then the ratio of Milk and Water is 1 : 2

⇒ (2x – 24) : (3x – 36 + 60) = 1 : 2

⇒ (2x – 24)/(3x + 24) = 1 : 2

⇒ 2(2x – 24) = 1(3x + 24)

⇒ 4x – 48 = 3x + 24

⇒ 4x – 3x = 24 + 48

⇒ x = 72

⇒ Total capacity of container = 2x + 3x

⇒ Total capacity of container = 5x

⇒ Total capacity of container = 5 × 72

⇒ Total capacity of container =  360 liters

∴ Total capacity of the container is 360 liters.

Shortcut Trick 

In a vessel, a mixture of milk and water is in ratio 8 : 7, while in another vessel mixture of milk and water is in ratio 7 : 9. In what ratio mixture of both the vessels should be mixed together so that in the resultant mixture ratio of water and milk becomes 9 : 8?

  1. 135 : 256
  2. 256 : 135
  3. 265 : 129
  4. 129 : 265

Answer (Detailed Solution Below)

Option 1 : 135 : 256

Mixture Problems Question 11 Detailed Solution

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Given:

The ratio of milk and water in the first vessel = 8 : 7

The ratio of milk and water in the second vessel = 7 : 9

The ratio of water and milk in the resultant mixture = 9 : 8

Calculation:

Let x litre of the first mixture and y litre of the second mixture are mixed.

Quantity of milk in x litre of the first mixture = 8x/15

Quantity of milk in y litre of the second mixture = 7y/16

Total quantity of the resultant mixture = (x + y)

Quantity of milk in (x + y) litre of the resultant mixture = 8(x + y)/17

8x/15 + 7y/16 = 8(x + y)/17

⇒ 8x/15 + 7y/16 = 8x/17 + 8y/17

⇒ 8x/15 – 8x/17 = 8y/17 – 7y/16

⇒ (136x – 120x)/15 × 17 = (128y – 119y)/17 × 16

⇒ 16x/15 = 9y/16

⇒ 256x = 135y

⇒ x/y = 135/256

 The required ratio is 135 : 256

Alternative Method:

The concentration of milk in the first mixture = 8/15

The concentration of milk in the second mixture = 7/16

The concentration of milk in the resultant mixture = 8/17

By the rule of Allegation,

⇒ 9/272 : 16/255

⇒ 9 × 255 : 16 × 272

⇒ 9 × 15 : 16 × 16

⇒ 135 : 256     

 The required ratio is 135 : 256.

A dairy farmer's can contains 6 litres of milk. His wife adds some water to it such that milk and water are in the ratio 4 ∶ 1. How many litres of milk should the farmer add so that the milk and water are in the ratio 5 ∶ 1?

  1. 1.5
  2. 1.2
  3. 1.0
  4. 1.8

Answer (Detailed Solution Below)

Option 1 : 1.5

Mixture Problems Question 12 Detailed Solution

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Given:

A dairy farmer's can contains 6 litres of milk.

His wife adds some water to it such that milk and water are in the ratio 4 ∶ 1.

Calculation:

Milk : Water = 4 : 1

Let the quantity of milk and water be 4x and x.

Quantity of Milk = 4x = 6 litres

⇒ x = 1.5 litres

Quantity of Water = x = 1.5 litres

According to the question,

 = 

⇒ 6 + x = 7.5

⇒ x = 7.5 - 6 = 1.5 litres

Alternate Method 

A solution contains a mixture of acid and base in the ratio 17 : 3. How much fraction of the mixture must be drawn off and substituted by the base so that the ratio of acid and base in the resultant mixture in the solution becomes 1 : 1?

  1. 1/17
  2. 7/17
  3. 5/17
  4. 2/17

Answer (Detailed Solution Below)

Option 2 : 7/17

Mixture Problems Question 13 Detailed Solution

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Given:

Initial ratio of acid and base = 17 : 3

Final mixture of acid and base = 1 : 1

Calculation:

Let the acid and base be 17x litres and 3x litres resp

⇒ Total mixture = 20x

Let the drawn part of the mixture be 'y' liters

Acid in (20 - y) litres mixture

⇒ (20x - y) × (17/20) = (340x - 17y)/20         ----(i)

Now adding 'y' litres of base to the mixture

Base in the resultant mixture

⇒ (3/20) × (20x - y) + y = (60x + 17y)/20      ----(ii)

According to the question, ratio of acid and base in resultant mixture is 1:1

Thus, equating Equations (1) and (2)

(340x - 17y)/20 = (60x + 17y)/20

⇒ 340x - 17y = 60x + 17y 

⇒ 34y = 280x

⇒ y/x = 280/34

⇒ y/x = 140/17

Total mixture = 20x = (20 × 17) liters

Mixture to be removed and replaced = y = 140 liters

⇒ Required fraction = (140)/(20 × 17) = 7/17

∴ 7/17 fraction of the mixture must be drawn off and substituted by the base so that the ratio of acid and base in the resultant mixture in the solution becomes  1:1
 Shortcut Trick

Let us remove some quantity of the mixture from the solution.

After that

                   Acid         Base

Initial ratio     17    :       3

Final ratio       1    :       1

We are adding the Base so the quantity of Acid will remain the same.

So multiply the second ratio by 17.

                   Acid         Base

Initial ratio     17    :       3

Final ratio     17    :      17

So Base added = 17 - 3 = 14 units

Here note that the initial quantity of mixture = Final quantity of mixture

So

Initial quantity of mixture = 17 + 17 = 34 units

the required ratio = 14/34 = 7/17

The ratio of sugar to water in solution A is 1  4 and the ratio of salt to water in solution B is 1 26. To make an ORS solution, A and B are mixed in 2 3. Find the ratio of sugar to salt in ORS.

  1. 45 ∶ 16
  2. 52 15 
  3. 18 5
  4. 12 7

Answer (Detailed Solution Below)

Option 3 : 18 5

Mixture Problems Question 14 Detailed Solution

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Given:

The ratio of sugar to water in solution A = 1  4

The ratio of salt to water in solution B = 1  26

Calculation:

First, make the quantity of solution A and solution B same.

Total unit of sugar and water in solution A = 1 + 4 = 5 units

Total unit of salt and water in solution B = 1 + 26 = 27 units

Now, multiply the ratio of solution A by 27 and multiply the ratio of solution B by 5.

The ratio of sugar to water in solution A = 1 × 27 ∶ 4 × 27 = 27 : 108

The ratio of salt to water in solution B = 1 × 5 ∶ 26 × 5 = 5 : 130

Now, mix the solution 2 : 3.

Therefore, multiply the new ratio of solution A by 2 and multiply the new ratio of the solution B by 3.

The new required ratio of solution A = 54 : 216

The new required ratio of solution B = 15 : 390

The ratio of sugar, salt and water in ORS = 54 : 15 : 606

The ratio of sugar and salt = 54 : 15 = 18 : 5

Therefore, "18 : 5" is the required answer.

Shortcut Trick 

In what ratio should water be mixed with wine, that costs Rs. 60 per liter, so that the price of the resultant mixture is Rs.40 per litre?

  1. 2 ∶ 3
  2. 3 ∶ 4
  3. 1 ∶ 2
  4. 4 ∶ 5

Answer (Detailed Solution Below)

Option 3 : 1 ∶ 2

Mixture Problems Question 15 Detailed Solution

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Given: 

The cost price of wine = Rs. 60 per litre

The cost price of water = Rs. 0 per litre

The cost price of mixture = Rs. 40 per litre

Calculations:

Let the quantity of wine and water added in final mixture be x and y respectively.

According to the question:

60 × x + 0y = (x + y) × 40

⇒ 60x  = 40x + 40y

⇒ 60x - 40x = 40y

⇒ 20x = 40y

⇒ x : y = 2 : 1

∴ The ratio in which water and wine should be mixed is 1 : 2.

Alternate Method

Given: 

The cost price of wine = Rs. 60 per litre

The cost price of water = Rs. 0 per litre

The cost price of mixture = Rs. 40 per litre

Concept used:

If two indigents are mixed, then

Calculation:

Using alligation, 

Ratio of wine and water = 40 : 20 = 2 :1

∴ The ratio in which water and wine should be mixed is 1 : 2.

Important Points

 

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