To Make a Mixture from Two Mixtures is an essential sub-topic of Ratio and proportion, which is a topic of Mathematics. Make a Mixture from Two Mixtures questions hold quite a weightage in distinct government examinations and is known for testing the aspirants’ aptitude. Testbook presents Make a Mixture from Two Mixtures MCQs Quiz to help you in analysing your preparation for this section. Nevertheless, to assist you, even more, solutions to all the questions and explanations for the same are also stated.

Option 3 : 20/39

**Given:**

The percentage of alcohol in the bottle of rum = 68%

The percentage of water in another bottle of rum = 71%

The percentage of alcohol after replacement = 48%

**Formula Used:**

Rule of Alligation

If two ingredients are mixed, then

**Calculation:**

The required ratio = 19 : 20

Replaced portion = 20 unit

Total part = (19 + 20) = 39 unit

Replaced part = 20/39

**∴ The quantity of alcohol replaced is 20/39.**

The ratio of milk and water in a given mixture is 4:9. If 20 litre of water is added then the ratio of the mixture becomes 1:3. Then the initial quantity of water in the mixture?

Option 4 : 60 L

**Given:**

The initial and the final ratio of milk and water in a given mixture are 4∶9 and 1∶3. 20 L of water is added to this mixture.

**Calculation:**

Let the initial quantity of milk and water in the mixture be 4x, 9x

After adding 20 L of water the ratio becomes 1:3

⇒ (4x)/(9x + 20) = 1/3

⇒ 12x = 9x + 20

⇒ 3x = 20, x = 20/3

Now the initial quantity of water in the mixture = 9×(20/3) = 60 L

**∴ Initial quantity of water in the mixture is 60L.**

Option 2 : 39 litre

**Given**

Quantity of A : Quantity of B = 3 : 2

**Calculation**

Let initial quantity of A and B be 3x and 2x.

Total 5 lt of mixture is drown out.

Hence, quantity of A drown out = 5 × (3/5) = 3 lt

And, quantity of B drown out = 5 × (2/5) = 2 lt

According to question

(3x - 3) - (2x - 2) = 12

⇒ x - 1 = 12

⇒ x = 13

∴ Initial quantity of A = 3x = 3 × 13 = 39 litres.

Option 4 : 47.50%

**Given:**

In a 15 litre solution 40% is acid and 60% is water and in 25 litre solution 60% is acid and 40% is water

**Concept Used:**

Concept of percentage.

The percentage is calculated based on 100

For example, 4 out of 10 means 40 out of 100 i.e. 40% of 10

**Calculation:**

In 15 litre solution, 40% is acid and 60% water.

That means, quantity of acid = (40/100) × 15

⇒ 6 litres

60% water that means (60/100) × 15

⇒ 9 litres

In 25 litre acid is 60% and water is 40%

That means, quantity of acid = (60/100) × 25

⇒ 15 litres

40% water means quantity of water = (40/100) × 25

⇒ 10 litre

From the above, acid in resultant mixture = (6 + 15) = 21 litters

Water in resultant mixture = (9 + 10) = 19 litters

Total resultant mixture = (15 + 25) = 40 litters

Percentage of water in resultant mixture = (19/40) × 100

⇒ 47.50%

**∴**** Percentage of water in resultant mixture is 47.50%**

Option 4 : 4 : 5

**Given:**

First vessel A contain spirit and water in the ratio 2 : 7

Second vessels B contain spirit and water in the ratio 5: 4

**Calculation:**

Let x letre of first mixture and 2x of second mixture is taken of,

As we know that if an amount of mixture is taken of then it is also in same ratio.

According to the question:

Spirit = (x × 2/9) = 2x/9 in first mixture

Water = (x × 7/9) = 7x/9 in first mixture

Spirit = (2x × 5/9) = 10x/9 in second mixture

Water = (2x × 4/9) = 8x/9 in second mixture

Now,

Total sprit = 2x/9 + 10x/9 = 12x/9

Total water = 7x/9 + 8x/9 = 15x/9

Ratio of spirit and water = 12x/9 : 15x/9 = 4 : 5

**∴ The ratio of spirit and water in the new mixture is 4 : 5.**

In 84 ml solution of beer and wine, 3/7 part of solution is beer and rest is wine, now 'x' ml of mixture is taken out 8 ml of beer is added now the concentration of beer and wine in the solution is same. Find the value of 'x'.

Option 1 : 28 ml

Quantity of beer in the solution = 3/7 × 84 = 36 ml

Quantity of wine in the solution = 4/7 × 84 = 48 ml

The quantity of solution replaced will be removed in the same ratio as the contents in the solution

Quantity of beer left after removal = 36 – (x × 36/84) = 36 – 3x/7

Quantity of wine left after removal = 48 – (x × 48/84) = 48 – 4x/7

Now, 8 ml of beer is added in the mixture and the concentration of beer and wine in the solution is same

∴ 36 – 3x/7 + 8 = 48 – 4x/7

⇒ x/7 = 4

⇒ x = 28

**∴ Quantity of mixture removed is 28 ml**

Option 4 : 43 ∶ 47

GIVEN :

Two equal vessels are filled with a mixture of water and glycerine in the ratio 2 ∶ 3 and 5 ∶ 4 respectively.

CONCEPT :

Mixture Problems

CALCULATION :

Let the initial quantity of both the mixtures be:

LCM of [(2 + 3), (5 + 4)] = 45 litres

So, the amount of water in the first vessel = (2/5) × 45 = 18 litres

The amount of glycerine in the first vessel = (3/5) × 45 = 27 litres

So, the amount of water in the second vessel = (5/9) × 45 = 25 litres

The amount of glycerine in the second vessel = (4/9) × 45 = 20 litres

When the two mixtures are mixed together,

The resulting quantity of water = 18 + 25 = 43 litres

The resulting quantity of glycerine = 27 + 20 = 47 litres

**∴ The resulting ratio of water and glycerine is 43 ∶ 47**

A = (2 ∶ 3) × (5 + 4) = 18 ∶ 27 ---(1)

B = (5 ∶ 4) × (2 + 3) = 25 ∶ 20 ---(2)

**On adding (1) and (2), we get the ratio of the Water and glycerine in the third vessel = 43 ∶ 47 **

Option 2 : 35 ml

Let the quantity of milk and water in the mixture be x and 56 – x

Now, 16ml of the solution is is removed

Quantity of solution replaced will be removed in the same ratio as the contents in the solution

∴ Quantity of milk left after removal = x – (16 × x/56) = 5x/7

Quantity of water left after removal = 56 – x – {16 × (56 – x)/56} = 40 – 5x/7

Ratio of milk and water after adding 5 ml of water is 5 ∶ 4

⇒ (5x/7)/(40 + 5 – 5x/7) = 5/4

⇒ 20x = 45 × 7 × 5 – 25x

⇒ x = 7 × 5 = 35

**∴ Concentration of milk in the mixture initially was 35 ml**

Option 3 : Rs. 46.57

**Given:**

10 kg sugar at price of Rs. 32 per kg

7 kg precious sugar of price Rs. 38 per kg

**Calculation:**

__Shortcut Trick__

Let the price of precious sugar per kg be Rs. ‘x’

⇒ 6/(x - 38) = 7/10

⇒ 60 = 7x – 266

⇒ 7x = 326

⇒ x = 46.57

∴ The price of precious sugar = Rs. 46.57

__Alternate Method__

Let the price of precious sugar be Rs. 'x'

Then, (32 × 10 + x × 7)/(10 + 7) = 38

⇒ 320 + 7x = 38 × 17

⇒ 7x = 646 - 320

⇒ x = 326/7 = 46.57

∴ The price of precious sugar = Rs. 46.57

Option 5 : 12 liter

1/ 4^{th} of the mixture is taken out

Means quantity of milk has taken out

⇒ 1/ 4^{th} of 44 = 11 liter

And the quantity of water has taken out

⇒ 1/ 4^{th} of 20 = 5 liter

Remaining Quantities of milk and water is 33 liter of milk and 15 liters of water

Now we have to make it in the ratio of 3:1 by adding milk

⇒ 33 + x/ 15 = 3/1

⇒ x = 12 liter

Option 4 : 90 L

**GIVEN:**

Salt in initial mixture = 15%

Salt in final mixture = 25%

Final mixture quantity = 102 L

**CALCULATION:**

Let the initial Quantity of mixture be "x"

Initial Ratio of Salt to Water = 15 : 85 = 3 : 17

Final Ratio of Salt to Water = 25 : 75 = 1 : 3

According to question, the quantity of water remains the same

Thus, initial and final ratios become

Initial Ratio of Salt to Water = 9 : 51

Final Ratio of Salt to Water = 17 : 51

Total units in final mixture = 68 units = 102 L

Initial units of mixture = 9 + 51 = 60 units

Initial Quantity of mixture = 60 × (102/68)

∴ Initial Quantity of mixture = 90 L

Option 1 : 52.5%

**Given: **

The ratio of rice and wheat in the first mixture is 2 : 3

The ratio of rice and wheat in the second mixture is 3 : 2

New mixture is a mixture of old mixtures in the ratio of 3 : 5.

**Calculation: **

⇒ Percentage of rice in first mixture = (2/5) × 100

= 40%

⇒ Percentage of rice in second mixture = (3/5) × 100

= 60%

⇒ Percentage of rice in new mixture = (3 × 40 + 5 × 60) / (3 + 5)

⇒ 420/8

⇒ 52.5%

Option 1 : 7 : 17

**GIVEN:**

1/3^{rd} and 1/4^{th} parts of two cans A and B of equal volume are filled with milk.

**CONCEPT:**

LCM: It is a number that is a multiple of two or more numbers.

**CALCULATION:**

Suppose the capacity of cans = 12 L (LCM of 3 and 4)

Amount of milk in can A = 12/3 = 4 L

⇒ Amount of water in can A = 12 – 4 = 8 L

Amount of milk in can B = 12/4 = 3 L

⇒ Amount of water in can B = 12 – 3 = 9 L

When both the cans are poured into a new pot:

∴ Ratio of Milk and Water = (4 + 3) : (8 + 9) = 7 : 17Option 1 : 3 : 5

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

Quantity of spirit in x litre of the first mixture = 5x/6

Quantity of spirit in y litre of second mixture = 3y/10

The total quantity of the resultant mixture = (x + y)

Quantity of spirit in (x + y) litre of the resultant mixture = (x + y) /2

5x/6 + 3y/10 = (x + y) /2

⇒ (25x + 9y) /30 = (x + y) /2

⇒ 25x + 9y = 15 × (x + y)

⇒ 25x + 9y = 15x + 15y

⇒ 10x = 6y

⇒ x/y = 3/5

∴ required ratio = 3 : 5

**Alternative method : **

Concentration of spirit in the first mixture = 5/6

Concentration of spirit in the second mixture = 3/10

Concentration of spirit in the resultant mixture = 1/2

By rule of allegation,

∴ Required ratio = (1/5) : (1/3) = 3 : 5

Option 2 : 30 litres

Ratio of milk and water in a mixture is = 4x : 3x

According to the question

4x/(3x + 2) = 8/7

⇒ 4x × 7 = 8 (3x + 2)

⇒ 28x = 24x + 16

⇒ 28x – 24x = 16

⇒ 4x = 16

⇒ x = 16/4

⇒ x = 4

Quantity of mixture in the last = (4x + 3x) = 7x = 7 × 4 = 28 litres

Quantity of mixture in the last = 28 + 2 = 30 litresOption 2 : 50 litres

**Given:**

Initial Mixture = 170liters

The ratio milk to water = 12 : 5

Initial quantity = 170 liters

After 20% mixture was taken out

Quantity = 170 – 20% of 170 = 136 liters

Milk : water = 12 : 5

Water = 136 × 5/17 = 40 liter

After adding 10 liters more

∴ Required quantity = 40 + 10 = 50 liters

Calculation:

Initial Mixture = 170 liters

Initial quantity of milk = 12/(12 + 5 )× 170 = 120 liters

Initial quantity of water = 170 - 120 = 50 liters

Final quantity of water = Initial quantity of water - taken out quantity of water + added quantity of water

Final quantity of water = 50 - 20/100 × 50 + 10 = 50 - 10 + 10 = 50 liters

**∴ The quantity of water in final mixture is 50 liters**

Option 4 : 17 : 35

Let the two alloys be mixed in the ratio of x:y

∴ Brass and Aluminum content in mixture from Alloy X will be (8x/17) and (9x/17) respectively.

Similarly, Brass and Aluminum content in mixture from alloy Y will be (y/7) and (6y/7) respectively.

⇒ Total Brass content in mixture = (8x/17) + (y/7)

⇒ Total Aluminum content in mixture = (9x/17) + (6y/7)

According to the condition given in the problem, Brass content in mixture = 25%

∴ Aluminum content in mixture = 100 – 25 = 75%

Ratio of Brass to Aluminium in the final mixture = 25 : 75 = 1:3

∴ [(8x/17) + (y/7)] / [(9x/17) + (6y/7)] = 1 / 3

⇒ 3 × [(8x/17) + (y/7)] = [(9x/17) + (6y/7)]

⇒ [(24x/17) + (3y/7)] = [(9x/17) + (6y/7)]

⇒ x × [(24/17) – (9/17)] = y × [(6/7) – (3/7)]

⇒ (15/17)x = (3/7)y

⇒ x/y = 17/35

∴ x ∶ y = 17 ∶ 35

__Alternate Method__

**1.**

Let the quantity of the first mixture be x and the second mixture be y.

Quantity of brass in the first mixture = 8x/17

Quantity of brass in the second mixture = y/7

Quantity of mixture in resultant mixture = (1/4) × (x + y)

Then, 8x/17 + y/7 = (1/4) × (x + y)

Or, 15x/(17 × 4) = 3y/(7 × 4)

Then, x/y = 17/35.

**2.**

Let us try allegation method to solve the question.

The ratio in which the alloys are mixed = 3/(7 × 4) ∶ 15/(17 × 4)

⇒ 1/7 ∶ 5/17 = 17 ∶ 35

Option 1 : 20 kg

**Given**

Per kg price wheat_{1} = Rs. 6

Per kg price wheat_{2} = Rs. 3

Total wheat = 40 kg

Total price 40kg wheat is = Rs. 180

**Concept Used:**

Mixture of two quantities.

**Calculation:**

Let the required quantity of the first type of wheat = x kg

So, the quantity of second type = (40 - x) kg

According to the question,

⇒ 6x + 3(40 - x) = 180

⇒ 6x + 120 - 3x = 180

⇒ 3x = 60

⇒ x = 20

**∴ The required quantity of the first type of wheat is 20 kg**

Option 4 : 27%

**Given:**

The mixture A contains milk and water in the ratio of 3 : 7.

The mixture B contains milk and water in the ratio of 1 : 3.

**Calculation:**

Let the quantity of mixture C be 100.

The ratio of mixture A to the mixture B in the mixture C = 4 : 6 = 2 : 3

The quantity of milk of mixture A poured in mixture C = 3/(3 + 7) × 2/(2 + 3) = 3/10 × 2/5 = 3/25

The quantity of milk of mixture B poured in mixture C = 1/(1 + 3) × 3/(2 + 3) = 1/4 × 3/5 = 3/20

The quantity of milk in mixture C = 3/25 + 3/20 = (12 + 15)/100 = 27/100

The percentage of milk in the mixture C = 27/100 × 100 = 27%

**∴ The percentage of milk in the mixture C is 27%.**

Option 3 : 4 : 5

By rule of allegation,

∴ Required ratio = 1/30 : 1/24 = 4 : 5