# Mixture and Alligation Questions and Answers updated daily – Aptitude

Mixture and Alligation Questions: Solved 187 Mixture and Alligation Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Mixture and Alligation Questions

61. How many kg of sugar worth Rs. 25 per kg must be blended with 30 kg of sugar worth Rs. 30 per kg so that by selling the blended variety at Rs. 30 per kg there should be gain of 10%?

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Correct Ans:36 kg

Explanation:

CP of mixture of sugar = SP*[100/(100 + profit)]

= 30*[100/(100 + 10)]

= 300/11 rupee/kg.

According to allegation of mixture,

Quantity of Cheaper/Quantity of Dearer = (CP of Dearer - Mean price)/(Mean price - CP of Cheaper)

Let sugar worth Rs. 25 per kg be Sugar1 and sugar worth Rs. 30 per kg be Sugar2.

Sugar1/Sugar2 = (30/11)/(25/11) = 30/25 = 6/5

Therefore, quantity of sugar1= (6/5)*30 = 36 kg.

Workspace

62. A container was full of champagne. A person used to draw out 20% of the champagne from the container and replaced it with sugar solution. He has repeated the same process 3 more times and thus there was only 512 gm of champagne left in the container, the rest part of the container was filled with the sugar solution. The initial amount of the champagne in the jar was:

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Correct Ans:1.25 kg

Explanation:

Let the initial amount of milk be x kg.

Champagne left in container after n operations= x(1- (y/x))â¿

Here, number operations = 4

Quantity of champagnedrawn out, y = 20% = 1/5

Champagne left in container = 521 gm

(512/1000) = x[1 - (1/5)]â´

(512/1000) = x(4/5)â´

(512/1000) = (256x/625)

x = (512*625)/(1000*256)

x = 1.25 kg

Champagne left in container after n operations= x(1- (y/x))â¿

Here, number operations = 4

Quantity of champagnedrawn out, y = 20% = 1/5

Champagne left in container = 521 gm

(512/1000) = x[1 - (1/5)]â´

(512/1000) = x(4/5)â´

(512/1000) = (256x/625)

x = (512*625)/(1000*256)

x = 1.25 kg

Workspace

63. From 10 kgs of sunflower seeds, 2.5 litres of oil can be extracted. A person has 45 kg of sunflower seeds but he mixes the oil extracted from them with coconut oil in the ratio of 2 : 3 respectively then how many litres of coconut oil will be required to make the mixture of sunflower oil and coconut oil?

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Correct Ans:16.875 litres

Explanation:

From 10 kgs of sunflower seeds, 2.5 litres of oil can be extracted.

Therefore, from 45 kgs of sunflower seeds,

10 kgs - 2.5 litres

45 kgs - ?

The quantity of oil = (2.5*45)/10 = 11.25 litres

In the mixture, let the quantity of sunflower oil be 2x litres.

2x = 11.25

x = 5.625 litres

Therefore, the quantity of coconut oil = 3x = 3*5.625

= 16.875 litres.

Therefore, from 45 kgs of sunflower seeds,

10 kgs - 2.5 litres

45 kgs - ?

The quantity of oil = (2.5*45)/10 = 11.25 litres

In the mixture, let the quantity of sunflower oil be 2x litres.

2x = 11.25

x = 5.625 litres

Therefore, the quantity of coconut oil = 3x = 3*5.625

= 16.875 litres.

Workspace

64. Copper and Zinc are in the ratio 2 : 3 in 200 gms of an alloy. The quantity (in grams) of copper to be added to it to make the ratio 3 : 2 is:

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Correct Ans:100 gm

Explanation:

Given that quantity of alloy = 200 gm.

Ratio of copper and zinc = 2 : 3

Then quantity of copper = (2/5)*200 = 80 gm

Quantity of zinc = (3/5)*200 = 120 gm

Let assume, the quantity of copper to be added as x.

(80 + x)/120 = 3/2

160 + 2x = 360

2x = 200

x =100 gm

Ratio of copper and zinc = 2 : 3

Then quantity of copper = (2/5)*200 = 80 gm

Quantity of zinc = (3/5)*200 = 120 gm

Let assume, the quantity of copper to be added as x.

(80 + x)/120 = 3/2

160 + 2x = 360

2x = 200

x =100 gm

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65. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:

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Correct Ans:Rs. 175.50

Explanation:

Since first and second varieties are mixed in equal proportions.

So, their average price = Rs. (126 + 135) / 2

=

So, the mixture is formed by mixing two varieties, one (variety 1 and 2) at Rs. 130.50 per kg and the other (variety 3) at say, Rs. x per kg in the ratio (1 + 1): 2 ie., 2 : 2 => 1:1

We have to find the

By the rule of alligation we have,

Therefore, Quantity of 1st kind / Quantity of 2nd kind = 1 / 1

=> (x - 153) / 22.50 = 1/1

=>(x - 153) / 22.50 = 1

=> x = 22.50 + 153

=>

So, their average price = Rs. (126 + 135) / 2

=

**Rs. 130.50**So, the mixture is formed by mixing two varieties, one (variety 1 and 2) at Rs. 130.50 per kg and the other (variety 3) at say, Rs. x per kg in the ratio (1 + 1): 2 ie., 2 : 2 => 1:1

We have to find the

**price of the third variety per kg ie., "x"**By the rule of alligation we have,

Therefore, Quantity of 1st kind / Quantity of 2nd kind = 1 / 1

=> (x - 153) / 22.50 = 1/1

=>(x - 153) / 22.50 = 1

=> x = 22.50 + 153

=>

**x = 175.50**

Thus,price of the third variety per kg is Rs.175.50Thus,price of the third variety per kg is Rs.175.50

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66. Vijay purchased two different kinds of alcohol. In the first mixture, the ratio of alcohol to water is 3 : 4 and the second mixture it is 5 : 6. If he mixes, the two given mixtures and makes a third mixture of 18 litres in which the ratio of alcohol to water is 4 : 5, the quantity of the first mixture (whose ratio is 3 : 4) that is required to make 18 litres of the third kind of mixture is:

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Correct Ans:7

Explanation:

Given

Workspace

67. An alloy contains only zinc and copper. One such alloy weighing 15 gm contains zinc and copper in the ratio of 2 : 3 by weight. If 10 gm of zinc is added then find what amount of copper has to be removed from the alloy such that the final alloy has zinc and copper in the ratio of 4 : 1 by weight?

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Correct Ans:5 gm

Explanation:

Given

1st Alloy Zinc

= (2 / 5 ) * 15

=> 6

Copper => (3 / 5 ) * 15

=> 9

We get

6 + 10 / (9 - x) = 4 / 1

=> 6 + 10 = 36 - 4x

=> 16 - 36 = - 4x

=> -20 = -4x

=> x = 5gms.

1st Alloy Zinc

= (2 / 5 ) * 15

=> 6

Copper => (3 / 5 ) * 15

=> 9

We get

6 + 10 / (9 - x) = 4 / 1

=> 6 + 10 = 36 - 4x

=> 16 - 36 = - 4x

=> -20 = -4x

=> x = 5gms.

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68. A vessel is filled with 120 litres of Chemical solution, Acid “A”. Some quantity of Acid “A” was taken out and replaced with 23 litres of Acid “B” in such a way that the resultant ratio of the quantity of Acid “A” to Acid “B” is 4:1. Again 23 litres of the mixture was taken out and replaced with 28 litre of Acid “B”. What is the ratio of the Acid “A” to Acid “B” in the resultant mixture?

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Correct Ans:46 : 29

Explanation:

In 23 litre mixture, Quantity of Acid “B” = 23 * 1/5 = 4.6 litre

Acid “A” in the mixture = 23 – 4.6 = 18.4 litre

120 – x / 23 = 4 / 1

x = 28

Ratio = 92-18.4 : 18.4 + 28

Ratio = 46 : 29

Acid “A” in the mixture = 23 – 4.6 = 18.4 litre

120 – x / 23 = 4 / 1

x = 28

Ratio = 92-18.4 : 18.4 + 28

Ratio = 46 : 29

Workspace

69. There are two containers on a table. A and B. A is half full of wine, while B,which is twice A's size, is one quarter full of wine. Both containers are filled with water and the contents are poured into a third container C. What portion of container C's mixture is wine?

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Correct Ans:33.33%

Explanation:

Let the size of container A is "x"

then B's size will be "2x"

A is half full of wine ⇒x2

So remaining "x2 " of A contains water B is quarter full of win

⇒2x4

⇒x2

So remaining ⇒2x–x2=3x2

So 3x2 of B contains water.

Totally C has A's content + B's Content

= x + 2x = 3x

Wine portion in C = x2 of "A" + x2 of

"B" x portion of wine Water portion in C

= x2 of "A" + 3x2 of "B"

⇒4x2

⇒2x portion of water

So portion of wine in C is

x3x=13 portion of wine

if 1/3 expressed in % 13×100 = 33.33%

Ans : 33.33% of wine.

then B's size will be "2x"

A is half full of wine ⇒x2

So remaining "x2 " of A contains water B is quarter full of win

⇒2x4

⇒x2

So remaining ⇒2x–x2=3x2

So 3x2 of B contains water.

Totally C has A's content + B's Content

= x + 2x = 3x

Wine portion in C = x2 of "A" + x2 of

"B" x portion of wine Water portion in C

= x2 of "A" + 3x2 of "B"

⇒4x2

⇒2x portion of water

So portion of wine in C is

x3x=13 portion of wine

if 1/3 expressed in % 13×100 = 33.33%

Ans : 33.33% of wine.

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70. How many kgs of wheat costing Rs.24/- per kg must be mixed with 30 kgs of wheat costing Rs.18.40/- per kg so that 15% profit can be obtained by selling the mixture at Rs.23/- per kg?

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Correct Ans:12

Explanation:

Given

S.P. of 1 kg mixture = Rs.23

Gain = 15%

= {100/ (100 + 15)} x 23

= {100 / 115} x 23

= 20

Hence,

Let the total quantity of wheat costing Rs.24/- per kg is

Then, total quantity of wheat in resultant mixture = (x + 30) kgs

Now,

=

Total C.P of variety 1 (ie., x kg of wheat costing Rs.24/- per kg) = x * 24

Total C.P of variety 2 (ie., 30 kg of wheat costing Rs.18.40/- per kg) = 30 * 18.40

= 552

Hence, Total C.P of variety 1 + Total C.P of variety 2 = Total C.P of resultant mixture

But We have --> Total C.P of resultant mixture = 20 * (x + 30)

Then, Total C.P of variety 1 + Total C.P of variety 2 = 20 * (x + 30)

=> (x * 24) + 552 = 20 * (x + 30)

=> 24x + 552 = 20x + 600

=> 4x = 600 - 552

=> 4x = 48

=>

Thus, the required quantity of wheat costing Rs.24/- per kg = 12 kg

S.P. of 1 kg mixture = Rs.23

Gain = 15%

__-__**Formula:****C.P. of 1 kg mixture = {100/ (100 + Gain %)} x S.P.**= {100/ (100 + 15)} x 23

= {100 / 115} x 23

= 20

Hence,

**C.P. of 1 kg mixture = Rs.20**Let the total quantity of wheat costing Rs.24/- per kg is

**x kgs**Then, total quantity of wheat in resultant mixture = (x + 30) kgs

Now,

**Total C.P of resultant mixture**= C.P. of 1 kg mixture * total quantity of wheat in resultant mixture=

**20 * (x + 30)**Total C.P of variety 1 (ie., x kg of wheat costing Rs.24/- per kg) = x * 24

Total C.P of variety 2 (ie., 30 kg of wheat costing Rs.18.40/- per kg) = 30 * 18.40

= 552

Hence, Total C.P of variety 1 + Total C.P of variety 2 = Total C.P of resultant mixture

But We have --> Total C.P of resultant mixture = 20 * (x + 30)

Then, Total C.P of variety 1 + Total C.P of variety 2 = 20 * (x + 30)

=> (x * 24) + 552 = 20 * (x + 30)

=> 24x + 552 = 20x + 600

=> 4x = 600 - 552

=> 4x = 48

=>

**x = 12 kg**Thus, the required quantity of wheat costing Rs.24/- per kg = 12 kg

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71. The average marks of 10 students of a class is 72 and the average marks of another 12 students of the same class is 75, find the overall average of the class.

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Correct Ans:73.64

Explanation:

**Given**

Average marks of 10 students = 72

**Average marks = Total marks / No of students**

**Total marks of 10 students = ( Average marks of 10 students ) * ( No of students )**

= 10 x 72 = 720

Given , Average marks of

**another**12 students = 75

Total marks of

**another**12 students = 12 x 75 = 900

Total marks of the class = 900 + 720 = 1620

Total number of students in the class = 10 + 12 = 22

**Overall average of the class = (Total marks of the class) / (Total number of students in the class)**

= 1620 / 22

**= 73.64**

Workspace

72. A 20 liters mixture of milk and water comprising 60% pure milk is mixed with "x" liters of pure milk. The new mixture comprises 80% milk. What is the value of "x"?

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Correct Ans:20

Explanation:

Original mixture comprises 20 liters of milk and water.

Out of the 20 liters, 60% is pure milk.

=> ( 60 / 100 ) x 20 = pure milk

=>

In 20 liters mixture remaining 8 liters = water

When "x" liters of pure milk is added to 20 liters of mixture

New mixture = ( 20 + x ) liters

Milk in new mixture = ( 12 + x ) liters

Given milk in new mixture = 80% of ( 20 + x )

=> 12 + x = ( 80 / 100 ) * ( 20 + x )

=> 12 + x = ( 4 / 5 ) * ( 20 + x )

=> 5 ( 12 + x ) = 4 ( 20 + x )

=> 60 + 5 x = 80 + 4 x

=> 5 x - 4 x = 80 - 60

Out of the 20 liters, 60% is pure milk.

=> ( 60 / 100 ) x 20 = pure milk

=>

**12 liters = pure milk**In 20 liters mixture remaining 8 liters = water

When "x" liters of pure milk is added to 20 liters of mixture

New mixture = ( 20 + x ) liters

Milk in new mixture = ( 12 + x ) liters

Given milk in new mixture = 80% of ( 20 + x )

=> 12 + x = ( 80 / 100 ) * ( 20 + x )

=> 12 + x = ( 4 / 5 ) * ( 20 + x )

=> 5 ( 12 + x ) = 4 ( 20 + x )

=> 60 + 5 x = 80 + 4 x

=> 5 x - 4 x = 80 - 60

**=> x = 20 liters**
Workspace

73. A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?

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Correct Ans:50

Explanation:

**Solution is:**

Given , Zoo had either pigeons or horses

Heads of the animals in Zoo = 80

=> pigeons + horses = 80

Let p = number of pigeons

h = number of horses

=> p = 80 - h

Given , Legs of the animals = 260

Each pigeon has 2 legs and each horse has 4 legs

=> 2p + 4h = 260

Substitute

**p = 80 - h**

=> 2 ( 80 - h ) + 4h = 260

=> 160 - 2h + 4h = 260

=> 2h = 260 - 160

=> 2h = 100

**h = 50**

So , number of horses in the Zoo =

**50**

Workspace

74. How many liters of a 12 litre mixture containing milk and water in the ratio of 2 : 3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion?

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Correct Ans:2 liters

Explanation:

The mixture contains 40% milk and 60% water in it.

Given , Mixture ( milk + water ) = 12 liters

milk ( m ) / water ( w ) = 2 / 3

=> m / w = 2x liters / 3x liters

so, 2x + 3x = 12 liters

5x = 12

x = 12 / 5

milk , m = 2x liters

= 2 * 2.4 liters

water , w = 3x liters

= 3 * 2.4 liters

That is 4.8 liters of milk and 7.2 liters of water.

Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.

That is wewill end up with 6 litres of milk and 6 litres of water.

=> Water gets reduced by 1.2 litres.

=> To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove (1.2 / 0.6) litres of the mixture =

Given , Mixture ( milk + water ) = 12 liters

milk ( m ) / water ( w ) = 2 / 3

=> m / w = 2x liters / 3x liters

so, 2x + 3x = 12 liters

5x = 12

x = 12 / 5

**x = 2.4**milk , m = 2x liters

= 2 * 2.4 liters

**m = 4.8 liters**water , w = 3x liters

= 3 * 2.4 liters

**w = 7.2 liters**That is 4.8 liters of milk and 7.2 liters of water.

Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%.

That is wewill end up with 6 litres of milk and 6 litres of water.

=> Water gets reduced by 1.2 litres.

=> To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove (1.2 / 0.6) litres of the mixture =

**2 litres.**
Workspace

75. 48 liters of a mixture has 75% alcohol. How much water must be added to it to get 60% alcohol concentration ? (in liters)

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Correct Ans:12

Explanation:

**Solution is**

Given In 48 litres => Alcohol = 75 %

=> Alcohol (in litres) = ( 75 / 100 ) x 48

= 36000 / 100

= 36 litres

Let x litres of water is added to this mixture to get 60 % alcohol concentration

So Total quantity of new mixture = ( 48 + x ) litres

Workspace

76. 40 liters of mixture has 65% milk. How much milk should be added to the mixture to make it 85% pure?(in liters)

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Correct Ans:53

Explanation:

Original mixture comprises 40 liters of milk and water.

Out of the 40 liters, 65% is pure milk.

=> ( 65 / 100 ) x 40 = pure milk

=>

In 40 liters mixture, remaining 14 liters = water

When "

New mixture = ( 40 + x ) liters

Milk in new mixture = (26 + x) liters

Given milk in new mixture = 85% of (40 + x)

=> 26 + x = (85/ 100) * (40 + x)

=> 100 * (26 + x) = 85 *(40 + x)

=> 2600 + 100x = 3400 + 85x

=> 100x - 85x = 3400 - 2600

=> 15x = 800

=> x = 800/15

=> x = 53.33

=>

Hence we need to "

Out of the 40 liters, 65% is pure milk.

=> ( 65 / 100 ) x 40 = pure milk

=>

**26 liters = pure milk**In 40 liters mixture, remaining 14 liters = water

When "

**x**" liters of pure milk is added to 40 liters of mixture, to make new mixture 85% pure milk.New mixture = ( 40 + x ) liters

Milk in new mixture = (26 + x) liters

Given milk in new mixture = 85% of (40 + x)

=> 26 + x = (85/ 100) * (40 + x)

=> 100 * (26 + x) = 85 *(40 + x)

=> 2600 + 100x = 3400 + 85x

=> 100x - 85x = 3400 - 2600

=> 15x = 800

=> x = 800/15

=> x = 53.33

=>

**x= 53 (approximately)**Hence we need to "

**add 53 liters of pure milk" to the 40 liters of mixtureto make it 85 % pure.**
Workspace

77. If the average income of a family of ‘x’ members is Rs. 25000 while average income of another family of 6 members is Rs. 32000, then find the value of ‘x’ if the average income of both families is Rs. 28000.

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Correct Ans:8

Explanation:

**Given**

Avg income of family one = 25

Avg income of family two = 32

Avg income of both the families = 28

32 - 28 = 4 ; 28 - 25 = 3

Required ratio is 4 : 3

Ratio of the number members in family 1 and 2 is 4 : 3

=>number of members in family 1 = 4x

and, number of members in family 2 = 3x

Given, In 2nd family there are 6 members => 3x = 6

=>x = 2

In family 1, no. of people = 4x= 4 x 2 =

**8**

Therefore, the value of x (ie., number of members in 1st Family) = 8

Therefore, the value of x (ie., number of members in 1st Family) = 8

Workspace

78. Two liquids A and B are mixed together in the ratio 2 : 3. The average cost of liquid B is Rs 30 per liter and the average cost of the mixture is Rs. 25 per liter, then find the average cost of liquid A per liter.

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Correct Ans:17.5

Explanation:

**Solution is**

Given liquids A and B = 2 : 3

Average cost of mixture = Rs. 25 / liter

Average cost of liquid B = Rs. 30 / liter

Let Average cost of liquid A = x per liter

According to the alligation equation,

( 30 - 25 ) / ( 25 - x ) = 2 / 3

=> 5 / ( 25 - x ) = 2 / 3

=> 5 x 3 / 2 = ( 25 - x )

=> 15 / 2 = 25 - x

=> 7.5 = 25 - x

=> x = 25 - 7.5

**x = 17.5**

So, Average cost of liquid A =

**17.5 per liter**

Workspace

79. Two liquids A and B are mixed together in the ratio 1 : 3. The average cost of liquid B is Rs 30 per liter and the average cost of the mixture is Rs. 25 per liter, then find the average cost of liquid A per liter.

SHOW ANSWER

Correct Ans:10

Explanation:

Given liquids A and B = 1 : 3

Average cost of mixture = Rs. 25 / liter

Average cost of liquid B = Rs. 30 / liter

Let Average cost of liquid A = x per liter

According to the alligation equation,

( 30 - 25 ) / ( 25 - x ) = 1 / 3

=> 5 / ( 25 - x ) = 1 / 3

=> 5 x 3 = ( 25 - x )

=> 15 = 25 - x

=> x = 25 - 15

=>

So,

Average cost of mixture = Rs. 25 / liter

Average cost of liquid B = Rs. 30 / liter

Let Average cost of liquid A = x per liter

According to the alligation equation,

( 30 - 25 ) / ( 25 - x ) = 1 / 3

=> 5 / ( 25 - x ) = 1 / 3

=> 5 x 3 = ( 25 - x )

=> 15 = 25 - x

=> x = 25 - 15

=>

**x = 10**So,

**Average cost of liquid A =10 per liter**
Workspace

80. Two liquids A and B are mixed together in the ratio 6 : 7. The average cost of liquid A is Rs 65 per liter and the average cost of the mixture is Rs. 60 per liter, then find the average cost of liquid B per liter.

SHOW ANSWER

Correct Ans:Rs. 64.29

Explanation:

Given liquids A and B = 6 : 7

Average cost of mixture = Rs. 60/ liter

Average cost of liquid A = Rs. 65 / liter

Let

According to the alligation equation,

=> (x - 60) / (65 - 60) = 6 / 7

=> (x - 60) / 5 = 6/7

=> (x - 60) = (6/7) * 5

=> (x - 60) = 30 /7

=> x = (30 /7) + 60

=> x = (30 + 420) /7

=> x = 450 /7

=>

So,

Average cost of mixture = Rs. 60/ liter

Average cost of liquid A = Rs. 65 / liter

Let

**Average cost of liquid B = x per liter**According to the alligation equation,

=> (x - 60) / (65 - 60) = 6 / 7

=> (x - 60) / 5 = 6/7

=> (x - 60) = (6/7) * 5

=> (x - 60) = 30 /7

=> x = (30 /7) + 60

=> x = (30 + 420) /7

=> x = 450 /7

=>

**x = 64.29**So,

**Average cost of liquid B = Rs. 64.29per liter.**
Workspace

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