# 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

41. A mixture contains water and milk in the ratio of 3 : 5. If 4 litres of water and 5 litres of milk are added to the mixture, the ratio of water and milk becomes 2 : 3. What is the quantity of water in the final mixture?

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Correct Ans:10 liters

Explanation:

Let the quantity of water and milk be 3x and 5x respectively.

It is given that, (3x + 4) / (5x + 5) = 2 / 3

---> 3 * (3x + 4) = 2 * (5x + 5)

---> 9x + 12 = 10x + 10

---> 10x - 9x = 12 - 10

--->

Therefore, the

It is given that, (3x + 4) / (5x + 5) = 2 / 3

---> 3 * (3x + 4) = 2 * (5x + 5)

---> 9x + 12 = 10x + 10

---> 10x - 9x = 12 - 10

--->

**x = 2**Therefore, the

**quantity of water in the final mixture**= (3x + 4) = (3 * 2 + 4) litres =**10 litres**.
Workspace

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

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Correct Ans:20 litres

Explanation:

The quantity of the initial mixture of milk and water is 20 litres.

Out of the 20 litres, 60% is pure milk. i.e., 12 litres is milk and the remaining 8 litres is water.

When "x" litres of pure milk is added, we will have (12 + x) litres of milk in the new mixture.

And there will be 20 + x litres of the new mixture.

The question states that (12 + x) litres of milk accounts for 80% of (20 + x) litres of the mixture.

Therefore, 12 + x = 80/100 * (20 + x)

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

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

60 + 5x = 80 + 4x

x = 20 litres

Out of the 20 litres, 60% is pure milk. i.e., 12 litres is milk and the remaining 8 litres is water.

When "x" litres of pure milk is added, we will have (12 + x) litres of milk in the new mixture.

And there will be 20 + x litres of the new mixture.

The question states that (12 + x) litres of milk accounts for 80% of (20 + x) litres of the mixture.

Therefore, 12 + x = 80/100 * (20 + x)

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

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

60 + 5x = 80 + 4x

x = 20 litres

Workspace

43. Two vessels contain mixtures of honey and water in the ratio of 8: 1 and 1: 5 respectively. The contents of both of these are mixed in a specific ratio into a third vessel. How much mixture must be drawn from the second vessel to fill the third vessel (capacity 36 gallons) completely in order that the resulting mixture may be half honey and half water?

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Correct Ans:14 gallons

Explanation:

Vessel 1 = honey : water = 8 : 1

Vessel 2 = honey : water = 1 : 5

Required amount = 7/18 * 36 = 14 gallons.

Vessel 2 = honey : water = 1 : 5

Required amount = 7/18 * 36 = 14 gallons.

Workspace

44. A quantity of eight litres is drawn from a cask full of wine and then cask is filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16 : 65. How much wine did the cask hold originally?

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Correct Ans:24 litres

Explanation:

Let initial quantity of wine = x litre

After a total of 4 operations, quantity of wine = x[1-(y/x)

Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

16/(16+65) = x[1-(8/x)

16/81 = [1-(8/x)

(2/3)

2/3 = 1 - (8/x)

8/x = 1/3

x = 24 litres.

After a total of 4 operations, quantity of wine = x[1-(y/x)

^{n}] = x[1-(8/x)^{4}]Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

16/(16+65) = x[1-(8/x)

^{4}] / x16/81 = [1-(8/x)

^{4}](2/3)

^{4}= [1-(8/x)^{4}]2/3 = 1 - (8/x)

8/x = 1/3

x = 24 litres.

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45. Jar A has x litre milk and jar B has y litre water. 80% milk and 20% water was taken out from the respective jars and were mixed in jar C. The respective ratio between milk and water in jar C was 2 : 1. When 28 litre pure milk was added to jar C, the total quantity of mixture in jar C became 76 litre. What was the value of x ?

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Correct Ans:40

Explanation:

Case I,

Quantity of milk in jar C = 80% of x = 4x/5 litres

Quantity of water = 20% of y = y/5 litres

According to the questions,

(4x/5) / (y/5) = 2/1

4x/y = 2

2x = y ......(i)

Case II,

On adding 28 litres of milk,

(4x/5) + (y/5) + 28 = 76

(4x/5) + (2x/5) = 76 - 28

6x/5 = 48

x = (48*5)/6 = 40 litres.

Quantity of milk in jar C = 80% of x = 4x/5 litres

Quantity of water = 20% of y = y/5 litres

According to the questions,

(4x/5) / (y/5) = 2/1

4x/y = 2

2x = y ......(i)

Case II,

On adding 28 litres of milk,

(4x/5) + (y/5) + 28 = 76

(4x/5) + (2x/5) = 76 - 28

6x/5 = 48

x = (48*5)/6 = 40 litres.

Workspace

46. A vessel contains a mixture of milk and water in the respective ratio of 10 : 3. Twenty-six litres of this mixture was taken out and replaced with 10 litres of water. If the respective ratio of milk and water in the resultant mixture was 5 : 2, what was the initial quantity of mixture in the vessel ? (In litres)

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Correct Ans:156

Explanation:

In 26 litres of mixture,

milk : water = 10 : 3

So, mixture contains,

Milk = (10/13)*26 = 10*2 = 20 litres

Water = (3/13)*26 = 3*2 = 6 litres

According to the question,

(10x - 20) / (3x - 6 + 10) = 5/2

(2x - 4) / (3x + 4) = 1/2

4x - 8 = 3x + 4

x = 8 + 4 = 12

Therefore, initial quality of mixture

= 13x = 13*12 = 156 litres.

milk : water = 10 : 3

So, mixture contains,

Milk = (10/13)*26 = 10*2 = 20 litres

Water = (3/13)*26 = 3*2 = 6 litres

According to the question,

(10x - 20) / (3x - 6 + 10) = 5/2

(2x - 4) / (3x + 4) = 1/2

4x - 8 = 3x + 4

x = 8 + 4 = 12

Therefore, initial quality of mixture

= 13x = 13*12 = 156 litres.

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47. A 24 litres of milk and water mixture contains milk and water in the ratio 3 : 5. What litres of the mixture should be taken out and replaced with pure milk so that the final mixture contains milk and water in equal proportions?

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Correct Ans:24/5 L

Explanation:

In 24 L of mixture, milk = 3/8 * 24 = 9 L, so water = 15 L

Now since the mixture is to be replaced with pure milk, the amount of mixture will remain same after replacement too.

In 24 L mixture, to have 12 L water and 12 L milk, 3 L of water should be taken out, since we are only adding milk.

Let x L of mixture taken out. So 5/8 * x = 3,

Solve, x = 24/5 L

Now since the mixture is to be replaced with pure milk, the amount of mixture will remain same after replacement too.

In 24 L mixture, to have 12 L water and 12 L milk, 3 L of water should be taken out, since we are only adding milk.

Let x L of mixture taken out. So 5/8 * x = 3,

Solve, x = 24/5 L

Workspace

48. There are two mixtures of wine and water, the quantity of wine in them being 25% and 75% of the mixture. If two gallons of the first are mixed with three gallons of the second, what will be the ratio of wine to water in the new mixture?

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Correct Ans:11:09

Explanation:

In first mixture contains 25% of wine so the remaining 75% is the water.

In second mixture contains 75% of wine so the remaining 25% is the water.

The required ratio = wine/water

= [(25% * 2) + (75% * 3)]/ [(75% * 2) + (25% * 3)]

= [(1/2) + (9/4)]/ [(3/2) + (3/4)]

= (11/4) / (9/4) = 11/9

Hence, option (b) is the correct answer.

In second mixture contains 75% of wine so the remaining 25% is the water.

The required ratio = wine/water

= [(25% * 2) + (75% * 3)]/ [(75% * 2) + (25% * 3)]

= [(1/2) + (9/4)]/ [(3/2) + (3/4)]

= (11/4) / (9/4) = 11/9

Hence, option (b) is the correct answer.

Workspace

49. A bucket contains some quantity of milk and water, in the ratio of water and milk is 3:5. 40 litres of mixture is drawn out and replaced with water and the ratio of milk and water becomes 5:11 then find the initial quantity of milk

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Correct Ans:50 litres

Explanation:

Given that, Milk : Water = 5 : 3

WKT, 40 litres of mixture is drawn out and replaced with water.

5x + 3x = 40

8x = 40

x = 40/8 = 5

Milk : Water = 5*5 : 3*5 = 25 : 15

So, 40 litres of mixture contains 25 L of milk and 15 L of water is drawn out and after replaced with 40 L of water the ratio of milk and water becomes 5:11.

(5x - 25)/(3x - 15 + 40) = 5/11

55x - 275 = 15X + 125

=> x = 10

litres Initial quantity of milk = 5*10 = 50 litres.

WKT, 40 litres of mixture is drawn out and replaced with water.

5x + 3x = 40

8x = 40

x = 40/8 = 5

Milk : Water = 5*5 : 3*5 = 25 : 15

So, 40 litres of mixture contains 25 L of milk and 15 L of water is drawn out and after replaced with 40 L of water the ratio of milk and water becomes 5:11.

(5x - 25)/(3x - 15 + 40) = 5/11

55x - 275 = 15X + 125

=> x = 10

litres Initial quantity of milk = 5*10 = 50 litres.

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50. A milkman purchases milk at Rs 20/litre and mixes 4 litres of water in it. By selling the resultant mixture at the rate of Rs 20/litre, he earns a profit of 40%. The amount of mixture he had with him to sell was:

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Correct Ans:14 litres

Explanation:

Given: CP of milk = Rs. 20/litre

SP of milk = Rs.20/litre;

Profit = 40%

To earn 40% profit, CP of mixture per litre = 20[100/(100 + 40)]

= (20 x 100)/140

=

Note: There is no cost price for water so it will be zero.

Solving by allegation rule,

Therefore, Milk : Water = 100/7 : 40/7

=

Hence, there is 5x of milk and 2x of water in the resultant mixture.

Given that, 4 litres of water is added.

2x = 4

x = 2

Number of litres of milk = 5x = 5(2)

=

Therefore,

SP of milk = Rs.20/litre;

Profit = 40%

**WKT, CP = SP[100/(100 + Profit)]**To earn 40% profit, CP of mixture per litre = 20[100/(100 + 40)]

= (20 x 100)/140

=

**Rs. 100/7**Note: There is no cost price for water so it will be zero.

Solving by allegation rule,

Therefore, Milk : Water = 100/7 : 40/7

=

**5 : 2**Hence, there is 5x of milk and 2x of water in the resultant mixture.

Given that, 4 litres of water is added.

2x = 4

x = 2

Number of litres of milk = 5x = 5(2)

=

**10 litres**Therefore,

**amount of mixture he had with him = 10 + 4 = 14 litres.**
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51. One type of mixture contains 25% of milk the another type of mixture contains 30% of milk. A container is filled with 6 parts of the first mixture and 4 parts of the second mixture. The percentage of milk in the mixture is,

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Correct Ans:27%

Explanation:

Mixture 1 = 25% Milk

Mixture 2 = 30% Milk

= {(1/100) (150 + 120) / 10} * 100

= {270 / (100 x 10)}* 100

=

Mixture 2 = 30% Milk

**Required Percentage**= {[(25/100) x 6 + (30/100) x 4] / (6 + 4)} * 100= {(1/100) (150 + 120) / 10} * 100

= {270 / (100 x 10)}* 100

=

**27%**
Workspace

52. A bottle is full of Dettol. One-third of it is taken out and then an equal amount of water is poured into the bottle to fill it. The operation is done four times. Find the final ratio of Dettol and water in the bottle.

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Correct Ans:16 : 65

Explanation:

The bottle originally contains Dettol only.

Let the bottle contain 1 litre of dettol originally, (say

In first operation, amount of Dettol taken out = 1/3 (say

(after first operation, amount of Dettol removed is replaced by equal amount of water, so, before second operation, Dettol in the bottle gets mixed with the some amount of water.. Hence, in second operation, we cannot take only Dettol from the bottle, but we take out a mixture of Dettol and water).

If this operation of taking out and replacing with equal amount of water is repeated 'n' times, then

= (1 - [1/3])

= (2/3)

= (16/81)/ {1 - (16/81)}

= (16/81)/ {65/81}

=

Let the bottle contain 1 litre of dettol originally, (say

**a =1**)In first operation, amount of Dettol taken out = 1/3 (say

**x = 1/3**)(after first operation, amount of Dettol removed is replaced by equal amount of water, so, before second operation, Dettol in the bottle gets mixed with the some amount of water.. Hence, in second operation, we cannot take only Dettol from the bottle, but we take out a mixture of Dettol and water).

If this operation of taking out and replacing with equal amount of water is repeated 'n' times, then

**Final ratio of Dettol and water in the bottle**= Amount of dettol left/Amount of water left =**(1 - x/a)**^{n}/{1 - (1 - x/a)^{n}}**Final Dettol/Final water**= (1 - [1/3]/1)^{4}/{1 - (1 - [1/3]/1)^{4}}= (1 - [1/3])

^{4}/{1 - (1 - [1/3])^{4}}= (2/3)

^{4}/{1 - (2/3)^{4}}= (16/81)/ {1 - (16/81)}

= (16/81)/ {65/81}

=

**16/65**
Workspace

53. The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kg of wheat so that the percentage of low quality wheat becomes 5%?

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Correct Ans:150 kg

Explanation:

Given, Total quantity of Wheat = 150 kg

Quantity of

= (10/100) * 150

=

Then, Quantity of

= 150 - 15

=

Let

Then, the percentage of

Now, in the

Quantity of Good quality wheat/Quantity of Low quality wheat = Percentage of Good quality wheat/ Percentage of Low quality wheat

---> (135 + x)/ 15 = 95/5

---> (135 + x)/ 15 = 19/1

---> (135 + x)/ 15 = 19

---> (135 + x) = 19 * 15

---> (135 + x) = 285

---> x = 285 - 135

--->

Hence, 150 kg of good quality wheat needs to be added to make the low quality wheat be 5%.

Quantity of

**Low quality wheat**= 10% of 150 kg= (10/100) * 150

=

**15 kg**Then, Quantity of

**Good quality wheat**= Total quantity of Wheat - Quantity of Low quality wheat= 150 - 15

=

**135 kg**Let

**'x'**be the amount of good quality wheat to be added to make the percentage of low quality wheat to 5%.Then, the percentage of

**good quality wheat**in new mixture becomes**95%**Now, in the

**new mixture**of 150 kg Wheat,Quantity of Good quality wheat/Quantity of Low quality wheat = Percentage of Good quality wheat/ Percentage of Low quality wheat

---> (135 + x)/ 15 = 95/5

---> (135 + x)/ 15 = 19/1

---> (135 + x)/ 15 = 19

---> (135 + x) = 19 * 15

---> (135 + x) = 285

---> x = 285 - 135

--->

**x = 150 kg**Hence, 150 kg of good quality wheat needs to be added to make the low quality wheat be 5%.

Workspace

54. A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the mixture is taken out and 10 litres of liquid B is poured into the jar, the ratio becomes 2 : 3. How many litres of liquid A was contained in the jar?

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Correct Ans:16 litres

Explanation:

Given ratio of liquids A and B = 4 : 1

Let the Quantity of liquid A = 4x liters

and Quantity of liquid B = x liters

Total part of the mixture initially = 5 parts

When 10 litres of the mixture is taken out, then

Final Quantity of A becomes 4x - 10 * (4/5)

=

Quantity of B becomes x - 10 * (1/5)

= x - 2

When 10 litres of liquid B is poured into the jar, the Final quantity of B becomes (x - 2) + 10

=

Given resultant ratio is 2 : 3, So

Final Quantity of A/ Final Quantity of B = 2 / 3

--> [4x - 8] / [x + 8] = 2/3

---> 3 * [4x - 8] = 2 * [x + 8]

---> 12x - 24 = 2x + 16

---> 12x - 2x = 16 + 24

---> 10x = 40

--->

Hence, quantity of liquid A which was contained in the jar (initially)= 4x = 4 * 4 = 16 liters

Let the Quantity of liquid A = 4x liters

and Quantity of liquid B = x liters

Total part of the mixture initially = 5 parts

When 10 litres of the mixture is taken out, then

Final Quantity of A becomes 4x - 10 * (4/5)

=

**4x - 8**Quantity of B becomes x - 10 * (1/5)

= x - 2

When 10 litres of liquid B is poured into the jar, the Final quantity of B becomes (x - 2) + 10

=

**x + 8**Given resultant ratio is 2 : 3, So

Final Quantity of A/ Final Quantity of B = 2 / 3

--> [4x - 8] / [x + 8] = 2/3

---> 3 * [4x - 8] = 2 * [x + 8]

---> 12x - 24 = 2x + 16

---> 12x - 2x = 16 + 24

---> 10x = 40

--->

**x = 4**Hence, quantity of liquid A which was contained in the jar (initially)= 4x = 4 * 4 = 16 liters

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55. A vessel has 80 litre mixture of water and milk. In it the water content is 25%. If the milkman takes out 8 litres of mixture and adds 8 litres of pure water in the vessel. What is the percentage of water in the resulting mixture?

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Correct Ans:32.50%

Explanation:

Given:

Total quantity of mixture = 80 litres

Quantity of water in 80 liters of mixture = 25%

Therefore, quantity of water in 80 liters of mixture = 25% of 80

= (25/100)*80

= 20 litres

So, quantity of milk in 80 litres of mixture = 80 - 20 = 60 litres

Quantity of water in 8 litres of mixture = 25% of 8

= (25/100)*8

= 2 liters

Therefore, quantity of water in the new mixture = 20 - 2 + 8

= 26 litres

Percentage of water in new mixture,

x% of 80 = 26

80x/100 = 26

x = (26 * 100)/80 = 32.5%

Required percentage = 32.5%.

Total quantity of mixture = 80 litres

Quantity of water in 80 liters of mixture = 25%

Therefore, quantity of water in 80 liters of mixture = 25% of 80

= (25/100)*80

= 20 litres

So, quantity of milk in 80 litres of mixture = 80 - 20 = 60 litres

Quantity of water in 8 litres of mixture = 25% of 8

= (25/100)*8

= 2 liters

Therefore, quantity of water in the new mixture = 20 - 2 + 8

= 26 litres

Percentage of water in new mixture,

x% of 80 = 26

80x/100 = 26

x = (26 * 100)/80 = 32.5%

Required percentage = 32.5%.

Workspace

56. In a mixture of 25 L, the ratio of acid to water is 4 : 1. Another 3 L of water is added to the mixture. The ratio of acid to water in the new mixture is ?

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Correct Ans:5 : 2

Explanation:

Given, acid : water = 4 : 1

Total mixture = 25 L

Now in that mixture,

Acid = (4/5) * 25

= 20 L

Water = (1/5) * 25

= 5 L

After the addition of 3 L of water,

New ratio = 20 : 8 = 5 : 2

Total mixture = 25 L

Now in that mixture,

Acid = (4/5) * 25

= 20 L

Water = (1/5) * 25

= 5 L

After the addition of 3 L of water,

New ratio = 20 : 8 = 5 : 2

Workspace

57. The ratio, in which tea costing Rs.192 per kg is to be mixed with tea costing Rs. 150 per kg so that the mixed tea, when sold for Rs. 194.40 per kg gives a profit of 20% is

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Correct Ans:2:5

Explanation:

Given, SP of mixed tea = Rs. 194.40

CP of mixed tea = [100/(100 + Profit)]*SP

= [100/(100 + 20)]*194.40

= (100/120)*194.40

= Rs. 162/kg

By Alligation rule,

Therefore, required ratio = 12/30

= 2/5

Required ratio = 2 : 5.

CP of mixed tea = [100/(100 + Profit)]*SP

= [100/(100 + 20)]*194.40

= (100/120)*194.40

= Rs. 162/kg

By Alligation rule,

Therefore, required ratio = 12/30

= 2/5

Required ratio = 2 : 5.

Workspace

58. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 5 : 3 and 5 : 11, respectively. Equal quantities of these alloys are melted to form a third alloy C. The ratio of gold and copper in the alloy C is

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Correct Ans:15:17

Explanation:

Let 1kg of each of the alloys A and B be mixed together.

In alloy A, ratio of gold and copper is 5 : 3.

Quantity of gold = 5/8 kg

Quantity of copper = 3/8 kg

In alloy B, ratio of gold and copper is 5 : 11.

Quantity of gold = 5/16 kg

Quantity of copper = 11/16 kg

Therefore, required ratio = Gold : Copper

= [(5/8) + (5/16)] : [(3/8) + (11/16)]

= [(10/16) + (5/16)] : [(6/16) + (11/16)]

= (15/16) : (17/16)

Required ratio = 15 : 17

In alloy A, ratio of gold and copper is 5 : 3.

Quantity of gold = 5/8 kg

Quantity of copper = 3/8 kg

In alloy B, ratio of gold and copper is 5 : 11.

Quantity of gold = 5/16 kg

Quantity of copper = 11/16 kg

Therefore, required ratio = Gold : Copper

= [(5/8) + (5/16)] : [(3/8) + (11/16)]

= [(10/16) + (5/16)] : [(6/16) + (11/16)]

= (15/16) : (17/16)

Required ratio = 15 : 17

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59. A trader mixes 14 kg rice of variety A which costs Rs. 60/kg with 18 kg of quantity of type B rice. He sells the mixture at Rs. 65/Kg and earns a profit of 100/3%. Then what was the cost price of type B rice.

SHOW ANSWER

Correct Ans:40

Explanation:

Let cost price of mixture = y

So, 4/3y = 65

y = 48.75

From mixture and allegation (refer the image)

7/9 = (x-48.75)/48.75-60

341.25 âˆ’ 420 = 9ð‘¥ âˆ’ 438.75

360 = 9ð‘¥

ð‘¥ = 40 Rs./kg

So, 4/3y = 65

y = 48.75

From mixture and allegation (refer the image)

7/9 = (x-48.75)/48.75-60

341.25 âˆ’ 420 = 9ð‘¥ âˆ’ 438.75

360 = 9ð‘¥

ð‘¥ = 40 Rs./kg

Workspace

60. A vessel contains milk and water in which 20% is water. 20 liters of mixture was taken out and replaced by water and the ratio becomes 12:13. Find the initial quantity of milk in the vessel?

SHOW ANSWER

Correct Ans:40 liters

Explanation:

Milk and water ratio = 4:1

Given,

(4x-16)/(x-4+20) =12/13

13x-52 = 3x+48

= 10x=100

x=10

Initial quantity of milk in the vessel= 40 liters

Given,

(4x-16)/(x-4+20) =12/13

13x-52 = 3x+48

= 10x=100

x=10

Initial quantity of milk in the vessel= 40 liters

Workspace

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