# 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

1. In a mixture, the ratio of the alchohol and water is 6 : 5. When 22 litre mixture are replaced by water, the ratio becomes 9 : 13. Find the quantity of water after replacement.

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Correct Ans:52 litre

Explanation:

-----> Let alchohol = 6x, water = 5x

-----> According to the question,

-----> 6x â€“ 22 * (6/11) : 5x â€“ 22 * (5/11) + 22 = 9 : 13

-----> 6x â€“ 12 : 5x â€“ 10 + 22 = 9 : 13

-----> 13 (6x â€“ 12) = 9 (5x + 12)

-----> 78x â€“ 156 = 45x + 108

-----> 78x â€“ 45x = 156 + 108

-----> 33x = 264

-----> x = 8

-----> Water after replacement = 5 * 8 â€“ 10 + 22 = 40 + 22 = 52 litre

-----> According to the question,

-----> 6x â€“ 22 * (6/11) : 5x â€“ 22 * (5/11) + 22 = 9 : 13

-----> 6x â€“ 12 : 5x â€“ 10 + 22 = 9 : 13

-----> 13 (6x â€“ 12) = 9 (5x + 12)

-----> 78x â€“ 156 = 45x + 108

-----> 78x â€“ 45x = 156 + 108

-----> 33x = 264

-----> x = 8

-----> Water after replacement = 5 * 8 â€“ 10 + 22 = 40 + 22 = 52 litre

**Hence, option D is correct.**
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2. The cost of Type 1 rice is Rs. 30 per kg and Type 2 rice is Rs. 40 per kg. if both Type 1 and Type 2 are mixed in the ratio of 1 : 4, then what will be the price per kg of the mixed variety of rice?

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Correct Ans:38 per kg

Explanation:

-----> Let the price of the mixed variety be x per kg.

cost of 1 kg of type 1 rice cost of 1 kg of type 2 rice

Rs . 30 Rs. 40

Mean Price

x

40 - x x - 30

-----> (40 - x/x - 30 )

-----> 160 â€“ 4x = x â€“ 30

-----> x = 38

----->

cost of 1 kg of type 1 rice cost of 1 kg of type 2 rice

Rs . 30 Rs. 40

Mean Price

x

40 - x x - 30

-----> (40 - x/x - 30 )

-----> 160 â€“ 4x = x â€“ 30

-----> x = 38

----->

**So, price of the mixture is Rs. 38 per kg.**
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3. A alcohol solution has 35% alcohol. Another solution has 25% alcohol. How many liters of the second solution must be added to 18 liters of the first solution to make a solution of 30% alcohol?

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Correct Ans:18

Explanation:

---> From Alligation:

---> 1st Solution 2nd Solution

---> 35% 25%

---> 30%

---> (30-25)=5 (35-30)=5

---> HENCE

---> Ratio of solution 1 : solution 2 = (5:5) = (1:1)

---> As 1 = 18 litres

---> Therefore we add 18 liter of 2nd solution

---> 1st Solution 2nd Solution

---> 35% 25%

---> 30%

---> (30-25)=5 (35-30)=5

---> HENCE

---> Ratio of solution 1 : solution 2 = (5:5) = (1:1)

---> As 1 = 18 litres

---> Therefore we add 18 liter of 2nd solution

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4. A bucket contains 60 litres of pure wine, in which x litres drawn off and replaced with water. This process is repeated for two times. Find the value of x if the final ratio of wine to water is 81 : 19

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Correct Ans:6 litres

Explanation:

Given,

Wine and water in final process (60/100*81 = 48.6) and (60/100*19 = 11.4) respectively

48.6 = 60 (1 - x/60)

48.6/60 = (1 - x/60)

WKT (48.6/60 = 81/100)

81/100 = (1 - x/60)

9/10 = 1 - x/60

x/60 = 1 - 9/10

x/60 = 1/10

x = 6 litres

Wine and water in final process (60/100*81 = 48.6) and (60/100*19 = 11.4) respectively

48.6 = 60 (1 - x/60)

^{2}48.6/60 = (1 - x/60)

^{2}WKT (48.6/60 = 81/100)

81/100 = (1 - x/60)

^{2}9/10 = 1 - x/60

x/60 = 1 - 9/10

x/60 = 1/10

x = 6 litres

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5. Two equal vessels A and B contain 60% of sugar and 40% of sugar respectively and the remaining Rava. In which 40 kg of mixture is taken out from vessel A and replaced into vessel B. Find the initial quantity of vessel if the final ratio of sugar and Rava in vessel B is 16 : 19

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Correct Ans:100 litres

Explanation:

Vessel A Sugar and Rava ratio = 3 : 2

Vessel B sugar and Rava ratio = 2 : 3

Given,

(2x + 24)/(3x + 16) = 16/19

38x + 456 = 48x + 256

10x = 200

x = 20 litres

Initial quantity = x*5 = 100 litres

Vessel B sugar and Rava ratio = 2 : 3

Given,

(2x + 24)/(3x + 16) = 16/19

38x + 456 = 48x + 256

10x = 200

x = 20 litres

Initial quantity = x*5 = 100 litres

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6. In a mixture of milk and water of volume 30 litres, the ratio of milk and water is 7 : 3. How much quantity of water is to be added to the mixture to make the ratio of milk and water 1 : 2?

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Correct Ans:33 liters

Explanation:

In 30 liter of mixture quantity of milk = (7*30)/10 = 21 liter

Hence quantity of water = 30 - 21 = 9 liter

Now let x liter of water is added in mixture to make the ratio of milk and water 1 : 2.

Hence now quantity of water = (9 + x) L

Milk = 21 L

Therefore (21 : (9 + x)) = 1 : 2

21/(9 + x) = 1/2

42 = 9 + x

x = 42 - 9 = 33

Hence 33 liters of water is added in the mixture.

Hence quantity of water = 30 - 21 = 9 liter

Now let x liter of water is added in mixture to make the ratio of milk and water 1 : 2.

Hence now quantity of water = (9 + x) L

Milk = 21 L

Therefore (21 : (9 + x)) = 1 : 2

21/(9 + x) = 1/2

42 = 9 + x

x = 42 - 9 = 33

Hence 33 liters of water is added in the mixture.

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7. A mixture of milk and water in a jar contains 32 liters milk and 10 liters water. X liter of milk and X liter of water is added to a new mixture. If 40 % of new mixture is 24 liters, then find the value of X (In liters)?

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Correct Ans:9

Explanation:

(40/100) * New mixture = 24

New mixture = 24 * (5/2) = 60 liters

Given,

32 + X + 10 + X = 60

42 + 2X = 60

2X = 18

X = 9

New mixture = 24 * (5/2) = 60 liters

Given,

32 + X + 10 + X = 60

42 + 2X = 60

2X = 18

X = 9

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8. A container contains pure milk. From these 4 litres of milk taken out and replaced by water. This process is repeated for one more time and the remaining milk in the container is 12.8 litres. What is the initial quantity of milk in the container?

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Correct Ans:20 litres

Explanation:

Let us take initial quantity of a container be x

Remaining milk = Initial (1 - Replaced/Initial)

12.8 = x (1 - 4/x)

12.8 = x (1 + 16/x

12.8 = x [(x

12.8x = x

5x

Simplify the above equation, we get x= 20 and 0.8 (Eliminate)

Remaining milk = Initial (1 - Replaced/Initial)

^{n}12.8 = x (1 - 4/x)

^{2}12.8 = x (1 + 16/x

^{2}- 8x)12.8 = x [(x

^{2}+ 16 - 8x) / x^{2}]12.8x = x

^{2}+ 16 - 8x5x

^{2}- 104x + 80 = 0Simplify the above equation, we get x= 20 and 0.8 (Eliminate)

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9. There are two containers P and Q. P contains 56 kg of salt and Q contains 36 kg of sugar. From P 24 kg of salt is taken out and poured into Q. Then 20kg of the mixture from Q is taken out and poured into P. Find the ratio of final quantity of salt to sugar in container P.

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Correct Ans:10/3

Explanation:

Initially, the Amount of sugar in Q = 36kg

Now, 24kg of salt is poured in Q,

Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture)

The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3

Now, the Amount of salt in P = 56 - 24 = 32 kg

Again, 20kg of the mixture is taken out from Q and poured into P

Therefore, quantity of salt and sugar in P becomes

= [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar

= (32 + 8) kg of salt + 12 kg of sugar

= 40 kg of salt + 12 kg of sugar

Required ratio = (40/12) = (10/3)

Now, 24kg of salt is poured in Q,

Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture)

The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3

Now, the Amount of salt in P = 56 - 24 = 32 kg

Again, 20kg of the mixture is taken out from Q and poured into P

Therefore, quantity of salt and sugar in P becomes

= [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar

= (32 + 8) kg of salt + 12 kg of sugar

= 40 kg of salt + 12 kg of sugar

Required ratio = (40/12) = (10/3)

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10. A mixture contains 200 litres Milk and 40 litres water, ______ litres of mixture are removed and ________ litres of pure water were added to it. If the final quantity of milk is 124 litres more than the final quantity of water. The values given in which of the following options will fill the blanks in the same order in which it is given to make the above statement true:

A) 30, 20

B) 18, 24

C) 24, 20

D) 36, 16

A) 30, 20

B) 18, 24

C) 24, 20

D) 36, 16

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Correct Ans:C and B only

Explanation:

---> Milk = 200

---> Water = 40

---> Ratio = (5: 1)

---> Let â€˜xâ€™ litres be removed

---> So milk removed = (5/6) * x

---> Water removed = (1/6) * x

---> Milk remaining = 200 - (5/6) * x

---> Water remaining = 40 - (1/6) * x

---> Let â€˜yâ€™ litres of water is added

---> So, total water = 40 - (1/6) * (x + y)

---> Given, 200 - (5/6) * x = 124 + 40 - (1/6) (x+y)

---> 36 - (2/3) * x = y

---> For option a : 30, 20

---> 36 - (2/3) * x = y

---> x = 30

---> 36 - 20 = 16

---> So, option a canâ€™t be the answer

---> For option b: 18, 24

---> 36 - (2/3) * x = y

---> x = 18

---> 36 - 12 = 24

---> So, option b can be the answer

---> For option c : 24, 20

---> 36 - (2/3) * x = y

---> x = 24

---> 36 - 16 = 20

---> So, option c can be the answer

---> For option d: 36, 16

---> 36 - (2/3) * x = y

---> x = 36

---> 36 - 24 = 12

---> So, option d canâ€™t be the answer

--->

---> Water = 40

---> Ratio = (5: 1)

---> Let â€˜xâ€™ litres be removed

---> So milk removed = (5/6) * x

---> Water removed = (1/6) * x

---> Milk remaining = 200 - (5/6) * x

---> Water remaining = 40 - (1/6) * x

---> Let â€˜yâ€™ litres of water is added

---> So, total water = 40 - (1/6) * (x + y)

---> Given, 200 - (5/6) * x = 124 + 40 - (1/6) (x+y)

---> 36 - (2/3) * x = y

---> For option a : 30, 20

---> 36 - (2/3) * x = y

---> x = 30

---> 36 - 20 = 16

---> So, option a canâ€™t be the answer

---> For option b: 18, 24

---> 36 - (2/3) * x = y

---> x = 18

---> 36 - 12 = 24

---> So, option b can be the answer

---> For option c : 24, 20

---> 36 - (2/3) * x = y

---> x = 24

---> 36 - 16 = 20

---> So, option c can be the answer

---> For option d: 36, 16

---> 36 - (2/3) * x = y

---> x = 36

---> 36 - 24 = 12

---> So, option d canâ€™t be the answer

--->

**So option (b) is the correct answer.**
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11. The respective ratio of zinc and copper in the first and second alloy are 1 : 2 and 2 : 3 respectively. In what ratio should these two alloys be mixed to form a new alloy in which the two elements are in ratio 5 : 8?

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Correct Ans:(3: 10)

Explanation:

---> Let the amount of mixture taken from 1st allow = x

---> And, the amount of mixture from the second allow = y

---> so = [(x/3)+(2y/5)/(2x/3)+(3y/5)] = (5/8)

---> = ((5x+6y)/(10x+9y)) = (5/8)

---> 40x + 48y = 50x + 45y

---> 10x = 3y; (x/y) = (3/10)

---> And, the amount of mixture from the second allow = y

---> so = [(x/3)+(2y/5)/(2x/3)+(3y/5)] = (5/8)

---> = ((5x+6y)/(10x+9y)) = (5/8)

---> 40x + 48y = 50x + 45y

---> 10x = 3y; (x/y) = (3/10)

**So, Second option is the correct answer.**
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12. In a container containing a mixture of 60 litres, the ratio of water to spirit is 4: 1. How much spirit should be added to make the ratio 3: 2?

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Correct Ans:20 litres

Explanation:

Amount of mixture in the container = 60 litres

Given, the ratio of water to spirit is 4: 1

Therefore, Amount of water = 4/5 * 60 = 48

Amount of spirit = 1/5 * 60 = 12

Let the amount of spirit added be x

Therefore, Amount of water/Amount of spirit = 3/2

48/(12 + x) = 3/2

96 = 36 + 3x

3x = 96 - 36 = 60

x = 60/3 = 20

Given, the ratio of water to spirit is 4: 1

Therefore, Amount of water = 4/5 * 60 = 48

Amount of spirit = 1/5 * 60 = 12

Let the amount of spirit added be x

Therefore, Amount of water/Amount of spirit = 3/2

48/(12 + x) = 3/2

96 = 36 + 3x

3x = 96 - 36 = 60

x = 60/3 = 20

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13. A company is creating a new signature drink. They are using two alcoholic ingredients in the drink vodka and gin. They are using two non-alcoholic ingredients in the drink: orange juice and cranberry juice. The alcoholic ingredients contain 50% alcohol. The non- alcoholic ingredients contain no alcohol. How many litres of non-alcoholic ingredients must be added to 5 litres of alcoholic ingredients to produce a mixture that is 20% alcohol?

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Correct Ans:7.5

Explanation:

A company is creating a new signature drink. They are using two alcoholic ingredients in the drink vodka and gin. They are using two non-alcoholic ingredients in the drink: orange juice and cranberry juice. The alcoholic ingredients contain 50% alcohol. The non- alcoholic ingredients contain no alcohol. How many litres of non-alcoholic ingredients must be added to 5 litres of alcoholic ingredients to produce a mixture that is 20% alcohol:

From the reference :

Find the value:

----> (5+X) (20 %) = 5 (50 %) + X (0 %)

----> (5+X) (0.2) = 5 (0.5) + 0

----> 1 + 0.2 X = 2.5

----> X = ((2.5 -1) / 0.2)

----> X = (1.5 /0.2 ) = 7.5

From the reference :

Find the value:

----> (5+X) (20 %) = 5 (50 %) + X (0 %)

----> (5+X) (0.2) = 5 (0.5) + 0

----> 1 + 0.2 X = 2.5

----> X = ((2.5 -1) / 0.2)

----> X = (1.5 /0.2 ) = 7.5

**Hence the answer is : 7.5**
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14. Rani mixes 70kg of sugar worth Rs.28.50 per kg with 100kg of sugar worth Rs.30.50 per kg. At what rate shall he sell the mixture to gain 10% ?

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Correct Ans:Rs.32.64

Explanation:

Rani mixes 70kg of sugar worth Rs.28.50 per kg with 100kg of sugar worth Rs.30.50 per kg. At what rate shall he sell the mixture to gain 10% :

From the Reference:

Find the value:

----> (70/100) = 100 X - (30.50(110)/28.50(110))

----> 0.7 = 100 X - (3355/3135) - 100 X

----> 2194.5 - 70 X = 100 X - 3355

----> 170 X = 5549.5

----> X = 32.64

From the Reference:

Find the value:

----> (70/100) = 100 X - (30.50(110)/28.50(110))

----> 0.7 = 100 X - (3355/3135) - 100 X

----> 2194.5 - 70 X = 100 X - 3355

----> 170 X = 5549.5

----> X = 32.64

**Hence the answer is : Rs.32.64**
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15. In a 70 litres mixture of milk and water, % of water is 30%. The milkman gave 20 litres of this mixture to a customer and then added 20 litres of water to the remaining mixture. What is the % of milk in the final mixture ?

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Correct Ans:50%

Explanation:

In a 70 litres mixture of milk and water, % of water is 30%. The milkman gave 20 litres of this mixture to a customer and then added 20 litres of water to the remaining mixture. What is the % of milk in the final mixture

From the reference:

From the value :

----> 20 litre given remaining = 70-20 = 50 litre

---> Quantity of milk = (( 50*70)/100 ) = 35 litre

---> Quantity of water = 50 - 35 = 15 litre

----> 20 litres of water added = 50+20 = 70

----> % of milk = (35*(100/70)) = 50%

From the reference:

From the value :

----> 20 litre given remaining = 70-20 = 50 litre

---> Quantity of milk = (( 50*70)/100 ) = 35 litre

---> Quantity of water = 50 - 35 = 15 litre

----> 20 litres of water added = 50+20 = 70

----> % of milk = (35*(100/70)) = 50%

**Hence the answer is : 50%**
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16. A mixture contains A and B in the ratio 2 : 5. 7 litres of mixture is replaced by 18 litres of A and the new ratio becomes 8 : 11 respectively. What is the amount of A present in the original mixture?

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Correct Ans:24 litres

Explanation:

Reference:

A mixture contains A and B in the ratio = 2 : 5

-----> A = 2x

-----> B = 5x

Find A And B in the mixture ,After 7 litres draw out:

-----> A = 2x - (2/7)*7 lit = (2x - 2)

-----> B = 5x - (5/7)*7 lit = (5x -5)

Find A in the new mixture, After 18 litres of A put back

-----> A = (2x - 2) + 18 = (2x +16)

-----> B = (5x - 5)

compare with new ratio

------> ((2x +16)/(5x - 5)) = (8 : 11)

------> (2x +16) * 11 = (5x - 5) * 8

------> Solve, x = 12

Find A present in the original mixture:

------> A = 2x

------> A = (2 * 12)

------> A = 24 litres

A mixture contains A and B in the ratio = 2 : 5

-----> A = 2x

-----> B = 5x

Find A And B in the mixture ,After 7 litres draw out:

-----> A = 2x - (2/7)*7 lit = (2x - 2)

-----> B = 5x - (5/7)*7 lit = (5x -5)

Find A in the new mixture, After 18 litres of A put back

-----> A = (2x - 2) + 18 = (2x +16)

-----> B = (5x - 5)

compare with new ratio

------> ((2x +16)/(5x - 5)) = (8 : 11)

------> (2x +16) * 11 = (5x - 5) * 8

------> Solve, x = 12

Find A present in the original mixture:

------> A = 2x

------> A = (2 * 12)

------> A = 24 litres

**Hence the answer is: 24 litres**
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17. A can contains milk and water in the ratio 3:1. A part of this mixture is replaced with milk, and now the new ratio of milk to water is 15:4.What proportion of original mixture had been replaced by milk?

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Correct Ans:(3/19)

Explanation:

Reference:

A can contains milk and water in the ratio 3:1

-----> Let total original quantity = x litres

-----> Mixture of milk and water

-----> (3/4) * x = (1/4) * x

-----> Let y litres replaced in both

-----> ((3/4) * x - (3/4) * y) = ((1/4) * x - (1/4) * y )

-----> After y litres of milk poured in can, Milk becomes :

-----> (3/4) * x - (3/4) * y + y

------> solve, so we get:

------> (3/4) * x - (1/4) * y

Compare with new ratio:

------> (( (3/4) * x - (1/4) * y )/((1/4) * x - (1/4) * y )) = (15 /4)

------> solve ,we get:

-----> y = (3/19)

The original mixture had been replaced by milk is (3/19)

**Hence the answer is : (3/9)**

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18. A 132 litres of mixture contains milk and water in the ratio 5 : 7. How much milk need to be added to this mixture so that the new ratio is 13 : 11 respectively?

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Correct Ans:36 litres

Explanation:

Reference:

A 132 litres of mixture contains milk and water in the ratio 5 : 7.

Let Find the Milk in original mixture :

------> (5/12) * 132 = 55 Lit

Find the Water in original mixture :

------> (7/12) * 132 = 77 Lit

Let x Litres of milk to be added

-----> so,((55 +x) : 77)

-----> compare with new ratio(13 :11)

------> ((55 +x) / 77) = (13 /11)

------> solve, x = 36 Litres

so,The 36 Litres milk need to be added to this new mixture.

A 132 litres of mixture contains milk and water in the ratio 5 : 7.

Let Find the Milk in original mixture :

------> (5/12) * 132 = 55 Lit

Find the Water in original mixture :

------> (7/12) * 132 = 77 Lit

Let x Litres of milk to be added

-----> so,((55 +x) : 77)

-----> compare with new ratio(13 :11)

------> ((55 +x) / 77) = (13 /11)

------> solve, x = 36 Litres

so,The 36 Litres milk need to be added to this new mixture.

**Hence the answer is : 36 Litres**
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19. A vessel contains mixture of milk and water mixed in the ratio of 9 : 4 respectively. 78 litres of the mixture is taken out of the vessel and replaced with 44 litres of water, so that the ratio of the water to milk in the vessel becomes 8 : 7 respectively. Find the final quantity of mixture in the vessel.

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Correct Ans:135 litres

Explanation:

Given:

Ratio of mixture of milk and water = 9 : 4

Let the initial quantity of milk be 9X and water be 4X.

Quantity of mixture taken out = 78 litres

Quantity of milk from the mixture taken out = (9/13)*78 = 54 litres

Quantity of water from the mixture taken out = 78 - 54 = 24 litres

When 44 litres of water is replaced, ratio of milk and water becomes 8 : 7

(9X - 54)/(4X - 24 + 44) = 7/8

(9X - 54)/(4X + 20) = 7/8

72X - 432 = 28X + 140

44X = 572

X = 13

So, initial quantity of mixture = 13X = 13*13 = 169 litres

Final quantity of mixture = 169 - 78 + 44 = 135 litres.

Ratio of mixture of milk and water = 9 : 4

Let the initial quantity of milk be 9X and water be 4X.

Quantity of mixture taken out = 78 litres

Quantity of milk from the mixture taken out = (9/13)*78 = 54 litres

Quantity of water from the mixture taken out = 78 - 54 = 24 litres

When 44 litres of water is replaced, ratio of milk and water becomes 8 : 7

(9X - 54)/(4X - 24 + 44) = 7/8

(9X - 54)/(4X + 20) = 7/8

72X - 432 = 28X + 140

44X = 572

X = 13

So, initial quantity of mixture = 13X = 13*13 = 169 litres

Final quantity of mixture = 169 - 78 + 44 = 135 litres.

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20. A container has a mixture of milk and water in the ratio 7: 4, respectively. 88 litres of mixture is taken out of the container and is replaced with another mixture, which contains milk and water in the ratio 3: 2, such that the ratio of milk to water in the container becomes 5: 3, respectively. Find the difference between quantity of milk and water present initially in the container if quantity of milk in final mixture is 120 litres.

SHOW ANSWER

Correct Ans:60 litres

Explanation:

Given:

Initial mixture of milk and water in the ratio - 7: 4

Final mixture of milk and water in the ratio - 5 : 3

Another mixture, ratio of milk and water - 3: 2

Let the quantity of milk and water present initially in the container be '7x' litres and '4x' litres, respectively.

Let the quantity of milk and water present in another mixture be '3y' litres and '2y' litres, respectively.

Given that quantity of milk in final mixture = 120 litres

So, quantity of water in final mixture = (3/5)*120 = 72 litres.

88 litres of mixture is taken out and another mixture is added then the quantity of milk in the final mixture is

7x + 3y - (7/11)*88 = 120

7x + 3y - 56 = 120

7x + 3y = 176

y = (176 - 7x)/3

Similarly, quantity of water in the final mixture is

4x + 2y - (4/11)*88 = 72

4x + 2[(176 - 7x)/3] - 32 = 72

12x + 352 - 14x - 96 = 216

2x = 40

x = 20

Required difference between quantity of milk and water present initially = 7x - 4x = 3x = 3(20) = 60 litres.

Initial mixture of milk and water in the ratio - 7: 4

Final mixture of milk and water in the ratio - 5 : 3

Another mixture, ratio of milk and water - 3: 2

Let the quantity of milk and water present initially in the container be '7x' litres and '4x' litres, respectively.

Let the quantity of milk and water present in another mixture be '3y' litres and '2y' litres, respectively.

Given that quantity of milk in final mixture = 120 litres

So, quantity of water in final mixture = (3/5)*120 = 72 litres.

88 litres of mixture is taken out and another mixture is added then the quantity of milk in the final mixture is

7x + 3y - (7/11)*88 = 120

7x + 3y - 56 = 120

7x + 3y = 176

y = (176 - 7x)/3

Similarly, quantity of water in the final mixture is

4x + 2y - (4/11)*88 = 72

4x + 2[(176 - 7x)/3] - 32 = 72

12x + 352 - 14x - 96 = 216

2x = 40

x = 20

Required difference between quantity of milk and water present initially = 7x - 4x = 3x = 3(20) = 60 litres.

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