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
21. There are two barrels - A & B. Barrel - A contains 80 litres mixture of petrol and kerosene oil in the ratio of 7 : 3 and barallel B contains petrol and kerosene oil in the ratio 5 : 9. 20 litres mixture of barallel A is poured into barallel B, due to which ratio of petrol and kerosene oil in barallel B becomes 11 : 15. Then, find the initial quantity of mixture in barallel B.
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Correct Ans:84 litres
Explanation:
Given:
Ratio of petrol and kerosene oil in barallel A = 7 : 3
Ratio of petrol and kerosene oil in barallel B = 5 : 9
Let the initial quantity of petrol and kerosene oil in barallel B be 5X and 9X respectively.
As per the question,
[5X + (20*(7/10))]/[9X + (20*(3/10))] = 11/15
[5X + 14]/[9X + 6] = 11/15
75X + 210 = 99X + 66
24X = 144
X = 6
So, initial quantity of oil in barallel B = 5X + 9X = 14X
= 14(6)
= 84 litres.
Ratio of petrol and kerosene oil in barallel A = 7 : 3
Ratio of petrol and kerosene oil in barallel B = 5 : 9
Let the initial quantity of petrol and kerosene oil in barallel B be 5X and 9X respectively.
As per the question,
[5X + (20*(7/10))]/[9X + (20*(3/10))] = 11/15
[5X + 14]/[9X + 6] = 11/15
75X + 210 = 99X + 66
24X = 144
X = 6
So, initial quantity of oil in barallel B = 5X + 9X = 14X
= 14(6)
= 84 litres.
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22. Rice worth Rs. 120 per kg and Rs. 130 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 170 per kg, the price of the third variety per kg will be?
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Correct Ans:160
Explanation:
Let Mixture of Rice worth Rs. 120 per kg and Rs. 130 per kg
Mixed with a third variety in the ratio = 1 : 1 : 2.
The mixture is worth Rs. 170 per kg,
Let x be the price of third variety per kg.
The price of the third variety
cost of three varieties = cost of the mixture
120*1+130*1+2*x = 170*4
The price of the third variety x = 160
Mixed with a third variety in the ratio = 1 : 1 : 2.
The mixture is worth Rs. 170 per kg,
Let x be the price of third variety per kg.
The price of the third variety
cost of three varieties = cost of the mixture
120*1+130*1+2*x = 170*4
The price of the third variety x = 160
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23. An alloy contains Brass, Iron and Zinc In the ratio 2:3:1 and another contains Iron, zinc and lead In the ratio 5:4:3.lf equal weights of both alloys are melted together to form a third alloy, then what will be the weight of lead per kg in new alloy?
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Correct Ans:(1/8)
Explanation:
B:I:Z = 2:3:1
Similarly,In the second alloy, ratio of Iron, zinc and lead is given as,
I:Z:L = 5:4:3
The trick here is to arrive at a quantity where calculation becomes easy.
To do that, we take LCM of6(=2+3+1),taken as 2kg Brass, 3kg Iron and 1Kg Zinc), and12(=5+4+3), taken as 5 kg Iron, 4 kg zinc and 3 Kg Lead, which is12.
Thus we assume that both the alloys are being mixed at 12 Kgs each.
Alloys are mixed together to form third alloy. Then the ratio of content in it,
Brass:Iron:Zinc:Lead
B:I:Z:L = 4:(6+5):(2+4):3
Weight of the third alloy= 12+1 2 = 24Kg.
So, weight of the Lead,
Weight of lead per kg in new alloy = (3/24) = 1/8
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24. A vessel is filled with liquid, 4 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
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Correct Ans:1/10
Explanation:
Suppose the vessel initially contains 9 litres of liquid
Let x litres of this liquid be replaced with water.
(4-(4x/9) +x) = (5-(5x/9))
x = (9/10)
So, part of the mixture replaced = (9/10) * (1/9) = (1/10)
Let x litres of this liquid be replaced with water.
(4-(4x/9) +x) = (5-(5x/9))
x = (9/10)
So, part of the mixture replaced = (9/10) * (1/9) = (1/10)
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25. A jar full of vinegar contains 48% alcohol. A part of this vinegar is replaced by another containing 16% alcohol and now the percentage of alcohol was found to be 26%. The quantity of vinegar replaced is
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Correct Ans:(11/16)
Explanation:
Concentration of alcohol in 1st jar = 48%
Concentration of alcohol in 2nd jar = 16%
After the mixing, concentration of alcohol in the mixture = 26%
By alligation rule,
The ratio of 1st and 2nd quantity = 10 : 22
= 5 : 11
The quantity of vinegar replaced is = 11/(5 + 11)
= 11/16 parts
Concentration of alcohol in 2nd jar = 16%
After the mixing, concentration of alcohol in the mixture = 26%
By alligation rule,
The ratio of 1st and 2nd quantity = 10 : 22
= 5 : 11
The quantity of vinegar replaced is = 11/(5 + 11)
= 11/16 parts
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26. There are two containers of equal capacity. The ratio of water to alcohol in the first container is 6 : 7, and in the second container is 4 : 5. If they are mixed up, the ratio of water to alcohol in the mixture will be?
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Correct Ans:53 : 64
Explanation:
Sum of the ratios = 6 + 7 = 13 and 4 + 5 = 9
As both the containers have equal volume, each container's volume
=> 13*9 = 117 liters
The ratio in the first container becomes
6*9 : 7*9 = 54 : 63
In the second container the ratio will be,
13*4 : 13*5 = 52 : 65
Adding both, we get
Water : Alcohol => (54+52) : (63+65) = 106 : 128 = 53 : 64
As both the containers have equal volume, each container's volume
=> 13*9 = 117 liters
The ratio in the first container becomes
6*9 : 7*9 = 54 : 63
In the second container the ratio will be,
13*4 : 13*5 = 52 : 65
Adding both, we get
Water : Alcohol => (54+52) : (63+65) = 106 : 128 = 53 : 64
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27. One test tube contains some acid and another test tube contains an equal quantity of water. To prepare a solution, 20 L of the acid is poured into the second test tube. Then, two-thirds of the so formed solution is poured from the second test tube into the first. If the fluid in the first test tube is four times that in second, what quantity of water was taken initially.
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Correct Ans:100 litres
Explanation:
Initially, let x litres of water and Acid was taken.
In first process,
20 litres of the acid is poured into the second test tube
2nd test tube = (x + 20) litres
1st test tube = (x - 20) litres
In second process,
2/3 of the so formed solution is poured from the second test tube into the first.
1st test tube = (x - 20) + [(x + 20)*2/3]
= x - 20 + 2x/3 + 40/3 = 5x/3 - 20/3 = (5x - 20)/3
2nd test tube = (x + 20)*(1/3) = (x + 20)/3
If the fluid in the first test tube is four times that in second then,
(5x - 20)/3 = 4 * [(x + 20)/3]
5x - 20 = 4 * (x + 20)
5x - 20 = 4x + 80
x = 100 litres
In first process,
20 litres of the acid is poured into the second test tube
2nd test tube = (x + 20) litres
1st test tube = (x - 20) litres
In second process,
2/3 of the so formed solution is poured from the second test tube into the first.
1st test tube = (x - 20) + [(x + 20)*2/3]
= x - 20 + 2x/3 + 40/3 = 5x/3 - 20/3 = (5x - 20)/3
2nd test tube = (x + 20)*(1/3) = (x + 20)/3
If the fluid in the first test tube is four times that in second then,
(5x - 20)/3 = 4 * [(x + 20)/3]
5x - 20 = 4 * (x + 20)
5x - 20 = 4x + 80
x = 100 litres
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28. A dishonest hairdresser uses a mixture having 5 parts after-shave lotion and 2 parts water. After taking out some portion of the mixture, he adds an equal amount of water to the remaining portion of the mixture such that the amount of after-shave lotion and water becomes equal. Find the part of mixture taken out.
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Correct Ans:3/10
Explanation:
Let quantity of after-shave lotion = 5 ltr
Quantity of water = 2 ltr
Total quantity = 5 + 2 = 7 ltr
Let quantity of mixture taken out be x
Quantity of after-shave lotion removed = 5x/7
Quantity of water removed = 2x/7
Amount of after-shave lotion and water becomes equal
5 - (5x/7) = 2 - (2x/7) + x
Where some amount of water is added after some portion of mixture is removed.
5 - 2 = (5x/7) - (2x/7) + x
3 = (10x/7)
x = 21/10
Therefore the part of mixture taken out = (21/10) / 7
= 21/(10*7)
= 3/10
= 3/10 of mixture is taken out.
Quantity of water = 2 ltr
Total quantity = 5 + 2 = 7 ltr
Let quantity of mixture taken out be x
Quantity of after-shave lotion removed = 5x/7
Quantity of water removed = 2x/7
Amount of after-shave lotion and water becomes equal
5 - (5x/7) = 2 - (2x/7) + x
Where some amount of water is added after some portion of mixture is removed.
5 - 2 = (5x/7) - (2x/7) + x
3 = (10x/7)
x = 21/10
Therefore the part of mixture taken out = (21/10) / 7
= 21/(10*7)
= 3/10
= 3/10 of mixture is taken out.
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29. A solution of salt and water contains 30% of salt by weight. If its 20 ltr water evaporates and the solution contains 45% of salt ,the original quantity of solution was:
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Correct Ans:60 ltr
Explanation:
Let mixture was X initially.
Initial Salt = 30% of X = 0.3X
Initial Water = 70% of X = 0.7X
20 liters of water evaporated.
So, Final water = (0.7X - 20)
Now, Total Mixture = (0.7X - 20 + 0.3X)
Now, salt rises to 45% of the mixture, it means as water dried up salt rises.
Final salt = 45% of (0.7X - 20 + 0.3X)
As, Initial Amount of salt = Final amount of salt
0.3X = 45% (1.0X - 20)
0.3X = 0.45X - 9
0.45X - 0.3X = 9
0.15X = 9
(15/100)X = 9
X = (3*100)/5 = 60 ltr.
Initial Salt = 30% of X = 0.3X
Initial Water = 70% of X = 0.7X
20 liters of water evaporated.
So, Final water = (0.7X - 20)
Now, Total Mixture = (0.7X - 20 + 0.3X)
Now, salt rises to 45% of the mixture, it means as water dried up salt rises.
Final salt = 45% of (0.7X - 20 + 0.3X)
As, Initial Amount of salt = Final amount of salt
0.3X = 45% (1.0X - 20)
0.3X = 0.45X - 9
0.45X - 0.3X = 9
0.15X = 9
(15/100)X = 9
X = (3*100)/5 = 60 ltr.
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30. Three vessels contain equal mixtures of acid and water in the ratio 6 : 1, 5 : 2 and 3 : 1 respectively. If all the solutions are mixed together, the ratio of acid to water in the final mixture will be
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Correct Ans:65 : 19
Explanation:
All vessels contains equal mixtures of acid and water in different ratios. And ratios are given in question as:
And the Ratios of Acid to Water for 3 vessels:
Hence the final ratio of mixture = Total Acid content / Total Water content
= [(6/7) + (5/7) + (3/4)] / [(1/7) + (2/7) + (1/4)]
= [(65/28) / (19/28)]
= (65/19) or 65 :19
Ist vessel |
IInd vessel |
IIIrd vessel |
6 : 1 |
5 : 2 |
3 : 1 |
And the Ratios of Acid to Water for 3 vessels:
Ist vessel |
IInd vessel |
IIIrd vessel |
|||
Acid |
Water |
Acid |
Water |
Acid |
Water |
6/7 |
1/7 |
5/7 |
2/7 |
3/4 |
1/4 |
Hence the final ratio of mixture = Total Acid content / Total Water content
= [(6/7) + (5/7) + (3/4)] / [(1/7) + (2/7) + (1/4)]
= [(65/28) / (19/28)]
= (65/19) or 65 :19
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31. A vessel which contains a mixture of acid and water in ratio 13 : 4. 25.5 litres of mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of acid is added to the mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture in the vessel before the replacement in litres?
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Correct Ans:68
Explanation:
Given: Ratio of syrup and water = 13 : 4
Let the quantity of syrup be '13x' and quantity of water be '4x'.
Total quantity of mixture = 17x
Quantity of resultant mixture = 17x - 25.5 + 2.5 + 5 = 17x - 18
Quantity of resultant water = 4x - 25.5(4/17) + 2.5
= 4x - 3.5
Given that resultant mixture contain 25% of water,
25% of (17x - 18) = 4x - 3.5
(25/100)*(17x -18) = 4x - 3.5
(1/4)*(17x -18) = (4x - 3.5)
17x - 18 = 16x - 14
x = 4
Therefore, initial quantity of mixture = 17x = 17(4) = 68 liters.
Let the quantity of syrup be '13x' and quantity of water be '4x'.
Total quantity of mixture = 17x
Quantity of resultant mixture = 17x - 25.5 + 2.5 + 5 = 17x - 18
Quantity of resultant water = 4x - 25.5(4/17) + 2.5
= 4x - 3.5
Given that resultant mixture contain 25% of water,
25% of (17x - 18) = 4x - 3.5
(25/100)*(17x -18) = 4x - 3.5
(1/4)*(17x -18) = (4x - 3.5)
17x - 18 = 16x - 14
x = 4
Therefore, initial quantity of mixture = 17x = 17(4) = 68 liters.
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32. There are two mixtures such that they contain 75% milk and 80% milk respectively. Some amount from first mixture is mixed with twice the same amount of second mixture. Find the percentage of milk in the resultant mixture?
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Correct Ans:78.30%
Explanation:
Let the amount from first mixture be 'X' and amount from the second mixture be '2X'.
Milk from first mixture = (75/100)*X = 75X/100
And milk from second mixture = (80X/100)*2X = 160X/100
So, milk in the resultant mixture = (75X/100) + (160X/100)
= 235X/100
= 2.35X
Total resultant mixture = X + 2X = 3X
Therefore, percentage of milk in resultant mixture = (2.35X/3X)*100
= 78.3%
Milk from first mixture = (75/100)*X = 75X/100
And milk from second mixture = (80X/100)*2X = 160X/100
So, milk in the resultant mixture = (75X/100) + (160X/100)
= 235X/100
= 2.35X
Total resultant mixture = X + 2X = 3X
Therefore, percentage of milk in resultant mixture = (2.35X/3X)*100
= 78.3%
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33. What is the minimum quantity of milk in liters that should be mixed in a mixture of 60 liters in which the initial ratio of milk to water is 1:4 such that the resulting mixture has 15 % milk?
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Correct Ans:Not possible
Explanation:
Given: Quantity of mixture = 60L
Quantity of milk = (1/5)*60 = 12 L
Quantity of water = 60 - 12 = 48L
Percentage of the milk in the mixture = (12/60)*100 = 20%
Therefore, when milk is added to the mixture, the percentage of the milk will be increased. But in question its given that it is 15% which is lesser than the original milk percentage.
So the correct answer is not possible.
Quantity of milk = (1/5)*60 = 12 L
Quantity of water = 60 - 12 = 48L
Percentage of the milk in the mixture = (12/60)*100 = 20%
Therefore, when milk is added to the mixture, the percentage of the milk will be increased. But in question its given that it is 15% which is lesser than the original milk percentage.
So the correct answer is not possible.
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34. A vessel contains 72 litres of mixture of milk and water. The respective ratio of milk and water in the vessel is 5 : 4. If 18 litres of mixtures is taken out from that vessel and then 10 litres of water is added, what will be the percentage of milk in that mixture?
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Correct Ans:46(7/8)
Explanation:
If 18 litres of mixtures is taken out,
Remaining mixture = 72 - 18 = 54 litres of mixtures.
Quantity of milk = (5/9)*54 = 30 litres
Quantity of water = (4/9)*54 = 24 litres
On adding 10 litres of water to the mixture,
Final quantity of mixture = 54 + 10 = 64 litres
Percentage of milk in final mixture = (30/64)*100
= 375/8
= 46(7/8)%
Remaining mixture = 72 - 18 = 54 litres of mixtures.
Quantity of milk = (5/9)*54 = 30 litres
Quantity of water = (4/9)*54 = 24 litres
On adding 10 litres of water to the mixture,
Final quantity of mixture = 54 + 10 = 64 litres
Percentage of milk in final mixture = (30/64)*100
= 375/8
= 46(7/8)%
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35. In a mixture of 60 L the ratio of the spirit and water is 2: 1. If the ratio of spirit and water is to be 1: 2, then the amount of water to be added in the mixture is?
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Correct Ans:60 L
Explanation:
Given: Mixture total quantity - 60L; Ratio of the spirit and water = 2: 1
Quantity of Spirit = (2/3)*60 = 40 L
Quantity of Water = (1/3)*60 = 20 L
Let the amount of water to be added be X.
If the ratio of spirit and water is to be 1: 2,
40/(20 + X) = 1/2
80 = 20 + X
X = 60
Therefore, the amount of water to be added in the mixture is 60 L.
Quantity of Spirit = (2/3)*60 = 40 L
Quantity of Water = (1/3)*60 = 20 L
Let the amount of water to be added be X.
If the ratio of spirit and water is to be 1: 2,
40/(20 + X) = 1/2
80 = 20 + X
X = 60
Therefore, the amount of water to be added in the mixture is 60 L.
Workspace
36. In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 per kg?
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Correct Ans:(7 : 3)
Explanation:
By the rule of alligation:
Therefore, Required ratio = 3.50 : 1.50
(Multiply by 100 on both sides)
= 350 : 150
= 35 : 15
(Divide by 5 on both sides)
= 7 : 3
Therefore, Required ratio = 3.50 : 1.50
(Multiply by 100 on both sides)
= 350 : 150
= 35 : 15
(Divide by 5 on both sides)
= 7 : 3
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37. In what ratio,water must be mixed with fruit juice costing Rs.24 per litre so that the juice would be worth of Rs.20 per litre?
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Correct Ans:(1 : 5)
Explanation:
Cost of 1 litre of water = Rs.0 = cheaper quantity.
Cost of 1 litre of juice = Rs.24 = dearer quantity.
And, the mean price = m = Rs.20
Applying the rule of alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (Cost of dearer – Mean Price):(Mean Price – Cost of cheaper)
Therefore, Quantity of Water : Juice = (24 - 20) : (20 - 0)
= 4 : 20
= 1 : 5
Cost of 1 litre of juice = Rs.24 = dearer quantity.
And, the mean price = m = Rs.20
Applying the rule of alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (Cost of dearer – Mean Price):(Mean Price – Cost of cheaper)
Therefore, Quantity of Water : Juice = (24 - 20) : (20 - 0)
= 4 : 20
= 1 : 5
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38. A certain quantity of water is mixed with milk priced at Rs 12 per litre. The price of mixture is Rs 8 per litre. Find out the ratio of water and milk in the new mixture.
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Correct Ans:(1 : 2)
Explanation:
Let, Cost Price of 1 litre of water = Rs 0.
Given, Cost Price of 1 litre of milk = Rs 12.
Mean Price of Mixture = Rs 8.
According to the Rule of Alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (CP of dearer â€“ Mean Price):(Mean Price â€“ CP of cheaper)
Therefore, Ratio of Quantity of Water : Milk = (12-8):(8-0)
= 4:8
= 1:2.
Given, Cost Price of 1 litre of milk = Rs 12.
Mean Price of Mixture = Rs 8.
According to the Rule of Alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (CP of dearer â€“ Mean Price):(Mean Price â€“ CP of cheaper)
Therefore, Ratio of Quantity of Water : Milk = (12-8):(8-0)
= 4:8
= 1:2.
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39. A shopkeeper deals in milk and 45 litre mixture is to be distributed in Milk & Water in the ratio of 4 : 1. If 4 litre milk & 3 litre water will be added in the mixture then what will be the new ratio of water and milk?
SHOW ANSWER
Correct Ans:(3 : 10)
Explanation:
As per the question, Milk : Water = 4 : 1
---> Milk = 4x liters
Water = x liters
Mixture = 45 liters
---> 4x + x = 45
---> 5x = 45
---> x = 9
Thus, Milk = 4 * 9 = 36 liters
Water = 9 liters
If 4 litre milk & 3 litre water is added in the mixture then,
Milk = 36 + 4 = 40 liters
Water = 9 + 3 = 12 liters
New Ratio becomes ---> Water : Milk = 12 : 40
= 3 : 10
---> Milk = 4x liters
Water = x liters
Mixture = 45 liters
---> 4x + x = 45
---> 5x = 45
---> x = 9
Thus, Milk = 4 * 9 = 36 liters
Water = 9 liters
If 4 litre milk & 3 litre water is added in the mixture then,
Milk = 36 + 4 = 40 liters
Water = 9 + 3 = 12 liters
New Ratio becomes ---> Water : Milk = 12 : 40
= 3 : 10
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40. How many litres of water should be added to a 60 litres mixture containing milk and water in the ratio of 2:1 such that the resultant mixture has 50% milk in it?
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Correct Ans:20 litres
Explanation:
Total mixture = 60 litres
Ratio of Milk and water = 2 : 1
---> Milk = 2x litres and water = x litres
---> 2x + x = 60
---> 3x = 60
---> x = 20
Hence, Milk = 2x = 2 * 20 = 40 litres
water = x = 20 litres
Given, resultant mixture has 50% milk, which means the new ratio of Milk and water = 50% : 50% = 1:1
So, x litres of water should be added to make the ratio 1:1
----> Milk/water = 40/(20+x) = 1/1
----> 40 = 20+x
----> x = 20 litres
Hence, 20 litres of water should be added.
Ratio of Milk and water = 2 : 1
---> Milk = 2x litres and water = x litres
---> 2x + x = 60
---> 3x = 60
---> x = 20
Hence, Milk = 2x = 2 * 20 = 40 litres
water = x = 20 litres
Given, resultant mixture has 50% milk, which means the new ratio of Milk and water = 50% : 50% = 1:1
So, x litres of water should be added to make the ratio 1:1
----> Milk/water = 40/(20+x) = 1/1
----> 40 = 20+x
----> x = 20 litres
Hence, 20 litres of water should be added.
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