A 30-litre bucket contains a mixture of milk and water in the ratio 2:

Consider volume of milk = M and volume of water = W

Situation 1:
M/W=2/1
Let x denote the common factor: 2x+x=30
M = 20 and W=10

Situation 2:
Let us consider that "a" litres of solution has been replaced.
In "a" liters --> 2a/3 = milk and a/3 = water
M1= 20-2a/3
W1=10-(a/3)+a=10+(2a/3)
Note that we have replaced "a" liters of solution with water, hence "a" liters of water should be added
M1/W1=8/7
Solving this equation we get a=6
M1 = 16 and W1=14

Situation 3:
M2/W2=3/1
Let y be common factor: 3y+y=20
M2= 15 and W2=5

Total M = 16+15 = 31
Total W=14+5=19

Total M/Total(m+W) = 31/50 =62%

drakell thank you for noticing the mistake.

In situation 2, how is it possible for M to be the same as in the initial scenario? Volume of the milk is removed, and replaced with only water.

let m=percentage of milk in new mixture
(8/15)(30)+(3/4)(20)=50m➡
m=31/50=62%
D

1). we have 20 liters of milk & 10 liters of water
2). 20-x/10+x = 8/7 , we can find x = 4. so now we have 16 liters of milk and 14 liters of water
3). we pour another 20 liters in a ratio 3:1, so we have another 15 liters milk and 5 liters water. Combining with the previous numbers:

16+15/ 14+5 = 31/19 -> looking for % of total: 31/50 or 62/ 100 -> 62%

Several litres of acid were drawn off a 54 L vessel full of acid and an equal amount of water added. Again, the same volume of the mixture was drawn off and replaced by water. As a result the vessel contained 24 L of pure acid. How much of the acid was drawn off initially?

Please solve this one?

Is there a shorter way to solve this problem?

Given:
1. A 30-litre bucket contains a mixture of milk and water in the ratio 2:1.
2. A certain quantity of this mixture is replaced by water such that the ratio of milk and water becomes 8:7.
3. This resultant mixture is added to 20 litres of another mixture containing milk and water in the ratio 3:1.
Asked: What is the percentage of milk in the new mixture?

1. A 30-litre bucket contains a mixture of milk and water in the ratio 2:1.
Milk = 20 litres
Water = 10 litres

2. A certain quantity of this mixture is replaced by water such that the ratio of milk and water becomes 8:7.
Milk = 16 litres
Water = 14 litres

3. This resultant mixture is added to 20 litres of another mixture containing milk and water in the ratio 3:1.
Milk in another mixture = 15 litres
Water in another mixture = 5 litres

What is the percentage of milk in the new mixture?
Milk = 16 + 15 = 31 litres
Water = 14 + 5 = 19 litres
Milk % = 31/50 * 100 % = 62%

IMO D

for the given 30 ltrs of mix 2:1 milk and water
milk ; 20 and water = 10
later certain qty of mix is replaced by water so new ratio of water ; 7/15 and milk 8/15
qty of water ; 14 and milk ; 16

now 20 liters has milk ; 3/4 *20 ; 15 liters and water ; 5 liters
net qty ; 30+20 ; 50 ltrs
net milk ; 15+16 ; 31 liters
31/50 ; 62%
OPTION D

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