Conc of milk in mixture (C1) = 2/5
Conc of milk in pure milk (C2) = 1
Conc of milk in new mix (Cavg) = 1/2

w1/w2 = (1 - 1/2) / (1/2 - 2/5) = (1/2) / (1/10) = 5/1

So original mixture was 5 parts and pure milk was 1 part to get equal proportions of milk and water. Total 6 parts is actually 60 litres which means the 5 parts are 50 litres and 1 part is 10 litres.
Hence 10 litres of original mix was replaced with pure milk.

Answer (B)

Check this post: https://www.veritasprep.com/blog/2012/0 ... -mixtures/
_________________

A great weighted average method above by VeritasKarishma..
But let me give you another method..
M:W=2:3 means M = \(\frac{2}{5}\) and W=\(\frac{3}{5}\)
Total 60 liters means M = \(\frac{2}{5}*60=24\) and W=\(\frac{3}{5}*60=36\)
To be equal both should be 30 each, so a decrease of 6 liters of water from initial 36 liters...

Now if reduction of 1 liter would mean reduction of \(\frac{3}{5}\) of water, reduction of x liter would mean \(\frac{3}{5}*x\), but this should be equal to 6..
thus \(\frac{3}{5}*x=6.....x=\frac{5}{3}*6=10\)

B
_________________

Since milk and water is in a ratio of 2 to 3, we can create the following equation:

2n + 3n = 60

5n = 60

n = 12

So originally, the container has 24 litres of milk and 36 liters of water. We are told that x liters of the mixture are removed. Note that these x liters that are removed will consist of 3x/5 liters of water and 2x/5 liters of milk. . The x liters of liquid that were removed are replaced by x liters of pure milk. Therefore, we can create the equation:

(24 - 2x/5 + x)/(36 - 3x/5) = 1

24 + 3x/5 = 36 - 3x/5

120 + 3x = 180 - 3x

6x = 60

x = 10

Answer: B
_________________

Quantity of milk in original = 24L
Quantity of milk in final =30L
Quantity of milk removed = 2x/5 L
Quantity of milk added = x L
Therefore, 24-2x/5 +x = 30
x=10