Answer: (2)

Solution: (2)

In 20 litres of mixture

Milk $=\frac{3}{5} \times 20=12$ litres

Water $=8$ litres

In 10 litres of mixture,

Milk $=6$ litres

Water $=4$ litres

On adding 10 litres of milk,

Milk $\Rightarrow 12-6+10=16$ litres

Water $\Rightarrow 8-4=4$ litres

Again, in 10 litres of mixture,

Milk $=\frac{4}{5} \times 10=8$ litres

Water $=2$ litres

On adding 10 litres of milk,

Milk $=16-8+10=18$ litres

Water $=2$ litres

Required ratio $=18: 2=9: 1$

Aliter:

Remaining quantity of water $=$ Initial quantity of water $\left(1-\frac{\text { quantity taken out }}{\text { total capacity of vessel }}\right)^{n}$

[Note : Here, we are taking quantity of water, because mixture is replaced by pure milk]

$\Rightarrow$ Remaining quantity of water $=\frac{2}{5} \times 20\left(1-\frac{10}{20}\right)^{2}$ $=8\left(1-\frac{1}{2}\right)^{2}=8 \times\left(\frac{1}{2}\right)^{2}$ $=8 \times \frac{1}{4}=2$ Ltr.

$\therefore$ Ratio of milk and water in resultant mixture $=(20-2): 2=18: 2=9: 1$