From a container, full of pure milk, 20% is replaced by water and this process is repeated three times. At the end of third operation, the quantity of pure milk reduces to:

Solution(By Examveda Team)

Let pure milk was 100L. So,
Water is replaced 20% in per process = 20% of 100 = 20L.
Now, we use short-cut formula for it.
Quantity of Milk reduced to,
$$\eqalign{ & = {\text{X}} \times {\left[ {1 - {\frac{{\text{Y}}}{{\text{X}}}} } \right]^{\text{n}}} \cr & = 100 \times {\left[ {1 - {\frac{{20}}{{100}}} } \right]^3} \cr & = \frac{{100 \times 64}}{{125}} \cr & = 51.2\,{\text{%}} \cr} $$

Here,
X = Initial quantity of milk.
Y = Replaced water in per process.
n = No. of process repeated.

Note:
The formula used in above problem is quite similar to depreciation formula or Compound interest formula.

Alternatively,
Let pure milk be 100 litres initially.
After third operation, milk will be
100 == 20%↓(- 20L) ⇒ 80 == 20%↓(- 16L) ⇒ 64 == 20%↓(-12.8L) ⇒ 51.2 %