Question

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:

Answer 1

each time , 20% of the content is replaced by water.

Intially, it was 100% milk.

After first operation, pure milk = 100 - (100 × 20/100) = 100 - 20 = 80%

After second operation, pure milk = 80 - (80× 20/100) = 80 - 16 = 64%

After third operation, pure milk = 64 - (64× 20/100) = 64 - 12.8 = 51.2%

So, after third operation, pure milk = 51.2%
_________________________

fraction of Pure milk after 3rd operation = (1 - 20/100)^3 = 0.8^3 = 0.512

percentage = 0.512 × 100 = 51.2%

Answer 2

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each time , 20% of the content is replaced by water.

Intially, it was 100% milk.

After first operation, pure milk = 100 - (100 × 20/100) = 100 - 20 = 80%

After second operation, pure milk = 80 - (80× 20/100) = 80 - 16 = 64%

After third operation, pure milk = 64 - (64× 20/100) = 64 - 12.8 = 51.2%

So, after third operation, pure milk = 51.2%

_________________________

fraction of Pure milk after 3rd operation = (1 - 20/100)^3 = 0.8^3 = 0.512

percentage = 0.512 × 100 = 51.2%

HOPE SO IT WILL HELP..........