A bottle contains 81 litres of pure milk. fra | vedclass

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(D) Initial quantity of milk $(x)$ = $81$ litres.Amount replaced in each operation $(y)$ = $\frac{1}{3}$ of $81 = 27$ litres.Number of operations $(n)

Vedclass · Accepted Answer

$4:5$

Vedclass · Answer

(D) Initial quantity of milk $(x)$ = $81$ litres.Amount replaced in each operation $(y)$ = $\frac{1}{3}$ of $81 = 27$ litres.Number of operations $(n)$ = $2$.The quantity of milk remaining after $n$ operations is given by the formula: $Milk_{final} = x \left(1 - \frac{y}{x}ight)^n$.$Milk_{final} = 81 \left(1 - \frac{27}{81}ight)^2 = 81 \left(1 - \frac{1}{3}ight)^2 = 81 	imes \left(\frac{2}{3}ight)^2$.$Milk_{final} = 81 	imes \frac{4}{9} = 9 	imes 4 = 36$ litres.Quantity of water in the final mixture = Total volume - Quantity of milk = $81 - 36 = 45$ litres.Ratio of milk to water = $36 : 45$.Dividing both by $9$,we get $4 : 5$.

$A$ bottle contains $81$ litres of pure milk. $\frac{1}{3}$ of the milk is replaced by the same amount of water. Again,$\frac{1}{3}$ of the mixture is replaced by the same amount of water. The ratio of the milk and water in the new mixture is

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