A 12 text{ litres} cylinder contains a mixtur | vedclass

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(C) Let $y$ litres of the mixture be released each time.Initially,the volume of oxygen is $40 \%$ of $12 	ext{ litres} = 0.40 	imes 12 = 4.8 	ext{

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$6$

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(C) Let $y$ litres of the mixture be released each time.Initially,the volume of oxygen is $40 \%$ of $12 	ext{ litres} = 0.40 	imes 12 = 4.8 	ext{ litres}$.The formula for the remaining amount of a substance after $n$ operations of replacing a volume $y$ from a total volume $V$ is: $A_{final} = A_{initial} 	imes (1 - \frac{y}{V})^n$.Here,$A_{initial} = 0.40$,$A_{final} = 0.10$,$V = 12$,and $n = 2$.Substituting the values: $0.10 = 0.40 	imes (1 - \frac{y}{12})^2$.Dividing both sides by $0.40$: $\frac{0.10}{0.40} = (1 - \frac{y}{12})^2$.$\frac{1}{4} = (1 - \frac{y}{12})^2$.Taking the square root of both sides: $\frac{1}{2} = 1 - \frac{y}{12}$.$\frac{y}{12} = 1 - \frac{1}{2} = \frac{1}{2}$.$y = 12 	imes \frac{1}{2} = 6 	ext{ litres}$.

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