In a 120 litre mixture of milk and water,wate | vedclass

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(C) Initial mixture $= 120$ $litre$,water $= 25 \%$.Water $= 0.25 	imes 120 = 30$ $litre$.Milk $= 120 - 30 = 90$ $litre$.After selling $20$ $litre$ o

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$24$

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(C) Initial mixture $= 120$ $litre$,water $= 25 \%$.Water $= 0.25 	imes 120 = 30$ $litre$.Milk $= 120 - 30 = 90$ $litre$.After selling $20$ $litre$ of the mixture,the remaining mixture $= 120 - 20 = 100$ $litre$.In the remaining $100$ $litre$ mixture,the ratio of milk to water remains the same ($75 \%$ milk and $25 \%$ water).Water in $100$ $litre$ mixture $= 0.25 	imes 100 = 25$ $litre$.Milk in $100$ $litre$ mixture $= 0.75 	imes 100 = 75$ $litre$.After adding $16.2$ $litre$ of milk and $3.8$ $litre$ of water:New milk quantity $= 75 + 16.2 = 91.2$ $litre$.New water quantity $= 25 + 3.8 = 28.8$ $litre$.Total new mixture volume $= 91.2 + 28.8 = 120$ $litre$.Percentage of water $= (28.8 / 120) 	imes 100 = 24 \%$.

In a $120$ $litre$ mixture of milk and water,water is only $25 \%$. The milkman sold $20$ $litre$ of this mixture and then added $16.2$ $litre$ of pure milk and $3.8$ $litre$ of pure water to the remaining mixture. What is the percentage of water in the final mixture (in $\%$)?

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