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(B) Initial volume of mixture = $15$ litres.Quantity of alcohol = $20 \%$ of $15 = \frac{20}{100} 	imes 15 = 3$ litres.Quantity of water = $15 - 3 =

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$16 \frac{2}{3} \%$

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(B) Initial volume of mixture = $15$ litres.Quantity of alcohol = $20 \%$ of $15 = \frac{20}{100} 	imes 15 = 3$ litres.Quantity of water = $15 - 3 = 12$ litres.When $3$ litres of water is added,the new quantity of water = $12 + 3 = 15$ litres.The quantity of alcohol remains the same,i.e.,$3$ litres.Total volume of the new mixture = $3 + 15 = 18$ litres.Percentage of alcohol in the new mixture = $(\frac{	ext{Quantity of alcohol}}{	ext{Total volume}}) 	imes 100 = (\frac{3}{18}) 	imes 100 = \frac{1}{6} 	imes 100 = \frac{50}{3} \% = 16 \frac{2}{3} \%$.

$15$ litres of mixture contains $20 \%$ alcohol and the rest water. If $3$ litres of water be mixed with it,the percentage of alcohol in the new mixture would be:

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