In a mixture of milk and water,the proportion | vedclass

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(D) Initial weight of the mixture $= 60 	ext{ gm}$.Weight of water in the initial mixture $= 60 	imes \frac{75}{100} = 45 	ext{ gm}$.When $15 	ext

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$80$

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(D) Initial weight of the mixture $= 60 	ext{ gm}$.Weight of water in the initial mixture $= 60 	imes \frac{75}{100} = 45 	ext{ gm}$.When $15 	ext{ gm}$ of water is added,the new weight of water $= 45 + 15 = 60 	ext{ gm}$.The new total weight of the mixture $= 60 + 15 = 75 	ext{ gm}$.Required percentage of water $= \left( \frac{	ext{New weight of water}}{	ext{New total weight of mixture}} ight) 	imes 100 = \left( \frac{60}{75} ight) 	imes 100 = 0.8 	imes 100 = 80 \%$.

In a mixture of milk and water,the proportion of water by weight was $75 \%$. If $15 \text{ gm}$ of water is added to $60 \text{ gm}$ of this mixture,what will be the new percentage of water by weight (in $\%$)?

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