729 , ml of a mixture contains milk and water | vedclass

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Question

(D) Total volume of the mixture $= 729 \, ml$.Ratio of milk to water $= 7:2$.Sum of ratio parts $= 7 + 2 = 9$.Value of one part $= 729 / 9 = 81 \, ml$

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(D) Total volume of the mixture $= 729 \, ml$.Ratio of milk to water $= 7:2$.Sum of ratio parts $= 7 + 2 = 9$.Value of one part $= 729 / 9 = 81 \, ml$.Quantity of milk $= 7 	imes 81 = 567 \, ml$.Quantity of water $= 2 	imes 81 = 162 \, ml$.We need to add water such that the new ratio becomes $7:3$. Since the quantity of milk remains constant $(567 \, ml)$,the new ratio $7:3$ implies that $7$ parts correspond to $567 \, ml$.Value of one part in the new ratio $= 567 / 7 = 81 \, ml$.Required quantity of water $= 3 	imes 81 = 243 \, ml$.Water to be added $= 243 \, ml - 162 \, ml = 81 \, ml$.

$729 \, ml$ of a mixture contains milk and water in the ratio $7:2$. How much more water is to be added to get a new mixture containing milk and water in the ratio $7:3$? (in $ml$)

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