85 , L of a mixture contains milk and water i | vedclass

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(A) Total volume of the mixture = $85 \, L$.The ratio of milk to water is $27: 7$.Sum of ratio terms = $27 + 7 = 34$.Value of $1$ unit = $\frac{85}{34

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Total volume of the mixture = $85 \, L$.The ratio of milk to water is $27: 7$.Sum of ratio terms = $27 + 7 = 34$.Value of $1$ unit = $\frac{85}{34} \, L = 2.5 \, L$.Initial amount of milk = $27 	imes 2.5 = 67.5 \, L$.Initial amount of water = $7 	imes 2.5 = 17.5 \, L$.Let $x$ liters of water be added to the mixture.The new ratio of milk to water becomes $3: 1$.So,$\frac{67.5}{17.5 + x} = \frac{3}{1}$.$67.5 = 3 	imes (17.5 + x)$.$67.5 = 52.5 + 3x$.$3x = 67.5 - 52.5 = 15$.$x = \frac{15}{3} = 5 \, L$.Therefore,$5 \, L$ of water must be added.

$85 \, L$ of a mixture contains milk and water in the ratio $27: 7$. How much more water (in $L$) is to be added to get a new mixture containing milk and water in the ratio $3: 1$?

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