Two vessels A and B contain milk and water mi | vedclass

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(D) In vessel $A$,the fraction of milk is $\frac{8}{13}$.In vessel $B$,the fraction of milk is $\frac{5}{7}$.The required percentage of milk in the ne

Vedclass · Accepted Answer

$2:7$

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(D) In vessel $A$,the fraction of milk is $\frac{8}{13}$.In vessel $B$,the fraction of milk is $\frac{5}{7}$.The required percentage of milk in the new mixture is $69 \frac{3}{13} \% = \frac{900}{13} \% = \frac{900}{13 	imes 100} = \frac{9}{13}$.Using the rule of alligation:Milk in $A$ = $\frac{8}{13}$,Milk in $B$ = $\frac{5}{7}$,Mean value = $\frac{9}{13}$.Difference for $B$ = $\frac{8}{13} - \frac{9}{13} = \frac{1}{13}$ (Wait,using absolute difference: $|\frac{8}{13} - \frac{9}{13}| = \frac{1}{13}$).Difference for $A$ = $|\frac{5}{7} - \frac{9}{13}| = |\frac{65 - 63}{91}| = \frac{2}{91}$.Ratio of $A$ to $B$ = $\frac{2}{91} : \frac{1}{13} = \frac{2}{91} : \frac{7}{91} = 2:7$.

Two vessels $A$ and $B$ contain milk and water mixed in the ratio $8: 5$ and $5: 2,$ respectively. The ratio in which these two mixtures should be mixed to get a new mixture containing $69 \frac{3}{13} \%$ milk is

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