Acid and water are mixed in a vessel A in the | vedclass

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Question

(A) In vessel $A$,the fraction of acid is $\frac{5}{7}$.In vessel $B$,the fraction of acid is $\frac{8}{13}$.The required mixture should have an acid

Vedclass · Accepted Answer

$7:2$

Vedclass · Answer

(A) In vessel $A$,the fraction of acid is $\frac{5}{7}$.In vessel $B$,the fraction of acid is $\frac{8}{13}$.The required mixture should have an acid fraction of $\frac{9}{13+4} = \frac{9}{13}$.Using the rule of alligation:Ratio of quantities = (Difference between acid in $B$ and mean) : (Difference between mean and acid in $A$)Ratio = $|\frac{9}{13} - \frac{8}{13}| : |\frac{5}{7} - \frac{9}{13}|$Ratio = $\frac{1}{13} : |\frac{65-63}{91}| = \frac{1}{13} : \frac{2}{91}$To simplify,multiply both sides by $91$:Ratio = $(\frac{1}{13} 	imes 91) : (\frac{2}{91} 	imes 91) = 7 : 2$.

Acid and water are mixed in a vessel $A$ in the ratio of $5:2$ and in the vessel $B$ in the ratio $8:5$. In what proportion should quantities be taken out from the two vessels so as to form a mixture in which the acid and water will be in the ratio of $9:4$?

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