A can contains a mixture of two liquids A and | vedclass

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(B) Let the initial quantity of liquid $A$ be $4x$ and liquid $B$ be $7x$. The total quantity is $11x$.When $6 	ext{ litres}$ of the mixture are remo

Vedclass · Accepted Answer

$12.46$

Vedclass · Answer

(B) Let the initial quantity of liquid $A$ be $4x$ and liquid $B$ be $7x$. The total quantity is $11x$.When $6 	ext{ litres}$ of the mixture are removed,the amount of $A$ removed is $\frac{4}{11} 	imes 6 = \frac{24}{11}$ and the amount of $B$ removed is $\frac{7}{11} 	imes 6 = \frac{42}{11}$.After adding $6 	ext{ litres}$ of $B$,the new quantity of $A$ is $4x - \frac{24}{11}$ and the new quantity of $B$ is $7x - \frac{42}{11} + 6$.The new ratio is given as $3:7$,so:$\frac{4x - \frac{24}{11}}{7x - \frac{42}{11} + 6} = \frac{3}{7}$$\frac{44x - 24}{77x - 42 + 66} = \frac{3}{7}$$7(44x - 24) = 3(77x + 24)$$308x - 168 = 231x + 72$$77x = 240 \Rightarrow x = \frac{240}{77}$The initial amount of liquid $A$ was $4x = 4 	imes \frac{240}{77} = \frac{960}{77} \approx 12.467 	ext{ litres}$.

$A$ can contains a mixture of two liquids $A$ and $B$ in the ratio $4:7$. When $6 \text{ litres}$ of the mixture are drawn off and the can is filled with liquid $B$,the ratio of $A$ and $B$ becomes $3:7$. How many litres of liquid $A$ were contained in the can initially?

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