A tub contains a mixture of two liquids P and | vedclass

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(D) Let the initial amount of liquid $P$ be $4x$ and liquid $Q$ be $x$.When $10 \, L$ of the mixture is removed,the amount of $P$ removed is $\frac{4}

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) Let the initial amount of liquid $P$ be $4x$ and liquid $Q$ be $x$.When $10 \, L$ of the mixture is removed,the amount of $P$ removed is $\frac{4}{5} 	imes 10 = 8 \, L$ and the amount of $Q$ removed is $\frac{1}{5} 	imes 10 = 2 \, L$.After removing the mixture,the remaining amount of $P$ is $4x - 8$ and the remaining amount of $Q$ is $x - 2$.Then,$10 \, L$ of liquid $Q$ is added,so the new amount of $Q$ is $x - 2 + 10 = x + 8$.The new ratio is given as $2:3$,so $\frac{4x - 8}{x + 8} = \frac{2}{3}$.Cross-multiplying gives $3(4x - 8) = 2(x + 8)$,which simplifies to $12x - 24 = 2x + 16$.$10x = 40$,so $x = 4$.The initial quantity of liquid $P$ was $4x = 4 	imes 4 = 16 \, L$.

$A$ tub contains a mixture of two liquids $P$ and $Q$ in the ratio of $4:1$. When $10 \, L$ of the mixture is taken out and $10 \, L$ of liquid $Q$ is poured into the tub,the ratio becomes $2:3$. How many litres of liquid $P$ were contained in the tub initially?

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