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A container has two liquids $A$ and $B$ in certain ratio. $10\%$ of the mixture is taken out and replaced with liquid $B$. Now , the ratio of $A$ and $B$ become $2:3$. What was the initial ratio of $A$ and $B$ in the mixture?

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Let us assume there is 2 units of A and 3 units of B. If 10% of A was there previously, then there was $\frac1{1-0.1}\cdot2$ units of A (We take the inverse of the percentage of how much there is now compared to originally). This comes to $2\frac29$. So $2\frac29-2=\frac29$ of liquid B was added to replace the $\frac29$ units of A missing. That means there were originally $3-\frac29=2\frac79$ units of liquid B. Therefore, the ratio between A and B is $2\frac29:2\frac79=20:25=4:5$

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  • $\begingroup$ Thanks for answering. Could you elaborate it in a little detail? $\endgroup$ – user261207 Dec 15 '17 at 6:21

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