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(A) Given that the sum of $6\%$ of $A$ and $4\%$ of $B$ is $\frac{2}{3}$ of the sum of $6\%$ of $A$ and $8\%$ of $B$.Mathematically,this can be writte

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$2:3$

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(A) Given that the sum of $6\%$ of $A$ and $4\%$ of $B$ is $\frac{2}{3}$ of the sum of $6\%$ of $A$ and $8\%$ of $B$.Mathematically,this can be written as:$\frac{6}{100}A + \frac{4}{100}B = \frac{2}{3} \left( \frac{6}{100}A + \frac{8}{100}B \right)$Multiply both sides by $100$ to simplify:$6A + 4B = \frac{2}{3}(6A + 8B)$Multiply by $3$ to remove the fraction:$3(6A + 4B) = 2(6A + 8B)$$18A + 12B = 12A + 16B$Rearrange the terms to group $A$ and $B$:$18A - 12A = 16B - 12B$$6A = 4B$Therefore,the ratio $A:B$ is:$\frac{A}{B} = \frac{4}{6} = \frac{2}{3}$

Two numbers $A$ and $B$ are such that the sum of $6\%$ of $A$ and $4\%$ of $B$ is two-thirds of the sum of $6\%$ of $A$ and $8\%$ of $B$. Find the ratio of $A:B$.

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