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(A) Given that the sum of $5 \%$ 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 wr

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$4:3$

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(A) Given that the sum of $5 \%$ 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:$0.05A + 0.04B = \frac{2}{3} (0.06A + 0.08B)$Multiply both sides by $3$ to eliminate the fraction:$3(0.05A + 0.04B) = 2(0.06A + 0.08B)$$0.15A + 0.12B = 0.12A + 0.16B$Rearrange the terms to group $A$ and $B$:$0.15A - 0.12A = 0.16B - 0.12B$$0.03A = 0.04B$$\frac{A}{B} = \frac{0.04}{0.03} = \frac{4}{3}$Therefore,the ratio $A:B$ is $4:3$.

Two numbers $A$ and $B$ are such that the sum of $5 \%$ of $A$ and $4 \%$ of $B$ is $\frac{2}{3}$ of the sum of $6 \%$ of $A$ and $8 \%$ of $B$. The ratio $A:B$ is

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