Two numbers are in the ratio 3: 5. If 9 is su | vedclass

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(B) Let the two numbers be $3k$ and $5k$,where $k$ is a constant.According to the problem,if $9$ is subtracted from each,the ratio becomes $12: 23$.So

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$33$

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(B) Let the two numbers be $3k$ and $5k$,where $k$ is a constant.According to the problem,if $9$ is subtracted from each,the ratio becomes $12: 23$.So,$\frac{3k - 9}{5k - 9} = \frac{12}{23}$.Cross-multiplying,we get: $23(3k - 9) = 12(5k - 9)$.Expanding the terms: $69k - 207 = 60k - 108$.Rearranging the terms to solve for $k$: $69k - 60k = 207 - 108$.$9k = 99$.$k = 11$.The smaller number is $3k = 3 	imes 11 = 33$.

Two numbers are in the ratio $3: 5$. If $9$ is subtracted from each,the new numbers are in the ratio $12: 23$. The smaller number is:

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