The sum of the digits of a two-digit number i | vedclass

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(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.According to the problem,the sum of the digits is $x + y = 9$ (Equat

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$36$

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(B) Let the tens digit be $x$ and the units digit be $y$. The number is $10x + y$.According to the problem,the sum of the digits is $x + y = 9$ (Equation $1$).When $27$ is added to the number,the digits are reversed,so the new number is $10y + x$.Thus,$(10x + y) + 27 = 10y + x$.Rearranging the terms: $10x - x + y - 10y = -27$,which simplifies to $9x - 9y = -27$.Dividing by $9$,we get $x - y = -3$ (Equation $2$).Adding Equation $1$ and Equation $2$: $(x + y) + (x - y) = 9 + (-3)$,which gives $2x = 6$,so $x = 3$.Substituting $x = 3$ into Equation $1$: $3 + y = 9$,so $y = 6$.The number is $10(3) + 6 = 36$.

The sum of the digits of a two-digit number is $9$. If $27$ is added to it,the digits of the number get reversed. The number is

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