In a two-digit number,the digit at the ten's | vedclass

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(D) Let the digit at the unit's place be $x$ and the digit at the ten's place be $y$.According to the problem,the digit at the ten's place is double t

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(D) Let the digit at the unit's place be $x$ and the digit at the ten's place be $y$.According to the problem,the digit at the ten's place is double the digit at the unit's place,so $y = 2x$.The original number can be expressed as $10y + x$.Substituting $y = 2x$,the original number is $10(2x) + x = 20x + x = 21x$.When the digits are interchanged,the new number becomes $10x + y$.Substituting $y = 2x$,the new number is $10x + 2x = 12x$.According to the problem,the new number is $27$ less than the original number,so $21x - 12x = 27$.$9x = 27$,which gives $x = 3$.Since $y = 2x$,we have $y = 2(3) = 6$.The original number is $10(6) + 3 = 63$.

In a two-digit number,the digit at the ten's place is double the digit at the unit's place. If the digits of the number are interchanged,the new number so formed is $27$ less than the original number. Find the original number.

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