If the digit in the unit's place of a two-dig | vedclass

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(B) Let the two-digit number be $10x + y$,where $x$ is the digit in the ten's place and $y$ is the digit in the unit's place.According to the problem,

Vedclass · Accepted Answer

Digit in the unit's place is twice the digit in the ten's place

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(B) Let the two-digit number be $10x + y$,where $x$ is the digit in the ten's place and $y$ is the digit in the unit's place.According to the problem,if the unit's digit is halved $(y/2)$ and the ten's digit is doubled $(2x)$,the new number is $10(2x) + y/2$.This new number is equal to the number obtained by interchanging the digits,which is $10y + x$.So,$10(2x) + y/2 = 10y + x$.$20x + y/2 = 10y + x$.Multiply the entire equation by $2$ to eliminate the fraction: $40x + y = 20y + 2x$.Rearranging the terms: $40x - 2x = 20y - y$.$38x = 19y$.Dividing both sides by $19$,we get $2x = y$.This means the digit in the unit's place $(y)$ is twice the digit in the ten's place $(x)$.

If the digit in the unit's place of a two-digit number is halved and the digit in the ten's place is doubled,the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?

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