In a two-digit number,the digit in the unit's | vedclass

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(B) Let the digit in the ten's place be $x$ and the digit in the unit's place be $y$. The original number is $10x + y$.According to the problem,$y = 2

Vedclass · Accepted Answer

$37$

Vedclass · Answer

(B) Let the digit in the ten's place be $x$ and the digit in the unit's place be $y$. The original number is $10x + y$.According to the problem,$y = 2x + 1$ (Equation $1$).When the digits are interchanged,the new number is $10y + x$.The difference between the new number and the original number is $(10y + x) - (10x + y) = 9y - 9x$.According to the problem,this difference is $1$ less than the original number:$9y - 9x = (10x + y) - 1$$8y - 19x = -1$ (Equation $2$).Substituting $y = 2x + 1$ into Equation $2$:$8(2x + 1) - 19x = -1$$16x + 8 - 19x = -1$$-3x = -9$$x = 3$.Now,find $y$ using Equation $1$:$y = 2(3) + 1 = 7$.The original number is $10(3) + 7 = 37$.

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