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

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Question

(C) Let the digit at the ten's place be $x$ and the digit at the unit's place be $y$ in the original two-digit number.Then,the original number $= 10x

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(C) Let the digit at the ten's place be $x$ and the digit at the unit's place be $y$ in the original two-digit number.Then,the original number $= 10x + y$.The sum of the digits is $x + y = 9$  ......... $(1)$The number obtained by interchanging the digits is $10y + x$.According to the problem,the new number exceeds the original number by $45$:$10y + x = (10x + y) + 45$$10y + x - 10x - y = 45$$-9x + 9y = 45$Dividing by $9$,we get $-x + y = 5$  ......... $(2)$Adding equations $(1)$ and $(2)$:$(x + y) + (-x + y) = 9 + 5$$2y = 14$$y = 7$Substituting $y = 7$ into equation $(1)$:$x + 7 = 9$$x = 2$Therefore,the original number $= 10x + y = 10(2) + 7 = 27$.

The sum of the digits of a two-digit number is $9$. The number obtained by interchanging the digits exceeds the original number by $45$. Find the original number.

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