Question

The sum of a two digit number and the number obtained by interchanging the digits is 143. If the digit at the units place is 3 more than the digit at tens place , find the original number.​

Answer 1

Answer:

The original number is 58.

Step-by-step-explanation:

Let the digit at the units place be x.

And the digit at the tens place be y.

∴ The original number = 10 y + x

The number obtained by interchanging the digits

= 10 y + x

From the first condition,

Two digit number + The number obtained by interchanging the digits = 143

∴ ( 10 y + x ) + ( 10 x + y ) = 143

∴ 10y + x + 10x + y = 143

∴ 11 x + 11 y = 143

∴ $\boxed{x + y = 13}$ ... [ Dividing

each term by 11. ] ( 1 )

Now, from the second condition,

The digit at units place = The digit at tens place + 3

x = y + 3

∴ $\boxed{x - y = 3}$ .... ( 2 )

Adding equations ( 1 ) and ( 2 ),

x + y = 13 ... ( 1 )

+ x - y = 3 ... ( 2 )

__________

2x = 16

∴ $\boxed{x = 8}$

Now, by substituting x = 8 in equation ( 1 ),

∴ x + y = 13

∴ 8 + y = 13

∴ y = 13 - 8

∴ $\boxed{y = 5}$

The original number

= 10y + x

= ( 10 × 5 ) + 8

= 50 + 8

∴ $\boxed{The\:original\:number \:= 58}$

Ans.: The original number is 58.