Question

The sum of the digits of a two-digit number is 12 number obtained by interchanging the two digits exceeds the given number by 18 find the number

Answer 1

$\bold{GIVEN}$

Sum of two digits number = 12

Let number be at tenth place x

and at unith place be y.

$\mathsf{Number = 10 × x + y}$

$\mathsf{= 10x + y}$

$\mathsf{x+ y = 12 -----(1)}$

Number formed after interchanging digits,

New number will be

$\mathsf{= 10 ×y + x}$

$\mathsf{=10y + x }$

ACCORDING TO QUESTION,

If digits are interchanged the formed number wil be 18 more than real one.

$\mathsf{10x + y + 18 = 10y + x }$

$\mathsf{18 = 10y - y + x - 10}$

$\mathsf{18 = 9y - 9x----(2)}$

On multiplying eq(1) by 9

$\mathsf{x+ y = 12 -----(1)}$ ×9

$\mathsf{9x+ 9y = 108}$

On subtracting eq(2) from eq(1)

$\mathsf{ 9x+ 9y = 108}$

$\mathsf{-9x+9y = 18}$

______________________________

18y = 90

______________________________

$\mathsf{\dfrac{90}{18} = y}$

$\boxed{\mathsf{5 = y}}$

For x,

$\mathsf{x+ 5 = 12}$

$\boxed{ \mathsf{x = 7}}$

NUMBER =

$\mathsf{7 \times 10 + 5}$

$\boxed { \mathsf{75}}$

Answer 2

Solution:-

Let the Two Digit Number be x & y.

Where,

x is at tens place & y is at once place.

=> Original Number = 10x + y.

CONDITION | ,

x + y = 12 ________(1)

CONDITION ||,

After Interchanging the Digit, we get

10y + x

A.T.Q.

10y + x = 10x + y + 18

=> 9x - 9y = 18._________(2).

By Elimination Method,

9 ( x + y ) = 9 (12)

=> 9x + 9y = 108__________(3).

Subtracting eq (2) from (3). we get,

9x + 9y = 108

-9x + 9y = - 18

_______________

18y = 90

=> y = 5.

Substituting y=5 in eq (1). we get,

x = 7.

Hence,

The Original Number is 75.