Question

the sum of thedigits of a two digit number is 12 the number obtained by interchanging the digits exceeds the originalnumber by 54 . find the original number​

Answer 1

Let the number be 10x+y(expanded form)

So,the actual number can be written as xy

That means x+y=12

Now, 10y+x=10x+y+54

So,9y-9x=54

Divide the equation by 9

y-x=6

Now,x+y=y+x=12

Now use elimination method to solve the equation

So the result will be 2y=18

y=9

Substitute the value of y in any of one equation

So,9+x=12(used 1st equation)

x=3

So the required number is 39

Answer 2

GIVEN :

• The sum of the digits of a two digit number is 10.

• The number interchanging the digit exceeds the original number by 54.

TO FIND :

• The original number = ?

STEP - BY - STEP EXPLANATION :

NOTE :---- If the digit of a two digit number are x (ones) and Y (tens) then the required number is 10y + x example as -----» 10 × Tens digit + Ones digit

[ RECALL THAT

=> 25 = 10 × 2 + 5

=> 36 = 10 × 3 + 6 etc.]

Since, the required number is a two digit number so we will have to find its ones digit and its tens digit.

=>Let the digit at ones place be 'X'

it is given that the sum of digit of the number is 12.

Hence, the digit at ones place = 12-x

Thus, the original number

= 10 × ( 12 - x ) + x

= 120 - 10x + x

= 120 - 9x

onenter changing the digits of the given number the digit at ones place becomes (12x - x) and the digit at tens place becomes x.

New number = 10x + ( 12-x )

= 9x + 12

it is given that the new numbers exceeds the original number by 54.

=> (9x + 12) – (120–9x) = 54

=> 9x + 12 – 120 + 9x = 54

=> 18x – 108 = 54

=> 18x = 54 + 108

=> 18x = 162

$= > x = \frac{162}{18} = 9$

The digit at ones place = 9

The digit at tens place = (12 –9) = 3

Hence, original number = 39