Question

The sum of digits of a two-digit number
is 7. If the digits are reversed, the new
number decreased by 2 equals twice the
original number. Find the number.
​

Answer 1

Let the digits be x and y... and so, the original number be 10x + y ( since x in tenth place and y in unit place)

so, given : sum of the digits = x + y = 7 ---------> (A)

                 

reversing the digits means y in tenth place and x in unit place.. so, the reversed number be 10y + x

        

  given: reversed number is decreased by 2 = twice original number

 

         so, 10y + x - 2 = 2(10x + y) 

          simplifying :   19x - 8y = -2       ------------> (B)

solving 2 eqns (A) & (B) ,

 we get x = 2 and y = 5

so the original number is 10x + y = 10(2) + 5 = 25

  & the reversed number is 10y + x = 10(5) + 2 = 52

   & the reversed number 52  when decreased by 2 is 50 which is twice the original number 25 

Ans: the original number is 25

Answer 2

Answer:

Let the digits be x and y... and so, the original number be 10x + y ( since x in tenth place and y in unit place)

so, given : sum of the digits = x + y = 7 ---------> (A)

                

reversing the digits means y in tenth place and x in unit place.. so, the reversed number be 10y + x

       

 given: reversed number is decreased by 2 = twice original number

 

        so, 10y + x - 2 = 2(10x + y) 

         simplifying :   19x - 8y = -2       ------------> (B)

solving 2 eqns (A) & (B) ,

 we get x = 2 and y = 5

so the original number is 10x + y = 10(2) + 5 = 25

 & the reversed number is 10y + x = 10(5) + 2 = 52

  & the reversed number 52  when decreased by 2 is 50 which is twice the original number 25 

Ans: the original number is 25

Step-by-step explanation:

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