Question

A number consists of two digits such that the digit
in the unit's place is more by 2 than the digit in the
ten's place. Three times the number added to
times the number obtained by reversing the digits
equals 108. Then the number is:​

Answer 1

Step-by-step explanation:

step 1 :

let digit in tens place=x

digit in units place=x+2

then the number becomes= 10x+(X+2)

=11x+2----------(1)

step 2:

number obtained by reversing the digits:

digit in tens place becomes=X+2

digit in units place becomes=x

then the number obtained by reversing the digits becomes= 10(X+2)+x

= 11x+20----------(2)

step 3:

it is given in the question that

3 times the number added to the number obtained by reversing the digits=108

therefore,

3(11x+2)+(11x+20)=108

33x+6+11x+20=108

44x+26=108

44x=82

I think the question is wrong!!