Question

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 54 more than the original number. Find the original number.​

Answer 1

Solution:

Let say ‘x' is digit on one's place so 3x is digit on tens' place

So number is 10*3x+x=31x

Now if reverse the position of digit new number is

1. 10*x+3x=13x

As given new number is more than by 54 from old number

    2. 31x+54=13x

• -18x=54 x=-3  ----> Putting x= -3 in $Eq_{1}$

So number is 10*3(-3)+(-3)=-93.

Hence, we get

-93 is the original number.