Question 35727
let the tens digit of the original whole number be x the according to the problem the ones digit is one more than twice x hence the ones digit is 2x + 1

the value of original number is obtained as follows
any number with A as unit digit and B as tens digit can we written as
10*B+ A*1 since the place value of B is 10 and that of A is 1

similarly for the number with 2x+1 as ones digit and x tens digit
we have the value of the number as 10x + 2x+ 1 = 12x + 1

if the digits are reversed then the new numbers unit digit is x and the tens digit is 2x+1

hence the value of the new number is 10*(2x+1)+x = 21x + 10

it is given the new number is 4 less than twice the original number

therefore we have 21x + 10 = 2*(12x+1) - 4

21x + 10 = 24x + 2 - 4
3x = 12
x = 4
the original number is 12x+1 = 49