The value of a two-digit number exceeds twice the sum of its digits by 19. The tens digit is two less than the units digit. Find the number.

Answer with Step-by-step explanation:

Let the ones digit of the number be y

and tens digit of the number be x.

The tens digit is two less than the units digit.

i. e. x=y-2

The value of the number exceeds twice the sum of its digits by 19.

i. e. 10x+y=2 (x+y) + 19

10 (y-2) + y=2 (y-2+y) + 19

10y-20+y=2 (2y-2) + 19

11y-20=4y-4+19

11y-4y=20-4+19

7y=35

dividing both sides by 7, we get

y=5

Putting value of y in x=y-2, we get

x=3

10x+y=10*3+5

=35

Hence, the two digit number is:

35