Question

The sum of the digits of a two-digit number is 8. If the number formed by reversing the digits, is less than the original number by 18, find the number.​

Answer 1

Answer:

Let the digit at tens place =x

and the digit at ones place =y

so, original number =10x+y

so,the number formed by reversing it's digit =x+10y

By given data,

x+y=8 -----------eq-1

Also,

x+10y = 10x+y-18 -----------eq-2

x+10y-10x-y = -18

-9x+9y = -18

-9(x-y) = -9×2

x-y = 2 ---------eq-3

Now,add eq-1 and eq-3

x+y = 8

x-y = 2

------------------

2x = 10

x = 5

Now, put the value of x in eq-3

x-y =2

5-y =2

-y = 2-5

-y = -3

y = 3

Therefore,the number is = 10x+y

= 10(5)+3

= 50+3

= 53