Question

the digit of a two digit number differ by 18 the digit are interchanged and the resulting number is added to the original number we get 99 find the original number

Answer 1

Let one digit be x

Therefore the other digit is x-4

Let x be the in tenths digit

Thus,

10x + x - 4 = 11x - 4

On reversing the digits,

10(x-4) + x = 11x - 40

On subtracting the two we get

11x - 4 -(11x - 40)

= 40-4= 36

Answer 2

Answer:

let the ones place digit be y and tens place digit be x

so original no. becomes 10x+y

so interchanged number become 10y+x

according to ques..

x-y=18

and ,

{10y+x }+{10x + y }= 99

so we get our two equations

as

x-y = 18.

{10y +x }+{10x+y}= 99

from 1st equation

x=18+y

putting the value of x in eq .

{10y+18+y}+{10×(18+y)+y}=99

{11y+18}+{180+10y +y}=99

11y+18+180+11y=99

22y+198=99

22y=99-198

22y=99

y=9/2

putting the value of y in equation

x-y =18

x-9/2=18

{2x-9}/2=18

after solving this

we get x = 45/2

so the values r

x =45/2

y=9/2