Question

The sum of a two-digit number and the number obtained by interchanging the digits is 110. If the digits of the number
differ by 4, then find the original number. [Assume, digit at tens place is less than the digit at unit place.]

Answer 1

Given units digit is x and tens digit is y
Hence the two digit number = 10y + x
Number obtained by reversing the digits = 10x + y
Given that sum of a two digit number and the number obtained by reversing the order of its digits is 121.
Hence (10y + x) + (10x + y) = 121
⇒ 11x + 11y = 121
∴ x + y = 11
Thus the required linear equation is x + y = 11.

Answer 2

Let the digits of the number be x and x+4

Then, the two digit number would be 10x+ x+4
As it is tens place

By interchanging the digits, we get 10(x+4)+ x

That is 10x+40+x

Their sum= 110

10x+x+4 + 10x+ 40+ x = 110
22x+44=110

22x= 66
x=3

x+4= 7

The number is 37 as the tens place is smaller than unit digit as mentioned above...

Hope it helps

Plz mark as brainliest